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https://research.tue.nl/en/publications/interlace-polynomials | [
"# Interlace polynomials\n\nM. Aigner, H. Holst, van der\n\nResearch output: Contribution to journalArticleAcademicpeer-review\n\n41 Citations (Scopus)\n1 Downloads (Pure)\n\n## Abstract\n\nIn a recent paper Arratia, Bollobás and Sorkin discuss a graph polynomial defined recursively, which they call the interlace polynomial q(G,x). They present several interesting results with applications to the Alexander polynomial and state the conjecture that |q(G,-1)| is always a power of 2. In this paper we use a matrix approach to study q(G,x). We derive evaluations of q(G,x) for various x, which are difficult to obtain (if at all) by the defining recursion. Among other results we prove the conjecture for x=-1. A related interlace polynomial Q(G,x) is introduced. Finally, we show how these polynomials arise as the Martin polynomials of a certain isotropic system as introduced by Bouchet.\nOriginal language English 11-30 Linear Algebra and Its Applications 377 https://doi.org/10.1016/j.laa.2003.06.010 Published - 2004"
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https://reasonablypolymorphic.com/blog/freer-yet-too-costly/ | [
"# Freer, yet Too Costly Higher-order Effects\n\nI’m throwing in the towel. For now at least.\n\nAs of today I have free, higher-order effects working. Unfortunately, they are not fast. I don’t think this is a fundamental limitation, merely that whatever code I’ve written isn’t amenable to GHC’s optimization process.\n\nI’ve been hammering on this for about 50 hours now. It’s been driving me slowly crazy, and promised myself I’d stop if I hadn’t solved it by now. That being said, before putting this project to rest, I wanted to do a quick write-up detailing what I’ve learned, how everything fits together, and where I’m hoping someone will pick up the ball. Here’s the repository.\n\n## Higher Order Effects🔗\n\nIn the freer-simple model, effects are first-order—meaning they are unable to embed Eff computations inside of them. This is occasionally annoying, primarily when trying to write bracket-like effects.\n\nYou can sort of work around the problem by encoding your scoped computation as an interpretation of an effect, but this often comes at the cost of fixing the interpretations of the other effects you’re dealing with.\n\nThis fundamental limitation comes from the fact that freer-simple effects have kind * -> *. There’s nowhere in here to stick the Eff stack you’re working in. You can kind of hack it in, but it never plays nicely.\n\nThe solution is given in the paper Effect Handlers in Scope, and implemented in the fused-effects package. The idea is to parameterize each of your effects with a monad—ie. they have kind (* -> *) -> * -> *. This parameter gets instantiated at the entire Eff stack, as seen by the type of send :: Member e r => e (Eff r) a -> Eff r a\n\nWhile it’s an obvious insight, actually getting everything to play nicely is tricky. The primary issue is how do you push interpretations through these additional monadic contexts? For example, let’s consider a bracket-esque effect:\n\ndata Bracket m x where\nBracket :: m a -- ^ allocate\n-> (a -> m ()) -- ^ deallocate\n-> (a -> m x) -- ^ use\n-> Bracket m x\n\nAssume we want to push a State effect through these ms. What are the correct semantics for how the state is accumulated? Should any state changed in the deallocate block count outside of the bracket? What should happen in the use case if an exception is thrown and the rest of the block is short-circuited?\n\nNot only are we concerned with the semantics, but also the actual mechanism of propagating this state throughout the computation.\n\nEffect Handlers in Scope introduces weave in a typeclass, which is responsible for this state-propagation behavior. Statefulness for an effect is encoded as some arbitrarily chosen functor f, and weave describes how the effect should move that state through the effect. Behold:\n\nclass Effect e where\nweave\n:: Functor f\n=> f ()\n-> (forall x. f (m x) -> n (f x))\n-> e m a\n-> e n (n (f a))\n\nThe f () parameter is the current “state of the world”, and the rank-2 thing is this distribution law. You can intuit the m parameter being an effect stack with all of the effects present, and the n parameter as being the same effect stack—but with the top of it taken off. To clarify, we could monomorphize it thusly:\n\nweave\n:: Functor f\n=> f ()\n-> (forall x. f (Eff (g ': r) x) -> Eff r (f x))\n-> e (Eff (g ': r)) a\n-> e (Eff r) (Eff r (f a))\n\nThis janky return type: e (Eff r) (Eff r (f a)) comes from the fact that Effect Handlers in Scope describes a traditional “free” (as opposed to freer) approach to a free monad. The last parameter of an effect is actually a continuation for the next piece of the computation. By mangling it, we’re ensuring the caller (which will be library code) respects the evaluation semantics.\n\nweave implicitly defines the evaluation semantics of an effect—it pins how state propagates through them, which in turn defines which pieces of the effect are observable to the outside.\n\n## Freeing the Higher Orders🔗\n\nWarning: This next section describes a janky-ass solution to the problem. It’s clearly a hack and clearly not the right answer. But maybe by writing out what I did, someone with a keen eye can point out where I went wrong.\n\nSo this is all well and good. It works, but requires a lot of boilerplate. As presented in the paper, a new effect requires:\n\n• A Functor instance\n• An MFunctor instance (providing hoist :: forall x. (f x -> g x) -> e f a -> e g a)\n• An Effect instance as above, and an additional method not described here\n\nIf you’re following in the fused-effects tradition, for each interpretation you additionally need a new Carrier type, with its own Functor, Applicative and Monad instances, and then another typeclass tying the effect to its carrier.\n\nfused-effects improves the $O(n^2)$ MTL instance problem to $O(n)$—albeit with a big constant factor :(\n\nThis is a huge amount of work! I’ve said it before and I’ll say it again: ain’t nobody got time for that. If it feels like too much work, people aren’t going to do it. A solution that depends on humans not being lazy isn’t one that’s going to take off.\n\nSo wouldn’t it be nice if we could just all of this effect stuff for free?\n\nHere’s where I admittedly went a little off the rails. The first step towards getting a freer Functor-less Monad instance for Eff is to define it in terms of its final encoding. I made the obvious changes to last time without thinking too much about it:\n\nnewtype Freer f a = Freer\n{ runFreer\n:: forall m\n=> (forall x. f (Freer f) x -> m x)\n-> m a\n}\n\nI have no idea if this is right, but at least it gives a Monad instance for free. One limitation you’ll notice is that the continuation in runFreer is a natural transformation, and thus it’s unable to change its return type.\n\nThat means interpretations like runError :: Eff (Error e ': r) a -> Eff r (Either e a) are surprisingly difficult to implement. More on this later—I just wanted to point out this flaw.\n\nFrom here I followed Oleg and implemented Eff as a type synonym, making sure to tie the knot and instantiate the m parameter correctly:\n\ntype Eff r = Freer (Union r (Eff r))\n\nBut how can we get a free implementation for weave?\n\nIt’s this mental thing I came up with, which is sort of like Coyoneda but for weave:\n\ndata Yo e m a where\nYo :: (Monad n, Functor f)\n=> e m a\n-> f ()\n-> (forall x. f (m x) ~> n (f x))\n-> (f a -> b)\n-> Yo e n b\n\nIn retrospect, I would not spend any more time on this approach—I’d just make people give an instance of weave for higher-order effects, and machinery to derive it automatically for first-order ones.\n\nBut then how can we get an MFunctor instance for free? You can’t just derive it generically—lots of higher-order effects want existentials, and thus can’t have Generic instances.\n\nThis Yo thing mirrors the definition of weave pretty closely. The idea is that it can accumulate arbitrarily many weaves into a single Yo, and then dispatch them all simultaneously.\n\nSome interesting points to note are that the state functor f is existentialized, and that there is this final f a -> b parameter to make it play nicely with the Union (more on this in a second.) We can implement weave now by replacing the existing state functor with a Compose of the new one and the old one.\n\nweave\n=> f ()\n-> (forall x. f (m x) -> n (f x))\n-> Union r m a\n-> Union r n (f a)\nweave s' distrib (Union w (Yo e s nt f)) =\nUnion w $Yo e (Compose$ s <$s') (fmap Compose . distrib . fmap nt . getCompose) (fmap f . getCompose) We can also use Yo to get a free MFunctor instance: instance MFunctor (Yo e) where hoist f (Yo e s nt z) = Yo e s (f . nt) z OK, all of this works I guess. But what’s with this weird f a -> b thing in Yo that I mentioned earlier? Well recall the type of runFreer, when instantiated at Union: runFreer :: forall m . Monad m => (forall x. Union r (Eff r) x -> m x) -> m a The only way we can produce an m a is via this rank-2 thing, which is a natural transformation from Union r (Eff r) to m. In other words, it’s not allowed to change the type. We can’t just stuff the f into the result and return an m (f a) instead—this thing doesn’t form a Monad! Fuck! All of this comes to a head when we ask ourselves how to actually get the state out of such a contraption. For example, when we call runState we want the resulting state at the end of the day! The trick is the same one I used in the last post—we’re able to instantiate the m inside runFreer at whatever we like, so we just choose StateT s (Eff r) and then run that thing afterwards. Again, this is very clearly a hack. Because weave is given freely, interpretations must eventually actually decide what that thing should look like. Some combinators can help; for example, this is the interface I came up with for implementing runBracket: runBracket :: Member (Lift IO) r => (Eff r ~> IO) -> Eff (Bracket ': r) a -> Eff r a runBracket finish = deep$ \\start continue -> \\case\nBracket alloc dealloc use -> sendM $X.bracket (finish$ start alloc)\n(finish . continue dealloc)\n(finish . continue use)\n\nThe deep combinator gives you start and continue (which are the cleverly disguised results of weave), and asks you to give a natural transformation from your effect into Eff r.\n\nThe actual implementation of deep isn’t going to win any awards in understandability, or in inline-ability:\n\ndeep\n:: (∀ m f y\n. Functor f\n=> (forall x. m x -> (Eff r (f x)))\n-> (∀ i o. (i -> m o) -> f i -> Eff r (f o))\n-> e m y\n-> Eff r (f y)\n)\n-> Eff (e ': r) a\n-> Eff r a\ndeep transform (Freer m) = m $\\u -> case decomp u of Left x -> liftEff$ hoist (deep transform) x\nRight (Yo eff state nt f) -> fmap f $transform (deep transform . nt . (<$ state))\n(\\ff -> deep transform . nt . fmap ff)\neff\n\nNotice that whoever implements transform needs to give an equivalent of an implementation of weave anyway. Except that instead of only writing it once per effect, they need to write it per interpretation.\n\nWe can also give an implementation for runState in terms of a StateT s:\n\nrunState :: forall s r a. s -> Eff (State s ': r) a -> Eff r (s, a)\nrunState s = flip runStateT s . go\nwhere\ngo :: forall x. Eff (State s ': r) x -> StateT s (Eff r) x\ngo (Freer m) = m $\\u -> case decomp u of -- W T F Left x -> StateT$ \\s' ->\nliftEff . weave (s', ())\n(uncurry (flip runStateT))\n$hoist go x Right (Yo Get state nt f) -> fmap f$ do\ns' <- get\ngo $nt$ pure s' <$state Right (Yo (Put s') state nt f) -> fmap f$ do\nput s'\ngo $nt$ pure () <$state This is also completely insane. Notice the aptly-named section W T F, where for no reason I can discern other than satisfying the typechecker, we convert from a StateT s to a (s, ()) and back again. But why?? Because this is what weave wants—and we need to satisfy weave because it’s the only way to change the type of a Union—and we need to do that in order to reinterpret everything as a StateT s. There is a similarly WTF implementation for runError. But worse is the combinator I wrote that generalizes this pattern for running an effect in terms of an underlying monad transformer: shundle :: ∀ a f t e r . ( MonadTrans t , ∀ m. Monad m => Monad (t m) , Functor f ) => (∀ x. Eff r (f x) -> t (Eff r) x) -> (∀ x. t (Eff r) x -> Eff r (f x)) -> (∀ x. f (t (Eff r) x) -> Eff r (f x)) -> f () -> (∀ m tk y . Functor tk => (∀ x. f () -> tk (m x) -> Eff r (f (tk x))) -> tk () -> e m y -> t (Eff r) (tk y) ) -> Eff (e ': r) a -> Eff r (f a) shundle intro finish dist tk zonk = finish . go where go :: ∀ x. Eff (e ': r) x -> t (Eff r) x go (Freer m) = m$ \\u ->\ncase decomp u of\nLeft x -> intro . liftEff . weave tk dist $hoist go x Right (Yo e sf nt f) -> fmap f$\nzonk (\\r -> shundle intro finish dist r zonk . nt) sf e\n\nDon’t ask why it’s called shundle or why it has a variable called zonk. It was funny to me at the time and I needed whatever smiles I could get to continue making progress. Believe it or not, this abomination does indeed generalize runState:\n\nstatefully\n:: (∀ m. e m ~> StateT s (Eff r))\n-> s\n-> Eff (e ': r) a\n-> Eff r (s, a)\nstatefully f s =\nshundle\n(StateT . const)\n(flip runStateT s)\n(uncurry $flip runStateT) (s, ())$ \\_ tk -> fmap (<$tk) . f runState :: s -> Eff (State s ': r) a -> Eff r (s, a) runState = statefully$ \\case\nGet -> get\nPut s' -> put s'\n\nrunState is actually quite reasonable! At this point I was willing to concede “worse is better”—that most of the time people only really care about StateT-esque state in their effects. And really the only thing they’ve been missing is bracket. And we now have bracket, and an easy way of doing StateT-esque state.\n\nSo I was going to hand it in to one library or another—if not for full marks, then at least for the participation trophy.\n\nBut then I ran my benchmark, and saw that it performs 10x slower than freer-simple. Even for effect stacks that don’t need to weave.\n\nSHAME.\n\nSORROW.\n\nAnd I guess this is where I leave it. The machinery is clearly wrong, but amazingly it actually does what it says on the tin. Unfortunately it does it so slowly that I think the people who complain about the performance of free monads might actually have a point this time.\n\nI’ve got my heart crossed that someone will swoop in here and say “here’s what you’ve done wrong” and it will be a minor change and then everything will optimize away and we will make merry and the freer monad revolution will be complete.\n\nI’m very tired.\n\n## What a Real Solution Would Look Like🔗\n\nI think I have enough familiarity with the problem at this point to know what a solution would look like—even if I can’t find it myself:\n\n• runFreer could produce a f a if you asked for it\n• Weave would be a typeclass that effects authors would need to implement. But they could derive it for free if the m parameter was unused. This would be the only instance necessary to implement by hand.\n• The library would provide a handleRelayS-esque interface for defining interpreters.\n• By the time the user’s code ran for the interpretation, every monadic value in their effect would be bindable without any further effort—ala cata.\n\nI’d also be happy if an existing solution (probably fused-effects) were given the tools to satisfy the above list. In particular, the handleRelayS thing."
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http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/ruby-talk/283626 | [
"```Shoot, I'll give it a whirl, it beats the meta discussion. I added a\ncheck for branch contents, I think that qualified as an aspect per the\nquiz rules.\n\n- donald\n\navl_tree.rb:\n\n#!/usr/bin/env ruby -wKU\n\nclass AVLTree\n\ndef initialize\n@contents =3D []\nend\n\ndef empty?\[email protected]?\nend\n\ndef include?(obj)\[email protected]?(obj)\nend\n\ndef <<(obj)\n@contents << obj\nend\n\ndef height\nMath::log(@contents.length).ceil\nend\n\ndef remove(obj)\[email protected](obj)\nend\n\nend=20\n\n#!/usr/bin/env ruby -wKU\n\nrequire \"test/unit\"\n\nrequire \"avl_tree\"\n\nclass TestAVLTree < Test::Unit::TestCase\ndef setup\n@tree =3D AVLTree.new\nend\n\ndef test_tree_membership\nassert_equal(true, @tree.empty?)\nassert_equal(false, @tree.include?(3))\n\n@tree << 3\n\nassert_equal(false, @tree.empty?)\nassert_equal(true, @tree.include?(3))\nend\n\ndef test_tree_should_allow_more_than_one_element\n@tree << 3\n@tree << 4\n\nassert(@tree.include?(4))\nassert(@tree.include?(3))\nend\n\n#disabled this test case as it is order dependent and not valid # def\ntest_tree_height_of_one_or_two_nodes_is_N\n# @tree << 5\n# assert_equal 1, @tree.height\n# @tree << 6\n# assert_equal 2, @tree.height #changed from 1\n# end\n\ndef test_tree_height_of_three_nodes_should_be_greater_than_1\n@tree << 5\n@tree << 6\n@tree << 7\nassert(@tree.height > 1, \"Tree appears to have stunted growth.\") end\n\ndef test_tree_growth_limit_is_1pt44_log_N\n(1..10).each{|i|\n@tree << i\nlimit =3D (1.44 * Math::log(i)).ceil+1\nassert( @tree.height <=3D limit, \"Tree of #{i} nodes is too tall =\nby\n#{@tree.height - limit}\")\n}\nend\n\ndef test_remove_node\n@tree << 314\[email protected](314)\nassert([email protected]?(314))\nend\n\ndef test_has_branches\n[50, 17, 72].each {|n| @tree << n}\nassert(50 =3D=3D @tree.value)\nassert(17 =3D=3D @tree.left.value)\nassert(72 =3D=3D @tree.right.value)\nend\n\nend\n\n```"
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https://jp.mathworks.com/matlabcentral/cody/problems/2640-find-similar-sequences | [
"Cody\n\n# Problem 2640. Find similar sequences\n\nAnother problem inspired by a question on the answers forum.\n\nGiven a matrix of positive integer numbers, find all the rows that are similar to the first rows and return these rows as a new matrix.\n\nRows are considered similar if the numbers common to both rows are in the exact same order with no other numbers in between. 0s in a row are always ignored and only occur at the end of the row.\n\nFor example:\n\n``` [3 1 5 0 0] and [4 2 1 5 0] are similar (1 5 are the common numbers and occur in the same order)\n[3 1 5 0 0] and [3 4 1 5 0] are not similar (3 1 5 are the common numbers, there's a 4 in between)```\n\n### Solution Stats\n\n30.19% Correct | 69.81% Incorrect\nLast solution submitted on Aug 30, 2019"
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https://www.fineconverter.com/days-to-minutes/18-day-to-min.html | [
"day to min\n Days day Minutes min\n\n# 18 Days to Minutes (d to min)\n\nUse 18 days to minutes converter to find how many minutes in 18 Day. To convert 18 days to minutes enter the time in days and get the value converted into minutes.\n\n## Formula to convert 18 days to minutes:\n\nmin = d × 1440\n\nWhere,\n\nmin = Time in minutes and,\n\nd = Time in days\n\n## How to convert days to minutes? (d into min)\n\nWhen you are searching for days to minutes, you are indirectly searching for day to min.\n\nBelow we will show you how to convert days to minutes.\n\nTo convert days into min use above conversion formula.\n\n01 day is equal to 1440 minutes. (i.e 1 day = 1440 minutes).\n\nTo convert your days to minutes, multiply days by 1440 to get the result.\n\n### Convert 18 days to minutes (18 days to Min)\n\n01 day is equal to 1440 minutes.\n\nTo convert 18 days in min, multiply days by 1440 to get the result.\n\nmin = (18 d × 1440)\n\nTherefore, the answer to 18 days to minutes is 25920 minutes, which can be written as follows:\n18 days = 25920 minutes"
]
| [
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.80437654,"math_prob":0.93896115,"size":946,"snap":"2021-43-2021-49","text_gpt3_token_len":243,"char_repetition_ratio":0.2760085,"word_repetition_ratio":0.07608695,"special_character_ratio":0.3012685,"punctuation_ratio":0.10294118,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98783654,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-20T16:21:55Z\",\"WARC-Record-ID\":\"<urn:uuid:03cb30aa-fa72-42b9-808e-90f83597229e>\",\"Content-Length\":\"6129\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:febea481-3a27-4b48-8713-864926768914>\",\"WARC-Concurrent-To\":\"<urn:uuid:f00c3c9d-76cf-4656-be88-030a2f7f9bb6>\",\"WARC-IP-Address\":\"104.21.66.141\",\"WARC-Target-URI\":\"https://www.fineconverter.com/days-to-minutes/18-day-to-min.html\",\"WARC-Payload-Digest\":\"sha1:EBTZOLICJBRBVQ2GUW4S5CLX4YFNQ5W4\",\"WARC-Block-Digest\":\"sha1:L7CZGDJIFSBGK22NKHI3O7Q6W454ZLEC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585322.63_warc_CC-MAIN-20211020152307-20211020182307-00475.warc.gz\"}"} |
https://www.exceldemy.com/excel-trim-function/ | [
"# How to use TRIM function in Excel (7 Examples)\n\nExcel provides several text functions to help you perform your desired tasks easily and swiftly. Today we are going to show you how to use a text function called: TRIM. For this session, we are using Excel 2019, feel free to use yours (at least 2003).",
null,
"## TRIM Function\n\n### 1. Basics of TRIM\n\nThe Excel TRIM function is categorized under the TEXT functions. It removes the extra spaces from a text string.",
null,
"#### Summary\n\nRemoves all spaces from a text string except for single spaces between words.\n\n#### Syntax\n\n`TRIM (text)`\n\n#### Arguments\n\ntext: The text string from which to eradicate unnecessary spaces.\n\n#### Versions\n\nWorkable from Excel 2003.\n\n### 2. Uses of TRIM\n\n#### I. Remove Extra Spaces\n\nThe description of the TRIM function might have let you understand that the prime task of this function is to remove any extra spaces from a string. Let’s explore with examples.\n\nOur example dataset contains few movie names.",
null,
"Here we have intentionally put extra spaces in the beginning, middle, or end of the names. To remove these extra spaces, we need to provide the Cell Reference of the text within TRIM.\n\n`=TRIM(B4)`\n\nB4 is the Cell Reference for the first row of the Movie column.",
null,
"You can see the spaces between Few and Good have been eradicated (only one space remains).\n\nA similar formula (change in Cell Reference) will remove the spaces from the beginning, middle, and end for the rest of the rows.",
null,
"#### II. Remove Spaces and Clean String\n\nSometimes merely using TRIM may not be useful to clean data. The function removes spaces, but if our dataset has a string where texts are separated into different lines?\n\nWe have brought the dataset of movie names and their respective releasing year. Here movie names and the release year are in different lines.",
null,
"There are extra spaces also. To eradicate the issues and make the data an organized one, we are going to use a function called CLEAN along with TRIM.\n\nThe CLEAN function converts text to be cleaned of line breaks and other non-printable characters.\n\nFor the first row, our formula will be\n\n`=TRIM(CLEAN(B4))`",
null,
"The formula provided the cleaned data, no extra spaces, no line breaks.",
null,
"The TRIM function removes all the extra spaces from every part. But it can be also used for removing the leading spaces only.\n\nTo show you examples, we have introduced a dataset of several area codes.",
null,
"The area codes have spaces at the beginning as well as in between the words. We aim to remove the spaces from the beginning only.\n\nWe will use MID, FIND, and LEN along with TRIM. To know more about these functions, visit these articles: MID, FIND, LEN.\n\nAnd the formula will be:\n\n`=MID(B4,FIND(MID(TRIM(B4),1,1),B4),LEN(B4))`",
null,
"The combination of FIND, MID, and TRIM calculates the position of the first text character in a string.\n\nAnd then, supply that number to the outer MID function so that it returns the entire text string starting at the position of the first text character.",
null,
"#### IV. Concatenate with TRIM\n\nWe can combine the values from different cells using the TRIM function.\n\nOur example dataset consists of a few random items, we are set to combine them together.",
null,
"To be honest we don’t need TRIM to combine. We will only use TRIM when the combining values have extra spaces.\n\nSo, our formula will be\n\n`=TRIM(B4&\" \"&C4&\" \"&D4) `\n\n`& `will concatenate the values. And the TRIM functions will remove all extra spaces.\n\nTRIM automatically strips space at the start and end of a given string and leaves just one space between all words inside the string. It takes care of extra spaces caused by empty cells.\n\nWe can replace the space between two items using the SUBSTITUTE function. Visit this SUBSTITUTE article for further information.\n\n`=SUBSTITUTE(TRIM(B5&\" \"&C5&\" \"&D5),\" \",\", \")`",
null,
"SUBSTITUTE is used to replace each space `(\" \")` with a comma and space `(\", \")`.",
null,
"#### V. Count Spaces in a String\n\nThe TRIM function can be helpful to count the number of spaces from a string. We are using the movie dataset for showing you examples.",
null,
"To find the extra spaces in the string, we are going to use the formula written below.\n\n`=LEN(B4)-LEN(TRIM(B4))`` `",
null,
"LEN provides the length of the string. To know more about the function visit the LEN article.\n\nLEN(B4) provided the full length of the string of cell B4 and LEN(TRIM(B4)) provided the length after trimming.\n\nSubtraction of these two will provide the total number of extra spaces.",
null,
"#### VI. Count Words in a String\n\nWe can count the number of words present in a string using the TRIM function though we need to get help from other functions also.",
null,
"We need to use the LEN, and SUBSTITUTE functions along with TRIM.\n\nOur formula will be\n\n`=(LEN(TRIM(B4))-LEN(SUBSTITUTE(B4,\" \",\"\")))+1`",
null,
"We have found the number of words from this string.\n\nSUBSTITUTE removes all spaces from the string, then LEN calculates the length without spaces.\n\nThis number is then subtracted from the length of the text with spaces. We have used TRIM to eradicate any spaces at the beginning or end of the string.\n\nAnd finally, 1 is added to the result, since the number of words is the number of spaces + 1.\n\nThe same formula will provide the number of words for the rest of the texts.",
null,
"#### VII. Numbers TRIM\n\nYou can provide numbers within the TRIM function. To show you examples we have picked several numbers randomly.",
null,
"Use the trim function as below\n\n`=TRIM(B4)`",
null,
"We have found the trimmed number.",
null,
"But are they really numbers? Let’s check by summing them together.",
null,
"Oh, dear! The sum of the value is 0.",
null,
"This is because the digits you are seeing are not numbers, they are in Text format. To convert these into numbers after TRIM we will use a function called VALUE.\n\nAnd the formula will be\n\n`=VALUE(TRIM(B4)) `",
null,
"The first-row value became Number and you can see the sum result changed as well. For the rest of the values, we are using the AutoFill feature.",
null,
"All the values are now Numbers and they can be added.\n\n### 3.Quick Notes\n\n1. You can directly insert the text into the TRIM function.",
null,
"You will find the trimmed text.",
null,
"1. The TRIM function only removes the space character from the text.",
null,
"Here only spaces have been removed. The line breaks remain as it is.",
null,
"## Conclusion\n\nThat’s all for today. We have tried showing how you can use the TRIM function. You can use the function to remove extra spaces from a string as well as several advanced operations like words or spaces count. Hope you will find this helpful.\n\nFeel free to comment if anything seems difficult to understand. Let us know any of your TRIM function-related scenarios where you have stuck, we are ready to help.",
null,
"",
null,
""
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.85839015,"math_prob":0.94293654,"size":6526,"snap":"2021-43-2021-49","text_gpt3_token_len":1467,"char_repetition_ratio":0.16467342,"word_repetition_ratio":0.008818342,"special_character_ratio":0.2195832,"punctuation_ratio":0.10703364,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9508039,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-09T05:39:04Z\",\"WARC-Record-ID\":\"<urn:uuid:b9234a96-3b6d-42be-b603-9b28a480cd0f>\",\"Content-Length\":\"277761\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:baf98216-fef0-4935-b13e-be3a4b6ed431>\",\"WARC-Concurrent-To\":\"<urn:uuid:7c481e41-652b-4a29-a838-3c76c889fc06>\",\"WARC-IP-Address\":\"104.21.92.91\",\"WARC-Target-URI\":\"https://www.exceldemy.com/excel-trim-function/\",\"WARC-Payload-Digest\":\"sha1:LQJQXS7RS7B5E4UUMTBQW4GBO35ZLFPD\",\"WARC-Block-Digest\":\"sha1:4AK2A2B4DEO4RSEFCY4YPNYODI6Z43BE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363659.21_warc_CC-MAIN-20211209030858-20211209060858-00161.warc.gz\"}"} |
https://community.jmp.com/t5/JMP-Wish-List/Summary-CV-make-a-percent-column/idi-p/107532 | [
"BookmarkSubscribe\n\nSummary CV make a percent column\n\nWhen I make a CV Column with Summary\n\nNames default to here(1);\ndt = open(\"\\$SAMPLE_DATA\\Big Class.jmp\");\ndt_sum = dt << Summary(\nGroup( :age ),\nMean( :height ),\nStd Dev( :height ),\nCV( :height ),\nFreq( \"None\" ),\nWeight( \"None\" ),\noutput table name( \"Summary\" )\n);\n\nThe Column is already multiplied by 100 to make it a percent. So if I try to make it a percent format I have to first divide by 100. I think it would make more sense and be more intuitive if it wasn't multiplied by 100 but just formatted to be a percentage. That way the percent sign is there and it seems more intuitituve to me. This is probably a thing that goes deeper than summary, and this may cause problems."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.91087645,"math_prob":0.9396994,"size":677,"snap":"2019-43-2019-47","text_gpt3_token_len":171,"char_repetition_ratio":0.11589896,"word_repetition_ratio":0.0,"special_character_ratio":0.28655836,"punctuation_ratio":0.14184397,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.982275,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-17T06:09:43Z\",\"WARC-Record-ID\":\"<urn:uuid:9c66d717-f2ca-42c2-ae31-d9ef8e7010f8>\",\"Content-Length\":\"94077\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a82adc64-d5b4-4d41-ac26-44dcdd608113>\",\"WARC-Concurrent-To\":\"<urn:uuid:4783481f-28e4-4b8a-b33e-9df9c6851d5a>\",\"WARC-IP-Address\":\"208.74.205.23\",\"WARC-Target-URI\":\"https://community.jmp.com/t5/JMP-Wish-List/Summary-CV-make-a-percent-column/idi-p/107532\",\"WARC-Payload-Digest\":\"sha1:4P7MK5C46NMQ3OLF7ZB2M6CGC27W3D64\",\"WARC-Block-Digest\":\"sha1:EOMVZAQ7NZS7M6MWFQ2WYHOLL5T4QAYQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986672723.50_warc_CC-MAIN-20191017045957-20191017073457-00302.warc.gz\"}"} |
https://www.ijert.org/the-bounds-of-crossing-number-in-complete-bipartite-graphs | [
"# The Bounds of Crossing Number in Complete Bipartite Graphs\n\nText Only Version\n\n#### The Bounds of Crossing Number in Complete Bipartite Graphs\n\nM. Malathi Department of Mathematics Saradha Gangadharan College Pondicherry, India 605 004.\n\nDr. J. Ravi Sankar\n\nSchool of Advanced Sciences VIT University\n\nVellore, India 632 014.\n\nDr. N. Selvi Department of Mathematics\n\nADM College for Women, Nagapattinam, India-611001\n\nAbstractWe compare the lower bound of crossing number of bipartite and complete bipartite graph with Zarankiewicz conjecture and we illustrate the possible upper bound by a\n\nII A MODIFIED ZARANKIEWICZ CONJECTURE:\n\nFor any complete bipartite graphs with n vertices,\n\nmodified Zarankiewicz conjecture.\n\n1 n 2 n n\n\nn n\n\nKeywordscomplete bipartite graphs, crossing numbers\n\nZ (n, n)\n\n2 2\n\n2 2 1 2 2 1\n\n1. INTRODUCTION\n\nLet G = (V,E) be a simple connected graph with vertex set V(G) and edge set E(G).\n\nThe crossing number of a graph G, denoted by Cr(G), is the minimum number of crossings in a drawing of G in the plane[2,3,4].\n\nThe crossing number of the complete bipartite graph was first introduced by Paul Turan, by his brick factory problem.\n\nIn 1954, Zarankiewicz conjectured that,\n\nBy using this conjecture, we can get all the possible number of crossings between every vertices for a given n, without any good drawing D. This reverse way of finding the crossings facilitates for large n by without drawing the graph, we can get all possible crossings between every edges.\n\nThe best known lower bound on general case for all m,n N which was proved by D.J.Kleitman in the following theorem. That is,\n\nTheorem1:\n\nn n 1\n\nm m 1 n n 1\n\ncr(K5,n ) 4 2 2\n\nZ(m,n) = 2\n\n2 2\n\n2\n\ncr(K\n\n) 6 n n 1\n\nWhere m and n are vertices.\n\nLater, Richard Guy shown that the conjecture doesnot holds for all m,n. Then in 1970 D.J.Kleitman proved that\n\n6,n\n\nFrom this he deduced that\n\n2 2\n\nZarankiewicz conjecture holds for Min(m,n) 6.\n\ncr(K\n\n) 1 m(m 1) n n 1\n\nA good drawing of a graph G is a drawing where the edges are non-self-intersecting in which any two edges have atmost one point in common other than end vertex. That is, a crossing is a point of intersection of two edges and no three edges\n\nm,n 5\n\nTheorem2:\n\n2 2\n\nintersect at a common point. So a good drawing is a crossing free drawing by arriving at a planar graph.\n\nThe crossing number is an important measure of the non- planarity of a graph. Therefore this application can be widely\n\nFor m > n, cr Z (m, n) cr Km,n cr Kn,m .\n\nProof:\n\nFrom theorem 1,\n\napplied in all real time problems.\n\ncr(K\n\n) 1 m(m 1) n n 1\n\nm,n 5\n\n2 2\n\nBy definition,\n\nZ(m,n) = m m 1 n n 1\n\ncr Z (5, 4) 2.2.2.1 8\n\n5\n\n5\n\ncr K 1 .5.4.2.1 8\n\n2 2 2 2\n\n5,4\n\nWe can prove the theorem byinduction. Since in\n\ncr K\n\n1 .4.3.2.2 48 9.8\n\ncr Km,n , there are c K subgraphs of K with the\n\nm\n\n5 5,n\n\n5 5,n\n\nm,n\n\n4,5 5 5\n\npartite with n vertices in K5,n . So we shall obtain the lower\n\ncr Z (5, 4) cr K5,4 cr K4,5\n\nbound of cr Km,n for m 5 and n 3. Case(i): Let n=3.\n\nSubcase(ii): m=6,\n\ncr Z (6, 4) 3.2.2.1 12\n\nSubcase(i): m=5,\n\ncr Z (5, 3) 2.2.1.1 4 1\n\ncr K\n\ncr K\n\n6,4\n\n1 .6.5.2.1 12\n\n5\n\n1 .4.3.3.2 72 14.4\n\ncr K\n\n5,3\n\n.5.4.1.1 4\n\n5\n\n4,6 5 5\n\ncr K3,5\n\n1 .3.2.2.2 24 4.8\n\n5 5\n\ncr Z (6, 4) cr K6,4 cr K4,6\n\nSubcase(ii): m=7,\n\ncr Z (5, 3) cr K5,3 cr K3,5\n\ncr Z (7, 4) 3.3.2.1 18\n\nSubcase(ii): m=6,\n\ncr Z (6, 3) 3.2.1.1 6\n\ncr K\n\n7,4\n\n1 .7.6.2.1 84 16.8\n\n5 5\n\n1 108\n\ncr K6,3\n\n1 .6.5.1.1 6\n\ncr K4,7 .4.3.3.3\n\n21.6\n\n5\n\n5\n\n5\n\n5\n\n5 cr Z (7, 4) cr K7,4 cr K4,7\n\n3,6\n\n3,6\n\n1 36\n\ncr K .3.2.3.2 7.2\n\n5 5\n\ncr Z (m, 4) cr K\n\nm,4 cr K4,m\n\ncr Z (6, 3) cr K\n\n6,3 cr K3,6\n\nIn general,\n\ncr Z (m, n) cr K\n\nm,n cr Kn,m .\n\nSubcase(ii): m=7,\n\ncr Z (7, 3) 3.3.1.1 9\n\nWe also observe that the following inequality ,\n\n2cr Z (m, n 1) 2cr Km,n1 2cr Kn1,m\n\ncr K\n\n7,3\n\n1 .7.6.1.1 42 8.4\n\n5 5\n\nAlso holds good for the above cases. Hence the proof.\n\ncr K\n\n3,7\n\n1 .3.2.3.3 54 10.8\n\n5 5\n\nTheorem 2:\n\ncr Z (7, 3) cr K\n\ncr K\n\nFor m = n, cr Z (m, n) cr Km,n .That is,\n\n7,3 3,7\n\n1 n 2 n n n n\n\ncr Z (m, 3) cr Km,3 cr K3,m\n\n2 2 2 2 1 2 2 1\n\nCase(ii): Let n=4.\n\nn 4 1 n n 1\n\nSubcase(i): m=5,\n\n2\n\nProof:\n\nn(n 1)\n\n5 2 2\n\nWhen m = n, Z (n, n) is a complete bipartite graph.\n\nBy theorem 1,\n\ncr K\n\n1 n(n 1) n n 1\n\nm,n 5\n\n2\n\n2\n\nWe shall prove the theorem for large sufficiently larger n and hence deducing the result for subsequent small n.\n\nCase(i):\n\ncr K9,9 256\n\n1 4(4)2 3(4)2 3(4)2 3(4)2 3(4)2\n\n2 4(4)2 4(4)2 4(4)2 4(4)2\n\ncr K11,11 625\n\n1 5(5)2 4(5)2 4(5)2 4(5)2 4(5)2 4(5)2 1 (4)2 5.4 4.3\n\n2\n\n2 5(5)2 5(5)2 5(5)2 5(5)2 5(5)2\n\n1 n 2 n n\n\nn n\n\n1 (5)2 6.5 5.4\n\n2 2\n\n2 2 1 2 2 1\n\n2\n\n3\n\n1 n 2 n n n n\n\n1 n 2 n\n\n1 1\n\n2 2\n\n2\n\n2 2\n\n3\n\n3\n\n1 n\n\n2 2\n\n2 n\n\n2 2\n\nn 4\n\n2\n\n2 2 2\n\nn 4\n\n2\n\ncr Z (9,9) cr K\n\nIn general,\n\n9,9\n\ncr Z (11,11) 550 1 .11.10.5.5\n\n5\n\n1 .11(111) 11 111\n\n5 2 2\n\ncr Z (m, n) cr Km,n\n\nHence the proof.\n\nIII CONCLUSION\n\nWe have given an alternate way of finding crossings in complete bipartite graphs. We also proved in bipartite\n\ngraphs, the best lower bound of cr Km,n will always be a\n\n1 .n(n 1) n n 1\n\nlower bound until m and n are altered.\n\n5 2 2\n\ncr Z (11,11) cr K11,11\n\nCase(ii):\n\nREFERENCES\n\ncr Z (9, 9) 256\n\n1\n\n1 .9.8.4.4\n\n5\n\n9 9 1\n\n1. Daniel J. Kleitman, The crossing number of K5,n, J. Combinatorial Theory 9(1970),315323.\n\n2. D.R. Woodall, Cyclic-order graph and Zarankiewicz's crossing number conjecture, J.Graph Theory. 17 (1994), 657-671.\n\n5 9(9 1) 2 2\n\n1 .n(n 1) n n 1\n\n5 2 2\n\n3. P. Erdos, R.P. Guy, Crossing number problem, American Mathematical Monthly 80 (1973), 52-58.\n\n4. R.K. Guy, The decline and fall of Zarankiewicz's theorem, Proof techniques in Graph Theory, F. Harary, ed. Academic Press, New York(1969),63-69.\n\n5. R.K. Guy and T. Jenkins, The toroidal crossing number of Km;n, J. CombinatorialTheory.6(1969),235-250.\n\n6. N.H. Nahas, On the crossing number of Km;n. Electron. J. Combin, 10:N8,(2003).\n\n7. R. Bruce Richter, Carsten Thomassen, Relations between crossing numbers of com-plete and complete bipartite graphs,The American Mathematical Monthly104 (1997),131137.\n\n8. K. Zarankiewicz, On a problem of P. Turan concerning graphs, Fund. Math. 41 (1954),137-145."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.69330186,"math_prob":0.99396247,"size":6104,"snap":"2020-45-2020-50","text_gpt3_token_len":2536,"char_repetition_ratio":0.119672135,"word_repetition_ratio":0.0518755,"special_character_ratio":0.4125164,"punctuation_ratio":0.19157088,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9864076,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-05T17:59:32Z\",\"WARC-Record-ID\":\"<urn:uuid:1f551ad5-c6c0-41ab-982d-d31ea153ce44>\",\"Content-Length\":\"283042\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:29af72cf-903f-43ce-9788-c0281ccb86ab>\",\"WARC-Concurrent-To\":\"<urn:uuid:1482380a-c923-4ae5-a224-fb3014ec8937>\",\"WARC-IP-Address\":\"149.129.130.163\",\"WARC-Target-URI\":\"https://www.ijert.org/the-bounds-of-crossing-number-in-complete-bipartite-graphs\",\"WARC-Payload-Digest\":\"sha1:GJ5JMOWECZLFCCJ6H7LNFLORDU2SJYRN\",\"WARC-Block-Digest\":\"sha1:ONLPPPHTIKXGI3FL5KQZXPVCIFIINDNU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141748276.94_warc_CC-MAIN-20201205165649-20201205195649-00603.warc.gz\"}"} |
https://www.geeksforgeeks.org/root-to-leaf-path-product-equal-to-a-given-number/?ref=rp | [
"# Root to leaf path product equal to a given number\n\n• Difficulty Level : Medium\n• Last Updated : 13 Sep, 2021\n\nGiven a binary tree and a number, the return is true if the tree has a root-to-leaf path such that the product of all the values along that path equals the given number. The return is false if no such path can be found.",
null,
"Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.\n\nIn case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.\n\nFor example, in the above tree, there exist three roots to leaf paths with the following products.\n\n• 240 –> 10 – 8 – 3\n• 400 –> 10 – 8 – 5\n• 40 –> 10 – 2 – 2\n\nApproach: The idea is to start traversing the tree recursively and divide the current node’s value from the product if it is divisible when recurring down, and check to see if the product is 1 when you reach the leaf node of the current path of the tree.\n\nBelow is the implementation of the above approach:\n\n## C++\n\n `// C++ program to check if there exists``// a root to leaf path product with``// given product`` ` `#include ``using` `namespace` `std;`` ` `// Binary Tree Node``struct` `node {`` ``int` `data;`` ``struct` `node* left;`` ``struct` `node* right;``};`` ` `// Function to check if there exists a path``// with given product`` ` `// Strategy: divide the node value from the product``// if it is divisible when recurring down, and check``// to see if the product is 1 when you reach leaf``// node of the current path out of tree.``bool` `hasPathProduct(``struct` `node* node, ``int` `prod)``{`` ``// return true if we run out`` ``// of tree and prod==1`` ``if` `(node == NULL) {`` ``return` `(prod == 1);`` ``}`` ``else` `{`` ``bool` `ans = 1;`` ` ` ``// Check if product is divisible by`` ``// current node, if not we are on wrong path`` ``if` `(prod % (node->data))`` ``return` `false``;`` ` ` ``// otherwise check both subtrees`` ``int` `subProduct = prod / node->data;`` ` ` ``// If we reach a leaf node and prod`` ``// becomes 1 then return true`` ``if` `(subProduct == 1 && node->left == NULL`` ``&& node->right == NULL)`` ``return` `1;`` ` ` ``if` `(node->left)`` ``ans = hasPathProduct(node->left, subProduct);`` ``if` `(node->right)`` ``ans = hasPathProduct(node->right, subProduct);`` ` ` ``return` `ans;`` ``}``}`` ` `/* UTILITY FUNCTIONS */``// Helper function that allocates``// a new node with the given data``// and NULL left and right pointers``struct` `node* newnode(``int` `data)``{`` ``struct` `node* newNode = ``new` `node();`` ``newNode->data = data;`` ``newNode->left = NULL;`` ``newNode->right = NULL;`` ` ` ``return` `(newNode);``}`` ` `// Driver Code``int` `main()``{`` ``int` `prod = 400;`` ` ` ``/* Constructed binary tree is `` ``10 `` ``/ \\ `` ``8 2 `` ``/ \\ / `` ``3 5 2 `` ``*/`` ` ` ``struct` `node* root = newnode(10);`` ``root->left = newnode(8);`` ``root->right = newnode(2);`` ``root->left->left = newnode(3);`` ``root->left->right = newnode(5);`` ``root->right->left = newnode(2);`` ` ` ``if` `(hasPathProduct(root, prod))`` ``cout << ``\"YES\"``;`` ``else`` ``cout << ``\"NO\"``;`` ` ` ``return` `0;``}`\n\n## Java\n\n `// Java program to check if there exists ``// a root to leaf path product with ``// given product ``class` `GFG``{`` ` ` ``// Binary Tree Node `` ``static` `class` `node `` ``{`` ` ` ``int` `data;`` ``node left;`` ``node right;`` ``};`` ` ` ``// Function to check if there exists a path `` ``// with given product `` ``// Strategy: divide the node value from the product `` ``// if it is divisible when recurring down, and check `` ``// to see if the product is 1 when you reach leaf `` ``// node of the current path out of tree. `` ``static` `boolean` `hasPathProduct(node node, ``int` `prod)`` ``{`` ``// return true if we run out `` ``// of tree and prod==1 `` ``if` `(node == ``null``)`` ``{`` ``return` `(prod == ``1``);`` ``} `` ``else` ` ``{`` ``boolean` `ans = ``true``;`` ` ` ``// Check if product is divisible by `` ``// current node, if not we are on wrong path `` ``if` `(prod % (node.data) == ``1``)`` ``{`` ``return` `false``;`` ``}`` ` ` ``// otherwise check both subtrees `` ``int` `subProduct = prod / node.data;`` ` ` ``// If we reach a leaf node and prod `` ``// becomes 1 then return true `` ``if` `(subProduct == ``1` `&& node.left == ``null`` ``&& node.right == ``null``) `` ``{`` ``return` `true``;`` ``}`` ` ` ``if` `(node.left != ``null``) `` ``{`` ``ans = hasPathProduct(node.left, subProduct);`` ``}`` ``if` `(node.right != ``null``)`` ``{`` ``ans = hasPathProduct(node.right, subProduct);`` ``}`` ` ` ``return` `ans;`` ``}`` ``}`` ` ` ``/* UTILITY FUNCTIONS */`` ``// Helper function that allocates `` ``// a new node with the given data `` ``// and null left and right pointers `` ``static` `node newnode(``int` `data)`` ``{`` ``node node = ``new` `node();`` ``node.data = data;`` ``node.left = ``null``;`` ``node.right = ``null``;`` ` ` ``return` `(node);`` ``}`` ` ` ``// Driver Code `` ``public` `static` `void` `main(String[] args)`` ``{`` ``int` `prod = ``400``;`` ` ` ``/* Constructed binary tree is `` ``10 `` ``/ \\ `` ``8 2 `` ``/ \\ / `` ``3 5 2 `` ``*/`` ``node root = newnode(``10``);`` ``root.left = newnode(``8``);`` ``root.right = newnode(``2``);`` ``root.left.left = newnode(``3``);`` ``root.left.right = newnode(``5``);`` ``root.right.left = newnode(``2``);`` ` ` ``if` `(hasPathProduct(root, prod))`` ``{`` ``System.out.println(``\"Yes\"``);`` ``}`` ``else` ` ``{`` ``System.out.println(``\"No\"``);`` ``}`` ``}``}`` ` `// This code is contributed by Princi Singh`\n\n## Python3\n\n `# Python3 program to check if there exists ``# a root to leaf path product with ``# given product `` ` `\"\"\" UTILITY FUNCTIONS \"\"\"``# Helper function that allocates ``# a new node with the given data ``# and None left and right pointers ``class` `newnode: `` ``def` `__init__(``self``, data): `` ``self``.data ``=` `data `` ``self``.left ``=` `self``.right ``=` `None`` ` `# Function to check if there exists ``# a path with given product `` ` `# Strategy: divide the node value from ``# the product if it is divisible when ``# recurring down, and check to see if ``# the product is 1 when you reach leaf ``# node of the current path out of tree. ``def` `hasPathProduct(node, prod) :`` ` ` ``# return true if we run out `` ``# of tree and prod==1 `` ``if` `(node ``=``=` `None``) :`` ``return` `(prod ``=``=` `1``) `` ` ` ``else` `:`` ` ` ``ans ``=` `1`` ` ` ``# Check if product is divisible by `` ``# current node, if not we are on wrong path `` ``if` `(prod ``%` `(node.data)) :`` ``return` `False`` ` ` ``# otherwise check both subtrees `` ``subProduct ``=` `prod ``/``/` `node.data `` ` ` ``# If we reach a leaf node and prod `` ``# becomes 1 then return true `` ``if` `(subProduct ``=``=` `1` `and` ` ``node.left ``=``=` `None` `and` ` ``node.right ``=``=` `None``) :`` ``return` `1`` ` ` ``if` `(node.left) :`` ``ans ``=` `hasPathProduct(node.left, `` ``subProduct) `` ``if` `(node.right) :`` ``ans ``=` `hasPathProduct(node.right, `` ``subProduct) `` ` ` ``return` `ans `` ` `# Driver Code ``if` `__name__ ``=``=` `'__main__'``:`` ``prod ``=` `400`` ` ` ``\"\"\" Constructed binary tree is `` ``10 `` ``/ \\ `` ``8 2 `` ``/ \\ / `` ``3 5 2 `` ``\"\"\"`` ``root ``=` `newnode(``10``) `` ``root.left ``=` `newnode(``8``) `` ``root.right ``=` `newnode(``2``) `` ``root.left.left ``=` `newnode(``3``) `` ``root.left.right ``=` `newnode(``5``) `` ``root.right.left ``=` `newnode(``2``) `` ` ` ``if` `(hasPathProduct(root, prod)) :`` ``print``(``\"YES\"` `)`` ``else``:`` ``print``(``\"NO\"``)`` ` `# This code is contributed``# by SHUBHAMSINGH10`\n\n## C#\n\n `// C# program to check if there exists ``// a root to leaf path product with ``// given product``using` `System;`` ` `class` `GFG``{`` ` ` ``// Binary Tree Node `` ``public` `class` `node `` ``{`` ` ` ``public` `int` `data;`` ``public` `node left;`` ``public` `node right;`` ``};`` ` ` ``// Function to check if there exists a path `` ``// with given product `` ``// Strategy: divide the node value from the product `` ``// if it is divisible when recurring down, and check `` ``// to see if the product is 1 when you reach leaf `` ``// node of the current path out of tree. `` ``static` `bool` `hasPathProduct(node node, ``int` `prod)`` ``{`` ``// return true if we run out `` ``// of tree and prod==1 `` ``if` `(node == ``null``)`` ``{`` ``return` `(prod == 1);`` ``} `` ``else`` ``{`` ``bool` `ans = ``true``;`` ` ` ``// Check if product is divisible by `` ``// current node, if not we are on wrong path `` ``if` `(prod % (node.data) == 1)`` ``{`` ``return` `false``;`` ``}`` ` ` ``// otherwise check both subtrees `` ``int` `subProduct = prod / node.data;`` ` ` ``// If we reach a leaf node and prod `` ``// becomes 1 then return true `` ``if` `(subProduct == 1 && node.left == ``null`` ``&& node.right == ``null``) `` ``{`` ``return` `true``;`` ``}`` ` ` ``if` `(node.left != ``null``) `` ``{`` ``ans = hasPathProduct(node.left, subProduct);`` ``}`` ``if` `(node.right != ``null``)`` ``{`` ``ans = hasPathProduct(node.right, subProduct);`` ``}`` ` ` ``return` `ans;`` ``}`` ``}`` ` ` ``/* UTILITY FUNCTIONS */`` ``// Helper function that allocates `` ``// a new node with the given data `` ``// and null left and right pointers `` ``static` `node newnode(``int` `data)`` ``{`` ``node node = ``new` `node();`` ``node.data = data;`` ``node.left = ``null``;`` ``node.right = ``null``;`` ` ` ``return` `(node);`` ``}`` ` ` ``// Driver Code `` ``public` `static` `void` `Main(String[] args)`` ``{`` ``int` `prod = 400;`` ` ` ``/* Constructed binary tree is `` ``10 `` ``/ \\ `` ``8 2 `` ``/ \\ / `` ``3 5 2 `` ``*/`` ``node root = newnode(10);`` ``root.left = newnode(8);`` ``root.right = newnode(2);`` ``root.left.left = newnode(3);`` ``root.left.right = newnode(5);`` ``root.right.left = newnode(2);`` ` ` ``if` `(hasPathProduct(root, prod))`` ``{`` ``Console.WriteLine(``\"Yes\"``);`` ``}`` ``else`` ``{`` ``Console.WriteLine(``\"No\"``);`` ``}`` ``}``}`` ` `// This code is contributed by Princi Singh`\n\n## Javascript\n\n ``\n\nOutput\n\n`YES`\n\nTime Complexity : O(n)\nAuxiliary Space: O(n)\n\nMy Personal Notes arrow_drop_up"
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"https://media.geeksforgeeks.org/wp-content/uploads/sum_property_tree1.gif",
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https://admiralmarkets.com.au/products/margin-calculation-examples | [
"We use cookies to give you the best possible experience on our website. By continuing to browse this site, you give consent for cookies to be used. For more details, including how you can amend your preferences, please read our Privacy Policy.\n\n# Margin calculation examples\n\n#### Example 1: Buying a rolling spot FX product\n\nAssuming your account type is Trade.MT4 and its deposit currency is USD, the leverage on major Forex instruments is provided as per table below and calculated as follows:\n\nNotional Position Value, USD Leverage Rate\nUp to 7,500,000 1:500\n7,500,000 — 10,000,000 1:200\n10,000,000 — 12,500,000 1:50\nOver 12,500,000 1:10\n\nLet`s open a position: Buy 10 lots EURUSD at 1.04440.\n\nThe notional position value in the account`s currency (USD) is 10 lots x 100,000 x 1.0444 = 1,044,400 USD, which is less than the first tier of 7,500,000 USD.\n\nTherefore, a leverage of 1:500 is applied to this position and the margin requirements are calculated as 1,044,400 / 500 = 2,088.8 USD.\n\n#### Example 2: Buying a cash index CFD product\n\nAssuming your account type is Trade.MT4 and its deposit currency is USD, the leverage on cash index CFDs is provided as per table below and calculated as follows:\n\nNotional Position Value, USD Leverage Rate\nUp to 500,000 1:500\n500,000 — 3,500,000 1:200\n3,500,000 — 4,700,000 1:50\nOver 4,700,000 1:10\n\nLet`s open a position: Buy 100 lots on [DAX30] at 11,467.88 while the EURUSD rate in MetaTrader 4 is 1.04440.\n\n[DAX30] is quoted in EUR, so the notional position value in the account`s currency (USD) is 100 lots x 11,467.88 x 1.04440 = 1,197,705.39 USD.\n\nThe above value is more than the first tier of 500,000 USD, but less than the second tier of 3,500,000 USD.\n\nTherefore, a leverage of 1:500 is applied to the first 500,000 USD of this position and a leverage of 1:200 is applied to the remainder. So, the margin requirements are calculated as 500,000 / 500 + 697,705.39 / 200 = 4,488.53 USD.\n\n#### Example 3: Selling a metal CFD product and increasing the open position on the same product\n\nAssuming your account type is Trade.MT4 and its deposit currency is GBP, the leverage on metal CFDs is provided according to the table below as shown in the following two examples.\n\nNotional Position Value, GBP Leverage Rate\nUp to 400,000 1:500\n400,000 — 2,500,000 1:200\n2,500,000 — 3,300,000 1:50\nOver 3,300,000 1:10\n##### Example 3.1\n\nLet`s open a position: Sell 25 lots GOLD at 1158.15 while the GBPUSD rate in MetaTrader 4 is 1.22462.\n\nGOLD is quoted in USD, so the notional position value in the account`s currency (GBP) is 25 lots x 100 oz x 1158.15 / 1.22462 = 2,364,304.85 GBP\n\nThe above value is more than the first tier of 400,000 GBP, but less than the second tier of 2,500,000 GBP.\n\nTherefore, a leverage of 1:500 is applied to the first 400,000 GBP of this position and a leverage of 1:200 is applied to the remainder. So, the margin requirements are calculated as 400,000 / 500 + 1,964,304.85 / 200 = 10,621.52 GBP.\n\n##### Example 3.2\n\nLet`s open an additional position on the same instrument, for example: Sell 5 lots GOLD at 1158.15.\n\nGOLD is quoted in USD, so the notional position value in the account`s currency (GBP) is 5 lots x 100 oz x 1158.15 / 1.22462 = 472,860.97 GBP.\n\nThe summary notional value of both positions in the account`s currency (GBP) is 2,364,304.85 + 472,860.97 = 2,837,165.82 GBP, which is more than the second tier of 2,500,000 GBP, but less than the third tier of 3,300,000 GBP.\n\nTherefore, a leverage of 1:500 is applied to the first 400,000 GBP of the summary notional value of both positions, while a leverage of 1:200 is applied to the next 2,100,000 GBP and a leverage of 1:50 is applied to the remainder. So, the margin requirements for both positions are calculated as 400,000 / 500 + 2,100,000 / 200 + 337,165.82 / 50 = 18,043.32 GBP.\n\n#### Example 4: Buying a rolling spot FX product within an hour before the close of the Friday trading session\n\nAssuming your account type is Trade.MT4 and its deposit currency is USD, the margin terms applied to FX majors are the same as shown in Example 1.\n\nAdditionally, there is a term which concerns all highly leveraged products according to which positions opened within an hour of the close of the trading session on Fridays will receive a leverage of 1:50—with the exception of positions that open with a lower leverage (e.g. 1:10).\n\nLet`s open a position: Buy 100 lots USDJPY at 117.311 on Friday 23:35 (EET).\n\nAccording to the Contract Specifications, the trading schedule of USDJPY in EET is 00:05 Mon - 23:59 Fri, therefore our position is subject to the pre-close margining term.\n\nUSDJPY is quoted in USD, so the notional position value in the account`s currency (USD) is 100 lots x 100,000 x 1 = 10,000,000 USD, which is less than the fourth tier of 12,500,000 USD—so there is no part of the position to which a leverage of 1:10 should be applied.\n\nTherefore, a leverage of 1:50 is applied to the entire position, with the margin requirements calculated as 10,000,000 / 50 = 200,000 USD."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8733825,"math_prob":0.9874778,"size":5105,"snap":"2021-04-2021-17","text_gpt3_token_len":1517,"char_repetition_ratio":0.15153891,"word_repetition_ratio":0.30248308,"special_character_ratio":0.365524,"punctuation_ratio":0.17699836,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95919216,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-18T07:56:18Z\",\"WARC-Record-ID\":\"<urn:uuid:c71d010f-93bf-4a3e-9b66-8585743e79c4>\",\"Content-Length\":\"138711\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fd51432e-e19d-492f-9434-c9c9e17aafb7>\",\"WARC-Concurrent-To\":\"<urn:uuid:2500a7f2-123b-4326-ba05-af29d21d8e0f>\",\"WARC-IP-Address\":\"35.176.49.55\",\"WARC-Target-URI\":\"https://admiralmarkets.com.au/products/margin-calculation-examples\",\"WARC-Payload-Digest\":\"sha1:RCNEBN4DA7JDG36YLKXGPVDU7VCT3XPT\",\"WARC-Block-Digest\":\"sha1:NW2CQWQSXBNRAN26HFNH3BW4D6QPDP6K\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703514423.60_warc_CC-MAIN-20210118061434-20210118091434-00778.warc.gz\"}"} |
http://support.sas.com/documentation/cdl/en/statug/63962/HTML/default/statug_surveymeans_a0000000259.htm | [
"Example 88.1 Stratified Cluster Sample Design\n\nConsider the example in the section Stratified Sampling. The study population is a junior high school with a total of 4,000 students in grades 7, 8, and 9. Researchers want to know how much these students spend weekly for ice cream, on the average, and what percentage of students spend at least \\$10 weekly for ice cream.\n\nThe example in the section Stratified Sampling assumes that the sample of students was selected using a stratified simple random sample design. This example shows analysis based on a more complex sample design.\n\nSuppose that every student belongs to a study group and that study groups are formed within each grade level. Each study group contains between two and four students. Table 88.4 shows the total number of study groups for each grade.\n\nTable 88.4 Study Groups and Students by Grade\n\nNumber of Study Groups\n\nNumber of Students\n\n7\n\n608\n\n1,824\n\n8\n\n252\n\n1,025\n\n9\n\n403\n\n1,151\n\nTotal\n\n617\n\n4,000\n\nIt is quicker and more convenient to collect data from students in the same study group than to collect data from students individually. Therefore, this study uses a stratified clustered sample design. The primary sampling units, or clusters, are study groups. The list of all study groups in the school is stratified by grade level. From each grade level, a sample of study groups is randomly selected, and all students in each selected study group are interviewed. The sample consists of eight study groups from the 7th grade, three groups from the 8th grade, and five groups from the 9th grade.\n\nThe SAS data set IceCreamStudy saves the responses of the selected students:\n\n```data IceCreamStudy;\nif (Spending < 10) then Group='less';\nelse Group='more';\ndatalines;\n7 34 7 7 34 7 7 412 4 9 27 14\n7 34 2 9 230 15 9 27 15 7 501 2\n9 230 8 9 230 7 7 501 3 8 59 20\n7 403 4 7 403 11 8 59 13 8 59 17\n8 143 12 8 143 16 8 59 18 9 235 9\n8 143 10 9 312 8 9 235 6 9 235 11\n9 312 10 7 321 6 8 156 19 8 156 14\n7 321 3 7 321 12 7 489 2 7 489 9\n7 78 1 7 78 10 7 489 2 7 156 1\n7 78 6 7 412 6 7 156 2 9 301 8\n;\n```\n\nIn the data set IceCreamStudy, the variable Grade contains a student’s grade. The variable StudyGroup identifies a student’s study group. It is possible for students from different grades to have the same study group number because study groups are sequentially numbered within each grade. The variable Spending contains a student’s response regarding how much he spends per week for ice cream, in dollars. The variable GROUP indicates whether a student spends at least \\$10 weekly for ice cream. It is not necessary to store the data in order of grade and study group.\n\nThe SAS data set StudyGroup is created to provide PROC SURVEYMEANS with the sample design information shown in Table 88.4:\n\n```data StudyGroups;\ndatalines;\n7 608\n8 252\n9 403\n;\n```\n\nThe variable Grade identifies the strata, and the variable _TOTAL_ contains the total number of study groups in each stratum. As discussed in the section Specification of Population Totals and Sampling Rates, the population totals stored in the variable _TOTAL_ should be expressed in terms of the primary sampling units (PSUs), which are study groups in this example. Therefore, the variable _TOTAL_ contains the total number of study groups for each grade, rather than the total number of students.\n\nIn order to obtain unbiased estimates, you create sampling weights by using the following SAS statements:\n\n```data IceCreamStudy;\nset IceCreamStudy;\nWeight=1/Prob;\nrun;\n```\n\nThe sampling weights are the reciprocals of the probabilities of selections. The variable Weight contains the sampling weights. Because the sampling design is clustered and all students from each selected cluster are interviewed, the sampling weights equal the inverse of the cluster (or study group) selection probabilities.\n\nThe following SAS statements perform the analysis for this sample design:\n\n```title1 'Analysis of Ice Cream Spending';\nproc surveymeans data=IceCreamStudy total=StudyGroups;\ncluster StudyGroup;\nvar Spending Group;\nweight Weight;\nrun;\n```\n\nOutput 88.1.1 provides information about the sample design and the input data set. There are three strata in the sample design, and the sample contains 16 clusters and 40 observations. The variable Group has two levels, 'less' and 'more.'\n\nOutput 88.1.1 Data Summary and Class Information\n Analysis of Ice Cream Spending\n\nThe SURVEYMEANS Procedure\n\nData Summary\nNumber of Strata 3\nNumber of Clusters 16\nNumber of Observations 40\nSum of Weights 3162.6\n\nClass Level Information\nClass Variable Levels Values\nGroup 2 less more\n\nOutput 88.1.2 displays information for each stratum. Since the primary sampling units in this design are study groups, the population totals shown in Output 88.1.2 are the total numbers of study groups for each stratum or grade. This differs from Output 88.3, which provides the population totals in terms of students since students were the primary sampling units for that design. Output 88.1.2 also displays the number of clusters for each stratum and analysis variable.\n\nOutput 88.1.2 Stratum Information\nStratum Information\nStratum\nIndex\nGrade Population Total Sampling Rate N Obs Variable Level N Clusters\n1 7 608 1.32% 20 Spending 20 8\nGroup less 17 8\nmore 3 3\n2 8 252 1.19% 9 Spending 9 3\nGroup less 0 0\nmore 9 3\n3 9 403 1.24% 11 Spending 11 5\nGroup less 6 4\nmore 5 4\n\nOutput 88.1.3 displays the estimates of the average weekly ice cream expenditure and the percentage of students spending at least \\$10 weekly for ice cream.\n\nOutput 88.1.3 Statistics\nStatistics\nVariable Level N Mean Std Error of Mean 95% CL for Mean\nSpending 40 8.923860 0.650859 7.51776370 10.3299565\nGroup less 23 0.561437 0.056368 0.43966057 0.6832130\nmore 17 0.438563 0.056368 0.31678698 0.5603394"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8436383,"math_prob":0.82991594,"size":5766,"snap":"2019-26-2019-30","text_gpt3_token_len":1544,"char_repetition_ratio":0.15012148,"word_repetition_ratio":0.032160804,"special_character_ratio":0.30575788,"punctuation_ratio":0.10425717,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95734704,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-17T16:51:34Z\",\"WARC-Record-ID\":\"<urn:uuid:b9242f11-53e7-40a7-8f1f-1a1337d72e07>\",\"Content-Length\":\"28425\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c2175342-5f64-42fb-ac4d-febacc3d41b9>\",\"WARC-Concurrent-To\":\"<urn:uuid:420f2e1b-78ba-4dfb-9276-468c95e17afb>\",\"WARC-IP-Address\":\"149.173.160.38\",\"WARC-Target-URI\":\"http://support.sas.com/documentation/cdl/en/statug/63962/HTML/default/statug_surveymeans_a0000000259.htm\",\"WARC-Payload-Digest\":\"sha1:PD55UIN6EL6FGC24PN2FW6KNZRDZTVVX\",\"WARC-Block-Digest\":\"sha1:C63ZKXNESJRMKRFSSC5USH2UCODX45FY\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998513.14_warc_CC-MAIN-20190617163111-20190617185111-00078.warc.gz\"}"} |
https://huaweicloud.csdn.net/63806accdacf622b8df87753.html | [
"## Python调用C语言\n\nWindows下Demo亲测一波，话不多说，上图上代码。\nfoo.c:\n\n``````# include<stdio.h>\n\nint foo(){\nint i,k,m;\nfor(i=0;i<1000;i++){\nfor(k=0;k<1000;k++){\nfor(m=0;m<1000;m++)\t{\n}\n}\n}\nreturn 0;\n}\n``````\n\npython代码\n\n``````from ctypes import cdll\nfrom time import time\n\n#.c-->.so\nstart=time()\ndll.foo()\nprint(f'c,so:{time()-start}')\n\n#.c-->.dll\nstart=time()\ndll.foo()\nprint(f'c,dll:{time()-start}')\n\n#Python\nstart=time()\nfor i in range(1000):\nfor k in range(1000):\nfor m in range(1000):\npass\nprint(f'python:{time()-start}')\n``````",
null,
"1、提高性能，弥补了python性能上的不足。\n2、相对安全，对dll文件或so文件进行加密，有效防止逆解析保证代码安全\n3、结合Python强大的生态环境，也满足了高效快速的开发。",
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| {"ft_lang_label":"__label__zh","ft_lang_prob":0.5789481,"math_prob":0.9821537,"size":1242,"snap":"2023-40-2023-50","text_gpt3_token_len":714,"char_repetition_ratio":0.09854604,"word_repetition_ratio":0.0,"special_character_ratio":0.32930756,"punctuation_ratio":0.22304833,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97179574,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-22T17:15:17Z\",\"WARC-Record-ID\":\"<urn:uuid:58606df6-1680-4c60-a132-8d0eb00062b4>\",\"Content-Length\":\"101260\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:64abbfbe-3610-4070-8a6b-36f00c9d6322>\",\"WARC-Concurrent-To\":\"<urn:uuid:59ae2c9a-c2a0-462a-9bc2-4e00f2e95bc4>\",\"WARC-IP-Address\":\"121.36.68.44\",\"WARC-Target-URI\":\"https://huaweicloud.csdn.net/63806accdacf622b8df87753.html\",\"WARC-Payload-Digest\":\"sha1:LZD5BQXLPRS6HSJRR4Z3K53KXENJQ3LX\",\"WARC-Block-Digest\":\"sha1:WPMQJITQPZYAORUIRXDIXTIKXXCAJZLG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506421.14_warc_CC-MAIN-20230922170343-20230922200343-00379.warc.gz\"}"} |
https://social.microsoft.com/Forums/en-US/963b271e-103e-4510-a5cf-66317a55466a/how-to-index?forum=Offtopic | [
"#",
null,
"How to index?",
null,
"• ### Question\n\n•",
null,
"Hello,\n\nI am trying to index a cumulative moving average so i can create more than one cma in my chart.\n\nAs you can see in the chart the blue line is the result of the code and the yellow is what it should look like. The blue lines all start at the 4th candle when actually it should create a cma that starts at every candle and cumulate the period from left to right. The system refuse to let me insert an image. sorry\n\n```if (CurrentBar < (ChartBars.ToIndex-Period))\nreturn;\n// test if is there enough bars in the chart.\n\nint count = 0;\ndouble sum01 = 0;\ndouble sum02 = 0;\ndouble cma = 0;\n\nPrint(\"***** Bar \" + CurrentBar + \" CMA Values: *****\");\n\nfor (int i = 0; i <= Period-1; i++)\n{\n// return periods to index from bar 4 to 1, the opposite is also possible (int i = Period-1; i >= 0; i--)\n\ndouble privol =\tVolume[i] * Input[i];\n// input returns all prices in the series and [i] is the indexation\n\nsum01 += privol;\nsum02 += Volume[i];\n\ncma = sum01 / sum02;\n\nValues[count] = cma;\n// Values return the plots to create the object\n// int count = 0;\nfor (int i = 0; i <= Period-1; i++)\n{\nAddPlot(Brushes.Aqua, \"MyPlot\"+ count);\ncount++;\n} //\n\nPrint(\"Bar \" + count + \" of Period CMA value: \" + cma);\n\ncount++;\n\nPrint(\"i\"+i);\nPrint(\"count\" +count);\n\n// this part list the result of the cma//\n\nfor(int n = 0; n < Values.Count(); n++)\n{\ndouble value = Values[n];\n\nPrint(value);\n}\n\n}\n```\n\nthanks\n\nWednesday, January 8, 2020 2:31 PM\n\n### All replies\n\n•",
null,
"Ninjatrader related, its a charting software.\nWednesday, January 8, 2020 3:27 PM\n•",
null,
"Hi,\nThis forum is discussing and asking questions about the Windows Forms.\nIf your question is related to the winform, please provide more details about it.\nIf it is related to Ninjatrader that you said, please contact Ninjatrader developer.\nBest Regards,\nDaniel Zhang\n\nMSDN Community Support\nPlease remember to click \"Mark as Answer\" the responses that resolved your issue, and to click \"Unmark as Answer\" if not. This can be beneficial to other community members reading this thread. If you have any compliments or complaints to MSDN Support, feel free to contact [email protected].\n\nThursday, January 9, 2020 6:00 AM"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7342164,"math_prob":0.8900143,"size":1370,"snap":"2021-04-2021-17","text_gpt3_token_len":400,"char_repetition_ratio":0.11566618,"word_repetition_ratio":0.04597701,"special_character_ratio":0.34525546,"punctuation_ratio":0.14590748,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9532781,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-12T23:50:51Z\",\"WARC-Record-ID\":\"<urn:uuid:551f86fc-179b-4f37-9862-62e678e03498>\",\"Content-Length\":\"47379\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f4f02088-065b-45ef-a0be-300c1e094ef3>\",\"WARC-Concurrent-To\":\"<urn:uuid:3d270838-1811-4623-985e-54e80bb547b7>\",\"WARC-IP-Address\":\"23.208.48.203\",\"WARC-Target-URI\":\"https://social.microsoft.com/Forums/en-US/963b271e-103e-4510-a5cf-66317a55466a/how-to-index?forum=Offtopic\",\"WARC-Payload-Digest\":\"sha1:LHKCA4VIS7G7EFVKWUJ37LD36M77WBNI\",\"WARC-Block-Digest\":\"sha1:3TCPYGP5L5DGZBC6X7H5FWQKBGVSLOVB\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038069267.22_warc_CC-MAIN-20210412210312-20210413000312-00012.warc.gz\"}"} |
https://help.altair.com/hwsolvers/ms/topics/solvers/ms/curve_api.htm | [
"# Curve\n\nModel ElementCurve defines a parametric curve in 3D space.\n\nCurve\n\n## Description\n\nA parametric curve has one free parameter, u, and three associated functions f(u), g(u) and h(u). For every value, u=u*, f(u*), g(u*), and h(u*) represent the coordinates of a point in space. As u is swept from its minimum to maximum value, a curve in 3D space is generated. For additional information on the definition of Curve, please see the Description and the Comments section of Reference_ParamCurve.\n\n## Attribute Summary\n\nName Property Modifiable by command? Designable?\nid Int ()\nlabel Str ()\nmatrix Reference (\"Matrix\") Yes FD Only\nxyx Nodes (None, count=0)\ncurve_points Bool () Yes\nclosed Bool (False) Yes\nminpar Double (-1.0)\nmaxpar Double (1.0)\nfunction Function (\"CURSUB\") FD Only\nroutine Routine () or Str () FD Only\nscript Script ()\n\n## Usage\n\nCurves are available in the following forms:\n#1: Curve data provided in a Matrix\nCurve(matrix=objMatrix, optional_attributes)\n\n#2. Curve data provided as xyz points\nCurve(xyz=xyz_points, optional_attributes)\n\n#3. Curve specified in a compiled user-written subroutine\nCurve(function=userString, routine=string, optional_attributes)\n\n#4. Curve specified in a Python function\nCurve(function=userString, routine=functionPointer, curve_points=boolean, optional_attributes)\n\n## Attributes\n\nCurve data provided in a Matrix.\nmatrix\nReference to an existing matrix object.\nSpecifies the Matrix that contains the curve data. The matrix should contain only the x-, y-, and z- data for the individual curve points. For example: [x1,y1,z1,x2,y2,z2,…].\nNote: For a closed curve, the first and last data points must be the same.\nCurve data provided as xyz points.\nxyz\nNodes\nList of [x,y,z] points used by the curve. For example: [[x1,y1,z1],[x2,y2,z2],…]\nNote: For a closed curve, the first and last data points must be the same.\nCurve specified in a compiled user-written subroutine.\nfunction\nString\nThe list of parameters that are passed from the data file to the user-defined subroutine.\nroutine\nString\nSpecifies an alternative name for the user subroutine. The name consists of two pieces of information, separated by \"::\". The first is the pathname to the shared library containing the function that computes the response of the user-defined Surface. The second is the name of the function in the shared library that does the computation.\nAn example is: routine=\"/staff/Altair/engine.dll::myCurve\"\n• \"/staff/Altair/engine.dll is the DLL.\n• \"myCurve\" is the function within this DLL that performs the calculations.\nThe attribute routine is optional.\nWhen not specified, routine defaults to CURSUB.\nminpar\nDouble\nSpecifies the minimum value of the curve parameter.\nThe minpar attribute is optional. When not specified, it defaults to -1.\nNote: When specified, minpar < maxpar.\nmaxpar\nDouble\nSpecifies the maximum value of the curve parameter.\nThe maxpar attribute is optional. When not specified, it defaults +1.\nNote: When specified, minpar < maxpar.\nCurve specified in a Python function.\nfunction\nString\nThe list of parameters that are passed from the data file to the user-defined function.\nroutine\nCallable function in Python or string.\nAn example is: routine=myCurve or routine=”myCurve”.\n• If script is not used, then myCurve is a Python function or method defined within the namespace.\n• If script is used, then “myCurve” is a string that names the function in an external written script.\nThe attribute routine is optional.\nWhen not specified, routine defaults to CURSUB.\nscript\nSpecifies the path and name of the user-written script that contains the function specified by routine.\nThe attribute script is optional.\nWhen not specified, MotionSolve looks for a function within the namespace.\nminpar\nDouble\nSpecifies the minimum value of the curve parameter.\nThe minpar attribute is optional. When not specified, it defaults to -1.\nNote: When specified, minpar < maxpar.\nmaxpar\nDouble\nSpecifies the maximum value of the curve parameter.\nThe maxpar attribute is optional. When not specified, it defaults +1.\nNote: When specified, minpar < maxpar.\nOptional attributes: Available to all description methods.\nid\nInteger\nSpecifies the element identification number. This number must be unique among all the Surface objects in the model.\nThis attribute is optional. MotionSolve automatically creates an ID when one is not specified.\nRange of values: id > 0\nlabel\nString\nSpecifies the name of the Surface object.\nThis attribute is optional. When not specified, MotionSolve will create a label for you.\nclosed\nString\nSelect from \"OPEN\" or \"CLOSED\".\nNote: For a closed curve, the first and last data points must be the same.\nThe type attribute is mandatory.\ncurve_points\nBoolean\nSelect from TRUE or FALSE.\n• TRUE means that the curve passes through the points.\n• FALSE means that the curve (B-spline) stays close to the points, but does not pass through them in general.\nWhen not specified, curve_points defaults to False.\n\n## Example\n\nA cycloid is defined in parametric space as:\nx = a * (t - sin(t))\ny = a * (1-cos(t))\n\"t\" is the independent parameter. The cycloid is shown in the image below.\n1. Define a parametric curve using a matrix.\nThe statement defining the cycloid described above has the following curve attributes:\n• The Curve is open\n• The independent parameter t starts at zero\n• The independent parameter t ends at 5\n• The curve points are specified in the Matrix object mat23\nThe corresponding MotionSolve Python specification is:\n# Note, since curve_points=True, MINPAR=-1 and MAXPAR=+1.\n# This range in u will accommodate all the data provided in the matrix.\ncurve1 = Curve (type=\"OPEN\", curve_points=True, matrix=mat23)\n2. Define a parametric curve in a python function.\nThis example shows how to specify exactly the same for implementation in a user-defined subroutine written in Python:\n# The cycloid function saved in file: cycloid.py\nfrom msolve import sin, cos\ndef cycloid (id, par, npar, t, iord, iflag):\na = par\nif iord == 0:\nx = a * (t - sin(t))\ny = a * (1 - cos(t))\nz = 0.0\nelif iord == 1:\nx = a * (1 - cos(t))\ny = a * sin(t)\nz = 0.0\nelse:\nx = a * sin(t)\ny = a * cos(t)\nz = 0.0\n\nreturn [x, y, z]\nThe corresponding MotionSolve Python specification is:\ncurve2 = Curve(function=\"USER(1)\", routine=cycloid, type=\"OPEN\", minpar=0, maxpar=6)"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6465643,"math_prob":0.893621,"size":6232,"snap":"2023-40-2023-50","text_gpt3_token_len":1561,"char_repetition_ratio":0.15382145,"word_repetition_ratio":0.22758621,"special_character_ratio":0.25369063,"punctuation_ratio":0.15325342,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9862402,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-21T11:20:57Z\",\"WARC-Record-ID\":\"<urn:uuid:8845fda2-4de5-4c4f-ae1e-fb8e17a94595>\",\"Content-Length\":\"136338\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0082403b-3668-417e-a361-4ccc66109135>\",\"WARC-Concurrent-To\":\"<urn:uuid:019e1912-b722-45d0-b7ed-66da6b4cc5be>\",\"WARC-IP-Address\":\"173.225.177.121\",\"WARC-Target-URI\":\"https://help.altair.com/hwsolvers/ms/topics/solvers/ms/curve_api.htm\",\"WARC-Payload-Digest\":\"sha1:BMSVTGVWGSGE2QCR5QDWK3T2F53WB5G5\",\"WARC-Block-Digest\":\"sha1:BV7JZC7TW64DDIIGN2SBGO6ZDMN3S3JL\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506027.39_warc_CC-MAIN-20230921105806-20230921135806-00505.warc.gz\"}"} |
https://alex.state.al.us/cr_view.php?res_id=3642&res_id=3642&res_type=CR | [
"# ALEX Classroom Resource\n\n## Algebra I Module 2, Topic C: Categorical Data on Two Variables\n\nClassroom Resource Information\n\nTitle:\n\nAlgebra I Module 2, Topic C: Categorical Data on Two Variables\n\nURL:\n\nhttps://www.engageny.org/resource/algebra-i-module-2-topic-c-overview\n\nContent Source:\n\nEngageNY\nType: Lesson/Unit Plan\n\nOverview:\n\nIn Module 2, Topic C, students reconnect with previous work in Grade 8 involving categorical data. Students use a two-way frequency table to organize data on two categorical variables. Students calculate the conditional relative frequencies from the frequency table. They explore a possible association between two categorical variables using differences in conditional, relative frequencies. Students also come to understand the distinction between association between two categorical variables and a causal relationship between two variables. This provides a foundation for work on sampling and inference in later grades.\n\nContent Standard(s):\n Mathematics MA2015 (2016) Grade: 9-12 Algebra I 44 ) Summarize categorical data for two categories in two-way frequency tables. Interpret relative frequencies in the context of the data (including joint, marginal, and conditional relative frequencies). Recognize possible associations and trends in the data. [S-ID5] Alabama Alternate Achievement Standards AAS Standard: M.AAS.SP.HS.44- Calculate the mean of a given data set (number of data points limited to fewer than five, values of less than 10).\nTags: associations, categorical data, causation, correlation, relative frequencies, trends, twoway frequency tables"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.89687264,"math_prob":0.8100428,"size":831,"snap":"2022-27-2022-33","text_gpt3_token_len":141,"char_repetition_ratio":0.11729141,"word_repetition_ratio":0.016949153,"special_character_ratio":0.16847172,"punctuation_ratio":0.11510792,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9682669,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-19T11:37:13Z\",\"WARC-Record-ID\":\"<urn:uuid:5e57e187-99a5-417f-be7c-50191163a9ed>\",\"Content-Length\":\"16461\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7ea8a7d2-cea2-4a6c-b5db-6538b6212f8f>\",\"WARC-Concurrent-To\":\"<urn:uuid:488fc561-bfe7-4cf0-afc6-decb2e768583>\",\"WARC-IP-Address\":\"207.157.1.17\",\"WARC-Target-URI\":\"https://alex.state.al.us/cr_view.php?res_id=3642&res_id=3642&res_type=CR\",\"WARC-Payload-Digest\":\"sha1:TK2YR2ZQFRIUQ6NCUPAH6YYDIOUY3WHJ\",\"WARC-Block-Digest\":\"sha1:P3YGBBLJHUG3NK5HQUML2HUZMLBTOKHY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573667.83_warc_CC-MAIN-20220819100644-20220819130644-00132.warc.gz\"}"} |
https://www.onlinestudy.xyz/2019/08/conditional-statements-in-programming.html | [
"## Conditional Statements In Programming Languages\n\nIn every programming language to specify the order of execution of particular statements we need to make decisions on the basis of the nature of some particular expressions that if they are true(1) or false(0).\nTo do so we have following conditional statements: -\nIf statement\nIf-else statement\nNested if-else statement\nSwitch statement\nThe above listed statements have different scopes and manner of usage which we shall discuss below in detail.\nThese statements work till the condition is true and when the condition is false then the compiler simply skips the statements and passes the command to the statement written outside these conditional statements.\n\n### If statement: -\n\nThe single if statement is used when the given statement is true. And if the statement is false then the compiler skips the statement under ‘if’ condition.\n\n#### Syntax: -\n\nif(expression)\n{\nStatements //codes to be executed\n}\n\n### Working: -\n\nIf the expression is true then the statements written within parenthesis are executed.\nIf the expression is false then the statements outside the parenthesis are executed.\n\n### Example for if statement: -\n\n#include<stdio.h>\n#include<conio.h>\nvoid main()\n{\nInt a;\nclrscr();\nprintf(“enter the number::”);\nscanf(“%d”,&a);\nif(a%2==0)\n{\nprintf(“%d is even”,a);\n}\ngetch();\n}\n\n### Output:\n\nenter the number::8\n8 is even\n\n### If-else statement: -\n\nThe if-else statement is used when there are possibilities of the condition of getting true or false. We may also denote this as two way selection statement.\n\n### Syntax: -\n\nifexpression)\n{\nStatements //codes to be executed if expression is true\n}\nelse\n{\nStatements //codes to be executed if expression is false\n}\n\n### Working: -\n\nIf the expression is true then the if block will be executed.\nIf the expression is false then the control will be passed to else block and it will be executed.\n\n### Example for if-else statement: -\n\n#include<stdio.h>\n#include<conio.h>\nvoid main()\n{\nint a;\nclrscr();\nprintf(“enter a number::”);\nscanf(“%d”,&a);\nif(a<0)\n{\nprintf(“number is negative”);\n}\nelse\n{\nprintf(“number is positive”);\n}\ngetch();\n}\n\n### Output:\n\nenter a number::67\nnumber is positive\n\n### Nested if-else statement: -\n\nThe nested form of if-else statement is used when there are more than one expression/ condition to be checked and operated. This statement could also be denoted as multi-way selection statement.\n\n### Syntax: -\n\nif(expression 1)\n{\nif(expression 2) //if expression 1 is true\n{\nStatements //codes to be executed if expression 2 is true\n}\nelse\n{\nStatements //codes to be executed if expression 2 is false\n}\n}\nelse\n{\n\nStatements //codes to be executed if expression 1 is false\n}\n\n### Working: -\n\nIf the condition is true then the control enters the if block and executes the sub conditions under it (further execution will be same as of if-else statement).\nIf the condition is false then the compiler executes the else block.\n\n### Example for nested if-else statement: -\n\n#include<stdio.h>\n#include<conio.h>\nvoid main()\n{\nint a,b,c;\nclrscr();\nprintf(“enter 3 numbers::\\n”);\nscanf(“%d%d%d”,&a,&b,&c);\nif(a>b)\n{\nIf(a>c)\n{\nprintf(“a is greatest”);\n}\nelse\n{\nprintf(“c is greatest”);\n}\n}\nelse\n{\nIf(b>c)\n{\nprintf(“b is greatest”);\n}\nelse\n{\nprintf(“c is greatest”);\n}\n}\ngetch();\n}\n\n### Output:\n\nenter 3 numbers::\n6\n5\n7\nc is greatest\n\nA chain of if-else-if statement where each if statement is associated with an else-if condition is used when we have to execute one code from multiple conditions.\nThe last statement is always an else statement.\n\n### Syntax: -\n\nif(expression 1)\n{\nStatements //codes to be executed if expression 1 is true\n}\nelse if(expression 2)\n{\nstatements //codes to be executed if expression 2 is true\n}\nelse\n{\nStatements //codes to be executed if all the expressions are false\n}\n\n### Working: -\n\nif the expression is true then the if block will be executed.\nIf the expression is false then the control passes to else if statement and if the expression (within else-if) is true then the respective block will be executed and so on.\nIf the expression (within else-if) is false then the compiler will execute the else block\n\n#### Note: -\n\nThe compiler will check all the expressions in each and every else-if block inspite of their numbers.\n\n### Example for if-else-if ladder: -\n\n#include<stdio.h>\n#include<conio.h>\nvoid main()\n{\nint a;\nclrscr();\nprintf(“enter marks::”);\nscanf(“%d”,&a);\nif(a>=90)\n{\n}\nelse if((a>=70)&&(a<90))\n{\n}\nelse if((a>=50)&&(a<70))\n{\n}\nelse\n{\nprintf(“Fail!!”);\n}\ngetch();\n}\n\nenter marks::59\n\n### Switch statement: -\n\nSometimes our source code become complex due to long if-else-if ladders and hence reduces the effectiveness of the program.\nTo solve this problem C-programming language provide us the concept of switch statements.\nA switch statement is the substitute to the if-else-if ladder.\n\n### Syntax:-\n\nswitch(expression)\n{\ncase 1:\n{\nStatements //codes to be executed if expression satisfies case 1\nbreak;\n}\ncase 2:\n{\nStatements //codes to be executed if expression satisfies case 2\nbreak;\n}\ncase 3:\n{\nStatements //codes to be executed if expression satisfies case 3\nbreak;\n}\ndefault\n{\nStatements //codes to be executed if expression doesn’t satisfy any case\n}\n}\n\n### Working: -\n\nAccording to the expression the respective case statements are executed and at the end of each case statement there is a break statement which breaks the loop and passes the control out of the switch statement.\nIf any of the case expression is not true then the compiler will execute the default statement.\n\n#### Note: -\n\nThere should be a colon (:) following the case label.\nCase label should be constants.\nThere must be a default statement.\nBreak statement is optional.\n\n### Example for switch statement: -\n\n#include<stdio.h>\n#include<conio.h>\nvoid main()\n{\nchar ch;\nint a,b,c;\nprintf(“enter the numbers\\n”);\nscanf(“%d%d”,&a,&b);\nprintf(“enter the operation to be performed\\n”);\nscanf(“ %c”,&ch);\nswitch(ch)\n{\ncase '+':\nc=a+b;\nbreak;\ncase '-':\nc=a-b;\nprintf(“sub=%d”,c);\nbreak;\ncase '*':\nc=a*b;\nprintf(“mul=%d”,c);\nbreak;\ncase '/':\nc=a/b;\nprintf(“div=%d”,c);\nbreak;\ncase '%':\nc=a%b;\nprintf(“mod=%d”,c);\nbreak;\ndefault:\nprintf(“invalid selection”);\n}\ngetch();\n}\n\n### Output: -\n\nenter the numbers::\n2\n3\nenter the operation to be performed::\n+"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.66328967,"math_prob":0.9138831,"size":8723,"snap":"2021-43-2021-49","text_gpt3_token_len":2375,"char_repetition_ratio":0.21596514,"word_repetition_ratio":0.54024655,"special_character_ratio":0.2972601,"punctuation_ratio":0.1668552,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9634418,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-29T09:27:44Z\",\"WARC-Record-ID\":\"<urn:uuid:ec634b86-5f0b-4523-a466-95e1756446a9>\",\"Content-Length\":\"263576\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9b9812e0-75c3-4428-b379-171b0bf97c04>\",\"WARC-Concurrent-To\":\"<urn:uuid:19a0e065-37e1-4e70-aae7-bb9c11eb0775>\",\"WARC-IP-Address\":\"142.250.188.51\",\"WARC-Target-URI\":\"https://www.onlinestudy.xyz/2019/08/conditional-statements-in-programming.html\",\"WARC-Payload-Digest\":\"sha1:FVZC3NQBTQFIV6YXSWEUD6GSF2X2MWLC\",\"WARC-Block-Digest\":\"sha1:24UN467LMFF3BK3PE2NAG5N7AAWPCHNH\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358702.43_warc_CC-MAIN-20211129074202-20211129104202-00089.warc.gz\"}"} |
https://www.wyzant.com/resources/answers/229213/find_the_function_f_that_satisfies_the_following_differential_equations | [
"Setayesh A.\n\nfind the function F that satisfies the following differential equations.\n\nf'''(x)=4x, f''(0)=0, f'(0)=1, f(0)=3\n\nBy:",
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"Tutor\n3 (1)\n\nMath major grad: Can tutor various math courses and computer science\n\nStill looking for help? Get the right answer, fast.\n\nGet a free answer to a quick problem.\nMost questions answered within 4 hours.\n\nOR\n\nChoose an expert and meet online. No packages or subscriptions, pay only for the time you need."
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null,
"https://www.wyzant.com/resources/answers/229213/find_the_function_f_that_satisfies_the_following_differential_equations",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7602228,"math_prob":0.8842432,"size":605,"snap":"2022-05-2022-21","text_gpt3_token_len":257,"char_repetition_ratio":0.10981697,"word_repetition_ratio":0.0,"special_character_ratio":0.4231405,"punctuation_ratio":0.012903226,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95818627,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-22T09:33:07Z\",\"WARC-Record-ID\":\"<urn:uuid:6770f97e-d434-4d4a-8cbb-0b363cb7a1a4>\",\"Content-Length\":\"66717\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:97bd06a1-96de-43d3-8043-934cb56b9b99>\",\"WARC-Concurrent-To\":\"<urn:uuid:d1364601-17fb-41aa-898d-ed655174c35b>\",\"WARC-IP-Address\":\"34.117.195.90\",\"WARC-Target-URI\":\"https://www.wyzant.com/resources/answers/229213/find_the_function_f_that_satisfies_the_following_differential_equations\",\"WARC-Payload-Digest\":\"sha1:H4HXX3GRCLZTG6PG5EM7R6WRQ34WPMSL\",\"WARC-Block-Digest\":\"sha1:GHAKRH6HSL4GXPOKTMLGJ7TFQA4TKPVL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320303779.65_warc_CC-MAIN-20220122073422-20220122103422-00335.warc.gz\"}"} |
https://javatutoring.com/armstrong-number-in-java/ | [
"# Armstrong Number In Java Program – 5 Simple Ways\n\nArmstrong number in Java. Here we have written the code in four different ways standard, using for loop, recursion, while loop and also with different examples as like: between 100 and 999, between 1 to 1000 and between 1 to 500 with sample outputs and online execution tool embedded.\n\nWhat is Armstrong Number?\n\nA: An Armstrong number of three digits is an integer, where the sum of the cubes of its digits is equal to the number itself.\n\n[table id=16 /]\n\nHere is the example for you :\n\nConsider the example: 371\n\nA: 3^3 + 7^3 + 1^3 = 371 ( If you add those all numbers, the final digit should be same as given number ).\n\nHere is the list of Java programs with different methods in different ways. Do check out the table of contents so that you will get an idea.",
null,
"1. Using static method Between 100 and 999\n\nHere is the first sample program using the static method with sample output as well. Here we used the method ‘ Static ‘ and taken numbers between 100 to 999. Once you are done with the execution, it automatically displays the Armstrong numbers between 100 and 999. Check out the sample output.\n\nOutput:\n\n2. Using while Loop ( 1 to 500 )\n\nThere you go another method using while loop. Java, while loop is nothing but executing a set of statements repeatedly as long as condition, is true – here is the complete guide on while loop in Java with examples. In this case, we have taken the example of numbers from 1 to 500. Upon execution, you will get to know what are Armstrong numbers. Check out the output below so that you will get an idea.\n\nOutput:\n\n3. Using Recursion ( Between 1 to 1000 )\n\nAnother method using recursion: A function that calls itself is called recursion. Here is the complete code with the sample program and the output of the program. Here we have taken the example of numbers from 1 to 1000. Check it out.\n\nOutput:\n\nThere you another method using bufferedreader. Do check it out.\n\nOutput:\n\n5. Using for loop ( 100 to 500 )\n\nAnother sample program to print Armstrong number using for loop. Check out the program along with the sample output.\n\nThat’s it. If you have any doubts related to the above code. Just leave a comment here at the end of the post. We do try to help you out related to ant query asap.\n\nMore programs for you:\n\nx\n\n## Rhombus Star Pattern Program In Java – Patterns\n\nJava program to print Rhombus star pattern program. We have written the below print/draw Rhombus ..."
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"https://javatutoring.com/wp-content/uploads/2017/04/Armstrong-Number-In-Java-1024x576.jpg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6649627,"math_prob":0.940813,"size":4761,"snap":"2022-40-2023-06","text_gpt3_token_len":1467,"char_repetition_ratio":0.14820264,"word_repetition_ratio":0.2606635,"special_character_ratio":0.36189875,"punctuation_ratio":0.15173675,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9872777,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-07T12:30:20Z\",\"WARC-Record-ID\":\"<urn:uuid:302e814e-ceff-4f87-862e-e90afe7fb093>\",\"Content-Length\":\"175825\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7c24c71a-1043-490b-955f-c06ed3b5ec38>\",\"WARC-Concurrent-To\":\"<urn:uuid:e0d9c956-a16c-4b8b-8304-f29e83de0305>\",\"WARC-IP-Address\":\"139.59.82.24\",\"WARC-Target-URI\":\"https://javatutoring.com/armstrong-number-in-java/\",\"WARC-Payload-Digest\":\"sha1:4CP4DCEBWIT7MQTUNDDVXQVGBNLVX7DB\",\"WARC-Block-Digest\":\"sha1:R3AZHEKYPU3FCDE2QAYZ3FJISTCYASHH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500456.61_warc_CC-MAIN-20230207102930-20230207132930-00593.warc.gz\"}"} |
https://www.turito.com/learn/math/quadratic-functions-in-standard-form-grade-10 | [
"#### Need Help?\n\nGet in touch with us\n\nThe component learnSearchBar has not been created yet.\n\n# Quadratic Functions in Standard Form\n\n## Key Concepts\n\n• Define the standard form of a quadratic function.\n• Find the Y-intercept of a quadratic function.\n• Find the axis of symmetry of a quadratic function.\n• Find the vertex of a quadratic function.\n\nA function f defined by f(x) = ax2+bx+c, where a, b and c are real numbers and a≠0, is called a quadratic function\n\n• The graph of a quadratic function is a curve called a parabola\n• The quadratic parent function is f(x) = x2\n• For 0<|a|<1, the shape of the parabola is wider than the parent function.\n• For |a|>1, the shape of the parabola is narrower than the parent function.\n• f(x) = ax2 is the reflection of f(x) = −ax2 over the x-axis.\n\n### Vertex form of the quadratic function\n\n• The function f(x) = a(x−h)2+k, where a≠0 is called the vertex form of the quadratic function\n• The vertex of the graph g is (h, k).\n• The graph of f(x) = ax−h2+k is a translation of the function f(x) = ax2 that is translated h units horizontally and k units vertically.\n• The value of a does not affect the location of the vertex.\n\n### Graph of g(x) = x2+k\n\nThe value of k in g(x) = x2+k translates the graph of parent function f, vertically k units. The value of k does not affect the axis of symmetry.\n\n### Graph of g(x) = (x−h)2\n\n• The value of h in g(x) = (x−h)2 translates the graph of parent function f, horizontally h units.\n• The vertex of the graph g is (0, h).\n• The value of h translates the axis of symmetry.\n\n### A standard form of a quadratic function\n\n• The standard form of a quadratic function is ax2+bx+c=0, where a≠0\n• The axis of symmetry of a standard form of quadratic function f(x) = ax2+bx+c is the line x=−b/2a.\n• The y-intercept of f(x) is c.\n• The x-coordinate of the graph of f(x) = ax2+bx+c is –b/2a.\n\nThe vertex of f(x) = ax2+bx+c is (-b/2a, f(-b/2a))\n\n## Exercise\n\n• Find the intercept, x intercept, axis of symmetry, and vertex of the graph of f(x) = 3x2 + 6x + 3\n• Graph g(x) = x2 + 2x – 7\n\n### What have we learned\n\n• The standard form of a quadratic function is ax2+bx+c=0, where a≠0\n\n### Concept Map\n\n#### Composite Figures – Area and Volume\n\nA composite figure is made up of simple geometric shapes. It is a 2-dimensional figure of basic two-dimensional shapes such as squares, triangles, rectangles, circles, etc. There are various shapes whose areas are different from one another. Everything has an area they occupy, from the laptop to your book. To understand the dynamics of composite […]",
null,
"#### Special Right Triangles: Types, Formulas, with Solved Examples.\n\nLearn all about special right triangles- their types, formulas, and examples explained in detail for a better understanding. What are the shortcut ratios for the side lengths of special right triangles 30 60 90 and 45 45 90? How are these ratios related to the Pythagorean theorem? Right Angle Triangles A triangle with a ninety-degree […]",
null,
"#### Ways to Simplify Algebraic Expressions\n\nSimplify algebraic expressions in Mathematics is a collection of various numeric expressions that multiple philosophers and historians have brought down. Talking of algebra, this branch of mathematics deals with the oldest concepts of mathematical sciences, geometry, and number theory. It is one of the earliest branches in the history of mathematics. The study of mathematical […]",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.86763215,"math_prob":0.99811363,"size":3696,"snap":"2022-40-2023-06","text_gpt3_token_len":946,"char_repetition_ratio":0.16440953,"word_repetition_ratio":0.090076335,"special_character_ratio":0.25487012,"punctuation_ratio":0.09638554,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997066,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-29T23:28:20Z\",\"WARC-Record-ID\":\"<urn:uuid:cb3119b3-90ba-4fc0-96fe-73d795c771dc>\",\"Content-Length\":\"117682\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fb66e9f3-f5f9-4344-855e-b990b671ef9e>\",\"WARC-Concurrent-To\":\"<urn:uuid:4e599686-d0ac-49fd-9f0b-8ffb8be4f664>\",\"WARC-IP-Address\":\"65.1.150.125\",\"WARC-Target-URI\":\"https://www.turito.com/learn/math/quadratic-functions-in-standard-form-grade-10\",\"WARC-Payload-Digest\":\"sha1:3FROLLTR4MIHOCT4YWJJPWXO2EB5I24J\",\"WARC-Block-Digest\":\"sha1:XRWMIOKF3CUSZTPEJOR52FGKHLKMCDAT\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499768.15_warc_CC-MAIN-20230129211612-20230130001612-00101.warc.gz\"}"} |
https://blogs.msdn.microsoft.com/ericgu/2003/11/14/generics-algorithms/ | [
"Generics Algorithms\n\nA few weeks ago, Anders walked into the C# design meeting and said, \"I think I have\na way to get generic algorithms to work with our current support\", and then proceeded\nto outline a scheme on the whiteboard. I dutifully copied it into our design notes\n(after being away from the design process for a couple of years, keeping the notes\nup to date is my big contribution).\n\nToday, I was home taking care of my sick daughter, so I spent some time playing around\nwith his idea.\n\nThe basic idea is to create a generic class that defines the operations that can be\nperformed on a data type, and then create specialized implementations. Something like:\n\npublic abstract class Calculator<T>\n{\npublic abstract T Add(T a, T b);\n\n}\n\nand then derive a specific class from it:\n\nnamespace Int32\n{\npublic class Calculator: Calculator<int>\n{\npublic override int Add(int a, int b)\n{\nreturn a + b;\n}\n}\n\n}\n\nWe'll be using Int32.Calculator to perform operations on ints. We could extend thte\ncalculator class to perform more operations, and create implementations to deal with\ndifferent types. So that's the first part of the scheme. It's somewhat reminiscent\nof template specializations in C++, but it's not\n\nWe can now define a library to use it:\n\nclass AlgorithmLibrary<T> where T: new()\n\n{\nCalculator<T> calculator;\n\npublic AlgorithmLibrary(Calculator<T> calculator)\n{\nthis.calculator = calculator;\n}\n\npublic T Sum(List<T> items)\n{\nT sum = new T();\n\nfor (int i = 0; i < items.Count; i++)\n{\nitems[i]);\n}\n\nreturn sum;\n}\n}\n\nThis class encapsulates algorithms inside of it. When we want to use it, we can write:\n\nAlgorithmLibrary library = new AlgorithmLibrary<int>(new Int32.Calculator());\n\nI should perhaps explain how this works. When we construct library with int as the\ntype argument, that means that constructor is looking for a Calculator<int>\nrather than a Calculator<T>, which is convenient, since that's the class we\npassed into it.\n\nUltimately, we end up with a version of Sum that uses the int calculator to do its\noperations.\n\nThat gives us a workable scheme, though it is a bit ugly because we have to create\nthe calculator and pass it in. The neat way that Anders came up to get around this\nis to use a pattern-based approach, and use reflection. We first look for the calculator\nclass off of the type, and if we don't find it, we look in the standard place. This\nmeans that you can have the AlgorithmLibrary constructor find the right calculator\nfor us.\n\nThis all worked out fairly well. I'd show you the whole code, but I'm not quite happy\nwith it, and I'm not sure it would run on the PDC bits, so you'll probably have to\nwait for a while. Here's a sample of using it:\n\nList<double> ld = new List<double>();\n\nAlgorithmLibrary<double> cd = new AlgorithmLibrary<double>();\n\nConsole.WriteLine(\"Sum is {0}\", cd.Sum(ld));\nConsole.WriteLine(\"Mean is {0}\", cd.Mean(ld));\nConsole.WriteLine(\"SD is {0}\", cd.StandardDeviation(ld));\n\nOne concern with this approach is performance. I did a few quick benchmarks, and for\nlists, this approach takes about 50% longer than coding it by hand. For arrays, it's\nabout twice as slow. I think the major problem is the virtual function, but there\nmay be a way to make that faster.\n\nTags\n\n1.",
null,
"Olivier says:\n\nYou can avoid virtual with:\npublic interface ICalculator<T>\n{\nT Add(T a, T b);\n}\n\nclass AlgorithmLibrary<T, IC>\nwhere T: new()\nwhere CT: ICalculator<T>\n{\nCT calculator;\n\npublic AlgorithmLibrary(CT calculator)\n{\nthis.calculator = calculator;\n}\n\npublic T Sum(List<T> items)\n{\nT sum = new T();\n\nfor (int i = 0; i < items.Count; i++)\n{\nsum = calculator.Add(sum, items[i]);\n}\n\nreturn sum;\n}\n}\nnamespace Int32\n{\npublic class Calculator: ICalculator<int>\n{\npublic int Add(int a, int b)\n{\nreturn a + b;\n}\n}\n}\nbut i don’t yet have received Whidbey to compare performance.\n\n2.",
null,
"Nicholas Paldino says:\n\nWhile the solution does not strike me as anything incredibly new (think STL), the only way you will get it to work (and I mean work in the sense of acceptance) is if you create your own algorithm library, with skeletons required (abstract classes and interfaces) to inject our own custom types into the algorithms.\n\nI don’t remember if it was you or someone else from MS, but it was said that the only way that something like this is effective is if there is universal coverage, which has more potential to lead to universal adoption.\n\nBased on that, my question would be, are you going to create such an algorithm library, along the lines of STL, and if so, what kinds of algorithms are you going to create?\n\n3.",
null,
"Eric says:\n\nThanks, Olivier. I had thought about trying interfaces, but hadn’t gotten around to it yet.\n\n4.",
null,
"Dilip says:\n\n5.",
null,
"(luKa) says:\n\nWouldn’t be possible to have a constraints for a type parameter to specifie a required member (a method, an operator, ecc) just like this?\n\nnamespace Int32\n{\npublic class Calculator<T>\nwhere T: T Add(T, T)\n{\npublic T Add(T a, T b)\n{\nreturn a + b;\n}\n}\n}\n\nTIA (luKa)\n\n6.",
null,
"(luKa) says:\n\nOps… I was meaning this\n\nclass AlgorithmLibrary<T>\n….where T: new(), T Add(T, T)\n{\n….// …\n}\n\n7.",
null,
"Anonymous says:\n8.",
null,
"andrew queisser says:\n\nluKa, I think you’re pointing out the fundamental problem with generic algorithms in C#. There’s no way to do anything inside a generic algorithm unless you specify an exact type or interface. That puts the burden on the caller of the generic algorithm to pass in a the type needed by the algorithm. That’s the exact oppositie of a \"generic\" algorithm. Looks like several people are thinking about how to get around this problem (see Eric’s post above).\n\nAndrew Queisser\n\n9.",
null,
"RebelGeekz says:\n10.",
null,
"Generic Algorithms in C# « The Wandering Glitch 2 says:\n11.",
null,
"Snapping says:\n\nRef:\n\nhttp://www.codeproject.com/csharp/genericnumerics.asp\n\nIntroduction\n\nThecurrentimplementatio…\n\n12.",
null,
"Employment Wages » Eric Gunnerson’s C# Compendium : Generics Algorithms says:\n13.",
null,
"Eric Gunnerson s C Compendium Generics Algorithms | Weak Bladder says:\n\n14.",
null,
"Eric Gunnerson s C Compendium Generics Algorithms | Quick Diets says:\n\nPingBack from http://quickdietsite.info/story.php?id=14591\n\n15.",
null,
"Eric Gunnerson s C Compendium Generics Algorithms | internet marketing tools says:\n16.",
null,
"Miron says:\n\nAny progress?\n\nI guess what ticks me off is that applicability of operator(s) to any given type is foreknown during compile time.\nAll that is asked is a bit of syntactic sugar from the compiler to avoid writing something like this:\nI’d think if instead of writing compiler code while tending to child the author was tending to child, he will have time to write proper compiler code when he writes compiler code.\n\n1.",
null,
"Miron says:\n\nI meant to say like this:\nhttps://github.com/MironAtHome/MathFun"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.85037446,"math_prob":0.87738156,"size":7041,"snap":"2019-43-2019-47","text_gpt3_token_len":1811,"char_repetition_ratio":0.14153759,"word_repetition_ratio":0.09073359,"special_character_ratio":0.2579179,"punctuation_ratio":0.16945606,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9578551,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-14T18:35:01Z\",\"WARC-Record-ID\":\"<urn:uuid:c244ba84-e65c-4ed9-b2be-1fa6144390c4>\",\"Content-Length\":\"69269\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c61e517e-5ab5-4795-ab30-7c8c403dd1e7>\",\"WARC-Concurrent-To\":\"<urn:uuid:be778b0c-e790-4df3-950b-5dad36d26f23>\",\"WARC-IP-Address\":\"104.112.30.208\",\"WARC-Target-URI\":\"https://blogs.msdn.microsoft.com/ericgu/2003/11/14/generics-algorithms/\",\"WARC-Payload-Digest\":\"sha1:RVSLM73NHJ5S53FB3F633APRXTY62X6S\",\"WARC-Block-Digest\":\"sha1:F7S3AJFB5FIQUONE3QYLXWSL7FEA7K6G\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986654086.1_warc_CC-MAIN-20191014173924-20191014201424-00250.warc.gz\"}"} |
https://www.examgyani.com/question-answer/solve-for-x-given-the-equation-square-root-of-x-plus-9-%E2%88%92-4-1/ | [
"solve for x, given the equation square root of x plus 9 − 4 = 1.\n\nHi, I have a question and I hope anyone could answer it:\n\nSolve for x, given the equation Square root of x plus 9 − 4 = 1.\n\nx = 16, solution is not extraneous\n\nx = 16, solution is extraneous\n\nx = 34, solution is not extraneous\n\nx = 34, solution is extraneous\n\n• 0\nShare"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9156502,"math_prob":0.9999225,"size":746,"snap":"2023-14-2023-23","text_gpt3_token_len":172,"char_repetition_ratio":0.1212938,"word_repetition_ratio":0.609375,"special_character_ratio":0.22654155,"punctuation_ratio":0.10606061,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9980435,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-21T07:37:30Z\",\"WARC-Record-ID\":\"<urn:uuid:37e500fd-e1e8-4205-81f1-033fd29fb1fb>\",\"Content-Length\":\"284727\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:72067ab0-5bd3-4d1f-a9d0-d9b236f94a0d>\",\"WARC-Concurrent-To\":\"<urn:uuid:de4b36db-c977-48f9-a779-fd539ed1a01a>\",\"WARC-IP-Address\":\"104.21.1.142\",\"WARC-Target-URI\":\"https://www.examgyani.com/question-answer/solve-for-x-given-the-equation-square-root-of-x-plus-9-%E2%88%92-4-1/\",\"WARC-Payload-Digest\":\"sha1:LOYSVEX5IUAPEQRPTVNEDJW6F2HKCXYC\",\"WARC-Block-Digest\":\"sha1:35XHEQMUYIGOYIZ2XEXZB2EGXBQGI7NU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296943637.3_warc_CC-MAIN-20230321064400-20230321094400-00540.warc.gz\"}"} |
https://hess.copernicus.org/articles/25/3539/2021/ | [
"Hydrol. Earth Syst. Sci., 25, 3539–3553, 2021\nhttps://doi.org/10.5194/hess-25-3539-2021\nHydrol. Earth Syst. Sci., 25, 3539–3553, 2021\nhttps://doi.org/10.5194/hess-25-3539-2021\n\nResearch article 23 Jun 2021\n\nResearch article | 23 Jun 2021",
null,
"# Probabilistic modeling of field-scale CO2 generation by carbonate–clay reactions in sedimentary basins\n\nProbabilistic modeling of field-scale CO2 generation by carbonate–clay reactions in sedimentary basins\nGiulia Ceriotti1, Claudio Geloni2, Matilde Dalla Rosa2, Alberto Guadagnini1,3, and Giovanni Porta1 Giulia Ceriotti et al.\n• 1Department of Civil and Environmental Engineering, Politecnico di Milano, Piazza L. Da Vinci 32, 20133 Milan, Italy\n• 2Eni S.p.A.-Upstream and Technical Services, via Emilia, 1 20097 San Donato Milanese (MI), Italy\n• 3Department of Hydrology and Atmospheric Sciences, University of Arizona, Tucson, Arizona, USA\n\nCorrespondence: Giulia Ceriotti ([email protected])\n\nAbstract\n\nThis work explores a probabilistic modeling workflow and its implementation targeting CO2 generation rate and CO2 source location by the occurrence of carbonate–clay reactions (CCRs) in three-dimensional realistic sedimentary basins. We ground our study on the methodology proposed for a one-dimensional case study and a single CCR formulation by which includes a framework to account for thermodynamic parameter uncertainties. This methodology is here extended to a realistic three-dimensional sedimentary basin setting and transferred to encompass different types of CCRs, including two newly formulated CCRs which account for minerals typically observed in sedimentary environments. While testing the ability of the selected procedure to model diverse CCRs in three-dimensional realistic subsurface sedimentary systems, we quantitatively compare the impact of CCR formulation on the spatial distribution of CO2 source location, temperature and pressure compatible with CO2 gaseous generation, and CO2 generation rate in three-dimensional environments characterized by complex and non-uniform stratigraphy. The application of the procedure to various types of CCRs enables us to provide an insight into the impact of mineralogical composition on the activation temperature and pressure and the amount of CO2 released by the different CCR mechanisms. Finally, we show the implementation of the proposed probabilistic framework to define scenarios associated with various levels of probability to be used as the input and boundary conditions for CO2 migration and transport models in the subsurface.\n\nShare\n1 Introduction\n\nCarbon dioxide (CO2) is ubiquitously found in gaseous mixtures accumulated across sedimentary basins together with hydrocarbon (e.g., CH4) and other non-hydrocarbon components (e.g., N2Feng et al.2016). With reference to gas reservoirs, it might constitute up to 90 % of the total gas volume , its presence often being the cause of dilution of the gas mixture and (sometimes marked) hampering of its energy content . Possible sources of such large amounts of CO2 in a sedimentary basin are associated with transformation of organic carbon, carbonate mineral dissolution, inorganic chemical equilibrium of the feldspar–clay–carbonate mineral system, and magma degassing . Processes of CO2 generation and accumulation may be of interest to the characterization of flow and geochemistry in deep subsurface systems as well as flow in relatively shallow groundwater bodies, as large CO2 accumulations may trigger vertical flow and transport processes . Characterization and understanding of the key mechanisms that control the natural formation of carbon dioxide are not completely explored and are still a subject of research. A considerable body of studies during the past 40 years has suggested that the interaction between carbonates, clays/aluminosilicates, and pore water might play a major role in controlling CO2 partial pressure in geologic systems. A relevant influence of mineral rock interactions on dissolved CO2 has been observed also in groundwater and surface water bodies . Considering a given sedimentary rock containing carbonates and clays/aluminosilicates, the amount of dissolved CO2 in the pore water is regulated by the chemical equilibrium among all mineral phases, such a buffering mechanism being typically denoted as carbonate–clay reaction (CCR; Hutcheon et al.1990b). The CCR buffering mechanism involves a complex system of geochemical reactions which is typically condensed in terms of a schematic reaction of the kind\n\n$\\begin{array}{ll}\\mathrm{CCR}\\mathrm{1}:& \\mathrm{5}\\mathrm{Dolomite}+\\mathrm{Kaolinite}+\\mathrm{Silica}+\\mathrm{2}{\\mathrm{H}}_{\\mathrm{2}}\\mathrm{O}\\\\ & ⇌\\mathrm{5}{\\mathrm{CO}}_{\\mathrm{2}}+\\mathrm{Clinochlore}+\\mathrm{5}\\mathrm{Calcite}.\\end{array}$\n\nThe possibility of occurrence of such a reaction in real (field-scale) sedimentary environments is supported by various studies . Several representations/formulations for carbonate–clay reactions that have been proposed share a reaction structure similar to CCR1 and differ in terms of the carbonate and aluminosilicate phases included therein . Each of the available carbonate–clay reactions can be characterized by an equilibrium constant that quantifies the relative partitioning between reactants and products and the amount of CO2 released in the porewater when the system attains thermodynamic equilibrium.\n\nSince the carbonate–clay buffering system is a reversible process, the CCR mechanism may act either as a CO2 sink or source depending on local temperature (T) and pressure (P), these quantities directly controlling the value of the equilibrium constant associated with the CCRs.",
null,
"Figure 1Outline of the two possible alternative scenarios according to the conceptual approach proposed in . At location A (characterized by shallow location and moderate TP values), the geochemical system is at equilibrium, the total gas pressure given by sum of all gases species partial pressures (Pgas) is smaller than porewater pressure (P), and the CO2 exists only as dissolved species. At location B (characterized by deep location where high TP values are expected), the total gas pressure is larger than porewater pressure. Then a CO2-rich gaseous phase is formed, which migrates upwards, and a disequilibrium is promoted, leading to continuous release of CO2 until one of the reactants of the CCR is exhausted.\n\nThe study of suggests that the equilibrium constant (KCCR1) characterizing CCR1 can take values larger than 1 for temperatures higher than 100–120 C, thus favoring release of CO2 which can then be found as a dissolved species in porewater. Starting from this analysis, propose a simple conceptual model which distinguishes two possible alternative CCR behaviors. These are exemplified in the depiction of Fig. 1, where we consider two points A and B located at different depths in a sedimentary environment, as described in the following.\n\n• Location A in Fig. 1 corresponds to shallow depths. Here, CO2 and all chemical species dissolved in the porewater are at equilibrium with the mineral assemblage. CO2 and other gaseous species (e.g., CH4) can appear only as dissolved phases. The moderate temperature and pressure typically associated with these shallow depths do not promote formation of large amounts of CO2. Thus, the sum of the partial pressures of all gaseous species attains values that are smaller than the porewater pressure (i.e., Pgas < P).\n\n• Location B corresponds to large depths. High temperature values that are expected to take place at such locations tend to remarkably shift the equilibrium towards the right-hand side of the CCR reaction, a high amount of CO2 being then released into the porewater. In this scenario, the partial pressure of CO2 (summed to other gaseous compounds, including, e.g., aqueous vapor) is typically higher than porewater pressure (i.e., Pgas > P). A CO2-rich gaseous phase is then separated and tends to migrate upwards through rock matrix fractures due to buoyancy effects. A disequilibrium between the rock mineral phases and the porewater is then promoted, and generation of CO2 takes place until at least one of the reactants of the CCR system is exhausted. The occurrence of the conditions corresponding to location B is hereafter denoted as CCR mechanism activation, implying that the geochemical disequilibrium and the formation of a separate CO2-rich gaseous phase have been triggered.\n\nprovide a first implementation of the above-described conceptual approach by relying on the linear TP trend proposed by , i.e., P (bar) =6 (T (C) −25), and using as a reference three CCR buffering mineral assemblages, corresponding to (i) calcite–laumontite–kaolinite–quartz, (ii) siderite–daphnite–kaolinite–quartz, and (iii) magnesite–daphnite–kaolinite–quartz. Results of their analysis (a) suggest that the formation of a separate CO2-rich gaseous phase is feasible for temperature higher than 330 C and (b) represent the first quantitative estimation of the temperature and pressure of CCR activation as source of gaseous CO2 in a sedimentary environment. However, it should be noted that these results cannot be readily transferred to a generic realistic sedimentary basin scenario because they are associated with (i) mineral phases that are rarely observed in real sediments (e.g., laumontite and daphnite) and (ii) a linear simplified TP relationship.\n\nOtherwise, T and P evolution in real sedimentary basins often displays complex patterns, each scenario being characterized by site-specific TP spatial and temporal distributions. These are a result of the diagenetic processes of rocks and the burial history of the sedimentary basin itself and should then be appropriately included in a CCR-based assessment of gaseous CO2 generation.\n\nThese aspects are fully recognized by , who combine a one-dimensional burial model with a geochemical model formulated according to the conceptual approach suggested in . A key point of novelty introduced by is the reliance on a probabilistic framework to propagate uncertainty of thermodynamic parameters associated with reaction CCR1 to target modeling goals (i.e., CO2 source location and CO2 generation rate). Such a stochastic modeling framework allows assessment of the probability distribution of (i) the depth at which the source of gases is located, (ii) the amount of CO2 generated (conditional on a given mineralogy of the sediments involved in the basin formation process), and (iii) the range of TP combinations associated with gaseous CO2 generation.\n\nIn this context, an appraisal of this probabilistic approach considering a fully three-dimensional scenario with the ensuing quantification of the amount of CO2 that can be realistically released by CCR reactions is still lacking. This is precisely the key goal of this study, which is geared to (i) providing a modeling workflow conducive to estimating the spatial distributions of CO2 sources and the associated generation rates in realistic three-dimensional sedimentary basins and (ii) assessing differences in the activation temperature and pressure characterizing various possible formulations of the CCR mechanism. We note that the evaluation of all these quantities is still a major element of study and debate in the literature . Modeling of CO2 generation and accumulation in large-scale geological systems is typically prone to considerable uncertainties, chiefly due to paucity of information and to the remarkably large spatial and temporal scales involved. In this context, we illustrate a modeling framework that leads to a probabilistic quantification of the generation of CO2 by a specific class of reactive processes (i.e., CCRs). As such, our study fills a knowledge gap by providing a methodology to support quantitative investigations of spontaneous CO2 generation in large-scale geological systems, these being otherwise typically based on mostly qualitative analyses. While we consider a simple geochemical model based on thermodynamic equilibrium, our probabilistic framework of analysis is flexible and can include treatment of model uncertainty (e.g., Walker et al.2003; Neuman2003) as an additional element. Setting a given model structure is simply a convenient choice to limit computational and conceptual complexity while at the same time considering a mathematical model that can be characterized by information that is typically available in field-scale settings (in terms of, e.g., mineral composition, pressure, and temperature distributions). Values of equilibrium constants are here considered uncertain because temperature and pressure values observed in sedimentary systems lie outside the range of conditions where such parameters are usually characterized . In this work we investigate the propagation of this parametric uncertainty in the presence of various (alternative) CCR formulations by focusing on a three-dimensional scenario. When considering the framework proposed by , our work allows combining uncertainty in model parameters (equilibrium thermodynamic constants) with input uncertainty, i.e., uncertainty in the description of the reference system. The latter type of uncertainty is reflected by our choice of considering diverse mineral assemblages leading to the occurrence of differing CCRs. Note that our approach is geared towards quantification on the space–time location and intensity of the CO2 source. This information can then be used as input to quantify scenario uncertainties by delineating the spatial and temporal extent of CO2 influx. Transport and accumulation of CO2 across the subsurface can then be analyzed through approaches such as those described, e.g., in . From an operational standpoint, our approach could be applied to enhance our knowledge of the degree of compatibility of CO2 concentrations observed in field-scale systems with the occurrence of CCR, as opposed to the action of other processes which might be considered in a large-scale transport model of choice.\n\nThe study is structured as follows. Section 2 is devoted to the presentation of the three-dimensional sedimentary setting and of the CCR formulations we consider. These include a typically employed formulation and two additional models involving clay and silicate minerals (such as beidellite and illite) that are frequently observed in sedimentary basins. Section 3 summarizes the modeling and uncertainty quantification workflow and procedures employed. Results are presented and discussed in Sect. 4. Finally, concluding remarks are provided in Sect. 5.\n\n2 Sedimentary setting and CCR formulations\n\nThe reference system considered in this study is a three-dimensional realistic sedimentary basin with a deposition history spanning a temporal window of 135 Ma (millions of years before present) and characterized by the deposition sequence listed in Table 1.\n\nTable 1Sequence of sediments deposited during the 135 Ma of basin deposition history and sediment density.",
null,
"The basin stratigraphy at the present time (which is taken as t=0 Ma) is depicted in Fig. 2 and comprises six layers (corresponding to four carbonate and two shale rock systems). The planar surface of the basin covers an area of about 177.5 km × 155 km, the maximum depth (below sea level) reached by the volume filled by sediments being approximately 8 km.",
null,
"Figure 2Sketch of the three-dimensional sedimentary basin setting considered at the present time, i.e., t= 0 Ma.\n\nThe geo-history of the basin is reconstructed using the widely tested and documented burial model E-SIMBA™ (for details see, e.g., Grigo et al.1993; Dalla Rosa et al.2015; Zattin et al.2016) which allows estimation of the three-dimensional dynamic evolution of stratigraphy as well as temperature, pressure and porosity distributions. These variables are here taken as input data.\n\nFigure 3 depicts the spatial distribution of temperature (T [C] in panel a) and pressure (P [bar] – in panel b) along two perpendicular vertical cross sections located at x=32 km and y= 105 km, respectively (see the reference system indicated in Figs. 23). Note that the z axis points downwards; i.e., the value of z increases with depth. Each cell of the spatial mesh used to describe the evolution of T, P and porosity has a uniform size of 2500 m × 2500 m × 200 m ($x×y×z$). Temperature and pressure display an overall increasing trend with depth which yields values of T and P close to those typically observed in real sedimentary basins (e.g., Colombo et al.2017, 2018). The largest temperature and pressure values (corresponding to 330 C and 800 bar, respectively) are observed at the deepest locations in the basin.",
null,
"Figure 3Evolution of temperature (T, C, panel a) and pressure (P, bar, panel b) simulated at the present time, i.e., t= 0 Ma, along two perpendicular planar sections at x= 32 km and y= 105 km.\n\nConsidering the reference geological setting described above, we investigate separately three differing CCR formulations which can be considered at the basis of CO2 generation. These include the classical reaction CCR1 (illustrated in Sect. 1 and recalled in the following) along with two alternative CCR models (labeled CCR2 and CCR3).\n\n$\\begin{array}{ll}& \\mathrm{CCR}\\mathrm{1}:\\mathrm{5}\\mathrm{Dolomite}+\\mathrm{Kaolinite}+\\mathrm{Silica}+\\mathrm{2}{\\mathrm{H}}_{\\mathrm{2}}\\mathrm{O}⇌\\mathrm{5}{\\mathrm{CO}}_{\\mathrm{2}}\\\\ & \\phantom{\\rule{0.25em}{0ex}}\\phantom{\\rule{0.25em}{0ex}}+\\mathrm{Clinochlore}+\\mathrm{5}\\mathrm{Calcite}\\\\ & \\mathrm{CCR}\\mathrm{2}:\\mathrm{0.33}\\mathrm{Dolomite}+\\mathrm{1.13}\\mathrm{Microcline}+\\mathrm{Beidellite}\\\\ & \\phantom{\\rule{0.25em}{0ex}}\\phantom{\\rule{0.25em}{0ex}}+\\mathrm{0.33}{\\mathrm{H}}_{\\mathrm{2}}\\mathrm{O}⇌\\mathrm{0.33}{\\mathrm{CO}}_{\\mathrm{2}}+\\mathrm{1.33}\\mathrm{Illite}+\\mathrm{1.5}\\mathrm{Quartz}\\\\ & \\phantom{\\rule{0.25em}{0ex}}\\phantom{\\rule{0.25em}{0ex}}+\\mathrm{0.33}\\mathrm{Albite}+\\mathrm{0.33}\\mathrm{Calcite}\\\\ & \\mathrm{CCR}\\mathrm{3}:\\mathrm{0.33}\\mathrm{Dolomite}+\\mathrm{1.13}\\mathrm{Microcline}+\\mathrm{Beidellite}\\\\ & \\phantom{\\rule{0.25em}{0ex}}\\phantom{\\rule{0.25em}{0ex}}+\\mathrm{0.66}{\\mathrm{H}}_{\\mathrm{2}}\\mathrm{O}⇌\\mathrm{0.33}{\\mathrm{CO}}_{\\mathrm{2}}+\\mathrm{1.33}\\mathrm{Illite}+\\mathrm{1.8}\\mathrm{Quartz}\\\\ & \\phantom{\\rule{0.25em}{0ex}}\\phantom{\\rule{0.25em}{0ex}}+\\mathrm{0.33}\\mathrm{Analcime}+\\mathrm{0.33}\\mathrm{Calcite}\\end{array}$\n\nCCR2 and CCR3 are here proposed based on laboratory tests aimed at investigating the role of different types of clay in sedimentary environments . These formulations include mineral phases (such as beidellite, analcime and microcline) which can be considered a proxy of clays and feldspars that have been observed promoting the release of CO2 by dolomite in laboratory experiments. The ability of CCR2 and CCR3 to interpret field ${P}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ data is further discussed in Sect. 3.2.\n\nDepending on the CCR investigated, we consider a given mineralogical composition of sediments, as listed in Table 2. These mineralogies (termed M1 for CCR1 and M2–3 for CCR2 and CCR3) are selected to maximize the mass of CO2 that can potentially be generated by a unit mass of sediment (${m}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ – kg${}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ kg${}_{\\mathrm{sed}}^{-\\mathrm{1}}$) when a prescribed CCR mechanism is activated. Details on the computation of ${m}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ are reported in the Supplement of . For simplicity, we assume here that the four carbonate rocks forming the sedimentary basin described in Fig. 2 are characterized by the same uniformly distributed mineralogical composition, i.e., M1 or M2–3 when modeling CO2 generation by CCR1 or CCR2 and CCR3, respectively. Otherwise, the shale rocks are assumed to be characterized by a negligible carbonate content, which is in turn incompatible with the occurrence of CCR (i.e., we assume that ${m}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ associated with shale layers is zero).\n\nTable 2Composition of the mineralogical scenarios used for the investigation of the three CCRs considered. The mass of CO2 released by a unit mass of sediment (${m}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ – kg${}_{{\\mathrm{CO}}_{\\mathrm{2}}}$, kg${}_{\\mathrm{sed}}^{-\\mathrm{1}}$) when the gaseous CO2 generation is activated is also listed for each setting.",
null,
"3 CCR modeling under uncertainties\n\nOur study relies on a given model structure, thus neglecting uncertainty in the latter. We rest on the equilibrium-based approach employed by . Thus, we consider pure mineral phases while neglecting other factors which would eventually influence the model structure (e.g., the occurrence of other mineral transformations or effects associated with salinity of brine). Consistent with this model structure, we consider the equilibrium constant of speciation reactions as the key source of parametric uncertainty. We note that this choice is motivated by the observation that temperature and pressure values observed in sedimentary systems lie outside the range of conditions where thermodynamic equilibrium constants are usually characterized . In addition to parametric uncertainty, we also consider input uncertainty, defined as the uncertainty related to the description of the system ; i.e., we assume that diverse CCRs may take place depending on the mineralogical assemblage. These two sources of uncertainty are propagated to the quantities of interest, i.e., the CO2 source location, the CO2 generation rate, and the temperature and pressure of CCR activation. Note that, as detailed in Sect. 2, we consider a uniform mineral composition across the domain, a setting corresponding to an upper limit condition for each of the considered CCRs. While it would be interesting in principle to investigate the impact of a spatially heterogeneous mineralogic composition, doing so would require having access to a suitable dataset and would increase complexity. Yet it is worth emphasizing that the proposed methodological framework and modeling approach are fully compatible with the presence of a spatially variable mineralogical composition, which can be accommodated in the presence of appropriate data to characterize it. As such, our approach can also be employed to assess the impact of uncertainties associated with spatially heterogeneous arrangements of mineral and sediment composition on CCR-based CO2 generation. The latter could be tackled by relying on appropriate techniques, such as, e.g., functional compositional kriging (see, e.g., Menafoglio et al.2016, and references therein). Analyzing this aspect is, however, beyond the scope of the present study.\n\n## 3.1 Speciation reactions and uncertainty characterization\n\nGiven a generic mineral, aqueous or gaseous phase (Ph), it is always possible to describe the speciation in water of mineral phase Ph upon relying on a set of aqueous basis species (Anderson2005). A speciation reaction can then be characterized by an equilibrium constant (KS,Ph), whose value depends on the system temperature and pressure. Following , we assume that the equilibrium constant driving speciation of Ph can be expressed as\n\n$\\begin{array}{}\\text{(1)}& \\mathrm{log}{\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{S},\\mathrm{Ph}}=\\stackrel{\\mathrm{̃}}{A}+B\\cdot T+\\frac{C}{T}+\\stackrel{\\mathrm{̃}}{D}\\cdot \\mathrm{log}T+\\frac{E}{{T}^{\\mathrm{2}}},\\end{array}$\n\nwhere T (K) is temperature and the symbol $\\stackrel{\\mathrm{̃}}{\\phantom{\\rule{0.125em}{0ex}}\\phantom{\\rule{0.125em}{0ex}}}$ denotes (uncertain) random variables (to distinguish these from deterministic quantities). Note that this formulation holds for a given pressure of P=1 bar. The format of Eq. (1) resembles the one characterizing the expression of a temperature-dependent equilibrium constant derived from the Maier–Kelley heat capacity assumption which is typically used in thermodynamic databases . The key difference between Eq. (1) and the classical expression for the (temperature-dependent) equilibrium constant is that the two parameters $\\stackrel{\\mathrm{̃}}{A}$ and $\\stackrel{\\mathrm{̃}}{D}$ are not considered deterministic effective parameters and are here interpreted as bivariate Gaussian random variables. We follow the approach of to define the mean values (μA and μD for $\\stackrel{\\mathrm{̃}}{A}$ and $\\stackrel{\\mathrm{̃}}{D}$, respectively) and the entries of the covariance matrix Ψ characterizing the bivariate Gaussian distribution of $\\stackrel{\\mathrm{̃}}{A}$ and $\\stackrel{\\mathrm{̃}}{D}$. Characterization of the mean and covariance of bivariate Gaussian variables is grounded on data taken from . Given the structure of Eq. (1), it then follows that, for a given temperature value log${\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{S},\\mathrm{Ph}}$ is described by a normal distribution with parameters related to the statistical moments of $\\stackrel{\\mathrm{̃}}{A}$ and $\\stackrel{\\mathrm{̃}}{D}$. Details about the characterization of $\\stackrel{\\mathrm{̃}}{A}$ and $\\stackrel{\\mathrm{̃}}{D}$ for all mineral, liquid, and gaseous phases appearing in this study are reported in the Supplement. Uncertainties associated with the characterization of $\\stackrel{\\mathrm{̃}}{A}$ and $\\stackrel{\\mathrm{̃}}{D}$ can be propagated to the Ph speciation equilibrium constant through Eq. (1). It then follows that ${\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{S},\\mathrm{Ph}}$ is not a deterministic quantity but rather an uncertain variable described by a probability density function (pdf).\n\n## 3.2 CO2 partial pressure computation\n\nWe introduce here a generalized CCR formulation in the form of\n\n$\\begin{array}{ll}& \\mathrm{CCR}:{\\mathit{\\alpha }}_{\\mathrm{1}}{\\mathrm{Ph}}_{\\mathrm{1}}+\\mathrm{\\dots }+{\\mathit{\\alpha }}_{i}{\\mathrm{Ph}}_{i}+\\mathrm{\\dots }+{\\mathit{\\alpha }}_{I}{\\mathrm{Ph}}_{I}\\\\ & \\phantom{\\rule{0.25em}{0ex}}\\phantom{\\rule{0.25em}{0ex}}⇌{\\mathit{\\alpha }}_{I+\\mathrm{1}}{\\mathrm{Ph}}_{I+\\mathrm{1}}+\\mathrm{\\dots }+{\\mathit{\\alpha }}_{I+J}{\\mathrm{Ph}}_{I+J}+{\\mathit{\\alpha }}_{\\mathrm{0}}{\\mathrm{CO}}_{\\mathrm{2}},\\end{array}$\n\nwhere the symbol Phi indicates the ith mineral phase (with $i=\\mathrm{1}\\mathrm{\\dots }I+J$) appearing in the CCR, the term αi representing the stoichiometric coefficient of the mineral phase i; I and J quantify the number of CCR reactants and products, respectively, with the exception of CO2, which is explicitly accounted for on the right-hand side of the CCR with its stoichiometric coefficient, α0. Each of the mineral and gaseous phases involved in the CCR is associated with a speciation reaction and an uncertain speciation equilibrium constant, as described in Sect. 3.1. Note that the proposed generic CCR formulation can be readily recast into CCR1, CCR2, or CCR3.\n\nWe can express the equilibrium constant of the CCR (${\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{CRR}}$) as (Anderson2005)\n\n$\\begin{array}{}\\text{(2)}& \\begin{array}{rl}\\mathrm{log}{\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{CCR}}\\left(T\\right)& =\\sum _{i=\\mathrm{1}}^{I}{\\mathit{\\alpha }}_{i}\\mathrm{log}{\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{S},{\\mathrm{Ph}}_{i}}-\\sum _{i=I+\\mathrm{1}}^{J+I}{\\mathit{\\alpha }}_{i}\\mathrm{log}{\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{S},{\\mathrm{Ph}}_{i}}\\\\ & -{\\mathit{\\alpha }}_{\\mathrm{0}}\\mathrm{log}{\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{S},{\\mathrm{CO}}_{\\mathrm{2}}}.\\end{array}\\end{array}$\n\nThe quantities ${\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{S},{\\mathrm{CO}}_{\\mathrm{2}}}$ and ${\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{S},{\\mathrm{Ph}}_{i}}$ correspond to the speciation equilibrium constants associated with CO2 and the ith mineral phase contributing to the CCR, respectively. The uncertain variables ${\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{S},{\\mathrm{Ph}}_{i}}$ and ${\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{S},{\\mathrm{CO}}_{\\mathrm{2}}}$ are evaluated through Eq. (1) as a function of temperature. The value of ${\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{CCR}}$ resulting from Eq. (2) is then temperature-dependent and affected by uncertainty. The effect of pressure on ${\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{CCR}}$ is considered through a correction term (Millero1982)\n\n$\\begin{array}{}\\text{(3)}& \\begin{array}{rl}\\mathrm{log}{\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{CCR}}\\left(T,P\\right)& =\\mathrm{log}{\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{CCR}}\\left(T,P=\\mathrm{1}\\right)\\\\ & -\\frac{\\mathrm{\\Delta }{V}^{\\circ }}{\\mathrm{2.303}{R}_{\\mathrm{g}}T}\\cdot \\left(P-\\mathrm{1}\\right),\\end{array}\\end{array}$\n\nwhere ${\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{CCR}}\\left(T,P\\right)$ is the CCR equilibrium constant computed for a generic value of T and P; ${\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{CCR}}\\left(T,P=\\mathrm{1}\\right)$ is the CCR equilibrium constant computed for a generic value of T and pressure P=1 bar as resulting from Eq. (2); ΔV (m3 mol−1) represents the change in the molar volume associated with the CCR, and Rg is the ideal gas constant.\n\nThe partial pressure of CO2 (${\\stackrel{\\mathrm{̃}}{P}}_{{\\mathrm{CO}}_{\\mathrm{2}}}$) associated with the CCR can then be evaluated as\n\n$\\begin{array}{}\\text{(4)}& \\mathrm{log}{\\stackrel{\\mathrm{̃}}{P}}_{{\\mathrm{CO}}_{\\mathrm{2}}}\\left(P,T\\right)=\\frac{\\mathrm{log}{\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{CCR}}\\left(P,T\\right)}{{\\mathit{\\alpha }}_{\\mathrm{0}}}.\\end{array}$\n\nEquation (4) rests on the assumption that the CO2 fugacity coefficient is set to unity . Equations (2)–(4) allow computation of the partial pressure of CO2 as a function of basin temperature and pressure, yielding a three-dimensional distribution of ${\\stackrel{\\mathrm{̃}}{P}}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ as a function of basin stratigraphy and burial history. To provide a preliminary assessment, Fig. 4 reports the mean values of log ${\\stackrel{\\mathrm{̃}}{P}}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ associated with CCR1, CCR2 and CCR3 as a function of temperature, assuming that P (bar) = 6 × (T (C) −22) . The log ${\\stackrel{\\mathrm{̃}}{P}}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ trends are compared against values of ${P}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ sampled in different sedimentary basins obtained from the literature . We note that, on the one hand, the mean log ${\\stackrel{\\mathrm{̃}}{P}}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ trend associated with CCR1 provides a good interpretation of data observed at temperatures higher than 100 C (specifically for Norway, Texas and Thailand basins). On the other hand, the log ${\\stackrel{\\mathrm{̃}}{P}}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ mean trend resulting from CCR2 and CCR3 formulations appears to explain data observed at lower temperatures, ranging between 50 and 100 C (Norway, Paris Basin and Arkansas). This is consistent with the considerations already provided by , who suggest that CCR formulations accounting for complex clay phases (such as illite) can feasibly interpret low-temperature ${P}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ trends. We can conclude that the three formulations considered in this work are compatible with data observed in real sedimentary environments.",
null,
"Figure 4Evolution of the mean log ${\\stackrel{\\mathrm{̃}}{P}}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ trend as a function of temperature computed for CCR1, CCR2 and CCR3 assuming that P and T are described by P (bar) = 6× (T (C) −22) suggested by (suggested by Smith and Ehrenberg1989). As a term of comparison, ${P}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ is measured in various sedimentary basins labeled Norway, Texas, Alberta, Arkansas, Madison, Paris Basin, and Thailand (reported in Coudrain-Ribstein et al.1998).\n\n## 3.3 CO2 source location\n\nAccording to the conceptual model of , the CCR mechanism activates when the sum of the partial pressures of all gaseous species is higher than the porewater pressure (see Sect. 1). Here, we assume that only CO2 and aqueous vapor partial pressure might contribute to the formation of a CO2-rich separate gas phase, while the effect of other gas species (e.g., hydrocarbon gases) is neglected.\n\nFor a selected observation time ($t=\\stackrel{\\mathrm{^}}{t}$) and location (identified by the coordinates $x=\\stackrel{\\mathrm{^}}{x}$ and $y=\\stackrel{\\mathrm{^}}{y}$) on the planar surface of the sedimentary basin, we define the quantity $\\stackrel{\\mathrm{̃}}{R}\\left(\\stackrel{\\mathrm{^}}{t},\\stackrel{\\mathrm{^}}{x},\\stackrel{\\mathrm{^}}{y},z\\right)$ as\n\n$\\begin{array}{}\\text{(5)}& \\stackrel{\\mathrm{̃}}{R}\\left(\\stackrel{\\mathrm{^}}{t},\\stackrel{\\mathrm{^}}{x},\\stackrel{\\mathrm{^}}{y},z\\right)=\\frac{{\\stackrel{\\mathrm{̃}}{P}}_{{\\mathrm{CO}}_{\\mathrm{2}}}\\left(\\stackrel{\\mathrm{^}}{t},\\stackrel{\\mathrm{^}}{x},\\stackrel{\\mathrm{^}}{y},z\\right)+{P}_{\\mathrm{v}}\\left(\\stackrel{\\mathrm{^}}{t},\\stackrel{\\mathrm{^}}{x},\\stackrel{\\mathrm{^}}{y},z\\right)}{P\\left(\\stackrel{\\mathrm{^}}{t},\\stackrel{\\mathrm{^}}{x},\\stackrel{\\mathrm{^}}{y},z\\right)}.\\end{array}$\n\nHere, the symbol Pv denotes the aqueous vapor partial pressure, which we evaluate according to the procedure described by . The variable $\\stackrel{\\mathrm{̃}}{R}$ is affected by uncertainty because it depends on the random variable ${\\stackrel{\\mathrm{̃}}{P}}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ and on elevation z and can undertake values equal to or larger than unity when a location is compatible with the activation of a CCR mechanism. The CO2 source (${\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}\\left(\\stackrel{\\mathrm{^}}{t},\\stackrel{\\mathrm{^}}{x},\\stackrel{\\mathrm{^}}{y}\\right)$) is then evaluated as the position corresponding to the shallowest vertical coordinate z where $\\stackrel{\\mathrm{̃}}{R}\\left(\\stackrel{\\mathrm{^}}{t},\\stackrel{\\mathrm{^}}{x},\\stackrel{\\mathrm{^}}{y},z\\right)\\ge \\mathrm{1}$. Application of this procedure for all combinations of x and y coordinates enables us to delineate a CCR activation surface in the three-dimensional basin as the collection of points with coordinates (x, y, z) with $z={\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}\\left(x,y\\right)$, i.e., where the CCR mechanism is activated.\n\n## 3.4 CO2 generation rate\n\nWe provide an estimate of the rate of CO2 generated by the CCR mechanism activation per unit area of the CCR activation surface (${\\stackrel{\\mathrm{̃}}{F}}_{{\\mathrm{CO}}_{\\mathrm{2}}}\\left(t,x,y\\right)$ – kg${}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ Ma−1 m−1) as\n\n$\\begin{array}{}\\text{(6)}& {\\stackrel{\\mathrm{̃}}{F}}_{{\\mathrm{CO}}_{\\mathrm{2}}}\\left(t,x,y\\right)={m}_{{\\mathrm{CO}}_{\\mathrm{2}}}\\cdot \\left[\\mathrm{1}-\\mathit{\\varphi }\\right]\\cdot {v}_{\\mathrm{b}}\\cdot \\mathit{\\rho },\\end{array}$\n\nwhere ${m}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ (kg${}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ kgsed) is the mass of CO2 released by a unit mass of sediment upon activation of CCR, which depends on the CCR formulation and mineral composition (see Sect. 2); ϕ (–) and ρ (kgsed m${}_{\\mathrm{sed}}^{-\\mathrm{3}}$) are the sediment porosity and density, respectively, and vb (m Ma−1) is the burial velocity of sediments, a quantity governing the rate at which the sediments reach the location of the source. The CO2 generation rate in Eq. (6) rests on the hypothesis that the water–rock system located at a certain depth attains equilibrium before being buried to a deeper level. As opposed to porosity, the density of a given sediment type can be taken as a constant, its value being listed in Table 1 for each type of rock. The quantity ${\\stackrel{\\mathrm{̃}}{F}}_{{\\mathrm{CO}}_{\\mathrm{2}}}\\left(t,x,y\\right)$ depends indirectly on the activation depth ${\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}\\left(x,y\\right)$ since both ϕ and vb are a function of z, their value in Eq. (6) being related to the depth of the CO2 source. Outputs of the burial model employed in this study (i.e., E-SIMBA™; see Sect. 2) do not include the space–time evolution of vb across the basin. Results from a series of preliminary investigations (not shown here) performed with a one-dimensional burial model (STREAM; see, e.g., Formaggia et al.2013) at various planar locations of the three-dimensional basin investigated suggest that the burial velocity of sediments does not significantly vary along depth for z>2 km, where the CCR activation is more likely to occur. The value of vb in these regions is approximately equal to 40 m Ma−1. We take this as a representative value for vb in Eq. (6) in our analyses, thus disregarding the vertical variation of burial velocity.\n\n4 Results and discussion\n\nWe tackle probabilistic modeling of the CCRs introduced in Sect. 1 upon relying on a numerical Monte Carlo (MC) approach. Parameters $\\stackrel{\\mathrm{̃}}{A}$ and $\\stackrel{\\mathrm{̃}}{D}$ associated with each mineral phase Phi appearing in a given CCR are sampled N times (for a total of N=105 Monte Carlo replicates for each CCR mechanism) to yield N arrays\n\n${V}_{n}=\\left|\\begin{array}{ccccc}{A}_{\\mathrm{1}}& \\mathrm{\\cdots }& {A}_{i}& \\mathrm{\\cdots }& {A}_{I+J}\\\\ {D}_{\\mathrm{1}}& \\mathrm{\\cdots }& {D}_{i}& \\mathrm{\\cdots }& {D}_{I+J}\\end{array}\\right|,$\n\nwhere Vn indicates the nth sampled array (with n= 1, , N) and quantities Ai and Di represent the nth values sampled from the bivariate Gaussian distribution of $\\stackrel{\\mathrm{̃}}{A}$ and $\\stackrel{\\mathrm{̃}}{D}$ associated with the ith mineral phase (i.e., Phi) appearing in the generalized CCR formulation (see Sect. 3.2). The modeling approach detailed in Sect. 3 is applied for each sample Vn to yield N Monte Carlo realizations of the CCR mechanism occurrence as a function of space and time. The results presented and discussed in this section are all associated with t=0 Ma, i.e., the present time, corresponding to the setting when the basin structure reaches the largest depths and the highest temperature and pressure are observed (see Fig. 3, Sect. 2). Note that the modeling approach can be applied to any time level across the basin burial history.\n\n## 4.1 Source location, activation temperature and pressure\n\nBy relying on the N MC realizations of our model, we compute the frequency at which the activation of the CCR mechanism is observed at each location in the sedimentary space (${C}_{A}\\left(x,y,z\\right)$, i.e., the number of times of activation of a CCR at a given spatial location (x, y, z)). We start by focusing on the quantity\n\n$\\begin{array}{}\\text{(7)}& f\\left({\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}\\right)=\\frac{{C}_{A}\\left(x,y,z\\right)}{N},\\end{array}$\n\nwhich quantifies the three-dimensional distribution of the relative frequency of the source location.\n\nFigure 5 displays $f\\left({\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}\\right)$ for CCR1 (A), CCR2 (B), and CCR3 (C) evaluated at t=0 Ma using Eq. (7) along the two cross sections of the basin depicted in Fig. 3. While the three CCRs analyzed yield similar qualitative patterns of $f\\left({\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}\\right)$, some key quantitative differences can be noted. The spatial region associated with non-zero probability to observe activation of the CCR mechanism (i.e., $f\\left({\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}\\right)>\\mathrm{0}$) is broadest for CCR2. Moreover, Fig. 5 suggests that values of $f\\left({\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}\\right)$ do not increase monotonically with depth and attain their largest values at different depths, depending on the considered CCR. These maximum values are located approximately at 7 km for CCR1, at depths ranging between 5 and 6 km for CCR2 and at 6.5 km for CCR3. The documented peak in $f\\left({\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}\\right)$ and the ensuing decreasing trend observed for very large depths are consistent with the assumptions underlying our conceptual model, according to which the CO2 source is positioned in the shallowest point where a combination of temperature and pressure compatible with CO2 generation is first attained.",
null,
"Figure 5Spatial distribution of $f\\left({\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}\\right)$ computed at t= 0 Ma along two planar perpendicular sections of the basin case study located at x= 32 km and y= 105 km for CCR1 (a), CCR2 (b) and CCR3 (c).\n\nFurthermore, our results show that the three CCRs examined yield markedly different ranges of values of $f\\left({\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}\\right)$, the largest observed value for CCR1 being 0.1 (i.e., the probability of activation of CCR1 at a given location can be as high as 10 %) while being considerably lower for CCR2 and CCR3 (corresponding to 5 % and 3 %, respectively). CO2 generation by CCR1 is associated with a high frequency in the thin layer of sediment located at 7 km depth. Otherwise, CCR2 and CCR3 display a smooth spatial distribution of $f\\left({\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}\\right)$, displaying a smaller maximum value of $f\\left({\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}\\right)$ if compared with CCR1.\n\nThe differences observed in $f\\left({\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}\\right)$ indicate that (i) the CO2 generation occurrence is sensitive to the selected buffering CCR mechanism and that (ii) relevant shifts in the source location, characteristic temperature and pressure of activation may be expected as a function of the CCR considered. This element is further explored through the analysis of the pdfs of ${\\stackrel{\\mathrm{̃}}{T}}_{\\mathrm{act}}$ and ${\\stackrel{\\mathrm{̃}}{P}}_{\\mathrm{act}}$ and their comparison against the pdf of ${\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}$. The latter is evaluated as\n\n$\\begin{array}{}\\text{(8)}& \\mathrm{pdf}\\left({\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}\\right)=\\frac{{\\int }_{x}{\\int }_{y}{C}_{A}\\left(x,y,z\\right)\\mathrm{d}x\\mathrm{d}y}{{\\int }_{x}{\\int }_{y}{\\int }_{z}{C}_{A}\\left(x,y,z\\right)\\mathrm{d}x\\mathrm{d}y\\mathrm{d}z}.\\end{array}$\n\nThe pdfs of temperature (${\\stackrel{\\mathrm{̃}}{T}}_{\\mathrm{act}}$) and pressure of activation (${\\stackrel{\\mathrm{̃}}{P}}_{\\mathrm{act}}$) of the CCR mechanism are evaluated from the three-dimensional distribution of ${C}_{A}\\left(x,y,z\\right)$ and the temperature and pressure computed in the burial basin model.\n\nFigure 6 depicts the sample pdfs obtained for CCR1, CCR2, and CCR3. While these pdfs are characterized by a seemingly similar shape, each of them embeds the signature of the corresponding CCR mechanism, as seen in terms of spread, mean, and mode (see also Table 3). For example, the mean and mode of the activation temperature are lowest for CCR2, the highest values being associated with CCR1. This can be explained upon observing that the mean of log${\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{CCR}\\mathrm{2}}$ is more sensitive to temperature than log${\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{CCR}\\mathrm{1}}$ and log${\\stackrel{\\mathrm{̃}}{K}}_{\\mathrm{CCR}\\mathrm{3}}$ (see Fig. 4). This implies that, on average, CCR2 is activated at lower temperatures than CCR1 and CCR3. For the three considered CCRs, the mean temperature of activation is comprised between 246 and 287 C, values which are significantly lower than the threshold of 330 C reported by . The standard deviation (σ) of the distribution of ${\\stackrel{\\mathrm{̃}}{T}}_{\\mathrm{act}}$ depends on the CCR mechanism considered (Table 3). The combination of higher spread and lower mean characterizing the sample pdf of ${\\stackrel{\\mathrm{̃}}{T}}_{\\mathrm{act}}$ for CCR2 yields non-zero probability values even for quite low values of temperature (i.e., 159 C; see Table 3) as compared to the results of the preliminary assessments of .",
null,
"Figure 6Probability density functions (pdfs) of ${\\stackrel{\\mathrm{̃}}{T}}_{\\mathrm{act}}$ (a), ${\\stackrel{\\mathrm{̃}}{P}}_{\\mathrm{act}}$ (b) and ${\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}$ (c) evaluated for CCR1 (solid blue line), CCR2 (solid red line) and CCR3 (solid black line) at t= 0 Ma.\n\nSimilar observations can be made from the sample pdf of ${\\stackrel{\\mathrm{̃}}{P}}_{\\mathrm{act}}$ depicted in Fig. 6b. While pressure is known to have a limited impact on equilibrium constants of reactions, our results reveal its major role in the activation of the CCR mechanism. This is related to the observation that pore water pressure sets the threshold that is required to be exceeded so that a separate gas phase can be found in the system. As such, the key statistics of ${\\stackrel{\\mathrm{̃}}{P}}_{\\mathrm{act}}$ depend on the CCR mechanism investigated (Table 3).\n\nFigure 6 depicts the sample pdf of ${\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}$ for the three CCR mechanisms analyzed. The behavior of these results mirrors the one displayed by the distributions of ${\\stackrel{\\mathrm{̃}}{T}}_{\\mathrm{act}}$ in Fig. 6a. According to our probabilistic modeling framework, the distribution of ${\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}$ and the associated main statistics (Table 3) suggest that CCR2 is the activation mechanism which tends to take place at the shallowest depths. Indeed, while the mode and the mean of ${\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}$ are larger than 6.60 km for CCR1 and CCR3, the source depth with the highest probability is found at about 5.78 km for CCR2. A similar behavior is shown for the mean of ${\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}$. The higher spread associated with the population of sampled ${\\stackrel{\\mathrm{̃}}{T}}_{\\mathrm{act}}$ values for CCR2 is mirrored by the behavior of ${\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}$. As a consequence, the shallowest depth where CO2 generation might take place under the action of CCR2 corresponds to 3.2 km from the sea level, which is about 1.4 km smaller than that observed for CCR1. Note that the distributions of ${\\stackrel{\\mathrm{̃}}{T}}_{\\mathrm{act}}$ and ${\\stackrel{\\mathrm{̃}}{P}}_{\\mathrm{act}}$ collected in Fig. 6 provide a first quantitative assessment of the temperature and pressure of activation of CO2 generation characterizing CCR1, CCR2, and CCR3. Thus, results of this kind can be used to perform preliminary probabilistic evaluations of CCR activation as a CO2 source.\n\nTable 3Mean (μ), standard deviation (σ), mode and minimum value associated with the sample pdfs of ${\\stackrel{\\mathrm{̃}}{T}}_{\\mathrm{act}}$, ${\\stackrel{\\mathrm{̃}}{P}}_{\\mathrm{act}}$ and ${\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}$. Statistics are computed for CCR1, CCR2 and CCR3. The maximum values of ${\\stackrel{\\mathrm{̃}}{T}}_{\\mathrm{act}}$, ${\\stackrel{\\mathrm{̃}}{P}}_{\\mathrm{act}}$ and ${\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}$ are not listed as they correspond to the maximum temperature, pressure and depth observed in the selected setting, independent of the target CCR mechanism.",
null,
"The extent of the impact of the CCR formulations considered on the occurrence of CO2 generation can also be assessed by analyzing the relative frequency of activation associated with each point of the basin planar surface (fA (x, y)). The latter is depicted in Fig. 7a and has been estimated as\n\n$\\begin{array}{}\\text{(9)}& {f}_{A}\\left(x,y\\right)=\\sum _{z=\\mathrm{0}}^{z={Z}_{T}\\left(x,y\\right)}f\\left({\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}\\right),\\end{array}$\n\nwhere ZT is the maximum depth attained for each pair of coordinates (x, y) in the basin at t= 0 Ma. Figure 7b, c, and d depict the spatial distribution of fA(x, y) for CCR1, CCR2, and CCR3, respectively. These results indicate that the frequency of activation of the CCR mechanism is spatially heterogeneous. Its distribution shows a pattern that is closely dependent on the maximum depth attained by the sediments (see Fig. 7), being linked to the burial history of the considered basin. For all CCRs explored, the highest relative frequency of activation is observed at a location x and y where the basin stratigraphy is thickest (i.e., 8 km in our setting). This is consistent with the observation that sediments reaching deeper locations experience higher temperatures, thus leading to an overall increase in the probability that activation of CCR will be observed for a given location (x, y). We note that both CCR1 and CCR2 are characterized by a maximum value of fA(x, y) equal to 0.7; i.e., there is a planar location in the system where activation of these CCRs takes place in 70 % of the N MC realizations. On the other hand, the largest values of fA(x, y) for CCR3 attain values that are about 0.3, i.e., significantly smaller than those recorded for CCR1. This result is consistent with the observation that CC3 is less likely to activate than CCR1 at large depths, as suggested by the spatial distributions of $f\\left({\\stackrel{\\mathrm{̃}}{Z}}_{\\mathrm{act}}\\right)$ reported in Fig. 5a and c.",
null,
"Figure 7Maximum depth attained at each point of the basin ZT (a) and spatial distribution of fA(x,y), i.e., the total frequency of CCR activation for each combination of x and y coordinates for the corresponding column of sediments, throughout the planar surface of the basin domain at t= 0 Ma associated with CCR1 (b), CCR2 (c), and CCR3 (d).\n\nOur probabilistic workflow documents that the characteristic temperature and pressure associated with the activation of the CCR mechanism are driven by (a) the considered CCR formulation and (b) the mineralogical assemblage constituting the buffering systems. Thus, the probability of CO2 generation taking place at some depth in a sedimentary basin is markedly dependent on the three-dimensional temperature and pressure distribution as well as the selected buffering system.\n\nThe probabilistic delineation of the source location may profoundly depend on the CCR mechanism employed in the modeling workflow. This result is of key relevance in light of a subsequent analysis involving modeling of transport, migration, and accumulation of the generated CO2. Shallow sources are typically associated with a reduced traveling path of gaseous CO2 and a decreased possibility of CO2 re-mineralization. Therefore, a location of the CO2 source at relatively shallow depths may increase the probability of observing large accumulation in reservoirs of interest for oil and gas exploration as well as the probability that CO2 migration may influence vertical flow processes capable of influencing shallow groundwater bodies.\n\n## 4.2 Implications for a scenario-based CO2 migration modeling\n\nWhen dealing with subsurface CO2 migration modeling, a key step is the design of the input scenario, i.e., the definition of a location of the CO2 source (i.e., activation surface in a three-dimensional setting) and the CO2 generation rate. Our probabilistic framework can assist the design of multiple scenarios. In practice, this can be obtained through the following steps:\n\n1. the solution of Eqs. (1)–(4) for all N Monte Carlo samples yields the pdf characterizing ${\\stackrel{\\mathrm{̃}}{P}}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ at each spatial location of the three-dimensional sedimentary basin, such a pdf being conditional to the T and P values rendered by the burial model;\n\n2. starting from the cumulative probability distribution of ${\\stackrel{\\mathrm{̃}}{P}}_{{\\mathrm{CO}}_{\\mathrm{2}}}$, we obtain scenarios of CO2 partial pressure (ppw(CO2)), each associated with a given percentile (pw);\n\n3. a given scenario ppw(CO2) constitutes the input to the system of Eqs. (5)–(6) for the evaluation of the spatial distribution of Zact and ${F}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ associated with the percentile (or probability level) w.\n\nFor the considered time t = 0 Ma, we exemplify the types of scenarios which can be used as input for CO2 transport modeling by (a) selecting the 25th, 50th, 75th and 99th percentiles of the sample pdf of ${\\stackrel{\\mathrm{̃}}{P}}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ at each point in the sedimentary basin and (b) building corresponding three-dimensional scenarios of CO2 partial pressure distribution (ppw(CO2), with w=25, 50, 75, and 99) for each of the CCRs investigated in this study.\n\nFigure 8 depicts the spatial location of the activation source associated with the 50th, 75th and 99th percentiles of the distributions stemming from CCR1 (a), CCR2 (b), and CCR3 (c). Note that, regardless of the selected CCR formulation, when considering the 25th percentile of the CO2 partial pressure pdfs (corresponding to the pp25(CO2) scenarios), none of the locations in the basin satisfies the criterion of CCR mechanism activation. Thus, an activation surface is not observed for pp25(CO2) scenarios. The same reasoning underpins the lack of activation surfaces associated with pp50(CO2) and pp75(CO2) for CCR3 in Fig. 8c.",
null,
"Figure 8Three-dimensional illustration of activation surfaces yielded by pp50(CO2) (dark blue), pp75(CO2) (light blue) and pp99(CO2) (red) for CCR1 (a), CCR2 (b) and CCR3 (c).\n\nComparison of Fig. 8a and b indicates that the scenarios corresponding to the 50th and 75th percentiles yield activation surfaces with a similar extent and average depth for CCR1 and CCR2. Otherwise, the scenario associated with the 99th percentile displays markedly different features across the CCRs analyzed, the activation surface characterizing CCR2 being located at considerably shallower depths (and hence being more extended) than its counterparts in CCR1 or CCR3 (Fig. 8). This result derives from the differences between the CCRs observed for the probability densities of the activation mechanism at relatively low temperature, i.e., Tact < 250C, as discussed in Sect. 4.1 and illustrated in Fig. 6. Therefore, the extent and location of the activation surface may deeply change, depending on the selected CCR as well as its characteristic activation temperature. This aspect is further elucidated in the detailed depiction of Fig. 9, which juxtaposes the activation surfaces associated with pp99(CO2) for CCR1 and CCR2, the color scale quantifying the CO2 generation rate per unit square meter (${F}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ – kg CO2 m−2 Ma−1). While CCR2 yields an activation surface with a larger spatial extent, CCR1 is characterized by a higher specific CO2 generation rate. Values of the mean $\\mathit{\\mu }\\left({F}_{{\\mathrm{CO}}_{\\mathrm{2}}}\\right)$ and standard deviation $\\mathit{\\sigma }\\left({F}_{{\\mathrm{CO}}_{\\mathrm{2}}}\\right)$ of ${F}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ observed for both activation surfaces displayed in Fig. 9 are listed in Table 4. These results show that $\\mathit{\\mu }\\left({F}_{{\\mathrm{CO}}_{\\mathrm{2}}}\\right)$ is almost 1 order of magnitude larger for the CCR1 activation surface than for CCR2. This is consistent with the values of ${m}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ associated with CCR1 (0.182 kg${}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ kg${}_{\\mathrm{sed}}^{-\\mathrm{1}}$) and for CCR2 (0.02 kg${}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ kg${}_{\\mathrm{sed}}^{-\\mathrm{1}}$) (see Table 2). Because we assume here a constant burial velocity in Eq. (6), ${m}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ is the main quantity affecting ${F}_{{\\mathrm{CO}}_{\\mathrm{2}}}$, which varies mildly across the activation surface for both CCRs (see values of standard deviations in Table 4), a result which is in line with the modest spatial variability of porosity resulting from the burial model.",
null,
"Figure 9Three-dimensional illustration of activation surface yielded by pp99(CO2) for CCR1 and CCR2 and the corresponding CO2 generation rate for each point of the activation surfaces (kg CO2 m−2 Ma−1).\n\nThe overall estimated CO2 rates of emission from the two surfaces depicted in Fig. 9 are equal to 3.42 × 104 and 1.47 × 104 kg${}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ yr−1 for CCR1 and CCR2, respectively. Even as the activation surface associated with CCR2 is characterized by a remarkably smaller specific rate of emission, the order of magnitude of the ensuing overall gas generation is similar to the one of CCR1. Moreover, the shape of the activation surface (in both Figs. 89) is significantly influenced by the basin structure, which may lead to discontinuities in the spatial structure of the CO2 sources. The basin structure and stratigraphy are then key factors driving the amount of CO2 potentially generated by CCR mechanisms. As such, while the methodological framework we present is general, the results are case-specific, and an appropriate quantification of the uncertainty associated with the geological setting is always required to constrain modeling results.\n\nTable 4Mean(μ) and standard deviation (σ) computed for ${F}_{{\\mathrm{CO}}_{\\mathrm{2}}}$ computed for activation surfaces depicted in Fig. 9.",
null,
"5 Conclusions\n\nWe rely on a probabilistic modeling framework to model CO2 generation by considering the effect of a variety of carbonate–clay reactions in a realistic large-scale three-dimensional sedimentary setting. Our work is grounded on the probabilistic approach proposed by to treat carbonate–clay reactions (CCRs). Such an approach embeds quantification of parametric uncertainty associated with the thermodynamic equilibrium constants driving CCR and has been showcased by these authors in a one-dimensional set-up. In summary, the methodological approach we propose and the ensuing results can contribute to enhancing our understanding of the strength of the controls of diverse geochemical mechanisms on CO2 dynamics in subsurface environments, with potential implications for several fields of practical interest, including, e.g., carbon capture and storage (CCS, Metz et al.2005), large-scale groundwater flow modeling and enhanced oil recovery practices.\n\nHere, we consider a three-dimensional system with a diagenetic history feasibly encountered in a real geological setting. We analyze the impact of three different CCR formulations and mineral assemblage on (i) the probability of CCR activation as a function of temperature and pressure; (ii) the frequency of activation as a function of depth; and (iii) the shape and extent of the surface delimiting the three-dimensional CO2 source. Our study leads to the following major conclusions.\n\n1. The temperature and pressure of activation depend on the CCR considered. Modifying the reference CCR can lead to a markedly different scenario in terms of depth of the source and extent of the activation surface. We rely on geochemical equilibrium and quantify uncertainty associated with model parameters and inputs, the latter source of uncertainty corresponding to the uncertainty in the information required to describe the reference system (i.e., input uncertainty; Walker et al.2003). The presence of input uncertainties implies the possibility that diverse CCRs may occur and lead to differing degrees of importance of parametric uncertainty on CO2 generation. Our stochastic framework allows quantification of the (spatially and temporally dependent) probability distribution of the activation temperature and pressure associated with a given CCR. With reference to the depositional setting here analyzed, non-zero probabilities of CO2 generation are associated with temperature and pressure equal to 159 C and 569 bar, respectively. These values are considered to be small if compared to those typically observed in sedimentary basins and support the potential of CCR mechanisms to act as a CO2 source in diagenetic environments. Notably, activation of CCR in our showcase scenario might be feasible even at a depth of 3.2 km, i.e., at a location compatible with the average depths of a typical gas extraction well (i.e., 2.5 km). This result is of particular interest because the occurrence of shallow CO2 sources reduces the CO2 migration path towards hydrocarbon reservoirs, thus increasing the probability that the CO2 generated by CCR might reach the shallow cap rock without being precipitated as newly formed carbonates, diluted or re-dissolved in water.\n\n2. We quantify the way the considered input and parametric uncertainty propagates onto estimates of generated mass of CO2 in a three-dimensional system. This allows description of the extent and the shape of the CO2-generating source together with the associated CO2 generation rate. These are the two key elements contributing to the estimation of the amount of CO2 generated by a given CCR mechanism. Scenarios characterized by different surface-specific rates and source areas might lead to a similar overall amount of CO2 generated per unit of time. We document the benefits resulting from the implementation of a three-dimensional probabilistic quantification of the main features of CCR activation temperature and pressure and set the grounds for a quantitative stochastic appraisal of CO2 accumulation in subsurface systems.\n\n3. We show that the shape of the CO2-generating source is closely dependent on the basin structure and stratigraphy. Thus, the overall amount of CO2 generated in a sedimentary basin requires a site-specific assessment, fully embedding uncertainty quantification. In this context, our modeling approach and probabilistic framework are readily transferable to other cases of interest to design site-specific studies.\n\nOur methodology considers a single type of uncertainty source, i.e., the system thermodynamic parameters. As a future development, one can envision exploring the effects of multiple sources of uncertainty, including model and parametric uncertainties. Key points of interest include the study of (i) the impact of qualitatively and quantitatively different mineralogical compositions and heterogeneous spatial arrangement on CCR activation and CO2 generation rate, (ii) the joint occurrence of CCR and other processes, and (iii) the contribution to CCR characteristic activation temperature and pressure of uncertainties associated with parameters and factors embedded into the burial model (e.g., burial model boundary conditions, sediment thermal and mechanical properties).\n\nSupplement\n\nAuthor contributions\n\nGC performed numerical simulations, contributed to designing the research methodology, analyzed data, created figures, and wrote the first draft; CG contributed to designing the research methodology and contributed data; MDR provided research funding and contributed data; AG supervised the research, contributed to designing the research methodology, discussed the results, and contributed to the writing; GP supervised the research, contributed to designing the research methodology, discussed the results, and contributed to the writing.\n\nCompeting interests\n\nAlberto Guadagnini is a member of the editorial board of the journal.\n\nAcknowledgements\n\nWe are grateful to Brian Berkowitz and three anonymous reviewers for their constructive reviews and comments.\n\nFinancial support\n\nThe study is financed by the Eni SpA R&D project “Gas Systems – Basin Gas Systems Risk Analysis”.\n\nReview statement\n\nThis paper was edited by Brian Berkowitz and reviewed by three anonymous referees.\n\nReferences\n\nAllis, R., Chidsey, T., Gwynn, W., Morgan, C., White, S., Adams, M., and Moore, J.: Natural CO2 reservoirs on the Colorado Plateau and southern Rocky Mountains: Candidates for CO2 sequestration, in: Proceedings of the First National Conference on Carbon Sequestration, US Department of Energy, National Energy Technology Laboratory Washington, DC., 14–17, 2001. a\n\nAnderson, G. 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J., 25, 385–403, 2017. a\n\nColombo, I., Nobile, F., Porta, G., Scotti, A., and Tamellini, L.: Uncertainty Quantification of geochemical and mechanical compaction in layered sedimentary basins, Comput. Meth. Appl. M., 328, 122–146, 2018. a\n\nCoudrain-Ribstein, A. and Gouze, P.: Quantitative study of geochemical processes in the Dogger aquifer, Paris Basin, France, Appl. Geochem., 8, 495–506, 1993. a\n\nCoudrain-Ribstein, A., Gouze, P., and de Marsily, G.: Temperature-carbon dioxide partial pressure trends in confined aquifers, Chem. Geol., 145, 73–89, 1998. a, b, c, d, e, f, g, h\n\nDalla Rosa, M., Ruffo, P., Grigo, D., Caldiero, L., Dolci, D., et al.: Hydrocarbon Retention: A New Way to Evaluate Source Rock and Unconventional Resource Potential With Petroleum System Modelling Applications, in: Offshore Mediterranean Conference and Exhibition, Offshore Mediterranean Conference, Ravenna, Italy, March 2015, 2015. a\n\nDelany, J. and Lundeen, S.: The LLNL thermochemical database, Lawrence Livermore National Laboratory Report UCRL-21658, 150, 1990. a\n\nFeng, Z., Liu, D., Huang, S., Gong, D., and Peng, W.: Geochemical characteristics and genesis of natural gas in the Yan’an gas field, Ordos Basin, China, Org. Geochem., 102, 67–76, 2016. a\n\nFischer, M., Botz, R., Schmidt, M., Rockenbauch, K., Garbe-Schönberg, D., Glodny, J., Gerling, P., and Littke, R.: Origins of CO2 in permian carbonate reservoir rocks (Zechstein, Ca2) of the NW-German Basin (Lower Saxony), Chem. Geol., 227, 184–213, 2006. a\n\nFormaggia, L., Guadagnini, A., Imperiali, I., Lever, V., Porta, G., Riva, M., Scotti, A., and Tamellini, L.: Global sensitivity analysis through polynomial chaos expansion of a basin-scale geochemical compaction model, Comput. Geosci., 17, 25–42, 2013. a\n\nGiggenbach, W. F.: Geothermal gas equilibria, Geochim. Cosmochim. Ac., 44, 2021–2032, 1980. a\n\nGrigo, D., Maragna, B., Arienti, M., Fiorani, M., Parisi, A., Marrone, M., Sguazzero, P., and Uberg, A.: Issues in 3D sedimentary basin modelling and application to Haltenbanken, offshore Norway, Basin Modelling: Advances and Applications, Norwegian Petroleum Society (NPF), Special Publication, 3, 455–468, 1993. a\n\nHutcheon, I. and Abercrombie, H.: Carbon dioxide in clastic rocks and silicate hydrolysis, Geology, 18, 541–544, 1990. a, b, c\n\nHutcheon, I., Oldershaw, A., and Ghent, E. D.: Diagenesis of Cretaceous sandstones of the Kootenay Formation at Elk Valley (southeastern British Columbia) and Mt Allan (southwestern Alberta), Geochim. Cosmochim. Ac., 44, 1425–1435, 1980. a\n\nHutcheon, I., Abercrombie, H. J., Putnam, P., Gardner, R., and Krouse, H. R.: Diagenesis and sedimentology of the Clearwater Formation at Tucker Lake, B. Can. Petrol. Geol., 37, 83–97, 1989. a\n\nHutcheon, I., Abercrombie, H. J., and Krouse, H.: Inorganic origin of carbon dioxide during low temperature thermal recovery of bitumen: Chemical and isotopic evidence, Geochim. Cosmochim. Ac., 54, 165–171, 1990a. a\n\nHutcheon, I., Spencer, R. J., and Chou, I.: Clay carbonate reactions in the venture area, Scotian Shelf, Nova Scotia, Canada, Geo. Soc. S. P., 2, 199–212, 1990b. a, b, c, d, e\n\nHutcheon, I., Shevalier, M., and Abercrombie, H. J.: pH buffering by metastable mineral-fluid equilibria and evolution of carbon dioxide fugacity during burial diagenesis, Geochim. Cosmochim. Ac., 57, 1017–1027, 1993. a\n\nImbus, S. W., Katz, B. J., and Urwongse, T.: Predicting CO2 occurrence on a regional scale: Southeast Asia example, Org. Geochem., 29, 325–345, 1998. a\n\nJarvie, B. and Jarvie, D.: Thermal decomposition of various carbonates: Kinetics results and geological temperatures of conversion: 23rd International Meeting on Organic Geochemistry (IMOG) 2007, Torquay, UK, 311–312, 2007. a\n\nKissinger, A., Noack, V., Knopf, S., Konrad, W., Scheer, D., and Class, H.: Regional-scale brine migration along vertical pathways due to CO2 injection – Part 2: A simulated case study in the North German Basin, Hydrol. Earth Syst. Sci., 21, 2751–2775, https://doi.org/10.5194/hess-21-2751-2017, 2017. a, b\n\nKotarba, M. J. and Nagao, K.: Composition and origin of natural gases accumulated in the Polish and Ukrainian parts of the Carpathian region: Gaseous hydrocarbons, noble gases, carbon dioxide and nitrogen, Chem. Geol., 255, 426–438, 2008. a\n\nLi, M., Wang, T., Liu, J., Lu, H., Wu, W., and Gao, L.: Occurrence and origin of carbon dioxide in the Fushan depression, Beibuwan Basin, South China Sea, Mar. Petrol. Geol., 25, 500–513, 2008. a\n\nMaier, C. G. and Kelley, K.: An equation for the representation of high-temperature heat content data, J. Am. Chem. Soc., 54, 3243–3246, 1932. a\n\nMarín-Moreno, H., Bull, J. M., Matter, J. M., Sanderson, D. J., and Roche, B. J.: Reactive transport modelling insights into CO2 migration through sub-vertical fluid flow structures, Int. J. Greenh. Gas Con., 86, 82–92, https://doi.org/10.1016/j.ijggc.2019.04.018, 2019. a\n\nMenafoglio, A., Guadagnini, A., and Secchi, P.: Stochastic simulation of soil particle-size curves in heterogeneous aquifer systems through a Bayes space approach, Water Resour. Res., 52, 5708–5726, 2016. a\n\nMetz, B., Davidson, O., De Coninck, H., Loos, M., and Meyer, L.: IPCC special report on carbon dioxide capture and storage, Tech. rep., Intergovernmental Panel on Climate Change, Geneva (Switzerland), Working Group III, 2005. a\n\nMillero, F. J.: The effect of pressure on the solubility of minerals in water and seawater, Geochim. Cosmochim. Ac., 46, 11–22, 1982. a\n\nNeuman, S. P.: Maximum likelihood Bayesian averaging of uncertain model predictions, Stoch. Env. Res. Risk A., 17, 291–305, 2003. a\n\nPanariti, N. and Previde Massara, E.: Relazione tecnica: Generazione di CO2 da rocce carbonatiche, Eni SpA, Tech. rep., 2000. a\n\nParkhurst, D. L. and Appelo, C.: Description of input and examples for PHREEQC version 3: a computer program for speciation, batch-reaction, one-dimensional transport, and inverse geochemical calculations, Tech. rep., US Geological Survey, 2013. a\n\nShin, W. J., Chung, G. S., Lee, D., and Lee, K. S.: Dissolved inorganic carbon export from carbonate and silicate catchments estimated from carbonate chemistry and δ13CDIC, Hydrol. Earth Syst. Sci., 15, 2551–2560, https://doi.org/10.5194/hess-15-2551-2011, 2011. a\n\nSmith, J. and Ehrenberg, S.: Correlation of carbon dioxide abundance with temperature in clastic hydrocarbon reservoirs: relationship to inorganic chemical equilibrium, Mar. Petrol. Geol., 6, 129–135, 1989. a, b, c, d, e, f\n\nvan Berk, W., Schulz, H.-M., and Fu, Y.: Controls on CO2 fate and behavior in the Gullfaks oil field (Norway): How hydrogeochemical modeling can help decipher organic-inorganic interactions, AAPG Bull., 97, 2233–2255, 2013. a\n\nWalker, W., Harremoës, P., Rotmans, J., van der Sluijs, J., van Asselt, M., Janssen, P., and von Krauss, M. K.: Defining Uncertainty: A Conceptual Basis for Uncertainty Management in Model-Based Decision Support, Integrat. Ass., 4, 5–17, https://doi.org/10.1076/iaij.4.1.5.16466, 2003. a, b, c, d\n\nWycherley, H., Fleet, A., and Shaw, H.: Some observations on the origins of large volumes of carbon dioxide accumulations in sedimentary basins, Mar. Petrol. Geol., 16, 489–494, 1999. a\n\nXu, T. and Pruess, K.: On fluid flow and mineral alteration in fractured caprock of magmatic hydrothermal systems, J. Geophys. Res.-Sol. Ea., 106, 2121–2138, 2001. a\n\nZattin, M., Andreucci, B., de Toffoli, B., Grigo, D., and Tsikalas, F.: Thermochronological constraints to late Cenozoic exhumation of the Barents Sea Shelf, Mar. Petrol. Geol., 73, 97–104, 2016. a\n\nZhang, S., FitzGerald, J. D., and Cox, S. F.: Reaction-enhanced permeability during decarbonation of calcite quartz = wollastonite carbon dioxide, Geology, 28, 911–914, 2000. a"
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https://dokumen.pub/an-introduction-to-algebraic-geometry-and-algebraic-groups-1nbsped-0198528310-9780198528319-9780199676163-019967616x.html | [
"Content: 1. Algebraic sets and algebraic groups\n2. Affine varieties and finite morphisms\n3. Algebraic representations and Borel subgroups\n4. Frobenius maps and finite groups of Lie type\nBibliography\nIndex\n\n##### Citation preview\n\nOXFORD GRADUATE TEXTS IN MATHEMATICS Series Editors R. COHEN S.K. DONALDSON S. HILDEBRANDT T.J. LYONS M.J. TAYLOR\n\nBooks in the series 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23.\n\nKeith Hannabuss: An Introduction to Quantum Theory Reinhold Meise and Dietmar Vogt: Introduction to Functional Analysis James G. Oxley: Matroid Theory N.J. Hitchin, G.B. Segal, and R.S. Ward: Integrable Systems: Twistors, Loop Groups, and Riemann Surfaces Wulf Rossmann: Lie Groups: An Introduction through Linear Groups Qing Liu: Algebraic Geometry and Arithmetic Curves Martin R. Bridson and Simon M. Salamon (eds): Invitations to Geometry and Topology Shmuel Kantorovitz: Introduction to Modern Analysis Terry Lawson: Topology: A Geometric Approach Meinolf Geck: An Introduction to Algebraic Geometry and Algebraic Groups Alastair Fletcher and Vladimir Markovic: Quasiconformal Maps and Teichm¨ uller Theory Dominic Joyce: Riemannian Holonomy Groups and Calibrated Geometry Fernando Villegas: Experimental Number Theory P´eter Medvegyev: Stochastic Integration Theory Martin A. Guest: From Quantum Cohomology to Integrable Systems Alan D. Rendall: Partial Differential Equations in General Relativity Yves F´elix, John Oprea, and Daniel Tanr´e: Algebraic Models in Geometry Jie Xiong: Introduction to Stochastic Filtering Theory Maciej Dunajski: Solitons, Instantons, and Twistors Graham R. Allan: Introduction to Banach Spaces and Algebras James Oxley: Matroid Theory, Second Edition Simon Donaldson: Riemann Surfaces Clifford Henry Taubes: Differential Geometry: Bundles, Connections, Metrics and Curvature\n\nAn Introduction to Algebraic Geometry and Algebraic Groups Meinolf Geck Department of Mathematics, University of Stuttgart.\n\n1\n\n3\n\nGreat Clarendon Street, Oxford OX2, 6DP, United Kingdom Oxford University Press is a department of the University of Oxford. It furthers the Universitys objective of excellence in research, scholarship, and education by publishing worldwide. Oxford is a registered trade mark of Oxford University Press in the UK and in certain other countries c Oxford University Press, 2003 \u0001 The moral rights of the author have been asserted First published 2003 First published in paperback 2013 Impression: 1 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior permission in writing of Oxford University Press, or as expressly permitted by law, by licence or under terms agreed with the appropriate reprographics rights organization. Enquiries concerning reproduction outside the scope of the above should be sent to the Rights Department, Oxford University Press, at the address above You must not circulate this work in any other form and you must impose this same condition on any acquirer British Library Cataloguing in Publication Data Data available ISBN 978–0–19–852831–9 (Hbk.) ISBN 978–0–19–967616–3 (Pbk.) Printed in Great Britain by Biddles Ltd, www.biddles.co.uk\n\nPreface Algebraic geometry, in its classical form, is the study of algebraic sets in affine or projective space. By definition, an algebraic set in kn (where k is a field) is the set of all common zeros of a collection of polynomials in n variables. Algebraic groups are both groups and algebraic sets, where the group operations are given by polynomial functions. For example, the special linear group SLn (k) consisting of all n × n matrices with determinant 1 is an algebraic group. Historically, these groups were first studied in an analytic context, where the ground field k is R or C. This is the classical theory of ‘Lie groups’ (see Chevalley (1946) or Rossmann (2002), for example), which plays an important role in various branches of mathematics. Through the fundamental work of Borel and Chevalley in the 1950s, it is known that this theory also makes sense over an arbitrary algebraically closed field. This book contains an introduction to the theory of ‘groups of Lie type’ over such a general ground field k; consequently, the main flavour of the exposition is purely algebraic. In the last chapter of this book, we will even exclusively study the case where k is an algebraic closure of a finite field of characteristic p > 0. Then the corresponding algebraic groups give rise to various families of finite groups. By the classification of the finite simple groups, every non-abelian finite simple group arises from an algebraic group over a field of characteristic p > 0, except for the alternating groups and 26 sporadic simple groups; see Gorenstein et al. (1994). This is one reason why algebraic groups over fields of positive characteristic also play an important role. Although large parts of this book are developed in a general setting, the choice of the material and the orientation is towards groups over fields of positive characteristic and the corresponding ‘finite groups of Lie type’. Many research articles in this area begin with a statement like ‘Let G be a connected reductive algebraic group over Fq ; we assume that G is defined over Fq , with corresponding Frobenius map F ’\n\nvi\n\nPreface\n\nPreface\n\nvii\n\nLie type. And last but not least, I imagine someone who just wishes to study some basic aspects in a beautiful area of mathematics, where algebraic geometry and finite groups (which are, a priori, quite unrelated areas!) get mixed. In the elaboration of this book, I have used a number of sources. As far as algebraic geometry and commutative algebra are concerned, my favourite references are Atiyah and Macdonald (1969), Mumford (1988), and Shafarevich (1994). The standard textbooks on algebraic groups are Borel (1991), Humphreys (1991), and Springer (1998), which contain much more material than we can cover here. Our treatment owes a lot to the lecture notes of Steinberg (1974) and the article by Steinberg (1977), in which Steinberg gave new and, in many cases, more elementary proofs for standard results on algebraic groups. For the more specific theory of finite groups of Lie type and their characters (following Deligne–Lusztig), the standard reference is Carter (1985). Finally, the work of George Lusztig on characters of finite groups of Lie type has had the most important influence on the development of my own understanding of all these matters. For the experts, let me try to mention some points which I consider to be the distinctive features of this book. (1) The running examples in this book are the ‘classical groups’, that is, groups of matrices which leave a non-degenerate bilinear or quadratic form invariant. We will not discuss the classification of such forms, for which one can consult Dieudonn´e (1971) or Grove (2002), for example. Here, we introduce three families of groups (which are precisely the ‘groups of Lie type’ Bm , Cm , Dm ) corresponding to three fixed choices of a symmetric or alternating form. We will give a complete treatment of these groups in the framework of so-called ‘groups with a BN -pair’, a most useful concept introduced by Tits in the 1960s. In this approach, for example, the Dickson pseudodeterminant for orthogonal groups in characteristic 2 is not needed at all; see Section 1.7. (2) In this book, we exclusively work in the framework of affine varieties. There is no mention at all of general (quasi-projective) varieties, sheaves of functions, or such things. We do introduce projective space, but this will only play an auxiliary role, as a topological space on which an algebraic group acts. All the properties of projective varieties, grassmannian varieties, and flag varieties that are needed can be quite naturally obtained by working directly with\n\nviii\n\nPreface\n\nclosed cones in the underlying affine spaces. This is the approach taken by Steinberg (1977), and it will be used to establish the main results on Borel subgroups in Section 3.4. (3) The general structure theory of connected reductive groups shows that these groups always have a BN -pair in which B is a Borel subgroup. This important result is achieved as the culmination of a long series of arguments (see §14.15 of Borel (1991), for example), and we will not endeavour to prove it. In this book, we turn these matters upside down: We shall consider algebraic groups G with a BN -pair satisfying an additional condition which guarantees that G is reductive and that B is a Borel subgroup. As mentioned in (1), we establish the existence of such BN -pairs for all groups of classical type. In this way, we can avoid various classification results which form a tricky part in the general theory (for example, the classification of algebraic groups of dimension one). (4) There is a general theory of fields of definition and rationality properties for algebraic varieties and groups. In this book, we only discuss this in the case of fields of positive characteristic where the rational structure is determined by a Frobenius map. As a highlight, we obtain the general order formula for finite groups of Lie type; see Section 4.2. Finally, following Carter (1985), we present some basic properties of the theory of \u0001-adic cohomology in the form of axioms, on which a number of concrete applications and explicit computations can be based. This will be illustrated in detail for the example of the finite Suzuki groups. Most of the material in the first two chapters has been used as a basis for my undergraduate courses on elementary algebraic geometry at the University of Lyon 1 in 2000/2001/2002. The first five sections of Chapter 1 might actually be used as the core of a short course in algebraic geometry. For example, one could say more on algorithmic aspects using the excellent text of Cox et al. (1992); or treat in more detail algebraic curves in two-dimensional space, using Fischer (1994) or Reid (1988), for example; or one could extend the discussion of tangent spaces and Lie algebras in Section 1.4, by including more examples concerning classical groups, following Goodman and Wallach (1998). I used parts of Chapters 3 and 4 for a recent graduate course on algebraic groups over a field of characteristic p > 0 in Lyon and, much earlier, in 1995, in a two-semester course on algebraic geometry and algebraic groups at the RWTH Aachen.\n\nPreface\n\nix\n\nFinally, I wish to thank David H´ezard, Gerhard Hiss, and Gunter Malle, who carefully read various chapters and sent me detailed lists of corrections. For the typesetting of this text, the ‘LATEXCompanion’ by M. Goossens, F. Mittelbach, and A. Samarin (Addison–Wesley 1994) was an indispensible tool. Lyon, February 2003\n\nContents\n\n1\n\n2\n\n3\n\n4\n\nAlgebraic sets and algebraic groups 1.1 The Zariski topology on affine space 1.2 Groebner bases and the Hilbert polynomial 1.3 Regular maps, direct products, and algebraic groups 1.4 The tangent space and non-singular points 1.5 The Lie algebra of a linear algebraic group 1.6 Groups with a split BN -pair 1.7 BN -pairs in symplectic and orthogonal groups 1.8 Bibliographic remarks and exercises Affine varieties and finite morphisms 2.1 Hilbert’s nullstellensatz and abstract affine varieties 2.2 Finite morphisms and Chevalley’s theorem 2.3 Birational equivalences and normal varieties 2.4 Linearization and generation of algebraic groups 2.5 Group actions on affine varieties 2.6 The unipotent variety of the special linear groups 2.7 Bibliographic remarks and exercises\n\n1 1 10 23 36 48 60 74 87 102 103 111 121 132 140 151 158\n\nAlgebraic representations and Borel subgroups 3.1 Algebraic representations, solvable groups, and tori 3.2 The main theorem of elimination theory 3.3 Grassmannian varieties and flag varieties 3.4 Parabolic subgroups and Borel subgroups 3.5 On the structure of Borel subgroups 3.6 Bibliographic remarks and exercises\n\n165\n\nFrobenius maps and finite groups of Lie type 4.1 Frobenius maps and rational structures 4.2 Frobenius maps and BN -pairs\n\n224 225 235\n\n166 174 184 195 205 216\n\nxii\n\nContents 4.3 4.4 4.5 4.6 4.7\n\nFurther applications of the Lang–Steinberg theorem Counting points on varieties over finite fields The virtual characters of Deligne and Lusztig An example: the characters of the Suzuki groups Bibliographic remarks and exercises\n\n245 257 267 278 289\n\nIndex\n\n305\n\n1 Algebraic sets and algebraic groups In this first chapter, we introduce the basic objects: algebraic sets in affine space, the corresponding affine algebras, and algebraic groups. We proceed in a completely elementary way and assume only some very basic knowledge about groups, polynomial rings, quotient rings, and fields. The general theory will be mainly illustrated with examples concerning hypersurfaces and algebraic groups. In Section 1.1, we give the basic definitions concerning algebraic sets and prove a version of Hilbert’s nullstellensatz for hypersurfaces. As a tool for working out non-trivial examples, we introduce Groebner bases and some of their applications in Section 1.2. The dimension of an algebraic set will be defined using the Hilbert polynomial of an ideal. In Section 1.3, we introduce regular maps, direct products, and algebraic groups. As an important class of examples, we define symplectic and orthogonal groups. In Section 1.4, we prove the existence of non-singular points in irreducible algebraic sets. This theme will be continued in Section 1.5, where we introduce the Lie algebra of an algebraic group. As an application, we determine the dimensions of the orthogonal and symplectic groups. The final two sections contain a self-contained introduction to the theory of groups with a BN -pair. This concept has been introduced by Tits, and it has turned out to be extremely useful. It does not only apply to algebraic groups, but also to the finite groups arising as the fixed-point set under a Frobenius map. (The latter aspect will be studied in more detail in Chapter 4.) We shall describe such BN pairs for the symplectic and orthogonal groups. This will allow us to determine exactly which of these groups are connected.\n\n1.1 The Zariski topology on affine space In this section, we will define the basic objects of our study: algebraic sets in affine space kn . Before we do this, we establish Hilbert’s fundamental ‘basis theorem’. Throughout, let k be a field and consider\n\n2\n\nAlgebraic sets and algebraic groups\n\nthe polynomial ring k[X1 , . . . , Xn ], where the Xi are independent variables. For any subset S ⊆ k[X1 , . . . , Xn ], we denote by (S) the ideal generated by S. Thus, (S) consists of all finite sums of terms of the form hf where h ∈ k[X1 , . . . , Xn ] and f ∈ S. Recall that a commutative ring A with 1 is called a noetherian ring if every ideal of A is generated by a finite number of elements. (Note that, if I is an ideal in a noetherian ring A generated by S ⊆ I, then we can always choose a finite generating set for I from the elements of S.) 1.1.1 Theorem (Hilbert’s basis theorem) If A is noetherian and X is an indeterminate over A, then A[X] is also noetherian. In particular, if k is a field, then k[X1 , . . . , Xn ] is noetherian. Proof The second statement follows from the first by induction on n, starting with the fact that a field k certainly is noetherian. Now let I ⊆ A[X] be a non-zero ideal and assume that I is not finitely generated. Let d1 \u0001 0 be the minimum of all deg(f) (where deg(f) denotes the degree of f) for f ∈ I0 := I \\ {0}; choose f1 ∈ I0 with deg(f1 ) = d1 . Since I is not finitely generated, I1 := I \\ (f1 ) is not empty. Let d2 \u0001 d1 be the minimum of all deg(f) for f ∈ I1 ; choose f2 ∈ I1 with deg(f2 ) = d2 . Since I is not finitely generated, I2 := I \\ (f1 , f2 ) is not empty. Going on as above, we obtain a sequence of polynomials f1 , f2 , . . . ∈ I, of degrees d1 \u0002 d2 \u0002 . . ., such that fi+1 ∈ I \\ (f1 , . . . , fi ) for all i. Now let ai ∈ A be the leading coefficient of fi . Then the ideal I \u0002 = (ai | i \u0001 1) ⊆ A is finitely generated by assumption, and so there exists some h\u0001 \u0001 1 such that I \u0002 = (a1 , . . . , ah ). Then ah+1 ∈ I \u0002 means that ah+1 = hi=1 xi ai with xi ∈ A. We set g := fh+1 −\n\nh \u0002\n\nxi X dh+1 −di fi ∈ I \\ (f1 , . . . , fh ).\n\ni=1\n\nBy definition, we have deg(g) \u0002 dh+1 , but the coefficient of X dh+1 in \u0001 g is ah+1 − hi=1 xi ai = 0. So we have deg(g) < dh+1 , contradicting the choice of fh+1 . \u0003 We shall fix the following notation. Given x = (x1 , . . . , xn ) ∈ kn , there exists a unique k-algebra homomorphism εx : k[X1 , . . . , Xn ] → k such that εx (Xi ) = xi for all i. To simplify notation, we shall write f(x) = f(x1 , . . . , xn ) = εx (f) for f ∈ k[X1 , . . . , Xn ]. In particular,\n\nThe Zariski topology on affine space\n\n3\n\nany f ∈ k[X1 , . . . , Xn ] defines a function f˙ : kn → k, x \u0004→ f(x). Note ˙ However, this is that, in general, f is not uniquely determined by f. certainly the case if |k| = ∞; see Exercise 1.8.2. 1.1.2 Definition Let S ⊆ k[X1 , . . . , Xn ] be any subset. Then V(S) := {x ∈ kn | f(x) = 0 for all f ∈ S} is called the algebraic set defined by S. A subset of kn is called algebraic if it is of the form V(S) for some subset S ⊆ k[X1 , . . . , Xn ]. It is readily checked that V(S) = V(I), where I = (S) is the ideal generated by S. Then, using Hilbert’s basis theorem, we see that there exist f1 , . . . , fr ∈ S such that V(S) = V({f1 , . . . , fr }). Conversely, for any subset V ⊆ kn , the ideal I(V ) := {f ∈ k[X1 , . . . , Xn ] | f(x) = 0 for all x ∈ V } is called the vanishing ideal of V . Note that I(V ) is an ideal of k[X1 , . . . , Xn ]. The quotient A[V ] := k[X1 , . . . , Xn ]/I(V ) is called the affine algebra of V . 1.1.3 Example (a) Consider the constant 1 ∈ k[X1 , . . . , Xn ]. Then, clearly, V({1}) = ∅. On the other hand, V({0}) = kn . Thus, ∅ and kn are algebraic sets. We certainly have I(∅) = k[X1 , . . . , Xn ]. Similarly, one might expect that I(kn ) = {0}, but this will only be true if |k| = ∞; see Exercise 1.8.2. (b) Let x = (x1 , . . . , xn ) ∈ kn . Then {x} = V({X1 −x1 , . . . , Xn − xn }) ⊆ kn , and so the singleton set {x} is an algebraic set in kn . (c) Consider the case n = 1. Then an algebraic set V ⊆ k is the set of zeros of a collection of polynomials in k[X1 ]. Since every non-zero polynomial in one variable has at most finitely many zeros, we conclude that |V | < ∞ or V = k. Conversely, assume that we have a finite subset V = {v1 , . . . , vm } ⊆ k. Then V = V({f}), where f = (X1 − v1 ) · · · (X1 − vm ) ∈ k[X1 ]. Thus, we see that the algebraic sets in k are precisely the finite sets and k itself. (d) Assume that k is infinite. Let I ⊆ k[X1 , X2 , X3 ] be the ideal generated by f = X2 − X12 and g = X3 − X13 . The algebraic set C = V(I) ⊆ k3 is called the twisted cubic. Explicitly, we have C = {(x, x2 , x3 ) | x ∈ k} ⊆ k3 . It is already a non-trivial task to find I(C). Using Exercise 1.8.4(a), we obtain that I(C) = (f, g).\n\n4\n\nAlgebraic sets and algebraic groups\n\n1.1.4 Remark The following elementary properties are easily proved. (a) For subsets S ⊆ S \u0002 ⊆ k[X1 , . . . , Xn ], we have V(S \u0002 ) ⊆ V(S). (b) For any S ⊆ k[X1 , . . . , Xn ], we have S ⊆ I(V(S)). Similarly, we also have the following two further properties. (c) For subsets V ⊆ V \u0002 ⊆ kn , we have I(V \u0002 ) ⊆ I(V ). (d) For any subset V ⊆ kn , we have V ⊆ V(I(V )). Let {Vλ }λ∈Λ be a family of subsets of kn . Then it is clear that \u0003 \u0005 \u0004 \u0006 I Vλ = I(Vλ ). λ∈Λ\n\nλ∈Λ\n\nNow let {Sλ }λ∈Λ be a family of subsets of k[X1 , . . . , Xn ]. Then it is clear that \u0003 \u0005 \u0004 \u0006 V Sλ = V(Sλ ). λ∈Λ\n\nλ∈Λ\n\nIn particular, the intersection of an arbitrary number of algebraic sets in kn is again an algebraic set. Note that infinite unions of algebraic sets need not be algebraic. For example, assume that |k| = ∞ and let Z \u0002 k be a subset which is still infinite. Then the union of the algebraic sets {z} ⊆ k, for z ∈ Z, is not algebraic; see Example 1.1.3(c). However, we have the following result. 1.1.5 Lemma For any subsets S, S \u0002 ⊆ k[X1 , . . . , Xn ], we have V({f · g | f ∈ S, g ∈ S \u0002 }) = V(S) ∪ V(S \u0002 ). If S, S \u0002 are ideals in k[X1 , . . . , Xn ], then we also have V(S ∩ S \u0002 ) = V(S) ∪ V(S \u0002 ). Proof Clearly, we have V(S) ∪ V(S \u0002 ) ⊆ V({f · g | f ∈ S, g ∈ S \u0002 }). To prove the reverse inclusion, let x ∈ kn be such that f(x)g(x) = 0 for all f ∈ S and g ∈ S \u0002 . Assume, if possible, that x \b∈ V(S) and x \b∈ V(S \u0002 ). Then f(x) \b= 0 for some f ∈ S and g(x) \b= 0 for some g ∈ S \u0002 . Hence, f(x)g(x) \b= 0, a contradiction. Finally, assume that S, S \u0002 are ideals. Then we have V(S) ⊆ V(S∩ \u0002 S ) and V(S \u0002 ) ⊆ V(S ∩ S \u0002 ). Furthermore, for f ∈ S and g ∈ S \u0002 , we have f · g ∈ S ∩ S \u0002 , and so V(S ∩ S \u0002 ) ⊆ V({f · g | f ∈ S, g ∈ S \u0002 )} = V(S) ∪ V(S \u0002 ). \u0003\n\nThe Zariski topology on affine space\n\n5\n\n1.1.6 The Zariski topology We have seen in Remark 1.1.4 that arbitrary intersections of algebraic sets in kn are again algebraic. It follows from Lemma 1.1.5 that finite unions of algebraic sets in kn are again algebraic. Finally, by Example 1.1.3, the empty set ∅ and kn itself are algebraic. Thus, the algebraic sets in kn form the closed sets of a topology on kn , which is called the Zariski topology. A subset X ⊆ kn is open if kn \\ X is closed (that is, algebraic). Now every algebraic set V ⊆ kn is itself a topological space, with the induced topology. Explicitly, a subset Z ⊆ V is closed (algebraic) if Z is the intersection of V and an algebraic set in kn . Thus, by Remark 1.1.4, a closed subset of V is nothing but an algebraic set in kn which is contained in V . 1.1.7 The closure of a set Given any subset V ⊆ kn (not necessarily algebraic), we denote by V¯ ⊆ kn its closure in the Zariski topology on kn . Thus, V¯ is the intersection of all algebraic subsets of kn which contain V . We have V(I(V )) = V¯ . Indeed, by Remark 1.1.4, we have V ⊆ V(I(V )) and so V¯ ⊆ V(I(V )). Conversely, assume that W ⊆ kn is any algebraic set containing V . We write W = V(S) for some S ⊆ k[X1 , . . . , Xn ]. Then, again by Remark 1.1.4, we have S ⊆ I(V(S)) = I(W ) ⊆ I(V ) and so V(I(V )) ⊆ V(S) = W , as required. The above result shows, in particular, that we have V(I(V )) = V for any algebraic set V ⊆ kn . Hence the operator I : {V ⊆ kn | V algebraic} → {I ⊆ k[X1 , . . . , Xn ] | I ideal} is injective. The following example shows that I is not surjective. 1.1.8 Example Let f ∈ k[X1 , . . . , Xn ] be non-constant. Then the algebraic set Hf := V ({f}) ⊆ kn is called a hypersurface. If n = 2, then Hf is called a plane curve. Note that, with no further assumption, it may happen that Hf = ∅. (For example, if k = R and f = X12 + 1.) However, if k is algebraically closed and n \u0001 2, then |Hf | = ∞; see Exercise 1.8.2(b). Now assume that m \u0001 2. We claim that I(V ) \b= (f m ) for any algebraic set V ⊆ kn . Indeed, if we had I(V ) = (f m ), then we would also have V = V(I(V )) = V({f m }) = V({f}), and so f ∈ I(V ) = (f m ), which is impossible.\n\n6\n\nAlgebraic sets and algebraic groups\n\nThe following result describes the vanishing ideal of a hypersurface. 1.1.9 Theorem (Hilbert’s nullstellensatz for hypersurfaces) Let k be an algebraically closed field. Let f ∈ k[X1 , . . . , Xn ] be nonconstant and ∅ \b= Hf ⊆ kn the corresponding hypersurface; see Example 1.1.8. Write f = f1n1 · · · frnr , where f1 , . . . , fr are irreducible and pairwise coprime to each other. Then we have Hf = Hf1 ∪ · · · ∪ Hfr\n\nand\n\nI(Hf ) = (f1 · · · fr ),\n\nIn particular, if f is irreducible (or, more generally, separable), then I(Hf ) = (f). Proof Assume first that f is irreducible. Relabelling the variables Xi if necessary, we may assume that Xn occurs in f, i.e. we have f = al + al−1 Xn + · · · + a0 Xnl ,\n\nwhere ai ∈ k[X1 , . . . , Xn−1 ], a0 \b= 0, l \u0001 1.\n\nNow let 0 \b= g ∈ I(Hf ). Assume, if possible, that f does not divide g. Setting A := k[X1 , . . . , Xn−1 ], this means that f ∈ A[Xn ] is a nonconstant irreducible polynomial which does not divide g ∈ A[Xn ]. By Gauss’ lemma, this statement remains true if we regard f and g as polynomials in K[Xn ], where K is the field of fractions of A. Hence, since K[Xn ] is a principal ideal domain, we can write 1 = ˜ + F˜ g, where F˜ , G ˜ ∈ K[Xn ]. So there exist F, G ∈ A[Xn ] such Gf that 0 \b= d := Gf + Fg ∈ A = k[X1 , . . . , Xn−1 ]. Since |k| = ∞, there exists some x\u0002 = (x1 , . . . , xn−1 ) ∈ kn−1 with a0 (x\u0002 )d(x\u0002 ) \b= 0; see Exercise 1.8.2. Now consider the polynomial f˜ := al (x\u0002 ) + al−1 (x\u0002 )Xn + · · · + a0 (x\u0002 )Xnl ∈ k[Xn ]. Since a0 (x\u0002 ) \b= 0 and l \u0001 1, this is a non-constant polynomial. Hence, since k is algebraically closed, there exists some xn ∈ k such that ˜ n ) = 0. Now, set x = (x1 , . . . , xn−1 , xn ) ∈ kn . Then we have f(x ˜ n ) = 0 and so x ∈ Hf . Since g ∈ I(Hf ), we also have f(x) = f(x g(x) = 0 and, consequently, d(x\u0002 ) = d(x) = G(x)f(x)+ F (x)g(x) = 0, a contradiction. Thus, f divides g and so g ∈ (f). Now consider the general case and write f = f1n1 · · · frnr , where the fi are irreducible and pairwise coprime. Then, clearly, we have\n\nThe Zariski topology on affine space\n\n7\n\nHf = V({f1 · · · fr }) and, by Lemma 1.1.5, we have Hf = Hf1 ∪ · · · ∪ Hfr . This yields I(Hf ) = I(Hf1 ) ∩ · · · ∩ I(Hfr ) = (f1 ) ∩ · · · ∩ (fr ), where the first equality holds by Remark 1.1.4 and the second equality holds since each fi is irreducible. Finally, we have (f1 )∩· · ·∩(fr ) = (f1 · · · fr ) since f1 · · · fr is the least common multiple of f1 , . . . , fr . \u0003 We now discuss some topological notions concerning algebraic sets; in particular, we will discuss consequences of the fact that k[X1 , . . . , Xn ] is noetherian. 1.1.10 Definition Let Z \b= ∅ be a topological space. We say that Z is reducible if we can write Z = Z1 ∪ Z2 , where Z1 , Z2 ⊆ Z are non-empty closed subsets with Z1 \b= Z, Z2 \b= Z. Otherwise, we say that Z is an irreducible topological space. A subset Y ⊆ Z is called irreducible if Y is irreducible with the induced topology. (The empty set will not be considered as irreducible.) Furthermore, we say that Z is a noetherian topological space if every chain of closed sets Z1 ⊇ Z2 ⊇ · · · in Z becomes stationary, i.e., if there exists some m \u0001 1 such that Zm = Zm+i for all i \u0001 1. 1.1.11 Proposition Let Z \b= ∅ be a noetherian topological space. Then there are only finitely many maximal closed irreducible subsets in Z; they are called the irreducible components of Z. If these are Z1 , . . . , Zr , we have Z = Z1 ∪ · · · ∪ Zr . Proof First we show that Z can be written as a finite union of closed irreducible subsets. Assume, if possible, that this is not the case. Then, in particular, Z itself cannot be irreducible, and so we can write Z = Z1 ∪ Z1\u0002 , where Z1 , Z1\u0002 are proper closed subsets of Z. Now at least one of these two subsets, Z1 say, is not irreducible, and so we can write Z1 = Z2 ∪ Z2\u0002 , where Z2 , Z2\u0002 are proper closed subsets of Z1 . Going on this way, we find a strictly decreasing chain of closed subsets Z1 \u0003 Z2 \u0003 Z3 \u0003 · · · in Z, which is impossible since Z is assumed to be noetherian. Thus, our assumption was wrong and we can write Z = Z1 ∪ · · · ∪ Zr , where Z1 , . . . , Zr are closed irreducible subsets of Z. We may assume that Zi is not contained in Zj , for all i \b= j.\n\n8\n\nAlgebraic sets and algebraic groups\n\nNext we show that each closed irreducible subset Y ⊆ Z is contained in some Zi . Indeed, we have Y = Y ∩Z = Y ∩(Z1 ∪· · ·∪Zr ) = (Y ∩ Z1 ) ∪ · · · ∪ (Y ∩ Zr ). Now each Y ∩ Zi is closed in Y . So, since Y is irreducible, we must have Y ∩ Zi = Y for some i and, hence, Y ⊆ Zi . Thus, the Zi are precisely the maximal closed irreducible subsets of X. \u0003 1.1.12 Proposition Let V ⊆ kn be a non-empty algebraic set. (a) V is noetherian with respect to the Zariski topology. Thus, we have V = V1 ∪ · · · ∪ Vr , where the Vi are the maximal closed irreducible subsets of V . (b) V is irreducible (in the Zariski topology) if and only if I(V ) is a prime ideal. (Note that I(V ) \b= k[X1 , . . . , Xn ] since V \b= ∅.) Proof (a) Let V1 ⊇ V2 ⊇ V3 ⊇ · · · be a chain of closed (i.e algebraic) sets in V . Then we obtain a corresponding \u0007 chain I(V1 ) ⊆ I(V2 ) ⊆ · · · of ideals in k[X1 , . . . , Xn ]. Let I := i\u00011 I(Vi ). Then I is an ideal in k[X1 , . . . , Xn ] and so, by Hilbert’s basis theorem 1.1.1, I is finitely generated. A finite generating set for I will lie in I(Vm ) for some m \u0001 1. Then we have I(Vm ) = I(Vm+1 ) = · · · . By §1.1.7, we have Vi = V(I(Vi )) for all i, and so Vm = Vm+1 = · · · . (b) Assume first that V is irreducible, and let f, g ∈ k[X1 , . . . , Xn ] be such that fg ∈ I(V ). Then V ⊆ Hfg = Hf ∪ Hg (see Lemma 1.1.5 for the last equality). So we obtain V = V ∩ Hfg = (V ∩Hf )∪(V ∩Hg ). Since V is irreducible, we must have V ∩Hf = V or V ∩ Hg = V , which means that f ∈ I(V ) or g ∈ I(V ), as desired. Conversely, assume that I(V ) is a prime ideal and consider a decomposition V = V1 ∪ V2 , where V1 , V2 are closed subsets. Assume, if possible, that V1 \b= V and V2 \b= V . Using 1.1.7, this implies I(V ) \u0002 I(V1 ) and I(V ) \u0002 I(V2 ). So there exist some f ∈ I(V1 ) \\ I(V ) and some g ∈ I(V2 ) \\ I(V ). But, since V = V1 ∪ V2 , we also have fg ∈ I(V1 ) ∩ I(V2 ) = I(V ), which is impossible. \u0003 1.1.13 Example (a) A singleton set is obviously irreducible. Thus, if k is finite, then any algebraic set V ⊆ kn is the finite union of its points, which are all closed and irreducible. Now assume that k is infinite. Then, by Exercise 1.8.2, we have I(kn ) = {0}, and this is a prime ideal. Thus, kn is irreducible. (b) Assume that |k| = ∞ and consider the twisted cubic C ⊆ k3 introduced in Example 1.1.3(d). We have I(C) = (X2 −X12 , X3 −X13 ),\n\nThe Zariski topology on affine space\n\n9\n\nand this is a prime ideal by Exercise 1.8.4. Thus, C is an irreducible algebraic set. (c) Assume that k is algebraically closed and consider the hypersurface Hf ⊆ kn defined by a non-constant polynomial f ∈ k[X1 , . . . , Xn ]. By Theorem 1.1.9, we have Hf = Hf1 ∪ · · · ∪ Hfr , where the fi are irreducible and pairwise coprime. We have I(Hfi ) = (fi ) for each i, and so Hfi is irreducible by Proposition 1.1.12(b). Furthermore, we have Hfi \b⊆ Hfj for i \b= j, since the fi are coprime. Thus, Hf1 , . . . , Hfr are the irreducible components of Hf . 1.1.14 Principal open subsets Let V ⊆ kn be a non-empty algebraic set. Let f ∈ k[X1 , . . . , Xn ] be such that f \b∈ I(V ), and set Vf := {v ∈ V | f(v) \b= 0} ⊆ V. Then Vf is a non-empty open set in V and is called a principal open set. Any open set in V is a finite union of principal open sets. Indeed, let U ⊆ V be open. Then V \\ U is closed and so there exists an algebraic set W ⊆ kn such that V \\U = W ∩V . Now, by Hilbert’s basis theorem, W ∩ V is defined by a finite collection of polynomials, f1 , . . . , fr say. Then we have U = Vf1 ∪ · · · ∪ Vfr . There is another reason why principal open sets are important: in a suitable sense, they can be regarded as algebraic sets in their own right. (This will play a role, for example, in Section 2.2). To make this more precise, consider the polynomial ring k[X1 , . . . , Xn , Y ]. We identify kn × k = kn+1 . Then we define V˜f := {(v, y) ∈ V × k | f(v)y = 1} ⊆ kn+1 . The set V˜f is algebraic in kn+1 and the projection map πf : V˜f → V , (v, y) \u0004→ v, defines a bijection between V˜f and Vf . 1.1.15 Lemma\n\nIn the above setting, we have\n\nI(V˜f ) = (I, fY − 1) and\n\n¯ − 1), A[V˜f ] ∼ = (A[V ])[Y ]/(fY\n\nwhere I := I(V ) ⊆ k[X1 , . . . , Xn ] and f¯ denotes the image of f in A[V ]. Thus, in the notation of Exercise 1.8.11, A[V˜f ] ∼ = A[V ]f¯ is the ¯ localization of A[V ] in f. Proof We certainly have (I, fY − 1) ⊆ I(V˜f ). To prove the reverse inclusion, let g ∈ k[X1 , . . . , Xn , Y ] be such that g(v, y) = 0 for\n\n10\n\nAlgebraic sets and algebraic groups\n\nmust show that g ∈ (I, fY − 1). To see this, all (v, y) ∈ V˜f . We\u0001 r i let us\u0001write g = i=0 gi Y , where gi ∈ k[X1 , . . . , Xn ]; we set r r−i \u0002 g := i=0 f gi ∈ k[X1 , . . . , Xn ]. Then we have f r+1 g ≡ fg\u0002 mod (fY − 1) and so fg\u0002 ∈ I(V˜f ). Now, since g\u0002 ∈ k[X1 , . . . , Xn ], we have g\u0002 (v) = g\u0002 (v, y) = 0 for all v ∈ Vf (where y ∈ k is such that f(v)y = 1). On the other hand, f(v) = 0 for all v ∈ V \\ Vf , and so fg\u0002 ∈ I = I(V ). Thus, we have shown that f r+1 g ∈ (I, fY − 1). Writing g = (1 − (fY )r+1 )g + Y r+1 (f r+1 g) and noting that 1 − (fY )r+1 is divisible by 1 − fY in k[X1 , . . . , Xn , Y ], we conclude that g ∈ (I, fY − 1). The statement about A[V˜f ] is then clear. \u0003\n\n1.2 Groebner bases and the Hilbert polynomial The aim of this section is to develop some basic tools for working with ideals in k[X1 , . . . , Xn ]. More precisely, we introduce Groebner basis and the Hilbert polynomial of an ideal. For this purpose, let us fix some notation. Let Zn\u00010 := {α = (α1 , . . . , αn ) ∈ Zn | αi \u0001 0 for all i}. For α = (α1 , . . . , αn ) ∈ Zn\u00010 , we denote the corresponding monomial by X α := X1α1 · · · Xnαn . Then any non-zero f ∈ k[X1 , . . . , Xn ] is expressed uniquely as \u0002 f= aα X α , where α ∈ Zn\u00010 and aα ∈ k, α\n\nwith aα = 0 for all but finitely many α. We write |α| = α1 + · · · + αn . The maximum of all |α| such that aα \b= 0 is called the degree of f and denoted by deg(f). (If f = 0, we set deg(f) = −∞.) 1.2.1 Definition A total order on Zn\u00010 is called a monomial order if the following conditions hold. (a) We have (0, . . . , 0) α for all α ∈ Zn\u00010 . (b) For any α, β, γ ∈ Zn\u00010 we have: α β ⇒ α + γ β + γ. \u0001 Now let 0 \b= f = α aα X α ∈ k[X1 , . . . , Xn ]. The expressions aα X α , where aα \b= 0, are called the terms of f. Let α0 be maximal (with respect to a given monomial order ) such that aα0 \b= 0. Then LT(f) := aα0 X α0 is called the leading term and LM(f) := X α0 is called the leading monomial of f.\n\nGroebner bases and the Hilbert polynomial\n\n11\n\nGiven α, β ∈ Zn\u00010 , we also write X α X β if α β. Note that X α | X β ⇒ α β. 1.2.2 Example (a) The lexicographic order LEX: Let α, β ∈ Zn\u00010 , α \b= β. Let i ∈ {1, . . . , n} be minimal such that αi \b= βi . Then we write α β if αi < βi . (b) The graded lexicographic order GR-LEX: Let α, β ∈ Zn\u00010 , α \b= β. If |α| \b= |β|, we write α β if |α| < |β|. On the other hand, if |α| = |β|, we write α β if αi < βi where, as before, i is minimal such that αi \b= βi . It is easily verified that these two order relations indeed are monomial orders. 1.2.3 Lemma Let I ⊆ k[X1 , . . . , Xn ] be an ideal which is generated by a set G of monomials. Then a polynomial f ∈ k[X1 , . . . , Xn ] lies in I if and only if every term of f is divisible by some g ∈ G. Proof If every term of f is divisible by some g ∈ G,\u0001then f certainly lies in I. Now assume that f ∈ I and write f = ri=1 hi gi , where hi ∈ k[X1 , . . . , Xn ] and gi ∈ G. Expressing each hi as a k-linear combination of monomials and expanding the products, we see that f is a k-linear combinations of terms of form X α gi , for various i and α ∈ Zn\u00010 , as required. \u0003 1.2.4 Lemma Let (gi )i\u00011 be a sequence of monomials in k[X1 , . . . , Xn ] such that g1 \u000e g2 \u000e g3 \u000e · · · for some monomial order . Then there exists some r \u0001 1 such that gr = gr+1 = gr+2 = · · · . Proof Let I be the ideal generated by all gi (i \u0001 1). By Hilbert’s basis theorem, I is generated by the first, r say, elements g1 , . . . , gr . Now let i > r; then gr \u000e gi . On the other hand, since gi ∈ I, there exists some j ∈ {1, . . . , r} such that gj divides gi ; see Lemma 1.2.3. Then gi \u000e gj \u000e gr and so gi = gr , as desired. \u0003 1.2.5 Proposition (division algorithm) Let us fix a monomial order and let f, f1 , . . . , fs ∈ k[X1 , . . . , Xn ] be non-zero. Then we have f = h1 f1 +· · ·+hs fs +r, where r, h1 , . . . , hs ∈ k[X1 , . . . , Xn ] are such that LT(hi fi ) LT(f) for all i with hi \b= 0, and where r = 0 or no term of r is divisible by LT(fi ) for any 1 \u0002 i \u0002 s.\n\n12\n\nAlgebraic sets and algebraic groups\n\nProof Since every non-empty collection of monomials in k[X1 , . . . , Xn ] has a minimal element by Lemma 1.2.4, we can proceed by downward induction on LT(f) with respect to . If f is constant, we can take hi = 0 and r = f. Now assume that f is non-constant. If some LT(fi ) divides LT(f), we set LT(f) f˜ := f − fi . LT(fi ) ˜ ≺ LT(f) and we can If f˜ = 0, we are done. Otherwise, we have LT(f) ˜ ˜ 1 f1 + · · · + apply induction to f. Thus, we have an expression f˜ = h ˜ hs fs + r satisfying the above conditions. Then f = h1 f1 + · · · + ˜ j for j \b= i and hi = h ˜ i + LT(f)/ LT(fi ), hs fs + r, where hj = h and all the required conditions are satisfied. On the other hand, if no such i exists, we set f˜ := f − LT(f). If f˜ = 0, we are done. Otherwise, we can apply induction to f˜ and obtain an expression f˜ = h1 f1 + · · · + hs fs + r˜ satisfying the above conditions. Then we have f = h1 f1 + · · · + hs fs + r, where r = r˜ + LT(f), and all the required conditions are satisfied. \u0003 1.2.6 Example Let us consider the order LEX in k[X, Y ] with X \u000e Y . Let f = XY 2 − X and f1 = XY + 1, f2 = Y 2 − 1. Then we have f = Y · f1 + 0 · f2 + (−X − Y )\n\nand f = 0 · f1 + X · f2 + 0.\n\nBoth expressions satisfy all the conditions in Proposition 1.2.5. So the ‘remainder’ r is not uniquely determined. Even worse, the second expression shows that f ∈ (f1 , f2 ), while this is not obvious from the first expression. 1.2.7 Definition (Buchberger) Let I ⊆ k[X1 , . . . , Xn ] be a nonzero ideal and be a monomial order on Zn\u00010 . A finite subset G ⊆ I \\ {0} is called a Groebner basis of I if the monomials LT(g) (g ∈ G) generate the ideal (LT(I)) := (LT(f) | 0 \b= f ∈ I) ⊆ k[X1 , . . . , Xn ]. By Hilbert’s basis theorem, every non-zero ideal has a Groebner basis. 1.2.8 Theorem Let I ⊆ k[X1 , . . . , Xn ] be a non-zero ideal and be a monomial order on Zn\u00010 . Let G be a Groebner basis of I. Then\n\nGroebner bases and the Hilbert polynomial\n\n13\n\nwe have I = (G), and a k-basis of k[X1 , . . . , Xn ]/I is given by the residue classes of X α , where α runs over the elements in C(I) := {α ∈ Zn\u00010 | X α is not divisible by LT(g) for any g ∈ G}. Proof Let G = {f1 , . . . , fs } and 0 \b= f ∈ k[X1 , . . . , Xn ]. By Proposition 1.2.5, we can write f = h1 f1 + · · · + hs fs + r, where r = 0 or no term of r is divisible by LT(fi ) for any i. Note that this means that r is a k-linear combination of monomials X α , where α ∈ C(I). This already shows that the residue classes of the monomials X α (α ∈ C(I)) span k[X1 , . . . , Xn ]/(G). Now let f \u0001∈ I and assume, if possible, that r \b= 0. Then 0 \b= r = f − si=1 hi fi ∈ I and so LT(r) ∈ (LT(f1 ), . . . , LT(fs )). By Lemma 1.2.3, some LT(fi ) must divide LT(r), contradicting the conditions on r in Proposition 1.2.5. Thus, we have r = 0. This shows, in particular, that I = (G) and that the residue classes of X α (α ∈ C(I)) are linearly independent in k[X1 , . . . , Xn ]/I. \u0003 1.2.9 Example (a) Let 0 \b= f ∈ k[X1 , . . . , Xn ] and I = (f). Choose any monomial order on Zn\u00010 . Then LT(f) divides LT(g) for any non-zero g ∈ I. (Indeed, write g = hf with h ∈ k[X1 , . . . , Xn ]. Using Definition 1.2.1(b), we see that LT(g) = LT(h) LT(f).) Hence, {f} is a Groebner basis of I. (b) Consider the ideal I = (X2 − X12 , X3 − X13 ) ⊆ k[X1 , X2 , X3 ]; see Example 1.1.3(d). If we choose a lexicographic order with X1 X2 X3 , then Exercise 1.8.4 shows that {X2 − X12 , X3 − X13 } is a Groebner basis of I. On the other hand, choosing a graded lexicographic order with X3 X2 X1 , it turns out that {X12 − X2 , X1 X2 − X3 , X1 X3 − X22 } is a Groebner basis of I. (c) Let {0} \b= I ⊆ k[X1 , . . . , Xn ] be an ideal which is generated by a finite set of monomials, G say. Then G is a Groebner basis for any monomial order . Indeed, we clearly have G ⊆ (LT(I)). Conversely, let 0 \b= f ∈ I. Then LT(f) is divisible by some g ∈ G (see Lemma 1.2.3) and so LT(f) ∈ (G). 1.2.10 Remark There are algorithms for computing Groebner bases of any non-zero ideal in k[X1 , . . . , Xn ]. Without giving the proof, let us just describe a simple version of Buchberger’s algorithm, following §2.7 of Cox et al. (1992). First, we shall need the following definition. Let f, g ∈ k[X1 , . . . , Xn ] be non-zero. We write LM(f) = X α and LM(g) = X β and set γ = (γ1 , . . . , γn ) where\n\n14\n\nAlgebraic sets and algebraic groups\n\nγi = max{αi , βi } for all i. Then the S-polynomial of f and g is defined by Xγ Xγ ·f − · g, S(f, g) := LT(f) LT(g) Now let I = (f1 , . . . , fs ) be an ideal in k[X1 , . . . , Xn ] where fi \b= 0 for all i. Then a Groebner basis G is constructed by the following steps: Initialize G\u0002 := {f1 , . . . , fs }. For each pair p \b= q in G\u0002 , compute S(p, q) and divide S(p, q) by G\u0002 , leaving a remainder. Add all nonzero remainders that you obtain in this way to G\u0002 and repeat the above procedure. After a finite number of iterations, you will find that, for any pair p \b= q in G\u0002 , the remainder of S(p, q) under division by G\u0002 is zero. Then G = G\u0002 is a Groebner basis for I. For practical issues (computer implementation etc.) see, for example, the SINGULAR system described by Greuel and Pfister (2002). We also remark that, in general, Groebner bases are not unique. A Groebner basis G is called reduced if all g ∈ G satisfy the following conditions. (a) The coefficient of LT(g) is 1. (b) No term of g is divisible by LT(g\u0002 ) for any g\u0002 ∈ G, g\u0002 \b= g. Given a Groebner basis G, it is easy to construct a new Groebner basis which is reduced; furthermore, such a reduced Groebner basis is uniquely determined. For more details on all of this, see Cox et al. (1992). Our next aim is to define an invariant which measures the size of an ideal I ⊆ k[X1 , . . . , Xn ]. For this purpose, we study the ‘growth’ of k[X1 , . . . , Xn ]/I. 1.2.11 Definition Let I ⊆ k[X1 , . . . , Xn ] be an ideal. For any integer s \u0001 0, we set k[X1 , . . . , Xn ]\u0002s := {f ∈ k[X1 , . . . , Xn ] | deg(f) \u0002 s} and I\u0002s := I ∩ k[X1 , . . . , Xn ]\u0002s . Then k[X1 , . . . , Xn ]\u0002s is a finitedimensional vector space over k and I\u0002s is a subspace of it. Thus, we can define a function a\n\nHFI : Z\u00010 → Z\u00010 ,\n\ns \u0004→ dimk (k[X1 , . . . , Xn ]\u0002s /I\u0002s ).\n\nThis function is called the(affine) Hilbert function of I.\n\nGroebner bases and the Hilbert polynomial\n\n15\n\n1.2.12 Example (a) Let I = k[X1 , . . . , Xn ]. Then a HFI (s) = 0 for all s \u0001 0. (b) Let I = {0}. Then a HFI (s) = dimk (k[X1 , . . . , Xn ]\u0002s ), and this is the number of all monomials X1α1 · · · Xnαn such that |α| \u0002 s. Thus, we have to count all n-tuples of non-negative integers whose sum is \u0002 s. It is well-known\b that the number of such tuples is given by the binomial coefficient s+n s . So we have\n\n1 s+n s+n a = = (s + n)(s + n − 1) · · · (s + 1) HF{0} (s) =\n\n\u000e \u000f n s n! n factors\n\nn \u0010\u0002 1 n 1 n + 1 n−1 1\u0011 s s + 1. = s + + ··· + 2 n! n! i\n\ni=1\n\n(c) Let I = (f), where 0 \b= f ∈ k[X1 , . . . , Xn ]. Then a k-vectorspace basis of I is given by all products X1α1 · · · Xnαn f, where α = (α1 , . . . , αn ) ∈ Zn\u00010 ; the degree of such a product is |α| + d, where d = deg(f). Now assume that s \u0001 d. Then dimk (I\u0002s ) is the number of all α such that |α| + d \u0002 s or, equivalently, |α| \u0002 s − d. Using the formulas in (b), we find that\n\ns−d+n s+n s−d+n s+n a − = − HF(f) (s) = n n s−d s d = sn−1 + linear combination of lower powers of s. (n − 1)! Note that the above formula \b is only valid for s \u0001 d; if s < d, we have = a HF{0} (s) = s+n n . In the above examples, the Hilbert function is given by a polynomial in s. Our aim is to show that this is true for any ideal in k[X1 , . . . , Xn ]. a HF (s) (f)\n\n1.2.13 Lemma (Macaulay) Let be a graded lexicographic monomial order. For an ideal I ⊆ k[X1 , . . . , Xn ], we have a HFI (s) = a HF (LT(I)) (s) for all s \u0001 0. Proof If I = {0}, there is nothing to prove. So let us assume now that I \b= {0} and let s \u0001 0. Then, clearly, there are only finitely many monomials occurring in the elements of I\u0002s . In particular, we can write {LM(f) | 0 \b= f ∈ I\u0002s } = {LM(f1 ), . . . , LM(fm )},\n\n(†)\n\n16\n\nAlgebraic sets and algebraic groups\n\nwhere fi ∈ I\u0002s and LM(fm ) \u0004 · · · \u0004 LM(f1 ). We shall show that (a) {f1 , . . . , fm } is a k-basis of I\u0002s ; (b) {LM(f1 ), . . . , LM(fm )} is a k-basis of (LT(I))\u0002s . (Note that, once (a) and (b) are established, we are done since then I\u0002s and (LT(I))\u0002s have the same dimension.) First, we check that the sets in (a) and (b) are linearly independent. This is clear in (b) since any set of distinct monomials\u0001 is linearly independent. Now assume that we have a relation f = m i=1 ai fi , where not all ai ∈ k are 0. We must show that f \b= 0. Let i0 be minimal such that ai0 \b= 0. Then, since LM(fi ) \u0004 LM(fi0 ) for all i > i0 , we have LM(f) = ai0 LM(fi0 ) \b= 0, as desired. Next, we check that the sets in (a) and (b) are generating sets. We begin with (b). Let\u0001g ∈ (LT(I))\u0002s . This means that we can write g in the form g = j hj LM(gj ), where hj ∈ k[X1 , . . . , Xn ] and gj ∈ I. Writing each hj as a sum of its terms and expanding the product, we see that we may assume without loss of generality that each hj lies in k. Then, since we have chosen a graded lexicographic order, this implies that gj ∈ I\u0002s for all j. Using (†), we conclude that g is a k-linear combination of {LM(f1 ), . . . , LM(fm )}. Finally, consider (a). Let 0 \b= f ∈ I\u0002s . Then, by (†), we have LM(f) = LM(fi ) for some i, and so LT(f) = a LT(fi ), where 0 \b= a ∈ k. Now set f \u0002 := f − afi . If f \u0002 = 0, we are done. If f \u0002 \b= 0, we repeat the whole argument. Since LT(f \u0002 ) \u0004 LT(f), this process stops after a finite number of iterations, by Lemma 1.2.4. \u0003 1.2.14 Theorem For any ideal I ⊆ k[X1 , . . . , Xn ], there exists a unique polynomial a HPI (t) ∈ Q[t] (where t is an indeterminate) and s0 \u0001 0 such that a\n\nHPI (s) = a HFI (s) = dimk (k[X1 , . . . , Xn ]\u0002s /I\u0002s )\n\nfor all s \u0001 s0 .\n\nIf I = k[X1 , . . . , Xn ], we have a HPI (t) = 0. If I \b= k[X1 , . . . , Xn ], then a HPI (t) is a non-zero polynomial with the following properties. (a) The degree of a HPI (t) is the largest d with the property that there exist indices 1 \u0002 i1 < · · · < id \u0002 n such that I ∩ k[Xi1 , . . . , Xid ] = {0}. (b) Let d = deg a HPI (t). Then we have a HPI (t) = ad td + · · · + a1 t + a0 with d! ai ∈ Z for all i and d! ad > 0.\n\nGroebner bases and the Hilbert polynomial The polynomial nomial of I.\n\na HP (t) I\n\n17\n\nis called the (affine) Hilbert poly-\n\nProof First note that the polynomial a HPI (t) is unique, once we know that it exists. If I = {0} or I = k[X1 , . . . , Xn ], then everything follows from the explicit computations in Example 1.2.12. So let us now assume that {0} \b= I \b= k[X1 , . . . , Xn ]. Let G be a Groebner basis of I (with respect to a graded lexicographic order) and write {LM(g) | g ∈ G} = {X β | β ∈ M }. We set C(I) = {α ∈ Zn\u00010 | X α is not divisible by X β for any β ∈ M }. For any s \u0001 0, let C(I)\u0002s be the set of all α ∈ C(I) such that |α| \u0002 s. We claim that a HFI (s) = |C(I)\u0002s | for all s \u0001 0. Indeed, by Lemma 1.2.13, we have a HFI (s) = a HF(LT(I)) (s). Furthermore, by Example 1.2.9(c), {X β | β ∈ M } is a Groebner basis of (LT(I)). Hence, by Lemma 1.2.3, we have that dim LT(I)\u0002s equals the number of all α such that |α| \u0002 s and X β | X α for some β ∈ M . Thus, we have a HFI (s) = |C(I)\u0002s | as required. Now we continue in four steps. Step 1. Given a subset J ⊆ {1, . . . , n} and a function τ : J → Z\u00010 , we define C(J, τ ) := {(α1 , . . . , αn ) ∈ Zn\u00010 | αj = τ (j) for all j ∈ J}. Now we claim that there exists a finite collection J of pairs (J, τ ) as above such that \u0004 C(I) = C(J, τ ). (∗) (J,τ )∈J\n\nThis is seen as follows. For β = (β1 , . . . , βn ) ∈ Zn\u00010 , we define C(β) n β α to be the set of all \u0012 α ∈ Z\u00010 such that X does not divide X\u0002 . \u0002Then we have C(I) = β∈M C(β). Now note that if (J, τ ) and (J , τ ) are two pairs as above, then ⎧ ⎪ if τ (j) \b= τ \u0002 (j) for some ⎨∅ C(J, τ ) ∩ C(J \u0002 , τ \u0002 ) = j ∈ J ∩ J \u0002, ⎪ ⎩ C(J ∪ J \u0002 , τ0 ) otherwise, where τ0 is defined by τ0 (j) = τ (j) if j ∈ J and τ0 (j) = τ \u0002 (j) if j ∈ J \u0002 . Thus, it remains to show that C(β) (for any β ∈ Zn\u00010 ) is a finite union of sets of the form C(J, τ ). But, we have (α1 , . . . , αn ) ∈ C(β) if\n\n18\n\nAlgebraic sets and algebraic groups\n\nand only if αi < βi for some 1 \u0002 i \u0002 n. Thus, as required, we have C(β) =\n\nn β\u0004 i −1 \u0004\n\n{(α1 , . . . , αn ) ∈ Zn\u00010 | αi = ti }\n\ni=1 ti =0\n\n=\n\nn β\u0004 i −1 \u0004\n\nC({i}, τ : i \u0004→ ti ).\n\ni=1 ti =0\n\nStep 2. We claim that, for any pair (J, τ ) as in Step 1, there exists a polynomial FJ,τ (t) ∈ Q[t] such that deg FJ,τ (t) = n − |J| and FJ,τ (s) = |C(J, τ )\u0002s | \u0001 for all s \u0001 |τ | := j∈J τ (j). Indeed, let γ = (γ1 , . . . , γn ) ∈ Zn\u00010 be such that γj = 0 for j \b∈ J and γj = τ (j) for j ∈ J. Then C(J, τ ) is the set of all α + γ, where α = (α1 , . . . , αn ) ∈ Zn\u00010 is such that αj = 0 for j ∈ J. Thus, |C(J, τ )\u0002s | is the number of monomials in the variables Xj (j \b∈ J) such that, when multiplied with X γ , the total degree is at most s. This is the same as the number of monomials in d := n − |J| variables of total degree at most s − |γ| (note that |τ | = |γ|). Hence, as in Example 1.2.12(b), we find that\n\n1 d + s − |τ | = (sd + combination of lower |C(J, τ )\u0002s | = d d! powers of s). Step 3. If A1 , . . . , Am are finite subsets of some set, then we have |A1 ∪ · · · ∪ Am | =\n\nm \u0002\n\n\u0002\n\n(−1)r−1 |Ai1 ∩ · · · ∩ Air |;\n\nr=1 1\u0002i1 0, and consider the regular map ϕ : k → k, x \u0004→ xp . Then ϕ is bijective but ϕ∗ : k[X] → k[X], X \u0004→ X p , is not surjective. Thus, ϕ is not an isomorphism! 1.3.6 Example Let f1 , . . . , fm ∈ k[X1 , . . . , Xn ] and define ϕ : kn → km by ϕ(x) = (f1 (x), . . . , fm (x)). As we have seen in the above example, the image of ϕ need not be closed. So let us set Vϕ :=\n\n26\n\nAlgebraic sets and algebraic groups\n\nof R := k[X1 , . . . , Xn , Y1 , . . . , Ym ] and consider the ideal I = Iˆ ∩ k[Y1 , . . . , Ym ]\n\nwhere Iˆ := (Y1 − f1 , . . . , Ym − fm ) ⊆ R.\n\nˆ ⊆ kn × km = kn+m and Then consider the algebraic set Vˆ = V(I) n m m the projection πm : k ×k → k . We claim that, if |k| = ∞, then πm (Vˆ ) = ϕ(kn )\n\nand Vϕ = V(I) ⊆ km .\n\nProof For any (x, y) ∈ kn × km , we have (x, y) ∈ Vˆ if and only if y = ϕ(x). Thus, we see that πm (Vˆ ) = ϕ(kn ), and it remains to show that V(I) = πm (Vˆ ). Now, if (x, y) ∈ Vˆ and f ∈ I, then f(y) = f(x, y) = 0, where the first equality holds since f only involves the variables Y1 , . . . , Ym ˆ Thus, and the second equality holds since f ∈ Iˆ and (x, y) ∈ V(I). we have πm (Vˆ ) ⊆ V(I). Conversely, let f ∈ I(πm (Vˆ )). Since f only involves the variables ˆ by Exercise 1.8.4. Y1 , . . . , Ym , we have f ∈ I(Vˆ ) and so f ∈ I, ˆ Thus, f ∈ I ∩ k[Y1 , . . . , Ym ] = I, and it follows that Vϕ = V(I) ⊆ \u0003 V(I(πm (Vˆ ))) = πm (Vˆ ) using §1.1.7. For example, consider the map ϕ : k → k3 , x \u0004→ (x, x2 , x3 ). Then the ideal Iˆ is generated by {Y1 − X, Y2 − X 2 , Y3 − X 3 } ⊆ k[X, Y1 , Y2 , Y3 ]. A Groebner basis with respect to LEX with Y1 Y2 Y3 X is given by {Y2 −Y12 , Y3 −Y13 , X−Y1 }. By Exercise 1.8.5, I = Iˆ ∩ k[Y1 , Y2 , Y3 ] is generated by Y2 − Y12 and Y3 − Y13 . Thus, we have recovered the defining equations of the twisted cubic. In the above setting, there are examples which show that the image of a closed subset Vˆ ⊆ kn × km under the projection map πm need not be closed; see Exercise 1.8.9. We will see later in Section 3.2 that this phenomenon can not occur when we consider algebraic sets in projective space.\n\n1.3.7 Direct products of algebraic sets Let V ⊆ kn and W ⊆ km be (non-empty) algebraic sets, defined by sets S ⊆ k[X1 , . . . , Xn ] and T ⊆ k[Y1 , . . . , Ym ], respectively. Identifying kn × km with kn+m and k[X1 , . . . , Xn ], k[Y1 , . . . , Ym ] with subrings of\n\nRegular maps, direct products, and algebraic groups\n\n27\n\nk[X1 , . . . , Xn , Y1 , . . . , Ym ], we see that \u0018 \u0019 \u0017 \u0018 n+m \u0018 f(v) = 0 for all f ∈ S, V × W = (v, w) ∈ k \u0018 g(w) = 0 for all g ∈ T = V(S ∪ T ) ⊆ kn+m , where we use the identities f(v) = f(v, w) for all f ∈ S and g(w) = g(v, w) = 0 for all g ∈ T . Thus, the direct product V × W is an algebraic set in kn+m . Since (I(V ), I(W )) ⊆ I(V × W ), we have a well-defined k-bilinear map ¯ g¯) \u0004→ f¯ × g¯ := fg + I(V × W ). A[V ] × A[W ] → A[V × W ], (f, So we get an induced k-linear map A[V ] ⊗k A[W ] → A[V × W ]. 1.3.8 Proposition Let V ⊆ kn and W ⊆ km be non-empty algebraic sets. (a) If V and W are irreducible, then so is V × W . (b) We have I(V × W ) = (I(V ), I(W )) ⊆ k[X1 , . . . , Xn , Y1 , . . . , Ym ]; the map A[V ] ⊗k A[W ] → A[V × W ], f¯ ⊗ g¯ \u0004→ f¯ × g¯, is an isomorphism. (c) We have dim(V × W ) = dim V + dim W . Proof (a) We must show that I(V ×W ) ⊆ k[X1 , . . . , Xn , Y1 , . . . , Ym ] is a prime ideal; see Proposition 1.1.12. So let f1 , f2 ∈ k[X1 , . . . , Xn , Y1 , . . . , Ym ] be such that f1 f2 ∈ I(V × W ). For a fixed w ∈ W and i = 1, 2, we define an algebraic set by Vi (w) := {v ∈ V | fi (v, w) = 0} ⊆ V . Since f1 f2 ∈ I(V × W ), we have V = V1 (w) ∪ V2 (w). Since V is irreducible, we conclude that V = V1 (w) or V = V2 (w). Consequently, we have W = W1 ∪ W2 , where Wi := {w ∈ W | Vi (w) = V } for i = 1, 2. Now note that Wi = {w ∈ W | fi (v, w) = 0 for all v ∈ V } and so Wi is algebraic. Since W is irreducible, we may therefore conclude that W = W1 or W = W2 . In the first case, we have f1 (v, w) = 0 for all v ∈ V and w ∈ W , and so f1 ∈ I(V × W ). Similarly, in the second case, we have f2 ∈ I(V × W ). Thus, we have shown that I(V × W ) is a prime ideal, as desired. (b) Since (I(V ), I(W )) ⊆ I(V × W ), we have a canonical surjection B := k[X1 , . . . , Xn , Y1 , . . . , Ym ]/(I(V ), I(W )) −→ A[V × W ].\n\n(∗)\n\nWe have to show that it is also injective. For this purpose, let {f¯α | α ∈ I} be a k-basis of A[V ] (where fα ∈ k[X1 , . . . , Xn ]) and\n\n28\n\nAlgebraic sets and algebraic groups\n\n{¯ gβ | β ∈ J } be a k-basis of A[W ] (where gβ ∈ k[Y1 , . . . , Ym ]). For example, we could take the bases provided by Theorem 1.2.8. Then the residue classes in B of the products fα gβ span B over k. Hence it will be enough to show that all the products fα gβ are linearly modulo I(V × W ). So let xα,β ∈ k be such \u0001 independent \u0001 that α∈I β∈J xα,β fα gβ ∈ I(V × W ), that is, we have \u0002\u0002 xα,β fα (v)gβ (w) = 0 for all v ∈ V , w ∈ W . α∈I β∈J\n\nWe must show that xα,β = 0 for all α, β. To see this, fix w ∈ W and set \u0002 yα (w) := xα,β gβ (w) ∈ k. \u0001\n\nβ∈J\n\nThen we have α∈I yα (w)fα (v) = 0 for all v ∈ V . Since the polynomials {fα } are linearly independent modulo I(V ), we conclude that yα (w) = 0 for all α. This holds for all w ∈ W and so, since the polynomials {gβ } are linearly independent modulo I(W ), we conclude that xα,β = 0 for all α, β, as desired. Finally, consider the map A[V ] ⊗k A[W ] → A[V × W ], which certainly is surjective. To prove injectivity, consider the substitution homomorphism from k[X1 , . . . , Xn , Y1 , . . . , Ym ] into A[V ] ⊗k A[W ] ¯ i ⊗ 1 and Yj to 1 ⊗ Y¯j for all i and j. defined by sending Xi to X The kernel of that map contains I(V ) and I(W ). Hence we get an induced map B → A[V ] ⊗k A[W ]. The composition of that map with A[V ] ⊗k A[W ] → A[V × W ] is easily seen to be the map (∗) and, hence, is an isomorphism. Consequently, A[V ] ⊗k A[W ] → A[V × W ] also is an isomorphism. (c) By Theorem 1.2.14, there are subsets I ⊆ {1, . . . , n} and J ⊆ {1, . . . , m} such that dim(V × W ) = |I| + |J | and k[Xi , Yj | i ∈ I, j ∈ J ]∩I(V ×W ) = {0}. Using (b), this implies k[Xi | i ∈ I]∩I(V ) = {0} and k[Yj | j ∈ J ]∩I(W ) = {0} and so dim(V ×W ) \u0002 dim V +dim W . Conversely, let I ⊆ {1, . . . , n} be such that dim V = |I| and k[Xi | i ∈ I] ∩ I(V ) = {0}. Furthermore, let J ⊆ {1, . . . , m} be such that dim W = |J | and k[Yj | j ∈ J ] ∩ I(W ) = {0}. It is easily seen (for example, using an argument analogous to that in the proof of (b)) that then we also have k[Xi , Yj | i ∈ I, j ∈ J ] ∩ I(V × W ) = {0}. Hence, by Theorem 1.2.14, we have dim(V × W ) \u0001 |I| + |J |, as required. \u0003 Now we are ready to define algebraic monoids and algebraic groups.\n\nRegular maps, direct products, and algebraic groups\n\n29\n\n1.3.9 Definition Consider Mn (k) = kn×n as an algebraic set 2 (under the identification kn×n = kn ). Let µ : Mn (k) × Mn (k) → Mn (k) be the usual matrix multiplication. Thus, for A = (aij ) and B = (bij ) in Mn (k), we have µ(A, B) = (cij ) where cij = \u0001 n l=1 ail blj . This formula shows that µ is a regular map. Then Mn (k) together with µ is called the general linear algebraic monoid of degree n; the identity element is the n × n identity matrix In . Setting A[Mn (k)] = k[Xij | 1 \u0002 i, j \u0002 n]/I(Mn (k)), the algebra homomorphism µ∗ : A[Mn (k)] → A[Mn (k)] ⊗k A[Mn (k)] is given by ∗\n\n¯ ij ) = µ (X\n\nn \u0002\n\n¯ il ⊗ X ¯ lj X\n\nfor all i, j ∈ {1, . . . , n}.\n\nl=1\n\n(Of course, we have I(Mn (k)) = {0} if k is an infinite field.) A linear algebraic monoid is an algebraic subset G ⊆ Mn (k) such that In ∈ G and µ(A, B) ∈ G for all A, B ∈ G. A homomorphism of algebraic monoids is a regular map ϕ : G → H between two linear algebraic monoids G ⊆ Mn (k) and H ⊆ Mm (k) such that ϕ(AB) = ϕ(A)ϕ(B) for all A, B ∈ G and ϕ(In ) = Im . If, moreover, every element A ∈ G has an inverse and the map ι : G → G, A \u0004→ A−1 , is regular, then G is called a linear algebraic group. 1.3.10 Example Let SLn (k) = {A ∈ Mn (k) | det(A) = 1}. Then SLn (k) is closed under multiplication. Furthermore, let \u0002 det := sgn(σ)X1σ(1) · · · Xnσ(n) ∈ k[Xij | 1 \u0002 i, j \u0002 n], σ∈Sn\n\nwhere Sn is the symmetric group of degree n. Then SLn (k) = V({det −1}) ⊆ Mn (k) is algebraic, and so SLn (k) is a linear algebraic monoid. Furthermore, for any invertible matrix A ∈ Mn (k), we have A−1 = det(A)−1 A˜tr , where A˜ is the matrix of cofactors of A; the entries of A˜ are given by the determinants of various submatrices of A of size n − 1. Thus, ι : SLn (k) → SLn (k), A \u0004→ A−1 , is regular and so SLn (k) is a linear algebraic group, which is called the special linear group. Finally, if k is algebraically closed, then SLn (k) is an irreducible hypersurface with I(SLn (k)) = (det −1). For this purpose, by Theorem 1.1.9, one has to check that det −1 ∈ k[Xij | 1 \u0002 i, j \u0002 n] is irreducible. (We leave this an exercise; a different proof that SLn (k)\n\n30\n\nAlgebraic sets and algebraic groups\n\nis irreducible will be given in Remark 1.6.11.) In particular, we see that dim SLn (k) = n2 − 1 by Example 1.2.16(b). (e) More generally, let e \u0001 1 and set SLn (k) = {A ∈ Mn (k) | (e) det(A)e = 1}. Then, as above, we see that SLn (k) is a linear (e) algebraic monoid. Furthermore, for any A ∈ SLn (k), we have det(A)e = 1 and so det(A)−1 = det(A)e−1 . Thus, A−1 can also be (e) expressed by polynomials in the entries of A. Consequently, SLn (k) is a linear algebraic group. 1.3.11 Example Let G ⊆ Mn (k) be an algebraic set which is a (e) subgroup of SLn (k) for some e \u0001 1. Then G is a linear algebraic group. (a) Let Un (k) be the set of all upper unitriangular matrices in Mn (k), that is, we have A = (aij ) ∈ Un (k) if and only if aii = 1 for all i and aij = 0 for all i > j. Then Un (k) is an algebraic subset of Mn (k) and a subgroup of SLn (k). Thus, Un (k) is a linear algebraic group. As an algebraic set, we have Un (k) ∼ = kn(n−1)/2 and so Un (k) is irreducible of dimension n(n − 1)/2 (if |k| = ∞). If n = 2, then \u001a \u001b 1 a k → U2 (k), a \u0004→ , 0 1 is an isomorphism of algebraic sets. Furthermore, the multiplication in U2 (k) simply corresponds to the addition in k. Thus, the additive group of k may be regarded as a linear algebraic group, which we denote by Ga (k). (b) Let G be any finite group. Then, by Cayley’s theorem, G may be regarded as a subgroup of Sn for some n \u0001 1. Using the representation of Sn by permutation matrices, we obtain an embedding (2) of G into the linear algebraic group SLn (k). Thus, being a finite set, G itself is an algebraic group. 1.3.12 Remark Let G ⊆ Mn (k) be a linear algebraic group. For a fixed x ∈ G, we define a map λx : G → G by λx (g) := µ(x, g). Thus, λx is the composition of µ with δ : G → G × G, g \u0004→ (x, g). The map δ is certainly regular; the same holds for µ by definition. Hence λx is regular. Furthermore, the associativity of the multiplication in G implies that λx ◦ λy = λxy for all x, y ∈ G. Since every element of G has an inverse, we conclude that λx is an isomorphism, the inverse being given by λx−1 . Similarly, we define\n\nRegular maps, direct products, and algebraic groups\n\n31\n\na map ρx : G → G by ρx (g) := µ(g, x). As above, we see that ρx is an isomorphism. Furthermore, we have ρx ◦ ρy = ρyx for all x, y ∈ G. 1.3.13 Proposition Let G ⊆ Mn (k) be a linear algebraic group. Let G◦ ⊆ G be an irreducible component containing 1. Then the following hold. (a) G◦ is a closed normal subgroup of G of finite index, and the cosets of G◦ are precisely the irreducible components of G. In particular, the irreducible components of G are disjoint, and G◦ is uniquely determined. (b) Every closed subgroup of G of finite index contains G◦ . (c) G is irreducible if and only if G is connected as a topological space (i.e. G cannot be expressed as the disjoint union of two nonempty open subsets). Proof (a) Consider the multiplication map µ : G × G → G. Let X ⊆ G be an irreducible component with 1 ∈ X. Then X · G◦ = µ(X × G◦ ) ⊆ G is a closed irreducible subset of G (see Remark 1.3.2 and Proposition 1.3.8). Since 1 ∈ X and 1 ∈ G◦ , we have X, G◦ ⊆ X · G◦ ⊆ X · G◦ . Since X, G◦ are maximal closed irreducible subsets, we must have X = X · G◦ and G◦ = X · G◦ . In particular, we see that X = G◦ and so G◦ is uniquely determined. The above argument also shows that G◦ · G◦ ⊆ G◦ · G◦ = G◦ and so G◦ is closed under multiplication. Now consider the inversion map ι : G → G. Since this is an isomorphism of algebraic sets, (G◦ )−1 = ι(G◦ ) is also an irreducible component containing 1, and hence must be equal to G◦ . Thus, G◦ is closed under inversion and so is a subgroup of G. Now let x ∈ G. We have seen in Remark 1.3.12 that ρx is an isomorphism of algebraic sets and so it maps irreducible components to irreducible components. Thus, the coset G◦ x = ρx (G◦ ) is an irreducible component of G. Conversely, let X ⊆ G be any irreducible component and let x ∈ G be such that x−1 ∈ X. Then ρx (X) = Xx is an irreducible component containing 1 and so Xx = G◦ . Hence X is equal to the coset G◦ x−1 . Thus, the cosets of G◦ are precisely the irreducible components of G. Consequently, G◦ has finite index in G. Furthermore, for any x ∈ G, the two cosets G◦ x = ρx (G◦ ) and xG◦ = λx (G◦ ) are irreducible components with non-empty intersection; hence they must be equal. This shows that G◦ is a normal subgroup. (b) Let H ⊆ G be any closed subgroup of finite index. Let g1 , . . . , gr ∈ G (with g1 = 1) be such that G is the disjoint union\n\n32\n\nAlgebraic sets and algebraic groups\n\nr\n\ni=1 Hgi . As above we see that each \u0007 coset Hgi is a closed subset of G. It follows that G◦ = G◦ ∩ G = ri=1 (G◦ ∩ Hgi ). Since G◦ is irreducible, we must have G◦ ∩ Hgi = G◦ and so G◦ ⊆ Hgi for some i. Since 1 ∈ G◦ , it follows that i = 1 and so G◦ ⊆ H, as desired. (c) It is clear that if G is a disjoint union of two open subsets, then each of these open subsets is also closed and so G is not irreducible. Conversely, if G is not irreducible, we write G as the union of its irreducible components. By (a), this union is disjoint and so each irreducible component is also open. \u0003\n\n1.3.14 Corollary All irreducible components of a linear algebraic group G ⊆ Mn (k) have the same dimension. In particular, we have dim G = dim G◦ . Proof We have seen above that the irreducible components of G are the cosets of G◦ . Such a coset is of the form G◦ x = ρx (G◦ ) for some x ∈ G, where ρx : G → G is an isomorphism of algebraic sets. Hence we have dim(G◦ x) = dim G◦ . The fact that dim G = dim G◦ follows from Proposition 1.2.17. \u0003 1.3.15 Example We shall now introduce an important class of groups: the so-called classical groups. Let Q ∈ Mn (k) be an invertible matrix, and set Γn (Q, k) := {A ∈ Mn (k) | Atr Q A = Q}. First note that taking the determinant of Atr QA = Q and using that det(Q) \b= 0, we obtain det(A) = ±1. Next, if A, B ∈ Γn (Q, k), then we also have AB and A−1 ∈ Γn (Q, k). Finally, writing out the equation Atr QA = Q for all matrix entries, we see that Γn (Q, k) (2) is a closed subset of Mn (k). Thus, Γn (Q, k) ⊆ SLn (k) is a linear algebraic group, called a classical group. If Q\u0002 ∈ Mn (k) is another invertible matrix, we say that Q, Q\u0002 are equivalent if there exists some invertible matrix R ∈ Mn (k) such that Q\u0002 = Rtr QR. In this case, we have A ∈ Γn (Q\u0002 , k) if and only if RAR−1 ∈ Γn (Q, k). Thus, the map ϕR : Γn (Q\u0002 , k) → Γn (Q, k),\n\nA \u0004→ RAR−1 ,\n\nis an isomorphism of algebraic groups. Now Q defines a bilinear form βQ : kn × kn → k by βQ (v, w) = vtr Qw (where v, w are regarded as column vectors).\n\nRegular maps, direct products, and algebraic groups\n\n33\n\n(Sym)\n\nWe say that Q is symmetric if char(k) \b= 2 and βQ (v, w) = βQ (w, v) for all v, w ∈ kn . The latter condition certainly is equivalent to Q = Qtr .\n\n(Alt)\n\nWe say that Q is alternating if βQ (v, v) = 0 for all v ∈ kn . This certainly implies that Q = −Qtr . (If k has characteristic \b= 2, the latter condition is also necessary.)\n\nNow, without any further assumption on k, there may be many pairwise inequivalent matrices Q as above. However, if k is algebraically closed, the picture simplifies drastically. In this case, there is only one symmetric Q and one alternating Q up to equivalence; see §2.10 and §4.4 of Grove (2002). Hence, since we will mainly be interested in the case where k is algebraically closed, it will be enough to study the groups Γn (Q, k) for the following special choices of the matrix Q. (The motivation for these choices will be become clear in Section 1.7 where we construct a BN -pair for these groups.) Assume first that we are in the case (Sym) where char(k) \b= 2. Then ⎤ ⎡ 0 ··· 0 1 ⎥ ⎢ .. . . . . . . 0 ⎥ ⎢ . On (k) := Γn (Qn , k), where Qn := ⎢ ⎥ ∈ Mn (k), ⎣ 0 1 . . . ... ⎦ 1 0 ··· 0 is called the orthogonal group of degree n. Note that On (k) is not irreducible: indeed, the determinant defines a group homomorphism ε : On (k) → {±1}, and there exist matrices A ∈ On (k) with determinant −1. (Check this!) Thus, On (k) is the union of two cosets of the closed normal subgroup SOn (k) := On (k) ∩ SLn (k), which is called the special orthogonal group of dimension n. For n = 1, SO1 (k) is the trivial group and O1 (k) = {±1}. For n = 2, we obtain \u001b\u0018 \u0017\u001a \u0019 \u001a \u001b a 0 0 1 \u0018 SO2 (k) = 0 = \b a ∈ k and ∈ O2 (k)\\SO2 (k); \u0018 1 0 0 a−1 see Exercise 1.8.19 for the case n = 3. Now assume that we are in the case (Alt) where k may have any characteristic. Then n = 2m must be even, and ⎡ ⎤ Q 0 m − ⎣ ⎦∈ M2m (k), Sp2m (k) := Γ2m (Q− 2m , k), where Q2m := −Qm 0\n\n34\n\nAlgebraic sets and algebraic groups\n\nis called the symplectic group of degree 2m. Note that, for m = 1, we obtain Sp2 (k) = SL2 (k). We shall see in Theorems 1.7.4 and 1.7.8 that SOn (k), Sp2m (k) are connected. 1.3.16 Orthogonal groups and quadratic forms When dealing with even-dimensional orthogonal groups, it is actually more convenient to work with a slightly different description, using quadratic forms. In particular, this will also give rise to a definition of orthogonal groups in characteristic 2. Let n = 2m for some m \u0001 1 and consider the matrix Q2m defined above. Now let P = (pij ) ∈ M2m (k) be such that Q2m = P + P tr . Then the corresponding polynomial fP :=\n\n2m \u0002\n\npij Xi Xj ∈ k[X1 , . . . , Xn ]\n\ni,j=1\n\ndefines in the usual way a function on k2m . We set Γ2m (fP , k) := {A ∈ M2m (k) | fP (Ax) = fP (x) for all x ∈ k2m }, where x ∈ k2m is regarded as a column vector. We claim that Γ2m (fP , k) is a linear algebraic group. Indeed, first note that Γ2m (fP , k) is closed in M2m (k) since the condition fP (Ax) = fP (x) yields a system of polynomial equations in the coefficients of A. Furthermore, Γ2m (fP , k) certainly is closed under multiplication. To see that every element in Γ2m (fP , k) is invertible, we note that fP (x + y) − fP (x) − fP (y) =\n\n2m \u0002\n\n(pij + pji )xi yj = βQ2m (x, y),\n\n(∗)\n\ni,j=1\n\nwhere x = (x1 , . . . , x2m ), y = (y1 , . . . , y2m ), and βQ2m is defined as in §1.3.15. Consequently, any matrix A ∈ Γ2m (fP , k) also satisfies the equation Atr Q2m A = Q2m , and so det(A) = ±1. Hence, we have Γ2m (fP , k) ⊆ Γ2m (Q2m , k). We remark that Γ2m (fP , k) is defined by the polynomial equations defining Γ2m (Q2m , k) plus the 2m equations fP (Aer ) = fP (er ), where {er | 1 \u0002 r \u0002 2m} is the standard basis of k2m . (This easily follows using the relation (∗).) These 2m\n\nRegular maps, direct products, and algebraic groups\n\n35\n\nequations yield the relations (Atr PA)rr = Prr . Thus, we have Γ2m (fP , k) = {A ∈ Γ2m (Q2m , k) | (Atr PA)rr = Prr for 1 \u0002 r \u0002 2m}. If fP \u0002 is another polynomial as above, we say that fP , fP \u0002 are equivalent if there exists an invertible R ∈ M2m (k) such that Q2m = Rtr Q2m R and fP (Rv) = fP \u0002 (v) for all v ∈ k2m . Hence, we have A ∈ Γ2m (fP \u0002 , k) if and only if RAR−1 ∈ Γ2m (fP , k). Thus, as before, the map A \u0004→ RAR−1 defines an isomorphism of algebraic groups Γ2m (fP \u0002 , k) → Γ2m (fP , k). Now, if k is algebraically closed, then there is (up to equivalence) a unique quadratic polynomial fP as above; see Theorems 4.4 and 12.9 of Grove (2002). We make the following definite choice: ⎡ ⎤ m \u0002 0 Qm ⎦. f2m := Xi X2m+1−i with matrix P2m := ⎣ 0 0 i=1 The corresponding group Γ2m (f2m , k) will be denoted by O+ 2m (k). Now, if char(k) \b= 2, then 2f2m (x) = βQ2m (x, x) for x ∈ k2m . Thus, we have O+ 2m (k) = O2m (k)\n\n(char(k) \b= 2);\n\nso we don’t obtain anything new here. In order to have a uniform notation, we shall also set SO+ 2m (k) := SO2m (k) in this case. Now assume that char(k) = 2. Then we have Γ2m (Q2m , k) = Sp2m (k). Writing out the relations (Atr PA)rr = Prr (1 \u0002 r \u0002 2m) for the above special choice of P = P2m yields # \\$ m \u0018\u0002 \u0018 aij a2m+1−i,j = 0 for 1 \u0002 j \u0002 2m ; O+ 2m (k) = (aij ) ∈ Sp2m (k)\u0018 i=1\n\nthis will be called the orthogonal group in characteristic 2. For m = 1, we obtain \u001b\u0018 \u0019 \u0017\u001a \u0017\u001a \u001b\u0018 \u0019 0 a \u0018 a 0 \u0018 + O2 (k) = \u0018 0 \b= a ∈ k ∪ \u0018 0 \b= a ∈ k . a−1 0 0 a−1 Thus, O+ 2 (k) \u0002 Sp2 (k) certainly is not connected. We shall set + ◦ SO+ 2m (k) := O2m (k)\n\n(char(k) = 2).\n\nIn Theorem 1.7.8 it will be shown that SO+ 2m (k) has index 2 in + O2m (k).\n\n36\n\nAlgebraic sets and algebraic groups Table 1.1\n\nSeries of groups of classical type\n\nType\n\nGroup\n\nRemarks\n\nAm−1 Bm Cm Dm\n\nSLm (k) SO2m+1 (k) Sp2m (k) SO+ 2m (k)\n\nany any any any\n\nm \u0001 1, m \u0001 1, m \u0001 1, m \u0001 1,\n\nany characteristic char(k) \b= 2 any characteristic any characteristic\n\nIn Lie notation, SLn (k), SOn (k), Sp2m (k), SO+ 2m (k) are groups of classical type A, B, C, D. The precise correspondence is given in Table 1.1; see Chapter 1 and §11.3 of Carter (1972) for more details. Note that we could have also defined SO2m+1 (k) in characteristic 2, as in §1.3.15. However, if char(k) = 2, then there exists a bijec∼ tive homomorphism of algebraic groups SO2m+1 (k) −→ Sp2m (k); see Theorem 14.2 of Grove (2002).\n\n1.4 The tangent space and non-singular points The aim of this section is to introduce the tangent space of an algebraic set at a point and to prove the existence of non-singular points. This involves a process of ‘linearization’ of the polynomials defining an algebraic set. On a more formal level, this requires a further basic notion, that of derivations. 1.4.1 Definition Let A be a k-algebra and M be an A-module; we denote the action of A on M by (a, m) \u0004→ a.m (a ∈ A, m ∈ M ). A k-linear map D : A → M is called a derivation if we have the product rule D(ab) = a.D(b) + b.D(a) for all a, b ∈ A. The set of all derivations of A into M is denoted by Derk (A, M ). This set naturally is an A-module. Indeed, if c ∈ A and D ∈ Derk (A, M ), then we define c.D : A → M by (c.D)(a) := c.D(a) for all a ∈ A. It is readily checked that c.D ∈ Derk (A, M ). 1.4.2 Example (a) Let us take M = A, where the action is given by multiplication. Then Derk (A, A) carries an additional structure, that of a Lie algebra. Recall that a Lie algebra is a vector space L\n\nThe tangent space and non-singular points\n\n37\n\nequipped with a bilinear product [ , ] : L × L → L which satisfies [x, x] = 0, for all x ∈ L, and the Jacobi identity [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0\n\nfor all x, y, z ∈ L.\n\nFor example, Mn (k) is a Lie algebra with product [A, B] := AB − BA. Now, Derk (A, A) is a Lie algebra with product [D, D\u0002 ] := D ◦ D\u0002 − D\u0002 ◦ D. Indeed, it is clear that [D, D] = 0; the Jacobi identity is checked by a straightforward verification. Moreover, we have the following Leibniz rule: n\n\nD (ab) =\n\nn \u0002 n i=0\n\ni\n\nDi (a)Dn−i (b)\n\nfor all a, b ∈ A.\n\nIf the characteristic of k is a prime p > 0, this shows that Dp is also a derivation. (b) Let R = k[X1 , . . . , Xn ] where Xi are indeterminates. We define Di : R → R by Di (f) = ∂f/∂Xi (the usual partial derivative with respect to Xi ). Then Di ∈ Derk (R, R). Furthermore, we have D=\n\nn \u0002\n\nD(Xi ) Di\n\nfor any D ∈ Derk (R, R).\n\ni=1\n\nTo check this identity, we note that D is uniquely determined by its values on a set of algebra generators of R. (This follows from the product rule for derivations.) Hence it is enough to verify the identify on Xj for 1 \u0002 j \u0002 n, in which case it is trivial. Since D1 , . . . , Dn certainly are linearly independent, we conclude that Derk (R, R) is a free R-module with basis D1 , . . . , Dn . 1.4.3 Lemma Let A = k[X1 , . . . , Xn ]/I where I ⊆ k[X1 , . . . , Xn ] is an ideal. Let M be an A-module, and set n \u0018\u0002 % & \u0018 Di (f).vi = 0 for all f ∈ I , TA,M := (v1 , . . . , vn ) ∈ M n \u0018 i=1\n\nwhere Di denotes partial derivative with respect to Xi and the bar denotes the canonical map k[X1 , . . . , Xn ] → A. Then, for any v = (v1 , . . . , vn ) ∈ TA,M , we have a well-defined derivation\n\n38\n\nAlgebraic sets and algebraic groups\n\nDv ∈ Derk (A, M ) given by ¯ = Dv (f)\n\nn \u0002\n\nDi (f).vi\n\nfor f ∈ k[X1 , . . . , Xn ].\n\ni=1\n\nThe map Φ : TA,M → Derk (A, M ), v \u0004→ Dv , is an A-module isomorphism. Proof Let R := k[X1 , . . . , Xn ] and consider the canonical map ¯ with kernel I. Then we can also regard M as an π : R → A, f \u0004→ f, R-module via π. First we must\u0001 check that Dv is well-defined. For this purpose we must check that ni=1 Di (f).vi = 0 for any f ∈ I, which is true by the definition of TA,M . Next we show that Dv ∈ Derk (A, M ). Indeed, for any f, g ∈ R, we have Dv (f¯ g¯) =\n\nn \u0002\n\nDi (fg).vi =\n\ni=1\n\nn \u0010 \u0002\n\n\u0011 gDi (f).vi + fDi (g).vi\n\ni=1\n\n¯ + f¯ Dv (¯ g), = g¯ Dv (f) as required. Thus, we have a well-defined A-linear map Φ : TA,M → Derk (A, M ). We have Dv (X¯j ) = vj for 1 \u0002 j \u0002 n, and so Φ is injective. It remains to show that Φ is also surjective. So let D ∈ ˜ = D ◦ π is a derivation in Derk (A, M ). Then the composition D Derk (R, M ). As in Example 1.4.2(b), one sees that ˜ D(f) =\n\nn \u0002\n\n˜ i) Di (f) D(X\n\nfor all f ∈ R.\n\ni=1\n\n˜ i ) ∈ M for 1 \u0002 i \u0002 n. We claim that v = Now set vi := D(X (v1 , . . . , vn ) ∈ TA,M . Indeed, we have ¯ = D(f) ˜ D(f) =\n\nn \u0002 i=1\n\nDi (f).vi =\n\nn \u0002\n\nDi (f).vi\n\nfor all f ∈ R.\n\ni=1\n\nThus, if f ∈ I, then f¯ = 0 and so v ∈ TA,M as claimed. ¯ j ) = vj = D(X ¯ j ), we see that Dv = D. Thus, Φ is Since Dv (X surjective. \u0003\n\nThe tangent space and non-singular points\n\n39\n\n1.4.4 Remark In the setting of Lemma 1.4.3, assume that the ideal I is generated by f1 , . . . , fm ∈ k[X1 , . . . , Xn ]. Then we claim that TA,M\n\nn \u0018\u0002 % & n\u0018 = (v1 , . . . , vn ) ∈ M \u0018 Di (fj ).vi = 0 for 1 \u0002 j \u0002 m . i=1\n\nIndeed, the inclusion ‘⊆’ is clear. To prove the reverse inclusion, assume that (v1 , . . . , vn ) ∈ M n lies in the right-hand side \u0001mof the above identity. Now let f ∈ I. Then we can write f = j=1 hj fj with hj ∈ k[X1 , . . . , Xn ], and it follows that Di (f) =\n\nm \u0002\n\nDi (hj fj ) =\n\nj=1\n\nHence we also have\n\nm \u0010 \u0002\n\nm \u0011 \u0002 hj Di (fj ) + fj Di (hj ) = hj Di (fj ).\n\nj=1\n\n\u0001n\n\ni=1 Di (f).vi\n\nj=1\n\n= 0, as required.\n\nWe now obtain a new characterization of the transcendence degree of a finitely generated field extension K ⊇ k. Recall that k is called a perfect field if either k has characteristic 0 or k has characteristic p > 0 and k = {xp | x ∈ k}. For example, finite fields or algebraically closed fields are perfect. 1.4.5 Proposition Assume that k is a perfect field, and let A be a finitely generated k-algebra which is an integral domain. (a) Let K be the field of fractions of A. Then ∂k (A) = dimK Derk (K, K). (b) If A = k[X1 , . . . , Xn ]/(f1 , . . . , fm ), where fi ∈ k[X1 , . . . , Xn ], then \u0011 \u0010 ∂k (A) = n − rankK Di (fj ) 1\u0001i\u0001n . 1\u0001j\u0001m\n\nProof We begin with the following observations. Let R ⊆ K be any k-subalgebra such that K is the field of fractions of R. We regard K as an R-module (by left multiplication). Now consider the restriction map ρ : Derk (K, K) → Derk (R, K),\n\nD \u0004→ D|R .\n\nSince R generates K, it is clear that ρ is injective. On the other hand, if D ∈ Derk (R, K), then we can extend D to a derivation\n\n40\n\nAlgebraic sets and algebraic groups\n\n˜ ∈ Derk (K, K) by setting D 1 ˜ D(f/g) = 2 (gD(f) − fD(g)) for f, g ∈ R, g \b= 0. g ˜ is well-defined and lies in It is straightforward to check that D Derk (K, K). (This is just similar to the rule for the derivative of ˜ = D, we see that ρ is sura quotient of two functions.) Since ρ(D) jective. Also note that Derk (R, K) still is a K-vector space and ρ is K-linear. Thus, we have dimK Derk (K, K) = dimK Derk (R, K). Furthermore, let us assume that R can be written as a quotient of a polynomial ring. Then the R-module TR,K defined in Lemma 1.4.3 certainly is a K-vector space and the map Φ is seen to be K-linear. Thus, we have in fact dimK Derk (K, K) = dimK TR,K .\n\n(∗)\n\nTo prove (a), let d = ∂k (A). Then we also have d = ∂k (K) by Proposition 1.2.18. So, since k is perfect, we can apply Exercise 1.8.15 and conclude that there exist z1 , . . . , zd+1 ∈ K such that K = k(z1 , . . . , zd+1 ), {z1 , . . . , zd } are algebraically independent over k and zd+1 is separable algebraic over k(z1 , . . . , zd ). We now apply the above discussion to the subalgebra R := k[z1 , . . . , zd+1 ] ⊆ K. Since {z1 , . . . , zd+1 } are not algebraically independent, there exists some 0 \b= f ∈ k[Y1 , . . . , Yd+1 ] such that f(z1 , . . . , zd+1 ) = 0. Since K is a field, we may assume that f is irreducible. Now consider the map α : k[Y1 , . . . , Yd+1 ] → K,\n\ng \u0004→ g(z1 , . . . , zd+1 ).\n\nWe claim that ker(α) = (f). Indeed, it is clear that α(f) = 0. Conversely, let g ∈ k[Y1 , . . . , Yd+1 ] be such that α(g) = 0, and assume that f does not divide g. As in the proof of Theorem 1.1.9, there exist F, G ∈ k[Y1 , . . . , Yd+1 ] such that 0 \b= d := Gf + Fg ∈ k[Y1 , . . . , Yd ]; note that Yd+1 occurs in some term of f. Then we obtain d(z1 , . . . , zd ) = α(Gf + Fg) = 0, contradicting the fact that z1 , . . . , zd are algebraically independent. So we have ker(α) = (f) as claimed. On the other hand, the image of α is just R. Hence we have R∼ = k[Y1 , . . . , Yd+1 ]/(f). It follows that TR,K is a subspace of K d+1 defined by one linear equation; see Remark 1.4.4. That linear equation is non-zero since some partial derivative of f will be non-zero; see Exercise 1.8.14. So we have dimK TR,K = d and (∗) yields (a). Finally, (b) follows from (a) and by applying (∗) to R = A. \u0003\n\nThe tangent space and non-singular points\n\n41\n\n1.4.6 Corollary Assume that k is a perfect field. Let V ⊆ kn be an irreducible algebraic set and assume that I(V ) ⊆ k[X1 , . . . , Xn ] is generated by f1 , . . . , fm . Then we have \u0010 \u0011 dim V = n − rank Di (fj ) 1\u0001i\u0001n , 1\u0001j\u0001m\n\nwhere Di denotes the partial derivative with respect to Xi and the bar denotes the canonical map k[X1 , . . . , Xn ] → A[V ]; the rank is taken over the field of fractions of A[V ]. Proof See Proposition 1.2.18 and Proposition 1.4.5.\n\n\u0003\n\nThe above results were developed into a purely algebraic setting. We now translate them into geometric properties of algebraic sets. 1.4.7 Taylor expansion Let V ⊆ kn be a non-empty algebraic set with vanishing ideal I(V ) ⊆ k[X1 , . . . , Xn ]. Let us fix a point p = (p1 , . . . , pn ) ∈ V . For f ∈ k[X1 , . . . , Xn ], we define a linear polynomial dp (f) :=\n\nn \u0002\n\nDi (f)(p) Xi ∈ k[X1 , . . . , Xn ],\n\ni=1\n\nwhere Di denotes the usual partial derivative with respect to Xi . We have the usual rules for computing derivates: dp (f+g) = dp (f)+dp (g) and dp (fg) = dp (f) g(p) + f(p) dp (g) for all f, g ∈ k[X1 , . . . , Xn ]. Furthermore, we have dp (f) = f − f(0) if f is linear. For any v = (v1 , . . . , vn ) ∈ kn , we now consider the ‘line through p in direction v’ defined by Lv := {p + tv | t ∈ k} ⊆ kn . Then, given any f ∈ k[X1 , . . . , Xn ], we consider the Taylor expansion of f(p + tv) (as a function in t) around t = 0: f(p + tv) = f(p) +\n\nn \u0010\u0002\n\n\u0011 Di (f)(p) vi t + combination of higher powers of t. \u000f i=1 \u000e =dp (f)(v)\n\nWe say that Lv is a tangent line at p ∈ V if dp (f)(v) = 0 for all f ∈ I(V ). 1.4.8 Definition Let V ⊆ kn be a non-empty algebraic set. For fixed p ∈ V , let us regard k as an A[V ]-module where f¯ ∈ A[V ]\n\n42\n\nAlgebraic sets and algebraic groups\n\nacts by multiplication with f(p). We denote this A[V ]-module by kp . Then we define Tp (V ) := TA[V ],kp = {v ∈ kn | dp (f)(v) = 0 for all f ∈ I(V )} = {v ∈ kn | Lv is a tangent line at p ∈ V } to be the tangent space at p ∈ V . Thus, Tp (V ) is at the same time a linear subspace and an algebraic subset of kn . Assume that |k| = ∞. Then, by Exercise 1.8.6, Tp (V ) is irreducible and we may unambiguously define dim Tp (V ) as in Definition 1.2.15 or as the dimension of a vector space. 1.4.9 Remark (a) Assume that the vanishing ideal I(V ) is generated by the polynomials f1 , . . . , fm ∈ k[X1 , . . . , Xn ]. Then Remark 1.4.4 shows that Tp (V ) = V({dp (f1 ), . . . , dp (fm )}) ⊆ kn . (b) By Lemma 1.4.3, we have a k-linear isomorphism Φ : Tp (V ) → Derk (A[V ], kp ), v \u0004→ Dv , ¯ = dp (f)(v). In what follows, we will where Dv is defined by Dv (f) often identify Tp (V ) = Derk (A[V ], kp ) using that isomorphism. 1.4.10 Example Assume that k is an infinite field. (a) Let V be a linear subspace of kn . Then V is defined by linear polynomials f1 , . . . , fm ∈ k[X1 , . . . , Xn ] with constant term zero. By Exercise 1.8.6, we have I(V ) = (f1 , . . . , fm ). Now dp (f) = f for all such polynomials, and so Tp (V ) = V for all p ∈ V . In particular, we have dim V = dim Tp (V ) for all p ∈ V . This applies, in particular, to the case where V = kn . Then we have I(V ) = {0} by Exercise 1.8.2, and so Tp (kn ) = kn for all p ∈ kn . (b) Let V = {p} ⊆ kn be a singleton set. Then, as in Exercise 1.2.19(a), we see that I(V ) = (X1 − p1 , . . . , Xn − pn ), and this yields Tp ({p}) = {0}. (c) Consider the twisted cubic C = V(f1 , f2 ) ⊆ k3 , where f1 = X2 − X12 and f2 = X3 − X13 . By Exercise 1.8.4(a), we have I(C) = (f1 , f2 ). Let p = (x, x2 , x3 ) ∈ C. Then we have dp (f1 ) = −2xX1 +X2 and dp (f2 ) = −3x2 X1 + X3 , and so Tp (C) = \u0013(1, 2x, 3x2 )\u0014k ⊆ k3 . Thus, Tp (C) is a one-dimensional subspace for each p ∈ C.\n\nThe tangent space and non-singular points\n\n43\n\n1.4.11 Theorem Assume that k is an infinite perfect field, and let V ⊆ kn be an irreducible (and, hence, non-empty) algebraic set. (a) We have dim Tp (V ) \u0001 dim V for all p ∈ V . Furthermore, the set of all p ∈ V with dim Tp (V ) = dim V is non-empty and open. (b) Let p ∈ V be such that dim Tp (V ) = dim V . Then there exists some f ∈ k[X1 , . . . , Xn ] with f(p) \b= 0 and regular maps ψj : V˜f → kn such that q ), . . . , ψd (˜ q )\u0014k ⊆ kn Tπf (˜q) (V ) = \u0013ψ1 (˜\n\nfor all q˜ ∈ V˜f ,\n\nwhere πf : V˜f → V is defined as in §1.1.14. We will say that a point p ∈ V is non-singular if dim Tp (V ) = dim V ; otherwise, p is called singular. If all points of V are non-singular, we call V non-singular. Later, we will give another characterization of non-singular points; see Proposition 2.3.8. Proof (a) Assume that I(V ) = (f1 , . . . , fm ). Since V is irreducible, A[V ] is an integral domain; let K be its field of fractions. Then, by Corollary 1.4.6, we have dim V = n − rankK M,\n\nwhere M := (Di (fj )) 1\u0001i\u0001n ∈ A[V ]n×m . 1\u0001j\u0001m\n\n\u0001n Now note that dp (fj ) = i=1 Di (fj )(p)Xi for any p ∈ V . Hence, by Remark 1.4.9(a), we have dim Tp (V ) = n − rankk (M (p)). Since the rank of a matrix is given by the vanishing and non-vanishing of certain minors, it is clear that rankk M (p) \u0002 rankK (M ) for all p ∈ V , with equality for at least some p. Thus, we have dim Tp (V ) \u0001 dim V for all p ∈ V , where equality holds for some p. Now set r = rankK M . Then the set of all p ∈ V where dim Tp (V ) = dim V is the complement of the set of all p ∈ V where all (r+1)×(r+1) minors of M (p) vanish. Since these minors are given by polynomial expressions in the coefficients of M (p) and, hence, by polynomials in the coordinates of p, we see that the set of all p ∈ V with dim Tp (V ) = dim V is open. (b) Let d = dim V . Since dim Tp (V ) = d, there exist f1 , . . . , fn−d ∈ I(V ) such that Tp (V ) = V({dp (f1 ), . . . , dp (fn−d )}). We shall assume that the labelling of the variables is chosen\n\n44\n\nAlgebraic sets and algebraic groups\n\nsuch that\n\n\u0010 \u0011 det Dj (fi )(p)\n\n1\u0002i,j\u0002n−d\n\n\b= 0.\n\n(1)\n\nNow consider the matrix (where i is the row index and j is the column index) \u0010 \u0011 Mq := Dj (fi )(q) 1\u0001i\u0001n−d where q ∈ V . 1\u0001j\u0001n\n\nConsider the elements of kn as column vectors and let v ∈ kn , with components v1 , . . . , vn . Then the ith component of the product Mq v ∈ kn is given by (Mq v)i =\n\nn \u0002\n\nDj (fi )(q) vj = dq (fi )(v)\n\nwhere q ∈ V .\n\n(2)\n\nj=1\n\n\b Our assumption (1) means that, if we write Mq = Aq Bq with Aq of size (n−d)×(n−d) and Bq of size (n−d)×d, then det Ap \b= 0. Since the determinant of a matrix is given by a polynomial expression in the coefficients, there exists a non-zero polynomial f ∈ k[X1 , . . . , Xn ] such that det Aq = f(q) for all q ∈ V . Since f(p) = det Ap \b= 0, we have p ∈ Vf ⊆ V ; furthermore, d = dim V({dq (f1 ), . . . , dq (fn−d )})\n\nfor all q ∈ Vf .\n\nFor any q ∈ Vf , we have that Tq (V ) is contained in the set on the right-hand side of the above identity (by the definition of Tq (V )). On the other hand, we have dim Tq (V ) \u0001 d by (a). So we conclude that Tq (V ) = V({dq (f1 ), . . . , dq (fn−d )})\n\nfor all q ∈ Vf .\n\n(3)\n\nNow consider V˜f = {(q, y) ∈ V × k | f(q)y = 1}. For 1 \u0002 j \u0002 d, we define \u001b \u001a −Aq−1 Bq n ˜ ψj : Vf → k , (q, y) \u0004→ jth column of . Id We claim that ψj is a regular map. Indeed, using the formula for the inverse of a matrix and recalling the definition of f, we see that there exist polynomials hj1 , . . . , hjn ∈ k[X1 , . . . , Xn ] such that the ith component of ψj (q, y) is given by hji (q)/f(q) = yhji (q) for all (q, y) ∈ V˜f . Thus, ψj certainly is regular.\n\nThe tangent space and non-singular points\n\n45\n\nWorking out the product Mq ψj (q, y), we see that the first n − d components of the result are 0. By (2), this means that dq (fi )(ψj (q, y)) = 0 for 1 \u0002 i \u0002 n − d and (q, y) ∈ V˜f . Furthermore, {ψ1 (q, y), . . . , ψd (q, y)} are certainly linearly independent. Thus, using (3), we see that Tq (V ) = \u0013ψ1 (q, y), . . . , ψd (q, y)\u0014k for all (q, y) ∈ V˜f . \u0003 1.4.12 Example Assume that k is algebraically closed and f ∈ k[X1 , . . . , Xn ] is a non-constant and irreducible polynomial. Then the corresponding hypersurface Hf ⊆ kn is irreducible, and we have I(Hf ) = (f); see Theorem 1.1.9. Thus, using Remark 1.4.9, we see that Tp (Hf ) = V({dp (f)}) ⊆ kn for any p ∈ Hf that is, Tp (Hf ) is defined by one linear equation. We conclude that \u0017 n − 1 if Di (f)(p) \b= 0 for some i, dim Tp (Hf ) = n if Di (f)(p) = 0 for all i. Using Exercise 1.8.14, it is easily checked that the set U := {p ∈ Hf | Di (f)(p) = 0 for all i} is a proper closed subset of V . Hence we have dim Tp (Hf ) = n−1 for all p ∈ Hf \\U ; note that dim Hf = n−1 by Example 1.2.16(b). Now consider the special case where n = 2 so that Hf ⊆ k2 is an affine curve; see Example 1.2.19(b). We claim that all but finitely many points of Hf are non-singular. Indeed, let U ⊆ Hf be the set of singular points. We have seen above that U \b= Hf . So we have dim U < dim Hf = 1 by Proposition 1.2.20. Thus, dim U = 0, and so U is a finite set by Exercise 1.8.10(a) and Proposition 1.1.11. 1.4.13 The differential of a regular map Let V ⊆ kn and W ⊆ km be non-empty algebraic sets and ϕ : V → W be a regular map. Let p ∈ V and q := ϕ(p) ∈ W . The map dp ϕ : Derk (A[V ], kp ) → Derk (A[W ], kq ),\n\nD \u0004→ D ◦ ϕ∗ ,\n\nwhere ϕ∗ : A[W ] → A[V ] is the algebra homomorphism in §1.3.3, is called the differential of ϕ. If V = W and ϕ = idV , then dp (id) is the identity map on Derk (A[V ], kp ) for all p ∈ V . Furthermore, if Z ⊆ kl is another algebraic set and ψ : W → Z is a regular map, then we have dp (ψ ◦ ϕ) = dq ψ ◦ dp ϕ.\n\n46\n\nAlgebraic sets and algebraic groups\n\nIn particular, this shows that if ϕ : V → W is an isomorphism, then the differential dp ϕ : Derk (A[V ], kp ) → Derk (A[W ], kq ) is a vector-space isomorphism. Finally, on the level of Tp (V ) ⊆ kn and Tq (W ) ⊆ km , the differential dp ϕ is given as follows. Suppose that ϕ is defined by f1 , . . . , fm ∈ k[X1 , . . . , Xn ]. Then, for v ∈ Tp (V ), we have dp ϕ(Dv ) = Dw , where w = (w1 , . . . , wn ) ∈ Tq (W ) is given by wi = Dw (Y¯i ) = (Dv ◦ ϕ∗ )(Y¯i ) = Dv (f¯i ) for all j. Thus, the linear map dp ϕ : Tp (V ) → Tq (W ) is given by multiplication with the m × n matrix \u0010 \u0011 Jp (ϕ) := Dj (fi )(p)\n\n1\u0001i\u0001m 1\u0001j\u0001n\n\n.\n\nThe following result shows that the tangent space at a point p ∈ V only depends on an open neighbourhood of that point. 1.4.14 Lemma Let V ⊆ kn be a non-empty algebraic set and 0 \b= f¯ ∈ A[V ]. As in §1.1.14, let V˜f = {(v, y) ∈ V × k | f(v)y = 1} ⊆ kn+1 . Then the map πf : V˜f → V,\n\n(v, y) \u0004→ v,\n\nis regular, and d(p,x) πf : T(p,x) (V˜f ) → Tp (V ) is an isomorphism for (p, x) ∈ V˜f . Proof By §1.4.13, d(p,x) (πf ) is given by the matrix J(p,x) (ϕ) which, in our case, is an n × (n + 1)-matrix whose first n columns form the (n × n)-identity matrix and whose last column is zero. Thus, d(p,x) (πf ) : T(p,x) (V˜f ) → Tp (V ) is given by projecting onto the first n coordinates. On the other hand, we know that I(V˜f ) = (I(V ), fY − 1) ⊆ k[X1 , . . . , Xn , Y ] by Lemma 1.1.15. So we have T(p,x) (V˜f ) = {(v, y) ∈ kn+1 | v ∈ Tp (V ) and d(p,x) (fY − 1)(v, y) = 0} = {(v, y) ∈ kn+1 | v ∈ Tp (V ) and y = −xdp (f)(v)/f(p)}, where the last equality follows from d(p,x) (fY −1)(v, y) = xdp (f)(v)+ f(p)y. Thus, for each (v, y) ∈ T(p,x) (V˜f ), we have v ∈ Tp (V ), and y is uniquely determined by v. Consequently, d(πf )(p,x) is an isomorphism. \u0003\n\nThe tangent space and non-singular points\n\n47\n\n1.4.15 Proposition (differential criterion for dominance) Let k be an infinite perfect field, V ⊆ kn and W ⊆ km be irreducible algebraic sets, and ϕ : V → W be a regular map. Assume that dp ϕ : Tp (V ) → Tϕ(p) (W ) is surjective, where p ∈ V is non-singular and ϕ(p) ∈ W is non-singular. Then ϕ is dominant; that is, we have ϕ(V ) = W . Proof We begin by showing that there is an open set U ⊆ V containing p such that the above hypotheses are satisfied for all x ∈ U , that is, x is non-singular, ϕ(x) is non-singular, and dx ϕ is surjective. We shall obtain U as the intersection of three open subsets. First, let U1 be the open set of all non-singular points in V . Next, let U2 be the preimage under ϕ of the open set of all nonsingular points in W . Finally, we claim that there exists an open set U3 ⊆ V such that p ∈ U3 and the image of Tx (V ) under dx ϕ has dimension \u0001 dim W for all x ∈ U3 . This is seen as follows. By Theorem 1.4.11(b), there exists some f ∈ k[X1 , . . . , Xn ] with f(p) \b= 0 and regular maps ψj : V˜f → kn (1 \u0002 j \u0002 d := dim V ) such that Tx (V ) = \u0013ψ1 (˜ x), . . . , ψd (˜ x)\u0014k ⊆ kn\n\nfor all x ∈ Vf .\n\nLet Ψ(x) be the n × d-matrix whose columns are given by ψ1 (˜ x), . . . , ψd (˜ x). Now dx ϕ : Tx (V ) → Tϕ(x) (W ) is the linear map given by v \u0004→ Jx (ϕ)v (v ∈ Tx (V )) where Jx (ϕ) is the m × n matrix defined in §1.4.13. Consequently, the image of Tx (V ) under dx ϕ is the k-span of the columns of Jx (ϕ)Ψ(x). As usual, the condition that the rank of Jx (ϕ)Ψ(x) be \u0001 dim W can be expressed by the non-vanishing of certain determinants in the coordinates of x. But, for p ∈ V , we know that this condition is satisfied. Hence, the set of all x ∈ Vf for which Jx (ϕ)Ψ(x) has rank \u0001 dim W is open. This yields the desired set U3 . Now set U := U1 ∩ U2 ∩ U3 . First note that we have p ∈ U by our hypothesis. Now let x ∈ U . Then x and ϕ(x) are non-singular since x ∈ U1 ∩ U2 . Furthermore, the image of Tx (V ) under dx ϕ is contained in Tϕ(x) (W ) and, hence, has dimension at most dim W (since ϕ(x) ∈ U2 ). Since we also have x ∈ U3 , that dimension must be equal to dim W , and so dx ϕ is surjective. Now we can complete the proof as follows. Let Z := ϕ(V ) ⊆ W . Then Z is closed irreducible (Remark 1.3.2) and we must show that dim Z = dim W ; see Proposition 1.2.20. This is seen as follows. We can factor ϕ as\n\n48\n\nAlgebraic sets and algebraic groups\n\nϕ = ι ◦ ϕ\u0002 , where ϕ\u0002 : V → Z is dominant and ι : Z \u0011→ W is the inclusion. Now consider the open set S of all non-singular points in Z. Since ϕ\u0002 (V ) = ϕ(V ) is dense in Z, we have S ∩ ϕ(V ) \b= ∅. Hence (ϕ\u0002 )−1 (S) is a non-empty open set in V . This open set has non-empty intersection with U , and so there exists some x ∈ U such that ϕ(x) is non-singular in Z. Now dx ϕ is the composition of dx ϕ\u0002 : Tx (V ) → Tϕ(x) (Z) with the inclusion Tϕ(x) (Z) ⊆ Tϕ(x) (W ); see Exercise 1.8.16. The map dx ϕ is surjective since x ∈ U . Hence we must have Tϕ(x) (Z) = Tϕ(x) (W ). Since ϕ(x) ∈ Z is non-singular, this yields dim Z = Tϕ(x) (Z) = dim Tϕ(x) (W ) \u0001 dim W , as desired. \u0003\n\n1.5 The Lie algebra of a linear algebraic group The purpose of this section is to study the tangent spaces of a linear algebraic group G ⊆ Mn (k). We shall show that the tangent space at the identity element of G carries an additional structure, that of a Lie algebra. (The definition of a Lie algebra is recalled in Example 1.4.2.) As examples, we will determine the Lie algebras of SLn (k) and the symplectic and orthogonal groups. 1.5.1 Remark Let G ⊆ Mn (k) be a linear algebraic group, as in Definition 1.3.9. For x ∈ G, consider the map λx : G → G, λx (g) := µ(x, g). By Remark 1.3.12, λx is an isomorphism of algebraic sets. So, using §1.4.13, we conclude that dg (λx ) : Tg (G) → Txg (G) is an isomorphism for all g ∈ G. In particular, taking x = 1, we see that dim Tg (G) = dim T1 (G) for all g ∈ G. 1.5.2 Proposition Let G ⊆ Mn (k) be a linear algebraic group, where k is an infinite perfect field. Then we have dim Tg (G) = dim G for all g ∈ G. In particular, if G is irreducible then G is non-singular. Proof Let G◦ ⊆ G be the irreducible component containing the identity element 1 ∈ G. By Exercise 1.8.16, we have Tg (G◦ ) ⊆ Tg (G) for any g ∈ G◦ . We prove that we have in fact equality. This \u001cis seen as follows. There exist 1 = g1 , g2 , . . . , gr ∈ G such that G = ri=1 G◦ gi . ◦ By Proposition 1.3.13, the cosets G \u0007rgi are◦ precisely the irreducible ◦ components of G. Thus, G \\ G = i=2 G gi ⊆ G is a closed subset of G. Since, by §1.1.7, the operator I is injective, there exists some\n\nThe Lie algebra of a linear algebraic group\n\n49\n\nf ∈ I(G \\ G◦ ) such that f \b∈ I(G). Let p ∈ G◦ be such that f(p) \b= 0. Now let v ∈ Tp (G). This means that dp (h)(v) = 0 for all h ∈ I(G). Furthermore, for h\u0002 ∈ I(G◦ ), we have h\u0002 f ∈ I(G) and h\u0002 (p) = 0, and so 0 = dp (h\u0002 f)(v) = dp (h\u0002 )(v)f(p) + h\u0002 (p) dp (f)(v) = dp (h\u0002 )(v)f(p) for any v ∈ Tp (G). Since f(p) \b= 0, we conclude that dp (h\u0002 )(v) = 0. This holds for all h\u0002 ∈ I(G◦ ), and so v ∈ Tp (G◦ ). Thus, we have shown that Tp (G◦ ) = Tp (G) for some p ∈ G◦ . By Remark 1.5.1, the tangent spaces at all elements of G◦ have the same dimension. Since there exist non-singular points by Theorem 1.4.11, we conclude that dim G◦ = dim Tp (G◦ ) = dim Tp (G). Finally, we have dim G = dim G◦ by Corollary 1.3.14, and so dim G = dim Tp (G). It remains to apply once more Remark 1.5.1 to G and its tangent spaces. \u0003 The following result shows that the tangent space at 1 ∈ G carries naturally the structure of a Lie algebra. In the following result, we use the classical fact that Mn (k) is a Lie algebra with product given by [A, B] = AB − BA. 1.5.3 Theorem Let G ⊆ Mn (k) be a linear algebraic group and T1 (G) ⊆ Mn (k) be the corresponding tangent space at 1 ∈ G. (a) For A, B ∈ T1 (G), we have [A, B] := AB − BA ∈ T1 (G). Thus, T1 (G) is a Lie subalgebra of Mn (k). (b) If H ⊆ Mm (k) is another linear algebraic group and ϕ : G → H is a homomorphism of algebraic groups, then d1 ϕ : T1 (G) → T1 (H) is a homomorphism of Lie algebras. Proof (a) We begin by introducing a product on Derk (A[G], k1 ). This will then be transported to T1 (G), using the isomorphism in Remark 1.4.9(b). For D ∈ Derk (A[G], k1 ) and f ∈ A[G], we define f \u0012 D ∈ A[G] by (f \u0012 D)(x) = D(λ∗x (f)) for x ∈ G.\n\n(∗)\n\nHere, λ∗x : A[G] → A[G] is the algebra homomorphism induced by the map λx : G → G, g \u0004→ xg; see Remark 1.3.12. First we claim that fg \u0012 D = f(g \u0012 D) + g(f \u0012 D) for all f, g ∈ A[G]. Indeed, for any x ∈ G, we have (fg \u0012 D)(x) = D(λ∗x (f)λ∗x (g)) = λ∗x (f)(1)D(λ∗x (g)) +λ∗x (g)(1)D(λ∗x (f)) = f(x)(g \u0012 D)(x) + g(x)\n\n50\n\nAlgebraic sets and algebraic groups\n\n(f \u0012 D)(x), as required. Now, for D, D\u0002 ∈ Derk (A[G], k1 ), we define a map [D, D\u0002 ] : A[G] → k by [D, D\u0002 ](f) = D(f \u0012 D\u0002 ) − D\u0002 (f \u0012 D)\n\nfor f ∈ A[G].\n\n(∗\u0002 )\n\nA straightforward computation, using (∗), yields that [D, D\u0002 ](fg) = f(1)(D(g \u0012 D\u0002 ) − D\u0002 (g \u0012 D)) + g(1)(D(f \u0012 D\u0002 ) − D\u0002 (f \u0012 D)) = f(1)[D, D\u0002 ](g) + g(1)[D, D\u0002 ](f)\n\nfor all f, g ∈ A[G].\n\nThus, we have [D, D\u0002 ] ∈ Derk (A[G], k1 ). We will now transport this operation to T1 (G), using the isomorphism Φ : T1 (G) → Derk (A[G], k1 ), P = (pij ) \u0004→ DP ; see Remark 1.4.9(b). We have ¯ ij denote the image of Xij in A[G] = k[Xij | 1 \u0002 i, j \u0002 n]/I(G). Let X ¯ ij ) = pij for 1 \u0002 i, j \u0002 n. To A[G]. Then DP is determined by DP (X complete the proof of (a), we must identify the matrix in T1 (G) corresponding to the derivation [DP , DP \u0002 ] where P, P \u0002 ∈ T1 (G). Thus, we must show the following identity for all D, D\u0002 ∈ Derk (A[G], k1 ): n \u0002 \b\n\n¯ lj ) − D\u0002 (X ¯ il )D(X ¯ lj ) . ¯ il )D\u0002 (X D(X\n\n¯ ij ) = [D, D\u0002 ](X\n\nl=1\n\nUsing the defining formula for [D, D\u0002 ], we see that it is enough to show that ¯ ij \u0012 D = X\n\nn \u0002\n\n¯ il D(X ¯ lj ) X\n\nfor all D ∈ Derk (A[G], k1 ).\n\nl=1\n\n¯ ij \u0012 To prove this, we argue as follows. Let x ∈ G. Then (X ∗ ¯ D)(x) \u0001 = D(λx (Xij )), and evaluating the right-hand side at x ¯ il (x)D(X ¯ lj ). Thus, the above identity will hold if the yields nl=1 X following identity holds: ¯ ij ) λ∗x (X\n\n=\n\nn \u0002\n\n¯ il (x)X ¯ lj X\n\nfor all x ∈ G.\n\nl=1\n\nTo see this, let us evaluate both sides on y ∈ G. Then the left-hand ¯ ij )(y) = X ¯ ij (xy) = X ¯ ij (µ(x, y)) = µ∗ (X ¯ ij )(x, y). side becomes λ∗x (X ∗ ¯ ij ) = But, the formula for µ in Definition 1.3.9, we have µ∗ (X \u0001n using ¯ ¯ l=1 Xil ⊗ Xlj , and this yields the desired identity. Note that,\n\nThe Lie algebra of a linear algebraic group\n\n51\n\nsince the Jacobi identity is known to hold for the Lie product on Mn (k), it now automatically follows that the product introduced on Derk (A[G], k1 ) also satisfies the Jacobi identity. (b) Again, we work with Derk (A[G], k1 ) and the Lie product defined by (∗\u0002 ). First we claim that we have the following identity: ϕ∗ (h) \u0012 D = ϕ∗ (h \u0012 d1 ϕ(D)) for all D ∈ Derk (A[G], k1 ) and h ∈ A[H].\n\n(†)\n\nIndeed, for any x ∈ G, we have (ϕ∗ (h) \u0012 D)(x) = D(λ∗x (ϕ∗ (h))) and ϕ∗ (h \u0012 d1 ϕ(D))(x) = (h \u0012 d1 ϕ(D))(ϕ(x)) = D(ϕ∗ (λ∗ϕ(x) (h))). Hence it is enough to check that we have λ∗x (ϕ∗ (h)) = ϕ∗ (λ∗ϕ(x) (h)). Evaluating both sides at y ∈ G, the left-hand side becomes λ∗x (ϕ∗ (h))(y) = ϕ∗ (h)(xy) = h(ϕ(xy)) and the right-hand side becomes ϕ∗ (λ∗ϕ(x) (h))(y) = (λ∗ϕ(x) (h))(ϕ(y)) = h(ϕ(x)ϕ(y)) = h(ϕ(xy)), where the last equality holds since ϕ is a group homomorphism. Thus, (†) is proved. Now let D, D\u0002 ∈ Derk (A[G], k1 ) and h ∈ A[H]. Then we have [d1 ϕ(D), d1 ϕ(D\u0002 )](h) = D(ϕ∗ (h \u0012 d1 ϕ(D\u0002 ))) − D\u0002 (ϕ∗ (h \u0012 d1 ϕ(D))) = D(ϕ∗ (h) \u0012 D\u0002 ) − D\u0002 (ϕ∗ (h) \u0012 D) \u0002\n\n(by (†))\n\n\u0002\n\n= [D, D ](ϕ (h)) = d1 ϕ([D, D ])(h). Thus, d1 ϕ is a homomorphism of Lie algebras.\n\n\u0003\n\n1.5.4 Remark Identifying T1 (G) = Derk (A[G], k1 ), we have seen in the above proof that the Lie product on T1 (G) can be characterized entirely in terms of the affine algebra A[G]. Indeed, for D, D\u0002 ∈ Derk (A[G], k1 ), we have [D, D\u0002 ](f) = D(f \u0012 D\u0002 ) − D\u0002 (f \u0012 D)\n\nfor all f ∈ A[G],\n\nwhere f \u0012 D ∈ A[G] is defined by (f \u0012 D)(x) = D(λ∗x (f)) for x ∈ G and λ∗x : A[G] → A[G] is the algebra homomorphism induced by left translation with x.\n\n52\n\nAlgebraic sets and algebraic groups\n\n1.5.5 Example Let k be an infinite perfect field. (a) Let Un (k) be the group of all upper triangular matrices with 1 on the diagonal; see Example 1.3.11. Then we have T1 (Un (k)) = {(aij ) ∈ Mn (k) | aij = 0 for 1 \u0002 j \u0002 i \u0002 n} = {A − In | A ∈ Un (k)}. Indeed, Un (k) is defined by the polynomials Xij (1 \u0002 j < i \u0002 n) and Xii − 1 (1 \u0002 i \u0002 n). Computing d1 (f) for these polynomials, we obtain the inclusion ‘⊆’. On the other hand, as an algebraic set, we have Un (k) ∼ = kn(n−1)/2 and so T1 (Un (k)) has dimension n(n − 1)/2 (see Example 1.4.10). Thus, the inclusion must be an equality. By a similar argument, we see that T1 (Un\u0002 (k)) = {A − In | A ∈ Un\u0002 (k)}, where Un\u0002 (k) is the group of all lower triangular matrices with 1 on the diagonal. (1) (b) Let Tn (k) be the group of all diagonal matrices in Mn (k) with determinant 1. Then, by a similar argument, we see that T1 (Tn(1) (k)) = {A ∈ Mn (k) | A diagonal and trace(A) = 0}. (1)\n\nIndeed, Tn is defined by the polynomials Xij (i \b= j) and X11 · · · Xnn − 1. Computing d1 (f) for these polynomials, we obtain the inclusion ‘⊆’. To prove the inclusion ‘⊇’, it is enough to show (1) that Tn is connected of dimension n − 1 (see Proposition 1.5.2). Now we note that the following map is an isomorphism: n−1 '\n\n{(xi , yi ) ∈ k2 | xi yi = 1}\n\n(1)\n\n−→ Tn (k)\n\ni=1\n\n((x1 , y1 ), . . . , (xn−1 , yn−1 ))\n\n\u0004→\n\n.\n\ndiag(x1 , . . . , xn−1 , y1 · · · yn−1 )\n\nSince {(x, y) ∈ k2 | xy = 1} is connected of dimension 1 (why?), (1) we conclude that Tn (k) is connected of dimension n − 1 (by Proposition 1.3.8). We now establish some differentiation formulas which will turn out to be useful tools for computing tangent spaces; see Example 1.5.8.\n\nThe Lie algebra of a linear algebraic group\n\n53\n\n1.5.6 Lemma Let G ⊆ Mn (k) be a linear algebraic group; then T1 (G) ⊆ Mn (k). Let µ : G × G → G and ι : G → G be the regular maps defining the multiplication and the inversion in G, respectively. (a) We have d(1,1) µ(A, B) = A + B for all A, B ∈ T1 (G), where we identify T(1,1) (G × G) = T1 (G) ⊕ T1 (G); see Exercise 1.8.16. (b) We have d1 ι(A) = −A for all A ∈ T1 (G). Proof (a) Let x = (xij ) and y = (yij ) be two matrices in G. Then \u0001 µ(x, y) is the matrix with (i, j)-coordinate µij := nl=1 xil ylj . So we obtain (x) (y) Drs (µij )(In , In ) = δir δsj , (µij )(In , In ) = Drs (x)\n\n(y)\n\nwhere Drs and Drs denote partial derivatives with respect to the x-variable and to the y-variable, respectively, which is indexed by (r, s). This yields d(In ,In ) µij (A, B) =\n\nn \u0002\n\nδir δsj (ars + brs ) = aij + bij\n\nr,s=1\n\nwhere A = (aij ) and B = (bij ) are in T1 (G). (b) For\u0001 x = (xij ) ∈ G, let us denote ι(x) = (˜ xij ). Thus, we have δij = nl=1 xil x ˜lj for 1 \u0002 i, j \u0002 n. Denoting by Drs the partial derivative with respect to the variable indexed by (r, s), we obtain 0=\n\n=\n\nn \u0002 \b\n\n˜lj + xil Drs (˜ xlj ) (In ) Drs (xil ) x\n\nl=1 n \u0002\n\n˜lj (In ) + δil Drs (˜ xlj )(In ) = δir δsj + Drs (˜ xij )(In ). δir δls x\n\n\b\n\nl=1\n\nxij )(In ) = −δir δsj and so d1 ι(A)ij Thus, \u0001 we have Drs (˜ − nr,s=1 δir δsj ars = −aij for all A = (aij ) ∈ T1 (G).\n\n= \u0003\n\n1.5.7 Lemma Let G ⊆ Mn (k) be a linear algebraic group. Let ϕ : G → Mn (k) be a regular map with ϕ(1) = In , and let B ∈ Mn (k) be fixed. Then ϕ˜B : G → Mn (k),\n\ng \u0004→ gBϕ(g),\n\nis a regular map and d1 ϕ˜B : T1 (G) → TB (Mn (k)) = Mn (k) is given by d1 ϕ˜B (A) = AB + B d1 ϕ(A)\n\nfor all A ∈ T1 (G).\n\n54\n\nAlgebraic sets and algebraic groups\n\nProof First note that, if B = b1 B1 + b2 B2 , where bi ∈ k and Bi ∈ Mn (k), then ϕ˜B = b1 ϕ˜B1 +b2 ϕ˜B2 , and so d1 ϕ˜B = b1 d1 ϕ˜B1 +b2 d1 ϕ˜B2 . Thus, it will be enough to consider the case where B is an elementary matrix, that is, there exist 1 \u0002 u, v \u0002 n such that the (u, v)-entry in B is 1 and all other entries are zero. In this case, the (i, j)-coefficient of ϕ˜B (g) ∈ Mn (k) is given by giu ϕ(g)vj . Denoting by Drs the partial derivative with respect to the (r, s)-variable, we obtain Drs (giu ϕ(g)vj )(In ) = δri δus ϕ(g)vj + giu Drs (ϕ(g)vj )(In ). Given A = (ars ) ∈ T1 (G), this yields d1 ϕ˜B (A)ij = aiu δvj + δiu d1 ϕvj (A) = (AB)ij + (B d1 ϕ(A))ij , as required. \u0003 1.5.8 Example Let k be an infinite perfect field and G = SLn (k). With the notation of Example 1.5.5, consider the regular map ϕ : Un (k) × Tn(1) (k) × Un\u0002 (k) → G,\n\n(u, h, u\u0002 ) \u0004→ uhu\u0002 .\n\nSince uhu\u0002 = µ(u, µ(h, u\u0002 )), Lemma 1.5.6 shows that d1 ϕ : T1 (Un (k)) ⊕ T1 (Tn(1) (k)) ⊕ T1 (Un\u0002 (k)) → T1 (G) is given by (x, h, x\u0002 ) \u0004→ x + h + x\u0002 . Now T1 (Un (k)) consists of strictly (1) upper triangular matrices, T1 (Tn (k)) consists of diagonal matrices and T1 (Un\u0002 (k)) consists of strictly lower triangular matrices. It follows that d1 ϕ is injective and so dim T1 (G) \u0001 n2 − 1. On the other hand, we certainly have T1 (G) ⊆ {A ∈ Mn (k) | d1 (det)(A) = 0}. Now, denoting by Drs the partial derivative with respect to the variable indexed by (r, s), we obtain d1 (det) =\n\n\u0002\n\nsgn(π)\n\n\u0002\n\nsgn(π)\n\n\u0002 π∈Sn\n\nn \u0010 \u0002\n\nδsπ(r)\n\nr,s=1\n\nπ∈Sn\n\n=\n\nDrs (X1π(1) X2π(2) · · · Xnπ(n) )(In ) Xrs\n\nr,s=1\n\nπ∈Sn\n\n=\n\nn \u0002\n\nsgn(π)\n\nn \u0010\u0002 r,s=1\n\nn '\n\n\u0011 Xiπ(i) (In ) Xrs\n\ni=1 i\u0003=r\n\nδsπ(r)\n\nn '\n\n\u0011 δiπ(i) Xrs .\n\ni=1 i\u0003=r\n\n( Now note that the expression i\u0007=r δiπ(i) is zero unless π ∈ Sn fixes n − 1 points. This automatically implies that π must be the identity,\n\nThe Lie algebra of a linear algebraic group\n\n55\n\n\u0001 and so the above expression simplifies to d1 (det) = nr,s=1 δsr Xrs = \u0001n r=1 Xrr . Thus, we have d1 (det)(A) = trace(A) and this gives an upper bound for dim T1 (G). Combining this with dim T1 (G) \u0001 n2 −1, we finally conclude that T1 (SLn (k)) = {A ∈ Mn (k) | trace(A) = 0}. By Theorem 1.5.3, the Lie product on T1 (SLn (k)) is given by [A, A\u0002 ] = AA\u0002 − A\u0002 A. The Lie algebra T1 (SLn (k)) is usually denoted by sln (k). The above discussion also implies that d1 ϕ is surjective and so the map ϕ is dominant, by Theorem 1.4.15. Our next aim is to determine the Lie algebras of the symplectic and orthogonal groups that we introduced in Section 1.3. The method is roughly the same as that in Example 1.5.8. For this purpose, we need the following preliminary results about some subgroups. Recall the definition of Un (k) from Example 1.5.5. 1.5.9 Lemma The group U := U2m (k) ∩ Sp2m (k) (m \u0001 1) consists of all matrices of the form ⎡ ⎤ A AQm S ⎦, u(A, S) := ⎣ −1 tr Qm (A ) Qm 0 where A ∈ Um (k) and S ∈ Mm (k) is such that S = S tr . We have dim U = m2 and U is connected if |k| = ∞. Furthermore, a diagonal matrix belongs to Sp2m (k) if and only if −1 −1 the diagonal entries are t1 , t2 , . . . , tm−1 , tm , t−1 m , tm−1 , . . . , t1 , where 0 \b= ti ∈ k. Proof The statements concerning the matrices are proved by a straightforward computation (which we leave as an exercise to the reader). Now assume that k is infinite. Then we note that the map Um (k) × {S ∈ Mm (k) | S = S tr } → U,\n\n(A, S) \u0004→ u(A, S),\n\nis regular and bijective. Furthermore, the inverse is easily seen to be a regular map, too. Thus, the above map is an isomorphism of algebraic sets. Now Um (k) and {S ∈ Mm (k) | S = S tr } are linear subspaces of Mn (k). Hence these sets are irreducible, and their\n\n56\n\nAlgebraic sets and algebraic groups\n\ndimension as an algebraic set equals the dimension as a vector space; see Exercise 1.8.6. This yields the desired formula and the fact U is irreducible, using Proposition 1.3.8(a) and (c). \u0003 1.5.10 Lemma Assume that char(k) \b= 2. Then the group U := U2m+1 (k) ∩ O2m+1 (k) (m \u0001 1) consists of all matrices of the form ⎤\n\n⎡ ⎢ ⎢ u(A, S, w) := ⎢ ⎢ ⎣\n\nA\n\nv\n\nAQm S\n\n0\n\n1\n\nw\n\n0\n\n0\n\nQm (A−1 )tr Qm\n\n⎥ ⎥ ⎥ (v := −AQm wtr ), ⎥ ⎦\n\nwhere A ∈ Um (k), S ∈ Mm (k), and w ∈ km (row vector) are such that S + S tr = −wtr w. We have dim U = m2 and U is connected if |k| = ∞. Furthermore, a diagonal matrix belongs to O2m+1 (k) if and only −1 −1 if the diagonal entries are t1 , t2 , . . . , tm−1 , tm , ±1, t−1 m , tm−1 , . . . , t1 where 0 \b= ti ∈ k. Proof This is similar to the proof of the previous lemma. Now the map Um (k) × {(w, S) ∈ km ×Mm (k) | S+S tr = −wtr w} → U, given by sending (A, w, S) to u(A, S, w), is an isomorphism of algebraic sets. To compute the dimension of these sets, it remains to note that the map km ×{S ∈ Mm (k) | S+S tr = 0} → {(w, S) ∈ km ×Mm (k) | S+S tr = −wtr w}, given by sending (w, S) to (w, S − wtr w/2), is an isomorphism.\n\n\u0003\n\nFinally, we consider the even-dimensional orthogonal groups, where we use the description using a suitable quadratic form as in §1.3.16.\n\nThe Lie algebra of a linear algebraic group\n\n57\n\n1.5.11 Lemma The group U := U2m (k)∩O+ 2m (k) (m \u0001 1) consists of all matrices of the form ⎡ ⎤ A AQm S ⎦, u(A, S) := ⎣ 0 Qm (A−1 )tr Qm where A ∈ Um (k) and S ∈ Mm (k) is such that S + S tr = 0 and all diagonal entries of S are zero. We have that dim U = m2 − m and U is connected if |k| = ∞. Furthermore, a diagonal matrix belongs to O+ 2m (k) if and only if −1 −1 the diagonal entries are t1 , t2 , . . . , tm−1 , tm , t−1 , m tm−1 , . . . , t1 , where 0 \b= ti ∈ k. Proof Again, this is similar to the proof of Lemma 1.5.9. Just note that, if char(k) \b= 2, then the condition S + S tr = 0 automatically implies that all diagonal entries of S are zero. If char(k) = 2, the fact that all diagonal entries of S are zero follows from the additional equations in the description of O+ 2m (k) in §1.3.16. Now we have an isomorphism of algebraic sets \u0019 \u0017 \u0018 \u0018 S + S tr = 0 and → U, Um (k) × S ∈ Mm (k)\u0018 sii = 0 for 1 \u0002 i \u0002 m given by sending (A, S) to u(A, S). The dimension is computed as before. \u0003 1.5.12 Remark Let G ⊆ Mn (k) be one of the classical groups: ⎧ ⎨SO2m+1 (k), n = 2m + 1, char(k) \b= 2, Sp2m (k), n = 2m, any characteristic, G= ⎩ SO+ (k), n = 2m, any characteristic; 2m see Example 1.3.15 and §1.3.16. We set \u0017 − Q2m if n = 2m and G = Sp2m (k), n0 := Qn otherwise. This matrix has the property that conjugation by n0 transforms an upper triangular matrix into a lower triangular matrix. We have A ∈ G ⇒ n0 An0−1 ∈ G\n\n(note that n02 = ±In ).\n\nThis is easily shown by an explicit computation which we leave as an exercise. Thus, the results in Lemmas 1.5.9–11 also hold for the\n\n58\n\nAlgebraic sets and algebraic groups\n\ngroups of lower unitriangular matrices in G. Now set U := Un (k) ∩ G\n\nand\n\nH := {A ∈ G | A diagonal}.\n\nAssume that k is infinite. Then we have already remarked in Lemmas 1.5.9–11 that U is connected; similarly, U \u0002 := n0 Un0−1 is connected. Finally, H is connected and dim H = m. This is seen by an argument similar to that which was used to deal with the group of diagonal matrices in SLn (k); see Example 1.5.5. 1.5.13 Theorem Let k be an infinite perfect field. Then we have T1 (SOn (k)) = {A ∈ Mn (k) | Atr Qn + Qn A = 0}\n\n(char(k) \b= 2),\n\n− T1 (Sp2m (k)) = {A ∈ M2m (k) | Atr Q− 2m + Q2m A = 0}\n\n(any characteristic), where Qn and Q− 2m are defined as in §1.3.15. Furthermore, if char(k) = 2, then ⎫ ⎧ \u0018 tr ⎨ \u0018A Q2m + Q2m A = 0 and ⎬ \u0018 tr T1 (SO+ 2m (k)) = ⎩A ∈ M2m (k)\u0018\u0018 (A P2m + P2m A)rr = 0 ⎭ , for 1 \u0002 r \u0002 2m where P2m and SO+ 2m (k) are defined as in §1.3.16. Proof Let Q ∈ Mn (k) be invertible. Writing out the equation Atr QA = Q for all matrix entries, we get a system of n2 quadratic polynomials in n2 variables. By a computation similar to that in Example 1.5.8, we obtain dIn (f) for each of these polynomials. This yields that T1 (Γn (Q, k)) ⊆ {A ∈ Mn (k) | Atr Q + QA = 0}. Now assume that char(k) = 2 and that Q ∈ M2m (k) \u0001 is alternating. tr As in §1.3.16, write Q = P + P , where f = ij pij Xi Xj . Let {e1 , . . . , e2m } be the standard basis of k2m . For each i, the identity f(Aei ) = f(ei ) yields a polynomial equation for the coefficients of A.\n\nThe Lie algebra of a linear algebraic group\n\n59\n\nComputing dIn for these polynomials, we obtain \u0017 \u0019 \u0018 Atr Q + QA = 0 and \u0018 + T1 (Γ2m (f, k)) ⊆ A ∈ M2m (k)\u0018 tr . (A P + PA)rr = 0 for 1 \u0002 r \u0002 2m Thus, in each case, we have already established one inclusion. To show that these inclusions are equalities, we use a dimensional argument. Let us give the details for G = Sp2m (k). (The proof for the other cases is completely analogous.) Let U := U2m (k) ∩ Sp2m (k)\n\nand H := {A ∈ Sp2m (k) | A diagonal}.\n\nNow note that the map γ : A \u0004→ n0 An0−1 defines an isomorphism of Sp2m (k); see Remark 1.5.12. Let U \u0002 := γ(U ) and consider the regular map ϕ : U × H × U \u0002 → Sp2m (k),\n\n(u, h, u\u0002 ) \u0004→ uhu\u0002 .\n\nSince uhu\u0002 = µ(u, µ(h, u\u0002 )), Lemma 1.5.6 shows that d1 ϕ : T1 (U )⊕T1 (H)⊕T1 (U \u0002 ) → T1 (Sp2m (k)),\n\n(x, u, x\u0002 ) \u0004→ x+u+x\u0002 .\n\nNow, we certainly have T1 (U ) ⊆ T1 (U2m (k)), and so Example 1.5.5(a) shows that T1 (U ) consists of strictly upper triangular matrices. Similarly, T1 (H) consists of diagonal matrices. Finally, note that γ transforms an upper triangular matrix into a lower triangular matrix. Consequently, T1 (U \u0002 ) consists of strictly lower triangular matrices. It follows that d1 ϕ is injective and, hence, dim T1 (Sp2m (k)) \u0001 dim(T1 (U ) ⊕ T1 (H) ⊕ T1 (U \u0002 )) = 2 dim T1 (U ) + dim T1 (H), where the last equality holds since U and U \u0002 are isomorphic. Using Lemma 1.5.9, Remark 1.5.12, and Proposition 1.5.2, we conclude that dim T1 (Sp2m (k)) \u0001 2m2 + m. On the other hand, we have − already seen that T1 (Sp2m (k)) ⊆ {A ∈ M2m (k) | Atr Q− 2m+Q2m A = 0} and the space on the right hand side has dimension 2m2 + m. Thus, we must have dim T1 (Sp2m (k)) = 2m2 + m, as required. \u0003 1.5.14 Corollary Let k be an infinite perfect field. Then we have dim SO2m+1 (k) = 2m2 + m\n\n(any m \u0001 1, char(k) \b= 2),\n\ndim Sp2m (k) = 2m2 + m\n\n(any m \u0001 1, any characteristic),\n\n= 2m − m\n\n(any m \u0001 1, any characteristic).\n\ndim SO+ 2m (k)\n\n2\n\n60 Proof\n\nAlgebraic sets and algebraic groups This follows from Proposition 1.5.13 and Proposition 1.5.2. \u0003\n\n1.5.15 Groups of Lie type For readers familiar with the basic theory of semisimple complex Lie algebras (see, for example, Humphreys (1972)), we just give a very rough sketch of Chevalley’s construction of groups of Lie type. Let g be a semisimple Lie algebra over C and let h ⊆ g be a Cartan subalgebra. Then we have a corresponding Cartan decomposition of g; see §8 of Humphreys (1972). Thus, there exists a finite subset Φ ⊆ Hom(h, C) \\ {0} and a basis B = {h1 , . . . , hn ; eα , α ∈ Φ} of g where h1 , . . . , hn form a basis of h and [hi , eα ] = α(hi )eα for all i and all α ∈ Φ. Now Chevalley has shown that B can be chosen such that the structure constants of g with respect to B lie in Z; see §25.2 of Humphreys (1972). Hence, for any field k, we obtain a Lie algebra over k by setting gk := k ⊗Z \u0013B\u0014Z . Now, given α ∈ Φ and t ∈ C, then t ad(eα ) is a nilpotent endomorphism of g, and so we have exp(t ad(eα )) ∈ SL(g); see §2.3 of Humphreys (1972). Moreover, it can be shown that the entries of the matrix of exp(t ad(eα )) with respect to B can be written as polynomials in t with coefficients in Z. This makes it possible to define an automorphism xα (t) ∈ SL(gk ), by transferring the above exponential construction to k; see §25.5 of Humphreys (1972). Then we define G(k) := \u0013xα (t) | α ∈ Φ, t ∈ k\u0014 ⊆ SL(gk ), and call this the Chevalley group or the group of Lie type over k associated with g. Now Chevalley’s construction shows that, for a fixed α ∈ Φ, the map xα : k → SL(gk ), t \u0004→ xα (t), is a regular map; see §3 of Steinberg (1967). This implies, using a general criterion which we shall establish later in Theorem 2.4.6, that G(k) is a closed connected subgroup of SL(gk ) when k is algebraically closed. Then one can also show that gk indeed is the Lie algebra of G(k), in the sense of Theorem 1.5.3. For more details see Steinberg (1967) and Carter (1972).\n\n1.6 Groups with a split BN -pair We now study an important class of groups, the groups with a socalled BN -pair or Tits system. As mentioned in the introduction of\n\nGroups with a split BN -pair\n\n61\n\nthis chapter, this concept has turned out to be extremely useful. Our aim is to determine such BN -pairs for all the groups of classical type introduced in §1.3.15 and §1.3.16. As an immediate application, we can determine exactly which of these groups are connected. In Chapter 3, we will exhibit a deep property of the subgroup B involved in a BN -pair: it is a Borel subgroup in the sense of Definition 3.4.1. 1.6.1 Definition (Tits) Let G be an abstract group. We say that G is a group with a BN -pair or that G admits a Tits system if there are subgroups B, N ⊆ G such that the following conditions are satisfied. (BN1)\n\nG is generated by B and N .\n\n(BN2)\n\nH := B ∩ N is normal in N , and the quotient W := N/H is a group generated by a set S of elements of order 2.\n\n(BN3)\n\nns Bns \b= B if s ∈ S and ns is a representative of s in N .\n\n(BN4)\n\nns Bn ⊆ Bns nB ∪ BnB for any s ∈ S and n ∈ N . \u0006 nBn−1 = H.\n\n(BN5)\n\nn∈N\n\nThe group W is called the Weyl group of G. (BN5) is called the saturation axiom. Sometimes, it is omitted from the definition of a BN -pair, but we shall find it convenient to include this condition here. Before we give some examples, we shall establish some elementary properties of a group with a BN -pair. For any w ∈ W , we set C(w) := Bnw B, where nw ∈ N is a representative of w in N . Since any two representatives of w lie in the same coset of H ⊆ B, we see that C(w) does not depend on the choice of the representative. The sets C(w) are called Bruhat cells. We define a length function on W as follows. We set l(1) = 0. If w \b= 1, we define l(w) to be the length of a shortest possible expression of w as a product of generators in S. (Note that we don’t have to take into account inverses, since s2 = 1 for all s ∈ S.) Thus, any w ∈ W can be written in the form w = s1 · · · sp , where p = l(w) and si ∈ S for all i. Such an expression (which is by no means unique) will be called a reduced expression for w. Note that, given a non-unit w ∈ W , we can always find some s ∈ S such that l(sw) < l(w). (Take, for example, the first factor in a reduced expression for w.) This will be used frequently in proofs by induction on l(w).\n\n62\n\nAlgebraic sets and algebraic groups\n\n1.6.2 Lemma Let J ⊆ S and define WJ ⊆ W to be the subgroup generated by J . Let NJ ⊆ N be the preimage of WJ under the natural map N → W . Then PJ := BNJ B is a subgroup of G. Proof The set PJ is closed under inversion, so it only remains to show that it is closed under multiplication. Now, we certainly have BPJ ⊆ PJ . Hence it will be enough to show that NJ PJ ⊆ PJ . Since NJ = \u0013H; ns , s ∈ J \u0014, it will be enough to show that ns PJ ⊆ PJ for all s ∈ J . But this is clear by (BN4). \u0003 1.6.3 Proposition (Bruhat decomposition) We double-coset decomposition , G= Bnw B.\n\nhave\n\nthe\n\nw∈W\n\nFurthermore, given s ∈ S and w ∈ W , we have l(sw) = l(w) ± 1 and \u0017 if l(sw) = l(w) + 1, Bns nw B Bns B.Bnw B = Bns nw B ∪ Bnw B if l(sw) = l(w) − 1. Proof \u0007 Taking J = S in Lemma 1.6.2, we see that G = BNB and so G = w∈W Bnw B. Now let y, w ∈ W . Then we must show that Bny B = Bnw B implies y = w. We will prove this by induction on min{l(y), l(w)}. We may assume without loss of generality that l(y) \u0002 l(w). Now, if l(y) = 0, then y = 1 and so B = Bny B = Bnw B. This yields nw ∈ B ∩N = H and so w = 1, as required. Now suppose that l(y) > 0. Then we can write y = sx, where s ∈ S and x ∈ W are such that l(y) = l(x) + 1. Then we have ns nx B ⊆ Bny B = Bnw B and so, using (BN4): nx B ⊆ ns Bnw B ⊆ Bns nw B ∪ Bnw B. Hence Bnx B = Bns nw B or Bnx B = Bnw B. By induction, we have x = sw or x = w. The latter case is impossible since l(x) < l(y) \u0002 l(w). Thus, we must have x = sw, and so y = sx = w, as required. Now consider the multiplication rule. Let s ∈ S and w ∈ W . First note that we certainly have l(w) − 1 \u0002 l(sw) \u0002 l(w) + 1. Now assume first that l(sw) \u0001 l(w). We show by induction on l(w) that Bns B.Bnw B = Bns nw B. If l(w) = 0, then w = 1 and so there is nothing to prove. Now assume that l(w) > 0 and write w = yt, where t ∈ S and y ∈ W are such that l(w) = l(y) + 1. By (BN4),\n\nGroups with a split BN -pair\n\n63\n\nwe already know that Bns B.Bnw B = Bns nw B or ns Bnw ∩Bnw B \b= ∅. Assume, if possible, that we were in the second case. Then ns Bnw ∩Bnw B \b= ∅ and so ns Bny ∩Bnw Bnt \b= ∅. Now l(sy) \u0001 l(y) and so, by induction, we have ns Bny ⊆ Bns ny B. We conclude that Bns ny B ∩ Bnw Bnt \b= ∅. Now, by inverting the relation in (BN4), we see that there is also a ‘right-handed’ version of (BN4). This yields that Bns ny B ∩ Bnw nt B \b= ∅ or Bns ny B ∩ Bnw B \b= ∅. Thus, by the first part of the proof, we have sy = wt or sy = w. The former equation gives sy = yt2 = y and so s = 1, contradicting (BN3). The latter gives sw = y and so l(sw) = l(y) < l(w), a contradiction. Thus, our assumption was wrong, and so we must have Bns B.Bnw B = Bns nw B, as required. Now assume that l(sw) \u0002 l(w). By (BN4), we have ns Bns ⊆ B ∪ Bns B. Using (BN3), this implies that ns Bns ∩ Bns B \b= ∅. Consequently, we also have ns B ∩Bns Bns \b= ∅ and ns Bnw ∩Bns Bns nw \b= ∅. Now, we have l(s(sw)) = l(w) \u0001 l(sw) and so Bns Bns nw ⊆ Bnw B by the previous case. This means ns Bnw ∩ Bnw B \b= ∅. On the other hand, we certainly have ns nw ∈ Bns B.nw B. So (BN4) shows that Bns B.Bnw B = Bns nw B ∪ Bnw B. \u0003 The sharpened multiplication rules for Bruhat cells in Proposition 1.6.3 are extremely strong statements. We shall give a number of applications. 1.6.4 Corollary\n\nWe have NG (B) = B.\n\nProof Let g ∈ NG (B). Then, by the Bruhat decomposition, we can write g = bnw b\u0002 , where b, b\u0002 ∈ B and w ∈ W . This yields nw ∈ NG (B). Now let s ∈ S be such that l(sw) < l(w). By Proposition 1.6.3, we have ns Bnw ∩ Bnw B \b= ∅ and so ns ∈ Bnw Bn−1 w B ⊆ NG (B), contradicting (BN3). Thus, we have w = 1, as required. \u0003 1.6.5 Corollary We have S = {1 \b= w ∈ W | B ∪ Bnw B ⊆ G is a subgroup}. Proof Let s ∈ S. Then P{s} = B ∪ Bns B is a subgroup by Lemma 1.6.2. Conversely, let 1 \b= w ∈ W be such that B ∪ Bnw B is a subgroup. Let s ∈ S be such that l(sw) < l(w). Then, as in the proof of Corollary 1.6.4, we have ns ∈ Bnw Bn−1 w B ⊆ B ∪ Bnw B and so s = w by Proposition 1.6.3. \u0003\n\n64\n\nAlgebraic sets and algebraic groups\n\n1.6.6 Corollary Let G be a group with a BN -pair, where W is finite. Let n0 ∈ N be such that the image of n0 has maximal length in W . Then we have G = \u0013B, n0 Bn0−1 \u0014. Proof Let w0 ∈ W denote the image of n0 . Then, for any s ∈ S, we have l(sw0 ) \u0002 l(w0 ) and so, using Proposition 1.6.3, n0 ∈ Bns B · Bn0 B. This yields ns ∈ Bn0 Bn0−1 B ⊆ \u0013B, n0 Bn0−1 \u0014 and so G = \u0013B, N \u0014 ⊆ \u0013B, n0 Bn0−1 \u0014. \u0003 1.6.7 Remark Assume that G is an affine algebraic group and that G has a BN -pair, where W is finite and B is a closed connected subgroup of G. Then B ⊆ G◦ and n0 Bn0−1 ⊆ G◦ , and so G = G◦ by Corollary 1.6.6. This will be one of our tools for determining which of the classical groups are connected. 1.6.8 Corollary (Exchange condition) Let w ∈ W and write w = s1 · · · sp , where si ∈ S and p = l(w). Let s ∈ S and assume that l(sw) < l(w). Then sw = s1 · · · si−1 si+1 · · · sp\n\nfor some 1 \u0002 i \u0002 p.\n\n−1 B = Proof As in the proof of Corollary 1.6.4, ns ∈ Bnw Bnw −1 C(w).C(w ). Hence, by Exercise 1.8.22(a), there exists some 0 \u0002 q \u0002 p such that\n\ns = xw−1 ,\n\nwhere x = si1 · · · siq and 1 \u0002 i1 < · · · < iq \u0002 p.\n\nNow consider a reduced expression for x, and evaluate the product xw−1 by multiplying one by one the factors in that expression for x; at each step, the length either increases or decreases by 1. As a result, we see that l(xw−1 ) \u0001 l(w) − l(x). Since l(x) \u0002 q and s = xw−1 , we deduce that 1 \u0001 p − q and so q \u0001 p − 1. If we had q = p, then x = w and so s = 1, which is absurd. Hence we have q = p − 1, and so there exists some 1 \u0002 i \u0002 p such that x = s1 · · · si−1 si+1 · · · sp . This yields the desired assertion. \u0003 1.6.9 Example Let k be a field and G = GLn (k). We set Bn (k) := subgroup of all upper triangular matrices in GLn (k), Nn (k) := subgroup of all monomial matrices in GLn (k). (A matrix is called monomial if it has exactly one non-zero entry in each row and and each column.) Our aim will be to show that the\n\nGroups with a split BN -pair\n\n65\n\ngroups Bn (k) and Nn (k) form a BN -pair in G. For this purpose, we introduce some special matrices and subgroups. We have Tn (k) := Bn (k) ∩ Nn (k) = subgroup of all diagonal matrices in GLn (k) and Bn (k) = Un (k).Tn (k), where Un (k) is the group of all upper unitriangular matrices, as in Example 1.3.11. Note that Un (k) ∩ Tn (k) = {1} and Un (k) ⊆ Bn (k) is a normal subgroup. For 1 \u0002 i \u0002 n − 1, let ni be the permutation matrix which is obtained by interchanging the ith and the (i + 1)th row in In . More generally, if w ∈ Sn is any permutation, let nw be the matrix which is obtained by permuting the rows of In as specified by w. (Thus, the ith row of In is the w(i)th row of nw .) Then we have Nn (k) = {hnw | h ∈ Tn (k), w ∈ Sn } and so W := Nn (k)/Tn (k) ∼ = Sn\n\nand ni ∈ Nn (k) ↔ σi ∈ Sn ,\n\nwhere σi denotes the basic transposition σi = (i, i + 1). Next, for 1 \u0002 i, j \u0002 n, let Eij be the ‘elementary’ matrix with coefficient 1 at the position (i, j) and 0 otherwise. We define Xij := {In + λEij | λ ∈ k},\n\nwhere 1 \u0002 i, j \u0002 n and i \b= j.\n\nWe write Xi := Xi,i+1 and X−i := Xi+1,i for 1 \u0002 i \u0002 n − 1. These are all subgroups of G. Furthermore, let Vi := group of all matrices in Un (k) with 0 at position (i, i + 1). By straightforward matrix computations, it is readily checked that the following relations hold. (a) Un (k) = Xi .Vi = Vi .Xi and Vi ∩Xi = {1} for 1 \u0002 i \u0002 n−1. (b) nw Xij n−1 w = Xw(i),w(j) for all w ∈ Sn and i \b= j. −1 (c) ni Xi n−1 i = X−i and ni Vi ni = Vi for any 1 \u0002 i \u0002 n − 1. 1.6.10 Proposition In the above setting, the groups Bn (k) and Nn (k) form a BN -pair in G = GLn (k), with Weyl group W ∼ = Sn . Proof We must show that the five axioms in Definition 1.6.1 are satisfied. Let B := Bn (k), N := Nn (k), H := Tn (k) = B ∩ N , and U := Un (k). (BN1): Let g ∈ G and choose b ∈ B to maximize the total number of zeros at the beginnings of all the rows of bg. These beginnings\n\n66\n\nAlgebraic sets and algebraic groups\n\nmust all be of different lengths since otherwise we could subtract a multiple of some row from an earlier row, that is, modify b, and increase the total number of zeros. It follows that, for some w ∈ Sn , we have nw bg ∈ B, as required. (BN2): The group H is normal in N , and W = N/H ∼ = Sn is generated by the involutions ni (1 \u0002 i \u0002 n − 1). (BN3): Let 1 \u0002 i \u0002 n − 1. By Example 1.6.9(a), we have B = Xi .Vi .Tn (k). Conjugating by ni and using Example 1.6.9(c) yields −1 −1 that ni Bni = ni Bn−1 i = ni Xi ni .ni Vi ni .H = X−i .Vi .H \b⊆ B. (BN4): First we show that ni Bni ⊆ B ∪ Bni B. For this purpose, we begin with the following identity in GL2 (k), where 0 \b= t ∈ k: \u001a \u001b \u001a \u001b \u001b \u001a \u001b \u001a \u001b\u001a 0 1 0 0 1 1 −t t 1 t−1 = . · · t 1 1 0 0 1 0 1 0 −t−1 Now let 1 \u0002 i \u0002 n − 1. Then we have an embedding ⎡ ⎤ Ii−1 0 0 ⎢ ⎥ A 0 ⎦. ϕi : GL2 (k) \u0011→ GLn (k), A \u0004→ ⎣ 0 0 0 In−i−1 Applying ϕi to the above identity, we obtain X−i ⊆ {1} ∪ Xi ni Xi H.\n\n(∗)\n\nNow we write B = Vi .Xi .H. Then we have ni Bni = ni Bn−1 = i −1 Vi .ni Xi ni .H = Vi .X−i .H by Example 1.6.9. Using (∗), this yields ni Bni ⊆ Vi .({1} ∪ Xi ni Xi H).H ⊆ Vi .H ∪ U ni Xi H ⊆ B ∪ Bni B. Next, let w ∈ W and 1 \u0002 i \u0002 n − 1. Assume first that w−1 (i) < w−1 (i + 1). Then, using Example 1.6.9(b), we have n−1 w X i nw = Xw−1 (i),w−1 (i+1) ⊆ U . Furthermore, by Example 1.6.9(c), we also have ni Vi ⊆ Vi ni . Thus, we obtain ni Bnw ⊆ ni Vi Xi Hnw ⊆ Vi ni nw .n−1 w X i nw H ⊆ V i ni nw U H ⊆ Bni nw B. Finally, assume that w−1 (i) > w−1 (i + 1) and set y := σi w. Then we have y −1 (i) = w−1 (i + 1) < w−1 (i) = y −1 (i + 1). Hence, by the previous argument, we have ni Bny ⊆ Bni ny B = Bnw B. This\n\nGroups with a split BN -pair\n\n67\n\nyields that ni Bnw = ni Bni ny ⊆ (B ∪ Bni B)ny ⊆ Bny ∪ Bni Bny ⊆ Bni nw B ∪ Bnw B. Combining the above statements, we obtain that ni Bnw ⊆ Bnw B ∪ Bni nw B for all w ∈ Sn and all i ∈ {1, . . . , n − 1}, as required. (BN5): Let B− ⊆ G be the subgroup of all lower triangular matrices. Then we have B− = n0 Bn0−1 , where n0 ∈ N is the permutation matrix \u0012 corresponding to (1, n)(2, n−1)(3, n−2) · · · ∈ Sn . Thus, we have n∈N nBn−1 ⊆ B ∩ B− = H. \u0003 1.6.11 Remark Let G\u0002 = SLn (k) and set B\u0002 := Bn (k) ∩ G\u0002 and N \u0002 := Nn (k) ∩ G\u0002 . Since G\u0002 is a normal subgroup of GLn (k) and Un (k) ⊆ G\u0002 , we can use Exercise 1.8.23 to conclude that B\u0002 and N \u0002 form a BN -pair in G\u0002 , with Weyl group W ∼ = Sn . Furthermore, we (1) (1) have Bn (k) = Un (k).Tn (k) with Tn (k) as in Example 1.5.5(b). Now assume that k is an infinite field. As we have seen in (1) Example 1.5.5, the groups Un (k) and Tn (k) are connected. Hence (1) so is Bn (k) = Un (k).Tn (k). (Indeed, the natural map Un (k) × (1) Tn (k) → Bn (k) given by multiplication is surjective; so it remains to use Proposition 1.3.8(a) and Remark 1.3.2.) Thus, Remark 1.6.7 gives yet another proof of the fact that SLn (k) is connected. 1.6.12 Remark Consider once more the BN -pair for G = GLn (k). We have already remarked in Example 1.6.9 that Bn (k) = Un (k).Tn (k), where Un (k) is normal and Un (k) ∩ Tn (k) = {1}. Thus Bn (k) is a semidirect product of Tn (k) with Un (k). Actually, something stronger holds: Let g ∈ G and u ∈ Un (k) be such that gug−1 ∈ Bn (k). Then gug−1 is a triangular matrix all of whose eigenvalues are 1. Thus, we must have gug−1 ∈ Un (k). A similar argument also applies to the BN -pair in G\u0002 = SLn (k). Thus, we see that these BN -pairs are split in the sense of the following definition. 1.6.13 Definition Let G be a group with a BN -pair. We say that this is a split BN -pair if there exists a normal subgroup U ⊆ B such that the following conditions are satisfied. (a) Let H = B ∩ N as in Definition 1.6.1. Then we have B = UH and U ∩ H = {1}. Thus, B is the semidirect product of H with U . (b) For any n ∈ N , we have n−1 Un ∩ B ⊆ U .\n\n68\n\nAlgebraic sets and algebraic groups\n\nWe mention here that all BN -pairs that we will encounter in this book are split. We shall now establish another important property of these groups, namely, a sharp form of the Bruhat decomposition. This will require a number of preparations. For the remainder of this section, we shall assume that G is a group with a BN -pair where W is finite. 1.6.14 Proposition (the longest element) There exists a unique element w0 in W of maximal length. We have w02 = 1, and w0 ∈ W is characterized by the property that l(sw0 ) < l(w0 ) for all s ∈ S. Proof Since W is finite, there exists some element w0 ∈ W such that l(w) \u0002 l(w0 ) for all w ∈ W . We claim that l(ww0 ) = l(w0 ) − l(w)\n\nfor all w ∈ W .\n\n(∗)\n\nWe prove this by induction on l(w). If l(w) = 0, then w = 1 and there is nothing to prove. Now assume that l(w) = p + 1 where p \u0001 0 and write w = s1 · · · sp s, where s, si ∈ S. Then q := l(s1 · · · sp w0 ) = l(w0 ) − p by induction. Now take a reduced expression s1 · · · sp w0 = t1 · · · tq , where tj ∈ S. Then w0 = sp · · · s2 s1 t1 t2 · · · tq is a reduced expression for w0 . Now, since l(sw0 ) < l(w0 ), the exchange condition in Corollary 1.6.8 shows that there are two possibilities: (1) sw0 = sp · · · si+1 si−1 · · · s2 s1 t1 t2 · · · tq for some i ∈ {1, . . . , p} or (2) sw0 = sp · · · s2 s1 t1 t2 · · · tj−1 tj+1 · · · tq for some j ∈ {1, . . . , q}. If (1) holds, then ssp · · · s2 s1 = sp · · · si+1 si−1 · · · s1 and so l(w) = l(w−1 ) \u0002 p − 1, a contradiction. Thus, (2) must hold and this yields ww0 = t1 t2 · · · tj−1 tj+1 · · · tq . Hence, we have l(ww0 ) = q − 1 = l(w0 ) − p − 1 = l(w0 ) − l(w), as desired. Thus, (∗) is proved. Setting w = w0 , we see that l(w02 ) = 0 and so 2 w0 = 1. Now let w ∈ W be another element such that l(w) = l(w0 ). Then l(ww0 ) = 0 by (∗) and so w = w0−1 = w0 , as required. Finally,\n\nGroups with a split BN -pair\n\n69\n\nassume that w ∈ W is such that l(sw) < l(w) for all s ∈ S. Assume, if possible, that l(w) < l(w0 ). Then we have l(ww0 ) = l(w0 ) − l(w) > 0 by (∗) and so l(sww0 ) < l(ww0 ) for some s ∈ S. Using (∗) once more, we obtain l(sw) > l(w), a contradiction. \u0003 1.6.15 Lemma l(w) + 1. Then\n\nLet s ∈ S and w ∈ W be such that l(sw) = −1 ns Bns ∩ Bnw Bnw ⊆ B.\n\nProof Assume this to be false. Then, since ns Bns ⊆ B ∪ Bns B, −1 \b= ∅ and so Bn Bn ∩ Bn B \b= ∅. But we have Bns B ∩ Bnw Bnw s w w we also have Bns Bnw ⊆ Bns nw B ∪ Bnw B, and so we must have Bns Bnw \b⊆ Bns nw B, contradicting Lemma 1.6.3. \u0003 It will now be convenient to introduce the following notation. We set −1 −1 Bw := nw Bnw and Bw := Bw0 w = B ∩ nw Bnw0 w 0w\n\nfor any w ∈ W . (Note that Bw , Bw do not depend on the choice of the representative nw ∈ N .) 1.6.16 Lemma Then\n\nLet y, w ∈ W be such that l(yw) = l(y) + l(w). B ∩ Byw ⊆ B ∩ Bw .\n\nFurthermore, we have B ∩ Bw0 = H. Proof We proceed by induction on l(y). If l(y) = 0, then y = 1 and there is nothing to prove. Next assume that l(y) = 1; then y = s ∈ S. By Lemma 1.6.15, we have −1 −1 −1 −1 nw (B ∩ Bsw )nw = nw (B ∩ nw ns Bns nw )nw −1 = nw Bnw ∩ ns Bns ⊆ B\n\nand so B ∩Bsw ⊆ B ∩Bw , as required. Now assume that l(y) > 1 and choose s ∈ S such that l(ys) < l(y). We set y\u0002 := ys and w\u0002 := sw. Then yw = y\u0002 sw = y\u0002 w\u0002 and l(y\u0002 w\u0002 ) = l(y\u0002 ) + l(w\u0002 ). So, by induction, \u0002 we have B ∩ Byw ⊆ B ∩ Bw = B ∩ Bsw . By the case l(y) = 1, we have B ∩ Bsw ⊆ B ∩ Bw , as required. Now consider the statement concerning B ∩ Bw0 . Let y ∈ W . By relation (∗) in the proof of Proposition 1.6.14, we can write w0 = xy with x ∈ W such that\n\n70\n\nAlgebraic sets and algebraic groups\n\n0 = B ∩ Bxy\u0012⊆ By . This holds l(w0 ) = l(x) + l(y). Then B ∩ Bw\u0012 w for all y ∈ W and so B ∩ B 0 ⊆ y∈W By ⊆ n∈N n−1 Bn = H, using (BN5). \u0003\n\n1.6.17 Lemma Let s ∈ S and w ∈ W be such that l(ws) = l(w) + 1. Then the following hold. (a) B = Bs .Bw0 s = Bw0 s .Bs and H \u0002 Bs . (b) Bs ⊆ Bw0 w . s = B s .B . (c) Bws = Bs .Bw w s Proof (a) Let y := w0 s; then we have l(ys) = l(y) + 1 and so l(sy−1 ) = l(y−1 ) + 1. By Proposition 1.6.3, this yields ns Bny−1 ⊆ Bns ny−1 B and so B ⊆ ns−1 Bns .ny−1 Bny = Bs By . Let b ∈ B. Then b = b\u0002 b\u0002\u0002 with b\u0002 ∈ Bs and b\u0002\u0002 ∈ By . Now b\u0002 ∈ ns Bns ∩ Bny−1 Bny ⊆ B by Lemma 1.6.15, and so b\u0002 , b\u0002\u0002 both lie in B. Hence, B = (B ∩ Bs ).(B ∩ By ) = Bw0 s .Bs . Taking inverses also yields B = Bs .Bw0 s , as required. Furthermore, it is clear that H ⊆ Bs . If we had Bs = H, then B = Bw0 s = B ∩ Bs and so ns Bns = B, contradicting (BN3). (b) By relation (∗) in the proof of Proposition 1.6.14, we can write w0 = yws, where y ∈ W is such that l(w0 ) = l(y) + l(ws) = l(y)+l(w)+1. Then w0 s = yw, where l(w0 s) = l(w0 )−1 = l(y)+l(w). So Lemma 1.6.16 yields Bs = B ∩ Bw0 s = B ∩ Byw ⊆ B ∩ Bw = Bw0 w . (c) We have B = Bs .Bw0 s by (a). Now note that l(w0 wss) = l(w0 ) − l(w) = l(w0 ) − l(ws) + 1 = l(w0 ws) + 1 and so Bs ⊆ Bws , using (b). This yields Bws = Bws ∩B = Bws ∩Bs Bw0 s = Bs Bws ∩Bs Bw0 s = Bs (Bws ∩Bw0 s ). s . First note that It remains to show that Bws ∩ Bw0 s = Bw\n\nBws ∩ Bw0 s = B ∩ Bw0 ws ∩ Bs = ns−1 (Bs ∩ Bw0 w ∩ B)ns = ns−1 (Bs ∩ Bw )ns . Now l(w0 ws) = l(w0 ) − l(w) − 1 = l(w0 w) − 1, and so we can write w0 w = xs where x ∈ W is such that l(xs) = l(x) + 1. By\n\nGroups with a split BN -pair\n\n71\n\nLemma 1.6.16, this implies that Bw = B ∩ Bw0 w ⊆ Bs . Hence we have Bws ∩ Bw0 s = ns−1 Bw ns as required. \u0003 1.6.18 Proposition We have B = Bw .Bw0 w = Bw0 w .Bw for all w ∈ W. Proof We begin with the following statement. Bsw = Bw .Bsw\n\nfor s ∈ S and w ∈ W such that l(sw) = l(w) + 1. (∗)\n\nWe will prove this by induction on l(w). If l(w) = 0, then w = 1 and so B1 = B ∩ Bw0 = H by Lemma 1.6.16. Thus, we have Bs = B1 .Bs as required. Now let l(w) > 0 and choose t ∈ S such that l(wt) < l(w). We set y := wt. Then sw = syt and l(sy) = l(y) + 1. So we obtain t Bsw = Bsyt = Bsy .Bt\n\n(by Lemma 1.6.17(c))\n\n= nt−1 (By .Bsy )nt Bt = (Bsy .By )t Bt =\n\nBsyt .Byt .Bt\n\n=\n\nBsyt .Byt\n\n=\n\nBsw .Bw\n\n(by induction) (by Lemma 1.6.17(c)),\n\nas required. Now we prove the factorization B = Bw .Bw0 w . Again, we will do this by induction on l(w). If l(w) = 0, then w = 1 and Bw0 = B. Thus, we certainly have B = B1 .Bw0 . Now let l(w) > 0 and write w = sy, where s ∈ S and y ∈ W are such that l(sy) = l(y) + 1. Then, using (∗), we obtain Bw .Bw0 w = Bsy .Bw0 sy = By .Bsy .Bw0 sy . sy Now note that Bw0 sy = B ∩ Bsy = Bw −1 . This yields 0 (sy) \b sy −1 s Bs .Bw Bw .Bw0 w = By .Bsy .Bw −1 ny −1 = By .ny 0 (sy) 0 (sy) y = By .ny−1 Bw0 (sy)−1 s ny = By .Bw −1 0y y by Lemma 1.6.17(c). Now, we note again that Bw = Bw0 y . −1 0y Thus, by induction, the right-hand side of the above identity equals By .Bw0 y = B. \u0003\n\nThe following result will turn out to be most useful. For example, it leads to an order formula for finite groups with a split BN -pair; see Exercise 1.8.21(c) for an example. Another application will be given in the following section, where we construct BN -pairs for the symplectic and orthogonal groups.\n\n72\n\nAlgebraic sets and algebraic groups\n\n1.6.19 Theorem (sharp form of the Bruhat decomposition) Let G be a group with a split BN -pair, and assume that W is finite. We set −1 Uw := U ∩ nw Unw0 w for w ∈ W 0w (Note again that Uw does not depend on the choice of the representatives.) Then every element g ∈ Bnw B is uniquely expressible in the form g = bnw u, where b ∈ B, w ∈ W , and u ∈ Uw . Thus, we have , B nw Uw with uniqueness of expressions. G= w∈W\n\nFurthermore, we have Uw \b= {1} for all non-unit w ∈ W . −1 Un for all w ∈ W . First we Proof We shall write U w := nw w show that Bw = Uw H for all w ∈ W . This is seen as follows. Since H is normal in N , and since B = UH, we have B ∩ Bw = UH ∩ U w H ⊇ (U ∩ U w )H. We must show that the reverse inclusion also −1 un h = u\u0002 h\u0002 , holds. So let g ∈ B ∩ Bw . Then we can write g = nw w \u0002 \u0002 where u, u ∈ U and h, h ∈ H. Since g ∈ B, we conclude that −1 un ∈ n−1 Un ∩ B ⊆ U by condition (b) in Definition 1.6.13. nw w w w −1 un = u\u0002 and h = h\u0002 . Thus, using condition (a), we conclude that nw w w This yields g ∈ (U ∩ U )H as required. Replacing w by w0 w, we see that Bw = Uw H. Now Proposition 1.6.18 yields the factorization U = Uw .Uw0 w = Uw0 w .Uw for all w ∈ W . Further note that −1 −1 −1 −1 nw Uw0 w nw = nw (U ∩ nw Unw )nw = nw Unw ∩ U ⊆ U,\n\nand so Bnw B = Bnw HU = Bnw U = Bnw Uw0 w Uw = Bnw Uw0 w −1 n U = Bn U . Thus, we have nw w w w w , Bnw Uw , G= w∈W\n\nand it remains to prove the uniqueness of expressions. Thus, let g ∈ Bnw Uw and write g = bnw u = b\u0002 nw u\u0002 , where b, b\u0002 ∈ B and u, u\u0002 ∈ Uw . Then we have −1 −1 nw u(u\u0002 )−1 nw = b−1 b\u0002 ∈ nw (U ∩ U w0 w )nw ∩ B ⊆ B ∩ U w0 .\n\nThus, it will be enough to show that B ∩ U w0 = {1}. Now, by Lemma 1.6.16, we have B ∩ U w0 ⊆ B ∩ Bw0 = H. But, by Definition 1.6.13(b), we also have B ∩ U w0 ⊆ U , and so that intersection must be trivial, as desired.\n\nGroups with a split BN -pair\n\n73\n\nFinally, assume that Uw = {1}. Then we have Bw = H. If w \b= 1, then there exists some s ∈ S such that l(ws) = s ⊇ B \u0003 H, a l(w) − 1. By Lemma 1.6.17, we have Bw = Bs .Bws s contradiction. \u0003 1.6.20 Remark In the above proof, we have seen that there is a factorization U = Uw .Uw0 w = Uw0 w .Uw\n\nfor all w ∈ W .\n\nThus, the map Uw × Uw0 w → U given by multiplication is surjective. We claim that it is also injective. Indeed, we have −1 −1 −1 (nw0 Unw0 ∩ U )nw ⊆ nw Hnw = H, Uw ∩ Uw0 w ⊆ nw\n\nwhere we used Lemma 1.6.16. Thus, since the left hand side lies in U , we conclude that Uw ∩ Uw0 w = {1} and this yields the injectivity of the above map. 1.6.21 Coxeter groups Finally, we mention the fact that W = N/H is always a Coxeter group; that is, we have the following presentation of W : W = \u0013s ∈ S | s2 = 1; (st)mst = 1 (s \b= t in S with mst < ∞)\u0014, where mst \u0001 2 denotes the order of st ∈ W . This is deduced from the fact that the exchange condition in Lemma 1.6.8 holds in W . Since this result will not be used explicitly in this book, we refer to Chapter IV, §1, Theorem 1 of Bourbaki (1968) for the proof. Furthermore, we remark that there is a complete classification of the finite Coxeter groups. To describe the result, we encode the above presentation in a graph, called the Coxeter graph of W . It has vertices labelled by the elements of S, and two vertices labelled by s \b= t are joined by an edge if mst \u0001 3. Moreover, if mst \u0001 4, we label the edge by mst . If the graph is connected, we say that W is an irreducible Coxeter group. The graphs corresponding to the irreducible finite Coxeter groups are given in Table 1.2; the orders of the corresponding groups are given as follows. Type Order\n\nAn (n+1)!\n\nBn 2n n!\n\nDn 2n−1 n!\n\nI2 (m) 2m\n\nH3 120\n\nType Order\n\nH4 14400\n\nF4 1152\n\nE6 51840\n\nE7 2903040\n\nE8 696729600\n\n74\n\nAlgebraic sets and algebraic groups\n\nTable 1.2 An n\u00011\n\nBn n\u00012\n\nI2 (m)\n\nCoxeter graphs of irreducible finite Coxeter groups\n\n1s 1 s\n\n4\n\n2s\n\n3s\n\np p p\n\nns\n\n2s\n\n3s\n\np p p\n\nns\n\n1s m 2s\n\nF4\n\n1s\n\nDn n\u00014\n\n2s\n\n4\n\n1s\n\n5\n\n3s\n\n4s\n\n1s\n\nE7\n\n1s\n\n5\n\n2s\n\n3s\n\n3s\n\n4s\n\ns2\n\nH4\n\n2s\n\n3s\n\n5s\n\n6s\n\n7s\n\nE8\n\n4\n\ns\n\np p p\n\nn\n\n3s\n\n4s\n\n5s\n\n6s\n\n7s\n\n8s\n\ns\n\ns2 E6\n\nm\u00015\n\nH3\n\ns1 @ 3 @s\n\n1s\n\n4s\n\ns2 1s\n\n3s\n\n4s\n\n5s\n\n6s\n\ns2\n\nThe numbers on the vertices correspond to a chosen labelling of the elements of S.\n\nIn general, if Γ1 , . . . , Γr are the connected components of the Coxeter graph of W , we have a corresponding direct-product decomposition W = W1 × · · · × Wr , where Wi = \u0013Si \u0014 and Si ⊆ S is the subset labelling the vertices in Γi ; furthermore, each Wi is a Coxeter group with generating set Si . For more details, see Chapter VI, no. 4.1, of Bourbaki (1968). For example, the Weyl group of GLn (k) is Sn , whose Coxeter graph is that of type An−1 in Table 1.2, where the vertex labelled by i corresponds to the generator σi = (i, i + 1). We shall say that GLn (k) has a BN -pair of type An−1 .\n\n1.7 BN -pairs in symplectic and orthogonal groups In this section, we will determine BN -pairs for the symplectic and orthogonal groups introduced in §1.3.15 and §1.3.16. For the groups SO2m+1 (k) (char(k) \b= 2) and Sp2m (k), we obtain a BN -pair by a general construction from the BN -pair in GLn (k). For the evendimensional orthogonal groups SO+ 2m (k), we verify the BN -pair axioms along the lines of the proof of Proposition 1.6.10. As an application, we obtain the result that all these groups are connected. 1.7.1 Groups with a split BN -pair and automorphisms Let G be a group with a split BN -pair having a finite Weyl group W ,\n\nBN -pairs in symplectic and orthogonal groups\n\n75\n\nand let ϕ : G → G be a bijective homomorphism of (abstract) groups satisfying the following conditions. (BNϕ 1)\n\nWe have ϕ(U ) = U, ϕ(H) = H and ϕ(N ) = N , where B = U.H and H = B ∩ N as in Definition 1.6.13.\n\n(BNϕ 2)\n\nEvery coset Hn (where n ∈ N ) such that ϕ(Hn) ⊆ Hn contains an element which is fixed by ϕ.\n\nSince ϕ is a group homomorphism, the fixed point set Gϕ := {g ∈ G | ϕ(g) = g} is a subgroup of G. Similarly, the fixed point sets Bϕ , N ϕ are subgroups of Gϕ . We would like to show that Bϕ , N ϕ form a BN -pair in Gϕ . First, since B = UH is a semidirect product, we conclude that we have a similar decomposition of Bϕ : Bϕ = U ϕ .H ϕ\n\nand H ϕ = Bϕ ∩ N ϕ .\n\n(1)\n\nThen it is also clear that n−1 U ϕ n ∩ Bϕ ⊆ U ϕ for all n ∈ N ϕ . Thus, the splitness conditions in Definition 1.6.13 are satisfied. Next, consider W = N/H. Since N and H are invariant under ϕ, we obtain an induced homomorphism ϕ¯ : W → W, Hn \u0004→ Hϕ(n). Since ϕ(N ) = N , we note that ϕ¯ is surjective and, hence, an isomorphism (since W is finite). Then ϕ induces a permutation of the Bruhat cells; we have ϕ(C(w)) = C(ϕ(w)) ¯ for all w ∈ W .\n\n(2)\n\n= ϕ(B ∪ C(s)) is a subgroup of G Now let s ∈ S. Then B ∪ C(ϕ(s)) ¯ and so ϕ(s) ¯ ∈ S; see Corollary 1.6.5. Thus, ϕ¯ leaves S invariant. Furthermore, we have a natural injective map N ϕ /H ϕ → W = N/H, H ϕ n \u0004→ Hn. By (BNϕ 2), the image of this map is precisely the group of fixed points W ϕ¯ = {w ∈ W | ϕ(w) ¯ = w}. We shall identify W ϕ¯ with N ϕ /H ϕ using the above map. Thus, we have W ϕ¯ = N ϕ /H ϕ\n\nand l(ϕ(w)) ¯ = l(w) for all w ∈ W .\n\n(3)\n\n¯ we consider Let S¯ be the set of orbits of ϕ¯ on S. For each J ∈ S, the subgroup WJ ⊆ W generated by J . By Exercise 1.8.24(b), there is a unique longest element wJ ∈ WJ . The invariance of S under ¯ ϕ¯ and the uniqueness of wJ imply that wJ ∈ W ϕ¯ for all J ∈ S.\n\n76\n\nAlgebraic sets and algebraic groups\n\nFurthermore, by Exercise 1.8.25, we have ¯ W ϕ¯ = \u0013wJ | J ∈ S\u0014. Thus, we have all the ingredients for a BN -pair: the subgroups Bϕ , ¯ for the group W ϕ¯ = N ϕ , and the set of generators {wJ | J ∈ S} ϕ ϕ N /H . 1.7.2 Proposition \u0013Bϕ , N ϕ \u0014 and Gϕ =\n\n, w∈W\n\nIn the set-up of §1.7.1, we have Gϕ =\n\nU ϕ H ϕ nw Uwϕ\n\nwith uniqueness of expressions,\n\nϕ ¯\n\nwhere nw ∈ N ϕ for any w ∈ W ϕ¯ . For J ∈ S¯ and w ∈ W ϕ¯ , we have nJ B ϕ n w ⊆ B ϕ n J n w B ϕ ∪ B ϕ n w B ϕ ,\n\nwhere nJ := nwJ ∈ W ϕ¯ .\n\n¯ then Bϕ and N ϕ form a split Moreover, if UwϕJ \b= {1} for all J ∈ S, ϕ BN -pair in G , with Weyl group W ϕ¯ . Proof We have already remarked in §1.7.1 that the splitness conditions in Definition 1.6.13 hold. We will now show that all axioms for a BN -pair are satisfied, except possibly (BN3) in Definition 1.6.1. The assumption that UwϕJ \b= {1} for all J ∈ S¯ will be needed to show that (BN3) holds. Now, by eqn 1.7.1(3), every w ∈ W ϕ¯ has a representative nw ∈ ϕ N . In combination with eqn 1.7.1(2) and the Bruhat decomposition for G, this yields that , \b ϕ Bnw B . Gϕ = w∈W ϕ¯\n\nNow let us take into account the sharp form of the Bruhat decomposition. Since the length function is invariant under ϕ, ¯ the longest element w0 ∈ W certainly is fixed by ϕ. ¯ Consequently, we have ϕ(Uw ) ⊆ Uw\n\nfor any w ∈ W ϕ¯ .\n\nThus, given g ∈ Gϕ and w ∈ W ϕ¯ such that g ∈ Bnw B, we can write g = uhnw u\u0002 , where u ∈ U, h ∈ H, and u\u0002 ∈ Uw . Since U, H, and\n\nBN -pairs in symplectic and orthogonal groups\n\n77\n\nUw are all invariant under ϕ, the uniqueness of the above expression implies that u, h, and u\u0002 are all fixed by ϕ. This yields the decomposition , \b ϕ U ϕ H ϕ nw Uwϕ and Bnw B = U ϕ H ϕ nw Uwϕ Gϕ = w∈W ϕ¯\n\n= B ϕ nw B ϕ .\n\n(†)\n\nWe have uniqueness of expressions since this already holds in G. Furthermore, we now see that Gϕ = \u0013Bϕ , N ϕ \u0014 as required. Next, we have already seen in §1.7.1 that Bϕ = U ϕ H ϕ where H ϕ = Bϕ ∩ N ϕ . Furthermore, we have \u0006 nBϕ n−1 ⊆ (B ∩ n0−1 Bn0 )ϕ = H ϕ . Hϕ ⊆ n∈N ϕ\n\nThus, (BN5) is satisfied. Next, we have N ϕ /H ϕ = W ϕ¯ , and this group is generated by elements of order 2; thus (BN2) holds. Let us now consider (BN4). Let J ∈ S¯ and w ∈ W ϕ¯ . Writing wJ = s1 · · · sn where si ∈ J , and applying the rule in Exercise 1.8.22(a), we obtain \u0004 C(wJ ).C(w) ⊆ C(w\u0002 w). w\u0002 ∈WJ\n\nNow let us take fixed points under ϕ. We have C(w\u0002 w)ϕ = ∅ if ϕ(w ¯ \u0002 w) \b= w\u0002 w. Since w ∈ W ϕ¯ , we have ϕ(w ¯ \u0002 w) = w\u0002 w if and only if \u0002 \u0002 ϕ(w ¯ ) = w . But, by Exercise 1.8.25, we have WJϕ¯ = {1, wJ } and so \u0004 \b ϕ C(w\u0002 w)ϕ nJ Bϕ nw ⊆ C(wJ ).C(w) ⊆ w\u0002 ∈WJ\n\n= C(w)ϕ ∪ C(wJ w)ϕ = Bϕ nJ nw Bϕ ∪ Bϕ nw Bϕ as required, where the last equality holds again by (†). Thus, all the required assertions are proved. It remains to check that (BN3) holds under the given hypothesis. So let J ∈ S¯ and assume, if possible, that nJ Bϕ nJ = Bϕ . Since wJ2 = 1, we can write this as nJ Bϕ nJ−1 = Bϕ . Now, since UwϕJ ⊆ Bϕ , we also have nJ UwϕJ nJ−1 ⊆ Bϕ . On the other hand, recalling the definition of UwJ from Theorem 1.6.19, we see that nJ UwJ nJ−1 ⊆ nJ UnJ−1 ∩ U w0 ⊆ U w0 . Thus, we have nJ UwϕJ nJ−1 ⊆ Bϕ ∩ U w0 ⊆ B ∩ Bw0 = H, where the last equality holds by Lemma 1.6.16. Since nJ normalizes H, we conclude that UwϕJ ⊆ H and so UwϕJ = {1}, contrary to our assumption. \u0003\n\n78\n\nAlgebraic sets and algebraic groups\n\n1.7.3 Example We have a split BN -pair in GLn (k) (where k may be any field) by taking the groups Bn (k) = Un (k).Tn (k) ⊆ GLn (k) and Nn (k) ⊆ GLn (k); see Proposition 1.6.10. Let n0 be the matrix defined in Remark 1.5.12; this matrix is used to define the groups On (k) (if char(k) \b= 2) or Sp2m (k) (if n = 2m is even). To place ourselves in the setting of §1.7.1, we consider the map ϕ : GLn (k) → GLn (k),\n\nA \u0004→ n0−1 (Atr )−1 n0 .\n\nThen ϕ is a bijective homomorphism of algebraic groups, and we have ⎧ ⎨ Sp2m (k) if n = 2m is even (any characteristic), O2m (k) if char(k) \b= 2 and n = 2m, GLn (k)ϕ = ⎩ O2m+1 (k) if char(k) \b= 2 and n = 2m + 1. We wish to show that the two conditions in §1.7.1 are satisfied. It is readily checked that ϕ(Un (k)) = Un (k), ϕ(Tn (k)) = Tn (k), and ϕ(Nn (k)) = Nn (k). Now let us identify the Weyl group of GLn (k) with Sn ; we have Sn = \u0013σ1 , . . . , σn−1 \u0014,\n\nwhere σi = (i, i + 1) for 1 \u0002 i \u0002 n − 1. (a)\n\nThen the induced automorphism ϕ¯ : Sn → Sn is given by conjugation with the permutation w0 := (1, n)(2, n−1)(3, n−2) · · · ∈ Sn . Thus, we are in the setting of Exercise 1.8.26 which shows that ¯ Sϕ n = \u0013t, s1 , . . . , sm−1 \u0014, where \u0017 si = σm−i σm+i and t = σm if n = 2m, (b) si = σm−i σm+i+1 and t = σm σm+1 σm if n = 2m+1. ¯ Now, for each element in Sϕ n , we have to find a representative in ϕ Nn (k) . It will actually be enough to do this for the above generators ¯ of Sϕ n . One checks that such a representative for the generator t is given by ⎤ ⎡ 0 0 Im−1 ⎥ ⎢ J 0 ⎦ ∈ GLn (k)ϕ , ⎣ 0 0 0 Im−1\n\nwhere J is one of the matrices \u001a \u001b \u001a 0 1 0 or −1 0 1\n\n\u001b 1 0\n\n⎤ 0 0 1 ⎣0 −1 0⎦ 1 0 0 ⎡\n\nor\n\naccording to whether GLn (k)ϕ is Sp2m (k) or O2m (k) or O2m+1 (k), respectively. Furthermore, representatives for the generators si can\n\nBN -pairs in symplectic and orthogonal groups\n\n79\n\nbe found in the image of the embedding GLm (k) \u0011→ GLϕ n defined by sending A ∈ GLm (k) to ⎡ ⎤ . 0 A 0 A 0 ⎢ ⎥ or ⎣ 0 1 0 ⎦ , where A\u0002 = Qm (A−1 )tr Qm , A\u0002 0 0 0 A\u0002 according to whether n is even or odd. (We leave the details of these calculations as an exercise to the reader.) Thus, the two conditions in §1.7.1 are satisfied. 1.7.4 Theorem In the above setting, let G := GLn (k)ϕ and assume that G = Sp2m (k) (any characteristic) or G = O2m+1 (k) (char(k) \b= 2). Then B := Bn (k) ∩ G\n\nand\n\nN := Nn (k) ∩ G\n\nform a split BN -pair in G, where W ϕ is a Coxeter group of type Bm ; see Table 1.2. We have B = UH, where H = B ∩ N and U = Un (k) ∩ G. Now assume that k is an infinite field. Then the following hold: (a) If G = Sp2m (k), then B is a closed connected subgroup. (b) If G = O2m+1 (k), let B\u0002 := B ∩ SO2m+1 (k) and N \u0002 := N ∩ SO2m+1 (k). Then B\u0002 and N \u0002 form a split BN -pair in SO2m+1 (k). The group B\u0002 is a closed connected subgroup of SO2m+1 (k). (c) The groups Sp2m (k) and SO2m+1 (k) are connected. Proof We have already checked in Example 1.7.3 that the assumptions (BNϕ 1) and (BNϕ 2) in §1.7.1 are satisfied. Furthermore, using the description of the groups Uw in Exercise 1.8.21, one checks that ¯ J . Thus, all the hypotheses of ProposiUwϕJ ∼ = k for every ϕ-orbit tion 1.7.2 are satisfied. This yields that B and N form a split BN -pair in G. The identification of W ϕ is given by Exercise 1.8.26(a). Now assume that k is an infinite field. If G = Sp2m (k), then the groups U and H are closed and connected by Remark 1.5.12. Hence so is B (by an argument similar to that in Remark 1.6.11). Thus, (a) is proved. Next assume that G = O2m+1 (k) and let G\u0002 := SO2m+1 (k). Then G\u0002 is a closed normal subgroup of G containing U . Thus, by Exercise 1.8.23, the groups B\u0002 and N \u0002 form a split BN -pair in G\u0002 . The descriptions of U and H in Lemma 1.5.10 show that U and\n\n80\n\nAlgebraic sets and algebraic groups\n\nH \u0002 := H ∩ G\u0002 are closed and connected. Hence so is B\u0002 = U.H \u0002 (by the same argument as above). This proves (b). Finally, (c) is a consequence of Remark 1.6.7. \u0003 We now turn to the even-dimensional orthogonal groups, defined in §1.3.16: O+ 2m (k) = {A ∈ M2m (k) | f2m (Av) = f2m (v)\n\nfor all v ∈ k2m }.\n\nIf the characteristic of k is not 2, then we could in fact apply a similar argument as in Theorem 1.7.4 to exhibit a BN -pair. However, this methods fails in characteristic 2 since then O+ 2m (k) is not defined as the fixed-point set under an automorphism. Instead, we shall proceed as in the proof of Proposition 1.6.10, by verifying directly the BN -pair axioms. Incidentally, this works simultaneously for the case of odd characteristic as well. One of the key problems will be ◦ to identify the group O+ 2m (k) . Recall from §1.3.15 and §1.3.16 that we define \u0017 + O2m (k) ∩ SL2m (k) if char(k) \b= 2, + SO2m (k) := ◦ O+ if char(k) = 2. 2m (k) In the subsequent discussion, the following matrix will play a special role: ⎤ ⎡ Im−1 0 0 ⎥ ⎢ 0 1 ⎥ ⎢ tm := ⎢ 0 0 ⎥ ∈ O+ 2m (k). 1 0 ⎦ ⎣ 0 Im−1 0 1.7.5 An intermediate group We place ourselves in the setting of Example 1.7.3 where n = 2m. Then GL2m (k)ϕ is given by O2m (k) (if char(k) \b= 2) or Sp2m (k) (if char(k) = 2). Thus, if we set Gm := ϕ O+ 2m (k), then we have Gm ⊆ GL2m (k) in both cases. Now let Bm := B2m (k) ∩ Gm ,\n\nNm := N2m (k) ∩ Gm ,\n\nUm := U2m (k) ∩ Gm .\n\nUsing Lemma 1.5.11, we see that Tm := T2m (k)ϕ ⊆ G in both cases. Furthermore, we have Bm = Um Tm and Tm = Bm ∩ Nm . Con¯ sidering the generators of Sϕ 2m in Example 1.7.3, we also see that\n\nBN -pairs in symplectic and orthogonal groups\n\n81\n\nN2m (k)ϕ ⊆ Gm in both cases, and so Nm = N2m (k)ϕ . Thus, we have ϕ ¯ Wm := Nm /Tm ∼ = S = \u0013t, s1 , . . . , sm−1 \u0014. 2m\n\n\u0002 Wm\n\nNow let ⊆ Wm be the subgroup of index 2 defined in Exer\u0002 (under the ˜m ⊆ Nm be the preimage of Wm cise 1.8.26(c), and let N map Nm → Wm ). We set ˜ m := \u0013Bm , N ˜ m \u0014 ⊆ Gm . G We have the following two properties: (a) The matrix tm (as defined above) lies in Nm , and it normalizes Bm . ˜ where n0 is the matrix defined in (b) We have n0 Bm n0−1 ⊆ G, Example 1.7.3. Indeed, (a) is checked by a straightforward computation, using the descriptions of Um , Tm in Lemma 1.5.11. Now consider (b). It is readily checked that n0 ∈ N2m (k)ϕ ⊆ Nm . Now, if m is even, then ˜m ⊆ G ˜ m and so we trivially have n0 Bm n−1 ⊆ G ˜ m . On the n0 ∈ N 0 ˜m but n0 tm ∈ N ˜m . Then, using other hand, if m is odd, then n0 \b∈ N −1 −1 −1 ˜ (a), we have n0 Bm n0 = n0 tm Btm n0 ⊆ Gm as required. ˜m form 1.7.6 Lemma In the above setting, the groups Bm and N \u0002 ˜ a split BN -pair in Gm , with Weyl group Wm . Furthermore, we have ˜m. tm \b∈ G Proof First note that the splitness conditions in Definition 1.6.13 are satisfied (by the same argument as in Remark 1.6.12). So we must verify the axioms in Definition 1.6.1. The condition (BN1) is clear by definition, and (BN2) holds by Exercise 1.8.26(c): \u0002 Wm = \u0013u, s1 , . . . , sm−1 \u0014,\n\nwhere u := ts1 t.\n\nFurthermore, we have seen in the proof of §1.7.5(b) that there exists ˜m such that n0 Bm n−1 = n some n ˜ ∈ N ˜ Bm n ˜ −1 . Then we have 0 \u0012 −1 −1 ⊆ Bm ∩ n0 Bm n0 = Tm ; thus (BN5) holds. It ˜m nBm n n∈N remains to verify (BN3) and (BN4). For this purpose, we work with ˜ m given by the embedding GLm (k) \u0011→ G . A 0 A \u0004→ Aˆ := , where A\u0002 = Qm (A−1 )tr Qm . A\u0002 0 ˜m for 1 \u0002 i \u0002 m − 1, where we regard ˆm−i ∈ N We set ni = σ ˜m is σ1 , . . . , σm−1 as generators for Sm ⊆ GLm (k). Then ni ∈ N\n\n82\n\nAlgebraic sets and algebraic groups\n\n\u0002 . Note also that n := t n t ∈ N ˜m is a representative of si ∈ Wm u m 1 m a representative of u. Furthermore, using Lemma 1.5.11, it is easily checked that\n\nˆm (k) = U ˆm (k).Vm Um = Vm .U\n\nand\n\nni Vm ni = Vm for 1 \u0002 i \u0002 m − 1, where\n\n#-\n\nVm :=\n\n(1)\n\n. \\$ Qm S \u0018\u0018 S = (sij ) ∈ Mm (k) such that . \u0018 S = −S tr and sii = 0 for all i Im\n\nIm 0\n\nNow consider (BN3). Assume, if possible, that ni Bm ni = B for some i ∈ {1, . . . , m − 1}. Since ni normalizes Tm and Vm , this implies that ˆm (k)ni = U ˆm (k) and so σm−i Um (k)σm−i = Um (k), contradicting ni U (BN3) for GLm (k); see Proposition 1.6.10. Thus, we have ni Bm ni \b= Bm for 1 \u0002 i \u0002 m − 1. Since tm normalizes Bm , we also have nu Bm nu = tm n1 tm Bm tm n1 tm \b= Bm . Thus, (BN3) holds. Finally, in order to verify (BN4), we proceed as in the proof of Proposition 1.6.10. For 1 \u0002 i \u0002 m − 1, we define Xsi := {I2m + λ(Em−i,m−i+1 − Em+i,m+i+1 ) | λ ∈ k} ⊆ Um , Vsi := {(aij ) ∈ Um | am−i,m−i+1 = am+i,m+i+1 = 0}. It is readily checked that these are subgroups such that Um = ˆ m−i Xsi .Vsi = Vsi .Xsi and Xsi ∩ Vsi = {1}. In fact, we have Xsi = X and Vsi = Vˆm−i Vm with Xm−i , Vm−i ⊆ GLm (k) as in Example 1.6.9. First we claim that ni Vsi ni = Vsi\n\nand ni Xsi ni ⊆ {1} ∪ Xsi ni Xsi Tm\n\nfor 1 \u0002 i \u0002 m − 1.\n\n(2)\n\nIndeed, using the above descriptions of Xsi and Vsi , the proof of (2) can be reduced to the analogous statements for Xm−i , Vm−i ⊆ GLm (k), where they are verified in Example 1.6.9; note also that ni Vm ni = Vm . Using (2), we obtain ni Bm ni = ni Xsi Vsi Tm ni = ni Xsi ni Vsi Tm ⊆ ni Xsi ni Bm ⊆ ({1} ∪ Xsi ni Xsi Tm ) Bm ⊆ Bm ∪ Bm ni Bm . \u0002 Next we wish to show that ni Bm nw ⊆ Bm ni nw Bm , where w ∈ Wm is such that l(σm−i w) > l(w); here, l is the usual length function\n\nBN -pairs in symplectic and orthogonal groups\n\n83\n\non S2m . For this purpose, we first prove: −1 X si nw ⊆ U m nw\n\n\u0002 such that l(σ for all w ∈ Wm m−i w) > l(w).\n\n(3) This is seen as follows. By the invariance of the length function under ϕ, ¯ we also have l(σm+i w) > l(w). Using Example 1.6.9(b), we obtain −1 nw Xsi nw ⊆ {I2m +λ(Ew−1 (m−i),w−1 (m−i+1)\n\n− Ew−1 (m+i),w−1 (m+i+1) ) | λ ∈ k}, and this is contained in U2m (k) by the characterization of the length −1 X n ⊆ U condition in Exercise 1.8.21(a). Thus, we have nw si w 2m (k)∩ Gm = Um as claimed. \u0002 be Now we can proceed as follows. Let 1 \u0002 i \u0002 m and w ∈ Wm such that l(σm−i w) > l(w). Then, using (2) and (3), we obtain −1 X si nw ni Bm nw = ni Vsi Xsi Tm nw ⊆ Vsi ni nw nw\n\n⊆ Vsi ni nw Um Tm ⊆ Bm ni nw Bm . \u0002 be such that l(σ Next, let w ∈ Wm m−i w) < l(w). Then set y := si w ∈ \u0002 Wm . If we had l(σm−i y) < l(y), then we could apply the argument \u0002 is in Exercise 1.8.25(∗) and conclude that y = si y\u0002 , where y\u0002 ∈ Wm \u0002 \u0002 such that l(y) = l(si ) + l(y ). But then we automatically have y = w and so l(si w) > l(w). This in turn would imply l(σm−i w) > l(w), contrary to our assumption. Thus, we must have l(σm−i y) > l(y). So we can apply the previous discussion and conclude that\n\nni Bm nw = ni Bm ni ny ⊆ Bm ny ∪ Bm ni Bm ny ⊆ Bm ny ∪ Bm ni ny Bm = Bm ni nw Bm ∪ Bm nw Bm . Thus, we have shown (BN4) for all the generators si . Finally, using \u0002 : the fact that tm normalizes Bm , we also have for any w ∈ Wm nu Bm nw = tm n1 tm Bm nw ⊆ tm (n1 Bm ntwt )tm ⊆ tm (Bm n1 ntwt Bm ∪ Bm ntwt Bm )tm = tm Bm n1 ntwt Bm tm ∪ tm Bm ntwt Bm tm = Bm tm n1 tm nw Bm ∪ Bm nw Bm = Bm nu nw Bm ∪ Bm nw Bm . Thus, we have verified all the axioms for a BN -pair. To show that ˜ m , we can now argue as follows. If we had tm ∈ G ˜ m , then tm \b∈ G\n\n84\n\nAlgebraic sets and algebraic groups\n\ntm ∈ NG˜ m (Bm ) = Bm by Corollary 1.6.4 and §1.7.5(a). This is absurd, since tm evidently is not a triangular matrix. \u0003 ˜m The next step consists of clarifying the relation between G ◦ and Gm . 1.7.7 Lemma Assume that k is an infinite field. Then we have ◦ =G ˜ m , and this group has index 2 in Gm . Gm ◦ ⊆ G \u0002 · ˜ m . To see this, set X := Um Proof First we show that Gm \u0002 are connected (see Remark 1.5.12), we Tm · Um . Since Um , Tm , Um \u0002 ◦ . Conversely, let g ∈ G◦ and consider have X ⊆ \u0013Um , Tm , Um \u0014 ⊆ Gm m ◦ ◦ is open. Hence the subset gX ⊆ Gm . By Exercise 1.8.27, X ⊆ Gm so is gX (since left multiplication by g is an isomorphism of affine ◦ is irreducible, this implies X ∩ gX \b= ∅ and so varieties). Since Gm −1 g ∈ X · X. Thus, we have shown that ◦ \u0002 Gm = \u0013Um , Tm , Um \u0014. \u0002 ⊆ G ˜ m . We On the other hand, by §1.7.5(b), we have Um , Tm , Um ◦ ˜ conclude that Gm ⊆ Gm , as claimed. Using Lemma 1.7.6, we obtain ◦ ˜ m \u0002 Gm . Gm ⊆G\n\nHence, in order to complete the proof, it will be enough to show that ◦ has index at most 2 in G . Since G◦ is a normal subgroup, it Gm m m will be enough to show that ◦ ◦ ◦ Gm = \u0013Gm , tm \u0014 = Gm ∪ tm Gm .\n\nWe will do this by induction on m. For m = 1, this holds by the explicit description of O+ 2 (k) in §1.3.15 and §1.3.16. Now let m > 1. The group Gm acts naturally on k2m (column vectors). Let g ∈ Gm and set x := g.e1 where e1 ∈ k2m is the first vector of the standard basis of k2m . Then x is a non-zero vector of k2m such that f2m (x) = f2m (g.e1 ) = f2m (e1 ) = 0. We claim that there exists some ◦ , t \u0014 such that g .e = x. To see this, assume first that x g1 ∈ \u0013Gm m 1 1 has the property that x1 \b= 0. Then we can find a suitable diagonal ◦ such that the first component of h.x is 1. So matrix h ∈ Tm ⊆ Gm 1 we may assume without loss of generality that x1 = 1. Then consider\n\nBN -pairs in symplectic and orthogonal groups\n\n85\n\nthe matrix ⎡\n\n··· 0\n\n0\n\n1 x2 .. .\n\n⎢ ⎢ ⎢ u(x) := ⎢ ⎢ ⎢ ⎣x2m−1 x2m\n\n0 0 .. .\n\nI2(m−1) · · · x2\n\nx2m−1\n\n⎥ ⎥ ⎥ ⎥. ⎥ ⎥ 0 ⎦ 1\n\nIt is readily checked that u(x) ∈ Gm . (To see this, one needs the fact \u0002 ⊆ G◦ and u(x).e = x, as that f2m (x) = 0.) We have u(x) ∈ Um 1 m ◦ , t \u0014 such desired. Now, if x1 = 0, then there exists some g\u0002 ∈ \u0013Gm m that the first component of g\u0002 .x is non-zero. To see this, first note that we have a natural embedding\n\nSLm (k) →\n\nO+ 2m (k),\n\nA A \u0004→ Aˆ := 0\n\n. 0 ∈ Gm , A\u0002\n\nwhere A\u0002 = Qm (A−1 )tr Qm . Since SLm (k) is connected (see Example ◦ for all A ∈ SL (k). Thus, if x \b= 0 1.6.11), this implies that Aˆ ∈ Gm m i ◦ in the image of the for some i \u0002 m, we can find some g\u0002 ∈ Gm above embedding such that the first component of g\u0002 .xi is non-zero. Otherwise, we have xi \b= 0 for some i > m and so there exists some ◦ in the image of the above embedding such that the (m + 1)g \u0002 ∈ Gm component of g\u0002 .x is non-zero. Then the mth component of tm g\u0002 .x will be non-zero, and we can apply the previous discussion. In any ◦ ,t \u0014 case, as desired, we have shown that there exists some g1 ∈ \u0013Gm m −1 such that g1 .e1 = x = g.e1 . Then g1 g = (aij ) ∈ Gm is a matrix such that ai1 = 0 for all i > 1. It is easily checked that then we automatically have a2m,j = 0 for all j < 2m, and so ⎡ g1−1 g\n\n⎢ ⎢ ⎢ =⎢ ⎢ ⎣\n\n1 0 .. .\n\ng2\n\n0 0\n\n0 ··· 0\n\n⎥ ⎥ ⎥ ∗ ⎥, ⎥ ⎦ 1\n\nwhere g2 ∈ Gm−1 .\n\n86\n\nAlgebraic sets and algebraic groups\n\nNow we have a natural embedding ⎡\n\nGm−1 \u0011→ Gm ,\n\n⎢ ⎢ ⎢ ⎢ ˜ A \u0004→ A := ⎢ ⎢ ⎢ ⎢ ⎣\n\n1 0 .. .\n\n0 ··· 0\n\n0\n\nA\n\n0 .. .\n\n0 0\n\n0 ··· 0\n\n⎥ ⎥ ⎥ ⎥ ⎥. ⎥ 0 ⎥ ⎥ ⎦ 1\n\n˜◦ Under this embedding, we certainly have t˜m−1 = tm and G m−1 ⊆ −1 −1 ◦ ◦ Gm . This yields g˜2 g1 g ∈ Um ⊆ Gm . Now, we have g2 ∈ ◦ ◦ , t \u0014. This yields \u0013Gm−1 , tm−1 \u0014 by induction, and so g˜2 ∈ \u0013Gm m ◦ g ∈ \u0013Gm , tm \u0014 as required. \u0003 \u0002 := SO+ (k) 1.7.8 Theorem Assume that k is infinite, and let Gm 2m (m \u0001 1) be the even-dimensional orthogonal group, as defined in §1.3.16. Then \u0002 \u0002 := B2m (k) ∩ Gm Bm\n\nand\n\n\u0002 \u0002 Nm := N2m (k) ∩ Gm\n\n\u0002 , where W ∼ W \u0002 is of type D ; see form a split BN -pair in Gm = m m \u0002 \u0002 T \u0002 , where T \u0002 = B \u0002 ∩ N \u0002 and U \u0002 := Table 1.2. We have Bm = Um m m m m m \u0002 is closed and connected. Furthermore, U2m (k) ∩ G\u0002 . The group Bm \u0002 = O+ (k)◦ , and this group has index 2 in O+ (k). Gm 2m 2m ◦ has index 2 in G . This implies that Proof By Lemma 1.7.7, Gm m \u0002 ◦ Gm = Gm . Indeed, this is clear by definition if char(k) = 2; on the \u0002 is a closed subgroup of index 2 other hand, if char(k) \b= 2, then Gm ◦ ⊆ G\u0002 by Proposition 1.3.13. Since G◦ has index 2, in Gm and so Gm m m ◦ = G\u0002 . we conclude that Gm m \u0002 = G◦ = G ˜ m . FurtherUsing Lemma 1.7.7, we also obtain Gm m \u0002 \u0002 . Thus, ˜m and Bm = Bm more, it is easily checked that Nm = N \u0002 and N \u0002 form a split BN -pair in G\u0002 . Lemma 1.7.6 yields that Bm m m \u0002 The fact that Bm is closed and connected is shown as in the proof of Theorem 1.7.4. \u0003\n\nBibliographic remarks and exercises\n\n87\n\n1.7.9 Summary: BN -pairs in classical groups Let k be an infinite field and G ⊆ GLn (k) be one of the classical groups: ⎧ any n, any characteristic; SLn (k), ⎪ ⎪ ⎨ SO2m+1 (k), n = 2m + 1, char(k) \b= 2; G= Sp2m (k), n = 2m, any characteristic; ⎪ ⎪ ⎩ + SO2m (k), n = 2m, any characteristic; see §1.3.15 and §1.3.16. Let B := Bn (k) ∩ G and N = Nn (k) ∩ G, where Bn (k) and Nn (k) are defined in Example 1.6.9. Furthermore, let U := Un (k) ∩ G and H := Tn (k) ∩ G. Then the following hold. (a) The groups B and N form a split BN -pair in G; we have B = UH. (b) The group U is closed, connected, and normal in B; we have dim U = l(w0 ). The group H is closed and connected, and consists of diagonal matrices. (c) The group B is closed and connected. Furthermore, G is connected. For the proofs, see once more Remark 1.6.11 and Theorems 1.7.4 and 1.7.8. The fact that dim U = l(w0 ) follows from Exercises 1.8.21 and 1.8.26. It is also easily checked that U is nilpotent and B is solvable; we will come back to this point in Example 2.4.9 and Definition 3.4.5. See also Example 4.2.6 where we will describe the finite classical groups.\n\n1.8 Bibliographic remarks and exercises The material in the first five sections on algebraic sets, tangent spaces, and so on is more or less standard. The main differences to standard text books on algebraic geometry (e.g. Mumford (1988) or Shafarevich (1994)), are (1) that we tried to keep the prerequisites to a minimum (in particular, we only prove a version of Hilbert’s nullstellensatz for hypersurfaces), (2) that we address algorithmic questions (Groebner bases), and (3) that we illustrate the general theory from an early point on by examples from the theory of algebraic groups. For further reading on Groebner bases and algorithmic aspects, see Cox et al. (1992) and Greuel and Pfister (2002). Example 1.3.6\n\n88\n\nAlgebraic sets and algebraic groups\n\nand Exercise 1.8.9 are taken from Chapter 3, §3 of Cox et al. (1992). The discussion of the Hilbert polynomial and the dimension follows Chapter 9 of Cox et al. (1992). This polynomial plays an important role in the general dimensional theory of noetherian rings; see Chapter 11 of Atiyah and Macdonald (1969). The results on algebraic groups are all given by Borel (1991), Humphreys (1991), or Springer (1998). The proof of Theorem 1.4.11(b) is taken from §A.3 of Goodman and Wallach (1998). For more on classical groups, see Dieudonn´e (1971), Taylor (1992) (especially for finite ground fields), Goodman and Wallach (1998) (over ground fields of characteristic 0), Chapter 7 of Aschbacher (2000), and Grove (2002). Here, we just develop this material to the point where we can show the existence of BN -pairs in these groups; see Section 1.7. The general construction of groups of Lie type in §1.5.15 was discovered by C. Chevalley in the 1950s; see Humphreys (1972) for an introduction and Steinberg (1967) and Carter (1972) for detailed expositions of this work. BN -pairs were introduced by Tits (1962). A survey on groups with a BN -pair, associated geometric structures (‘buildings’), and Tits’ classification of such groups (where W is finite and |S| > 2) can be found in Chapter 15 of Carter (1972). The discussion of the sharp form of the Bruhat decomposition follows §2.5 of Carter (1985); here, we have tried to reduce the required prerequisites to a minimum. Our definition of split BN -pairs is slightly weaker than the usual one; see, for example, §30 of Gorenstein et al. (1994). Proposition 1.7.2 is adapted from §13.5.4 of Carter (1972). The argument for the proof of (BN1) in Proposition 1.6.10 is taken from p. 36 of Steinberg (1967). For the determination of the BN -pairs in the symplectic and orthogonal groups (except in characteristic 2), we followed §2.14 of Steinberg (1974). For a more direct approach, see Chapter IV, §2, Exercise 20 of Bourbaki (1968). The problem with orthogonal groups in characteristic 2 is that, in some + ◦ way, one has to show that O+ 2m (k) has index 2 in O2m (k). Usually, this is based on the so-called Dickson invariant, which is the group homomorphism D:\n\nO+ 2m (k)\n\n→ Z/2Z ⊆ k,\n\n(aij ) \u0004→\n\nm \u0002\n\nai,2m+1−j a2m+1−i,j ;\n\ni,j=1\n\nsee §II.10 of Dieudonn´e (1971). The proof uses the Clifford algebra associated with O+ 2m (k); see also Grove (1982) and Exercise 14.5 of\n\nBibliographic remarks and exercises\n\n89\n\nAschbacher (2000). A different approach is offered by Dye (1977), who shows that D(A) ≡ rankk (A − I2m )(mod2) for all A ∈ O+ 2m (k). Our argument in Theorem 1.7.8 provides yet an alternative approach, based on the theory of groups with a BN -pair. For the existence of BN -pairs in groups of Lie type in general, see §3 of Steinberg (1967) and §8.2 of Carter (1972). 1.8.1 Exercise Let A be a commutative noetherian ring (with 1). Show that every submodule of a finitely generated A-module is again finitely generated. [Hint. Every finitely generated A-module is a quotient of the free A-module An for some n \u0001 1. So it is enough to consider An . If you need further help, see Proposition 6.5 of Atiyah and Macdonald (1969).] 1.8.2 Exercise Let k be any field. (a) Assume that |k| = ∞, and let f ∈ k[X1 , . . . , Xn ]. Show that f = 0 if and only if f(x) = 0 for all x ∈ kn . (b) Assume that k is algebraically closed. Show that |k| = ∞. Let n \u0001 2 and let f ∈ k[X1 , . . . , Xn ] be non-constant. Show that |V ({f})| = ∞. 1.8.3 Exercise Assume that k is an infinite field. (a) Let f, g ∈ k[X, Y ] be non-constant polynomials which have no common irreducible factor. Then show that V({f, g}) is a finite set. (b) Assume that V \u0002 k2 is an irreducible algebraic set. Show that either V is a singleton set or there exists an irreducible polynomial f ∈ k[X, Y ] such that V = V({f}). The statement in (a) is a first step toward B´ezout’s theorem on the intersection of curves in projective space; see Fischer (1994) or Chapter 5 of Fulton (1969). 1.8.4 Exercise We consider the polynomial ring R = k[X1 , . . . , Xn , Y1 , . . . , Ym ], where k is an infinite field. Let g1 , . . . , gn ∈ k[Y1 , . . . , Ym ] ⊆ R and consider the ideal I = (X1 − g1 , . . . , Xn − gn ) ⊆ R. (a) Show that every f ∈ R is of the form f = g + h, where g ∈ I and h ∈ k[Y1 , . . . , Ym ]. Deduce from this that R/I ∼ = k[Y1 , . . . , Ym ] and I(V(I)) = I.\n\n90\n\nAlgebraic sets and algebraic groups\n\n(b) Given the LEX order with Ym · · · Y1 Xn · · · X1 , show that (LT(f) | 0 \b= f ∈ I) is generated by G = {X1 , . . . , Xn }, and that G is a Groebner basis of I. 1.8.5 Exercise Let I ⊆ k[X1 , . . . , Xn ] and choose a lexicographic order such that Xn Xn−1 · · · X1 . Let G be a Groebner basis of I. Then show that, for any i ∈ {1, . . . , n}, we have I ∩ k[Xi , Xi+1 , . . . , Xn ] = (G ∩ k[Xi , Xi+1 , . . . , Xn ]). 1.8.6 Exercise Assume that |k| = ∞. Let us regard kn as a k-vector space in the usual way, and let V ⊆ kn be a linear subspace of kn . (a) Show that V = V({f \u0001 1 , . . . , fm }), where each fi is a polynomial of the form fi = nj=1 aij Xj with aij ∈ k. Show that V is irreducible. (b) Using Gaussian elimination, show that we can assume the polynomials fi to be of the following special form. We have m \u0002 n, and there exist 1 \u0002 j1 < j2 < · · · < jm \u0002 n such that fi = Xji + hi , where hi is a linear polynomial (with constant term zero) in the variables Xj (j > ji and j \b∈ {j1 , . . . , jm }). (c) Use Exercise 1.8.4 to show that I(V ) = (f1 , . . . , fm ) and that A[V ] is a polynomial ring in n−m variables. Deduce that the dimension of V as an algebraic set equals the dimension of V as a vector space. 1.8.7 Exercise Let k be algebraically closed, and consider the algebraic set V = V({X12 − X2 X3 , X1 X3 − X1 }) ⊆ k3 . Show that V has three irreducible components. Determine their affine algebras and show that dim V = 1. 1.8.8 Exercise Let V ⊆ kn and W ⊆ km be non-empty algebraic sets. Consider the direct product V × W ⊆ kn+m ; see §1.3.7. Show that the projection maps pr1 : V × W → V\n\nand\n\npr2 : V × W → W\n\nare open maps, that is, they send open sets to open sets. [Hint. It is enough to prove this for a principal open set. / Now a principal open subset of V × 0 W is a subset of the form (v, w) ∈ \u0001 V × W | ri=1 fi (v)gi (w) \b= 0 , where f1 , . . . , fr ∈ k[X1 , . . . , Xn ] and g1 , . . . , gr ∈ k[Y1 , . . . , Ym ]. Then check explicitly that images of such a set under pr1 and pr2 are open.]\n\nBibliographic remarks and exercises\n\n91\n\n1.8.9 Exercise We place ourselves in the setting of §1.3.6, where k ⊆ C. (a) Let ϕ : k2 → k3 be defined by ϕ(u, v) = (uv, v, u2 ) for u, v ∈ k, and set Wϕ = ϕ(k2 ) ⊆ k3 , the so-called Whitney umbrella. Consider the ideal Iˆ := (X − UV, Y − V, Z − U 2 ) ⊆ k[U, V, X, Y, Z]. Compute a Groebner basis of Iˆ with respect to the order LEX with U \u000e V \u000e X \u000e Y \u000e Z. Then use Exercise 1.8.5 to show that I := Iˆ ∩ k[X, Y, Z] = (X 2 − Y 2 Z). By §1.3.6, we have V(I) = Wϕ . Show that ϕ(k2 ) = Wϕ if k = C, and that ϕ(k2 ) \u0002 Wϕ if k = R; describe the ‘missing’ points for k = R. (b) Now let ϕ : k2 → k3 be defined by ϕ(u, v) = (uv, uv2 , u2 ) for u, v ∈ k, and set Vϕ = ϕ(k2 ) ⊆ k3 . Use the same technique as in (a) to show that Vϕ = V(X 4 − Y 2 Z) ⊆ k3 . Show that ϕ(k2 ) \u0002 Vϕ , and determine the ‘missing’ points in the cases where k = R or k = C. 1.8.10 Exercise Assume that k is algebraically closed, and let V ⊆ kn be an irreducible algebraic set. Show that the following hold. (a) We have dim V = 0 if and only if V is a singleton set. (b) We have dim V = n − 1 if and only if V is a hypersurface. (c) We have dim V = n if and only if V = kn . [Hint. (a) By Example 1.2.19(a), we have dim V = 0 if |V | = 1. To prove the converse, use Proposition 1.2.20. (b) By Example 1.2.16(b), a hypersurface in kn has dimension in n − 1. Conversely, assume that dim V = n − 1. This means that ∂k (A[V ]) < ∂k (k[X1 , . . . , Xn ]) = n and so I(V ) \b= {0}. Let 0 \b= f ∈ I(V ). Since I(V ) is a prime ideal, we may assume that f is irreducible. Thus, we have V = V(I(V )) ⊆ Hf , where Hf ⊆ kn is irreducible. Now use Proposition 1.2.20. (c) By Example 1.2.16(a), we have dim kn = n. The reverse implication follows from Proposition 1.2.20.] 1.8.11 Exercise This exercise is concerned with a special case of localization. (In fact, this is all we shall need to know about localization.) Let A be a commutative ring (with 1) and suppose 0 \b= f ∈ A. Let X be an indeterminate over A, and set Af := A[X]/(fX − 1). Composing the natural inclusion A ⊆ A[X] with the canonical projection onto Af , we obtain a ring homomorphism ι : A → Af .\n\n92\n\nAlgebraic sets and algebraic groups\n\n(a) Show that ker(ι) = {a ∈ A | af n = 0 for some n \u0001 0} (where we set f 0 = 1). Thus, ι is injective if and only if f is not a zero divisor in A. Furthermore, we have that f is nilpotent if and only if 1 = 0 in Af . If f is not nilpotent, then ι(f) ∈ Af is invertible, with inverse being given by the residue class of X in Af . (b) Assume that f is not nilpotent. Then show that we have the following universal property: if α : A → B is any homomorphism into a commutative ring B (with 1) such that α(f) ∈ B is invertible, then there exists a unique ring homomorphism α ˜ : Af → B such that α=α ˜ ◦ ι. (c) Let I ⊆ A be an ideal such that f n \b∈ I for all n \u0001 0. Let f¯ be the image of f in A/I and If be the ideal generated by ι(I) in Af . Show that (A/I)f¯ ∼ = Af /If . Furthermore, if I is a prime ideal and f¯ is not invertible in Af , then If also is a prime ideal. (d) Assume that A is an integral domain with field of fractions K. Let α : A \u0011→ K be the inclusion. Show that α ˜ : Af → K is an injective ring homomorphism with image given by {a/f n | a ∈ A, n \u0001 0} ⊆ K. [Hint. (a) Let a ∈ A and write out explicitly the condition that a ∈ (fX −1). Apply this also to a = 1. (b) Given α, we have a unique ring homomorphism α\u0002 : A[X] → B such that α\u0002 (X) = α(f)−1 and the restriction of α\u0002 to A is α. Since fX − 1 ∈ ker(α\u0002 ), we get an induced map α ˜ : Af → B as desired. (c) Consider the canonical map π : A → A/I; check that we have an induced map π\u0002 : Af → (A/I)f¯ and consider its kernel. (d) To show that α ˜ is injective, check that the ring A\u0002 := {a/f n | a ∈ A, n \u0001 0} ⊆ K (together with the inclusion A ⊆ A\u0002 ) also satisfies the universal property in (b).] 1.8.12 Exercise This exercise contains some standard results about algebraic field extensions and the transcendence degree. Let k be a field and A be a finitely generated k-algebra. Assume that A is an integral domain, and let K be its field of fractions. Then K ⊇ k is a finitely generated field extension, and ∂k (K) is also called the transcendence degree of K over k. (a) Show that, if A = k[a1 , . . . , am ], then ∂k (A) \u0002 m and one can select an algebraically independent subset of size ∂k (A) from {a1 , . . . , am }. (b) Assume that z1 , . . . , zd ∈ K are algebraically independent and that every element of K is algebraic over k(z1 , . . . , zd ). Then d = ∂k (K).\n\nBibliographic remarks and exercises\n\n93\n\n(c) Let F ⊆ K be a subfield containing k. Then ∂k (K) = ∂k (F )+ ∂F (K). [Hint. (a) Assume that A = k[a1 , . . . , am ]. Then we have a canonical surjection k[X1 , . . . , Xm ] → A such that Xi \u0004→ ai for all i; let I be the kernel of that map. Then use Proposition 1.2.18. The results (b) and (c) are standard; see for example Chapter X of Lang (1984).] 1.8.13 Exercise Let k be an infinite field and V ⊆ k n be an irreducible algebraic set. Let f ∈ k[X1 , . . . , Xn ] be such that f \b∈ I(V ). Show that V˜f is irreducible and dim V˜f = dim V . [Hint. Use Exercise 1.8.11 and Proposition 1.2.18.] 1.8.14 Exercise Denote by ∂f /∂Xi the usual partial derivative of a polynomial f ∈ k[X1 , . . . , Xn ] with respect to Xi . Show that the following holds. (a) If the characteristic of k is 0 and ∂f /∂Xi = 0 for all i, then f is constant. (b) Assume that k is perfect of characteristic p > 0. If ∂f /∂Xi = 0 for all i, then there exists some g ∈ k[X1 , . . . , Xn ] such that f = gp. (c) Assume that k is perfect and 0 \b= f ∈ k[X1 , . . . , Xn ] is nonconstant irreducible. Conclude from (a) and (b) that ∂f /∂Xi \b= 0 for some i. 1.8.15 Exercise This exercise is used in the proof of Proposition 1.4.5. Let k be a perfect field and K be a finitely generated extension field. Let d = ∂k (K). Then show that there exist z1 , . . . , zd+1 ∈ K such that: (a) K = k(z1 , . . . , zd+1 ); (b) {z1 , . . . , zd } are algebraically independent over k; (c) zd+1 is separable algebraic over k(z1 , . . . , zd ). Proceed as follows. Let us write K = k(a1 , . . . , an ) with ai ∈ K. By Exercise 1.8.12, we have d \u0002 n and we may assume that a1 , . . . , ad are algebraically independent. We use induction on n − d. If n = d, then K = k(a1 , . . . , ad ) and we may take ad+1 = 1. Then all the conditions are satisfied. Now assume that d < n and proceed by the following steps. Step 1. The set {a1 , . . . , ad+1 } cannot be algebraically independent, so there exists some non-zero F ∈ k[X1 , . . . , Xd+1 ] such that\n\n94\n\nAlgebraic sets and algebraic groups\n\nF (a1 , . . . , ad+1 ) = 0. Show that we can assume that F is irreducible and that ∂f/∂Xi \b= 0 for some i. Step 2. Show that a1 , . . . , ai−1 , ai+1 , . . . , ad+1 are algebraically independent. Then let L ⊆ K be the subfield generated by these elements. Show that ai is separable algebraic over L. Thus, we are done if n = d + 1. Step 3. Now assume that n > d + 1. Then ai and ad+2 are algebraic over L and ai is separable. So the ‘primitive-element theorem’ shows that there exists some b ∈ L(ai , ad+2 ) with L(b) = L(ai , ad+2 ). Thus, K = k[a1 , . . . , ai−1 , ai+1 , . . . , ad+1 , b, ad+3 , . . . , an ], that is, we have decreased the number of generators by 1. For more details see Appendix 5, Proposition 1 of Shafarevich (1994). 1.8.16 Exercise\n\nLet V ⊆ kn and W ⊆ km be algebraic sets.\n\n(a) Let ϕ : V → W be a regular map. Let Z := ϕ(V ) ⊆ W, p ∈ V , and q := ϕ(p) ∈ Z. Then show that dp ϕ(Tp (V )) ⊆ Tq (Z) ⊆ Tq (W ). (b) Let p ∈ V and q ∈ W . Identifying kn+m = kn × km as usual, show that T(p,q) (V × W ) = Tp (V ) ⊕ Tq (W ). [Hint. (a) We have a canonical factorization ϕ = ι ◦ ϕ\u0002 , where ι : Z → W is the inclusion and ϕ\u0002 : V → W, v \u0004→ ϕ(v). Check that dq ι : Tq (Z) → Tq (W ) is just the inclusion. (b) Use Proposition 1.3.8(b).] 1.8.17 Exercise field.\n\nAssume that k is an algebraically closed\n\n(a) Compute all non-singular points of V = V({f}) ⊆ k2 for the following polynomials in k[X, Y ]: Y 2 − X 2 (X + 1), Y 2 − X(X 2 + 1), Y 2 − X 3 . The same for f = X 2 − Y 2 Z ∈ k[X, Y, Z] and the corresponding algebraic set V = V({f}) ⊆ k3 , where k = C; see also Exercise 1.8.9(a). (b) Let V = V({ZX, YZ}) ⊆ k3 . Show that there exist points p ∈ V with dim Tp (V ) < dim V . How do you interpret this result? 1.8.18 Exercise Assume that k is perfect and infinite. Let V ⊆ kn be an irreducible algebraic set. Assume that dim V = d. Then show\n\nBibliographic remarks and exercises\n\n95\n\nthat there exist polynomials f1 , . . . , fn−d ∈ k[X1 , . . . , Xn ] such that V is an irreducible component of V({f1 , . . . , fn−d }). [Hint. By Theorem 1.4.11, there exists some non-singular p ∈ V . Since Tp (V ) is a linear subspace of kn , this means that there exist f1 , . . . , fn−d ∈ I(V ) such that Tp (V ) = V({dp (f1 ), . . . , dp (fn−d )}) ⊆ kn . Now we have V ⊆ V({f1 , . . . , fn−d }) and so V is contained in some irreducible component V \u0002 of V({f1 , . . . , fn−d }). Then use Proposition 1.2.20 and Theorem 1.4.11 to conclude that V = V \u0002 .] 1.8.19 Exercise Let k be an algebraically closed field of characteristic \b= 2 and consider the orthogonal group SO3 (k) as defined in §1.3.15. Let ω ∈ k be such that ω2 = −2. Then show that we have a unique homomorphism of algebraic groups ϕ : SL2 (k) → SO3 (k) such that ⎡ ⎤ ⎡ ⎤ \u001a \u001b \u001a \u001b 1 0 0 1 ωt t2 1 t 1 0 1 0⎦ ϕ: \u0004→ ⎣0 1 −ωt⎦ and \u0004→ ⎣ωt 0 1 t 1 2 0 0 1 t −ωt 1 for all t ∈ k. Show that ϕ is surjective and ker(ϕ) = {±I2 }. [Hint. First check that SL2 (k) is generated by the matrices specified above. To show the existence of ϕ, consider the natural action of SL2 (k) on the polynomial ring k[X, Y ]; we have \u001a \u001b a b g.X = aX + cY , where g = ∈ SL2 (k). g.Y = bX + dY c d (See, for example, Example V.5.13 of Huppert (1983).) The subspace V2 ⊆ k[X, Y ] of homogeneous polynomials of degree 2 is invariant under this action. Now express the action of g ∈ SL2 (k) on V2 with respect to the basis {ω−1 X 2 , XY, ω−1 Y 2 }.] 1.8.20 Exercise Let k be an algebraically closed field and G = O+ 4 (k), as defined in §1.3.16. Show that the following maps are welldefined injective homomorphisms of algebraic groups: ⎤ ⎡ a 0 b 0 \u001a \u001b ⎢0 a 0 −b⎥ a b ⎥, ϕ1 : SL2 (k) → G, \u0004→ ⎢ ⎣c 0 c d d 0⎦ 0 −c 0 d \u001b \u001a A 0 ϕ2 : SL2 (k) → G, , where A\u0002 = Q2 (A−1 )tr Q2 . A \u0004→ 0 A\u0002\n\n96\n\nAlgebraic sets and algebraic groups\n\nLet G1 := ϕ1 (SL2 (k)) and G2 := ϕ2 (SL2 (k)). Show that G1 and G2 are closed connected normal subgroups of G◦ and that we have G◦ = G1 .G2 where G1 ∩ G2 = {±I4 }. Furthermore, we have g1 g2 = g2 g1 for all g1 ∈ G1 and g2 ∈ G2 . 1.8.21 Exercise Let k be a field and let G = GLn (k). By Proposition 1.6.10 and Remark 1.6.12, we know that the groups Bn (k) and Nn (k) form a split BN -pair in G, with Weyl group W ∼ = Sn . (a) Let 1 \u0001 i \u0001 n − 1 and w ∈ Sn . Then show that l(σi w) > l(w)\n\nw−1 (i) < w−1 (i + 1).\n\nDeduce that l(w) = |{(i, j)|1 \u0001 i < j \u0001 n and w(i) > w(j)}| and that w0 = (1, n)(2, n − 1)(3, n − 2) · · · is the longest element in Sn . (b) Let w ∈ Sn and show that Uw (see Theorem 1.6.19) is given by Uw = \u0005Xij | 1 \u0001 i < j \u0001 n and w(i) > w(j)\u0006. (c) Assume that k = Fq is a finite field with q elements. Show that |Uw | = q l(w) . Then use the sharp form of the Bruhat decomposition to show that \u0001 | GLn (Fq )| = q n(n−1)/2 (q − 1)n q l(w) . w∈Sn\n\n[Hint. (a) In the proof of Proposition 1.6.10, we have seen that, if w−1 (i) < w−1 (i + 1), then ni Bnw ⊆ Bni nw B. Now use Proposition 1.6.3 to conclude that this implies l(σi w) > l(w). Conversely, if w−1 (i) > w−1 (i + 1), then set y := σi w and apply the previous argument. (b) Let Xi be defined as in Example 1.6.9. Show that Xi = Uσi . Then use induction on l(w) and relation (∗) in the proof of Proposition 1.6.18. For a more direct approach, see also Theorem 5.10 of Taylor (1992).] 1.8.22 Exercise Let G be a group with a BN -pair. The purpose of this exercise is to establish some multiplication rules for Bruhat cells. Let us write C(w) := Bnw B for w ∈ W . (Recall that Bnw B\n\nBibliographic remarks and exercises\n\n97\n\ndoes not depend on the choice of the representative nw ∈ N .) We certainly have C(1) = B,\n\nC(ww\u0002 ) ⊆ C(w).C(w\u0002 )\n\nand C(w−1 ) = C(w)−1 .\n\n(a) Let s1 , . . . , sp ∈ S. Then show that \u0004 C(si1 · · · siq w). C(s1 · · · sp ).C(w) ⊆ 1\u0001i1 1,\n\nsi si+1 si = si+1 si si+1 for i \u0002 1.\n\nNow show the following statements. m (a) We have Sϕ n = Wm and |Wm | = 2 m!. Conclude that Wm is a Coxeter group of type Bm (see Table 1.2). Show that w0 also is the longest element in Wm , and that its length (with respect to the generators of Wm ) is m2 . (b) Show that there exists a unique group homomorphism εt : Wm → {±1} such that εt (t) = −1 and εt (si ) = 1 for 1 \u0001 i \u0001 m − 1. \u0002 := ker(ε ) ⊆ W ; then W \u0002 has index 2 in W . (c) Let Wm t m m m \u0002 = \u0005ts t, s , . . . , s \u0002 Show that Wm 1 1 m−1 \u0006 if m \u0002 2. Conclude that Wm is a Coxeter group of type Dm . Show that the length of the longest \u0002 is m(m − 1). element in Wm\n\n[Hint. The relations are checked by a straightforward computation. (a) To show Sϕ n = Wm , apply Exercise 1.8.25. On the other hand, we have Wm = CSn (w0 ) and it is easy to compute the order of that centralizer. To obtain an expression for w0 in terms of the generators of Wm , define t0 := t and ti := si ti−1 si for 1 \u0001 i \u0001 m − 1. Then check that w0 = t0 t1 · · · tm−1 . (b) Restrict the sign homomorphism \u0002 is the set of all elements of W which ε : Sn → {±1} to Wm . (c) Wm m can be written as a product of the generators t, s1 , . . . , sm−1 with an\n\n100\n\nAlgebraic sets and algebraic groups\n\neven number of occurrences of t. Use the above relations to rearrange \u0002 is generated as desired. To such a product in order to see that Wm \u0002 obtain the length of the longest element, note that either w0 ∈ Wm \u0002 .] or tw0 ∈ Wm 1.8.27 Exercise The principal minors of a matrix A = (aij ) ∈ Mn (k) are the determinants ⎤ ⎡ a11 . . . a1m ⎢ .. ⎥ for 1 \u0002 m \u0002 n. dm := det ⎣ ... . ⎦ am1 . . . amm Let δ ∈ k[Xij | 1 \u0002 i, j \u0002 n] be the polynomial which gives the product of the principal minors of A ∈ Mn (k) for 1 \u0002 m \u0002 n. Now let G ⊆ GLn (k) be one of the classical groups defined in §§1.3.15–16; thus, we have ⎧ SLn (k), any n, ⎪ ⎪ ⎨ SO2m+1 (k), n = 2m + 1, char(k) \b= 2, G= Sp2m (k), n = 2m, any characteristic, ⎪ ⎪ ⎩ + O2m (k), n = 2m, any characteristic. Let U and U \u0002 be the subgroups of all upper unitriangular matrices and all lower unitriangular matrices in G, respectively. Let T be the subgroup of all invertible diagonal matrices in G. Show that U \u0002 · T · U = {g ∈ G | δ(g) \b= 0} is an open subset of G. [Hint. First note that the principal minors of a matrix A ∈ Mn (k) do not change if we multiply A on the left by a lower unitriangular matrix or on the right by an upper unitriangular matrix. Thus, U · T · U \u0002 ⊆ {g ∈ G | δ(g) \b= 0}. To prove the reverse inclusion, first consider G = SLn (k). Let A = (aij ) be a matrix in G such that δ(A) \b= 0. We have to solve the following system of equations: ⎤⎡ ⎡ ⎤⎡ ⎤ 1 0 ··· 0 1 u21 · · · un1 h1 0 · · · 0 ⎤ ⎡ · · · a a ⎥ ⎢ \u0002 . . . 11 1n . . . ⎢ ⎥ ⎢ ⎥ . . .. ⎥ ⎢ 0 h . . .. ⎥ ⎢0 1 . . .. ⎥ ⎢ ⎢ u21 1 .. ⎥ . 2 ⎥⎢ ⎢ ⎥ ⎢. . . ⎥ = ⎣ .. . ⎦ ⎥ ⎣ .. . . . . ⎢ .. . . . . ⎦ ⎣ ⎦ . . . ⎣ . . . 0⎦ . . . 0 . . . un,n−1 a · · · a n1 nn \u0002 · · · u\u0002 un1 0 · · · 0 hn 0 ··· 0 1 n,n−1 1 \u0002 and h . We can solve this system recursively. with unknowns uij , uij i The fact that all principal minors of A are non-zero is used in the\n\nBibliographic remarks and exercises\n\n101\n\nfollowing way. First, we have h1 = a11 \b= 0. Then, for i > 1, we \u0002 h = a and so we determine u\u0002 . Similarly, for j > 1, we have ui1 1 i1 i1 have h1 uj1 = a1j and so we can determine uj1 . Now assume that the first j − 1 rows and columns of the above matrices have already been determined, where h1 , . . . , hj−1 are all non-zero. We have an \u0002 \u0002 h u equation hj + uj,j−1 hj−1 uj,j−1 + · · · + uj1 1 j1 = ajj , which can be solved uniquely for hj . In particular, we have now determined all coefficients which belong to the first j rows and columns. To show that hj \b= 0, we consider the subsystem of equations made up of these rows and columns; this subsystem looks like the original system written in matrix form above, with n replaced by j. The righthand side has a non-zero determinant by assumption. Hence all the coefficients h1 , . . . , hj must be non-zero. Thus, the assertion holds for G = SLn (k). Next note that, given g ∈ G such that g = u\u0002 hu, where u ∈ U , u\u0002 ∈ U \u0002 , and h ∈ T , then the factors u, u\u0002 ,and h are unique. Now use Example 1.7.3 to deduce the desired assertion for SO2m+1 (k) (char(k) \b= 2), Sp2m (k) (any characteristic), and O2m (k) = O+ 2m (k) (char(k) \b= 2). It remains to consider the case where G = O+ 2m (k) ˜ and char(k) = 2. We have G ⊆ G = Sp2m (k). Let g ∈ G be such that δ(g) \b= 0. Then we can write g = u ˜\u0002 h˜ u, where u ˜\u0002 and u ˜ are lower and ˜ ˜ Next upper unitriangular in G, respectively, and h is diagonal in G. note that h ∈ G by Lemmas 1.5.9 and 1.5.11. Furthermore, check that u ˜ = vu, where u ∈ G and . Im Qm S v= , with S ∈ Mm (k) diagonal. Im 0 Similarly, we have u ˜\u0002 = u\u0002 v\u0002 , where u\u0002 ∈ G. This yields v\u0002 hv ∈ G. Now check explicitly, using the defining equations for G, that this implies v = v\u0002 = 1.]\n\n2 Affine varieties and finite morphisms\n\nIn this chapter, we come to those results on affine varieties and algebraic groups which rely in an essential way on the hypothesis that the ground field k is algebraically closed. We will see that, under this hypothesis, the image of a morphism between affine varieties is a relatively ‘thick’ subset, more precisely, it contains an open subset of its closure. This fact will turn out to be a key tool. The basic algebraic notion which underlies these results is that of an integral extension of rings; see Section 2.1. As a first application, we obtain the general form of Hilbert’s Nullstellensatz which will reveal the geometric meaning of many of the purely algebraic constructions in Chapter 1. Before we go on to study morphisms, we address another point. One disadvantage of the set-up in Chapter 1 is that it always refers to an embedding of an algebraic set in some affine space kn . Instead, one would just like to work with an intrinsic description of an affine variety and the algebra of regular functions on it. In Section 2.1, following Steinberg, we will give such an abstract definition of affine varieties. Then, in Section 2.2, we prove Chevalley’s theorem on the image of a morphism. Section 2.3 is concerned with the problem of finding criteria which guarantee that a bijective morphism between affine varieties actually is an isomorphism. In general, this is a very difficult and subtle problem, especially over fields of positive characteristic. We shall prove some weak versions of ‘Zariski’s main theorem’ which are sufficient for many applications to algebraic groups. In Section 2.4, we come to first applications to algebraic groups. We show that a group generated by a family of closed irreducible subsets is itself irreducible. This is an important tool for proving the connectedness of a group. In Section 2.5, we establish some general results concerning algebraic group actions on affine varieties. One of the main results shows that there are always closed orbits for the action of an algebraic group. Finally, in Section 2.6, we illustrate\n\nHilbert’s Nullstellensatz and abstract affine varieties\n\n103\n\nthese ideas by studying in detail the conjugation action of SLn (k) on unipotent elements.\n\n2.1 Hilbert’s Nullstellensatz and abstract affine varieties In the previous chapter, we have introduced various notions related to an algebraic set V ⊆ kn : the dimension, regular maps, and the tangent space. However, we have also seen characterizations entirely in terms of the algebra A[V ]: see Proposition 1.2.18 for the dimension, Proposition 1.3.4 for regular maps, and Remark 1.4.9(b) for the tangent space. All this suggests that there is a purely axiomatic description of algebraic sets in terms of algebras of functions. Our aim here is to give a precise sense to this idea. First we establish Hilbert’s nullstellensatz, whose proof relies on some properties of integral ring extensions. Let A and B be commutative rings (with 1) such that A ⊆ B. We say that b ∈ B is integral over A, if there is an equation of ‘integral dependence’ bn +an−1 bn−1 +· · ·+a1 b+a0 = 0,\n\nwhere n \u0002 1 and ai ∈ A for all i.\n\nWe say that B is integral over A if every element of B is integral over A. 2.1.1 Lemma equivalent.\n\nLet b ∈ B. Then the following conditions are\n\n(a) b is integral over A. (b) The subring A[b] ⊆ B generated by b is a finitely generated A-module. (c) There exists a subring C ⊆ B such that A[b] ⊆ C and C is finitely generated as an A-module. Proof ‘(a) ⇒ (b)’: We have A[b] = {f(b) | f ∈ A[X]}, where X is an indeterminate. By assumption, there exists a monic polynomial g : = a0 + a1 X + · · · + an−1 X n−1 + X n ∈ A[X] with n \u0002 1 such that g(b) = 0. Then we can divide any f ∈ A[X] by g, leaving a remainder r such that f = qg + r, where q, r ∈ A[X] and r = 0 or deg(r) < n. Consequently, f(b) = q(b)g(b) + r(b) = r(b), and every element in A[b] is an A-linear combination of 1, b, b2 , . . . , bn−1 . ‘(b) ⇒ (c)’: This is trivial (set C : = A[b]).\n\n104\n\nAffine varieties and finite morphisms\n\n\u0003n ‘(c) ⇒ (a)’: Let c1 , . . . , cn ∈ C be such that C = i=1 Aci . Since b ∈ C and C is a ring, we have bc ∈ C for all i. So there i \u0003n exist aij ∈ A such that bci = a c for all i. Setting M = j=1 ij j (aij )1\u0002i,j\u0002n ∈ Mn (A), we can write these equations in matrix form: Mv = bv, where v ∈ An is the column vector whose ith component is ci . Thus, we have (M − bIn )v = 0, where In is the n × n identity matrix. By Cramer’s rule, this implies that ci det(M − bIn ) = 0 for all i. \u0003 Now, we have 1 ∈ C and so there exist ai ∈ A such that n 1 = i=1 ai ci . Multiplying this equation with det(M − bIn ), we obtain det(M − bIn ) = 0. Expanding the determinant yields the required equation of integral dependence for b. \u0003 2.1.2 Corollary\n\nLet A ⊆ B be commutative rings.\n\n(a) If B = A[b1 , . . . , bn ], where each bi ∈ B is integral over A[b1 , . . . , bi−1 ], then B is a finitely generated A-module and B is integral over A. (b) The set AB : = {b ∈ B | b is integral over A} is a subring of B and is called the integral closure of A in B. (c) Let C ⊆ B be a subring such that A ⊆ C. If C is integral over A, and B is integral over C, then B is integral over A. (d) If B is a field, and B is integral over A, then A also is a field. Proof (a) This immediately follows from Lemma 2.1.1, using induction on n. (b) Let b, b\u0002 ∈ B be integral over A, and consider the subring A[b, b\u0002 ] ⊆ B. By (a), A[b, b\u0002 ] is a finitely generated A-module. So every element in A[b, b\u0002 ] is integral over A. In particular, b ± b\u0002 and bb\u0002 are integral over A. (c) Let b ∈ B. Since b is integral over C, there exists an equation m b + cm−1 bm−1 + · · · + c1 b + c0 = 0, where m \u0002 1 and ci ∈ C. Using (a), we see that the subring C \u0002 : = A[b, c0 , c1 , . . . , cm−1 ] ⊆ B is a finitely generated A-module. Hence, by Lemma 2.1.1, b ∈ C \u0002 is integral over A, as desired. (d) Let 0 = a ∈ A. Since B is a field, there exists some b ∈ B such that 1 = ab. Now B is integral over A, and so we have an equation bn + an−1 bn−1 + · · · + a1 b + a0 = 0, where n \u0002 1 and ai ∈ A. Multiplying with an−1 and using ab = 1, we obtain b = −(an−1 + an−2 a + · · · + a0 an−1 ) ∈ A, as required. \u0003\n\nHilbert’s Nullstellensatz and abstract affine varieties\n\n105\n\n2.1.3 Example We say that A is integrally closed in B if AB = A. For example, if A is a factorial domain with field of fractions K, then A is integrally closed in K. (We leave the proof as an exercise to the reader.) In particular, the polynomial ring k[X1 , . . . , Xn ] is integrally closed in its field of fractions. 2.1.4 Theorem (Noether normalization) Let A be a finitely generated k-algebra.1 Then there exist algebraically independent elements a1 , . . . , ad ∈ A such that A is integral over the subring k[a1 , . . . , ad ]. Proof Let us write A = k[a1 , . . . , an ] with ai ∈ A. We now proceed by an induction on n. If n = 0, then A = k and there is nothing to prove. Now assume that n > 0. If a1 , . . . , an are algebraically independent, we are done. So let us assume that a1 , . . . , an are not algebraically independent. Then there exists some non-constant F ∈ k[X1 , . . . , Xn ] such that F (a1 , . . . , an ) = 0. Renumbering the variables if necessary, we may assume that some term of F involves \u0003 the variable Xn . Writing F = α aα X α , let r be an integer which is bigger than any\u0003component of any α such that aα = 0. Then, n−1 i setting N (α) : = i=0 αn−i r for any α = (α1 , . . . , αn ), we have N (α1 ) = N (α2 ) for any α1 = α2 such that aα1 = 0 and aα2 = 0. (Check this!) We now proceed as follows. We set ri : = rn−i for 1 \u0001 i \u0001 n − 1, and define Yi : = Xi − Xnri for 1 \u0001 i \u0001 n − 1. We claim that, if we set N = max{N (α) | aα = 0}, then F has the form F =\n\nλXnN\n\n+\n\nN −1 \u0001\n\ngi Xni ,\n\nwhere 0 = λ ∈ k and gi ∈ k[Y1 , . . . , Yn−1 ].\n\ni=0\n\nThis follows from the fact that all N (α) for aα = 0 are different and that X α = X1α1 · · · Xnαn = (Y1 + Xnr1 )α1 · · · (Yn−1 + Xnrn−1 )αn−1 Xnαn \u0001\n\nN (α)−1\n\n=\n\nXnN (α)\n\n+\n\nhi Xni\n\nwith hi ∈ k[Y1 , . . . , Yn−1 ].\n\ni=0\n\nFurthermore, note that N is the degree of F in Xn . Now set yi : = ai − arni for 1 \u0001 i \u0001 n − 1, and consider the subring 1\n\nRecall that the term k-algebra means a commutative associative k-algebra with 1.\n\n106\n\nAffine varieties and finite morphisms\n\nR : = k[y1 , . . . , yn−1 ] ⊆ A. We define g : = F (y1 , . . . , yn−1 , Xn ) ∈ R[Xn ]. Then g = 0 and g(an ) = 0. Dividing g by λ, we see that an is integral over R. Furthermore, each ai = yi + arni (1 \u0001 i \u0001 n) is also integral over R and so A is integral over R by Corollary 2.1.2(a). By induction, we can find r1 , . . . , re ∈ R which are algebraically independent and such that R is integral over the subring k[r1 , . . . , re ]. Then A is integral over k[r1 , . . . , re ] by Corollary 2.1.2(c). \u0003 2.1.5 Theorem (Hilbert’s nullstellensatz, weak form) Assume that k is algebraically closed. Then the maximal ideals in k[X1 , . . . , Xn ] are precisely the ideals of the form (X1 − v1 , . . . , Xn − vn ), where vi ∈ k. More generally, if A is any finitely generated k-algebra, then A/m ∼ = k for every maximal ideal m in A. Proof Since every finitely generated k-algebra is a quotient of a polynomial ring, it is enough to prove the statement about k[X1 , . . . , Xn ]. Now, first note that an ideal of the form I = (X1 − v1 , . . . , Xn − vn ) is maximal, since k[X1 , . . . , Xn ]/I ∼ = k. Conversely, let I ⊆ k[X1 , . . . , Xn ] be any maximal ideal, and consider A : = k[X1 , . . . , Xn ]/I. Then A is a finitely generated k-algebra which is in fact a field. Let R = k[a1 , . . . , ad ] ⊆ A be a Noether normalization, as in Theorem 2.1.4. Now Corollary 2.1.2(d) implies that R must also be a field. This is only possible if d = 0, that is, we have R = k. Consequently, A is a finite algebraic extension of k. Since k is algebraically closed, we have A = k. Thus, for each i, there exists some vi ∈ k such that Xi − vi ∈ I, and so I = (X1 − v1 , . . . , Xn − vn ). \u0003 We are now ready to give the abstract definition of affine varieties. 2.1.6 Definition Let X be any (non-empty) set and k be a field. We consider the k-algebra Maps(X, k) of all maps X → k, with pointwise defined operations. For any x ∈ X, we have the evaluation map εx which sends a function f : X → k to its value f (x). Now let A be a k-subalgebra of Maps(X, k). Then the pair (X, A) is called an affine variety (over k) if the following conditions hold. (a) A is finitely generated as an algebra over k, and we have 1 ∈ A. (b) For x = y in X, there exists some f ∈ A such that f (x) = f (y).\n\nHilbert’s Nullstellensatz and abstract affine varieties\n\n107\n\n(c) For every k-algebra homomorphism λ : A → k, there exists some x ∈ X such that λ = εx . (By convention, we also consider (∅, {0}) as an affine variety.) We call A the algebra of regular functions on X and also write A = A[X] to indicate the relation with X. Note that (b), (c) mean that the assignment x → εx is a bijection between X and the set of k-algebra homomorphisms A → k. The condition (a) means that A is a quotient of a polynomial ring in finitely many indeterminates over k. Hence, by Hilbert’s basis theorem 1.1.1, A is a noetherian ring. If (X, A) and (Y, B) are affine varieties over k, then a map ϕ : X → Y is called a morphism of affine varieties if g ◦ ϕ ∈ A for all g ∈ B. In this case, we have an induced k-algebra homomorphism ϕ∗ : B → A, g → g ◦ ϕ. As in §1.3.3 and Proposition 1.3.4, one checks that the assignment ϕ → ϕ∗ is functorial and sets up a bijection between morphisms X → Y and k-algebra homomorphisms B → A. Given a homomorphism α : B → A, the unique morphism ϕ : X → Y such that ϕ∗ = α is determined by the condition that εx ◦ α = εϕ(x)\n\nfor any x ∈ X.\n\nIndeed, by Definition 2.1.6(c), there exists some y ∈ Y such that εx ◦ α = εy . By Definition 2.1.6(b), y is uniquely determined, and so we can set ϕ(x) : = y. In this way, we obtain a map ϕ : X → Y satisfying the condition ϕ∗ = α. 2.1.7 The abstract Zariski topology Let (X, A) be an affine variety over k. For any subset S ⊆ A, we define VX (S) : = {x ∈ X | f(x) = 0 for all f ∈ S}. By exactly the same argument as in §1.1.6, we see that the sets VX (S) (S ⊆ A) form the closed sets of a topology on X which is called the Zariski topology. Furthermore, for any subset Y ⊆ X, we define IA (Y ) : = {f ∈ A | f(y) = 0 for all y ∈ Y }. Then IA (Y ) is an ideal in A, the vanishing ideal of Y . As in §1.1.7, we see that VX (IA (Y )) = Y¯ = closure of Y in the Zariski topology on X. Thus, the operators VX and IA have the same formal properties as the operators V and I introduced in Section 1.1.\n\n108\n\nAffine varieties and finite morphisms\n\nArguing as in the proof of Proposition 1.1.12(a), we see that X is a noetherian topological space. Furthermore, as in Proposition 1.1.12(b), we see that X is irreducible if and only if A is an integral domain. Let ϕ : X → Y be a morphism of affine varieties. Then the following hold. (a) ϕ is continuous. (b) If X is irreducible, then ϕ(X) ⊆ Y is also irreducible. This follows by exactly the same arguments as in Remark 1.3.2. 2.1.8 Example Let V ⊆ kn be a non-empty algebraic set and A[V ] its affine algebra, regarded as the algebra of regular functions on V , as in §1.3.3. Then (V, A[V ]) is an affine variety in the sense of Definition 2.1.6. Indeed, we have A[V ] = k[X1 , . . . , Xn ]/I(V ), and this is a finitely generated k-algebra. Furthermore, for v = (v1 , . . . , vn ) ∈ V , we have ¯ i (v) = vi for 1 \u0001 i \u0001 n. Thus, given v = w in V , there exists X ¯ i (v) = X ¯ i (w). It remains to show that every some i such that X k-algebra homomorphism λ : A[V ] → k is an evaluation map. To ¯ i ) for 1 \u0001 i \u0001 n and v : = (v1 , . . . , vn ) ∈ kn . prove this, set vi : = λ(X We claim that v ∈ V and λ = εv . Indeed, consider the canonical map π : k[X1 , . . . , Xn ] → A[V ]. Then λ ◦ π : k[X1 , . . . , Xn ] → k is a k-algebra homomorphism whose kernel contains the ideal I : = (X1 − v1 , . . . , Xn − vn ). This ideal is maximal (since k[X1 , . . . , Xn ]/I ∼ = k) and so ker(λ ◦ π) = I. We have I(V ) ⊆ I and so {v} = V(I) ⊆ ¯ i ) = vi = X ¯ i (v) = V(I(V )) = V ; see §1.1.7. Finally, we have λ(X ¯ εv (Xi ) for all i and so λ = εv , as required. To state the next result, we need one more √ definition. Given any ideal I in a commutative ring A, the set I : = {a ∈ √ I | m a ∈ I for some m \u0002 1} is√ called the radical of I. Note that I itself is an ideal: if a, b ∈ \u0003 I, we \u0004have\u0005 ar ∈ I and bs ∈ I for some r+s i (r−i)+s r+s r, s \u0002 1 and so (a + b) = r+s ∈ I. i=0 i a b 2.1.9 Theorem (Hilbert’s nullstellensatz, strong form) Let k be an algebraically closed field and (X, A) be an affine variety over k. For any ideal I ⊆ A and any closed subset Y ⊆ X, we have √ IA (VX (I)) = I and VX (IA (Y )) = Y.\n\nHilbert’s Nullstellensatz and abstract affine varieties\n\n109\n\nFurthermore, the maximal ideals in A are the ideals ker(εp ) for p ∈ X, where εp : A → k denotes the evaluation homomorphism. Proof By the same proof as in §1.1.7, we have VX (IA (Y )) = Y for any closed √ set Y ⊆ X. Now let I ⊆ A be any ideal. Then we certainly have I ⊆ IA (V √X (I)). To prove the reverse inclusion, let f ∈ A be such that f ∈ I. We shall show that there exists some x ∈ VX (I) such that f(x) = 0. This is done as follows. Consider the localization Af and √ the natural map ι : A → Af as √defined in Exercise 1.8.11. √ generated by ι( I) in Af . √ Now let I f be the ideal I, the image of f in A/ I is not nilpotent and so Since f ∈\n\n√ √ Af / I f = {0} by Exercise 1.8.11(c). Now, since A / I is a finitely f f √ generated k-algebra, the quotient of Af / I f by any maximal ideal is isomorphic to k, by the weak form of Hilbert’s nullstellensatz. Hence, there exists a k-algebra homomorphism λ : Af → k with √ I f ⊆ ker(λ). By Definition 2.1.6(c), we have εx = λ ◦ ι : A → k for some x ∈ X. Since I ⊆ ker(εx ), we have x ∈ VX (ker(εx )) ⊆ VX (I). Furthermore, since ι(f) ∈ Af is invertible, we have εx (f) = λ(ι(f)) = 0, as required. Finally, consider the statement concerning the maximal ideals in A. Every ideal ker(εp ) certainly is maximal. Conversely, if m ⊆ A is a maximal ideal, then using Theorem 2.1.5, we see that A/m ∼ = k. Thus, m is the kernel of an evaluation homomorphism by Definition 2.1.6(c). \u0003 By Example 2.1.8, the above result applies to all algebraic sets in kn . Next we show that every abstract affine variety is isomorphic to an algebraic set. 2.1.10 Proposition Let (X, A) be an affine variety over k, where k is algebraically closed. Since A is finitely generated, there exists some n \u0002 1 and a surjective algebra homomorphism π : k[T1 , . . . , Tn ] → A (where Ti are indeterminates). Then the following hold: (a) Define ϕ : X → kn by ϕ(x) = (π(T1 )(x), . . . , π(Tn )(x)). Then ϕ is a morphism of affine varieties and ϕ(X) = V(ker(π)) is closed. (b) The restricted map X → ϕ(X), x → ϕ(x), is an isomorphism where the closed subset ϕ(X) ⊆ kn is regarded as an affine variety as in Example 2.1.8.\n\n110\n\nAffine varieties and finite morphisms\n\nProof Set I : = ker(π) ⊆ k[T1 , . . . , Tn ] and V : = V(I) ⊆ kn . First note that ϕ is a morphism of affine varieties such that ϕ∗ = π. (Indeed, given f ∈ k[T1 , . . . , Tn ], we have f ◦ ϕ = f(π(T1 ), . . . , π(Tn )) = π(f)√∈ A.) Next note that I certainly is a radical ideal, that is we have I = I. (Indeed, if f m ∈ ker(π) for some f ∈ k[T1 , . . . , Tn ] and some m \u0002 1, then π(f)m = 0 in A. But A is an algebra of k-valued functions and so π(f) = 0.) Thus, Theorem 2.1.9 shows that I = I(V ). Consequently, we have A[V ] = k[T1 , . . . , Tn ]/I and so π induces a k-algebra isomorphism π1 : A[V ] → A. Hence, by the remarks following Definition 2.1.6, there exists an isomorphism of affine varieties ϕ1 : X → V such that ϕ∗1 = π1 . Let ι : V → kn be the inclusion. Then ι∗ is the natural map k[T1 , . . . , Tn ] → A[V ] and so ϕ∗ = π = π1 ◦ ι∗ = ϕ∗1 ◦ ι∗ = (ι ◦ ϕ1 )∗ . This implies ϕ = ι ◦ ϕ1 . Thus, we see that ϕ1 is the restricted map X → V , x → ϕ(x). \u0003 Thus, if k is algebraically closed, then everything that we proved for algebraic sets in Chapter 1 will remain valid for an abstract affine variety (X, A) as above. 2.1.11 Definition Let (X, A) be an affine variety over k. We define the dimension of X by dim X = ∂k (A). Furthermore, for p ∈ X, we define the tangent space by Tp (X) = Derk (A, kp ). If X is irreducible, we say that p ∈ X is non-singular if dim X = dim Tp (X). We shall now show that the usual constructions involving algebraic sets all have a natural interpretation in the above abstract setting. 2.1.12 Closed subvarieties Let (X, A) be an affine variety and Y ⊆ X be a closed subset. Then the algebra homomorphism A → Maps(Y, k) given by restricting functions from X to Y has kernel I : = IA (Y ). Thus, we may regard A/I as a k-algebra of functions on Y . We claim that (Y, A/I) is an affine variety and the inclusion ι : Y → X is a morphism, where ι∗ is the natural map A → A/I. Indeed, B : = A/I certainly is finitely generated. Furthermore, condition (b) in Definition 2.1.6 is clearly satisfied. Finally, if λ : B → k is a k-algebra homomorphism, then we can lift λ to A and obtain an element x ∈ X such that λ(f|B ) = f(x) for all f ∈ A.\n\nFinite morphisms and Chevalley’s theorem\n\n111\n\nLet S ⊆ A be such that Y = VX (S). Then, for any f ∈ S, we have f|B = 0 and so f(x) = 0, that is, x ∈ Y . 2.1.13 Direct products of affine varieties Let (X, A) and (Y, B) be affine varieties over k. Then we have a natural map A ⊗k B → Maps(X × Y, k) given by sending f ⊗ g to the function X × Y → k, (x, y) → f(x)g(y). As in Proposition 1.3.8(b), one sees that this map is in fact injective. Thus, we may regard A ⊗k B as a finitely generated subalgebra of Maps(X × Y, k). It is easily checked that the remaining conditions in Definition 2.1.6 are satisfied, and so (X × Y, A ⊗k B) is an affine variety. 2.1.14 Affine open subvarieties Let (X, A) be an affine variety and 0 = f ∈ A. Consider the open set Xf : = {x ∈ X | f(x) = 0}, and let Af be the k-subalgebra of Maps(Xf , k) generated by 1/f (restricted to Xf ) and the functions in A (restricted to Xf ). Then Af is finitely generated, and it is easily checked that the remaining conditions in Definition 2.1.6 are also satisfied. Thus, (Xf , Af ) is an affine variety. Using an argument similar to that in the proof of Lemma 1.1.15, we see that Af = A[X]/(fX − 1) is the localization of A in f; see Exercise 1.8.11. The embedding ι : Xf → X is a morphism of affine varieties, where the corresponding algebra homomorphism ι∗ : A → Af is the natural map.\n\n2.2 Finite morphisms and Chevalley’s theorem Whenever we consider affine varieties, we will assume from now on that the ground field k is algebraically closed. The purpose of this section is to introduce an important class of morphisms: the so-called finite morphisms. We shall see (1) that these morphisms have many special properties and (2) that many questions about arbitrary morphisms can be reduced to questions about finite morphisms. We will usually denote the algebra of regular functions on an affine variety X by A[X] and just write X instead of (X, A[X]).\n\n112\n\nAffine varieties and finite morphisms\n\nWe begin by establishing some general results on morphisms. For this purpose, we introduce the following notation. Let ϕ : X → Y be a morphism between affine varieties. We say that ϕ is dominant if ϕ(X) = Y . We say that ϕ is a closed embedding if the image ϕ(X) ⊆ Y is a closed subset and the restricted map ϕ1 : X → ϕ(X), x → ϕ(x), is an isomorphism of affine varieties. Here, we regard ϕ(X) ⊆ Y as an affine variety as in §2.1.12. 2.2.1 Proposition Let ϕ : X → Y be a morphism between affine varieties. Then the following hold. (a) ϕ is dominant if and only if ϕ∗ : A[Y ] → A[X] is injective. (b) ϕ is a closed embedding if and only if ϕ∗ : A[Y ] → A[X] is surjective. Proof (a) Let Z : = ϕ(X) ⊆ Y . Assume first that Z \u0002 Y . Since the operator IY is injective, there exists some non-zero g ∈ IY (Z). Then we have ϕ∗ (g)(x) = g(ϕ(x)) = 0 for all x ∈ X. This means g ∈ ker(ϕ∗ ), and so ϕ∗ is not injective. Conversely, if ϕ∗ is not injective, there exists some non-zero g ∈ A[Y ] such that g(ϕ(x)) = ϕ∗ (g)(x) = 0 for all x ∈ X, and so g ∈ IY (ϕ(X)). Consequently, we have ϕ(X) ⊆ VY ({g}) and, since the latter set is closed, ϕ(X) ⊆ VY ({g}). Since g = 0, this shows VY ({g}) \u0002 Y and so ϕ(X) \u0002 Y . (b) If ϕ is a closed embedding, then we have a factorization ϕ = ι ◦ ϕ1 , where ι is the inclusion ϕ(X) ⊆ Y , and so ϕ∗ = ϕ∗1 ◦ ι∗ . Since ϕ∗1 is an isomorphism and ι∗ is surjective, we conclude that ϕ∗ is surjective. Conversely, assume that ϕ is a morphism such that ϕ∗ is surjective. First we claim that ϕ(X) is a closed subset of Y ; more precisely: ϕ(X) = VY (ker(ϕ∗ )) ⊆ Y. To see this, let x ∈ X and g ∈ ker(ϕ∗ ). Then g(ϕ(x)) = ϕ∗ (g)(x) = 0 and so ϕ(x) ∈ VY (ker(ϕ∗ )). Conversely, let y ∈ Y be such that g(y) = 0 for all g ∈ ker(ϕ∗ ). Consider the evaluation map εy : A[Y ] → k. Since ϕ∗ is surjective and g(y) = 0 for all g ∈ ker(ϕ∗ ), we have a well-defined k-algebra homomorphism λ : A[X] → k such that εy = λ ◦ ϕ∗ . By Definition 2.1.6(c), there exists some x ∈ X such that λ = εx and so εy (g) = εx (ϕ∗ (g)) = ϕ∗ (g)(x) = g(ϕ(x)) = εϕ(x) (g) for all g ∈ A[Y ]. Thus, we have y = ϕ(x) ∈ ϕ(X), as desired.\n\nFinite morphisms and Chevalley’s theorem\n\n113\n\n\u0006 As in the proof of Proposition 2.1.10, one sees that ker(ϕ∗ ) = ker(ϕ∗ ). Hence, Theorem 2.1.9 implies that IA[Y ] (ϕ(X)) = ker(ϕ∗ ). We can now argue as in the proof of Proposition 2.1.10 to conclude that the restricted map ϕ1 : X → ϕ(X) is an isomorphism of affine varieties, as required. \u0003 2.2.2 Definition Let ϕ : X → Y be a morphism of affine varieties. We say that ϕ is a finite morphism if A[X] is integral over ϕ∗ (A[Y ]). The simplest possible example is given as follows. Consider k as an affine variety, where k[X] is the algebra of regular functions on k. Let e \u0002 1 and consider the morphism ϕe : k → k, x → xe . Then X certainly is integral over ϕ∗e (k[X]) = k[X e ], and so ϕe is a finite morphism. A general class of examples is given by the Noether normalization theorem. Let X be any affine variety over k and k[a1 , . . . , ad ] ⊆ A[X] be as in Theorem 2.1.4. Since a1 , . . . , ad are algebraically independent, we may regard k[a1 , . . . , ad ] as the algebra of regular functions on kd . Then the embedding k[a1 , . . . , ad ] ⊆ A[X] is the algebra homomorphism corresponding to a dominant finite morphism ϕ : X → kd . 2.2.3 Lemma Let ϕ : X → Y be a finite morphism of affine varieties over k. Then there exists a constant c > 0 such that |ϕ−1 (q)| < c for all q ∈ Y . Proof By Proposition 2.1.10, we may assume without loss of generality that X ⊆ kn and Y ⊆ km are algebraic sets and that A[X] = k[x1 , . . . , xn ] and A[Y ] = k[y1 , . . . , ym ], where xi and yj are the corresponding coordinate functions. Let q ∈ ϕ(X) and p ∈ X be such that ϕ(p) = q. Then we have p = (x1 (p), . . . , xn (p)), and it is enough to show that the number of possible values for xi (p) (1 \u0001 i \u0001 n) is bounded independently of q. Now let us fix some i. Since A[X] is integral over ϕ∗ (A[Y ]), we can write ∗ ∗ xli +ϕ∗ (bl−1 )xl−1 i +· · ·+ϕ (b1 )xi +ϕ (b0 ) = 0, where l \u0002 1 and bj ∈ A[Y ]. Evaluating at p and using ϕ∗ (bj )(p) = bj (q), we see that xi (p) is a zero of T l + bl−1 (q)T l−1 + · · · + b1 (q)T + b0 (q) ∈ k[T ] (where T is an indeterminate). So there are only finitely many possibilities for\n\n114\n\nAffine varieties and finite morphisms\n\nxi (p). Since this works for all i, we also obtain a global bound c > 0 for the number of preimages of q. \u0003 More precise results on finite morphisms will be a consequence of the following basic result, which is in fact a special case of the ‘goingup theorem’ of Cohen–Seidenberg; see, for example, Corollary 5.8 and Theorem 5.10 of Atiyah and Macdonald (1969). 2.2.4 Lemma (‘Going up’, weak form) Assume that A ⊆ B are commutative rings with 1 such that B is finitely generated and integral over A. Let m \u0002 A be a maximal ideal. Then there exists a maximal ideal q \u0002 B such that q ∩ A = m. Proof By Corollary 2.1.2(a), we can write B = Ab1 + · · · + Abn , where bi ∈ B. Now consider the ideal Bm ⊆ B generated by m. Assume, if possible, that Bm = B. Then, by Nakayama’s lemma (see Exercise 2.7.3), there exists some f ∈ 1 + m such that fB = {0}. But we have 1 ∈ B and so f = 0, that is, 1 ∈ m, a contradiction. Thus, we must have Bm \u0002 B. So then there exists a maximal ideal q \u0002 B such that Bm ⊆ q. Then we have m ⊆ q ∩ A \u0002 A. Since m is \u0003 maximal, we deduce that m = q ∩ A, as required. 2.2.5 Proposition varieties.\n\nLet ϕ : X → Y be a finite morphism of affine\n\n(a) Let X \u0002 ⊆ X and Y \u0002 ⊆ Y be closed subsets such that ϕ(X \u0002 ) ⊆ Y \u0002 . Then the restricted morphism ϕ\u0002 : X \u0002 → Y \u0002 is also finite. (b) The map ϕ is closed (that is, maps closed subsets to closed subsets). In particular, if ϕ is dominant then ϕ is surjective. Proof We set A = A[X] and B = A[Y ]. (a) Let I : = IA (X \u0002 ) ⊆ A and J : = IB (Y \u0002 ). By Example 2.1.12, the algebras of regular functions on X \u0002 and Y \u0002 can be identified with A/I and B/J , respectively. Then ϕ\u0002 : X \u0002 → Y \u0002 corresponds to the induced algebra homomorphism (ϕ\u0002 )∗ : B/J → A/I. Now, let a ∈ A and consider an equation of integral dependence for a over ϕ∗ (B). Then we also get an equation of integral dependence for the image of a in A/I over (ϕ\u0002 )∗ (B/J ), as desired. (b) First we show that ϕ(X) ⊆ Y is closed. For q ∈ Y , consider the evaluation homomorphism εq : B → k and set Iq : = ker(εq ). We\n\nFinite morphisms and Chevalley’s theorem claim that\n\n115\n\nϕ(X) = {q ∈ Y | ker(ϕ∗ ) ⊆ Iq }.\n\nIndeed, first let p ∈ X and q = ϕ(p) ∈ Y . Then, for any g ∈ ker(ϕ∗ ), we have g(q) = g(ϕ(p)) = ϕ∗ (g)(p) = 0 and so ker(ϕ∗ ) ⊆ Iq . Conversely, let q ∈ Y be such that ker(ϕ∗ ) ⊆ Iq . Then ϕ∗ (Iq ) ⊆ ϕ∗ (B) is a maximal ideal. So, by Lemma 2.2.4 (‘going-up’), there exists a maximal ideal m ⊆ A such that m ∩ ϕ∗ (B) = ϕ∗ (Iq ). By Theorem 2.1.9, we have m = ker(εp ) for some p ∈ X. Then εp ◦ ϕ∗ = εq and so ϕ(p) = q. Thus, the claim is proved. Now note that, for any q ∈ Y , we have f(q) = εq (f) = 0 for all f ∈ ker(ϕ∗ ) if and only if ker(ϕ∗ ) ⊆ Iq . Thus, VY (ker(ϕ∗ )) = {q ∈ Y | ker(ϕ∗ ) ⊆ Iq } is a closed set. Now consider any closed subset Z ⊆ X. By (a), the restriction ϕ|Z : Z → Y is also finite, and so ϕ(Z) ⊆ Y is closed. \u0003 2.2.6 Lemma Let R and S be finitely generated k-algebras, where k is any field. Assume that R and S are integral domains and that R ⊆ S. Let K be the field of fractions of S and F ⊆ K be the field of fractions of R. Then K is a finitely generated field extension of F , and we set d : = ∂k (K) − ∂k (F ) \u0002 0. Then there exists some non-zero f ∈ R and s1 , . . . , sd ∈ S such that (a) Sf is integral over Rf [s1 , . . . , sd ], (b) s1 , . . . , sd ∈ S are algebraically independent over F . Here, we have Rf = {a/f n | a ∈ R, n \u0002 0} as in Exercise 1.8.11, where the quotients are taken in the field of fractions of R. Proof We consider the subring S ∗ : = {s/r ∈ K | s ∈ S, 0 = r ∈ R} ⊆ K. We have F ⊆ S ∗ ⊆ K, and S ∗ is a finitely generated F -algebra (since S is finitely generated over k); furthermore, K is also the field of fractions of S ∗ . So, using Lemma 1.2.18, we have ∂F (S ∗ ) = ∂F (K) = ∂k (K) − ∂k (F ) = d. Now, by the Noether normalization theorem, there exist y1 , . . . , yd ∈ S ∗ which are algebraically independent over F and such that S ∗ is integral over F [y1 , . . . , yd ]. We can write yi = si /r with si ∈ S and 0 = r ∈ R. Then s1 , . . . , sd are also algebraically independent over F , and so these elements satisfy the condition in (b). We now have a subring R[s1 , . . . , sd ] ⊆ S and, for any non-zero f ∈ R, we also have Rf [s1 , . . . , sd ] ⊆ Sf . We claim that there exists some f such that (a) holds. To see this, consider any s ∈ S. Then s is integral over F [y1 , . . . , yd ], and\n\n116\n\nAffine varieties and finite morphisms\n\nso we have sm + pm−1 sm−1 + · · · + p1 s + p0 = 0 for some m \u0002 1 and pi ∈ F [y1 , . . . , yd ]. Writing each coefficient of each pi as a quotient with denominator in R and using the expressions yi = si /r as above, there exists some 0 = g ∈ R such that gpi ∈ R[s1 , . . . , sd ] for all i. Then the above relation shows that s is integral over Rg [s1 , . . . , sd ]. We apply this construction to a finite set of k-algebra generators of S. For each of these generators, we obtain a corresponding element g. Taking the product of all these elements g, we see that there exists some non-zero f ∈ R such that every algebra generator of S is integral over Rf [s1 , . . . , sd ]. Thus, by Corollary 2.1.2, every element in S is integral over Rf [s1 , . . . , sd ]. Finally, this implies that also every element in Sf is integral over that subring. Indeed, let s/f n ∈ Sf and sm + pm−1 sm−1 + · · · + p1 s + p0 = 0 be an equation of integral dependence for s, where m \u0002 1 and pi ∈ Rf [s1 , . . . , sd ]. Dividing this equation by f nm , we obtain an equation of integral dependence for s/f n over Rf [s1 , . . . , sd ]. \u0003 We now translate the above result to the context of affine varieties. For this purpose, recall from §2.1.14 that, if (X, A) is an affine variety and 0 = f ∈ A, then we have an affine open subvariety (Xf , Af ), where Xf = {x ∈ X | f(x) = 0} and the embedding Xf ⊆ X is a morphism. 2.2.7 Theorem Let ϕ : X → Y be a dominant morphism of irreducible affine varieties. In particular, ϕ∗ : A[Y ] → A[X] is injective and d : = dim X − dim Y \u0002 0. Then we have a factorization ϕ ˜\n\npr\n\n1 ϕ|Xϕ∗ (g) : Xϕ∗ (g) −→ Yg × kd −→ Yg\n\nfor some 0 = g ∈ A[Y ],\n\nwhere ϕ˜ is a finite dominant morphism and pr1 is the first projection. Proof To simplify notation, let us write A = A[X] and B = A[Y ], and denote b∗ = ϕ∗ (b) for b ∈ B. Since X and Y are irreducible, A and B are integral domains and we have B∗ ⊆ A. Let K be the field of fractions of A, and F ⊆ K be the field of fractions of B∗ . Since K is finitely generated over F and ϕ∗ is injective, we have ∂k (K) − ∂k (F ) = ∂k (A) − ∂k (B) = dim X − dim Y = d. We now apply Lemma 2.2.6 to the rings R = B∗ and S = A. So there exist elements a1 , . . . , ad ∈ A and a non-zero element g ∈ B\n\nFinite morphisms and Chevalley’s theorem\n\n117\n\nsuch that (a) Ag∗ is integral over Bg∗∗ [a1 , . . . , ad ], (b) a1 , . . . , ad ∈ A are algebraically independent over F . Now consider the homomorphism ϕ∗ : B → B∗ ⊆ A. We can canonically extend it to a ring homomorphism ψ : Bg → Bg∗∗ ⊆ Ag∗ , by setting ψ(b/gm ) = b∗ /(g∗ )m for any b ∈ B and m \u0002 0. (Check this!) Now, denoting by T1 , . . . , Td independent indeterminates over Bg , we can extend ψ to a unique ring homomorphism ψ˜ : Bg [T1 , . . . , Td ] → Bg∗∗ [a1 , . . . , ad ] ⊆ Ag∗ such that Ti → ai . Since a1 , . . . , ad are algebraically independent, ψ˜ is an isomorphism onto its image. Now Bg and Ag∗ can be identified with the algebras of regular functions on Yg and Xg∗ , respectively; see §2.1.14. Furthermore, Bg [T1 , . . . , Td ] can be identified with the algebra of regular functions on Yg × kd ; see §2.1.13. Then ψ˜ is the algebra homomorphism corresponding to a dominant regular map ϕ˜ : Xg∗ → Yg × kd ; furthermore, by (a), we have that ϕ˜ is a finite morphism. Finally, the embedding Bg ⊆ Bg [T1 , . . . , Td ] corresponds to the projection map pr1 : Yg × kd → Yg , and the composition of Bg ⊆ Bg [T1 , . . . , Td ] with ψ˜ is nothing but the canonical extension of ϕ∗ to the homomorphism Bg → Ag∗ , b/gm → b∗ /(g∗ )m . The latter homomorphism corresponds to the restriction of ϕ to Xg∗ ; see once more §2.1.14. Thus, we have a factorization of that restriction as desired. \u0003 2.2.8 Corollary Let ϕ : X → Y be a dominant morphism of irreducible affine varieties. Then the image of any non-empty open subset U ⊆ X contains a non-empty open subset of Y . Proof First we show that ϕ(X) itself contains some non-empty open subset of Y . For this purpose, consider a factorization of ϕ as in Theorem 2.2.7. Since ϕ˜ is a finite morphism, it is surjective by Proposition 2.2.5. Hence, since pr1 also is surjective, we have ϕ(Xϕ∗ (g) ) = Yg . Thus, ϕ(X) contains the open set Yg . Now let U ⊆ X be any non-empty open subset. Since, by §1.1.14, U is a finite union of affine open sets, it is enough to prove the assertion in the case where U = Xf for some non-zero f ∈ A[X]. Then the inclusion Xf → X corresponds to the embedding A[X] → A[X]f . Consequently, the restriction ϕ|Xf : Xf → Y is still a dominant morphism between irreducible affine varieties, and so ϕ(Xf ) contains a non-empty open subset of Y . \u0003\n\n118\n\nAffine varieties and finite morphisms\n\n2.2.9 Corollary Let ϕ : X → Y be a dominant morphism of irreducible affine varieties. Then there exists a non-empty open subset U ⊆ Y with U ⊆ ϕ(X) and such that dim ϕ−1 (y) = dim X − dim Y for all y ∈ U . In particular, if |ϕ−1 (ϕ(x))| < ∞ for all x ∈ X, then dim X = dim Y . Proof We set d : = dim X − dim Y . Consider a factorization of ϕ as in Theorem 2.2.7 and set U : = Yg . Now let y ∈ Yg . Since ϕ−1 (Yg ) = Xϕ∗ (g) , we also have ϕ−1 (y) = ϕ˜−1 ({y} × kd ). Now {y} × kd is irreducible of dimension d; see Example 1.2.16(a). Let Z1 , . . . , Zm be the irreducible components of ϕ−1 (y). Then each restriction ϕ| ˜ Zi : Zi → {y} × kd is also finite by Proposition 2.2.5(a), and so we have dim Zi \u0001 d (since A(Zi ) is algebraic over the field of fractions of the image of A[{y} × kd ] in A[Zi ]). On the other hand, ϕ(Z ˜ i ) ⊆ {y} × kd is closed and irreducible for all i. Thus, since {y} × kd is irreducible, there exists some i0 such ϕ(Z ˜ i0 ) = {y} × kd , and so we have dim Zi0 = d. \u0003 2.2.10 Example Consider the regular map ϕ : k2 → k2 , (x, y) → (xy, y), introduced in Example 1.3.5(a). We have already seen that ϕ(k2 ) = {(0, 0)} ∪ {(x, y) ∈ k2 | y = 0}. Thus, ϕ(k2 ) is dense in k2 , and so ϕ is dominant but not surjective. In particular, ϕ cannot be a finite morphism; see Proposition 2.2.5. We have ϕ−1 ({0, 0}) = {(x, 0) | x ∈ k}, and this is 1-dimensional. On the other hand, for any (x, y) ∈ k2 with y = 0, we have ϕ−1 (x, y) = {(x/y, y)}, and this is 0-dimensional. Furthermore, the set of all (x, y) ∈ k2 with y = 0 is open in k2 . To state the next result, we need the following definition. Let Z be any topological space. A subset Y ⊆ Z is called locally closed if Y is the intersection of an open subset and a closed subset of Z. (For example, all open subsets are locally closed, and so are all closed subsets.) We say that Y ⊆ Z is constructible if Y is a finite union of locally closed subsets. For some basic properties of constructible subsets, see Exercise 2.7.6. 2.2.11 Theorem (Chevalley) Let X and Y be affine varieties (not necessarily irreducible) and ϕ : X → Y be a morphism (not necessarily dominant). Then the image of any constructible subset of X is a constructible subset of Y .\n\nFinite morphisms and Chevalley’s theorem\n\n119\n\nProof We proceed in two steps. Step 1. First assume that X is irreducible. Then we claim that ϕ(X) ⊆ Y is constructible. We prove this by induction on dim Y . If dim Y = 0, then Y is a finite set and so every subset of Y is constructible. Now assume that dim Y > 0. Since X is irreducible, Y \u0002 = ϕ(X) is a closed irreducible subset of Y and we may regard ϕ as a dominant morphism ϕ : X → Y \u0002 . Furthermore, if ϕ(X) is constructible in Y \u0002 then it is also constructible in Y . Thus, we may assume without loss of generality that Y \u0002 = Y . Applying Corollary 2.2.8, we obtain a non-empty open subset U ⊆ Y such that U ⊆ ϕ(X). Let Z \u0002 be an irreducible component of Y \\ U and X \u0002 be an irreducible component of ϕ−1 (Z \u0002 ). Restricting ϕ to X \u0002 , we obtain a morphism ϕ\u0002 : X \u0002 → Z \u0002 . Since U = ∅, we have dim Z \u0002 < dim Y by Proposition 1.2.20. So, by induction, ϕ(X \u0002 ) is constructible in Z \u0002 and, hence, also in Y . We conclude that U \u0002 : = ϕ(ϕ−1 (Y \\ U )) is constructible in Y . Consequently, ϕ(X) = U ∪ U \u0002 is constructible in Y , as required. Step 2. Now let X be arbitrary and \u0007 Z ⊆ X be any constructible subset. By definition, we can write Z = ri=1 (Oi ∩Ci ), where Oi ⊆ X is open and Ci ⊆ X is closed. Now each closed set in X can be written as a finite union of irreducible closed sets, and each open set in X can be written as a finite union of principal open sets; see §2.1.14. Thus, we may assume without loss of generality that each Ci is irreducible and each Oi is a principal open set. Then we have \u0007 ϕ(Z) = i ϕ(Oi ∩ Ci ), and so it is enough to consider the restriction of ϕ to Oi ∩ Ci . Thus, we may assume without loss of generality that X is irreducible and Z ⊆ X is a principal open subset, that is, we have Z = Xf for some non-zero f ∈ A[X]. Since Xf also is an irreducible affine variety and the inclusion Xf ⊆ X is a morphism, we can apply Step 1 to the restriction ϕ|Xf : Xf → Y and conclude that its image is constructible, as desired. \u0003 In the case of homomorphisms between algebraic groups, the above results take a particularly simple form. Let us first give an ‘abstract’ definition of algebraic groups, in the spirit of Definition 2.1.6. 2.2.12 Affine algebraic groups Let (G, A) be an affine variety and µ : G × G → G be a morphism. We say that G is an affine algebraic monoid if the conditions in Definition 1.3.9 are satisfied, that is, µ is associative and there exists an identity element\n\n120\n\nAffine varieties and finite morphisms\n\n1 ∈ G. If, moreover, every element g ∈ G has an inverse and the map ι : G → G, g → g−1 , is a morphism, then G is called an affine algebraic group. A homomorphism of affine algebraic groups is a group homomorphism which is also a morphism of the underlying affine varieties. By Example 2.1.8 and the characterization of morphisms in Proposition 1.3.4, it is clear that every linear algebraic group in the sense of Definition 1.3.9 also is an affine algebraic group in the above abstract sense. (The converse will be proved in Corollary 2.4.4.) We remark that the statement of Proposition 1.3.13 remains true, with the same proof, word by word. Thus, given an affine algebraic group G, there is a unique irreducible component G◦ containing the identity element of G. Furthermore, G◦ is a closed normal subgroup whose cosets are the irreducible components of G. In particular, we have dim G = dim G◦ . 2.2.13 Lemma Let G be an affine algebraic group. If H ⊆ G is a subgroup (not necessarily closed), then H ⊆ G also is a subgroup. If H is any constructible subset of G such that H is a subgroup, then H = H · H. In particular, if H is a subgroup which is a constructible subset of G, then H is closed. Proof Let H ⊆ G be a subgroup. Since inversion is a homeomorph−1 ism (with respect to the Zariski topology), we have H = H −1 = H. Similarly, left multiplication by any x ∈ H is a homeomorphism, and so xH = xH = H, that is, we have H · H ⊆ H. Consequently, if x ∈ H, we have Hx ⊆ H and so Hx = Hx ⊆ H. Thus, we see that H is a subgroup. Now, if H is any constructible subset, there exists some dense open set U ⊆ H such that U ⊆ H. Now assume also that H is a subgroup. Since inversion is a homeomorphism, U −1 is also dense and open in H. Since left multiplication by any h ∈ H is a homeomorphism, hU −1 is also dense and open in H. Thus, we must have U ∩hU −1 = ∅ and so h ∈ U ·U . Hence, H = U ·U ⊆ H ·H = H. \u0003 2.2.14 Proposition algebraic groups.\n\n˜ → G be a homomorphism of affine Let ϕ : G\n\n˜ and the image ϕ(G) ˜ ⊆ G are closed (a) The kernel ker(ϕ) ⊆ G ˜ = dim ϕ(G) ˜ + dim ker(ϕ). subgroups. We have dim G ◦ ◦ ˜ ˜ (b) We have ϕ(G ) = ϕ(G) .\n\nBirational equivalences and normal varieties\n\n121\n\nProof First note that ker(ϕ) = ϕ−1 (1) is a closed normal subgroup ˜ is a subgroup. The fact that ϕ(G) ˜ is closed follows and that ϕ(G) from Lemma 2.2.13 and Chevalley’s Theorem 2.2.11. This proves (a) except for the statement concerning the dimensions. Next, con˜ ◦ ) ⊆ ϕ(G) ˜ ◦ . On the other sider (b). By Remark 1.3.2, we have ϕ(G ◦ ˜ ) is a closed subgroup hand, by the part of (a) already shown, ϕ(G ˜ and so it contains ϕ(G) ˜ ◦. of finite index in ϕ(G) It remains to prove the statement in (a) concerning the dimensions. For this purpose, we may now assume that ϕ is surjective. ˜ ◦ ) = G◦ . Denote by ϕ◦ : G ˜ ◦ → G◦ Then, by (b), we also have ϕ(G ◦ the restriction of ϕ. Then ϕ is a surjective morphism between irreducible affine varieties. The preimage of any g ∈ G◦ is a coset of ker(ϕ◦ ). Thus, all these preimages have the same dimension and ˜ ◦ = dim ker(ϕ◦ ) + dim G◦ , by Corollary 2.2.9. But all so dim G ˜ have the same dimension, irreducible components of G (or of G) ◦ ˜ ˜ ◦ . Finally, ker(ϕ◦ ) = and so dim G = dim G and dim G = dim G ◦ ˜ ker(ϕ) ∩ G is a closed subgroup of finite index in ker(ϕ) and so contains ker(ϕ)◦ . Hence dim ker(ϕ◦ ) = dim ker(ϕ). \u0003\n\n2.3 Birational equivalences and normal varieties In this section, we study bijective morphisms between irreducible affine varieties. We have already encountered examples which show that such a morphism need not be an isomorphism. We would like to know: In addition to being bijective, what is required for a morphism to be an isomorphism? We shall see that there is a satisfactory answer to this question. This will involve two basic notions: birational equivalences and normal varieties. Again, finite morphisms will play a key role. Let us first consider the following example. 2.3.1 Example Let C : = {(x, y) ∈ k2 | x3 = y2 }. Then C ⊆ k2 is irreducible and we have a bijective morphism ϕ : k → C, t → (t2 , t3 ). (Check this!) We have ϕ∗ (A[C]) = k[T 2 , T 3 ] \u0002 k[T ], and so ϕ is not an isomorphism. On the other hand, note the following three facts: (1) The field of fractions of k[T 2 , T 3 ] certainly equals k(T ). Thus ϕ at least induces an isomorphism on the level of fields of fractions. (2) It is not difficult to check that I(C) = (X 3 − Y 2 ) ⊆ k[X, Y ]. This yields T(0,0) (C) = k2 and T(x,y) (C) ∼ = k for all (x, y) ∈ C \\ {(0, 0)}. Thus,\n\n122\n\nAffine varieties and finite morphisms\n\n(0, 0) is the unique singular point of C. (3) The variable T is integral over ϕ∗ (A[C]) and so A[C] is not integrally closed in its field of fractions. We shall see that the failure of being an isomorphism is related to facts (2) and (3). 2.3.2 Definition Let X and Y be irreducible affine varieties and ϕ : X → Y be a dominant morphism. Since A[X] and A[Y ] are integral domains, we have corresponding fields of fractions denoted by A(X) and A(Y ), respectively. Then ϕ∗ induces a k-algebra homomorphism A(Y ) → A(X) which we denote by the same symbol. We say that ϕ is a birational equivalence if ϕ∗ induces a field isomorphism A(Y ) ∼ = A(X). Clearly, if ϕ : X → Y is an isomorphism, then ϕ is a birational equivalence. The above example shows that the converse need not be true. In Theorem 2.3.10 below, we shall establish a differential criterion for birational equivalence. 2.3.3 Definition Let X be an irreducible affine variety. We say that X is normal if A[X] is integrally closed in its field of fractions. In Proposition 2.3.11, we will see that there always exists an affine open subvariety in X which is normal. Furthermore, in Example 2.3.14 below, we will see that all connected affine algebraic groups are normal. 2.3.4 Remark Let us give an illustration of how the above notions can be combined to yield a criterion for a morphism to be an isomorphism. So, let ϕ : X → Y be a morphism between irreducible affine varieties and assume that (a) ϕ is a finite morphism; (b) ϕ is a birational equivalence; (c) Y is normal. Then ϕ is an isomorphism. Indeed, by (a), ϕ∗ (A[Y ]) ⊆ A[X] is an integral extension. By (b), ϕ∗ (A[Y ]) and A[X] have the same field of fractions. Hence, by (c), we must have ϕ∗ (A[Y ]) = A[X]. Thus, ϕ is an isomorphism as claimed. This is the simplest possible case of a much deeper result: Zariski’s main theorem. In one formulation, this asserts that condition (a) can be replaced by the substantially weaker condition\n\nBirational equivalences and normal varieties\n\n123\n\nthat ϕ is surjective and all fibres of ϕ are finite —but we will not prove this here. See §5 of Dieudonn´e(1974), §5.2 of Springer (1998), and §III.9 of Mumford (1988), for further details. In any case, as far as applications are concerned, the trouble is that the assumptions (b) and (c) are rather strong and may be difficult to verify. Our next aim is to develop critera for checking these conditions. This will involve some properties of the local ring of an affine variety at a point. 2.3.5 The local ring of a point Let X be an irreducible affine variety over k, and fix a point p ∈ X. We consider the evaluation homomorphism εp : A[X] → k, f → f(p). Since X is irreducible, A[X] is an integral domain. Let A(X) be the field of fractions of A[X]. Then it is easily checked that Op : = {x ∈ A(X) | x = a/b with a, b ∈ A[X] and εp (b) = 0} ⊆ A(X) is a subring of A(X) containing A[X]. We can extend εp canonically to a ring homomorphism ε˜p : Op → k by setting ε˜p (x) : = εp (a)/εp (b) for x = a/b ∈ Op . The ring Op is local, that is, it has a unique maximal ideal which is given by εp ) ⊂ Op . mp : = ker(˜ (Indeed, if x ∈ Op is such that ε˜p (x) = 0, then, writing x = a/b, we see that εp (a) = 0 and εp (b) = 0 and so x−1 ∈ Op . Thus, every non-invertible element of Op lies in mp .) We call Op the local ring of the point p ∈ X. 2.3.6 Lemma\n\nIn the above set-up, the following hold. \b (a) We have A[X] = p∈X Op , where the intersection is taken inside A(X). (b) The ring Op is noetherian. \b Proof (a) Let f ∈ p∈X Op and consider the ideal I = {a ∈ A[X] | af ∈ A[X]}. Suppose that I = A[X]. Then there exists some p ∈ X such that I ⊆ ker(εp ). On the other hand, since f ∈ Op , there exists some b ∈ I such that εp (b) = 0, a contradiction. So we have 1 ∈ I and hence f ∈ A[X].\n\n124\n\nAffine varieties and finite morphisms\n\n(b) Let I ⊆ Op be an ideal and set J : = I ∩ A[X]. Assume that J is generated by a1 , . . . , ad ∈ A[X] as an ideal in A[X]. Let f ∈ I and g ∈ A[X] be such that\u0003g(p) = 0 and gf ∈ A[X]. Then gf ∈ A[X] ∩ I = J and so gf = ri=1 ri ai with ri ∈ A[X]. Consequently, I is generated by a1 , . . . , ad as an ideal in Op . \u0003 2.3.7 Lemma Let ϕ : X → Y be a surjective morphism of irreducible affine varieties such that ϕ∗ (Oϕ(p) ) = Op for all p ∈ X. Then ϕ is an isomorphism. Proof Since ϕ is dominant, the map ϕ∗ : A[Y ] → A[X] is injective. We must show that it is also surjective. So let f ∈ A[X] and consider the ideal I = {g ∈ A[Y ] | ϕ∗ (g)f ∈ ϕ∗ (A[Y ])}. Assume, if possible, that I = A[Y ]. Then there exists some q ∈ Y such that I ⊆ ker(εq ). Since ϕ is surjective, there exists some p ∈ X such that ϕ(p) = q. Now, by assumption, we have f ∈ Op = ϕ∗ (Oq ). So there exist some g, h ∈ A[Y ] such that h(q) = 0 and ϕ∗ (g/h) = f. This implies that ϕ∗ (h)f ∈ ϕ∗ (A[Y ]) and so h ∈ I ⊆ ker(εq ), a contradiction. Thus, I = A[Y ] and so f ∈ ϕ∗ (A[Y ]). \u0003 Now we can give a new characterization of tangent spaces and non-singular points, as promised in Section 1.4. In the following statement, note that mp /m2p is naturally a vector space over k, via the identification k = Op /mp . (See also Exercise 2.7.4.) 2.3.8 Proposition p ∈ X.\n\nLet X be an irreducible affine variety and\n\n(a) There is a well-defined k-linear isomorphism δp : Homk (mp /m2p , k) → Derk (A[X], kp ),\n\nµ → Dµ ,\n\nsuch that Dµ (a) = µ(a + m2p ) for all a ∈ A[X] with a(p) = 0. (b) p ∈ X is non-singular if and only if mp is generated by dim X elements. Proof (a) Let µ : mp /m2p → k be k-linear and define a map Dµ : A[X] → k by Dµ (a) = µ((a − a(p)) + m2p ) for a ∈ A[X]. We check that Dµ ∈ Derk (A[X], kp ). We have (a − a(p))(b − b(p)) ∈ m2p\n\nBirational equivalences and normal varieties\n\n125\n\nand so ab − a(p)b(p) ≡ a(p)(b − b(p)) + b(p)(a − a(p)) mod m2p . Applying µ to this equation yields that Dµ is a derivation as required. Thus, we obtain a k-linear map δp : Homk (mp /m2p , k) → Derk (A[X], kp ), where δp (µ) = Dµ . Since µ is k-linear, the map Dµ is uniquely determined by its values on functions a ∈ A[X] such that a(p) = 0. To show that δp is an isomorphism, we construct an inverse map. To do this, first note that we can extend uniquely any ˜ ∈ Derk (Op , kp ), by setting D ∈ Derk (A[X], kp ) to a derivation D ˜ D(f) =\n\n1 (a(p)D(b) − b(p)D(a)) b(p)2\n\nfor f = a/b ∈ Op .\n\n(This is just analogous to the rule for the derivative of the quotient of ˜ ˜ ˜ two functions.) Then we have D(ab) = a(p)D(b) + b(p)D(a) = 0 for all a, b ∈ mp , and so D induces a k-linear function λD : mp /m2p → k, ˜ Thus, we obtain a k-linear map a + m2p → D(a). λp : Derk (A[X], kp ) → Homk (mp /m2p , k),\n\nD → λD .\n\nIt is readily checked that λp and δp are inverse to each other. (b) By Remark 1.4.9, we can identify Tp (X) = Derk (A[X], kp ). Thus, by (a), we have dimk Tp (X) = dimk (mp /m2p ). Hence the assertion follows by combining this with Exercise 2.7.4 and Theorem 1.4.11. \u0003 2.3.9 Remark The isomorphism in Proposition 2.3.8(a) is functorial, in the following sense. Let ϕ : X → Y be a dominant morphism between irreducible affine varieties. Let p ∈ X and q : = ϕ(p) ∈ Y . Then the corresponding map ϕ∗ : A[Y ] → A[X] induces a homomorphism A(Y ) → A(X) which we denote by the same symbol. With this convention, we can apply ϕ∗ to Oq ⊂ A(Y ) and obtain a subring ϕ∗ (Oq ) ⊆ A(X). It is readily checked that ϕ∗ (Oq ) ⊆ Op ⊆ A(X). Consequently, we also have ϕ∗ (mq ) ⊆ mp , and so ϕ induces a k-linear map ϕ¯∗ : mq /m2q → mp /m2p ,\n\na + m2q → ϕ∗ (a) + m2p .\n\n126\n\nAffine varieties and finite morphisms\n\nThen we get an induced k-linear map on the dual spaces, and this makes the following diagram commutative: µ → µ ◦ ϕ¯∗ - Hom (m /m2 , k) Hom (m /m2 , k) k\n\np\n\nq\n\nk\n\np\n\nδp\n\nq\n\nδq . ?\n\nDerk (A[X], kp )\n\ndp ϕ\n\n-\n\n?\n\nDerk (A[Y ], kq ).\n\nWith the above preparations at hand, we can now establish the following basic result which provides a criterion for checking condition (b) in Remark 2.3.4. 2.3.10 Theorem (differential criterion for birational equivalence) Let X and Y be irreducible affine varieties. Assume that ϕ : X → Y is a dominant injective morphism such that dp ϕ : Tp (X) → Tϕ(p) (Y )is surjective for some non-singular p ∈ X. (∗) Then ϕ is a birational equivalence. More precisely, there exists some x ∈ X such that ϕ∗ (Oϕ(x) ) = Ox , where ϕ∗ also denotes the induced map A(Y ) → A(X). Proof First note that, since ϕ is injective, we have dim X = dim Y ; see Corollary 2.2.9. Hence, (∗) implies that dim Tϕ(p) (Y ) \u0001 dim Tp (X) = dim X = dim Y . Thus, by Theorem 1.4.11, ϕ(p) ∈ Y is non-singular and so dp ϕ is an isomorphism. We now proceed in two steps. Step 1. First we prove the desired assertion under the additional assumption that ϕ is a finite morphism. As in Remark 2.3.9, ϕ∗ induces a homomorphism A(Y ) → A(X) which we denote by the same symbol. We want to show that this induced map is an isomorphism. For this purpose, consider the local rings Op ⊂ A(X) and Oϕ(p) ⊂ A(Y ). As in Remark 2.3.9, we have ϕ∗ (Oϕ(p) ) ⊆ Op . Then it will be enough to prove that we have in fact ϕ∗ (Oq ) = Op ,\n\nwhere q : = ϕ(p).\n\n(†)\n\nTo prove this, we consider Op as a ϕ∗ (Oq )-module. By the discussion in Remark 2.3.9, we conclude that the k-linear map ϕ¯∗ : mq /m2q → mp /m2p ,\n\na + m2q → ϕ∗ (a) + m2p ,\n\nBirational equivalences and normal varieties\n\n127\n\nis an isomorphism. Thus, if mq is generated by a1 , . . . , ad , then mp /m2p is spanned by the images of ϕ∗ (a1 ), . . . , ϕ∗ (ad ). So Exercise 2.7.4 shows that mp is generated by ϕ∗ (a1 ), . . . , ϕ∗ (ad ), that is, we have mp = ϕ∗ (mq )Op . Since, clearly, we have Op = mp +ϕ∗ (Oq ) (note that 1 ∈ mp ), we can now deduce that Op = ϕ∗ (mq )Op + ϕ∗ (Oq ). Hence (†) would follow if we could apply Nakayama’s lemma (see Exercise 2.7.3). Now, by Lemma 2.3.6, ϕ∗ (Oq ) is noetherian. So it will be enough to show that Op is contained in a submodule of a finitely generated ϕ∗ (Oq )-module (see Exercise 1.8.1). To prove this, let x ∈ Op and write x = a/b, where a, b ∈ A[X] and b(p) = 0. Then consider the set Y \u0002 : = ϕ(VX ({b})) ⊆ Y . Since ϕ is a finite morphism, Y \u0002 is a closed subset of Y . Furthermore, we necessarily have q ∈ Y \u0002 . (Indeed, if q ∈ Y \u0002 , then q = ϕ(v) for some v ∈ X such that b(v) = 0. Since ϕ is injective, this would imply p = v and so b(p) = 0, a contradiction.) So there exists some g ∈ A[Y ] such that g(q) = 0 and g(y) = 0 for all y ∈ Y \u0002 . Then ϕ∗ (g)(x) =\u0006g(ϕ(x)) = 0 for all x ∈ VX ({b}), and so ϕ∗ (g) ∈ IA[X] (VX ({b})) = (b), where the last equality holds by Hilbert’s nullstellensatz. Hence we can write ϕ∗ (gm ) = ϕ∗ (g)m = cb for some m \u0002 1 and some c ∈ A[X]. It follows that x = (ca)/(cb) = ca/ϕ∗ (gm ) ∈ A[X]ϕ∗ (Oq ). Thus, we have shown that Op ⊆ A[X]ϕ∗ (Oq ). Now, since ϕ is a finite morphism, we have that ϕ∗ (A[Y ]) ⊆ A[X] is an integral extension. So, since A[X] is a finitely generated k-algebra, there exist u1 , . . . , ul ∈ A[X] which generate A[X] as a ϕ∗ (A[Y ])-module. Then we also have Op ⊆\n\nl \u0001\n\nϕ∗ (Oq )ui .\n\ni=1\n\nThus, Op is a submodule of a finitely generated ϕ∗ (Oq )-module, as required. Step 2. Now we show that the general case can be reduced to the special case considered in Step 1. So let ϕ : X → Y be a dominant injective morphism between irreducible affine varieties and assume that (∗) holds. Now, as in the proof of Proposition 1.4.15, one shows that there is an open subset U ⊆ X containing p such that the above conditions are satisfied for all x ∈ U , that is, x is non-singular, ϕ(x) is non-singular, and dx ϕ is bijective.\n\n128\n\nAffine varieties and finite morphisms\n\nWe can now proceed as follows. Since dim X = dim Y , there exists some non-zero g ∈ A[Y ] such that ϕ restricts to a finite bijective morphism ϕg : Xϕ∗ (g) → Yg ; see Theorem 2.2.7. Now let us choose x ∈ U ∩Xϕ∗ (g) . By Lemma 1.1.14, we may identify Tx (Xϕ∗ (g) ) with Tx (X) and Tϕ(x) (Yg ) with Tϕ(x) (Y ). Then (∗) also holds for the restricted morphism ϕg . So, by Step 1, ϕg is a birational equivalence. It remains to note that the algebra homomorphism ϕ∗g is the natural map A[Y ]g → A[X]ϕ∗ (g) induced by ϕ∗ . Hence ϕ is a birational equivalence. \u0003 2.3.11 Proposition Let X be an irreducible affine variety. Then there is a non-empty open set U ⊆ X such that Op is integrally closed in A(X) for all p ∈ U . Proof Let K be the field of fractions of A[X]. We begin by showing that there exists some 0 = f ∈ A[X] and a finitely generated k-subalgebra B ⊆ K such that the following conditions hold. (a) B is integrally closed in its field of fractions F ⊆ K. (b) We have B ⊆ A[X]f , and A[X]f is integral over B. (c) K is a finite separable extension of F . This is proved as follows. By Exercise 1.8.15, there exist algebraically independent elements z1 , . . . , zd ∈ K such that K is a finite separable algebraic extension of F : = k(z1 , . . . , zd ). We will construct B as a certain localization of the ring R : = k[z1 , . . . , zd ]. Since K is the field of fractions of A[X], we can write zi = ai /h, where 0 = h ∈ A[X] and ai ∈ A[X] for 1 \u0001 i \u0001 d. Thus, we have R ⊆ A[X]h . We claim that there exists some non-zero g ∈ R such that every element in A[X]h is integral over Rg . To see this, let a ∈ A[X]h . Then a is algebraic over F . Since F is the field of fractions of R, this implies that a is integral over Rs for some non-zero s ∈ R. Applying this to a finite set of algebra generators of A[X]h , we see that there exists some non-zero g ∈ R such that all elements of A[X]h are integral over Rg , as claimed. Then we have Rg ⊆ A[X]gh , and A[X]gh will be integral over Rg (see the argument in the proof of Lemma 2.2.6). Now set B : = Rg and f : = gh. Then (b) holds. By construction, (c) also holds. Finally, since R is a polynomial ring, R is integrally closed in F . Then Rg will also be integrally closed in F (see Exercise 2.7.7) and so (a) holds.\n\nBirational equivalences and normal varieties\n\n129\n\nNow let S be the integral closure of B in K. We claim that K is the field of fractions of S; more precisely: there exists an F -basis {u1 , . . . , un } of K such that all ui lie in S. (1) Indeed, let u ∈ K. Then u is algebraic over F and, hence, satisfies an equation of the form bm um + bm−1 um−1 + · · · + b0 = 0, where m−1 , m \u0002 1, bi ∈ B, and bm = 0. Multiplying this equation by bm we see that bm u is integral over B. Thus, given any F -basis of K, we may multiply the basis elements by suitable elements of B to get an F -basis consisting of elements in S. Next we claim that there exist z1 , . . . , zn in K such that S ⊆ Bz1 + · · · + Bzn .\n\n(2)\n\nTo prove this, we use the trace map TK/F introduced in Exercise 2.7.5. Since K ⊇ F is separable, the symmetric F -bilinear form K × K → F , (x, y) → TK/F (xy), is non-degenerate; see Exercise 2.7.5(d). Let {z1 , . . . , zn } be the F -basis of K dual to an F -basis {u1 , . . . , un } ⊆ S as in (1), that is, we have TK/F (z\u0003 j ui ) = δij n for all i, j. Now let s ∈ S. Then we can write s = j=1 xj zj with xj ∈ F . Since ui ∈ S, we have sui ∈ S for all i and so TK/F (sui ) ∈ B, see Exercise 2.7.5(c). It follows that TK/F (sui ) = \u0003 n j=1 xj TK/F (zj ui ) = xi ∈ B. Thus, every element of S is contained Bz1 + · · · + Bzn , as claimed. Now, by (2), S is contained in a finitely generated B-module. Since B is noetherian, S is finitely generated as a B-module (see Exercise 1.8.1) and so S is a finitely generated k-algebra.\n\n(3)\n\nLet us write S = k[s1 , . . . , sl ], where si ∈ S. Since S ⊆ K and K is the field of fractions of A[X], there exists some non-zero c ∈ A[X] and ai ∈ A[X] such that si = ai /c for 1 \u0001 i \u0001 l. Thus, we have S ⊆ A[X]c . We claim that this implies that Op is integrally closed in K for all p ∈ U : = Xc . Indeed, let x ∈ K be integral over Op where p ∈ U . Then we have an equation of the form xm + (am−1 /bm−1 )xm−1 + · · · + (a1 /b1 )x + a0 /b0 = 0, where m \u0002 1 and ai , bi ∈ A[X] are such that bi (p) = 0. Now set d : = b0 b1 · · · bm−1 and multiply the above equation by dm . Then one obtains an equation of integral dependence of dx over A[X].\n\n130\n\nAffine varieties and finite morphisms\n\nHence dx is integral over A[X]f ⊇ A[X] and, hence, also over B. Thus, we have dx ∈ S ⊆ A[X]c and so we can write dx = a/cl , where a ∈ A[X] and l \u0002 0. This yields that x = a/(dcl ) ∈ Op , as required. \u0003 2.3.12 Remark Let X be an irreducible affine variety. Then one can show that the following implication holds: p ∈ X is non-singular\n\nOp is a factorial ring;\n\nin particular, if p ∈ X is non-singular, then Op is integrally closed in A(X). Here, we will not give the proof of this result; for further details see §III.7 of Mumford (1988) and §II.3 of Shafarevich (1994). The above Proposition 2.3.11 in combination with Theorem 1.4.11 shows that, at least, there exist non-singular points p ∈ X such that Op is integrally closed in A(X). As in the previous section, one may expect that the above results admit a simpler formulation when we consider algebraic groups. This is indeed the case. 2.3.13 Algebraic group actions Let G be an affine algebraic group and X be an affine variety. We say that X is a G-variety if there exists a morphism of affine varieties α : G × X → X,\n\n(g, x) → g.x,\n\nwhich defines an abstract operation of G on the set X; that is, we have 1.x = x and (gh).x = g.(h.x) for all g, h ∈ G and x ∈ X. We say that X is a homogeneous G-variety if the action of G on X is transitive. Note the following simple facts. For each g ∈ G, the map πg : X → X, x → g.x, is obtained by composing α with the morphism X → G × X, x → (g, x). Hence πg is a morphism. Furthermore, we have π1 = idX and πg ◦ πh = πgh for all g, h ∈ G. Consequently, each πg is an isomorphism of affine varieties of X, with inverse πg−1 . 2.3.14 Example Let X be an irreducible affine variety which is a homogeneous G-variety for some affine algebraic group G. Then X is normal. Indeed, let K be the field of fractions of A[X]. By Proposition 2.3.11, there exists some p ∈ X such that Op is integrally closed in K. Now let q ∈ X and choose g ∈ G such that p = g.q. Then\n\nBirational equivalences and normal varieties\n\n131\n\nconsider the isomorphism πg : X → X, x → g.x. The corresponding algebra homomorphism πg∗ : A[X] → A[X] also is an isomorphism, and this induces a field automorphism of K. It follows that Oq = πg∗ (Og.q ) will also be integrally closed in K. Since this holds for all q ∈ X, we see that all local rings are integrally closed in K. Hence X is normal (see Exercise 2.7.7). The above discussion applies, in particular, to the case where G is a connected affine algebraic group. Then we may regard G as a homogeneous G-variety via left multiplication. Hence, G is normal. 2.3.15 Proposition Let G be an affine algebraic group and X, Y be irreducible affine varieties which are homogeneous G-varieties. Let ϕ : X → Y be a G-equivariant bijective morphism such that dp ϕ : Tp (X) → Tϕ(p) (Y ) is surjective for some non-singular p ∈ X. Then ϕ is an isomorphism. Proof By Theorem 2.3.10, there exists some x ∈ X such that ϕ∗ (Oϕ(x) ) = Ox . Since ϕ is G-equivariant, this implies that ϕ∗ (Oϕ(g−1 .x) ) = ϕ∗ (Og−1 .ϕ(x) ) = ϕ∗ (πg∗ (Oϕ(x) )) = πg∗ (ϕ∗ (Oϕ(x) )) = πg∗ (Ox ) = Og−1 .x\n\nfor all g ∈ G.\n\nThus, since X is a homogeneous G-variety, we have ϕ∗ (Oϕ(x\u0002 ) ) = Ox\u0002 for all x\u0002 ∈ X. It remains to apply Lemma 2.3.7. \u0003 ˜ → G be a bijective homomorph2.3.16 Example Let ϕ : G ism between connected affine algebraic groups. Assume that ˜ → T1 (G) is surjective. Then ϕ is an isomorphism. d1 ϕ : T1 (G) ˜ ˜ ˜ acts on itself Indeed, G and G are homogeneous G-varieties, where G ˜ by left multiplication and on G via G×G → G, (˜ g, h) → ϕ(˜ g)h. Then ϕ is G-equivariant and the assertion follows from Theorem 2.3.15. We remark that, if k has characteristic 0, then it can be shown that the assumption on d1 ϕ is in fact unnecessary; see Corollary 11.1.3 of Goodman and Wallach (1998). However, in characteristic p > 0, that assumption cannot be omitted. Take, for example, the additive group G = Ga (k), where k is an algebraic closure of the field with p elements, and let ϕ : G → G, x → xp . Then ϕ is a bijective homomorphism, but ϕ−1 certainly is not a morphism of affine varieties; see Example 1.3.5(c).\n\n132\n\nAffine varieties and finite morphisms\n\n2.4 Linearization and generation of algebraic groups Our first aim in this section is to show that every ‘abstract’ affine algebraic group, as defined in §2.2.12, is isomorphic to a linear algebraic group in the sense of Definition 1.3.9. Then we establish a basic connectedness criterion and give some applications. Throughout, k is an algebraically closed field. 2.4.1 Example Consider the general linear monoid Mn (k) with algebra of regular functions given by A = k[Xij | 1 \u0001 i, j \u0001 n]; see Definition 1.3.9. Let \u0001 det : = sgn(σ) X1σ(1) X2σ(2) · · · Xnσ(n) ∈ A. σ∈Sn\n\nSince det(In ) = 1, we certainly have that det is not nilpotent in A. Thus, using the notation in §2.1.14 (see also Exercise 1.8.11) we have an affine variety (GLn (k), Adet ). The matrix multiplication in Mn (k) defines a multiplication map µ : GLn (k) × GLn (k) → GLn (k) and an inversion map ι : GLn (k) → GLn (k). We claim that (GLn (k), Adet ) is an affine algebraic group. So, we must check that µ and ι are morphisms. First, using the formula for the inverse of a matrix, we see that ι is a morphism. Next, we have µ∗ (Xij ) =\n\nn \u0001\n\nfor all i, j ∈ {1, . . . , n}.\n\nl=1\n\nSo it remains to consider µ∗ (1/ det). For x, y ∈ G, we have (1/ det ◦µ)(x, y) = µ(ι∗ (det)(x), ι∗ (det)(y)). Now ι is a morphism and so ι∗ (det) ∈ Adet . Thus, we also see that µ∗ (1/ det) ∈ Adet ⊗ Adet , as required. Note also that the group GL1 (k) is nothing but the multiplicative group of k, which we denote by Gm (k). 2.4.2 Lemma Let (G, A) be an affine algebraic group over k, and assume that ϕ : G → GLn (k) is a homomorphism of abstract groups. For each x ∈ G, let us write ϕ(x) = (aij (x)), where aij (x) ∈ k for all i, j. Then ϕ is a homomorphism of algebraic groups if all the maps aij : G → k belong to A.\n\nLinearization and generation of algebraic groups\n\n133\n\nProof We use the notation in Example 2.4.1. For 1 \u0001 i, j \u0001 n, we have Xij ◦ ϕ = aij ∈ A by assumption. Next, we have (1/ det ◦ ϕ)(x) = det(ϕ(x))−1 = det(ϕ(x−1 )) = det(ϕ(ι(x))) for all x ∈ G. Thus, we obtain 1/ det ◦ϕ = det ◦ϕ ◦ ι. We certainly have det ◦ϕ ∈ A. Finally, since ι is a morphism, we also have (det ◦ϕ) ◦ ι ∈ A, as required. \u0003 As a first application, we shall prove that an affine algebraic group (G, A) can always be embedded in some general linear group. We first need the following result which is a basic tool in the study of algebraic groups. Let µ : G × G → G be the multiplication map. For any x ∈ G, we define λx : G → G,\n\ny → xy,\n\nand ρx : G → G,\n\ny → yx.\n\nAs in §1.3.12, we see that λx and ρx are isomorphisms of affine varieties. Since ρx is an isomorphism, the corresponding k-algebra homomorphism ρx∗ : A[G] → A[G] also is an isomorphism, where ∗ . Hence, we have a group homomorphism ρ∗ : G → ρx∗ ◦ ρy∗ = ρxy Autk (A[G]), where Autk (A[G]) denotes the group of all k-algebra automorphisms of A[G]. Note that the analogous construction with λx yields an anti-homomorphism λ∗ : G → Autk (A[G]), x → λ∗x . 2.4.3 Theorem Let (G, A) be an affine algebraic group, with multiplication map µ : G × G → G. Let F ⊂ A be any finite subset. Then the subspace E : = \u0005ρx∗ (f) | f ∈ F, x ∈ G\u0006k ⊆ A has finite dimension. If {e1 , . . . , en } is a basis of E, then we have ∗\n\nµ (ej ) =\n\nn \u0001\n\nei ⊗ aij\n\nfor all j ∈ {1, . . . , n}, where aij ∈ A.\n\ni=1\n\nFurthermore, the map ϕ : G → GLn (k), x → (aij (x)), is a homomorphism of affine algebraic groups. Proof Fix some non-zero f ∈ F and consider the subspace Ef = \u0005ρx∗ (f) | \u0003 x ∈ G\u0006k ⊆ A. Then choose m \u0002 1 minimal such that ∗ µ (f) = m i=1 gi ⊗ hi where gi , hi ∈ A. Now, for x, y ∈ G, we have\n\n134\n\nAffine varieties and finite morphisms\n\nρy∗ (f)(x) = f(xy) = µ∗ (f)(x, y) = ρy∗ (f) =\n\nm \u0001\n\n\u0003m\n\ni=1 gi (x)hi (y)\n\nhi (y)gi\n\nand so\n\nfor all y ∈ G.\n\n(1)\n\ni=1\n\nSimilarly, we have µ∗ (ρx∗ (f))(y, z) = f(yzx) = f(yρx (z)) = µ∗ (f)(y, ρx (z)) for any y, z ∈ G, and so µ∗ (ρx∗ (f)) =\n\nm \u0001\n\ngi ⊗ ρx∗ (hi )\n\nfor all x ∈ G.\n\n(2)\n\ni=1\n\nNow, the minimality of m shows that {h1 , . . . , hm } is linearly independent. Hence there exist elements y1 , . . . , ym ∈ G such that the matrix (hi (yj ))1\u0002i,j\u0002m \u0003 is invertible. Consequently, we can ‘invert’ the equations ρy∗j (f) = m i=1 hi (yj )gi and find that each gi is a linear ∗ combination of ρyj (f) (1 \u0001 j \u0001 m). We conclude that gi ∈ Ef for all i. On the other hand, (1) shows that any ρy∗ (f) is a linear combination of g1 , . . . , gm . Hence {g\u0003 1 , . . . , gm } spans Ef . Applying this to each f ∈ F , we see that E = f∈F Ef has finite dimension. Now let {e1 , . . . , en } be a basis of E. By writing all of the above expressions in terms of this basis, we see that (2) yields relations of the form ∗\n\nµ (ej ) =\n\nn \u0001\n\nei ⊗ aij\n\nfor all j ∈ {1, . . . , n}, where aij ∈ A. (3)\n\ni=1 ∗ (f) ∈ E, and For any x, y ∈ G and f ∈ F , we have ρy∗ (ρx∗ (f)) = ρyx ∗ so E is invariant under all ρy (y ∈ G). Thus, for any y ∈ G, we have an element ϕy ∈ GL(E) given by ϕy (f) : = ρy∗ (f) (f ∈ E). Then the map\n\nϕ : G → GLn (k), y → My = matrix of ϕy with respect to {e1 , . . . , en }, is a group homomorphism. Furthermore, for any x ∈ G, we have ϕy (ej )(x) = ρy∗ (ej )(x) = ej (µ(x, y)) = µ∗ (ej )(x, y) =\n\nm \u0001\n\naij (y)ei (x)\n\ni=1\n\n\u0003n\n\nfor all x ∈ G and so ϕy (ej ) = i=1 aij (y)ei . Thus, the coefficients of My are given by the regular functions aij ∈ A evaluated at y. Hence ϕ is a homomorphism of affine algebraic groups by Lemma 2.4.2. \u0003\n\nLinearization and generation of algebraic groups\n\n135\n\n2.4.4 Corollary Let (G, A) be an affine algebraic group over k. Then there exists some n \u0002 1 and a closed embedding of algebraic groups ϕ : G → GLn (k); that is, ϕ(G) ⊆ GLn (k) is a closed subgroup and the restricted map G → ϕ(G), g → ϕ(g), is an isomorphism of algebraic groups. Proof Since A is a finitely generated k-algebra, we have A = k[f1 , . . . , fm ], where fj ∈ A. We set F = {f1 , . . . , fm } and apply Theorem 2.4.3. This yields a finite-dimensional subspace E ⊂ A containing F with a basis {e1 , . . . , en } such that ∗\n\nµ (ej ) =\n\nn \u0001\n\nei ⊗ aij\n\nfor all 1 \u0001 j \u0001 n, where aij ∈ A.\n\ni=1\n\nFurthermore, ϕ : G → GLn (k), x → (aij (x)), is a homomorphism of affine algebraic groups. By Proposition 2.2.1, it remains to show that ϕ∗ : Adet → A is surjective. Now, E contains all fi and, hence, we have A = k[e1 , . . . , en ]. So it is enough to show that all ej lie in the image of ϕ∗ . First note that all aij lie in that image since ϕ∗ (Xij ) = aij . Now, using the above relation, we obtain ej (y) = ej (µ(1, y)) = µ∗ (ej )(1, y) =\n\nn \u0001\n\nei (1)aij (y) for all y ∈ G\n\ni=1\n\nand so ej =\n\n\u0003n\n\ni=1 ei (1)aij .\n\nHence we have ej ∈ ϕ∗ (Adet ), as required. \u0003\n\n2.4.5 Remark Let ϕ : G → GLn (k) be a closed embedding as above. Furthermore, note that we also have a closed embedding\n\nA 0 ι : GLn (k) → SLn+1 (k), . A → 0 det(A)−1 Thus, the composition ι ◦ ϕ : G → SLn+1 (k) yields an isomorphism of G with a closed subgroup of SLn+1 (k). So everything that we proved for a linear algebraic group in the sense of Definition 1.3.9 remains valid for an abstract affine group (G, A) as defined above. For example, Proposition 1.3.13 (concerning the irreducible components of an algebraic group) remains true without any change. Furthermore, we have dim Tg (G) = dim T1 (G) = dim G for all g ∈ G (see Proposition 1.5.2) and the tangent space T1 (G) carries in a\n\n136\n\nAffine varieties and finite morphisms\n\nnatural way the structure of a Lie algebra, where the Lie product is given by the formula in Remark 1.5.4. 2.4.6 Theorem Let G be an affine algebraic group and {(Yλ , Bλ )}λ∈Λ be a family of irreducible affine varieties. We assume that there exist morphisms ϕλ : Yλ → G such that 1 ∈ Xλ : = ϕλ (Yλ ) for all λ ∈ Λ. Let H be the subgroup of G generated by all Xλ . Then H is closed and irreducible and we have H = Xλε11 · · · Xλεnn\n\nfor some n \u0002 0, where λi ∈ Λ and εi ∈ {±1}.\n\nIn particular, a collection of closed irreducible subgroups in G generates a subgroup which is itself closed and irreducible. Proof We may assume that, for each λ, there exists some λ\u0002 such that Xλ\u0002 = Xλ−1 . Now let n \u0002 0 and λ = (λ1 , . . . , λn ) with λi ∈ Λ, and consider the morphism ϕλ : Yλ1 × · · · × Yλn → G,\n\n(y1 , . . . , yn ) → ϕλ1 (y1 ) · · · ϕλn (yn ).\n\nDenote by Xλ = Xλ1 · · · Xλn the image of ϕλ . Since each Yλ is irreducible, the above direct product is irreducible and, hence, X λ is an irreducible closed subset of G. Now let ν = (ν1 , . . . , νm ) for some m \u0002 0, where νi ∈ Λ. Denote by (λ, ν) the concatenation of λ and ν. Then Xλ · Xν = X(λ,ν) and we claim that X λ · X ν ⊆ X (λ,ν) .\n\n(a)\n\nIndeed, for a fixed x ∈ Xν , the map Xλ → X(λ,ν) , y → yx, is continuous, and so X λ x ⊆ X (λ,ν) . Thus, we have X λ · Xν ⊆ X (λ,ν) . On the other hand, for a fixed y ∈ X λ , the map Xν → X (λ,ν) , x → yx, is continuous, and so yX ν ⊆ X (λ,ν) . Thus, (a) is proved. Now choose λ as above such that dim X λ is maximal. Then, for any ν, we have X λ ⊆ X λ · 1 ⊆ X λ · X ν ⊆ X (λ,ν) (since 1 ∈ Xµ for µ ∈ Λ). By the maximality of λ and Proposition 1.2.20, we have X λ = X (λ,ν) and, hence, using a similar argument, also X ν ⊆ X (λ,ν) = X λ . In particular, this implies that Xλ · Xλ ⊆ Xλ\n\n−1\n\nand X λ ⊆ X λ ,\n\n(b)\n\nand so X λ ⊆ G is a subgroup. Now, Xλ is constructible by Theorem 2.2.11, and so X λ = Xλ · Xλ by Lemma 2.2.13. Finally,\n\nLinearization and generation of algebraic groups\n\n137\n\nsince X ν ⊆ X λ for all ν, we have H = X λ , and the sequence (λ, λ) has the required properties. \u0003 2.4.7 Example Let G be an affine algebraic group. For g, h ∈ G, the commutator is defined by [g, h] : = g −1 h−1 gh ∈ G. If U, H ⊆ G are subgroups, the corresponding commutator subgroup is defined to be the subgroup [U, H] : = \u0005[u, h] | u ∈ U, h ∈ H\u0006 ⊆ G. Now, if U and H are closed and U or H is connected, then [U, H] is closed and connected. Indeed, assume for example that H is connected. For u ∈ U , we consider the morphism ϕu : H → G, h → [u, h]. We certainly have 1 ∈ ϕu (H). Thus, all the conditions in Theorem 2.4.6 are satisfied, and so [U, H] = \u0005ϕu (H) | u ∈ U \u0006 is a closed connected subgroup, as claimed. 2.4.8 Definition Let G be an affine algebraic group. We define a chain of subgroups G ⊇ G(1) ⊇ G(2) ⊇ · · · as follows. We set G(0) : = G and G(i+1) : = [G(i) , G(i) ] for i \u0002 0. It is well known (and easy to check) that the subgroups G(i) are normal and that the quotients G(i) /G(i+1) are abelian. The group G is called solvable if G(r) = {1} for some r \u0002 0. Now assume that G is connected. Then, by Example 2.4.7, all subgroups G(i) are closed and connected. Similarly, we can define a descending chain of normal subgroups by K0 (G) : = G and Ki+1 (G) : = [G, Ki (G)] for i \u0002 0. We say that G is nilpotent if Kr (G) = {1} for some r \u0002 1. Clearly, nilpotent groups are solvable. As before, we see that, if G is connected, then the groups Ki (G) are closed and connected. 2.4.9 Example Let Bn (k) ⊆ GLn (k) be the closed subgroup consisting of all upper triangular invertible matrices, that is, all matrices A = (aij )1\u0001i,j\u0001n such that aij = 0 for 1 \u0001 j < i \u0001 n. As abstract groups, we have Bn (k) = Un (k)Tn (k), where Un (k) is the group of all upper triangular matrices with 1 on the diagonal (see Example 1.3.11) and Tn (k) is the group of all invertible diagonal matrices. Since Un (k) and Tn (k) are connected (see Example 1.5.5)\n\n138\n\nAffine varieties and finite morphisms\n\nand the natural map given by multiplication Un (k) × Tn (k) → Bn (k) is surjective, we conclude that Bn (k) is connected. Furthermore, Un (k) is normal in Bn (k), and we have Bn (k)/Un (k) ∼ = Tn (k) (as abstract groups). Hence, since Tn (k) is abelian, we have [Bn (k), Bn (k)] ⊆ Un (k). Now it is easily checked that there exists an element t ∈ Tn (k) whose centralizer in Bn (k) is just Tn (k). So, by Exercise 2.7.13, we can even conclude that [Bn (k), Bn (k)] = Un (k). Now consider the group Un (k). A matrix calculation shows that Kr (Un (k)) = {(aij ) ∈ Un (k) | aij = 0 if 1 \u0001 j − i \u0001 r} for r \u0002 1; (see, for example, Propostion III.16.3 of Huppert (1983). Thus, Kn (Un (k)) = {1}, and so Un (k) is nilpotent; consequently, Bn (k)(n) ⊆ Kn (Un (k)) = {1}, and so Bn (k) is solvable. The group Bn (k) provides a model for the further study of connected solvable groups. In Section 3.5, we will obtain a generalization of the decomposition Bn (k) = Un (k)Tn (k) for an arbitrary connected solvable group. 2.4.10 Example Consider the algebraic group SLn (k). Let Un (k) and Un\u0002 (k) be the subgroups of SLn (k) consisting of all upper unitriangular matrices and all lower unitriangular matrices, respectively, as in Example 1.5.5. These are closed connected subgroups in SLn (k). We claim that SLn (k) = [SLn (k), SLn (k)] = \u0005Un (k), Un\u0002 (k)\u0006. This is seen as follows. By the Gauss elimination algorithm, we know that SLn (k) is generated by Un (k) and Un\u0002 (k), together with the set of all monomial matrices in SLn (k). (Recall that a matrix is monomial if there is precisely one non-zero coefficient in each row and each column.) Let us now check that every monomial matrix in SLn (k) lies in \u0005Un (k), Un\u0002 (k)\u0006. For this purpose, we use the following computations in SL2 (k): 1 0 0 a 1 a 1 a · = · n0 (a) : = 0 1 0 1 −a−1 1 −a−1 0 for any non-zero a ∈ k. This also shows that the diagonal matrix a 0 = n0 (a)n0 (−1) 0 a−1\n\nLinearization and generation of algebraic groups\n\n139\n\ncan be written as a product of upper and lower unitriangular matrices in SL2 (k). Now let 1 \u0001 i \u0001 n − 1. Then we have an embedding ⎤ ⎡ 0 Ii−1 0 ⎥ ⎢ ⎥. A → ⎢ ϕi : SL2 (k) → SLn (k), 0 A 0 ⎦ ⎣ 0 0 In−i−1 (Thus, the four coefficients of A are placed at positions (i, i), (i, i+1), (i + 1, i), and (i + 1, i + 1) in the bigger matrix.) It is then easy to show (using the fact that every permutation of 1, . . . , n can be written as a product of transpositions permuting two consecutive numbers) that every monomial matrix in SLn (k) can be expressed as a product ϕi1 (A1 ) · · · ϕir (Ar ), for various ij ∈ {1, . . . , n − 1} and various monomial matrices Aj ∈ SL2 (k). Thus, we have shown that SLn (k) = \u0005Un (k), Un\u0002 (k)\u0006. Furthermore, by Example 2.4.9, we know that Un (k) = [Bn (k), Bn (k)] ⊆ [SLn (k), SLn (k)]. A similar argument also applies to Un\u0002 (k). Thus, we conclude that SLn (k) = [SLn (k), SLn (k)]. 2.4.11 Example Let G be one of the classical groups defined in §1.3.15 and §1.3.16; thus, we have ⎧ any m \u0002 1, char(k) = 2, ⎨ SO2m+1 (k), Sp2m (k), any m \u0002 1, any characteristic, G= ⎩ SO+ any m \u0002 2, any characteristic. 2m (k), Let us set n = 2m + 1 if G = SO2m+1 (k), and n = 2m otherwise. Then G is a closed subgroup of GLn (k). Furthermore, by Theorems 1.7.4 and 1.7.8, we know that G is connected and that there is a split BN -pair in G. We claim that G = [G, G] = \u0005U, U \u0002 \u0006\n\nand\n\nZ(G) = {± id},\n\nwhere U and U \u0002 are the subgroups of all upper unitriangular matrices and all lower unitriangular matrices in G, respectively. We sketch the proof but leave some details as an exercise to the reader. Let T be the subgroup of all invertible diagonal matrices in G. By Remark 1.5.12, the groups U, U \u0002 , and T are connected. Using Exercise 1.8.27 and Lemma 2.2.13, we conclude that G = \u0005U, T, U \u0002 \u0006. Now recall from Lemmas 1.5.9–11 the explicit description of the matrices in T . Using that explicit description, one finds some t ∈ T such that CG (t) = T . Then, by Exercise 2.7.13, we may conclude that\n\n140\n\nAffine varieties and finite morphisms\n\nU, U \u0002 ⊆ [G, G]. So it will be enough to show that T ⊆ \u0005U, U \u0002 \u0006. Now, in each case, we have an embedding SLm (k) → G, A → A, defined similarly as in the proof of Lemma 1.7.6. (If G = SO2m+1 (k), one places \u0002 (k)\u0006, 1 at position (m + 1, m + 1).) Thus, since SLm (k) = \u0005Um (k), Um one already finds a big portion of T in \u0005U, U \u0002 \u0006. In fact, it remains to show that the diagonal matrices with entries 1, . . . , 1, a, a−1 , 1, . . . , 1 lie in \u0005U, U \u0002 \u0006 for all non-zero a ∈ k (where a and a−1 are at the positions m and m + 1, respectively; if G = SO2m+1 (k), one has to insert a coefficient 1 between a and a−1 ). Thus, the problem can be reduced to an explicit computation in Sp2 (k) = SL2 (k), SO3 (k), or SO+ 4 (k); to deal with these groups, use Exercises 1.8.19 and 1.8.20. To determine Z(G), we argue as follows. Since Z(G) certainly normalizes \bB = U T , we have Z(G) ⊆ B by Lemma 1.6.4, and hence Z(G) ⊆ n∈N nBn−1 = T . Thus, it remains to find all diagonal matrices which commute with all elements of U and U \u0002 . It is easily checked that ± id are the only such matrices. 2.4.12 Remark Let G be a group with a split BN -pair. Assume that G = [G, G],\n\nB is solvable, and W is irreducible. \b Then it can be shown that Z : = g∈G gBg −1 ⊆ H is the centre of G and that G/Z is a simple group. See Chapter IV, §2, no. 7 of Bourbaki (1968) of §11.1 of Carter (1972). This criterion can be applied to show that G/Z is simple, where G = SLn (k) or G is one of the classical groups as in Example 2.4.11. More generally, this also works for groups of Lie type as in §1.5.15; see §11.1 of Carter (1972) and §4 of Steinberg (1967).\n\n2.5 Group actions on affine varieties The aim of this section is to establish some basic results concerning algebraic group actions on affine varieties. The two main results are Proposition 2.5.2, which shows that closed orbits always exist, and Theorem 2.5.5, which shows that orbit maps are open maps. At the end of this section, we will also discuss the problem of defining ‘affine quotients’. In general, this is a very difficult problem, but we will at least show that quotients by finite groups always exist. Let G be an affine algebraic group and X be an affine variety (both over an algebraically closed field k). Recall from §2.3.13 that\n\nGroup actions on affine varieties\n\n141\n\nwe call X is a G-variety if there exists a morphism of affine varieties α : G × X → X,\n\n(g, x) → g.x,\n\nwhich defines an abstract operation of G on the set X. We have already noted that, for each g ∈ G, the map πg : X → X, x → g.x, is an isomorphism of affine varieties. Similarly, for each x ∈ X, the orbit map ϕx : G → X, g → g.x, is obtained by composing α with the morphism G → G × X, g → (g, x). Hence ϕx is a morphism. The orbit of x is the set Ox : = {g.x | g ∈ G}. For any subsets H ⊆ G and Y ⊆ X, we write H.Y : = {h.y | h ∈ H, y ∈ Y }. 2.5.1 Lemma\n\nLet X be a G-variety. Then the following hold.\n\n(a) Let Y, Z ⊆ X, with Z closed. Then the transporter TranG (Y, Z) : = {g ∈ G | g.Y ⊆ Z} is a closed subset of G. In particular, for x ∈ X, the stabilizer StabG (x) = TranG ({x}, {x}) is a closed subgroup. (b) The identity component G◦ leaves each irreducible component of X invariant. Proof (a) Let y ∈ Y and consider the orbit map ϕ X. \by : G → −1 (Z) Then ϕ−1 (Z) ⊆ G is closed. But then Tran (Y, Z) = ϕ G y y∈Y y also is closed. (b) Let X \u0002 ⊆ X be an irreducible component and G\u0002 = TranG (X \u0002 , X \u0002 ) be its stabilizer in G. By (a), G\u0002 is a closed subset in G, and it is a subgroup. Since G permutes the finitely many irreducible components of X, we have that G\u0002 is a closed subgroup of finite index in G, and so G◦ ⊆ G\u0002 by Proposition 1.3.13. \u0003 2.5.2 Proposition\n\nLet X be a G-variety.\n\n(a) Let O be a G-orbit. Then O is open in O, and O is a union of orbits. \u0002 (b) For G-orbits O, O\u0002 ⊆ X, we write O \u0015 O\u0002 if O ⊆ O (= Zariski closure of O\u0002 ). This defines a partial order on the set of G-orbits in X. (c) There exist minimal elements in the partial order defined in (b) and each of these is a closed orbit. Proof (a) First we show that O is a union of orbits. For g ∈ G, consider the map πg : X → X, y → g.y. Since πg is continuous for\n\n142\n\nAffine varieties and finite morphisms\n\neach g ∈ G, we actually have that πg is a homeomorphism (with inverse given by πg−1 ). Thus, we have πg (O) ⊆ O, and so O is a union of orbits. Now let O = Ox , where x ∈ X. Since O is the image of the morphism ϕx : G → X, g → g.x, we know by Theorem 2.2.11 that O is constructible. So there exists a subset U ⊆ O such that U is dense open in O (see Exercise 2.7.6). Let g0 ∈ G be such that g0 .x ∈ U . Then x \u0007 ∈ g0−1 .U and so g.x ∈ gg0−1 .U ⊆ O for all g ∈ G. It follows that O = g∈G g.U , which is open in O (since g.U = πg (U ) and πg is a homeomorphism). (b) To show that \u0015 is a partial order, it only remains to check \u0002 that if O, O\u0002 are G-orbits such that O = O , then O = O\u0002 . Now, by (a), O is open in O. Hence, if O\u0002 = O is a G-orbit contained in O, \u0002 then O\u0002 is contained in the closed set O \\ O and so O ⊆ O \\ O. (c) For x ∈ X we set Sx : = Ox \\Ox . By (a), this is a closed subset of X. We can now construct a decreasing chain of closed subsets in X as follows. Let x1 ∈ X and assume that Sx1 = ∅. Then let x2 ∈ Sx1 . By (a), the orbit Ox2 is still contained in Sx1 , and we have Sx2 \u0002 Sx1 . If Sx2 = ∅, we choose x3 ∈ Sx2 and consider Ox3 . Thus, we obtain a chain Sx1 \u0003 Sx2 \u0003 Sx3 \u0003 · · · of closed subvarieties. Since X is noetherian, there exists some n \u0002 1 such that Sxn = ∅. This means that Oxn is a closed subset of X. \u0003 2.5.3 Proposition Let X be a G-variety and x ∈ X. Denote by Ox◦ the orbit of x under the action of the identity component G◦ . Then we have dim Ox = dim Ox◦ , dim StabG (x) = dim StabG◦ (x), and dim G = dim Ox + dim StabG (x). \u0016 Proof Write G = ti=1 gi G◦ , where gi ∈ G. Then we have Ox = \u0007t \u0007t ◦ ◦ i=1 gi Ox and so Ox = i=1 gi Ox . Since left multiplication by any element of G is an isomorphism of affine varieties, gi Ox◦ is a closed irreducible subset, and we have dim gi Ox◦ = dim Ox◦ for all i. This shows that dim Ox = dim Ox◦ . Furthermore, since G◦ is a normal subgroup, the product G◦ StabG (x) is a subgroup of G, and we have the following isomorphism of abstract groups: StabG (x)/StabG◦(x) = StabG (x)/(G◦ ∩StabG (x)) ∼ = G◦ StabG (x)/G◦ . Hence, since G◦ has finite index in G, we conclude that StabG◦ (x) has finite index in StabG (x), and so dim StabG (x) = dim StabG◦ (x). To prove the remaining statement, we may now assume without loss of generality that G is connected.\n\nGroup actions on affine varieties\n\n143\n\nConsider the orbit map ϕx : G → Ox ⊆ X. Then ϕx is a dominant morphism between irreducible affine varieties. By Corollary 2.2.9, there exists a non-empty open subset U ⊆ Ox such that dim ϕ−1 x (y) = dim G − dim Ox for all y ∈ U . Ox by Proposition 2.5.2, we conclude that Ox ∩ U = ∅. Now let y ∈ Ox ∩ U and g ∈ G be such that y = g.x. Then we have ϕ−1 x (y) = {h ∈ G | h.x = y} = {h ∈ G | g −1 h ∈ StabG (x)} = g StabG (x). Since left multiplication with g is an isomorphism of affine varieties, we have dim ϕ−1 x (y) = dim StabG (x), and so dim G = dim O x + \u0003 dim StabG (x). 2.5.4 Example Let G be an affine algebraic group. Then there are several ways to see G itself as a G-variety. For example, we may consider the action given by left multiplication. Another example is the conjugation action G × G → G, (g, x) → gxg −1 . For x ∈ G, the orbit Ox = {gxg −1 | g ∈ G} is nothing but the conjugacy class of x in G, and the stabilizer of x ∈ G is the centralizer CG (x) : = {g ∈ G | gx = xg}. Thus, Proposition 2.5.3 shows that dim G = dim Ox + dim CG (x). See Section 2.6 for worked examples in the case where G = SLn (k); in particular, Theorem 2.6.5 gives an explicit combinatorial characterization of the partial order on conjugacy classes of unipotent elements. Furthermore, let S ⊆ G be a closed subset. Then NG (S) : = {g ∈ G | gsg −1 ∈ S for all s ∈ S} is a closed subgroup in G, which is called the normalizer of S. To prove this, just note that NG (S) = TranG (S, S) and use Lemma 2.5.1. 2.5.5 Theorem (Steinberg) Let X1 and X2 be G-varieties and ϕ : X1 → X2 be a dominant G-equivariant morphism. Assume that X1 consists of a single G-orbit. Then ϕ is an open map, that is, it sends open sets to open sets.\n\n144\n\nAffine varieties and finite morphisms\n\nProof We begin with some reductions. First we show that it is enough to consider the special case where X1 = G and G acts by left multiplication. Indeed, let us fix x1 ∈ X1 and set x2 = ϕ(x1 ) ∈ X2 . Then we have the orbit maps ϕi : G → Xi , g → g.xi (i = 1, 2). Since ϕ2 (g) = g.x2 = g.ϕ(x1 ) = ϕ(g.x1 ) = ϕ(ϕ1 (g)) for all g ∈ G, we have ϕ2 = ϕ ◦ ϕ1 . Now, since ϕ1 is assumed to be surjective, we see that the openness of ϕ would follow if we knew that ϕ2 was open. So let us now change the notation and consider a G-variety X, let x ∈ X and consider the orbit map ϕx : G → X. Assume that this map is dominant (that is, the G-orbit of x is dense in X). Then we must show that ϕx is an open map. We claim that it is enough to show that ϕx |U : U → X is an open map for some non-empty open U ⊆ G. (∗) Indeed, if we have such an open subset U , we can proceed as follows. Since left multiplication by any g ∈ G is a\u0007homeomorphism, the map ϕx |gU : gU → X is also open. Since G = g∈G gU , this implies that ϕx is open. Let us now prove (∗) in the special case where G is connected. Since ϕx is assumed to be dominant, this also forces X to be irreducible. Let d : = dim G − dim X. By Theorem 2.2.7, there exists a non-empty affine open subset U ⊆ G and a non-empty affine open subset V ⊆ X such that we have a factorization ϕ ˜\n\npr\n\n1 ϕx |U : U −→ V × kd −→ V,\n\nwhere ϕ˜ is a dominant finite morphism; in particular, we have ϕx (U ) = V . We claim that U has the desired property. Since V ⊆ X is open, it is enough to show that the restricted map ϕx |U : U → V (which is still a dominant morphism between irreducible affine varieties) is open. To prove this, we argue as follows. By Lemma 2.2.5, ϕ˜ is a closed map. Hence, by taking complements, one sees that ϕ(U ˜ 1 ) is open for any open subset U1 ⊆ U which consists of complete fibres of ϕ, ˜ that is, we have U1 = ϕ˜−1 (ϕ(U ˜ 1 )). Since the projection map pr1 is open by Exercise 1.8.8, we conclude that ϕx (U1 ) ⊆ V is open for any open U1 ⊆ U such that U1 = ϕ˜−1 (ϕ(U ˜ 1 )). Now, given any open subset U1 ⊆ U , we set ˜1 : = U1 H ∩ U ⊇ U1 , U\n\nwhere H : = StabG (x).\n\nGroup actions on affine varieties\n\n145\n\n˜1 is open since U1 H = \u0007 Then U h∈H U1 h is open. Furthermore, ˜ ˜ ˜1 = we have ϕx (U1 ) = U1 .x = U1 .x = ϕx (U1 ). We claim that U −1 ˜ ϕ˜ (ϕ( ˜ U1 )). The inclusion ‘⊆’ is clear. Now let u ∈ U be such ˜1 . Composing with pr1 , we that ϕ(u) ˜ = ϕ(u ˜ 1 ) for some u1 ∈ U obtain u.x = ϕx (u) = ϕx (u1 ) = u1 .x and so u1−1 u.x = x. This ˜1 , as required. Thus, we conclude that implies that u ∈ u1 H ∩ U ⊆ U ˜1 ) is open, as claimed. ϕx (U1 ) = ϕx (U So (∗) is proved in the special case where G is irreducible. Now consider the general case. Then (∗) holds for the restricted orbit map ◦ ϕ◦x : G◦ → X ◦ , where X ◦ = Ox . So there is a non-empty open set U ⊆ G◦ such that ϕx |U : U → X ◦ is open. Now, by Exercise 2.8.10, Ox◦ is open in Ox and, hence, open in X = Ox . Consequently, ϕx |U : U → X is an open map. \u0003 2.5.6 Example (a) Let X be a G-variety and Ox be the G-orbit of some x ∈ X. Then the orbit map ϕx : G → Ox , g → g.x, is an open map. (Apply Theorem 2.5.5 where X1 = G and G acts by left multiplication.) (b) Let ϕ : G1 → G2 be a surjective homomorphism of affine algebraic groups. Then ϕ is an open map. (Apply Theorem 2.5.5, where G = G1 acts on G1 by left multiplication and on G2 by g.g2 = ϕ(g)(g2 ).) In particular, a bijective homomorphism between affine algebraic groups is a homeomorphism. 2.5.7 Corollary With the same assumptions as in Theorem 2.5.5, let Y be any affine variety. Then the map ϕ × id : X1 × Y → X2 × Y is open. Proof Let us regard Y as an algebraic set in some affine space kn . Then Xi × Y is a closed subset of Xi × kn for i = 1, 2. Hence, if U ⊆ X1 × Y is open, we have U = (X1 × Y ) ∩ U \u0002 for some open set U \u0002 ⊆ X1 × kn and (ϕ × id)(U ) = (X2 × Y ) ∩ (ϕ × id)(U \u0002 ). Thus, it will be enough to prove the assertion in the special case where Y = kn . In this case, we use the fact that kn is a kn -variety, where we regard kn as an affine algebraic group under addition. Then Xi × kn are (G × kn )-varieties, with action given by (g, v).(xi , v\u0002 ) := (g.xi , v + v\u0002 ) (where g ∈ G, xi ∈ Xi , and v, v\u0002 ∈ kn ). Furthermore, X1 ×kn is a single orbit and ϕ×id is a (G×kn )-equivariant dominant morphism. Thus, ϕ × id is open by Theorem 2.5.5. \u0003\n\n146\n\nAffine varieties and finite morphisms\n\nGiven a G-variety X, we can consider the set of orbits of G on X. A natural question is whether this set of orbits might be itself regarded as an affine variety. This leads us to the following notion. 2.5.8 Affine quotients Let G be an affine algebraic group and X be a G-variety, as in §2.3.13. For any g ∈ G, consider the corresponding morphism πg : X → X. This morphism is in fact an isomorphism, with inverse being given by πg−1 . Hence πg∗ : A[X] → A[X] is a k-algebra isomorphism. Explicitly, πg∗ is given by the formula πg∗ (f)(x) = f(g.x), where f ∈ A[X] and x ∈ X. We set A[X]G : = {f ∈ A[X] | f(g.x) = f(x) for all g ∈ G, x ∈ X}. Then A[X]G is a k-subalgebra of A[X]. By definition, an affine quotient of X by G is a pair (X0 , ϕ) consisting of an affine variety X0 over k and a morphism of affine varieties ϕ : X → X0 such that the following conditions are satisfied: (a) For each x0 ∈ X0 , the set ϕ−1 (x0 ) is a G-orbit. In particular, as a set, X0 is in bijection with the G-orbits on X. (b) The map ϕ is open, that is, π(U ) ⊆ X0 is open for any open U ⊆ X. (c) We have ϕ∗ (A[X0 ]) = A[X]G . If it exists, an affine quotient is unique up to unique isomorphism. This follows from the following universal property. Assume that (X0 , ϕ) is an affine quotient, and let ψ : X → Y be any morphism of affine varieties which is constant on the orbits of G. Then there exists a unique morphism of affine varieties ψ¯ : X0 → Y such that ψ = ψ¯ ◦ ϕ. ¯ Indeed, define ψ(ϕ(x)) : = ψ(x) for x ∈ X. (Note that this is well-defined and the unique possibility.) We must check that ψ¯ is a morphism of affine varieties. So let f ∈ A[Y ]. Since ψ is constant on the orbits of G, we have ψ∗ (f) ∈ A[X]G . By condition (c), there exists a unique f0 ∈ A[X0 ] such that ψ∗ (f) = ϕ∗ (f0 ). Now let x0 ∈ X0 and choose x ∈ X with ϕ(x) = x0 . Then we have ¯ 0 ) = f(ψ(x)) = ψ∗ (f)(x) = ϕ∗ (f0 )(x) = f0 (ϕ(x)) = f0 (x0 ), (f ◦ ψ)(x that is, f ◦ ψ¯ = f0 ∈ A[X0 ]. So ψ¯ is a morphism. 2.5.9 Example Let X be a G-variety, where X is irreducible. Assume that an affine quotient (X0 , ϕ) exists. Then, since ϕ is surjective, X0 must be irreducible, too. The following example shows\n\nGroup actions on affine varieties\n\n147\n\nthat affine quotients do not always exist. Let G = GLn (k) and X = kn , where G acts on X in the natural way. Then there are only two orbits: {0} and kn \\ {0}. If an affine quotient existed, it would have only two elements and, hence, would not be irreducible. In fact, the question of whether an affine quotient exists or not is rather subtle. With the methods that we have at hand right now, we can give a complete answer in the special case where G is a finite group. This will be used, for example, in Chapter 4 where we study the finite fixed-point set of an algebraic group under a Frobenius map. Note that a finite group always is an affine algebraic group in a natural way. (The algebra of regular functions is given by the algebra of all k-valued functions on the group.) 2.5.10 Theorem (quotients by finite groups) Let X be a G-variety, where G is a finite group. Let X/G be the set of G-orbits on X and π : X → X/G be the natural map. We set A[X/G] = {f¯: X/G → k | f¯ ◦ π ∈ A[X]}. Then (X/G, A[X/G]) is an affine variety, and the pair (X/G, π) is an affine quotient; furthermore, π : X → X/G is a finite morphism. Proof First note that A[X/G] is a k-subalgebra of Maps(X/G, k), and the map π∗ : A[X/G] → A[X], f → f ◦ π, is an injective k-algebra homomorphism; furthermore, π∗ (A[X/G]) = A[X]G . We now proceed in several steps. Step 1. First we show that A[X]G is a finitely generated k-algebra and that A[X] ⊇ A[X]G is an integral extension. This is done as follows. Since A[X] is finitely generated, we can write A[X] = k[f1 , . . . , fr ], where fi ∈ A[X]. Now set Fi : =\n\n\u0017\u0018\n\n\u0019\n\nT − πg∗ (fi )\n\nfor 1 \u0001 i \u0001 r,\n\ng∈G\n\nwhere T is an indeterminate. Thus, Fi is a monic polynomial with coefficients in A[X]. By construction, since the product runs over all elements of G, the coefficients are even invariant under πg∗ for all g ∈ G. So the coefficients of Fi lie in A[X]G . Thus, since Fi (fi ) = 0, we have found an equation of integral dependence of fi over A[X]G .\n\n148\n\nAffine varieties and finite morphisms\n\nThis already shows that A[X] is integral over A[X]G . Now write \u0003|G| Fi = j=0 aij T j (aij ∈ A[X]G ) and consider the k-algebra A0 : = k[aij | 1 \u0001 i \u0001 r, 0 \u0001 j \u0001 |G|] ⊆ A[X]G . Then A0 is finitely generated, and every fi is integral over A0 . Consequently, by Corollary 2.1.2(a), A[X] is a finitely generated A0 -module. So A[X]G will also be a finitely generated A0 -module; see Exercise 1.8.1. This implies that A[X]G is finitely generated as a k-algebra. Step 2. We now show that (X/G, A[X/G]) is an affine variety. This means that we have to verify the conditions in Definition 2.1.6. By Step 1 and the fact that π∗ : A[X/G] → A[X]G is an isomorphism, we know that A[X/G] is a finitely generated k-algebra. Now let x, y ∈ X be such that the G-orbits Ox and Oy are disjoint. We must ¯ x ) = f(O ¯ y ). show that there exists some f¯ ∈ A[X/G] such that f(O G In other words, we must show that there exists some f ∈ A[X] such that f(x) = f(y). Now, since G is a finite group, Ox is a finite set and, hence, closed. So there exists some a ∈ A[X] such that a(g.x) = 0 for all g ∈ G and a(y) = 1. Hence, setting a\u0002 : = a − 1 ∈ A[X], \u0002 we have = −1 for all g ∈ G, and a\u0002 (y) \u001a a (g.x) \u001a = 0. Now set ∗ \u0002 f : = g∈G πg (a ) ∈ A[X]G . Then we have f(x) = g∈G πg∗ (a\u0002 )(x) = \u001a \u001a \u001a \u0002 ∗ \u0002 \u0002 g∈G a (g.x) = ±1 and f(y) = g∈G πg (a )(y) = g∈G a (g.y) = 0, as required. Finally, let λ : A[X/G] → k be any k-algebra homomorphism. We must show that λ is given by evaluation at a point in X/G. Now, ker(λ) is a maximal ideal in A[X/G]. Since A[X] ⊇ A[X]G = π∗ (A[X/G]) is an integral extension, Lemma 2.2.4 (‘going up’) shows that there exists a maximal ideal m ⊆ A[X] such that π∗ (ker(λ)) = A[X]G ∩ m. Now, by Theorem 2.1.9, we have m = ker(εx ) for some x ∈ X. Let Ox be the G-orbit of x. Then it is readily checked that λ is the evaluation at Ox . Step 3. It remains to check that the three conditions in §2.5.8 are satisfied. This is clear for (a) and (c). To verify (b), let U ⊆ X be an \u0007 open subset. We must show that π(U ) ⊆ X/G is open. Now U \u0002 = g∈G g.U also is open, and we have π(U ) = π(U \u0002 ). Thus, we may assume without loss generality that U = π−1 (π(U )). But then we have π(X \\ U ) = (X/G) \\ π(U ). Since π is a closed map (see Lemma 2.2.5), this shows that π(U ) is open, as desired. \u0003 2.5.11 Example Let G ⊆ GLn (k) be a finite subgroup. Then we may regard kn as a G-variety in a natural way; the algebra\n\nGroup actions on affine varieties\n\n149\n\nof regular functions on kn is just given by the polynomial ring R = k[X1 , . . . , Xn ]. By Theorem 2.5.10, the ring of invariants RG is finitely generated; let us write RG = k[f1 , . . . , fm ], where F : = {f1 , . . . , fm } ⊆ RG . Now consider the ideal IF : = {h ∈ k[Y1 , . . . , Ym ] | h(f1 , . . . , fm ) = 0} ⊆ k[Y1 , . . . , Ym ] and set VF : = V(IF ) ⊆ km . We have a corresponding morphism ϕ : kn → VF , x → (f1 (x), . . . , fm (x)), which is constant on the G-orbits on kn . Hence, by the universal property in §2.5.8, we obtain an induced morphism ϕ¯ : kn /G → VF . We claim that this map is an isomorphism. Indeed, the map k[Y1 , . . . , Ym ] → RG , h → h(f1 , . . . , fm ), is a surjective k-algebra homomorphism with kernel IF . Since RG is the algebra of regular functions on kn /G, √ it has no nilpotent elements except 0. Thus, we have IF = IF and so IF = I(VF ) by Hilbert’s nullstellensatz (Theorem 2.1.9). Hence, we obtain an induced isomorphism A[VF ] → RG , which is easily seen to be equal to ϕ¯∗ , as required. Now, assuming that we have found F = {f1 , . . . , fm }, we can explicitly find generators for IF . Indeed, by Example 1.3.6, we have IF = JF ∩ k[Y1 , . . . , Ym ], where JF : = (f1 − Y1 , . . . , fm − Ym ) ⊆ k[X1 , . . . , Xn , Y1 , . . . , Ym ]. Hence, choosing the LEX order with X1 \u0016 · · · \u0016 Xn \u0016 Y1 \u0016 · · · \u0016 Ym , we have IF = (G ∩ k[Y1 , . . . , Ym ]), where G is a Groebner basis for JF . 2.5.12 Quotients by finite normal subgroups Let G be an affine algebraic group and Z be a finite closed normal subgroup of G. We consider G as a Z-variety, where z ∈ Z acts by left multiplication on G. Then G/Z = {Zg | g ∈ G} is the set of Z-orbits on G, and this is an abstract group. Since Z is finite, we also have a structure of G/Z as an affine variety by Theorem 2.5.10. We claim that G/Z is an affine algebraic group. To see this, we must check that multiplication and inversion in G/Z are given by morphisms of affine varieties. Let π : G → G/Z, g → Zg, be the natural map and ι : G → G be inversion in G. Then π ◦ι : G → G/Z is a morphism of affine varieties\n\n150\n\nAffine varieties and finite morphisms\n\nwhich is constant on the cosets of Z. Hence, by the universal property of the affine quotient, we obtain an induced map ¯ι : G/Z → G/Z which is a morphism of affine varieties and which gives exactly the inversion in G/Z. Now consider the multiplication map µ : G × G → G. The map π ◦ µ : G × G → G/Z is a morphism of affine varieties which is constant on the cosets of the finite closed normal subgroup Z × Z ⊆ G × G. Hence, applying the universal property of the affine quotient of G × G by Z × Z and using Exercise 2.7.11, we obtain an induced map µ ¯ : (G/Z) × (G/Z) → G/Z which is a morphism of affine varieties and which gives exactly the multiplication map for G/Z. 2.5.13 Example Consider the group G = SLn (k), and let Z ⊆ G be the centre of G. Then Z = {aIn | an = 1} is a finite group. So, by §2.5.12, the factor group PSLn (k) = SLn (k)/Z can be naturally regarded as an affine algebraic group. Similarly, let G be one of the classical groups introduced in §1.3.15 and §1.3.16, that is, G equals SO2m+1 (k) (any m \u0002 1, char(k) = 2), Sp2m (k), or SO+ 2m (k) (any m \u0002 1, any characteristic). + If G = SO2m (k), we assume that m \u0002 2. Then, by Example 2.4.11, we have Z(G) = {±1}. Hence, we obtain affine algebraic groups G/Z(G) which we denote by PSO2m+1 (k), PSp2m (k), and PSO+ 2m (k), respectively. We close this section with the following result which provides a criterion for checking if a morphism of affine varieties is a quotient morphism as in §2.5.8. It relies on the results on birational equivalences and normal varieties in Section 2.3. (If we had shown the general form of Zariski’s main theorem, then the assumption that ϕ be finite in the following result would be unnecessary.) 2.5.14 Proposition Let X be an irreducible G-variety, where G is a finite group. Let ϕ : X → X0 be a dominant finite morphism such that the following conditions are satisfied. (a) X0 is normal. (b) For each x0 ∈ X0 , the set ϕ−1 (x0 ) ⊆ X is a G-orbit. (c) dp ϕ : Tp (X) → Tϕ(p) (X0 ) is surjective for some p ∈ X. Then the pair (X0 , ϕ) is an affine quotient of X by G. Proof Let (X/G, π) be the affine quotient constructed in Theorem 2.5.8. Since ϕ is constant on the G-orbits on X, the\n\nThe unipotent variety of the special linear groups\n\n151\n\nuniversal property in 2.5.8 implies that there exists a unique morphism ϕ¯ : X/G → X0 such that ϕ = ϕ¯ ◦ π. We shall show that ϕ¯ is an isomorphism. To prove this, we use the criterion in Remark 2.3.4 (the very weak form of Zariski’s main theorem). First note that ϕ¯ is injective by (b). Next, X0 is normal by (a). So it only remains to show that ϕ¯ is a birational equivalence. By Theorem 2.3.10, it will be enough to find some non-singular q ∈ X/G such that dq ϕ¯ is surjective. To show the existence of q, first note that dim X = dim X0 . Hence (c) implies that dim Tϕ(p) (X) \u0001 dim Tp (X) = dim X = dim X0 . Thus, by Theorem 1.4.11, ϕ(p) ∈ X0 is non-singular and so dp ϕ is an isomorphism. As in the proof of Theorem 1.4.15, there exists an open set U ⊆ X containing p such that the above properties hold for all x ∈ U , that is, x is non-singular, ϕ(x) is non-singular, and dx ϕ is an isomorphism. Now, by §2.5.8(b), π(U ) ⊆ X/G is a non-empty open set. Hence, by Theorem 1.4.11, there exist some non-singular point x ∈ U such that q : = π(x) ∈ π(U ) is non-singular. Now consider the factorization of differentials dx ϕ = dq ϕ¯ ◦ dx π. Since dx ϕ is an isomorphism, we conclude that dq ϕ¯ is surjective, as required. \u0003\n\n2.6 The unipotent variety of the special linear groups This section contains a detailed study of the conjugation action of SLn (k) on the variety of unipotent matrices. This will illustrate many of the ideas in the previous sections. We assume that the reader is familiar with the theory of Jordan normal forms for matrices. Throughout, let G = SLn (k),\n\nwhere k is an algebraically closed field.\n\nA matrix g ∈ G is called unipotent if all eigenvalues of g are 1. Note that, since k is algebraically closed, we have GLn (k) = ZG, where Z is the group of all scalar matrices in GLn (k). Thus, two elements of G are conjugate in G if and only if they are conjugate in GLn (k), and the latter condition holds if and only if the two elements have the same Jordan normal form. In particular, an element g ∈ G is unipotent if and only if g is conjugate to a matrix in Un (k), where Un (k) is the group of all upper triangular matrices with 1 on the diagonal; see Example 1.3.11(a). Recall that the conjugation action\n\n152\n\nAffine varieties and finite morphisms\n\nof G on itself is given by the regular map G × G → G,\n\n(g, x) → gxg−1 .\n\nThe unipotent variety is defined by Un : = {g ∈ G | g unipotent}. We claim that Un is a closed irreducible subset of G (invariant v under conjugation). This is seen as follows. A matrix g ∈ G is unipotent if and only if its characteristitic polynomial is of the form (X − 1)n (where X is an indeterminate). Hence, using the Cayley–Hamilton theorem, we have Un = {g ∈ G | (g − In )n = 0}, and so we see that Un is closed. Now we consider the morphism ϕ : G × Un (k) → Un , (g, u) → gug−1 . By the above remarks, ϕ is surjective. Furthermore, G and Un (k) are irreducible and so G × Un (k) is irreducible by Proposition 1.3.8. Hence, Remark 1.3.2 implies that Un is irreducible, as claimed. We want to understand how Un decomposes into conjugacy classes. The Jordan normal form of an element in Un consists of Jordan blocks corresponding to the eigenvalue 1, and such a block is given by ⎤ ⎡ 1 1 0 ··· 0 . .. ⎢ . .. ⎥ 1 ⎥ ⎢ 0 1 ⎥ ⎢ ∈ Mm (k). Jm = ⎢ .. . . . . . . . . . 0 ⎥ ⎥ ⎢ . ⎣ 0 0 1 1 ⎦ 0 ··· 0 0 1 Thus, we only need to specify how many blocks of a given size there are. Hence, the conjugacy classes in Un are parametrized by the partitions of n. Recall that a partition of n is a finite and weakly decreasing sequence of \u0003non-negative integers λ = (λ1 \u0002 λ2 \u0002 · · · \u0002 λr \u0002 0) such that ri=1 λi = n. We will not distinguish between partitions which have different length but the same non-zero parts. We write λ \u0017 n if λ is a partition of n and denote by Oλ the corresponding conjugacy class in Un . A representative uλ ∈ Oλ is given by the block diagonal matrix with r Jordan blocks of sizes λ1 , . . . , λr on the diagonal. The following result gives some information on the centralizer of a unipotent element. For any x ∈ G, the centralizer is defined by CG (x) = {g ∈ G | gxg−1 = x}.\n\nThe unipotent variety of the special linear groups 2.6.1 Proposition and we have\n\n153\n\nLet λ \u0017 n. Then CG (uλ ) is a closed subgroup\n\ndim CG (uλ ) = n − 1 + 2n(λ)\n\nwhere\n\nn(λ) : =\n\nr \u0001\n\n(i − 1)λi .\n\ni=1\n\nProof We set Cλ : = {A ∈ Mn (k) | Auλ = uλ A}. Then Cλ is a k-subspace of Mn (k) and, hence, a closed irreducible subset of Mn (k); see Exercise 1.8.6. Furthermore, the dimension of Cλ as an algebraic set is the same as the dimension as a k-subspace of Mn (k). Now note that CG (uλ ) is the intersection of Cλ with the zero set of the polynomial given by the determinant. Thus, CG (uλ ) is a hypersurface in Cλ , and so dim CG (uλ ) = dim Cλ − 1 by Example 1.2.16(b). It remains to determine dimk Cλ . For this purpose, write λ = (λ1 \u0002 · · · \u0002 λr > 0). Let A ∈ Mn (k) and partition A into blocks as follows: ⎡ ⎤ A11 A12 · · · A1r ⎢ A21 A22 · · · A2r ⎥ ⎢ ⎥ λ ×λ A = ⎢ .. .. ⎥ , where Aij ∈ k i j . .. ⎣ . . ⎦ . Ar1 Ar2 · · · Arr Then the condition that uλ A = Auλ is equivalent to the condition Jλi Aij = Aij Jλj\n\nfor all i, j ∈ {1, . . . , r}.\n\nHence our task is reduced to finding the set of matrices commuting with two Jordan blocks of possibly different size. A direct computation shows that these are precisely the matrices of the form ⎤ ⎡ a1 a2 · · · aλj ⎤ ⎢ 0 a1 a2 · · · aλj −1 ⎥ ⎡ ⎥ ⎢ . 0 · · · 0 a1 a2 · · · aλj . . . ⎥ ⎢ ⎥ ⎥ ⎢ .. . . . . . . ⎢ 0 · · · 0 0 a1 a2 · · · aλ j−1 ⎥ ⎥ ⎢ ⎢ a 0 · · · 0 a ⎥ ⎥ ⎢ ⎢ .. . . 1 2 .. .. . . . . . . . . . Aij = ⎢ . ⎥ ⎥ or ⎢ 0 · · · 0 0 a ⎥ ⎥ ⎢ ⎢ 1 ⎥ ⎣ 0 · · · 0 0 · · · 0 a1 a2 ⎦ ⎢ 0 · · · 0 0 0 ⎥ ⎢ ⎥ ⎢ 0 · · · 0 0 · · · 0 0 a1 .. .. .. ⎦ ⎣ ... . . . 0 ··· 0 0 0 according to whether λi \u0001 λj or λi > λj . Thus, the dimension of the space of all such matrices Aij is just min{λi , λj } and so\n\n154\n\nAffine varieties and finite morphisms\n\n\u0003 dimk Cλ = i,j min{λi , λj }. Since λi \u0002 λj if i \u0001 j, we can evaluate this sum as follows. dimk Cλ =\n\nr \u0001\n\n\u0001\n\nmin{λi , λj } = n + 2\n\nλj = n + 2n(λ),\n\n1\u0001i 0, and p : = 1 if char(k) = 0. Then CG (uλ )/CG (uλ )◦ is a cyclic group of order gcd(n\u0002 , λ1 , λ2 , . . .), where λ1 , λ2 , . . . are the non-zero parts of λ and n\u0002 is the largest divisor of n which is prime to p. In fact, we have CG (uλ ) = Z(G) CG (uλ )◦ , where Z(G) is the centre of G, and so CG (uλ )/CG (uλ )◦ is seen to be a quotient of Z(G). We leave the proof of these assertions as an exercise to the reader. In order to study some geometric properties of the conjugacy classes of unipotent elements in G, we need a convenient characterization of the Jordan normal form of a unipotent matrix. This is provided by the following result. 2.6.3 Lemma Then we have A ∈ Oλ\n\nLet A ∈ Un be any unipotent matrix and λ \u0017 n.\n\nt \u0001\n\nλ∗i = n − rt (A)\n\nfor 1 \u0001 t \u0001 n,\n\ni=1\n\nwhere we set rt (A) = rankk (A − In )t for all t. Here, the numbers λ∗i are defined as follows. If λ has parts λ1 \u0002 λ2 \u0002 · · · , we set λ∗i = |{j | λj \u0002 i}| for i = 1, 2, . . .. Then λ∗ : = (λ∗1 \u0002 λ∗2 \u0002 · · · λ∗n \u0002 0) is called the conjugate partition (which also is a partition of n). Note that λ and λ∗ determine each other, since (λ∗ )∗ = λ. Proof Assume that A ∈ Oλ . For 1 \u0001 i \u0001 n, let li \u0002 0 be the number of parts of λ which are equal to i. (In other words, li is the number of Jordan blocks of size i in the Jordan normal form of A.)\n\nThe unipotent variety of the special linear groups\n\n155\n\nWith this notation, we have λ∗i = |{j | λj \u0002 i}| = li + li+1 + · · · + ln\n\nfor 1 \u0001 i \u0001 n.\n\nNow the numbers li satisfy the following relations: l1 +2l2 +3l3 l2 +2l3\n\n· · · +(n−1)ln−1 +nln · · · +(n−2)ln−1 +(n−1)ln .. .. . . ln−1 +2ln ln\n\n= rankk (A−In )0 = n = rankk (A−In )1 .. .. . . = rankk (A−In )n−2 = rankk (A−In )n−1 .\n\n(This easily follows from the fact that rankk (Jl − Il )i = l − i for 0 \u0001 i \u0001 l.) The above relations form a system of linear equations for the quantities li . Since the matrix of that system is triangular with an all-unity diagonal, we see that the numbers li are uniquely determined by the vector (r1 (A), . . . , rn (A)). Furthermore, using the above expression for λ∗i , we see that rt (A) = λ∗t+1 + λ∗t+2 + · · · + λ∗n , and so n − rt (A) = λ∗1 + · · · + λ∗t , as required. \u0003 To state the following result, we introduce an order relation on the partitions of n. Assume that λ = (λ1 \u0002 · · · \u0002 λr \u0002 0) and µ = (µ1 \u0002 · · · \u0002 µr \u0002 0) are partitions of n. Then we write def\n\nλ \u0004 µ if and only if\n\nt \u0001 i=1\n\nλi \u0001\n\nt \u0001\n\nµi\n\nfor 1 \u0001 t \u0001 r.\n\ni=1\n\nThen \u0004 is a partial, but in general not total, order. It is called the dominance order on the set of partitions of n. We have λ \u0004 µ if and only if µ∗ \u0004 λ∗ ; see §I.1.11 of Macdonald (1995). \u0007 2.6.4 Lemma For any partition λ \u0017 n, the set C˜λ : = µ\u0002λ Oµ is closed in G. Proof Let µ \u0017 n and take A, B ∈ Un with A ∈ Oλ and B ∈ Oµ . We have already remarked above that µ \u0004 λ ⇔ λ∗ \u0004 µ∗ . Furthermore, by Lemma 2.6.3, the latter condition is equivalent to rt (B) \u0001 rt (A) for all t. Thus, we have shown that C˜λ = {B ∈ Un | rt (B) \u0001 rt (A) for 1 \u0001 t \u0001 n} (where A ∈ Oλ ). (∗) Now, the condition rt (B) \u0001 rt (A) means that the determinants of all square submatrices of B of size strictly bigger than rt (A) are 0.\n\n156\n\nAffine varieties and finite morphisms\n\nSince these determinants are given by polynomials in the coefficients \u0003 of B, we see that C˜λ is closed. As in Proposition 2.5.2, given partitions λ, µ \u0017 n, we write Oµ \u0015 Oλ if Oµ ⊆ Oλ . Thus, by Lemma 2.6.4, we already know that Oµ \u0015 Oλ ⇒ µ \u0004 λ. The following result shows that the reverse implication also holds. 2.6.5 Theorem if µ \u0004 λ.\n\nLet λ, µ \u0017 n. Then we have Oµ ⊆ Oλ if and only\n\nProof By Lemma 2.6.4 we already know that Oµ \u0015 Oλ ⇒ µ \u0004 λ. To prove the reverse implication, let λ1 \u0002 · · · \u0002 λr > 0 be the nonzero parts of λ, and consider the subgroup Uλ (k) ⊆ G consisting of all block matrices of the form ⎡ ⎤ B1r Iλ1 B12 · · · .. .. ⎢ ⎥ . . Iλ2 ⎢ 0 ⎥ B=⎢ . with Bij ∈ kλi ×λj , ⎥ .. .. ⎣ .. ⎦ . . Br−1,r 0 ··· 0 Iλr where the diagonal blocks are the identity matrices of sizes λ1 , . . . , λr . It is readily checked that, for any matrix B as above, we have rt (B) = rankk (B − In )t \u0001 n −\n\nt \u0001\n\nλi\n\nfor 1 \u0001 t \u0001 n.\n\n(†)\n\ni=1\n\nThus, working with the conjugate partition λ∗ and using Lemma 2.6.3, we have rt (B) \u0001 rt (A\u0002 ) for all t, where A\u0002 ∈ Oλ∗ . So, by relation (∗) in the proof of Lemma 2.6.4, we have (†\u0002 )\n\nUλ (k) ⊆ C˜λ∗ .\n\nWe can in fact find some element in Uλ (k) for which equality holds in (†) for all t = 1, . . . , n. Indeed, let uλ\u0002 ∈ Uλ (k) be the matrix ⎡\n\nIλ1 E12 0\n\n··· .. . .. .\n\n0 .. .\n\n⎢ ⎢ 0 Iλ2 E23 ⎢ \u0001 uλ = ⎢ .. . . . . . . ⎢ . 0 ⎣ 0 · · · 0 Iλr−1 Er−1,r 0 ··· 0 0 Iλr\n\n⎢ ⎢ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ where Ei,i+1 = ⎢ ⎢ ⎥ ⎢ ⎦ ⎢ ⎢ ⎣\n\n1\n\n0\n\n0 .. .\n\n1 ..\n\n0 0 .. . 0\n\n. ··· ··· ···\n\n··· .. . .. . 0 0 .. . 0\n\n⎤ 0 .. ⎥ . ⎥ ⎥ ⎥ 0 ⎥ ⎥ 1 ⎥. 0 ⎥ ⎥ .. ⎥ . ⎦ 0\n\nThe unipotent variety of the special linear groups\n\n157\n\n(Note that Ei,i+1 has λi rows and λi+1 columns, where λi \u0002 λi+1 .) \u0003 Then a direct computation shows that rt (uλ\u0002 ) = n − ti=1 λi for all t, and so uλ\u0002 ∈ Oλ∗ by Lemma 2.6.3. Now we claim that Uλ (k) ⊆ Oλ∗ . For this purpose, we consider the morphism ϕλ : G × Uλ (k) → G, (g, u) → gug−1 . We denote by Uλ the image of ϕλ . For example, if λ = (1, 1, . . . , 1), then Uλ (k) = Un (k) and Uλ = Un is the unipotent variety of G. We will now try to prove results about Uλ which are similar to those obtained for Un . The subgroup Uλ (k) certainly is closed and irreducible. (Indeed, since we may place any element of k in the blocks above the diagonal, we have that Uλ (k) is isomorphic to some affine space kN as an affine variety.) Consequently, Uλ is irreducible and, hence, so is Uλ . Furthermore, Uλ is invariant under conjugation by elements of G, that is, Uλ is a union of unipotent classes of G. Hence the same also holds for Uλ . Writing the irreducible closed set Uλ as a union of finitely many conjugacy classes (and then taking the closure of these classes), we see that there exists some λ0 \u0017 n such that Uλ = Oλ0 . Hence it will be enough to show that λ0 = λ∗ . Since uλ\u0002 ∈ Uλ (k) ∩ Oλ∗ , we have Oλ∗ ⊆ Uλ = Oλ0 and so Oλ∗ \u0015 Oλ0 . We have already seen that this implies λ∗ \u0004 λ0 . On the other hand, we have Uλ ⊆ C˜λ∗ by (†\u0002 ) and so Oλ0 ⊆ C˜λ∗ . By definition, this means that λ0 \u0004 λ∗ . Thus, we must have λ0 = λ∗ , as claimed. So we have reached the conclusion that Uλ (k) ⊆ Oλ∗ . Since this holds for all λ, we also have Uλ∗ (k) ⊆ Oλ . Now it will be enough to show that, for each µ \u0017 n with µ \u0004 λ, we can find a representative of Oµ in the subgroup Uλ∗ (k). We will first show how this works in the special case where λ = (r \u0002 s \u0002 0)\n\nand µ = (r − 1 \u0002 s + 1 \u0002 0),\n\nwhere r + s = n.\n\nNow the conjugate partition λ∗ has s parts equal to 2 and the remaining non-zero parts are equal to 1. Similarly, µ∗ has s + 1 parts equal to 2 and the remaining non-zero parts are equal to 1. Then it is obvious that Uµ∗ (k) ⊆ Uλ∗ (k). In particular, the representative uµ\u0002 ∗ ∈ Oµ defined above lies in Uλ∗ (k). Thus, the implication µ \u0004 λ ⇒ Oµ \u0015 Oλ is proved in the above special case. Finally, we show that the general case can be reduced to this special case. Indeed, in order to show that Oµ \u0015 Oλ for all µ \u0017 n with µ = λ and µ \u0004 λ, it is clearly enough to do this for those\n\n158\n\nAffine varieties and finite morphisms\n\npartitions µ which are maximal with respect to these properties. It is readily checked that such a partition µ has the following form: µ = (λ1 , . . . , λi−1 , λi − 1, λi+1 , . . . , λj−1 , λj + 1, λj+1 , . . . , λr \u0002 0), where 1 \u0001 i < j \u0001 r and λi − 1 \u0002 λi+1 ; see §I.1.16 of Macdonald (1995). Now we can find representatives u\u0002λ ∈ Oλ and u\u0002µ ∈ Oµ such that ⎤ ⎤ ⎡ ⎡ 0 0 ··· 0 Jλi 0 0 · · · 0 Jλi −1 ⎢ 0 Jλ 0 · · · 0 ⎥ ⎢ 0 Jλ +1 0 · · · 0 ⎥ j j ⎥ ⎥ ⎢ ⎢ ⎥ ⎥ ⎢ 0 0 ⎢ \u0002 \u0002 0 uλ = ⎢ ⎥ , uµ = ⎢ 0 ⎥, ⎥ ⎥ ⎢ .. ⎢ .. .. .. ⎦ ⎦ ⎣ . ⎣ . . . C C 0\n\n0\n\n0\n\n0\n\nwhere C is a unipotent matrix. Setting m = λi + λj , we consider SLm (k) as a subgroup of SLn (k), by sending a matrix A in SLm (k) to the block diagonal matrix in SLn (k) with diagonal blocks A and In−m . Then it will be enough to show that the conjugacy class in SLm (k) which is labelled by the partition (λi \u0002 λj ) is contained in the Zariski closure of the class labelled by (λi − 1 \u0002 λj + 1). Thus, we are reduced to the special case considered above. \u0003 2.6.6 Example There is a unique maximal partition with respect to the dominance order; this is the partition (n). The corresponding unipotent element u(n) is just the Jordan block of size n. By Theorem 2.6.5, all unipotent classes are contained in the closure of O(n) . Hence, O(n) is dense and open in Un . For this reason, the elements in O(n) are called regular unipotent elements. We claim dim Un = dim O(n) = n(n − 1). Indeed, by Proposition 2.6.1, we have dim CG (u(n) ) = n − 1. Thus, the above assertion follows from Proposition 2.5.3, using that dim G = n2 − 1.\n\n2.7 Bibliographic remarks and exercises For the proof of the strong form of Hilbert’s nullstellensatz, we followed Chapter 7, Exercise 14, of Atiyah and Macdonald (1969). The abstract definition of affine varieties is taken from Steinberg (1974). The results on the images of a morphism can all be found in §I.8\n\nBibliographic remarks and exercises\n\n159\n\nof Mumford (1988). A classical application of Corollary 2.2.8 in combination with the differential criterion for dominance (see Proposition 1.4.15) is the proof that any two Cartan subalgebras of a finite-dimensional complex semisimple Lie algebra are conjugate by an inner automorphism; see §D.22 of Fulton and Harris (1991). The discussion of birational equivalences and normal varieties partly follows Springer (1998) and Shafarevich (1994). The proof of Theorem 2.3.10 essentially appears in §II.5.4 of Shafarevich (1994). The proof of Proposition 2.3.11 follows §5.17 of Atiyah and Macdonald (1969) and §5.2.11 of Springer (1998). In fact, the argument in that proof can be generalized to show that, given an irreducible affine variety X over k, then the integral closure of A[X] in its field of fractions is a finitely generated k-algebra; see §II.5.2, Theorem 4, of Shafarevich (1994). The examples in Section 2.4 are taken from p. 31 and p. 277 of Mumford (1988). For a thorough discussion of the geometric content of Zariski’s main theorem, see §III.9 of Mumford (1988). See also §III.11 of Hartshorne (1977) for a ‘higher point of view’, involving the cohomology theory of schemes. The connectedness criterion Theorem 2.4.6 can be found in any book on algebraic groups; see, for example, §7.5 of Humphreys (1991). The proof of Theorem 2.5.5 and its corollary is adapted from the Appendix to §2.11 of Steinberg (1974). For a different argument, see 5.3.2 of Springer (1998). The construction of the affine quotient by a finite group in Theorem 2.5.10 appears in §5.52 of Fogarty (1969). More generally, it is known that if H is closed subgroup of an affine algebraic group G, then one can define on G/H a natural structure as a ‘quasi-projective’ variety. This fact can be proved using the general form of Zariski’s main theorem (for this approach, see §5.5, of Springer (1998)), or one can proceed as in Chapter 12 of Humphreys (1991), using a purely algebraic result due to Chevalley; see §5.3A of Humphreys (1991). See also Chapter 2 of Borel (1991) for a more general discussion of quotients. For more on invariants of finite groups as in Example 2.5.11, see Chapter 7 of Cox et al. (1992) and Sturmfels (1993). The results on unipotent classes in SLn (k) in Section 2.6 have been generalized to symplectic and orthogonal groups and, more generally, to groups of Lie type; see Spaltenstein (1982). In the exercises, all affine varieties and algebraic groups are defined over an algebraically closed ground field k.\n\n160\n\nAffine varieties and finite morphisms\n\n2.7.1 Exercise Let V be a finite dimensional k-vector space. Let A[V ] be the k-algebra generated by all elements of the dual space Homk (V, k): A[V ] : = k[λ | λ ∈ Hom(V, k)] ⊆ Maps(V, k). Show that (V, A[V ]) is an affine variety. Furthermore, if {e1 , . . . , en } is a basis of V and {λ1 , . . . , λn } is the dual basis of Homk (V, k), then the map V → kn , v → (λ1 (v), . . . , λn (v)), is an isomorphism of affine varieties. 2.7.2 Exercise Let A = {0} be a finitely generated k-algebra with 1 which has no nilpotent elements except 0. Let X be the set of all maximal ideals in A. For every x ∈ X, there is a unique k-algebra homomorphism λx : A → k with kernel x. For f ∈ A, we define a ˙ function f˙ : X → k by f(x) : = λx (f) for x ∈ X. Show that the assignment f → f˙ is injective. Thus, we may regard A as a k-subalgebra of Maps(X, k). Then show that (X, A) is an affine variety. [Hint. If λx (f) = 0 for all x ∈ X, then f lies in the intersection of all maximal ideals of A. To see that this intersection is 0, one can argue as follows. Write A = k[X1 , . . . , Xn ]/I for some ideal I and define V√: = V(I) ⊆ kn . Then, by Hilbert’s nullstellensatz, we have I(V ) = I = I, thanks to the assumption on A. Hence (V, A) is an affine variety. Now use the injectivity of the operator I.] 2.7.3 Exercise Let A be a commutative ring with 1, and let I ⊆ A be an ideal and M be a finitely generated A-module. Prove the following version of Nakayama’s lemma: If M = I.M , then there exists some f ∈ 1 + I such that f.M = {0}. Furthermore, assume that A is local and I ⊂ A is the unique maximal ideal. Show that, if N ⊆ M is a submodule with M = I.M + N , then M = N . [Hint. Use a ‘determinantal trick’ as in the proof of Lemma 2.1.1.] 2.7.4 Exercise Let A be a commutative local ring (with 1). Let m be the unique maximal ideal of A, and denote by F = A/m the residue field. Then we may consider m/m2 as a F -vector space where\n\nBibliographic remarks and exercises\n\n161\n\nthe operation of F is given by (x + m).(a + m2 ) : = xa + m2\n\nfor x ∈ A and a ∈ m.\n\nAssume that m is finitely generated, and let d \u0002 0 be minimal such that m can be generated by d elements. Show that d = dimF (m/m2 ). [Hint. First check that, if m is generated as an ideal by a1 , . . . , ad , then the images of a1 , . . . , ad generate m/m2 as a F -vector space. Conversely, assume that dimF (m/m2 ) = l, and let {b1 + m2 , . . . , bl + m2 } be an F -basis, where bi ∈ m. To show that m is generated by b1 , . . . , bl , consider the left A-module M : = m/(b1 , . . . , bl ) and check that mM = M . Then apply Exercise 2.7.3.] 2.7.5 Exercise Let R ⊆ A be integral domains such that A is integral over R. Let K be the field of fractions of A and F ⊆ K be the field of fractions of R. Assume that K is a finite-dimensional F -vector space. The purpose of this exercise is to establish some basic properties of the trace map. (This is used in the proof of Proposition 2.3.11.) For any x ∈ K, we define αx : K → K by αx (y) = xy. Then αx is F -linear, and the trace of x is defined as TK/F (x) : = Trace(αx ) ∈ F. Prove the following statements. (a) The map α : K → EndF (K), x → αx , is an F -algebra homomorphism. For x ∈ K, the minimal polynomial of x over F equals the minimal polynomial of αx . (b) TK/F (ax + by) = aTK/F (x) + bTK/F (y) for all x, y ∈ K and a, b ∈ F . (c) Assume that R is integrally closed in F . Then we have TF/K (x) ∈ R for all x ∈ K which are integral over A. (d) If K ⊇ F is separable, there exists some x ∈ K such that TK/F (x) = 0. Deduce that the bilinear form K × K → F , (x, y) → TK/F (xy), is non-degenerate. [Hint. (c) Let x ∈ K be integral over A. Then x is integral over R, and so there exists a monic g ∈ R[X] such that g(x) = 0. Let f ∈ F [X] be the minimal polynomial of x over F . Then f divides g, and so all eigenvalues of αx are integral over R, in some finite extension of K. Hence Trace(αx ) ∈ F is integral over R and so Trace(αx ) ∈ R since R is integrally closed in F . (d) This is a standard\n\n162\n\nAffine varieties and finite morphisms\n\nresult on separable field extensions; see §VIII.5 of Lang (1984), for example.] 2.7.6 Exercise This exercise is concerned with some properties of constructible sets. Let X be any topological space. (a) Show that finite unions and finite intersections of constructible subsets are constructible. Show that the complement of a constructible subset is constructible. (b) Assume that X is noetherian and that Y ⊆ X is a constructible subset. Show that there exists a subset U ⊆ Y which is dense and open in Y . [Hint.\u0007(b) First assume that Y is irreducible. Then we can write Y = ni=1 (Oi ∩ Fi ), where each Oi ⊆ X is open, each Fi ⊆ X is closed, and Fi \u0002 F1 = Y for i > 1. Now we can take U : = O1 ∩F1 . In the general case, apply this argument to the irreducible components of Y .] 2.7.7 Exercise Let X be an irreducible affine variety. Let K be the field of fractions of A[X]. Show that the following conditions are equivalent: (a) A[X] is integrally closed in K; (b) A[X]f is integrally closed in K for all non-zero f ∈ A[X]; (c) Op is integrally closed in K for all p ∈ X. [Hint. ‘(a) ⇒ (b)’ Let x ∈ K be integral over A[X]f . Write down an equation for integral dependence of x over A[X]f . Multiplying that equation by a sufficiently large power of f yields an equation of integral dependence of xf m over A[X], for some m \u0002 0. Hence, we have xf m ∈ A[X] and so x ∈ A[X]f . ‘(b) ⇒ (c)’ Similarly. ‘(c) ⇒ (a)’ Use Lemma 2.3.6.] 2.7.8 Exercise The purpose of this exercise is to show that there exist normal varieties which have singular points. Let V be one of the following algebraic sets. (a) V = {(x, y, z) ∈ k3 | x2 + y2 = z 2 }, where char(k) = 2. (b) V = {(x, y, z) ∈ k3 | xy = z 2 }.\n\nBibliographic remarks and exercises\n\n163\n\nShow that V is irreducible with dim V = 2. Furthermore, V is normal and (0, 0, 0) is a singular point of V . The first example is taken from p. 277 of Mumford (1988); the second one appears (with hints) in §II.5.1 of Shafarevich (1994). 2.7.9 Exercise Mn (k) is dense.\n\nShow that the set of all diagonalizable matrices in\n\n[Hint. Let Sn (k) ⊆ Mn (k) be the set of diagonal matrices, and consider the morphism ϕ : GLn (k) × Sn (k) → Mn (k), (g, t) → g−1 tg. Determine the dimension of the closure of the image of ϕ by looking at the fibres ϕ−1 (y) for suitable diagonal matrices and using Corollary 2.2.9.] 2.7.10 Exercise Let G be an affine algebraic group and X be a G-variety. Let x ∈ X and Ox be the corresponding G-orbit. Let G◦ be the connected component of 1 ∈ G and Ox◦ be the G◦ -orbit. (By Proposition 1.3.13, G◦ is a closed subgroup of finite index in G.) Show that Ox◦ is open and closed in Ox . [Hint. Since [G : G◦ ] < ∞, it is clear that Ox is a finite union of G◦ -orbits, and if Oy◦ is such an orbit, then there exists some g ∈ G such g.x = y. So there exist g1 , . . . , gr ∈\u0007G such that Ox = \u0016r that ◦ r ◦ , where g = 1. This implies O = O 1 x i=1 gi .x i=1 Ogi .x . Since all ◦ ◦ Ogi .x have the same dimension, we have Og◦i .x ⊆ Ogj .x for i = j, and ◦ so Ogi .x ∩ Ox = Og◦i .x . Thus, Og◦i .x is closed in Ox .] 2.7.11 Exercise The purpose of this exercise is to show that the notion of affine quotients in §2.5.8 is compatible with direct products. For i = 1, 2, let Xi be a Gi -variety and assume that an affine quotient (X0i , πi ) exists. (a) Show that X1 ×X2 is a (G1 ×G2 )-variety, where the operation is given by the formula (g1 , g2 ).(x1 , x2 ) = (g1 .x1 , g2 .x2 ). (b) Consider the morphism ϕ : X1 ×X2 → X01 ×X02 which sends (x1 , x2 ) to (π1 (x1 ), π2 (x2 )). Show that the pair (X01 × X02 , ϕ) is an affine quotient of X1 × X2 by G1 × G2 . [Hint. Consider the three conditions in §2.5.8. To check (b), note that it is enough to consider a subset U ⊆ X01 × X02 which is of the form U = U1 × U2 with Ui ⊆ X0i . Furthermore, U is open if and only if U1 and U2 are open.]\n\n164\n\nAffine varieties and finite morphisms\n\n2.7.12 Exercise Let G = SL2 (k). Show that every element of G is conjugate to one of the following matrices: x 0 1 1 1 0 , (x = ±1). ± , ± 0 1 0 1 0 x−1 Show that the centralizers of these matrices are respectively G, ZU , and H, where Z is the centre of G, U is the group of all upper triangular matrices with an all-unity diagonal, and H is the group of diagonal matrices in G. Determine explicitly which of these centralizers are connected. 2.7.13 Exercise Let B be a connected affine algebraic group. Assume that there exist closed connected subgroups U, T ⊆ B such that U is normal in B, T is abelian, B = U.T , and U ∩ T = {1}. Furthermore, assume that CB (t) = T for some t ∈ T . Then show that U = [B, B]. [Hint. First note that [B, B] ⊆ U . Now, since CB (t) = T , we have t−1 ut = u for all non-unit u ∈ U , and so the morphism ϕ : U → U , u → [t, u] = t−1 u−1 tu, is injective. Hence Corollary 2.2.9 implies that ϕ is dominant. Now, we have ϕ(U ) ⊆ [B, B] ⊆ U . Finally, use the fact that [B, B] is closed by Example 2.4.7.]\n\n3 Algebraic representations and Borel subgroups One of the principal aims of this chapter is to prove the conjugacy (and further basic properties) of Borel subgroups, that is, maximal closed connected solvable subgroups in an affine algebraic group. This will require various preparations concerning algebraic representations, which certainly is an interesting topic in its own right. To begin with, in Section 3.1, we study linear actions of algebraic groups on vector spaces. We prove the Lie–Kolchin theorem which shows that a solvable closed connected group of invertible matrices is always conjugate to a group of triangular matrices. At the end of this section, we give the general definition of semisimple and reductive algebraic groups. Typical examples are SLn (k) and GLn (k), respectively. The linear action of an algebraic group on a vector space V will lead us to consider the induced action on projective space P(V ). This is no longer an affine variety, but it carries a topology defined by the vanishing of homogeneous polynomials. One of the main properties of that topology is the so-called ‘completeness of projective space’; this will be discussed in Section 3.2. In Section 3.3, we introduce the related grassmannian and flag varieties. In Section 3.4, by combining practically everything that we did so far, we establish the main properties of Borel subgroups in algebraic groups. As a highlight, we obtain the combinatorial characterization of the Bruhat–Chevalley order in groups with a BN -pair in Theorem 3.4.8. This will be completed in Section 3.5, where we obtain some detailed structure results on connected solvable groups.\n\n166\n\nAlgebraic representations and Borel subgroups\n\n3.1 Algebraic representations, solvable groups, and tori In this section, we obtain the first substantial results on solvable algebraic groups. All this will be based on some notions related to algebraic representations. Throughout, k is algebraically closed. 3.1.1 Definition Let V be a finite-dimensional k-vector space and G be an affine algebraic group over k. Let ρ : G → GL(V ) be a homomorphism of abstract groups. Fixing a basis {e1 , . . . , en } of V , we have equations ρ(g)(ej ) =\n\nn \u0001\n\naij (g) ei ,\n\nwhere aij (g) ∈ k.\n\ni=1\n\nLet CF(V, ρ) be the subspace of Maps(G, k) which is spanned by the set of functions aij : G → k (1 \u0001 i, j \u0001 n). Since any two bases of V can be transformed into each other by a linear transformation, CF(V, ρ) is independent of the choice of the basis; it is called the coefficient space of (V, ρ). We say that V is an algebraic G-module, or that ρ is an algebraic representation of G, if CF(V, ρ) ⊆ A[G]. By Lemma 2.4.2, the map G → GLn (k), g → (aij (g)), is a homomorphism of affine algebraic groups. Conversely, given a homomorphism ρ : G → GLn (k) of affine algebraic groups, then V = kn is an algebraic G-module. 3.1.2 Invariant subspaces Assume that V is an algebraic G-module, and let U ⊆ V be a G-invariant subspace, that is, we have ρ(g)(U ) ⊆ U for all g ∈ G. Then U and V/U (with the induced actions of G) also are algebraic G-modules. This is seen as follows. Assume that 0 < m = dim U < dim V = n and let {e1 , . . . , en } be a basis of V such that {e1 , . . . , em } is a basis of U . Then the matrices A(g) := (aij (g)) ∈ GLn (k) have a block-triangular shape\n\nA1 (g) ∗ A(g) = , 0 A2 (g) where A1 (g) ∈ GLm (k) describes the representation of G on U (with respect to the basis {e1 , . . . , em }) and A2 (g) ∈ GLn−m (k) describes the representation of G on V/U (with respect to the basis given by the images of {em+1 , . . . , en } in V/U ). Now it may happen that V = {0}\n\nAlgebraic representations, solvable groups, and tori\n\n167\n\ndoes not have any non-trivial proper G-invariant subspaces at all. In this case, we say that V is irreducible. Otherwise, we can always find a chain of G-invariant subspaces {0} = V0 \u0002 V1 \u0002 V2 \u0002 · · · \u0002 Vr = V such that Vi /Vi−1 is irreducible for all i. Choosing a basis of V which is adapted to this chain, we have ⎡ ⎤ ∗ ··· ∗ A1 (g) .. ⎥ . ⎢ A2 (g) . . . ⎥ ⎢ 0 A(g) = ⎢ . for all g ∈ G, ⎥ . . ⎣ .. .. .. ∗ ⎦ 0 ··· 0 Ar (g) where Ai (g) describes the (irreducible) representation of G on Vi /Vi−1 . 3.1.3 Definition A character of G is a homomorphism of algebraic groups from G into the multiplicative group Gm = GL1 (k). Since Gm is a principal open subset of k, every character can be regarded as an element of A[G]. We will use frequently the fact that pairwise different characters are linearly independent in A[G]. This follows from Dedekind’s theorem which says (in a completely general setting) that pairwise different homomorphisms of a group into the multiplicative group of a field are linearly independent (see Chapter VIII, §4, of Lang (1984)). The set of all characters of G is a group (under pointwise defined multiplication), called the character group of G and denoted by X(G). We write this group additively, that is, we have (χ1 +χ2 )(g) = χ1 (g) χ2 (g) for all g ∈ G. The identity element of X(G) is given by the unit character 1G : G → Gm , g → 1. 3.1.4 Example Consider the group G = Gm itself. For any m ∈ Z, we have an element χm ∈ X(G) given by χm (x) = xm for x ∈ Gm . We claim that X(G) = {χm | m ∈ Z} ∼ = Z. Indeed, first note that A[G] = k[X ±1 ] is the ring of Laurent polynomials in one variable X. Thus, we have χm = X m under the identification X(G) ⊆ A[G]. Now let χ ∈ X(G). Then we can find ±1 ] such that χ(x) = f(x) for all x ∈ G. Now some non-zero \u0003r f ∈ k[X m i write f = i=1 a\u0003 i X , where ai ∈ k and mi ∈ Z. Then we obtain the relation χ = ri=1 ai χmi . So Dedekind’s theorem shows that we must have χ = χmi for some i, and the above claim is proved.\n\n168\n\nAlgebraic representations and Borel subgroups\n\nMore generally, let n \u0002 1 and define Dn (k) = Gm × · · · × Gm (n factors). Then we have X(Dn (k)) ∼ = Zn , and A[Dn (k)] is spanned by X(Dn (k)) over k. This follows from the above statement and the fact that, if we have a direct product G1 ×G2 , then there is a natural isomorphism X(G1 × G2 ) ∼ = X(G1 ) × X(G2 ). (We leave this as an exercise for the reader.) Thus, we have X(Dn (k)) = \u0005χ1 , . . . , χn \u0006, where χi ∈ X(Dn (k)) is the projection onto the ith factor. 3.1.5 Lemma We define\n\nLet ρ : G → GL(V ) be an algebraic representation.\n\nVχ := {v ∈ V | ρ(g)(v) = χ(g)v for all g ∈ G} ⊆ V for any χ ∈ X(G). There are only finitely many χ ∈ X(G) such that Vχ = {0}, and the sum of these non-zero subspaces is direct. These subspaces are the weight spaces of G on V . Proof Let χ1 , . . . , χr ∈ X(G) be pairwise different characters such that Vχi = 0 for all i. We claim that the sum of the subspaces Vχi is direct. We proceed by induction on r. If r = 1, there is nothing to prove. Now let r \u0002 2 and vi ∈ Vχi be such that v1 + · · · + vr = 0. We must show that vi = 0 for all i. Let i \u0002 2 and gi ∈ G be such that χ1 (gi ) = χi (gi ). Then we have two equations: χ1 (gi )v1 + χ2 (gi )v2 + · · · + χr (gi )vr = ρ(gi )(v1 + · · · + vr ) = 0, χ1 (gi )v1 + χ1 (gi )v2 + · · · + χ1 (gi )vr = χ1 (gi )(v1 + · · · + vr ) = 0. Subtracting these equations, we obtain a linear relation among v2 , . . . , vr , where the coefficient of vi is non-zero. So, by induction, we conclude that vi = 0. This holds for 2 \u0001 i \u0001 r and, consequently, also for i = 1. Thus, the above claim is proved. Since dim V < ∞, it is now clear that {χ | Vχ = {0}} is finite. \u0003 The next result is a technical tool which will be used in the proof of Theorem 3.1.7 and later again in the proof of Theorem 3.4.2. 3.1.6 Lemma Let G be a connected affine algebraic group and V be an algebraic G-module. Let H ⊆ G be a closed normal subgroup, and assume that there exists some ψ ∈ X(H) such that Vψ = {0}. Then Vψ is G-invariant. If, furthermore, H = [G, G], then ψ is the trivial character.\n\nAlgebraic representations, solvable groups, and tori\n\n169\n\n\u0003 Proof We set W := χ∈X(H) Vχ ⊆ V . Since Vψ = {0}, we have W = {0}. Now, by Lemma 3.1.5, there are only finitely many χ ∈ X(H) such that Vχ = {0}, and their sum is direct. Let X1 ⊆ X(H) be the finite set of all χ ∈ X(H) such that Vχ = {0}. For g ∈ G, the map γg : H → H, h → ghg−1 , is an isomorphism of algebraic groups. Hence, for each χ ∈ X(H), we also have χg := χ ◦ γg ∈ X(H), and a straightforward computation shows that ρ(g)(Vχ ) = Vχg . Thus, each g ∈ G induces a permutation of the elements in X1 . Then, since X1 is finite, we see that StabG (ψ) = {g ∈ G | ψ(gxg−1 ) = ψ(x) for all x ∈ G} is a closed subgroup of G of finite index. Hence, since G is connected, the above stabilizer must be all of G and so Vψ is a G-invariant subspace. Now assume that H = [G, G], and consider the induced action of G on Vψ . Let ρ : G → GL(Vψ ) be the corresponding algebraic representation. Then we have ρ(h) = ψ(h)idVψ for all h ∈ H. Furthermore, for all x, y ∈ G, we have det(ρ(x−1 y−1 xy)) = 1, and so ψ(h)r = det(ρ(h)) = 1 for all h ∈ H, where r = dimk Vψ . Hence the image of ψ : H → Gm is finite. Since H is connected by Example 2.4.7, the image of ψ is irreducible. Consequently, the image is a singleton set, and so ψ(h) = 1 for all h ∈ H. \u0003 3.1.7 Theorem (Lie–Kolchin) Assume that G is a connected solvable affine algebraic group, and let V be an algebraic G-module. Then there exists a basis {e1 , . . . , en } of V such that the corresponding matrices (aij (g)) ∈ GLn (k) all lie in Bn (k). In particular, every closed connected solvable subgroup of GLn (k) is conjugate to a subgroup of Bn (k). Proof First we show that there exists some non-zero e1 ∈ V such that \u0005e1 \u0006 is a G-invariant subspace of V . We do this by induction on dim G. If dim G = 0, then G = {1} and there is nothing to prove. Now assume that dim G > 0. Let G\u0002 = [G, G]; by Example 2.4.7, this is a closed connected subgroup of G. Since G is solvable, we have dim G\u0002 < dim G. So, by induction, there exists some G\u0002 -invariant 1-dimensional subspace of V . This means that we have Vψ = {0} for some ψ ∈ X(G\u0002 ). By Lemma 3.1.6, Vψ is G-invariant and ψ = 1G\u0002 . So G\u0002 acts trivially on Vψ . For each g ∈ G, denote by ρg : Vψ → Vψ the induced linear map. Then the fact that G\u0002 = [G, G] acts trivially on Vψ means that the operators ρg commute with each other. Hence, by\n\n170\n\nAlgebraic representations and Borel subgroups\n\na standard result in linear algebra (note also that k is algebraically closed), there exists a non-zero common eigenvector, that is, a nonzero e1 ∈ V such that \u0005e1 \u0006 is a G-invariant subspace of V . Thus, the above claim is proved. Now we consider the algebraic G-module V /\u0005e1 \u0006. If this is nonzero, then we can find some e2 ∈ V \\ \u0005e1 \u0006 such that e2 defines a 1-dimensional G-invariant subspace in V /\u0005e1 \u0006. Going on in this way, we obtain a basis {e1 , . . . , en } of V such that the matrices (aij (g)) are all upper triangular, as desired. \u0003 3.1.8 Remark The assumption that G is connected is essential in the above result. For example, let G = S3 be the symmetric group on three letters. Then −1 0 1 1 , , (2, 3) → (1, 2) → 1 1 0 −1 defines an irreducible representation ρ : G → GL2 (C). We shall now study diagonalizable groups and tori. Recall that Tn (k) is the closed subgroup of all invertible diagonal matrices in GLn (k). This group clearly is isomorphic to Dn (k) := Gm × · · · × Gm (n factors). 3.1.9 Proposition Let G be a connected affine algebraic group over k. We set p := char(k) if char(k) > 0, and p := 1 if char(k) = 0. Let Gfin := {g ∈ G | g d = 1 for some d \u0002 1 such that gcd(p, d) = 1}. Then the following conditions are equivalent. (a) G is abelian and G = Gfin . (b) For every homomorphism of algebraic groups ρ : G → GLn (k), there exists some T ∈ GLn (k) such that T ρ(G)T −1 ⊆ Tn (k). (c) There exists a closed embedding G ⊆ GLn (k) such that G ⊆ Tn (k). (d) G is isomorphic to Dd (k) for some d \u0002 1. A group G satisfying the above conditions will be called a torus. Proof ‘(a) ⇒ (b)’ Let ρ : G → GLn (k) be a homomorphism. For g ∈ Gfin , we have g d = 1, where d is coprime to p. This implies that the minimal polynomial of ρ(g) ∈ GLn (k) has distinct roots,\n\nAlgebraic representations, solvable groups, and tori\n\n171\n\nand so ρ(g) is diagonalizable. Since G is abelian, we can even find some P ∈ GLn (k) such that P ρ(g)P −1 ∈ Tn (k) for all g ∈ Gfin . (This is a standard result in linear algebra.) Thus, we may assume without loss of generality that ρ(Gfin ) ⊆ Tn (k). Since ρ is continuous and Tn (k) is closed in GLn (k), we conclude that ρ(G) = ρ(Gfin ) ⊆ ρ(Gfin ) ⊆ Tn (k). ‘(b) ⇒ (c)’ By Corollary 2.4.4, there exists a closed embedding ρ : G → GLn (k). Thus, by hypothesis (b), there exists some T ∈ GLn (k) such that T ρ(G)T −1 ⊆ Tn (k). Hence we obtain a closed embedding G → Tn (k). ‘(c) ⇒ (d)’ By assumption, there exists a closed embedding ϕ : G → Dn (k). Consider the corresponding comorphism ϕ∗ : A[Dn (k)] → A[G]. First note that ϕ∗ induces a group homomorphism ϕ∗ : X(Dn (k)) → X(G). We claim that this induced homomorphism is surjective. This is seen as follows. Let ψ ∈ X(Dn (k)) ⊆ A[G]. Since ϕ is a closed embedding, ϕ∗ : A[Dn (k)] → A[G] is surjective; see Proposition 2.2.1. So there exists some f ∈ A[Dn (k)] such that ϕ∗ (f ) = ψ. By the discussion in Example \u0003r3.1.4, f can be written as a linear combination of characters: f = i=1 ai ψi say, where ai ∈ k. \u0003r ∗ ∗ Then we obtain ψ = ϕ (f ) = i=1 ai ϕ (ψi ). By Dedekind’s theorem, we must have ψ = ϕ∗ (ψi ) for some i, as required. Thus, as claimed, ϕ∗ : X(Dn (k)) → X(G) is surjective, and A[G] is spanned by X(G). Now X(G) is a finitely generated abelian group, since X(Dn (k)) ∼ = Zn . Furthermore, we claim that X(G) is free abelian. Indeed, if this were not the case, then—as is well-known—there would exist some element 1G = χ ∈ X(G) of finite order, e \u0002 2 say. But then χ(G) ⊆ {x ∈ Gm | xe = 1} would be a finite set, contradicting the assumption that G is connected. Thus, X(G) is a free abelian group of finite rank. Let {ε1 , . . . , εd } be a basis of X(G), and consider the homomorphism ψ : G → Dd (k), g → (ε1 (g), . . . , εd (g)). Then ψ ∗ will be surjective, since X(G) spans A[G]. Hence ψ is a closed embedding by Proposition 2.2.1. In particular, ψ is injective and so dim G \u0001 d. On the other hand, the fact that X(G) is free abelian means that all monomials in ε1 , . . . , εd are linearly independent in A[G]. Hence we have dim G = ∂k (A[G]) \u0002 d and so, finally, ψ is an isomorphism. (d) ⇒ (a) By assumption, G certainly is abelian. Furthermore, every element of finite order in G has order prime to p. It remains to show that the set of these elements is dense in G. For this purpose, it is actually enough to consider the case d = 1. But it is easily checked that Gm contains infinitely many roots of unity of order prime to p.\n\n172\n\nAlgebraic representations and Borel subgroups\n\nThe closure of that set will be an algebraic subset of Gm of dimension \u0002 1 and, hence, must be equal to Gm . \u0003 3.1.10 Proposition (Rigidity of tori) Let G be an affine algebraic group and T ⊆ G be a closed subgroup which is a torus. Then NG (T )◦ = CG (T )◦ . Proof It is clear that CG (T ) ⊆ NG (T ) and so CG (T )◦ ⊆ NG (T )◦ . To prove the reverse inclusion, let N := NG (T )◦ and consider the closed set X := {t ∈ T | ntn−1 = t for all n ∈ N }. We must show that X = T . To prove this, it will be enough to show that X contains a dense subset of T . Now, by Proposition 3.1.9(a), the set of elements of finite order in T is dense. So it will be enough to show that any t ∈ T of finite order lies in X. Assume that t has order d. Then consider the morphism ψt : N → T , n → ntn−1 . We have ψt (n)d = (ntn−1 )d = ntd n−1 = 1 for all n ∈ N . Thus, the subset ψt (N ) of T consists of elements whose order divides d. Now, by Proposition 3.1.9, T is isomorphic to Dn (k) for some n \u0002 1. It is easily checked that, in Dn (k), there are only finitely many elements of a given order (since this is the case in Gm ). Hence, ψt (N ) is a finite set. Since N is irreducible, ψt (N ) actually is a singleton set. Thus, we have ntn−1 = t for all n ∈ N , as required. \u0003 3.1.11 The unipotent radical Let G be a connected solvable affine algebraic group. The unipotent radical of G is defined as the subgroup \u001b Ru (G) := ker(χ). χ∈X(G)\n\n(For example, if G = Bn (k), then we have Ru (G) = Un (k).) Clearly, Ru (G) is a closed normal subgroup of G. Furthermore, since G is a noetherian topological space, there exist finitely many char\b acters χ1 , . . . , χn ∈ X(G) such that Ru (G) = ni=1 ker(χi ). Then we obtain a homomorphism ϕ : G → Dn (k) by sending g ∈ G to (χ1 (g), . . . , χn (g)), such that ker(ϕ) = Ru (G). 3.1.12 Lemma Let G be connected and solvable. Then the unipotent radical Ru (G) is nilpotent. Furthermore, if ρ : G → GLn (k) is any homomorphism of algebraic groups, then the eigenvalues of ρ(g) (g ∈ Ru (G)) are all 1.\n\nAlgebraic representations, solvable groups, and tori\n\n173\n\nProof Let ρ : G → GLn (k) be any homomorphism of algebraic groups. By Theorem 3.1.7, we may assume that ρ(G) ⊆ Bn (k). Now recall that Bn (k) = Un (k)Tn (k). For 1 \u0001 i \u0001 n, let εi ∈ X(Bn (k)) be the character which assigns to a matrix in Bn (k) its ith diagonal coefficient. Then we have εi ◦ ρ ∈ X(G) for all i. Now, if g ∈ Ru (G), then εi (ρ(g)) = 1 for all i, and so the diagonal coefficients of ρ(g) are 1, as claimed. To show that Ru (G) is nilpotent, choose ρ to be a closed embedding. Then Ru (G) is isomorphic (as an abstract group) to a subgroup of Un (k) and, hence, Ru (G) is nilpotent (since Un (k) is). \u0003 3.1.13 Semisimple and reductive groups Let G be a connected affine algebraic group. Then the radical of G is defined as \u001d \u001c \u001e \u001d S is a closed connected R(G) := S ⊆ G \u001d\u001d ⊆ G. solvable normal subgroup It is clear that R(G) is a normal subgroup. By Theorem 2.4.6, R(G) is a closed connected subgroup of G; furthermore, there are finitely many closed connected solvable normal subgroups S1 , . . . , Sn such that R(G) = S1 · · · Sn . Now it is a general fact about (abstract) groups that the product of two normal solvable subgroups of a given group is again a normal solvable subgroup. Thus, we conclude that R(G) is solvable. In fact, R(G) is the unique maximal closed connected solvable normal subgroup of G. We say that G is a semisimple group if R(G) = {1}. We say that G is a reductive group if R(G) is a torus. 3.1.14 Lemma Let G be a connected reductive affine algebraic group. Then R(G) = Z(G)◦ , where Z(G) is the centre of G. Furthermore, [G, G] is a connected semisimple group and we have |R(G) ∩ [G, G]| < ∞. Proof Since R(G) is a torus, we can apply Proposition 3.1.10 and conclude that G = NG (R(G)) = CG (R(G)). So we have R(G) ⊆ Z(G)◦ , while the reverse inclusion is clear anyway (since Z(G)◦ is a closed connected abelian normal subgroup). Now, [G, G] certainly is connected by Example 2.4.7. We also note that the radical of [G, G] is contained in R(G). (Indeed, the radical of [G, G] is also a closed connected solvable normal subgroup in G.) Hence, it will be enough to show that |R(G) ∩ [G, G]| < ∞, For this purpose, we choose a closed embedding G ⊆ GLn (k). Then V = kn is an algebraic\n\n174\n\nAlgebraic representations and Borel subgroups\n\nr G-module and, by Proposition 3.1.9, we can write V = i=1 Vi , where Vi := Vχi and χi ∈ X(R(G)). Since R(G) lies in the centre of G, we have g.Vi ⊆ Vi for all g ∈ G and all i; see Lemma 3.1.6. Thus, as in §3.1.2, we actually obtain an injective representation ρ : G → GL(V1 ) × · · · × GL(Vr ),\n\ng → (ρ1 (g), . . . , ρr (g)),\n\nwhere ρi (z) = χi (z) idVi for all z ∈ R(G) and all i. Now, if we take an element g ∈ [G, G], then we can write g as a product of commutators, and so ρi (g) is seen to have determinant 1. If g also lies in R(G), then ρi (g) will be a scalar multiple of the identity for each i. But there are only finitely many scalar matrices whose determinant is 1. Thus, |R(G) ∩ [G, G]| < ∞, as required. \u0003 3.1.15 Example The group GLn (k) is reductive and SLn (k) is semisimple. Indeed, by Theorem 3.1.7, the radical of GLn (k) is contained in Bn (k). But Bn (k) is conjugate to the subgroup Bn− (k) of all lower triangular invertible matrices, and so the radical will be contained in Bn (k) ∩ Bn− (k) = Tn (k). Thus, if g ∈ R(GLn (k)), then xgx−1 must be a diagonal matrix for all x ∈ GLn (k). Taking for x suitable upper unitriangular matrices, we can conclude that g must be a scalar matrix. Hence, we have R(GLn (k)) = Z(G) ∼ = Gm and so this is a torus. Now consider SLn (k). Since det : GLn (k) → k× is a group homomorphism, we have [GLn (k), GLn (k)] ⊆ SLn (k). On the other hand, by Example 2.4.10, we have SLn (k) = [SLn (k), SLn (k)] and so we must have SLn (k) = [GLn (k), GLn (k)]. It remains to apply Lemma 3.1.14. In Definition 3.4.5, we will introduce a general class of reductive groups.\n\n3.2 The main theorem of elimination theory In the previous sections, we have studied actions of algebraic groups on affine varieties. The linear action of an algebraic group G on an algebraic G-module V will now lead us to consider the induced action on the projective space P(V ). We will see that this basically amounts\n\nThe main theorem of elimination theory\n\n175\n\nto working with affine algebraic sets which are defined by homogeneous polynomials. A fundamental result about such sets is the main theorem of elimination theory,1 which we establish in Theorem 3.2.7. 3.2.1 Example Let G be an affine algebraic group, and assume that ρ : G → GL(V ) is an algebraic representation, as in Definition 3.1.1. Then, regarding V as an affine variety as in Exercise 2.7.1, we see that V is a G-variety, where the action is given by g.v = ρ(g)(v) for g ∈ G and v ∈ V . We can go one step further. Consider the multiplicative group Gm of k. We have a natural action of Gm on V , simply given by scalar multiplication. We can combine this action of Gm with that of G to obtain an action (Gm × G) × V → V,\n\n((λ, g), v) → λρ(g)(v).\n\nSince the above map is still a morphism of affine varieties, we see that V is a Gm × G-variety. The set of non-zero Gm -orbits on V is the projective space P(V ) := {\u0005v\u0006 | 0 = v ∈ V }. We have an induced action of G on P(V ) given by g.\u0005v\u0006 = \u0005g.v\u0006, where g ∈ G and \u0005v\u0006 ∈ P(V ). Then we see that we have a bijection {non-zero (Gm × G)-orbits on V }\n\n←→\n\n{G-orbits on P(V )}.\n\n3.2.2 The Zariski topology on P(V ) Let V be a finitedimensional k-vector space, and consider the corresponding projective space P(V ), as defined above. Using the natural map π : V \\{0} → P(V ), v → \u0005v\u0006, we could define a topology on P(V ) by declaring a subset U ⊆ P(V ) to be open if π −1 (U ) is open in the induced topology on V \\ {0}. However, it is possible to define this topology on P(V ) in a more direct way, using zero sets of polynomials as in the case of affine algebraic sets. For this purpose, we choose a basis {e1 , . . . , en } of V and identify V = kn using that basis. Let T1 , . . . , Tn be indeterminates. A non-zero polynomial f ∈ k[T1 , . . . , Tn ] is called homogeneous of degree d \u0002 0 if f is a linear combination of monomials which all have total degree d. For such a polynomial f, we have f(av1 , . . . , avn ) = ad f(v1 , . . . , vn ) for all a ∈ k and vi ∈ k. Thus, 1\n\nFor the relation with ‘elimination theory’ see e.g. Chapter 8, §5, of Cox et al. (1992).\n\n176\n\nAlgebraic representations and Borel subgroups\n\nfor \u0005v\u0006 ∈ P(V \u0003 ), the condition f(v1 , . . . , vn ) = 0 is well-defined, where v = i vi ei . For any subset H ⊆ k[T1 , . . . , Tn ] consisting of homogeneous polynomials, we define Vp (H) := {\u0005v1 e1 + · · · + vn en \u0006 ∈ P(V ) | f(v1 , . . . , vn ) = 0 for all f ∈ H}. By exactly the same arguments as in §1.1.6, we see that the sets Vp (H) form the closed sets of a topology on P(V ), which we call the Zariski topology. Note that this topology does not depend on the choice of the basis {e1 , . . . , en }. This follows from the fact that any two bases of V can be transformed into each other by a linear transformation of V , and such a linear transformation also transforms homogeneous polynomials into homogeneous polynomials. We summarize the above discussion in the following result. 3.2.3 Lemma\n\nWe set V := V \\ {0}. Then the map π : V → P(V ),\n\nv → \u0005v\u0006,\n\nis continuous and open. We have π(Z \\ {0}) = π(Z) for any Gm invariant subset Z ⊆ V . If Z is closed, then π(Z) also is closed. Proof To check that π is continuous, we must check that π−1 (Z) is closed for any closed subset Z ⊆ P(V ). Let H ⊆ k[T1 , . . . , Tn ] be a set of homogeneous polynomials such that Z = Vp (H). But then it is clear that π−1 (Z) = V(H) \\ {0}, where V(H) denotes the usual affine algebraic set in V = kn defined by H. To show that π is open, we must check that π(U ) is open for any open subset U ⊆ V . Since π(U ) = π(π−1 (π(U ))), it is enough to do this for all open sets which are unions of complete fibres of π, that is, which are Gm -invariant. But this is equivalent to checking that π(C) is closed for all Gm -invariant closed subsets C ⊆ V . Now, by Exercise 3.6.4, such a subset C is of the form V(H) \\ {0} for some set H ⊆ k[T1 , . . . , Tn ] of homogeneous polynomials. But then we see that π(C) = Vp (H) is closed. Finally, let Z ⊆ V be any Gm -invariant subset. By Exercise 3.6.4 and §1.1.7, we have Z = V(H), where H ⊆ k[T1 , . . . , Tn ] is a set of homogeneous polynomials. In particular, Z is Gm -invariant, and so π(Z \\ {0}) = Vp (H) by the previous discussion. We certainly have π(Z) ⊆ Vp (H) and so π(Z) ⊆ π(Z \\ {0}). On the other hand, π is continuous and so π(Z \\ {0}) ⊆ π(Z). \u0003\n\nThe main theorem of elimination theory\n\n177\n\n3.2.4 Remark In Chapter 1, we defined the dimension of an algebraic set V ⊆ kn using the Hilbert polynomial of an ideal. In a similar way, one can also define the dimension of a projective algebraic set in P(V ), using the ‘projective’ Hilbert polynomial of a homogeneous ideal; see Exercise 3.6.6. However, we shall not need this here; see Chapter 9 of Cox et al. (1992) for more details. 3.2.5 Proposition Let V be an algebraic G-module, and consider the (abstract) operation of G on P(V ) given by g.\u0005v\u0006 = \u0005g.v\u0006, where g ∈ G and \u0005v\u0006 ∈ P(V ). This operation has the following properties: (a) For any non-zero v ∈ V , the subgroup StabG (\u0005v\u0006) ⊆ G is closed. (b) Let O ⊆ P(V ) be a G-orbit. Then O is open in O, and O is a union of orbits. (c) Every non-empty G-invariant closed subset of P(V ) contains a closed orbit. In particular, closed orbits exist. Proof (a) Let \u0005v\u0006 ∈ P(V ). Then we have StabG (\u0005v\u0006) = TranG (\u0005v\u0006, \u0005v\u0006), and so this is closed by Lemma 2.5.1(a). (b) Let O be the G-orbit of \u0005v\u0006 ∈ P(V ). As remarked in §3.2.2, we have π−1 (O) = C, where C is the (Gm × G)-orbit of v in\n\nV . By Lemma 3.2.3, we have π(C ) = O, where the superscript \u0011 denotes intersection with V . Now we can argue as follows. By Proposition 2.5.2(a), C is in fact a union of (Gm × G)-orbits, and C \\ C is closed. Let us write C = C ∪ R, where R is a closed (Gm × G)-invariant subset and C ∩ R = ∅. The Gm -invariance implies that O ∩ π(R ) = π(C ) ∩ π(R ) = ∅. Hence, we have\n\nO = π(C ) = O ∪ π(R ), where the union is disjoint and π(R ) is closed (again, by Lemma 3.2.3). Hence O is open in O. (c) Let C be a G-invariant closed subset of P(V ). Then C is a union of orbits which contains the closure of each of these orbits. The existence of a closed orbit in C follows from the fact that P(V ) is noetherian (see Exercise 3.6.3), by exactly the same argument as in the proof of Proposition 2.5.2(c). \u0003 We now turn to another important property of projective space, namely, what is usually called the ‘main theorem of elimination theory’ or the ‘completeness of projective space’. To prove this, we first need the following ‘homogeneous’ version of Hilbert’s nullstellensatz.\n\n178\n\nAlgebraic representations and Borel subgroups\n\n3.2.6 Lemma Recall our standing assumption that k is algebraically closed. Let H ⊆ k[T1 , . . . , Tn ] be a set of homogeneous polynomials. Then Vp (H) = ∅ if and only if there exists some s \u0002 1 such that all monomials of degree s lie in the ideal generated by H. Proof First assume that there exists some s \u0002 1 such that all monomials of degree s lie in (H). Then H contains Tis for 1 \u0001 i \u0001 n, and so Vp (H) ⊆ Vp ({T1s , . . . , Tns }) = ∅. Conversely, assume that Vp (H) = ∅. Then we consider the algebraic set in kn defined by p (H) = ∅ means that H, that is, V(H) ⊆ kn . Then the condition V\u0006 V(H) ⊆ {0} and so I({0}) ⊆ I(V(H)) = (H), where the last equality holds by Hilbert’s nullstellensatz; see Theorem \u0006 2.1.5. On the other hand, we certainly have Ti ∈ I({0}), and so Ti ∈ (H) for 1 \u0001 i \u0001 n. Hence there exist some mi \u0002 1 such that Timi ∈ (H) for all i. Now set m := max{mi } and s := nm \u0002 1. Then, if T1α1 · · · Tnαn is any monomial of degree s, there exists some i such that αi \u0002 m \u0002 mi . Since Timi ∈ (H), we conclude that T1α1 · · · Tnαn ∈ (H), as desired. \u0003 To state the following result, we introduce the following notation. Let (X, A) be an affine variety over k and V be a finite-dimensional k-vector space. Let {e1 , . . . , en } be a basis of V, and consider the polynomial ring A[T1 , . . . , Tn ], where T1 , . . . , Tn are indeterminates. Then, for any x ∈ X, the evaluation homomorphism εx : A → k canonically extends to a k-algebra homomorphism ε˜x : A[T1 , . . . , Tn ] → k[T1 , . . . , Tn ], simply by applying εx to the coefficients of polynomials. Thus, for any subset H ⊆ A[T1 , . . . , Tn ] consisting of homogeneous polynomials and any x ∈ X, we can define ˜ V(H) := {(\u0005v\u0006, x) ∈ P(V ) × X | ε˜x (f)(v) = 0 for all f ∈ H}. ˜ As in §1.1.6, it is readily checked that the sets V(H) form the closed subsets of a topology on P(V ) × X. 3.2.7 Theorem (Main theorem of elimination theory) Recall that k is algebraically closed. In the above setting, the second projection map pr2 : P(V ) × X → X is a closed map, that is, it sends closed subsets to closed subsets. Proof Let H ⊆ A[T1 , . . . , Tn ] be a collection of homogeneous poly˜ nomials. We must show that pr2 (V(H)) ⊆ X is a closed set. First\n\nThe main theorem of elimination theory\n\n179\n\nnote that we may assume without loss of generality that H is a finite set and that all polynomials f ∈ H are homogeneous of the same degree. Indeed, since A is a noetherian ring, the same is also true for A[T1 , . . . , Tn ] (by Hilbert’s basis theorem) and so there exist finitely many homogeneous polynomials f1 , . . . , fr ∈ H such that the ideal (H) ⊆ A[T1 , . . . , Tn ] is generated by f1 , . . . , fr . Then it ˜ ˜ is straightforward and easy to check that V(H) = V({f 1 , . . . , fr }). Assume that each fi is homogeneous of degree di \u0002 0, say. Then set d := max {di }, and define gij := Tjd−di fi ∈ A[T1 , . . . , Tn ] for 1 \u0001 i \u0001 r and 1 \u0001 j \u0001 n. Again, it is straightforward and easy ˜ ˜ to check that V(H) = V({g ij | 1 \u0001 i \u0001 r, 1 \u0001 j \u0001 n}). Thus, we are reduced to the case where H consists of finitely many polynomials which are all homogeneous of the same degree, d say. If d = 0, then all f ∈ H are constant and there is nothing to prove. So let us assume that d \u0002 1. ˜ Let us fix x ∈ X. Then we have x ∈ pr2 (V(H)) if and only if p Vx (H) := V ({˜ εx (f) | f ∈ H}) = ∅. We begin by trying to find a more convenient characterization of the condition Vx (H) = ∅. By Lemma 3.2.6, we have Vx (H) = ∅ if and only if, for some s \u0002 1, all monomials of degree s lie in the ideal (˜ εx (f) | f ∈ H) ⊆ k[T1 , . . . , Tn ]. As in Section 1.2, we write Ms := {α = (α1 , . . . , αn ) ∈ Zn\u00030 | α1 + · · · + αn = s}, and denote the monomial T1α1 · · · Tnαn simply by T α . With this notation, we see that the condition Vx (H) = ∅ is equivalent to the condition that there exists some s \u0002 1 such that, for any α ∈ Ms , we have \u0001 Tα = hαf ε˜x (f), where hαf ∈ k[T1 , . . . , Tn ]. f∈H\n\nExpressing all hαf as linear combinations of monomials, and noting that all ε˜x (f) are homogeneous of degree d (or 0) and T α is just a monomial of degree s, we see that the above condition is equivalent to the condition that there exists some s \u0002 d such that, for any α ∈ Ms , we have \u0001 \u0001 α α ∈ k. cf,β T β ε˜x (f), where cf,β Tα = f∈H β∈Ms−d (s)\n\nTo simplify notation, denote by Rx the subspace of k[T1 , . . . , Tn ] spanned by the homogeneous polynomials {T β ε˜x (f) | β ∈ Ms−d ,\n\n180\n\nAlgebraic representations and Borel subgroups\n\nf ∈ H}. Then the above condition can be rephrased by saying that (s) Rx is the whole space of homogeneous polynomials of degree s, for some s \u0002 d. Thus, we have: Vx (H) = ∅\n\ndimk Rx(s) = |Ms | for some s \u0002 d. \b Hence we have {x ∈ X | Vx (H) = ∅} = s\u0003d {x ∈ X | dimk (s)\n\nRx < |Ms |}. So it will be enough to show that, for a fixed s \u0002 d, (s) the set of all x ∈ X with dimk Rx\u0003< |Ms | is closed. This is seen as follows. For f ∈ H, we write f = γ∈Md fγ T γ , where fγ ∈ A. Then we have \u0001 T βf = fα−β T α for any β ∈ Ms−d α∈Ms\n\n(where we set fα−β = 0 if some coefficient in α − β is negative) and so \u0001 T β ε˜x (f) = fα−β (x) T α for any f ∈ H. α∈Ms\n\nNow let us arrange the coefficients fα−β (for various f, α, β) in a matrix, where the rows are labelled by the various pairs (f, β) (f ∈ H, β ∈ Ms−d ) and the columns are labelled by the various (s) α ∈ Ms . Then the condition dimk Rx < |Ms | is equivalent to the condition that the rank of that matrix is less than |Ms |. The latter condition can be expressed by the vanishing of certain minors of that matrix, that is, by determinantal expressions in the coefficients fα−β . These determinantal expressions yield a collection of regular functions in A. The above discussion shows that the closed set of X defined by that collection is precisely the set of all x ∈ X with (s) dimk Rx < |Ms |, as desired. \u0003 3.2.8 Example Consider the polynomials F1 := T1 + T2 Y , F2 := T1 + T1 Y in k[T1 , T2 , Y ]. These polynomials are homogeneous in T1 and T2 , and so the set \u001d \u0002 \u001d F1 (x1 , x2 , y) = 0 2 ˜ \u001d V({F 1 , F2 }) = (\u0005x1 , x2 \u0006, y) ∈ P(k ) × k \u001d F2 (x1 , x2 , y) = 0 is well-defined. It is easily checked that the above set just contains the two elements (\u00050, 1\u0006, 0) and (\u00051, 1\u0006, −1). Thus, we have ˜ pr2 (V({F 1 , F2 })) = {0, −1}. ˜ We conclude that pr2 (V({F 1 , F2 })) = V({Y (Y + 1)}). It is a natural question to ask if there is a general procedure for finding defining\n\nThe main theorem of elimination theory\n\n181\n\n˜ See Exercise 3.6.5 for an answer to this polynomials for pr2 (V(H)). question. 3.2.9 Lemma Let V = V \\ {0} and X be an affine variety. Consider the map π × id : V × X → P(V ) × X,\n\n(v, x) → (\u0005v\u0006, x).\n\nLet C ⊆ V ×X be a subset satisfying the following two conditions. (a) C is closed in the induced topology on V × X as a subset of V × X. (b) For any (v, x) ∈ C and any 0 = t ∈ k, we have (tv, x) ∈ C. Then (π × id)(C) ⊆ P(V ) × X is closed. Proof Using a basis {e1 , . . . , en } of V , we can identify A[V × X] with the polynomial ring A[T1 , . . . , Tn ]. Let C be the closure of C in V × X. By §1.1.7, we have C = V(I(C)). Now, by (b) and Exercise 3.6.4, I(C) is generated by homogeneous polynomials. Thus, we have C = V(H), where H ⊆ A[T1 , . . . , Tn ] is a set of polynomials which are homogeneous in T1 , . . . , Tn . We claim that ˜ (π × id)(C) = V(H) ⊆ P(V ) × X. Indeed, the inclusion ‘⊆’ is clear by definition. Conversely, let ˜ (\u0005v\u0006, x) ∈ V(H). Then F (v, x) = 0 for all F ∈ H, and so (v, x) ∈ C = V(H) ⊆ V × X. Since v = 0, we also have (v, x) ⊆ V × X, and so (v, x) ∈ C, using (a). \u0003 We now describe applications of the above results to algebraic groups. 3.2.10 Definition Let G be an affine algebraic group. A closed subgroup P ⊆ G is called a parabolic subgroup if there exists an algebraic G-module V and a non-zero vector v ∈ V such that (a) P is the stabilizer of \u0005v\u0006 ∈ P(V ) and (b) the orbit {\u0005g.v\u0006 | g ∈ G} ⊆ P(V ) is closed. Here, we consider the induced action of G on P(V ), as in Proposition 3.2.5. For the time being, this is a rather technical condition. In Theorem 3.4.3, we will obtain a purely group-theoretic characterization of parabolic subgroups. For example, P = G always is a parabolic subgroup. (To see this, we simply take the trivial G-module V = k, where each g ∈ G acts\n\n182\n\nAlgebraic representations and Borel subgroups\n\nas the identity.) A similar argument also shows that, if P ⊆ G is a parabolic subgroup and H is any affine algebraic group, then H × P is a parabolic subgroup of H × G. Now we can prove a basic property of parabolic subgroups, which is essential in many applications; see, for example, the two results at the end of Section 3.4. 3.2.11 Theorem Let X be an affine variety, G an affine algebraic group, and P ⊆ G a parabolic subgroup. Assume that Z ⊆ G × X is closed and that (g, x) ∈ Z\n\n(gp, x) ∈ Z\n\nfor all p ∈ P .\n\n(∗)\n\nThen pr2 (Z) ⊆ X is closed, where pr2 : G × X → X is the second projection. Proof The statement already bears some similarity to that in Theorem 3.2.7. To establish the link, we proceed as follows. By definition, there exists an algebraic G-module V and some non-zero v0 ∈ V such that the G-orbit C := O\bv0 ⊆ P(V ) is closed, and we have P = StabG (\u0005v0 \u0006). Then we also have pr2 (Z) = pr2 ((ϕ\bv0 × id)(Z)), where ϕ\bv0 : G → P(V ), g → \u0005g.v0 \u0006. Hence, by Theorem 3.2.7, it will be enough to show that the image of Z under ϕ\bv0 × id : G × X → P(V ) × X is closed. For this purpose, we begin with the following ˜ = Gm × G. As we have seen in §3.2.1, preparations. Let us set G ˜ ˜ we may also regard V as an algebraic G-module, where (t, g) ∈ G ˜ → V , (t, g) → tg.v0 , be the acts by v → tg.v (v ∈ V ). Let ϕ˜v0 : G corresponding orbit map. Then we have ˜ (ϕ\bv0 × id)(Z) = (π × id)((ϕ˜v0 × id)(Z)), where π × id is as in Lemma 3.2.9 and ˜ × X | t ∈ Gm , (g, x) ∈ Z}. Z˜ = {((t, g), x) ∈ G ˜ ⊆ V ×X Thus, it will be sufficient to show that the set (ϕ˜v0 ×id)(Z) satisfies the two conditions in Lemma 3.2.9. By construction, it is clear that condition (b) is satisfied. Hence it remains to show that ˜ is closed in V × X ⊆ V × X. (ϕ˜v0 × id)(Z)\n\n(†)\n\nTo see this, consider the canonical map π : V → P(V ), v → \u0005v\u0006, and let CV = π−1 (C) ∪ {0}. As we have seen in §3.2.2, CV is\n\nThe main theorem of elimination theory\n\n183\n\n˜ ⊆ CV , it will be enough a closed subset in V . Hence, since ϕ˜v0 (G) to show that ˜ is closed in C × X ⊆ CV × X, (ϕ˜v0 × id)(Z) V\n\n(†\u0002 )\n\nwhere CV := CV \\ {0}. To prove this, first note that Z˜ is a ˜ × X. Furthermore, by Corollary 2.5.7, we already closed subset of G know that ˜ × X → CV × X ϕ˜v0 × id : G is an open map. Hence, in order to prove (†\u0002 ), it is enough to show ˜ × X consists of complete fibres of the map ϕ˜v0 × id. To that Z˜ ⊆ G ˜ ×X check the latter property, let ((t, g), x) ∈ Z˜ and ((t\u0002 , g\u0002 ), x\u0002 ) ∈ G \u0002 \u0002 \u0002 be such that (tg.v0 , x) = (t g .v0 , x ). Then we certainly have x = x\u0002 and g−1 g\u0002 .v0 ∈ \u0005v0 \u0006. This implies g−1 g\u0002 ∈ StabG (\u0005v0 \u0006) = P , and so g\u0002 = gp for some p ∈ P . The fact that Z satisfies condition (∗) now ˜ as required. Thus, (†\u0002 ) is implies that we also have ((t\u0002 , g\u0002 ), x\u0002 ) ∈ Z, proved. \u0003 3.2.12 Corollary Let P be a parabolic subgroup of G. (a) Let X be a G-variety and Y ⊆ X be a P -invariant closed subset. Then G.Y = {g.y | g ∈ G, y ∈ Y } ⊆ X is closed. (b) Let V be an algebraic G-module and C ⊆ P(V ) be a P invariant closed subset. Then G.C = {\u0005g.v\u0006 | g ∈ G, \u0005v\u0006 ∈ C} ⊆ P(V ) is closed. Proof (a) Consider the direct product G × X, and set Z := {(g, x) ∈ G × X | g −1 .x ∈ Y }. Then Z is closed. Furthermore, let (g, x) ∈ Z and p ∈ P . Then we have (gp)−1 .x = p−1 (g−1 .x) ∈ Y since Y is P -invariant. Thus, all the assumptions of Theorem 3.2.11 are satisfied, and so G.Y = pr2 (Z) is closed. ˜ = Gm × G and (b) As in the proof of Theorem 3.2.11, let G ˜ consider V as a G-variety. Furthermore, we set CV = π−1 (C) ∪ {0}, where π : V → P(V ) is the canonical map. By §3.2.2, CV is a closed ˜ V ⊆ V is also closed. By the remarks following set. We claim that G.C ˜ Since CV is Definition 3.2.10, Gm × P is a parabolic subgroup in G. ˜ V is invariant under Gm × P , the claim follows from (a). Since G.C ˜ Gm -invariant, G.C = π(G.CV ) is closed, as required. \u0003\n\n184\n\nAlgebraic representations and Borel subgroups\n\n3.2.13 Example Let P be a parabolic subgroup of G and U ⊆ P be a closed normal subgroup. Then ! xU x−1 is a closed subset of G. x∈G\n\nTo see this, we consider G as a G-variety via conjugation. Then U is a P -invariant closed subset of G, and we have G.U = {xux−1 | \u0007 x ∈ G, u ∈ U } = x∈G xU x−1 . By Corollary 3.2.12, this is a closed subset of G.\n\n3.3 Grassmannian varieties and flag varieties The purpose of this section is to describe various constructions of G-varieties arising from the linear action of an affine algebraic group G on a vector space. This will lead to the important examples of grassmannian varieties and flag varieties. We assume that the reader has some familiarity with the basic facts concerning tensor products and exterior powers of vector spaces; see, for example, Appendix B of Fulton and Harris (1991). As a first application, we prove a result due to Chevalley which asserts the existence of certain algebraic representations. 3.3.1 Partial flag varieties Let V be an algebraic G-module. Then G acts naturally on various sets constructed from V . Indeed, let 0 < n1 < n2 < · · · < nd \u0001 dim V be a sequence of strictly increasing integers. Then the corresponding partial flag variety is defined by \u001d \u0002 \u001dE ⊆ V subspace, dimk Er = nr . Fn1 ,...,nd (V ) := (E1 , . . . , Ed ) \u001d\u001d r for all r, and E1 ⊆ · · · ⊆ Ed The given action of G on V certainly induces an action on Fn1 ,...,nd (V ). Just note that, for g ∈ G and a subspace E ⊆ V , the set g.E = {g.v | v ∈ E} ⊆ V is a subspace with dimk E = dimk g.E (since πg : V → V is a bijective linear map). We note the following special cases of the above construction. Let n = dim V . Taking the sequence 0 < 1 < 2 < · · · < n − 1 < n, the corresponding set (a) F(V ) := F1,2,...,n (V )\n\nGrassmannian varieties and flag varieties\n\n185\n\nis called the flag variety. For 0 \u0001 r \u0001 n, the corresponding set Gr (V ) := Fr (V ) = {E ⊆ V | E subspace with dimk E = r}\n\n(b)\n\nis called a grassmannian variety. For r = 1, we just obtain the projective space P(V ) := {\u0005v\u0006 | 0 = v ∈ V }. However, note that the induced actions on the above sets are only defined on a purely abstract level. The aim of this section is to show that the sets Fn1 ,...,nd (V ) can be identified with some closed subsets in a suitably chosen projective space. We begin by showing that the notion of algebraic G-modules is compatible with the usual linear algebra constructions: tensor products and exterior powers. 3.3.2 Tensor products Let V and W be algebraic G-modules. Then the tensor product V ⊗ W also is an algebraic G-module, with G-action given by g.(v ⊗ w) = (g.v) ⊗ (g.w),\n\nwhere g ∈ G, v ∈ V , and w ∈ W .\n\nIndeed, first note that the above formula defines an (abstract) action of G on V ⊗ W which is linear. Now, by assumption, V has a basis {e1 , . . . , en } such that the action of G on V is given by the formula in §3.1.1, with regular functions aij ∈ A[G]. Similarly, there is a basis {f1 , . . . , fm } of W such that the action of G on W is given by regular functions brs ∈ A[G]. Now it is well-known that {ei ⊗ fr | 1 \u0001 i \u0001 n, 1 \u0001 r \u0001 m} is a basis of V ⊗ W . Then we have the formula g.(ej ⊗ fs ) =\n\nm n \u0001 \u0001\n\naij (g)brs (g) ei ⊗ fr ,\n\ni=1 r=1\n\nwhich shows that the action of G on V ⊗ W is given by the functions aij brs ∈ A[G]. More generally, if V1 , . . . , Vd are algebraic G-modules, then the tensor product V1 ⊗ · · · ⊗ Vd is an algebraic G-module, with G-action given by g.(v1 ⊗ · · · ⊗ vd ) = (g.v1 ) ⊗ · · · ⊗ (g.vd ), where g ∈ G and vi ∈ Vi for all i.\n\n186\n\nAlgebraic representations and Borel subgroups\n\n3.3.3 Lemma (the Segre embedding) braic G-modules. Then the map σ:\n\nLet V1 , . . . , Vd be alge-\n\nP(V1 ) × · · · × P(Vd ) → P(V1 ⊗ · · · ⊗ Vd ) (\u0005v1 \u0006, . . . , \u0005vd \u0006) → \u0005v1 ⊗ · · · ⊗ vd \u0006\n\nis well-defined, injective, and G-equivariant, and its image is closed. Here, the G-action on the left-hand side is given by g.(\u0005v1 \u0006, . . . , \u0005vd \u0006) = (\u0005g.v1 \u0006, . . . , \u0005g.vd \u0006). See also Exercise 3.6.7 where we give an explicit description of the subsets of P(V1 ) × · · · × P(Vd ) which correspond to closed subsets under the Segre embedding. Proof By induction on d, it is enough to prove this in the case where d = 2. Let {e1 , . . . , en } be a basis of V = V1 and {f1 , . . . , fm } be a basis of W = V2 . Then {e \u0003i ⊗fj | 1 \u0001 i \u0001 n, 1 \u0001 j \u0001 m} is\u0003a basis of V ⊗ W . If we write v = i vi ei with vi ∈ k, and w = j wj fj with wj ∈ k, then v⊗w=\n\nm n \u0001 \u0001\n\nvi wj (ei ⊗ fj ).\n\ni=1 j=1\n\nThus, if v = 0 and w = 0, then vi = 0 for some i, and wj = 0 for some j, and so v ⊗ w = 0. Furthermore, if we replace v by av for 0 = a ∈ k, and w by bw for some 0 = b ∈ k, then v ⊗ w is replaced by ab(v ⊗ w). This shows that the map σ is well-defined. The above formula for v ⊗ w also shows how the coordinates of v and w can be recovered from the coordinates of v ⊗ w (up to scalar multiples). Thus, σ is injective. It is also clear that σ is G-equivariant. Now let us prove that the image of σ is closed. With respect to the basis {ei ⊗ fj } of V ⊗ W , the closed sets in P(V ⊗ W ) are given by the zero sets of homogeneous polynomials in the nm variables Yij (1 \u0001 i \u0001 n, 1 \u0001 j \u0001 m). We claim that σ(P(V ) × P(W )) = Vp (H) ⊆ P(V ⊗ W ), where H := {Yij Yrs − Yis Yrj | 1 \u0001 i, r \u0001 n, 1 \u0001 j, s \u0001 m}. \u0003 To see this, let 0 = z = i,j cij ei ⊗ fj ∈ V ⊗ W . If z = v ⊗ w for some \u0003 \u0003 v = i vi ei , and w = j wj fj , then cij = vi wj , and so f(cij ) = 0 for all f ∈ H. Conversely, assume that f(cij ) = 0 for all f ∈ H. Now, there exist some i0 and j0 such that ci0 j0 = 0; we may even\n\nGrassmannian varieties and flag varieties\n\n187\n\nassume that ci0 j0 = 1. Considering Yij0 Yi0 j − Yij Yi0 j0 ∈ H shows that cij0 ci\u0003 = cij ci0 j0 = cij \u0003 for all i and j, and so z = v ⊗ w, where 0j v := i cij0 ei and w := j ci0 j fj . \u0003 3.3.4 Exterior powers Let V be an algebraic G-module. We set T r (V ) := V · · ⊗ V% \" ⊗ ·#\\$\n\nfor any r \u0002 0, where T 0 (V ) = k.\n\nr times\n\nBy §3.3.2, T r (V ) also is an algebraic G-module. Recall that if {e1 , . . . , en } is a basis of V , then {ej1 ⊗ · · · ⊗ ejr | 1 \u0001 j1 , . . . , jr \u0001 n} is a basis of T r (V ); in particular, we have dim T r (V ) = nr . Now consider the subspace I r (V ) = \u0005v1 ⊗ · · · ⊗ vr | vi = vj for some i = j\u0006k ⊆ T r (V ). & Clearly, I r (V ) is G-invariant, and so the exterior power r V := T r (V )/I r (V & ) is an algebraic G-module. As usual, the natural map T r (V ) → r V is written as v1 ∧ · · · ∧ vr := v1 ⊗ · · · ⊗ vr + I r (V ), and then the G-action is given by g.(v1 ∧· · ·∧vr ) = (g.v1 )∧· · ·∧(g.vr ). It is well known that if {e1 , . . . , en } is a basis of V , then {ej1 ∧ · · · ∧ ejr | 1 \u0001 j1 < · · · < jr \u0001 n} &r &r is a basis of V . In particular, we have V = {0} for r \u0002 n + 1, \u0004 \u0005 &r n while dim \u0003 V = r for 0 \u0001 r \u0001 n. Given v1 , . . . , vr ∈ V and writing vj = ni=1 aij ei , where A = (aij ) ∈ kn×r , we have \u0001 pi1 i2 ...ir (A) ei1 ∧ · · · ∧ eir , (a) v1 ∧ · · · ∧ vr = 1\u0001i1"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.87705916,"math_prob":0.99769247,"size":363709,"snap":"2023-40-2023-50","text_gpt3_token_len":128389,"char_repetition_ratio":0.19830357,"word_repetition_ratio":0.1456247,"special_character_ratio":0.35240537,"punctuation_ratio":0.16530254,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.99982756,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-10T09:49:08Z\",\"WARC-Record-ID\":\"<urn:uuid:36c9e7d6-5c03-447d-bf76-fe0fee31a044>\",\"Content-Length\":\"432778\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0a652fc7-de4b-4ed4-8624-c3ff2b1ce653>\",\"WARC-Concurrent-To\":\"<urn:uuid:3e584025-aebf-47d5-ac97-cf83415617ef>\",\"WARC-IP-Address\":\"104.21.62.192\",\"WARC-Target-URI\":\"https://dokumen.pub/an-introduction-to-algebraic-geometry-and-algebraic-groups-1nbsped-0198528310-9780198528319-9780199676163-019967616x.html\",\"WARC-Payload-Digest\":\"sha1:IB4J3HWGVXGIK3BRGI5Z4UAP22EYE2FV\",\"WARC-Block-Digest\":\"sha1:FLRRD2LOLLA5WZQGEATQABCXX26ICZI7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679101779.95_warc_CC-MAIN-20231210092457-20231210122457-00256.warc.gz\"}"} |
https://solvedlib.com/n/find-cos0-csce-and-tan0-where-0-is-the-angle-shown-in-the,885214 | [
"# Find cos0, csce, and tan0, where 0 is the angle shown In the figureGive exact values; not decimal approximations_cos8csc0tan0\n\n###### Question:\n\nFind cos0, csce, and tan0, where 0 is the angle shown In the figure Give exact values; not decimal approximations_ cos8 csc0 tan0",
null,
"",
null,
"#### Similar Solved Questions\n\n##### 1.9 Use Fermat' Inethod of tangents to find the derivative function of V9\n1.9 Use Fermat' Inethod of tangents to find the derivative function of V9...\n##### Joule - What is the difference between static mic, d al wees for fund flow? How...\nJoule - What is the difference between static mic, d al wees for fund flow? How can we measure these types of pees? - Water flows in a pipe network between two w a factory is a certains part of the pipe system the flow is 295 There is where we changes in the pipe as illustrated in the next figure. T...\n##### 2: For five of the following eight series, determine if the series converges O1 diverges (only do five of the eight) . There are five different tests listed below . In order to receive credit for this question, YOu must use each of the tests OnCC . Write your work on your paper: When JOU are finished with the question; enter \"Done\" in the answer box. Divergence Test Integral Test Limit Comparison Test Ratio Test Root Test 2n 7n2 + n b) In? (n +2)! 1)(n + 2)6n - 3)(n +4) n=1 \"=1 n=\n2: For five of the following eight series, determine if the series converges O1 diverges (only do five of the eight) . There are five different tests listed below . In order to receive credit for this question, YOu must use each of the tests OnCC . Write your work on your paper: When JOU are finishe...\n##### Exercise 13-9 Net Present Value Analysis and Simple Rate of Return [LO13-2, LO13-6] Derrick Iverson is...\nExercise 13-9 Net Present Value Analysis and Simple Rate of Return [LO13-2, LO13-6] Derrick Iverson is a divisional manager for Holston Company. His annual pay raises are largely determined by his division’s return on investment (ROI), which has been above 20% each of the last three years. Der...\n##### Question 4: [13 marks] For function f(t) = 4cos\"(0) + 2sin(4t) - cos(t). Find the period T of f(t) [3 marks]Using the period T in (a), compute the Fourier series of the function f(t) and list all the Fourier coefficients. [10 marks]Page\nQuestion 4: [13 marks] For function f(t) = 4cos\"(0) + 2sin(4t) - cos(t). Find the period T of f(t) [3 marks] Using the period T in (a), compute the Fourier series of the function f(t) and list all the Fourier coefficients. [10 marks] Page...\n##### Chapter 3, Section 3.5, Question 12 Find the solution of the given initial value problem. 01...\nChapter 3, Section 3.5, Question 12 Find the solution of the given initial value problem. 01 -25 x(0)= 10) x, Click here to enter or edit your answer The solution is given by x(t) -...\n##### Solve the given system of differential equations by systematic elimination. \\begin{aligned} &\\frac{d x}{d t}=-x+z\\\\ &\\frac{d y}{d t}=-y+z\\\\ &\\frac{d z}{d t}=-x+y \\end{aligned}\nSolve the given system of differential equations by systematic elimination. \\begin{aligned} &\\frac{d x}{d t}=-x+z\\\\ &\\frac{d y}{d t}=-y+z\\\\ &\\frac{d z}{d t}=-x+y \\end{aligned}...\n##### Find the principal values of the following: $\\cot ^{-1}(\\sqrt{3})$\nFind the principal values of the following: $\\cot ^{-1}(\\sqrt{3})$...\n##### Whats the approximate value of 0, 7 decimal places in theinterval [0,pi/2] that defines each as truesin theta= 0.9872cot theta= 4.2669sec theta= 3.302\nwhats the approximate value of 0, 7 decimal places in the interval [0,pi/2] that defines each as true sin theta= 0.9872 cot theta= 4.2669 sec theta= 3.302...\nDiscounted payback period) Assuming an appropriate discount rate of 11 percent, what is the discounted payback period on a project with an initial outlay of $100,000 and the following cash flows? Year 1$30,000 Year 2 $35,000 Year 3$25,000 Year 4 $25,000 Year 5$30,000 Year 6 $20,000 The project\u0003... 1 answer ##### Harbour Company makes two models of electronic tablets, the Home and the Work. Basic production information... Harbour Company makes two models of electronic tablets, the Home and the Work. Basic production information follows: Home Work Direct materials cost per unit$ 39 $70 Direct labor cost per unit 24 37 Sales price per unit 360 565 Expected production per month 600 units 310 un... 1 answer ##### 1) Which reaction will shift to the left in response to a decrease in volume? A)... 1) Which reaction will shift to the left in response to a decrease in volume? A) 2 SO3 (8) 2 SO2(g) + O2(8) B) N2(g) + 3H2(g) + 2NH3(g) C) 2H1 (8) H2(g) + 12 (8) D) 4 Fe (8) +32(g) 2 Fe2O3 (s) E) H2 (8) + Cl2(g) 2 HCl (g) Explain 2) Based on Le Châtelier's principle, increasing pressure at... 1 answer ##### Which type of blood cells are the most abundant in a healthy human body? Which type of blood cells are the most abundant in a healthy human body?... 5 answers ##### 12. Write the IUPAC name of the compound formed when PPh;*-C HCH; reacts with 3- pentanone. Write the complete chemical equation for the same_ 12. Write the IUPAC name of the compound formed when PPh;*-C HCH; reacts with 3- pentanone. Write the complete chemical equation for the same_... 1 answer ##### Two charges of magnitude |?| = 10 ?? are separated by a distance of 10 ??... Two charges of magnitude |?| = 10 ?? are separated by a distance of 10 ?? along a horizontal axis. What is the potential as well as the magnitude and direction of the electric field at the midpoint between the two charges if a) both charges are positive and b) the left charge is positive and the rig... 5 answers ##### Question 32What Is the initiator triplet In both prokaryotes and eukaryotes, and what amina scid [ recrulted by this uiplen? UAA or UGA\" arginineAUG; methionineUAA methlonineUAA; no amino acid recruited Question 32 What Is the initiator triplet In both prokaryotes and eukaryotes, and what amina scid [ recrulted by this uiplen? UAA or UGA\" arginine AUG; methionine UAA methlonine UAA; no amino acid recruited... 1 answer ##### Pl) Frori Rogauclim eacioI6? LactcIA V(z,y) I8vProblemsHroblcm$ Zroblem 2 Jrblcm *Necloruracicntal(Uso SYiribolz rolaltnIraclizngha FcECedDisplay OptiongEvJllaleInc Intearal)or 1 <0 <2MutnA(Use s7 boic WalonHraciceetem necaccaen nuclnneHota: Ycucan Gan 'Bori 'GreuronthsAnco\npl) Frori Rogauclim eacioI6? LactcIA V(z,y) I8v Problems Hroblcm \\$ Zroblem 2 Jrblcm * Neclor uracicntal (Uso SYiribolz rolaltn Iraclizngha FcECed Display Optiong EvJllale Inc Intearal )or 1 <0 <2 MutnA (Use s7 boic Walon Hraciceetem necacc aen nuclnne Hota: Ycucan Gan 'Bori 'Greuront...\n##### 4. Swor et al. (A-9) looked at the effectiveness of cardiopul- monary resuscitation (CPR) training in...\n4. Swor et al. (A-9) looked at the effectiveness of cardiopul- monary resuscitation (CPR) training in people over 55 years old. They compared the skill retention rates of subjects in this age group who completed a course in traditional CPR instruction with those who received chest-compression only c...\n##### Saved Help Save & Exi Required information [The following information applies to the questions displayed below)...\nSaved Help Save & Exi Required information [The following information applies to the questions displayed below) Warnerwoods Company uses a perpetual inventory system. It entered into the following purchases and sales transactions for March Date Activities Units Acquired at Cost Units sold at Ret...\n##### 9 ~H3~B K2~H KS~B a Denying the antecedent b Invalid. C Pure hypothetical syllogism. d Constructive dilemma. C_ Destructive dilemma.\n9 ~H3~B K2~H KS~B a Denying the antecedent b Invalid. C Pure hypothetical syllogism. d Constructive dilemma. C_ Destructive dilemma....\n##### 1 ['84 S 1 1 0 1 1 1 1 L\n1 ['84 S 1 1 0 1 1 1 1 L...\n##### 1 . What are the three regions Of a cell? 2. What is the structure of the plasma membrane? of the endomembrane system? Are the mitochondria included? Which organelles are part centrioles, cytoskeleton; endoplasmic reticulum; Golgi = apparat What are the functions of the mitochondria? What does the nucleus consist of? which are passive: - facilitated diffusio use of energy and involves the 6. Which of the following and simple diffusion? osmosis, phagocytosis, the phases of the cell cycle? order;\n1 . What are the three regions Of a cell? 2. What is the structure of the plasma membrane? of the endomembrane system? Are the mitochondria included? Which organelles are part centrioles, cytoskeleton; endoplasmic reticulum; Golgi = apparat What are the functions of the mitochondria? What does the n...\n##### A particular first-order reaction has rate If Eas 85.6 kJ/mol? constant of 1.35 x 102 s-1at25C What is the magnitude of k at75C A) 670 B) 3.47 * 104 C) 3.85 106 D) 1.93 x 104 E) 1.36 x 102\nA particular first-order reaction has rate If Eas 85.6 kJ/mol? constant of 1.35 x 102 s-1at25C What is the magnitude of k at75C A) 670 B) 3.47 * 104 C) 3.85 106 D) 1.93 x 104 E) 1.36 x 102..."
]
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null,
"https://i.imgur.com/RKfkaAV.jpeg",
null,
"https://i.imgur.com/2TQgtSt.jpeg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.79688084,"math_prob":0.9539498,"size":14996,"snap":"2023-14-2023-23","text_gpt3_token_len":4457,"char_repetition_ratio":0.101520814,"word_repetition_ratio":0.47067901,"special_character_ratio":0.28034142,"punctuation_ratio":0.14667925,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98476166,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-24T22:26:52Z\",\"WARC-Record-ID\":\"<urn:uuid:4ff0a0d2-6eb2-444b-a874-d94f2e957abb>\",\"Content-Length\":\"89099\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3e95f6b3-7166-467a-a1da-71582a813650>\",\"WARC-Concurrent-To\":\"<urn:uuid:6f164830-7ad5-4dfb-89fc-70e0c5692458>\",\"WARC-IP-Address\":\"104.21.12.185\",\"WARC-Target-URI\":\"https://solvedlib.com/n/find-cos0-csce-and-tan0-where-0-is-the-angle-shown-in-the,885214\",\"WARC-Payload-Digest\":\"sha1:U22MTYBF5SAIHKXUFRRUEQIZRLCGAEVP\",\"WARC-Block-Digest\":\"sha1:WICSQ3WWFHJVIAQUWFDZB56233CWK5FJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945289.9_warc_CC-MAIN-20230324211121-20230325001121-00411.warc.gz\"}"} |
https://tex.stackexchange.com/questions/330070/alignment-of-chemfig-molecules-in-a-reaction | [
"# Alignment of chemfig molecules in a reaction\n\nI'm using chemfig for my chemistry course this period, and facing problems with the alignment of the molecules, arrows, + signs and also the names. Basically everything is messed up haha.\n\nCan someone explain me how to do it right? I would like to have the molecules' center on one horizontal line, the arrows and + signs aligning that line and the names aligned horizontally just below the biggest molecule.\n\n \\begin{figure}[width=\\textwidth]\n\\scriptsize\n\\begin{scheme}\n\\schemestart\n\\chemname{\\chemfig{OH-[:210,,1]-[:270]=_[:210]-[:150]=_[:90]-[:30](=_[:330])-[:90](-[:30,,,1]OH)=[:150]O}}{salicylic acid}\n\\+{2em}\n\\chemname{\\chemfig{-[:30](=[:90]O)-[:330]O-[:30](-[:330])=[:90]O}}{acetic anhydride}\n\\arrow{->}\n\\chemname{\\chemfig{OH-[:210,,1](=[:150]O)-[:270]=_[:330](-[:30]O-[:330](=[:270]O)-[:30])-[:270]=_[:210]-[:150]=_[:90](-[:30])}}{acetylsalicylic acid}\n\\+{2em}\n\\chemname{\\chemfig{-[:30](-[:330,,,1]OH)=[:90]O}}{acetic acid}\n\\schemestop\n\\end{scheme}\n\\caption{Reaction for synthesizing acetylsalicylic acid} \\label{fig:reaction}\n\\end{figure}",
null,
"The main secret is starting the molecules with the right atom: the first atom of a formula determines the baseline of the corresponding molecule. The rest is only shifting the arrow a bit using its optional argument as described in the manual:\n\n\\documentclass{article}\n\\usepackage{chemfig}\n\n\\begin{document}\n\n\\begin{center}\n\\setatomsep{1.8em}\n\\small\n\\schemestart\n\\chemname{%\n\\chemfig{\nOH-[:210,,1]-[:270]=_[:210]-[:150]\n=_[:90]-[:30](=_[:330])\n-[:90](-[:30,,,1]OH)=[:150]O}%\n}{salicylic acid}\n\\+\n\\chemname{%\n\\chemfig{\n(-[:-150])(=[:90]O)-[:330]O-[:30](-[:330])=[:90]O}%\n}{acetic anhydride}\n\\arrow{->[][][9pt]}\n\\chemname{%\n\\chemfig{\n{\\vphantom{H}}-[:-150](=[:-90]O)-[:150]\nO-[:210,,1]-[:270]=_[:210]-[:150]\n=_[:90]-[:30](=_[:330])\n-[:90](-[:30,,,1]OH)=[:150]O}%\n}{acetylsalicylic acid}\n\\+\n\\chemname{%\n\\chemfig{(-[:-150])(-[:330,,,1]OH)=[:90]O}%\n}{acetic acid}\n\\schemestop\n\\end{center}\n\n\\end{document}",
null,
"• Thanks clemens, this definitely works out. Can you explain me what the \\vphantom part does? How does it work?\n– Lisa\nSep 17, 2016 at 18:50\n• The \\vphantom reserves the vertical space that its argument needs. In this case it ensures that the molecule has the same baseline as the first which starts with an OH group. Sep 17, 2016 at 22:47"
]
| [
null,
"https://i.stack.imgur.com/e2fWm.png",
null,
"https://i.stack.imgur.com/db7cn.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.6244427,"math_prob":0.9830935,"size":1068,"snap":"2023-40-2023-50","text_gpt3_token_len":366,"char_repetition_ratio":0.1231203,"word_repetition_ratio":0.0,"special_character_ratio":0.40355805,"punctuation_ratio":0.23744293,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99248,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-11T23:19:44Z\",\"WARC-Record-ID\":\"<urn:uuid:b3a8efc1-d59c-4de8-93a3-e7ae53bd7ad7>\",\"Content-Length\":\"144804\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:65677219-d878-4dd3-86ca-998e4ac972f3>\",\"WARC-Concurrent-To\":\"<urn:uuid:b2ec7076-b691-496e-add9-d97bfc8dea4b>\",\"WARC-IP-Address\":\"172.64.144.30\",\"WARC-Target-URI\":\"https://tex.stackexchange.com/questions/330070/alignment-of-chemfig-molecules-in-a-reaction\",\"WARC-Payload-Digest\":\"sha1:OE2LYZPUV5CZP64HCH6E56FPRLZENPLR\",\"WARC-Block-Digest\":\"sha1:UO35SAHRSUTLFVCXKDWJZMTI2VXCUXGO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679518883.99_warc_CC-MAIN-20231211210408-20231212000408-00471.warc.gz\"}"} |
http://litung-hk.com/portfolio-item/learners-a-comprehensive-course-polynomial-and-equation-9789627915362/ | [
"# Learner’s – A Comprehensive Course -Polynomial and Equation9789627915362\n\n 圖書名稱 Learner's - A Comprehensive Course-Polynomial and Equation 編著者 C.S.Lee 條碼 9789627915362 出版社 Learner's 頁數 168 幀 裝 平裝 開度 19 x 26.5cm 出版日期 2019/04 類別 課本參考書 每本價錢 HKD\\$\\$105.00\n\nPolynomial and Equation (圖書描述以英語為準)：\n\nPolynomial and Equation:\nThis volume is on the elementary theory of polynomials and polynomial equations and of rational functions, including partial fractions. Roots of unity is discussed in complex numbers."
]
| [
null
]
| {"ft_lang_label":"__label__zh","ft_lang_prob":0.62344384,"math_prob":0.9775459,"size":481,"snap":"2020-10-2020-16","text_gpt3_token_len":335,"char_repetition_ratio":0.16142558,"word_repetition_ratio":0.0,"special_character_ratio":0.14968815,"punctuation_ratio":0.06349207,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9815582,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-24T19:49:51Z\",\"WARC-Record-ID\":\"<urn:uuid:c63b8f86-e92f-4553-a131-66727e3ea3e6>\",\"Content-Length\":\"94789\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cbf292ce-a52a-45ea-8f1a-c16b2479e21c>\",\"WARC-Concurrent-To\":\"<urn:uuid:95e48cd1-6dd8-4f85-a537-244a1ccc5cc8>\",\"WARC-IP-Address\":\"148.66.54.130\",\"WARC-Target-URI\":\"http://litung-hk.com/portfolio-item/learners-a-comprehensive-course-polynomial-and-equation-9789627915362/\",\"WARC-Payload-Digest\":\"sha1:CYXZU6O4LA7D5KKNDJHHS6CATTROOGVI\",\"WARC-Block-Digest\":\"sha1:MQQXQRIO3A6ATXRWSFHELSBPIWIZRZG7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875145981.35_warc_CC-MAIN-20200224193815-20200224223815-00152.warc.gz\"}"} |
https://mammothmemory.net/maths/geometry/area-of-a-triangle/slanting-triangles.html | [
"",
null,
"# Slanting triangles\n\nEven for triangles shaped as below, the area of a triangle = base x height and half it.",
null,
"The proof (but you don’t need it) is as follows:\n\nRedraw the above diagram",
null,
"Area of triangle ABC=Area triangle ADB-Area triangle ADC\n\nArea of triangle ABC=1/2timesDBtimesAD-1/2timesDCtimesAD\n\nYou can remove 1/2AD\n\nArea of triangle ABC=1/2AD(DB-DC)\n\nAnd because DB-DC=CB\n\nWe can now say:\n\nArea of triangle ABC=1/2ADtimesCB\n\nNOTE:\n\nSlanting triangles are rare and it’s still best to think of areas of triangles in a rectangle."
]
| [
null,
"https://mammothmemory.net/images/mobile-home.svg",
null,
"https://mammothmemory.net/images/user/base/Maths/Geometry/Area of a Triangle/even-a-slanting-triangles-area-is-half-base-times-height.67871e8.jpg",
null,
"https://mammothmemory.net/images/user/base/Maths/Geometry/Area of a Triangle/proof-of-area-of-a-slanting-triangle.2f49894.jpg",
null
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https://www.teachercreated.com/lessons/174 | [
"## Iditarod: Statistics and Rates\n\nMathematics\n\n### Objective\n\nStudents analyze statistical data and work with rates.\n\n### Directions\n\nIntroduce the Exploration\nAsk the students if they have ever heard of the Iditarod. Let them talk about it or tell them yourself that the Iditarod is a race of over 1,000 miles that mushers and their dogs run in Alaska. Tell them that they will learn about this race in this exploration, and they will have an opportunity to work with data from the race.\nActivity 1: Prize Money\nHave students use the website listed below to answer the following questions. How much more does the first place winner make than the last place winner? Express your answer in dollars, and then show the increase as a percentage. You may need to review with your students how to calculate the percentage increase.\nActivity 2: The Red Lantern Musher\nHave students find the most recent race that has data on the finishing time of the winner and the finishing time of the last place finisher, also known as the Red Lantern Musher. They should calculate their rates of travel in miles per day to the nearest hundredth. What is the difference in their rate of travel? You may need to review with students how to calculate a rate.\nActivity 3: Winning Rates\nHave students complete the following activities. You may need to review with your students how to calculate a rate. The equation should be in the form t = d / r (where t = time, d = distance, and r = rate).\nFind the times of the winner in 1973 and the most recent winner. Calculate their rates of travel in miles per day to the nearest hundredth. What is the difference in their rate of travel?\nImagine that the most recent winner can always run his or her dogs at the same rates no matter what the distance. Write an equation that shows how to calculate the time to run a given distance.\nImagine that the most recent winner ran his or her dogs for a distance of 1,300 miles at the rate you calculated in Part A. Use the equation you wrote in Part B to calculate how much time it would take to go 1,300 miles.\nActivity 4: A Sample of Mushers\nHave students complete the following activities. You may need to review the topic of probability with your students. Help them understand that if their sample size was 10 and five out of the 10 mushers were between the ages of 40 and 49, then the probability of a musher being 40 to 49 in age is 5/10 or 1/2. Discuss the results with your students. If they came fairly close in the predictions they should feel successful in their ability to predict based on looking at a sample.\nLook at the page that names the mushers. You are going to try to predict the age of a musher. First, you will look at the ages of many mushers. How many mushers and their ages do you think you should look at in order to make an accurate prediction? Once you decide how many mushers you need age information on (sample size), click on that number of mushers to collect the age data. Use the chart on the work sheet to assign probabilities various age categories.\nMake a prediction on the ages of a randomly selected group of 10 mushers. Use the probabilities you created in Part A.\nSelect 10 mushers that you did not select in Part A. Collect the data on their ages. How accurate was the prediction you made in Part B?\n\n### Resources\n\n• copies of the activity sheet (see the link below)\n• computer access"
]
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https://www.spicum.com/uncle-boonmee-nddfma/83e7bb-flash-forward-in-a-sentence | [
"Square Root Function f(x) = x 7. Get access risk-free for 30 days, A parent function is the simplest function that still satisfies the definition of a certain type of function. Namely, if y'(x)=0 for all real numbers x, then y is a constant function.. 2.) Already registered? Sociology 110: Cultural Studies & Diversity in the U.S. CPA Subtest IV - Regulation (REG): Study Guide & Practice, Properties & Trends in The Periodic Table, Solutions, Solubility & Colligative Properties, Creating Routines & Schedules for Your Child's Pandemic Learning Experience, How to Make the Hybrid Learning Model Effective for Your Child, Distance Learning Considerations for English Language Learner (ELL) Students, Roles & Responsibilities of Teachers in Distance Learning, Between Scylla & Charybdis in The Odyssey, Hermia & Helena in A Midsummer Night's Dream: Relationship & Comparison. Well, that's not exactly right; however, there are some similarities that we can observe between our own parents and parent functions. c For functions between preordered sets, constant functions are both order-preserving and order-reversing; conversely, if f is both order-preserving and order-reversing, and if the domain of f is a lattice, then f must be constant. Earn Transferable Credit & Get your Degree, Translating & Reflecting Graphs of Linear Functions, Graphing Square Root & Cube Root Functions, How to Graph an Absolute Value and Do Transformations, Graphing Absolute Value Equations: Dilations & Reflections, Cubic Function: Definition, Formula & Examples, Identify Where a Function is Linear, Increasing or Decreasing, Positive or Negative, Zeroes, Roots & X-Intercepts: Definitions & Properties, Transformation of Exponential Functions: Examples & Summary, Absolute Value Function: Definition & Examples, Horizontal & Vertical Shifts of Linear Functions, What is Function Notation: Definition & Examples, Analyzing the Graph of a Rational Function: Asymptotes, Domain, and Range, Introduction to Statistics: Tutoring Solution, Introduction to Statistics: Help and Review, Holt McDougal Algebra I: Online Textbook Help, DSST Principles of Statistics: Study Guide & Test Prep, SAT Subject Test Mathematics Level 2: Tutoring Solution, Common Core ELA - Literature Grades 11-12: Standards, Common Core ELA - Writing Grades 11-12: Standards. Th, A company needs 400,000 items per year. When you hear the term parent function, you may be inclined to think of two functions who love each other very much creating a new function. All other trademarks and copyrights are the property of their respective owners. Decreasing(left, right) D: (-∞,∞ Range: y values How low and high does the graph go? ... certain pieces of the function have specific behavior. With the information in this lesson, we should now be familiar with what parent functions are and how to identify them. Cubic functions form a family of functions. Let's consider the family of functions that are exponential functions with base 2. To unlock this lesson you must be a Study.com Member. Example: The function y(x) = 2 or just y = 2 is the specific constant function where the output value is c = 2. The sign function sgn(x), which is −1 for negative numbers and +1 for positive numbers, and is the simplest non-constant step function. 0 Take a look! These transformations include horizontal shifts, stretching or compressing vertically or horizontally, reflecting over the x or y axes, and vertical shifts. This would be y= x^3. If f(x)=x+\\sqrt {3-x} and g(u)=u+\\sqrt {3-u}, is it true that f=g? In the same way that we share similar characteristics, genes, and behaviors with our own family, families of functions share similar algebraic properties, have similar graphs, and tend to behave alike. Parent Functions 1. For example, when we think of the linear functions which make up a family of functions, the parent function would be y = x. Related to CCSS standard F.IF.7. This would give us y = 2|x| + 3. Read Also: What Are Agents of Socialization? Inverse Function (Reciprocal of … Algebraically, these transformations correspond to adding or subtracting terms to the parent function and to multiplying by a constant. This is often written: The production costs are $250 to prepare for a production run and$7 for each item produced. Quadratic Function f(x) = x 2 5. Log in here for access. For example, the function y = 2x^2 + 4x can be derived by taking the parent function y = x^2, multiplying it by the constant 2, and then adding the term 4x to it. 's' : ''}}. A special type of linear function is the constant function, a function whose output value has the same result for every input value. As mentioned above, each family of functions has a parent function. A lot of notes here concern defining the __DIR__ magic constant for PHP versions not supporting the feature. Notice that the simplest exponential function in the above family is y = 2^x. This is the parent function of the family of functions. Its graph is the x-axis in the plane.. credit-by-exam regardless of age or education level. Observe that they all have the same shape as the parent function and that they can all be derived by performing the transformations previously mentioned to the parent function. 1.) Let us start with a function, in this case it is f(x) = x 2, but it could be anything: f(x) = x 2. courses that prepare you to earn Select a subject to preview related courses: Let's look at some more examples to further our understanding of parent functions. Log in or sign up to add this lesson to a Custom Course. On the other hand, the polynomial f(x) = 0 is the identically zero function. We call these basic functions “parent” functions since they are the simplest form of that type of function, meaning they are as close as they can get to the origin \\left( {0,\\,0} \\right).The chart below provides some basic parent functions that you should be familiar with. The next transformation occurs when we add a constant c to the input of the parent function $f\\left(x\\right)={b}^{x}$ giving us a horizontal shift c units in the opposite direction of the sign. Constant Function f(x) = a where a = any # 2. Learn the definition of a function and see the different ways functions can be represented. Create an account to start this course today. It is the (trivial) constant function and every x is a root. As a member, you'll also get unlimited access to over 83,000 Cubic Function f(x) = x 3 6. You can test out of the Let f(x,y) = e^{xy}x^2. // CONST_PARENT ';?> to the left of (0,0) you graph (-1,(-1) 3 )=(-1,-1), (-2,(-2) 3 )=(-2,-8), etc.. and career path that can help you find the school that's right for you. The similarities don't end there! An error occurred trying to load this video. A:: getParent (). ' Find h(-4) and h'(-4). // CONST_A parent CONSTANT: '. Just as all of us share certain characteristics with our parents, graphs in a family of functions all share certain characteristics with their parent function. Algebraically, these transformations correspond to adding or subtracting terms to the parent function and to multiplying by a constant. lessons in math, English, science, history, and more. flashcard set{{course.flashcardSetCoun > 1 ? A:: getSelf (). ' The independent variable x does not appear on the right side of the function expression and so its value is \"vacuously substituted\". In mathematics, the maximum and minimum of a function (known collectively as extrema)are the largest and smallest value that a function takes at a point either within a given neighborhood (local or relative extremum ) or within the function domain in its entirety (global or absolute extr… For example, the function y(x) = 4 is a constant function because the value of y(x) is 4 regardless of the input value x (see image). someone help me in pre-cal. A parent function is the simplest function that still satisfies the definition of a certain type of function. You can't go through algebra without learning about functions. For example, in the above graph, we see that the graph of y = 2x^2 + 4x is the graph of the parent function y = x^2 shifted one unit to the left, stretched vertically, and shifted down two units. x Exponential functions are functions of the form y = ab^x, where a and b are both positive (greater than zero), and b is not equal to one. Get the unbiased info you need to find the right school. An example of a family of functions are the quadratic functions. This is the parent function of exponential functions with base b. In mathematics, a constant function is a function whose (output) value is the same for every input value. Study.com has thousands of articles about every Lv 4. These transformations correspond algebraically to adding or subtracting terms to the function, or multiplying by a constant. Then, write a complete approach for f(x). Characteristics will vary for each piecewise function. For example, the function y = 2x^2 + 4x can be derived by taking the parent function y = x^2, multiplying it by the constant 2, and then adding the term 4x to it. As a real-valued function of a real-valued argument, a constant function has the general form y(x) = c or just y = c., The graph of the constant function y = c is a horizontal line in the plane that passes through the point (0, c). the graph of a constant function is symmetric with respect to the y-axis. The number b is the base of the exponential function. Learn Math in the Blogosphere: 10 Top Math Blogs, Universities with Master's Degrees in Math: How to Choose, White House Announces New Math and Science Achievement Campaign, Register for the 2010 American Math Challenge, Tau Day Generates Controversy Among Math Scholars, How to Become a Sports Radio Personality: Step-by-Step Career Guide, Political Media Strategist: Job Description & Salary, Online Barber Courses, Classes and Training Programs, Online MBA Vs PhD Programs in Business Whats the Difference, Online Epidemiology Course and Class Information, Acting Manager Vs Interim Manager Comparison Pay Benefits, Top Make-Up Artistry Schools Information and Overview, ACT English - Section Overview: Tutoring Solution, ACT English - Punctuation: Tutoring Solution, ACT English - Grammar and Usage: Tutoring Solution, ACT English - Sentence Structure: Tutoring Solution, ACT English - Rhetorical Strategy: Tutoring Solution, ACT English - Organization: Tutoring Solution, ACT Math - Pre-Algebra: Tutoring Solution, ACT Math - Algebraic Expressions: Tutoring Solution, ACT Math - Linear Equations: Tutoring Solution, Parent Function in Math: Definition & Examples, ACT Math - Absolute Value: Tutoring Solution, ACT Math - Inequalities: Tutoring Solution, ACT Math - Probability: Tutoring Solution, ACT Math - Data and Statistics: Tutoring Solution, ACT Math - Polynomials and Quadratics: Tutoring Solution, ACT Math - Rational Equations: Tutoring Solution, ACT Math - Complex Numbers: Tutoring Solution, ACT Math - Exponentials and Logarithms: Tutoring Solution, ACT Math - Coordinate Geometry: Tutoring Solution, ACT Math - Conic Sections: Tutoring Solution, ACT Math - Plane Geometry: Tutoring Solution, ACT Math - Logic in Mathematics: Tutoring Solution, ACT Math - Trigonometry: Tutoring Solution, ACT Science Reasoning - Overview: Tutoring Solution, ACT Science Reasoning - Fundamentals: Tutoring Solution, ACT Reading - Overview: Tutoring Solution, ACT Reading - Question Types: Tutoring Solution, ACT Reading - Understanding Passages: Tutoring Solution, ACT Reading - Literary Terms: Tutoring Solution, ACT Reading - Practice: Tutoring Solution, ACT Writing - Overview: Tutoring Solution, ACT Writing - Essay Skills: Tutoring Solution, ACT Writing - Essay Parts: Tutoring Solution, ACT Writing - Planning: Tutoring Solution, ACT Writing - Advanced Skills: Tutoring Solution, Praxis World & U.S. History - Content Knowledge (5941): Practice & Study Guide, FTCE General Knowledge Test (GK) (082): Study Guide & Prep, CSET Science Subtest II Life Sciences (217): Practice & Study Guide, Praxis Chemistry (5245): Practice & Study Guide, CSET Business Subtest II (176): Practice & Study Guide, Praxis Social Studies - Content Knowledge (5081): Study Guide & Practice, CSET Business Subtest I (175): Practice & Study Guide, Praxis Business Education - Content Knowledge (5101): Practice & Study Guide, ILTS Business, Marketing, and Computer Education (171): Test Practice and Study Guide, FTCE School Psychologist PK-12 (036): Test Practice & Study Guide, SAT Subject Test US History: Practice and Study Guide, The Colour Out of Space by HP Lovecraft: Summary & Analysis, A Descent into the Maelstrom: Summary & Analysis, The Facts in the Case of M. Valdemar: Summary & Analysis, The Hop Frog by Edgar Allan Poe: Summary & Analysis, Quiz & Worksheet - Writing Clear Technical Documents, Quiz & Worksheet - Organizing Technical Communication for Clarity, Quiz & Worksheet - Grammatical & Contextual Correctness in Technical Communication, Quiz & Worksheet - Observation, Focus Groups and Other Research Methods, Quiz & Worksheet - Parallel Structure in Technical Writing, AP Chemistry: Experimental Laboratory Chemistry, AP Chemistry: The Periodic Table of Elements, California Sexual Harassment Refresher Course: Supervisors, California Sexual Harassment Refresher Course: Employees. This is the simplest linear function. Find the, Find all the local maxima, local minima and saddle points of the function f(x,y) = y \\sin x. Basically, exponential functions are functions with the variable in the exponent. Find b(x), the end-behavior function of f(x) = \\frac{x^3 + 3x^2 - 4x -1}{x^2 - 1} . Harold’s Parent Functions “Cheat Sheet” 6 November 2019 Function Name Parent Function Graph Characteristics Algebra Constant ( T)= Domain: (− ∞, ) Range: [c, c] Inverse Function: Undefined (asymptote) Restrictions: c is a real number Odd/Even: Even General Form: + =0 Linear or = Identity ( T) T Domain: (−∞, ∞)"
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https://bestfitpetsit.com/prime-and-composite-numbers/ | [
"",
null,
"# Prime and composite numbers",
null,
"In this article we will study prime/composite numbers and what is multiples. First, we will give definitions of prime and composite numbers and give examples. Then we will prove that there are infinitely many prime numbers. Then we will write down a table of prime numbers, and consider methods of compiling a table of prime numbers, focusing especially on the method called Eratosthenes’ sieve. Finally, we will highlight the main points to consider when proving that a given number is prime or composite.\n\nPrime and composite numbers – definitions\n\nThe concepts of prime and composite numbers refer to positive integers that are greater than one. Such integers, depending on the number of their positive divisors, are divided into prime and composite numbers. Thus, to understand the definitions of prime and composite numbers, you need to have a good idea of what divisors and multiples are.",
null,
"Definition.\n\nPrime numbers are integers larger than one that have only two positive divisors, namely themselves and 1.\n\nDefinition.\n\nCompound numbers are integers greater than ones that have at least three positive divisors.\n\nSeparately, note that the number 1 refers to neither prime nor composite numbers. One has only one positive divisor, which is the number 1 itself. This distinguishes the number 1 from all other positive integers, which have at least two positive divisors.\n\nGiven that positive integers are natural numbers, and that one has only one positive divisor, we can give other formulations of the sounded definitions of prime and composite numbers. They have been repeatedly confirmed by experts within the framework of the https://argoprep.com/blog/textual-evidence-writing-engaging-essays/.\n\nDefinition.\n\nPrime numbers are natural numbers that have only two positive divisors.\n\nDefinition.\n\nComposite numbers are natural numbers that have more than two positive divisors.\n\nNote that every positive integer greater than one is either a prime or a composite number. In other words, there is no integer that is neither prime nor composite. This follows from the property of divisibility, which states that the numbers 1 and a are always divisors of any integer a.\n\nFrom the information in the previous paragraph,",
null,
"we can give the following definition of composite numbers.\n\nDefinition.\n\nNatural numbers that are not prime numbers are called composite numbers.\n\nHere are examples of prime and composite numbers.\n\nFor example, the numbers 2, 3, 11, 17, 131, 523 are prime numbers. Undoubtedly, this is far from obvious. But all our attempts to find any positive divisor of any of these numbers, other than one and the numbers themselves, will fail. This shows that the numbers written down are prime numbers. In the last paragraph of this article, we will talk in more detail about proving the simplicity of a given number.\n\nAs examples of composite numbers, let’s take 6, 63, 121, and 6,697. This assertion also needs explanation. Number 6 has positive divisors 1 and 6 as well as divisors 2 and 3, because 6=2-3, so 6 is indeed a composite number. The positive divisors of 63 are numbers 1, 3, 7, 9, 21, and 63. The number 121 is equal to the product of 11-11, so its positive divisors are 1, 11, and 121. And number 6,697 is composite, because its positive divisors in addition to 1 and 6,697 are 37 and 181.\n\nIn conclusion, we would like to point out that prime numbers and mutually prime numbers are not the same thing.\n\n## How do you properly negotiate on the phone?How do you properly negotiate on the phone?\n\nThe 21st century is the age of capitalism, the age of new technologies, mass digitalization and general globalization. It would seem that everything is getting easier every day: it is",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9348668,"math_prob":0.9875414,"size":3414,"snap":"2021-43-2021-49","text_gpt3_token_len":736,"char_repetition_ratio":0.23870967,"word_repetition_ratio":0.03057554,"special_character_ratio":0.21558289,"punctuation_ratio":0.13273002,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9898103,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,9,null,4,null,3,null,3,null,2,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-24T00:23:36Z\",\"WARC-Record-ID\":\"<urn:uuid:743df010-9042-4db8-9232-fad05d5a17c6>\",\"Content-Length\":\"29666\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f4cc1955-4f6a-4def-ad43-86142994f158>\",\"WARC-Concurrent-To\":\"<urn:uuid:7991e6f4-0d0f-4ad6-85c9-3284887d5f94>\",\"WARC-IP-Address\":\"172.67.217.236\",\"WARC-Target-URI\":\"https://bestfitpetsit.com/prime-and-composite-numbers/\",\"WARC-Payload-Digest\":\"sha1:7GQVUIVKA52VMW7JSWUFSBF6NLFAR2IR\",\"WARC-Block-Digest\":\"sha1:PV7L4R2W25BB4Y4L4CXEARMHYHVVQFEU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585828.15_warc_CC-MAIN-20211023224247-20211024014247-00079.warc.gz\"}"} |
https://dynesty.readthedocs.io/en/stable/crashcourse.html | [
"# Crash Course¶\n\ndynesty requires three basic ingredients to sample from a given distribution:\n\n• the likelihood (via a loglikelihood() function),\n• the prior (via a prior_transform() function that transforms samples from the unit cube to the target prior), and\n• the dimensionality of the parameter space.\n\nAs an example, let’s define our likelihood to be a 3-D correlated multivariate Normal (Gaussian) distribution and our prior to be uniform in each dimension from [-10, 10):\n\nimport numpy as np\n\n# Define the dimensionality of our problem.\nndim = 3\n\n# Define our 3-D correlated multivariate normal likelihood.\nC = np.identity(ndim) # set covariance to identity matrix\nC[C==0] = 0.95 # set off-diagonal terms\nCinv = np.linalg.inv(C) # define the inverse (i.e. the precision matrix)\nlnorm = -0.5 * (np.log(2 * np.pi) * ndim +\nnp.log(np.linalg.det(C))) # ln(normalization)\n\ndef loglike(x):\n\"\"\"The log-likelihood function.\"\"\"\n\nreturn -0.5 * np.dot(x, np.dot(Cinv, x)) + lnorm\n\n# Define our uniform prior.\ndef ptform(u):\n\"\"\"Transforms samples u drawn from the unit cube to samples to those\nfrom our uniform prior within [-10., 10.) for each variable.\"\"\"\n\nreturn 10. * (2. * u - 1.)\n\n\nEstimating the evidence and posterior is as simple as:\n\nimport dynesty\n\n# \"Static\" nested sampling.\nsampler = dynesty.NestedSampler(loglike, ptform, ndim)\nsampler.run_nested()\nsresults = sampler.results\n\n# \"Dynamic\" nested sampling.\ndsampler = dynesty.DynamicNestedSampler(loglike, ptform, ndim)\ndsampler.run_nested()\ndresults = dsampler.results\n\n\nCombining the results from multiple (independent) runs is easy:\n\nfrom dynesty import utils as dyfunc\n\n# Combine results from \"Static\" and \"Dynamic\" runs.\nresults = dyfunc.merge_runs([sresults, dresults])\n\n\nWe can visualize our results using several of the built-in plotting utilities. For instance:\n\nfrom dynesty import plotting as dyplot\n\n# Plot a summary of the run.\nrfig, raxes = dyplot.runplot(results)\n\n# Plot traces and 1-D marginalized posteriors.\ntfig, taxes = dyplot.traceplot(results)\n\n# Plot the 2-D marginalized posteriors.\ncfig, caxes = dyplot.cornerplot(results)\n\n\nWe can post-process these results using some built-in utilities. For instance:\n\nfrom dynesty import utils as dyfunc\n\n# Extract sampling results.\nsamples = results.samples # samples\nweights = np.exp(results.logwt - results.logz[-1]) # normalized weights\n\n# Compute 10%-90% quantiles.\nquantiles = [dyfunc.quantile(samps, [0.1, 0.9], weights=weights)\nfor samps in samples.T]\n\n# Compute weighted mean and covariance.\nmean, cov = dyfunc.mean_and_cov(samples, weights)\n\n# Resample weighted samples.\nsamples_equal = dyfunc.resample_equal(samples, weights)\n\n# Generate a new set of results with statistical+sampling uncertainties.\nresults_sim = dyfunc.simulate_run(results)"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.6471183,"math_prob":0.9815017,"size":2742,"snap":"2021-04-2021-17","text_gpt3_token_len":693,"char_repetition_ratio":0.1311176,"word_repetition_ratio":0.03183024,"special_character_ratio":0.25091174,"punctuation_ratio":0.18126273,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9948244,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-18T06:23:26Z\",\"WARC-Record-ID\":\"<urn:uuid:e5eca83d-01c9-46b4-8d7b-642b0dc42051>\",\"Content-Length\":\"20331\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c5f41ede-ee43-4439-b58e-07e4ad4c9587>\",\"WARC-Concurrent-To\":\"<urn:uuid:fcb94ef7-68fe-4c42-aa07-4f85e130e5ae>\",\"WARC-IP-Address\":\"104.17.33.82\",\"WARC-Target-URI\":\"https://dynesty.readthedocs.io/en/stable/crashcourse.html\",\"WARC-Payload-Digest\":\"sha1:H7NFIC4YZCMDOWICPUY53VGIFYE6TJN3\",\"WARC-Block-Digest\":\"sha1:6TW2KXW2ZB5LUGA4WTSD6KWM3RV7IDJ3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038468066.58_warc_CC-MAIN-20210418043500-20210418073500-00612.warc.gz\"}"} |
https://researchers.unab.cl/es/publications/genetic-algorithm-in-the-wavelet-domain-for-large-p-small-n-regre | [
"# Genetic algorithm in the wavelet domain for large p small n regression\n\nEylem Deniz Howe, Orietta Nicolis\n\nResultado de la investigación: Contribución a una revistaArtículorevisión exhaustiva\n\n4 Citas (Scopus)\n\n## Resumen\n\nMany areas of statistical modeling are plagued by the \"curse of dimensionality,\" in which there are more variables than observations. This is especially true when developing functional regression models where the independent dataset is some type of spectral decomposition, such as data from near-infrared spectroscopy. While we could develop a very complex model by simply taking enough samples (such that n > p), this could prove impossible or prohibitively expensive. In addition, a regression model developed like this could turn out to be highly inefficient, as spectral data usually exhibit high multicollinearity. In this article, we propose a two-part algorithm for selecting an effective and efficient functional regression model. Our algorithm begins by evaluating a subset of discrete wavelet transformations, allowing for variation in both wavelet and filter number. Next, we perform an intermediate processing step to remove variables with low correlation to the response data. Finally, we use the genetic algorithm to perform a stochastic search through the subset regression model space, driven by an information-theoretic objective function. We allow our algorithm to develop the regression model for each response variable independently, so as to optimally model each variable. We demonstrate our method on the familiar biscuit dough dataset, which has been used in a similar context by several researchers. Our results demonstrate both the flexibility and the power of our algorithm. For each response variable, a different subset model is selected, and different wavelet transformations are used. The models developed by our algorithm show an improvement, as measured by lower mean error, over results in the published literature.\n\nIdioma original Inglés 1144-1157 14 Communications in Statistics: Simulation and Computation 44 5 https://doi.org/10.1080/03610918.2013.809101 Publicada - 1 ene. 2015"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9179498,"math_prob":0.8822148,"size":4406,"snap":"2022-27-2022-33","text_gpt3_token_len":907,"char_repetition_ratio":0.11585643,"word_repetition_ratio":0.79576397,"special_character_ratio":0.20789832,"punctuation_ratio":0.10843374,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97786427,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-20T05:41:34Z\",\"WARC-Record-ID\":\"<urn:uuid:6d16528a-3241-4ca1-bee9-7f317876c7c2>\",\"Content-Length\":\"59552\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6824bc4d-6a56-4b4d-9589-3270865f0b85>\",\"WARC-Concurrent-To\":\"<urn:uuid:6fb956e6-68a1-459b-a7ba-872470fd82b9>\",\"WARC-IP-Address\":\"104.18.5.125\",\"WARC-Target-URI\":\"https://researchers.unab.cl/es/publications/genetic-algorithm-in-the-wavelet-domain-for-large-p-small-n-regre\",\"WARC-Payload-Digest\":\"sha1:CVYDVKPLV2UENI4773HG5N25XMP6EUSX\",\"WARC-Block-Digest\":\"sha1:S3Y6KAA7CBS6URCGULGMYVAI5TWBXIEO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573908.30_warc_CC-MAIN-20220820043108-20220820073108-00471.warc.gz\"}"} |
https://m.scirp.org/papers/38844 | [
"Solution of Laplace’s Differential Equation and Fractional Differential Equation of That Type\nABSTRACT\nIn a preceding paper, we discussed the solution of Laplace’s differential equation by using operational calculus in the framework of distribution theory. We there studied the solution of that differential equation with an inhomogeneous term, and also a fractional differential equation of the type of Laplace’s differential equation. We there considered derivatives of a function",
null,
"on",
null,
", when",
null,
"is locally integrable on",
null,
", and the integral",
null,
"converges. We now discard the last condition that",
null,
"should converge, and discuss the same problem. In Appendices, polynomial form of particular solutions are given for the differential equations studied and Hermite’s differential equation with special inhomogeneous terms.\n\nCite this paper\nT. Morita and K. Sato, \"Solution of Laplace’s Differential Equation and Fractional Differential Equation of That Type,\" Applied Mathematics, Vol. 4 No. 11, 2013, pp. 26-36. doi: 10.4236/am.2013.411A1005.\nReferences\n K. Yosida, “The Algebraic Derivative and Laplace’s Differential Equation,” Proceedings of the Japan Academy, Vol. 59, Ser. A, 1983, pp. 1-4.\n\n K. Yosida, “Operational Calculus,” Springer-Verlag, New York, 1982, Chapter VII.\n\n J. Mikusiński, “Operational Calculus,” Pergamon Press, London, 1959.\n\n T. Morita and K. Sato, “Remarks on the Solution of Laplace’s Differential Equation and Fractional Differential Equation of That Type,” Applied Mathematics, Vol. 4, No. 11A, 2013, pp. 13-21.\n\n T. Morita and K. Sato, “Solution of Fractional Differential Equation in Terms of Distribution Theory,” Interdisciplinary Information Sciences, Vol. 12, No. 2, 2006, pp. 71-83.\n\n T. Morita and K. Sato, “Neumann-Series Solution of Fractional Differential Equation,” Interdisciplinary Information Sciences, Vol. 16, No. 1, 2010, pp. 127-137.\n\n M. Abramowitz and I. A. Stegun, “Handbook of Mathematical Functions with Formulas, Graphs and Mathematical Tables,” Dover Publ., Inc., New York, 1972, Chapter 13.\n\n M. Magnus and F. Oberhettinger, “Formulas and Theorems for the Functions of Mathematical Physics,” Chelsea Publ. Co., New York, 1949, Chapter VI.\n\n T. Morita and K. Sato, “Liouville and Riemann-Liouville Fractional Derivatives via Contour Integrals,” Fractional Calculus and Applied Analysis, Vol. 16, No. 3, 2013, pp. 630-653.\n\n L. Levine and R. Maleh, “Polynomial Solutions of the Classical Equations of Hermite, Legendre and Chebyshev,” International Journal of Mathematical Education in Science and Technology, Vol. 34, 2003, pp. 95-103.\n\n F. Riesz and B. Sz.-Nagy, “Functional Analysis,” Dover Publ., Inc., New York, 1990, p. 146.\n\nTop"
]
| [
null,
"https://m.scirp.org/papers/Edit_e43f170b-2a6b-4d14-a65d-8890cd706f8e.png",
null,
"https://m.scirp.org/papers/Edit_c3fa0992-39d2-4da2-aea9-a5140e612ab6.png",
null,
"https://m.scirp.org/papers/Edit_9a1c6a5d-fd83-4617-8142-6db97e475ec2.png",
null,
"https://m.scirp.org/papers/Edit_99741e6f-b5bb-44d2-aaa5-e88d13d71788.png",
null,
"https://m.scirp.org/papers/Edit_339bcd8c-75d2-404f-81f8-a69195ff2722.png",
null,
"https://m.scirp.org/papers/Edit_f6b6fb63-83fa-471d-b865-5933b5d0359c.png",
null
]
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https://brilliant.org/practice/arithmetic-progressions-level-3-challenges/?subtopic=arithmetic-and-geometric-progressions&chapter=arithmetic-progressions | [
"",
null,
"Algebra\n\n# Arithmetic Progressions: Level 3 Challenges\n\n$\\left (1 + 3 + 5 + \\ldots + p^\\text{th} \\right ) \\\\ + \\\\ \\left (1 + 3 + 5 + \\ldots + q^\\text{th} \\right ) \\\\ = \\\\ \\left (1 + 3 + 5 + \\ldots + r^\\text{th} \\right )$\n\nGiven the above equation for positive integers $p,q,r$ with $p^\\text{th}$ term greater than 39.\n\nWhat is the smallest possible value of the expression below?\n\n$\\large p^\\text{th} \\ \\text{term } + q^\\text{th} \\ \\text{term } + r^\\text{th} \\ \\text{term }$\n\n$\\dfrac{a(q-r)}{p} + \\dfrac{b(r-p)}{q} + \\dfrac{c(p-q)}{r}$\n\nIn an arithmetic progression, the sum of the first $p, q, r$ terms are $a, b, c$ respectively. Compute the expression above.\n\nConsider an arithmetic progression with 2 and 101 as its first term and last term respectively. If the sum of the first 5 terms of this arithmetic progression is 40, find the sum of the last 5 terms of this progression.\n\n$\\begin{array}{llllll} A_1 : & 2, & 9, & 16, \\ldots , & 2 + (1000-1) \\times 7 \\\\ A_2: & 3, & 12 , & 21, \\ldots, & 3 + (1000-1) \\times 9? \\\\ \\end{array}$\n\nHow many integers appear in both of the following arithmetic progressions above?\n\nDetails and assumptions\n\nSince 2 appears in $A_1$ but not in $A_2$, it does not appear in both of the arithmetic progressions.\n\n$\\large \\frac ab \\ , \\ ab \\ , \\ a -b \\ , \\ a+b$\n\nAbove shows real numbers that belong to an arithmetic progression in order. Find the next term of this sequence.\n\n×"
]
| [
null,
"https://ds055uzetaobb.cloudfront.net/brioche/chapter/Arithmetic%20Progressions-zAiar9.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9164439,"math_prob":0.99991405,"size":1152,"snap":"2019-43-2019-47","text_gpt3_token_len":242,"char_repetition_ratio":0.17334495,"word_repetition_ratio":0.20707071,"special_character_ratio":0.2126736,"punctuation_ratio":0.14102565,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99998546,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-19T01:29:02Z\",\"WARC-Record-ID\":\"<urn:uuid:9ec4d210-8941-4cd0-929d-46a5a3b94b9d>\",\"Content-Length\":\"101530\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:81be771c-c737-444e-a508-ef880ba4a142>\",\"WARC-Concurrent-To\":\"<urn:uuid:1f00c97d-0a46-4d3b-b4ef-1bea57eaffab>\",\"WARC-IP-Address\":\"104.20.35.242\",\"WARC-Target-URI\":\"https://brilliant.org/practice/arithmetic-progressions-level-3-challenges/?subtopic=arithmetic-and-geometric-progressions&chapter=arithmetic-progressions\",\"WARC-Payload-Digest\":\"sha1:J5UYTYYMIWYC7GGN2WO2COOIYXB5OWHI\",\"WARC-Block-Digest\":\"sha1:RGT7MUIJVVTOEKD3FIPHYSBKW6R4YLX7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496669868.3_warc_CC-MAIN-20191118232526-20191119020526-00267.warc.gz\"}"} |
http://bioconductor.statistik.tu-dortmund.de/packages/3.10/bioc/vignettes/EMDomics/inst/doc/EMDomics.html | [
"# Welcome\n\nWelcome to the EMDomics package! This vignette will explain the functionality of the package through the creation and analysis of a toy data set.\n\n# Earth Mover’s Distance\n\nEMDomics analyzes differences in genomics data between groups of samples. Typically the data will be gene expression levels from array- or sequence-based experiments, but other scenarios are possible. In a real two-class experiment, the groups might be test vs. control, sensitive vs. resistant, etc. In a multi-class experiment (i.e., more than two groups of patients), groups may be associated with patients (in the case of single cell measurements) or disease subtypes. Typically you’ll be analyzing differences across multiple genes, but we’ll start with a single gene to get a feel for how the Earth Mover’s Distance (EMD) algorithm works. Note also that this package includes functionality for Komolgorov-Smirnov (K-S) and Cramer von Mises (CVM) distribution comparison tests. To access these tests, use calculate_ks or calculate_cvm. The input and output syntax is the same as calculate_emd, with “emd” being replaced with “ks” or “cvm” when accessing output values.\n\nBecause this package is EMDomics we will go through functionality with calculations for EMD, but K-S and CVM can be accessed with ease by replacing the function name.\n\nWe’ll create a vector of expression data for 100 samples. We’ll assign the first 50 to group “A,” the next 20 to group “B,” and the final 30 to group “C.” We will create a vector of group labels that describes which group each of the samples is a part of. Note that the vector of labels must have names corresponding to the sample identifiers in the data:\n\nexp_data <- rnorm(100)\nnames(exp_data) <- paste(\"sample\", 1:100)\n\ngroupA.labels <- rep(\"A\",50)\ngroupB.labels <- rep(\"B\",20)\ngroupC.labels <- rep(\"C\",30)\n\nlabels <- c(groupA.labels, groupB.labels, groupC.labels)\nnames(labels) <- names(exp_data)\n\nWe’ll take a quick look at the three distributions using ggplot:\n\nlibrary(ggplot2)\n\ndf <- as.data.frame(exp_data)\ndf$group[1:50] <- \"A\" df$group[51:70] <- \"B\"\ndf$group[71:100] <- \"C\" ggplot(df, aes(exp_data, fill=group)) + geom_density(alpha=0.5)",
null,
"We shouldn’t expect the three groups to look too different, since we’re just sampling from the normal distribution. Intuitively, the “work” required to transform any one distribution into another should be low. We can calculate the EMD score for this single gene using the function calculate_emd_gene: library(EMDomics) calculate_emd_gene(exp_data, labels, names(exp_data)) ## 2.2 Now we’ll modify the expression data for group A and see how the EMD score changes. We’ll randomly add or subtract 2 from each data point in group A: exp_data2 <- exp_data mod_vec <- sample(c(2,-2), 50, replace=TRUE) exp_data2[1:50] <- exp_data2[1:50] + mod_vec Let’s again visualize the distributions and calculate the EMD score: df <- as.data.frame(exp_data2) df$group[1:50] <- \"A\"\ndf$group[51:70] <- \"B\" df$group[71:100] <- \"C\"\n\nggplot(df, aes(exp_data2, fill=group)) + geom_density(alpha=0.5)",
null,
"calculate_emd_gene(exp_data2, labels, names(exp_data2))\n## 6.82\n\nThe EMD score is larger, reflecting the increased work needed to transform one distribution into another. Note that since we have three classes defined, we aren’t able to tell from the EMD score alone which two groups (or, potentially, all three groups) demonstrate differences in gene behavior. The composite EMD score in a multi-class analysis is the average of all the pairwise EMD scores. The pairwise EMD scores are computed by comparing all possible combinations of two of the classes. More information on multi-class analysis is in the next section.\n\nNote that in a two-class analysis, a greater EMD score is directly indicative of a greater difference between the measurement distributions of the two classes.\n\n# Analyzing Significance\n\nThe EMD score increases as the distributions become increasingly dissimilar, but we have no framework for estimating the significance of a particular EMD score. EMDomics uses a permutation-based method to calculate a q-value that is interpreted analogously to a p-value. To access the full functionality of the package, we’ll use the function calculate_emd.\n\nWe’ll first create a matrix of gene expression data for 100 samples (tumors, patients, etc.) and 100 genes. We’ll just sample from the normal distribution for now. The first 50 samples will be our “group A,” second 20 will be “group B,” and the final 30 will be “group C.” Just as before, we will store these sample labels in a named vector associating group with sample identifier:\n\ndata <- matrix(rnorm(10000), nrow=100, ncol=100)\nrownames(data) <- paste(\"gene\", 1:100, sep=\"\")\ncolnames(data) <- paste(\"sample\", 1:100, sep=\"\")\n\ngroupA.labels <- rep(\"A\",50)\ngroupB.labels <- rep(\"B\",20)\ngroupC.labels <- rep(\"C\",30)\n\nlabels <- c(groupA.labels, groupB.labels, groupC.labels)\nnames(labels) <- colnames(data)\n\nNow we can call calculate_emd. We’ll only use 10 permutations for the purposes of this vignette, but in actual experiments using at least 100 permutations is advised. For this example we will turn off parallel processing, but in general it should be enabled.\n\nresults <- calculate_emd(data, labels, nperm=10, parallel=FALSE)\n\nMost of the time, you’ll be interested in the emd matrix returned as a member of the return object:\n\nemd <- results$emd head(emd) ## emd q-value ## gene1 1.880000 1.0000000 ## gene2 2.116667 0.5806452 ## gene3 1.716667 1.0000000 ## gene4 2.120000 0.5806452 ## gene5 2.450000 0.1500000 ## gene6 3.060000 0.0000000 This matrix lists the emd score and the q-value for each gene in the data set. Because we’re not analyzing many genes and the data is randomly generated, there may be some significant q-values in the results simply by chance. We can order the emd matrix by q-value: emd2 <- emd[(order(emd[,\"q-value\"])),] head(emd2) ## emd q-value ## gene6 3.060000 0 ## gene13 3.110000 0 ## gene31 3.283333 0 ## gene37 3.000000 0 ## gene38 4.866666 0 ## gene48 3.026667 0 Note the correlation of significant q-values with relatively large EMD scores. In a multi-class analysis, it may not be enough to know that a gene behaves differently somehow among the defined classes. We may be interested in finding which two classes display a greater difference in gene behavior, or if all three classes are somehow different. The differences between each of the classes is defined in the pairwise.emd.table. Note that EMD is not directional, so all possible combinations, not permutations, of the class labels are used in the pairwise EMD score calculations. Each of the columns represents a pairwise comparison (e.g. Group A vs Group B), each row represents a gene, and the cell content is the EMD score quantifying the work required to transform the distribution of one group into the other. emd.pairwise <- results$pairwise.emd.table\nhead(emd.pairwise)\n## A vs B A vs C B vs C\n## gene1 1.78 1.8800000 1.0666666\n## gene2 1.61 1.0066668 2.1166666\n## gene3 1.55 1.0133333 1.7166666\n## gene4 2.12 1.2066667 1.4000001\n## gene5 2.45 2.1800001 0.7500001\n## gene6 3.06 0.9333333 2.5333333\n\n# Visualization\n\nEMDomics includes a few visualization functions. The function plot_density will display the density distributions of each of the groups for a given gene, along with the EMD score. We can compare the gene with the largest EMD score and the gene with the smallest EMD score, for example:\n\nemd3 <- emd[(order(emd[,\"emd\"])),]\nsmallest_gene <- rownames(emd3)\nbiggest_gene <- rownames(emd3)[nrow(emd3)]\n\nplot_emd_density(results, smallest_gene)",
null,
"plot_emd_density(results, biggest_gene)",
null,
"Note that the EMD score is the average of the each of the pairwise EMD scores. This means that the smallest and largest EMD scores may have ambiguous meanings in a multi-class analysis. To understand how each class compares to the others, the pairwise.emd.table provides pairwise comparisons of gene behavior. These pairwise EMD scores will lend more insight into how the gene is similar or different across classes.\n\nIn a two-class analysis, the smallest score represents the gene that demonstrates the most similar behavior in both classes, and the largest score represents the gene that demonstrates the most different behavior in both classes.\n\nWe can plot a histogram of all the calculated EMD scores with the function plot_emdperms:\n\nplot_emdperms(results)",
null,
"This plot can help intuitively understand the relative significance of an EMD score. For example, almost all the randomly permuted EMD scores are smaller than the largest calculated EMD score plotted above.\n\nIn a similar vein, the function plot_emdnull plots the null distribution (the median of the permuted EMD scores) for each gene vs. the calculated EMD score (the line x=y is superimposed in red):\n\nplot_emdnull(results)",
null,
"# Wrapping Up\n\nThis concludes the EMDomics vignette. For additional information, please consult the reference manual.\n\n# Session Info\n\n## R version 3.6.1 (2019-07-05)\n## Platform: x86_64-pc-linux-gnu (64-bit)\n## Running under: Ubuntu 18.04.3 LTS\n##\n## Matrix products: default\n## BLAS: /home/biocbuild/bbs-3.10-bioc/R/lib/libRblas.so\n## LAPACK: /home/biocbuild/bbs-3.10-bioc/R/lib/libRlapack.so\n##\n## locale:\n## LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C\n## LC_TIME=en_US.UTF-8 LC_COLLATE=C\n## LC_MONETARY=en_US.UTF-8 LC_MESSAGES=en_US.UTF-8\n## LC_PAPER=en_US.UTF-8 LC_NAME=C\n## LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C\n##\n## attached base packages:\n## stats graphics grDevices utils datasets methods base\n##\n## other attached packages:\n## EMDomics_2.16.0 ggplot2_3.2.1\n##\n## loaded via a namespace (and not attached):\n## Rcpp_1.0.2 pillar_1.4.2 compiler_3.6.1\n## tools_3.6.1 emdist_0.3-1 digest_0.6.22\n## preprocessCore_1.48.0 evaluate_0.14 tibble_2.1.3\n## gtable_0.3.0 pkgconfig_2.0.3 rlang_0.4.1\n## yaml_2.2.0 parallel_3.6.1 xfun_0.10\n## withr_2.1.2 dplyr_0.8.3 stringr_1.4.0\n## knitr_1.25 grid_3.6.1 tidyselect_0.2.5\n## glue_1.3.1 R6_2.4.0 BiocParallel_1.20.0\n## rmarkdown_1.16 purrr_0.3.3 magrittr_1.5\n## scales_1.0.0 htmltools_0.4.0 matrixStats_0.55.0\n## CDFt_1.0.1 assertthat_0.2.1 colorspace_1.4-1\n## labeling_0.3 stringi_1.4.3 lazyeval_0.2.2\n## munsell_0.5.0 crayon_1.3.4"
]
| [
null,
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",
null,
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",
null,
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",
null,
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",
null,
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o4J+gKLMOxLQw1OqdgoIKgHfqyCVi5kUUHXMjb7033vjWUjSuJq8ErFRgFBPfCjFPTQ0rnDWFTQ0AQ2U3kZ794R7IVYHW6p2CggqAd+lIJuUQ/emqBFjKk/UPI0Gxd7Gzm3VGwUENQDP0pBq3bv3l0UFXQFy1Jf9PY5Y7FfZuaWio0CgnrgRymoQklU0KdZrpooZXEX9txSsVFAUA9AUGkOW6AmGjLZuuh6c+mxyZMnv1EnREgS286OBn/DSSFfw4XqfQ3XEPY1nM+TVy8lGq3+c3ceBL2fPaOVjWZvRNebS0tmzJjx9xoh6iSx7WryzajFDYLhbJDqfQ1XX+druFDY13A+T16tlGi01X4Iqr+uaBSLvUmWWyp2HBA+xD9mRi3GId4DjfIQr75TXzmYr42u55aKjQKCegCCSovYw2riBGOxp0bcUrFRQFAPQFDpFTZevdG5jbHi6HpuqdgoIKgHIKj0BWM7I4kl+q1P21KxUUBQD0BQKTROvaNUOZbF/aQYt1RsFBDUAxBUkt5nbPHB+u05bGSpkl0+Y0aZtRSCugSCGvBDUOk5xthgxoarD9/zGCuxlkJQl0BQA74IKm2anTVkYqH204C6oMZSCOoSCGqgSf/JBwSFoBAUgnoAgkJQCApBfQ0HQQ1AUM9AUHEgKASFoBDU13AQ1AAE9QwEFQeCQlAICkF9DQdBDUBQz0BQcSAoBIWgENTXcBDUAAT1DAQVB4JCUAgKQX0NB0ENQFDPQFBxICgEhaAQ1NdwENQABPUMBBUHgkJQCApBfQ0HQQ1AUM9AUHEgqP+CWoIWzOe2IgoENQBBPQeFoOJAUAgKQSEoBBUFgkJQCApBIagoEBSCQtAkqBCiRqoU23C+GbU4JBbNLmhBPrcVUWqqvW1voq7B13AeJ89MlZRotFUpFbREiEqpVGzDPDNqcZ1YNLugBfO5rYhSVeFtexM1Db6G8zh5Zsqk8gQ1ylMqqNhxAId4D+AQD0EhqDAQFIJCUAgKQUWBoBAUgkJQCCoKBIWgEBSCQlBRICgEhaAQFIKKAkEhKASFoBBUFAgKQSEoBIWgokBQCApBISgEFQWCQlAICkEhqCgQFIJCUAgKQUWBoBAUgkJQCCoKBIWgELQxC2pTI6mgiQW1bOLUDAQ1AEEhaFJAUCsQ1AMQFIJCUGEgKASFoBAUgooCQSEoBIWgEFQUCApBIWg2i3F3rLg4WpgHQSEon3QKugmC2m4CQTVSIujWTzQ2TGKvx4rfZFO08h0QFILySe056DtsTjiWK2QLLTXERgFBPQBBo+wZemdZXDaXrYKgNptAUI1UClozJfOb+PwEtlGSaiEobxMIqpFKQV9iT8VnazPZxoIslpX7EQSFoHakUNDDQ4cdj8/vZSxTvYjPrVTyh/r16/dsWAxJcLsCM5ZwNjWSCpo4hosqJw1JdPLs4qU4XJ1fghaw1wz5jYzdueFQxY58xhYp+Yply5Z9JvZzZCfxl+ZsaiQVNPEvzVk2cWqmtia5QSYAvzSn8g0bazzfLFr0nPrpvISxb/VCseMADvEewCFeJY+9ZLOmZgh7C4JCUC4pE/TQ4MxDduvui109iY0CgnoAgkZYymbbrpvGlkNQCMolVYLWjmYbjCXhu4au1tYNZeshKATlkipBN7HMClPRUjb2SCSxhI0shaAQlEuqBC1kD8Qyy2fMKJOkw2PYxPeLD255lLGPoqvERgFBPQBBFSawJbFMHmMl8mLHMPU+vX6oh6AQ1EKKBN3D2L8sgkplL07LGvFQ4cG4imKjgKAegKDJIDYKCOoBCApBIagwEBSCQlAICkFFgaAQFIJCUAgqCgT1LKhVLgjqHxAUgkJQCApBRYGgEBSCQlAIKgoEhaAQFIJCUFEgKASFoBAUgooCQSEoBIWgEFQUCApBISgEhaCiQFAICkEhKAQVBYJCUAgKQSGoKBAUgkJQCApBRYGgEBSCQlAIKgoEhaAQFIJCUFEgKASFoElQJ0RIcLu6BWbU4nDCGk4xLBQIxHAabUhwtHwawonrJBXP12j1Un2CGjUpFVTsvxk+QT3QND9B9+6ti8ud2Ps9BIWgKsEQlKgoLvc8XQJBIahKEAVdRB0hKARVSbugyxSI5i6LUtATgkJQnbQLSjyuh6AQVCWQgnb6BIJCUJW0C/qeAtFf3ouxrtTGNwgKQb3hz0WSf4iNorEJ6gKHTmuCJrOJE01T0AceOABBbWpAUB8J2qNOsVFAUAgKQSGoO4Ii6PcvZf8ujj0QFIJGCIigb3Qy3mj6NwSFoBGCIejuU013QsshKASNEAxB76TmD20+fDSGP35CUAhqRFTQy2mBT0pCUAjqgKigbU6pgqA2NSCoj4gKmtHtpPgJQSGoEVFBr6NjENSmBgT1EVFBl9DzENSmBgT1EVFBQzeetgmC8mtAUB8RvlFfmtnirs8rICgENRMMQQcNGtiaiM7ooQNBIahKMAS1fKcegkJQlWAI2tcMBIWgKsEQ9GQhNgoICkEhKAR1BwS1AkEhKASFoO4IhqBPm9gIQSGoSjAENd9lyoGgEFQliIK2aDETgkJQlWAIukHn3SdGNafsBmftiplOnqF885ysIZMKD0JQCGqHHxdJO66gKc6CbuIL+oJaNuxLCApBbfDlKn736fQPR0HfZFM+UdkRV7qWsdmf7ntvLBtRAkEhKB9/bjNNotsdBS1kC62FoQlsZr283DuCvQBBISgffwRdTOc4CprLVlkLixjbGkk8zcaFISgE5eKPoAuptaOgE9hGSao1Fa5gWeq11eeMRd9EJjYKCApBHQW9iX7q5GdtJttYkMWycj+KL32a5aqJUsa+khd1X3/99aESISqlUrEN88yoxXUJazjFsFAwP3GdhM3EjbbCZc/cUdMguKENdYmrJEGZVJ6ghv7KEHtBQ3OJbnUSdC9jmeoFe25lrHQO0/60viGTrZMXB/v06fOUU5iTQIEZP2r4gh99/3FQry2NgvaP0bcLEb3rFGIjY3duOFSxI5+xRbHS+9kzWmo0e0P+t/bTTz/dVypEpXTCVb15CVHr1TtsIhC0ID9xnYTNxKiqdNkzd9Q0CG5oQ33iKklQLlUkqKH/3ZHjo84xjo4XLXpOPVFYwti3cYIu1lKjYtdQYicqbs9B3Z764RzUA8E4B82I4+ybn3P0M0bNEPZWNDOHaQcl+RC/FoK6tw2CGvD163b3sdhZ5iL2sJo4wVj0WZLYKCBogk2cgKBxTGPLo+lX2Hj19uc2xoohqHvbIKgB74KG7xq6Wk3VDmXro8VfMLYzklii3xCFoK5sg6AGnATdveGddTvDnBVGlrKxRzQVR8Z+USk0Tr3PVDmWxV6jIzYKCJpgEyearKBf33NG5Aq+84TtCQQ9PIZNfL/44JZHGftIyS+fMaNMXrzP2OKD9dtz4q0VGwUETbCJE01V0AUtozeZWv4pgaE7hqn36bVDfR5jke8vPScXDWZs+NZYTbFRQNAEmzjRRAUtkMVse8Ok2ffc2F5Ocb6sZKDsxWlZIx7Sv5qsCyptmp01ZGLhobiKYqOAoAk2caJpCrrrVGqeo55YHstpTq2+S2CoW8RGAUETbOJE0xQ0m+JeUp/fWP5ozq0XENQDwRD0Cro0lglfSldCUHsgqDiigp5OE+JyE6kzBLUHgoojKuiphoN6LrWCoPZAUHFEBe1Ot8TlBtHZENQeCCqOqKADqf3eaGZ/R8qEoPZAUHFEBV1K1EdfU/YLomUQ1B4IKo6ooPU/I+qct7VcKv9qfhein4UgqD0QVBzhJ0nfdos85WwX+bfrtxbTIGgMCCqO+LP47zOb6c/iBx0wewZB44Gg4nj5ut3X8zKvvuTqzLyvfdMTgrqxDYIaaEpvWHbrBQT1QIAEVb+qvA+CJgCCiiMuaP1TA3sqyxBdNLMGgjoBQcURFnT/L4m6qYISXbkbgjoAQcURFTR8FVGLwZHU3d2IfgVBHYCg4ogK+grRL/dr6crRRK9BUHsgqDjCv3ZMHY9EMzXn0HAIag8EFUdU0ItpbFxuMl0OQe2BoOKICtqaHo7LzaHTIKg9EFQcUUG70si43CjqCkHtgaDiiAraj3pURTPV5/h2GS82ioigLnaYWzvykzEncbhGJKgPQYMhaCHRiGhmAtGfIag9EFQcUUFrLiS6arXyuwj1H/yK6KwKi2oQNAoEFUf4SdI25cXfzc7qc15zedl6vcU0CBoDgooj/ix+16+i72a6eLNffkJQ22ZiQFAD9t9mWj/lik7N2l4w5nW//t4Dgjo0EwOCGmg03wd1Mbdu7YCgHoJCUCsQNMEmTkDQZKgTIqRst8CMtZ6lig0FDusEgjqFc91M3GhD/HbF5q6uIew0GIF4gv3gUy/VJ6ihfxkZn6A2UROHwyeoOEH7BBUbBQRNsIkTEBSCQlBhICgPCOohKAS1AkETbOIEBIWgEFQYCMoDgnoICkGtQNAEmzgBQSEoBBUGgvKAoB6CQlArEDTBJk5AUAgKQYWBoDwgqIegENQKBE2wiRMQFIJCUGEgKA8I6iEoBLUCQRNs4gQEhaAQVBgIygOCeggKQa1A0ASbOAFBISgEFQaC8oCgHoJCUCsQNMEmTkBQCApBhYGgPJwEFUBEUAfmzeOXi80dBIWgEFQcCMoDgnoICkGtQFAFsbmDoBAUgooDQXlAUA9BIagVCKogNncQFIJCUHFSJej6RyffMSV/k6GsmOnkQdAkgKAG/BC0fJZm4pzquNJNEFQICGrAD0HnsczCnce2L8xkC+NK32RTPlHZAUGTAIIa8EHQ/YwtjyRWMrYzVlxo0BWCugWCGvBB0A/Z7eqhPXQHezNWnMtWQVABIKgBHwRdyu7TUlPZ4ljxBLZRkmohaLJAUAM+CLpr87dqomIoeytaWpvJNhZksazcjyBoUkBQA37eB32S3XE8mtnLWKZ6EZ9bqeQrV69e/VW5ENVSRXn5PDPWepYqNixwW9EdBfm+hsu3CSc2d+V1DU4zlHy8esF+8KmUqhLW8EvQ0nzG3ollNzJ254ZDFTvk0kVK/mCfPn2e8hC+wIyLKk0KD3OXzCQGjXpt6VXQqleHs9vfiisoWvSc+um8hDHlFKDhxIkTNUeFKJeOHT1qOTpZ67k9iAb7ED/f5hAvNndHq0NOM5R8vFrBfvApkU4kqFHmj6BrsxibtY+7qmZI7MxU7EQF56AKYnOHc1CFQzMZy9lqt/Y+Fj20i40CgiqIzR0EVfycwEZ8bL96mnYfH4K6AoIa8EHQ0D0s+5C5MHzX0NVqqnYoWw9B3QNBDfgg6Do2usxaupSNPRJJLGEjSyGoeyCoAR8EzWU5/9IplvPLZ8yQhT08hk18v/jglkcZ+yhaVWwUEFRBbO4gqCSNZzFek/N5jJXIix3D1CL9UA9B3QFBDXgXtDaTL6hU9uK0rBEPFR6Mqys2CgiqIDZ3EDQpxEYBQRXE5g6CQlAIKg4E5QFBPQSFoFYgqILY3EFQCApBxYGgPCCoh6AQ1AoEVRCbOwiaNkHFaZyCWnA3eRAUgkJQYSAoDwiaZIx4IKgVCMrF3eRBUAgKQYWBoDwgaJIx4oGgViAoF3eTB0EhKAQVBoLygKBJxogHglqBoFzcTR4EhaAQVBgIygOCJhkjHghqBYJycTd5EBSCQlBhICgPCJpkjHggqBUIysXd5EFQCApBhYGgPCBokjHigaBWICgXd5MHQSEoBBUGgvKAoEnGiAeCWoGgXNxNHgSFoBBUmKAJelyICqnk+PE8/1jgYyyZgvm+hps/z2VFd5NX0xCXEYwRT13ymzhwQipLUMOnn6FxSbUQdVJNdXW+fyzwMZZMgb/xXEdzN3mhcFxGMEY8Dclv4kCNVJuoSkoFFTsO4BDPxd3k4RAPQSGoMBCUBwRNMkY8ENQKBOXibvIgKASFoMJAUB4QNMkY8UBQKxCUi7vJg6C+COq8ywoE9rMDTVVQbq35C7x1x9yIP4Lq0fMK5tkMRgeCegaCJo8eHYLygKBJYG4EglqBoFzcTSIEhaAQNGn06BCUBwRNAnMjENQKBOXibhIhKASFoEmjR4egPCBoEpgbgaBWICgXd5MIQSEoBE0aPToE5QFBk8DcCAS1AkG5uJtECApBIWjS6NEhKA8ImgTmRiCoFQjKxd0kQlAICkGTRo8OQXlA0CQwNwJBrUBQLu4mEYJCUAiaNHp0CMoDgiaBuREIagWCcnE3iRAUgkLQpNGjQ1AeEDQJzI1AUCsQlIu7SYSgCpvnZA2ZVHgwYWnCDnOBoFzcTSIElXmBRRj2ZaLShB3mAkG5uJtECCpJaxmb/em+98ayESUJShN2mAsE5eJuEiGoFJrAZtbLy70j2AsJShN2mAsE5eJuEiGoVMTY1kjiaTYu7FyasMNcICgXd5MIQaUVLKshkvicsQPOpQk7zAWCcnE3iRBU/ojMVROljH3lXJqww1wgKBd3kwhBpTlsgZpoyGTrbEtP5OXlrbX9MRznXw4qSP7Xhhzj+RsuML+TxA/ncfLMjfjzO0nR8AU2zcTwLuj97BktNZq9YVt6NCsra2W9ECFJbDs7wv6Gkxp8DdcQ8jecz6P1N1xISjTaWj8EXaylRrFVzqVix4PIK8B9xN+3WB8JV/garvyEr+EMrwD3gUb4CvA5rEBNyAfztc6lYqOAoB6AoNIi9rCaOMHYl86lYqOAoB6AoNIrbLx6o3MbY8XOpWKjgKAegKDSF4ztjCSW6Lc+bUvFRgFBPQBBpdA49Y5S5Vj2fIJSsVFAUA9AUEl6n7HFB+u357CRpUp2+YwZZdZSCOoSCGrAl6/bPccYG8zYcPXhex5jJdZSCOoSCGrAny8sb5qdNWRi4SHJIKixFIK6BIIaCMiffDgCQT0AQSEoBBUGgvKAoB6AoFYgqAcgKASFoMJAUB4Q1ANNW1Axdj5Vk+4uOFG4Jd09cGLdi+nugRPHnipOXClCkAX9e58TiSulj+uWp7sHTjx+e7p74MR/+nzmsiYEFQaCigNBUwAEFadpCFpZ3JC4Uvo4UJbuHjhR+kO6e+BEXbHby4sgCwoABAXBBoKCQANBQaAJtKAr41yYA1gAAAeXSURBVF6pEzDWPzr5jin5m9LdDRtqV84aP3zakwcS10wbR0azSjf1gixo6O6gClo+S31JL5tTnbhy6tl3t9q7295Jd09sCT3EGr2glQtZUAWdxzILdx7bvjCTLUx3VziEp7NRf993+F+/Z7d9m+6+2LGUNXJBDy2dO4wFVdD9jKk36Vfqf2MdKD7V5q18LHsy3X2x4bPMzEYu6Bb1IBVMQT9kt6uH9tAd7M0094XDK+w+NfEX9kB6e2LH0TGZSxq5oFW7d+8uCqqgS3UDpKnRV6UFiHz9DZgvsknp7YkNDTls+ZZGLqhCSVAF3bVZO7erGMreSm9XeITqQpFl+A9sXpq7wucl9lAIgqaCJ9kdx9PdBxtqjxUtYMP+k+5u8Pgic9RhCYKefErzGQvqjZyjyhn8tEBexB/PYv+UIOhJp+rV4ez2AB7gVY7eNv7PeVkvh9LdDyvhWexZCYKedNZmMTZrX7p74cz77PF0d8HKCnZ/nQRBTzKHZjKWszVxvXTQUFKifXDWDmZ709sXK99lDtlWIrOOseISF19Ih6AiHJrARnyc7k7YUZ3JNqup8FDlbC9YFLE47k5cH4IKELqHZR9KXC1dTGEr1ERxAB90QdAUsI6NDvKfezzNRquvZp3HRgTyyywKOAc9ieSynH/puP377hRybAS7c+13P2yWz5M/THdfbIGgJ5HxcUep19LdGQ4bhmtft3sp3T2xB4KePGozAy6odHxxdtbI7EUB/HSP0hQEBQCCgmADQUGggaAg0EBQEGggKAg0EBQEGggKAg0EBYEGgoJAA0FTSS/qle4uNDYgaCqBoEkDQVMJBE0aCJpKIGjSQNBUAkGTBoKmEp8ELX31iSC/m9ZXIKggGyZd2KbLNdn6n/X2p0HSkZmXtOnc92U5t2bQma16jtih1903/fIO7S+f9QNH0HWj+nTMuGbSN3q+9pkburW+JHNFWMuHlmf2OLXzldP3aPm+NFJak0H0Aa8TTREIKkT1GFJp/ohaIAv6ZQ/qeJpcNC08leiMU4haqhZJy9uodTPWmQWtHKCHeUwt2H6RVnD1/kh+Vy8t30p7BYMs6PpWpApq6URTBIKK0NBf1u+m7KnXyHZMiZT0p6t6XPlZuOGvHYluoAnFUuV0oosin4MrmxF1vnVa/7Z0+hkmQYcRtRvz8IN9iZqtUfK7uxB1H56b1YHouno5f7AbUfsBM8ZdLLdTENmiL91yLp0/es4BXieaIhBUhHyi3pH3irx9JtH/KglZlp6RP/H9b1mXqZFK/Yj2yIsjnYgGK6/QKO4jrzIIuo3o8qNKopCIKctfEw1X/lLn8C+JnpKXg4l+rsQIz29OLSNnDLLLND9k04mmCAQVoCqDOmmvZVrbjK5RlrKgb0QKdsifeepfpRcQ/UNe5BH9Wj2jrOhmEvQ5omWRRLgLdZcX64muUtfsb0ED5VjNqLP2dpgconuUpSzoANtONEUgqABvEs3W0zfSKcorQmVB1TeFlsk+qmtWECnvB5c/Nz/R6j5uElT+4HwsPj+FaLWWHNytZ1iaR5Sn5Q+3pa7KUhb0I9tONEUgqADZRNE3M+USKW9H6E+nq/kKoglqamVE0Orm1EOvu98k6Cb5JHJy3G8tXUHNDL+xmkkUfUHZDeoJgyxoqW0nmiIQVIDhZEA5tvenDHWdLKj2YnhVUFnK3+jbhVubLpLuUzb/6bhntBchd6KzDKuvJoq+YmcS0QZJEbSNfSeaIhBUgJuNbrwoOQj6DdGY6IYXmO+Dvv4zNUSv55XT1GbUx7D2EmoXTc8mUt7m3JfOtu9EUwSCCjBMlqUohnIhbivo90T9ohu2tT5J+nZhZoYi2MCwsvp8wzr5E7RcT99LtE5SBNVPGDidaIpAUAHkI/MaU5GtoHWt6By90g/Ef9S55XcdiF5SPmBb1GtF4VAocg4afffPjUTKqxRjgnI60RSBoAK8SjRHT7+9OHJ0thVU+i8i/S2yz5oEfXnxC1rqb5F77aOJ1moFdxPtU+5Q5Wv5Yx2os9JOTFBOJ5oiEFSAkrbU4Xs1ufdU5Yalk6CPR4/xtT8xCdqPaLua2kw0UZJWEV2v5qu60UWStJ0oQzvG52g3B2KCcjrRFIGgImQT/SLy/Y79vYgiv/RhL+ixzkRDlYvxw9ebnyQ9RHSz8oMCUtWtRM9LUqgn0Z3K86ga+bP0D/JyENG1kYfyC5pTy4jLMUE5nWiKQFARqnsTtWIz/jisFdHvIyX2gkqrmhGdMXg6O506DDQ96mxBdPbkubPHyw73rJILNrUm+smYuVPOIeqt3BE90JWoI8u967K4Z/FRQa2daIpAUCGO36jd3Wl5f0OkwEFQaZn2baYuH+aYLpJWdNDvEl37XaRgXVct33d3JP/vS/VvM/1Z3SJOUGsnmiIQVJB3R53buut1U3dpWSdBI98HbdMzu1hak2P6ffmSR39z4Wldrx4TfSZUVnBdRuvL7lih5+v/Z1D3Uzv1nq5/5TNeUEsnmiIQ9KQRPvLVsXT3ofEDQUGggaAg0EBQEGggKAg0EBQEGggKAg0EBYEGgoJAA0FBoIGgINBAUBBoICgINBAUBBoICgLN/wNgTUeCcFTpGgAAAABJRU5ErkJggg==",
null,
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",
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https://ie.boun.edu.tr/tr/publications?page=4&%3Bs=author&%3Bf%5Bag%5D=R&s=title&o=desc&f%5Bauthor%5D=264 | [
"51 sonucu aktar:\nYazar [ Başlık",
null,
"] Tür Yıl\nSüzgeçler: Yazar: Ekim, Tinaz [Clear All Filters]\nG\nDiscrete Mathematics and Theoretical Computer Science, vol. 20, pp. 1b–1b, 2018.\nDiscrete Applied Mathematics, vol. 215, pp. 47–60, 2016.\nTheoretical Computer Science, vol. 629, pp. 40-50, 2016.\nE\nDiscrete Mathematics, vol. 339, pp. 2964–2969, 2016.\nDiscrete Mathematics, vol. 341, pp. 2859-2871, 2018.\nDemange, M., and T. Ekim, Information Processing Letters, vol. 114, no. 1: Elsevier, pp. 66–71, 2014.\nDeniz, Z., and T. Ekim, Discrete Applied Mathematics, vol. 261, pp. 136-147, 2019.\n17th Haifa Workshop on Interdisciplinary Applications of Graphs – CRI, 2017.\nD\nŞeker, O., T. Ekim, and Z. C. Taşkın, Networks, vol. 73, pp. 145-169, 2019.\nBodur, M., T. Ekim, and Z. C. Taşkın, Networks, vol. 62, no. 4, pp. 273–287, 2013.\nC\nde Werra, D., T. Ekim, and C. Raess, Discrete Applied Mathematics, vol. 154, no. 1: North-Holland, pp. 47–58, 2006.\nGeinoz, A., T. Ekim, and D. de Werra, Operations Research Letters, vol. 36, no. 3: North-Holland, pp. 279–282, 2008.\nLATIN 2012: Theoretical Informatics: Springer Berlin Heidelberg, pp. 279–290, 2012.\nGraphs and Combinatorics, vol. 33, pp. 595–615, 2017.\nDiscrete Mathematics, vol. 343, pp. 111665, 2020.\nB\nEkim, T., and A. Erey, RAIRO-Operations Research, vol. 48, no. 04: EDP Sciences, pp. 497–507, 2014.\nA\nEkim, T., and V. Th Paschos, International Journal of Computer Mathematics, vol. 81, no. 5: Taylor & Francis Group, pp. 569–582, 2004.\nDemange, M., T. Ekim, D. de Werra, and , J. Graph Algorithms Appl., vol. 10, no. 2, pp. 297–315, 2006.\nDiscrete Mathematics & Theoretical Computer Science, vol. 20, 2018.\nAkdemir, A., and T. Ekim, Discrete Optimization, vol. 16, pp. 62-69, 2015."
]
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"https://ie.boun.edu.tr/sites/all/modules/biblio/misc/arrow-asc.png",
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https://uqworld.org/t/vectorized-mfile-based-model-with-an-if-statement/1199 | [
"# Vectorized mfile-based model with an if statement\n\nDear UQLab Community\n\nI have a mfile-based model with this if statement that needs to be true so I can evaluate a performance function according to different metamodel previously established.\n\nIt is my understanding that the mfile is vectorized, improving the performance of the code by several orders of magnitude, since it allows the evaluation of N different realizations in a single call.\n\nNevertheless, I notice that the if statement that I am using the mfile is never true, but it should be for some runs. When the mfile was not vectorized I could confirm this.\n\nIs this something that can be done? If so, please let me know what I am missing.\nPlease find enclosed a short example of the code in the file I am using.\nmfilebased_model.m (308 Bytes)\n\nFurthermore, I should say that I am using subset simulation as the reliability analysis method.\n\nI thank you in advance for your time and support\nBest Regards,\nNeryvaldo Galvão\n\nDear Neryvaldo Galvão,\n\nSince your code is short, I take the liberty of posting it below to make it more convenient for other users.\n\nThe fact that your IF statement never evaluates to true after you vectorized your code makes me wonder if your condition is correct. Your current condition is equivalent to:\n`all(X(:,18) <= X(:,19))`\nHowever, you probably wanted the following condition:\n`any(X(:,18) <= X(:,19))`\nCan you confirm that your condition is correct?\n\nBest regards,\nStyfen\n\n``````function Y = myHRAEventTree(X)\n\nload(Metamodel_1)\nload(Metamodel_2)\nload(Metamodel_3)\n\nif X(:,18)<=X(:,19)\nY = uq_evalModel(Metamodel_1,X(:,1:16));\nelseif X(:,18)<=X(:,19)+X(:,20)\nY = uq_evalModel(Metamodel_2,X(:,1:16));\nelse\nY = uq_evalModel(Metamodel_3,X(:,1:16));\nend\n\nend\n``````\n\nDear Styfen Schaer\n\nThank you for your support\n\nall(X(:,18) <= X(:,19)) is not what I meant to do",
null,
". Nevertheless, I not sure if any(X(:,18) <= X(:,19)) will work as well",
null,
".\n\nWhat I need is: for any realization N of a single call, if X(18)<=X(19) the Metamodel_1 should be used to get the output of interest i.e., Y for that single realization where the condition “if” is verified.\n\nThe non-vectorized code that I had before was something like the following and it was working fine, but the performance was terrible.",
null,
"``````function Y = myHRAEventTree(X)\n\nload(Metamodel_1)\nload(Metamodel_2)\nload(Metamodel_3)\n\nfor i=1:size(X,1)\nif X(i,18)<=X(i,19)\nY(i) = uq_evalModel(Metamodel_1,X(i,1:16));\nelseif X(i,18)<=X(i,19)+X(i,20)\nY(i) = uq_evalModel(Metamodel_2,X(i,1:16));\nelse\nY(i) = uq_evalModel(Metamodel_3,X(i,1:16));\nend\nend\n\nend\n``````\n\nHow can I convert this code to a vectorized mfile-based model?",
null,
"Looking forward to hearing from you.\n\nThank you in advance for your support.\nBest regards,\nNeryvaldo Galvão\n\nDear Neryvaldo Galvão;\n\nany() will not work in this case. However, I think you’re looking for something like below:\n\nBest regards,\nStyfen\n\n``````function Y = myHRAEventTree(X)\n\nload(Metamodel_1)\nload(Metamodel_2)\nload(Metamodel_3)\n\ncond1 = X(:,18) <= X(:,19);\nX1 = X(cond1,16);\n\ncond2 = X(:,18) <= X(:,19) + X(:,20);\nX2 = X(cond2,16);\n\ncond3 = ~cond1 & ~cond2;\nX3 = X(cond3,16);\n\nY1 = uq_evalModel(Metamodel_1, X1);\nY2 = uq_evalModel(Metamodel_2, X2);\nY3 = uq_evalModel(Metamodel_3, X3);\n\nY = zeros(size(X,1),1);\nY(cond1) = Y1;\nY(cond2) = Y2;\nY(cond3) = Y3;\n\nend\n``````\n1 Like\n\nDear Styfen Schaer\n\nThank you very much",
null,
". It looks like the code you suggested is working fine",
null,
".\n\nNevertheless, I need to understand a little more about it. Where can I read more about the approach you suggesting?\n\nFurthermore, lets say I need to do the following multiplication X(i,6) = X(i,17) * X(i,15) when cond1 = X(:,18) <= X(:,19) is true. How can I do this in this vectorized approach you proposing? I have tried a couple of things, but sadly it is not working.\n\nMany thanks for your support\nBest Regards,\nNeryvaldo Galvão\n\nDear Neryvaldo Galvão,\n\nI’m glad it worked.\nThe proposed approach is not directly related to UQLab but to Matlab.\nHowever, the following link shows some examples of how to use logical operators to select submatrizes:\n\nhttps://ch.mathworks.com/help/matlab/matlab_prog/find-array-elements-that-meet-a-condition.html\n\nRegarding your problem you may have missed to use the elementwise operator.\nThe following should work:\n\n``````cond1 = X(:,18) <= X(:,19);\nX(cond1,6) = X(cond1,17) .* X(cond1,15);\n``````\n\nBest regards,\nStyfen\n\n1 Like\n\nDear Styfen Schaer\n\nOnce again thank you very much. Almost everything is working fine",
null,
".\n\nBut it seems to me that these lines:\n\n``````Y(cond1) = Y1;\nY(cond2) = Y2;\nY(cond3) = Y3;\n``````\n\nare overwriting each other. At the end Y(cond3)=Y3 overwrites the values provided by Y(cond2) = Y2 and this last one overwrite the Y(cond1) = Y1. Making my efforts to separate the estimation of Y with different metamodel inconsequential",
null,
".\n\n``````Y1 = uq_evalModel(Metamodel_1, X1);\nY2 = uq_evalModel(Metamodel_2, X2);\nY3 = uq_evalModel(Metamodel_3, X3);\n``````\n\nPlease let me know what are your thoughts on this.\n\nBest Regards,\nNeryvaldo Galvão\n\nWell, I think that’s a tiny mistake on my part.",
null,
"The second condition should read:\n\n``````cond2 = X(:,18) <= X(:,19) + X(:,20) & ~cond1;\n``````\n\nYou should double check that all conditions are correct now.\n\nBest regards,\nStyfen\n\n1 Like\n\nGreat!! Looks like that everything is working fine. I still have to run a few more tests though.\n\nThank you very much for the super support",
null,
"",
null,
". It is great to know I have you on the other side.\n\nRegards,\nNeryvaldo Galvão"
]
| [
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"https://uqworld.org/images/emoji/google/frowning_face.png",
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"https://uqworld.org/images/emoji/google/thinking.png",
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"https://uqworld.org/images/emoji/google/pray.png",
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"https://uqworld.org/images/emoji/google/grinning.png",
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"https://uqworld.org/images/emoji/google/+1.png",
null,
"https://uqworld.org/images/emoji/google/grinning.png",
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"https://uqworld.org/images/emoji/google/cold_sweat.png",
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"https://uqworld.org/images/emoji/google/wink.png",
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"https://uqworld.org/images/emoji/google/muscle.png",
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"https://uqworld.org/images/emoji/google/muscle.png",
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https://answers.everydaycalculation.com/compare-fractions/4-3-and-6-5 | [
"Solutions by everydaycalculation.com\n\n## Compare 4/3 and 6/5\n\n1st number: 1 1/3, 2nd number: 1 1/5\n\n4/3 is greater than 6/5\n\n#### Steps for comparing fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 3 and 5 is 15\n\nNext, find the equivalent fraction of both fractional numbers with denominator 15\n2. For the 1st fraction, since 3 × 5 = 15,\n4/3 = 4 × 5/3 × 5 = 20/15\n3. Likewise, for the 2nd fraction, since 5 × 3 = 15,\n6/5 = 6 × 3/5 × 3 = 18/15\n4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction\n5. 20/15 > 18/15 or 4/3 > 6/5\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:"
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"https://answers.everydaycalculation.com/mathstep-app-icon.png",
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https://finstanon.com/articles/34-liquidity-of-short-term-assets-and-liquidity-ratio-analysis | [
"# Liquidity of Short-Term Assets and Liquidity Ratio Analysis\n\n## General understanding of liquidity\n\nSince the short-term debt-paying ability is a very important indicator of the enterprise stability, the liquidity ratio analysis becomes a useful method of analyzing firm’s performance. The ability to pay current obligations means there is a higher chance company can also maintain a long-term debt-paying ability and not find itself bankrupt because of not being able to meet its obligations to short-term creditors. Having systematic troubles meeting its short-term obligations means a higher risk of firm’s bankruptcy, thus calculating the liquidity ratios and analyzing the results is highly important both for company owners and for potential investors.\n\n## Liquidity ratio calculation and analysis\n\n### Current Ratio\n\nThe formula for the current ratio is as follows:\n\nCurrent Ratio = Current Assets ÷ Current Liabilities\n\nThe current ratio indicates a firm's ability to pay its current liabilities from its current assets. This is the basic indicator of the company’s liquidity. Higher numbers are better, meaning that the current assets amount of a firm is higher comparing to current liabilities and thus, company has the ability to easily pay off its short-term debt. Generally, the normal value for this ratio is 2 or more, however, the comparison with other similar companies should necessarily be made, because for some industries values below 2 are adequate.\n\n### Quick Ratio (Acid Test Ratio)\n\nThe purpose of calculating the quick ratio (also referred as acid test ratio) is to measure how well a company can meet its short-term obligations with its most liquid assets:\n\nQuick Ratio = (Cash Equivalents + Marketable Securities + Net Receivables) ÷ Current Liabilities\n\nThis formula can be used for the most conservative ratio calculation when one needs to exclude items that don’t represent current cash flow from current assets. The normal value for this ratio is 1, but as with current ratio, the comparison with similar companies should be made, because in some industries firms with quick ratio below 1 still have normal liquidity.\n\nA common alternative formula for acid test ratio is:\n\nQuick Ratio = (Cash + Marketable Securities + Accounts and Notes Receivable) ÷ Current Liabilities\n\nThis formula provides you an indication of the business liquidity by comparing the amount of cash, marketable securities and accounts and notes receivable to current liabilities. Should also be mentioned, that the quick ratio does not include inventory and prepaid expenses in the calculation, which is the main difference between the current ratio and the quick ratio.\n\nFinally, another formula for calculating quick ratio exists, which is more general:\n\nQuick Ratio = (Current assets - Inventory) ÷ Current liabilities\n\nThe quick ratio allows to focus on quick assets (those that could be quickly converted to cash), that’s why it keeps inventories out of equation. Once again, the normal value for this ratio is 1 or more, meaning that for every dollar of company’s current liabilities the firm has at least 1 dollar of very liquid assets to cover those immediately, if needed.\n\n### Cash Ratio\n\nA way of company’s cash and equivalents amount estimation in terms of liquidity is the calculation of the cash ratio. That can be done using the following formula:\n\nCash Ratio = (Cash Equivalents + Marketable Securities) ÷ Current Liabilities\n\nThe cash ratio is the most conservative indicator of firm’s liquidity, indicating its immediate liquidity. Having calculated the cash ratio one can see how well a company can pay off its current liabilities with only cash and cash equivalents. However, it is not realistic to expect the company to have enough cash and equivalents to cover all the current liabilities, because if this occurs, it means that the company’s usage of cash is not efficient as cash should rather be put to work in the operations of the firm. Considering this, the detailed knowledge of the business is required to have a chance to draw a conclusion based on the cash ratio calculation. Most commonly, big values of cash ratio would mean inappropriate usage of cash by the company, while cash ratio lesser than 0.2 would mean that the firm might face an immediate problem with paying bills.\n\n### Working Capital\n\nCompany’s working capital indicates its short-run solvency and financial sustainability. The calculation formula for the working capital as follows:\n\nWorking Capital = Current Assets - Current Liabilities\n\nThe working capital amount is a value that should be compared with past periods of time within the same company to determine its reasonability, while comparing the working capital amount of different companies is pointless due to the different sizes of firms. Current assets are assets that are expected to be turned into cash by a company within one year, or one business cycle. Analogically, current liabilities are liabilities that are expected to be paid by a firm within a year, or one business cycle. Financially sustainable business would be able to pay its current liabilities with its current assets.\n\n### Sales to Working Capital (Working Capital Turnover)\n\nA part of liquidity ratio analysis is the calculation of sales to working capital (also referred as working capital turnover). The formula for doing that is as follows:\n\nSales to Working Capital = Sales ÷ Average Working Capital\n\nThe sales to working capital ratio indicates how much cash is needed to generate a certain level of sales. In other words, it measures dollars of sales generated by a dollar of working capital investment. That’s why a low working capital turnover ratio most likely indicates an unprofitable use of working capital. In other words, sales are not adequate in relation to the available working capital. As with many other ratios, before drawing a conclusion based on sales to working capital ratio, one should make a comparison with other similar companies, industry averages, to compare the dynamics of this ratio comparing to past periods of time. Seeing the increase of sales to working capital in dynamics over some period would witness the overall increase of liquidity of the firm.\n\n## Tools and Software to Analyze Liqudity\n\nYou can use Financial Statement Analysis Desctop App to calculate ratios and generate conclusion.\n\nExcel for GAAP US\n\nExcel for IFRS\n\n## Summary\n\nThe liquidity of short-term assets and the short-term debt-paying ability of the company can be measured by the liquidity ratio analysis, including calculation of the following indicators:\n\nCurrent Ratio = Current Assets ÷ Current Liabilities\n\nQuick ratio (Acid Test Ratio) = (Cash Equivalents + Marketable Securities + Net Receivables) ÷ (Current Liabilities)\n\nQuick Ratio = (Cash + Marketable Securities + Accounts and Notes Receivable) ÷ Current Liabilities\n\nQuick Ratio = (Current assets - Inventory) ÷ Current liabilities\n\nCash Ratio = (Cash Equivalents + Marketable Securities) ÷ Current Liabilities\n\nWorking Capital = Current Assets - Current Liabilities\n\nSales to Working Capital (Working Capital Turnover) = Sales ÷ Average Working Capital"
]
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https://pointclouds.org/documentation/classpcl_1_1_point_representation.html | [
"Point Cloud Library (PCL) 1.13.1-dev\npcl::PointRepresentation< PointT > Class Template Referenceabstract\n\nPointRepresentation provides a set of methods for converting a point structs/object into an n-dimensional vector. More...\n\n`#include <pcl/point_representation.h>`",
null,
"Inheritance diagram for pcl::PointRepresentation< PointT >:\n\n## Public Types\n\nusing Ptr = shared_ptr< PointRepresentation< PointT > >\n\nusing ConstPtr = shared_ptr< const PointRepresentation< PointT > >\n\n## Public Member Functions\n\nvirtual ~PointRepresentation ()=default\nEmpty destructor. More...\n\nvirtual void copyToFloatArray (const PointT &p, float *out) const =0\nCopy point data from input point to a float array. More...\n\nbool isTrivial () const\nReturns whether this point representation is trivial. More...\n\nvirtual bool isValid (const PointT &p) const\nVerify that the input point is valid. More...\n\ntemplate<typename OutputType >\nvoid vectorize (const PointT &p, OutputType &out) const\nConvert input point into a vector representation, rescaling by alpha. More...\n\nvoid setRescaleValues (const float *rescale_array)\nSet the rescale values to use when vectorizing points. More...\n\nint getNumberOfDimensions () const\nReturn the number of dimensions in the point's vector representation. More...\n\n## Protected Attributes\n\nint nr_dimensions_ = 0\nThe number of dimensions in this point's vector (i.e. More...\n\nstd::vector< float > alpha_\nA vector containing the rescale factor to apply to each dimension. More...\n\nbool trivial_ = false\nIndicates whether this point representation is trivial. More...\n\n## Detailed Description\n\n### template<typename PointT> class pcl::PointRepresentation< PointT >\n\nPointRepresentation provides a set of methods for converting a point structs/object into an n-dimensional vector.\n\nNote\nThis is an abstract class. Subclasses must set nr_dimensions_ to the appropriate value in the constructor and provide an implementation of the pure virtual copyToFloatArray method.\n\nDefinition at line 59 of file point_representation.h.\n\n## ◆ ConstPtr\n\ntemplate<typename PointT >\n using pcl::PointRepresentation< PointT >::ConstPtr = shared_ptr >\n\nDefinition at line 79 of file point_representation.h.\n\n## ◆ Ptr\n\ntemplate<typename PointT >\n using pcl::PointRepresentation< PointT >::Ptr = shared_ptr >\n\nDefinition at line 78 of file point_representation.h.\n\n## ◆ ~PointRepresentation()\n\ntemplate<typename PointT >\n virtual pcl::PointRepresentation< PointT >::~PointRepresentation ( )\nvirtualdefault\n\nEmpty destructor.\n\n## ◆ copyToFloatArray()\n\ntemplate<typename PointT >\n virtual void pcl::PointRepresentation< PointT >::copyToFloatArray ( const PointT & p, float * out ) const\npure virtual\n\n## ◆ getNumberOfDimensions()\n\ntemplate<typename PointT >\n int pcl::PointRepresentation< PointT >::getNumberOfDimensions ( ) const\ninline\n\nReturn the number of dimensions in the point's vector representation.\n\nDefinition at line 172 of file point_representation.h.\n\n## ◆ isTrivial()\n\ntemplate<typename PointT >\n bool pcl::PointRepresentation< PointT >::isTrivial ( ) const\ninline\n\nReturns whether this point representation is trivial.\n\nIt is trivial if and only if the following conditions hold:\n\n• the relevant data consists only of float values\n• the vectorize operation directly copies the first nr_dimensions_ elements of PointT to the out array\n• sizeof(PointT) is a multiple of sizeof(float) In short, a trivial point representation converts the input point to a float array that is the same as if the point was reinterpret_casted to a float array of length nr_dimensions_ .\n\nDefinition at line 98 of file point_representation.h.\n\n## ◆ isValid()\n\ntemplate<typename PointT >\n virtual bool pcl::PointRepresentation< PointT >::isValid ( const PointT & p ) const\ninlinevirtual\n\nVerify that the input point is valid.\n\nParameters\n p The point to validate\n\nDefinition at line 104 of file point_representation.h.\n\n## ◆ setRescaleValues()\n\ntemplate<typename PointT >\n void pcl::PointRepresentation< PointT >::setRescaleValues ( const float * rescale_array )\ninline\n\nSet the rescale values to use when vectorizing points.\n\nParameters\n [in] rescale_array The array/vector of rescale values. Can be of any type that implements the [] operator.\n\nDefinition at line 165 of file point_representation.h.\n\n## ◆ vectorize()\n\ntemplate<typename PointT >\ntemplate<typename OutputType >\n void pcl::PointRepresentation< PointT >::vectorize ( const PointT & p, OutputType & out ) const\ninline\n\nConvert input point into a vector representation, rescaling by alpha.\n\nParameters\n [in] p the input point [out] out The output vector. Can be of any type that implements the [] operator.\n\nDefinition at line 144 of file point_representation.h.\n\n## ◆ alpha_\n\ntemplate<typename PointT >\n std::vector pcl::PointRepresentation< PointT >::alpha_\nprotected\n\nA vector containing the rescale factor to apply to each dimension.\n\nDefinition at line 65 of file point_representation.h.\n\n## ◆ nr_dimensions_\n\ntemplate<typename PointT >\n int pcl::PointRepresentation< PointT >::nr_dimensions_ = 0\nprotected\n\nThe number of dimensions in this point's vector (i.e.\n\nthe \"k\" in \"k-D\")\n\nDefinition at line 63 of file point_representation.h.\n\n## ◆ trivial_\n\ntemplate<typename PointT >\n bool pcl::PointRepresentation< PointT >::trivial_ = false\nprotected\n\nIndicates whether this point representation is trivial.\n\nIt is trivial if and only if the following conditions hold:\n\n• the relevant data consists only of float values\n• the vectorize operation directly copies the first nr_dimensions_ elements of PointT to the out array\n• sizeof(PointT) is a multiple of sizeof(float) In short, a trivial point representation converts the input point to a float array that is the same as if the point was reinterpret_casted to a float array of length nr_dimensions_ . This value says that this representation can be trivial; it is only trivial if setRescaleValues() has not been set.\n\nDefinition at line 75 of file point_representation.h.\n\nThe documentation for this class was generated from the following file:"
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https://export.arxiv.org/abs/2101.09444?context=math.OA | [
"math.OA\n\n# Title: On the anti-commutator of two free random variables\n\nAbstract: Let $(\\kappa_n(a))_{n\\geq 1}$ denote the sequence of free cumulants of a random variable $a$ in a non-commutative probability space $(\\mathcal{A},\\varphi)$. Based on some considerations on bipartite graphs, we provide a formula to compute the cumulants $(\\kappa_n(ab+ba))_{n\\geq 1}$ in terms of $(\\kappa_n(a))_{n\\geq 1}$ and $(\\kappa_n(b))_{n\\geq 1}$, where $a$ and $b$ are freely independent. Our formula expresses the $n$-th free cumulant of $ab+ba$ as a sum indexed by partitions in the set $\\mathcal{Y}_{2n}$ of non-crossing partitions of the form\n$\\sigma=\\{B_1,B_3,\\dots, B_{2n-1},E_1,\\dots,E_r\\}, \\quad \\text{with }r\\geq 0,$\nsuch that $i\\in B_{i}$ for $i=1,3,\\dots,2n-1$ and $|E_j|$ even for $j\\leq r$. Therefore, by studying the sets $\\mathcal{Y}_{2n}$ we obtain new results regarding the distribution of $ab+ba$. For instance, the size $|\\mathcal{Y}_{2n}|$ is closely related to the case when $a,b$ are free Poisson random variables of parameter 1. Our formula can also be expressed in terms of cacti graphs. This graph theoretic approach suggests a natural generalization that allows us to study quadratic forms in $k$ free random variables.\n Subjects: Operator Algebras (math.OA); Combinatorics (math.CO); Probability (math.PR) MSC classes: 46L54, 05A18 Cite as: arXiv:2101.09444 [math.OA] (or arXiv:2101.09444v1 [math.OA] for this version)\n\n## Submission history\n\nFrom: Daniel Perales Anaya [view email]\n[v1] Sat, 23 Jan 2021 07:26:22 GMT (38kb,D)\n\nLink back to: arXiv, form interface, contact."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7199675,"math_prob":0.99938303,"size":1557,"snap":"2021-21-2021-25","text_gpt3_token_len":487,"char_repetition_ratio":0.098519,"word_repetition_ratio":0.0,"special_character_ratio":0.29736674,"punctuation_ratio":0.12852664,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998429,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-18T13:56:41Z\",\"WARC-Record-ID\":\"<urn:uuid:48461b3a-b736-49b3-a7f2-1a232c9a16ab>\",\"Content-Length\":\"16284\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:12e03352-0c17-4b03-9b6e-180bd4ae41f4>\",\"WARC-Concurrent-To\":\"<urn:uuid:ddfd6b59-cf63-487d-8398-2566cd60c0ef>\",\"WARC-IP-Address\":\"128.84.21.203\",\"WARC-Target-URI\":\"https://export.arxiv.org/abs/2101.09444?context=math.OA\",\"WARC-Payload-Digest\":\"sha1:AMYRL5UCWXPDRXQA4MTHQDS43CB6PETC\",\"WARC-Block-Digest\":\"sha1:2F3XWTBFCXRI3GKDFRKDIITILXJS3WHY\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243989637.86_warc_CC-MAIN-20210518125638-20210518155638-00532.warc.gz\"}"} |
https://link.springer.com/article/10.2478/s11533-013-0216-x | [
"Central European Journal of Mathematics\n\n, Volume 11, Issue 5, pp 972–983\n\nApproximate multiplication in adaptive wavelet methods\n\nResearch Article\n\nAbstract\n\nCohen, Dahmen and DeVore designed in [Adaptive wavelet methods for elliptic operator equations: convergence rates, Math. Comp., 2001, 70(233), 27–75] and [Adaptive wavelet methods II¶beyond the elliptic case, Found. Comput. Math., 2002, 2(3), 203–245] a general concept for solving operator equations. Its essential steps are: transformation of the variational formulation into the well-conditioned infinite-dimensional l 2-problem, finding the convergent iteration process for the l 2-problem and finally using its finite dimensional approximation which works with an inexact right-hand side and approximate matrix-vector multiplication. In our contribution, we pay attention to approximate matrix-vector multiplication which is enabled by an off-diagonal decay of entries of the wavelet stiffness matrices. We propose a more efficient technique which better utilizes actual decay of matrix and vector entries and we also prove that this multiplication algorithm is asymptotically optimal in the sense that storage and number of floating point operations, needed to resolve the problem with desired accuracy, remain proportional to the problem size when the resolution of the discretization is refined.\n\nMSC\n\n65T60 65F99 65N99\n\nReferences\n\n1. \nČerná D., Finěk V., Construction of optimally conditioned cubic spline wavelets on the interval, Adv. Comput. Math., 2011, 34(2), 219–25\n2. \nCohen A., Dahmen W., DeVore R., Adaptive wavelet methods for elliptic operator equations: convergence rates, Math. Comp., 2001, 70(233), 27–75\n3. \nCohen A., Dahmen W., DeVore R., Adaptive wavelet methods II¶beyond the elliptic case, Found. Comput. Math., 2002, 2(3), 203–245\n4. \nDahmen W., Wavelet and multiscale methods for operator equations, In: Acta Numer., 6, Cambridge University Press, Cambridge, 1997, 55–228Google Scholar\n5. \nDeVore R.A., Nonlinear approximation, In: Acta Numer., 7, Cambridge University Press, Cambridge, 1998, 51–150Google Scholar\n6. \nDijkema T.J., Schwab Ch., Stevenson R., An adaptive wavelet method for solving high-dimensional elliptic PDEs, Constr. Approx., 2009, 30(3), 423–455\n7. \nStevenson R., Adaptive solution of operator equations using wavelet frames, SIAM J. Numer. Anal., 2003, 41(3), 1074–1100\n8. \nStevenson R., On the compressibility operators in wavelet coordinates, SIAM J. Math. Anal., 2004, 35(5), 1110–1132",
null,
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]
| [
null,
"https://link.springer.com/track/controlled/article/denied/10.2478/s11533-013-0216-x",
null
]
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https://www.arxiv-vanity.com/papers/0808.1316/ | [
"# Growth factor parametrization and modified gravity\n\nYungui Gong College of Mathematics and Physics, Chongqing University of Posts and Telecommunications, Chongqing 400065, China\n###### Abstract\n\nThe growth rate of matter perturbation and the expansion rate of the Universe can be used to distinguish modified gravity and dark energy models in explaining the cosmic acceleration. The growth rate is parametrized by the growth index . We discuss the dependence of on the matter energy density and its current value for a more accurate approximation of the growth factor. The observational data, including the data of the growth rate, are used to fit different models. The data strongly disfavor the Dvali-Gabadadze-Porrati model. For the dark energy model with a constant equation of state, we find that and . For the CDM model, we find that . For the Dvali-Gabadadze-Porrati model, we find that .\n\n###### pacs:\n95.36.+x; 98.80.Es; 04.50.-h\npreprint: arXiv:0808.1316\n\n## I Introduction\n\nThe discovery of late time cosmic acceleration acc1 challenges our understanding of the standard models of gravity and particle physics. Within the framework of Friedmann-Robertson-Walker cosmology, an exotic energy ingredient with negative pressure, dubbed dark energy, is invoked to explain the observed accelerated expansion of the Universe. One simple candidate of dark energy which is consistent with current observations is the cosmological constant. Because of the many orders of magnitude discrepancy between the theoretical predication and the observation of vacuum energy, other dynamical dark energy models were proposed rev . By choosing a suitable equation of state for dark energy, we can recover the observed expansion rate and the luminosity distance redshift relation . Current observations are unable to distinguish many different dark energy models which give the same , and the nature of dark energy is still a mystery. Many parametric and nonparametric model-independent methods were proposed to study the property of dark energy astier01 ; huterer ; par2 ; par3 ; lind ; alam ; jbp ; par4 ; par1 ; par5 ; gong04 ; gong05 ; gong06 ; gong08 ; sahni .\n\nBear in mind that the only observable effect of dark energy is through gravitational interaction; it is also possible that the accelerated expansion is caused by modification of gravitation. One example of the alternative approach is provided by the Dvali-Gabadadze-Porrati (DGP) brane-world model dgp , in which gravity appears four dimensional at short distances but is modified at large distances. The question one faces is how to distinguish such an alternative approach from the one involving dark energy. One may answer the question by seeking a more accurate observation of the cosmic expansion history, but this will not break the degeneracies between different approaches of explaining the cosmic acceleration. Recently, it was proposed to use the growth rate of large scale in the Universe to distinguish the effect of modified gravity from that of dark energy. While different models give the same late time accelerated expansion, the growth of matter perturbation they produce differ jetp . To linear order of perturbation, at large scales the matter density perturbation satisfies the following equation:\n\n ¨δ+2H˙δ−4πGeffρδ=0, (1)\n\nwhere is the matter energy density and denotes the effect of modified gravity. For example, for the Brans-Dicke theory boiss and for the DGP model lue , the dimensionless matter energy density . In terms of the growth factor , the matter density perturbation Eq. (1) becomes\n\n f′+f2+(˙HH2+2)f=32GeffGΩ, (2)\n\nwhere . In general, there is no analytical solution to Eq. (2), and we need to solve Eq. (2) numerically; it is very interesting that the solution of the equation can be approximated as peebles ; fry ; lightman ; wang and the growth index can be obtained for some general models. The approximation was first proposed by Peebles for the matter dominated universe as peebles ; then a more accurate approximation, , for the same model was derived in fry ; lightman . For the Universe with a cosmological constant, the approximation can be made lahav . For a dynamical dark energy model with slowly varying and zero curvature, the approximation was given in wang . For the DGP model, linder07 . Therefore, instead of looking for the growth factor by numerically solving Eq. (2), the growth index may be used as the signature of modified gravity and dark energy models. It was found that with and with for the dynamical dark energy model in flat space linder05 ; linder07 . Recently, the use of the growth rate of matter perturbation in addition to the expansion history of the Universe to differentiate dark energy models and modified gravity attracted much attention huterer07 ; sereno ; knox ; ishak ; yun ; polarski ; sapone ; balles ; gannouji ; berts ; laszlo ; kunz ; kiakotou ; porto ; wei ; ness .\n\nThe dependence of on the equation of state has received much attention in the literature; we discuss the dependence of on and for a more accurate approximation in this paper. We discuss a more accurate approximation for the dark energy model with constant in Sec. II. Then we discuss the DGP model in Sec. III. In Sec. IV, we apply the Union compilation of type Ia supernovae (SNe) data union , the baryon acoustic oscillation (BAO) measurement from the Sloan Digital Sky Survey (SDSS) sdss6 , the shift parameter measured from the Wilkinson Microwave Anisotropy Probe 5 yr data (WMAP5) wmap5 , the Hubble parameter data hz1 ; hz2 , and the growth factor data porto ; ness ; guzzo to constrain the models. We also use the growth factor data alone to find out the constraint on the growth index . We conclude the paper in Sec. V.\n\n## Ii ΛCDM model\n\nWe first review the derivation of given in wang . For the flat dark energy model with constant equation of state , we have\n\n ˙HH2=−32[1+w(1−Ω)]. (3)\n\nThe energy conservation equation tells us that\n\n Ω′=3wΩ(1−Ω). (4)\n\nSubstituting Eqs. (3) and (4) into Eq. (2), we get\n\n 3wΩ(1−Ω)dfdΩ+f2+[12−32w(1−Ω)]f=32Ω. (5)\n\nPlugging into Eq. (5), we get\n\n 3wΩ(1−Ω)lnΩdγdΩ−3wΩ(γ−1/2)+Ωγ−32Ω1−γ+3wγ−32w+12=0. (6)\n\nExpanding Eq. (6) around , to the first order of , we get wang\n\n γ=3(1−w)5−6w+3125(1−w)(1−3w/2)(1−6w/5)2(1−Ω). (7)\n\nIn order to see how well the approximation fits the growth factor , we need to solve Eq. (6) numerically with the expression of . The dimensionless matter density is\n\n Ω=Ω0Ω0+(1−Ω0)(1+z)3w. (8)\n\nSince does not change very much, we first use to approximate the growth factor. For convenience, we choose , and the result is shown in Fig. 1.",
null,
"Figure 1: The relative difference between the growth factor f and Ωγ∞ with γ∞=3(1−w)(5−6w) for the dark energy model with constant w.\n\nFrom Fig. 1, we see that approximates better than 2%.\n\nFor the CDM model, , so the growth index becomes\n\n γ∞=3(1−w)5−6w=611, (9)\n\nand\n\n γ1=−dγdΩ=3125(1−w)(1−3w/2)(1−6w/5)3=151331. (10)\n\nFor the CDM model, the matter density is\n\n Ω=Ω0(1+z)3Ω0(1+z)3+1−Ω0. (11)\n\nSubstituting Eq. (11) into Eq. (6) and solving the equation numerically, we compare the numerical result with the analytical approximation . The results are shown in Figs. 2 and 3. Since is very small, does not change much, and we can take the approximation . In Fig. 2, we compare with for different values of . From Fig. 2, we see that approximates very well; the smaller , the larger the error. When the Universe deviates farther from the matter dominated era, the error becomes larger. For , the approximation overestimates by only 2%, or underestimates . To get a better approximation, we expand to the first order of and use . In Fig. 3, we plot the relative difference between and . From Fig. 3, we see that using approximates the growth factor much better; now the error is only 0.6% for .",
null,
"Figure 2: The relative difference between the growth factor f and Ωγ∞ with γ∞=6/11.",
null,
"Figure 3: The relative difference between the growth factor f and Ωγ with γ=γ∞+γ1(1−Ω) for the ΛCDM model.\n\n## Iii DGP model\n\nFor the flat DGP model, we have\n\n GeffG=2(1+2Ω2)3(1+Ω2). (12)\n\nThe Friedmann equation tells us that\n\n ˙HH2=−3Ω1+Ω. (13)\n\nThe energy conservation equation tells us that\n\n Ω′=−3Ω(1−Ω)1+Ω. (14)\n\nThe matter energy density is given by\n\n Ω=Ω0(1+z)3[(1−Ω0)/2+√Ω0(1+z)3+(1−Ω0)2/4]2. (15)\n\nSubstituting Eqs. (12), (13) and (14) into Eq. (2), we get\n\n −3Ω(1−Ω)1+ΩdfdΩ+f2+2−Ω1+Ωf=Ω(1+2Ω2)1+Ω2. (16)\n\nPlugging into Eq. (16), we get\n\n −3Ω(1−Ω)lnΩ1+ΩdγdΩ−3(1−Ω)γ1+Ω+Ωγ+2−Ω1+Ω−Ω1−γ(1+2Ω2)1+Ω2=0. (17)\n\nExpanding Eq. (17) around , to the first order of , we get\n\n γ=1116+75632(1−Ω). (18)\n\nSo and . The change of is very small because is very small; we first approximate by and the result is shown in Fig. 4. From Fig 4, we see that the error becomes larger when the Universe deviates farther from the matter domination. When , underestimates by 4.6%. So the growth factor is overestimated. If we use the first order approximation, the error becomes larger because . Linder and Cahn give the approximation linder07\n\n γ≈7+5Ω+7Ω2+3Ω3(1+Ω2)(11+5Ω)≈1116+7256(1−Ω). (19)\n\nAgain, to the first order approximation of , the error becomes larger than that with . In wei , the author found the approximation\n\n γ=1116−316(1−Ω)+116(1−Ω)2. (20)\n\nIn this approximation, the correction to is too big because is too large even though the sign is correct. To get better than 1% approximation, we first assume is a constant and solve Eq. (17) at to get ; we then approximate . The values of and for different are listed in table 1. The difference between with and is shown in Fig. 5. As promised, the error is under 1%.",
null,
"Figure 4: The relative difference between the growth factor f and Ωγ∞ with γ∞=11/16 for the DGP model.",
null,
"Figure 5: The relative difference between the growth factor f and Ωγ with γ=γ∞+γ1(1−Ω) for the DGP model.\n\n## Iv Observational Constraints\n\nNow we use the observational data to fit the dark energy model with constant and the DGP model. The parameters in the models are determined by minimizing . For the SNe data, we use the reduced Union compilation of 307 SNe union . The SNe compilation includes the Supernova Legacy Survey astier and the ESSENCE Survey riess ; essence , the older observed SNe data, and the extended dataset of distant SNe observed with the Hubble space telescope. To fit the SNe data, we define\n\n χ2sn=307∑i=1[μobs(zi)−μ(zi)]2σ2i, (21)\n\nwhere the extinction-corrected distance modulus , is the total uncertainty in the SNe data, and the luminosity distance is\n\n dL(z)=H−10(1+z)∫z0dz′E(z′). (22)\n\nThe dimensionless Hubble parameter for the dark energy model with constant and for the DGP model. The nuisance parameter is marginalized over using a flat prior.\n\nTo use the BAO measurement from the SDSS data, we define , where the distance parameter sdss6 ; wmap5\n\n A=√Ω00.35⎡⎣0.35E(0.35)(∫0.350dzE(z))2⎤⎦1/3=0.469(0.96/0.98)−0.35±0.017. (23)\n\nTo use the shift parameter measured from the WMAP5 data, we define , where the shift parameter wmap5\n\n R=√Ω0∫10890dzE(z)=1.715±0.021. (24)\n\nSimon, Verde, and Jimenez obtained the Hubble parameter at nine different redshifts from the differential ages of passively evolving galaxies hz1 . Recently, the authors in hz2 obtained and by taking the BAO scale as a standard ruler in the radial direction. To use these 11 data, we define\n\n χ2h=11∑i=1[Hobs(zi)−H(zi)]2σ2hi, (25)\n\nwhere is the uncertainty in the data. We also add the prior km/s/Mpc given by Freedman et al. freedman .\n\nFor the growth factor data, we define\n\n χ2f=12∑i=1[fobs(zi)−Ωγ(zi)]2σ2fi, (26)\n\nwhere is the uncertainty in the data. For reference, we compile the available data porto ; ness ; guzzo in Table 2. The data are obtained from the measurement of the redshift distortion parameter , where the bias factor measures how closely galaxies trace the mass density field. Note that some of the measured redshift distortion parameter is obtained by fitting with given by the CDM model, and some analyses tried to account for extra distortions due to the geometric Alcock-Paczynski effect guzzo . With these caveats in mind, it is still worthwhile to apply the data to fit the models. As discussed in the previous sections, we can use within the accuracy of a few percent. For the CDM model, we use and given by Eq. (11). For the DGP model, we use and given by Eq. (15).\n\nBy fitting the dark energy model with constant to the combined data, we get , , and . The , , and contours of and are shown in Fig. 6. From Fig. 6, we see that the CDM model is consistent with the current observation. By fitting the CDM model to the combined data, we get and . By fitting the DGP model to the combined data, we get and .",
null,
"Figure 6: The 1σ, 2σ, and 3σ contours of Ω and w by fitting the dark energy model with constant w to the combined data. The point with + denotes the best fit value.\n\nIf we fit to the growth factor data alone, we can get a constraint on the growth index . For the CDM model with the best fit value , we find that and . The theoretical value is consistent with the observation at the level. For the DGP model with the best fit value , we find and which is barely consistent with the theoretical value at the level. The best fit curves and the observational data are shown in Fig. 7.",
null,
"Figure 7: The growth factor data and the best fit curves. The solid line is for the dark energy model with Ω0=0.272 and w=−0.97. The dashed line is for the ΛCDM model with Ω0=0.273 and γ=0.64. The dotted line is for the DGP model with Ω0=0.278 and γ=0.55.\n\n## V Discussions\n\nThe simple analytical formulas can be used to approximate the growth rate . The value of provides useful information about the dark energy model and the modification of gravity. For the dark energy model within general relativity, . For the DGP model, . If the accuracy of the growth factor data is in the range of a few percent, we can use a constant to approximate . For example, for constant wang , for dynamical models linder05 , and for the DGP model. The value of is obtained by approximating the differential equation of around . This approximation is reasonably good at high redshift () when . However, at lower redshift the dark energy or the effect of modified gravity starts to dominate and deviates from 1; we expect the approximation to break down. Therefore, to get a better than 1% fit, the dependence of needs to be considered. We show that can approximate to better than 1%. The value of is very small compared with and usually depends on the value of . For the CDM model, . For the DGP model, we give a prescription of how to find ; the values of are listed in Table 1 for some values of .\n\nTo distinguish different dark energy models and modified gravity, we use the observational data to fit the models. Fitting the combined SNe, SDSS, WMAP5, , and data to the dark energy model with constant , we find that , , and . For the CDM model, we find that and . For the DGP model, we find that and . The results suggest that the data strongly disfavor the DGP model. If we fit the SNe data alone to the DGP model, we get and . This result is consistent with the results in song ; lue1 ; song05 ; koyama ; roy06 ; gong ; zhu . If the same data were fitted to the CDM model, we would get . If we fit the CDM model to the combined SNe and data, we get . For the DGP model, we get and . These results show that both the CDM model and the DGP model give almost the same expansion rate. If we fit the CDM model to the combined SNe, , and data, we get . For the DGP model, we get and . With the addition of data to the expansion data, the DGP model is readily distinguishable from the CDM model. As mentioned above, the best fit value of in the DGP model tends to be lower, i.e., , when we fit the model to the SNe, , and data. On the other hand, we get and if we fit the model to the distance parameter and the shift parameter . If the two sets of data are combined together, takes the value in the middle and we get a large value of . When we fit the models to the combined SNe, the distance parameter , the shift parameter , and data, we get and for the CDM model and and for the DGP model. This result is consistent with the analysis by Song et al song . In song , they found that the flat DGP model is excluded at about by the combined SNe, 3 yr WMAP, and Hubble constant data.\n\nThe observational data of can be used to provide information on the growth index and the modified gravity. As discussed above, is almost a constant; we can use with a constant to approximate . For the CDM model, we find that , which is consistent with the theoretical value 0.55. This result is also consistent with that in ness ; porto . For the DGP model, we find that . The theoretical value lies on the upper limit of the error. From Fig. 7, we see that the growth rates for the DGP model and the CDM model are distinguishable even with the best fitting values of . If the theoretical values of are used, the difference will be larger. The approximation of to is good for theories with and . If the observational data for are larger than 1 at high redshift (), then the approximation is broken and the effect of modified gravity is explicit. For the Brans-Dicke theory, during the matter domination, and porto ; here is the Brans-Dicke constant. In this case, modification of the approximation is needed. This will be discussed in future work. In conclusion, more precise future data on along with the SNe data will differentiate dark energy models from modified gravity.\n\n###### Acknowledgements.\nThis research was supported in part by the Project of Knowledge Innovation Program (PKIP) of the Chinese Academy of Sciences and by NNSFC under Grant No. 10605042. The author thanks the hospitality of Kavli Institute for Theoretical Physics China and the Abdus Salam International Center for Theoretical Physics where part of the work was done."
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https://www.systutorials.com/docs/linux/man/3-ql-y0/ | [
"## NAME\n\nQuantLib::G2Process - G2 stochastic process\n\n## SYNOPSIS\n\n#include <ql/processes/g2process.hpp>\n\nInherits QuantLib::StochasticProcess.\n\n### Public Member Functions\n\nG2Process (Real a, Real sigma, Real b, Real eta, Real rho)\n\nReal x0 () const\n\nReal y0 () const\n\nReal a () const\n\nReal sigma () const\n\nReal b () const\n\nReal eta () const\n\nReal rho () const\n\nStochasticProcess interface\n\nSize size () const\nreturns the number of dimensions of the stochastic process\nDisposable< Array > initialValues () const\nreturns the initial values of the state variables\nDisposable< Array > drift (Time t, const Array &x) const\nreturns the drift part of the equation, i.e., \\$ mu(t, mathrm{x}_t) \\$\nDisposable< Matrix > diffusion (Time t, const Array &x) const\nreturns the diffusion part of the equation, i.e. \\$ igma(t, mathrm{x}_t) \\$\nDisposable< Array > expectation (Time t0, const Array &x0, Time dt) const\n\nDisposable< Matrix > stdDeviation (Time t0, const Array &x0, Time dt) const\n\nDisposable< Matrix > covariance (Time t0, const Array &x0, Time dt) const\n\n## Detailed Description\n\nG2 stochastic process\n\n## Member Function Documentation\n\n### Disposable<Array> expectation (Time t0, const Array & x0, Time dt) const [virtual]\n\nreturns the expectation \\$ E(mathrm{x}_{t_0 + Delta t} | mathrm{x}_{t_0} = mathrm{x}_0) \\$ of the process after a time interval \\$ Delta t \\$ according to the given discretization. This method can be overridden in derived classes which want to hard-code a particular discretization.\n\nReimplemented from StochasticProcess.\n\n### Disposable<Matrix> stdDeviation (Time t0, const Array & x0, Time dt) const [virtual]\n\nreturns the standard deviation \\$ S(mathrm{x}_{t_0 + Delta t} | mathrm{x}_{t_0} = mathrm{x}_0) \\$ of the process after a time interval \\$ Delta t \\$ according to the given discretization. This method can be overridden in derived classes which want to hard-code a particular discretization.\n\nReimplemented from StochasticProcess.\n\n### Disposable<Matrix> covariance (Time t0, const Array & x0, Time dt) const [virtual]\n\nreturns the covariance \\$ V(mathrm{x}_{t_0 + Delta t} | mathrm{x}_{t_0} = mathrm{x}_0) \\$ of the process after a time interval \\$ Delta t \\$ according to the given discretization. This method can be overridden in derived classes which want to hard-code a particular discretization.\n\nReimplemented from StochasticProcess.\n\n## Author\n\nGenerated automatically by Doxygen for QuantLib from the source code."
]
| [
null
]
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https://scialert.net/fulltext/?doi=jas.2012.440.449&org=11#ref | [
"Research Article\n\n# An Analytical Closed Form Solution Around Underground Openings Using Different Methods",
null,
"O. Saeidi, A. Elyasi, S. Maleki and A. Fegh\n\nABSTRACT\n\nThis study presents a closed form solution around underground openings based on the four rock failure criteria. The results of the closed form solution are compared to the numerical method to establish the solution. The Tresca, Mohr-Coulomb, Mogi-Coulomb and generalized Hoek-Brown failure criteria are used to determine stress distribution and plastic zone around the circular space. The solutions are implemented in three dimensional distinct element code (3DEC) and assessed numerically. Also, the parametric study using these criteria is presented. Results show that the Tresca failure criterion does not fit in rock mass problems. Because of the employing axial stress, the Mogi-Coulomb failure criterion fitted the numerical results, appropriately. It was observed that increasing the axial stress using Mogi-Coulomb failure criterion decreases the stress distribution and the plastic zone, significantly.\n\n Services Related Articles in ASCI Search in Google Scholar View Citation Report Citation Science Alert\n\n How to cite this article: O. Saeidi, A. Elyasi, S. Maleki and A. Fegh, 2012. An Analytical Closed Form Solution Around Underground Openings Using Different Methods. Journal of Applied Sciences, 12: 440-449. DOI: 10.3923/jas.2012.440.449 URL: https://scialert.net/abstract/?doi=jas.2012.440.449\n\nReceived: January 24, 2012; Accepted: February 20, 2012; Published: April 16, 2012\n\nINTRODUCTION\n\nDuring excavation in rock mass far-filed in situ stresses are redistributed and cause a plastic deformation around the openings such as tunnels, shafts, wellbores etc. Analytical solution in this case involves elastoplasticity which have been used in previous texts (Carranza-Torres and Fairhurst, 1999; Chen et al., 1999; Hoek and Brown, 1980; Li and Michel, 2009; Detournay, 1986; Carranza-Torres, 1998). Alani (2002) and Alani and Nasser (2001) developed a new cubic macro-element and quadratic analyses for plate with a hole under bending and compared with closed form solution, respectively. Exact determination of stress field in elastoplastic solution requires appropriate rock failure criterion. The medium in elasto-plastic solution for circular space presumed homogenous, compressive and isotropic far-filed stress that subjected to internal pressure, Pi that applied in plane strain condition (Carranza-Torres and Fairhurst, 2000; Muhlhaus, 1985; Hoek, 1998; Malvern, 1969). Closed form solution of GRC implementing in convergence-confinement method by elasto-plastic model are among the most broadly used for general design evaluation, especially regarding excavations and support design. Taha et al. (2009) applied the Mohr-Coulomb material and simulated stress distribution around a pile in cohesion less soil. It was found that dry soil condition gives more resistance than others. Stress concentration analysis around a wellbore showed that in addition to rock-mud interaction drilling string vibration could cause many problems (Ibrahim et al., 2004). Macro element analysis and closed form solution of stress distribution around cavities in plate bending were modeled and had excellent results with regards to conventional finite element solutions (Al-Ani, 2010).\n\nThe Mohr-Coulomb criterion (M-C) was the most common criterion that has been used in elasto-plastic solution of stress state around openings (Florence and Schwer, 1978). However, The Hoek-Brown criterion (H-B) could find wide practical application as a method of describing the stress condition in rock mass surrounded the opening. Hassani et al. (2008) presented a 3D finite element analysis of Siah Bishe tunnels by using ABAQUS software. It was observed maximum displacement occurs in the roof of the tunnels. Stress concentration was intensified in transition zone of tunnel and shaft. The rock mass condition under which Hoek-Brown criterion can be applied is only intact rock or heavily jointed rock masses that can be considered homogenous and isotropic (Hoek et al., 1998). The Classic Tresca criterion related the difference between maximum and minimum principal stresses to the cohesion without friction, like Von Mises (Hill, 1950). Stress analysis and hydro mechanical behavior of the Bisotun epigraph showed that heterogeneity is one of the most significant factors on hydraulic and mechanical properties of rock mass (Karimnia and Shahkarami, 2011). Fatigue behavior of a cylindrical hole in piston was studied by Rahman et al. (2009) using the Tresca and Von Mises materials. It was observed that more conservative prediction to use Signed Tresca parameter and Signed von Mises stress gives the result that lie between the absolute maximum principal stress and signed Tresca results.\n\nMost of the cited failure criteria which applied in rock mechanics were extended before the function of the intermediate principal stress was evident. According to experimental data has shown that the intermediate principal stress has a substantial-although slight-influence on the strength of several rock classes (Colmenares and Zoback, 2002; Mogi, 1967). The Mogi-Coulomb Criterion (MG-C) clearly showed the impact of intermediate principal stress that was based on linear Mogi criterion in terms of first and second stress invariants (Al-Ajmi and Zimmerman, 2006). A mathematical model for couple coal/rock mass visco-elastic deformation was presented by Sun (2006). It could properly show the gas leak flow in these mediums.\n\nThis study concerns analytical solution of stress distribution about an underground circular space via four rock failure criteria includes the generalized Hoek-Brown, the Mohr-Coulomb, the Mogi-Coulomb and the Tresca criterion which implemented in 3DEC by means of a FISH program. The aim of this paper was to compare elasto-plastic solution of the rock failure criteria numerically in the 3DEC. Advantages and deficient of each criterion is presented. In addition, parametric study of the rock failure criteria in elasto-plastic solution has been carried out.\n\nFOUR ROCK FAILURE CRITERIA\n\nThe Tresca criterion: After a series of experiments, Tresca achieved that the material will failed when a critical amount of shear stress is reached (Tresca, 1868):",
null,
"(1)\n\nwhere, C is the cohesion and τmax is the maximum shear stress of the material. Notice that the Tresca criterion can be considered as a particular type of the M-C criterion, with φ = 0.\n\nThe generalized Hoek-Brown criterion: The generalized H-B criterion concerns the maximum principal stress, σ1 to the minimum principal stress, σ3 via Eq. 2:",
null,
"(2)\n\nwhere, σc is the Uniaxial Compressive Strength (UCS) of intact rock, m, s and a are constants which depend on the rock mass properties:",
null,
"(3)",
null,
"(4)",
null,
"(5)\n\nwhere, mi is the value of m for intact rock and can be obtained from laboratory tests. While, D is the disturbance factor which varies from 0.0 for undisturbed rock masses to 1.0 for very disturbed rock mass e.g., by stress release and drill-blast. The Geological Strength Index (GSI) introduced by Hoek indicates the characteristic of the rock mass (Hoek, 1994).\n\nThe Mohr-Coulomb criterion: A more general and frequently used criterion is the Mohr-Coulomb failure criterion. Failure will occur when in any (failure) plane the shear stress, τ reaches the failure shear stress, τmax which is given by a functional relation of the form:",
null,
"(6)\n\nwhere, c is the cohesion of the rock mass, φ is the internal friction angle of the rock mass and σn is the normal stress working on the individual failure plane. The M-C criterion can be written with regard to the maximum and minimum principal stresses as follows (Benz and Schwab, 2008):",
null,
"(7)\n\nThe Mogi-Coulomb failure criterion: All three rock failure criteria considered above, did not take into account the influence of intermediate principal stress and determined from triaxial tests. Mogi’s experimental attempts revealed that rock strength varied with the intermediate principal stress, σ2 which was obtained from polyaxial (True-Triaxial) tests (Mogi, 1971). He related the octahedral shear stress at failure to the sum of the minimum and maximum principal stresses",
null,
"(8)\n\nwhere, f is a monotonically increasing function. The MG-C criterion can be stated as:",
null,
"(9)\n\nAccording to Al-Ajmi and Zimmerman (2006) the linear Mogi parameters a and b can be related to the Coulomb shear strength parameters c and φ then can be extended as follows:",
null,
"(10)\n\nSTRESSES AROUND CIRCULAR SPACE BY ELASTO-PLASTIC SOLUTION\n\nElasto-plastic solution around circular space using the Tresca failure criterion: The elasto-plastic analytical solution is commonly carries out for simplified models. Consequently as shown in Fig. 1a circular space is utilized in plane strain condition which subjected to isotropic stresses, σh and σv at infinity and internal pressure Pi. Then, Ri is the primary radius of the underground space; Re is the plastic zone radius; R and θ are the cylindrical coordinate of an assumed location, while σR and σθ are the related radial and tangential stresses, respectively.\n\nAccording to equilibrium equation (Jaeger et al., 2007):",
null,
"(11)\n\nAnd it is assumed that σθ, σr will be σ1 and σ3, respectively. The Tresca failure criterion (Sec. 1.1) then requires:",
null,
"(12)\n\nin the plastic zone (Ri<R<Re). Introducing Eq. 12 into Eq. 11 the stresses at boundary condition are given by:",
null,
"(13)",
null,
"(14)\n\nAs the Tresca criterion does not comprise the intermediate principal stress the hydrostatic ground pressure σv = σh is assumed to solve the problem. Therefore, the induced stresses in the elastic region can be found in Hiramatsu and Oka (1968) as follows:",
null,
"(15)",
null,
"Fig. 1: The circular space subjected to isotropic stresses, σh and σv; internal pressure, Pi",
null,
"(16)\n\nwhere, σRe is the radial stress at elastic-plastic interface i.e. (R = Re). From Eq. 15 and 16 at this interface it can be obtained:",
null,
"(17)\n\nSubstituting Eq. 17 into Eq. 12 the induced radial stress can be determined as follows:",
null,
"(18)\n\nThe plastic zone radius is determined from Eq. 18 and 13 as:",
null,
"(19)\n\nAssumed the circular space with Ri = 1 m and σv = σh = 30 MPa. Figure 2 illustrates the effects of internal pressure Pi on the plastic zone radius around the circular space by the Tresca criterion for three types of rocks. The plastic zone radius decreases by increasing of UCS, while internal pressure exceeds ground pressure, the results become vice versa. For an invariable UCS, increasing the internal pressure Pi decreases the plastic zone radius Re.",
null,
"Fig. 2: The effects of internal pressure on plastic zone radius at different compressive strengths",
null,
"Fig. 3: The effects of hydrostatic ground pressure on the plastic zone radius at different compressive strengths\n\nFigure 3 displays the effects of ground pressure, σv = σh on the plastic zone radius by the Tresca criterion for three types of rocks where internal pressure Pi is 10 MPa. An increase in ground pressure leads to increase in the plastic zone radius in an invariable UCS. It can be seen that increasing UCS tends to decrease in the plastic zone radius.\n\nElasto-plastic solution around circular space using the generalized H-B failure criterion: In the plastic region, radial and tangential stresses can be found.\n\nThus, induced stresses by introducing Eq. 2 into Eq. 11 are as follows:",
null,
"(20)",
null,
"(21)\n\nThe radius of plastic zone can be determined from Eq. 17 and 20 that will be expressed as:",
null,
"(22)\n\nAll the parameters in this paper were assumed, where the ground pressure P0 is equal to hydrostatic stresses σv = σh. Figure 4 shows the effects of GSI and D on the plastic zone radius by the H-B failure criterion around circular space for different rock types where, mi = 4, Pi = 10 MPa, σv = σh = 30 MPa and σc = 20 MPa. An increase in the D factor for the certain GSI leads to increasing of the plastic zone radius. Whereas increasing the GSI decreases the plastic zone radius.\n\nFigure 5 shows the effects of internal pressure on the radius of plastic zone by H-B failure criterion for different rock types where, σv = σh = 60 MPa, mi = 4, σc = 20 MPa and D = 0.1. It can be understood with increasing of internal pressure the plastic zone radius will decrease in a certain GSI. Figure 6 displays the influence of the GSI on the radial stress around circular space for different rock types with σv = σh = 50 MPa, mi = 4, σc = 20 MPa. It can be observed the radial stress around circular space will decrease when GSI increases.\n\nThe influence of ground pressure on tangential stress around circular space has been showed in Fig. 7 by the H-B failure criterion with Pi = 0, σc = 20 MPa, D = 05 and GSI = 20. It can be noticed that for a certain ground pressure the increasing of mi will decrease the tangential stress. On the other hand when the ground pressure increases for a fixed mi the tangential stress raises.\n\nElasto-plastic solution around circular space using the Mogi-Coulomb failure criterion: As pointed out in Sec. 1.4 the strengthening effect of the intermediate principal stress can be taking into account by utilizing the Mogi-Coulomb formula. In terms of first and second stress invariants I1 and I2 defined by Al-Ajmi and Zimmerman (2006):",
null,
"(23)",
null,
"Fig. 4: Variation of the plastic zone radius with disturbance factor at different GSI using the H-B failure criterion",
null,
"Fig. 5: Variation of the radius of plastic zone under increasing of GSI at different internal pressures by the H-B failure criterion",
null,
"Fig. 6: The influence of D on the radial stress at different GSI around circular space using the H-B failure criterion",
null,
"Fig. 7: The influence of ground pressure on the tangential stress at different mi around circular space using the H-B failure criterion\n\nSince the maximum and minimum principal stresses, σ1 and σ2 are corresponded to σθ and θr around the underground circular spaces then intermediate principal stress will be σ2 = σz along the circular space axis. In plane strain condition (εz = 0) the axial stress, σz can be determined as follows:",
null,
"(24)\n\nwhere, σa is the axial in situ stress, v is the Poisson’s ratio. In hydrostatic ground pressure around the circular space, σv = σh = σa the Eq. 15 can be rewritten as:",
null,
"(25)\n\nThe radial stress in this case can be determined through Eq. 23 and 11 as follows:",
null,
"(26)",
null,
"Fig. 8: The influence of axial stress on the plastic zone radius at different internal pressures around the circular space by the MG-C failure criterion",
null,
"Fig. 9: The influence of compressive strength on the plastic zone radius at different Poisson’s ratios around circular space using MG-C criterion\n\nSubstituting Eq. 26 and 25 into Eq. 23 the tangential stress around circular space using MG-C criterion can be found as:",
null,
"(27)\n\nAs stated in previous sections the plastic zone radius is obtained using continues equations (σRe = σR) at elastic-plastic interface. Figure 8 represents the influence of intermediate principal stress (axial stress) on the plastic zone radius around the circular space based on the MG-C failure criterion. The features of the circular space and rock mass assumed as Ri = 1 and cohesion C = 3.45 MPa, v = 0.23 and φ = 30°. It can be understood that an increase in intermediate principal stress in a certain internal pressure, causes decreasing the plastic zone radius.",
null,
"Fig. 10: Variation of the plastic zone radius under increasing rock cohesion at different friction using the MG-C criterion\n\nFigure 9 indicates the effects of the Poisson’s ratio and uniaxial compressive strength on the plastic zone radius around the circular space by using the MG-C failure criterion for different rock strengths. In this case, the internal pressure Pi is 10 MPa, ground pressure σh = σv is 30 MPa and the initial radius of underground space Ri is 1 m. It can be obtained that the plastic zone radius increases by increasing of the Poisson’s ratio in a particular UCS. Also in a fixed Poisson’s ratio an increase in UCS decreases the plastic zone radius. As illustrated in Fig. 10 when cohesion of the rock around the circular space increases, in a fixed friction angle the plastic zone radius decreases. On the other hand by increasing friction angle the plastic zone radius decreases subsequently.\n\nNUMERICAL ANALYSIS\n\nAs a common criterion the elasto-plastic solution for the circular space using M-C failure criterion has been given in Salencon (1969). For this reason the solution has not mentioned in the previous section. Here, the analytical solution of the Tresca, generalized H-B, M-C and MG-C failure criteria implemented in 3DEC using a FISH program (ITASCA, 2003). Because of axisymmetric and plain strain conditions only one-fourth of the sketch in the Fig. 1 has been modeled.\n\nAll the parameters in these solutions were assumed, where the initial radius of the underground circular space Ri is 1 m, the in situ stress σh = σv is 30 MPa and the internal pressure Pi is 5 MPa. The compressive strength of rock mass σc is 50 MPa, constant parameter m is 4.5, the Geological Strength Index GSI is 40, the Poisson’s ratio v is 0.22, cohesion C is 3.45 MPa and internal friction φ is 30°. Figure 11 compares the results of the generalized H-B and Tresca failure criteria with the 3DEC.",
null,
"Fig. 11: The comparison of the radial and tangential stresses using the Tresca, the generalized H-B criteria with the 3DEC around the circular space",
null,
"Fig. 12: The comparison of the radial and tangential stresses using the MG-C, the generalized H-B criteria with the 3DEC around the circular space",
null,
"Fig. 13: The comparison of the radial and tangential stresses using the MG-C and M-C criteria with the 3DEC around the circular space",
null,
"Fig. 14: The influence of intermediate principal stress using MG-C criterion (σ2 = 10 and σ2 = 15) on the stress distribution around the circular space\n\n Table 1: The plastic zone radius around the circular space using four rock failure criteria compared with 3DEC",
null,
"It can be seen that stress distribution around the circular space using the generalized H-B criterion (the tangential stress in turquoise line and the radial stress in blue line) is close to the 3DEC (the tangential stress in red line and the radial stress in green line) than the Tresca criterion. Because the Tresca criterion does not concern with internal friction of the rock mass then it cannot be used in rock properties determination and the elasto-plastic solution around the underground spaces.\n\nThe results of the generalized H-B and MG-C failure criteria, compared with the 3DEC have shown in Fig. 12. The stress distribution around the circular space using the MG-C criterion (the tangential stress in turquoise line and the radial stress in brown line) is more similar to the 3DEC than the generalized H-B criterion.\n\nIn Fig. 13, the comparison of stress distribution using the MG-C and the M-C criteria has represented. The both results of criteria are similar to the 3DEC, completely. The effect of intermediate principal stress (axial stress in this case) is illustrated in Fig. 14. It can be discerned that an increase in intermediate principal stress from σ2 = 10 MPa (the tangential stress in turquoise line and the radial stress in brown line) to σ2 = 15 MPa (the tangential stress in cyan line and the radial stress in red line) increases the stress distribution around the circular space using the Mogi-Coulomb (MG-C) failure criterion.\n\nThe plastic zone radius around the underground circular space using four rock failure criteria are given in Table 1.\n\nIt can be found that the Tresca failure criterion overestimated the plastic zone radius more than other criteria and an increase in axial stress using MG-C failure criterion tends to decrease of the plastic zone radius. The generalized H-B failure criterion evaluated the least amount with regard to other criteria and numerical analysis in 3DEC.\n\nCONCLUSIONS\n\nElasto-plastic analytical solution of stress distribution around the underground circular space using four rock failure criteria has been given. The following consequences attained:\n\n • The Tresca failure criterion overestimated the stress distribution and radius of the plastic zone around the underground circular space. Because of neglecting internal friction, this criterion does not answer the rock mass problems • Increasing of uniaxial compressive strength, GSI, mi and internal pressure Pi decreases the plastic zone radius and the induced stresses while an increase in ground pressure P0, disturbance factor D and Poisson’s ratio v increases the induced stresses and the plastic zone radius around the underground circular space • Numerical analysis of elasto-plastic solution using four rock failure criteria has been carried out using 3DEC. The criterion that has closest fitting to 3DEC was the Mohr-Coulomb, Mogi-Coulomb and generalized Hoek-Brown failure criterion, respectively • The advantage of Mogi-Coulomb failure criterion to the others is that comprising the axial stress (intermediate principal stress) in this criterion. It was declared that an increase in axial stress increases the stress distribution while decrease the plastic zone radius • The generalized H-B failure criterion estimated the smallest amount of the plastic zone radius vis-à-vis other criteria and 3DEC\n\nREFERENCES\n\n1: Al-Ajmi, A.M. and R.W. Zimmerman, 2006. Stability analysis of vertical boreholes using the Mogi-Coulomb failure criterion. Int. J. Rock Mechanics Min. Sci., 43: 1200-1211.\n\n2: Alani, H.R.A. and I.B. Nasser, 2001. Development of quadratic plate bending macro-elements. Pak. J. Applied Sci., 1: 417-425.\n\n3: Alani, H.R.A., 2002. Cubic macro-element for plates under bending. J. Applied Sci., 2: 1084-1091.\n\n4: Al-Ani, H.R.A., 2010. Beams and plate bending macro-elements. Asian J. Applied Sci., 3: 109-134.\nCrossRef |\n\n5: Benz, T. and R. Schwab, 2008. A quantitative comparison of six rock failure criteria. Int. J. Rock Mechanics Min. Sci., 45: 1176-1186.\nCrossRef |\n\n6: Carranza-Torres, C. and C. Fairhurst, 1999. The elasto plastic response of underground excavations in rock masses that satisfy the Hoek Brown failure criterion. Int. J. Rock Mech. Min. Sci., 36: 777-809.\nCrossRef |\n\n7: Carranza-Torres, C. and C. Fairhurst, 2000. Application of the convergence-confinement method of tunnel design to rock masses that satisfy the hoek-brown failure criterion. Tunnelling Underground Space Technol., 15: 187-213.\nCrossRef |\n\n8: Carranza-Torres, C., 1998. Self-similarity analysis of the elasto-plastic response of underground openings in rock and effects of practical variables. Ph.D. Thesis, University of Minnesota.\n\n9: Chen, X., C.P. Tan and C.M. Haberfield, 1999. Solutions for the deformations and stability of elastoplastic hollow cylinders subjected to boundary pressures. Int. J. Numerical Anal. Methods Geomech., 23: 779-800.\nCrossRef |\n\n10: Colmenares, L.B. and M.D. Zoback, 2002. A statistical evaluation of intact rock failure criteria constrained by polyaxial test data for five different rocks. Int. J. Rock Mech. Min. Sci., 39: 695-729.\n\n11: Detournay, E., 1986. Elastoplastic model of a deep tunnel for a rock with variable dilatancy. Rock Mech. Rock Eng., 19: 99-108.\nCrossRef |\n\n12: Florence, A.L. and L.E. Schwer, 1978. Axisymmetric compression of a Mohr-Coulomb medium around a circular hole. Int. J. Numerical Anal. Methods Geomech., 2: 367-379.\nCrossRef |\n\n13: Hassani, H., S. Arshadnejad, H. Khodadadi and N. Goodarzi, 2008. 3D numerical modeling of a couple of power intake shafts and head race tunnels at vicinity of a rock slope in siah bishe pumped storage dam, North of Iran. J. Applied Sci., 8: 4294-4302.\n\n14: Hill, R., 1950. The Mathematical Theory of Plasticity. Oxford University Press, Oxford\n\n15: Hiramatsu, Y. and Y. Oka, 1968. Determination of the stress in rock unaffected by boreholes or drifts, from measured strains or deformations. Int. J. Rock Mech. Min. Sci. Geomechanics Abst., 5: 337-353.\nCrossRef |\n\n16: Hoek, E., P.K. Kaiser and W.F. Bawden, 1998. Underground Excavations in Hard Rock. A.A. Balkema, Rotterdam, Brookfield\n\n17: Hoek, E. and E.T Brown, 1980. Underground Excavations in Rock. MAA Publishing Company, London\n\n18: Hoek, E., 1994. Strength of rock and rock masses. ISRM News J., 2: 4-16.\n\n19: Hoek, E., 1998. Reliability of Hoek-Brown estimates of rock mass properties and their impact on design. Int. J. Rock Mech. Min. Sci. Geomech. Abstr., 35: 63-68.\n\n20: Ibrahim, A.A., T.A. Musa and A.M. Fadoul, 2004. Drilling mechanics: Consequences and relevance of drill string vibration on wellbore stability. J. Applied Sci., 4: 106-109.\n\n21: ITASCA, 2003. 3 Dimensional Distinct Element Code (3DEC 3.0). Itasca Consulting Group Inc., Minnesota\n\n22: Jaeger, J.C., N.G.W. Cook and R.W. Zimmerman, 2007. Fundamentals of Rock Mechanics. 4th Edn., Wiley- Blackwell, Oxford\n\n23: Karimnia, M. and A. Shahkarami, 2011. Hydro mechanical coupling behavior effects on the Bisotun epigraph damaging progress. J. Applied Sci., 11: 2764-2772.\n\n24: Li, L. and A. Michel, 2009. An elastoplastic evaluation of the stress state around cylindrical openings based on a closed multiaxial yield surface. Int. J. Numer. Anal. Methods Geomech., 33: 193-213.\n\n25: Malvern, L.E., 1969. Introduction to the Mechanics of a Continuous Medium. Prentice-Hall, New Jersey\n\n26: Mogi, K., 1967. Effect of intermediate principal stress on rock failure. J. Geophys. Res., 72: 5117-5131.\nCrossRef |\n\n27: Mogi, K., 1971. Fracture and flow of rocks under high triaxial compression. J. Geophys. Res., 76: 1255-1269.\nCrossRef |\n\n28: Muhlhaus, H.B., 1985. Lower bound solutions for circular tunnels in two and three dimensions. Rock Mech. Rock Eng., 18: 37-52.\nCrossRef |\n\n29: Rahman, M.M., A.K. Ariffin, M.R.M. Rejab, K. Kadirgama and M.M. Noor, 2009. Multiaxial fatigue behavior of cylinder head for a free piston linear engine. J. Applied Sci., 9: 2725-2734.",
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https://compete.etutor.co/neet/resource/1st-year/chemistry/structure-of-atom/nature-of-electromagnetic-radiation-and-photoelectric-effect | [
"",
null,
"# Nature of electromagnetic radiation and Photoelectric effect\n\n### Nature of Light : Different Theories\n\nNewton's Corpuscular Theory :\nLight comes out of the source as small particles called corpuscles. It traces in straight lines.\n\nHuygen's Wave Theory :\nAccording to Huygen, light has wave character. He compared light with mechanical waves.",
null,
"Maxwell's Electromagnetic Theory :\nA reaction which consists of both electric and magnetic components which are perpendicular to each other and perpendicular to direction of propagation of light",
null,
"Wave length (λ) : Units cm/s ; m/sec\n\nFrequency(ν) : \\tt \\nu = \\frac{C}{\\lambda}\nunits Hz (or) Cps (or) S-1\neg : λ400nm = 400 × 10-9 = 4 × 10-7 m\n\nwave number ( \\overline{\\nu}) :\n\\overline{\\nu} = \\frac{1}{\\lambda}\nunits = cm-1 (or) m-1\n\nVelocity of light (c) :\nC = 3 × 108 m/s (or) 3 × 1010 cm/sec",
null,
"Amplitude (A) : I ∝ a2\nI = Intensity\n\nElectromagnetic spectrum :",
null,
"Photo electric effect : The emission of electrons by a metal surface when a radiation having same frequency\nh\\nu = W + KE \\ \\ \\ \\ \\ \\ \\ \\ KE = \\frac{1}{2}mv^{2}\nW → Work function (The amount of energy required to free the electron from the metal surface)\nW = hν0 → The minimum frequency\nν0 = Threshold frequency → The minimum frequency required to free the electron.",
null,
"### Part2: View the Topic in this video From 0:40 to 43:40\n\nDisclaimer: Compete.etutor.co may from time to time provide links to third party Internet sites under their respective fair use policy and it may from time to time provide materials from such third parties on this website. These third party sites and any third party materials are provided for viewers convenience and for non-commercial educational purpose only. Compete does not operate or control in any respect any information, products or services available on these third party sites. Compete.etutor.co makes no representations whatsoever concerning the content of these sites and the fact that compete.etutor.co has provided a link to such sites is NOT an endorsement, authorization, sponsorship, or affiliation by compete.etutor.co with respect to such sites, its services, the products displayed, its owners, or its providers.\n\n1. Frequency = \\tt V=\\frac{C}{\\lambda}\n\n2. Wavelength = \\tt \\lambda=\\frac{C}{V}\n\n3. Wave number = \\tt \\overline{v}=\\frac{1}{\\lambda}, v=\\frac{c}{\\lambda}=c\\overline{v}\n\n4. Time period \\tt T=\\frac{1}{\\upsilon}\n\n5. No. of waves \\tt n=\\frac{2\\pi r}{\\lambda}\\left(where \\ \\lambda=\\frac{h}{m\\upsilon}\\right)\n\n6. No. of revolutions of e- per second is =\\tt \\frac{\\upsilon}{2\\pi r}"
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https://la.mathworks.com/help/comm/ref/ssbmod.html | [
"# ssbmod\n\nSingle sideband amplitude modulation\n\n## Syntax\n\n```y = ssbmod(x,Fc,Fs) y = ssbmod(x,Fc,Fs,ini_phase) y = ssbmod(x,fc,fs,ini_phase,'upper') ```\n\n## Description\n\n`y = ssbmod(x,Fc,Fs)` uses the message signal `x` to modulate a carrier signal with frequency `Fc` (Hz) using single sideband amplitude modulation in which the lower sideband is the desired sideband. The generated output `y` is a single side band signal with a suppressed carrier. The carrier signal and `x` have sample frequency `Fs` (Hz). The modulated signal has zero initial phase.\n\n`y = ssbmod(x,Fc,Fs,ini_phase)` specifies the initial phase of the modulated signal in radians.\n\n`y = ssbmod(x,fc,fs,ini_phase,'upper')` uses the upper sideband as the desired sideband.\n\n## Examples\n\ncollapse all\n\nSet the sample rate to 100 Hz. Create a time vector 100 seconds long.\n\n```fs = 100; t = (0:1/fs:100)';```\n\nSet the carrier frequency to 10 Hz. Generate a sinusoidal signal.\n\n```fc = 10; x = sin(2*pi*t);```\n\nModulate `x` using single- and double-sideband AM.\n\n```ydouble = ammod(x,fc,fs); ysingle = ssbmod(x,fc,fs);```\n\nCreate a spectrum analyzer object to plot the spectra of the two signals. Plot the spectrum of the double-sideband signal.\n\n```sa = dsp.SpectrumAnalyzer('SampleRate',fs, ... 'PlotAsTwoSidedSpectrum',false, ... 'YLimits',[-60 40]); sa(ydouble)```",
null,
"Plot the single-sideband spectrum.\n\n`sa(ysingle)`",
null,
"## Version History\n\nIntroduced before R2006a"
]
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http://www2.econ.iastate.edu/classes/econ355/choi/ho1.htm | [
"Stolper-Samuelson Theorem\n\n 1. The Relationship between commodity and factor prices Commodity prices and factor prices The SS theorem links commodity prices and factor prices. Competitive markets If a country suddenly opens up to trade, how will factor prices change? FPE theorem presupposes that all markets are perfectly competitive. Perfect competition implies that in the long run profits are zero in every industry. Specifically, Π1 = p1y1 - wL1 - rK1 = 0. Π2 = p2y2 - wL2 - rK2 = 0. Output prices and input prices Per unit profits are also zero. Dividing the two profit functions by the respective outputs, we get the relationship between prices and factor prices. price = unit labor cost + unit capital cost. P1 = aL1 w + aK1 r. P2 = aL2 w + aK2 r. Ricardo vs. HO (i) Output price includes not only wage but also rent. (ii) Input-output coefficients (e.g., aL2 ) now depend on input prices (w,r).\n\n 2. Iso-unit cost curve/Factor Price Frontier It is well known that the unit cost function g1(w,r) = aL1 w + aK1 r is concave in factor prices, w and r. That is, as either factor price rises, unit cost rises at a decreasing rate. Moreover, the iso-unit cost contour of (w,r) is convex to the origin, as shown by two curves in Figure 12. Iso-unit cost curve of p1 An iso-unit cost curve, also known as factor price frontier, is a locus of factor price combinations along which the unit cost of a good remains constant. Paul Samuelson first considered this notion and called it factor price frontier. These are derived from the above two unit cost equations. r = p1/aK1 - w(aL1/aK1). Intuitively, p1/aK1 is the vertical intercept, where w = 0, and aL1/aK1 is the slope of the curve, except that its slope is a function of factor prices (w,r). its slope Recall that the input-output coefficients are not fixed, but are actually functions of factor prices, i.e., aij = aij(w,r). The slope of the isoprice curve p1 is aL1/aK1 = L1/y1 ÷ K1/y1 = L1/K1 = 1/k1 , where k1= K1/L1. Thus, a more capital-intensive industry has a flatter curve. Recall that the slope is changing as w or r changes. That is why the iso-unit cost curves are not linear. A pair of output prices (p1,p2) results in a unique combination of factor prices, (w,r)e, an equilibrium set of wage and rental. Equilibrium wage and rental",
null,
"Note that k2 > k1 implies that p2 is flatter than p1 everywhere.\n\n 3. The Stolper-Samuelson Theorem The SS Theorem An increase in the price of the capital-intensive good increases the return to capital and decreases the return to the other factor (labor). Likewise, an increase in the price of the labor-intensive good increases wage and reduces rent. (L and the labor-intensive industry are friends, L and the capital-intensive industry are enemies.)",
null,
"When does the price of a good rise? Prices of importable goods rise and those of the exportable goods fall dramatically during the war. Price changes are moderate when tariffs are imposed.\n\n 4. Magnification Effect What is it? The magnification effect states that an increase in the price of a capital-intensive good increases the return to capital more than proportionately. Proof p2 = aL2w + aK2r. Δp2 = aL2Δw + aK2Δr = aL2w(Δw/w) + aK2r(Δr/r). Here, Δ reads \"change in\" and the percentage change in x is written as x with a hat. That is, ^x = Δx/x. Divide both sides by p2: ^p2 = θL2^w + θK2^r = θL2^w + (1 - θL2)^r for example = 75% × ^w + 25% × ^r labor share, capital share The percentage change in product price is a weighted average of percentage change factor price changes.) Note ^w = Δw/w is the \"percentage change in\" w, and θL2 = aL2w/p2 = wL2/p2y2 is the share of labor in industry 2. Moreover, θL2 + θK2 = 1 (For example, labor share 75% + capital share 25% = 100%). That is, the sum of labor and capital shares is unity in every industry. Intuition Intuitive Reason: If both the rental rate and wage were to double, the product price must also double to breakeven. Recall that if the labor share is 75%, then the capital share is 25%. If the wage rate rises by 20% and the interest rate by 10%, then the product price rises by 20% × 3/4 + 10% × 1/4 = 17.5%. If the rental rate were to remain constant, then a 10% increase in output price must be accompanied by 10%/.75 = 13.33%. However, by the SS Theorem, we know that a 10% rise in the output price (of a capital-intensive good) results in a reduction in the wage rate, and hence the interest rate must rise even faster than 13.33% (a magnification effect on the interest rate) to more than offset the negative effect of the falling wage rate. Similarly, a rise in the price of a labor-intensive good reduces the interest rate and hence increases the wage rate more than proportionately. This is the magnification effect on the wage rate. An increase in the price of the labor (capital) intensive good increases wages (rent) more than proportionately. (The labor-intensive good and labor are friends.) Price and Factor Intensities An increase in the price of a capital-intensive good raises (r/w), and hence all industries become less capital intensive to minimize costs."
]
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"http://www2.econ.iastate.edu/classes/econ355/choi/images/h/ho20.gif",
null,
"http://www2.econ.iastate.edu/classes/econ355/choi/images/h/ho21.gif",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9006441,"math_prob":0.9888904,"size":5105,"snap":"2019-43-2019-47","text_gpt3_token_len":1380,"char_repetition_ratio":0.14840227,"word_repetition_ratio":0.030735455,"special_character_ratio":0.26699314,"punctuation_ratio":0.11339199,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9977084,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-22T13:38:03Z\",\"WARC-Record-ID\":\"<urn:uuid:4b24f678-68e3-4657-b4e0-073280a52642>\",\"Content-Length\":\"14132\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3d5e9c6e-0789-43c7-a4dc-0a4d4c728ce4>\",\"WARC-Concurrent-To\":\"<urn:uuid:c7e4a8d9-a290-42ac-9665-79f37ffd5c39>\",\"WARC-IP-Address\":\"129.186.133.13\",\"WARC-Target-URI\":\"http://www2.econ.iastate.edu/classes/econ355/choi/ho1.htm\",\"WARC-Payload-Digest\":\"sha1:W5ACAS5OEFMKDMOQLF4HLLKENQ2NUXAM\",\"WARC-Block-Digest\":\"sha1:HBWRHOKB3OJLULTGSJNANFSB6XBBAG7B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987822098.86_warc_CC-MAIN-20191022132135-20191022155635-00508.warc.gz\"}"} |
https://www.gradesaver.com/textbooks/math/calculus/thomas-calculus-13th-edition/chapter-5-integrals-section-5-3-the-definite-integral-exercises-5-3-page-275/7 | [
"## Thomas' Calculus 13th Edition\n\n--$\\int^0_\\frac{-\\pi}{4}(sec x)dx$\nInterpreting Limits of sums as Integrals -$\\lim\\limits_{||p|| \\to 0}$ $\\Sigma^n_{k=1} (sec c_k) Δx_k$ where P is a partition of $[\\frac{-\\pi}{4},0]$ as --$\\int^0_\\frac{-\\pi}{4}(sec x)dx$"
]
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null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5980098,"math_prob":0.999995,"size":247,"snap":"2021-21-2021-25","text_gpt3_token_len":104,"char_repetition_ratio":0.16049382,"word_repetition_ratio":0.0,"special_character_ratio":0.4331984,"punctuation_ratio":0.019607844,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999778,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-14T00:14:45Z\",\"WARC-Record-ID\":\"<urn:uuid:2b5465ca-9c86-491f-8b2a-b4bebe694d93>\",\"Content-Length\":\"77888\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e66974a3-cac8-4447-a82c-e17a926ee6e5>\",\"WARC-Concurrent-To\":\"<urn:uuid:d06a59b2-8c9c-4f72-a62e-242cca31cae8>\",\"WARC-IP-Address\":\"54.83.50.97\",\"WARC-Target-URI\":\"https://www.gradesaver.com/textbooks/math/calculus/thomas-calculus-13th-edition/chapter-5-integrals-section-5-3-the-definite-integral-exercises-5-3-page-275/7\",\"WARC-Payload-Digest\":\"sha1:LAMC6BLMXF6WFGESVVKIHSL72ZRWBJE7\",\"WARC-Block-Digest\":\"sha1:JQZZHLMYKPEW5CZEWHOBRZH6RTRXCI6U\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243989616.38_warc_CC-MAIN-20210513234920-20210514024920-00605.warc.gz\"}"} |
https://lang-ship.com/reference/unofficial/M5StickC/Class/ESP32/BLECharacteristicMap/ | [
"# BLECharacteristicMap¶\n\nA data mapping used to manage the set of BLE characteristics known to the server.\n\n## メンバー¶\n\n### setByUUID()¶\n\n```void BLECharacteristicMap::setByUUID(BLECharacteristic *pCharacteristic, const char *uuid)\n```\n\n• BLECharacteristic* `pCharacteristic`\n• constchar * `uuid`\n\n### setByUUID()¶\n\nSet the characteristic by UUID.\n\n```void BLECharacteristicMap::setByUUID(BLECharacteristic *pCharacteristic, BLEUUID uuid)\n```\n\n• BLECharacteristic* `pCharacteristic`\n• BLEUUID `uuid` The uuid of the characteristic.\n\n### setByHandle()¶\n\nSet the characteristic by handle.\n\n```void BLECharacteristicMap::setByHandle(uint16_t handle, BLECharacteristic *pCharacteristic)\n```\n\n• uint16_t `handle` The handle of the characteristic.\n• BLECharacteristic* `pCharacteristic`\n\n### getByUUID()¶\n\nReturn the characteristic by UUID.\n\n```BLECharacteristic * BLECharacteristicMap::getByUUID(const char *uuid)\n```\n\n• constchar * `uuid`\n\nBLECharacteristic* The characteristic.\n\n### getByUUID()¶\n\nReturn the characteristic by UUID.\n\n```BLECharacteristic * BLECharacteristicMap::getByUUID(BLEUUID uuid)\n```\n\n• BLEUUID `uuid`\n\nBLECharacteristic* The characteristic.\n\n### getByHandle()¶\n\nReturn the characteristic by handle.\n\n```BLECharacteristic * BLECharacteristicMap::getByHandle(uint16_t handle)\n```\n\n• uint16_t `handle` The handle to look up the characteristic.\n\nBLECharacteristic* The characteristic.\n\n### getFirst()¶\n\nGet the first characteristic in the map.\n\n```BLECharacteristic * BLECharacteristicMap::getFirst()\n```\n\nBLECharacteristic* The first characteristic in the map.\n\n### getNext()¶\n\nGet the next characteristic in the map.\n\n```BLECharacteristic * BLECharacteristicMap::getNext()\n```\n\nBLECharacteristic* The next characteristic in the map.\n\n### toString()¶\n\nReturn a string representation of the characteristic map.\n\n```std::string BLECharacteristicMap::toString()\n```\n\nstd::string A string representation of the characteristic map.\n\n### handleGATTServerEvent()¶\n\nPass the GATT server event onwards to each of the characteristics found in the mapping\n\n```void BLECharacteristicMap::handleGATTServerEvent(esp_gatts_cb_event_t event, esp_gatt_if_t gatts_if, esp_ble_gatts_cb_param_t *param)\n```\n\n• esp_gatts_cb_event_t `event`\n• esp_gatt_if_t `gatts_if`\n• esp_ble_gatts_cb_param_t* `param`"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5243978,"math_prob":0.71629786,"size":2164,"snap":"2019-51-2020-05","text_gpt3_token_len":544,"char_repetition_ratio":0.3523148,"word_repetition_ratio":0.11790393,"special_character_ratio":0.18160814,"punctuation_ratio":0.16785714,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9797205,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-29T18:43:52Z\",\"WARC-Record-ID\":\"<urn:uuid:416eb0d3-3833-4cc5-a1b0-eaa3e5cc7606>\",\"Content-Length\":\"210965\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b1ef6cb9-ec3c-4d76-ab73-9cd33f958941>\",\"WARC-Concurrent-To\":\"<urn:uuid:c203276e-3501-4fec-b708-fa1b13090f0c>\",\"WARC-IP-Address\":\"202.172.28.25\",\"WARC-Target-URI\":\"https://lang-ship.com/reference/unofficial/M5StickC/Class/ESP32/BLECharacteristicMap/\",\"WARC-Payload-Digest\":\"sha1:LCJZD2B4PZB5BSUCHRSOXSYSA2HANZOA\",\"WARC-Block-Digest\":\"sha1:MWIECHRGFLOLWJ527COMBINGWNIZ46K6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251801423.98_warc_CC-MAIN-20200129164403-20200129193403-00518.warc.gz\"}"} |
https://jeopardylabs.com/play/enter-title11974 | [
"Multiplication: 2 digits by 1 digit\nMultiplication: 3 digits by 1 digit\nMultiplication: 4 digits by 1 digit\nMultiplication with Zeros\nWrite an equation\n100\n23 x 6\n138\n100\n225 x 4\n900\n100\n3,446 x 3\n10,338\n100\n208 x 4\n832\n100\nKris organized his baseball cards into rows. He had 6 rows with 56 cards in each row. A. Write an equation to show how many cards Kris had in all. B. Solve the equation.\nA. 6 x 56 = c B. c = 336 cards\n200\n48 x 7\n336\n200\nFind the value of n. 456 x n = 2280\nn=5\n200\n5,457 x 6\n32,742\n200\n3,080 x 7\n21,560\n200\nKelsey had 7 rows of 34 stickers. She got 13 more from Meghan. A. Write an equation to show how many stickers Kelsey has in all. B. Solve the equation.\nA. (7 x 34) + 13 = s B. 251 stickers\n300\n72 x 9\n648\n300\n736 x 7\n5,152\n300\n6,778 x 5\n33,890\n300\n\\$14.05 x 8\n\\$112.40\n300\nRobert makes \\$24 an hour. Write an equation to show how much money Robert will make in 6 hours and then solve the equation.\n24 x 6 = m; m = \\$144\n400\n39 x 6\n234\n400\n858 x 4\n3,432\n400\n4,989 x 8\n39,912\n400\n40,046 x 9\n360,414\n400\nRyan had \\$18. If he bought 6 video games for \\$2.50 each, how much money did he have left over?\n\\$18 - (6 x \\$2.50) = \\$18 - \\$15 = \\$3\n500\n98 x 4\n392\n500\n289 x 6\n1,734\n500\n5,972 x 4\n23,888\n500\nJessica earns \\$27.04 per hour. How much will she earn in 6 hours?\n\\$162.24\n500\nTrevor had \\$35. He bought 1 football for \\$15.30 and 2 pairs of socks for \\$3.98 each. How much money does Trevor have left?\n\\$35 - \\$15.30 = \\$19.70, and \\$19.70 - \\$7.96 = \\$11.74, so Trevor had \\$11.74 left.\nClick to zoom"
]
| [
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.90851235,"math_prob":0.9979823,"size":1835,"snap":"2022-05-2022-21","text_gpt3_token_len":709,"char_repetition_ratio":0.11414528,"word_repetition_ratio":0.04511278,"special_character_ratio":0.4746594,"punctuation_ratio":0.13721414,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97712004,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-18T19:56:27Z\",\"WARC-Record-ID\":\"<urn:uuid:ab111630-dfd9-4796-811a-5ad09b36f1d3>\",\"Content-Length\":\"53554\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ccc1c5be-6c36-4411-a92e-54f0ba750528>\",\"WARC-Concurrent-To\":\"<urn:uuid:1d97c44f-4d0c-466c-843e-c2632a0f9348>\",\"WARC-IP-Address\":\"198.100.157.237\",\"WARC-Target-URI\":\"https://jeopardylabs.com/play/enter-title11974\",\"WARC-Payload-Digest\":\"sha1:H2DNEB4JO2TTTEVKHR4WDL3QSWGLDTXI\",\"WARC-Block-Digest\":\"sha1:AIGJE3VPP3LNZLUDZ4OY3Q4Z5L5MAZEY\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662522309.14_warc_CC-MAIN-20220518183254-20220518213254-00298.warc.gz\"}"} |
https://research.vu.nl/en/publications/bipartite-networks | [
"Bipartite Networks\n\nF. Agneessens, C. Moser\n\nResearch output: Chapter in Book / Report / Conference proceedingChapterAcademic\n\nAbstract\n\nBipartite networks refer to a specific kind of network in which the nodes (or actors) can be partitioned into two subsets based on the fact that no links exist between actors within each subset, but only between the two subsets.\n\nDue to the partition of actors in two sets and the absence of relations within the parts, bipartite networks form a specific type of complete network, which differs from classic complete networks in a number of ways. One particular property includes the absence of three-cycles, or any higher order cycles of uneven number. As a result the smallest closed configuration is a four-cycle, i.e., a configuration where two nodes of one set (i and j) are both connected to two nodes of another set (k and l).\n\nIn practice, bipartite networks seem to be similar to two-mode networks, and sometimes bipartite networks are used interchangeably with two-mode or affiliation networks. However, theoretically bipartite and two-mode networks are clearly distinct, since in the former the focus of the research question is on all the actors in the two subsets, whereas in the two-mode network the focus is often on one of the two modes (subsets). Affiliation (two-mode) networks a priori define two clearly distinct kinds of actors (such as events and persons), and the relations involve persons attending these events, such as theater attendees, researchers working together on projects, or collaboration on articles. These are represented in an actor-by-event matrix, and the interest is generally on one of the two kinds of actors (modes).\n\nBipartite networks generally refer to one type of actor, which can be distinguished based on some characteristic or, alternatively, ad hoc based on the structure. The focus often lies on both subsets of actors at the same time. According to Stephen Borgatti and Martin Everett, “In general, bipartite networks are distinct from two-mode networks in the reasons for their structure, not in their resulting properties.” Hence, these are represented in an actor-by-actor matrix, where both subsets of actors are presented in both the columns and the rows. Nevertheless, two-mode networks can also be translated into a bipartite format by putting actors and events behind each other. That is why some will treat two-mode networks as a subset of all bipartite networks (or a representation of bipartite), which has the same structure but is only theoretically distinct.\nOriginal language English Encyclopedia of Social Networks G.A. Barnett Thousand Oaks Sage 75-77 9781412979115 Published - 2011\n\nevent\nhuman being\ntheater\n\nCite this\n\nAgneessens, F., & Moser, C. (2011). Bipartite Networks. In G. A. Barnett (Ed.), Encyclopedia of Social Networks (pp. 75-77). Thousand Oaks: Sage.\nAgneessens, F. ; Moser, C. / Bipartite Networks. Encyclopedia of Social Networks. editor / G.A. Barnett. Thousand Oaks : Sage, 2011. pp. 75-77\n@inbook{61acdfb4bc124409911c8d4612110011,\ntitle = \"Bipartite Networks\",\nabstract = \"Bipartite networks refer to a specific kind of network in which the nodes (or actors) can be partitioned into two subsets based on the fact that no links exist between actors within each subset, but only between the two subsets.Due to the partition of actors in two sets and the absence of relations within the parts, bipartite networks form a specific type of complete network, which differs from classic complete networks in a number of ways. One particular property includes the absence of three-cycles, or any higher order cycles of uneven number. As a result the smallest closed configuration is a four-cycle, i.e., a configuration where two nodes of one set (i and j) are both connected to two nodes of another set (k and l).In practice, bipartite networks seem to be similar to two-mode networks, and sometimes bipartite networks are used interchangeably with two-mode or affiliation networks. However, theoretically bipartite and two-mode networks are clearly distinct, since in the former the focus of the research question is on all the actors in the two subsets, whereas in the two-mode network the focus is often on one of the two modes (subsets). Affiliation (two-mode) networks a priori define two clearly distinct kinds of actors (such as events and persons), and the relations involve persons attending these events, such as theater attendees, researchers working together on projects, or collaboration on articles. These are represented in an actor-by-event matrix, and the interest is generally on one of the two kinds of actors (modes).Bipartite networks generally refer to one type of actor, which can be distinguished based on some characteristic or, alternatively, ad hoc based on the structure. The focus often lies on both subsets of actors at the same time. According to Stephen Borgatti and Martin Everett, “In general, bipartite networks are distinct from two-mode networks in the reasons for their structure, not in their resulting properties.” Hence, these are represented in an actor-by-actor matrix, where both subsets of actors are presented in both the columns and the rows. Nevertheless, two-mode networks can also be translated into a bipartite format by putting actors and events behind each other. That is why some will treat two-mode networks as a subset of all bipartite networks (or a representation of bipartite), which has the same structure but is only theoretically distinct.\",\nauthor = \"F. Agneessens and C. Moser\",\nyear = \"2011\",\nlanguage = \"English\",\nisbn = \"9781412979115\",\npages = \"75--77\",\neditor = \"G.A. Barnett\",\nbooktitle = \"Encyclopedia of Social Networks\",\npublisher = \"Sage\",\n\n}\n\nAgneessens, F & Moser, C 2011, Bipartite Networks. in GA Barnett (ed.), Encyclopedia of Social Networks. Sage, Thousand Oaks, pp. 75-77.\n\nBipartite Networks. / Agneessens, F.; Moser, C.\n\nEncyclopedia of Social Networks. ed. / G.A. Barnett. Thousand Oaks : Sage, 2011. p. 75-77.\n\nResearch output: Chapter in Book / Report / Conference proceedingChapterAcademic\n\nTY - CHAP\n\nT1 - Bipartite Networks\n\nAU - Agneessens, F.\n\nAU - Moser, C.\n\nPY - 2011\n\nY1 - 2011\n\nN2 - Bipartite networks refer to a specific kind of network in which the nodes (or actors) can be partitioned into two subsets based on the fact that no links exist between actors within each subset, but only between the two subsets.Due to the partition of actors in two sets and the absence of relations within the parts, bipartite networks form a specific type of complete network, which differs from classic complete networks in a number of ways. One particular property includes the absence of three-cycles, or any higher order cycles of uneven number. As a result the smallest closed configuration is a four-cycle, i.e., a configuration where two nodes of one set (i and j) are both connected to two nodes of another set (k and l).In practice, bipartite networks seem to be similar to two-mode networks, and sometimes bipartite networks are used interchangeably with two-mode or affiliation networks. However, theoretically bipartite and two-mode networks are clearly distinct, since in the former the focus of the research question is on all the actors in the two subsets, whereas in the two-mode network the focus is often on one of the two modes (subsets). Affiliation (two-mode) networks a priori define two clearly distinct kinds of actors (such as events and persons), and the relations involve persons attending these events, such as theater attendees, researchers working together on projects, or collaboration on articles. These are represented in an actor-by-event matrix, and the interest is generally on one of the two kinds of actors (modes).Bipartite networks generally refer to one type of actor, which can be distinguished based on some characteristic or, alternatively, ad hoc based on the structure. The focus often lies on both subsets of actors at the same time. According to Stephen Borgatti and Martin Everett, “In general, bipartite networks are distinct from two-mode networks in the reasons for their structure, not in their resulting properties.” Hence, these are represented in an actor-by-actor matrix, where both subsets of actors are presented in both the columns and the rows. Nevertheless, two-mode networks can also be translated into a bipartite format by putting actors and events behind each other. That is why some will treat two-mode networks as a subset of all bipartite networks (or a representation of bipartite), which has the same structure but is only theoretically distinct.\n\nAB - Bipartite networks refer to a specific kind of network in which the nodes (or actors) can be partitioned into two subsets based on the fact that no links exist between actors within each subset, but only between the two subsets.Due to the partition of actors in two sets and the absence of relations within the parts, bipartite networks form a specific type of complete network, which differs from classic complete networks in a number of ways. One particular property includes the absence of three-cycles, or any higher order cycles of uneven number. As a result the smallest closed configuration is a four-cycle, i.e., a configuration where two nodes of one set (i and j) are both connected to two nodes of another set (k and l).In practice, bipartite networks seem to be similar to two-mode networks, and sometimes bipartite networks are used interchangeably with two-mode or affiliation networks. However, theoretically bipartite and two-mode networks are clearly distinct, since in the former the focus of the research question is on all the actors in the two subsets, whereas in the two-mode network the focus is often on one of the two modes (subsets). Affiliation (two-mode) networks a priori define two clearly distinct kinds of actors (such as events and persons), and the relations involve persons attending these events, such as theater attendees, researchers working together on projects, or collaboration on articles. These are represented in an actor-by-event matrix, and the interest is generally on one of the two kinds of actors (modes).Bipartite networks generally refer to one type of actor, which can be distinguished based on some characteristic or, alternatively, ad hoc based on the structure. The focus often lies on both subsets of actors at the same time. According to Stephen Borgatti and Martin Everett, “In general, bipartite networks are distinct from two-mode networks in the reasons for their structure, not in their resulting properties.” Hence, these are represented in an actor-by-actor matrix, where both subsets of actors are presented in both the columns and the rows. Nevertheless, two-mode networks can also be translated into a bipartite format by putting actors and events behind each other. That is why some will treat two-mode networks as a subset of all bipartite networks (or a representation of bipartite), which has the same structure but is only theoretically distinct.\n\nM3 - Chapter\n\nSN - 9781412979115\n\nSP - 75\n\nEP - 77\n\nBT - Encyclopedia of Social Networks\n\nA2 - Barnett, G.A.\n\nPB - Sage\n\nCY - Thousand Oaks\n\nER -\n\nAgneessens F, Moser C. Bipartite Networks. In Barnett GA, editor, Encyclopedia of Social Networks. Thousand Oaks: Sage. 2011. p. 75-77"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9491709,"math_prob":0.82002676,"size":7774,"snap":"2019-43-2019-47","text_gpt3_token_len":1642,"char_repetition_ratio":0.17129987,"word_repetition_ratio":0.88263667,"special_character_ratio":0.20517108,"punctuation_ratio":0.10496851,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9582941,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-22T04:38:42Z\",\"WARC-Record-ID\":\"<urn:uuid:b0908d43-6908-4d77-9322-1b5f8b7a0197>\",\"Content-Length\":\"36099\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b2a85272-45c2-478c-aaef-ecef19f56d69>\",\"WARC-Concurrent-To\":\"<urn:uuid:c6d20701-b2a1-4b11-96e8-56e0f2fc7792>\",\"WARC-IP-Address\":\"52.51.22.49\",\"WARC-Target-URI\":\"https://research.vu.nl/en/publications/bipartite-networks\",\"WARC-Payload-Digest\":\"sha1:ISXZIN2Y743XV7W2G3M6X2HINTZ4SDWF\",\"WARC-Block-Digest\":\"sha1:2PZYVXFKHYD6TL6PIGGVCJYC4F4FTIKP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987798619.84_warc_CC-MAIN-20191022030805-20191022054305-00345.warc.gz\"}"} |
https://ocw.ilas.nagoya-u.ac.jp/index.php?lang=en&mode=c&id=254&page_type=index | [
" Econometrics I | School of Economics / Graduate School of Economics | NU OCW\n•",
null,
"• Browse by Category\n•",
null,
"• Browse by School/\n\n# Econometrics I\n\nDepartment: School of Economics / Graduate School of Economics\n\n Class Time: 2010 Fall Monday Recommended for: Economics students\n\n##",
null,
"Course Overview\n\n### Course Overview\n\nWe focus on \"regression analysis,\" in which a variable of interest (for example, hourly wage of workers) is explained by other variables (for example, years of schooling, work experience, age, sex), and study a fundamental method called \"ordinary least squares\". We also study statistical inferences based on this method, in which we ask, for example, \"Does the hourly wage of workers increase with their years of schooling?\" or \"Does the hourly wage differ between male and female workers with the same years of schooling?\".\n\n### Key Features\n\nTo increase students' understanding, I try to keep calculations and proof in statistics to a minimum and explain the concepts and technical terms in econometrics using economic examples and illustrations. Due to a lack of Japanese textbooks containing a good balance of appropriate examples and statistical calculations, I instead prepare a handout based on a popular textbook of introductory econometrics which is used in other countries. The handout has some blank spaces left for figures, calculations and proofs which students fill in themselves to increase their understanding.\n\nClose Section\n\n##",
null,
"Syllabus\n\n### Course Aims\n\nWe focus on \"regression analysis,\" in which a variable of interest (for example, hourly wage of workers) is explained by other variables (for example, years of schooling, work experience, age, sex), and study a fundamental method called \"ordinary least squares\". We also study statistical inferences based on this method, in which we ask, for example, \"Does the hourly wage of workers increase with their years of schooling?\" or \"Does the hourly wage differ between male and female workers with the same years of schooling?\".\n\nNone.\n\n### Course Requirements\n\nIt is preferable for students to already know a basic statistical method.\n\nSession Contents\n1 What is econometrics? / The nature of economic data\n2 Review of basic statistics (1)\n3 Simple regression model (1)\n4 Simple regression model (2)\n5 Simple regression model (3)\n6 Multiple regression model: estimation (1)\n7 Multiple regression model: estimation (2)\n8 Review of basic statistics (2)\n9 Multiple regression model: statistical inference (1)\n10 Multiple regression model: statistical inference (2)\n11 Other fundamental tools in econometrics (1)\n12 Other fundamental tools in econometrics (2)\n13 Dummy variables\n14 Course Review\n15 Summary / Final exam"
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"https://ocw.ilas.nagoya-u.ac.jp/images/en/j_category.svg",
null,
"https://ocw.ilas.nagoya-u.ac.jp/images/en/j_school.svg",
null,
"https://ocw.ilas.nagoya-u.ac.jp/images/en/close_text.svg",
null,
"https://ocw.ilas.nagoya-u.ac.jp/images/en/close_text.svg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.90108603,"math_prob":0.6055905,"size":2933,"snap":"2023-40-2023-50","text_gpt3_token_len":642,"char_repetition_ratio":0.11061796,"word_repetition_ratio":0.36525613,"special_character_ratio":0.21581998,"punctuation_ratio":0.097165994,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95650506,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,5,null,5,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-29T16:23:36Z\",\"WARC-Record-ID\":\"<urn:uuid:8a60b72a-ac58-443b-a9bc-8ea54006ccee>\",\"Content-Length\":\"15776\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:22d241e3-0b17-4cc1-a653-b639fb771046>\",\"WARC-Concurrent-To\":\"<urn:uuid:4062d831-907f-4744-b1f4-57910ae194a8>\",\"WARC-IP-Address\":\"133.6.182.40\",\"WARC-Target-URI\":\"https://ocw.ilas.nagoya-u.ac.jp/index.php?lang=en&mode=c&id=254&page_type=index\",\"WARC-Payload-Digest\":\"sha1:G26776M6C4XCOGLPHD3WMVFTEIXE2ELH\",\"WARC-Block-Digest\":\"sha1:3WP5W3HY3ET2R5WAHL43L6YX5WWES5ZL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100112.41_warc_CC-MAIN-20231129141108-20231129171108-00054.warc.gz\"}"} |
http://semiticroots.net/index.php?r=root/search&%3BRoot_sort=rad2&%3BRoot_page=10&Root_sort=concept.desc&Root_page=19 | [
"Root Filter Search\n\nTo filter on radicals you must enter their index number, click the tabs below to see how the indexes map out in various scripts.\n\n 𐩱 = 0 𐩨 = 1 𐩩 = 2 𐩻 = 3 𐩴 = 4 𐩢 = 5 𐩭 = 6 𐩵 = 7 𐩹 = 8 𐩧 = 9 𐩸 = 10 𐩪 = 11 𐩦 = 12 𐩮 = 13 𐩳 = 14 𐩷 = 15 𐩼 = 16 𐩲 = 17 𐩶 = 18 𐩰 = 19 𐩤 = 20 𐩫 = 21 𐩡 = 22 𐩣 = 23 𐩬 = 24 𐩠 = 25 𐩥 = 26 𐩺 = 27 𐩯 = 28\n\nYou may optionally enter a comparison operator (<, <=, >, >=, <> or =) at the beginning of each of your search values to specify how the comparison should be done. To filter on more than one index for a radical, separate them with a comma (eg. 0,1,2)."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.57193923,"math_prob":0.9999441,"size":1035,"snap":"2022-05-2022-21","text_gpt3_token_len":608,"char_repetition_ratio":0.0814743,"word_repetition_ratio":0.0,"special_character_ratio":0.49371982,"punctuation_ratio":0.05394191,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9921377,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-27T00:02:53Z\",\"WARC-Record-ID\":\"<urn:uuid:fb6adedc-195a-4e31-9d09-a7718fd76909>\",\"Content-Length\":\"20787\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c7fa344b-6871-4cbd-ba40-66e54a34d3ea>\",\"WARC-Concurrent-To\":\"<urn:uuid:77207974-73af-4e9c-bc06-8a8ccd606efd>\",\"WARC-IP-Address\":\"69.163.217.228\",\"WARC-Target-URI\":\"http://semiticroots.net/index.php?r=root/search&%3BRoot_sort=rad2&%3BRoot_page=10&Root_sort=concept.desc&Root_page=19\",\"WARC-Payload-Digest\":\"sha1:ESURBDNFXU2QVGYHGZBHMPPNSFXAXUIP\",\"WARC-Block-Digest\":\"sha1:TBOO4ND5J7AN33SM5XZKYUKLSJX77RYH\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320305006.68_warc_CC-MAIN-20220126222652-20220127012652-00138.warc.gz\"}"} |
https://tightwadkitty.savingadvice.com/2010/08/ | [
"User Real IP - 3.228.220.31\n```Array\n(\n => Array\n(\n => 182.68.68.92\n)\n\n => Array\n(\n => 101.0.41.201\n)\n\n => Array\n(\n => 43.225.98.123\n)\n\n => Array\n(\n => 2.58.194.139\n)\n\n => Array\n(\n => 46.119.197.104\n)\n\n => Array\n(\n => 45.249.8.93\n)\n\n => Array\n(\n => 103.12.135.72\n)\n\n => Array\n(\n => 157.35.243.216\n)\n\n => Array\n(\n => 209.107.214.176\n)\n\n => Array\n(\n => 5.181.233.166\n)\n\n => Array\n(\n => 106.201.10.100\n)\n\n => Array\n(\n => 36.90.55.39\n)\n\n => Array\n(\n => 119.154.138.47\n)\n\n => Array\n(\n => 51.91.31.157\n)\n\n => Array\n(\n => 182.182.65.216\n)\n\n => Array\n(\n => 157.35.252.63\n)\n\n => Array\n(\n => 14.142.34.163\n)\n\n => Array\n(\n => 178.62.43.135\n)\n\n => Array\n(\n => 43.248.152.148\n)\n\n => Array\n(\n => 222.252.104.114\n)\n\n => Array\n(\n => 209.107.214.168\n)\n\n => Array\n(\n => 103.99.199.250\n)\n\n => Array\n(\n => 178.62.72.160\n)\n\n => Array\n(\n => 27.6.1.170\n)\n\n => Array\n(\n => 182.69.249.219\n)\n\n => Array\n(\n => 110.93.228.86\n)\n\n => Array\n(\n => 72.255.1.98\n)\n\n => Array\n(\n => 182.73.111.98\n)\n\n => Array\n(\n => 45.116.117.11\n)\n\n => Array\n(\n => 122.15.78.189\n)\n\n => Array\n(\n => 14.167.188.234\n)\n\n => Array\n(\n => 223.190.4.202\n)\n\n => Array\n(\n => 202.173.125.19\n)\n\n => Array\n(\n => 103.255.5.32\n)\n\n => Array\n(\n => 39.37.145.103\n)\n\n => Array\n(\n => 140.213.26.249\n)\n\n => Array\n(\n => 45.118.166.85\n)\n\n => Array\n(\n => 102.166.138.255\n)\n\n => Array\n(\n => 77.111.246.234\n)\n\n => Array\n(\n => 45.63.6.196\n)\n\n => Array\n(\n => 103.250.147.115\n)\n\n => Array\n(\n => 223.185.30.99\n)\n\n => Array\n(\n => 103.122.168.108\n)\n\n => Array\n(\n => 123.136.203.21\n)\n\n => Array\n(\n => 171.229.243.63\n)\n\n => Array\n(\n => 153.149.98.149\n)\n\n => Array\n(\n => 223.238.93.15\n)\n\n => Array\n(\n => 178.62.113.166\n)\n\n => Array\n(\n => 101.162.0.153\n)\n\n => Array\n(\n => 121.200.62.114\n)\n\n => Array\n(\n => 14.248.77.252\n)\n\n => Array\n(\n => 95.142.117.29\n)\n\n => Array\n(\n => 150.129.60.107\n)\n\n => Array\n(\n => 94.205.243.22\n)\n\n => Array\n(\n => 115.42.71.143\n)\n\n => Array\n(\n => 117.217.195.59\n)\n\n => Array\n(\n => 182.77.112.56\n)\n\n => Array\n(\n => 182.77.112.108\n)\n\n => Array\n(\n => 41.80.69.10\n)\n\n => Array\n(\n => 117.5.222.121\n)\n\n => Array\n(\n => 103.11.0.38\n)\n\n => Array\n(\n => 202.173.127.140\n)\n\n => Array\n(\n => 49.249.249.50\n)\n\n => Array\n(\n => 116.72.198.211\n)\n\n => Array\n(\n => 223.230.54.53\n)\n\n => Array\n(\n => 102.69.228.74\n)\n\n => Array\n(\n => 39.37.251.89\n)\n\n => Array\n(\n => 39.53.246.141\n)\n\n => Array\n(\n => 39.57.182.72\n)\n\n => Array\n(\n => 209.58.130.210\n)\n\n => Array\n(\n => 104.131.75.86\n)\n\n => Array\n(\n => 106.212.131.255\n)\n\n => Array\n(\n => 106.212.132.127\n)\n\n => Array\n(\n => 223.190.4.60\n)\n\n => Array\n(\n => 103.252.116.252\n)\n\n => Array\n(\n => 103.76.55.182\n)\n\n => Array\n(\n => 45.118.166.70\n)\n\n => Array\n(\n => 103.93.174.215\n)\n\n => Array\n(\n => 5.62.62.142\n)\n\n => Array\n(\n => 182.179.158.156\n)\n\n => Array\n(\n => 39.57.255.12\n)\n\n => Array\n(\n => 39.37.178.37\n)\n\n => Array\n(\n => 182.180.165.211\n)\n\n => Array\n(\n => 119.153.135.17\n)\n\n => Array\n(\n => 72.255.15.244\n)\n\n => Array\n(\n => 139.180.166.181\n)\n\n => Array\n(\n => 70.119.147.111\n)\n\n => Array\n(\n => 106.210.40.83\n)\n\n => Array\n(\n => 14.190.70.91\n)\n\n => Array\n(\n => 202.125.156.82\n)\n\n => Array\n(\n => 115.42.68.38\n)\n\n => Array\n(\n => 102.167.13.108\n)\n\n => Array\n(\n => 117.217.192.130\n)\n\n => Array\n(\n => 205.185.223.156\n)\n\n => Array\n(\n => 171.224.180.29\n)\n\n => Array\n(\n => 45.127.45.68\n)\n\n => Array\n(\n => 195.206.183.232\n)\n\n => Array\n(\n => 49.32.52.115\n)\n\n => Array\n(\n => 49.207.49.223\n)\n\n => Array\n(\n => 45.63.29.61\n)\n\n => Array\n(\n => 103.245.193.214\n)\n\n => Array\n(\n => 39.40.236.69\n)\n\n => Array\n(\n => 62.80.162.111\n)\n\n => Array\n(\n => 45.116.232.56\n)\n\n => Array\n(\n => 45.118.166.91\n)\n\n => Array\n(\n => 180.92.230.234\n)\n\n => Array\n(\n => 157.40.57.160\n)\n\n => Array\n(\n => 110.38.38.130\n)\n\n => Array\n(\n => 72.255.57.183\n)\n\n => Array\n(\n => 182.68.81.85\n)\n\n => Array\n(\n => 39.57.202.122\n)\n\n => Array\n(\n => 119.152.154.36\n)\n\n => Array\n(\n => 5.62.62.141\n)\n\n => Array\n(\n => 119.155.54.232\n)\n\n => Array\n(\n => 39.37.141.22\n)\n\n => Array\n(\n => 183.87.12.225\n)\n\n => Array\n(\n => 107.170.127.117\n)\n\n => Array\n(\n => 125.63.124.49\n)\n\n => Array\n(\n => 39.42.191.3\n)\n\n => Array\n(\n => 116.74.24.72\n)\n\n => Array\n(\n => 46.101.89.227\n)\n\n => Array\n(\n => 202.173.125.247\n)\n\n => Array\n(\n => 39.42.184.254\n)\n\n => Array\n(\n => 115.186.165.132\n)\n\n => Array\n(\n => 39.57.206.126\n)\n\n => Array\n(\n => 103.245.13.145\n)\n\n => Array\n(\n => 202.175.246.43\n)\n\n => Array\n(\n => 192.140.152.150\n)\n\n => Array\n(\n => 202.88.250.103\n)\n\n => Array\n(\n => 103.248.94.207\n)\n\n => Array\n(\n => 77.73.66.101\n)\n\n => Array\n(\n => 104.131.66.8\n)\n\n => Array\n(\n => 113.186.161.97\n)\n\n => Array\n(\n => 222.254.5.7\n)\n\n => Array\n(\n => 223.233.67.247\n)\n\n => Array\n(\n => 171.249.116.146\n)\n\n => Array\n(\n => 47.30.209.71\n)\n\n => Array\n(\n => 202.134.13.130\n)\n\n => Array\n(\n => 27.6.135.7\n)\n\n => Array\n(\n => 107.170.186.79\n)\n\n => Array\n(\n => 103.212.89.171\n)\n\n => Array\n(\n => 117.197.9.77\n)\n\n => Array\n(\n => 122.176.206.233\n)\n\n => Array\n(\n => 192.227.253.222\n)\n\n => Array\n(\n => 182.188.224.119\n)\n\n => Array\n(\n => 14.248.70.74\n)\n\n => Array\n(\n => 42.118.219.169\n)\n\n => Array\n(\n => 110.39.146.170\n)\n\n => Array\n(\n => 119.160.66.143\n)\n\n => Array\n(\n => 103.248.95.130\n)\n\n => Array\n(\n => 27.63.152.208\n)\n\n => Array\n(\n => 49.207.114.96\n)\n\n => Array\n(\n => 102.166.23.214\n)\n\n => Array\n(\n => 175.107.254.73\n)\n\n => Array\n(\n => 103.10.227.214\n)\n\n => Array\n(\n => 202.143.115.89\n)\n\n => Array\n(\n => 110.93.227.187\n)\n\n => Array\n(\n => 103.140.31.60\n)\n\n => Array\n(\n => 110.37.231.46\n)\n\n => Array\n(\n => 39.36.99.238\n)\n\n => Array\n(\n => 157.37.140.26\n)\n\n => Array\n(\n => 43.246.202.226\n)\n\n => Array\n(\n => 137.97.8.143\n)\n\n => Array\n(\n => 182.65.52.242\n)\n\n => Array\n(\n => 115.42.69.62\n)\n\n => Array\n(\n => 14.143.254.58\n)\n\n => Array\n(\n => 223.179.143.236\n)\n\n => Array\n(\n => 223.179.143.249\n)\n\n => Array\n(\n => 103.143.7.54\n)\n\n => Array\n(\n => 223.179.139.106\n)\n\n => Array\n(\n => 39.40.219.90\n)\n\n => Array\n(\n => 45.115.141.231\n)\n\n => Array\n(\n => 120.29.100.33\n)\n\n => Array\n(\n => 112.196.132.5\n)\n\n => Array\n(\n => 202.163.123.153\n)\n\n => Array\n(\n => 5.62.58.146\n)\n\n => Array\n(\n => 39.53.216.113\n)\n\n => Array\n(\n => 42.111.160.73\n)\n\n => Array\n(\n => 107.182.231.213\n)\n\n => Array\n(\n => 119.82.94.120\n)\n\n => Array\n(\n => 178.62.34.82\n)\n\n => Array\n(\n => 203.122.6.18\n)\n\n => Array\n(\n => 157.42.38.251\n)\n\n => Array\n(\n => 45.112.68.222\n)\n\n => Array\n(\n => 49.206.212.122\n)\n\n => Array\n(\n => 104.236.70.228\n)\n\n => Array\n(\n => 42.111.34.243\n)\n\n => Array\n(\n => 84.241.19.186\n)\n\n => Array\n(\n => 89.187.180.207\n)\n\n => Array\n(\n => 104.243.212.118\n)\n\n => Array\n(\n => 104.236.55.136\n)\n\n => Array\n(\n => 106.201.16.163\n)\n\n => Array\n(\n => 46.101.40.25\n)\n\n => Array\n(\n => 45.118.166.94\n)\n\n => Array\n(\n => 49.36.128.102\n)\n\n => Array\n(\n => 14.142.193.58\n)\n\n => Array\n(\n => 212.79.124.176\n)\n\n => Array\n(\n => 45.32.191.194\n)\n\n => Array\n(\n => 105.112.107.46\n)\n\n => Array\n(\n => 106.201.14.8\n)\n\n => Array\n(\n => 110.93.240.65\n)\n\n => Array\n(\n => 27.96.95.177\n)\n\n => Array\n(\n => 45.41.134.35\n)\n\n => Array\n(\n => 180.151.13.110\n)\n\n => Array\n(\n => 101.53.242.89\n)\n\n => Array\n(\n => 115.186.3.110\n)\n\n => Array\n(\n => 171.49.185.242\n)\n\n => Array\n(\n => 115.42.70.24\n)\n\n => Array\n(\n => 45.128.188.43\n)\n\n => Array\n(\n => 103.140.129.63\n)\n\n => Array\n(\n => 101.50.113.147\n)\n\n => Array\n(\n => 103.66.73.30\n)\n\n => Array\n(\n => 117.247.193.169\n)\n\n => Array\n(\n => 120.29.100.94\n)\n\n => Array\n(\n => 42.109.154.39\n)\n\n => Array\n(\n => 122.173.155.150\n)\n\n => Array\n(\n => 45.115.104.53\n)\n\n => Array\n(\n => 116.74.29.84\n)\n\n => Array\n(\n => 101.50.125.34\n)\n\n => Array\n(\n => 45.118.166.80\n)\n\n => Array\n(\n => 91.236.184.27\n)\n\n => Array\n(\n => 113.167.185.120\n)\n\n => Array\n(\n => 27.97.66.222\n)\n\n => Array\n(\n => 43.247.41.117\n)\n\n => Array\n(\n => 23.229.16.227\n)\n\n => Array\n(\n => 14.248.79.209\n)\n\n => Array\n(\n => 117.5.194.26\n)\n\n => Array\n(\n => 117.217.205.41\n)\n\n => Array\n(\n => 114.79.169.99\n)\n\n => Array\n(\n => 103.55.60.97\n)\n\n => Array\n(\n => 182.75.89.210\n)\n\n => Array\n(\n => 77.73.66.109\n)\n\n => Array\n(\n => 182.77.126.139\n)\n\n => Array\n(\n => 14.248.77.166\n)\n\n => Array\n(\n => 157.35.224.133\n)\n\n => Array\n(\n => 183.83.38.27\n)\n\n => Array\n(\n => 182.68.4.77\n)\n\n => Array\n(\n => 122.177.130.234\n)\n\n => Array\n(\n => 103.24.99.99\n)\n\n => Array\n(\n => 103.91.127.66\n)\n\n => Array\n(\n => 41.90.34.240\n)\n\n => Array\n(\n => 49.205.77.102\n)\n\n => Array\n(\n => 103.248.94.142\n)\n\n => Array\n(\n => 104.143.92.170\n)\n\n => Array\n(\n => 219.91.157.114\n)\n\n => Array\n(\n => 223.190.88.22\n)\n\n => Array\n(\n => 223.190.86.232\n)\n\n => Array\n(\n => 39.41.172.80\n)\n\n => Array\n(\n => 124.107.206.5\n)\n\n => Array\n(\n => 139.167.180.224\n)\n\n => Array\n(\n => 93.76.64.248\n)\n\n => Array\n(\n => 65.216.227.119\n)\n\n => Array\n(\n => 223.190.119.141\n)\n\n => Array\n(\n => 110.93.237.179\n)\n\n => Array\n(\n => 41.90.7.85\n)\n\n => Array\n(\n => 103.100.6.26\n)\n\n => Array\n(\n => 104.140.83.13\n)\n\n => Array\n(\n => 223.190.119.133\n)\n\n => Array\n(\n => 119.152.150.87\n)\n\n => Array\n(\n => 103.125.130.147\n)\n\n => Array\n(\n => 27.6.5.52\n)\n\n => Array\n(\n => 103.98.188.26\n)\n\n => Array\n(\n => 39.35.121.81\n)\n\n => Array\n(\n => 74.119.146.182\n)\n\n => Array\n(\n => 5.181.233.162\n)\n\n => Array\n(\n => 157.39.18.60\n)\n\n => Array\n(\n => 1.187.252.25\n)\n\n => Array\n(\n => 39.42.145.59\n)\n\n => Array\n(\n => 39.35.39.198\n)\n\n => Array\n(\n => 49.36.128.214\n)\n\n => Array\n(\n => 182.190.20.56\n)\n\n => Array\n(\n => 122.180.249.189\n)\n\n => Array\n(\n => 117.217.203.107\n)\n\n => Array\n(\n => 103.70.82.241\n)\n\n => Array\n(\n => 45.118.166.68\n)\n\n => Array\n(\n => 122.180.168.39\n)\n\n => Array\n(\n => 149.28.67.254\n)\n\n => Array\n(\n => 223.233.73.8\n)\n\n => Array\n(\n => 122.167.140.0\n)\n\n => Array\n(\n => 95.158.51.55\n)\n\n => Array\n(\n => 27.96.95.134\n)\n\n => Array\n(\n => 49.206.214.53\n)\n\n => Array\n(\n => 212.103.49.92\n)\n\n => Array\n(\n => 122.177.115.101\n)\n\n => Array\n(\n => 171.50.187.124\n)\n\n => Array\n(\n => 122.164.55.107\n)\n\n => Array\n(\n => 98.114.217.204\n)\n\n => Array\n(\n => 106.215.10.54\n)\n\n => Array\n(\n => 115.42.68.28\n)\n\n => Array\n(\n => 104.194.220.87\n)\n\n => Array\n(\n => 103.137.84.170\n)\n\n => Array\n(\n => 61.16.142.110\n)\n\n => Array\n(\n => 212.103.49.85\n)\n\n => Array\n(\n => 39.53.248.162\n)\n\n => Array\n(\n => 203.122.40.214\n)\n\n => Array\n(\n => 117.217.198.72\n)\n\n => Array\n(\n => 115.186.191.203\n)\n\n => Array\n(\n => 120.29.100.199\n)\n\n => Array\n(\n => 45.151.237.24\n)\n\n => Array\n(\n => 223.190.125.232\n)\n\n => Array\n(\n => 41.80.151.17\n)\n\n => Array\n(\n => 23.111.188.5\n)\n\n => Array\n(\n => 223.190.125.216\n)\n\n => Array\n(\n => 103.217.133.119\n)\n\n => Array\n(\n => 103.198.173.132\n)\n\n => Array\n(\n => 47.31.155.89\n)\n\n => Array\n(\n => 223.190.20.253\n)\n\n => Array\n(\n => 104.131.92.125\n)\n\n => Array\n(\n => 223.190.19.152\n)\n\n => Array\n(\n => 103.245.193.191\n)\n\n => Array\n(\n => 106.215.58.255\n)\n\n => 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Array\n(\n => 103.224.54.190\n)\n\n => Array\n(\n => 49.36.141.210\n)\n\n => Array\n(\n => 5.62.63.218\n)\n\n => Array\n(\n => 39.59.9.18\n)\n\n => Array\n(\n => 111.88.86.45\n)\n\n => Array\n(\n => 178.54.139.5\n)\n\n => Array\n(\n => 116.68.105.241\n)\n\n => Array\n(\n => 119.160.96.187\n)\n\n => Array\n(\n => 182.189.192.103\n)\n\n => Array\n(\n => 119.160.96.143\n)\n\n => Array\n(\n => 110.225.89.98\n)\n\n => Array\n(\n => 169.149.195.134\n)\n\n => Array\n(\n => 103.238.104.54\n)\n\n => Array\n(\n => 47.30.208.142\n)\n\n => Array\n(\n => 157.46.179.209\n)\n\n => Array\n(\n => 223.235.38.119\n)\n\n => Array\n(\n => 42.106.180.165\n)\n\n => Array\n(\n => 154.122.240.239\n)\n\n => Array\n(\n => 106.223.104.191\n)\n\n => Array\n(\n => 111.93.110.218\n)\n\n => Array\n(\n => 182.183.161.171\n)\n\n => Array\n(\n => 157.44.184.211\n)\n\n => Array\n(\n => 157.50.185.193\n)\n\n => Array\n(\n => 117.230.19.194\n)\n\n => Array\n(\n => 162.243.246.160\n)\n\n => Array\n(\n => 106.223.143.53\n)\n\n => Array\n(\n => 39.59.41.15\n)\n\n => Array\n(\n => 106.210.65.42\n)\n\n => Array\n(\n => 180.243.144.208\n)\n\n => Array\n(\n => 116.68.105.22\n)\n\n => Array\n(\n => 115.42.70.46\n)\n\n => Array\n(\n => 99.72.192.148\n)\n\n => Array\n(\n => 182.183.182.48\n)\n\n => Array\n(\n => 171.48.58.97\n)\n\n => Array\n(\n => 37.120.131.188\n)\n\n => Array\n(\n => 117.99.167.177\n)\n\n => Array\n(\n => 111.92.76.210\n)\n\n => Array\n(\n => 14.192.144.245\n)\n\n => Array\n(\n => 169.149.242.87\n)\n\n => Array\n(\n => 47.30.198.149\n)\n\n => Array\n(\n => 59.103.57.140\n)\n\n => Array\n(\n => 117.230.161.168\n)\n\n => Array\n(\n => 110.225.88.173\n)\n\n => Array\n(\n => 169.149.246.95\n)\n\n => Array\n(\n => 42.106.180.52\n)\n\n => Array\n(\n => 14.231.160.157\n)\n\n => Array\n(\n => 123.27.109.47\n)\n\n => Array\n(\n => 157.46.130.54\n)\n\n => Array\n(\n => 39.42.73.194\n)\n\n => Array\n(\n => 117.230.18.147\n)\n\n => Array\n(\n => 27.59.231.98\n)\n\n => Array\n(\n => 125.209.78.227\n)\n\n => Array\n(\n => 157.34.80.145\n)\n\n => Array\n(\n => 42.201.251.86\n)\n\n => Array\n(\n => 117.230.129.158\n)\n\n => Array\n(\n => 103.82.80.103\n)\n\n => Array\n(\n => 47.9.171.228\n)\n\n => Array\n(\n => 117.230.24.92\n)\n\n => Array\n(\n => 103.129.143.119\n)\n\n => Array\n(\n => 39.40.213.45\n)\n\n => Array\n(\n => 178.92.188.214\n)\n\n => Array\n(\n => 110.235.232.191\n)\n\n => Array\n(\n => 5.62.34.18\n)\n\n => Array\n(\n => 47.30.212.134\n)\n\n => Array\n(\n => 157.42.34.196\n)\n\n => Array\n(\n => 157.32.169.9\n)\n\n => Array\n(\n => 103.255.4.11\n)\n\n => Array\n(\n => 117.230.13.69\n)\n\n => Array\n(\n => 117.230.58.97\n)\n\n)\n```\nArchive for August, 2010: Tightwad Kitty's \\$\\$\\$ Stretching Blog\n << Back to all Blogs Login or Create your own free blog Layout: Blue and Brown (Default) Author's Creation\nHome > Archive: August, 2010",
null,
"",
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"",
null,
"# Archive for August, 2010\n\n## Put extra photos in Pic posts while it's working\n\nAugust 31st, 2010 at 02:12 am\n\nPut extra photos in Pic posts while it's working!\n\nI have added more photos to some of my posts that I have photos in.\n\nLook up pic for quick look.\n\nSome ramdom photos from Ekka",
null,
"Quail & Pheasant",
null,
"Bark painting",
null,
"Start of main Sideshow Ally",
null,
"Cermic plate",
null,
"wombat\n\n## Trip to a Garage Sale\n\nAugust 30th, 2010 at 07:07 pm\n\nFinding things that I can afford to buy in charity op shops now days is rare as they have put up their prices up to nearly new or prices you would pay on Ebay or Antique stores, in fact some Antiques store are cheaper than some of our charity shops for the same things. I don’t buy that much collectible but I do look at them. Now I have started to go back to looking at Garage/Yard Sales. Just last weekend I went to one, the lady was quite nice and had the prices at reasonable inexpensive level.\n\nI did buy from her as she was planning to leave the unsold stuff for the taking next day, I bought a black handbag hardly used, two money wallets in black hardly used, Gold Gomesh Keyring never used that I will put in my gift box for the future, tartan scarf, two teaspoons, one a collector item that will go with my collection of stuff from my old job. What else oh! Black & silver rolled neck jumper classic style that I can wear out at some time, Pair of cargo shorts in my size for summer, cookbook, and small brown ramekin dish as brown is not in at the moment. For me colour doesn’t count but size and what I can use it for does. I have a got a lot of bit & pieces in the way brown dishes so it will match anyway. This is how I can get things a lot cheaper as I don’t worry about matching colours nor get rid of things because the colour doesn’t match. It’s just a Rustic country style. Well anyway I just paid \\$7.00 for all the above and the real retail cost new would have been around \\$150 mark.\n\n## Financial summery of the past month\n\nAugust 28th, 2010 at 08:53 pm\n\nIt’s time to start again on reviewing my monthly daily living allowance.\n\nI have been spending a little more on some books and cookware over the past few months than I really should on my monthly allowance but it’s spent now so I will to have work around that and make up for it over time. I am still just under my budgeted allowance so I am on track.\n\nThese are the categories for this challenge.\n\nItems === Approx Budget===Spent\n\nEating out===\\$35.00=====\\$42.05\nGroceries===\\$135.50==== \\$109.77\nNon-Food===\\$17.00=====\\$4.29\nClothes ====\\$35.00=====\\$69.95\nGrooming===\\$15.00\nGas======\\$25.00\nPets====== \\$50.00=== ==\\$7.47\nHH items===\\$42.50======\\$117.39\nFares== ====\\$9.00 =====\\$20.00\nEntrainments= \\$12.00\nGarden=====\\$10.00=====\\$19.96\nMedical=====\\$15.00\nMisc======\\$9.00\nPersonal Allowance=\\$85.00 ==\\$75.39\nNote only one amount is no spend\n\nBudgeted for this period \\$530 spent \\$502.27\n\nNo Spending days over the 28 days = 14 days in total.\n\nI have to still line up some painters for quotes the outside of my house over the next year this should be my major expense if all goes to normal here. I think I just like having the money in the bank and not spending it. I do know that it will make me save a lot more when I do spend it as I don’t like not having money in the bank as safety blanket for me.\n\nAs for banking in the past month I have saved over \\$200 in cash so you still can save money on a pension but need to make choices on where you spend your money. As for saving money on things I will save that for another blog.\n\n## Rebuilding showground\n\nAugust 27th, 2010 at 08:25 pm\n\nI got very tired last time I was there, so I went again to look at some of the place that I missed then.\n\nI found out that Showbag pavilion, horse stables and Dog show pavilion are the ones getting pull down next to start their makeover it will take over 15 years to do and many areas will be moved around as they progress in this makeover. So each year some thing will be in a new place and it may be a temporary home only for a few years. I may never see it finished in my lifetime. They are planning to have open it all year not just for the 10 days at EKKA show time. The grounds has it’s own railway station which they are hoping have open all year around but it on a bypass ring loop so if they get a new line underground built that in the pipeline they may link in with it and its next door the major hospital which has a bus station running pass it. Major roads bypass and tunnels have been built near this site so it a site ready for big development but lucky for the show grounds, they are ones doing the development for next 100 years plan which includes the Show (EKKA) only 3km from CBD. It you take up and buying into any of the building sites planned its taken as granted that you can’t complain about any noise or people in the area as it’s in a special noise zone area. Parking will be another issue too with only 1.5 car spaces for each unit only.\n\nThey are planning to build on out side of the showgrounds and go up in the centre for the pavilions\n\nEnough rambling on for now until next year!\n\nSome pictures",
null,
"CBD in the background",
null,
"Planning\n\n## Photo Test Run didn’t work.\n\nAugust 14th, 2010 at 09:21 pm\n\nIt looks like I need to go back to writing without pictures full time. I did a test run with one lot and it down again this morning.\n\nWill try to put up new post today.\n\n## Some Fruit & Vegetables\n\nAugust 11th, 2010 at 05:32 am\n\nThese displays are some of interesting ones this year.",
null,
"",
null,
"",
null,
"## Iced Chocolate Cake\n\nAugust 10th, 2010 at 05:18 am\n\nFirst prize EKKA in under 15 - (A 11 year olds winner entry)\n1 layer cake\n\n90g butter (3oz), softened\n2 eggs\n1 cup self raising flour\n1 cup sugar\n1/2 cup milk\n2 tablespoons cocoa powder (8 teaspoons)\n\nGrease or line a 20cm (8 inch) ring tin with baking paper.\nBeat ingredients together for 4 minutes.\nCook at 190C (375F) for 35 to 40 minutes\n\nNB: Most likely beaten in a cake mixer or processer the recipe doesn't say.\nCup are 250ml and tablespoons are (20ml) or 4 tsp.\n\nSource - The Courier-Mail\n\n## Who said Kids can’t Cook!\n\nAugust 9th, 2010 at 08:14 pm\n\nWe have coming up a MasterChef series with children cooking soon! After watching three series of MasterChef, 2 full series and Celerity Masterchef now all the children wanted have ago too!\n\nIf these photos are anything to go by the next generation will be able to cook or not kept out of the kitchen and may be less obese children too!",
null,
"This one is a mini-masterclass for 5-10 year old. It’s a taken on day when primary school children were taken to the show as class lesson.",
null,
"This one is a picture of some cake icing in a new category for 6 -10 year olds. Only very best were shown and won by two 8 year olds who mothers are cake- icing teachers.",
null,
"The entries on the bottom of this picture were done by next age group 11 -15 year olds.\n\nHeres A few more Pictures\n\nChildern School Work Competions",
null,
"",
null,
"## Some more photos of the EKKA!\n\nAugust 8th, 2010 at 07:13 pm\n\nI went to the first day of\n\nEKKA as it’s commonly known around here for the last 135 years. The showground are on prime real estate, just a few kilometres from the CBD. They are changing it a bit and putting in a few apartment blocks, shopping area & restaurants which will become part of the EKKA when the show is on. They will have to put up with a full EKKA show each August as the EKKA is not moving at all and any extra events during the year as it’s hired out for events too! It will take time to make everyone happy but they will not kill the show. I didn’t see the new plans for the grounds as yet! I do know that I wouldn’t travel to another area if they did move it.",
null,
"",
null,
"This picture is of the show bag pavilion which was to be pulled down this year so it may not be around much longer.",
null,
"This one is of one the streets at show time. Only one main road is closed during the show. Others are more like side streets within the showgrounds.",
null,
"Looking up the closed Main Road",
null,
"This one is antique farm machines with the small children sideshow ally in the background.",
null,
"Start of Main Sideshow Ally\nThe main one takes up a quarter of the showgrounds and can be expensive. So far I haven’t ventured into that area and some years I never get that far!\n\n## What do these photos have in common?\n\nAugust 7th, 2010 at 09:21 am",
null,
"",
null,
"Sugar\n\nAnother go at keeping these photos here.\n\n## Is Canned Tuna Frugal?\n\nAugust 4th, 2010 at 09:34 pm\n\nI fine that canned tuna is one of my frugal proteins as I have posted recipes a few times. It's easy to use and an economical here. We can buy it in 3 sizes 95g, 180g & 420g most are just plain tuna in brine, springwater or oil in Chunky or flaked. There are numerous recipes printed for using canned tuna.\n\nPasta & Tuna - 1/2 can flavoured tuna, 1/4 cup tomato sauce mix- balance of tuna used for lunch with lavish crackers\n\nTuna Pasta\n\n1 can 95g Tuna - Flavoured Tomato & Onion or other flavour\n2 cups of spiral pasta\n1/2 jar of pasta sauce\nDried parsley\nSeasoning\n\nCook pasta as instructed\nMix tuna & pasta sauce into a microwave dish, add seasoning & dried parsley to taste and heat until hot.\nWhen pasta is cooked drain, and reserves a little pasta water to pour back over the pasta and add tuna pasta sauce mixture to pasta and serve on 2 spaghetti bowls.\n\nServe 2\nFor one serving divide in half.\n\nTuna Recipes posted so far here.\n\nEasy Tuna Dish\n\nUpdate on -Easy Tuna Dish\n\nTuna Mornay\n\nTuna with macaroni and peas in the comments section\n\nAugust 3rd, 2010 at 12:27 am\n\n200g -250g frozen fresh broad beans\n20g butter\n1 tsp herbes de Provence or other dried herbs\n2 tomatoes, peeled, seeded and diced or 1/2 can tomatoes drained and diced\nSalt and pepper\n\n1. Cook the broad beans in boiling salted water until they are just tender, about 8 minutes.\n2. Drain and refresh the beans under cold water. Peel off the outer skins, if preferred.\n3. Melt the butter in a large frying pan and add the beans, together with the herbes de Provence.\n4. Add the tomatoes, salt and pepper. Heat through, stirring continuously.\nServe immediately\n\nServes 3\n\nNote: Remove any yellow beans use only green ones, do a taste test.\nI used a Spicy Italian Mixed Herbs that I have in the cupboard and kept it the fridge for about 3 days after cooking also reheating the amount that I was eating that night in the microwave.\n\nHerbs de Provence\n\n1 teaspoon thyme\n1 teaspoon summer savory\n1/2 teaspoon lavender\n1/4 teaspoon rosemary\n1/2 teaspoon oregano or basil\n1/4 teaspoon sage\n\nHerbs de provence is best made with dried herbs as fresh herbs lose their flavour if the cooking is longer than about 20 minutes. This blend is excellent in soups, on potatoes, rice, pasta, fish, roasted vegetables or bread. Mix with 1/4 lb butter for a real treat.\n\n## Full Freezer again!\n\nAugust 2nd, 2010 at 09:09 am\n\nBoth my fridge and freezers are full as my friend as cleaned her freezer out and we went shopping today and she bought me a cooked chicken and 2 T-bone steaks as well buying her self the same. I got 4 x 1kg each bags of frozen fresh broad beans & large bag another kind frozen fresh bean. I did give two kilos away of broad beans and re-bag the balance into 200g portions which will give you about 3 servings when cooked. I found an interesting recipe that I liked for them. I will post the recipe later."
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https://sciencing.com/calculate-voltage-drop-across-resistors-6128036.html | [
"# How to Calculate a Voltage Drop Across Resistors",
null,
"••• LightFieldStudios/iStock/GettyImages\n\nA simple electrical circuit contains a source of voltage (a power supply, such as a battery, generator or the utility wires coming into your building), a wire to carry current in the form of electrons, and a source of electrical resistance. In reality, such circuits are rarely simple and include a number of branching and re-joining points.\n\n• Voltage (V) is measured in volts (the symbol is also V); current (I) is measured in amperes or \"amps\" (A); and resistance (R) is measured in ohms (Ω).\n\nAlong the branches, and sometimes along the main trunk of the circuit, items such as household appliances (lamps, refrigerators, television sets) are placed, each drawing current to keep itself going. But what exactly happens to the voltage and current within a given electrical circuit set-up from a physics standpoint when each resistor is encountered and the voltage \"drops\"?\n\n## Electrical Circuit Basics\n\nOhm's law states that current flow is voltage divided by resistance. This can apply to a circuit as a whole, an isolated set of branches or to a single resistor, as you'll see. The most common form of this law is written:\n\nV = IR\n\nCircuits can be arranged in two basic ways.\n\nSeries circuit: Here, current flows entirely along one path, through a single wire. Whatever resistances current encounters along the way simply add up to give the total resistance of the circuit as a whole:\n\nRS = R1 + R2 + ... + RN (series circuit)\n\nParallel circuit: In this case, a primary wire branches (shown as right angles) into two or more other wires, each with its own resistor. In this case, the total resistance is given by:\n\n1/RP = 1/R1 + 1/R2 + ... + 1/RN (parallel circuit)\n\nIf you explore this equation, you find that by adding the resistances of the same magnitude, you decrease the resistance of the circuit as a whole. (Picking 1 ohm, or 1 Ω, makes the math easier.) By Ohm's law, this actually increases the current!\n\nIf this seems counterintuitive, imagine the flow of cars on a busy highway served by a single tollbooth that backs up traffic for a mile, and then imagine the same scenario with four more tollbooths identical to the first. This will plainly increase the flow of cars despite technically adding resistance.\n\n## Voltage Drop: Series Circuit\n\nIf you want to find voltage drops across individual resistors in a series, you proceed as follows:\n\n1. Calculate the total resistance by adding the individual R values.\n2. Calculate the current in the circuit, which is the same across each resistor since there is only one wire in the circuit.\n3. Calculate the voltage drop across each resistor using Ohm's law.\n\nExample: A 24-V power source and three resistors are connected in series with R1= 4 Ω, R2= 2 Ω and R3 = 6 Ω. What is the voltage drop across each resistor?\n\nFirst, calculate total resistance: 4 + 2 + 6 = 12 Ω\n\nNext, calculate the current: 24 V/12 Ω = 2 A\n\nNow, use the current to calculate the voltage drop across each resistor. Using V = IR for each, the values of R1, R2 and R3 are 8 V, 4 V and 12 V.\n\n## Voltage Drop: Parallel Circuit\n\nExample: A 24-V power source and three resistors are connected in parallel with R1= 4 Ω, R2= 2 Ω and R3 = 6 Ω, as before. What is the voltage drop across each resistor?\n\nIn this case, the story is simpler: Regardless of the resistance value, the voltage drop across each resistor is the same, making the current the variable that differs across resistors in this case. This means that the voltage drop across each is just the total voltage of the circuit divided by the number of resistors in the circuit, or 24 V/3 = 8 V.\n\n## Resistor Voltage Drop Calculator\n\nSee the Resources for an example of an instance in which you can use an automatic tool to calculate the voltage drop in a kind of circuit arrangement called a voltage divider.\n\nDont Go!\n\nWe Have More Great Sciencing Articles!"
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https://kishorekaranam.tk/blog/oct12/ | [
"# Algorithms 101: Sorting II October 12, 2020\n\nThese three remaining sorting algorithms are better af than the previous three and more complicated af to implement. That said these three algorithms excel where the previous three didn't that is, in sorting the lists efficiently. Though the implementations I'm doing are too basic, these algorithms can be modified to accommodate the sorting involving huge data and huge in the sense millions and billions of items!!\n\nMerge Sort",
null,
"It's too easy to understand but a little complicated than the previous when it comes to the implementation. So, how merge sort works? It just breaks an array of elements into its components and then pairs them up by arranging them in their proper respective places and rejoins them into one sorted array. That said, the time complexity of merge sort is O(n log n) and this is the case at worst-case too thereby making this a powerful algorithm. This algorithm can kick others out when it comes to handling a large amount of data by performing significantly faster than the previous ones.\n\nPseudocode:\n``` ```\nmethod mergesort(arr)\nif(arr.length==1)\nreturn arr\narr1...arr1[n/2]\narr2[n/2+1]...arr2[n]\narr1 = mergesort(arr1)\narr2 = mergesort(arr2)\nreturn merge(arr1,arr2)\nend mergesort\nmethod merge(arr1, arr2)\ntake arr3\nwhile(arr1 and arr2 have elements)\nif(arr1 > arr2)\nadd arr2 to end of arr3\nremove arr2 from arr2\nelse\nadd arr1 to end of arr3\nremove arr1 from arr1\nend if\nend while\nwhile(arr1 has elements)\nadd arr1 to end of arr3\nremove arr1 from arr1\nend while\nwhile(arr2 has elements)\nadd arr2 to end of arr3\nremove arr2 from arr2\nend while\nreturn arr3\n```\n```\nImplementation in Java:\n\nQuickSort",
null,
"Quick sort is effective when it comes to its speed but in its base form, its not quite stable enough. It has the worst-case time complexity of O(n²) but worst-case is rarely the case for quick sort thereby making it faster than the previous algorithms. It's in general time complexity is O(n log n) thus it can be faster and more efficient when used in proper scenarios. That said, quick sort can be a little confusing to understand which is troublesome. In quick sort we first select a pivot element in the array and place the rest of the numbers accoring to this pivot element, that is smaller ones before the pivot and larger ones after. So, when this is done we have the array with pivot in middle and left small ones and right large ones, both of which are yet to be sorted. Now we chose the new pivot numbers in both ends and again repeat the process until the array is sorted.\n\nPseudocode:\n``` ```\nmethod quckSort(arr, low, high)\nif(low < high)\nmiddleElement = low+(high-low)/2\npivotElement = arr[middleElement]\nquickSort(arr, low, middleElement-1)\nquickSort(arr, middleElement+1, high)\nend\n```\n```\nImplementation in Java:\n\nLet's take an array consisting of elements {7,8,12,11,2,19,4,62}. The code will be as follows:\n\nHeapSort",
null,
"This guy is hard of all the algorithms but has the time complexity of O(n log n) at the worst-case meaning even at worst this guy kicks ass of the rest. That said, heap sort uses a heap which is a special kind of binary tree where the nodes are greater than their children making the root the largest of all. To make it simple basically The Heapsort algorithm involves preparing the list by first turning it into a max heap. The algorithm then repeatedly swaps the first value of the list with the last value, decreasing the range of values considered in the heap operation by one, and shifting the new first value into its position in the heap. This algorithm needs additional reading to understand what's what. So, the best would be to refer the gog.\n\nPseudocode:\n``` ```\nHeapsort(arr)\nBuildHeap(arr)\nfor i = n to 1\nswap(arr, arr[i])\nn = n - 1\nHeapify(arr, 1)\n\nBuildHeap(arr)\nn = elements_in(arr)\nfor i = floor(n/2) to 1\nHeapify(arr,i,n)\nHeapify(arr, i, n)\nleft = 2i\nright = 2i+1\nif (left <= n) and (A[left] > A[i])\nmax = left\nelse\nmax = i\nif (right<=n) and (A[right] > A[max])\nmax = right\nif (max != i)\nswap(arr[i], arr[max])\nHeapify(arr, max)\n```\n```\nImplementation in Java:\n\nLet's take an array consisting of elements {2,8,5,3,9,1}. The code will be as follows:\n\nSo, this winds up all the sorting algorithms!! These three algorithms are very hard but are much preferred by the developers all around whenever sorting is required. Though I've done the in general implementations of all these algorithms the main tricky part will be choosing them in the right scenario and using them according to that scenario.\n\nLater."
]
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"https://kishorekaranam.tk/blog/oct12/ms.gif",
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"https://kishorekaranam.tk/blog/oct12/qs.gif",
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"https://kishorekaranam.tk/blog/oct12/hs.gif",
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https://optimized-einsum.readthedocs.io/en/stable/reusing_paths.html | [
"# Reusing Paths¶\n\nIf you expect to use a particular contraction repeatedly, it can make things simpler and more efficient not to compute the path each time. Instead, supplying contract_expression() with the contraction string and the shapes of the tensors generates a ContractExpression which can then be repeatedly called with any matching set of arrays. For example:\n\n>>> my_expr = oe.contract_expression(\"abc,cd,dbe->ea\", (2, 3, 4), (4, 5), (5, 3, 6))\n>>> print(my_expr)\n<ContractExpression('abc,cd,dbe->ea')>\n1. 'dbe,cd->bce' [GEMM]\n2. 'bce,abc->ea' [GEMM]\n\n\nThe ContractExpression can be called with 3 arrays that match the original shapes without having to recompute the path:\n\n>>> x, y, z = (np.random.rand(*s) for s in [(2, 3, 4), (4, 5), (5, 3, 6)])\n>>> my_expr(x, y, z)\narray([[ 3.08331541, 4.13708916],\n[ 2.92793729, 4.57945185],\n[ 3.55679457, 5.56304115],\n[ 2.6208398 , 4.39024187],\n[ 3.66736543, 5.41450334],\n[ 3.67772272, 5.46727192]])\n\n\nNote that few checks are performed when calling the expression, and while it will work for a set of arrays with the same ranks as the original shapes but differing sizes, it might no longer be optimal.\n\n## Specifying Constants¶\n\nOften one generates contraction expressions where some of the tensor arguments will remain constant across many calls. contract_expression() allows you to specify the indices of these constant arguments, allowing opt_einsum to build and then reuse as many constant contractions as possible. Take for example the equation:\n\n>>> eq = \"ij,jk,kl,lm,mn->ni\"\n\n\nwhere we know that only the first and last tensors will vary between calls. We can specify this by marking the middle three as constant - we then need to supply the actual arrays rather than just the shapes to contract_expression():\n\n>>> # A B C D E\n>>> shapes = [(9, 5), (5, 5), (5, 5), (5, 5), (5, 8)]\n\n>>> # mark the middle three arrays as constant\n>>> constants = [1, 2, 3]\n\n>>> # generate the constant arrays\n>>> B, C, D = [np.random.randn(*shapes[i]) for i in constants]\n\n>>> # supplied ops are now mix of shapes and arrays\n>>> ops = (9, 5), B, C, D, (5, 8)\n\n>>> expr = oe.contract_expression(eq, *ops, constants=constants)\n>>> expr\n<ContractExpression('ij,[jk,kl,lm],mn->ni', constants=[1, 2, 3])>\n\n\nThe expression now only takes the remaining two arrays as arguments (the tensors with 'ij' and 'mn' indices), and will store as many reusable constant contractions as possible.\n\n>>> A1, E1 = np.random.rand(*shapes), np.random.rand(*shapes[-1])\n>>> out1 = expr(A1, E1)\n>>> out1.shap\n(8, 9)\n\n>>> A2, E2 = np.random.rand(*shapes), np.random.rand(*shapes[-1])\n>>> out2 = expr(A2, E2)\n>>> out2.shape\n(8, 9)\n\n>>> np.allclose(out1, out2)\nFalse\n\n>>> print(expr)\n<ContractExpression('ij,[jk,kl,lm],mn->ni', constants=[1, 2, 3])>\n1. 'jm,mn->jn' [GEMM]\n2. 'jn,ij->ni' [GEMM]\n\n\nWhere we can see that the expression now only has to perform two contractions to compute the output.\n\nNote\n\nThe constant part of an expression is lazily generated upon the first call (specific to each backend), though it can also be explicitly built by calling evaluate_constants().\n\nWe can confirm the advantage of using expressions and constants by timing the following scenarios, first setting A = np.random.rand(*shapes) and E = np.random.rand(*shapes[-1]).\n\n• contract from scratch:\n\n>>> %timeit oe.contract(eq, A, B, C, D, E)\n239 µs ± 5.06 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)\n\n• contraction with an expression but no constants:\n\n>>> expr_no_consts = oe.contract_expression(eq, *shapes)\n>>> %timeit expr_no_consts(A, B, C, D, E)\n76.7 µs ± 2.47 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n\n• contraction with an expression and constants marked:\n\n>>> %timeit expr(A, E)\n40.8 µs ± 1.22 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)\n\n\nAlthough this gives us a rough idea, of course the efficiency savings are hugely dependent on the size of the contraction and number of possible constant contractions.\n\nWe also note that even if there are no constant contractions to perform, it can be very advantageous to specify constant tensors for particular backends. For instance, if a GPU backend is used, the constant tensors will be kept on the device rather than being transferred each time."
]
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https://w3schoolsrus.github.io/php/func_string_fprintf.html | [
"THE WORLD'S LARGEST WEB DEVELOPER SITE\n\n# PHP fprintf() Function\n\n❮ PHP String Reference\n\n### Example\n\nWrite some text to a text file named \"test.txt\":\n\n<?php\n\\$number = 9;\n\\$str = \"Beijing\";\n\\$file = fopen(\"test.txt\",\"w\");\necho fprintf(\\$file,\"There are %u million bicycles in %s.\",\\$number,\\$str);\n?>\n\nThe output of the code above will be:\n\n40\n\nThe following text will be written to the file \"test.txt\":\n\nThere are 9 million bicycles in Beijing.\n\n## Definition and Usage\n\nThe fprintf() function writes a formatted string to a specified output stream (example: file or database).\n\nThe arg1, arg2, ++ parameters will be inserted at percent (%) signs in the main string. This function works \"step-by-step\". At the first % sign, arg1 is inserted, at the second % sign, arg2 is inserted, etc.\n\nNote: If there are more % signs than arguments, you must use placeholders. A placeholder is inserted after the % sign, and consists of the argument- number and \"\\\\$\". See example two.\n\nTip: Related functions: printf(), sprintf(), vprintf(), vsprintf() and vfprintf()\n\n## Syntax\n\nfprintf(stream,format,arg1,arg2,arg++)\nParameter Description\nstream Required. Specifies where to write/output the string\nformat Required. Specifies the string and how to format the variables in it.\n\nPossible format values:\n\n• %% - Returns a percent sign\n• %b - Binary number\n• %c - The character according to the ASCII value\n• %d - Signed decimal number (negative, zero or positive)\n• %e - Scientific notation using a lowercase (e.g. 1.2e+2)\n• %E - Scientific notation using a uppercase (e.g. 1.2E+2)\n• %u - Unsigned decimal number (equal to or greather than zero)\n• %f - Floating-point number (local settings aware)\n• %F - Floating-point number (not local settings aware)\n• %g - shorter of %e and %f\n• %G - shorter of %E and %f\n• %o - Octal number\n• %s - String\n• %x - Hexadecimal number (lowercase letters)\n• %X - Hexadecimal number (uppercase letters)\n\nAdditional format values. These are placed between the % and the letter (example %.2f):\n\n• + (Forces both + and - in front of numbers. By default, only negative numbers are marked)\n• ' (Specifies what to use as padding. Default is space. Must be used together with the width specifier. Example: %'x20s (this uses \"x\" as padding)\n• - (Left-justifies the variable value)\n• [0-9] (Specifies the minimum width held of to the variable value)\n• .[0-9] (Specifies the number of decimal digits or maximum string length)\n\nNote: If multiple additional format values are used, they must be in the same order as above.\n\narg1 Required. The argument to be inserted at the first %-sign in the format string\narg2 Optional. The argument to be inserted at the second %-sign in the format string\narg++ Optional. The argument to be inserted at the third, fourth, etc. %-sign in the format string\n\n## Technical Details\n\nReturn Value: Returns the length of the written string 5+\n\n## More Examples\n\n### Example\n\nWrite some text to a file:\n\n<?php\n\\$number = 123;\n\\$file = fopen(\"test.txt\",\"w\");\nfprintf(\\$file,\"%f\",\\$number);\n?>\n\nThe following text will be written to the file \"test.txt\":\n\n123.000000\n\n### Example\n\nUse of placeholders:\n\n<?php\n\\$number = 123;\n\\$file = fopen(\"test.txt\",\"w\");\nfprintf(\\$file,\"With 2 decimals: %1\\\\$.2f\n\\nWith no decimals: %1\\\\$u\",\\$number);\n?>\n\nThe following text will be written to the file \"test.txt\":\n\nWith 2 decimals: 123.00\nWith no decimals: 123\n\n### Example\n\nUsing printf() to demonstrate all possible format values:\n\n<?php\n\\$num1 = 123456789;\n\\$num2 = -123456789;\n\\$char = 50; // The ASCII Character 50 is 2\n\n// Note: The format value \"%%\" returns a percent sign\nprintf(\"%%b = %b <br>\",\\$num1); // Binary number\nprintf(\"%%c = %c <br>\",\\$char); // The ASCII Character\nprintf(\"%%d = %d <br>\",\\$num1); // Signed decimal number\nprintf(\"%%d = %d <br>\",\\$num2); // Signed decimal number\nprintf(\"%%e = %e <br>\",\\$num1); // Scientific notation (lowercase)\nprintf(\"%%E = %E <br>\",\\$num1); // Scientific notation (uppercase)\nprintf(\"%%u = %u <br>\",\\$num1); // Unsigned decimal number (positive)\nprintf(\"%%u = %u <br>\",\\$num2); // Unsigned decimal number (negative)\nprintf(\"%%f = %f <br>\",\\$num1); // Floating-point number (local settings aware)\nprintf(\"%%F = %F <br>\",\\$num1); // Floating-point number (not local settings aware)\nprintf(\"%%g = %g <br>\",\\$num1); // Shorter of %e and %f\nprintf(\"%%G = %G <br>\",\\$num1); // Shorter of %E and %f\nprintf(\"%%o = %o <br>\",\\$num1); // Octal number\nprintf(\"%%s = %s <br>\",\\$num1); // String\nprintf(\"%%x = %x <br>\",\\$num1); // Hexadecimal number (lowercase)\nprintf(\"%%X = %X <br>\",\\$num1); // Hexadecimal number (uppercase)\nprintf(\"%%+d = %+d <br>\",\\$num1); // Sign specifier (positive)\nprintf(\"%%+d = %+d <br>\",\\$num2); // Sign specifier (negative)\n?>\nTry it Yourself »\n\n❮ PHP String Reference"
]
| [
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.56961274,"math_prob":0.98260814,"size":3568,"snap":"2020-24-2020-29","text_gpt3_token_len":993,"char_repetition_ratio":0.17592593,"word_repetition_ratio":0.08239701,"special_character_ratio":0.3323991,"punctuation_ratio":0.19883041,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9877323,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-05T11:42:05Z\",\"WARC-Record-ID\":\"<urn:uuid:06f111bf-a2d6-4e14-9842-7264bc8d162e>\",\"Content-Length\":\"109133\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:233178a6-0ed1-4a90-a2f2-fc0b9b12d90d>\",\"WARC-Concurrent-To\":\"<urn:uuid:50798361-5476-47c4-bb15-54285aee766c>\",\"WARC-IP-Address\":\"185.199.111.153\",\"WARC-Target-URI\":\"https://w3schoolsrus.github.io/php/func_string_fprintf.html\",\"WARC-Payload-Digest\":\"sha1:Q7TOMH7AFS7LNA6ZVXDG2QFNOZMFMW2B\",\"WARC-Block-Digest\":\"sha1:ILGALNCMMQTVZV3TOAQBRTVFPD23DCC6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655887319.41_warc_CC-MAIN-20200705090648-20200705120648-00307.warc.gz\"}"} |
https://courses.lumenlearning.com/ivytech-collegealgebra/chapter/describe-local-behavior-of-polynomial-functions/ | [
"## Identifying Local Behavior of Polynomial Functions\n\nIn addition to the end behavior of polynomial functions, we are also interested in what happens in the “middle” of the function. In particular, we are interested in locations where graph behavior changes. A turning point is a point at which the function values change from increasing to decreasing or decreasing to increasing.",
null,
"Figure 10\n\nWe are also interested in the intercepts. As with all functions, the y-intercept is the point at which the graph intersects the vertical axis. The point corresponds to the coordinate pair in which the input value is zero. Because a polynomial is a function, only one output value corresponds to each input value so there can be only one y-intercept $\\left(0,{a}_{0}\\right).$ The x-intercepts occur at the input values that correspond to an output value of zero. It is possible to have more than one x-intercept.\n\n### A General Note: Intercepts and Turning Points of Polynomial Functions\n\nA turning point of a graph is a point at which the graph changes direction from increasing to decreasing or decreasing to increasing. The y-intercept is the point at which the function has an input value of zero. The x-intercepts are the points at which the output value is zero.\n\n### How To: Given a polynomial function, determine the intercepts.\n\n1. Determine the y-intercept by setting $x=0$ and finding the corresponding output value.\n2. Determine the x-intercepts by solving for the input values that yield an output value of zero.\n\n### Example 8: Determining the Intercepts of a Polynomial Function\n\nGiven the polynomial function $f\\left(x\\right)=\\left(x - 2\\right)\\left(x+1\\right)\\left(x - 4\\right),\\\\$ written in factored form for your convenience, determine the y– and x-intercepts.\n\n### Solution\n\nThe y-intercept occurs when the input is zero so substitute 0 for x.\n\n$\\begin{cases}f\\left(0\\right)=\\left(0 - 2\\right)\\left(0+1\\right)\\left(0 - 4\\right)\\hfill \\\\ \\text{ }=\\left(-2\\right)\\left(1\\right)\\left(-4\\right)\\hfill \\\\ \\text{ }=8\\hfill \\end{cases}\\\\$\n\nThe y-intercept is (0, 8).\n\nThe x-intercepts occur when the output is zero.\n\n$\\begin{cases}\\text{ }0=\\left(x - 2\\right)\\left(x+1\\right)\\left(x - 4\\right)\\hfill \\\\ x - 2=0\\hfill & \\hfill & \\text{or}\\hfill & \\hfill & x+1=0\\hfill & \\hfill & \\text{or}\\hfill & \\hfill & x - 4=0\\hfill \\\\ \\text{ }x=2\\hfill & \\hfill & \\text{or}\\hfill & \\hfill & \\text{ }x=-1\\hfill & \\hfill & \\text{or}\\hfill & \\hfill & x=4 \\end{cases}$\n\nThe x-intercepts are\n\n$\\left(2,0\\right),\\left(-1,0\\right), \\text{and} \\left(4,0\\right).\\\\$\n\nWe can see these intercepts on the graph of the function shown in Figure 11.",
null,
"Figure 11\n\n### Example 9: Determining the Intercepts of a Polynomial Function with Factoring\n\nGiven the polynomial function $f\\left(x\\right)={x}^{4}-4{x}^{2}-45,\\\\$ determine the y– and x-intercepts.\n\n### Solution\n\nThe y-intercept occurs when the input is zero.\n\n$\\begin{cases} \\\\ f\\left(0\\right)={\\left(0\\right)}^{4}-4{\\left(0\\right)}^{2}-45\\hfill \\hfill \\\\ \\text{ }=-45\\hfill \\end{cases}\\\\$\n\nThe y-intercept is $\\left(0,-45\\right).$\n\nThe x-intercepts occur when the output is zero. To determine when the output is zero, we will need to factor the polynomial.\n\n$\\begin{cases}f\\left(x\\right)={x}^{4}-4{x}^{2}-45\\hfill \\\\ =\\left({x}^{2}-9\\right)\\left({x}^{2}+5\\right)\\hfill \\\\ =\\left(x - 3\\right)\\left(x+3\\right)\\left({x}^{2}+5\\right)\\hfill \\end{cases}$\n$0=\\left(x - 3\\right)\\left(x+3\\right)\\left({x}^{2}+5\\right)\\\\$\n$\\begin{cases}x - 3=0\\hfill & \\text{or}\\hfill & x+3=0\\hfill & \\text{or}\\hfill & {x}^{2}+5=0\\hfill \\\\ \\text{ }x=3\\hfill & \\text{or}\\hfill & \\text{ }x=-3\\hfill & \\text{or}\\hfill & \\text{(no real solution)}\\hfill \\end{cases}\\\\$\n\nThe x-intercepts are $\\left(3,0\\right) \\text{and} \\left(-3,0\\right).$\n\nWe can see these intercepts on the graph of the function shown in Figure 12. We can see that the function is even because $f\\left(x\\right)=f\\left(-x\\right).\\\\$",
null,
"Figure 12\n\n### Try It 6\n\nGiven the polynomial function $f\\left(x\\right)=2{x}^{3}-6{x}^{2}-20x,\\\\$ determine the y– and x-intercepts.\n\nSolution\n\n## Comparing Smooth and Continuous Graphs\n\nThe degree of a polynomial function helps us to determine the number of x-intercepts and the number of turning points. A polynomial function of nth degree is the product of n factors, so it will have at most n roots or zeros, or x-intercepts. The graph of the polynomial function of degree n must have at most n – 1 turning points. This means the graph has at most one fewer turning point than the degree of the polynomial or one fewer than the number of factors.\n\nA continuous function has no breaks in its graph: the graph can be drawn without lifting the pen from the paper. A smooth curve is a graph that has no sharp corners. The turning points of a smooth graph must always occur at rounded curves. The graphs of polynomial functions are both continuous and smooth.\n\n### A General Note: Intercepts and Turning Points of Polynomials\n\nA polynomial of degree n will have, at most, n x-intercepts and n – 1 turning points.\n\n### Example 10: Determining the Number of Intercepts and Turning Points of a Polynomial\n\nWithout graphing the function, determine the local behavior of the function by finding the maximum number of x-intercepts and turning points for $f\\left(x\\right)=-3{x}^{10}+4{x}^{7}-{x}^{4}+2{x}^{3}.\\\\$\n\n### Solution\n\nThe polynomial has a degree of 10, so there are at most n x-intercepts and at most n – 1 turning points.\n\n### Try It 7\n\nWithout graphing the function, determine the maximum number of x-intercepts and turning points for $f\\left(x\\right)=108 - 13{x}^{9}-8{x}^{4}+14{x}^{12}+2{x}^{3}\\\\$\n\nSolution\n\n### Example 11: Drawing Conclusions about a Polynomial Function from the Graph\n\nWhat can we conclude about the polynomial represented by the graph shown in the graph in Figure 13 based on its intercepts and turning points?",
null,
"Figure 13\n\n### Solution",
null,
"Figure 14\n\nThe end behavior of the graph tells us this is the graph of an even-degree polynomial.\n\nThe graph has 2 x-intercepts, suggesting a degree of 2 or greater, and 3 turning points, suggesting a degree of 4 or greater. Based on this, it would be reasonable to conclude that the degree is even and at least 4.\n\n### Try It 8\n\nWhat can we conclude about the polynomial represented by Figure 15 based on its intercepts and turning points?",
null,
"Figure 15\n\nSolution\n\n### Example 12: Drawing Conclusions about a Polynomial Function from the Factors\n\nGiven the function $f\\left(x\\right)=-4x\\left(x+3\\right)\\left(x - 4\\right),\\\\$ determine the local behavior.\n\n### Solution\n\nThe y-intercept is found by evaluating $f\\left(0\\right).$\n\n$\\begin{cases}f\\left(0\\right)=-4\\left(0\\right)\\left(0+3\\right)\\left(0 - 4\\right)\\hfill \\hfill \\\\ \\text{ }=0\\hfill \\end{cases}\\\\$\n\nThe y-intercept is $\\left(0,0\\right).$\n\nThe x-intercepts are found by determining the zeros of the function.\n\n$\\begin{cases}0=-4x\\left(x+3\\right)\\left(x - 4\\right)\\\\ x=0\\hfill & \\hfill & \\text{or}\\hfill & \\hfill & x+3=0\\hfill & \\hfill & \\text{or}\\hfill & \\hfill & x - 4=0\\hfill \\\\ x=0\\hfill & \\hfill & \\text{or}\\hfill & \\hfill & \\text{ }x=-3\\hfill & \\hfill & \\text{or}\\hfill & \\hfill & \\text{ }x=4\\end{cases}\\\\$\n\nThe x-intercepts are $\\left(0,0\\right),\\left(-3,0\\right) \\text{and} \\left(4,0\\right).$\n\nThe degree is 3 so the graph has at most 2 turning points.\n\n### Try It 9\n\nGiven the function $f\\left(x\\right)=0.2\\left(x - 2\\right)\\left(x+1\\right)\\left(x - 5\\right),\\\\$ determine the local behavior.\n\nSolution"
]
| [
null,
"https://s3-us-west-2.amazonaws.com/courses-images-archive-read-only/wp-content/uploads/sites/924/2015/11/25201343/CNX_Precalc_Figure_03_03_0172.jpg",
null,
"https://s3-us-west-2.amazonaws.com/courses-images-archive-read-only/wp-content/uploads/sites/924/2015/11/25201345/CNX_Precalc_Figure_03_03_0182.jpg",
null,
"https://s3-us-west-2.amazonaws.com/courses-images-archive-read-only/wp-content/uploads/sites/924/2015/11/25201346/CNX_Precalc_Figure_03_03_0192.jpg",
null,
"https://s3-us-west-2.amazonaws.com/courses-images-archive-read-only/wp-content/uploads/sites/924/2015/11/25201347/CNX_Precalc_Figure_03_03_0202.jpg",
null,
"https://s3-us-west-2.amazonaws.com/courses-images-archive-read-only/wp-content/uploads/sites/924/2015/11/25201349/CNX_Precalc_Figure_03_03_0212.jpg",
null,
"https://s3-us-west-2.amazonaws.com/courses-images-archive-read-only/wp-content/uploads/sites/924/2015/11/25201350/CNX_Precalc_Figure_03_03_0224.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8502016,"math_prob":0.9987938,"size":5840,"snap":"2019-35-2019-39","text_gpt3_token_len":1485,"char_repetition_ratio":0.20630568,"word_repetition_ratio":0.15429235,"special_character_ratio":0.24589041,"punctuation_ratio":0.08729473,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99977154,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-19T02:17:30Z\",\"WARC-Record-ID\":\"<urn:uuid:7dbf731a-a67c-4752-b994-2e3e873851d3>\",\"Content-Length\":\"34828\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7aa4cc60-aeca-4d04-b16d-2e1af6e247df>\",\"WARC-Concurrent-To\":\"<urn:uuid:0bbd31f1-a105-438b-95e4-1ca6999cd997>\",\"WARC-IP-Address\":\"23.185.0.2\",\"WARC-Target-URI\":\"https://courses.lumenlearning.com/ivytech-collegealgebra/chapter/describe-local-behavior-of-polynomial-functions/\",\"WARC-Payload-Digest\":\"sha1:KUQU6QXBYHJUGMAPZKKU5EG2SXBPPXPY\",\"WARC-Block-Digest\":\"sha1:FV2TOKUSDYUMGBZPK5YQ6KDGNCUWJYJ7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573415.58_warc_CC-MAIN-20190919015534-20190919041534-00412.warc.gz\"}"} |
https://stat.ethz.ch/pipermail/r-help/2011-April/274773.html | [
"# [R] Adding text labels to lattice plots with multiple panels\n\nDeepayan Sarkar deepayan.sarkar at gmail.com\nTue Apr 12 08:50:54 CEST 2011\n\n```On Mon, Apr 11, 2011 at 12:49 AM, Jeff Stevens <stev0175 at gmail.com> wrote:\n> Many thanks, Peter. This works brilliantly, and I prefer to have the\n> labels assigned outside of panel function as well.\n\nYou could also consider using which.packet(). You haven't explicitly\ntold us how the labels are matched with the boxplots, but assuming\nthat the labels are in the order of plotting, you can do [using the\noriginal data]:\n\ndf <- data.frame(f1, f2, dv)\nlab <- c(1, 2, 3, 4, 5, 6)\n\nbwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),\ntext.labels = as.character(lab),\npanel = function(x, y, ..., text.labels) {\npanel.bwplot(x, y, ...)\nn <- nlevels(x)\ni <- seq_len(n) + (which.packet() - 1) * n\nprint(text.labels[i])\npanel.text(x = seq_len(n), y = 0.55,\nlabels = text.labels[i])\n})\n\n-Deepayan\n\n```"
]
| [
null
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https://socratic.org/questions/how-do-you-write-c-times-c-times-c-times-c-times-c-times-c-in-exponential-form | [
"# How do you write \"c (times) c (times) c (times) c (times) c (times) c\" in exponential form?\n\n$c \\times c \\times c \\times c \\times c \\times c = \\textcolor{red}{{c}^{6}}$\n${c}^{6}$ means $c$ multiplied by itself $6$ times."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.64039963,"math_prob":1.0000055,"size":289,"snap":"2021-04-2021-17","text_gpt3_token_len":75,"char_repetition_ratio":0.19298245,"word_repetition_ratio":0.11363637,"special_character_ratio":0.25259516,"punctuation_ratio":0.07692308,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99851817,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-23T11:19:11Z\",\"WARC-Record-ID\":\"<urn:uuid:0d26d63b-84a8-4172-8cfd-2446fd97e999>\",\"Content-Length\":\"32585\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aa20e4dd-431a-45bf-b812-98c55d96db52>\",\"WARC-Concurrent-To\":\"<urn:uuid:56f39f88-8e19-4a12-a12a-b15bed4bf98b>\",\"WARC-IP-Address\":\"216.239.38.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-write-c-times-c-times-c-times-c-times-c-times-c-in-exponential-form\",\"WARC-Payload-Digest\":\"sha1:EYABO43B5CRTJ7233OHDYJU77XDJC4J6\",\"WARC-Block-Digest\":\"sha1:WKLE3SNHVY346UYJXASJSBNDXKQQOKYM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703537796.45_warc_CC-MAIN-20210123094754-20210123124754-00551.warc.gz\"}"} |
https://web2.0calc.com/questions/what-is-the-value-of-a-b | [
"+0\n\nWhat is the value of a + b?\n\n0\n247\n1\n\nLet a and b real numbers such that x4+2x3-x2+ax+b = (Q(x))2 for some polynomial Q(x). What is the value of a + b?\n\nOct 16, 2018\n\n#1\n+2\n\n$$p(x)=x^4+2x^3-x^2+ax+b = (q(x))^2,~\\text{for some polynomial }q(x)$$\n\n$$\\text{we know q(x) will be of order 2 so write it as }\\\\ q(x)=q_2 x^2 + q_1 x + q_0 \\\\ (q(x))^2 = q_2^2 x^4+2 q_1 q_2 x^3+\\left(q_1^2+2 q_0 q_2\\right) x^2+2 q_0 q_1 x+q_0^2$$\n\n$$\\text{and we have equations }\\\\ q_2^2 = 1\\\\ 2q_1q_2 = 2\\\\ (q_1^2+2q_0q_2)=-1\\\\ 2q_0q_1=a\\\\ q_0^2=b$$\n\n$$\\text{clearly }q_2 = \\pm 1\\\\ \\text{suppose }q_2=1 \\\\ 2q_1 (1)=2,~q_1=1\\\\ (1+2q_0(1))=-1 \\\\ q_0=-1 \\\\ b=q_0^2 = 1 \\\\ a=2q_0 = 2$$\n\n$$\\text{now suppose }q_2=-1 \\\\ 2q_1(-1) = 2,~q_1=-1 \\\\ (1+2q_0(-1))=-1,~q_0=1 \\\\ b=q_0^2 = 1 \\\\ a = 2(-1)(-1) = 2$$\n\n$$\\text{so in both cases }a=2,~b=1 \\\\ \\text{and }a+b = 3$$\n\nI kind of suspect there is a simpler way to solve this.\n\n.\nOct 17, 2018\nedited by Rom Oct 17, 2018"
]
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null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7018776,"math_prob":1.0000099,"size":829,"snap":"2019-43-2019-47","text_gpt3_token_len":453,"char_repetition_ratio":0.12969697,"word_repetition_ratio":0.01459854,"special_character_ratio":0.59951746,"punctuation_ratio":0.044554457,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000094,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-16T08:16:06Z\",\"WARC-Record-ID\":\"<urn:uuid:b9b2b076-7782-4038-8852-c98ee795f58f>\",\"Content-Length\":\"22215\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4f367347-666e-4395-acb4-8b967188f0ff>\",\"WARC-Concurrent-To\":\"<urn:uuid:a8ff6a3c-e8e5-4ed6-8c88-51359750f3d5>\",\"WARC-IP-Address\":\"144.76.186.3\",\"WARC-Target-URI\":\"https://web2.0calc.com/questions/what-is-the-value-of-a-b\",\"WARC-Payload-Digest\":\"sha1:3AZN6NGOCJIBTOIP7GNFHP5LN2JLIONB\",\"WARC-Block-Digest\":\"sha1:LYJ7RFVYBN4C4KXFL3V7PXGEQQSLBBPZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986666467.20_warc_CC-MAIN-20191016063833-20191016091333-00555.warc.gz\"}"} |
https://math.stackexchange.com/questions/2084785/is-fracx2-2xx-a-polynomial | [
"# Is $\\frac{x^2 + 2x}{x}$ a polynomial? [duplicate]\n\nIs $$\\dfrac{x^2 + 2x}{x}$$ a polynomial?\n\n• One argument against it being a polynomial would be that the domain doesn't include zero, and a polynomial has always domain = $\\mathbb{R}$ – Guillermo Mosse Jan 5 '17 at 14:42\n• @SeñorBilly A polynomial does not \"always\" have domain $\\mathbb{R}$. Remember, domains are specified, not intrinsic. – MathematicsStudent1122 Jan 5 '17 at 14:46\n• @TonyK Definitely. If the expression is considered as an element of $\\mathbb{R}(x)$ (the field of fractions of the polynomial ring $\\mathbb{R}[x]$), then it is certainly a polynomial, because we can “cancel out the $x$”. As a function of a real variable, it is not a polynomial function, because its domain is not $\\mathbb{R}$. – egreg Jan 5 '17 at 14:51\n• @user1952009 You're comparing apples and oranges. – egreg Jan 5 '17 at 14:53\n• @user1952009 Not really: 9/3 = 3 is always true; $x^2/x=x$ is only correct if $x \\ne 0$; since the left-hand side is undefined for $x=0$. – StackTD Jan 5 '17 at 15:04\n\n## 2 Answers\n\nSome people would say that the rational number $7/1$ is not really equal to the integer $7$, but merely canonically identified with it. But (after reaching a certain level of sophistication) mathematicians say that $7/1$ and $7$ are indeed equal.\n\nSome people would say that the rational function $$\\frac{x^2+2x}{x} \\tag{*}$$ is not really equal to the polynomial $$x+2, \\tag{**}$$ but merely canonically identified with it. But (after reaching a certain level of sophistication) mathematicians say that (*) and (**) are indeed equal.\n\n• It may depend on the context, but there is some subtlety to this analogy (as pointed out in the comments). If you, after canonical identification, indeed would say they are equal, then what about (what happens at) $x=0$...? – StackTD Jan 5 '17 at 15:14\n\nWell, it's a polynomial in the variable $t=\\tfrac{x^2+2x}{x}$... But you probably mean a polynomial in the (real?) variable $x$. What precise definition of polynomial are you using?\n\nI would say no, because it is not of the form $$a_0+a_1x+a_2x^2+\\ldots+a_nx^n$$ for any $n \\in \\mathbb{N}$ and real numbers $a_i$ ($0 \\le i \\le n$).\n\nNote that you cannot just simplify $$\\frac{x^2+2x}{x} = x+2$$ as this equality is only valid for non-zero $x$, so not for all $x$."
]
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https://www.mathbits.com/MathBits/TISection/Trig/trigidentity.htm | [
"",
null,
"Verifying Trig Identities\n\nIf you need to determine if an equation is a trigonometric identity, you can use your graphing calculator. A \"formal proof\" of a trigonometric identity is done algebraically. The graphing calculator, however, can be used to \"check\" your work.\n\n Determine if cos2 x = 1 - sin2 x is a trig identity:\n\n Enter the left side in Y1. Enter the right side in Y2. Set the bubble animation for the second graph.",
null,
"Choose ZOOM #7 ZTRig",
null,
"If the bubble runs over the same graph as the first plotting, the equation is a trig identity.",
null,
"2nd GRAPH (TABLE)",
null,
"The TABLE can also be used to check to see if the results are the same from both graphs. Arrow up and down to see if the values listed under columns Y1 and Y2 remain the same. Adjust your table through 2nd WINDOW (TBLSET) to see more varied values.\n\nAnswer: Yes, cos2 x = 1 - sin2 x is a trig identity.",
null,
""
]
| [
null,
"https://www.mathbits.com/MathBits/TISection/Trig/logoTrig.gif",
null,
"https://www.mathbits.com/MathBits/TISection/Trig/idnetitypuc.jpg",
null,
"https://www.mathbits.com/MathBits/TISection/Trig/trigid1.gif",
null,
"https://www.mathbits.com/MathBits/TISection/Trig/idnetitypic2.jpg",
null,
"https://www.mathbits.com/MathBits/TISection/Trig/trigid2.gif",
null,
"https://www.mathbits.com/MathBits/TISection/Trig/tidivider.gif",
null
]
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https://discuss.codechef.com/t/meanprob-editorial/47454 | [
"",
null,
"# MEANPROB - Editorial\n\nPractice\n\nAuthor: Praveen Dhinwa\n\nTester: tncks0121\n\nDIFFICULTY:\n\nEasy\n\nPREREQUISITES:\n\nPROBLEM:\nYou are given a sequence a_1, \\ldots a_n. All the values are positive and \\leq 1000. For all i > n, a_i is equal to the floor of the mean of previous n elements of the sequence. Given q queries, each containing an index k, find a_k for each query.\n\nEXPLANATION:\nThe main idea here is that the sequence would become constant after i > 1000 * n.\n\nProof:\nClearly, if n consecutive values are equal (to t say), then all the following values must also be equal to t.\n\nDivide the sequence into chunks of size n. Say the maximum value in current chunk is m, and the values in the chunk are not equal. Then, each element in the next chunk must be < m, as the sum of n consecutive elements will always be < m n. This means that the maximum of next chunk is < m.\n\nSince the maximum can reduce \\leq 1000 times, the sequence must become constant after 1000n.\n\nSo, we precompute the first 1000n values in the sequence by maintaining sum of last n values. To answer a query k, we just return a_{\\min(1000n, k)}.\n\nAUTHOR’S and TESTER’S codes:\n\nThe author’s code can be found here\n\nThe tester’s code can be found here\n\n1 Like\n\nI didnt understand the explanation can any one here explain clearly please"
]
| [
null,
"https://s3.amazonaws.com/discourseproduction/original/3X/7/f/7ffd6e5e45912aba9f6a1a33447d6baae049de81.svg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.85422695,"math_prob":0.9901122,"size":1195,"snap":"2020-34-2020-40","text_gpt3_token_len":322,"char_repetition_ratio":0.12678422,"word_repetition_ratio":0.00913242,"special_character_ratio":0.26025105,"punctuation_ratio":0.13565892,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99054253,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-28T09:51:57Z\",\"WARC-Record-ID\":\"<urn:uuid:849fc844-7fae-437a-a6b1-f823abecc5a4>\",\"Content-Length\":\"16053\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:79805318-a752-46ac-8f1f-d935232c8313>\",\"WARC-Concurrent-To\":\"<urn:uuid:4d955c20-5f5b-45d5-b0d1-f63d207b2c26>\",\"WARC-IP-Address\":\"18.213.158.143\",\"WARC-Target-URI\":\"https://discuss.codechef.com/t/meanprob-editorial/47454\",\"WARC-Payload-Digest\":\"sha1:A5W6EYAT73T3RSK3DIOTETAP446C7RDO\",\"WARC-Block-Digest\":\"sha1:L7AAT7WSWSC3K33ZEMICAM4YTFRDPF3J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600401598891.71_warc_CC-MAIN-20200928073028-20200928103028-00699.warc.gz\"}"} |
https://stackoverflow.com/questions/64488529/swiftui-animated-shape-going-crazy | [
"# SwiftUI Animated Shape going crazy\n\nFirst off, my appologies for the unconventional title of this post, but I don't know of any way to describe the behavior any better than that.\n\nTo reproduce this problem, create a new project in xCode 12 specifying IOS App with any name and **Interface:**SwiftUI, Life Cycle: SwiftUI App, Language: Swift. Then replace all of Content View with the code listed here.\n\nIn either Live Preview, or when running the app, a click on the Poly will trigger the animation. It should move a fraction of a pixel, and do a 1/3 turn (as the angle is in radians). The problem, as noted on line 37 is that when I try to move or turn the Poly, it goes crazy, multiplying any move by much greater amounts. The color animates fine, but the Animatable properties in the shape do not. The location starts at 200,200 with an angle of 0, and if you try to move it very close, as the sample code does, it overreacts. If you try to move it only a few pixes (say 190,190) it will fly off the screen.\n\nI have not had this happen to any other animations I have done, and have no idea why this is behaving this way.\n\nIn trying to debug this, at one point I put print statements on the animatableData getters and setters, and can make no sense of what the animation engine is doing to the variables. It just seems to pick a number that is much further from the source that the value I am asking it to go to.\n\nI am confident that the trig in the path is correct, and suspect the issue lies in one of the following:\n\n• My declaration of animatableData somehow\n• The withAnimation function in the gesture\n• An issue with animating a CGFloat\n• SwiftUI is just going crazy\n\nI am running Xcode 12.1 (12A7403) and Swift 5. After many hours of trying to figure this out, I humbly present my problem here. Help me Obiwan Kenobi, you are my only hope...",
null,
"``````import SwiftUI\n\nlet twoPi:CGFloat = CGFloat.pi * 2\nlet pi = CGFloat.pi\n\nclass Poly: ObservableObject, Identifiable {\n\n@Published var location:CGPoint\n@Published var color:Color\n\nvar sides:CGFloat\nvar angle:CGFloat\n\ninit(at:CGPoint, color:Color, sides:CGFloat, radius:CGFloat, angle:CGFloat=0) {\nself.location = at\nself.color = color\nself.sides = sides\nself.angle = angle\n}\n}\n\nstruct ContentView: View {\n\n@ObservedObject var poly:Poly = Poly(at: CGPoint(x:200,y:200), color: .green, sides: 6, radius: 100)\n\nvar body: some View {\n\nPolyShape(poly: poly)\n.fill(poly.color)\n.gesture(\nDragGesture(minimumDistance: 0, coordinateSpace: .local)\n.onEnded { gesture in\nwithAnimation(.easeInOut(duration:15)) {\n//This is what doesn't work.\n//Try to nudge it a fraction of a pixel, and do only 1/3 of a turn, and it spins and moves much further.\npoly.location.x = 200.4\npoly.location.y = 200.2\npoly.angle = twoPi / 3\npoly.color = .red\n}\n}\n)\n}\n}\n\nstruct ContentView_Previews: PreviewProvider {\n\nstatic var previews: some View {\nGroup {\nContentView(poly: Poly(at: CGPoint(x:200,y:200), color: .blue, sides: 3, radius: 100))\n}\n}\n}\n\nstruct PolyShape:Shape {\n\nvar poly:Poly\n\npublic var animatableData: AnimatablePair<CGFloat, AnimatablePair<CGFloat,CGFloat>> {\nget { AnimatablePair(poly.angle, AnimatablePair(poly.location.x, poly.location.y))\n}\nset {\npoly.angle = newValue.first\npoly.location.x = newValue.second.first\npoly.location.y = newValue.second.second\n}\n}\n\nfunc path(in rect: CGRect) -> Path {\n\nvar path = Path()\n\nwhile radial < twoPi + 0.5 {\npath.move(to: CGPoint(x: newX, y: newY))\n} else {\n}\n}\nreturn path\n}\n}\n``````\n\nYou need to separate model from view model, because to have `PolyShape` correctly work in your case the input data have to a value.\n\nHere is tested solution (Xcode 12 / iOS 14)",
null,
"1. Separate model and view model\n``````class Poly: ObservableObject, Identifiable {\n@Published var data:PolyData\n\ninit(data: PolyData) {\nself.data = data\n}\n}\n\nstruct PolyData {\nvar location:CGPoint\nvar color:Color\n\nvar sides:CGFloat\nvar angle:CGFloat\n\ninit(at:CGPoint, color:Color, sides:CGFloat, radius:CGFloat, angle:CGFloat=0) {\nself.location = at\nself.color = color\nself.sides = sides\nself.angle = angle\n}\n}\n``````\n1. Make shape value dependent\n``````struct PolyShape:Shape {\n\nvar poly:PolyData // << here !!\n\n// ... other code no changes\n}\n``````\n1. Update dependent demo\n``````struct ContentView: View {\n\n@ObservedObject var poly:Poly = Poly(data: PolyData(at: CGPoint(x:200,y:200), color: .green, sides: 6, radius: 100))\n\nvar body: some View {\n\nPolyShape(poly: poly.data) // << pass data only here !!\n.fill(poly.data.color)\n.gesture(\nDragGesture(minimumDistance: 0, coordinateSpace: .local)\n.onEnded { gesture in\nwithAnimation(.easeInOut(duration:15)) {\npoly.data.location = CGPoint(x: 200.4, y: 200.2)\npoly.data.angle = twoPi / 3\npoly.data.color = .red\n}\n}\n)\n}\n}\n\nstruct ContentView_Previews: PreviewProvider {\n\nstatic var previews: some View {\nGroup {\nContentView(poly: Poly(data: PolyData(at: CGPoint(x:200,y:200), color: .blue, sides: 3, radius: 100)))\n}\n}\n}\n``````\n• Thank you very much! That did the trick, although I am not entirely why. I have seen other posts were users animate ObservableObject classes. One post even showed them doing that with a DispatchQueue Async call. Is it because Shape only wants to work with animation on a struct, or that AnimatableData only works with structs? – cdeerinck Oct 23 at 6:01\n\nCan you give the following a try?\n\n``````import SwiftUI\n\nlet twoPi:CGFloat = CGFloat.pi * 2\nlet pi = CGFloat.pi\n\nstruct ContentView: View {\n@State var location: CGPoint = CGPoint(x: 200, y: 200)\n@State var color: Color = .blue\n@State var angle: CGFloat = 0\n\nvar body: some View {\nPolyShape(location: location, color: color, angle: angle, sides: 3, vertexRadius: 100)\n.fill(color)\n.gesture(\nDragGesture(minimumDistance: 0, coordinateSpace: .local)\n.onEnded { gesture in\nwithAnimation(.easeInOut(duration:1)) {\n//This is what doesn't work.\n//Try to nudge it a fraction of a pixel, and do only 1/3 of a turn, and it spins and moves much further.\nlocation.x = 220\nlocation.y = 220\nangle = (CGFloat.pi * 2) / 3\ncolor = .red\n}\n}\n)\n}\n}\n\nstruct ContentView_Previews: PreviewProvider {\nstatic var previews: some View {\nGroup {\nContentView(location: CGPoint(x: 200, y: 200), color: .blue, angle: 0)\n}\n}\n}\n\nstruct PolyShape:Shape {\nvar location: CGPoint\nvar color: Color\nvar angle: CGFloat\nvar sides: CGFloat\n\npublic var animatableData: AnimatablePair<CGFloat, AnimatablePair<CGFloat,CGFloat>> {\nget {\nreturn AnimatablePair(angle, AnimatablePair(location.x, location.y))\n}\nset {\nangle = newValue.first\nlocation.x = newValue.second.first\nlocation.y = newValue.second.second\n}\n}\n\nfunc path(in rect: CGRect) -> Path {\nvar path = Path()\nwhile radial < twoPi + 0.5 {"
]
| [
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"https://i.stack.imgur.com/1GvoN.gif",
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"https://i.stack.imgur.com/wKMq8.gif",
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https://teamtreehouse.com/community/it-says-that-i-have-the-wrong-number-of-prints-i-am-confused-since-i-thought-there-is-two | [
"# It says that I have the wrong number of prints I am confused since I thought there is two.\n\nWhat is wrong with my code? Also do you see anything else that's wrong.\n\neven.py\n```import random\ndef even_odd(num):\nstart = 5\nwhile start == True:\nrandom.randint(1, 99)\nstart - 1\nif num % 2 == 0:\nprint(\"{} is even\").format(num)\nelse:\nprint(\"{} is odd\").format(num)\n\n# If % 2 is 0, the number is even.\n# Since 0 is falsey, we have to invert it with not.\nreturn not num % 2\n```",
null,
"MOD\n\nHi Annika,\n\nYou do have a few bugs in your code.\n\nFirst, and this should have been caught by the Challenge tester in task 2, is that you have declared your start variable inside your function, but the instructions tell you:\n\nMake a new variable, outside of the `even_odd` function, named `start` that's assigned to the number 5.\n\nYou're also getting a bit muddled by the rest of what code goes where. It's not totally obvious in the instructions but your code should comprise:\n\n• import random (task 1); then\n• a function called `even_odd` that returns whether the input value is odd or even (already created for you); then\n• setting a variable `start` to 5. (task 2); then lastly\n• a while loop that runs till `start` is falsey (task3)\n\nYou have a lot of code inside the even_odd function, but all the code you write should be outside that function.\n\nNow for the specific items in your code:\n\nYour while loop declaration has an error caused by a subtle difference between `True`/`False` and `Truthy`/`Falsey`. Basically, a value that is `Truthy` acts like it is true, but isn't exactly true. The main difference is that if you explicitly ask a Truthy value if it is true, it will reply that it is not:\n\n```>>> 5 == True\nFalse\n```\n\nBut as long as you don't expressly compare it to True, truthy value will behave like True:\n\n```my_number = 1 # all non-zero numbers are 'Truthy'\nif my_number:\nprint(\"my_number is acting like it is true!\")\nelse:\nprint(\"my_number is acting like it is false!\")\n```\n\nThis will print `\"my_number is acting like it is true!\"`. Now let's run the same code but with a falsey value:\n\n```my_number = 0 # all non-zero numbers are 'Truthy', zero is 'Falsey'\nif my_number:\nprint(\"my_number is acting like it is true!\")\nelse:\nprint(\"my_number is acting like it is false!\")\n```\n\nNow we get `\"my_number is acting like it is false!\"`.\n\nIn the above example, we used an `if` statement, but the logic holds true for every other situation where we compare values. In your code, `start` has a value of `5`, which is truthy, when it gets to 0, it will become falsey, so you can just write:\n\n```while start:\n```\n\nThis will behave like true as long as `start` isn't 0.\n\nYou're not doing anything with your `random.randint()` call. You should be assigning that result to a temporary variable and then using that variable when you call the `even_odd` function.\n\nNext, you have a bug here:\n\n```start - 1\n```\n\nThis just evaluates to 4 but doesn't actually do anything with that value. To change the value in the `start` variable you need to assign the new value to `start`. You can combine the assignment and deduction into a single operator: `-=`.\n\nYour use of the `.format()` method is also not quite right. Remember that `.format()` is a method on a string, therefore the `.` should directly connect the string and the method. For example:\n\n```name = \"Annika\"\nphrase = \"hello, {}\".format(name)\n```\n\nAlso, `print()` is a global function, so the parentheses should completely enclose the thing being printed.\n\n```print(phrase)\n```\n\nIs equivalent to:\n\n```print(\"hello, {}\".format(name))\n```\n\nI know this sounds like a lot, but you are on the right track. Feel free to ask if you have any specific questions about the above.\n\nCheers\n\nAlex"
]
| [
null,
"https://uploads.teamtreehouse.com/production/profile-photos/8418842/micro_IMG_1256__cropped_and_reduced_res_.jpg",
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https://www.mathphysicsbook.com/mathematics/algebraic-topology/counting-holes-that-arent-boundaries/calculating-homology-groups/ | [
"# Calculating homology groups\n\nIt should be kept in mind that although homology, like most of algebraic topology, is geometrically inspired, its algebraic constructions may or may not have ready geometric interpretations in odd situations or higher dimensions. Despite the “sensible” results above, counter-intuitive facts caution us to always deduce these and other measurements of a space with mathematical rigor. For example, the “hole” interpretation of homology does not easily extend to non-orientable surfaces:\n\n• $${H_{1}\\left(\\mathbb{R}\\textrm{P}^{2}\\right)=\\mathbb{Z}_{2}}$$, since if we view $${\\mathbb{R}\\textrm{P}^{2}}$$ as $${D^{2}}$$ with boundary $${S^{1}}$$ having antipodal points identified, the path connecting two antipodal points is a “loop,” but any two such loops together is homologous to a point\n• $${H_{2}\\left(\\mathbb{R}\\textrm{P}^{2}\\right)=0}$$, since any triangulation of $${\\mathbb{R}\\textrm{P}^{2}}$$ has a boundary, the real projective plane being non-orientable\n• In general, any non-orientable manifold $${M^{n}}$$ has $${H_{n}(M^{n})=0}$$\n\nAs one might guess from the examples considered so far, it is also a fact that the topology of 2-manifolds is completely determined by homology. This circumstance is certainly not true in higher dimensions, as we noted in the introduction to this chapter.\n\nThere are various relations that can help in calculating homology groups. For example, an immediate result is that if $${X}$$ has connected components $${X_{i}}$$, $${H_{n}(X)=\\bigoplus_{i}H_{n}\\left(X_{i}\\right)}$$. Another is the excision theorem, which states that for $${Z\\subset A\\subset X}$$, $${H_{n}\\left(X-Z,A-Z\\right)\\cong H_{n}\\left(X,A\\right)}$$. Here $${H_{n}\\left(X,A\\right)}$$ is a relative homology group, defined using $${n}$$-chains $${C_{n}\\left(X,A\\right)\\equiv C_{n}\\left(X\\right)/C_{n}(A)}$$ in place of $${C_{n}\\left(X\\right)}$$; this construction essentially ignores any holes in $${A\\subset X}$$. Thus the excision theorem states the intuitively expected fact that we can delete any portion of a subspace $${A}$$ without affecting $${H_{n}\\left(X,A\\right)}$$, which already ignores holes in $${A}$$."
]
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https://www.shivatutorials.com/2019/11/marksheet-program-solution-in-php.html | [
"# Marksheet program Solution in PHP\n\n0\nMarksheet program Solution in PHP:-\n\nWAP to create a mark sheet using five different subjects with the following conditions?\n\n1) all subject marks should be 0 to 100?\n\n2) If only one subject mark is <33 then the student will supply.\n\n3) If all subject marks are>33 then the student pass and display division with percentage?\n\n4) if the minimum of two subject marks is <33 then the student will fail.\n\n5) if the student is suppl and the mark is >28 then five bonus marks will be added then the student will pass by grace and grace subject name.\n\n6) display distinction subject name\n\n7) display suppl subject name\n\n<?php\n\n\\$m1=40;\n\\$m2=40;\n\\$m3=94;\n\\$m4=35;\n\\$m5=93;\n\\$s1=\"PHY\";\n\\$s2= \"Chem\";\n\\$s3=\"Maths\";\n\\$s4=\"English\";\n\\$s5=\"Hindi\";\n\nif((\\$m1>=0 && \\$m1<=100) && (\\$m2>=0 && \\$m2<=100) && (\\$m3>=0 && \\$m3<=100) && (\\$m4>=0 && \\$m4<=100) && (\\$m5>=0 && \\$m5<=100))\n{\n\n\\$c=0;\n\\$marks=0;\n\\$sub= \"\";\n\\$dist= \"\";\nif(\\$m1<33)\n{\n\\$c++;\n\\$marks=\\$m1;\n\\$sub=\\$sub.\\$s1;\n}\n\nif(\\$m2<33)\n{\n\\$c++;\n\\$marks=\\$m2;\n\\$sub=\\$sub.\\$s2;\n}\nif(\\$m3<33)\n{\n\\$c++;\n\\$marks=\\$m3;\n\\$sub=\\$sub.\\$s3;\n}\nif(\\$m4<33)\n{\n\\$c++;\n\\$marks=\\$m4;\n\\$sub=\\$sub.\\$s4;\n}\nif(\\$m5<33)\n{\n\\$c++;\n\\$marks=\\$m5;\n\\$sub=\\$sub.\\$s5;\n}\nif(\\$m1>=75)\n{\n\n\\$dist=\\$dist.\\$s1;\n}\n\nif(\\$m2>=75)\n{\n\n\\$dist=\\$dist.\\$s2;\n}\nif(\\$m3>=75)\n{\n\n\\$dist=\\$dist.\\$s3;\n}\nif(\\$m4>=75)\n{\n\n\\$dist=\\$dist.\\$s4;\n}\nif(\\$m5>=75)\n{\n\n\\$dist=\\$dist.\\$s5;\n}\nif(\\$c==0 || (\\$c==1 && \\$marks>=28))\n{\n\\$per= (\\$m1+\\$m2+\\$m3+\\$m4+\\$m5)/5;\nif(\\$per>33 && \\$per<45)\necho \"Congratulation you are pass in the third division with \\$per % \";\nelseif(\\$per<60)\necho \"Congratulation you are pass in the second division with \\$per % \";\nelse\necho \"Congratulation you are pass in first division with \\$per % \";\nif(\\$c==1)\necho \"You are passing by grace and grace mark is \",33-\\$marks,\" grace subject is \\$sub\";\nif(\\$dist!=\"\")\necho \"Distinction subject name is \",\\$dist;\n}\nelse if(\\$c==1)\necho \"Try again you are suppl in \\$sub\";\nelse\necho \"Sorry you have failed try again next year subject is \\$sub\";\n\n}\nelse\necho \"All Subject Marks should be 0 to 100\";\n\n?>\n\nTags"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6698683,"math_prob":0.9919008,"size":3795,"snap":"2023-14-2023-23","text_gpt3_token_len":1352,"char_repetition_ratio":0.14771828,"word_repetition_ratio":0.9484193,"special_character_ratio":0.44822136,"punctuation_ratio":0.15834348,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9990251,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-11T01:49:22Z\",\"WARC-Record-ID\":\"<urn:uuid:3cdccee5-eeb7-4af1-92c0-0ca8e80efebc>\",\"Content-Length\":\"249726\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b10a21f9-cef3-4519-8cb1-a84d4be5fecf>\",\"WARC-Concurrent-To\":\"<urn:uuid:f064c453-303c-4fe2-ab50-5e9176c41e67>\",\"WARC-IP-Address\":\"172.253.122.121\",\"WARC-Target-URI\":\"https://www.shivatutorials.com/2019/11/marksheet-program-solution-in-php.html\",\"WARC-Payload-Digest\":\"sha1:7SD5M5AVCGINVNMWGF6L3QEPGDEQDPZC\",\"WARC-Block-Digest\":\"sha1:BCFXUU56BJJ2JUJO6N54VA6IXXLKTVUK\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224646652.16_warc_CC-MAIN-20230610233020-20230611023020-00414.warc.gz\"}"} |
https://mathematica.stackexchange.com/questions/186732/rsolve-not-reducing-for-a-certain-recurrence-relation | [
"# RSolve not reducing for a certain recurrence relation\n\nI'm trying to use RSolve as follows to solve a recurrence relation:\n\nRSolve[{a == 1, a[2 n + 1] == a[2 n]*2, a[2 n] == a[2 n - 1] + 2}, a, n]\n\n\nI think the meaning of the relation is clear: for odd terms in the sequence, multiply the previous term by two, and for even terms, add two to the previous term.\n\nHowever, when evaluating this, Mathematica simply echos the input:",
null,
"rather than attempting to solve the recurrence.\n\nI can't find anything in the RSolve documentation which talks about cases where RSolve will do nothing, without any error message.\n\nHave I made a syntax error, or does this mean that Mathematica is not able to solve this type of recurrence relation? How can I change my input so that Mathematica does solve the recurrence, assuming it is possible?\n\n• Try replacing the next-to-last term in RSolve (a) with a[n]. Nov 26, 2018 at 20:30\n• It still simply echoes the input: RSolve[{a == 1, a[1 + 2 n] == 2 a[2 n], a[2 n] == 2 + a[-1 + 2 n]}, a[n], n] Nov 26, 2018 at 20:36\n• Welcome to Mathematica.SE! Interesting question(+1). I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise. Nov 27, 2018 at 0:28\n\nIt appears that RSolve cannot solve difference equations in which two different equations describe the same variable. If so, a work-around is to represent a[k] with k an even index as c[k], and with k an odd indix as b[k], so that there is only one equation per variable.\n\nFullSimplify[RSolveValue[{c == 1, b[k] == c[k - 1]*2, c[k + 2] == b[k + 1] + 2},\n{c[k], b[k]}, k] /. C -> 0]\n(* {-2 + 3 2^(-1 + k/2) (1 + (-1)^k), -4 - 3 2^(1/2 (-1 + k)) (-1 + (-1)^k)} *)\n\n\nThen, construct the desired a[k] as even-index terms of c[k] and odd-index terms of b[k].\n\nsol[k_] := If[EvenQ[k], -2 + 3 2^(k/2) , -4 + 3 2^((1 + k)/2) ]\nTable[sol[k], {k, 0, 10}]\n(* {1, 2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)\n\n\nas desired.\n\nWhen Mathematica returns the input without any error message, it is unable to evaluate the input.\n\nWith this particular recursion there is another approach. The recursion can be defined by\n\nClear[a, ar]\n\nar = 1; ar[n_?OddQ] := ar[n - 1]*2; ar[n_?EvenQ] := ar[n - 1] + 2;\n\n\nGenerating a sequence from this recursion\n\nseq = ar /@ Range\n\n(* {2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)\n\n\nUse FindSequenceFunction to find the closed form of the recursion\n\na[n_] = FindSequenceFunction[seq, n] // FullSimplify\n\n(* -3 + (-1)^n + 3 2^(-1 + n/2) (1 + Sqrt + (-1)^(1 + n) (-1 + Sqrt)) *)\n\n\nChecking equivalence outside of the range of seq\n\nAnd @@ Table[a[n] == ar[n], {n, 0, 100}]\n\n(* True *)\n\n\nEDIT: Verifying,\n\nSimplify[{a[2 n + 1] == a[2 n]*2,\na[2 n] == a[2 n - 1] + 2}, {Element[n, Integers], n >= 0}]\n\n(* {True, True} *)\n\n• Hi, thanks for the alternative solution. Do you know why Mathematica was unable to evaluate the input given? Nov 26, 2018 at 20:36\n• Don’t know. Presumably the algorithms used are not sufficiently robust to cover this case. Nov 26, 2018 at 20:52"
]
| [
null,
"https://i.stack.imgur.com/IWZmV.png",
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http://www.bom.gov.au/australia/radar/about/calculating_rainfall_accumulations.shtml | [
"# How Radar-Derived Rainfall Accumulations Are Calculated\n\nThere are three steps in creating the radar derived rainfall accumulation:\n1. Correct the raw radar reflectivity measurements.\n2. Convert the corrected radar reflectivity to a rainfall estimate.",
null,
"Figure 1. Flow chart illustrating how the radar derived rainfall accumulations are calculated.\n\n## Reflectivity measurement\n\nThe radar reflectivity measurements received from the radar are in a raw format in a 3D polar coordinate system. A polar coordinate system is where points (or pixels in this case) are given by an angle and a distance from an origin. In a radar scan, this means that each pixel has a reflectivity value in the range (distance), azimuth (angle) and elevation (height) axis.\n\nThese raw reflectivity values are assessed to see if they are within realistic thresholds. The reflectivity values are then quality checked, where clutter from the sea and ground is removed, as well as erroneous data from partial beam blocking and beam filling. A vertical profile bias adjustment is also completed as well as a scaling adjustment to compensate for the radar beam geometry errors. The data is then converted to Cartesian coordinates at a specified height above ground level for the next step.\n\n## Converting radar reflectivity to rainfall accumulation\n\nIn this step the newly corrected reflectivity information is classified into convective or stratiform rain (to lessen the rainfall estimation errors) depending on it's properties and behaviour. The reflectivity is then converted into a rain rate (mm/hour) using a power-law relationship (known as the Z-R relationship) with different constants for convective or stratiform rain. This rain rate is then converted to 1-minute accumulation data using the optical flow technique, to account for movement of the rain. These data are accumulated into maps of different durations."
]
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"http://www.bom.gov.au/australia/radar/images/create_radar_derived_rainfall_accumulation.png",
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8904109,"math_prob":0.9739383,"size":3241,"snap":"2022-05-2022-21","text_gpt3_token_len":603,"char_repetition_ratio":0.15971579,"word_repetition_ratio":0.0,"special_character_ratio":0.17803147,"punctuation_ratio":0.062271062,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97263616,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-23T11:46:01Z\",\"WARC-Record-ID\":\"<urn:uuid:e3d23149-efa1-45e4-84f4-0ab30d5b5762>\",\"Content-Length\":\"31537\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9a270a83-79ea-49ef-94bf-5e2402cf057d>\",\"WARC-Concurrent-To\":\"<urn:uuid:f8fa5fc2-149f-45db-9bf8-9824d0cf2d18>\",\"WARC-IP-Address\":\"23.220.129.39\",\"WARC-Target-URI\":\"http://www.bom.gov.au/australia/radar/about/calculating_rainfall_accumulations.shtml\",\"WARC-Payload-Digest\":\"sha1:WDTD5GLOFWNZ3YXLPSOPNZSFDFUYQGCI\",\"WARC-Block-Digest\":\"sha1:2F2PTHD2HBJYBLGYHO3EEX55LP237YYX\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662558015.52_warc_CC-MAIN-20220523101705-20220523131705-00329.warc.gz\"}"} |
https://www.fxsolver.com/browse/?like=500&p=8 | [
"'\n\n# Search results\n\nFound 1142 matches\nPower in a reference system(aerodynamic drag)\n\nIn fluid dynamics, drag (sometimes called air resistance, a type of friction, or fluid resistance, another type of friction or fluid friction) is a force ... more\n\nCentripetal Force - angular velocity\n\nCentripetal force (from Latin centrum “center” and petere “to seek”) is a force that makes a body follow a curved path: its ... more\n\nFlywheel energy storage (Energy density)\n\nA flywheel is a rotating mechanical device that is used to store rotational energy. Flywheel energy storage works by accelerating a rotor to a very high ... more\n\nWorksheet 290\n\nFind the terminal velocity of an 85-kg skydiver falling in a spread-eagle position.\n\nTerminal Velocity (without considering buoyancy)\nRectangle area\n\nwhere Vt is the terminal velocity, m is the mass of the skydiver, g is the acceleration due to gravity, Cd is the drag coefficient, ρ is the density of the fluid through which the object is falling, and A is the projected area of the object.\n\nReference : OpenStax College,College Physics. OpenStax College. 21 June 2012.\nhttp://openstaxcollege.org/textbooks/college-physics\n\nwhere h is skydiver height and w the width at “spread-eagle” position\n\nDrag coefficient\n\nDrag (sometimes called air resistance, a type of friction, or fluid resistance, another type of friction or fluid friction) refers to forces acting ... more\n\nDrag equation ( for fluids)\n\nDrag (sometimes called air resistance, a type of friction, or fluid resistance, another type of friction or fluid friction) refers to forces acting ... more\n\nPlanck temperature\n\nPlanck temperature, denoted by TP, is the unit of temperature in the system of natural units known as Planck units.\n\nIt serves as the ... more\n\nSpeed of Sound (air, ideal gases) - relative to molar mass\n\nThe speed of sound is the distance travelled per unit time by a sound wave propagating through an elastic medium. The SI unit of the speed of sound is the ... more\n\nIn the physical sciences, the wavenumber (also wave number) is the spatial frequency of a wave, either in cycles per unit distance or radians per unit ... more\n\nPower (aerodynamic drag)\n\nIn fluid dynamics, drag (sometimes called air resistance, a type of friction, or fluid resistance, another type of friction or fluid friction) is a force ... more\n\n...can't find what you're looking for?\n\nCreate a new formula\n\n### Search criteria:\n\nSimilar to formula\nCategory"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8966333,"math_prob":0.9516504,"size":2072,"snap":"2023-40-2023-50","text_gpt3_token_len":466,"char_repetition_ratio":0.16489361,"word_repetition_ratio":0.2670623,"special_character_ratio":0.2195946,"punctuation_ratio":0.17026378,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9728395,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-01T12:03:00Z\",\"WARC-Record-ID\":\"<urn:uuid:a2739fb1-c939-446b-8137-4a0d3e68a4b7>\",\"Content-Length\":\"122235\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:04a48d4d-0272-45ad-9999-22f8a88f2826>\",\"WARC-Concurrent-To\":\"<urn:uuid:4f121b5b-1fab-4520-bde8-518fe0780b0c>\",\"WARC-IP-Address\":\"178.254.54.75\",\"WARC-Target-URI\":\"https://www.fxsolver.com/browse/?like=500&p=8\",\"WARC-Payload-Digest\":\"sha1:6N7WQ5NXYPSXPDZFDR2OUMBWRWM6TNFV\",\"WARC-Block-Digest\":\"sha1:JVU5CZPKJFCWLDNY6KYBDEJ2AUUM3AEM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510888.64_warc_CC-MAIN-20231001105617-20231001135617-00766.warc.gz\"}"} |
http://blog.haoservice.cn/archives/1429 | [
"# golang教程（十八）：反射\n\n## 一、基本介绍\n\n1. 反射可以在运行时动态获取变量的各种信息, 比如变量的类型(type)，类别(kind)\n2. 如果是结构体变量，还可以获取到结构体本身的信息(包括结构体的字段、方法)\n3. 通过反射，可以修改变量的值，可以调用关联的方法。\n4. 使用反射，需要 import (“reflect”)\n\nType： 类型用来表示一个go类型。\n\nKind是一个大的分类，比如定义了一个Person类，它的Kind就是struct 而Type的名称是Person\n\nValue： 为go值提供了反射接口。\n\nValue类型的零值表示不持有某个值。零值的IsValid方法返回false，其Kind方法返回Invalid，而String方法返回””，所有其它方法都会panic。绝大多数函数和方法都永远不返回Value零值。如果某个函数/方法返回了非法的Value，它的文档必须显式的说明具体情况。",
null,
"",
null,
"",
null,
"## 二、基本用法及API\n\n• func ValueOf\n``````func ValueOf(i interface{}) Value\n``````\n\nValueOf返回一个初始化为i接口保管的具体值的Value，ValueOf(nil)返回Value零值。\n\n• func TypeOf\n``````func TypeOf(i interface{}) Type\n``````\n\nTypeOf返回接口中保存的值的类型，TypeOf(nil)会返回nil。\n\n• func (Value) Type\n``````func (v Value) Type() Type\n``````\n\n• func (Value) Elem\n``````func (v Value) Elem() Value\n``````\n\nElem返回v持有的接口保管的值的Value封装，或者v持有的指针指向的值的Value封装。如果v的Kind不是Interface或Ptr会panic；如果v持有的值为nil，会返回Value零值。\n\n• func (Type) Field\n``````func (Type) Field(i int) StructField\n``````\n\n// 返回索引序列指定的嵌套字段的类型，\n// 等价于用索引中每个值链式调用本方法，如非结构体将会panic\n\n``````type StructField struct {\n// Name是字段的名字。PkgPath是非导出字段的包路径，对导出字段该字段为\"\"。\n// 参见http://golang.org/ref/spec#Uniqueness_of_identifiers\nName string\nPkgPath string\nType Type // 字段的类型\nTag StructTag // 字段的标签\nOffset uintptr // 字段在结构体中的字节偏移量\nIndex []int // 用于Type.FieldByIndex时的索引切片\nAnonymous bool // 是否匿名字段\n}\n``````\n\n## 三、数值类型反射使用\n\n#### 基本类型的值修改\n\n``````package main\n\nimport (\n\"fmt\"\n\"reflect\"\n)\n\nfunc testBasic(i interface{}){\n//i实际是一个指针\nptrValueOfI := reflect.ValueOf(i)\nfmt.Println(ptrValueOfI.Kind())\n//获取指针指向的真正的值\nvalueOfI := ptrValueOfI.Elem()\n//为其更改值\nvalueOfI.SetInt(1000)\n}\n\nfunc main() {\nn:=1\ntestBasic(&n)\nfmt.Println(n)\n\n}\n``````\n\n``````ptr\n1000\n``````\n\n#### struct类型的属性修改\n\n``````package main\n\nimport (\n\"fmt\"\n\"reflect\"\n)\n\ntype Student struct {\nName string\nAge int\n}\n\nfunc test(i interface{}){\n//获取指针指向的真正的数值Value\nvalueOfI := reflect.ValueOf(i).Elem()\n//获取对应的Type这个是用来获取属性方法的\ntypeOfI := valueOfI.Type()\n//判断是否是struct\nif typeOfI.Kind()!=reflect.Struct{\nfmt.Println(\"except struct\")\nreturn\n}\n//获取属性的数量\nnumField := typeOfI.NumField()\n//遍历属性，找到特定的属性进行操作\nfor i:=0;i< numField;i++{\n//获得属性的StructField，次方法不同于Value中的Filed（这个返回的是Field）\nfield := typeOfI.Field(i)\n//获取属性名称\nfieldName := field.Name\nfmt.Println(fieldName)\n//找到名为Name的属性进行修改值\nif fieldName==\"Name\"{\n//改变他的值为jack\nvalueOfI.Field(i).SetString(\"jack\")\n}\n}\n}\n\nfunc main() {\nstu:=Student{Name:\"susan\",Age:58}\ntest(&stu)\nfmt.Println(stu.Name)\n}\n``````\n\n``````Name\nAge\njack\n``````\n\n## 四、引用类型修改\n\n##### slice\n\n``````package main\n\nimport (\n\"fmt\"\n\"reflect\"\n)\n\nfunc testSilce(i interface{}){\n//获取对应的Value\nvalueOfI := reflect.ValueOf(i)\n//获取Kind\nfmt.Println(\"kind = \",valueOfI.Kind())\n//获取slice长度\nfmt.Println(\"len = \",valueOfI.Len())\n//获取slice容量\nfmt.Println(\"cap = \",valueOfI.Cap())\n//获取指定位置的元素\nindexValue := valueOfI.Index(0)\nfmt.Println(\"index 0 = \",indexValue)\n//为指定位置的元素赋值\nindexValue.SetInt(300)\n}\n\nfunc main() {\ns:=make([]int,10,20)\ns=200\n//无需传入指针 因为其本身就是引用类型（slice map channel）\ntestSilce(s)\nfmt.Println(s)\n\n}\n``````\n\n``````kind = slice\nlen = 10\ncap = 20\nindex 0 = 200\n300\n\n``````\n\n##### map\n\n``````package main\n\nimport (\n\"fmt\"\n\"reflect\"\n)\n\nfunc testMap(i interface{}){\nvalueOfI := reflect.ValueOf(i)\n//获取长度\nlenOfI := valueOfI.Len()\nfmt.Println(\"len = \",lenOfI)\n//获取所有的keys\nkeys := valueOfI.MapKeys()\nfor i:=0;i<len(keys);i++{\nkey := keys[i]\n//获取对应的value\nvalue := valueOfI.MapIndex(key)\nfmt.Println(\"key=\",key,\"value=\",value)\nnum := value.Int()\n//更改数值\n//value.SetInt(num*num) 这种不行\nvalueOfI.SetMapIndex(key,reflect.ValueOf(int(num*num)))\n}\n\n}\n\nfunc main() {\nm:=make(map[string]int,10)\nm[\"a\"]=1\nm[\"b\"]=2\nm[\"c\"]=3\ntestMap(m)\nfmt.Println(m)\n}\n\n``````\n\n``````len = 3\nkey= a value= 1\nkey= b value= 2\nkey= c value= 3\nmap[a:1 b:4 c:9]\n``````\n\n## 结构体反射执行方法\n\n``````package main\n\nimport (\n\"fmt\"\n\"reflect\"\n)\n\ntype Student struct {\nName string\nAge int\n}\n\nfunc (s Student) Print() {\nfmt.Println(\"name=\", s.Name, \";age=\", s.Age)\n}\n\nfunc (Student) Plus(a int, b int) int {\nreturn a + b\n}\n\nfunc testMethod(i interface{}){\n//获取指针指向的值\nvalueOfI := reflect.ValueOf(i).Elem()\n//获取对应的Type\ntypeOfI := valueOfI.Type()\n//获取方法的数量\nmethodNum:= typeOfI.NumMethod()\nfmt.Println(\"methodNum=\",methodNum)\nfor i:=0;i<methodNum;i++{\n//获取对应位置的方法\nmethod := typeOfI.Method(i)\nmethodName := method.Name\nfmt.Println(methodName)\nif methodName ==\"Plus\"{\n//调用Plus\nm := valueOfI.MethodByName(methodName)\n//声明一个切片\nargs:=make([]reflect.Value,2)\nargs=reflect.ValueOf(78)\nargs=reflect.ValueOf(22)\nres := m.Call(args)\n//遍历结果\nfor j:=0;j<len(res);j++ {\nfmt.Println(methodName,\" ivoke result\",j+1,\" : \",res[j].Int())\n}\n\n}\n}\n}\n\nfunc main() {\nstu := Student{Name: \"susan\", Age: 58}\ntestMethod(&stu)\n}\n\n``````\n\n``````methodNum= 2\nPlus\nPlus ivoke result 1 : 100\nPrint``````\n\n### 觉得文章有用就打赏一下文章作者\n\n#### 支付宝扫一扫打赏",
null,
"#### 微信扫一扫打赏",
null,
"• QQ咨询\n• 回顶"
]
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"http://blog.haoservice.cn/wp-content/uploads/2020/08/20190302153202488.png",
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"http://blog.haoservice.cn/wp-content/uploads/2020/08/201903021532166.png",
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"http://blog.haoservice.cn/wp-content/uploads/2020/07/微信二维码.png",
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"http://blog.haoservice.cn/wp-content/uploads/2020/07/微信二维码.png",
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]
| {"ft_lang_label":"__label__zh","ft_lang_prob":0.5481234,"math_prob":0.9827627,"size":4924,"snap":"2023-14-2023-23","text_gpt3_token_len":2203,"char_repetition_ratio":0.15670732,"word_repetition_ratio":0.097510375,"special_character_ratio":0.27112103,"punctuation_ratio":0.18333334,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97656393,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,2,null,2,null,2,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-09T18:52:27Z\",\"WARC-Record-ID\":\"<urn:uuid:7402bc44-5ce3-4e84-8560-3152ef9dfc39>\",\"Content-Length\":\"71401\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:459e5cd6-2a82-428a-8eb6-86e8b4de2b9b>\",\"WARC-Concurrent-To\":\"<urn:uuid:1b805f54-8bca-4011-b841-a5c075861cea>\",\"WARC-IP-Address\":\"115.29.204.54\",\"WARC-Target-URI\":\"http://blog.haoservice.cn/archives/1429\",\"WARC-Payload-Digest\":\"sha1:BXNJE6C7DADGKHLNE775WI3VFRNJNG2T\",\"WARC-Block-Digest\":\"sha1:XJ62GH52WT37XAKFVFMA46NWY3HJ5KFK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224656788.77_warc_CC-MAIN-20230609164851-20230609194851-00760.warc.gz\"}"} |
https://www.colorhexa.com/6cfde9 | [
"# #6cfde9 Color Information\n\nIn a RGB color space, hex #6cfde9 is composed of 42.4% red, 99.2% green and 91.4% blue. Whereas in a CMYK color space, it is composed of 57.3% cyan, 0% magenta, 7.9% yellow and 0.8% black. It has a hue angle of 171.7 degrees, a saturation of 97.3% and a lightness of 70.8%. #6cfde9 color hex could be obtained by blending #d8ffff with #00fbd3. Closest websafe color is: #66ffff.\n\n• R 42\n• G 99\n• B 91\nRGB color chart\n• C 57\n• M 0\n• Y 8\n• K 1\nCMYK color chart\n\n#6cfde9 color description : Soft cyan.\n\n# #6cfde9 Color Conversion\n\nThe hexadecimal color #6cfde9 has RGB values of R:108, G:253, B:233 and CMYK values of C:0.57, M:0, Y:0.08, K:0.01. Its decimal value is 7142889.\n\nHex triplet RGB Decimal 6cfde9 `#6cfde9` 108, 253, 233 `rgb(108,253,233)` 42.4, 99.2, 91.4 `rgb(42.4%,99.2%,91.4%)` 57, 0, 8, 1 171.7°, 97.3, 70.8 `hsl(171.7,97.3%,70.8%)` 171.7°, 57.3, 99.2 66ffff `#66ffff`\nCIE-LAB 91.378, -43.636, -2.176 56.013, 79.317, 89.444 0.249, 0.353, 79.317 91.378, 43.69, 182.855 91.378, -59.234, 3.723 89.06, -43.591, 2.797 01101100, 11111101, 11101001\n\n# Color Schemes with #6cfde9\n\n• #6cfde9\n``#6cfde9` `rgb(108,253,233)``\n• #fd6c80\n``#fd6c80` `rgb(253,108,128)``\nComplementary Color\n• #6cfda1\n``#6cfda1` `rgb(108,253,161)``\n• #6cfde9\n``#6cfde9` `rgb(108,253,233)``\n• #6cc9fd\n``#6cc9fd` `rgb(108,201,253)``\nAnalogous Color\n• #fda16c\n``#fda16c` `rgb(253,161,108)``\n• #6cfde9\n``#6cfde9` `rgb(108,253,233)``\n• #fd6cc9\n``#fd6cc9` `rgb(253,108,201)``\nSplit Complementary Color\n• #fde96c\n``#fde96c` `rgb(253,233,108)``\n• #6cfde9\n``#6cfde9` `rgb(108,253,233)``\n• #e96cfd\n``#e96cfd` `rgb(233,108,253)``\n• #80fd6c\n``#80fd6c` `rgb(128,253,108)``\n• #6cfde9\n``#6cfde9` `rgb(108,253,233)``\n• #e96cfd\n``#e96cfd` `rgb(233,108,253)``\n• #fd6c80\n``#fd6c80` `rgb(253,108,128)``\n• #21fcde\n``#21fcde` `rgb(33,252,222)``\n• #3afce1\n``#3afce1` `rgb(58,252,225)``\n• #53fde5\n``#53fde5` `rgb(83,253,229)``\n• #6cfde9\n``#6cfde9` `rgb(108,253,233)``\n• #85fded\n``#85fded` `rgb(133,253,237)``\n• #9efef1\n``#9efef1` `rgb(158,254,241)``\n• #b7fef4\n``#b7fef4` `rgb(183,254,244)``\nMonochromatic Color\n\n# Alternatives to #6cfde9\n\nBelow, you can see some colors close to #6cfde9. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #6cfdc5\n``#6cfdc5` `rgb(108,253,197)``\n• #6cfdd1\n``#6cfdd1` `rgb(108,253,209)``\n• #6cfddd\n``#6cfddd` `rgb(108,253,221)``\n• #6cfde9\n``#6cfde9` `rgb(108,253,233)``\n• #6cfdf5\n``#6cfdf5` `rgb(108,253,245)``\n• #6cf9fd\n``#6cf9fd` `rgb(108,249,253)``\n• #6cedfd\n``#6cedfd` `rgb(108,237,253)``\nSimilar Colors\n\n# #6cfde9 Preview\n\nText with hexadecimal color #6cfde9\n\nThis text has a font color of #6cfde9.\n\n``<span style=\"color:#6cfde9;\">Text here</span>``\n#6cfde9 background color\n\nThis paragraph has a background color of #6cfde9.\n\n``<p style=\"background-color:#6cfde9;\">Content here</p>``\n#6cfde9 border color\n\nThis element has a border color of #6cfde9.\n\n``<div style=\"border:1px solid #6cfde9;\">Content here</div>``\nCSS codes\n``.text {color:#6cfde9;}``\n``.background {background-color:#6cfde9;}``\n``.border {border:1px solid #6cfde9;}``\n\n# Shades and Tints of #6cfde9\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000807 is the darkest color, while #f3fffd is the lightest one.\n\n• #000807\n``#000807` `rgb(0,8,7)``\n• #001b17\n``#001b17` `rgb(0,27,23)``\n• #012f28\n``#012f28` `rgb(1,47,40)``\n• #014239\n``#014239` `rgb(1,66,57)``\n• #01554a\n``#01554a` `rgb(1,85,74)``\n• #01695a\n``#01695a` `rgb(1,105,90)``\n• #027c6b\n``#027c6b` `rgb(2,124,107)``\n• #028f7c\n``#028f7c` `rgb(2,143,124)``\n• #02a38d\n``#02a38d` `rgb(2,163,141)``\n• #02b69d\n``#02b69d` `rgb(2,182,157)``\n• #03c9ae\n``#03c9ae` `rgb(3,201,174)``\n• #03ddbf\n``#03ddbf` `rgb(3,221,191)``\n• #03f0cf\n``#03f0cf` `rgb(3,240,207)``\n• #0bfcdb\n``#0bfcdb` `rgb(11,252,219)``\n• #1ffcdd\n``#1ffcdd` `rgb(31,252,221)``\n• #32fce0\n``#32fce0` `rgb(50,252,224)``\n• #45fce3\n``#45fce3` `rgb(69,252,227)``\n• #59fde6\n``#59fde6` `rgb(89,253,230)``\n• #6cfde9\n``#6cfde9` `rgb(108,253,233)``\n• #7ffdec\n``#7ffdec` `rgb(127,253,236)``\n• #93feef\n``#93feef` `rgb(147,254,239)``\n• #a6fef2\n``#a6fef2` `rgb(166,254,242)``\n• #b9fef5\n``#b9fef5` `rgb(185,254,245)``\n• #cdfef7\n``#cdfef7` `rgb(205,254,247)``\n• #e0fffa\n``#e0fffa` `rgb(224,255,250)``\n• #f3fffd\n``#f3fffd` `rgb(243,255,253)``\nTint Color Variation\n\n# Tones of #6cfde9\n\nA tone is produced by adding gray to any pure hue. In this case, #b1b8b7 is the less saturated color, while #6cfde9 is the most saturated one.\n\n• #b1b8b7\n``#b1b8b7` `rgb(177,184,183)``\n• #abbebb\n``#abbebb` `rgb(171,190,187)``\n• #a5c4c0\n``#a5c4c0` `rgb(165,196,192)``\n• #a0c9c4\n``#a0c9c4` `rgb(160,201,196)``\n• #9acfc8\n``#9acfc8` `rgb(154,207,200)``\n• #94d5cc\n``#94d5cc` `rgb(148,213,204)``\n• #8edbd0\n``#8edbd0` `rgb(142,219,208)``\n• #89e0d4\n``#89e0d4` `rgb(137,224,212)``\n• #83e6d8\n``#83e6d8` `rgb(131,230,216)``\n• #7decdd\n``#7decdd` `rgb(125,236,221)``\n• #77f2e1\n``#77f2e1` `rgb(119,242,225)``\n• #72f7e5\n``#72f7e5` `rgb(114,247,229)``\n• #6cfde9\n``#6cfde9` `rgb(108,253,233)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #6cfde9 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.5351698,"math_prob":0.7536834,"size":3709,"snap":"2019-35-2019-39","text_gpt3_token_len":1678,"char_repetition_ratio":0.12631579,"word_repetition_ratio":0.011111111,"special_character_ratio":0.5238609,"punctuation_ratio":0.23809524,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9845967,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-16T16:33:56Z\",\"WARC-Record-ID\":\"<urn:uuid:1d8f7e17-8998-4bcf-b374-08bb6067f76e>\",\"Content-Length\":\"36339\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:300f6a06-3148-434e-b6bf-f10d464aa9b6>\",\"WARC-Concurrent-To\":\"<urn:uuid:b41b3239-6943-487d-abf0-6626aee16a13>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/6cfde9\",\"WARC-Payload-Digest\":\"sha1:5J5KC3ZC7BX2I7FEAF7AEE6RP2E5ZAB6\",\"WARC-Block-Digest\":\"sha1:CIBTU5QWYD7XULXPVM7JZ3EDDQHA6SXZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514572879.28_warc_CC-MAIN-20190916155946-20190916181946-00537.warc.gz\"}"} |
http://freebsd.dk/hacks/HP5065A/20150828_c_pot/ | [
"# HP5065A C-field potentiometer¶\n\nMy attempt to reference the C-field constant current generator with my Fluke 732A voltage reference instead of the CR5 zener didn’t improve stability a lot, so the search for temperature coeffients continue.\n\nThe two most important resistors for C-field stability are R10 and R11 which sense the current through the C-field coil:",
null,
"The little asterix next to R10 indicates that it is “factory selected”, but the note doesn’t say what they select it for.\n\nMy guess is that it is selected for the voltage of the CR5 Zener, which does not have a particularly tight specification for voltage.\n\nIn my case, R10 is a 11k 1% resistor, giving a combined resistance of 680R for R10||R11.\n\nWhen I measured it by pulling A15 out, I found 677.4 Ohm and as the board cooled down, it rose to 677.8 Ohm over a few minutes.\n\nThat is a change of 600ppm for an unknown change in temperature.\n\nNot good.\n\nWhat is the criteria that R10||R11 and the CR4 zener voltage most fulfill?\n\nI measured the Q6[AB] base voltages as a function of the C-field potentiometer and calculated the resulting C-field current, based on a nominal 680R R10||R11 combination:",
null,
"This matches the box on page 8-64, which says that the current should be “2.5 to 6 mA”\n\nAs you can see the curve is not flat, this is deliberate, the Rb resonance does not depend linearly on the magnetic field, and the bend in this curve is what “calibrates” the C-field pot to be linear in frequency. (See TVB’s measurements of this here: HP 5065A Rubidium C-Field Resolution)\n\nThe line marked “Theory” is calculated as follows:\n\n```Lower_arm {Ohm} = (333 {R12, Ohm} + X {R6, Ohm})\n# Note X = C-field pot resistance in Ohms = reading on dial.\n\nTotal {Ohm} = Lower_arm + 1333 {R8, Ohm}\n\nV_q6a {Volt} = V_cr5 {Volt} * (Lower_arm {Ohm} / Total {Ohm})\n\nI_L1 {A} = V_q6a {Volt} / 680 {R10||R11, Ohm}\n# Note 680 Ohm = 725 Ohm (R11) parallel with 11000 Ohm (R10)\n```\n\nThat’s a pretty good match.\n\nLooking at the curve, and taking into account that the C-field setting 250 is used for the calibration, I think it is a pretty good guess that R10 is selected for 4mA C-field current when the C-field pot is set to 250.\n\nIf that is true, then the optimal value of R10||R11 is 76 times the reference voltage.\n\n## Various notes¶\n\nSomething bad clearly happens around 700 on the pot, looks like a loose connection but could also be instability in the circuit.\n\nI also measured this curve with the Fluke 732A reference, the higher 10V reference means that R10||R11 should be 770R to replicate these currents.\n\nThe offset between the two curves is Q6[AB] not being well matched. The difference is on the order of 45 mV which means replacing them with pretty much any op-amp would be an improvement.\n\nSo a recipe for a new circuit at this point sounds like: LT1021 voltage reference, decent op-amp, low temp-co resistor(s) to replace R10||R11.\n\nphk"
]
| [
null,
"http://freebsd.dk/_images/schem1.png",
null,
"http://freebsd.dk/_images/plot11.svg",
null
]
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https://enya.day/mathematics/question-956.html | [
"# I need help with this question.",
null,
"s = 6.3 in\n\nStep-by-step explanation:\n\nThe formula to find length of arc is\n\narc length = 2 x",
null,
"x r x (",
null,
")",
null,
"is the angle\n\nSo, arc length = 2 x",
null,
"x 8 x (",
null,
")\n\n= 6.3 in"
]
| [
null,
"https://enya.day/show_image.php",
null,
"https://tex.z-dn.net/",
null,
"https://tex.z-dn.net/",
null,
"https://tex.z-dn.net/",
null,
"https://tex.z-dn.net/",
null,
"https://tex.z-dn.net/",
null
]
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https://dict.woxikon.com/en-es/bisector | [
"Search term bisector has 3 results\nEN English ES Spanish\n(n) [geometry] bisector (n) {m} [geometry]\n(n) [A line or curve that bisects or divides a line segment, angle, or other figure into two equal parts] bisectriz (n) [A line or curve that bisects or divides a line segment, angle, or other figure into two equal parts]\nES Spanish EN English\n(n) [geometría] {m} bisector (n) [geometría]"
]
| [
null
]
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https://physics.stackexchange.com/questions/313680/calculating-force-through-icor-instantaneous-centre-of-rotation | [
"# Calculating force through ICOR (instantaneous centre of rotation)",
null,
"A purely rolling wheel of radius $R$, has its ICOR at the point $P$ (as per the figure).\n\nIf I calculate the centripetal force on a particle at the top most point of the rim, considering point $P$ as its center of rotation, I get the value of the force to be $F=2mR \\omega^2$, where $\\omega$ represents angular velocity.\n\nIf I calculate the force using the radius of curvature of cycloid, I end up with the value of $F=mR\\omega^2$, which is indeed true.\n\nMy concern is that why doesn't ICOR provide me with the correct answer.\n\n• Centripetal force is what keeps the particle in circular motion. The topmost particle is moving at a radius of $R$, why have you taken $2R$? Even in the frame of reference of the lowermost point, the particle is in uniform circular motion in a circle of radius $R$. – Yashas Feb 21 '17 at 14:08\n• @Yashas, in ground frame I would perceive the particle to move in a circle of radius 4 R(as 4R is the length of radius of curvature of the cycloid at top most point) and yes since it is force so the value won't change even if I see it from another inertial frame that is the centre of the circle which is moving along a straight line. But my concern here is that ICOR is also a static point thus it will make me feel as if the whole wheel is rotating about it(momentarily) then why doesn't it yield the correct answer. – Vasu Goyal Feb 21 '17 at 14:37\n• \"I would perceive the particle to move in a circle of radius $4R$ \" - that is the flaw in your reasoning. Consider a particle moving in projectile motion, at the highest point, try to calculate the radius. The radius is not the maximum height of the projectile. Rather, it is $v^2/g$. – Yashas Feb 21 '17 at 14:47\n• @Yashas, I appreciate your efforts but I think that I am not able to make myself clear here. What I mean(in a general sense) here is why can't I consider centripetal force on a body about its ICOR. And yes of course I am not getting confused in considering the bottom most point as point of rotation as you said in projectile (parabolic trajectory) example. – Vasu Goyal Feb 21 '17 at 14:55\n• The topmost particle is NOT rotating about the point P. The radius is not $2R$. It doesn't matter which frame of reference you are in, the topmost point is rotating in a circle of radius $R$. Distances don't change over frames of reference. – Yashas Feb 21 '17 at 14:58\n\nThe \"true\" acceleration of the rim point A is $a_A = 2 R \\omega^2$. How did you decide the true value is $R \\omega^2$?\n\nYou can look at the ICOR and arrive at $2 R \\omega^2$, or you can add the acceleration of A due to the rotation $R \\omega^2$ plus the acceleration of the center of mass $R \\omega^2$ to arrive at the same value.\n\nBTW Taking the acceleration of A and multiplying by the entire mass $m$ of the body is meaningless as a force quantity. Only the acceleration of the center of mass is important in deriving inertial forces. The is because linear momentum is defined as the mass of a body $m$ times the velocity of the center of mass $v_C = \\omega R$ and force is the time derivative of momentum.\n\n• firstly $m$ is the mass of a particle at top of the wheel, not the entire wheel. Secondly, if we take the same approach to find the acceleration of centre of the wheel, the acceleration comes to be $Rω²$, and is false as the trajectory of the centre is a straight line, thus the approach through ICOR method is incorrect without considering pseudo force. – Vasu Goyal Feb 24 '17 at 16:30\n\n$P$ is a non-inertial frame of reference which is why it does not provides us with correct answer. In order to obtain the correct result we need to write the relation between accelerations as $a$$real = a$$0$ +$a'$. Where $a$$real denotes acceleration of the particle(whose motion is considered) in inertial frame, a$$0$ denotes acceleration of non-inertial frame and $a'$ denotes acceleration of particle in non-inertial frame. $a$$0$ is equal to $Rω²$ directed towards centre of the wheel whereas $a'$ is the acceleration of the top most point equal to $2Rω²$ directed radially towards the center of the wheel.\n\n*Note: Since both points, ICOR and top most particle are diametrically opposite hence addition of acceleration vectors pointing towards centre produce a cancelation effect.\n\nAdding both vectors results in real acceleration equal to $Rω²$ pointing towards the centre of the wheel, which is the correct result.",
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"https://i.stack.imgur.com/7wr5O.png",
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"https://i.stack.imgur.com/ZoFdX.jpg",
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https://doc.arcgis.com/en/iot/analyze/generalized-linear-regression.htm | [
"# Generalized Linear Regression",
null,
"Available in big data analytics.\n\nThe Generalized Linear Regression tool",
null,
"performs Generalized Linear Regression (GLR) to generate predictions or to model a dependent variable in terms of its relationship to a set of explanatory variables. This tool can be used to fit Continuous (Gaussian), Count (Poisson), and Binary (Logistic) model types.\n\n## Workflow diagram",
null,
"## Example\n\nAs an analyst for a large city, you have access to past 911 call records and demographic information. You need to answer the following questions: What variables effectively predict 911 call volume? Given future projections, what is the expected demand for emergency response resources?\n\n## Usage notes\n\n• This tool can be configured to perform one of two operational methods:\n• Method 1—If only target (training) data is provided, the tool will fit a model to assess model performance. The tool then allows you to evaluate the performance of different models as you explore different explanatory variables and tool settings.\n• Method 2—Once you have identified a good model and explanatory variables, configure the model to also provide join (prediction) data. When join data is configured, the tool will predict values for the dependent variable for features in your join (prediction) data based on the mapped explanatory variables.\n• Use the Dependent variable parameter to select a field from the Target Input Layer (training data) representing the phenomena you are modeling. Use the Explanatory variables parameter to select one or more fields representing the explanatory variables from the Target Input Layer (training data). These fields must be numeric and have a range of values. Features that contain missing values in the dependent or explanatory variable will be excluded from the analysis. To modify null values, use the Calculate Field tool prior to updating values.\n• The Generalized Linear Regression tool also produces output features and diagnostics. Output feature layers automatically have a rendering scheme applied to model residuals. A full explanation of each output is provided below.\n• It is important to use the correct model type—Continuous (Gaussian), Count (Poisson), or Binary (Logistic)—for analysis to obtain accurate results of the regression analysis.\n• Model summary results and diagnostics are written to the analytic logs as well as the output feature layer item details page. These diagnostics include a summary of the Generalized Linear Regression model and statistical summaries that are used to assess whether a model is a good fit for the data. The diagnostics reported depend on the model type chosen. The three options for Model type are as follows:\n• Continuous (Gaussian)—Use if the dependent variable can take on a wide range of values such as temperature or total sales. Ideally, the dependent variable will be normally distributed.\n• Count (Poisson)—Use if the dependent variable is discrete and represents the number of occurrences of an event such as a count of crimes. Count models can also be used if the dependent variable represents a rate and the denominator of the rate is a fixed value such as sales per month or number of people with cancer per 10,000 in the population. A Count (Poisson) model type assumes the mean and variance of the dependent variable are equal and the values of the dependent variable cannot be negative or contain decimals.\n• Binary (Logistic)—Use if the dependent variable can take on one of two possible values such as success or failure, or presence or absence. The field containing the dependent variable must be numeric and contain only ones and zeros. There must be variation of the ones and zeros in the data.\n• The Dependent variable and Explanatory variable(s) parameters should be numeric fields containing a range of values. This tool cannot solve when variables have the same values (for example, if all the values for a field are 9.0).\n• Features with one or more null values or empty string values in prediction or explanatory fields will be excluded from the output. If necessary, modify values using the Calculate Field tool.\n• Visually inspect the over- and under-predictions evident in the regression residuals to see whether they provide clues about potential missing variables from your regression model.\n• Use the regression model that was created to make predictions for other features. Creating these predictions requires that each prediction feature (join dataset) has values for each of the explanatory variables specified. An explanatory variable mapping configuration is provided to map explanatory variable field names from the target (training) features and join (prediction) features. When matching the explanatory variable fields, the fields from the target (training data) and join (prediction data) features must be of the same type (for example, double fields must be matched with double fields).\n\n## Parameters\n\nParameterDescriptionData type\n\nTarget Input Layer (training data)\n\nThe training features used to generate a model.\n\nFeatures\n\nJoin Input Layer (prediction data)\n\n(Optional)\n\nThe prediction features for which the dependent variable will be predicted based on the specified explanatory variables and model type.\n\nThis parameter is optional. If not specified, the Generalized Linear Regression tool will fit a model to assess model performance based on the training data.\n\nFeatures\n\nModel type\n\nSpecifies the model type to use. The model type chosen depends on the type of data in the dependent variable field. Model type options include the following:\n\n• Continuous (Gaussian)—Choose if the dependent variable can take on a wide range of values such as temperature or total sales.\n• Count (Poisson)—Choose if the dependent variable is discrete and represents the number of occurrences of an event such as a count of crimes or a rate and the denominator of the rate is a fixed value.\n• Binary (Logistic)—Choose if the dependent variable can take on one of two possible values such as success or failure, or presence or absence.\n\nString\n\nDependent variable\n\nSpecifies the field representing the phenomena you are modeling.\n\nFieldName\n\nText to zero mapping\n\nFor theBinary (Logistic) model type, if a string field is specified for the Dependent variable, this parameter can be used to specify the string in the dependent variable to convert to zero.\n\nString\n\nText to one mapping\n\nFor the Binary (Logistic) model type, if a string field is specified for the Dependent variable, this parameter can be used to specify the string in the dependent variable to convert to one.\n\nString\n\nExplanatory variable(s)\n\nField or fields from the target schema to represent independent explanatory variables in the regression model.\n\nFieldNames\n\nExplanatory variable mapping (prediction only)\n\nMaps the selected explanatory variable field names in the target (training) schema to corresponding field names in join (predict) schema.\n\nThis parameter is optional. Explanatory variable mappings are only required to be specified if join (prediction) data is specified.\n\nExplanatoryVariableMappings\n\n## Output layer\n\nThe Generalized Linear Regression tool produces a variety of outputs. A summary of the Generalized Linear Regression model and statistical summaries are available on the item details page of the output feature layer or in the analytic logs.\n\nIf implementing Method 1 of this tool to simply fit a model to assess performance, the training data will be the output, as well as messages and diagnostics available in the item details of the output feature layer in addition to results in the analytic logs.\n\nIf implementing Method 2 of this tool to fit a model and predict values, the prediction data will be the output with predicted values appended, as well as messages and diagnostics available in the item details of the output feature layer in addition to results in the analytic logs.\n\nThe diagnostics generated depend on the model type of the input features and are described below.\n\n### Continuous (Gaussian)\n\n#### Interpret messages and diagnostics\n\n• AIC—This is a measure of model performance and can be used to compare regression models. Taking into account model complexity, the model with the lower AIC value provides a better fit to the observed data. AIC is not an absolute measure of goodness of fit but is useful for comparing models with different explanatory variables as long as they apply to the same dependent variable. If the AIC values for two models differ by more than 3, the model with the lower AIC value is considered more accurate.\n• AICc—AICc applies a bias correction to AIC for small sample sizes. AICc will approach AIC as the number of features in the input increase. See AIC above.\n• Multiple R-Squared—The R-Squared is a measure of goodness of fit. Its value varies from 0.0 to 1.0, with higher values being preferable. It may be interpreted as the proportion of dependent variable variance accounted for by the regression model. The denominator for the R-Squared computation is the sum of squared dependent variable values. Adding an extra explanatory variable to the model does not alter the denominator but does alter the numerator; this gives the impression of improvement in model fit that may not be real. See Adjusted R-Squared below.\n• Adjusted R-Squared—Because of the problem described above for the R-Squared value, calculations for the adjusted R-Squared value normalize the numerator and denominator by their degrees of freedom. This has the effect of compensating for the number of variables in a model, and consequently, the Adjusted R-Squared value is almost always less than the R-Squared value. However, in making this adjustment, you lose the interpretation of the value as a proportion of the variance explained. In Geographically Weighted Regression (GWR), the effective number of degrees of freedom is a function of the neighborhood used, so the adjustment may be quite marked in comparison to a global model such as GLR. For this reason, AICc is preferred as a means of comparing models.\n\n### Count (Poisson)\n\n#### Interpret messages and diagnostics\n\n• AIC—This is a measure of model performance and can be used to compare regression models. Taking into account model complexity, the model with the lower AIC value provides a better fit to the observed data. AIC is not an absolute measure of goodness of fit but is useful for comparing models with different explanatory variables as long as they apply to the same dependent variable. If the AIC values for two models differ by more than 3, the model with the lower AIC value is considered more accurate.\n• AICc—AICc applies a bias correction to AIC for small sample sizes. AICc will approach AIC as the number of features in the input increase. See AIC above.\n\n### Binary (Logistic)\n\n#### Interpret messages and diagnostics\n\n• AIC—This is a measure of model performance and can be used to compare regression models. Taking into account model complexity, the model with the lower AIC value provides a better fit to the observed data. AIC is not an absolute measure of goodness of fit but is useful for comparing models with different explanatory variables as long as they apply to the same dependent variable. If the AIC values for two models differ by more than 3, the model with the lower AIC value is considered more accurate.\n• AICc—AICc applies a bias correction to AIC for small sample sizes. AICc will approach AIC as the number of features in the input increase. See AIC above.\n\n## Considerations and limitations\n\nThe ArcGIS Velocity implementation of Generalized Linear Regression has the following limitations:\n\n• It is a global regression model and does not take the spatial distribution of data into account.\n• Analysis does not apply Moran's I test on the residuals.\n• Points, lines, polygons, and tables are supported as target (training data) dataset geometry.\n• You cannot classify values into multiple classes."
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"https://doc.arcgis.com/en/iot/analyze/GUID-6FBBEC30-E4BF-462A-86D9-E80C2842259F-web.png",
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https://lecatalogue.info/hewitt-the-role-of-compactness-in-analysis-49/ | [
"# HEWITT THE ROLE OF COMPACTNESS IN ANALYSIS PDF\n\nToposym 1. Edwin Hewitt. Some applications to harmonic analysis, and so clearly illustrate the importance of compactness, that they should be cited. The first. This paper traces the history of compactness from the original motivating questions E. Hewitt, The role of compactness in analysis, Amer. Compactness. The importance of compactness in analysis is well known (see Munkres, p). In real anal- ysis, compactness is a relatively easy property to.",
null,
"Author: Yole Meztigrel Country: South Africa Language: English (Spanish) Genre: Science Published (Last): 27 February 2006 Pages: 329 PDF File Size: 18.8 Mb ePub File Size: 3.32 Mb ISBN: 630-6-49877-995-9 Downloads: 71177 Price: Free* [*Free Regsitration Required] Uploader: Dagore",
null,
"In this topology we have less open sets which implies more compact sets and in particular, bounded sets are pre-compact sets. To prove your theorem without it: This relationship is a useful one because we now have a notion which is strongly related to boundedness which does generalise to topological spaces, unlike boundedness itself. As many have said, compactness is sort of a topological generalization jn finiteness. So, at least for closed sets, compactness and boundedness are the same.\n\n### general topology – Why is compactness so important? – Mathematics Stack Exchange\n\nThis list is far from over Compact spaces, being “pseudo-finite” in their nature are also well-behaved and we can prove interesting things about them. One reason is that boundedness doesn’t make sense in a general topological space. So why then compactness? The only theorems I’ve seen concerning it are the Heine-Borel theorem, and a proof continuous heitt on R from closed subintervals of R are bounded. Naalysis there a redefinition of discrete so this principle works for all topological spaces e.\n\nBy the way, as always, very nice to read your answers. Please, could you detail more your point of view to me? A locally compact abelian group is compact if and only if its Pontyagin dual is discrete.\n\nMathematics Stack Exchange compadtness best with JavaScript enabled. In addition, at least for Hausdorff topological spaces, compact sets are closed. I can’t think of a good example to make this more precise now, though. Here are some more useful things: Since there are a lot of theorems in real and complex analysis that uses Heine-Borel theorem, so the idea of compactness is too important.\n\nJOHN ZERZAN ELEMENTS OF REFUSAL PDF\n\nAnd this is true in a deep sense, because topology deals with open sets, and this means that we often “care about how something behaves on an open set”, and for compact spaces this means that there are only finitely many possible behaviors. Evan 3, 8 Consider the following Theorem:. Compactness is the next best thing to finiteness. If it helps answering, I am about to enter my third year of my undergraduate degree, and came to wonder this upon preliminary reading of introductory topology, where I first found the definition of compactness.\n\nAnd when one learns about first order logic, gets the feeling that compactness is, somehow, deduce information about an “infinite” object by deducing it from its “finite” or from a finite number of parts.\n\n## There was a problem providing the content you requested\n\nHonestly, discrete spaces come closer to my intuition for finite spaces than do compact spaces. In probability they use the term “tightness” for measures Hmm.",
null,
"It gives you convergent subsequences when working with arbitrary sequences that aren’t known to converge; the Arzela-Ascoli theorem hewit an important instance of this for snalysis space of continuous functions this point of view is the basis for various “compactness” methods in the theory of non-linear PDE. Every universal net in a compact set converges.\n\nSo they end up being useful for that reason. Every continuous function is Riemann integrable-uses Heine-Borel theorem. And yet, we work so much with these properties.\n\nThank you for the compliment. The concept of a “coercive” function was unfamiliar to me until I read your answer; I suspect the same will be true for many readers.",
null,
"FAT LAND BY GREG CRITSER PDF",
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""
]
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"x-raw-image:/6949928cfeaca6cb4a5d016c96753531ba87b0cfbe7501d749cf294bc34ef3de",
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"https://lecatalogue.info/download_pdf.png",
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"https://i1.rgstatic.net/publication/266019643_Optimization_Without_Compactness_and_Its_Applications/links/568aab3d08aebccc4e1a12a1/largepreview.png",
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"https://i1.rgstatic.net/publication/255612002_Volterra_Convolution_Operators_with_Values_in_Rearrangement_Invariant_Spaces/links/0f31753ba9b363ae99000000/largepreview.png",
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"https://i1.rgstatic.net/publication/258233532_Finite_tightness_and_applications_to_limits_of_ODE\\u0027_s_solutions/links/02e7e5277b8c8cb28d000000/largepreview.png",
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https://verimag.gricad-pages.univ-grenoble-alpes.fr/synchrone/sasa/_html/algo/Algo/index.html | [
"# Module `Algo`\n\n## The Algorithm programming Interface\n\nA SASA process is an instance of an algorithm defined via this module.\n\n### To be provided by users\n\nIn order to define a SASA algorithm, one need to provide an enable and a step function:\n\n• the enable function states which actions can be triggered;\n• the step function triggers an enable action.\n\nThe enable and the step functions should be of the type ` 's enable_fun ` and ` 's step_fun ` respectively.\n\n`type 's neighbor`\n`type action`` = string`\n`type 's enable_fun`` = 's -> 's neighbor list -> action list`\n`type 's step_fun`` = 's -> 's neighbor list -> action -> 's`\n\n`enable_fun` and `step_fun` have the same arguments in input:\n\n• The first argument holds the current state of the process. As it is polymorphic (`'s`), algorithms designers can put anything they need into this state (an integer, a structure, etc.). The only constraint is that all algorithms should use the same type.\n• The second argument holds the process neighbors (its successors in the graph). Note that SASA processes, that live in graph nodes, can only access to their immediate neighbors. From each neighbor, a process can access to various information (cf `state`, `reply`, and `weight` functions below).\n\n`enable_fun` returns the list of enable actions.\n\n`step_fun`, which also takes an action in input, returns the new value of the local state resulting from the execution of the action.\n\n`type 's state_init_fun`` = int -> string -> 's`\n\nThe initial value of the local state can be set using an initialization function that takes as input 2 arguments: the number of node neighbors, and the pid. Anonymous algorithms are not supposed to use the pid.\n\n### To be provided optionally\n\n#### Potential function\n\nLet the user define what the potential of a configuration is. Used to explore best/worst case daemons (--worst-daemon)\n\n`type pid`` = string`\n`type 's potential_fun`` = pid list -> (pid -> 's * ('s neighbor * pid) list) -> float`\n\n#### Legitimate Configurations\n\n`type 's legitimate_fun`` = pid list -> (pid -> 's * ('s neighbor * pid) list) -> bool`\n\nBy default, legitimate configurations (i.e., global states) are silent ones. But this is not true for all algorithms. Predicates of this type are used to redefine what's a legitimate configuration is.\n\n#### Fault Injection\n\n`type 's fault_fun`` = int -> string -> 's -> 's`\n\nThe fault function is called on each node to update their state each time a legitimate configuration is reached. It takes 3 arguments: the number of node neighbors, the pid, and the value of the current state.\n\n### Can be used in algorithms\n\n#### Process (local) Information\n\n`val state : 's neighbor -> 's`\n\nReturns the neighbor processes local state\n\n`val reply : 's neighbor -> int`\n\nReturns the neighbor channel number, that let this neighbor access to the content of the current process, if its neighbor can access it. The channel number is the rank, starting at 0, in the neighbors' list. Returns -1 if the neighbor can not access to the process, which may happen in directed graphs.\n\nAn algorithm that uses reply, and not the pid, is called semi-anonymous. It is called anonymous if it can access none.\n\n`val weight : 's neighbor -> int`\n\nReturns the weight of the edge from the current node to the neighbor. Note that \"weight\" is an edge (dot) attribute. 1 is returned if not weight is set in the graph.\n\n`val print_neighbor : 's neighbor -> unit`\n\nFor debugging\n\n`val fmt_print_neighbor : Stdlib.Format.formatter -> 's neighbor -> unit`\n\n#### Topological Information\n\nWhen writing algorithms, one can also have access to topological information (i.e., information that are relative to the graph)\n\n`val card : unit -> int`\n`val min_degree : unit -> int`\n`val mean_degree : unit -> float`\n`val max_degree : unit -> int`\n`val is_directed : unit -> bool`\n`val is_cyclic : unit -> bool`\n`val is_connected : unit -> bool`\n`val diameter : unit -> int`\n`val is_tree : unit -> bool`\n\n#### Trees\n\n`val is_in_tree : unit -> bool`\n`val is_out_tree : unit -> bool`\n`val sub_tree_size : string -> int`\n`val height : string -> int`\n\nmaps each node to its height in the tree\n\n`val parent : string -> int option`\n\nmaps each node to the channel number of its parent, and to None for the tree root.\n\n`val get_graph_attribute : string -> string`\n\nIt is possible to set some global parameters in the dot file using graph attributes. This function allows one the get their values.\n\n### Code Registration\n\nThe `register: 's to_register -> unit` function must be called once in the user code.\n\n`type 's algo_to_register`` = ``{`\n `algo_id : string;` `init_state : 's state_init_fun;` `enab : 's enable_fun;` `step : 's step_fun;`\n`}`\n`type 's to_register`` = ``{`\n `algo : 's algo_to_register list;` `state_to_string : 's -> string;` `state_of_string : (string -> 's) option;` `copy_state : 's -> 's;` `actions : action list;` Mandatory in custom daemon mode, or to use oracles `potential_function : 's potential_fun option;` Mandatory with Evil daemons `legitimate_function : 's legitimate_fun option;` `fault_function : 's fault_fun option;` called at legitimate configuration\n`}`\n• For the `state_to_string` field, the idea is to print the raw values contained in `'s`. If a value is omitted, one won't see it in the simulation outputs. If one prepend a value with \"some_id\", some_id will we used in the simulation outputs. Otherwise, an id will be invented by sasa.\n• For the `state_of_string` field, if some function is provided, sasa should be able to parse state init values in the dot.\n`val register : 's to_register -> unit`\n\nTo be called once\n\n## Automatic Code Registration\n\nOnce enable and step functions are defined, they need to be registered by calling the register function.\n\nAn alternative to writing this registration code is to let sasa generate it with the \"--gen-register\" (or \"-reg\") option. In this case, one needs to follow some naming conventions (w.r.t file and function names):\n\n(1) The internal state (i.e. the 's shared by all algos) should be defined in a file named \"state.ml\" that defines the following 5 entities:\n\n``````type t = define_me\nlet (to_string: t -> string) = fun _ -> \"define_me\"\nlet (of_string: string -> t option) = fun _ -> None\nlet (copy : t -> t) = fun x -> x\nlet actions = [\"action1\";\"action2\"]; ``````\n\nIf one does not want/need to provide the of_string state parser, returning None is fine. This is mandatory only if one wants to define initial values in the dot file.\n\nDefining a copy that is not the identity is necessary if the state is not functional (e.g., if it contains an array or an Hashtbl).\n\nIn the file \"state.ml\" does not exist in the current directory, a skeleton is generated.\n\n(2) The optional functions relative to the global configuration should be defined in a file named \"config.ml\" that defines the following 5 entities:\n\n``````let potential_function = None (** Mandatory with --worst-daemon (-wd) *);\nlet legitimate_function = None (** if different from silent ones *);\nlet fault_function = None (** called when a legitimate configuration is\nreached *) ``````\n\n(3) All the algos mentioned in the dot file should define the following functions:\n\n``````let (init_state: int -> string -> State.t) = xxx finishme\nlet (enable_f: State.t neighbor list -> State.t -> action list) = xxx\nlet (step_f : State.t neighbor list -> State.t -> action -> State.t) = xxx``````"
]
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null
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http://mathonline.wikidot.com/the-discrete-metric | [
"The Discrete Metric\n\n# The Discrete Metric\n\nRecall from the Metric Spaces page that if $M$ is a nonempty set then a function $d : M \\times M \\to [0, \\infty)$ is called a metric if for all $x, y, z \\in M$ we have that the following three properties hold:\n\n• $d(x, y) = d(y, x)$.\n• $d(x, y) = 0$ if and only if $x = y$.\n• $d(x, y) \\leq d(x, z) + d(z, y)$.\n\nFurthermore, the set $M$ with the metric $d$, denoted $(M, d)$ is called a metric space.\n\nWe will now look at a rather important metric of any nonempty set $M$ known as the discrete metric.\n\n Definition: Let $M$ be a nonempty set. The Discrete Metric on $M$ is defined to be the function $d : M \\times M \\to [0, \\infty)$ defined for all $x, y \\in M$ by $d(x, y) = \\left\\{\\begin{matrix} 0 & \\mathrm{if} \\: x = y\\\\ 1 & \\mathrm{if} \\: x \\neq y \\end{matrix}\\right.$.\n\nLet's verify that $d$ is indeed a metric. Let $x, y, z \\in M$.\n\nFor the first condition, suppose that $x = y$. Then $d(x, y) = 0$ and $d(y, x) = 0$, so $d(x, y) = d(y, x)$. Now suppose that $x \\neq y$. Then $d(x, y) = 1$ and $d(y, x) = 1$, so $d(x, y) = d(y, x)$ once again.\n\nFor the second condition, we have by the definition of $d$ we have that $d(x, y) = 0$ if and only if $x = y$.\n\nFor the third condition, suppose that $x = y$. Then $d(x, y) = 0$ and $d(x, z), d(z, y) \\geq 0$ so $d(x, y) \\leq d(x, z) + d(z, y)$. Now suppose that $x \\neq y$. Then $d(x, y) = 1$. The element $z$ cannot equal both $x$ and $y$ since $x \\neq y$, so $d(x, z) + d(z, y) \\geq 1$ so $d(x, y) \\leq d(x, z) + d(z, y)$.\n\nTherefore $d : M \\times M \\to [0, \\infty)$ is indeed a metric on $M$ and $(M, d)$ is a metric space."
]
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null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7488812,"math_prob":1.0000099,"size":1607,"snap":"2019-51-2020-05","text_gpt3_token_len":622,"char_repetition_ratio":0.15408608,"word_repetition_ratio":0.22157435,"special_character_ratio":0.41070318,"punctuation_ratio":0.1768868,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.00001,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-16T04:26:16Z\",\"WARC-Record-ID\":\"<urn:uuid:1e207eec-edb9-43b3-ad1c-fe91732442fe>\",\"Content-Length\":\"16653\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a04e7879-c53a-44a9-b104-01f2e817f7f8>\",\"WARC-Concurrent-To\":\"<urn:uuid:0843e471-79f2-4f66-b2d2-105812f93381>\",\"WARC-IP-Address\":\"107.20.139.176\",\"WARC-Target-URI\":\"http://mathonline.wikidot.com/the-discrete-metric\",\"WARC-Payload-Digest\":\"sha1:ATO4JC7QSQZ4M34BUUJRD7VVOE6U2HY3\",\"WARC-Block-Digest\":\"sha1:AI4B4566VOTQMNTYG7YVDZKMUGKJ4GYV\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575541317967.94_warc_CC-MAIN-20191216041840-20191216065840-00531.warc.gz\"}"} |
https://www.clear.rice.edu/comp212/08-spring/labs/12/ | [
"# Lab #12: Finite State Machine and the Cheap Calculator\n\nIn this lab we will create Graphical User Interface (\"GUI\") \"Cheap Calculator\" program which can do the four basic arithmetic operations: addition, subtraction, multiplication and division.\n\n### Demo\n\nTo see where we are going, download and run the demo As you can see this is a really cheap calculator: it misses a couple of digit buttons and does not support subtraction nor division. Nevertheless, it seems to calculate arithmetic expression in infix notation correctly. What do we mean by an \"infix\" expression? A binary operation is said to be in infix notation if the operator (e.g. +) is in between the two operands. There are also \"postfix\" and \"prefix\" expressions. The following table shows an example of each form of notation.\n\n Adding two numbers x, y together Infix Notation x + y Prefix Notation + x y Postfix Notation x y +\n\nNote that it is an Honor Code violation to attempt to decompile the demo.\n\n### Background Information\n\nI am interested in writing a GUI application that simulates the behavior of a \"cheap\" infix calculator. As the user interface, this calculator has on the outside:\n\nÀ 10 buttons labeled 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, respectively, to enter digit characters\n\nÀ one button labeled . to enter the decimal point\n\nÀ one button labeled +, to perform addition\n\nÀ one button labeled *, to perform multiplication\n\nÀ one button labeled =, to compute the result of the input so far\n\nÀ one button labeled C, to clear all entries in the calculator and reset the calculator to its pristine initial state.\n\nÀ and a display text field for me to view the input and the result of my desired computation.\n\nInside this calculator is perhaps some kind of electronic circuit board capable of carrying out the actual computations. The GUI provides the only way for me to interact with the internals of the calculator. How do I want the calculator to behave like? For a start, as I click on the digit buttons, I expect the calculator to accumulate the digits by concatenating them together and displaying the result on the display text field. But wait! What if I only click the digit '0' button? The calculator should not concatenate one '0' after another, should it? Run the demo and check it out yourself!\n\nNow, try this: click '3', then click '0' four times. This time around, clicking '0' results consistently in concatenating '0' to the existing display text. The calculator has changed its \"click '0'\" behavior after I click a digit button that is not a '0'. We say that the calculator has gone through a \"state change\" and behaves differently in each state.\n\nHere is another of example.\n\n• First click C to reset the calculator. The display should show '0'.\n\n• Now, click '3', then '3' again, then '3' again.\n\n• What do you see? The first time you click '3', the calculator replaces '0' with '3'. Then for each subsequent click on '3', the calculator simply appends '3' to the current display text field and update it with the resulting string. You should see \"333\" at this point.\n\n• Next, click '+': \"333\" is still being display. But now click '3'. What happens? '3' is no longer appended to the existing display string, isn't it? You are sending the same message to the calculator by clicking the same '3', but the calculator responds differently this time, doesn't it? It has dynamically changed its behavior after you clicked on '+'! Again what we see here is another example of \"state change\". Clicking the '+' causes the calculator to transition to another state.\n\nThe behaviors of my calculator can be modeled as what is called a \"finite state machine.\" What is a state? How many states are there? How do we go about describing the behavior of the calculator as it changes from state to state?\n\n### Finite State Machine (an informal discussion)\n\nWhen you click on a button of the calculator, you are in effect making a request to the calculator to perform certain task. At any point in time, the internals of the calculator are in some specific conditions with specific values that would cause the calculator to respond in a very specific manner to such a request. The condition in which the calculator is in at any point in time is called the state of the calculator at that point in time. A state of a system is defined by the behaviors of the system. The user can interact with the calculator in only five ways:\n\n1. click a digit button,\n2. click an operation button,\n3. click the equal button,\n4. click the clear button,\n5. click the point button.\n\nTo specify the behavior of the calculator, we must specify the calculator's response to each of the above five clicking events.\n\n#### Start State\n\nWhen you first turn on the calculator, imagine that it is in some initial condition called a start state. Suppose, from the start state, you...\n\n• click a digit button: what do you want to happen? It depends...\n• digit = '0': display '0' and not concatenating '0' on subsequent clicks on '0'. This implies the calculator can stay in the same start state when '0' is clicked.\n• digit != '0': display the digit and begin concatenating any digit (including '0') on subsequent digit clicks. This implies the existence of another state for the calculator to switch to after a non-zero digit is clicked for the first time. Let's call this new state the accumulate state. We shall analyze the behavior of this state later.\n• click an operation button, for instance, \"+\", which we will refer to as \"op\" for generalities sake. Among the many things we want to happens here, one thing for sure we want is for the calculator to remember op as a pending operation so that we can enter another number, click the equal button and get the correct result. A subtle problem arises here. Is there any current pending operation to be perform before updating op as the new pending operation? You may say, well, if we just turn on the calculator and the first thing you do is click an op button, then there is no pending operation to perform. The response to that is the calculator may transition form one state to another during its usage and may come back to the start state with some pending operation yet to be carried out. In any case, the calculator must have a way to remember a pending op, it is best to have the calculator remember a well-define pending op at all time, even if there is no pending op at all. This is where the concept of a no-op operation becomes useful. Define the no-op operation to be an operation that \"does nothing\" and use no-op as a pending operation when there is no operation to perform. In the design pattern language, this is called the \"Null Object\" pattern. The next question to ask is whether or not the calculator should change to a state different from both the start state and the accumulate state.\n• Can the calculator stay in the same start state? Say, in the beginning you start with the number 0 in the display, and you click '+' . This is tantamount to entering the expression 0 +. If you click the equal button next, then you are in essence asking the calculator to compute 0 +, which is an invalid infix expression. What do you want the calculator to do and display here? Examine a few real calculators of different brands and see how they behave. Also, examine a few different \"software\" calculators, if you have any. They do not all behave the same way, do they? The behavior here can be arbitrarily defined because the expression entered ( 0 + ) is not a valid infix expression. So we arbitrarily specify that our calculator display an error message, stop accepting any additional inputs and stop performing any additional operations. This implies the existence of a new state called the error state. With this new behavior specification, the calculator cannot stay in the same state as the start state because, as we shall see in the discussion of the click equal event that follows, the start state cannot display an error.\n• Can the new state be the same as the accumulate state? If we transition to the accumulate state after we click an op, then we should be accumulating '0' as we continually click on '0', shouldn't we? But is that want we want the calculator to behave like? Of course not. After we click an op button, we should be in a state where we can accept new digit inputs. If we click '0', we do not want to accumulate '0' at all. Since this new state does not have the same behavior as the accumulate state discovered in 1, it is a different state. Let's call this state the compute state. We get to a compute state after we click on an op button.\n• click the equal button: we want the calculator to compute the pending operation, which may be a no-op, update the pending operation to a no-op (since there is no more operation to perform), display the result and be ready to accept a brand new input. There is no need to transition to any state.\n• click the clear button: we want the calculator to be reset to the initial pristine condition as if it has just been turned on. Obviously, the calculator should stay in this same start state.\n• click the point button: we want to display \"0.\" to signify that we have just entered a decimal point in order to input a number less than one. Do we stay in the same start state? We can't, because if we click '0' next, we want to see \"0.\". Should we stay in the start state, we would be displaying \"0\" instead. The new state is obviously not the error state either. Let's call this new state the point state.\n• Can the point state be the compute state? Consider the following sequence of 4 clicks: '0', '.' , '0', '0'. After the first click, '0', the calculator stays in the same start state. If after the second click ('.'), we do not change state, then after the third click ('0'), and fourth click ('0'), the calculator would display \"0\" and not \"0.00\", as it should. Therefore the point state cannot be the compute state.\n• Can the point state be the accumulate state? It depends on what we want the calculator to behave like once we are in the point state and keep clicking '.'. The first '.' click will cause the calculator to display \"0.\". What do we want the computer to display if click '.' again? One reasonable behavior is to ignore the '.' click and do nothing. That is, once we are in the point state, we ignore all the point clicks. What should the accumulate state behave like when we click '.'? Can we ignore the point click once we are in the accumulate state? From the start state, we click '3' and change to the accumulate state as discussed in the above. Now, click '.'. If calculator ignores the point click and does nothing, then we cannot enter the decimal point at all. We can never enter something like \"3.5\". The accumulate state cannot ignore the first point click while the point state ignores all point clicks. Thus the point state and the accumulate state are not the same.\n\nBeginning with one state, the start state, by analyzing and specifying the behavior of the calculator in response to the five distinct click events, we have inferred the existence of four distinct other states: accumulate state, compute state, point state and error state. We can repeat the same analysis on each of these states for each of the click events. We will not discussed them in details here. Instead, we summarize the state behaviors and the state transition in the form of a diagram shown below.\n\n#### State Diagram",
null,
"### State Design Pattern\n\nTraditional procedural programming implements the behavior of finite state machines by using numbers, loops and conditional constructs such as if and switch statements. In OOP, each state is an object with behaviors. A finite state machine is an object that has states. The machine holds a reference to an abstract state and delegates all state-dependent behaviors to its current state. This is the gist of what is called the state design pattern. We will apply the state pattern to model the cheap calculator. Here is the UML diagram of the complete system that we will be building:",
null,
"## Step 0: Understand the problem\n\nStudy the above state transition diagram!\n\nStudy the code for InfixCalc to see how all state-dependent behaviors are delegated to the current state.\n\nNOTE: For the sake of expediency, we are \"hardwiring\" the model (InfixCalc) to its view (InfixGUI), instead of using adapters and a controller to connect the model and the view as usually done. We encourage that you modify the above design to make it a true MVC design.\n\n## Step 1: Complete the stub code\n\nThe stub code should compile. The main class is demo.AppLauncher.java. You should be able to run AppLauncher. However the calculator does not work as specified.\n\n• Imitate the code for StartState to complete the code for AccumState, PointState and CompState: use the above state diagram to write the code for all the methods of each of the concrete state classes, one class at a time.\n• Your code should compile and run like the demo applet.\n\n## Step 2: Add More Buttons and Event Listeners\n\n• Modify InFixGUI to add a '1' button and a '2' button. Add appropriate event listeners for these two buttons.\n• Add a subtraction binary operation and a division binary operation.\n• Modify InFixGUI to add a subtraction button and a division button. Add appropriate event listeners for these two buttons.\n• Compile and run and test at each of the above steps.\n• You now should have a fully functional \"cheap\" calculator.\n\n## Step 3: Here is the code of the demo (no peeking during the lab!)\n\nLast Revised Monday, 07-Apr-2008 11:48:58 CDT\n\n©2008 Stephen Wong and Dung Nguyen"
]
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null,
"https://www.clear.rice.edu/comp212/08-spring/labs/12/states.gif",
null,
"https://www.clear.rice.edu/comp212/08-spring/labs/12/InfixCalc.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.90070784,"math_prob":0.9393152,"size":7064,"snap":"2023-40-2023-50","text_gpt3_token_len":1518,"char_repetition_ratio":0.15864022,"word_repetition_ratio":0.0146818925,"special_character_ratio":0.2202718,"punctuation_ratio":0.11143482,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95887256,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-04T20:26:11Z\",\"WARC-Record-ID\":\"<urn:uuid:240ed999-1861-4acd-915d-3f0ec599d523>\",\"Content-Length\":\"20686\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dd7f2b5a-ddc9-4e59-8248-50eb9e332ff9>\",\"WARC-Concurrent-To\":\"<urn:uuid:681d5094-d058-4058-b781-855c1aa30f44>\",\"WARC-IP-Address\":\"128.42.124.176\",\"WARC-Target-URI\":\"https://www.clear.rice.edu/comp212/08-spring/labs/12/\",\"WARC-Payload-Digest\":\"sha1:FU5VUISRKYRO46KKSZWG6ONNAG2UZXUC\",\"WARC-Block-Digest\":\"sha1:JA3CHDAHRMGJ45TCPK54B67N53HFAF2H\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100534.18_warc_CC-MAIN-20231204182901-20231204212901-00891.warc.gz\"}"} |
https://www.semanticscholar.org/paper/The-path-to-recent-progress-on-small-gaps-between-Goldston-Pintz/479b35c7fb1384fb4285c1c4b39ecc5e27df4268 | [
"• Corpus ID: 7503161\n\n# The path to recent progress on small gaps between primes\n\n@article{Goldston2007ThePT,\ntitle={The path to recent progress on small gaps between primes},\nauthor={D. A. Goldston and Janos Pintz and C. Y. Yildirim},\njournal={arXiv: Number Theory},\nyear={2007}\n}\n• Published 19 December 2005\n• Economics\n• arXiv: Number Theory\nWe present the development of ideas which led to our recent find- ings about the existence of small gaps between primes.\n13 Citations\n\n### Strings of congruent primes in short intervals\n\n• T. Freiberg\n• Mathematics, Philosophy\nJ. Lond. Math. Soc.\n• 2011\nIt is proved that if (q, a )=1, then there are infinitely many pairs pr,p r+1 such that pr ≡ pr-1 ≡ a mod q and pr+1 − pr <� log pr.\n\n### TOWARDS A PROOF OF THE TWIN PRIME CONJECTURE 33\n\nPrime numbers differing by 2 are called twin primes. The twin prime conjecture states that the number of twin primes is infinite. Many attempts to prove or disprove this 2300-year old conjecture have\n\n### An Overview of Sieve Methods\n\nAn overview of the power of Sieve methods in number theory meant for the non-specialist is provided.\n\n### Levels of Distribution and the Affine Sieve\n\n• Alex Kontorovich\n• Mathematics\nAnnales de la Faculté des sciences de Toulouse : Mathématiques\n• 2014\nThis article is an expanded version of the author's lecture in the Basic Notions Seminar at Harvard, September 2013. Our goal is a brief and introductory exposition of aspects of two topics in sieve\n\n### Three topics in additive prime number theory\n\nThis is an expository article to accompany my two lectures at the CDM conference. I have used this an excuse to make public two sets of notes I had lying around, and also to put together a short\n\n### The Goldston-Pintz-Yildirim sieveand some applications\n\nThe twin prime conjecture - that there exist infinitely many pairs of \"twin primes\" p, p + 2 - is among the most famous problems in number theory. In 2005, Goldston, Pintz and Yildirim (GPY) made a\n\n### The Goldston-Pintz-Yıldırım sieve and some applications\n\n• Mathematics\n• 2013\nThe twin prime conjecture that there exist infinitely many pairs of “twin primes” p, p + 2 is among the most famous problems in number theory. In 2005, Goldston, Pintz and Yıldırım (GPY) made a major\n\n### Small gaps between primes\n\nAbstractCombining Goldston-Yildirim’s method on k-correlations of the truncated von Mangoldt function with Maier’s matrix method, we show that \\Xi _r : = \\lim \\inf _{n \\to \\infty } \\tfrac{{p_{n +\n\n### Towards a Proof of the Twin Prime Conjecture\n\nIntroduction: All primes greater than or equal to five are of the form 6j –1 or 6j +1. m =1 to n ∏ Pm = Jn is the product of the first n primes. The number of (6j –1, 6j +1) pairs with no factor less\n\n### Distribution of prime numbers by the modified chi-square function\n\nThe statistical distribution of prime numbers represents an open problem in number theory still nowadays. The methodology of experimental mathematics has not yet been attempted in this field, thus"
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https://quantumcomputing.stackexchange.com/questions/1323/quantum-bitcoin-subdivision | [
"# Quantum Bitcoin Subdivision\n\n## Background\n\nRecently I was reading the article \"Quantum Bitcoin: An Anonymous and Distributed Currency Secured by the No-Cloning Theorem of Quantum Mechanics\" which demonstrates how a quantum bitcoin could function. The article's conclusion states that:\n\nquantum bitcoins are atomic and there is currently no way to subdivide quantum bitcoin into smaller denominations, or merge them into larger ones.\n\nAs there is currently no way to subdivide or merge quantum bitcoins, you can not make change in a transaction. However, I could not understand why subdivision of a quantum bitcoin is not possible.\n\n## Question\n\nWhy can you not subdivide a quantum bitcoin?\n\n## Definitions\n\nA quantum bitcoin - like a regular bitcoin - is a currency with no central authority.\n\nThe main idea behind the implementation of a quantum bitcoin is the no-cloning theorem. The no-cloning theorem demonstrates how it is impossible to copy the arbitrary quantum state $\\left| \\varphi \\right>$.\n\n• Good question +1. If possible add the definition of \"quantum-bitcoin\". Secondly, always link to the abstract of a paper, rather than the pdf (I've edited the link, now). Mar 25, 2018 at 16:32\n• @Blue OK, I've updated the question and added the definition. I also added information on the no-cloning theorem as it is the main idea behind quantum bitcoins. Mar 25, 2018 at 17:00\n• Hmm, I do think this would be purely theoretical. I mean, all those qubits are very expensive. A single qubit-coin has to be very expensive to be more expensive than the qubit on which it is created! Mar 25, 2018 at 17:37\n• @Discretelizard I'd argue more longterm than theoretical.\n– auden\nMar 25, 2018 at 17:40\n• @heather That's a matter of semantics and optimism. I for one, don't see qubits becoming cheap in the next 50 years. But, who knows. Mar 25, 2018 at 18:03\n\nWhy can you not subdivide a quantum bitcoin?\n\nAnyone can create a Cryptocurrency, how it works is up to them, how well it is received is up to the public, generally it is decided by: Utility, Scarcity, Perceived Value.\n\nAs of today a Bitcoin is worth USD 7,073.54, A Bitcoin is 10$^8$ Satoshis which are 0.00000001 Bitcoins, so a Satoshi is worth: 7,073.54 * 0.00000001 = 7.07354 × 10$^{-5}$ USD or 0.00707354 pennies. In total there can be 21 million bitcoin units (2.1 quadrillion Satoshis). The creators of Bitcoin chose that it would be divided into Satoshis.\n\nIn Jogenfors' paper, which you cited, he decided that the Quantum Bitcoin protocol (not to be confused with the Qubitcoin (Q2C) a CPU and GPU based Cryptocurrency) will use the no-cloning theorem of quantum mechanics to construct quantum shards and two blockchains.\n\nThe answer to your question is, according to section 4.4 - Preventing the Reuse Attack:\n\n\"With Bitcoin the blockchain records all transactions and a miner therefore relinquishes control over the mined bitcoin as soon as it is handed over to a recipient. In Quantum Bitcoin, however, there is no record of who owns what, so there is no way to distinguish between the real and counterfeit quantum bitcoin.\n\nWe prevent the reuse attack by adding a secondary stage to the minting algorithm, where data is also appended to a new ledger $\\mathcal{L^′}$.\n\n...\n\nQuantum shard miners create and sell the quantum shards on a marketplace, another miner (called a quantum bitcoin miner) purchases $m$ quantum shards ${(s, ρi, σi)}1≤i≤m$ on the marketplace that, for all $1 ≤ i ≤ m$, fulfill the following conditions:\n\n• $\\mathsf{Verify}\\mathcal{_M} ((s, ρi, σi))$ accepts\n\n• The timestamp $T$ of the quantum shard in the Quantum Shard ledger $\\mathcal{L}$ fulfills $t − T ≤ T_{max}$, where $t$ is the current time\". See section \"A.2 The Reuse Attack\" for further proof.\n\nShards have a lifetime of $T_{max}$.\n\nWhile a shard is designed to have a short lifetime the Quantum Bitcoin is designed to have a great longevity (as long as it is intact, undivided), see section \"A.3 Quantum Bitcoin Longevity\":\n\n\"Theorem 4 (Quantum Bitcoin Longevity) The number of times a quantum bitcoin can be verified and reconstructed is exponentially large in $n$.\n\nProof Corollary 1 shows that the completeness error $\\varepsilon$ of $\\mathcal{Q}$ is exponentially small in $n$. When verifying a genuine quantum bitcoin \\$, the verifier performs the measurement$\\mathsf{Verify}\\mathcal{_Q}$(\\$) on the underlying quantum states $ρ$, which yields the outcome “Pass” with probability $1 − \\varepsilon$. Then lemma 2 shows that we can recover the underlying quantum states $\\widetilde{p}_{i}$ of \\$so that$ \\Vert \\widetilde{p}_{i} − p_i \\Vert _{tr} \\le \\sqrt\\varepsilon$. As$\\varepsilon$is exponentially small in$n$, the trace distance becomes exponentially small in$n$as well. Each time such a quantum bitcoin is verified and reconstructed, the trace distance between the “before” and “after” is exponentially small in$n$. Given any threshold after which we consider the quantum bitcoin “worn out”, the number of verifications it survives before passing this threshold is exponential in$n$. Theorem 4 shows that a quantum bitcoin \\$ can be verified and re-used many times before the quantum state is lost (assuming the absence of noise and decoherence). This is of course analogous to traditional, physical banknotes and coins which are expected to last for a large enough number of transactions before wearing out.\".\n\nSo, much like paper money, you need the whole bill (blockchain) undivided. Of course, that doesn't prevent you from exchanging one Quantum Bitcoin for however many Bitcoins and then offering Bitcoins (or fiat) as change."
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https://prepinsta.com/data-structures/minimum-number-of-merge-operations-to-make-an-array-palindrome/ | [
"# Minimum number of Merge Operations to make an Array Palindrome in C\n\n## Minimum number of merge operations\n\nHere, in this section we discuss code for finding Minimum number of merge operations to make array palindrome.\nWe are given with an array of positive integers. If array is not a palindrome, make merge operations and prints the number of merge operations. In each merge operation it will merge two adjacent elements. Here, merging two elements means replacing them with their sum.\n\nA palindrome is a word, phrase, or sequence that reads the same backwards as forwards.",
null,
"## Minimum Merge Operations To Make Array Palindrome\n\n### Algorithm :\n\n• Take the size of array from the user and store it in variable named n.\n• Declare an array of size n and take n integer values from the user.\n• Create two variables i,j. i will point to the start of the array and j to the end.\n• Run a loop till i less then or equal to j.\n• If arr[i] = arr[j], then there is no need to merge the elements. Increment i and decrement j\n• If arr[i] > arr[j], then do merge operation at index j ie, arr[j-1] = arr[j-1] + arr[j], decrement j and increment the no of merge operations count by 1\n• If arr[i] < arr[j], then do merge operation at index i ie, arr[i+1] = arr[i+1] + arr[i], increment i and increment the no of merge operations count by 1",
null,
"## Code in C, based on above algorithm\n\nRun\n```#include <stdio.h>\n\nint MinOps (int arr[], int n)\n{\nint ans = 0;\n\nfor (int i = 0, j = n - 1; i <= j;)\n{\nif (arr[i] == arr[j])\n{\ni++;\nj--;\n}\n\nelse if (arr[i] > arr[j])\n{\nj--;\narr[j] += arr[j + 1];\nans++;\n}\n\nelse\n{\ni++;\narr[i] += arr[i - 1];\nans++;\n}\n}\n\nreturn ans;\n}\n\nint main ()\n{\nint n;\nprintf (\"Enter the number of elements \");\nscanf (\"%d\", &n);\n\nint arr[n];\n\nprintf (\"Enter the elements \");\nfor (int i = 0; i < n; i++)\n{\nscanf (\"%d\", &arr[i]);\n}\nprintf (\"Count of minimum operations is %d\", MinOps (arr, n));\n\nreturn 0;\n}\n```\n\n### Output:\n\n```Enter the number of elements 4\nEnter the elements 1 2 3 1\nCount of minimum operations is 1```\n\n### Related Banners\n\nGet PrepInsta Prime & get Access to all 200+ courses offered by PrepInsta in One Subscription\n\n## Get over 200+ course One Subscription\n\nCourses like AI/ML, Cloud Computing, Ethical Hacking, C, C++, Java, Python, DSA (All Languages), Competitive Coding (All Languages), TCS, Infosys, Wipro, Amazon, DBMS, SQL and others\n\n## Checkout list of all the video courses in PrepInsta Prime Subscription\n\n### Traversals\n\n• Traversal in Trees\n• Tree Traversals: Breadth-First Search (BFS) :\nC | C++ | Java\n• Tree Traversals: Depth First Search (DFS) :\nC | C++ | Java\n• Construct a Binary Tree from Postorder and Inorder\n\n### AVL Trees\n\n• AVL Trees\n• AVL Trees: Introduction\n• AVL Tree Insertion :\nC | C++ | Java\n• AVL Tree Deletion :\nC | C++ | Java\n• Insertion in a Binary Tree (Level Order) –\nC | C++ | Java\n• Searching in Binary Tree –\nC | C++ | Java\n• Searching in a Binary Search Tree –\nC | C++ | Java\n\n### Complete Programs for Trees\n\n• Depth First Traversals –\nC | C++ | Java\n• Level Order Traversal –\nC | C++Java\n• Construct Tree from given Inorder and Preorder traversals –\nC | C++Java\n• Construct Tree from given Postorder and Inorder traversals –\nC | C++Java\n• Construct Tree from given Postorder and Preorder traversal –\nC | C++Java\n• Find size of the Binary tree –\nC | C++Java\n• Find the height of binary tree –\nC | C++Java\n• Find maximum in binary tree –\nC | C++Java\n• Check whether two tree are identical-\nCC++Java\n• Spiral Order traversal of Tree-\nCC++Java\n• Level Order Traversal Line by Line –\nC | C++Java\n• Hand shaking lemma and some Impotant Tree Properties.\n• Check If binary tree if Foldable or not.-\nCC++Java\n• check whether tree is Symmetric –\nC| C++Java.\n• Check for Children-Sum in Binary Tree-\nC|C++Java\n• Sum of all nodes in Binary Tree-\nCC++ | Java\n• Lowest Common Ancestor in Binary Tree-\nCC++ | Java\n\n### Traversals\n\n• Traversal in Trees\n• Tree Traversals: Breadth-First Search (BFS) : C | C++ | Java\n• Tree Traversals: Depth First Search (DFS) : C | C++ | Java\n• Construct a Binary Tree from Postorder and Inorder\n\n### AVL Trees\n\n• AVL Trees\n• AVL Trees: Introduction\n• AVL Tree Insertion : C | C++ | Java\n• AVL Tree Deletion : C | C++ | Java\n• Insertion in a Binary Tree (Level Order) – C | C++ | Java\n• Searching in Binary Tree – C | C++ | Java\n• Searching in a Binary Search Tree – C | C++ | Java\n\n### Complete Programs for Trees\n\n• Depth First Traversals – C | C++ | Java\n• Level Order Traversal – C | C++Java\n• Construct Tree from given Inorder and Preorder traversals – C | C++Java\n• Construct Tree from given Postorder and Inorder traversals – C | C++Java\n• Construct Tree from given Postorder and Preorder traversal – C | C++Java\n• Find size of the Binary tree – C | C++Java\n• Find the height of binary tree – C | C++Java\n• Find maximum in binary tree – C | C++Java\n• Check whether two tree are identical- CC++Java\n• Spiral Order traversal of Tree- CC++Java\n• Level Order Traversal LIne by Line – C | C++Java\n• Hand shaking lemma and some Impotant Tree Properties.\n• Check If binary tree if Foldable or not.- CC++Java\n• check whether tree is Symmetric C| C++Java.\n• Check for Children-Sum in Binary Tree- C|C++Java\n• Sum of all nodes in Binary Tree- CC++ | Java\n• Lowest Common Ancestor in Binary Tree. CC++ | Java"
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https://www.tangram-channel.com/the-eighth-book-of-tan-by-sam-loyd-page-1/the-eighth-book-of-tan-by-sam-loyd-page-17/ | [
"# The 8th Book of Tan by Sam Loyd\n\n## The unsolved problems\n\nThe Second Book of Tan, like the lost Euclid, was given to elementary problems, the correctness or fallacy in the construction of which was to be discovered by the student, who was required to decide between the correctness of two conflicting propositions. The squaring of the circle appears to have been treated according to the following kindergarten method, which will furnish food for reflection for the young folks as well as the brainy mathematicians.\n\nTo make it clear to all, it may be explained that the squaring of the circle is merely the problem of figuring out how many square feet are contained in a circle of a given diameter--viz., how many square feet of sod are there in a round grass-plot 100 feet in diameter?\n\n\"If the Pons Asinorum, as shown, is correct,\" says Tan, \"let us describe these semicircles upon the sides of the triangle A. Knowing that B is equal to C and D combined in Fig. 1, we will divide B by a straight cut from X to Z. Then the remaining half of B, as shown in Fig. 2, will be equal in area to C.\n\n\"Let us now, as shown in Fig. 3, continue the circle from X to Z and draw the vertical line from X to Y so as to cut the semicircular pieces D and D from both C and B. Those pieces which are removed being exactly similar, taken from forms of the same areas, prove that the remainders C and B must be of the same areas, although the one is a triangle and the other a crescent.\n\n\"The space A being of similar size and construction to B we proceed to place the pieces as shown in Fig. 4 and halve them by a straight cut from C to X, so as to have the remaining pieces as shown in Fig. 5, which have been proven by construction to be of the same areas.\n\n\"We will then cut the piece C by a diagonal line, and, as shown by the dotted lines, proceed to describe the circle enclosed within a square.\n\n\"Now let us show by analysis what has been proven: CC have been shown by Figs. 3, 4, and 5 to be equal to A.\n\n\"As BB added to CC would form a square of the same size as AD, then BB must be equal to D, as CC are equal to A. It shows that B must be equal to B and C equal to C, because BB equals A or BC equals A.\n\n\"Then if the segments B, B, C, and C are all equal, each piece represents the sixteenth part of the large square, and we find that twelve of these sixteen pieces are contained within the circle.\n\n\"Therefore, if the square is 100 in diameter, 100 x 100 = 10,000 square feet, of which our grass-plot, which was said to be 100 feet in diameter, would contain twelve-sixteenths, or just 7,500 square feet.\"\n\nAnd, as Tan says, \"There you are! the truth or fallacy of Pons Asinorum is submitted to the judgment of the students who will kindly pass in their examination papers to the professor.\"\n\n• Geometrical shapes\n\nTangram difficulty levels: Easy Medium Hard Expert\n\n• Letters, Numbers & Signs\n\nTangram difficulty levels: Easy Medium Hard Expert\n\n• People\n\nTangram difficulty levels: Easy Medium Hard Expert\n\n• Animals\n\nTangram difficulty levels: Easy Medium Hard Expert\n\n• Usual objects\n\nTangram difficulty levels: Easy Medium Hard Expert\n\n• Boats\n\nTangram difficulty levels: Easy Medium Hard Expert\n\n• Miscellaneous\n\nTangram difficulty levels: Easy Medium Hard Expert"
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https://immobilier-haute-garonne.info/and-relationship/voltage-relationship-in-star-connection-and-delta.php | [
"Voltage relationship in star connection and delta",
null,
"In the Star Connection, the similar ends (either start or finish) of the three windings are Relation Between Phase Voltage and Line Voltage in Star Connection. Comparison between Star and Delta Connections. difference between STAR (Y) and DELTA Delta Connection (Δ): 3 Phase Power, Voltage & Current Values. Three Phase Transformer Star and Delta Configurations Transformers for high voltage operation with the star connections has the advantage of Then the relationship between line and phase voltages and currents in a three-phase system.\n\nVoltage and current relationships in three-phase circuits.",
null,
"Star-Connected Balanced Load Phase current: The VI provides a visual aid to understanding the definitions of phase and line voltages and phase and line currents in the delta- and the star-connected ac systems that contain the loads as well as the ac supplies. In addition, the instantaneous voltage and currents are displayed in the front panel of the VI.\n\nStar Connection in a 3 Phase System\n\nShow that the line voltage Vline in the three-phase system is times the phase voltage Vphase, and verify the result by using the VI for a given phase voltage. Study the concept in question 1 this time for the line currents and the phase currents in the case of a delta-connected three-phase load.",
null,
"In question 2, find out the angles in degrees between the phase and the line quantities on the supply side and the load side. Use the single-phase equivalent circuit in each load configuration and calculate the phase currents for given values of the voltage and the load impedance. Three incandescent lamps rated 60 W, V rms are connected in the delta form.\n\nWhat line voltage is needed so that the lamps burn normally at rated conditions? What are the line and phase currents in the circuit?\n\nThree-phase Y and Delta Configurations | Polyphase AC Circuits | Electronics Textbook\n\nFirst calculate and set the resistance of the lamps using the controls provided. Three load resistors are connected in the delta form.",
null,
"The terms line current and phase current follow the same logic: Y-connected sources and loads always have line voltages greater than phase voltages, and line currents equal to phase currents.\n\nIf the Y-connected source or load is balanced, the line voltage will be equal to the phase voltage times the square root of 3: Take close notice of the polarity for each winding in Figure below. At first glance it seems as though three voltage sources like this would create a short-circuit, electrons flowing around the triangle with nothing but the internal impedance of the windings to hold them back.",
null,
"Due to the phase angles of these three voltage sources, however, this is not the case. If they do, then there will be no voltage available to push current around and around that loop, and consequently, there will be no circulating current. Starting with the top winding and progressing counter-clockwise, our KVL expression looks something like this: Indeed, if we add these three vector quantities together, they do add up to zero.\n\nVoltage and Currents in Star- and Delta-Connected Loads | Introduction to AC Circuits | InformIT\n\nAnother way to verify the fact that these three voltage sources can be connected together in a loop without resulting in circulating currents is to open up the loop at one junction point and calculate voltage across the break: Sure enough, there will be zero voltage across the break, telling us that no current will circulate within the triangular loop of windings when that connection is made complete. Conversely, because each line conductor attaches at a node between two windings, the line current will be the vector sum of the two joining phase currents.\n\nWith each load resistance receiving volts from its respective phase winding at the source, the current in each phase of this circuit will be So each line current in this three-phase power system is equal to The answer is no. With a Y-connected system, a neutral wire was needed in case one of the phase loads were to fail open or be turned offin order to keep the phase voltages at the load from changing. This is not necessary or even possible! With each load phase element directly connected across a respective source phase winding, the phase voltage will be constant regardless of open failures in the load elements.\n\nRelationship Between Line Quantities and Phase Quantities in Delta Connection - Three Phase Circuits"
]
| [
null,
"https://3.bp.blogspot.com/-KUYprke9aQc/V8wbYZ53EeI/AAAAAAAAB7o/6hUgSp2cYq4Kd45GVSdLUMbE7avrzxwBwCK4B/s1600/HT-and-LT-Motor-connection.jpg",
null,
"https://3.bp.blogspot.com/-Raj96onZf4Y/VNkehEtk2NI/AAAAAAAABLo/3MWduz7QCUI/s1600/Comparison+between+Star+and+Delta+Connections.png",
null,
"https://3.bp.blogspot.com/-_5oSxD8dgWA/TezXb22B6WI/AAAAAAAAAH4/4V9MrlRC2TM/s1600/Star Volts Amp Relation.jpg",
null,
"https://4.bp.blogspot.com/-lQorrgpFmeE/UlbdGlb_79I/AAAAAAAAAMw/FMGs8jIo6U4/s1600/Motor Windings Star Connection.jpg",
null,
"http://electricalacademia.com/wp-content/uploads/2018/05/Voltage-current-relationships-in-a-3-phase-delta-wiring-configuration.png",
null
]
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https://vivo.library.tamu.edu/vivo/display/n37415SE | [
"# Analytical and numerical studies of a singularly perturbed Boussinesq equation Academic Article",
null,
"•\n• Overview\n•\n• Research\n•\n• Identity\n•\n•\n• View All\n•\n\n### abstract\n\n• We study the singularly perturbed (sixth-order) Boussinesq equation recently introduced by Daripa and Hua [Appl. Math. Comput. 101 (1999) 159]. Motivated by their work, we formally derive this equation from two-dimensional potential flow equations governing the small amplitude long capillary-gravity waves on the surface of shallow water for Bond number very close to but less than 1/3. On the basis of far-field analyses and heuristic arguments, we show that the traveling wave solutions of this equation are weakly non-local solitary waves characterized by small amplitude fast oscillations in the far-field. We review various analytical and numerical methods originally devised to obtain this type of weakly non-local solitary wave solutions of the singularly perturbed (fifth-order) KdV equation. Using these methods, we obtain weakly non-local solitary wave solutions of the singularly perturbed (sixth-order) Boussinesq equation and provide estimates of the amplitude of oscillations which persist in the far-field. 2002 Elsevier Science Inc. All rights reserved.\n\n### published proceedings\n\n• APPLIED MATHEMATICS AND COMPUTATION\n\n### author list (cited authors)\n\n• Dash, R. K., & Daripa, P.\n\n• 18\n\n### complete list of authors\n\n• Dash, RK||Daripa, P\n\n### publication date\n\n• February 2002"
]
| [
null,
"https://vivo.library.tamu.edu/vivo/images/individual/uriIcon.gif",
null
]
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https://answers.everydaycalculation.com/divide-fractions/15-98-divided-by-1-90 | [
"Solutions by everydaycalculation.com\n\n## Divide 15/98 with 1/90\n\n15/98 ÷ 1/90 is 675/49.\n\n#### Steps for dividing fractions\n\n1. Find the reciprocal of the divisor\nReciprocal of 1/90: 90/1\n2. Now, multiply it with the dividend\nSo, 15/98 ÷ 1/90 = 15/98 × 90/1\n3. = 15 × 90/98 × 1 = 1350/98\n4. After reducing the fraction, the answer is 675/49\n5. In mixed form: 1338/49\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:"
]
| [
null,
"https://answers.everydaycalculation.com/mathstep-app-icon.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.67776567,"math_prob":0.93285817,"size":297,"snap":"2022-40-2023-06","text_gpt3_token_len":121,"char_repetition_ratio":0.17406143,"word_repetition_ratio":0.0,"special_character_ratio":0.44781145,"punctuation_ratio":0.073529415,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9669371,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-05T23:51:31Z\",\"WARC-Record-ID\":\"<urn:uuid:64b10bed-b2ec-433c-8812-12247de5dfbe>\",\"Content-Length\":\"7418\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5d796e53-8fba-4830-be69-40f417eeba86>\",\"WARC-Concurrent-To\":\"<urn:uuid:18f3b6be-ef68-42df-afa6-b69cbcda34ce>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/divide-fractions/15-98-divided-by-1-90\",\"WARC-Payload-Digest\":\"sha1:GAG5IXDAXVMEKJ4MVWWXSTHWTYD3HCW2\",\"WARC-Block-Digest\":\"sha1:GEHVNHM3SPDSU7SQBEETUQEHPULR54ZM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500294.64_warc_CC-MAIN-20230205224620-20230206014620-00189.warc.gz\"}"} |
https://docs.openvino.ai/latest/api/ie_python_api/_autosummary/openvino.runtime.opset8.proposal.html | [
"# openvino.runtime.opset8.proposal¶\n\nopenvino.runtime.opset8.proposal(class_probs: openvino.pyopenvino.Node, bbox_deltas: openvino.pyopenvino.Node, image_shape: Union[openvino.pyopenvino.Node, int, float, numpy.ndarray], attrs: dict, name: Optional[str] = None) → openvino.pyopenvino.Node\n\nFilter bounding boxes and outputs only those with the highest prediction confidence.\n\nParameters\n• class_probs – 4D input floating point tensor with class prediction scores.\n\n• bbox_deltas – 4D input floating point tensor with corrected predictions of bounding boxes\n\n• image_shape – The 1D input tensor with 3 or 4 elements describing image shape.\n\n• attrs – The dictionary containing key, value pairs for attributes.\n\n• name – Optional name for the output node.\n\n• base_size The size of the anchor to which scale and ratio attributes are applied.\n\nRange of values: a positive unsigned integer number Default value: None Required: yes\n\n• pre_nms_topn The number of bounding boxes before the NMS operation.\n\nRange of values: a positive unsigned integer number Default value: None Required: yes\n\n• post_nms_topn The number of bounding boxes after the NMS operation.\n\nRange of values: a positive unsigned integer number Default value: None Required: yes\n\n• nms_thresh The minimum value of the proposal to be taken into consideration.\n\nRange of values: a positive floating-point number Default value: None Required: yes\n\n• feat_stride The step size to slide over boxes (in pixels).\n\nRange of values: a positive unsigned integer Default value: None Required: yes\n\n• min_size The minimum size of box to be taken into consideration.\n\nRange of values: a positive unsigned integer number Default value: None Required: yes\n\n• ratio The ratios for anchor generation.\n\nRange of values: a list of floating-point numbers Default value: None Required: yes\n\n• scale The scales for anchor generation.\n\nRange of values: a list of floating-point numbers Default value: None Required: yes\n\n• clip_before_nms The flag that specifies whether to perform clip bounding boxes before\n\nnon-maximum suppression or not. Range of values: True or False Default value: True Required: no\n\n• clip_after_nms The flag that specifies whether to perform clip bounding boxes after\n\nnon-maximum suppression or not. Range of values: True or False Default value: False Required: no\n\n• normalize The flag that specifies whether to perform normalization of output boxes to\n\n[0,1] interval or not. Range of values: True or False Default value: False Required: no\n\n• box_size_scale Specifies the scale factor applied to logits of box sizes before decoding.\n\nRange of values: a positive floating-point number Default value: 1.0 Required: no\n\n• box_coordinate_scale Specifies the scale factor applied to logits of box coordinates\n\nbefore decoding. Range of values: a positive floating-point number Default value: 1.0 Required: no\n\n• framework Specifies how the box coordinates are calculated.\nRange of values: “” (empty string) - calculate box coordinates like in Caffe*\ntensorflow - calculate box coordinates like in the TensorFlow*\n\nObject Detection API models\n\nDefault value: “” (empty string) Required: no\n\nExample of attribute dictionary:\n\n# just required ones\nattrs = {\n'base_size': 85,\n'pre_nms_topn': 10,\n'post_nms_topn': 20,\n'nms_thresh': 0.34,\n'feat_stride': 16,\n'min_size': 32,\n'ratio': [0.1, 1.5, 2.0, 2.5],\n'scale': [2, 3, 3, 4],\n}\n\n\nOptional attributes which are absent from dictionary will be set with corresponding default. :return: Node representing Proposal operation."
]
| [
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https://www.coursehero.com/file/p6dauqg3/b-Find-the-dimensions-of-the-triangle-and-the-square-that-produce-a-maximum/ | [
"b Find the dimensions of the triangle and the square that produce a maximum\n\n# B find the dimensions of the triangle and the square\n\nThis preview shows page 11 - 15 out of 19 pages.\n\n(b) Find the dimensions of the triangle and the square that produce a maximum area. Justify your answer. (Area of an equilateral triangle = 2 3 , 4 x where x = side of triangle)",
null,
"Answers to Worksheet 2 on Optimization Note: Only the answers are shown below. Students must also justify their answer. See the article on AP Central called “On the Role of Sign Charts in the AP Calculus Exam for Justifying Local or Absolute Extrema” found at to see how justifications for absolute maximums and minimums should be written. 1. The maximum size of the population of bacteria is 387.298 (so 387 bacteria) and it occurs when t = 7.750 weeks. 2. The minimum velocity is 7, and it occurs when t = 3. The maximum velocity is 15.333, and it occurs when 4 . 3 t 3. The minimum cost is \\$330, and it occurs when the dimensions of the tank are 4 m x 3 m x 3 m. 4. The maximum area is 0.281, and it occurs when x = 0.860. The minimum area is 0.121, and it occurs when x = 0.25. 5. The minimum time is 1.733 hr when x = 1.5 mi. so he should land the boat 4.5 mi. from the house. 6. The minimum cost is \\$996,862.70 when the point P is 6.803 mi. from B . 7. The minimum area occurs when the side of the triangle is 1.883 and the side of the square is 1.087. The maximum area is occurs when the side of the triangle is 0 and the side of the square is 3.333.",
null,
"P CALCULUS AB WORKSHEET ON OPTIMIZATION AND PARTICLE MOTION Work the following on notebook paper. Use your calculator on 2 - 6, and give decimal answers correct to three decimal places. 1. A rectangle is bounded by the x -axis and the parabola 2 9 y x . What length and width should the rectangle have to that its area is a maximum? ___________________________________________________________________________________________ 2. A cylindrical container has a volume of 3 150cm . Find the radius and height of the cylinder so that its surface area will be a minimum. Use Calculus to find and to justify your answer. (Volume of a cylinder = 2 r h . Surface area of a cylinder = 2 2 2 rh r . ___________________________________________________________________________________________ 3. A drilling rig 12 miles offshore is to be connected to a refinery on shore, 20 miles down the coast from Rig the rig by using underwater pipe from the rig to point P and land-based pipe from point P to the refinery. If underwater pipe costs \\$5000 per mile and land-based 12 pipe costs \\$3500 per mile, how far should point P be from the refinery to minimize the cost? What will the cost be? Use Calculus to find and justify your answer. Refinery",
null,
"4. (Modified from 2010) A zoo sponsored a one-day contest to name a new baby elephant. Zoo visitors deposited entries into a box between noon ( t = 0) and 8 PM. At 8 PM, volunteers began to process the entries at a rate modeled by the function 2 3 30 298 976 P t t t t , where P t gives the number of hundreds of entries per hour for 8 12 t . At what time t , 8 12 t , were the entries being processed most quickly? Use Calculus to find and justify your answer. 5. The velocity of a particle at time t is given by 2 3 ln 2 for 0 6 v t t t . Find the acceleration of the particle each time its velocity is zero.",
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"",
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"#### You've reached the end of your free preview.\n\nWant to read all 19 pages?\n\n•",
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https://cxymm.net/article/starzhou/113832228 | [
"# 预训练模型transformers综合总结（一）\n\n# 加载本地模型须知\n\n1.使用transformers库加载预训练模型，99%的时间都是用于模型的下载。\n\n2.下载的模型通常会是\"模型名称-\"+\"config.json\"的格式例如(bert-base-cased-finetuned-mrpc-config.json)，但如果使用transformers库加载本地模型，需要的是模型路径中是config.json、vocab.txt、pytorch_model.bin、tf_model.h5、tokenizer.json等形式，为此在加载前，需要将把文件前面的模型名称，才能加载成功\n\n` `\n1. `#coding=utf-8`\n\n2. `import os`\n\n3. `import os.path`\n\n4. `# 模型存放路径`\n\n5. `rootdir = r\"H:\\code\\Model\\bert-large-uncased-whole-word-masking-finetuned-squad\"# 指明被遍历的文件夹`\n\n6.\n7. `for parent,dirnames,filenames in os.walk(rootdir):#三个参数：分别返回1.父目录 2.所有文件夹名字（不含路径） 3.所有文件名字`\n\n8. `for filename in filenames:#文件名`\n\n9. `# nameList=filename.split('.')`\n\n10. `# print(nameList)`\n\n11. `print(filename)`\n\n12. `# filenew=nameList+'.jpg'`\n\n13. `# print(filenew)`\n\n14. `#模型的名称`\n\n15. `newName=filename.replace('bert-large-uncased-whole-word-masking-finetuned-squad-','')`\n\n16. `os.rename(os.path.join(parent,filename),os.path.join(parent,newName))#重命名`\n\n1.使用管道\n\n` `\n1. `model_path=\"H:\\\\code\\\\Model\\\\bert-base-cased-finetuned-mrpc\\\\\"`\n\n2.\n3. `from transformers import pipeline`\n\n4. `#使用当前模型+使用Tensorflow框架，默认应该是使用PYTORCH框架`\n\n5. `nlp = pipeline(\"sentiment-analysis\",model=model_path, tokenizer=model_path, framework=\"tf\")`\n\n6. `result = nlp(\"I hate you\")`\n\n7. `print(f\"label: {result['label']}, with score: {round(result['score'], 4)}\")`\n\n8. `result = nlp(\"I love you\")`\n\n9. `print(f\"label: {result['label']}, with score: {round(result['score'], 4)}\")`\n\n2.直接使用模型\n\n` `\n1. `model_path=\"H:\\\\code\\\\Model\\\\bert-base-cased-finetuned-mrpc\\\\\"`\n\n2. `#pytorch框架`\n\n3.\n4. `from transformers import AutoTokenizer, AutoModelForSequenceClassification`\n\n5. `import torch`\n\n6. `tokenizer = AutoTokenizer.from_pretrained(model_path)`\n\n7. `model = AutoModelForSequenceClassification.from_pretrained(model_path)`\n\n8. `classes = [\"not paraphrase\", \"is paraphrase\"]`\n\n9. `sequence_0 = \"The company HuggingFace is based in New York City\"`\n\n10. `sequence_1 = \"Apples are especially bad for your health\"`\n\n11. `sequence_2 = \"HuggingFace's headquarters are situated in Manhattan\"`\n\n12. `paraphrase = tokenizer(sequence_0, sequence_2, return_tensors=\"pt\")`\n\n13. `not_paraphrase = tokenizer(sequence_0, sequence_1, return_tensors=\"pt\")`\n\n14. `paraphrase_classification_logits = model(**paraphrase).logits`\n\n15. `not_paraphrase_classification_logits = model(**not_paraphrase).logits`\n\n16. `paraphrase_results = torch.softmax(paraphrase_classification_logits, dim=1).tolist()`\n\n17. `not_paraphrase_results = torch.softmax(not_paraphrase_classification_logits, dim=1).tolist()`\n\n18. `# Should be paraphrase`\n\n19. `for i in range(len(classes)):`\n\n20. `print(f\"{classes[i]}: {int(round(paraphrase_results[i] * 100))}%\")`\n\n21. `# Should not be paraphrase`\n\n22. `for i in range(len(classes)):`\n\n23. `print(f\"{classes[i]}: {int(round(not_paraphrase_results[i] * 100))}%\")`\n\n24.\n25. `#tensorflow框架`\n\n26. `from transformers import AutoTokenizer, TFAutoModelForSequenceClassification`\n\n27. `import tensorflow as tf`\n\n28. `tokenizer = AutoTokenizer.from_pretrained(model_path)`\n\n29. `model = TFAutoModelForSequenceClassification.from_pretrained(model_path)`\n\n30. `classes = [\"not paraphrase\", \"is paraphrase\"]`\n\n31. `sequence_0 = \"The company HuggingFace is based in New York City\"`\n\n32. `sequence_1 = \"Apples are especially bad for your health\"`\n\n33. `sequence_2 = \"HuggingFace's headquarters are situated in Manhattan\"`\n\n34. `paraphrase = tokenizer(sequence_0, sequence_2, return_tensors=\"tf\")`\n\n35. `not_paraphrase = tokenizer(sequence_0, sequence_1, return_tensors=\"tf\")`\n\n36. `paraphrase_classification_logits = model(paraphrase)`\n\n37. `not_paraphrase_classification_logits = model(not_paraphrase)`\n\n38. `paraphrase_results = tf.nn.softmax(paraphrase_classification_logits, axis=1).numpy()`\n\n39. `not_paraphrase_results = tf.nn.softmax(not_paraphrase_classification_logits, axis=1).numpy()`\n\n40. `# Should be paraphrase`\n\n41. `for i in range(len(classes)):`\n\n42. `print(f\"{classes[i]}: {int(round(paraphrase_results[i] * 100))}%\")`\n\n43. `# Should not be paraphrase`\n\n44. `for i in range(len(classes)):`\n\n45. `print(f\"{classes[i]}: {int(round(not_paraphrase_results[i] * 100))}%\")`\n\n## 提取式问答\n\n1.使用管道\n\n` `\n1. `model_path=\"H:\\\\code\\\\Model\\\\bert-large-uncased-whole-word-masking-finetuned-squad\\\\\"`\n\n2.\n3. `from transformers import pipeline`\n\n4. `nlp = pipeline(\"question-answering\",model=model_path, tokenizer=model_path)`\n\n5. `context = r\"\"\"`\n\n6. `Extractive Question Answering is the task of extracting an answer from a text given a question. An example of a`\n\n7. `question answering dataset is the SQuAD dataset, which is entirely based on that task. If you would like to fine-tune`\n\n8. `a model on a SQuAD task, you may leverage the examples/question-answering/run_squad.py script.`\n\n9. `\"\"\"`\n\n10. `result = nlp(question=\"What is extractive question answering?\", context=context)`\n\n11. `print(f\"Answer: '{result['answer']}', score: {round(result['score'], 4)}, start: {result['start']}, end: {result['end']}\")`\n\n12. `result = nlp(question=\"What is a good example of a question answering dataset?\", context=context)`\n\n13. `print(f\"Answer: '{result['answer']}', score: {round(result['score'], 4)}, start: {result['start']}, end: {result['end']}\")`\n\n2.直接使用模型\n\n` `\n1. `model_path=\"H:\\\\code\\\\Model\\\\bert-large-uncased-whole-word-masking-finetuned-squad\\\\\"`\n\n2. `#使用pytorch框架`\n\n3. `from transformers import AutoTokenizer, AutoModelForQuestionAnswering`\n\n4. `import torch`\n\n5. `tokenizer = AutoTokenizer.from_pretrained(model_path)`\n\n6. `model = AutoModelForQuestionAnswering.from_pretrained(model_path)`\n\n7. `text = r\"\"\"`\n\n8. ` Transformers (formerly known as pytorch-transformers and pytorch-pretrained-bert) provides general-purpose`\n\n9. `architectures (BERT, GPT-2, RoBERTa, XLM, DistilBert, XLNet…) for Natural Language Understanding (NLU) and Natural`\n\n10. `Language Generation (NLG) with over 32+ pretrained models in 100+ languages and deep interoperability between`\n\n11. `TensorFlow 2.0 and PyTorch.`\n\n12. `\"\"\"`\n\n13. `questions = [`\n\n14. `\"How many pretrained models are available in Transformers?\",`\n\n15. `\"What does Transformers provide?\",`\n\n16. `\" Transformers provides interoperability between which frameworks?\",`\n\n17. `]`\n\n18. `for question in questions:`\n\n19. `inputs = tokenizer(question, text, add_special_tokens=True, return_tensors=\"pt\")`\n\n20. `input_ids = inputs[\"input_ids\"].tolist()`\n\n21. `text_tokens = tokenizer.convert_ids_to_tokens(input_ids)`\n\n22. `outputs = model(**inputs)`\n\n23. `answer_start_scores = outputs.start_logits`\n\n24. `answer_end_scores = outputs.end_logits`\n\n25. `answer_start = torch.argmax(`\n\n26. `answer_start_scores`\n\n27. `) # Get the most likely beginning of answer with the argmax of the score`\n\n28. `answer_end = torch.argmax(answer_end_scores) + 1 # Get the most likely end of answer with the argmax of the score`\n\n29. `answer = tokenizer.convert_tokens_to_string(tokenizer.convert_ids_to_tokens(input_ids[answer_start:answer_end]))`\n\n30. `print(f\"Question: {question}\")`\n\n31. `print(f\"Answer: {answer}\")`\n\n32.\n33. `#使用tensorflow框架`\n\n34. `from transformers import AutoTokenizer, TFAutoModelForQuestionAnswering`\n\n35. `import tensorflow as tf`\n\n36. `tokenizer = AutoTokenizer.from_pretrained(model_path)`\n\n37. `model = TFAutoModelForQuestionAnswering.from_pretrained(model_path)`\n\n38. `text = r\"\"\"`\n\n39. ` Transformers (formerly known as pytorch-transformers and pytorch-pretrained-bert) provides general-purpose`\n\n40. `architectures (BERT, GPT-2, RoBERTa, XLM, DistilBert, XLNet…) for Natural Language Understanding (NLU) and Natural`\n\n41. `Language Generation (NLG) with over 32+ pretrained models in 100+ languages and deep interoperability between`\n\n42. `TensorFlow 2.0 and PyTorch.`\n\n43. `\"\"\"`\n\n44. `questions = [`\n\n45. `\"How many pretrained models are available in Transformers?\",`\n\n46. `\"What does Transformers provide?\",`\n\n47. `\" Transformers provides interoperability between which frameworks?\",`\n\n48. `]`\n\n49. `for question in questions:`\n\n50. `inputs = tokenizer(question, text, add_special_tokens=True, return_tensors=\"tf\")`\n\n51. `input_ids = inputs[\"input_ids\"].numpy()`\n\n52. `text_tokens = tokenizer.convert_ids_to_tokens(input_ids)`\n\n53. `outputs = model(inputs)`\n\n54. `answer_start_scores = outputs.start_logits`\n\n55. `answer_end_scores = outputs.end_logits`\n\n56. `answer_start = tf.argmax(`\n\n57. `answer_start_scores, axis=1`\n\n58. `).numpy() # Get the most likely beginning of answer with the argmax of the score`\n\n59. `answer_end = (`\n\n60. `tf.argmax(answer_end_scores, axis=1) + 1`\n\n61. `).numpy() # Get the most likely end of answer with the argmax of the score`\n\n62. `answer = tokenizer.convert_tokens_to_string(tokenizer.convert_ids_to_tokens(input_ids[answer_start:answer_end]))`\n\n63. `print(f\"Question: {question}\")`\n\n64. `print(f\"Answer: {answer}\")`\n\n1.使用管道\n\n` `\n1. `model_path=\"H:\\\\code\\\\Model\\\\distilbert-base-cased\\\\\"`\n\n2. `from transformers import pipeline`\n\n3. `nlp = pipeline(\"fill-mask\",model=model_path, tokenizer=model_path, framework=\"tf\")`\n\n4. `from pprint import pprint`\n\n5. `pprint(nlp(f\"HuggingFace is creating a {nlp.tokenizer.mask_token} that the community uses to solve NLP tasks.\"))`\n\n2.直接使用模型\n\n` `\n1. `model_path=\"H:\\\\code\\\\Model\\\\distilbert-base-cased\\\\\"`\n\n2.\n3. `## 使用pytorch实现`\n\n4. `from transformers import AutoModelWithLMHead, AutoTokenizer`\n\n5. `import torch`\n\n6. `tokenizer = AutoTokenizer.from_pretrained(model_path)`\n\n7. `model = AutoModelWithLMHead.from_pretrained(model_path)`\n\n8. `sequence = f\"Distilled models are smaller than the models they mimic. Using them instead of the large versions would help {tokenizer.mask_token} our carbon footprint.\"`\n\n9. `input = tokenizer.encode(sequence, return_tensors=\"pt\")`\n\n10. `mask_token_index = torch.where(input == tokenizer.mask_token_id)`\n\n11. `token_logits = model(input).logits`\n\n12. `mask_token_logits = token_logits[0, mask_token_index, :]`\n\n13. `top_5_tokens = torch.topk(mask_token_logits, 5, dim=1).indices.tolist()`\n\n14. `for token in top_5_tokens:`\n\n15. `print(sequence.replace(tokenizer.mask_token, tokenizer.decode([token])))`\n\n16.\n17. `## 使用tensorflow实现`\n\n18. `from transformers import TFAutoModelWithLMHead, AutoTokenizer`\n\n19. `import tensorflow as tf`\n\n20. `tokenizer = AutoTokenizer.from_pretrained(model_path)`\n\n21. `model = TFAutoModelWithLMHead.from_pretrained(model_path)`\n\n22. `sequence = f\"Distilled models are smaller than the models they mimic. Using them instead of the large versions would help {tokenizer.mask_token} our carbon footprint.\"`\n\n23. `input = tokenizer.encode(sequence, return_tensors=\"tf\")`\n\n24. `mask_token_index = tf.where(input == tokenizer.mask_token_id)[0, 1]`\n\n25. `token_logits = model(input)`\n\n26. `mask_token_logits = token_logits[0, mask_token_index, :]`\n\n27. `top_5_tokens = tf.math.top_k(mask_token_logits, 5).indices.numpy()`\n\n28. `for token in top_5_tokens:`\n\n29. `print(sequence.replace(tokenizer.mask_token, tokenizer.decode([token])))`\n\n1.使用管道\n\n` `\n1. `model_path=\"H:\\\\code\\\\Model\\\\xlnet-base-cased\\\\\"`\n\n2. `from transformers import pipeline`\n\n3. `text_generator = pipeline(\"text-generation\",model=model_path, tokenizer=model_path, framework=\"tf\")`\n\n4. `print(text_generator(\"As far as I am concerned, I will\", max_length=50, do_sample=False))`\n\n2.直接使用模型\n\n` `\n1. `model_path=\"H:\\\\code\\\\Model\\\\xlnet-base-cased\\\\\"`\n\n2. `#使用pytorch版本`\n\n3. `from transformers import AutoModelWithLMHead, AutoTokenizer`\n\n4. `model = AutoModelWithLMHead.from_pretrained(model_path)`\n\n5. `tokenizer = AutoTokenizer.from_pretrained(model_path)`\n\n6. `# Padding text helps XLNet with short prompts - proposed by Aman Rusia in https://github.com/rusiaaman/XLNet-gen#methodology`\n\n7. `PADDING_TEXT = \"\"\"In 1991, the remains of Russian Tsar Nicholas II and his family`\n\n8. `(except for Alexei and Maria) are discovered.`\n\n9. `The voice of Nicholas's young son, Tsarevich Alexei Nikolaevich, narrates the`\n\n10. `remainder of the story. 1883 Western Siberia,`\n\n11. `a young Grigori Rasputin is asked by his father and a group of men to perform magic.`\n\n12. `Rasputin has a vision and denounces one of the men as a horse thief. Although his`\n\n13. `father initially slaps him for making such an accusation, Rasputin watches as the`\n\n14. `man is chased outside and beaten. Twenty years later, Rasputin sees a vision of`\n\n15. `the Virgin Mary, prompting him to become a priest. Rasputin quickly becomes famous,`\n\n16. `with people, even a bishop, begging for his blessing. <eod> </s> <eos>\"\"\"`\n\n17. `prompt = \"Today the weather is really nice and I am planning on \"`\n\n18. `inputs = tokenizer.encode(PADDING_TEXT + prompt, add_special_tokens=False, return_tensors=\"pt\")`\n\n19. `prompt_length = len(tokenizer.decode(inputs, skip_special_tokens=True, clean_up_tokenization_spaces=True))`\n\n20. `outputs = model.generate(inputs, max_length=250, do_sample=True, top_p=0.95, top_k=60)`\n\n21. `generated = prompt + tokenizer.decode(outputs)[prompt_length:]`\n\n22. `print(generated)`\n\n23.\n24. `#使用tensorflow版本`\n\n25. `from transformers import TFAutoModelWithLMHead, AutoTokenizer`\n\n26. `model = TFAutoModelWithLMHead.from_pretrained(model_path)`\n\n27. `tokenizer = AutoTokenizer.from_pretrained(model_path)`\n\n28. `# Padding text helps XLNet with short prompts - proposed by Aman Rusia in https://github.com/rusiaaman/XLNet-gen#methodology`\n\n29. `PADDING_TEXT = \"\"\"In 1991, the remains of Russian Tsar Nicholas II and his family`\n\n30. `(except for Alexei and Maria) are discovered.`\n\n31. `The voice of Nicholas's young son, Tsarevich Alexei Nikolaevich, narrates the`\n\n32. `remainder of the story. 1883 Western Siberia,`\n\n33. `a young Grigori Rasputin is asked by his father and a group of men to perform magic.`\n\n34. `Rasputin has a vision and denounces one of the men as a horse thief. Although his`\n\n35. `father initially slaps him for making such an accusation, Rasputin watches as the`\n\n36. `man is chased outside and beaten. Twenty years later, Rasputin sees a vision of`\n\n37. `the Virgin Mary, prompting him to become a priest. Rasputin quickly becomes famous,`\n\n38. `with people, even a bishop, begging for his blessing. <eod> </s> <eos>\"\"\"`\n\n39. `prompt = \"Today the weather is really nice and I am planning on \"`\n\n40. `inputs = tokenizer.encode(PADDING_TEXT + prompt, add_special_tokens=False, return_tensors=\"tf\")`\n\n41. `prompt_length = len(tokenizer.decode(inputs, skip_special_tokens=True, clean_up_tokenization_spaces=True))`\n\n42. `outputs = model.generate(inputs, max_length=250, do_sample=True, top_p=0.95, top_k=60)`\n\n43. `generated = prompt + tokenizer.decode(outputs)[prompt_length:]`\n\n44. `print(generated)`\n\n45.\n\n1.使用管道\n\n` `\n1. `model_path=\"H:\\\\code\\\\Model\\\\dbmdz\\\\bert-large-cased-finetuned-conll03-english\\\\\"`\n\n2. `tokenizer=\"H:\\\\code\\\\Model\\\\bert-base-cased\\\\\"`\n\n3. `from transformers import pipeline`\n\n4. `nlp = pipeline(\"ner\",model=model_path, tokenizer=tokenizer_path)`\n\n5. `sequence = \"Hugging Face Inc. is a company based in New York City. Its headquarters are in DUMBO, therefore very\"`\n\n6. `\"close to the Manhattan Bridge which is visible from the window.\"`\n\n7. `print(nlp(sequence))`\n\n2.直接使用模型\n\n` `\n1. `model_path=\"H:\\\\code\\\\Model\\\\dbmdz\\\\bert-large-cased-finetuned-conll03-english\\\\\"`\n\n2. `tokenizer_path=\"H:\\\\code\\\\Model\\\\bert-base-cased\\\\\"`\n\n3.\n4. `##使用pytorch格式`\n\n5.\n6. `from transformers import AutoModelForTokenClassification, AutoTokenizer`\n\n7. `import torch`\n\n8. `model = AutoModelForTokenClassification.from_pretrained(model_path)`\n\n9. `tokenizer = AutoTokenizer.from_pretrained(tokenizer_path)`\n\n10. `label_list = [`\n\n11. `\"O\", # Outside of a named entity`\n\n12. `\"B-MISC\", # Beginning of a miscellaneous entity right after another miscellaneous entity`\n\n13. `\"I-MISC\", # Miscellaneous entity`\n\n14. `\"B-PER\", # Beginning of a person's name right after another person's name`\n\n15. `\"I-PER\", # Person's name`\n\n16. `\"B-ORG\", # Beginning of an organisation right after another organisation`\n\n17. `\"I-ORG\", # Organisation`\n\n18. `\"B-LOC\", # Beginning of a location right after another location`\n\n19. `\"I-LOC\" # Location`\n\n20. `]`\n\n21. `sequence = \"Hugging Face Inc. is a company based in New York City. Its headquarters are in DUMBO, therefore very\" \\`\n\n22. `\"close to the Manhattan Bridge.\"`\n\n23. `# Bit of a hack to get the tokens with the special tokens`\n\n24. `tokens = tokenizer.tokenize(tokenizer.decode(tokenizer.encode(sequence)))`\n\n25. `inputs = tokenizer.encode(sequence, return_tensors=\"pt\")`\n\n26. `outputs = model(inputs).logits`\n\n27. `predictions = torch.argmax(outputs, dim=2)`\n\n28. `print([(token, label_list[prediction]) for token, prediction in zip(tokens, predictions.numpy())])`\n\n29.\n30. `##使用tensorflow格式`\n\n31. `from transformers import TFAutoModelForTokenClassification, AutoTokenizer`\n\n32. `import tensorflow as tf`\n\n33. `model = TFAutoModelForTokenClassification.from_pretrained(model_path)`\n\n34. `tokenizer = AutoTokenizer.from_pretrained(tokenizer_path)`\n\n35. `label_list = [`\n\n36. `\"O\", # Outside of a named entity`\n\n37. `\"B-MISC\", # Beginning of a miscellaneous entity right after another miscellaneous entity`\n\n38. `\"I-MISC\", # Miscellaneous entity`\n\n39. `\"B-PER\", # Beginning of a person's name right after another person's name`\n\n40. `\"I-PER\", # Person's name`\n\n41. `\"B-ORG\", # Beginning of an organisation right after another organisation`\n\n42. `\"I-ORG\", # Organisation`\n\n43. `\"B-LOC\", # Beginning of a location right after another location`\n\n44. `\"I-LOC\" # Location`\n\n45. `]`\n\n46. `sequence = \"Hugging Face Inc. is a company based in New York City. Its headquarters are in DUMBO, therefore very\" \\`\n\n47. `\"close to the Manhattan Bridge.\"`\n\n48. `# Bit of a hack to get the tokens with the special tokens`\n\n49. `tokens = tokenizer.tokenize(tokenizer.decode(tokenizer.encode(sequence)))`\n\n50. `inputs = tokenizer.encode(sequence, return_tensors=\"tf\")`\n\n51. `outputs = model(inputs)`\n\n52. `predictions = tf.argmax(outputs, axis=2)`\n\n53. `print([(token, label_list[prediction]) for token, prediction in zip(tokens, predictions.numpy())])`\n\n1.使用管道\n\n` `\n1. `model_path=\"H:\\\\code\\\\Model\\\\t5-base\\\\\"`\n\n2.\n3. `from transformers import pipeline`\n\n4. `summarizer = pipeline(\"summarization\",model=model_path, tokenizer=model_path)`\n\n5. `ARTICLE = \"\"\" New York (CNN)When Liana Barrientos was 23 years old, she got married in Westchester County, New York.`\n\n6. `A year later, she got married again in Westchester County, but to a different man and without divorcing her first husband.`\n\n7. `Only 18 days after that marriage, she got hitched yet again. Then, Barrientos declared \"I do\" five more times, sometimes only within two weeks of each other.`\n\n8. `In 2010, she married once more, this time in the Bronx. In an application for a marriage license, she stated it was her \"first and only\" marriage.`\n\n9. `Barrientos, now 39, is facing two criminal counts of \"offering a false instrument for filing in the first degree,\" referring to her false statements on the`\n\n10. `2010 marriage license application, according to court documents.`\n\n11. `Prosecutors said the marriages were part of an immigration scam.`\n\n12. `On Friday, she pleaded not guilty at State Supreme Court in the Bronx, according to her attorney, Christopher Wright, who declined to comment further.`\n\n13. `After leaving court, Barrientos was arrested and charged with theft of service and criminal trespass for allegedly sneaking into the New York subway through an emergency exit, said Detective`\n\n14. `Annette Markowski, a police spokeswoman. In total, Barrientos has been married 10 times, with nine of her marriages occurring between 1999 and 2002.`\n\n15. `All occurred either in Westchester County, Long Island, New Jersey or the Bronx. She is believed to still be married to four men, and at one time, she was married to eight men at once, prosecutors say.`\n\n16. `Prosecutors said the immigration scam involved some of her husbands, who filed for permanent residence status shortly after the marriages.`\n\n17. `Any divorces happened only after such filings were approved. It was unclear whether any of the men will be prosecuted.`\n\n18. `The case was referred to the Bronx District Attorney\\'s Office by Immigration and Customs Enforcement and the Department of Homeland Security\\'s`\n\n19. `Investigation Division. Seven of the men are from so-called \"red-flagged\" countries, including Egypt, Turkey, Georgia, Pakistan and Mali.`\n\n20. `Her eighth husband, Rashid Rajput, was deported in 2006 to his native Pakistan after an investigation by the Joint Terrorism Task Force.`\n\n21. `If convicted, Barrientos faces up to four years in prison. Her next court appearance is scheduled for May 18.`\n\n22. `\"\"\"`\n\n23. `print(summarizer(ARTICLE, max_length=130, min_length=30, do_sample=False))`\n\n2.直接使用模型\n\n` `\n1. `model_path=\"H:\\\\code\\\\Model\\\\t5-base\\\\\"`\n\n2.\n3. `#使用pytorch框架`\n\n4. `from transformers import AutoModelWithLMHead, AutoTokenizer`\n\n5. `model = AutoModelWithLMHead.from_pretrained(model_path)`\n\n6. `tokenizer = AutoTokenizer.from_pretrained(model_path)`\n\n7. `# T5 uses a max_length of 512 so we cut the article to 512 tokens.`\n\n8. `inputs = tokenizer.encode(\"summarize: \" + ARTICLE, return_tensors=\"pt\", max_length=512)`\n\n9. `outputs = model.generate(inputs, max_length=150, min_length=40, length_penalty=2.0, num_beams=4, early_stopping=True)`\n\n10. `# print(outputs)`\n\n11. `print(tokenizer.decode(outputs))`\n\n12.\n13. `#使用tensorflow框架`\n\n14. `from transformers import TFAutoModelWithLMHead, AutoTokenizer`\n\n15. `model = TFAutoModelWithLMHead.from_pretrained(model_path)`\n\n16. `tokenizer = AutoTokenizer.from_pretrained(model_path)`\n\n17. `# T5 uses a max_length of 512 so we cut the article to 512 tokens.`\n\n18. `inputs = tokenizer.encode(\"summarize: \" + ARTICLE, return_tensors=\"tf\", max_length=512)`\n\n19. `outputs = model.generate(inputs, max_length=150, min_length=40, length_penalty=2.0, num_beams=4, early_stopping=True)`\n\n1.使用管道\n\n` `\n1. `model_path=\"H:\\\\code\\\\Model\\\\t5-base\\\\\"`\n\n2. `from transformers import pipeline`\n\n3. `translator = pipeline(\"translation_en_to_de\",model=model_path, tokenizer=model_path)`\n\n4. `print(translator(\"Hugging Face is a technology company based in New York and Paris\", max_length=40))`\n\n2.直接使用模型\n\n` `\n1. `model_path=\"H:\\\\code\\\\Model\\\\distilbert-base-uncased\\\\\"`\n\n2. `#pytorch方式`\n\n3.\n4. `from transformers import DistilBertConfig, DistilBertTokenizer, DistilBertForSequenceClassification`\n\n5. `config = DistilBertConfig(n_heads=8, dim=512, hidden_dim=4*512)`\n\n6. `tokenizer = DistilBertTokenizer.from_pretrained(model_path)`\n\n7. `model = DistilBertForSequenceClassification(config)`\n\n8.\n9. `#tensorflow方式`\n\n10.\n11. `from transformers import DistilBertConfig, DistilBertTokenizer, TFDistilBertForSequenceClassification`\n\n12. `config = DistilBertConfig(n_heads=8, dim=512, hidden_dim=4*512)`\n\n13. `tokenizer = DistilBertTokenizer.from_pretrained(model_path)`\n\n14. `model = TFDistilBertForSequenceClassification(config)`\n\n15.\n\n` `\n1. `model_name=\"H:\\\\code\\\\Model\\\\distilbert-base-uncased\\\\\"`\n\n2.\n3. `#pytorch方式`\n\n4.\n5. `from transformers import DistilBertConfig, DistilBertTokenizer, DistilBertForSequenceClassification`\n\n6. `#model_name = \"distilbert-base-uncased\"`\n\n7. `model = DistilBertForSequenceClassification.from_pretrained(model_name, num_labels=10)`\n\n8. `tokenizer = DistilBertTokenizer.from_pretrained(model_name)`\n\n9.\n10. `#tensorflow方式`\n\n11. `from transformers import DistilBertConfig, DistilBertTokenizer, TFDistilBertForSequenceClassification`\n\n12. `#model_name = \"distilbert-base-uncased\"`\n\n13. `model = TFDistilBertForSequenceClassification.from_pretrained(model_name, num_labels=10)`\n\n14. `tokenizer = DistilBertTokenizer.from_pretrained(model_name)`\n\n` `\n1. `#模型保存`\n\n2. `##对模型进行微调后，可以通过以下方式将其与标记器一起保存：`\n\n3. `save_directory=\"./save/\"`\n\n4. `tokenizer.save_pretrained(save_directory)`\n\n5. `model.save_pretrained(save_directory)`\n\n6.\n7. `#模型加载`\n\n8. `##TensorFlow模型中加载已保存的PyTorch模型`\n\n9. `tokenizer = AutoTokenizer.from_pretrained(save_directory)`\n\n10. `model = TFAutoModel.from_pretrained(save_directory, from_pt=True)`\n\n11.\n12. `##PyTorch模型中加载已保存的TensorFlow模型`\n\n13. `tokenizer = AutoTokenizer.from_pretrained(save_directory)`\n\n14. `model = AutoModel.from_pretrained(save_directory, from_tf=True)`\n\n15.\n\n### QQ号码被盗如何申诉成功，我教您绝招!_qq被盗可以报警解决吗_ucshng的博客-程序员秘密\n\n这些信息在申诉中是非常重要的，比如登陆QQ官网后察看个人信息，就能看到您的昵称，姓名，联系电话等信息，有人说这些信息直接在QQ里就能察看啊，不是这样的，我认为90％的用户都用QQ软件，但很少用自己的QQ登陆官网，记得同步更新么？对，关键问题就在这里，由于您几乎从未登陆官网，官网里的个人信息资料可能就是很早以前的，甚至是您刚刚申请QQ时的资料，申诉时提供的资料可是越早越好的，所以，经常在官网同步\n\n### scrapy 教程_擒贼先擒王的博客-程序员秘密\n\n------------------------------------------------------------------------------------------scrapy中文文档 和 scrapy 英文文档参照看。因为中文文档比较老，英文文档是最新的。scrapy 英文文档：https://doc.scrapy.org/en/latestscrapy 中文文档：...\n\n### 20165211 2017-2018-2 《Java程序设计》课程总结_weixin_33831673的博客-程序员秘密\n\n20165211 2017-2018-2 《Java程序设计》课程总结一、每周作业及实验报告博客链接汇总预备作业1：我期望的师生关系预备作业2：学习基础和C语言调查预备作业3：Linux安装与学习第一周作业：Java入门及环境搭建第二周作业：基本数据类型与数，运算符、表达式和语句第三周作业：类与对象第四周作业：子类与继承，接口与实现第五...\n\n### RFC3920_weixin_30547797的博客-程序员秘密\n\nRFC3920 可扩展的消息和出席信息协议 (XMPP): 核心协议 关于本文的说明 本文为互联网社区定义了一个互联网标准跟踪协议，并且申请讨论协议和提出了改进的建议。请参照“互联网官方协议标准”的最新版本（STD 1）获得这个协议的标准化进程和状态。本文可以不受限制的分发。 版权声明 本文版权属于互联网社区 (C) The Internet Society (2004). 摘要 ..."
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https://ixtrieve.fh-koeln.de/birds/litie/document/35289 | [
"# Document (#35289)\n\nAuthor\nSong, F.W.\nTitle\nVirtual communities : bowling alone, online together\nImprint\nNew York : Lang\nYear\n2009\nPages\nXVII, 178 S\nIsbn\n978-1-433-10396-4 *\n978-1-433-10395-7\nSeries\nDigital formations ; Vol. 54\nAbstract\nDoes contemporary Internet technology strengthen civic engagement and democratic practice? The recent surge in online community participation has become a cultural phenomenon enmeshed in ongoing debates about the health of American civil society. But observations about online communities often concentrate on ascertaining the true nature of community and democracy, typically rehearsing familiar communitarian and liberal perspectives. This book seeks to understand the technology on its own terms, focusing on how the technological and organizational configurations of online communities frame our contemporary beliefs and assumptions about community and the individual. It analyzes key structural features of thirty award-winning online community websites to show that while the values of individual autonomy, egalitarianism, and freedom of speech dominate the discursive content of these communities, the practical realities of online life are clearly marked by exclusivity and the demands of commercialization and corporate surveillance. Promises of social empowerment are framed within consumer and therapeutic frameworks that undermine their democratic efficacy. As a result, online communities fail to revolutionize the civic landscape because they create cultures of membership that epitomize the commodification of community and public life altogether.\nContent\nInhalt: The virtual community -- A high-stakes battle : the context of virtual communities -- A cultural topography of virtual communities : the rough terrain of autonomy and control -- An alternative framework for understanding virtual communities -- The institutional landscape : the market of virtual communities -- The evolving landscape of virtual communities -- Technology, the self, and the market : eyeing the horizons of a brave new democracy -- Epilogue\nTheme\nInternet\nField\nKommunikationswissenschaften\nCOMPASS\nOnline social networks\nInternet / Social aspects\nCommunities / Philosophy\nComputers and civilization\nInformation technology / Social aspects\nDDC\n302.30285\nLCC\nHM742\n\n## Similar documents (author)\n\n1. Song, S.-f.: Rethinking of the development of reference service (1997) 4.06\n```4.0597486 = sum of:\n4.0597486 = weight(author_txt:song in 1860) [ClassicSimilarity], result of:\n4.0597486 = fieldWeight in 1860, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.119497 = idf(docFreq=34, maxDocs=43254)\n0.5 = fieldNorm(doc=1860)\n```\n2. Song, D.; Bruza, P.D.: Towards context sensitive information inference (2003) 4.06\n```4.0597486 = sum of:\n4.0597486 = weight(author_txt:song in 3429) [ClassicSimilarity], result of:\n4.0597486 = fieldWeight in 3429, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.119497 = idf(docFreq=34, maxDocs=43254)\n0.5 = fieldNorm(doc=3429)\n```\n3. 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Zhu, J.; Song, D.; Rüger, S.: Integrating multiple windows and document features for expert finding (2009) 3.04\n```3.0448115 = sum of:\n3.0448115 = weight(author_txt:song in 4756) [ClassicSimilarity], result of:\n3.0448115 = fieldWeight in 4756, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.119497 = idf(docFreq=34, maxDocs=43254)\n0.375 = fieldNorm(doc=4756)\n```\n\n## Similar documents (content)\n\n1. Berman, Y.; Phillips, D.: Information and social quality (2001) 0.19\n```0.19257398 = sum of:\n0.19257398 = product of:\n0.6877642 = sum of:\n0.11748599 = weight(abstract_txt:empowerment in 1829) [ClassicSimilarity], result of:\n0.11748599 = score(doc=1829,freq=1.0), product of:\n0.17008406 = queryWeight, product of:\n1.0504807 = boost\n8.841632 = idf(docFreq=16, maxDocs=43254)\n0.018312309 = queryNorm\n0.6907525 = fieldWeight in 1829, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.841632 = idf(docFreq=16, maxDocs=43254)\n0.078125 = fieldNorm(doc=1829)\n0.05208274 = weight(abstract_txt:aspects in 1829) [ClassicSimilarity], result of:\n0.05208274 = score(doc=1829,freq=2.0), product of:\n0.09888615 = queryWeight, product of:\n1.1327629 = boost\n4.76709 = idf(docFreq=999, maxDocs=43254)\n0.018312309 = queryNorm\n0.526694 = fieldWeight in 1829, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n4.76709 = idf(docFreq=999, maxDocs=43254)\n0.078125 = fieldNorm(doc=1829)\n0.058607638 = weight(abstract_txt:life in 1829) [ClassicSimilarity], result of:\n0.058607638 = score(doc=1829,freq=1.0), product of:\n0.13478836 = queryWeight, product of:\n1.3225055 = boost\n5.5655975 = idf(docFreq=449, maxDocs=43254)\n0.018312309 = queryNorm\n0.4348123 = fieldWeight in 1829, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.5655975 = idf(docFreq=449, maxDocs=43254)\n0.078125 = fieldNorm(doc=1829)\n0.03107141 = weight(abstract_txt:about in 1829) [ClassicSimilarity], result of:\n0.03107141 = score(doc=1829,freq=1.0), product of:\n0.101069614 = queryWeight, product of:\n1.4025786 = boost\n3.9350505 = idf(docFreq=2297, maxDocs=43254)\n0.018312309 = queryNorm\n0.30742583 = fieldWeight in 1829, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.9350505 = idf(docFreq=2297, maxDocs=43254)\n0.078125 = fieldNorm(doc=1829)\n0.1657575 = weight(abstract_txt:social in 1829) [ClassicSimilarity], result of:\n0.1657575 = score(doc=1829,freq=10.0), product of:\n0.15764312 = queryWeight, product of:\n2.0226643 = boost\n4.256064 = idf(docFreq=1666, maxDocs=43254)\n0.018312309 = queryNorm\n1.0514731 = fieldWeight in 1829, product of:\n3.1622777 = tf(freq=10.0), with freq of:\n10.0 = termFreq=10.0\n4.256064 = idf(docFreq=1666, maxDocs=43254)\n0.078125 = fieldNorm(doc=1829)\n0.09259498 = weight(abstract_txt:community in 1829) [ClassicSimilarity], result of:\n0.09259498 = score(doc=1829,freq=1.0), product of:\n0.24815395 = queryWeight, product of:\n2.837278 = boost\n4.7761307 = idf(docFreq=990, maxDocs=43254)\n0.018312309 = queryNorm\n0.3731352 = fieldWeight in 1829, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.7761307 = idf(docFreq=990, maxDocs=43254)\n0.078125 = fieldNorm(doc=1829)\n0.17016391 = weight(abstract_txt:communities in 1829) [ClassicSimilarity], result of:\n0.17016391 = score(doc=1829,freq=1.0), product of:\n0.39564133 = queryWeight, product of:\n3.9244857 = boost\n5.5052342 = idf(docFreq=477, maxDocs=43254)\n0.018312309 = queryNorm\n0.43009642 = fieldWeight in 1829, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.5052342 = idf(docFreq=477, maxDocs=43254)\n0.078125 = fieldNorm(doc=1829)\n0.28 = coord(7/25)\n```\n2. Mossberger, K.; Tolbert, C.J.; McNeal, R.S.: Digital citizenship : the internet, society, and participation (2007) 0.17\n```0.17427905 = sum of:\n0.17427905 = product of:\n0.6224252 = sum of:\n0.029462447 = weight(abstract_txt:aspects in 3973) [ClassicSimilarity], result of:\n0.029462447 = score(doc=3973,freq=1.0), product of:\n0.09888615 = queryWeight, product of:\n1.1327629 = boost\n4.76709 = idf(docFreq=999, maxDocs=43254)\n0.018312309 = queryNorm\n0.29794312 = fieldWeight in 3973, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.76709 = idf(docFreq=999, maxDocs=43254)\n0.0625 = fieldNorm(doc=3973)\n0.024857128 = weight(abstract_txt:about in 3973) [ClassicSimilarity], result of:\n0.024857128 = score(doc=3973,freq=1.0), product of:\n0.101069614 = queryWeight, product of:\n1.4025786 = boost\n3.9350505 = idf(docFreq=2297, maxDocs=43254)\n0.018312309 = queryNorm\n0.24594066 = fieldWeight in 3973, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.9350505 = idf(docFreq=2297, maxDocs=43254)\n0.0625 = fieldNorm(doc=3973)\n0.0455969 = weight(abstract_txt:technology in 3973) [ClassicSimilarity], result of:\n0.0455969 = score(doc=3973,freq=2.0), product of:\n0.12020805 = queryWeight, product of:\n1.5296204 = boost\n4.2914767 = idf(docFreq=1608, maxDocs=43254)\n0.018312309 = queryNorm\n0.37931654 = fieldWeight in 3973, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n4.2914767 = idf(docFreq=1608, maxDocs=43254)\n0.0625 = fieldNorm(doc=3973)\n0.12088951 = weight(abstract_txt:democratic in 3973) [ClassicSimilarity], result of:\n0.12088951 = score(doc=3973,freq=1.0), product of:\n0.253444 = queryWeight, product of:\n1.813478 = boost\n7.6317935 = idf(docFreq=56, maxDocs=43254)\n0.018312309 = queryNorm\n0.4769871 = fieldWeight in 3973, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.6317935 = idf(docFreq=56, maxDocs=43254)\n0.0625 = fieldNorm(doc=3973)\n0.0419337 = weight(abstract_txt:social in 3973) [ClassicSimilarity], result of:\n0.0419337 = score(doc=3973,freq=1.0), product of:\n0.15764312 = queryWeight, product of:\n2.0226643 = boost\n4.256064 = idf(docFreq=1666, maxDocs=43254)\n0.018312309 = queryNorm\n0.266004 = fieldWeight in 3973, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.256064 = idf(docFreq=1666, maxDocs=43254)\n0.0625 = fieldNorm(doc=3973)\n0.23949602 = weight(abstract_txt:civic in 3973) [ClassicSimilarity], result of:\n0.23949602 = score(doc=3973,freq=2.0), product of:\n0.31730613 = queryWeight, product of:\n2.029133 = boost\n8.5393505 = idf(docFreq=22, maxDocs=43254)\n0.018312309 = queryNorm\n0.75477904 = fieldWeight in 3973, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n8.5393505 = idf(docFreq=22, maxDocs=43254)\n0.0625 = fieldNorm(doc=3973)\n0.120189525 = weight(abstract_txt:online in 3973) [ClassicSimilarity], result of:\n0.120189525 = score(doc=3973,freq=5.0), product of:\n0.23436746 = queryWeight, product of:\n3.487787 = boost\n3.669478 = idf(docFreq=2996, maxDocs=43254)\n0.018312309 = queryNorm\n0.51282513 = fieldWeight in 3973, product of:\n2.236068 = tf(freq=5.0), with freq of:\n5.0 = termFreq=5.0\n3.669478 = idf(docFreq=2996, maxDocs=43254)\n0.0625 = fieldNorm(doc=3973)\n0.28 = coord(7/25)\n```\n3. Cao, X.; Wang, D.: ¬The role of online communities in reducing urban-rural health disparities in China (2018) 0.14\n```0.143956 = sum of:\n0.143956 = product of:\n0.71978 = sum of:\n0.040302347 = weight(abstract_txt:technology in 288) [ClassicSimilarity], result of:\n0.040302347 = score(doc=288,freq=1.0), product of:\n0.12020805 = queryWeight, product of:\n1.5296204 = boost\n4.2914767 = idf(docFreq=1608, maxDocs=43254)\n0.018312309 = queryNorm\n0.33527163 = fieldWeight in 288, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.2914767 = idf(docFreq=1608, maxDocs=43254)\n0.078125 = fieldNorm(doc=288)\n0.074129015 = weight(abstract_txt:social in 288) [ClassicSimilarity], result of:\n0.074129015 = score(doc=288,freq=2.0), product of:\n0.15764312 = queryWeight, product of:\n2.0226643 = boost\n4.256064 = idf(docFreq=1666, maxDocs=43254)\n0.018312309 = queryNorm\n0.47023308 = fieldWeight in 288, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n4.256064 = idf(docFreq=1666, maxDocs=43254)\n0.078125 = fieldNorm(doc=288)\n0.1603792 = weight(abstract_txt:community in 288) [ClassicSimilarity], result of:\n0.1603792 = score(doc=288,freq=3.0), product of:\n0.24815395 = queryWeight, product of:\n2.837278 = boost\n4.7761307 = idf(docFreq=990, maxDocs=43254)\n0.018312309 = queryNorm\n0.6462891 = fieldWeight in 288, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n4.7761307 = idf(docFreq=990, maxDocs=43254)\n0.078125 = fieldNorm(doc=288)\n0.1502369 = weight(abstract_txt:online in 288) [ClassicSimilarity], result of:\n0.1502369 = score(doc=288,freq=5.0), product of:\n0.23436746 = queryWeight, product of:\n3.487787 = boost\n3.669478 = idf(docFreq=2996, maxDocs=43254)\n0.018312309 = queryNorm\n0.6410314 = fieldWeight in 288, product of:\n2.236068 = tf(freq=5.0), with freq of:\n5.0 = termFreq=5.0\n3.669478 = idf(docFreq=2996, maxDocs=43254)\n0.078125 = fieldNorm(doc=288)\n0.29473257 = weight(abstract_txt:communities in 288) [ClassicSimilarity], result of:\n0.29473257 = score(doc=288,freq=3.0), product of:\n0.39564133 = queryWeight, product of:\n3.9244857 = boost\n5.5052342 = idf(docFreq=477, maxDocs=43254)\n0.018312309 = queryNorm\n0.74494886 = fieldWeight in 288, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n5.5052342 = idf(docFreq=477, maxDocs=43254)\n0.078125 = fieldNorm(doc=288)\n0.2 = coord(5/25)\n```\n4. Worrall, A.: \"Connections above and beyond\" : information, translation, and community boundaries in LibraryThing and Goodreads (2019) 0.13\n```0.12909131 = sum of:\n0.12909131 = product of:\n0.64545655 = sum of:\n0.032241877 = weight(abstract_txt:technology in 303) [ClassicSimilarity], result of:\n0.032241877 = score(doc=303,freq=1.0), product of:\n0.12020805 = queryWeight, product of:\n1.5296204 = boost\n4.2914767 = idf(docFreq=1608, maxDocs=43254)\n0.018312309 = queryNorm\n0.2682173 = fieldWeight in 303, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.2914767 = idf(docFreq=1608, maxDocs=43254)\n0.0625 = fieldNorm(doc=303)\n0.0838674 = weight(abstract_txt:social in 303) [ClassicSimilarity], result of:\n0.0838674 = score(doc=303,freq=4.0), product of:\n0.15764312 = queryWeight, product of:\n2.0226643 = boost\n4.256064 = idf(docFreq=1666, maxDocs=43254)\n0.018312309 = queryNorm\n0.532008 = fieldWeight in 303, product of:\n2.0 = tf(freq=4.0), with freq of:\n4.0 = termFreq=4.0\n4.256064 = idf(docFreq=1666, maxDocs=43254)\n0.0625 = fieldNorm(doc=303)\n0.10475925 = weight(abstract_txt:community in 303) [ClassicSimilarity], result of:\n0.10475925 = score(doc=303,freq=2.0), product of:\n0.24815395 = queryWeight, product of:\n2.837278 = boost\n4.7761307 = idf(docFreq=990, maxDocs=43254)\n0.018312309 = queryNorm\n0.42215428 = fieldWeight in 303, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n4.7761307 = idf(docFreq=990, maxDocs=43254)\n0.0625 = fieldNorm(doc=303)\n0.120189525 = weight(abstract_txt:online in 303) [ClassicSimilarity], result of:\n0.120189525 = score(doc=303,freq=5.0), product of:\n0.23436746 = queryWeight, product of:\n3.487787 = boost\n3.669478 = idf(docFreq=2996, maxDocs=43254)\n0.018312309 = queryNorm\n0.51282513 = fieldWeight in 303, product of:\n2.236068 = tf(freq=5.0), with freq of:\n5.0 = termFreq=5.0\n3.669478 = idf(docFreq=2996, maxDocs=43254)\n0.0625 = fieldNorm(doc=303)\n0.30439848 = weight(abstract_txt:communities in 303) [ClassicSimilarity], result of:\n0.30439848 = score(doc=303,freq=5.0), product of:\n0.39564133 = queryWeight, product of:\n3.9244857 = boost\n5.5052342 = idf(docFreq=477, maxDocs=43254)\n0.018312309 = queryNorm\n0.7693799 = fieldWeight in 303, product of:\n2.236068 = tf(freq=5.0), with freq of:\n5.0 = termFreq=5.0\n5.5052342 = idf(docFreq=477, maxDocs=43254)\n0.0625 = fieldNorm(doc=303)\n0.2 = coord(5/25)\n```\n5. Poole, A.H.: ¬The information work of community archives : a systematic literature review (2020) 0.13\n```0.12650812 = sum of:\n0.12650812 = product of:\n0.6325406 = sum of:\n0.09398879 = weight(abstract_txt:empowerment in 841) [ClassicSimilarity], result of:\n0.09398879 = score(doc=841,freq=1.0), product of:\n0.17008406 = queryWeight, product of:\n1.0504807 = boost\n8.841632 = idf(docFreq=16, maxDocs=43254)\n0.018312309 = queryNorm\n0.552602 = fieldWeight in 841, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.841632 = idf(docFreq=16, maxDocs=43254)\n0.0625 = fieldNorm(doc=841)\n0.12088951 = weight(abstract_txt:democratic in 841) [ClassicSimilarity], result of:\n0.12088951 = score(doc=841,freq=1.0), product of:\n0.253444 = queryWeight, product of:\n1.813478 = boost\n7.6317935 = idf(docFreq=56, maxDocs=43254)\n0.018312309 = queryNorm\n0.4769871 = fieldWeight in 841, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.6317935 = idf(docFreq=56, maxDocs=43254)\n0.0625 = fieldNorm(doc=841)\n0.05930321 = weight(abstract_txt:social in 841) [ClassicSimilarity], result of:\n0.05930321 = score(doc=841,freq=2.0), product of:\n0.15764312 = queryWeight, product of:\n2.0226643 = boost\n4.256064 = idf(docFreq=1666, maxDocs=43254)\n0.018312309 = queryNorm\n0.37618646 = fieldWeight in 841, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n4.256064 = idf(docFreq=1666, maxDocs=43254)\n0.0625 = fieldNorm(doc=841)\n0.22222795 = weight(abstract_txt:community in 841) [ClassicSimilarity], result of:\n0.22222795 = score(doc=841,freq=9.0), product of:\n0.24815395 = queryWeight, product of:\n2.837278 = boost\n4.7761307 = idf(docFreq=990, maxDocs=43254)\n0.018312309 = queryNorm\n0.8955245 = fieldWeight in 841, product of:\n3.0 = tf(freq=9.0), with freq of:\n9.0 = termFreq=9.0\n4.7761307 = idf(docFreq=990, maxDocs=43254)\n0.0625 = fieldNorm(doc=841)\n0.13613114 = weight(abstract_txt:communities in 841) [ClassicSimilarity], result of:\n0.13613114 = score(doc=841,freq=1.0), product of:\n0.39564133 = queryWeight, product of:\n3.9244857 = boost\n5.5052342 = idf(docFreq=477, maxDocs=43254)\n0.018312309 = queryNorm\n0.34407714 = fieldWeight in 841, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.5052342 = idf(docFreq=477, maxDocs=43254)\n0.0625 = fieldNorm(doc=841)\n0.2 = coord(5/25)\n```"
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https://www.mql5.com/en/articles/11200 | [
"Do you like the article?\nShare it with others —\nUse new possibilities of MetaTrader 5\n\n#### Similar articles",
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"",
null,
"# Data Science and Machine Learning (Part 06): Gradient Descent",
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"3 971",
null,
"0",
null,
"Premature optimization is the root of all evil in programming\n\n-Donald Knuth\n\n### Introduction\n\nAccording to Wikipedia gradient descent (also often called steepest descent) is a first-order iterative optimization algorithm for finding a local minimum of a differentiable function. The idea is to take repeated steps in the opposite direction of the gradient (or approximate gradient) of the function at the current point, because this is the direction of steepest descent. Conversely, stepping in the direction of the gradient will lead to a local maximum of that function; the procedure is then known as gradient ascent.\n\nBasically, the gradient descent is an optimization algorithm used to find the minimum of a function:",
null,
"The gradient descent is a very important algorithm in machine learning as it helps us find the parameters for the best model for our dataset. Let me first explain the term Cost Function.\n\n### Cost Function\n\nSome folks call this loss function, it is a metric for calculating how good or bad our model is at predicting the relationship between the values of x and y.\n\nThere are a lot of metrics that can be used to determine how the model is predicting but unlike all those, the cost function finds the average loss over the entire dataset, the larger the cost function the bad our model is at finding the relationships in our dataset.\n\nGradient Descent aims at minimizing the cost function because the model with the lowest cost function is the best model. For you to understand what I just explained let's see this following example.\n\nSuppose our cost function is the equation",
null,
"If we plot a graph of this function with python this is how it will look like this:",
null,
"The very first step we need to do to our cost function is to differentiate the cost function, using the Chain Rule:\n\nThe equation y= (x+5)^2 is a composite function (there is one function inside of another). The outer function being (x+5)^2 the Inner function being (x+5). To differentiate this let's apply the Chain rule, see the image:",
null,
"There is a video linked at the end of me doing the maths by hand, if you found this hard to understand. Ok so now this function that we just obtained is the Gradient. The process of finding the gradient of an equation is the most important step of all of them and I wish my mathematical teachers told me back in the day that the purpose of differentiating the function is so that we get the gradient of a function.\n\nThat is the first and most important step, below is the second step.\n\nStep 02:\n\nWe move in the negative direction of the gradient, here the question raises, how much should we move? This is where the learning rate comes into play.\n\n### Learning Rate\n\nBy definition, this is the step size at each iteration while moving toward a minimum of a loss function, take an example of a person stepping down the mountain, their steps are the learning rate, the smaller the steps the longer it will take them to reach the bottom of the mountain and vice versa.\n\nKeep the algorithm learning rate to smaller values but not very small like 0.0001 by doing so you are increasing the program execution time as it might take longer for the algorithm to reach the minimum values. In contrast, using big numbers for the learning rate will cause the algorithm to skip the minimum values which in the end may cause you to miss the minimum value targeted.\n\nThe default learning rate is 0.01.\n\nLet's perform the iteration to see how the algorithm works.\n\nFirst Iteration: We choose any random point as a starting point for our algorithm, I chose 0 as a the first value of x now, to update the values of x this is the formula",
null,
"By each iteration, we will descend toward the minimum value of the function and so is the name Gradient Descent. Making sense now?",
null,
"Let's see how this works in details. Now let's calculate the values manually on 2 iterations so that you get a solid understanding on what's happening:\n\n1st Iteration:\n\nformula: x1 = x0 - learning rate * ( 2*(x+5) )\n\nx1 = 0 - 0.01 * 0.01 * 2*(0+5)\n\nx1 = -0.01 * 10\n\nx1 = -0.1 (finally)\n\nNow the finally we update the values by assigning the new value to the old value and repeat the procedure for as much iterations until we reach the minimum of a function:\n\nx0 = x1\n\n2nd Iteration:\n\nx1 = -0.1 - 0.01 * 2*(-0.1+5)\n\nx1 = -0.198\n\nThen: x0 = x1\n\nIf we repeat this procedure a several times the output for 10 first iterations will be:\n\n```RS 0 17:15:16.793 gradient-descent test (EURUSD,M1) Gradient Descent CostFunction CUSTOM\nQQ 0 17:15:16.793 gradient-descent test (EURUSD,M1) 1 x0 = 0.0000000000 x1 = -0.1000000000 CostFunction = 10.0000000000\nES 0 17:15:16.793 gradient-descent test (EURUSD,M1) 2 x0 = -0.1000000000 x1 = -0.1980000000 CostFunction = 9.8000000000\nPR 0 17:15:16.793 gradient-descent test (EURUSD,M1) 3 x0 = -0.1980000000 x1 = -0.2940400000 CostFunction = 9.6040000000\nLE 0 17:15:16.793 gradient-descent test (EURUSD,M1) 4 x0 = -0.2940400000 x1 = -0.3881592000 CostFunction = 9.4119200000\nJD 0 17:15:16.793 gradient-descent test (EURUSD,M1) 5 x0 = -0.3881592000 x1 = -0.4803960160 CostFunction = 9.2236816000\nIG 0 17:15:16.793 gradient-descent test (EURUSD,M1) 6 x0 = -0.4803960160 x1 = -0.5707880957 CostFunction = 9.0392079680\nIG 0 17:15:16.793 gradient-descent test (EURUSD,M1) 7 x0 = -0.5707880957 x1 = -0.6593723338 CostFunction = 8.8584238086\nJF 0 17:15:16.793 gradient-descent test (EURUSD,M1) 8 x0 = -0.6593723338 x1 = -0.7461848871 CostFunction = 8.6812553325\nNI 0 17:15:16.793 gradient-descent test (EURUSD,M1) 9 x0 = -0.7461848871 x1 = -0.8312611893 CostFunction = 8.5076302258\nCK 0 17:15:16.793 gradient-descent test (EURUSD,M1) 10 x0 = -0.8312611893 x1 = -0.9146359656 CostFunction = 8.3374776213```\n\nLet's also see the other ten values of the algorithm when it is very close to the minimum of the function:\n\n```GK 0 17:15:16.800 gradient-descent test (EURUSD,M1) 1052 x0 = -4.9999999970 x1 = -4.9999999971 CostFunction = 0.0000000060\nIH 0 17:15:16.800 gradient-descent test (EURUSD,M1) 1053 x0 = -4.9999999971 x1 = -4.9999999971 CostFunction = 0.0000000059\nNH 0 17:15:16.800 gradient-descent test (EURUSD,M1) 1054 x0 = -4.9999999971 x1 = -4.9999999972 CostFunction = 0.0000000058\nQI 0 17:15:16.800 gradient-descent test (EURUSD,M1) 1055 x0 = -4.9999999972 x1 = -4.9999999972 CostFunction = 0.0000000057\nII 0 17:15:16.800 gradient-descent test (EURUSD,M1) 1056 x0 = -4.9999999972 x1 = -4.9999999973 CostFunction = 0.0000000055\nRN 0 17:15:16.800 gradient-descent test (EURUSD,M1) 1057 x0 = -4.9999999973 x1 = -4.9999999973 CostFunction = 0.0000000054\nKN 0 17:15:16.800 gradient-descent test (EURUSD,M1) 1058 x0 = -4.9999999973 x1 = -4.9999999974 CostFunction = 0.0000000053\nJO 0 17:15:16.800 gradient-descent test (EURUSD,M1) 1059 x0 = -4.9999999974 x1 = -4.9999999974 CostFunction = 0.0000000052\nJO 0 17:15:16.800 gradient-descent test (EURUSD,M1) 1060 x0 = -4.9999999974 x1 = -4.9999999975 CostFunction = 0.0000000051\nQL 0 17:15:16.800 gradient-descent test (EURUSD,M1) 1061 x0 = -4.9999999975 x1 = -4.9999999975 CostFunction = 0.0000000050\nQL 0 17:15:16.800 gradient-descent test (EURUSD,M1) 1062 x0 = -4.9999999975 x1 = -4.9999999976 CostFunction = 0.0000000049\nHP 0 17:15:16.800 gradient-descent test (EURUSD,M1) Local miminum found =-4.999999997546217```\n\nAfter 1062 (One thousand and sixty two) iterations the algorithm was able to reach the local minimum of this function.\n\nA thing to notice from this algorithm\n\nLooking at the values of Cost Function you will notice huge change in values at the beginning, but very tiny noticeable changes to the last values of a cost function.\n\nThe gradient descent takes larger steps when it is nowhere near the minimum of a function but, takes baby steps when is is near the minimum of the function, the same thing you will do when you are near the bottom of the mountain, so now you know that the gradient descent is pretty smart!\n\nIn the end the local minimum is\n\n`HP 0 17:15:16.800 gradient-descent test (EURUSD,M1) Local miminum found =-4.999999997546217`\n\nWhich is the accurate value because the minimum of this function is -5.0!\n\nThe Real Question\n\nHow does the Gradient know when to stop? See we can let the algorithm keep on iterating till infinity or at least the end of a computer's ability to calculate.\n\nWhen the cost function is Zero is when we know that the gradient descent has done it's job.\n\nNow let's code this whole operation in MQL5:\n\n``` while (true)\n{\niterations++;\n\nx1 = x0 - m_learning_rate * CustomCostFunction(x0);\n\nprintf(\"%d x0 = %.10f x1 = %.10f CostFunction = %.10f\",iterations,x0,x1,CustomCostFunction(x0));\n\nif (NormalizeDouble(CustomCostFunction(x0),8) == 0) { Print(\"Local minimum found =\",x0); break; }\n\nx0 = x1;\n} ```\n\nThe above block of code is the one that was able to get us the results we wanted but it's not alone in the class CGradientDescent. The Function CustomCostFunction is where our differentiated equation was being kept and calculated here it is\n\n```double CGradientDescent::CustomCostFunction(double x)\n{\nreturn(2 * ( x + 5 ));\n}```\n\n### What's the Purpose?\n\nOne might be asking themselves what's the purpose of all these calculations when you can just use the default linear model created by the previous libraries we discussed in this article series. Newsflash the model created using the default values isn't necessarily the best model so you need to let the computer learn the best parameters for the model with few errors (best model).\n\nWe are a few articles closer to building Artificial Neural Networks and for everyone to be able to understand how neural networks learn (teach themselves the patterns) in the process of back propagations and other techniques, gradient descent is the most popular algorithm that has made all that possible. Without a solid understanding of it you may never understand the process because things are about to get complicated.\n\n### Gradient Descent for a Regression Model",
null,
"Data visualization in Python:\n\n```import pandas as pd\nimport numpy as np\nimport matplotlib.pyplot as plt\n\nx = data[\"YearsExperience\"]\ny = data[\"Salary\"]\n\nplt.figure(figsize=(16,9))\nplt.title(\"Experience vs Salary\")\nplt.scatter(x,y,c=\"green\")\nplt.xlabel(xlabel=\"Years of Experience\")\nplt.ylabel(ylabel=\"Salary\")\nplt.show()```\n\nThis will be our graph:",
null,
"Looking at our dataset you can not help but notice that this dataset is for a regression problem, but we can have a million number of models to help us make the prediction or whatever it is we are trying to achieve.",
null,
"What is the Best model to use to make predictions of a person's experience and what their salary will be, that's what we are going to find out. But first let's derive the cost function for our regression model.\n\nTheory\n\nLet me take you back to Linear Regression.\n\nWe know for a fact that every linear model has errors associated with it. We also know that we can create a million lines in this graph and the best fit line is always the line with the least errors.\n\nThe cost function represents the error between our actual values and predicted values, we can write the formula for the cost function to be equal to:\n\nCost = Y Actual - Y Predicted\n\nSince we are seeing the magnitude of errors we raise the square, our formula now becomes",
null,
"But we are looking for Errors in our entire Dataset, we need to make the summation",
null,
"Finally, we divide the summation of errors by the m which is the number of items in the dataset :",
null,
"Here is the video on the whole mathematical procedures done by hand.\n\nNow that we have the cost function let's code for the gradient descent and find the best parameters for both. Coefficient of X(Slope) denoted as Bo and The Y Intercept denoted as B1\n\n```\ndouble cost_B0=0, cost_B1=0;\n\nif (costFunction == MSE)\n{\nint iterations=0;\nfor (int i=0; i<m_iterations; i++, iterations++)\n{\n\ncost_B0 = Mse(b0,b1,Intercept);\ncost_B1 = Mse(b0,b1,Slope);\n\nb0 = b0 - m_learning_rate * cost_B0;\nb1 = b1 - m_learning_rate * cost_B1;\n\nprintf(\"%d b0 = %.8f cost_B0 = %.8f B1 = %.8f cost_B1 = %.8f\",iterations,b0,cost_B0,b1,cost_B1);\n\nDBL_MAX_MIN(b0); DBL_MAX_MIN(cost_B0); DBL_MAX_MIN(cost_B1);\n\nif (NormalizeDouble(cost_B0,8) == 0 && NormalizeDouble(cost_B1,8) == 0) break;\n\n}\nprintf(\"%d Iterations Local Minima are\\nB0(Intercept) = %.5f || B1(Coefficient) = %.5f\",iterations,b0,b1);\n}```\n\nNotice a few things from the Gradient Descent code:\n\n• The process is still the same as the one we performed before but this time we are finding and updating the values twice at once the Bo and B1.\n• There is a restricted number of iterations, someone once said the best way to make an infinite loop is to use a while loop we do not use the while loop this time but instead we want to limit the number of times the algorithm will work out to find the coefficients for the best model.\n• DBL_MAX_MIN is a function for debugging purpose responsible to check and notify us if we have hit the mathematical limits of a computer.\n\nThis is the output of the operations of the algorithm. Learning Rate = 0.01 Iterations = 10000\n\n```PD 0 17:29:17.999 gradient-descent test (EURUSD,M1) 91738.0000 98273.0000 101302.0000 113812.0000 109431.0000 105582.0000 116969.0000 112635.0000 122391.0000 121872.0000\nRF 0 17:29:17.999 gradient-descent test (EURUSD,M1) 0 b0 = 1520.06000000 cost_B0 = -152006.00000000 B1 = 9547.97400000 cost_B1 = -954797.40000000\nOP 0 17:29:17.999 gradient-descent test (EURUSD,M1) 1 b0 = 1995.08742960 cost_B0 = -47502.74296000 B1 = 12056.69235267 cost_B1 = -250871.83526667\nLP 0 17:29:17.999 gradient-descent test (EURUSD,M1) 2 b0 = 2194.02117366 cost_B0 = -19893.37440646 B1 = 12707.81767044 cost_B1 = -65112.53177770\nQN 0 17:29:17.999 gradient-descent test (EURUSD,M1) 3 b0 = 2319.78332575 cost_B0 = -12576.21520809 B1 = 12868.77569178 cost_B1 = -16095.80213357\nLO 0 17:29:17.999 gradient-descent test (EURUSD,M1) 4 b0 = 2425.92576238 cost_B0 = -10614.24366387 B1 = 12900.42596039 cost_B1 = -3165.02686058\nGH 0 17:29:17.999 gradient-descent test (EURUSD,M1) 5 b0 = 2526.58198175 cost_B0 = -10065.62193621 B1 = 12897.99808257 cost_B1 = 242.78778134\nCJ 0 17:29:17.999 gradient-descent test (EURUSD,M1) 6 b0 = 2625.48307920 cost_B0 = -9890.10974571 B1 = 12886.62268517 cost_B1 = 1137.53974060\nDD 0 17:29:17.999 gradient-descent test (EURUSD,M1) 7 b0 = 2723.61498028 cost_B0 = -9813.19010723 B1 = 12872.93147573 cost_B1 = 1369.12094310\nHF 0 17:29:17.999 gradient-descent test (EURUSD,M1) 8 b0 = 2821.23916252 cost_B0 = -9762.41822398 B1 = 12858.67435081 cost_B1 = 1425.71249248\n\n<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<< Last Iterations >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>\n\nEI 0 17:29:48.247 gradient-descent test (EURUSD,M1) 6672 b0 = 25792.20019866 cost_B0 = -0.00000001 B1 = 9449.96232146 cost_B1 = 0.00000000\nNG 0 17:29:48.247 gradient-descent test (EURUSD,M1) 6673 b0 = 25792.20019866 cost_B0 = -0.00000001 B1 = 9449.96232146 cost_B1 = 0.00000000\nGD 0 17:29:48.247 gradient-descent test (EURUSD,M1) 6674 b0 = 25792.20019866 cost_B0 = -0.00000001 B1 = 9449.96232146 cost_B1 = 0.00000000\nPR 0 17:29:48.247 gradient-descent test (EURUSD,M1) 6675 b0 = 25792.20019866 cost_B0 = -0.00000001 B1 = 9449.96232146 cost_B1 = 0.00000000\nIS 0 17:29:48.247 gradient-descent test (EURUSD,M1) 6676 b0 = 25792.20019866 cost_B0 = -0.00000001 B1 = 9449.96232146 cost_B1 = 0.00000000\nRQ 0 17:29:48.247 gradient-descent test (EURUSD,M1) 6677 b0 = 25792.20019866 cost_B0 = -0.00000001 B1 = 9449.96232146 cost_B1 = 0.00000000\nKN 0 17:29:48.247 gradient-descent test (EURUSD,M1) 6678 b0 = 25792.20019866 cost_B0 = -0.00000001 B1 = 9449.96232146 cost_B1 = 0.00000000\nDL 0 17:29:48.247 gradient-descent test (EURUSD,M1) 6679 b0 = 25792.20019866 cost_B0 = -0.00000001 B1 = 9449.96232146 cost_B1 = 0.00000000\nRM 0 17:29:48.247 gradient-descent test (EURUSD,M1) 6680 b0 = 25792.20019866 cost_B0 = -0.00000001 B1 = 9449.96232146 cost_B1 = 0.00000000\nIK 0 17:29:48.247 gradient-descent test (EURUSD,M1) 6681 b0 = 25792.20019866 cost_B0 = -0.00000001 B1 = 9449.96232146 cost_B1 = 0.00000000\nPH 0 17:29:48.247 gradient-descent test (EURUSD,M1) 6682 b0 = 25792.20019866 cost_B0 = -0.00000001 B1 = 9449.96232146 cost_B1 = 0.00000000\nGF 0 17:29:48.247 gradient-descent test (EURUSD,M1) 6683 b0 = 25792.20019866 cost_B0 = -0.00000001 B1 = 9449.96232146 cost_B1 = 0.00000000\nMG 0 17:29:48.247 gradient-descent test (EURUSD,M1) 6684 b0 = 25792.20019866 cost_B0 = -0.00000000 B1 = 9449.96232146 cost_B1 = 0.00000000\nLE 0 17:29:48.247 gradient-descent test (EURUSD,M1) 6684 Iterations Local Minima are\nOJ 0 17:29:48.247 gradient-descent test (EURUSD,M1) B0(Intercept) = 25792.20020 || B1(Coefficient) = 9449.96232```\n\nIf we plot the graph using matplotlib",
null,
"B A M, The gradient Descent has been able to successfully get the best model out of 10000 models we tried out, great but there is one crucial step we are missing out that may cause our model to behave strangely and make us get the results we do not want.\n\n### Normalizing Linear Regression Input Variables Data\n\nWe know that for different datasets the best models can be found after different iterations some may take 100 iterations to reach the best models and some could take 10000 or up to a million iterations for the cost function to become zero not to mention that if we get the learning rate wrong values we may end up missing the local minima and if we miss that target we will end up hitting mathematical limits of a computer, let's see this in practice.\n\nLearning Rate = 0.1 Iterations 1000",
null,
"We just hit the maximum double value allowed by the system. Here are our logs.\n\n```GM 0 17:28:14.819 gradient-descent test (EURUSD,M1) Gradient Descent CostFunction MSE\nOP 0 17:28:14.819 gradient-descent test (EURUSD,M1) 0 b0 = 15200.60000000 cost_B0 = -152006.00000000 B1 = 95479.74000000 cost_B1 = -954797.40000000\nGR 0 17:28:14.819 gradient-descent test (EURUSD,M1) 1 b0 = -74102.05704000 cost_B0 = 893026.57040000 B1 = -512966.08473333 cost_B1 = 6084458.24733333\nNM 0 17:28:14.819 gradient-descent test (EURUSD,M1) 2 b0 = 501030.91374462 cost_B0 = -5751329.70784622 B1 = 3356325.13824362 cost_B1 = -38692912.22976952\nLH 0 17:28:14.819 gradient-descent test (EURUSD,M1) 3 b0 = -3150629.51591119 cost_B0 = 36516604.29655810 B1 = -21257352.71857720 cost_B1 = 246136778.56820822\nKD 0 17:28:14.819 gradient-descent test (EURUSD,M1) 4 b0 = 20084177.14287909 cost_B0 = -232348066.58790281 B1 = 135309993.40314889 cost_B1 = -1565673461.21726084\nOQ 0 17:28:14.819 gradient-descent test (EURUSD,M1) 5 b0 = -127706877.34210962 cost_B0 = 1477910544.84988713 B1 = -860620298.24803317 cost_B1 = 9959302916.51181984\nFM 0 17:28:14.819 gradient-descent test (EURUSD,M1) 6 b0 = 812402202.33122230 cost_B0 = -9401090796.73331833 B1 = 5474519904.86084747 cost_B1 = -63351402031.08880615\nJJ 0 17:28:14.819 gradient-descent test (EURUSD,M1) 7 b0 = -5167652856.43381691 cost_B0 = 59800550587.65039062 B1 = -34823489070.42410278 cost_B1 = 402980089752.84948730\nMP 0 17:28:14.819 gradient-descent test (EURUSD,M1) 8 b0 = 32871653967.62362671 cost_B0 = -380393068240.57440186 B1 = 221513298448.70788574 cost_B1 = -2563367875191.31982422\nMM 0 17:28:14.819 gradient-descent test (EURUSD,M1) 9 b0 = -209097460110.12799072 cost_B0 = 2419691140777.51611328 B1 = -1409052343513.33935547 cost_B1 = 16305656419620.47265625\nHD 0 17:28:14.819 gradient-descent test (EURUSD,M1) 10 b0 = 1330075004152.67309570 cost_B0 = -15391724642628.00976562 B1 = 8963022367351.18359375 cost_B1 = -103720747108645.23437500\nDP 0 17:28:14.819 gradient-descent test (EURUSD,M1) 11 b0 = -8460645083849.12207031 cost_B0 = 97907200880017.93750000 B1 = -57014041694401.67187500 cost_B1 = 659770640617528.50000000```\n\nThis signifies that if we got the learning rate wrong we may have a slim to none chance of finding the best model and chances are high that we will end up hitting the mathematical limit of a computer as you just saw the warning.\n\nBut if I try 0.01 for the learning rate in this dataset we will end up not having troubles though the training process will become much slower, but when I use this learning rate for this dataset I will end up hitting mathematical limits, so now you know that every dataset has it's learning rate but we may not have the chance to optimize for the learning rate because sometimes we have complex datasets with multiple variables and also this is an ineffective way of doing this whole process.\n\nthe solution to all these is to normalize the entire dataset so that it can be on the same scale, This improves readability when we plot the values on the same axis also it improves the training time because the normalized values are usually in the range of 0 to 1, Also we no longer have to worry about the learning rate because once we have only one learning rate parameter, we could use it for whatever dataset we face for example the learning rate of 0.01 read more about normalization here.\n\nLast but, not least\n\nAlso we know that our salary data values are from 39,343 to 121,782 mean while Years of Experience are from 1.1 to 10.5 if we keep the data this way, the values for salary are huge that they may make the model think that they are more important than any values so they will have a huge impact as compared to the years of experience, we need all the independent variables to have the same impacts as any other variables, now you see how important it is to normalize values.\n\n### (Normalization) Min-Max Scalar\n\nIn this approach we normalize the data to be within the range of 0 and 1. The formula is as given below:",
null,
"Converting this formula, into lines of code in MQL5 will become:\n\n```void CGradientDescent::MinMaxScaler(double &Array[])\n{\ndouble mean = Mean(Array);\ndouble max,min;\ndouble Norm[];\n\nArrayResize(Norm,ArraySize(Array));\n\nmax = Array[ArrayMaximum(Array)]; min = Array[ArrayMinimum(Array)];\n\nfor (int i=0; i<ArraySize(Array); i++)\nNorm[i] = (Array[i] - min) / (max - min);\n\nprintf(\"Scaled data Mean = %.5f Std = %.5f\",Mean(Norm),std(Norm));\n\nArrayFree(Array);\nArrayCopy(Array,Norm);\n}\n```\n\nThe function std() is just for letting us know the Standard Deviation after the data has been normalized. Here is its code:\n\n```double CGradientDescent::std(double &data[])\n{\ndouble mean = Mean(data);\ndouble sum = 0;\n\nfor (int i=0; i<ArraySize(data); i++)\nsum += MathPow(data[i] - mean,2);\n\nreturn(MathSqrt(sum/ArraySize(data)));\n}\n```\n\nNow let's call all this and print the output to see what happens:\n\n```void OnStart()\n{\n//---\nstring filename = \"Salary_Data.csv\";\n\ndouble XMatrix[];\ndouble YMatrix[];\n\nArrayPrint(\"Normalized X\",XMatrix);\nArrayPrint(\"Normalized Y\",YMatrix);\n\n}```\n\nOutput\n\n```OK 0 18:50:53.387 gradient-descent test (EURUSD,M1) Scaled data Mean = 0.44823 Std = 0.29683\nMG 0 18:50:53.387 gradient-descent test (EURUSD,M1) Scaled data Mean = 0.45207 Std = 0.31838\nMP 0 18:50:53.387 gradient-descent test (EURUSD,M1) Normalized X\nJG 0 18:50:53.387 gradient-descent test (EURUSD,M1) [ 0] 0.0000 0.0213 0.0426 0.0957 0.1170 0.1915 0.2021 0.2234 0.2234 0.2766 0.2979 0.3085 0.3085 0.3191 0.3617\nER 0 18:50:53.387 gradient-descent test (EURUSD,M1) 0.4043 0.4255 0.4468 0.5106 0.5213 0.6064 0.6383 0.7234 0.7553 0.8085 0.8404 0.8936 0.9043 0.9787 1.0000\nNQ 0 18:50:53.387 gradient-descent test (EURUSD,M1) Normalized Y\nIF 0 18:50:53.387 gradient-descent test (EURUSD,M1) [ 0] 0.0190 0.1001 0.0000 0.0684 0.0255 0.2234 0.2648 0.1974 0.3155 0.2298 0.3011 0.2134 0.2271 0.2286 0.2762\nIS 0 18:50:53.387 gradient-descent test (EURUSD,M1) 0.3568 0.3343 0.5358 0.5154 0.6639 0.6379 0.7151 0.7509 0.8987 0.8469 0.8015 0.9360 0.8848 1.0000 0.9939\n```\n\nThe graphs will now look like these:",
null,
"### Gradient Descent for Logistic Regression\n\nWe have seen how the linear side of the gradient descent, now let's see the logistic side.\n\nHere we do the same processes we just did on the linear regression part because the processes involved are the exact same only the process of differentiating the logistic regression gets more complex than that of a linear model, let's see the cost function first.\n\nAs discussed in the second article of the series about Logistic Regression the cost function of a logistic regression model is Binary Cross Entropy A.K.A Log Loss, given below.",
null,
"Now let's do the hard part first, differentiate this function to get it's gradient.\n\nAfter finding the derivatives",
null,
"Let's turn the formulas into MQL5 code inside the BCE function which stands for Binary Cross Entropy.\n\n```double CGradientDescent::Bce(double Bo,double B1,Beta wrt)\n{\ndouble sum_sqr=0;\ndouble m = ArraySize(Y);\ndouble x[];\n\nMatrixColumn(m_XMatrix,x,2);\n\nif (wrt == Slope)\nfor (int i=0; i<ArraySize(Y); i++)\n{\ndouble Yp = Sigmoid(Bo+B1*x[i]);\n\nsum_sqr += (Y[i] - Yp) * x[i];\n}\n\nif (wrt == Intercept)\nfor (int i=0; i<ArraySize(Y); i++)\n{\ndouble Yp = Sigmoid(Bo+B1*x[i]);\nsum_sqr += (Y[i] - Yp);\n}\nreturn((-1/m)*sum_sqr);\n}```\n\nSince we are dealing with the classification model, our dataset of choice is the Titanic dataset we used in logistic regression. Our independent variable is Pclass(Passenger class) our dependent variable on the other hand is Survived.",
null,
"Classified scatterplot",
null,
"Now we will call the class Gradient Descent but this time with the BCE (Binary Cross Entropy) as our cost function.\n\n``` filename = \"titanic.csv\";\n\nZeroMemory(XMatrix);\nZeroMemory(YMatrix);\n\n```\n\nLet's see the outcome:\n\n```CP 0 07:19:08.906 gradient-descent test (EURUSD,M1) Gradient Descent CostFunction BCE\nKD 0 07:19:08.906 gradient-descent test (EURUSD,M1) 0 b0 = -0.01161616 cost_B0 = 0.11616162 B1 = -0.04057239 cost_B1 = 0.40572391\nFD 0 07:19:08.906 gradient-descent test (EURUSD,M1) 1 b0 = -0.02060337 cost_B0 = 0.08987211 B1 = -0.07436893 cost_B1 = 0.33796541\nKE 0 07:19:08.906 gradient-descent test (EURUSD,M1) 2 b0 = -0.02743120 cost_B0 = 0.06827832 B1 = -0.10259883 cost_B1 = 0.28229898\nQE 0 07:19:08.906 gradient-descent test (EURUSD,M1) 3 b0 = -0.03248925 cost_B0 = 0.05058047 B1 = -0.12626640 cost_B1 = 0.23667566\nEE 0 07:19:08.907 gradient-descent test (EURUSD,M1) 4 b0 = -0.03609603 cost_B0 = 0.03606775 B1 = -0.14619252 cost_B1 = 0.19926123\nCF 0 07:19:08.907 gradient-descent test (EURUSD,M1) 5 b0 = -0.03851035 cost_B0 = 0.02414322 B1 = -0.16304363 cost_B1 = 0.16851108\nMF 0 07:19:08.907 gradient-descent test (EURUSD,M1) 6 b0 = -0.03994229 cost_B0 = 0.01431946 B1 = -0.17735996 cost_B1 = 0.14316329\nJG 0 07:19:08.907 gradient-descent test (EURUSD,M1) 7 b0 = -0.04056266 cost_B0 = 0.00620364 B1 = -0.18958010 cost_B1 = 0.12220146\nHE 0 07:19:08.907 gradient-descent test (EURUSD,M1) 8 b0 = -0.04051073 cost_B0 = -0.00051932 B1 = -0.20006123 cost_B1 = 0.10481129\nME 0 07:19:08.907 gradient-descent test (EURUSD,M1) 9 b0 = -0.03990051 cost_B0 = -0.00610216 B1 = -0.20909530 cost_B1 = 0.09034065\nJQ 0 07:19:08.907 gradient-descent test (EURUSD,M1) 10 b0 = -0.03882570 cost_B0 = -0.01074812 B1 = -0.21692190 cost_B1 = 0.07826600\n<<<<<< Last 10 iterations >>>>>>\n\nFN 0 07:19:09.725 gradient-descent test (EURUSD,M1) 6935 b0 = 1.44678930 cost_B0 = -0.00000001 B1 = -0.85010666 cost_B1 = 0.00000000\nPN 0 07:19:09.725 gradient-descent test (EURUSD,M1) 6936 b0 = 1.44678931 cost_B0 = -0.00000001 B1 = -0.85010666 cost_B1 = 0.00000000\nNM 0 07:19:09.726 gradient-descent test (EURUSD,M1) 6937 b0 = 1.44678931 cost_B0 = -0.00000001 B1 = -0.85010666 cost_B1 = 0.00000000\nKL 0 07:19:09.726 gradient-descent test (EURUSD,M1) 6938 b0 = 1.44678931 cost_B0 = -0.00000001 B1 = -0.85010666 cost_B1 = 0.00000000\nPK 0 07:19:09.726 gradient-descent test (EURUSD,M1) 6939 b0 = 1.44678931 cost_B0 = -0.00000001 B1 = -0.85010666 cost_B1 = 0.00000000\nRK 0 07:19:09.726 gradient-descent test (EURUSD,M1) 6940 b0 = 1.44678931 cost_B0 = -0.00000001 B1 = -0.85010666 cost_B1 = 0.00000000\nMJ 0 07:19:09.726 gradient-descent test (EURUSD,M1) 6941 b0 = 1.44678931 cost_B0 = -0.00000001 B1 = -0.85010666 cost_B1 = 0.00000000\nHI 0 07:19:09.726 gradient-descent test (EURUSD,M1) 6942 b0 = 1.44678931 cost_B0 = -0.00000001 B1 = -0.85010666 cost_B1 = 0.00000000\nCH 0 07:19:09.726 gradient-descent test (EURUSD,M1) 6943 b0 = 1.44678931 cost_B0 = -0.00000001 B1 = -0.85010666 cost_B1 = 0.00000000\nMH 0 07:19:09.727 gradient-descent test (EURUSD,M1) 6944 b0 = 1.44678931 cost_B0 = -0.00000001 B1 = -0.85010666 cost_B1 = 0.00000000\nQG 0 07:19:09.727 gradient-descent test (EURUSD,M1) 6945 b0 = 1.44678931 cost_B0 = -0.00000000 B1 = -0.85010666 cost_B1 = 0.00000000\nNG 0 07:19:09.727 gradient-descent test (EURUSD,M1) 6945 Iterations Local Minima are\nMJ 0 07:19:09.727 gradient-descent test (EURUSD,M1) B0(Intercept) = 1.44679 || B1(Coefficient) = -0.85011\n```\n\nWe don't normalize or scale classified data for logistic regression as we did in linear regression.\n\nThere you have it the gradient descent for the two and most important machine learning models, I hope it was easy to understand and helpful python code used in this article and the dataset is linked to this GitHub repo.\n\n### Conclusion\n\nWe have seen the gradient descent for one independent and one dependent variable, for multiple independent variables you need to use the vector/ matrices form of equations, I think this time it will become easy for anyone to try and find out themselves now that we have the Library for matrices released recently by MQL5, for any help on the matrices feel free to reach me out, I will be more than happy to help.\n\nBest regards\n\nAttached files |\n\n#### Other articles by this author",
null,
"Developing a trading Expert Advisor from scratch (Part 17): Accessing data on the web (III)\nIn this article we continue considering how to obtain data from the web and to use it in an Expert Advisor. This time we will proceed to developing an alternative system.",
null,
"Automated grid trading using limit orders on Moscow Exchange (MOEX)\nThe article considers the development of an MQL5 Expert Advisor (EA) for MetaTrader 5 aimed at working on MOEX. The EA is to follow a grid strategy while trading on MOEX using MetaTrader 5 terminal. The EA involves closing positions by stop loss and take profit, as well as removing pending orders in case of certain market conditions.",
null,
"Learn how to design a trading system by Chaikin Oscillator\nWelcome to our new article from our series about learning how to design a trading system by the most popular technical indicator. Through this new article, we will learn how to design a trading system by the Chaikin Oscillator indicator.",
null,
"Neural networks made easy (Part 16): Practical use of clustering\nIn the previous article, we have created a class for data clustering. In this article, I want to share variants of the possible application of obtained results in solving practical trading tasks."
]
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null,
"data:image/jpeg;base64,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",
null,
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"https://c.mql5.com/2/47/y_t_xc5_equation.png",
null,
"https://c.mql5.com/2/47/y_h_x95_squared_imageb.png",
null,
"https://c.mql5.com/2/48/chain_rule_to_y_x_gx35m3_2.png",
null,
"https://c.mql5.com/2/47/updating_the_values_in_gradient_descent.png",
null,
"https://c.mql5.com/2/47/making_sense_gif.gif",
null,
"https://c.mql5.com/2/48/dataset.jpg",
null,
"https://c.mql5.com/2/48/Years_of_experience_Vs_Salary.png",
null,
"https://c.mql5.com/2/48/random.png",
null,
"https://c.mql5.com/2/48/y_squared.png",
null,
"https://c.mql5.com/2/48/summation_cost.png",
null,
"https://c.mql5.com/2/48/finally_cost_function.png",
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"https://c.mql5.com/2/48/Best_model_gradient_descent.png",
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"https://c.mql5.com/2/48/math_limit.jpg",
null,
"https://c.mql5.com/2/48/normalization.png",
null,
"https://c.mql5.com/2/48/Normalize_price_graphs.png",
null,
"https://c.mql5.com/2/48/logloss_function.png",
null,
"https://c.mql5.com/2/48/logloss_derivatives.png",
null,
"https://c.mql5.com/2/48/Class_vs_survival_titanic_dataset.png",
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"https://c.mql5.com/2/48/Survived_vs_class.png",
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"https://c.mql5.com/2/47/development.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.74021804,"math_prob":0.9744778,"size":30098,"snap":"2023-40-2023-50","text_gpt3_token_len":9770,"char_repetition_ratio":0.2508141,"word_repetition_ratio":0.09790979,"special_character_ratio":0.4270051,"punctuation_ratio":0.17462073,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9991906,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58],"im_url_duplicate_count":[null,null,null,1,null,null,null,null,null,null,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,4,null,4,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-22T22:16:07Z\",\"WARC-Record-ID\":\"<urn:uuid:cd2fa7f8-1a95-4279-87b8-bc4aedbe1583>\",\"Content-Length\":\"147431\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:942bfc64-28ba-4a69-a814-4f48453f8fcd>\",\"WARC-Concurrent-To\":\"<urn:uuid:a678ccc6-ba2b-405c-8938-3226caa9f01e>\",\"WARC-IP-Address\":\"142.215.208.239\",\"WARC-Target-URI\":\"https://www.mql5.com/en/articles/11200\",\"WARC-Payload-Digest\":\"sha1:G4IQ6QYC5DQDS6WGKLTSTXF2A2UDLYBZ\",\"WARC-Block-Digest\":\"sha1:Y6USTTXFPDJSMEQ2B3U72E5C2FB4CE7C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506423.70_warc_CC-MAIN-20230922202444-20230922232444-00522.warc.gz\"}"} |
http://lisperator.net/pltut/parser/the-parser | [
"The parser\n\nThe parser creates AST nodes that are described in the AST section.\n\nThanks to the work we did in the tokenizer, the parser operates on a stream of tokens instead of dealing with individual characters. It still defines many helpers to keep complexity down. I'll discuss here the main functions that comprise the parser. Let's start with a high level one, the lambda parser:\n\nfunction parse_lambda() {\nreturn {\ntype: \"lambda\",\nvars: delimited(\"(\", \")\", \",\", parse_varname),\nbody: parse_expression()\n};\n}\n\nThis function will be invoked when the lambda keyword has already been seen and “eaten” from the input, so all it cares for is to parse the argument names; but they're in parentheses and delimited by commas. Rather than placing that code in parse_lambda, I preferred to write a delimited function that takes these arguments: the start token, the end token, the separator, and a function which parses whatever must be between those start/end tokens. In this case, it's parse_varname, which takes care to throw an error if it encounters anything which doesn't look like a variable. The body of the function is an expression, so we get it with parse_expression.\n\ndelimited is a bit lower-level:\n\nfunction delimited(start, stop, separator, parser) {\nvar a = [], first = true;\nskip_punc(start);\nwhile (!input.eof()) {\nif (is_punc(stop)) break;\nif (first) first = false; else skip_punc(separator);\nif (is_punc(stop)) break; // the last separator can be missing\na.push(parser());\n}\nskip_punc(stop);\nreturn a;\n}\n\nAs you can see, it uses more utilities: is_punc and skip_punc. The former will return true if the current token is the given punctuation sign (without “eating” it), while skip_punc will ensure that the current token is that punctuation (throws an error otherwise) and will discard it from the input.\n\nThe function that parses the whole program is probably the simplest:\n\nfunction parse_toplevel() {\nvar prog = [];\nwhile (!input.eof()) {\nprog.push(parse_expression());\nif (!input.eof()) skip_punc(\";\");\n}\nreturn { type: \"prog\", prog: prog };\n}\n\nSince we have no statements, we simply call parse_expression() and read expressions until we get to the end of the input. Using skip_punc(\";\") we demand semicolons between these expressions.\n\nAnother simple example: parse_if():\n\nfunction parse_if() {\nskip_kw(\"if\");\nvar cond = parse_expression();\nif (!is_punc(\"{\")) skip_kw(\"then\");\nvar then = parse_expression();\nvar ret = { type: \"if\", cond: cond, then: then };\nif (is_kw(\"else\")) {\ninput.next();\nret.else = parse_expression();\n}\nreturn ret;\n}\n\nIt skips over the if keyword with skip_kw (and this throws an error if the current token is not the given keyword), reads the condition using parse_expression(). Next, if the consequent branch doesn't start with a { we require the keyword then to be present (I feel like the syntax is too scarce without it). The branches are just expressions, so again we use parse_expression() for them. The else branch is optional so we need to check if the keyword is present before parsing it.\n\nHaving many small utilities helps a lot in keeping the code simple. We almost write the parser like we had a high level language dedicated for parsing. All these functions are “mutually recursive”, e.g.: there's a parse_atom() function which is the main dispatcher — based on the current token it calls other functions. One of them is parse_if() (called when the current token is if) and that in turn calls parse_expression(). But parse_expression() calls parse_atom(). The reason why there's no infinite loop is that at each step, one function or another will advance at least one token.\n\nThis kind of parser is called a recursive descent parser and it's probably the easiest kind to write manually.\n\nLower level: parse_atom() and parse_expression()\n\nparse_atom() does the main dispatching job, depending on the current token:\n\nfunction parse_atom() {\nreturn maybe_call(function(){\nif (is_punc(\"(\")) {\ninput.next();\nvar exp = parse_expression();\nskip_punc(\")\");\nreturn exp;\n}\nif (is_punc(\"{\")) return parse_prog();\nif (is_kw(\"if\")) return parse_if();\nif (is_kw(\"true\") || is_kw(\"false\")) return parse_bool();\nif (is_kw(\"lambda\") || is_kw(\"λ\")) {\ninput.next();\nreturn parse_lambda();\n}\nvar tok = input.next();\nif (tok.type == \"var\" || tok.type == \"num\" || tok.type == \"str\")\nunexpected();\n});\n}\n\nIf it sees an open paren, then it must be a parenthesized expression — thus, skip over paren, call parse_expression() and expect a closing paren. If it sees some keyword, it calls the appropriate parser function. If it sees a constant or an identifier, it's just returned as is. And if nothing works, unexpected() will throw an error.\n\nWhen an atomic expression is expected and it sees {, it calls parse_prog to parse a sequence of expressions. That's defined below. It will do some minor optimization at this point — if the prog is empty, then it just returns FALSE. If it has a single expression, it is returned instead of a \"prog\" node. Otherwise it returns a \"prog\" node containing the expressions.\n\n// we're going to use the FALSE node in various places,\n// so I'm making it a global.\nvar FALSE = { type: \"bool\", value: false };\n\nfunction parse_prog() {\nvar prog = delimited(\"{\", \"}\", \";\", parse_expression);\nif (prog.length == 0) return FALSE;\nif (prog.length == 1) return prog;\nreturn { type: \"prog\", prog: prog };\n}\n\nHere's the parse_expression() function. Contrary to parse_atom(), this one will extend an expression as much as possible to the right using maybe_binary(), which is explained below.\n\nfunction parse_expression() {\nreturn maybe_call(function(){\nreturn maybe_binary(parse_atom(), 0);\n});\n}\n\nThe maybe_* functions\n\nThese functions check what follows after an expression in order to decide whether to wrap that expression in another node, or just return it as is.\n\nmaybe_call() is very simple. It receives a function that is expected to parse the current expression. If after that expression it sees a ( punctuation token, then it must be a \"call\" node, which is what parse_call() makes (included below). Notice again how delimited() comes in handy for reading the argument list.\n\nfunction maybe_call(expr) {\nexpr = expr();\nreturn is_punc(\"(\") ? parse_call(expr) : expr;\n}\n\nfunction parse_call(func) {\nreturn {\ntype: \"call\",\nfunc: func,\nargs: delimited(\"(\", \")\", \",\", parse_expression)\n};\n}\n\nOperator precedence\n\nmaybe_binary(left, my_prec) is used to compose binary expressions like 1 + 2 * 3. The trick to parse them properly is to correctly define the operator precedence, so we'll start with that:\n\nvar PRECEDENCE = {\n\"=\": 1,\n\"||\": 2,\n\"&&\": 3,\n\"<\": 7, \">\": 7, \"<=\": 7, \">=\": 7, \"==\": 7, \"!=\": 7,\n\"+\": 10, \"-\": 10,\n\"*\": 20, \"/\": 20, \"%\": 20,\n};\n\nThis says that * binds tighter than +, so an expression like 1 + 2 * 3 must be read as (1 + (2 * 3)) instead of ((1 + 2) * 3), which would be the normal left-to-right order in which the parser operates.\n\nThe trick is to read an atomic expression (only the 1) and pass it to maybe_binary() (the left argument), along with the current precedence (the my_prec). maybe_binary will look at what follows. If it doesn't see an operator, or if it has a smaller priority, then left is returned as is.\n\nIf it's an operator that has a higher precedence than ours, then it wraps left in a new \"binary\" node, and for the right side it repeats the trick at the new precedence level (*):\n\nfunction maybe_binary(left, my_prec) {\nvar tok = is_op();\nif (tok) {\nvar his_prec = PRECEDENCE[tok.value];\nif (his_prec > my_prec) {\ninput.next();\nvar right = maybe_binary(parse_atom(), his_prec) // (*);\nvar binary = {\ntype : tok.value == \"=\" ? \"assign\" : \"binary\",\noperator : tok.value,\nleft : left,\nright : right\n};\nreturn maybe_binary(binary, my_prec);\n}\n}\nreturn left;\n}\n\nNote that before returning the binary expression we must also call maybe_binary at the old precedence level (my_prec), in order to wrap the expression in another one, should an operator with a higher precedence follow. If all this is confusing, read the code again and again (perhaps try to execute it mentally on some input expressions) until you get it.\n\nFinally, since my_prec is initially zero, any operator will trigger the building of a \"binary\" node (or \"assign\" when the operator is =).\n\nThere are a few more functions in the parser, so I'm including the whole parse function below. Click “Show code” to display it (about 150 lines).\n\nCredits\n\nThe moment I understood how to write a non-trivial parser occurred while studying Marijn Haverbeke's parse-js library (Common Lisp). The parser above, although for a much simpler language, is modeled after his code."
]
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https://math.stackexchange.com/questions/333213/why-does-the-power-rule-work | [
"# Why does the power rule work?\n\nIf $$f(x)=x^u$$ then the derivative function will always be $$f'(x)=u*x^{u-1}$$ I've been trying to figure out why that makes sense and I can't quite get there.\n\nI know it can be proven with limits, but I'm looking for something more basic, something I can picture in my head.\n\nThe derivative should be the slope of the function. If $$f(x)=x^3=x^2*x$$ then the slope should be $x^2$. But it isn't. The power rule says it's $3x^2$.\n\nI understand that it has to do with having variables where in a more simple equation there would be a constant. I'm trying to understand how that exactly translates into the power rule.\n\n• What if you write $x^3 = x*x*x$ and use the product rule. What do you get? What do you get with the product rule for $x^3 = x^2*x$? Do they all agree? – Amzoti Mar 17 '13 at 22:34\n• As per Amzoti's suggestion, if you want intuition think of the case where $u$ is an integer and use the product rule. So in the case of $x^3$, use the product rule and take the derivative of $x\\cdot x\\cdot x$. This is symmetric, so you will get $3\\cdot x^2$. – Lepidopterist Mar 17 '13 at 22:37\n• Since the definition of $f'(x)$ involves limits, you can't hope for \"something more basic\" that proves things about derivatives. – Henning Makholm Mar 17 '13 at 22:39\n• \"If $f(x)=x^3=x^2\\cdot x$ then the slope should be $x^2$\" - Why? (You are perhaps confusing it with rules that work to find slopes of lines of the form $y=ax+b$; you simply take the $a$ in front of the $x$ then.) – anon Mar 17 '13 at 22:52\n• @Amzoti, so that is just restating the question as.... why does the product rule work? – Pacerier Jan 6 '14 at 13:44\n\n## 4 Answers\n\nFirst let's try to understand why the derivative of the function $f$ given by $f(x) = x^2$ is equal to $2x$ and not to $x$. (The product rule and the power rule are both generalizations of this.)\n\nImagine that you have a square whose sides have length $x$. Now imagine what happens to its area if we increase the length of each side by a small amount $\\Delta x$. We can do this by adding three regions to the picture: two thin rectangles measuring $x$ by $\\Delta x$ (say one on the right of the square and another on the top) and one small square measuring $\\Delta x$ by $\\Delta x$ (say added in the top right corner.) So the change in the area $x^2$ is equal to $2x \\cdot \\Delta x + (\\Delta x)^2$. If we divide this by $\\Delta x$ and take the limit as $\\Delta x$ approaches zero, we get $2x$.\n\nSo geometrically what is happening is that the small square in the corner is too small to matter, but you have to count both rectangles. If you only count one of them, you will get the answer $x$; however, this only tells you what happens when you lengthen, say, the horizontal sides and not the vertical sides of your square to get a rectangle. This is a different problem than the one under consideration, which asks (after we put it in geometrical terms) how the area varies as we lengthen all the sides.\n\n• That's a really cool way to explain it. Like the OP, I too used to understand the math of the limits, but couldn't picture it. My high school calculus teacher explained it the same way as @Trevor, and it really helped me get my head around the concept visually. – David John Welsh Mar 18 '13 at 3:21\n• Hm, but this doesn't explain why power rule is valid in, say $y=\\sqrt x$, or $x^\\pi$. – Simply Beautiful Art Dec 17 '16 at 21:59\n\nIf you know the product rule, you can derive this for when $u$ is a positive integer, which should give you a basic intuitive understanding.\n\nSuppose, as an example, that $f(x) = x^2$. Equivalently, $f(x) = x*x$. By using the product rule, we have $$f(x) = (1)(x) + (x)(1)$$ $$= 2x$$\n\nMore generally, suppose that $f(x) = x^u$. Suppose for now that $u$ is a positive integer, which allows us to expand like this: $$f(x) = \\underbrace{(x)(x)...(x)(x)}_{u\\text{ terms}}$$ Using the product rule again, we can say $$f(x) = \\underbrace{\\underbrace{(1)(x)...(x)(x)}_{u \\text{ terms}} + (x)(1)...(x)(x) + ... + (x)(x)...(1)(x) + (x)(x)...(x)(1)}_{u\\text{ terms}}$$ which simplifies to $$f(x)=ux^{u-1}$$\n\nFor positive integers $n$, we can use the Binomial Theorem. Let $f(x)=x^n$. We want to find the slope of the tangent line to $y=f(x)$ at $x=a$. So take a very small $h$, and calculate $$\\frac{(a+h)^n-a^n}{h}.$$ By the Binomial Theorem, $(a+h)^n=a^n+na^{n-1}h +\\binom{n}{2}a^{n-2}h^2+\\cdots$.\n\nSince $h$ is tiny, $h^2$, $h^3$, and so on are negligible compared to $h$. Thus $$\\frac{(a+h)^n-a^n}{h}\\approx na^{n-1}.$$\n\n• Ok, this explanation is way too mathy – Pacerier Jan 6 '14 at 13:58\n• True, it is rather symbol-laden. – André Nicolas Jan 6 '14 at 15:55\n• I think it works. However, you should have used the limit as $x$ approaches $0$. @Pacerier have you dealt with binomial theorem? It is rather a beautiful thing. The trick to understanding this explanation lies in ignoring the $3rd,4th,5th,...$ term because when you set $h=0$, they all cancel. The $1st$ term is canceled by the $-f(x)$ in the derivative calculation and the second term's $h$ is removed by the $h$ in the denominator of the derivative quotient. Thus giving $na^{n-1}$. – Simply Beautiful Art Dec 21 '15 at 14:35\n\nThis may be too advanced for you right now, but knowing about the derivative of the log function can be very helpful.\n\nThe basic idea is that $(\\ln(x))' = 1/x$, where $\\ln$ is the natural log.\n\nApplying the chain rule, $(\\ln(f(x))' = f'(x)/f(x)$.\n\nFor this case, set $f(x) = x^n$. Then $\\ln(f(x)) = \\ln(x^n) = n \\ln(x)$. Taking derivatives on both sides, $(\\ln(x^n))' = f'(x)/f(x) = f'(x)/x^n$ and $(\\ln(x^n))' = (n \\ln(x))' = n/x$, so $f'(x)/x^n = n/x$ or $f'(x) = n x^{n-1}$.\n\nMore generally, if $f(x) = \\prod a_i(x)/\\prod b_i(x)$, then $\\ln(f(x)) = \\sum \\ln(a_i(x))-\\sum \\ln(b_i(x))$ so\n\n\\begin{align} f'(x)/f(x) &= (\\ln(f(x))'\\\\ &= \\sum (\\ln(a_i(x))'-\\sum (\\ln(b_i(x))'\\\\ &=\\sum (a_i(x))'/a_i(x)-\\sum (b_i(x))'/b_i(x)\\\\ \\end{align},\n\nso $f'(x) = f(x)\\left(\\sum (a_i(x))'/a_i(x)-\\sum (b_i(x))'/b_i(x)\\right)$.\n\nNote that this technique, called logarithmic differentiation, generalizes both the product and quotient rules for derivatives."
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https://www.antionline.com/showthread.php?254560-Adding-numbers-in-strings-in-C | [
"1. ##",
null,
"Adding numbers in strings in C++\n\nOk, I have been working on a project lately that required me to write a function that added together strings of numbers. Such as \"123\" + \"321\" = \"444\". It all needs to stay as a string. I wrote the function, and it works up until a certain point and I can't figure out what is wrong with it. The first while loop works perfectly, but then the second screws up and it's almost exactly the same as the first. Here is the code and the functions it itself uses. Why does it not work? or if you have a better function written, what is it?\nCode:\n```#include <string.h>\n#include <iostream.h>\n\nvoid cstradd(char *fnum, char *snum, char *lnum);\nvoid cstrrev(char *str);\n\nint main()\n{\nchar a = \"2242442245\"; //\"12345678910\";\nchar b = \"242\"; //\"1112131415\";\nchar c[] = \"\";\ncout << a << \" + \" << b << \" = \" << c << endl;\nreturn 0;\n}\n\nvoid cstradd(char fnum[], char snum[], char lnum[])\n{\ncstrrev(fnum);\ncstrrev(snum);\nint count = 0;\nchar carry = 0;\nwhile(fnum[count] != 0 && snum[count] != 0)\n{\nchar temp = \"0\";\ntemp = (fnum[count] - 48) + (snum[count] - 48) + carry;\ncarry = 0;\nif(temp > 9)\n{\ntemp = temp + 38;\ncarry = 1;\n}\nelse\n{\ntemp = temp + 48;\n}\nstrcat(lnum, temp);\ntemp = 0;\ncount++;\n}\n/* This is the while loop that fails, some how it is editing snum and just totally screws up, it's just like the first while loop but with only fnum, I don't know what is screwing up */\nwhile(fnum[count] != 0)\n{\nchar temp = \"0\";\ntemp = (fnum[count] - 48) + carry;\ncarry = 0;\nif(temp > 9)\n{\ntemp = temp + 38;\ncarry = 1;\n}\nelse\n{\ntemp = temp + 48;\n}\nstrcat(lnum, temp);\ntemp = 0;\ncount++;\n}\nwhile(snum[count] != 0)\n{\nchar temp = \"0\";\ntemp = (snum[count] - 48) + carry;\ncarry = 0;\nif(temp > 9)\n{\ntemp = temp + 38;\ncarry = 1;\n}\nelse\n{\ntemp = temp + 48;\n}\nstrcat(lnum, temp);\ntemp = 0;\ncount++;\n}\nif(carry == 1)\n{\nchar temp = \"1\";\nstrcat(lnum, temp);\n}\ncstrrev(lnum);\ncstrrev(fnum);\ncstrrev(snum);\n}\n\nvoid cstrrev(char str[]) // This works for reversing a string\n{\nint l=strlen(str)-1;\nfor(int x=0;x<l;x++,l--)\n{\nstr[x]^=str[l]; //triple XOR Trick\nstr[l]^=str[x]; //for not using a temp\nstr[x]^=str[l];\n}\n}```",
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"Reply With Quote\n\n2. alright... i've noticed two problems with your code... but neither one solves the bigger problem\n\nYou need to add a count=0; statement before each while loop. Otherwise the second and third loops will probably never be used.\n\nThe big problem comes from the strcat(str1, str2) statment. I've run some different numbers and each time snum is deleted at different times. *shrug* But i can say WITHOUT A DOUBT that the problem comes from the strcat(str1, str2) statement.\n\nHere's what i got from my msdn help system:\nStrCat\n\nAppends one string to another.\n\nLPTSTR StrCat(\nLPTSTR psz1,\nLPCTSTR psz2,\n);\n\nParameters\npsz1\n[in/out] Address of a null-terminated string to be appended to. It must be large enough to hold both strings.\npsz2\n[in] Address of the string to be appended to psz1.\nReturn Values\nReturns a pointer to psz1, which holds the combined strings.\n\ni'll keep working on this and when (if?) i can fix the problem i'll post my findings",
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"",
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"Reply With Quote\n\n3. Use stringstreams. They'll make the solution almost trivial.",
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"",
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"Reply With Quote\n\n4. ## ok\n\nOk, I looked at that one strcat in there and put some output around it and tested the values, your right. But what I don't get, is that it is doing the same way as in the first while loop. It's weird. As for string streams, I actually would rather find out why mine is failing, so I know for future reference.",
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"Reply With Quote\n\n5. ok, i fixed the program... how? by removing strcat of course!\n\nin place of strcat in the first loop, use this statement: lnum[count]=temp;\nstrcat is meant for much longer strings.. but this is just an add on.\n\nSecond... you've already done the addition, so the second loop is mostly useless. i just commented that whole thing out... Then i commented the Third loop out...\n\nSo, at this point you have the CHANGED numbers stored in lnum. All that's left is to fill in any missing values. Remember, if both fnum and snum had values, then lnum would have changed at that location.\n\nFinally, you need to keep the final carry==1 loop to fill in the top location (if needed)\n\nHere's how i have the final cstradd sub program\n\nvoid cstradd(char fnum[], char snum[], char lnum[])\n{\ncstrrev(fnum);\ncstrrev(snum);\nint count = 0;\nchar carry = 0;\nwhile(fnum[count] != 0 && snum[count] != 0)\n{\nchar temp = \"0\";\ntemp = (fnum[count] - 48) + (snum[count] - 48) + carry;\ncarry = 0;\nif(temp > 9)\n{\ntemp = temp + 38;\ncarry = 1;\n}\nelse\n{\ntemp = temp + 48;\n}\n//cout<<snum<<\"before\\n\";\nlnum [count] = temp;\n//cout<<snum<<\"after\\n\";\ntemp = 0;\ncount++;\n}\n//cout<<fnum<<\",\"<<snum<<\",\"<<lnum<<\", first loop out\\n\";\n\ncount = 0;\nwhile(fnum[count] != 0)\n{\nif (lnum[count] == 0)\n{\n//cout<<fnum[count]<<\",\"<<lnum<<\"\\n\";\nlnum[count]=fnum[count];\n}\ncount++;\n}\n\ncount = 0;\nwhile(snum[count] != 0)\n{\nif(lnum[count]==0)\n{\nlnum[count]=snum[count];\n}\ncount++;\n}\n\nif(carry == 1)\n{\nchar temp = \"1\";\nstrcat(lnum, temp);\n}\ncstrrev(lnum);\ncstrrev(fnum);\ncstrrev(snum);\n}\n\nNow, i've only tested this with a few numbers... you should test it with more numbers (don't forget about negatives)\n\nAlso, you could re-write the whole thing and use the VAL(str) function... at least i think that's the right one... i've not used it in a while.\n\nENJOY",
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