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https://mathoverflow.net/questions/23004 | 12 | Let R be a ring. I'm trying to understand when the categories of finitely presented R-modules and finitely generated R-modules can *fail* to be abelian categories.
Poking around on the internet has lead me to believe that these categories will agree and be abelian if R is a (left?) Noetherian ring, but I'm interested in the more general case. I've found warnings that these categories can fail to be abelian but haven't found much discussion of what goes wrong. I feel like there should be some really good illuminating examples.
I think it's not too hard to show that the cokernel of a map of finitely presented modules is again finitely presented. So one of the things I'm really looking for is an example of a map of between finitely presented modules in which the kernel is *not* finitely presented. It would be even better if the kernel was not finitely generated. Is this possible?
| https://mathoverflow.net/users/184 | Failure of Fin. Presented and Fin. Generated Modules to be Abelian Categories? | [Background to the following can be found in Lam's *Lectures on Modules and Rings*, section 4.G.]
**Definition:** A finitely generated (f.g.) right module $M\_R$ is *coherent* if every f.g. submodule of $M$ is finitely presented (f.p.).
Now let $M\_R$ be a finitely presented module that is not coherent. There exists some submodule $N\_R\subseteq M$ that is f.g. but not f.p. Pick a surjection $R^n\twoheadrightarrow N$, and consider the composition $f\colon R^n \twoheadrightarrow N\hookrightarrow M$. Because $N\cong R^n/\ker(f)$ is not f.p., it must be the case that $\ker(f)$ is not f.g.
So all we need is an explicit example of a f.p. module that is not coherent. For this, see Jack Schmidt's answer to this question. Using my approach in his case, instead of looking at a surjection $R\twoheadrightarrow R/I$, we would look at the composition $R^n\twoheadrightarrow I\hookrightarrow R$.
| 5 | https://mathoverflow.net/users/778 | 23025 | 15,166 |
https://mathoverflow.net/questions/23026 | 7 | Recall the notion of **groupoid** ([Wikipedia](http://en.wikipedia.org/wiki/Groupoid), [nLab](http://ncatlab.org/nlab/show/groupoid)). An important construction of groupoids is as "action groupoids" for group actions. Namely, let $X$ be a groupoid and $G$ a group, and suppose that $G$ acts on $X$ by groupoid automorphisms. Then we can form a new groupoid $X//G$, which has as objects the objects of $X$, but the morphisms include, in addition to the original morphisms of $X$, a morphism $x \overset g \to gx$ for each $g\in G$ and $x\in X$. The composition of morphisms is well-defined if the action is by groupoid automorphisms. (When $X$ is a set, then $X//G$ is equivalent to the skeletal groupoid whose objects are the elements of the "coarse" quotient $X/G$, and with ${\rm Aut}(\bar x) = {\rm Stab}\_G(x)$.)
(Probably there is a fancier construction, in which the conditions on the word "group action" be relaxed to an "action" up to specified natural isomorphism, and then $G$ could act on $X$ by autoequivalences, rather than autoisomorphisms, but this generalization won't concern me.)
Let $1$ denote the one-point set, thought of as a groupoid with only identity morphisms. Then any group $G$ acts uniquely on $1$, and so we have the groupoid $1//G$. In general, although $X\times 1 \cong X$, we do not have $X \times (1//G) \cong X//G$ for arbitrary $G$-actions on $X$ unless the action is trivial. (Here $\times$ denotes the **groupoid product**, which is just what you think it is.) However, the construction provides natural bijections between the objects of $X//G$ and the objects of $X \times (1//G)$, and between the morphisms of $X//G$ and the morphisms of $X \times (1//G)$.
>
> **Question:** Is there some sort of "semidirect" or "crossed" product of groupoids, which presumably depends on extra data, so that we do have $X//G \cong X \rtimes (1//G)$? By which I mean, what is the correct notion of "action" of a groupoid $Y$ on a groupoid $X$ and what is the corresponding correct notion of $X \rtimes Y$?
>
>
>
I see that the page [semidirect product in nLab](http://ncatlab.org/nlab/show/semidirect+product) defines $X \rtimes G$ as something closely related to $X//G$. But clearly this ought to be called $X\rtimes (1//G)$, but then I do not know what the right definition for $X\rtimes Y$ is, hence the question. And really I'd like to know about a "double crossed product" $X\bowtie Y$.
My motivation for this question is from my answer to [Do rational numbers admit a categorification which respects the following “duality”?](https://mathoverflow.net/questions/22860/do-rational-numbers-admit-a-categorification-which-respects-the-following-dualit/22879#22879).
| https://mathoverflow.net/users/78 | How can I understand the "groupoid" quotient of a group action as some sort of "product"? | Let $X$ and $Y$ be groupoids. An action of $Y$ on $X$ is a functor $\rho: Y \to B\operatorname{Aut}(X)$, where $B\operatorname{Aut}(X)$ is the one-object 2-groupoid such that $\operatorname{Hom}(\ast, \ast)$ is the 2-group of autoequivalences of $X$.
We define $X \rtimes Y$ as follows. Its objects are simply $\operatorname{Ob}(X) \times \operatorname{Ob}(Y)$. An element of $(X \rtimes Y)((x\_1, y\_1), (x\_2, y\_2))$ consists of a pair $(f, g)$, where $g \in Y(y\_1, y\_2)$, and $f \in X(x\_1, \rho(g)x\_2)$. Given $(f, g) \in (X \rtimes Y)((x\_1, y\_1), (x\_2, y\_2))$, and $(f', g') \in (X \rtimes Y)((x\_2, y\_2), (x\_3, y\_3))$, we define $(f', g') \circ (f, g) \in \operatorname{Hom}((x\_1, y\_1), (x\_3, y\_3))$ as $(\rho(g)f' \circ f, g' \circ g)$. It is straightforward to check that in the case that $Y$ is $1 // G$, $X \rtimes Y \cong X // G$.
| 11 | https://mathoverflow.net/users/396 | 23040 | 15,175 |
https://mathoverflow.net/questions/23029 | -3 | It is easy to show that similarity in matrices is an equivalence relation (two matrices A and B of same size being similar if there exists a matrix P such that B = PAP^(-1) )
Moreover, given a matrix, its equivalence class can be finite. E.g. The equivalence of nxn matrices containing the identity matrix I is singleton (i.e. it contains only the identity matrix itself).
But I do not know how many equivalence classes there are for matrices of a given size.
Thanks in advance for any comment.
| https://mathoverflow.net/users/5627 | Are there infinitely many equivalence classes of similar matrices? | [This is an easy question, but it doesn't feel like a homework question, so I will answer it. I have made the post community wiki to protect myself from unwanted votes, both upwards and downwards.]
For a positive integer $n$, consider the ring $M\_n(k)$ of $n \times n$ matrices with $k$-coefficients for $n \geq 1$.
If $k$ is finite, then $M\_n(k)$ is finite, so obviously there are only finitely many similarity classes.
If $k$ is infinite, then since the determinant map $M\_n(k) \rightarrow k$ is surjective and the determinant is a similarity invariant, there are infinitely many similarity classes.
One may ask a more precise question: what is the cardinality $S(n,k)$ of the set of
similarity classes of $n \times n$ matrices with coefficients in $k$?
When $k$ is infinite, it follows easily from the above that $S(n,k) = \# k$.
On the other hand, when $k \cong \mathbb{F}\_q$ is finite, it is a nice linear algebra exercise to give an explicit formula for $S(n,k)$ in terms of $q$ and $n$. It might (or might not) be appropriate to discuss how to derive such a formula here.
| 10 | https://mathoverflow.net/users/1149 | 23042 | 15,177 |
https://mathoverflow.net/questions/22990 | 22 | This question arose after reading the answers (and the comments to the answers) to [Why worry about the axiom of choice?](https://mathoverflow.net/questions/22927/why-worry-about-the-axiom-of-choice).
First things first. In *my intuitive conception of the hierarchy of sets*, the axiom of choice is obviously true. I mean, how can the product of a family of non-empty sets fail to be non-empty? I simply cannot fathom it. Now, I understand that there are people who disagree with me; a mathematician of a (more) constructive persuasion would reply that mathematical existence is constructive existence. Well, we can agree to disagree. And besides, the distinction between constructive and non-constructive proofs is very much worth having in mind. First, because constructive proofs usually give more information and second, there are many contexts where AC is not available (e.g. topoi).
A second (personal) reason for championing AC is a pragmatic one: it allows us to prove many things. And "many things" include things that physicists use without a blink. Analysis can hardly get off the ground without some form of choice. Countable choice (ACC) or dependent countable choice (ACDC) is enough for most elementary analysis and many constructivists have no problem with ACC or ACDC. For example, ACC and the stronger ACDC are enough to prove that the countable union of countable sets is countable or Baire's theorem but it is not enough to prove Hahn-Banach, Tychonoff or Krein-Milman (please, correct me if I am wrong). And this is where my question comes in. In one of the comments to the post cited above someone wrote (quoting from memory) that the majority of practicing mathematicians views countable choice as "true". I have seen this repeated many times, and the way I read this is that while the majority of practicing mathematicians views ACC as "obviously true", a part of this population harbours, in various degrees, some doubts about full AC. Assuming that I have not misread these statements, why in the minds of some people ACC is "unproblematic" but AC's validity is not? What is the intuitive explanation (or philosophical reason, if you will) why making countably infinite choices is "unproblematic" but making arbitrarily infinite choices is somehow "more suspicious" and "fraught with dangers"? I for one, cannot see any difference, but then again I freely confess my ignorance about these matters. Let me stress once again that I do not think for a moment that denying AC is "wrong" in some absolute sense of the word; I just would like to understand better what is the obstruction (to use a geometric metaphor) from passing from countably infinite choices to arbitrarily infinite ones.
Note: some rewriting and expansion of the original post to address some of the comments.
| https://mathoverflow.net/users/2562 | Choice vs. countable choice | Here is one explanation of why countable choice is not problematic in constructive mathematics.
For this discussion it is useful to formulate the axiom of choice as follows:
>
> $(\forall x \in X . \exists y \in Y . R(x,y)) \implies \exists f \in Y^X . \forall x \in X . R(x,f(x))$
>
>
>
This says that a total relation $R \subseteq X \times Y$ contains a function. The usual formulation of the axiom of choice is equivalent to the above one. Indeed, if $(S\_i)\_{i \in I}$ is a family of non-empty sets we take $X = I$, $Y = \bigcup\_i S\_i$ and $R(i,x) \iff x \in S\_i$ to obtain a choice function $f : I \to \bigcup\_i S\_i$. Conversely, given a total relation $R \subseteq X \times Y$, consider the the family $(S\_x)\_{x \in X}$ where $S\_x = \lbrace y \in Y \mid R(x,y)\rbrace$ and apply the usual axiom of choice.
One way of viewing sets in constructive mathematics is to imagine that they are collections together with *given* equality, i.e., some sort of "presets" equipped with equivalence relations. This actually makes sense if you think about how we implement abstract sets in computers: each element of the abstract set is represented by a finite sequence of bits, where each element may have many valid representations (and this is unavoidable in general). Let me give two specific examples:
* a natural number $n \in \mathbb{N}$ is represented in the usual binary system, and let us allow leadings zeroes, so that $42$ is represented by $101010$ as well as $0101010$, $00101010$, etc.
* a (computable) real $x \in \mathbb{R}$ is represented by machine code (a binary string) that computes arbitrarily good approximations of $x$. Specifically, a piece of code $p$ represents $x$ when $p(n)$ outputs a rational number that differs from $x$ by at most $2^{-n}$. Of course we only represent *computable* reals this way, and every computable real has many different representations.
Let me write $\mathbb{R}$ for the set of computable reals, because those are the only reals relevant to this discussion.
An essential difference between the first and the second example is that there is a computable canonical choice of representatives for elements of $\mathbb{N}$ (chop off the leading zeroes), whereas there is no such canonical choice for $\mathbb{R}$, for if we had it we could decide equality of computable reals and consequently solve the Halting problem.
According to the constructive interpretation of logic, a statement of the form
>
> $\forall x \in X . \exists y \in Y . R(x, y)$
>
>
>
holds if there is a program $p$ which takes as input a representative for $x \in X$ and produces a representative for $y \in Y$, together with a witness for $R(x,y)$. Crucially, $p$ need not respect equality of $X$. For example,
>
> $\forall x \in \mathbb{R} . \exists n \in \mathbb{N} . x < n$
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>
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is accepted because we can write a program which takes as input a representative of $x$, namely a program $p$ as described above, and outputs a natural number larger than $x$, for example $round(p(0)) + 1000$. However, the number $n$ will necessarily depend on $p$, and there is no way too make it depend only on $x$ (computably).
Let us have a look at the axiom of choice again:
>
> $(\forall x \in X . \exists y \in Y . R(x,y)) \implies \exists f \in Y^X . \forall x \in X . R(x,f(x))$
>
>
>
We accept this if there is a program which takes as input a $p$ witnessing totality of $R$ and outputs a representative of a choice function $f$, as well as a witness that $\forall x \in X. R(x, f(x))$ holds. This is probematic because $p$ need not respect equality of $X$, whereas a representative for $f$ must respect equality. It is not clear where we could get it from, and in specific examples we can show that there isn't one. Already the following fails:
>
> $(\forall x \in \mathbb{R} . \exists n \in \mathbb{N} . x < n) \implies \exists f \in \mathbb{N}^\mathbb{R} . \forall x \in \mathbb{R} . x < f(x)$.
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>
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Indeed, every computable map $f : \mathbb{R} \to \mathbb{N}$ is constant (because a non-constant one would allow us to write a Halting oracle).
However, if we specialize to *countable choice*
>
> $(\forall n \in \mathbb{N} . \exists y \in Y . R(n,y)) \implies \exists f \in Y^\mathbb{N} . \forall n \in \mathbb{N} . R(n,f(n))$
>
>
>
then we *can* produce the desired program. Given $p$ that witnesses totality of $R$, define the following program $q$ that represents a choice function: $q$ takes as input a binary representation of a natural number $n$, possibly with leading zeroes, chops of the leading zeroes, and applies $p$. Now, even if $p$ did not respect equality of natural numbers, $q$ does because it applies $p$ to canonically chosen representatives.
In general, we will accept choice for those sets $X$ that have computable canonical representatives for their elements. Ok, this was a bit quick, but I hope I got the idea accross.
Let me finish with a general comment. Most working mathematicians cannot imagine alternative mathematical universes because they were thoroughly trained to think about only one mathemtical universe, namely classical set theory. As a result their mathematical intuition has fallen a victim to classical set theory. The first step towards understanding why someone might call into question a mathematical principle which seems obviously true to them, is to broaden their horizon by studying other mathematical universes. On a smaller scale this is quite obvious: one cannot make sense of non-Euclidean geometry by interpreting points and lines as those of the Euclidean plane. Similarly, you cannot understand in what way the axiom of choice could fail by interpreting it in classical set theory. You *must* switch to a different universe, even though you think there isn't one... Of course, this takes some effort, but it's a real eye-opener.
| 44 | https://mathoverflow.net/users/1176 | 23043 | 15,178 |
https://mathoverflow.net/questions/23032 | 7 | The systems of the λ-cube have the axiom $\star:\square$.
I've listed a few meanings that the Curry-Howard isomorphism gives to $t : T$ below. What are the intuitive meanings of $\star$ and $\square$ in each interpretation?
$t : T : \star : \square$
**Programs:** t is a program of type T. (Possibility: T is a program of type $\star$?)
**Proofs:** t is a proof of theorem T. It's hard to see T as a proof of $\star$, though.
**Set elements:** t is a member of set T. (Possibility: T is a member of the universe $\star$ of sets. Then it seems difficult to assign a meaning to $\square$ that avoids the membership $\square : \star$.)
I'd like to fill out this table both vertically and horizontally, with both further interpretations and the missing descriptions of $\star$ and $\square$, and possibly meanings of $T : \square$ for $T \neq \star$.
Thank you!
| https://mathoverflow.net/users/5700 | What is the intuitive meaning of star and box in a pure type system? | $\star$ is a *kind*, which classifies *types*. $\square$ is a sort, and it classifies *kinds*. So this is a 4-layer deep classification. Once you get to have type-constructors, kinds get really useful. Eventually, you wish for kind-constructors too, and then you need sorts.
Turns out that you really rarely ever need to get deeper than that (even though Coq and Agda have infinitely many such levels). I am not sure I have ever read a good Curry-Howard explanation of kinds and sorts. I would hazard a guess that classical mathematics rarely worries about kinds/sorts, I would tend to dig into $n$-categories to find a good relation.
| 9 | https://mathoverflow.net/users/3993 | 23044 | 15,179 |
https://mathoverflow.net/questions/23019 | 8 | Historically, I've checked the quality of arXiv output by reading the PDF it creates; I learned today that this actually a serious mistake. If you look at the [PDF](http://arxiv.org/pdf/1001.2020.pdf) and [DVI](http://arxiv.org/dvi/1001.2020) for my most recent article, you'll see that the PDF looks fine, and the DVI is essentially unusable from a mathematical perspective. The tikz pictures don't come out at all, and there seem to be serious font issues.
I'll note that this file compiles fine as a DVI on my home computer.
>
> What's causing this? Is it a problem a user can fix by include font/style files with his/her upload? is there someone at the arXiv I should be telling to install something? Why is the arXiv less competent than a home installation of MikTeX?
>
>
>
---
**EDIT:** So, the take-away seems to be that your DVIs from the arXiv will only look right if you use Computer Modern. Hooray for all math papers looking the same!
| https://mathoverflow.net/users/66 | Why does the arXiv produce a messed-up DVI when the PDF is fine? | Apparently the source of the problem is that TikZ behaves differently when the TeX file
is compiled with pdftex/pdflatex instead of the standard tex/latex.
Instead of simply embedding PostScript into the DVI file, as any sane system would do,
it apparently tries to draw figures using characters from some special fonts.
In my opinion, TikZ authors should be crucified for this.
Mathematical papers are supposed to last a little longer than 10 or 15 years.
In 10 or 15 years PDF will be replaced by yet another format,
and all TeX files that are tailored specifically for PDF will become unusable.
So long for durability.
For this reason I urge everybody to ignore Andrew Stacey's advice (no offense meant).
One more reason to use METAPOST instead of TikZ. They seem to have
similar functionality, but METAPOST is much more powerful when it comes to solving
systems of equations, has more convenient syntax (no annoying backslashes),
and is also an order of magnitude more portable (for example, any document with METAPOST pictures looks
exactly the same in DVI, PostScript, and PDF formats).
As for the fonts, we should blame arXiv for this.
The DVI format *is* device independent.
However, a DVI file must always be accompanied by all fonts that it uses,
except for the standard Computer Modern family.
The arXiv simply gives you the DVI file without any fonts, which I find ridiculous.
Since the paper under discussion uses some weird PostScript fonts,
and the arXiv gives you a DVI file without any fonts,
it is natural that the DVI file is unreadable on many machines.
I apologize for somewhat strong language, but I feel very strongly when
it comes to the issues of portability and/or durability of mathematical papers.
For example, 10 or 15 year old LaTeX files are often no longer compilable on modern TeX systems
and require some voodoo magic on my part, because of the changes in the LaTeX
and its various packages.
For this reason I switched to Plain TeX about 10 years ago.
A Plain TeX file created in 1982 will compile without any errors and will produce
exactly the same document in 2012 or in 2042.
[If I am allowed to comment on the appearance of the paper, I must say
that it looks quite weird with the current fonts. Apparently, PostScript
fonts are used to typeset the main text, but all formulas are still typeset in Computer Modern.
Since the two fonts are obviously incompatible in style (for example, they have
different thickness), the end result is horrible. If I were to read this paper,
I would recompile it using the standard Computer Modern fonts.
Again, no offense meant.]
| 7 | https://mathoverflow.net/users/402 | 23046 | 15,181 |
https://mathoverflow.net/questions/22811 | 14 | Given natural numbers of special very composite form, like primorials or factorials, how to give some useful upper bound limit of continued fraction period length of their square roots?
I'm not a professional, any advices are welcome.
| https://mathoverflow.net/users/5539 | Upper bound of period length of continued fraction representation of very composite number square root | The continued fraction length is usually a small constant factor away from the regulator. A more precise version can also be achieved, but I don't remember a reference, so if anyone does...
Then, we know the regulator times the class number is usually a small constant factor away from the discriminant (of the order, not necessarily the field).
In addition, if a discriminant has $n$ prime factors in its squarefree part, the class number will be divisible by $2^{n-1}$.
Finally, most positive numbers of some size don't have many prime factors, and we suspect the real quadratic fields composed from these to have relatively small class number (look up Cohen-Lenstra).
Combining these facts and heuristics we get that primorials, and even factorials (still have large squarefree part), will have larger class number, hence small regulator, and therefore smaller continued fraction period length.
That said, we can dig even deeper. For factorials, there is a hefty squareful part. When we go from the maximal order of the field $\mathbb{Q}(\sqrt{n!})$ to the order $\mathbb{Z}(\sqrt{n!})$, the discriminant is enlarged hugely. Each prime (even) power $p^{2m}\ ||\ n!$ adds $p^{m-1}(p-(squarefree(n!)/p))$ to the original $h\_K\times R\_K$. So for each such factor, something goes into the class number of the smaller order, and something goes into the regulator of the smaller order.
Here comes the interesting bit. The factors that make up the new regulator tell us how far the unit group in the small order is from the unit group of the maximal order. Since we are in a real quadratic field the unit groups are of rank 1, so this distance is just the exponent to which the fundamental unit is powered by in order to enter the smaller order. Say $p\_1-1$ and $p\_2-1$ (say Legendre symbol is 1) have a large gcd. Once we power the fundamental unit to, say, $p\_1-1$ to get (locally) into the $p$-part of the smaller order, we can slack off when getting into the $p\_2$-part of the smaller order, because we've already done some of the work.
So in the factorial case, or in any number with many square factors (not dividing the squarefree part), since many of the above mentioned factors will have a large gcd, most of the factors will have to go into the class number of the small order - hence the very small regulator and continued fraction expansion.
---
We can write the above explicitly, but we need another notion. When a prime divides the discriminant to an even power greater than 2, the power of $p$ from the above mentioned factor that goes into the regulator measures how the fundamental unit is far p-adically from being 1. For most $p$, the fundamental unit will not be $1\ (mod\ p^2)$, so pretty much all of the $p^{m-1}$ goes into the regulator.
Hence, we expect:
$$ \frac{period-length(n!)}{\sqrt{n!}} \sim \frac{lcm(\{\ p-(p/squarefree(n!))\ |\ p^{2m}\ ||\ n!\ \}}{\prod\_{p^{2m}||n!} (p-(p/squarefree(n!))} $$
And how that factor behaves - I have no idea. Sounds like a combinatorical answer could exist.
| 11 | https://mathoverflow.net/users/2024 | 23054 | 15,186 |
https://mathoverflow.net/questions/23048 | 14 | Gromov-Hausdorff distance ([Wikipedia](https://en.wikipedia.org/wiki/Gromov%E2%80%93Hausdorff_convergence)) between two compact manifolds measures how far away the manifolds are from being isometric.
In many cases it is possible to do coarse estimates and conclude that a sequence of manifolds converges or diverges.
How does one usually go about calculating GH distance precisely?
Example: Take two spheres of different radii $r$ and $R$ with intrinsic (i.e the distance between two points is the length of the arc of a great circle that connects them) metrics obtained from standard embeddings into $\Bbb R^n$. What is the GH distance between them?
| https://mathoverflow.net/users/3375 | What are the tricks for computing/estimating Gromov-Hausdorff distance? | I misinterpreted the question at first, sorry. Here's my new answer:
**First, an answer to the wrong question**
For two $(n-1)$ dimensional spheres of radii r and R with the metrics induced from embedding in $\mathbb{R}^n$ (note, this is the "chord metric" not the "round metric" as Zarathustra desired), the Gromov-Hausdorff distance is $|r-R|$. We can achieve this as an upper bound by embedding the two spheres in a concentric fashion, and it's seen to be sharp by the inequality $d\_{GH}(X,Y)\geq \frac{1}{2}|\operatorname{diam}(X)-\operatorname{diam}(Y)|$.
See e.g. [Burago Burago and Ivanov, ex. 7.3.14](https://books.google.com/books?id=afnlx8sHmQIC&pg=PA255&dq=gromov+hausdorff+7.3.14&ei=3h7aS8ekIqeeygSR0OGtCQ&cd=1#v=onepage&q=gromov%20hausdorff%207.3.14&f=false) which is a good source in general.
**Now an answer to the correct question**
The answer is $\frac{\pi}{2}|R-r|$ for spheres with the round metric, as Anton more or less suggested. This follows easily from the discussion following [defn. 7.3.17 in BBI](https://books.google.com/books?id=afnlx8sHmQIC&pg=PA256&lpg=PA258&dq=burago+burago+ivanov+correspondence&source=bl&ots=JvJ19H2X98&sig=VVQSTNM35zJq_hecDlDC3Jpd1pk&hl=en&ei=6lHaS-u2KJPy9QS96ORF&sa=X&oi=book_result&ct=result&resnum=2&ved=0CBIQ6AEwAQ#v=onepage&q&f=false).
Out of laziness, I've written out some of the details here. A "correspondence" of metric spaces $X$ and $Y$ is defined to be a subset $\mathcal{R}$ of $X\times Y$ such that for every point $x\in X$ there is at least one point $(x,z)\in\mathcal{R}$ and for every $y\in Y$ there is at least one point $(w,y)\in\mathcal{R}$. From this one can prove Theorem 7.3.25 which states
$d\_{GH}(X,Y)=\frac{1}{2}\inf\_{\mathcal{R}}dis\mathcal{R}$
where the infimum is taken over all correspondences $\mathcal{R}$ and $dis\mathcal{R}$ is the distortion of $\mathcal{R}$, defined to be $\sup\{|d\_X(x,x')-d\_Y(y,y')|:(x,y),(x',y')\in\mathcal{R}\}$.
Take $\mathcal{R}$ to be the correspondence consisting of pairs $(x,y)$ with $x\in S^2\_{r}$ and $y\in S^2\_{R}$ if $x$ and $y$ lie on the same ray through the origin when the two spheres are embedded in $\mathbb{R}^3$. The distortion of this correspondence is $\pi|R-r|$ by taking $x$ and $x'$ to be antipodal points on one of the spheres. This gives an upper bound for $d\_{GH}$ of $\frac{\pi}{2}|R-r|$, and this is sharp again by the inequality above.
**A possibly useful reference for G-H distance for subsets of Euclidean space in general**
For the rest of your question, you might find interesting [a paper by Facundo Mémoli](https://people.math.osu.edu/memoli.2/papers/dgh-euclidean.pdf) which discusses the case when X and Y are subsets of Euclidean space. See also slides [here](https://people.math.osu.edu/memoli.2/talks/dgh-euclidean-cvpr.pdf).
| 10 | https://mathoverflow.net/users/353 | 23057 | 15,188 |
https://mathoverflow.net/questions/21607 | 5 | In a recent [question](https://mathoverflow.net/questions/21314/an-example-of-a-series-that-is-not-differentially-algebraic), we learned about the existence of functions that do not satisfy any algebraic differential equation.
One nice property of such equations is that there is a good way to enumerate a basis: we can produce the stream of "monomials" $\left(\prod\_i D^{\lambda\_i-1}f(x)\right)\_\lambda$, where $D$ is the differentiation operator and $\lambda=(\lambda\_1,\lambda\_2,\dots)$ runs over the integer partitions in lexicographic order:
$$
1, f(x), f(x)^2, f'(x), f(x)^3, f(x)f'(x), f''(x), f(x)^4, f(x)^2f'(x),\dots
$$
I'm wondering: is there a "natural" class of equations, more general than ADEs, that has a similar basis. (Natural meaning: equations that specify many functions occurring in "nature")? Or, alternatively, just another class of equations.
I realise this is vague, please bear with me...
edit: I should add that I'm aware of "algebraic recurrences" (i.e., shift instead of differentiation) and "Mahler-type functional equations" (i.e., $f(x^{k+1})$ instead of $f(x)^{(k)}$).
[Martin Klazar](http://kam.mff.cuni.cz/~klazar/bell.pdf) mentions that a few interesting sequences (eg. the ordinary generating function for Bell numbers) satisfy functional equations of the form
$$p\_1(x)f(x)=p\_2(x)+p\_3(x)f(\frac{x}{1-x}),$$
with polynomials p1, p2, p3 (and concludes that they are not differentially algebraic), but I'm not sure how common such equations are.
edit: the motivation for this question comes from the desire of being able to guess a formula (or recurrence, differential or functional equation) for a given sequence (of numbers or polynomials, etc.), as pioneered be GFUN, see also Section 7 in [my preprint with Waldek Hebisch](http://arxiv.org/abs/math/0702086).
For example, given the first few (say 100) terms of the sequence, we compute it's (truncated) generating function $f\_1 := f(x)$, and also $f\_2 := f(x)^2, f\_3 := f'(x), f\_4 := f(x)^3, f\_5 := f(x)f'(x), f\_6 := f''(x), f\_7 := f(x)^4, f\_8 := f(x)^2f'(x), \dots$. We fix the maximal degree, say $N$ of the coefficient polynomials $p\_1, p\_2, \dots, p\_m$, and then try to solve the linear system of equations obtained by equating coefficients in
$$
ord(p\_1 f\_1 + \dots + p\_m f\_m)\geq\sigma
$$
for large sigma. If we get a solution, and the given sequence is somehow naturally defined, chances are good that the equation holds for all terms of the sequence.
| https://mathoverflow.net/users/3032 | beyond differentially algebraic power series | I am not quite sure whether the question is about a "natural" graded algebra which is infinitely generated, or finitely generated algebras are fine as well. Because there are nice examples of algebras of multiple zeta values, but also of multiple polylogarithms and of finite multiple harmonic sums, as well as the algebra of classical modular forms. The latter gives rise to a certain structure which is presumably richer than the algebra of differential monomials, and I could probably try to explain this point.
The Eisenstein series $E\_2=1-24\sum\_{n=1}\sigma\_1(n)q^n$, $E\_4=1+240\sum\_{n=1}\sigma\_3(n)q^n$ and $E\_6=1-504\sum\_{n=1}\sigma\_5(n)q^n$, where $\sigma\_k(n)=\sum\_{d\mid n}d^k$, generate a differentially stable ring over $\mathbb Q$ with respect to the differentiation $D=q\dfrac{d}{dq}$ (a result usually attributed to Ramanujan). The weights $2,4,6$ are assigned to $E\_2,E\_4,E\_6$ respectively, and $D$ increases the weight by $2$. The graded ring $\mathbb Q[E\_2,E\_4,E\_6]$ possesses an additional structure coming from the functional equations for replacing $q$ by $q^k$ where $k$ is a positive integer, although it's very hard to write down the structure explicitly. Let me call the corresponding scale operators (substitutions $q^k$ for $q$) $D\_k$. They do not change weights.
The counterpart consists of the infinite family $F\_{2m+1}(q)=\sum\_{n=1}^\infty\sigma\_{2m}(n)q^n$, $m=0,1,2,\dots$, which are known to be linearly independent over $\mathbb Q$ and even over the field of meromorphic functions on $\mathbb C$. We can formally assign the weight $2m+1$ to each $F\_{2m+1}(q)$, although there could be reasons to normalize them in a way used for the Eisenstein series. Again, the differential operator $D$ increases weights by 2, and the open problem here is to show that the $F\_{2m+1}(q)$ are all algebraically differentially independent over $\mathbb C$ (or $\mathbb Q$). An expanded version of the problem is to show that the ring of all $D$- and $D\_k$-monomials have no nontrivial relations at all. In a sense this includes both the algebraic differential structure from the problem, as well as all kind of Mahler-type equations. If one restricts to considering $D$- and $D\_2$-monomials (or $D\_k$ monomials for a finite set of $k$'s), the corresponding set of monomials of finite weight will be finite.
| 5 | https://mathoverflow.net/users/4953 | 23059 | 15,189 |
https://mathoverflow.net/questions/23061 | 9 | Recall that a space is $\delta$-hyperbolic if there is some number $\delta$ with the property that every point on an edge of a geodesic triangle lies within $\delta$ of another edge. For example a tree is $0$-hyperbolic. One of the basic facts about standard hyperbolic space is that it is $\delta$-hyperbolic for some $\delta$, and I am looking for the smallest delta which makes this true.
Full disclosure: I stole this question from Dima Burago, who brought it up as an example of of a useless problem about which he is nevertheless a little curious. I haven't exactly burned the midnight oil, but I can't solve it.
| https://mathoverflow.net/users/4362 | It is well-known that hyperbolic space is delta-hyperbolic, but what is delta? | We can use the isometry group of $H^n$ to reduce to the case of an ideal triangle in the upper half plane, with vertices at -1, 1, and infinity. We want to find the distance between $i$ and the vertical geodesic with real part 1. To find the shortest geodesic, we reflect $i$ in the vertical line, and take half the distance between $i$ and $i+2$. The [distance formula](http://en.wikipedia.org/wiki/Poincare_metric) yields $\tanh^{-1}\left(\frac{|(i+2)-i|}{|(i+2) + i|} \right) = \tanh^{-1}(1/\sqrt2)$. This is about 0.8813735.
| 13 | https://mathoverflow.net/users/121 | 23064 | 15,192 |
https://mathoverflow.net/questions/22952 | 3 | In the K.Ueno and K.Takasaki's paper ``Toda Lattice hierarchy", advanced sdudies in pure mathematics 4(1984),pp1-95, they mentioned the Toda Lattice hierarchy of B and C type(we denote BTL and CTL hierarchies).In their paper, the bilinear relation for BTL and CTL hierarchies are almost the same as the one for Toda Lattice hierarchy except with the even time flows being zero( see eq.2.4.1 in reference).
My question is as follows:
1. Since the KP and TL hierarchies are almost the same (see Mark Adler et al's paper Comm.math.Phys.171,547-588(1995)), and considering the great difference between the bilinear relations for BKP and KP hierarchies(see E.Date,Jimbo,et al's paper,Transformation groups for soltion equations), why BTL and TL lattice hierarches are almost the same? Is there something wrong?
2. Can the BTL and CTL hierarchies have the fay like identities just as the KP and BKP hierarchies(see H.F.Shen and M.H.Tu's paper,arxiv:0811.1469)
3. why the literatures for the BTL and CTL hierarchies are so few? is there someting difficult in studying them?
thanks
| https://mathoverflow.net/users/5705 | the Toda Lattice hierarchy of B and C type | I suggest
<http://arxiv.org/abs/nlin.SI/0608018>
as a first step
| 5 | https://mathoverflow.net/users/3054 | 23076 | 15,201 |
https://mathoverflow.net/questions/19477 | 6 | Is there a good reference for the structure of the Chow ring of $\mathcal{A}\_g$, the moduli space of complex principally polarized abelian varieties? More generally, references for the intersection theory, enumerative geometry, and vector bundles on $\mathcal{A}\_g$ would be nice.
| https://mathoverflow.net/users/622 | Chow Ring of Moduli Space of Abelian Varieties | Van der Geer has written a paper computing what he calls the tautological subring of the chow ring of $\mathcal{A}\_g$. He also computes the tautological ring for a smooth toroidal compactification.
G. van der Geer, Cycles on the Moduli Space of Abelian Varieties, in "Moduli of Curves and Abelian Varieties (The Dutch Intercity Seminar on Moduli)", p. 65-89 (Carel Faber and Eduard Looijenga, editors), Aspects of Mathematics, Vieweg, Wiesbaden 1999.
It is available on the van der Geer's website [here](http://www.science.uva.nl/~geer/publications.html)
Regarding intersection theory, Erdenberger, Grushevsky, and Hulek have been working on this for the toroidal compactifications, mostly for small values of $g$. For example, see the following references.
C. Erdenberger, S. Grushevsky, K. Hulek, Intersection theory of toroidal compactifications of $\mathcal{A}\_4$. Bull. London Math. Soc. 38 (2006), no. 3, 396--400.
C. Erdenberger, S. Grushevsky, K. Hulek, Some intersection numbers of divisors on toroidal compactifications of $\mathcal{A}\_g$. J. Algebraic Geom. 19 (2010), no. 1, 99--132.
S. Grushevsky, Geometry of $\mathcal{A}\_g$ and its compactifications. Algebraic geometry---Seattle 2005. Part 1, 193--234, Proc. Sympos. Pure Math., 80, Part 1, Amer. Math. Soc., Providence, RI, 2009.
| 3 | https://mathoverflow.net/users/4910 | 23080 | 15,205 |
https://mathoverflow.net/questions/23009 | 15 | Let $K$ be a finite extension of the $p$-adic numbers. Suppose that $V$ and $W$ are two (finite dimensional, $p$-adic) continuous representations of $G\_K$. Suppose that $V \otimes W$ is crystalline. Is $V$ crystalline up to twist by a character of $G\_K$?
| https://mathoverflow.net/users/nan | If the tensor product of two representations are crystalline, are the original representations crystalline? | I'm indeed pretty sure that the answer is "yes". I'd prefer not to post the idea of the proof here because I asked one of my PhD students to write it down with all the details.
| 17 | https://mathoverflow.net/users/5743 | 23082 | 15,207 |
https://mathoverflow.net/questions/23083 | 2 | Let $k$ be a ring and $\overline{k} = k[\epsilon]/\epsilon^2$. For every $f \in k[t]$ there is a unique $f' \in k[t]$ such that $f(t+\epsilon)=f(t)+\epsilon f'(t)$ holds in $\overline{k}[t]$. It follows that $f \mapsto f'$ is a derivation. For example, the leibniz rule:
$(fg)(t+\epsilon)=f(t+\epsilon) g(t+\epsilon)=(f(t)+\epsilon f'(t)) (g(t)+\epsilon g'(t))$
$=f(t) g(t) + \epsilon (f'(t) g(t) + f(t) g'(t)) + \epsilon^2 f'(t) g'(t) = (fg)(t) + \epsilon (f'(t) g(t) + f(t) g'(t))$
Now what about the chain rule $(f \circ g)'(t)=g'(t) f'(g(t))$? Is there a proof for this just using the definition above? Of course, you could argue that it suffices to take $f=x^n$ and then use inductively the leibniz rule. Or you recoqnize $f'$ as the usual formal derivative of $f$ and make an explicit calculation. But I wonder if there is a direct proof in the sense that you just use the definition of $f'$ above. Perhaps this then makes sense in more general situations as well.
| https://mathoverflow.net/users/2841 | derivative in the ring k[e]/e², chain rule | The point is that you have the more general formula $f(g(t)+\epsilon h(t)) = f(g(t))+f'(g(t))h(t)\epsilon$. From that the chain rule follows:
$$(f\circ g)(t+\epsilon) = f(g(t+\epsilon)) = f(g(t)+g'(t)\epsilon) =
f(g(t))+f'(g(t))g'(t)\epsilon$$
The more general formula follows from the simpler "by substitution", i.e., by applying appropriate ring homomorphisms.
| 6 | https://mathoverflow.net/users/4008 | 23084 | 15,208 |
https://mathoverflow.net/questions/23085 | 3 | I would like to find a reference for the following fact:
every finite dimensional complex representation of a reductive Lie algebra is semisimple.
| https://mathoverflow.net/users/4821 | On the full reducibility of representations of reductive Lie algebras | The statement is false. The standard definition of "reductive" for a finite dimensional Lie algebra $\mathfrak{g}$ over an arbitrary field of characteristic 0 is given in a number of equivalent ways by Bourbaki in Chapter 1 (1960) of their treatise on Lie groups and Lie algebras: section 6, no. 4-5. By definition, $\mathfrak{g}$ is *reductive* provided its adjoint representation is semisimple (= completely reducible). Typical equivalent conditions: the derived algebra is semisimple; or $\mathfrak{g}$ is the direct sum of a semisimple and an abelian Lie algebra; or the solvable radical equals the center.
As a consequence, a finite dimensional representation of a reductive Lie algebra is semisimple iff the center acts by semisimple endomorphisms. (An abelian Lie algebra need not be represented in that way.)
Some of this is set up as an exercise at the end of Section 6 in my Springer graduate text (1972); see also Proposition 19.1.
| 20 | https://mathoverflow.net/users/4231 | 23093 | 15,212 |
https://mathoverflow.net/questions/23094 | 8 | Lets recall Platonic construction in plane geometry. It is impossible to square a circle using only ruler and callipers. But is also known that it is possible to do it with ruler which has a mark on it.
So we know by Gaois theory that we can't write expressions for the solutions to all algebraic equations using the four arithmetic operations and nth roots.
Question: is it possible to write expressions for the solutions to all algebraic equations using the four arithmetic operations and nth roots and something else in finite numbers of factors? For example adding logarithm to the set of operations? ( or other function, or class of functions...)?
If Yes: Do they form any kind of algebraic structure?
| https://mathoverflow.net/users/3811 | method of finding roots of polynominal equations with arithmetic operations and roots and other functions | The answer is no: you need to keep adding more and more operations. For degree $n=5$ you can use elliptic functions (or the Jacobi $\Theta$ function, or Bring radicals - see below), for $n=6,7$ the Lauricella functions are needed (they are a 2-variable version of hypergeometric functions), and after that you need more and more complicated 'new' functions. You can apparently use the elliptic Siegel functions (aka Siegel modular forms) for the general case, but I've never looked into that. Most of this was all worked out in gruesome detail by analysts for decades, with lots of research on this up until the early 20th century (and then generally forgotten!).
For trinomial equations, you can use hypergeometrics to solve all of them. The derivation of that is fun.
It is hard to find information about this on the web. The 'best' discussion of related ideas is the Wikipedia page on the [Bring radical](http://en.wikipedia.org/wiki/Bring_radical). There is a good [timeline](http://library.wolfram.com/examples/quintic/timeline.html) at Wolfram's site.
| 10 | https://mathoverflow.net/users/3993 | 23099 | 15,215 |
https://mathoverflow.net/questions/23104 | 1 | It is well known that e.g. $sin(1/x)$ is of [unbounded total variation](http://en.wikipedia.org/wiki/Bounded_variation#Examples) (in the interval [0,1] assuming $f(0)=0$). (Preliminary numerical tests suggest that) it is also of unbounded [quadratic variation](http://en.wikipedia.org/wiki/Quadratic_variation). $x\ sin(1/x)$ is of unbounded total variation too but (preliminary numerical tests suggest that) it is of zero quadratic variation.
**My question:** How to construct a deterministic function with unbounded total variation and bounded (non zero) quadratic variation along these lines? I don't want to have a function which is defined by a sum of terms (like the [Weierstrass function](http://en.wikipedia.org/wiki/Weierstrass_function)) but one which is defined straight forward like the two above mentioned examples. References (if available) would also be appreciated!
**Addendum:** If some of these conjectures are not true please tell me. And please tell me also if it is not possible to construct such a deterministic function (and why not).
| https://mathoverflow.net/users/1047 | Example for deterministic function with unbounded total variation and bounded quadratic variation | Function $x^a\sin(1/x)$ on $(0,1]$ has bounded variation iff $a>1$ and finite quadratic variation iff $a>1/2$. So for your example, take $1/2 < a \le 1$. It looks like more than numerical tests may be needed to decide questions like this...
I took "quadratic variation" to mean
$$
\sup \sum\_{j=1}^n |f(x\_j)-f(x\_{j-1})|^2
$$
with sup over all finite sequences $x\_0\lt x\_1\lt\cdots\lt x\_n$ in $(0,1]$. Perhaps you meant something else?
**addition**
On the other hand, maybe you mean
$$
\lim\_{\delta \to 0}\sup \sum\_{j=1}^n |f(x\_j)-f(x\_{j-1})|^2
$$
with sup over all finite sequences $x\_0\lt x\_1\lt\cdots\lt x\_n$ in $(0,1]$ such that $x\_j-x\_{j-1}<\delta$. In that case, on any inverval where $f$ is monotone it has quadratic variation $0$, so you won't find any such simple example.
| 4 | https://mathoverflow.net/users/454 | 23109 | 15,223 |
https://mathoverflow.net/questions/23113 | 29 | We know from elementary school that the triangle inequality holds in Euclidean geometry. Some where in High School or in Univ., we come across non-Euclidean geometries (hyperbolic and Riemannian) and Absolute geometry where in both the inequality holds.
I am curious whether the triangle inequality is made to hold in any geometry( from the beginning) or is a consequence of some axioms. Presumably, the denial of the inequality would create havoc in that conceivable geometry.
Thanks.
| https://mathoverflow.net/users/5627 | Is there any geometry where the triangle inequality fails? | There are people who seriously study quasi-normed spaces. The most natural examples are $\ell\_p$ spaces for p strictly between 0 and 1 (the "norm" given by the usual formula and the distance given by the norm of the difference). Although these spaces do not satisfy the triangle inequality, you get an inequality of the form $\|x+y\|\leq C(\|x\|+\|y\|)$.
| 32 | https://mathoverflow.net/users/1459 | 23121 | 15,232 |
https://mathoverflow.net/questions/23111 | 12 | Questions about continued fractions reminded me about a related diophantine problem. I am not quite sure that diophantine equations are still in fashion but
$$
1^k+2^k+\dots+(m-1)^k=m^k,
$$
the Erdős--Moser equation, is quite special. This seems to be the only known equation in two unknowns ($k$ and $m$ are assumed to be positive integers) for which it is not proven whether there are finitely many solutions. Presumably there is only one solution, $1^1+2^1=3^1$. In a joint work with Yves Gallos and Pieter Moree (the preprint is available from arxiv.org) we show that if a solution exists than $m>10^{10^9}$ "by showing that $2k/(2m-3)$ is a convergent of $\log 2$ and making an extensive continued fraction digits calculation of $(\log 2)/N$, with $N$ an appropriate integer." Trying to
build up some history on the use of continued fractions in diophantine equations we could not find any other example, besides the continued fractions of real quadratic irrationalities with their ultimate relation to Pell's equation (which in turn appears in studying some other equations). I am definitely interested if there are some, but also in other "serious" applications of "generic" (presumably pattern-free) continued fractions of mathematical constants (or functions) outside number theory. Please suggest "generic" cases where no rule for partial quotients is known.
| https://mathoverflow.net/users/4953 | Applications of pattern-free continued fractions | Continued fractions are used in the effective solution of Thue equations. Given such an equation, say $x^4-3x^3y+7x^2y^2-y^4 = 1$ to be specific, we can brute force find all small $x,y$ solutions, and linear forms in logarithms results can be used to eliminate solutions with $x,y$ astronomically large. In the large intermediate range (which typically can mean $x,y$ with between 50 and $10^{50}$ digits, depending on the specifics of the equation) LLL is often helpful. The small intermediate range (say $x,y$ having between 3 and 50 digits) is where continued fractions make their appearance.
An example of this is the following. The equation given above is the same as
$$r^4 - 3r^3 + 7 r^2 -1 = 1/y^4$$
where $r$ is the rational $x/y$. This factors (for some complex numbers $\alpha\_i$) as
$$(r-\alpha\_1)(r-\alpha\_2)(r-\alpha\_3)(r-\alpha\_4) = \frac{1}{y^4}.$$
Since the right hand side is small, one of the factors on the left must be small, say it's the first that is smallest. Letting $C=(\alpha\_1-\alpha\_2)(\alpha\_1-\alpha\_3)(\alpha\_1-\alpha\_4)/8$, our equation gives the inequality:
$$\left| \alpha\_1 - \frac xy \right| < \frac{2/(C y^2)}{2y^2}.$$
If $y$ is large enough (typically, say, 3 digits), then $2/(C y^2)<1$, and a theorem from a couple centuries ago states that if
$$\left| \alpha - \frac xy \right| < \frac{1}{2y^2},$$
then $x/y$ is a convergent of the continued fraction of $\alpha$. This means that every solution to the original equation (except for the small ones) is a convergent to one of the roots of the equation, and numerators and denominators of convergents grow exponentially. So, as a practical matter, we can find all solutions of a Thue equation with between 3 and 50 digits essentially instantaneously.
Disclaimer: all statements here about how many digits constitutes large or small depend on the specific equation (among other things, on the embeddings into $\mathbb{C}$ of the group of fundamental units of the relevant extension of $\mathbb{Q}$), and the numbers I've given here derive from my personal experience. Some years ago, I worked at Wolfram Research to help bring diophantine equation solving to Mathematica, it will solve the above equation if you input:
Reduce[x^4 - 3 x^3 y + 7 x^2 y^2 - y^4 == 1, {x, y}, Integers]
| 13 | https://mathoverflow.net/users/935 | 23138 | 15,241 |
https://mathoverflow.net/questions/23091 | 6 | I'm stuck on a technicality concerning singularities.
Basically, I have to show that the singularities of a $\mathbf{certain}$ normal projective variety over $\mathbf{C}$ are rational. (I won't bother you with the exact set-up. For, it's possible that I'm even wrong.)
My idea was to use a theorem of Viehweg and show that all singularities are quotient. But I don't have a clue of how to do the latter. What are the standard techniques used in such a proof? That is, say you want to show that all singularities are quotient. What would be your first idea to apply? Do problems like this become easier in low dimensions or does it really not matter?
**Note**. Forgive me if the question is ill-posed/vague.
**Added later**.
In view of Karl's remark, I decided to give the set-up.
Let $X$ be a smooth projective variety over $\mathbf{C}$ (of any dimension). Let $\pi:Y\longrightarrow X$ be a finite morphism, where $Y$ is a normal projective variety over $\mathbf{C}$. We also are given a flat morphism $h:X\longrightarrow C$, where $C$ is a smooth projective curve. Of course, it can happen that $Y$ has singularities that are not rational. (Example?) But I would like to show that the singularities of $Y$ are rational in the situation I will describe now.
Let $V\longrightarrow U$ be a connected finite etale covering of $U=X-D$, where $D$ is a simple normal crossings divisor on $X$. Define $Y$ to be the normalization of $X$ in the function field of $V$.
So the set-up is quite general, i.e., I don't have any equations. Maybe one could try to apply arguments based on fundamental groups to show that the singularities of $Y$ are quotient? I think I can show that the singularities of $Y$ occur in the inverse image under $\pi$ of $D^{sing}$, where $D^{sing}$ is the singular locus of $D$.
And Karl, what are the techniques in characteristic $p>0$ that you mention below?
| https://mathoverflow.net/users/4333 | Is there an obvious way for showing singularities are quotient? | Say $X=\mathbb A^n$ and $D\_1,\dots, D\_n$, the components of $D$, are the coordinate hyperplanes $x\_i=0$, for simplicity. You can assume that WLOG, because your $(X,D)$ is isomorphic to this one in étale topology.
$\pi\_1(X\setminus D) = \mathbb Z^n$. So the cover $V\to U$ corresponds to a finite quotient of $\mathbb Z^n$, which is a finite abelian group $G$. So the ring of regular functions on $V$ is generated by the *roots* of monomials in $x$. You can write these as $x^m$ for some $m\in \mathbb Q^n$. The lattice $H$ generated by $m\_i$ contains $\mathbb Z^n$, and the quotient $H/\mathbb Z^n$ is the dual abelian group $G^{\vee}$.
So what is $Y$ now? It is the normalization of the ring $k[x\_1,\dots,x\_n]$ in the bigger field $k(x^{m\_i})$. So it is a toric problem now. The normalization is generated by monomials in the lattice $H$ which lie in the cone $(\mathbb R\_{\ge0})^n$.
So $Y$ is toric and simplicial, and every such singularity is an abelian quotient singularity.
The condition $\pi\_1(X\setminus D) = \mathbb Z^n$ fails in char $p$ (indeed, the fundamental group in that case is huge), so this argument and the statement both fail in char $p$.
| 8 | https://mathoverflow.net/users/1784 | 23139 | 15,242 |
https://mathoverflow.net/questions/23107 | 15 | I would like to know how much of the recent results on the MMP (due to Hacon, McKernan, Birkar, Cascini, Siu,...) which are usually only stated for varieties over the complex numbers, extend to varieties over arbitrary fields of characteristic zero or to the equivariant case.
I assume that the basic finite generation results hold for any such field, by base extending to an algebraic closure, so I would guess that most results should extend without too much difficulty. The particular questions that I am really interested in are:
1) Given a smooth projective rationally connected variety X over a field k of charaterisitic zero, can we perform a finite sequence of divisorial contractions and flips to obtain a Mori fibre space?
and
2) The equivariant version of 1) for the action of a finite group on X.
| https://mathoverflow.net/users/519 | What is known about the MMP over non-algebraically closed fields | Both of these cases follow more or less automatically from the Minimal Model Program over an algebraically closed field $\bar k$. This is well, known, see for example the original Mori's paper *"Threefolds whose canonical bundles are not numerically effective"* or [Kollár's paper](http://www.math.princeton.edu/~kollar/FromMyHomePage/rat2.ps) on 3-folds over $\mathbb R$. Iskovskikh and Manin did both versions for surfaces in the 1970s, and their arguments still apply.
The point is that $K\_X$ is invariant under the action of any group $G\subset Aut(X)$ and $Gal(\bar k/k)$. So if $\bar C$ is a curve on $\bar X= X\otimes\_k \bar k$ with $K\_X . \bar C<0$ then $C= \sum\_{g\in G} g.\bar C$ (resp. the sum of the conjugates) also intersects $K\_X$ negatively.
So you can work with the $G$-invariant (resp. $Gal$-invariant) part of the Mori cone $NE(X)\subset N\_1(X)$. If $R$ is an extremal ray of $NE(X)^G$ then the supporting divisor $D$ can be chosen to be $G$-invariant, and it contracts a face of $NE(X)$ (instead of just a ray).
It is either a divisorial contraction over $k$, or a flipping contraction. In the latter case there is a flip defined over $k$, since it is an appropriate relative canonical model, and every canonical model is automatically $G$-equivariant, resp. $Gal$-equivariant, by its uniqueness.
So you just do the MMP over $k$. Unless $K\_X$ is pseudoeffective but not effective, MMP terminates by [BCHM], and you get either a minimal model (with $K\_{X\_{\rm min}}$ nef) of a Mori-Fano fibration.
Finally, if you started with a variety $X$ such that $\bar X$ is covered by rational curves then $K\_X$ is not pseudoeffective.
| 13 | https://mathoverflow.net/users/1784 | 23141 | 15,244 |
https://mathoverflow.net/questions/23136 | 1 | The regression depth of a line is the minimum number of points it has to cross to take it from its initial position to vertical. The undirected depth of a point is the minimum number of lines a ray originating at the point will cross before escaping. If we use projective duality, then regression depth is same as undirected depth. I cannot see how that would be.
My attempt is as follows: By duality, points will map to lines, and lines will map to points. Suppose I am trying to compute the undirected depth of a point. So there is a point such that all rays emanating from it will meet n/3 points or more in each direction. In the dual plane, the point will coincide with a line. But what will rays correspond to?
**Edit:** Quoting from paper by Nina Amenta, Marshall Bern et al, "Regression Depth and Center Points":
Geometrically, the regression depth of a hyperplane is the minimum number of points intersected by the hyperplane as it undergoes any continuous motion taking it from its initial position to vertical. In the dual setting of hyperplane arrangements, the undirected depth of a point in an arrangement is the minimum number of hyperplanes touched by or parallel to a ray originating at the point. Standard techniques of projective duality transform any statement about
regression depth to a mathematically equivalent statement about undirected depth and vice versa.
I was trying to come up with this duality but havent so far succeeded. In $\mathbb{R}^2$, when we consider lines and points, I was considering a point that has high undirected depth, but am unable to make a transformation that would show exactly how the undirected depth of the point will map to the regression depth of a line in the dual plane.
| https://mathoverflow.net/users/5553 | How does one map regression depth to undirected depth of a point? | (For some context: the question here is written in terms of points in a plane, and my answer is in the same terms, but the same duality works in higher dimensions as well. The paper in question is [arxiv:cs.CG/9809037](http://arxiv.org/abs/cs.CG/9809037).)
Let P be the plane in which you are measuring regression depth; regression depth is defined in a Euclidean plane, not a projective one, but we view P as being a Euclidean plane embedded as a subset of its projective completion P\*, so that a line is vertical iff it passes through the point at vertical infinity (the intersection point in P\* of two vertical lines).
Now take a dual projective plane Q\*, and again view Q\* as containing a Euclidean plane Q and a line at infinity; choose Q and Q\* in such a way that the line at infinity in Q\* is dual to the point at vertical infinity in P\*. (It's easy enough to do this with coordinates, if you prefer them to this sort of conceptual explanation.)
Then rotating a line in P until it is vertical is the same thing as moving the projectively dual point in Q until it reaches infinity, and the number of points crossed by the rotating line is the same as the number of lines crossed by the moving point.
| 0 | https://mathoverflow.net/users/440 | 23146 | 15,247 |
https://mathoverflow.net/questions/23128 | 5 | Let $B$ be an infinite-dimensional Banach space, and let $M\subset\mathbb{R}^n$ be a neighborhood of the origin in $\mathbb{R}^n$.
Suppose that $I:M\to B$ is a real-analytic function with $I(0)=0$ and such that the derivative of $I$ at $0$ has maximal rank.
Is it true that there exist neighborhoods $U,V\subset B$ of $0$ and a real-analytic diffeomorphism $\phi:U\to V$ such that $\phi\circ I$ is the restriction of a linear map $\mathbb{R}^n\to B$? If so, what is a good reference?
EDIT: I asked this question in the real-analytic setting, but might as well have done so in the complex-analytic case.
| https://mathoverflow.net/users/3651 | Local form of a real-analytic function taking values in a Banach space | I cannot give you a reference, but the answer ought to be yes.
To simplify notation, identify $\mathbb{R}^n$ with its image in $B$ under the derivative $I'(0)$. That image, being finite-dimensional, is the range $pB$ of a finite rank, bounded projection $p$. Replacing $M$ by a smaller neighbourhood if necessary, we can assume that $p\circ I$ is a diffeomorphism of $M$ onto a neighbourhood $N$ of $0$ in $pB$. Let $h\colon N\to M$ be its inverse, and define $\phi\colon p^{-1}(N)\to p^{-1}(M)$ by $$\phi(w)=x+w-I(x),\qquad x=h(pw).$$
If $w\in pB$ then $h(p(I(w)))=w$, so $\phi(I(w))=w+I(w)-I(w)=w$ – i.e., $\phi\circ I$ is the identity on $pB$.
To find the inverse of $\phi$, note that if $y=\phi(w)$ and $x=h(pw)\in pB$ then $py=x+pw-p(I(h(pw))=x$, so $w=y-x+I(x)=y-p(y)+I(py)$, i.e., $$\phi^{-1}(y)=y-py+I(py).$$
**Edit:** On second thought, it would have been more natural to define $\phi^{-1}$ first, with the requirement that its restriction to $pB$ be $I$. Letting its restriction to $(1-p)B$ be the identity is the simplest way to make it a diffeomorphism.
| 2 | https://mathoverflow.net/users/802 | 23148 | 15,249 |
https://mathoverflow.net/questions/22662 | 4 | Suppose $V\_1$ and $V\_2$ are two $(g,K)$ modules of some reductive group $G$ with maximal compact $K$. Let $P$ be the minimal parabolic of $G$, $U$ its unipotent part, and $u$ its Lie algebra. Suppose the quotients $V\_1/uV\_1$ and $V\_2/uV\_2$ are isomorphic as modules for the Levi component of $P$, then what else do we need to know to conclude that $V\_1$ and $V\_2$ are isomorphic as $G$ modules?
edit:Thanks for Kevin and Emerton's comments and sorry for the confusion about the base field. Here I'm assuming REAL reductive group.
| https://mathoverflow.net/users/1832 | what information of a representation was killed by Jacquet functor? | This is a comment, mostly on terminology that I think caused some confusion, not an answer, but I don't have enough "influence" to post this as a comment.
For real reductive groups, the Jacquet functor $V\mapsto J(V)$ (Jacquet-Casselman functor, Jacquet module, etc) is defined *differently* from the nonarchimedean case: it is $\varinjlim (V/n^iV)^\*$, which is dual to the n-adic completion of V. Thus J(V) is a p-finite g-module (n is the nilradical, p is the parabolic Lie subalgebra) and the target category of the functor J is parabolic category O for g. On the one hand, this provides more structure: we get a g-module and, for example, the infinitesimal character of V can be read off the infinitesimal character of J(V) (Casselman-Osborne); more generally, V and J(V) have the same annihilator in U(g). On the other hand, J(V) is morally a highest weight module, not a Harish-Chandra module, so some information is lost. The 0th n-homology that this problem is asking about, V/nV, is just the the top layer of J(V), so even more information is lost. By the way, unlike the p-adic Jacquet functor, n-homology is not exact (there may be higher homology), but V/nV is always non-zero (short proof was given by Beilinson and Bernstein).
I assume that the modules V1 and V2 in the formulation were presumed to be simple?
| 5 | https://mathoverflow.net/users/5740 | 23156 | 15,254 |
https://mathoverflow.net/questions/23154 | 6 | In [this](http://www.ams.org/journals/bull/1965-71-01/S0002-9904-1965-11265-4/S0002-9904-1965-11265-4.pdf) classic paper, Sakai proves the following Radon-Nikodym theorem:
>
> Let $M$ be a von Neumann algebra, and let $\phi$ and $\psi$ be two normal positive linear functionals on $M$. If $\psi \leq \phi$, then there is a positive operator $t\_0\in M$ such that $0 \leq t\_0 \leq 1$, and $\psi(x) = \phi(t\_0 x t\_0)$ for all $x \in M$.
>
>
>
The paper provides no uniqueness result. One would naively expect that any two such operators $t\_0$ and $t\_1$ would satisfy $\phi((t\_1-t\_0)^2)=0$. I can find no such statement in the literature. Is this true?
Please note that $\phi$ is not assumed to be faithful.
| https://mathoverflow.net/users/2206 | A non-commutative Radon-Nikodym derivative. | Such t\_0 is unique if its support is at most p, where p is the support of ϕ.
Note that we can replace t\_0 by pt\_0p and the support of pt\_0p is at most p.
Without this additional condition t\_0 is highly non-unique, because
we can replace t\_0 by t\_0 + q,
where q is an arbitrary self-adjoint element with support at most 1-p such that t\_0 + q ≥ 0.
Using simple algebraic manipulations one can show that all solutions can be obtained in this way.
See Lemma 15.4 (page 104) in
Takesaki's book “Tomita's theory of modular Hilbert algebras and its applications”. [Note that Takesaki implicitly assumes that φ\_0 is faithful, hence you need to introduce
an additional condition on the support of h.]
>
> One would naively expect that any two such operators t\_0 and t\_1 would satisfy ϕ((t\_1−t\_0)^2)=0.
>
>
>
This is a trivial corollary of the above statement characterizing all possible solutions.
| 6 | https://mathoverflow.net/users/402 | 23161 | 15,257 |
https://mathoverflow.net/questions/23153 | 12 | Given integers $a,b,c$ such that $\gcd(a,b,c) = 1$, it is well known that there exists only a finite set of numbers $n$ such that $n$ is not expressible as $ax+by+cz$ for non negative integers $x$,$y$,$z$.
It is also known that there exists a quadratic time algorithm for finding the maximal such $n$. However I was not able to spot the paper covering the algorithm.
Anybody happens to know the algorithm and/or a (free) reference to it?
| https://mathoverflow.net/users/1737 | Frobenius number for three numbers | Simple algoritm based on continued fractions was proposed by Rödseth, O. J. On a linear Diophantine problem of Frobenius J. Reine Angew. Math., 1978, 301, 171-178
All algorithm are described in Ramrez Alfonsn, J. L. The Diophantine Frobenius problem Oxford University Press, 2005
I think that Mathematica uses Rödseth's algoritm <http://demonstrations.wolfram.com/PositiveFrobeniusNumbersOfThreeArguments/>
| 10 | https://mathoverflow.net/users/5712 | 23165 | 15,258 |
https://mathoverflow.net/questions/23168 | 1 | Where can I found a description of the deformation theory for modules?Is it possible to deform a free module in such way that each fibre of the deformation is still free?
| https://mathoverflow.net/users/4821 | Deformations of free modules | A free module is rigid, because $Ext^1(E, E)=0$ for any free module $E.$
For deformation theory see for example ``Functors of Artin Rings''
by Michael Schlessinger.
| 2 | https://mathoverflow.net/users/2464 | 23170 | 15,260 |
https://mathoverflow.net/questions/23171 | 10 | This question has bugged me as I read McDuff-Salamon's book on pseudoholomorphic curves. I'll use their terminology.
Let $\Sigma$ be a compact surface possibly with boundary, $M$ an almost-complex manifold, and $L$ a totally real submanifold of $M$. A map
$u:(\Sigma, \partial\Sigma)\to(M, L)$
gives rise to a *bundle pair* over $\Sigma$: a complex vector bundle $u^\*TM$ over $\Sigma$, together with a totally real sub-bundle $u^\*TL$ over $\partial\Sigma$.
**Question:** Is there a nice description for the Maslov index of this bundle pair, in terms of a topological invariant of $u$? For instance, in terms of the homology class $u\_\*[\Sigma]\in H\_2(M, L)$, or in terms of the homotopy equivalence class of $u$?
**Motivating special case:** if $\partial\Sigma=\emptyset$, then the Maslov index of the bundle pair $(u^\*TM, \emptyset)$ is $2\langle c\_1(TM), u\_\*[\Sigma]\rangle$.
| https://mathoverflow.net/users/2819 | Maslov index of a pullback bundle | When you have a vector bundle on a manifold $X$ with boundary, trivialised over $\partial X$, there are characteristic classes valued in $H^\ast (X,\partial X)$. Here, when $L$ is orientable, the Maslov index is twice the first Chern class of $u^\ast TM$ relative to the trivialisation on the boundary induced by $L$, evaluated on $[\Sigma,\partial \Sigma]$.
When $\Sigma$ is closed, the Chern number of $u^\ast TM$ is the signed count of zeroes of a transversely-vanishing section $s$ of $u^\ast\Lambda^{max}\_{\mathbb{C}}TM$.
When there is an orientable boundary condition, the relative Chern number is the same thing, but you choose $s$ non-vanishing along the boundary and tangent to the real line sub-bundle $\Lambda^{max}\_{\mathbb{R}} TL$.
This doesn't make sense when $u^\*|\_{\partial \Sigma} TL \to \partial \Sigma$ is not orientable: its top exterior power then has no non-vanishing section. Besides, the Maslov index is odd in this case.
ADDED: Here's a proof using the method of Robbin's appendix to McDuff-Salamon ("$J$-holomorphic curves and symplectic topology"). Robbin characterises the boundary Maslov index as an invariant of bundle pairs (complex vector $E$ bundle over a surface, totally real sub-bundle $F$ over the boundary) which is additive under direct sum and under sewing boundaries and is suitably normalised for line bundles over the disc. The uniqueness proof, by "pair-of pants induction", still applies when $F$ is assumed orientable. The invariant "twice the relative Chern number" evidently satisfies the direct sum and sewing properties, and the section $z\mapsto z$ of the trivial line bundle over the disc satisfies the standard Maslov-index 2 boundary condition. Done!
| 10 | https://mathoverflow.net/users/2356 | 23176 | 15,264 |
https://mathoverflow.net/questions/23175 | 38 | This question is short but to the point: what is the "right" abstract framework where Mayer-Vietoris is just a trivial consequence?
| https://mathoverflow.net/users/5756 | Mathematically mature way to think about Mayer–Vietoris | The Mayer-Vietoris sequence is an upshot of the relationship between sheaf cohomology and presheaf cohomology (a.k.a. Cech cohomology).
Let $X$ be a topological space (or any topos), $\mathcal U$ a covering of $X$. Let $\mathop{\rm Sh}X$ be the category of sheaves on $X$ and $\mathop{\rm PreSh}X$ the category of presheaves. The embedding $\mathop{\rm Sh}X \subseteq \mathop{\rm PreSh}X$ is left-exact; its derived functors send a sheaf $F$ into the presheaves $U \mapsto \mathrm H^i(U, F)$. For any presheaf $P$, one can define Cech cohomology $\mathrm {\check H}^i(\mathcal U, P)$ of $P$ by the usual formulas (this is often done only for sheaves, but scrutinizing the definition, one sees that the sheaf condition is never used). One shows that the $\mathrm {\check H}^i(\mathcal U, -)$ are the derived funtors of $\mathrm {\check H}^0(\mathcal U, -)$; and of course for a sheaf $F$, $\mathrm {\check H}^0(\mathcal U, F)$ coincides with $\mathrm H^0(\mathcal U, F)$. The Grothendieck spectral sequence of this composition, in the case of a covering with two elements, gives the Mayer--Vietoris sequence.
There is also a spectral sequence for finite closed covers, which is obtained as in anonymous's answer.
I guess that this can also be interpreted as Tilman does, in a different language (I am not a topologist).
| 64 | https://mathoverflow.net/users/4790 | 23180 | 15,268 |
https://mathoverflow.net/questions/23182 | 1 | $S$ is a graded ring (over non-negative integers), $f \in S\_{+}$ is a homogeneous element of positive degree, $D(f)$ the elements of Proj $S$ not containing $f$. I don't see the bijection between $D(f)$ and Spec $S\_{(f)}$. Here $S\_{(f)}$ is the zero-degree part of $S\_{f}$ obtained from $S$ by inverting f. I see the bijection from $D(f)$ to the homogeneous primes in $S\_{f}$, but is there 1-1 correspondence between primes in $S\_{(f)}$ and homogeneous primes in $S\_{f}$?
| https://mathoverflow.net/users/5292 | Restriction of Proj S to D(f) is isomorphic to Spec S_{(f)} | The homeomorphism $D\_+(f) \to \text{Spec } S\_{(f)}$ is given by $\mathfrak{p} \mapsto \mathfrak{p} S\_f \cap S\_{(f)}$ with inverse map $\mathfrak{q} \mapsto \oplus\_n \{x \in S\_n : x^{|f|} / f^n \in \mathfrak{q}\}$. This can be checked by simple calcuations.
| 2 | https://mathoverflow.net/users/2841 | 23183 | 15,269 |
https://mathoverflow.net/questions/23192 | 2 | The question if there is an upper bound known for Brun's constant was discussed briefly here: <http://gowers.wordpress.com/2009/05/22/what-is-wolfram-alpha-good-for/> but no sure answer was given.
So I thought I'd ask the question here. Can one get any upper bound for the sum of the reciprocals of the twin primes?
| https://mathoverflow.net/users/2888 | Upper bound on Brun's constant | Crandall and Pomerance, "Prime numbers: a computational perspective" (Google books) says that Brun's constant B, the sum of the reciprocals of the twin primes, is known to be between 1.82 and 2.15.
**edited to add**: I'm aware that this isn't much of a citation. It would be nice if someone who has access to this book could give a better citation. I'd do it, but I'm not near a library today.
| 4 | https://mathoverflow.net/users/143 | 23194 | 15,274 |
https://mathoverflow.net/questions/23202 | 65 | In the following, I use the word "explicit" in the following sense: No choices of bases (of vector spaces or field extensions), non-principal ultrafilters or alike which exist only by Zorn's Lemma (or AC) are needed. Feel free to use similar (perhaps more precise) notions of "explicit", but reasonable ones! To be honest, I'm not so interested in a discussion about mathematical logic. If no example is there, well, then there is no example. ;-)
Can you give explicit large linearly independent subsets of $ \mathbb{R}$ over $\mathbb{Q}$? For example, $\{\ln(p) : p \text{ prime}\}$ is such a set, but it's only countable and surely is no basis. You can find more numbers which are linearly independent, but I cannot find uncountably many. AC implies $\dim\_\mathbb{Q} \mathbb{R} = |\mathbb{R}|$. Perhaps $ZF$ has a model in which every linearly independant subset of $ \mathbb{R}$ is countable?
The same question for algebraically independent subsets of $ \mathbb{R}$ over $\mathbb{Q}$? Perhaps the set above is such a subset? But anyway, it is too small.
Closely related problems: Can you give an explicit proper subspace of $ \mathbb{R}$ over $\mathbb{Q}$, which is isomorphic to $ \mathbb{R}$? If so, is the isomorphism explicit? Same question for subfields.
That would be *great* if there were explicit examples. :-)
| https://mathoverflow.net/users/2841 | explicit big linearly independent sets | Here is a linearly independent subset of $\mathbb{R}$ with size $2^{\aleph\_0}$.
Let $q\_0, q\_1, \ldots$ be an enumeration of $\mathbb{Q}$. For every real number $r$, let
$$T\_r = \sum\_{q\_n < r} \frac{1}{n!}$$
The proof that these numbers are linearly independent is similar to the usual proof that $e$ is irrational. (It's a cute problem; there's spoiler below.)
I think a similar trick might work for algebraic independence, but I don't recall having seen such a construction. Actually, John von Neumann showed that the numbers
$$A\_r = \sum\_{n=0}^\infty \frac{2^{2^{[nr]}}}{2^{2^{n^2}}}$$
are algebraically independent for $r > 0$. [[**Ein System algebraisch unabhängiger zahlen**](http://dx.doi.org/10.1007/BF01459089), *Math. Ann.* **99** (1928), no. 1, 134–141.] A more general result due to Jan Mycielski seems to go through in ZF + DC perhaps just ZF in some cases. [[**Independent sets in topological algebras**](http://matwbn.icm.edu.pl/ksiazki/fm/fm55/fm55112.pdf), *Fund. Math.* **55** (1964), 139–147.]
As for subspaces and subfields isomorphic to $\mathbb{R}$, the answer is no. (Since I'm not allowed to post any logic here, I'll refer you to [this answer](https://mathoverflow.net/questions/16666/does-conzf-imply-conzf-aut-c-z-2z/16683#16683) and let you figure it out.)
Well, I'll bend the rules a little... Consider a $\mathbb{Q}$-linear isomorphism $h:\mathbb{R}\to H$, where $H$ is a $\mathbb{Q}$-linear subspace of $\mathbb{R}$ (i.e. $h$ is an additive group isomorphism onto the divisible subgroup $H$ of $\mathbb{R}$). If $h$ Baire measurable then it must be continuous by an ancient theorem of Banach and Pettis. It follows that $h(x) = xh(1)$ for all $x \in \mathbb{R}$ and therefore $H = \mathbb{R}$. Shelah has produced a model of ZF + DC where all sets of reals have the [Baire property](http://en.wikipedia.org/wiki/Property_of_Baire), so any such $h$ in this model must be Baire measurable. A similar argument works if Baire measurable is replaced by Lebesgue measurable, but Solovay's model of ZF + DC where all sets of reals are Lebesgue measurable uses the existence of an inaccessible cardinal, and this hypothesis was shown necessary by Shelah.
---
***Spoiler***
Suppose for the sake of contradiction that $r\_1 > r\_2 > \cdots > r\_k$ and $a\_1,a\_2,\ldots,a\_k \in \mathbb{Z}$ are such that $a\_1T\_{r\_1} + a\_2T\_{r\_2} + \cdots + a\_kT\_{r\_k} = 0$. Choose a very large $n$ such that $r\_1 > q\_n > r\_2$. If $n$ is large enough that
$$(|a\_1| + |a\_2| + \cdots + |a\_k|) \sum\_{m=n+1}^\infty \frac{n!}{m!} < 1$$
then the tail terms of $n!(a\_1T\_{r\_1}+\cdots+a\_kT\_{r\_k}) = 0$ must cancel out, and we're left with
$$a\_1 = -\sum\_{m=0}^{n-1} \sum\_{q\_m < r\_i} a\_i \frac{n!}{m!} \equiv 0 \pmod{n}$$
If moreover $n > |a\_1|$, this means that $a\_1 = 0$. Repeat to conclude that $a\_1 = a\_2 = \cdots a\_k = 0$.
| 124 | https://mathoverflow.net/users/2000 | 23206 | 15,280 |
https://mathoverflow.net/questions/23193 | 21 | The real numbers can be axiomatically defined (up to isomorphism) as a Dedekind-complete ordered field.
What is a similar standard axiomatic definition of the integer numbers?
A commutative ordered ring with positive induction?
| https://mathoverflow.net/users/5761 | Axiomatic definition of integers | It's the unique commutative ordered ring whose positive elements are well-ordered.
| 28 | https://mathoverflow.net/users/5583 | 23212 | 15,284 |
https://mathoverflow.net/questions/23197 | 4 | **Question:** What are the connected components of the familiar spaces of functions between two (let's say compact and smooth, for simplicity) manifolds $M$ and $N$?
Specifically, I'm thinking of the Hölder spaces $\mathcal{C}^{k,\alpha}(M, N)$ and the Sobolev spaces $\mathcal{W}^{k,p}(M, N)$.
**Some comments:**
1. For a smooth function $f:M\to N$, it seems clear that, at least, all continuous functions homotopic to $f$ will be connected to it.
2. This question is inspired by the discussion of $\mathcal{W}^{k,p}(M, N)$ in McDuff-Salamon's book on $J$-holomorphic curves. There it's stated as an offhand remark that the connected components of $\mathcal{W}^{k,p}(M, N)$ (in the case of $M$ oriented & two-dimensional; I'm not sure if this makes a difference) are the completions of the sets {$f:M\to N \text{ smooth}: f\_\*[M]=A$}, for $A\in H\_{\dim M}(N)$.
3. If the McD-S factoid is true, there should exist sequences of smooth not-all-mutually-homotopic functions which converge in $\mathcal{W}^{k,p}(M, N)$. (This isn't too counterintuitive, since $\mathcal{W}^{k,p}(M, N)$ presumably contains functions which aren't continuous, & so don't themselves have a homotopy class). Can someone give me an example of this phenomenon?
Please feel free to re-tag -- I can't think of anything really appropriate.
| https://mathoverflow.net/users/2819 | Connected components of space of maps between two manifolds | Any continuous map from *M* to *N* is homotopic to a smooth map, and if two smooth maps are homotopic, then they are also smoothly homotopic. (More generally, two homotopic functions are homotopic through a homotopy that is smooth except at the endpoints.) The proof involves convolving with Gaussians, and is standard; I think you can find it in Milnor's *Topology from a Differentiable Viewpoint*, for instance. (It's also appeard on mathoverflow before, but I couldn't find it just now.) The hard issues for smooth vs. continuous functions arise only once you start demanding the maps be injective.
I can't say more about McDuff and Salomon without more context for the quote.
| 5 | https://mathoverflow.net/users/5010 | 23224 | 15,293 |
https://mathoverflow.net/questions/20263 | 8 | The ring of invariants $S^T$ of $k[a,b,c,d]$ under the algebraic torus action $T = k^{\*}$ with weights $(1,1,-1,-1)$
is $S = k[ac,ad,bc,bd]$ which has multigraded Hilbert series
$$
\frac{1 - abcd}{(1-ac)(1-ad)(1-bc)(1-bd)}.
$$
Is there a software package that can compute multigraded Hilbert series? Can it be computed using Macaulay2?
Alternatively, is there software that can compute the multigraded Hilbert series of a toric variety, specified by its fan?
For this example
$v\_1 = (0,0,1), v\_2 = (1,0,1), v\_3 = (1,1,1), v\_4 = (0,1,1)$ specify the vertices of the toric fan. The multigraded Hiblert series is given by the index which counts points in the dual cone $S\_{C^{\*}}$
$$
\sum\_{m \in S\_{C^{\*}}} q^m = \frac{(1 - q\_1)}{ (1 - q\_2)(1 - q\_3)(1 - q\_1 q\_2^{-1}) (1 - q\_1 q\_3^{-1}) }
$$
References:
Richard P. Stanley, "Linear Diophantine equations and local cohomology." Inventiones Mathematicae 68, 175-193 (1982), [MR0666158](http://www.ams.org/mathscinet-getitem?mr=MR0666158), [Zbl 0516.10009](https://zbmath.org/0516.10009).
Ezra Miller, Bernd Sturmfels, *Combinatorial commutative algebra.* Graduate Texts in Mathematics 227. New York, NY: Springer Verlag, pp. xiv, 417 (2005), ISBN 0-387-23707-0, [MR2110098](http://www.ams.org/mathscinet-getitem?mr=MR2110098), [Zbl 1090.13001](https://zbmath.org/1090.13001).
Dario Martelli, James Sparks, Shing-Tung Yau, "Sasaki-Einstein manifolds and volume minimisation." Communications in Mathematical Physics 280, No. 3, 611-673 (2008), [MR2399609](http://www.ams.org/mathscinet-getitem?mr=MR2399609), [Zbl 1161.53029](https://zbmath.org/1161.53029).
| https://mathoverflow.net/users/874 | Software for computing multi-graded Hilbert series | Macaulay 2 can do multigraded Hilbert series. Let's first assume that you have a presentation of your multigraded ring. I'll mention how to calculate this below. So for your $S = k[ac,ad,bc,bd]$, we'll write it as $S = k[x,y,z,w] / (xz - yw)$.
Assuming that each of $a,b,c,d$ has its own degree direction (so the grading is by ${\bf Z}^4$), we input $k[x,y,z,w]$ as
```
S = QQ[x,y,z,w, Degrees=> {{1,0,1,0}, {1,0,0,1}, {0,1,1,0}, {0,1,0,1}}]
```
where here QQ means the rationals. Then we want the Hilbert series of the ideal $(xz - yw)$, so we put
```
i8 : hilbertSeries ideal(y*z-x*w)
```
and the answer is:
```
1 - T T T T
0 1 2 3
o8 = ----------------------------------------
(1 - T T )(1 - T T )(1 - T T )(1 - T T )
1 3 1 2 0 3 0 2
```
If you also need to get the presentation, we can do this as follows. First, download normaliz:
<http://www.mathematik.uni-osnabrueck.de/normaliz/>
and the Macaulay 2 interface to normaliz (I think this is automatically there in Macaulay 2 1.3.1):
<http://www.math.uiuc.edu/Macaulay2/doc/Macaulay2-1.2/share/doc/Macaulay2/Normaliz/html/>
After loading the package with
```
loadPackage "Normaliz"
```
Set the path to normaliz in Macaulay 2 using the command
```
setNmzExecPath("path to the executables norm32 and norm64");
```
The [torusInvariants](http://www.math.uiuc.edu/Macaulay2/doc/Macaulay2-1.2/share/doc/Macaulay2/Normaliz/html/_torus__Invariants.html) command will give you generators for the subring of invariants. Finally, we can use the [ringmap](http://www.math.uiuc.edu/Macaulay2/doc/Macaulay2-1.3.1/share/doc/Macaulay2/Macaulay2Doc/html/___Ring__Map.html) command to define a surjection from a polynomial ring onto the subring to get the desired ideal.
| 11 | https://mathoverflow.net/users/321 | 23235 | 15,298 |
https://mathoverflow.net/questions/23240 | 7 | A commutative noetherian ring $R$ is Gorenstein if $R$ has finite injective dimension.
Obviously, if $R$ is Gorenstein, then $R$ localized at any prime ideal $P$ is also Gorenstein. But I don't know whether the converse holds?
| https://mathoverflow.net/users/5775 | Is there a non-Gorenstein ring but locally Gorenstein? | In all of the standard references I know, Gorenstein is either only defined for local rings or a not necessarily local ring is *defined* to be Gorenstein if all of its localizations at maximal ideals are Gorenstein (which implies that its localizations at all prime ideals are Gorenstein). So I am guessing you really want to ask: is there a commutative Noetherian ring $R$ all of whose localizations have finite injective dimension but such that $R$ itself does not?
I believe that the answer is "yes" and that a counterexample is given by Nagata's (in)famous example of a Noetherian ring of infinite Krull dimension. See (5.96) in Lam's *Lectures on modules and rings* for an explanation why Nagata's example is **regular**, hence locally Gorenstein, i.e., locally of finite injective dimension.
Furthermore, the proof of the Theorem at the bottom of p. 7 of
<http://www.math.hawaii.edu/~lee/homolog/Goren.pdf>
states that the injective dimension of a ring is the supremum of the injective dimensions of its local rings $R\_{\mathfrak{m}}$. In this case each $R\_{\mathfrak{m}}$ is regular, hence Gorenstein, hence its injective dimension is simply equal to its Krull dimension, i.e., to the height of $\mathfrak{m}$. It follows that the injective dimension of $R$ itself is infinite.
Conversely, Lee's handout contains a proof that the answer is "no" if $R$ is Noetherian of finite Krull dimension.
| 15 | https://mathoverflow.net/users/1149 | 23242 | 15,303 |
https://mathoverflow.net/questions/23204 | 7 | Suppose I have distinct real numbers $a\_i \in [-1,1]$, $i \in [k]$. I want to choose real numbers $b\_j, j\in [k]$ such that the matrix $(\arccos(a\_i b\_j))\_{i,j \in [k]}$ is nonsingular.
>
> Is this always possible? Equivalently, does $\arccos$ not satisfy a functional equation of the form $\det(\arccos(a\_i x\_j)) = 0$ for indeterminates $x\_j, j \in [k]$?
>
>
>
It seems like this should hold for any sufficiently ``non-algebraic'' function in place of $\arccos$. Is there a theory that handles such questions? Does functional analysis deal with these questions? I know very little functional analysis, so some general references may be helpful.
| https://mathoverflow.net/users/3318 | Nonexistence of determinantal functional equation for $\arccos$ | A quick counter-example to the question as stated is $a\_0=0$, $a\_1=1$ $a\_2=-1$. Since $2arccos(0)-arccos(b)-arccos(-b)=0$ for all $b$, we have $2M\_1-M\_2-M\_3=0$ where $M\_1, M\_2, M\_3$ are the rows of the matrix.
So it is better to assume that the $a\_i$ are nonnegative. In this case, the answer is yes. More generally, consider this problem for a function $f$. You want to choose $n$ linearly independent vectors from the set
$$
X := \{ (f(a\_1x), f(a\_2x),\dots, f(a\_nx)) \mid x\in\mathbb R \} \subset \mathbb R^n
$$
This is not possible if and only if $X$ lies in an $(n-1)$-dimensional subspace. This means that there are constants $c\_1,\dots,c\_n$ (not all zero) such that $c\_1v\_1+\dots+c\_nv\_n=0$ for all $v\in X$. Or, equivalently,
$$
c\_1 f(a\_1x) + \dots c\_n f(a\_nx) = 0
$$
for all $x\in\mathbb R$ (such that all $a\_ix$ belong to the domain of $f$). In other words, the functions $x\mapsto f(a\_ix)$ are linearly dependent over $\mathbb R$.
This cannot happen if $f=arccos$. Indeed, assuming that $a\_n$ is the maximum of $a\_i$ such that $c\_i\ne 0$, the above sum is well-defined and analytic on $[0,1/a\_n)$ but its derivative goes to infinity as $x$ approaches $1/a\_n$. Hence it is not constant on $[0,1/a\_n)$, and hence non-constant in any neighborhood of 0.
UPDATE. A similar argument shows that the answer is the same for any non-polynomial function analytic near the origin (assuming $a\_i>0$). Indeed, if $f(x)=\sum\_{j=1}^\infty q\_j x^{k\_j}$ where $q\_j\ne 0$, then Taylor expansion of the identity
$$
c\_1 f(a\_1x) + \dots c\_n f(a\_nx) = 0
$$
implies that $\sum\_i c\_i a\_i^{k\_j}=0$ for all $j$. This cannot happen because, if $a\_n$ is the maximum of $a\_i$, the term $a\_i^{k\_j}$ grows faster (or decays slower) than all other terms as $k\_j\to\infty$.
| 12 | https://mathoverflow.net/users/4354 | 23244 | 15,305 |
https://mathoverflow.net/questions/23246 | 13 | I've been going over the extremely interesting discussions about Axiom of Choice.
It looks to me like all the "weird" consequences of AC (Banach-Tarski etc) come from using it on uncountable collections of sets.
If, instead, we only believe the Axiom of Countable Choice, do we still get unintuitive consequences in the same sense ?
Apologies in advance if the question is vague.
| https://mathoverflow.net/users/4279 | Peculiar examples with Axiom of Countable Choice ? | If you assume the existence of suitable large cardinals, then $L(\mathbb{R})$ is a model of the Axiom of Determinacy $AD$ and the Axiom of Dependent Choice $DC$. In particular, since $DC$ is stronger than the Axiom of Countable Choice $AC\_{\omega}$, it follows that $AC\_{\omega}$ is also true in $L(\mathbb{R})$. Since $L(\mathbb{R})$ satisfies $AD$, all subsets $X \subseteq \mathbb{R}$ and maps $f: \mathbb{R} \to \mathbb{R}$ are measurable, etc. So it seems extremely unlikely that you will find any unintuitive consequences of $AC\_{\omega}$ in the more classical areas of mathematics.
| 15 | https://mathoverflow.net/users/4706 | 23249 | 15,306 |
https://mathoverflow.net/questions/23247 | 8 | Let $\mathrm{Hol}^d(\Sigma, \mathbb{C} \mathbb{P}^n)$ denote the space of holomorphic maps of degree $d$ from a Riemann surface $\Sigma$ to complex projective space of dimension $n$. Let $\mathrm{HolEmb}^d(\Sigma, \mathbb{C} \mathbb{P}^n)$ denote the subspace of those holomorphic maps which are also embeddings, so there is an inclusion
$$i : \mathrm{HolEmb}^d(\Sigma, \mathbb{C} \mathbb{P}^n) \longrightarrow \mathrm{Hol}^d(\Sigma, \mathbb{C} \mathbb{P}^n).$$
If we were discussing smooth maps, instead of holomorphic, Whitney's embedding theorem would say that this map is approximately $(\frac{n}{2}-2)$-connected. Is there a connectivity range for this map of spaces of holomorphic maps?
| https://mathoverflow.net/users/318 | Approximating holomorphic maps by holomorphic embeddings | I will assume that *d* is the degree of the pull-back of $\mathcal{O}(1)$ to $\Sigma$ and that it is sufficiently large with respect to the genus *g* of $\Sigma$. In this case, the dimension of the space of holomorphic maps of $\Sigma$ in $\mathbb{P}^n$ is
$$
D\_n := (n+1)d + n(1-g) ,
$$
while the dimension of the space of holomorphic maps that are not isomorphisms onto their image is
$$
(n+1)d + n(1-g) -n+2 = D\_n -(n-2) .
$$
In particular, since the inclusion that you are interested in has complement of (complex) codimension *n-2*, it follows that it is "quite connected", roughly *(2n-1)*-connected?
Unless I made some mistakes in my computations, the estimates for the dimensions above are only valid if *d* is sufficiently large, otherwise they should simply be lower bounds on the actual dimensions of the spaces. In the case of *n=2* you obviously must allow singularities in the image, but you can also prove that the locus where the morphism is not a local embedding (i.e. when the derivative is not injective somewhere) has codimension one.
| 4 | https://mathoverflow.net/users/4344 | 23251 | 15,307 |
https://mathoverflow.net/questions/23254 | 1 | Does it exist a Lie algebra $\mathfrak{g}$ that is reductive but if we consider the inclusion of Lie agebras $\mathfrak{g} \subset \mathfrak{h}$ then $\mathfrak{g}$ is not reductive in $\mathfrak{h}?$
| https://mathoverflow.net/users/4821 | Reductive Lie algebra | A reductive Lie algebra $L$ is the direct sum of a semisimple
Lie algebra $L\_1$ and an abelian Lie algebra $L\_2$. Let's consider
the case where $L\_2$ is one-dimensional.
We can embed $L$ into a larger Lie algebra $L=L\_1\oplus L\_2'$
by embedding $L\_2$ into $L\_2'$. Let $L\_2'$ be the two-dimensional
Lie subalgebra
$$\left(\begin{array}{cc}
\*& \*\\\
0& 0
\end{array}
\right)$$
of $\mathfrak{gl}(\mathbf{C})$
and
$$L\_2=\left(\begin{array}{cc}
0& \*\\\
0& 0
\end{array}
\right).$$
Then $L\_2$ does not act semisimply on $L\_2'$, so $L$ does not
act semisimply on $L'$.
| 4 | https://mathoverflow.net/users/4213 | 23255 | 15,308 |
https://mathoverflow.net/questions/22699 | 4 | I would like to understand the theory of pure complexes of (etale?) sheaves (of geometric origin?). In particular, I would like to understand which conditions are realy necessary in (part 1 of) Theorem 3.1.8 of Cataldo-Migliorini's survey <http://www.ams.org/journals/bull/2009-46-04/S0273-0979-09-01260-9/S0273-0979-09-01260-9.pdf>
(see page 33-567). Does the splitting of part 1 (only!) really requires $\overline{\mathbb{Q}\_l}$-coefficients? Which coefficients could be put here? Does the splitting exist over $X\_0$?
Below they explain that those extensions of mixed complexes $K\_0,L\_0$ of appropriate weights that come from $\mathbb{F}\_q$, become zero over $\mathbb{F}$. My main question is: does there exist a triangulated category of complexes of sheaves where the corresponding Ext-group is zero from the beginning (i.e. we consider $Ext^1$ in a single triangulated category instead of the image $Ext^1(K\_0.L\_0)\to Ext^1(K,L)$). Is there such a category over a (more or less) general base scheme $S$ (instead of $\mathbb{F}\_q$ or $\mathbb{F}$)? Again, which coefficient rings are possible here?
| https://mathoverflow.net/users/2191 | Morphisms between pure complexes of sheaves | Dear Mikhail,
I had been hoping someone else would attempt to answer this question, as I have been wondering very similar things lately. (In fact I drove myself crazy for about a month last year trying to work out some solution to what you are asking in the second paragraph.)
I can't answer everything but here is a start:
Write $K = \overline{\mathbb{Q}\_{\ell}}$. One already sees the problems with what you are asking for $X\_0 = Spec \mathbb{F}\_q$. Then the category of constructible $K$-sheaves on $X\_0$ is equivalent to the category of finite dimensional $K$-representations of $Gal(\mathbb{F}/\mathbb{F}\_q)$, the absolute Galois group of $\mathbb{F}\_q$. (The absolute Galois group is generated topologically by Frobenius, and so this is the same as giving a finite dimensional $K$-vector space together with an endomorphism.) [See BBD 5.1.11 for a statement, I think this is explained in Milne, but don't have it at the moment.]
Now, a pure sheaf on $X\_0$ is pure of weight $i$ if all the eigenvalues of Frobenius are algebraic integers all of whose complex conjugates over $\mathbb{C}$ have the same absolute value $q^{i/2}$.
Note that here we already see that over $X\_0$ a pure sheaf does not need to be semi-simple. Indeed, there is no reason why Frobenius should act semi-simply. (This is one example of what de Cataldo and Migliorini are talking about in Remark 3.1.9 after the Theorem 3.1.8.) I think it is part of the standard conjectures that Frobenius acts semi-simply on the $\ell$-adic cohomology of smooth projective varieties, which as I understand it, is still not known.
I don't know what you mean when you ask:
>
> Does the splitting of part 1 (only!) really requires $\overline{\mathbb{Q}\_{\ell}}$-coefficients?
>
>
>
As to your main question, I think that the above example shows that this is too much to hope. Without working over $X\_0$ one cannot define what it means to be mixed, and without going to $X\_0$ one can't expect the same ext vanishing.
I recently discovered your work on "weight structures" and found it very interesting. I guess you are asking the above, because you would like to argue that one gets a weight structure in the setting of $\overline{\mathbb{Q}\_{\ell}}$-sheaves.
There is one setting where I think that one really does get a weight structure. This is in the (at least formally) very similar world of "mixed Hodge modules" on complex varieties. There one has the desired ext vanishing from the outset.
| 3 | https://mathoverflow.net/users/919 | 23257 | 15,310 |
https://mathoverflow.net/questions/23229 | 23 | The Bohr-Mollerup theorem states that the Gamma function is the unique function that satisfies:
1) f(x+1) = x\*f(x)
2) f(1) = 1
3) ln(f(x)) is convex
The Gamma function is meant to interpolate the factorial function, so I can see the importance of the first two properties. But why is log convexity important? How does it affect the Gamma function's applicability in other areas of mathematics?
| https://mathoverflow.net/users/5768 | Importance of Log Convexity of the Gamma Function | First, let me mention that log convexity of a function is implied by an analytic property, which appears to be more natural than log convexity itself. Namely, if $\mu$ is a Borel measure on $[0,\infty)$ such that the $r$th moment
$$f(r)=\int\_{0}^{\infty}z^r d\mu(z)$$
is finite for all $r$ in the interval $I\subset \mathbb R$, then $\log f$ is convex on $I$.
Log convexity can be effectively used in derivation of various inequalities involving the gamma function (particularly, two-sided estimates of products of gamma functions). It is linked with the notion of [Schur convexity](http://en.wikipedia.org/wiki/Majorization) which is itself used in many applications.
**An appetizer.** Let $m=\max x\_i$, $s=\sum x\_i$, $x\_i > 0$, $i = 1,\dots,n$, then
$$[\Gamma(s/n)]^n\leq\prod\limits\_{1}^{n}\Gamma (x\_i)\leq \left[\Gamma\left(\frac{s-m}{n-1}\right)\right]^{n-1}\Gamma(m).\qquad\qquad\qquad (1)$$
(1) is trivial, of course, when all $x\_i$ and $s/n$ are integers, but in general the bounds do not hold without assuming log convexity.
**Edit added: a sketch of the proof**. Let $f$ be a continuous positive function defined on an interval $I\subset \mathbb R$. One may show that the function $\phi(x)=\prod\limits\_{i=1}^{n}f(x\_i)$, $x\in I^n$ is Schur-convex on $I^n$ if and only if $\log f$ is convex on $I$. Thus the function
$$\phi(x)=\prod\limits\_{i=1}^n \Gamma(x\_i),\quad x\_i>0,\qquad \quad\qquad\qquad\qquad\qquad\qquad\quad (2)$$
is Schur-convex on $I^n=(0,\infty)^n$. Since $x\_i\le m$, $i=1,\dots,n$, and $\sum x\_i=s$, it is easy to check that
$$x \prec \left(\frac{s-m}{n-1},\dots,\frac{s-m}{n-1},m\right).$$
The latter [majorization](http://en.wikipedia.org/wiki/Majorization) and the fact that $\phi(x)$ defined by (2) is Schur-convex imply the upper bound (1). The lower bound follows from the standard majorization $x\succ (s/n,\dots,s/n)$.
---
Have a look at [the recent short article](http://www.springerlink.com/content/k4285g1811744605/) by Marshall and Olkin concerning this and related inequalities for the gamma function.
| 15 | https://mathoverflow.net/users/5371 | 23262 | 15,312 |
https://mathoverflow.net/questions/23221 | 11 | Is the following result true: the Hilbert space $\ell^{2}\left(2^{\Gamma}\right)$ is a quotient of $\ell^{\infty}\left(\Gamma\right)$ for any
uncountable $\Gamma$ ? [I think it is, but cannot remember where I saw it, long time ago.] I would be very grateful for any (freely available, if possible) reference (Pelczynski ? Rosenthal ?).
| https://mathoverflow.net/users/2508 | Nonseparable Hilbert spaces as quotients of spaces of bounded functions | I don't know who first observed this (maybe Archimedes?) but it is true because $C(\{0,1 \}^\Gamma)$ is a quotient of $\ell\_1^\Gamma$ and hence $\ell\_1(2^\Gamma)$ embeds into $\ell\_\infty(\Gamma)$.
@Ady
Here is a more serious answer to your question. Take a quotient map $Q$ from $\ell\_1(2^\Gamma)$ onto $C([0,1]^{2^\Gamma})$ and extend to a norm one mapping $T$ from $\ell\_\infty(\Gamma)$ into some injective space $Z$ that contains $C([0,1]^{2^\Gamma})$ (you cannot extend $Q$ to an operator from $\ell\_\infty(\Gamma)$ into $C([0,1]^{2^\Gamma})$ because, e.g., $C([0,1]$ is not a quotient of $\ell\_\infty$). Use partitions of unity to get a net $(P\_a)$ of norm one finite rank projections on $Z$ taking values in $C([0,1]^{2^\Gamma})$ and whose restrictions to $C([0,1]^{2^\Gamma})$ converge strongly to the identity. A weak$^\*$ cluster point of $(P\_a^\* T^\*)$ gives an isometric embedding of the dual of $C([0,1]^{2^\Gamma})$ (which contains $L\_1([0,1]^{2^\Gamma})$) into the dual of $\ell\_\infty(\Gamma)$. Thus if $Y^\*$ is any reflexive subspace of $L\_1([0,1]^{2^\Gamma})$, such as $\ell\_2(2^\Gamma)$, then $Y$ is isometric to a quotient of $\ell\_\infty(\Gamma)$.
| 8 | https://mathoverflow.net/users/2554 | 23273 | 15,319 |
https://mathoverflow.net/questions/23268 | 38 | I'm the sort of mathematician who works really well with elements. I really enjoy point-set topology, and category theory tends to drive me crazy. When I was given a bunch of exercises on subjects like limits, colimits, and adjoint functors, I was able to do them, although I am sure my proofs were far longer and more laborious than they should have been. However, I felt like most of the understanding I gained from these exercises was gone within a week. I have a copy of MacLane's "Categories for the Working Mathematician," but whenever I pick it up, I can never seem to get through more than two or three pages (except in the introduction on foundations).
A couple months ago, I was trying to use the statements found in Hartshorne about glueing schemes and morphisms and realized that these statements were inadequate for my purposes. Looking more closely, I realized that Hartshorne's hypotheses are "wrong," in roughly the same way that it is "wrong" to require, in the definition of a basis for a topology that it be closed under finite intersections. (This would, for instance, exclude the set of open balls from being a basis for $\mathbb{R}^n$.) Working through it a bit more, I realized that the "right" statement was most easily expressed by saying that a certain kind of diagram in the category of schemes has a colimit. At this point, the notion of "colimit" began to seem much more manageable: a colimit is a way of gluing objects (and morphisms).
However, I cannot think of any similar intuition for the notion of "limit." Even in the case of a fibre product, a limit can be anything from an intersection to a product, and I find it intimidating to try to think of these two very different things as a special cases of the same construction. I understand how to show that they are; it just does not make intuitive sense, somehow.
For another example, I think (and correct me if I am wrong) that the sheaf condition on a presheaf can be expressed as stating that the contravariant functor takes colimits to limits. [This is not correct as stated. See Martin Brandenburg's answer below for an explanation of why not, as well as what the correct statement is.] It seems like a statement this simple should make everything clearer, but I find it much easier to understand the definition in terms of compatible local sections gluing together. I can (I think) prove that they are the same, but by the time I get to one end of the proof, I've lost track of the other end intuitively.
Thus, my question is this: Is there a nice, preferably geometric intuition for the notion of limit? If anyone can recommend a book on category theory that they think would appeal to someone like me, that would also be appreciated.
| https://mathoverflow.net/users/5094 | Geometric intuition for limits | I pick up your remarks about sheaves. Indeed, the sheaf condition is a very good example to get a geometric idea of a limit.
Assume that $X$ is a set and $X\_i$ are subsets of $X$ whose union is $X$. Then it is clear how to characterize functions on $X$: These are simply functions on the $X\_i$ which agree on the overlaps $X\_i \cap X\_j$. This can be formulated in a fancy way: Let $J$ be the category whose objects are the indices $i$ and pairs of such indices $(i,j)$. It should be a preorder and we have the morphisms $(i,j) \to i, (i,j) \to j$. Consider the diagram $J \to Set$, which is given by $i \mapsto X\_i, (i,j) \mapsto X\_i \cap X\_j$. What we have remarked above says exactly that $X$ is the colimit of this diagram! In a similar fashion, open coverings can be understood as colimits in the category of topological spaces, ringed spaces or schemes. It's all about gluing morphisms.
Now what about limits? I think it is important first to understand limits in the category of sets. If $F : J \to Set$ is a small diagram, then we can consider simply the set of "compatible elements in the image" of $F$, namely
$X = \{x \in \prod\_j F(j) : \forall i \to j : x\_j = F(i \to j)(x\_i)\}$.
A short definition would be $X = Cone(\*,F)$. Observe that we have projections $X \to F(j), x \mapsto x\_j$ and with these $X$ is the limit of $F$. Now the Yoneda-Lemma or just the definition of a limit tells you how you can think of a limit in an arbitrary category: That $X$ is a limit of a diagram $F : J \to C$ amounts to say that elements of $X$ .. erm we don't have any elements, so let's say morphisms $Y \to X$, naturally correspond to compatible elem... erm morphisms $Y \to F(i)$. In other words, for every $Y$, $X(Y)$ is the set-theoretic limit of the diagramm $F(Y)$. I hope that this makes clear that the concept of limits in arbitrary categories is already visible in the category of sets.
Now let $X$ be a topological space and $O(X)$ the category of open subsets of $X$; it's an preorder with respect to the inclusion. Thus a presheaf is just a functor $F$ from $O(X)^{op}$ to the category of sets (or which suitable category you like). Now open coverings can be described as certain limits in $O(X)^{op}$, i.e. colimits in $O(X)$, as above. Observe that $F$ is a sheaf if and only if $F$ preserves these limits: If $U$ is covered by $U\_i$, then $F(U)$ should be the limit of the $F(U\_i), F(U\_i \cap U\_j)$ with transition maps $F(U\_i) \to F(U\_i \cap U\_j), F(U\_j) \to F(U\_i \cap U\_j)$, i.e. $F(U)$ consists of compatible elements of the $F(U\_i)$, meaning that the elements of $F(U\_i)$ and $F(U\_j)$ restrict to the same element in $F(U\_i \cap U\_j)$. Thus we have a perfect geometric example of a limit: the set of sections on an open set is the limit of the set of sections on the open subsets of a covering.
Somehow this view takes over to the general case: Let $F : J \to Set$ be a functor. Regard it as a presheaf on $J^{op}$, and the map induced by $i \to j$ in $J^{op}$ as a restriction $F(j) \to F(i)$. Also call the elements of $F(i)$ sections on $i$. Then the limit of $F$ consists of compatible sections. Since I've been learning algebraic geometry, I almost always think of limits in this way.
Finally it is important to remember that limit is just the dual concept of colimit. And often algebra and geometry appear dually at once, for example sections and open subsets in sheaves. If $(X\_i,\mathcal{O}\_{X\_i})$ are ringed spaces and you want to find the colimit, well you can guess that you *have* to do: Take the colimit of the $X\_i$ and the limit of the $\mathcal{O}\_{X\_i}$ (pullbacked to the colimit).
>
> "...the sheaf condition on a presheaf can be expressed as stating that the contravariant functor takes colimits to limits"
>
>
>
This is not correct. The reason is that the index category can be rather wild and colimits in preorders don't care about that. In detail: Let $U : J \to O(X)^{op}$ be a small diagram. Then the limit is just the union $V$ of $U\_j$. Thus $F$ preserves this limit iff sections on $V$ are sections on the $U\_j$ which are compatible with respect to the restriction morphisms given by $U$. If $J$ is discrete and $U$ maps everything to the same open subset $V$ of $X$, then the compatible sections are $F(V)^J$, which is bigger than $F(V)$.
>
> "... I have a copy of MacLane's "Categories for the Working Mathematician," but whenever I pick it up, I can never seem to get through more than two or three pages (except in the introduction on foundations"
>
>
>
I think this book is still one of the best introductions into category theory. It can be hard to grasp all these abstract concepts and examples, but it gets easier as soon as you get input from other areas where category theoretic ideas are omnipresent. Your example about gluing morphisms illustrates this very well.
| 33 | https://mathoverflow.net/users/2841 | 23276 | 15,321 |
https://mathoverflow.net/questions/23278 | 10 | It seems the notion of tensor product of abelian categories exists naturally.
Does someone know the reference of the construction?
| https://mathoverflow.net/users/5082 | Tensor product of abelian categories | Deligne's article, 'Categories Tannakiennes,' section 5 would be a good place to look. It was published in the Grothendieck Festschrift, vol. 2.
| 11 | https://mathoverflow.net/users/373 | 23284 | 15,325 |
https://mathoverflow.net/questions/22533 | 13 | Let $\mathbf{Q}\_p$ denote the field of $p$-adic numbers.
Suppose that $K/\mathbf{Q}\_p$ is a finite extension, and let $O\_K$ denote the ring of integers of $K$. Suppose that $X$ is proper over $O\_K$, with smooth generic fibre. Consider the following three statements:
A. There exists a finite extension $L/K$ such that $X/L$ has a semi-stable model over $O\_L$.
B. There exists a finite extension $L/K$ such that, for every proper etale map $Y \rightarrow X$, the etale cohomology $H^i(Y/\overline{K},\mathbf{Q}\_p)$ is semi-stable when considered as a representation of $\mathrm{Gal}(\overline{K}/L)$.
C. There exists a finite extension $L/K$ such that $H^i(X/\overline{K},\mathbf{Q}\_p)$ is semi-stable when considered as a representation of $\mathrm{Gal}(\overline{K}/L)$.
In light of Tsuji's proof of the semi-stable conjecture of Fontaine and Jannsen, we know that $A \Rightarrow B \Rightarrow C$ (the point being that a semi-stable model for $X$ pulls back to one for $Y$). On the other hand, Tsuji also proved $C$ without proving $A$ by using de Jong's theory of alterations.
My question: Can one also use Tsuji's arguments to prove $B$? One could imagine some argument with alterations being compatible with taking etale covers, but this is not my field so I would rather ask an expert.
| https://mathoverflow.net/users/nan | Potential semi-stability of etale cohomology of etale covers. | Answered to my satisfaction. Well, except for the bit about typesetting $\mathbf{Q}\_p$, that still confuses me.
| 4 | https://mathoverflow.net/users/nan | 23289 | 15,329 |
https://mathoverflow.net/questions/23228 | 9 | Consider critical edge-percolation in the induced subgraph of the square grid with vertex set {$(i,j) \in Z \times Z:\ i+j \geq 0$}, and let $p\_n$ be the probability that the cluster containing $(0,0)$ has size $> n$. How quickly does $p\_n$ fall as $n \rightarrow \infty$?
| https://mathoverflow.net/users/3621 | half-plane percolation clusters | As Leandro suggested in the comments, this should follow a power-law decay in $n$. However, Hara and Slade's rigorous work using lace expansions is only valid for dimensions $\ge 19$. Much of the rigorous work on critical exponents for two-dimensional percolation has been done only recently, via connections to Schramm-Loewner Evolution (with $\kappa = 6$). A good starting place might be [this PowerPoint presentation by Oded Schramm](http://research.microsoft.com/en-us/um/people/schramm/memorial/victoria.pdf). Here is Page 23:
> Physicists have predicted some exponents describing asymptotics of critical percolation in 2D.
For example, Nienhuis conjectured that the probability that the origin is in a cluster of diameter $\ge R$ is $$R^{−5/48+o(1)}, \qquad R \to \infty$$ and Cardy conjectured that the probability that the origin is connected to distance $R$ within the upper half plane is $$R^{−1/3+o(1)}, \qquad R \to \infty.$$
| 4 | https://mathoverflow.net/users/238 | 23291 | 15,331 |
https://mathoverflow.net/questions/23285 | 1 | I am trying to plot the pdf of flipping heads when drawing from a bag of biased coins. Since I am interested in the % of heads flipped, not the number, I simulate 500K flips and group the results into buckets of 100, computing the % of heads in each bucket. I do this using two different methods, in method 1 I draw a coin per bucket, in method 2 I draw a coin per flip. They come up with the same mean, but the variance of the two methods are different. In fact, method 2 basically removes the variance, i.e. if I have a bag of coins with P of heads being [0.3, 0.35, 0.4, 0.45, 0.5] method 2 has simulation results that are equivalent to just having a bag with a coin with p 0.4 of flipping heads.
Why does method 2 (i.e. a draw from the bag per flip) remove the variance associated with the bag of biased coins?
See the following histograms of the results: <http://img208.imageshack.us/img208/111/biasedcoin.png> (I'm a new user and the site won't let me link this inline)
And the code that generated these:
```
#!/usr/bin/env python
import random as rand
import numpy as np
import matplotlib
matplotlib.use('TkAgg')
import matplotlib.mlab as mlab
import matplotlib.pyplot as plt
#biased_coin_bag = [0.4]
biased_coin_bag = [0.3, 0.35, 0.4, 0.45, 0.5]
Heads = 'H'
Tails = 'T'
def draw_coin():
return biased_coin_bag[rand.randint(0, len(biased_coin_bag)-1)]
def flip_coin(bc):
if bc >= rand.random():
return Heads
else:
return Tails
def histogram(iters=500000, bucket_size=100):
results_per_flip = []
heads_per_flip, tails_per_flip = 0, 0
results_per_bucket = []
heads_per_bucket, tails_per_bucket = 0, 0
bc_per_bucket = draw_coin()
for i in range(iters):
if i != 0 and i % bucket_size == 0:
#
# track the bucketized results when drawing per bucket
#
results_per_bucket.append(float(heads_per_bucket) /
float(heads_per_bucket + tails_per_bucket))
heads_per_bucket, tails_per_bucket = 0, 0
bc_per_bucket = draw_coin()
#
# track the bucketized results when drawing per flip
#
results_per_flip.append(float(heads_per_flip) /
float(heads_per_flip + tails_per_flip))
heads_per_flip, tails_per_flip = 0, 0
#
# flip the coins
#
if flip_coin(bc_per_bucket) == Heads:
heads_per_bucket += 1
else:
tails_per_bucket += 1
if flip_coin(draw_coin()) == Heads:
heads_per_flip += 1
else:
tails_per_flip += 1
#
# plot the draw per bucket results
#
plt.subplot(211)
mu_per_bucket = np.average(results_per_bucket)
sigma_per_bucket = np.std(results_per_bucket)
n, bins, patches = plt.hist(results_per_bucket, 40, normed=1, color='green',
alpha=0.75,
label=r'draw per bucket $\mathrm{\mu=%f\ \sigma=%f}$' %
(mu_per_bucket, sigma_per_bucket))
y = mlab.normpdf(bins, mu_per_bucket, sigma_per_bucket)
plt.plot(bins, y, 'r--', linewidth=1)
plt.legend()
plt.axis([0, 1, 0, 20])
#
# plot the draw per flip results
#
plt.subplot(212)
mu_per_flip = np.average(results_per_flip)
sigma_per_flip = np.std(results_per_flip)
n, bins, patches = plt.hist(results_per_flip, 40, normed=1, color='blue',
alpha=0.75,
label=r'draw per flip $\mathrm{\mu=%f\ \sigma=%f}$' %
(mu_per_flip, sigma_per_flip))
y = mlab.normpdf(bins, mu_per_flip, sigma_per_flip)
plt.plot(bins, y, 'r--', linewidth=1)
plt.legend()
plt.axis([0, 1, 0, 20])
plt.show()
histogram()
```
| https://mathoverflow.net/users/5783 | simulating chances of success when drawing from a bag of biased coins | You seem to answer the question yourself in the last sentence of your first paragraph. If you draw a new coin every flip, then the experiment is equivalent to a sequence of coin tosses with a single coin (with appropriately chosen probability).
Maybe helpful to think about independence. If you draw a new coin every flip, then the results of the different flips are all independent. So all that matters is the probability of a head on any given flip (and that is just a single number - 0.4 in your example).
If you reuse the same coin for several flips, then the results of these flips are no longer independent, and the form of the dependence between them is important.
Put another way - there are two contributions to the variance of the total number of heads. One comes from the variance of individual flips. The other comes from the covariance between pairs of flips. This comes from a simple formula like $Var (\sum\_{i=1}^n X\_i)
= \sum\_{i=1}^n Var (X\_i) + 2\sum\_{1\leq i<j\leq n} Cov(X\_i,X\_j)$.
The variance of any individual flip is the same in both your experiments. When you draw a new coin every time, the covariance between any pair of flips is 0. However, when you reuse the same coin for different flips, the covariance between two flips which use the same coin is positive. So the total variance is higher.
Try thinking about an extreme example. Suppose the bag contains two coins, one with probability 0 of heads and the other with probability 1 of heads. If you choose a coin from the bag and flip it N times, you will see either 0 or N heads. On the other hand, if you make N different draws and one flip after each of them, you are likely to see approximately N/2 heads. The variance is $N^2/4$ in the first case and $N/4$ in the second.
| 3 | https://mathoverflow.net/users/5784 | 23296 | 15,332 |
https://mathoverflow.net/questions/23295 | 2 | Hi all,
I have been looking at complex multiplication of elliptic curves for a course project in cryptography and the following question came up: Let $\mathcal{O}\_K$ be the maximal order in $K$ ($K$ is an imaginary quadratic field), let $h\_K (X)$ be the Hilbert class polynomial of $K$. Suppose that $\mathcal{O}$ is another order (say $\mathcal{O} =\mathbb{Z}[ \frac{1 + \sqrt{D}}{2}]$ and $\mathcal{O} = \mathbb{Z}[\sqrt{D}]$ for concreteness). Let $h\_\mathcal{O} (X)$ be the hilbert class polynomial of the order $\mathcal{O}$. Is there any relation between $h\_k(X)$ and $h\_\mathcal{O} (X)$? For example can one obtain $h\_\mathcal{O}(X)$ from $h\_K(X)$ and vice verse?
| https://mathoverflow.net/users/2917 | Relation between the Hilbert Class polynomial of $\mathcal{O}_K$ and an order. | As far as I know, the best relation between the two is the following: the field generated by the hilbert class polynomial $h\_\mathcal{O} (X)$ contains the field generated by $h\_K(X)$. This is implied by Proposition 25 of [this paper](http://alpha.math.uga.edu/~pete/torspaper_FINAL.pdf).
This implies among other things that $\deg(h\_K(X)) | \deg(h\_\mathcal{O}(X))$ (although this could be determined by simpler means).
Now as to your question about whether one can be generated from the other? No, unless you're in a very limited set of circumstances like $\deg(h\_\mathcal{O}(X)) =1$ or such a thing. In fact it's a celebrated theorem of Heilbronn that $\deg(h\_\mathcal{O}(X)) \to \infty$ as $|D| \to \infty$ where $D$ is the discriminant of $\mathcal{O}$.
| 3 | https://mathoverflow.net/users/3384 | 23302 | 15,334 |
https://mathoverflow.net/questions/23253 | 2 | Has there been a survey written on the model theory of first-order (non-)Euclidean geometry in the spirit of Hilbert and Tarski? I'm especially interested in two aspects of the model theory:
1. Countable models
2. Distinguished/canonical models.
I'm interested in 2 because I don't see why the standard ${\mathbb{R}}^2$ model is any special.
| https://mathoverflow.net/users/nan | Reference: Countable Models of (Non-)Euclidean Geometry | Marvin Jay Greenberg got very interested in the foundations for the fourth edition(2007) of his book. This led to a survey article in the March 2010 M.A.A. Monthly. Table of contents:
<http://www.maa.org/pubs/monthly_mar10_toc.html>
It does not seem to say explicitly on the link, volume 117, number 3, March 2010, pages 198-219
I have a pdf of the article if you have no better way to look at it, just email me. Marvin sent me a copy because my results are in it, however I was not doing genuine foundations.
The other book is by the well known R. Hartshorne, called Geometry:Euclid and Beyond (2000)
That's a pretty good start, article and two books. You may also want to look up Victor Pambuccian on MathSciNet
To summarize the bits I expect you find most interesting, Hilbert gave a recipe for defining the hyperbolic plane first and finding the underlying field second, this is called the "field of ends" in English. I prefer the "Euclidean" fields, if there is a field element $a > 0$ then $\sqrt a$ is also in the field. The smallest field that does everything I find interesting is the "constructible field" which is all numbers arrived at by starting with the rationals and taking a finite number of square roots, mixed with other field operation of course. This is a subfield of the algebraic numbers. Hilbert himself concentrated on the milder Pythagorean fields, if $a \in F$ then $ \sqrt{1 + a^2} \in F.$ So there may be positive field elements without square roots. As there is an intermediate step that usually requires interpreting the standard hyperbolic functions, Hartshorne wrote that entire section with a "multiplicative length" for segments, which corresponds to taking $e^x$ when $x$ is an ordinary length in a plane over the reals. Personally, I prefer $\sinh x$ because of the appearance of the (hyperbolic) Pythagorean Theorem and the integral triangles you get this way, but that is never going to be popular.
Well, I know a ton more about this at secondhand, you might say. Let me know if you need more info to get started.
| 1 | https://mathoverflow.net/users/3324 | 23304 | 15,336 |
https://mathoverflow.net/questions/23306 | 5 | Suppose you have a 2-dimensional polyhedral surface with specified lengths for the edges so that the vertices all have positive curvature. I believe this has a unique isometric embedding into 3-dimensional Euclidean space as the boundary of a convex polyhedron. Could someone confirm this? If so, is there a reasonable algorithm for finding the isometric embedding computationally?
| https://mathoverflow.net/users/613 | Isometric embedding of a positively curved polyhedral surface | Yes, this is the historically first version of the Alexandrov embedding theorem which was earlier discussed [here](https://mathoverflow.net/questions/22122/on-alexandrov-embedding-theorem). See the original Alexandrov's monograph (reviewed [here](http://www.math.cornell.edu/~connelly/alexandrov.pdf)), or see [my book](http://www.math.ucla.edu/~pak/book.htm) if you like downloadable stuff (at least for now).
The algorithmic issue is difficult. If you want the usual complexity, in my paper with Fedorchuk we [showed](http://www.math.ucla.edu/~pak/papers/pp16.pdf) that this requires finding roots of polynomials with exponential degrees. The practical algorithm (based on a new completely different proof of the Alexandrov theorem) was found by Bobenko and Izmestiev [here](http://arxiv.org/abs/math/0609447) and further analyzed [here](http://arxiv.org/abs/0812.5030).
| 11 | https://mathoverflow.net/users/4040 | 23309 | 15,338 |
https://mathoverflow.net/questions/23297 | 42 | I'm looking for basic examples that show the usefulness of spectral sequences even in the simplest case of spectral sequence of a filtered complex.
All I know are certain "extreme cases", where the spectral sequences collapses very early to yield the acyclicity of the given complex or some quasi-isomorphism to another easier complex (balancing tor, for example).
Is there an example of a useful filtration where one really computes something nontrivial also in the higher sheets?
The examples I have in mind come from topology. For example, the calculation of $H\_{\ast}(\Omega{\mathbb S}^n;{\mathbb Z})$ is simply beautiful using the Serre spectral sequence, and one needs to pass to the $n$-th sheet until something happens. Another more difficult example would be the computation of the rational cohomology of $K({\mathbb Z},n)$ by induction on $n$ (depending on the parity of $n$, we get a polynomial algebra or an exterior algebra, if I remember correctly).
Are there similar, but purely algebraic examples which could show the usefulness of spectral sequences to those seeing them the first time?
| https://mathoverflow.net/users/3108 | Simple examples for the use of spectral sequences | This isn't exactly what you asked, but its a very simple example that (to me) demonstrates some of the necessity of the complexities of spectral sequences. Consider the ring $R=\mathbb{C}[x,y]$, and consider the module $M=(Rx+Ry)\oplus R/x$. Then the double dual spectral sequence converges to the original module:
$$ Ext^{-i}\_R(Ext^j\_R(M,R),R) \Rightarrow M $$
The second page of this spectral sequence has
* $R$ in degree $(0,0)$
* $R/x$ in degree $(-1,1)$
* $R/(Rx+Ry)$ in degree $(-1,2)$
* A non-trivial knights-move map (differential on the second page) from $(0,0)$ to $(-1,2)$ which is the natural quotient map.
The spectral sequence collapses on the third page, with $(Rx+Ry)$ in degree $(0,0)$ and $R/x$ in degree $(-1,1)$. One shortcoming of this example is that you get the same module back, split apart into different components; rather than the associated graded of some interesting filtration. If memory serves, there was a way to tinker with this example to give it that property too, but it escapes me at the moment.
| 9 | https://mathoverflow.net/users/750 | 23316 | 15,344 |
https://mathoverflow.net/questions/23277 | 2 | While studying some class field theory there was a lot of talk on galois extensions. Of course. When talking about non-galois number fields, usually the text will quickly take the galois closure. At this point it occurred to me that this implies many properties are shared by number fields with the same galois closure.
So splitting of primes and ramification are controlled by the galois closure. But what other properties are shared?
Class group, unit group, higher cohomology and K-groups?
| https://mathoverflow.net/users/2024 | Properties shared by number fields with the same normal closure? | There's a huge amount of literature on this problem starting with
* F. Gassmann, *Über Beziehungen zwischen den Primidealen eines algebraischen Körpers und den Substitutionen seiner Gruppen*, Math. Z. 25, 661-675 (1926)
Gassmann constructed number fields with the same normal closure in which almost all primes split in the same way. Later, the question morphed into "do zeta functions determine the number field", for example in
* B. de Smit, Bart; R. Perlis,
*Zeta functions do not determine class numbers*,
Bull. Am. Math. Soc., New Ser. 31, No.2, 213-215 (1994)
There's actually a whole book out there on this problem:
* N. Klingen, *Arithmetical similarities. Prime decomposition and finite
group theory*, Oxford (1998)
where you can find the relevant literature up to the 1990s.
| 5 | https://mathoverflow.net/users/3503 | 23317 | 15,345 |
https://mathoverflow.net/questions/23312 | 8 | Recently, I'm tired of those theoretical parts on commuative algebra. So I hope that someone could recommend me some good textbooks on SINGULAR and Macaulay 2. And I'm wondering whether SINGULAR is better that Macaulay 2?
Thank you in advance!
| https://mathoverflow.net/users/5775 | Textbooks on SINGULAR and Macaulay 2 | Two remarks:
* Macauly 2 and Singular share the same computational engine (singular) so none if them is "better" in any real sense
* the best book+software combination I know of is COCOA plus the two volumes of "computational commutative algebra". My "issue" with the singular book is that it's too basic, and with the Macauly book that it's simply a compendium of articles, and not a real text book.
| 2 | https://mathoverflow.net/users/404 | 23319 | 15,347 |
https://mathoverflow.net/questions/23181 | 3 | I have n sectors, enumerated 0 to n-1 counterclockwise. The boundaries between these sectors are infinite branches (n of them).
The sectors live in the complex plane, and for n even,
sector 0 and n/2 are bisected by the real axis, and the sectors are evenly spaced.
These branches meet at certain points, called junctions. Each junction is adjacent to a subset of the sectors (at least 3 of them).
Specifying the junctions, (in pre-fix order, lets say, starting from junction adjacent to sector 0 and 1), and the distance between the junctions, uniquely describes the tree.
Now, given such a representation, how can I see if it is symmetric wrt the real axis?
For example, n=6, the tree (0,1,5)(1,2,4,5)(2,3,4) have three junctions on the real line,
so it is symmetric wrt the real axis.
If the distances between (015) and (1245) is equal to distance from (1245) to (234),
this is also symmetric wrt the imaginary axis.
The tree (0,1,5)(1,2,5)(2,4,5)(2,3,4) have 4 junctions, and this is never symmetric wrt either imaginary or real axis, but it has 180 degrees rotation symmetry if the distance between the first two and the last two junctions in the representation are equal.
Edit: Here are all trees with 6 branches, distances 1.
<http://www2.math.su.se/~per/files/allTrees.pdf>
So, given the description/representation, I want to find some algorithm to decide if it is symmetric wrt real, imaginary, and rotation 180 degrees. The last example have 180 degree symmetry.
Edit 2: If all length of the distances between the junctions were all the same, it is quite easy to find the reflection/rotation of a tree. The problem arises when the distances are of unequal length.
Notice that if I have a regular n-gon, with some non-intersecting chords, is sort of the dual to my trees. I use this in the drawing algorithm, for those that wonder.
That is, I create the n roots of unity (possible with some rotation), then the angle between junction (123) and (345) would be the same as for the mean of vertices 1,2,3 to the mean of vertices 3,4,5 in this n-gon.
The angles in the drawing is not really important, you may change the angles, but the order of the long branches should be the same, and you cannot rotate the tree.
EDIT 3:
Observe that there are many ways of drawing the trees. What I have is
an equivalence relation, T1 ~ T2 if the two trees have the same junction representation.
If S is an axis symmetry, or rotation by 180 degrees,
Then S(T1) ~ S(T2), so the notion of being the same tree is well-defined.
The question is therefore, how to determine if S(T1) ~ T1, or even better, compute S(T1).
By above, this is independent on how I draw the tree.
| https://mathoverflow.net/users/1056 | Find symmetries of a tree | I don't understand how the angles of the connecting finite segments are determined, so I'll assume the angles are set so that they don't break any symmetry. First observe that the reflection wrt the real axis sends sectors 0,1,2,3,4,5 to 0,5,4,3,2,1 respectively. So in your second example, tree
(0,1,5)(1,2,5)(2,4,5)(2,3,4) turns into
(0,5,1)(5,4,1)(4,2,1)(4,3,2)
which is different from the original tree (the original and transformed tree share only the first and last junction). So the transformation is not a symmetry of the tree. However, the same transformation sends the first example
(0,1,5)(1,2,4,5)(2,3,4) to
(0,5,1)(5,4,2,1)(4,3,2)
which is the same tree, represented in a non standard way because the junctions appear in the wrong order and the sectors of each junction are also in the wrong order.
Rotation of 180 degrees sends 0,1,2,3,4,5 to 3,4,5,0,1,2 (add 3 mod 6) so
(0,1,5)(1,2,5)(2,4,5)(2,3,4) turns into
(3,4,2)(4,5,2)(5,1,2)(5,0,1)
which is the same tree, again represented in a non standard way (the junctions appear in the inverse order, and each junction has its inciding sectors cycled).
So the recipe seems to be the following: Find out, for your transformation of the plane, which sectors goes to which, apply this permutation to the tree representation, and then reorder each tree representation (original and transformed) in a standard way that allows to compare if they are equal. If they are equal, then (assuming the angles are nice), the transformation of the plane is a symmetry of the tree. If they are not equal, then the transformation is not a symmetry of the tree.
| 1 | https://mathoverflow.net/users/4118 | 23327 | 15,350 |
https://mathoverflow.net/questions/23325 | 5 | I know that arithmetic sentences are conserved under the addition of the axiom of choice and the continuum hypothesis to ZF (i.e. ($ZF+AC \vdash \phi$ iff $ZF \vdash \phi$) and ($ZF+CH \vdash \phi$ iff $ZF \vdash \phi$) for arithmetic sentences $\phi$) Does this result extend to analytic sentences or other hyperarithmetic sentences? If so, how far does it extend up the hyperarithmetic hierarchy?
| https://mathoverflow.net/users/4085 | Conservation of Hyperarithmetic Sentences over AC and CH. | First, let me remark that in your question, you can combine AC and CH together, rather than having two separate conservation results as you did. In fact, you can ramp CH up to GCH and more, including such principles as $\Diamond$ or others, without any problem. That is, the conservation result is that ZFC + GCH proves $\varphi$ if and only if ZF proves $\varphi$, for a large class of statements $\varphi$, including the arithmetic statements, as you mentioned, but much more.
The phenomenon extends completely up the hyperarithmetic hierarchy and beyond, beyond even the analytic sentences up into the [projective hiearchary](http://en.wikipedia.org/wiki/Projective_hierarchy) at the level of $\Sigma^1\_2$. (In this hiearchy, the hyperarithmetic statements are $\Delta^1\_1$ and the analytic statements $\Sigma^1\_1$.)
This absoluteness result is the content of the [Shoenfield Absoluteness Theorem](http://en.wikipedia.org/wiki/Absoluteness#Shoenfield.27s_absoluteness_theorem), which asserts that any $\Sigma^1\_2$ statement is absolute between between any two transitive models of set theory $V\subset W$ having the same ordinals. In particular, a $\Sigma^1\_2$ statement holds in the universe if and only if it holds in the constructible universe $L$, where both AC and GCH hold.
Thus, the $\Sigma^1\_2$ statements provable in ZFC+GCH are the same as those provable in ZF.
| 7 | https://mathoverflow.net/users/1946 | 23332 | 15,353 |
https://mathoverflow.net/questions/23203 | 0 | Hi,
I've read this sentence but I can not understand what it means
[...] $\Phi'$ is the topological dual of some dense space $\Phi$ of $H\_{aux}$ [...] Notice that the choice of $\Phi$ is subject to the two conditions: [...] ,On the other hand it must be small enough so that its topological dual $\Phi$ is "sufficiently large" [...]
What the author means by "sufficiently large"? Is it the dimension? Knowing that the spaces under consideration are infinite dimensional
Edit: why does the dual becomes larger when the original space becomes smaller?
| https://mathoverflow.net/users/2597 | Topological dual and the notions of "smaller" and "larger" than... | To answer your final question: Let $\Phi \supset \Psi$. Consider $\Phi' \subset \Phi^\*$, the former is the continuous linear functionals on $\Phi$, and the latter is the set of all linear functionals on $\Phi$. Then the restriction of $\Phi'$ on $\Psi$ is obviously continuous, so $\Phi' \subset \Psi'\subset \Psi^\*$.
Therefore if you make a space smaller, you makes its dual bigger.
Intuitively speaking, elements of $\Psi'$ need to be continuous on fewer objects, and hence has fewer constraints; thus $\Psi'$ contains more objects.
For your original question: your interpretation is sort-of okay. The point is that infinite dimensional Hilbert spaces admit dense proper subspaces (hope I am getting the notation correct). And in particular you can have two dense subspaces of a Hilbert space with one strictly contained in the other. You may want to review volume 2 of Reed and Simon.
| 5 | https://mathoverflow.net/users/3948 | 23344 | 15,361 |
https://mathoverflow.net/questions/23322 | 6 | Let $C$ be a compact Riemann surface, and let $U$ be a Zariski open subset in $C.$ Let $L$ be a local system (with coefficients $\mathbb C$ or $\mathbb Q\_{\ell}$) on $U.$ For each point $z\_i\in C-U,$ let $M\_i$ be the monodromy matrix of $L$ around $z\_i.$ If we identify $L$ as a representation of $\pi\_1(U),$ and let $\gamma\_i$ be a small loop in $U$ around $z\_i,$ then $M\_i$ is the image of $\gamma\_i$ under the representation $\rho\_L:\pi\_1(U,a)\to GL(L\_a)$ (where $a$ is a base point and $L\_a$ is the fiber of $L$ at $a$).
Here's my question.
It seems to me that there should be some relation between (the traces of) these monodromies (as well as the monodromies around loops in $\pi\_1(C)$) and (Betti numbers of) the cohomology groups $H^i\_c(U,L)$ of $L$ with compact support, but I don't know the precise formulation or reference.
As an example, if $C$ has genus $g$ and $L=\mathbb C$ is constant of rank 1, then $H^1\_c(U,\mathbb C)$ has dimension $2g+n-1,$ where $n$ is the number of points at infinity. This number is the rank of $\pi\_1(U),$ and each ``canonical generator" of $\pi\_1(U)$ has trace 1.
| https://mathoverflow.net/users/370 | monodromy and global cohomology | Suppose that $C = \mathbb P^1$ and $U = \mathbb P^1\setminus \{0,\infty\}.$ Then
$\pi\_1(U)$ is cyclic, freely generated by a loop around $0$.
The local system $L$ is thus given by the vector space $L\_a$, equipped with an invertible operator,
call it $T$, corresponding to the action of the generator of $\pi\_1(U)$. This operator $T$ is the monodromy matrix.
Now $H^1\_c(U,L) = $ the space of $T$-invariants of $L\_a$, while $H^2\_c(U,L) = $ the space
of $T$-coinvariants of $L\_a$. If you think of the possible Jordan decompositions of
$T$, and the fact that the trace is insensitive to the unipotent part, but just depends on
the semi-simple part, you'll see that the it's going to be hard to find any interesting relation of the type that you want. (E.g. if $L\_a$ is $n$-dimensional, and $T$ acts by the
identity, or by a maximally non-trivial unipotent element, in both cases the trace of $T$ is equal to $n$, but in the first case the Betti numbers are also both equal to $n$,
while in the second, they are both equal to $1$.)
You might also wonder about the Euler characteristic $H^2\_c(U,L) - H^1\_c(U,L)$, but this is always
equal to (rank $L$) $\cdot \chi(U)$, and so is insensitive to the monodromy matrices.
| 10 | https://mathoverflow.net/users/2874 | 23348 | 15,363 |
https://mathoverflow.net/questions/23354 | 6 | This might turn out to be a silly question, but here goes.
Let $\mathcal{C}$ be a full additive subcategory of an abelian category $\mathcal{A}$. I'm wondering if the Grothendieck group $K(\mathcal{C})$ (defined below) depends on $\mathcal{A}$. I can't think of any examples, and I believe that it does not. Something with Yoneda embeddings (see Weibel) comes to mind, but I can't really get it precise.
**Idea**. Given two embeddings $\mathcal{C}\subset\mathcal{A}\_1$ and $\mathcal{C}\subset \mathcal{A}\_2$, I'm guessing it suffices to find a third one containing these.
**Def**.
Let $\textrm{Ob}(\mathcal{C})$ denote the class of objects in $\mathcal{C}$ and let $\textrm{Ob}(\mathcal{C})/\cong$ be the set of isomorphism classes. Let $F(\mathcal{C})$ be the free abelian group on $\textrm{Ob}(\mathcal{C})/\cong$. To any sequence $$(E) \ \ 0 \longrightarrow M^\prime \longrightarrow M \longrightarrow M^{\prime\prime} \longrightarrow 0 .$$ in $\mathcal{C}$, which is exact in $\mathcal{A}$, we associate the element $Q(E) = [M] - [M^\prime] - [M^{\prime\prime}]$ in $F(\mathcal{C})$. Let $H(\mathcal{C})$ be the subgroup generated by the elements $Q(E)$, where $E$ is a short exact sequence. We define the Grothendieck group, denoted by $K(\mathcal{C})$, as the quotient group $$K(\mathcal{C}) = F(\mathcal{C})/H(\mathcal{C}).$$
(All categories are assumed to be at least skeletally small in this definition, but ok.)
| https://mathoverflow.net/users/4333 | Does the Grothendieck group depend on the embedding? | I think that the Grothendieck group DOES depend on A. Indeed, any additive category C could be embedded (by the Yoneda embedding) into the abelian category of contravariant additive functors from C to abelian groups (some call this category the category of presheaves on C). For this embedding the only exact sequence of objects coming from C are those that are given by decompositions of direct sums (in C). On the other hand, you can take for C an arbitrary additive subcategory of an abelian A where there could be tons of (non-trivial) exact sequences.
| 11 | https://mathoverflow.net/users/2191 | 23356 | 15,366 |
https://mathoverflow.net/questions/23349 | 17 | This ought to be a simple one to answer. Does anyone know of, or can anyone provide, an accurate English translation of the marginal remarks in Goldbach's letter to Euler
<http://upload.wikimedia.org/wikipedia/commons/1/18/Letter_Goldbaxh-Euler.jpg>
in which a statement equivalent to the Goldbach conjecture is first stated?
| https://mathoverflow.net/users/5575 | Translation of Goldbach's 1742 letter to Euler | I have an English translation of the letter; you can find my email address on my homepage.
Here is the relevant part:
"By the way, I take it to be not useless to note down also such propositions
that are very probable, even if a real proof is lacking; for even if
afterwards they were found to be erroneous, they could all the same give
occasion for the discovery of a new truth. Thus Fermat's idea that all the
numbers $2^{2^{n-1}}+1$ yield a series of prime numbers cannot hold up, as
you already demonstrated, Sir; but it should still be remarkable if this
series were composed only of numbers that could be split into two squares in
a unique way. I should like to risk another conjecture of that kind: any number
composed from two primes is the sum of as many prime numbers (including $1$) as one wishes, right down to the sum that consists just of ones.
After reading this through again, I see that the conjecture can be proved
quite rigorously for the case $n+1$ if it holds for the case $n$ and
$n+1$ can be split into two prime numbers. The proof is very easy;
and at least it appears to be true that every number greater than
$2$ is the sum of three prime numbers."
The translation was done by Martin Mattmüller.
| 14 | https://mathoverflow.net/users/3503 | 23365 | 15,373 |
https://mathoverflow.net/questions/20007 | 9 | In Alain Roberts "Elliptic curves: notes from postgraduate lectures given in Lausanne 1971/72" page 11 (available on google books unless you already tried to read another chapter), there is a hand drawn picture of a real 2-dimensional torus and a real plane, which **topologically** represent the way a complex cubic (with two real components) and the real projective plane sit in the complex projective plane. Taking the picture on face value, one should be able to project an open subset of the complex projective plane to $\mathbb{R}^3$, so that there is some real line $L$ that passes through the "doughnut" defined by the image of the complex cubic.
I tried to reproduce this picture on a computer, using the map $\mathbb{CP}^2\to\mathbb{R}^7$ given by
$(z\_1:z\_2:z\_3)\mapsto(z\_2\overline{z\_3},z\_3\overline{z\_1},z\_1\overline{z\_2},|z\_1|^2-|z\_2|^2)/(|z\_1|^2+|z\_2|^2+|z\_3|^2)$,
projecting to various $\mathbb{R}^3$s, and looking for $L$ by trial and error; all in vain. Which brings me to....
Questions:
* Is there such a line (the map I used does not send the real projective plane to a plane, so it does not have to be the case even if Roberts picture is correct) ?
* Is there an algorithm to find such a line ?
* Is there a "better" way to project an open part of the complex projective plane to $\mathbb{R}^3$ ?
| https://mathoverflow.net/users/404 | Visualizing a complex plane cubic together with the real plane | I found this article:
"Visualizing Elliptic Curves"
by Donu Arapura
it is available at the following URL:
<http://www.math.purdue.edu/~dvb/graph/elliptic.pdf>
In it he discusses a projection that sends sends the real
part of $x$ to $x\_1$ and the real part of $y$ to $x\_3$ thus
it would seem to preserve the entire real plane and any line
in it. So this might be useful to you.
| 3 | https://mathoverflow.net/users/1098 | 23367 | 15,374 |
https://mathoverflow.net/questions/23334 | 6 | The generalized Korteweg-de Vries equation is
$u\_t + u\_{xxx} + (u^p)\_x=0$
for integer $p$. (The original Korteweg-de Vries equation is the case $p=2$.) I need to understand solutions for $p=1$, but I haven't been able to find this case addressed in the literature despite extensive searching. Is there a reason why $p=1$ isn't interesting? Does it reduce to a simpler form? Is it intractable?
If not, what the best way to learn about this case?
Thanks!
| https://mathoverflow.net/users/5789 | What are the interesting cases of the generalized Korteweg-de Vries equation? | For $p=1$ your equation indeed reduces to a simpler form. Put $X=x-t, T=t$. Then $u\_t=u\_T-u\_X$, $u\_x=u\_X$, and hence in the new independent variables $X,T$ your equation (with $p=1$) becomes a well-studied linear third-order equation
$$
u\_{T}+u\_{XXX}=0,
$$
which is, *inter alia*, a linearized version of the "standard" KdV equation with $p=2$,
see e.g. the discussion in [this book](http://books.google.com/books?id=6fr_gQfHna0C) by Ablowitz and Segur (and cf. also Willie's answer).
Note that the case of $p=3$ (the *modified* KdV equation) is also quite interesting. It is integrable by the inverse scattering transform just as the "usual" KdV and, in fact, is related to it through the Miura transformation (see e.g. the above book for details).
| 8 | https://mathoverflow.net/users/2149 | 23372 | 15,376 |
https://mathoverflow.net/questions/23368 | 0 | $A$ a local ring and $a\_{1}$, ..., $a\_{n}$ elements in its maximal ideal, $M$ a finitely generated $A$-module. In this case apparently the homologies from the Koszul complex are finitely generated as $A$-module. Is there a simple explanation for this? Or is it some deep result? Thanks!
| https://mathoverflow.net/users/5292 | Homology of koszul complex is finitely generated? | If your local rings are Noetherian it's obvious. The Koszul complex
consists of finitely generated free modules and the homology
modules are subquotients of it so also finitely generated.
For non-Noetherian rings it's not true, even when $n=1$. In
this case the $H\_1$ is the annihilator of $a\_1$ which may not
be finitely generated. As an example consider the quotient
of $k[x\_1,x\_2,\ldots]$ by the ideal generated by all the
$x\_i x\_j$. The annihilator of $x\_1$ is the
non-finitely generated maximal ideal.
| 8 | https://mathoverflow.net/users/4213 | 23373 | 15,377 |
https://mathoverflow.net/questions/23361 | 12 | I am delving a bit into category theory and something has me curious about opposite categories. I have checked several books and I can't seem to find an answer.
Given a category C, the opposite category is just the abstract category with the objects of C and with the arrows of C reversed. However, the opposite category can sometimes be realised (is equivalent to) a category where the objects are sets with additional structure, and the arrows are homomorphisms of these structures.
For instance one of the examples on Wikipedia is that the opposite category of commutative rings is equivalent to the category of affine schemes.
**Question. How would one in general, given a category C, find a category where the objects are mathematical structures with underlying sets that satisfy additional axioms, and the morphisms are homomorphisms of those structures, and which is the opposite of the original category C?**
For instance, if we take the category where the objects are groups, and the morphisms are group homomorphisms, what is its opposite category in the above sense? Is there some way to find this from the first-order axiomatization of groups?
| https://mathoverflow.net/users/5800 | Construction Of Opposite Category as a Structure | One way of formalizing the desired categories is given by concrete categories. A category is called concrete (more precise: "concretizable") if it has a faithful functor to the category of sets $Set$. Thus the question is: Is the dual of a concrete category concrete again? The answer is yes: Since a composition of faithful functors is faithful and a dual of a faithful functor is also faithful, it suffices to show that $Set^{op}$ is concrete. But it is not hard to see that the contravariant hom-functor $Hom(-,2)$ (i.e. the power set) yields the desired faithful functor $Set^{op} \to Set$.
However, this solution is somewhat useless. If we apply the proof to $Grp^{op}$, we get sets of the form $P(X)$, where $X$ is a group and morphisms $P(X) \to P(Y)$ should be induced by group morphisms $Y \to X$.
Perhaps we should demand of our concretization that it does not reuses the given category? But this seems to be hard to formalize. Anyway, in the category of groups it would be interesting ...
EDIT: What about the following: If $k$ is a field, then $(k-Vect)^{op}$ is equivalent to the category of pairs $(X,p)$, where $X$ is an affine $k$-scheme and $p$ is a rational point of $X$ such that the corresponding maximal ideal $a$ satisfies $a^2=0$. :-)
If $R$ is a ring, then $(R-Mod)^{op}$ is equivalent to the category of pairs $(X,p)$, where $X$ is an affine $R$-scheme, $p$ is a $R$-valued point of $X$ such that the closed image of $X \to Spec(R)$ is $Spec(R)$ and and the closed image of $p : Spec(R) \to X$ is cut out by an ideal $a \subseteq \mathcal{O}\_X(X)$ with $a^2=0$. What about dropping the affine-condition, do we get "global modules"? Abstract-Nonsense!
| 8 | https://mathoverflow.net/users/2841 | 23379 | 15,380 |
https://mathoverflow.net/questions/23092 | 8 | Let $K$ be a quadratic imaginary field, and $E$ an elliptic curve whose endomorphism ring is isomorphic to the full ring of integers of $K$. Let $j$ be its $j$-invariant, and $c$ an integral ideal of $K$. Consider the following tower:
$$K(j,E[c])\ \ /\ \ K(j,h(E[c]))\ \ /\ \ K(j)\ \ /\ \ K,$$
where $h$ here is any Weber function on $E$. (Note that $K(j)$ is the Hilbert class field of $K$).
We know that all these extensions are Galois, and any field has ABELIAN galois group over any smaller field, EXCEPT POSSIBLY THE BIGGEST ONE (namely, $K(j,E[c]) / K$).
Questions:
1. Does the biggest one have to be abelian? Give a proof or counterexample.
My suspicion: No, it doesn't. I've been trying an example with $K = \mathbf Q(\sqrt{-15})$, $E = C/O\_K$, and $c = 3$; it just requires me to factorise a quartic polynomial over $\bar{ \mathbf Q}$, which SAGE apparently can't do.
2. What about if I replace $E[c]$ in the above by $E\_{tors}$, the full torsion group?
| https://mathoverflow.net/users/5744 | Class Field Theory for Imaginary Quadratic Fields | Here is a case where it is non-Abelian. I use $K$ of class number 3. If I use the Gross curve, it is Abelian. If I twist in $Q(\sqrt{-15})$, it is Abelian for every one I tried, maybe because it is one class per genus. My comments are not from an expert.
```
> K<s>:=QuadraticField(-23);
> jinv:=jInvariant((1+Sqrt(RealField(200)!-23))/2);
> jrel:=PowerRelation(jinv,3 : Al:="LLL");
> Kj<j>:=ext<K|jrel>;
> E:=EllipticCurve([-3*j/(j-1728),-2*j/(j-1728)]);
> HasComplexMultiplication(E);
true -23
> c4, c6 := Explode(cInvariants(E)); // random twist with this j
> f:=Polynomial([-c6/864,-c4/48,0,1]);
> poly:=DivisionPolynomial(E,3); // Linear x Linear x Quadratic
> R:=Roots(poly);
> Kj2:=ext<Kj|Polynomial([-Evaluate(f,R[1][1]),0,1])>;
> KK:=ext<Kj2|Polynomial([-Evaluate(f,R[2][1]),0,1])>;
> assert #DivisionPoints(ChangeRing(E,KK)!0,3) eq 3^2; // all E[3] here
> f:=Factorization(ChangeRing(DefiningPolynomial(AbsoluteField(KK)),K))[1][1];
> GaloisGroup(f); /* not immediate to compute */
Permutation group acting on a set of cardinality 12
Order = 48 = 2^4 * 3
> IsAbelian($1);
false
```
This group has $A\_4$ and $Z\_2^4$ as normal subgroups, but I don't know it's name if any.
PS. 5-torsion is too long to compute most often.
| 8 | https://mathoverflow.net/users/5267 | 23390 | 15,387 |
https://mathoverflow.net/questions/23378 | 7 | Franel uses the convergence of
$ \frac{\zeta(s+1)}{\zeta(s)} = \sum \frac{c(n)}{n^s}$
as an equivalent to the Riemann hypothesis.
Does anybody have a citation for this result and/or hints for computing $c(n)$?
Thanks for any insight.
Cheers, Scott
| https://mathoverflow.net/users/4111 | $\zeta(s+1)/\zeta(s)$ | Since
$$\zeta(s+1) = \sum\_{n=1}^\infty \frac{1/n}{n^s}$$
and
$$\frac{1}{\zeta(s)} = \sum\_{n=1}^\infty \frac{\mu(n)}{n^s}$$
where $\mu$ is the [Möbius function](http://en.wikipedia.org/wiki/M%C3%B6bius_function), we have
$$c(n) = \sum\_{d \mid n} \frac{d}{n}\mu(d) = \frac{1}{n}\prod\_{p \mid n} (1-p)$$
using [Dirichlet convolution](http://en.wikipedia.org/wiki/Dirichlet_convolution).
| 11 | https://mathoverflow.net/users/2000 | 23395 | 15,388 |
https://mathoverflow.net/questions/23393 | 10 | From <http://en.wikipedia.org/wiki/Analytical_hierarchy>
"If the axiom of constructibility holds then there is a subset of the product of the Baire space with itself which is $\Delta^1\_2$ and is the graph of a well ordering of the Baire space. If the axiom holds then there is also a $\Delta^1\_2$ well ordering of Cantor space."
Can someone (give here or point me to) a (sketch or thorough description) of a $\Delta^1\_2$ set that does this for (Baire space or Cantor space)?
I can see how V=L implies there is a definable well order, but I can't see how it would be in the analytical hierarchy.
| https://mathoverflow.net/users/nan | Set Theory and V=L | In the constructible universe $L$, there is a definable well-ordering of the entire universe. This universe is built up in transfinite stages $L\_\alpha$, and the ordering has $x\lt\_L y$ when $x$ is constructed at an earlier stage, or else they are constructed at the same stage, but $x$ is constructed at that stage by an earlier definition, or with the same definition, but with earlier parameters. I also explain this in [this MO answer](https://mathoverflow.net/questions/6593/vl-and-a-well-ordering-of-the-reals/6600#6600).
One may extract from this definition a rather low-complexity definable well-ordering of the reals by capturing the countable pieces of the $L$ hieararchy by reals. That is, if $x$ is a real number of $L$, then it appears at some countable stage $L\_\alpha$ for a countable ordinal $\alpha$, and the entire structure $L\_\alpha$ is countable, and hence itself coded by a real. Here, we code a set by a real in any of the standard ways, for example, by coding a well-founded extensional relation on $\omega$ whose Mostowski collapse is the given set. Furthermore, the $L$-order is absolute to any $L\_\alpha$, since $L\_\alpha$ knows about the $L\_\beta$-heirarchy for $\beta<\alpha$. Also, if a countable structure is well-founded and thinks $V=L$, then it is $L\_\alpha$ for some $\alpha$. Note that if a real $z$ codes a first order structure $M$, then the question of whether $M$ satisfies a first order assertion is an arithmetic statement in $z$, since we need only quantify over the coded elements, which are coded by natural numbers.
Putting all this together, we get that the following are equivalent for any two reals $x$ and $y$:
* $x\lt\_L y$ in the $L$ order.
* There is some countable ordinal $\alpha$ such that $L\_\alpha$ satisfies $x\lt\_L y$.
* For every countable ordinal $\alpha$, if $x$ and $y$ are reals in $L\_\alpha$, then $L\_\alpha$ satisfies $x\lt\_L y$.
* There is a real $z$ coding a well-founded structure that thinks $V=L$ (and so this structure must be some $L\_\alpha$) in which $x$ and $y$ are reals and the structure satisfies $x\lt\_L y$.
* All reals $z$ coding well-founded structures $L\_\alpha$ in which $x$ and $y$ are reals satisfy $x\lt\_L y$.
The fourth statement has complexity $\Sigma^1\_2$, since being-well-founded is $\Pi^1\_1$. Similarly the fifth statement has complexity $\Pi^1\_2$, so overall the ordering is $\Delta^1\_2$.
The end result is that in the universe $L$, there is a low-complexity definable well-ordering of the reals. In this universe, therefore, all of the supposedly non-constructive applications of AC turn out to be completely definable.
| 16 | https://mathoverflow.net/users/1946 | 23399 | 15,389 |
https://mathoverflow.net/questions/23394 | 23 | I would like to understand at least one of the several existing approaches to algebraic geometry over $\mathbb{F}\_1$ (the field with one element). Is there an example of an "interesting" theorem that can be formulated purely in the language of ordinary schemes, but which can be proved using algebraic geometry over $\mathbb{F}\_1$?
Of course, the interpretation of the word "interesting" is entirely up to your own taste. An example in which the theorem cannot be proved using "classical" methods would be most desirable, but examples where (one of) the theories of schemes over $\mathbb{F}\_1$ gives an alternative proof of an already known result would also be very much appreciated.
[On a related note, perhaps there should be a tag "naive-question" for situations like this one.]
| https://mathoverflow.net/users/4384 | Applications of algebraic geometry over a field with one element | I'm confident that the answer to the original question is *no*. There are hardly any theorems at all in the subject, much less ones with external applications! In other words, if no further progress is ever made in any of the directions people have pursued, everything will likely be forgotten (which would not make it so unusual an area). What attracts people to these things is not a track record of existing applications but the possibility of exciting future ones. So investing time in the subject is something of a gamble---it might pay off if you're good at divining the future (or if you have insider information), or you might end up wasting a lot of time.
I don't mean to be too pessimistic. I for one have high hopes for certain directions (!), but I think it's best to see clearly what you'd be getting into.
| 15 | https://mathoverflow.net/users/1114 | 23418 | 15,400 |
https://mathoverflow.net/questions/23415 | 4 | Is there necessarily a CW structure on a space build out of cells without demanding them to be attached in "right" order?
More precisely, let $X$ be a topological space such that the map $\emptyset\to X$ factorizes as a transfinite composition of inclusions
$$
\emptyset\to\ldots\to X\_\beta\to X\_{\beta+1}\to\ldots\to X
$$
where every map $X\_\beta\to X\_{\beta+1}$ is a pushout
$$
\begin{array}{rcl}
S^{n-1} &\to& X\_\beta\\\
\downarrow && \downarrow\\\
D^{n} &\to& X\_{\beta+1}
\end{array}
$$
for some $n\geq 0$. Is $X$ necessarily a CW complex?
| https://mathoverflow.net/users/467 | CW structure on spaces obtained by attaching cells wildly | Fix an irrational number $\alpha$.
Let $X\_2 = [0,1]$ (built from attaching a 1-cell to two 0-cells) and, for each larger $n$, let $X\_n$ be built from $X\_{n-1}$ with a new a 1-cell by attaching the ends to $0$ and the fractional part of $n \alpha$. Take $X$ to be the union.
There are no embeddings from an open disc $D^n$ into $X$ for n greater than 1. If $X$ admitted a CW-complex structure, this would force it to be 1-dimensional. However, $X$ cannot be homeomorphic to a 1-dimensional CW-complex, for example because the set of points which have no neighborhood homeomorphic to $\mathbb{R}$ do not form a discrete subspace.
| 13 | https://mathoverflow.net/users/360 | 23425 | 15,403 |
https://mathoverflow.net/questions/23422 | 4 | Suppose we have, say, $n$ $2n$-dimensional linearly independent vectors over $\mathbb{F}\_2$. We do a projection on a random $d$-dimensional subspace. We are interested in probability that images of our vectors will be linearly independent (over $\mathbb{F}\_2$) too.
The question is as follows: how large $d - n$ should be if we want this probability to be, say, $1 - 1 / \mathrm{poly}(n)$?
| https://mathoverflow.net/users/3448 | Random projection and finite fields | Suppose the vectors are $e\_1,\dots,e\_n$. The kernel of projection onto a random subspace of dimension $n+r$ is a random subspace of dimension $n-r$, so you want the probability that such a subspace has trivial intersection with the span of $e\_1,\dots, e\_n$. Now just count the number of choices for a basis $v\_1,\dots, v\_{n-r}$ of such a space: $2^{2n} - 2^n$ for the first vector, then $2^{2n} - 2^{n+1}$ for the second, and so on. This is to be compared with $2^{2n} - 1$ choices for the first vector if one doesn't have an restriction, $2^{2n}-2$ for the second and so on.
So the probability of this happening is the ratio of these two quantities, which you need to find a good approximation for; a very brief back-of-an-envelope calculation suggested it's about $1 - c2^{-r}$, at least if $r$ is largeish. For your specific needs, then, $d - n$ should be about $C\log n$.
| 10 | https://mathoverflow.net/users/5575 | 23431 | 15,405 |
https://mathoverflow.net/questions/23398 | 5 | If $R$ is a commutative noetherian ring and $I$ is an ideal of $R$, $M$ is an $R$-module. Does $Tor\_i^R(R/I, M)$ is finitely generated for $i\ge 0$ imply $Ext^i\_R(R/I, M)$ is finitely generated for $i\ge 0$?
| https://mathoverflow.net/users/5775 | A problem on finiteness of Ext | In fact, the two are equivalent. My apologies for the length of this argument - if someone else has a shorter one, I'd be happy to hear it.
Let $(a\_1,\ldots,a\_k) = I$, and let $K\_\bullet$ be the [Koszul complex](http://en.wikipedia.org/wiki/Koszul_complex) associated to this set of generators. Note that its zero'th homology group is $R/I$, and all the homology groups are finitely generated $R/I$-modules because $R$ is Noetherian.
Let C be a Serre class of R-modules (i.e. one such that for any $0 \to A' \to A \to A'' \to 0$ exact, $A$ is in $C$ if and only if $A'$ and $A''$ are both in C). The result you ask is obtained by letting C be the class of finitely generated $R$-modules (which is only a Serre class because $R$ is Noetherian). We have that the following are equivalent for an R-module M:
1. $Tor\_i(R/I,M)$ is in C for all values of i.
2. $Tor\_i(N,M)$ is in C for all finitely generated $R/I$-modules $N$.
3. $H\_i(P \otimes\_R M)$ is in C for all bounded chain complexes $P$ of free $R$-modules whose homology groups are finitely generated $R/I$-modules.
4. $H\_i(K \otimes\_R M)$ is in C for all i.
The implication 1 => 2 follows inductively by writing $N$ in a short exact sequence $0 \to J \to \oplus R/I \to N \to 0$ and applying the long exact sequence of Tor.
The implication 2 => 3 follows from a [hyperhomology](http://en.wikipedia.org/wiki/Hyperhomology) spectral sequence.
The implication 3 => 4 is immediate from the definition of the Koszul complex.
The implication 4 => 1 is proved inductively. The hyperhomology spectral sequence
$$
E^2\_{p,q} = Tor\_p(H\_q(K), M) \Rightarrow H\_{p+q}(K \otimes\_R M)
$$
first shows $E^2\_{0,0} = Tor\_0(R/I,M)$ is in C. If $Tor\_i(R/I,M)$ is in C for $0 \leq i \leq m$, the above argument implies that $Tor\_i(N,M)$ is in C for all finitely generated $N$, which forces $E^2\_{p,q}$ to be in C for all $p \leq m$. As the abutment is in C, this forces $E^2\_{m+1,0} = Tor\_{m+1}(R/I,M)$ to be in C.
Now, there is an exactly analogous string of implications in Ext. The following are equivalent:
1. $Ext^i(R/I,M)$ is in C for all values of i.
2. $Ext^i(N,M)$ is in C for all finitely generated $R/I$-modules $N$.
3. $H^i(\underline{Hom}\_R(P,M))$ is in C for all bounded chain complexes $P$ of free $R$-modules whose homology groups are finitely generated $R/I$-modules.
4. $H^i(\underline{Hom}\_R(K,M))$ is in C for all i.
However, the Koszul complex has self-duality; the tensor product complex $K \otimes\_R M$ visibly has the same homology groups as a shift of the Hom-complex $\underline{Hom}\_R(K,M)$. Therefore, the two versions of statement (4) are equivalent.
| 9 | https://mathoverflow.net/users/360 | 23432 | 15,406 |
https://mathoverflow.net/questions/12810 | 36 | Here is a double series I have been having trouble evaluating:
$$S=\sum\_{k=0}^{m}\sum\_{j=0}^{k+m-1}(-1)^{k}{m \choose k}\frac{[2(k+m)]!}{(k+m)!^{2}}\frac{(k-j+m)!^{2}}{(k-j+m)[2(k-j+m)]!}\frac{1}{2^{k+j+m+1}}\text{.}$$
I am confident that $S=0$ for any $m>0$. In fact, I have no doubt. I have done lots of algebraic manipulation, attempted to "convert" it to a hypergeometric series, check tables (Gradshteyn and Ryzhik), etc., but I have not been able to get it into a form from which I can prove zero equivalence.
Here is another form of the sum (well, I hope at least) that might be easier to work with:
$$S=\sum\_{k=0}^{m}\sum\_{j=0}^{k+m-1}(-1)^{k}\frac{m!}{k!(m-k)!}\frac{(k+m-1-j)!}{(k+m)!}\frac{(k+m-1/2)!}{(k+m-1/2-j)!}\frac{1}{2^{k-j+m+1}}\text{.}$$
I have read Concrete Mathematics and $A=B$, and looked at Gosper's and Zeilberger's work for some hints, but no cigar.
Note: $0!=1$ and $n!=n(n-1)!$ for $n\in\mathbb{N}\cup\{0\}$. For $n\in\mathbb{R}^+$, $n!=n\Gamma(n)$ where $\Gamma\colon\: \mathbb{C}\to\mathbb{C}$ and, for $\Re z>0$ and $z\notin\mathbb{Z}^{-}$, $$\Gamma\colon\: z\mapsto \int\_0^\infty t^{z-1}\mathrm{e}^{-t}\,\mathrm{d}t\text{.}$$
which can be analytically extended to $\mathbb{C}$ via the recurrence $\Gamma(z+1)=z\Gamma(z)$.
| https://mathoverflow.net/users/556 | Help with a double sum, please | Olivier Gerard just told me about this wonderful website!
Regarding the question it can be done in one nano-second using the Maple
package
<http://www.math.rutgers.edu/~zeilberg/tokhniot/MultiZeilberger>
accopmaying my article with Moa Apagodu
<http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/multiZ.html>
Here is the command:
F:=(-1)^k\*binomial(m,k)\*(k+m-1-j)!/(k+m)!\*simplify((k+m-1/2)!/(k+m-1/2-j)!)/2^(k-j): lprint(MulZeil(F,[j,k],m,M,{})[1]);
and here is the output:
-1/4\*(2\*m+1)/(m+1)+M
(Note that I had to divide the summand by 1/2^(m+1) if you don't you get FAIL,
the prgram does not like extraneous factors)
Translated to humaneze we have that (my S(m) is hte original S(m) times 2^(m+1))
S(m+1)=(2m+1)/(m+1)S(m)
Since S(1)=0 (check!) This is a completely rigorous proof.
P.S. The proof can be gotten by finding the so-called multi-certificate
lprint(MulZeil(F,[j,k],m,M,{})[2]);
-Doron Zeilbeger
| 44 | https://mathoverflow.net/users/5822 | 23434 | 15,408 |
https://mathoverflow.net/questions/23426 | 13 | Let $G$ be algebraic group over $\mathbb(C)$(semisimple), $B$ be Borel subgroup. consider flag variety $G/B$. It is known that $G/B$ is projective variety. So one consider the homogeneous coordinate ring $ \mathbb{C}[G/B]$, it is known that it is a graded polynomial ring quotient by some homogeneous ideal. For example, consider flag variety of $sl\_2$, it is $\mathbb{P}^{1}$, so the coordinate ring is $\mathbb{C}[x\_1,x\_2]$ as graded ring.
I want to know for general case, i.e. flag variety of finite dimensional Lie algebra $g$($G/B$), how to compute $\mathbb{C}[G/B]$? But we know that from Borel-Weil, we can write it as
$\bigoplus\_{\lambda\in P\_+}$ $R\_\lambda$, where $R\_\lambda$ is irreducible highest dominant weight representations of $g$
My question
===========
How to compute $\mathbb{C}[G/B]$? explicitly?
How to build explicit ring isomorphism from $\bigoplus\_{\lambda\in P\_+}$ $R\_\lambda$ to some concrete ring?
This question might be elementary. Thanks anonymous and Ben pointing out the stupid mistake I made
| https://mathoverflow.net/users/1851 | How to Compute the coordinate ring of flag variety? | $G/B$ is most naturally a multi-projective variety, embedding in the product of projectivizations of fundamental representations: $\prod \mathbb{P}
(R\_{\omega\_i})$. So there is a multi-homogeneous coordinated ring on $G/B$. You mentioned that this ring is $\bigoplus\_{\lambda\in P\_+}$ $R\_\lambda$. This is correct, and the grading is also apparent. It's given by the weight lattice. (To be more canonical, you should take the duals of every highest weight representation, but what you've written down is isomorphic to that.)
So all that remains is giving the multiplication law on $\bigoplus\_{\lambda\in P\_+}$ $R\_\lambda$. You need to specify maps $R\_\lambda \otimes R\_\mu \mapsto R\_{\lambda+\mu}$. There's a natural candidate: If you decompose $R\_\lambda \otimes R\_\mu$ into a direct sum of irreducible representations, $R\_{\lambda+\mu}$ will appear exactly once. The multiplication law is simply projection onto this factor.
@Shizuo: For $sl\_n$, the situation is more explicit. The flag variety here is the set of flags $0 = V\_0 \subset V\_1 \subset \cdots \subset V\_{n-1} \subset V\_n = \mathbb{C}^n$ with $\dim V\_i = i$. So the flag variety is a closed subvariety of the product of Grassmannians $Gr(1,n) \times \cdots \times Gr(n,n) $. Each of these Grassmannians have a explicitly Plücker embedding into the projectivization of the exterior power of $\mathbb{C}^n$. In particular, the homogeneous ideal is explicitly given by the Plücker relations. So the multi-homogeneous coordinate ring of $Gr(1,n) \times \cdots \times Gr(n,n) $ is just the tensor product of the known homogeneous coordinate rings.
Finally, to get the multi-homogeneous coordinate ring for the flag variety, we need to specify an incidence locus inside $Gr(1,n) \times \cdots \times Gr(n,n) $. Namely, we need to specify those tuples of subspaces that form a flag. But this is easy: it corresponds to certain wedge products being zero. Just impose those additional relations, i.e. mod out by the corresponding multi-homogeneous ideal. Now you should have an explicit, albeit fairly long description of the homogeneous coordinate ring. The above answer for $SL\_3$ looks like a special case of this construction.
| 10 | https://mathoverflow.net/users/788 | 23446 | 15,415 |
https://mathoverflow.net/questions/23443 | 10 | I would like to know an example (not using the Gibbs measure Theory) of a sequence of measures $\mu\_n:\mathcal B\to[0,1]$ , where $\mathcal B$ is the $\sigma$-algebra of the borelians of a compact space $X$ such that :
1) $\mu\_n$ is ergodic, with respect to a fixed continuous function $T:X\to X$, for all $n\in\mathbb N$;
2) $\mu\_n\to \mu$ in the weak-$\*$ topology and $\mu$ is not ergodic.
| https://mathoverflow.net/users/2386 | On The Convergence of Ergodic Measures | Let $X=\{0,1\}^{\mathbb{N}}$ with the infinite product topology (which is metrisable). For each $n \geq 1$, define $x\_n$ to be the sequence given by $x\_i=0$ for $1 \leq i \leq n$, $x\_i=1$ for $n+1 \leq i \leq 2n$, and $x\_{2n+i}=x\_i$ for all $i$. Let $T \colon X \to X$ be the shift transformation $T[(x\_n)]= (x\_{n+1})$.
We have $T^{2n}x\_n=x\_n$ for every $n \geq 1$, so the measure $\mu\_n$ defined by $\mu\_n:=(2n)^{-1}\sum\_{j=0}^{2n-1}\delta\_{T^jx\_n}$ is an ergodic invariant Borel probability measure for $T$.
Let $\overline{0}$ denote the element of $X$ corresponding to an infinite sequence of zeroes, and similarly let $\overline{1}$ denote the infinite sequence of ones; we have $\lim\_{n \to \infty} \mu\_n = \frac{1}{2}(\delta\_{\overline{0}}+\delta\_{\overline{1}})$, and this limit is not ergodic (since the set containing only the point $\{\overline{0}\}$ has measure 1/2 but is invariant).
There is a nice paper by Parthasarathy - called, I think, "On the category of ergodic measures" - which shows that for this particular dynamical system and some of its generalisations, the set of all ergodic measures and the set of all non-ergodic measures are both weak-\* dense in the set of all invariant measures, so this phenomenon can actually happen quite a lot.
(Hmm, the definition of $X$ above is supposed to have curly set brackets in it, but I can't get them to appear for some reason. Anyway, it's supposed to be the set of all one-sided infinite sequences of zeroes and ones.)
| 17 | https://mathoverflow.net/users/1840 | 23449 | 15,416 |
https://mathoverflow.net/questions/23453 | 0 | It is well-known that in average $\varphi(n)$ behaves like $\frac1{\zeta(2)}n=\frac{6}{\pi^2}n$. But it looks that in some sense it is ``asymptotically larger''. In particular, the ratio
$$
\zeta(2)(1-t)^2\sum\_{n=1}^{\infty} \varphi(n)t^{n-1}=
\zeta(2)\frac{\sum\_{n=1}^{\infty} \varphi(n)t^{n-1}}{\sum\_{n=1}^{\infty} nt^{n-1}}
$$
seems to be greater then 1 when $t$ increases to 1 (my caclulations say so), and analogous things appear for other averaging procedures involving $\varphi(n)$.
Does it have some sense and/or explanation?
| https://mathoverflow.net/users/4312 | Asymptotic propeties of Euler function | Write $S(t)= \sum \varphi(n)t^n$. A standard calculation gives
$S(e^{-u})= \frac{1}{2\pi i}\int\_{(3)}\frac{\zeta(s-1)}{\zeta(s)}\Gamma(s)u^{-s} ds$,
so pushing the contour to $Re(s)=3/2$ (say) gives $S(e^{-u})=\zeta(2)^{-1} u^{-2}+O(u^{-\frac{3}{2}})$ as $u\to 1$. How many $n$ are you using in your numerical calculations?
| 3 | https://mathoverflow.net/users/1464 | 23458 | 15,420 |
https://mathoverflow.net/questions/23430 | 5 | I need this for a counterexample: the multiplication in the fundamental group $\pi\_1(\Sigma X\_+)$, when it is equipped with the topology inherited from $\Omega \Sigma X\_+$, fails to be continuous for the sort of space in the question, by a result from
>
> J. Brazas, The topological fundamental group and hoop earring spaces, 2009, arXiv:0910.3685
>
>
>
but the author doesn't supply an explicit example of such a space.
| https://mathoverflow.net/users/4177 | What is an example of a non-regular, totally path-disconnected Hausdorff space? | One of the easiest examples is the rational numbers with the subspace topology of the real line with the K-topology. Total path disconnectedness is not entirely necessary for multiplication of $\pi\_{1}(\Sigma X\_{+})$ to fail to be continuous. It just makes the path component space of $X$ equal to $X$, greatly simplifying complications.
| 6 | https://mathoverflow.net/users/5801 | 23461 | 15,423 |
https://mathoverflow.net/questions/23451 | 1 | I have constructed a hierarchical clustering of data using some proximity function. Now I would like to calculate a representative value of any given node of this clustering, such that it reflects the different contributions of the branch components.
I know that this is a very tricky and risky task, but still...
Can anyone point me to a simple method of averaging such branches?
| https://mathoverflow.net/users/5823 | Finding a representative of a branch in hierarchical clustering | If you do some sort of centroid linkage algorithm to construct your hierarchal clustering, you should be able to pick out the centroid values just before merging clusters as representatives at those levels. It'd require tweaking the algorithm to get the points to tag the corresponding tree, but shouldn't be too hard to write up.
Alternatively, reinterpret the hierarchal clustering as persistent homology and pick out representative cycles for the homology classes associated to each bar in the barcode. jPlex ( <http://comptop.stanford.edu/programs/jplex/index.html> ) or javaPlex ( <http://code.google.com/p/javaplex> ) should both be able to do just that if I remember correctly.
| 1 | https://mathoverflow.net/users/102 | 23467 | 15,428 |
https://mathoverflow.net/questions/23409 | 81 | My question is related to the question [Explanation for the Chern Character](https://mathoverflow.net/questions/6144/explanation-for-the-chern-character) to [this question about Todd classes](https://mathoverflow.net/questions/10630/why-do-todd-classes-appear-in-grothendieck-riemann-roch-formula), and to [this question about the Atiyah-Singer index theorem](https://mathoverflow.net/questions/1162/atiyah-singer-index-theorem).
I'm trying to learn the [Atiyah-Singer index theorem](https://en.wikipedia.org/wiki/Atiyah%E2%80%93Singer_index_theorem) from standard and less-standard sources, and what I really want now is some soft, heuristic, not-necessarily-rigourous intuitive explanation of why it should be true. I am really just looking for a mental picture, analogous somehow to the mental picture I have of [Gauss-Bonnet](https://en.wikipedia.org/wiki/Gauss%E2%80%93Bonnet_theorem): "increasing Gaussian curvature tears holes in a surface".
The Atiyah-Singer theorem reads $$\mathrm{Ind}(D)=\int\_{T^\ast M}\mathrm{ch}([\sigma\_m(D)])\smile \mathrm{Td}(T^\ast M \otimes \mathbb{C})$$
What I want to understand is what the Chern character cup Todd class is actually measuring (heuristically- it doesn't have to be precisely true), and why, integrated over the cotangent bundle, this should give rise to the index of a Fredholm operator. I'm not so much interested in exact formulae at this point as in gleaning some sort of intuition for what is going on "under the hood". The Chern character is beautifully interpreted in [this answer by Tyler Lawson](https://mathoverflow.net/questions/6144/explanation-for-the-chern-character/6188#6188), which, however, doesn't tell me what it means to cup it with the Todd class (I can guess that it's some sort of exponent of the logarithm of a formal group law, but this might be rubbish, and it's still not clear what that should be supposed to be measuring). Peter Teichner gives another, to my mind perhaps even more compelling [answer](https://mathoverflow.net/questions/6144/explanation-for-the-chern-character/18466#18466), relating the Chern character with looping-delooping (going up and down the n-category ladder? ), but again, I'm missing a picture of what role the Todd class plays in this picture, and why it should have anything to do with the genus of an elliptic operator. I'm also missing a "big picture" explanation of Fei Han's work, even after having read his thesis (can someone familiar with this paper summarize the conceptual idea without the technical details?). Similarly, [Jose Figueroa-O'Farrill's answer](https://mathoverflow.net/questions/1162/atiyah-singer-index-theorem/7683#7683) looks intriguing, but what I'm missing in that picture is intuitive understanding of why at zero temperature, the Witten index should have anything at all to do with Chern characters and Todd classes.
I know (at least in principle) that on both sides of the equation the manifold can be replaced with a point, where the index theorem holds true trivially; but that looks to me like an argument to convince somebody of the fact that it is true, and not an argument which gives any insight as to **why** it's true.
Let me add background about the Todd class, explained to me by Nigel Higson: "The Todd class is the correction factor that you need to make the Thom homomorphism commute with the Chern character." (I wish I could draw commutative diagrams on MathOverflow!) So for a vector bundle $V\longrightarrow E\longrightarrow X$, you have a Thom homomorphism in the top row $K(X)\rightarrow K\_c(E)$, one in the bottom row $H^\ast\_c(X;\mathbb{Q})\rightarrow H\_c^\ast(E;\mathbb{Q})$, and Chern characters going from the top row to the bottom row. This diagram doesn't commute in general, but it commutes modulo the action of $\mathrm{TD}(E)$. I don't think I understand why any of this is relevant.
In summary, my question is
>
> Do you have a soft not-necessarily-rigourous intuitive explanation of what each term in the Atiyah-Singer index theorem is trying to measure, and of why, in these terms, the Atiyah-Singer index theorem might be expected to hold true.
>
| https://mathoverflow.net/users/2051 | Intuitive explanation for the Atiyah-Singer index theorem | I don't think I can really give you the intuition that you seek because I don't think I quite have it yet either. But I think that understanding the relevance of Nigel Higson's comment might help, and I can try to provide some insight. (Full disclosure: most of my understanding of these matters has been heavily influenced by Nigel Higson and John Roe).
My first comment is that the index theorem should be regarded as a statement about K-theory, not as a cohomological formula. Understanding the theorem in this way suppresses many complications (such as the confusing appearance of the Todd class!) and lends itself most readily to generalization. Moreover the K-theory proof of the index theorem parallels the "extrinsic" proof of the Gauss Bonnet theorem, making the result seem a little more natural. The appearance of the Chern character and Todd class are explained in this context by the observations that the Chern character maps K-theory (vector bundles) to cohomology (differential forms) and that the Todd class measures the difference between the Thom isomorphism in K-theory and the Thom isomorphism in cohomology. I unfortunately can't give you any better intuition for the latter statement than what can be obtained by looking at Atiyah and Singer's proof, but in any event my point is that the Todd class arises because we are trying to convert what ought to be a K-theory statement into a cohomological statement, not for a reason that is truly intrinsic to the index theorem.
Before I elaborate on the K-theory proof, I want to comment that there is also a local proof of the index theorem which relies on detailed asymptotic analysis of the heat equation associated to a Dirac operator. This is analogous to certain intrinsic proofs of the Gauss-Bonnet theorem, but according to my understanding the argument doesn't provide the same kind of intuition that the K-theory argument does. The basic strategy of the local argument, as simplified by Getzler, is to invent a symbolic calculus for the Dirac operator which reduces the theorem to a computation with a specific example. This example is a version of the quantum-mechanical harmonic oscillator operator, and a coordinate calculation directly produces the $\hat{A}$ genus (the appropriate "right-hand side" of the index theorem for the Dirac operator). There are some slightly more conceptual versions of this proof, but none that I have seen REALLY explain the geometric meaning of the $\hat{A}$ genus.
So let's look at the K-theory argument. The first step is to observe that the symbol of an elliptic operator gives rise to a class in $K(T^\*M)$. If the operator acts on smooth sections of a vector bundle $S$, then its symbol is a map $T^\*M \to End(S)$ which is invertible away from the origin; Atiyah's "clutching" construction produces the relevant K-theory class. Second, one constructs an "analytic index" map $K(T^\*M) \to \mathbb{Z}$ which sends the symbol class to the index of $D$. The crucial point about the construction of this map is that it is really just a jazzed up version of the basic case where $M = \mathbb{R}^2$, and in that case the analytic index map is the Bott periodicity isomorphism. Third, one constructs a "topological index map" $K(T^\*M) \to \mathbb{Z}$ as follows. Choose an embedding $M \to \mathbb{R}^n$ (one must prove later that the choice of embedding doesn't matter) and let $E$ be the normal bundle of the manifold $T^\*M$. $E$ is diffeomorphic to a tubular neighborhood $U$ of $T^\*M$, so we have a composition
$K(T^\* M) \to K(E) \to K(U) \to K(T^\*\mathbb{R}^n)$
Here the first map is the Thom isomorphism, the second is induced by the tubular neighborhood diffeomorphism, and the third is induced by inclusion of an open set (i.e. extension of a vector bundle on an open set to a vector bundle on the whole manifold). But K-theory is a homotopy functor, so $K(T^\* \mathbb{R}^n) \cong K(\text{point}) = \mathbb{Z}$, and we have obtained our topological index map from $K(T^\*M)$ to $\mathbb{Z}$. The last step of the proof is to show that the analytic index map and the topological index map are equal, and here again the basic idea is to invoke Bott periodicity. Note that we expect Bott periodicity to be the relevant tool because it is crucial to the construction of both the analytic and topological index maps - in the topological index map it is hiding in the construction of the Thom isomorphism, which by definition is the product with the Bott element in K-theory.
To recover the cohomological formulation of the index theorem, just apply Chern characters to the composition of K-theory maps which defines the topological index. The K-theory formulation of the index theorem says that if you "plug in" the symbol class then you get out the index, and all squares with K-theory on top and cohomology on the bottom commute except for the "Thom isomorphism square", which introduces the Todd class. So the main challenge is to get an intuitive grasp of the K-theory formulation of the index theorem, and as I hope you can see the main idea is the Bott periodicity theorem.
I hope this helps!
| 42 | https://mathoverflow.net/users/4362 | 23469 | 15,430 |
https://mathoverflow.net/questions/23477 | 2 | I just used the following.
Lemma. Let $A$ be a $\mathbb{Z}\_p$-flat ring, of finite type over $\mathbb{Z}\_p$, and suppose that $A \otimes \mathbb{F}\_p$ is a domain. Then $A$ is a domain.
Proof: Suppose $ab = 0$ in $A$. Then one of $a, b$ must lie in $pA$, so we can write (without loss of generality) $a = p a\_1$. Then by flatness $a\_1 b=0$.
Continuing in this manner, we find that one of $a$ and $b$ must be infinitely divisible by $p$. But the finite type hypothesis implies that this is impossible unless one of $a$ or $b$ is in fact zero.
Given the statement, it seems like there should be a more conceptual reason why this should be true. Can anyone supply one? (A proof using more general facts in EGA counts as conceptual).
Edit: Kevin Buzzard gives a compelling reason why I have never seen this "fact" used before. Thank you both for your answers.
Edit 2: I suppose that replacing "finite type" with "p in the radical" would work (with an application of the Krull intersection theorem). In particular, the result is true as stated for a local $\mathbb{Z}\_p$-algebra.
| https://mathoverflow.net/users/1594 | Z_p flatness and irreducible components. | Your proof seems wrong to me. I might be misunderstanding some things you wrote, but surely $\mathbf{Q}{}\_p=\mathbf{Z}\_p[X]/(pX-1)$ is finite type over $\mathbf{Z}\_p$, and contains many elements which are infinitely divisible by $p$. Again if I've understood your definitions correctly, $\mathbf{Q}{}\_p\times\mathbf{Z}\_p$ is a counterexample to your assertion.
| 9 | https://mathoverflow.net/users/1384 | 23489 | 15,441 |
https://mathoverflow.net/questions/23391 | 31 | Let $S^{\lambda}$ be a Schur functor. Is there a known **positive** rule to compute the decomposition of $S^{\lambda}(\bigwedge^2 \mathbb{C}^n)$ into $GL\_n(\mathbb{C})$ irreps?
---
In response to Vladimir's request for clarification, the ideal answer would be a finite set whose cardinality is the multiplicity of $S^{\mu}(\mathbb{C}^n)$ in $S^{\lambda}(\bigwedge^2 \mathbb{C}^2)$. As an example, the paper [Splitting the square of a Schur function into its symmetric and anti-symmetric parts](http://www.ams.org/mathscinet-getitem?mr=1331743) gives such a rule for $\bigwedge^2(S^{\lambda}(\mathbb{C}^n))$.
Formulas involving evaluations of symmetric group characters, or involving alternating sums over stable rim hooks, are not good because they are not positive.
And, yes, it is easy to relate the answers for $\bigwedge^2 \mathbb{C}^n$ and $\mathrm{Sym}^2(\mathbb{C}^n)$, so feel free to answer with whichever is more convenient.
| https://mathoverflow.net/users/297 | What is known about this plethysm? | If I remember this correctly the cases $\mathrm{Sym}^k(\bigwedge^2 \mathbb{C}^n)$ and $\mathrm{Sym}^k(\mathrm{Sym}^2(\mathbb{C}^n))$ are known; and hence $\bigwedge^k(\bigwedge^2 \mathbb{C}^n)$ and $\bigwedge^k(\mathrm{Sym}^2(\mathbb{C}^n))$. I will look up the references tomorrow (if this is of interest).
**Edit** The result has now been stated. I learnt this from R.P.Stanley "Enumerative Combinatorics" Vol 2, Appendix 2. Specifically, A2.9 Example (page 449) which refers
to (7.202) on page 503. This gives as the original reference (11.9;4) of the 1950 edition of:
Littlewood, Dudley E.
"The theory of group characters and matrix representations of groups."
P.S. In the Notes at the end of 7.24 (bottom of page 404 in CUP 1999 edition)
it discusses the origin and the etymology of "plethysm". It says:
Plethysm was introduced in
MR0010594 (6,41c) Littlewood, D. E. Invariant theory, tensors and group characters.
Philos. Trans. Roy. Soc. London. Ser. A. 239, (1944). 305--365
The term "plethysm" was suggested to Littlewood by M. L. Clark after the Greek word
*plethysmos* $\pi\lambda\eta\theta\upsilon\sigma\mu\acute o\varsigma$ for "multiplication".
| 11 | https://mathoverflow.net/users/3992 | 23491 | 15,442 |
https://mathoverflow.net/questions/23476 | 2 | I was browsing in Plouffe's inverter and found that $\frac{\exp(\pi)-\ln(3)}{\ln(2)}$ is very nearly $\frac{159}{5}.$ The continued fraction is
[31, 1, 4, 12029125, ...].
Is this the same magic as $\exp(\pi \sqrt{163})$?
| https://mathoverflow.net/users/756 | Is this related to the j-function? | The trick with the modular invariant $j$ is for $\pi\sqrt D$ only (as $j(\pi\sqrt{-D})$ is rational). Your value is not the exponential of a CM-point in the upper halfplane, so nothing to do with modularity. This kind of experimental discoveries already exists in the literature; see, for example, [C. Schneider, R. Pemantle. When is 0.999... equal to 1?. Amer. Math. Monthly 114 (2007) 344-350.].
| 4 | https://mathoverflow.net/users/4953 | 23496 | 15,446 |
https://mathoverflow.net/questions/23502 | 11 | *Edit: I wrote the following question and then immediately realized an answer to it, and moonface gave the same answer in the comments. Namely, $\mathbb C(t)$, the field of rational functions of $\mathbb C$, gives a nice counterexample. Note that it is of dimension $2^{\mathbb N}$.*
The following is one statement of Schur's lemma:
>
> Let $R$ be an associative unital algebra over $\mathbb C$, and let $M$ be a simple $R$-module. Then ${\rm End}\_RM = \mathbb C$.
>
>
>
My question is: are there extra conditions required on $R$? In particular, how large can $R$ be?
In particular, the statement is true when $\dim\_{\mathbb C}R <\infty$ and also when $R$ is countable-dimensional. But I have been told that the statement fails when $\dim\_{\mathbb C}R$ is sufficiently large.
How large must $\dim\_{\mathbb C}R$ be to break Schur's lemma? I am also looking for an explicit example of Schur's lemma breaking for $\dim\_{\mathbb C}R$ sufficiently large?
| https://mathoverflow.net/users/78 | (When) does Schur's lemma break? | The idea behind Schur's Lemma is the following. The endomorphism ring of any simple $R$-module is a division ring. On the other hand, a finite dimensional division algebra over an algebraically closed field $k$ must be equal to $k$ (this is because any element generates a finite dimensional subfield over $k$, which must be equal to $k$).
Thus, when $R$ is an algebra over an algebraically closed field $k$, the endomorphism ring of a finite dimensional simple module is a finite dimensional division algebra over $k$ and hence is equal to $k$.
On the other hand, let $D$ be any division algebra over a field $k$, which we no longer assume to be algebraically closed. Then $D\_D$ is a simple module, and $\text{End}\_D(D)\cong D$. This allows us to break Schur's Lemma two different ways. If $D$ is infinite-dimensional and $k$ algebraically closed, the endomorphism ring $\text{End}\_D(D)$ will also be infinite dimensional over $k$, hence not isomorphic to $k$. We can easily come up with such $D$, even commutative examples. For instance, $k(x)$ will be an infinite dimensional division algebra over $k$. On the other hand, if $k$ is not algebraically closed, we can take $D$ to be a finite field extension, and we get a $\text{End}\_D(D)\cong D$ not isomorphic to $k$.
| 7 | https://mathoverflow.net/users/778 | 23503 | 15,452 |
https://mathoverflow.net/questions/23487 | 7 | A number theorist I know (who studies Galois representations) raised a question recently:
>
> Which finite groups can have an irreducible character of degree at least 2 having only $n=2, 3$, or 4 classes where the character takes nonzero values?
>
>
>
He has learned about a few special examples involving nonabelian groups which are very close to being abelian. My limited intuition about the question, based on finite groups of Lie type, suggests that the sort of group he is looking for will be far from simple. But maybe there is no reasonable characterization of these groups for a given $n \leq 4$? In any case, there should be some relevant literature out there. The question itself belongs to a familiar genre in finite group theory: What does the character table tell me about the structure of a group?
Appropriate references: S. , Gagola, Pacific J. Math 108 (1983), 363-385,
Berkovich-Zhmud', Characters of Finite Groups, Part 2, Chapter 21.
| https://mathoverflow.net/users/4231 | Finite groups with a character having very few nonzero values? | For the case of n=2 see Gagola's paper "Characters vanishing on all but two conjugacy classes" MR0721927. Some improvements on Gagola's results can be found in Kuisch and van der Waall's papers "Homogeneous character induction [I and II]" MR1172440/MR1302857 and in my paper ["Groups with a Character of Large Degree"](http://arxiv.org/abs/math/0603239).
See Theorem 4.1 in my paper for what I think is the strongest result in this direction (as far as I know):
>
> Gagola showed that the character of
> $V$ vanishes on all but one nontrivial
> conjugacy class if and only if there
> exists a subgroup $N$ of $G$ such that
> $N$ acts trivially on each simple
> $\mathbb{C}[G]$-module except $V$.
>
>
> Let
> $G$ be a group of order $n$ and $V$ a
> simple $\mathbb{C}[G]$-module of
> dimension $d$. Define $e$ such that
> $n = d(d+e)$. Assume that there
> exists a normal subgroup $N \neq
> > \{1\}$ such that $N$ acts trivially on
> each simple $\mathbb{C}[G]$-module
> other than $V$. Let $x$ be a
> nontrivial element of $N$ and $C$ be
> the centralizer of $x$ in $G$. Then
> there exist a prime number $p$, a
> positive integer $k$ and a
> non-negative integer $m$ such that:
>
>
> * $N$ is elementary abelian of order $p^k$,
> * $C$ has order $p^k e^2$, and $d = e(p^k-1)$, and $n = e^2 p^k (p^k-1)$,
> * $C$ is a Sylow $p$-subgroup of $G$ and $e = p^m$,
> * if $H$ is any group such that $N \subseteq H \subseteq C$ and $|H| >
> > p^{k+m}$, then $N \subseteq [H,H]$.
>
>
> The above conditions are also sufficient, see Thm 4.3
>
>
>
For n bigger than 2 there may be some results in MR1487363 Chillag "On zeroes of characters of finite groups."
| 8 | https://mathoverflow.net/users/22 | 23514 | 15,459 |
https://mathoverflow.net/questions/23250 | 7 | As I study category theory, I'm finding the use of universal mapping properties in defining basic concepts to be both simple and clever. Yet, the idea seems non-obvious enough that I expect quite a bit of mathematics had been done before the discovery of the technique.
> What is the chronologically earliest abstract definition given by a universal mapping property?
Note that this question is not intended to be restricted to category theory.
Thank you!
| https://mathoverflow.net/users/5700 | What is the earliest definition given by a universal mapping property? | I very much agree with Pete's comment that we will find incipient instances of the universal mapping property among many classical constructions in mathematics, even if the original users of those concepts would not describe the idea in those terms. Indeed, I believe that these instances will stretch back through the whole of mathematics, and for this reason, there may be no definitive answer to the question.
But let me anyway introduce a very early classical construction that we might agree has the hallmarks of a universal mapping property. To my mind, one of the fundamental essences of the UMP definitions is that they specify an object $U$ that relates to given objects $A,B,\ldots$ in a certain way, such that any other object $V$ relating to $A,B,\ldots$ in that same way then stands in a certain relation to $U$. This is the sense in which $U$ is universal with that property, and the particular details of the relations determine the nature of the universal property.
My proposal is to consider the ancient idea of [commensurability of line segments](http://en.wikipedia.org/wiki/Commensurability_%28mathematics%29), appearing in Euclid's Elements and used earlier. Commensurability is of course intimately connected with the concept of greatest common divisor arising in the [Euclidean algorithm](http://en.wikipedia.org/wiki/Euclidean_algorithm), appearing in Books VII and X of Euclid's Elements.
Specifically, two line segments $K$ and $L$ are *comensurable* when there is another line segment $R$ such that $K$ and $L$ are common multiples of $R$. The ancients knew not only that there was a largest such common measure $R$, but also that this largest common measure has a universal property: if $S$ is any other common measure of $K$ and $L$, then $R$ is a multiple of $S$. Thus, the largest common measure of commensurable line segments is characterized by a universal property, known in antiquity.
This fact is of course related to the fact, also known to the ancients, that the greatest common divisor $d$ of two natural numbers $a$ and $b$ is not only the greatest common divisor of $a$ and $b$, but has the universal property that any other common divisor of $a$ and $b$ is a divisor of $d$. Thus, the gcd of two integers is characterized by a universal property, also known in antiquity.
| 10 | https://mathoverflow.net/users/1946 | 23526 | 15,467 |
https://mathoverflow.net/questions/23337 | 24 | A morphism of schemes $f:X\to S$ is said to be quasi-compact if for every OPEN quasi-compact subset $K \subset S$ the subset $f^{-1}(K) \subset X$ is also quasi-compact (and open, of course!). The morphism $f:X\to S$ is said to be universally closed if for every morphism $T\to S$ the resulting base-changed morphism $X\_T \to T$ is closed. The title question (inspired by topology) is then:
**Question 1: If $f:X\to S$ is universally closed, does it follow that $f$ is quasi-compact?**
Here is a variant of this question, asking for a stronger conclusion :
**Question 2: If $f:X\to S$ is universally closed, does it follow that for every quasi-compact subset $K\subset S$, open or not,
$f^{-1}(K)$ is quasi-compact ?**
**REMARK 1** The converse of Question 1 is false: any morphism between affine schemes is quasi-compact but is not universally closed in general.
**REMARK 2** One might wonder whether $f$ proper implies $f$ quasi-compact. The answer is "yes" but for an irrelevant reason: proper is *defined* as separated, universally closed and of finite type. Since finite type already implies quasi-compact, proper obviously implies quasi-compact.
**REMARK 3** In topology "proper" is (or should be !) defined as universally closed; equivalently, closed with quasi-compact fibres. Topologically proper implies that every quasi-compact subset (open or not) of the codomain has quasi-compact inverse image. The converse is not true in general, but it is for locally compact spaces.
**REMARK 4** (edited).As BCnrd remarks in his comment below, it is not at all clear that the two questions are equivalent (I had stated they were in the previous version of this post, but I retract that claim ). Also, beware that in topology the notion of quasi-compact continuous map is so weak as to be essentially useless since decent topological spaces, the ones algebraic geometers never use :) , have so few open quasi-compact subsets.
| https://mathoverflow.net/users/450 | Is a universally closed morphism of schemes quasi-compact ? | Yes, a universally closed morphism is quasi-compact. (I haven't yet checked whether the same approach answers question 2.)
**Proof:** Without loss of generality, we may assume that $S=\operatorname{Spec} A$ for some ring $A$, and that $f$ is surjective. Suppose that $f$ is not quasi-compact. We need to show that $f$ is not universally closed.
Write $X = \bigcup\_{i \in I} X\_i$ where the $X\_i$ are affine open subschemes of $X$. Let $T=\operatorname{Spec} A[\{t\_i:i \in I\}]$, where the $t\_i$ are distinct indeterminates. Let $T\_i=D(t\_i) \subseteq T$. Let $Z$ be the closed set $(X \times\_S T) - \bigcup\_{i \in I} (X\_i \times\_S T\_i)$. It suffices to prove that the image $f\_T(Z)$ of $Z$ under $f\_T \colon X \times\_S T \to T$ is not closed.
There exists a point $\mathfrak{p} \in \operatorname{Spec} A$ such that there is no neighborhood $U$ of $\mathfrak{p}$ in $S$ such that $X\_U$ is quasi-compact, since otherwise we could cover $S$ with finitely many such $U$ and prove that $X$ itself was quasi-compact. Fix such $\mathfrak{p}$, and let $k$ be its residue field.
First we check that $f\_T(Z\_k) \ne T\_k$. Let $\tau \in T(k)$ be the point such that $t\_i(\tau)=1$ for all $i$. Then $\tau \in T\_i$ for all $i$, and the fiber of $Z\_k \to T\_k$ above $\tau$ is isomorphic to $(X - \bigcup\_{i \in I} X\_i)\_k$, which is empty. Thus $\tau \in T\_k - f\_T(Z\_k)$.
If $f\_T(Z)$ were closed in $T$, there would exist a polynomial $g \in A[\{t\_i:i \in I\}]$ vanishing on $f\_T(Z)$ but not at $\tau$. Since $g(\tau) \ne 0$, some coefficient of $g$ would have nonzero image in $k$, and hence be invertible on some neighborhood $U$ of $\mathfrak{p}$. Let $J$ be the finite set of $j \in I$ such that $t\_j$ appears in $g$. Since $X\_U$ is not quasi-compact, we may choose a point $x \in X - \bigcup\_{j \in J} X\_j$ lying above some $u \in U$. Since $g$ has a coefficient that is invertible on $U$, we can find a point $P \in T$ lying above $u$ such that $g(P) \ne 0$ and $t\_i(P)=0$ for all $i \notin J$. Then $P \notin T\_i$ for each $i \notin J$. A point $z$ of $X \times\_S T$ mapping to $x \in X$ and to $P \in T$ then belongs to $Z$. But $g(f\_T(z))=g(P) \ne 0$, so this contradicts the fact that $g$ vanishes on $f\_T(Z)$.
| 28 | https://mathoverflow.net/users/2757 | 23528 | 15,469 |
https://mathoverflow.net/questions/23479 | 4 | In R.O.Wells book "Differential Analysis on Complex Manifolds" p. 44 proof of Theorem 2.2 part b) the author claims that any two sections of an etale space which agree at a point agree in some neighborhood of that point (etale space is a (possibly non-Hausdorff) $Y$ with the surjection $p\colon Y \rightarrow X$ which is a local homeomorphism). Why is that true?
| https://mathoverflow.net/users/5838 | Sections of an etale space | First note that if $p\_1:Y\_1 \to X$ and $p\_2:Y\_2 \to X$ are two etale spaces over $X$ then
any morphism of etale spaces $f:Y\_1 \to Y\_2$ (where by morphism of etale spaces I mean that $f$ is continuous and that
$p\_2\circ f = p\_1$) makes $Y\_1$ an etale space over $Y\_2$.
(Proof: Let $y \in Y\_1$, and let $V$ be a n.h. of $f(y)$ in $Y\_2$ such that $p\_2:V \to
p\_2(V)$ is a homeo. Note that a local homeo. is open by definition, and so $p\_2(V)$
is also open in $X$. Now $f^{-1}(V)$ is an open subset of $Y\_1$ containing $y$,
and so we may find a n.h. $U$ of $y$ in $Y\_1$, contained in $f^{-1}(V)$, such that $p\_1: U \to p\_1(U)$ is a homeo.
Since $p\_1 = p\_2\circ f$, and since $f(U) \subset V$, we find that $f: U \to f(U)$ is
a homeo. and that $f(U)$ is open in $Y\_2$. Thus $f$ is a local homeo., as required.)
Now suppose that $f\_1,f\_2:Y\_1 \rightarrow Y\_2$ are two morphisms of etale spaces,
and that $f\_1(y) = f\_2(y)$ for some $y \in Y\_1$. Then $f\_1$ and $f\_2$ coincide in some
n.h. of $y$.
(Proof: Choose some sufficiently small n.h. $V$ of $f\_1(y) = f\_2(y)$ in $Y\_2$
such that $p\_2:V \to p\_2(V)$ is a homeo. Then choose a n.h. $U$ of $y$ contained
in $f\_1^{-1}(V) \cap f\_2^{-1}(V)$ such that $f\_i:U \to f\_i(U)$ is a homeo. for
both values of $i$. Since each $f\_i(U)$ is contained in $V$, we see
that $p\_2: f\_i(U) \to p\_2\circ f\_i(U) = p\_1(U)$ is a homeo. as well, and hence
that $p\_1 = p\_2\circ f\_i: U \to p\_1(U)$ is a homeo.
We then find that, on $U$, the map $f\_i$ (for either value of $i$) can be computed
as the composite of $p\_1:U \to p\_1(U)$ and $p\_2^{-1}: p\_1(U) \to f\_i(U) \subset V$.
Thus the two maps $f\_i$ coincide on $U$.)
Now if $p:Y \to X$ is an etale space and $\sigma$ is a section, we can think of $\sigma$
as a morphism from the etale space $X \to X$ (i.e. $X$ thought of as the trivial etale
space over itself) to the etale space $Y \to X$. The preceding result then shows that
if two sections coincide at a point, they coincide in a n.h. of that point.
| 8 | https://mathoverflow.net/users/2874 | 23536 | 15,475 |
https://mathoverflow.net/questions/23537 | 8 | The setup for my question is as follows: $k$ is a field, $K$ a Galois extension of $k$ with group $G$, $k^\prime$ an arbitrary extension of $k$, and $K^\prime/k^\prime$ another Galois extension of fields with group $G^\prime$ (the notation, as well as the setup, is essentially verbatim from Chapter X, Section 4 of Serre's *Local Fields*). Given a $k$-monomorphism $f:K\rightarrow K^\prime$, we get a homomorphism $\overline{f}:G^\prime\rightarrow G$ (sending $s^\prime\in G^\prime$ to the unique $s\in G$ with $f\circ s=s^\prime\circ f$), and if we have another $k$-monomorphism $g:K\rightarrow K^\prime$, then there is some unique $s\in G$ with $f\circ s=f^\prime$ (both these facts are consequences of the normality of $K$ over $k$, which implies that every $k$-monomorphism of $K$ into $K^\prime$ has the same image), so the homomorphism $\overline{g}:G^\prime\rightarrow G$ induced by $g$ is just that induced by $f$ followed by a conjugation of $G$.
Now, if $A$ is a commutative group scheme over $k$, then such an $f$ (as above) gives a map $A(K)\rightarrow A(K^\prime)$ that is compatible (in the sense of group cohomology) with $\overline{f}:G\rightarrow G^\prime$, so we get maps $H^n(G,A(K))\rightarrow H^n(G^\prime,A(K^\prime))$. The key fact which comes from the considerations in the previous paragraph is that if we use a different $k$-monomorphism $g:K\rightarrow K^\prime$, then the induced maps on cohomology differ by the map $H^n(G,A(K))\rightarrow H^n(G,A(K))$ induced by the pair $x\mapsto sxs^{-1}$ and $p\mapsto s^{-1}p$, which is the identity.
My question regards a particular case of this situation but with (what appears to me) to be a twist. $k$ is a global field, $k^\prime=k\_v$ is the completion of $k$ at some place $v$, $K=k^s$ is a separable closure of $k$, and $K^\prime=k\_v^s$ is a separable closure of $k$. Any two embeddings of $k^s$ into $k\_v^s$ over $k\rightarrow k\_v$ differ by an element of the absolute Galois group $G\_k=Gal(k^s/k)$ of $k$ (these are just two choices of places of $k^s$ extending $v$). As before, with such an embedding, I get a map $G\_{k\_v}=Gal(k\_v^s/k\_v)\rightarrow G\_k$ which is (by Krasner's lemma) an injection with image the decomposition group in $G\_k$ of the corresponding place. Any two such injections differ by conjugation in $G\_k$. So, if I have, for instance, an abelian variety $A$ over $k$, I get canonical "localization" maps $H^n(k,A(k^s))\rightarrow H^n(k\_v,A(k\_v^s))$, i.e., they are independent of the embedding of $k^s$ into $k\_v^s$. But now if I start with an arbitrary $G\_k$-module $M$, the homomorphism $G\_{k\_v}\rightarrow G\_{k}$ makes $M$ into a $G\_{k\_v}$-module and is compatible with the identity $M\rightarrow M$ (where the target is regarded as a module over the local Galois group), so I get a map on cohomology $H^n(k,M)\rightarrow H^n(k\_v,M)$. My question is **how can this map be independent of the embedding $k^s\rightarrow k\_v^s$**? It's true that any two embeddings give rise to conjugate injections of the local Galois group into the global one, but the action of the local group on $M$ is defined in terms of the map $G\_{k\_v}\rightarrow G\_k$. There is no canonical choice of $G\_{k\_v}$-action on $M$ as there is in the case of a commutative $k$-group scheme. As far as I can tell, if I use a different embedding, I get a conjugate homomorphism $G\_{k\_v}\rightarrow G\_k$, and this gives a DIFFERENT action of $G\_{k\_v}$ on $M$. So, strictly speaking, the map on cohomology induced by a second embedding shouldn't even have the same target as the one induced by the first. Even in dimension zero, shouldn't the Galois invariants of $G\_{k\_v}$ with respect to the different actions differ by an element of $G\_k$ (i.e. if $a$ is fixed by $G\_{k\_v}$ with respect to the first action then $sa$ is fixed by $G\_{k\_v}$ with respect to the second action, where $s\in G\_k$). I would greatly appreciate if somebody could clarify what I'm not getting right here. I know these maps have to be canonical, and everywhere this is stated it's attributed to the fact that conjugation induces the identity on cohomology, but when I've tried to apply this, it doesn't seem to work out if my Galois module isn't already endowed with an action of $G\_{k\_v}$.
I apologize for this post being so long. I suspect there's something simple I've overlooked that will warrant maybe a sentence or two, but I felt like I needed to try to motivate this question somewhat.
| https://mathoverflow.net/users/4351 | How canonical are localization maps in Galois cohomology? | As you note, if we choose two different embeddings $k^s \to k\_v^s$, say $\imath\_1$
and $\imath\_2$, then we get two different $G\_{k\_v}$-module structures on $M$, call
them $M\_1$ and $M\_2$, and two different restriction maps $r\_1:H^n(k,M) \to H^n(k\_v,M\_1)$
and $r\_2:H^n(k,M) \to H^n(k\_v,M\_2)$.
The point is that there will also be a canonical isomorphism
$i:H^n(k\_v,M\_1) \cong H^n(k\_v,M\_2)$ such that $i\circ r\_1 = r\_2,$
given by a formula analogous to the one you gave in the abelian variety context.
Namely, if $\imath\_2 = \imath\_1\circ g,$ then the isomorphism $i$ will be induced by
$m \mapsto g\cdot m$ (if I have things straight; you can easily check if this is correct,
of if I should have $g^{-1}$ instead). The fact that $i\circ r\_1 = r\_2$ will then depend
on the fact that the automorphism of $H^n(k,M)$ induced by conjugation by $g^{-1}$ and the
map $m\mapsto g\cdot m$ is the identity. So one does not have a literal independence of
the embedding, but rather, the restriction is defined up to a canonical isomorphism independent of the embedding, and this latter fact does depend upon conjugation inducing the
identity on cohomology (which is why people often summarize it in that way).
Note also that your abelian variety example is actually a special case of this, because
the $M$ is $A(k^s)$, and the $G\_{k\_v}$-action on $M$ *does* depend on the embedding
of $k^s$ in $k\_v^s$. But the natural map $A(k^s) \to A(k\_v^s)$ *also* depends on this embedding,
in such a way that, when you compose the restriction from $G\_k$ to $G\_{k\_v}$ (with coefficients in
$A(k^s)$) with the map on $G\_{k\_v}$-cohomology induced by the embedding $A(k^s) \hookrightarrow
A(k\_v^s)$, you *do* obtain a map on cohomology that is independent of the embedding.
But it is *not* that in this case $M$ has a well-defined action of $G\_{k\_v}$ independent
of the choice of embedding $k^s \hookrightarrow k\_v^s$. It is rather that $M$ embeds into
a bigger module $M\_v$ (in a way that also depends on the embedding) so that the
*composite* $H^n(k,M)\to H^n(k\_v,M) \to H^n(k\_v,M\_v)$ is independent of the embedding.
It is the embeddings $M \hookrightarrow M\_v$ that are missing in the more general context
(i.e. when $M$ is not of the form $A(k^s)$).
| 5 | https://mathoverflow.net/users/2874 | 23540 | 15,478 |
https://mathoverflow.net/questions/23534 | 7 | Suppose we have a normed vector space $V$ and its dual $V^\*$, and suppose that $X \subseteq V^\*$ has the property that for every $v \in V$, there is some $\phi \in X$ with $\Vert \phi \Vert = 1$ such that $\phi(v) = \Vert v \Vert$. Is $X$ dense in $V^\*$ (in the operator norm)? Note that this is a stronger property than $\Vert v \Vert = \sup\_{\phi\in X} \frac{\phi(v)}{\Vert \phi \Vert}$, since we are assuming that the supremum is realized.
I think the answer is probably "no." A nice example (passed to me originally made up by Terry Tao) showing that the second condition (the supremum over $X$ gives the norm) does not imply dense is the following: consider $l^1$ and $(l^1)^\* = l^\infty$. Then the space of eventually zero sequences in $l^\infty$ is sufficient for the norm: given $f\in l^1$, let $\phi\_n$ be a truncation of the sign function of $f$ to the first $n$ indices. Then $\lim\_{n\to \infty} \phi\_n(f) = \Vert f \Vert$. However (for $f$ with infinite support), there is no finite sequence $\phi$ with $\phi(f) = \Vert f \Vert$.
| https://mathoverflow.net/users/5852 | Is a subspace with a certain property dense in the dual of a vector space? | The answer is negative. Since the linear span of the Dirac masses is not a dense subspace of the dual of $C[0,1]$.
| 11 | https://mathoverflow.net/users/2508 | 23548 | 15,485 |
https://mathoverflow.net/questions/23547 | 42 | The motivation for this question comes from the novel *Contact* by Carl Sagan. Actually, I haven't read the book myself. However, I heard that one of the characters (possibly one of those aliens at the end) says that if humans compute enough digits of $\pi$, they will discover that after some point there is nothing but zeroes for a really long time. After this long string of zeroes, the digits are no longer random, and there is some secret message embedded in them. This was supposed to be a justification of why humans have 10 fingers and increasing computing power.
Anyway, apologies for the sidebar, but this all seemed rather dubious to me. So, I was wondering if it is known that $\pi$ does not contain 1000 consecutive zeroes in its base 10 expansion? Or perhaps it does? Of course, this question makes sense for any base and digit. Let's restrict ourselves to base 10. If $\pi$ does contain 1000 consecutive $k$'s, then we can instead ask if the number of consecutive $k$'s is bounded by a constant $b\_k$.
According to the [wikipedia page](http://en.wikipedia.org/wiki/%CE%A0), it is not even known which digits occur infinitely often in $\pi$, although it is conjectured that $\pi$ is a [normal number](http://en.wikipedia.org/wiki/Normal_number). So, it is theoretically possible that only two digits occur infinitely often, in which case $b\_k$ certainly exist for at least 8 values of $k$.
**Update.** As Wadim Zudilin points out, the answer is conjectured to be yes. It in fact follows from the definition of a normal number (it helps to know the correct definition of things). I am guessing that a string of 1000 zeroes has not yet been observed in the over 1 trillion digits of $\pi$ thus computed, so I am adding the open problem tag to the question. Also, Douglas Zare has pointed out that in the novel, the actual culprit in question is a string of 0s and 1s arranged in a circle in the base 11 expansion of $\pi$. See [here](http://en.wikipedia.org/wiki/Contact_%28novel%29) for more details.
| https://mathoverflow.net/users/2233 | Does pi contain 1000 consecutive zeroes (in base 10)? | Summing up what others have written, it is widely believed (but not proved) that every finite string of digits occurs in the decimal expansion of pi, and furthermore occurs, in the long run, "as often as it should," and furthermore that the analogous statement is true for expansion in base b for b = 2, 3, .... On the other hand, for all we are able to prove, pi in decimal could be all sixes and sevens (say) from some point on.
About the only thing we can prove is that it can't have a huge string of zeros too early. This comes from irrationality measures for pi which are inequalities of the form
$|\pi-(p/q)|>q^{-9}$ (see, e.g., Masayoshi Hata, Rational approximations to $\pi$ and some other numbers, Acta Arith. 63 (1993), no. 4, 335-349, MR1218461 (94e:11082)), which tell us that such a string of zeros would result in an impossibly good rational approximation to pi.
| 50 | https://mathoverflow.net/users/3684 | 23551 | 15,488 |
https://mathoverflow.net/questions/23512 | 1 | For a given integer $x>0$, I need to find all integers $a \in [0, 10^{15}]$ which have the following property: the digit sum of $a$ equals the digit sum of $x\cdot a$.
I found this link <http://mathworld.wolfram.com/CastingOutNines.html> which looks quite relevant to my task, but I can't figure out how to apply it.
Any ideas?
| https://mathoverflow.net/users/5846 | Finding integers whose sum of digits equals the sum of the digits of a multiple of them | If you wish to get the property $S(a)=S(xa)$ (where $S(\cdot)$ denotes the sum of decimal digits) valid for all positive integers $x$, then there are no such numbers $a$. Indeed, if such an $a$ existed, then concantination $\overline{aa}=a\cdot x$ with $x=10^n+1$, $n$ the number of digits in $a$, would give you a number $xa$ with sum of digits twice more than $S(a)$.
I put the answer because my comments on clarifying the problem were ignored.
| 1 | https://mathoverflow.net/users/4953 | 23563 | 15,497 |
https://mathoverflow.net/questions/23564 | 30 | In the [December 2009 issue](http://www.ems-ph.org/journals/newsletter/pdf/2009-12-74.pdf) of the [newsletter of the European Mathematical Society](http://www.ems-ph.org/journals/all_issues.php?issn=1027-488X) there is a very interesting interview with Pierre Cartier. In page 33, to the question
>
> What was the ontological status of categories for Grothendieck?
>
>
>
he responds
>
> Nowadays, one of the most interesting points in mathematics is that, although all categorical reasonings are formally contradictory, we use them and we never make a mistake.
>
>
>
Could someone explain what this actually means? (Please feel free to retag.)
| https://mathoverflow.net/users/394 | Is "all categorical reasoning formally contradictory"? | Note: I am not a historian. I'm just guessing as to what prompted the comments.
Here's my guess: if you do set theory naively, in the old-fashioned "anything is a set" way, then you run into Russell's paradox; the set consisting of all sets that aren't elements of themselves gives you trouble. So you then decide set theory needs formalising (I'm talking about 100 years ago here of course) and you write down some axioms, and the ones that "won" are ZFC, where only "small" collections of things are sets, and "big" things like "all groups" aren't sets. Of course there's nothing contradictory about considering all groups, or quantifying over groups (i.e. saying "every group has an identity element"), but you can't quantify over the *set* of all groups.
And now because it's the 50s or 60s and you want to do homological algebra and take derived functors and do spectral sequences and stuff in some abstract way, you are now feeling the pressure a bit, because you want to define "functions" from the category of all G-modules (G a group) to the category of abelian groups called "group cohomology", but "H^n(G,-)" isn't a function, because its domain and range aren't sets. So you call it a "functor", which is fine, and press on.
And as time goes on, and you start composing derived functors, you know in your heart that it's all OK.
And then Grothendieck comes along, and probably other people too, and raise the issue that one really should be a bit more careful, because we don't want another Frege (who wrote a huge treatise on set theory but allowed big sets and his axioms were contradictory because of Russell's paradox). So Grothendieck tried to tame these beasts and "go back to basics"---but in some sense he "failed"---or, more precisely, realised that there were fundamental problems if he really wanted to treat categories as sets. "Sod it all", thought Grothendieck, "this is not really the main point". So he said "let's just assume there's a universe, i.e. (basically) a set where all the axioms of set theory hold" (it was a bit more complicated than that but still). This assumption (a) fixed all his problems but was (b) unprovable from the axioms of ZFC (because of Goedel).
So there's a guess. Cartier is perhaps going over the top with "contradictory"---the statement "Russell's paradox is a paradox" is true but the statement "any mathematical manipulation with collections of objects that don't form a set is formally contradictory" is much stronger and surely false.
| 36 | https://mathoverflow.net/users/1384 | 23566 | 15,499 |
https://mathoverflow.net/questions/23561 | 14 | In his answer
[here](https://mathoverflow.net/questions/2022/definition-and-meaning-of-the-conductor-of-an-elliptic-curve)
Qing Liu mentioned "the 'discriminant' of X which measures the defect of a functorial isomorphism which involves powers of the relative dualizing sheaf of X/R."
Could somebody give a reference for or explain this?
I ask because there is a natural definition of conductor of a proper morphism to a normal scheme (for example as in Serre's "Algebraic groups and class fields"), and some relationship between the conductor and discriminant, but the definitions of discriminants I have seen always seem situation-specific and ad-hoc.
Thanks.
| https://mathoverflow.net/users/4710 | "Nice" definition of discriminant as alluded to in an answer of Qing Liu | The "magistral paper of Takeshi Saito" mentioned by Qing Liu is the following: 'Conductor, Discriminant, and the Noether formula of Arithmetic surfaces', Duke Math Journal, vol. 57, no. 1. See here if you have a subscription to Duke: projecteuclid.org/euclid.dmj/1077306852 The definition of the discriminant is at the top of the second page. Or see section three of Liu's notes which he referenced in his reply to which you provided a link.
But for what it's worth, here is a quick answer (basically copied from Saito's paper):
Let $T$ be a nice scheme (spectrum of a field is enough for us), and $g:Y\to T$ a relative curve (i.e. geometrically connected, proper, relative dimension one). When $Y$ is smooth over $T$, there is a functorial isomorphism $\Delta:\det Rg\_\*(\omega\_{Y/T}^{\otimes 2})\to(\det Rg\_\*\omega\_{Y/T})^{\otimes 13}$. This is due to Deligne (from [this unpublished letter to Quillen](http://publications.ias.edu/sites/default/files/Lettre-a-Quillen.pdf), link thanks to [D. Eriksson's answer](https://mathoverflow.net/a/178010/250)).
Now let $\mathcal{O}\_K$ be a Henselian discrete valuation ring with algebraically closed residue field, put $S=\mbox{Spec}\mathcal{O}\_K$,
and let $X\to S$ be a relative curve which is regular. We have an invertible $\mathcal{O}\_K$-module
$V=\mbox{Hom} (\det Rg\_\*(\omega\_{X/S}^{\otimes 2}),(\det Rg\_\*\omega\_{X/S})^{\otimes 13})$.
and Deligne's result, applied to the generic fibre of $X$, produces a natural element $\Delta\in V\otimes\_{\mathcal{O}\_K} K$. The discriminant of $X$ is defined to be the order of $\Delta$ (that is, the unique $d\in\mathbb{Z}$ such that $\mathcal{O}\_K\Delta=\mathfrak{p}^d V$).
I do not know if there is a similar notion of discriminant for higher dimensional $S$-schemes.
| 13 | https://mathoverflow.net/users/5830 | 23569 | 15,501 |
https://mathoverflow.net/questions/23571 | 17 | Are there non-isomorphic number fields (say of the same degree and signature) that have the same discriminant and regulator? I'm guessing the answer is no - why?
And focusing on fields of small degree (n=3 and n=4), what about a less restrictive question: can we find two such fields that have the same regulator (no discriminant restrictions)?
| https://mathoverflow.net/users/5860 | Number fields with same discriminant and regulator? | Yes, see e.g. the paper "Arithmetically equivalent number fields of small degree" (Google for it) by Bosma and de Smit.
In brief: two number fields $K$ and $K'$ are said to be **arithmetically equivalent** if they have the same Dedekind zeta function. A famous group-theoretic construction of Perlis (Journal of Number Theory, 1977) gives many nontrivial (i.e., non-isomorphic) pairs of arithmetically equivalent number fields. Remarkably, this construction works equally well to construct isospectral, non-isometric Riemannian manifolds, as was later shown by Sunada.
Arithmetically equivalent number fields necessarily share many of the simplest invariants, for instance they have equal discriminants.
As the aformentioned paper explains, for arithmetically equivalent $K$ and $K'$, comparing zeta functions gives
$h(K)r(K) = h(K')r(K')$,
where $h$ is the class number and $r$ is the regulator. Therefore, to get an affirmative answer to your question you want a nontrivial pair of arithmetically equivalent number fields $K$ and $K'$ with $h(K) = h(K')$. The paper by Bosma and de Smit gives such examples.
| 29 | https://mathoverflow.net/users/1149 | 23572 | 15,503 |
https://mathoverflow.net/questions/23145 | 6 | This might be a dumb question. If $C$ is an ordinary category, then for any $c \in C$ the covariant representable functor $\text{Hom}(c, -) : C \to \text{Set}$ preserves limits. However, it can happen that $c$ can be equipped with extra structure which in turn gives the morphisms out of $c$ extra structure, so that there is a "representable functor" $\text{Hom}(c, -) : C \to D$ where $D$ is a category equipped with a forgetful functor $F : D \to \text{Set}$ such that composing with the above gives the original representable functor.
In this situation, when does the functor into $D$ still preserve limits? How is this situation formalized? (Assume that $C$ is not enriched over $D$ in any obvious way.)
There are several examples of this coming from algebra, but the one that got me curious is the following. Let $C$ denote the homotopy category of pointed (path-connected?) topological spaces and let $S^1$ denote the circle with a distinguished point. I believe I am correct in saying that if the fundamental group functor $\pi\_1 : C \to \text{Grp}$ is composed with the forgetful functor $U : \text{Grp} \to \text{Set}$, then $S^1$ represents the resulting functor $U(\pi\_1(-))$. (The extra structure on $S^1$ that makes this possible is, if I'm not mistaken, a **cogroup** structure internal to $C$.) Can I conclude that $\pi\_1$ preserves limits?
---
**Edit:** I've been told that the above example is problematic, so here's a simpler one. Let $C = \text{Set}$ and suppose that $c \in C$ is equipped with a morphism $f : c \to c$. Then by precomposition $\text{Hom}(c, d)$ is also equipped with such a morphism, so $\text{Hom}(c, -)$ has values in the category of dynamical systems. Does it preserve limits? Another example is my attempted answer to [question #23188](https://mathoverflow.net/questions/23188/distributivity-of-unary-operations/23191#23191).
| https://mathoverflow.net/users/290 | When does a "representable functor" into a category other than Set preserve limits? | [Collecting my sporadic comments into one (hopefully) coherent answer.]
A more general question is as follows: For functors
$C\stackrel{F}{\to}D\stackrel{U}{\to}E$ and for an index category
$J$ such that $UF$ preserves $J$-limits, when does $F$ preserve $J$
limits?
A useful sufficient condition is that if $U$ creates $J$-limits, then
in the above situation $F$ preserves $J$-limits. Proof: Let $T\colon
J\to C$ be a functor, and suppose that $\tau\colon
\ell\stackrel{\cdot}{\to} T$ is a limiting cone in $C$. Since $UF$
preserves $J$-limits, $UF\tau\colon UF\ell\stackrel{\cdot}{\to} UFT$ is a limiting cone in
$E$. As $U$ creates $J$-limits, there is a unique lifting of $UF\tau$ to a cone in
$D$, and this cone is a limiting cone. But $F\tau\colon
F\ell\stackrel{\cdot}{\to} FT$ is such a lift, and hence we're done.
This condition is quite useful, because many forgetful functors are
[monadic](http://ncatlab.org/nlab/show/monadic+functor), and monadic functors create all limits (by combining their definition
on pp. 143--144 of Mac Lane and Ex. 6.2.2 on [p. 142](http://books.google.com/books?id=eBvhyc4z8HQC&lpg=PP1&dq=categories%20for%20the%20working%20mathematician&pg=PA142#v=onepage&q&f=false) of Mac Lane, or by
Proposition 4.4.1 on [p. 178](http://books.google.com/books?id=SGwwDerbEowC&lpg=PP1&dq=sheaves%20in%20geometry%20and%20logic&pg=PA178#v=onepage&q&f=false) of Mac Lane--Moerdijk, or really by a
[comment](https://mathoverflow.net/questions/9504/why-is-top-4-a-reflective-subcategory-of-top-3/9530#9530) of Tom Leinster from which I learned this :)).
For example, consider the category of all small algebraic systems of
some type. From the AFT, we know that the forgetful functor to
$\mathbf{Set}$ has a left adjoint, and it is the content of Theorem
6.8.1, p. 156 of Mac Lane that this forgetful functor is monadic.
Returning to the original question, this means that whenever the
category $D$ is one of $\mathbf{Grp}$, $\mathbf{Rng}$,
$\mathbf{Ab}$,... and $U\colon D\to \mathbf{Set}$ is the forgetful
functor, then for any $J$, $UF$ preserves $J$-limits implies $F$ preserves $J$
limits. In particular, if $UF$ is a representable functor
(and hence preserves all limits), then $F$ preserves all limits.
Next, let me try to comment on your motivating examples (the one from
Q. 23188 and the one from the 'Edit' part of the current question.)
Regarding your example in Q. 23188: Unfortunately I know nothing of
Hopf algebras, so I can't understand all the details of your
construction. If I understand correctly, you construct a functor
$F\colon\mathbf{Rng}\to\mathbf{Grp}$ whose composition with the
forgetful functor $U\colon \mathbf{Grp}\to \mathbf{Set}$ is
representable. If this is indeed the case, then by the above $F$
itself preserves all limits.
[EDIT: corrected the part concerning the last example.]
Finally, regarding your example in the edited question:
While I
know nothing of dynamical systems, from a quick glance at [Terence
Tao's blog](http://terrytao.wordpress.com/2008/01/10/254a-lecture-2-three-categories-of-dynamical-systems/)
it seems that the category of dynamical systems is the
category whose objects are pairs $\langle X,f\colon X\to X\rangle$ with
$X$ a (small) set and whose arrows
$\phi\colon\langle X, f\rangle\to\langle Y, g\rangle$ are those
functions $\phi\colon X\to Y$ with $g\circ\phi =\phi\circ f$.
To show that the above sufficient condition works in this case, we
would like to show that the forgetful functor to $\mathbf{Set}$ crates
limits. More generally, we will show that if $C$ is a
category and $D$ is the category whose objects are pairs $\langle
x,f\colon x\to x\rangle$ (where $x\in\operatorname{obj}(C)$,
$f\in\operatorname{arr}(C)$), and whose arrows $\phi\colon \langle
x,f\rangle\to \langle y,g\rangle$ are those arrows $\phi\colon x\to y$
with $g\circ\phi =\phi\circ f$, then the forgetful functor $U\colon
D\to C$ creates limits.
[I'm sure that this follows from some well-known result, but since I
don't see it, I'll just continue with a direct proof.]
So, let $J$ be an index category, let
$F\colon J\to D$ be a functor, and suppose that $\tau\colon
x\stackrel{.}{\to} UF$ is a limiting cone in $C$. We would like to
show that there exists a
unique cone $\sigma\colon L\stackrel{.}{\to} F$ in $D$ such that
$U\sigma=\tau$, and that this unique cone is a limiting cone.
For uniqueness, suppose that $\sigma\colon L\stackrel{.}{\to} F$
satisfies $U\sigma = \tau$. Write $F\_j:=\langle y\_j,f\_j\rangle$. Then
we must have for all $j$
$$
\sigma\_j=(x\stackrel{f}{\to}x)\stackrel{\tau\_j}{\to}(y\_j\stackrel{f\_j}{\to}y\_j)
$$
for some $f\colon x\to x$ (hence we immediately see that $\sigma$ is
determined up to $f$). Now, since by the above we see that $\tau\_j$
must be an arrow
$$
(x\stackrel{f}{\to}x)\stackrel{\tau\_j}{\to}(y\_j\stackrel{f\_j}{\to}y\_j)
$$
of $D$, the following diagram must be commutative for all $j$:
$$
\begin{matrix}
x & \stackrel{\tau\_j}{\longrightarrow} & y\_j =UF\_j\\
f\downarrow & & f\_j\downarrow\\
x&\stackrel{\tau\_j}{\longrightarrow} & y\_j = UF\_j.
\end{matrix} \quad \text{(Diagram 1)}
$$
Now we claim that the $\to\downarrow$ part of the above diagram
forms a cone to $UF$, that is, we claim that the family
$\{f\_j\tau\_j\}$ forms a cone $x\stackrel{.}{\to} UF$. Indeed, for an
arrow $g:j\to j'$ of $J$, consider the following diagram:
$$
\begin{matrix}
&&&&x\\
&&&\stackrel{\tau\_j}{\swarrow}&&\stackrel{\tau\_{j'}}{\searrow}\\
&&y\_j && \stackrel{UFg}{\longrightarrow} && y\_{j'}\\
&\stackrel{f\_j}{\swarrow} &&&&&&\stackrel{f\_{j'}}{\searrow}\\
y\_j&&&&\stackrel{UFg}{\longrightarrow}&&&&y\_{j'}
\end{matrix}
$$
The upper triangle is commutative because $\tau$ is a cone to the base
$UF$, and the lower trapezoid is commutative because $F$ is a
functor, and hence $Fg$ is an arrow $F\_j\to F\_{j'}$ in $D$. Hence the
outer triangle commutes, as required. From the universality of $\tau$,
it follows that there is a unique $f$ for which Diagram 1 is
commutative, and we have uniqueness.
For existence, we can take $f$ to be the unique arrow $x\to x$ for
which Diagram 1 is commutative, and we get a cone
$$
\sigma=\{\sigma\_j=\tau\_j\colon (x\stackrel{f}{\to}x)\to F\_j=(y\_j\stackrel{f\_j}{\to}y\_j)\}
$$
with $U\sigma=\tau$. We claim that this is a limiting cone.
To see this, let $\alpha\colon(z\stackrel{g}{\to}z)\stackrel{.}{\to}F$
be a cone, so that
for all $j$ the following diagram is commutative:
$$
\begin{matrix}
z & \stackrel{\alpha\_j}{\longrightarrow} & y\_j\\
g\downarrow & & f\_j\downarrow\\
z &\stackrel{\alpha\_j}{\longrightarrow} & y\_j.
\end{matrix} \quad\text{(Diagram 2)}
$$
Then $U\alpha$ is a cone $z\stackrel{.}{\to} UF$ in $C$, and by the
universality of $\tau$ there exists a unique arrow $h\colon z\to x$
for which the following diagram is commutative for all $j$:
$$
\begin{matrix}
z & \stackrel{\alpha\_j}{\longrightarrow} & y\_j\\
h\downarrow& \stackrel{\tau\_j}{\nearrow}\\
x&
\end{matrix}\quad\text{(Diagram 3)}
$$
If this $h$ is an arrow $(z\stackrel{g}{\to}z)\to
(x\stackrel{f}{\to}x)$ in $D$, then we're done. In other words, all
that remains to do is to show that the outer rectangle of the
following diagram is commutative:
$$
\begin{matrix}
z && \stackrel{h}{\longrightarrow} && x\\
& \stackrel{\alpha\_j}{\searrow} && \stackrel{\tau\_j}{\swarrow}\\
&& y\_j\\
g\downarrow&& \downarrow f\_j && \downarrow f\\
&& y\_j\\
& \stackrel{\alpha\_j}{\nearrow} && \stackrel{\tau\_j}{\nwarrow}\\
z && \stackrel{h}{\longrightarrow} && x\\
\end{matrix}
$$
Now, the left trapezoid is just Diagram 2, the upper
and lower triangles are just Diagram 3, and the
right trapezoid is commutative for all $j$ by the definition of $f$.
It follows that both paths of the outer rectangle have the same
composition with the limiting cone $\tau$, and hence the outer rectangle
is commutative, as required.
| 7 | https://mathoverflow.net/users/2734 | 23577 | 15,505 |
https://mathoverflow.net/questions/23583 | 15 | How many steps on average does a simple random walk in the plane take before it visits a vertex it's visited before?
If an exact formula does not exist (as seems likely), then I'm interested in good approximations.
I'd also like to know the standard nomenclature associated with the question, if any exists.
| https://mathoverflow.net/users/3621 | self-avoidance time of random walk | [I've corrected a stupid mistake below and added an upper bound... Please check the numerical values!]
Well, I doubt that an explicit expression exists. However, it should be possible to get good bounds. The lower bound is easy: observe that
$$
E(T) = \sum\_{k\geq 1} P(T> k-1) = 1+\sum\_{k\geq 1} c\_k 4^{-k},
$$
where $c\_k$ is the number of self-avoiding paths of length $k$ (and, of course, $4^k$ is the number of all paths of length $k$), so that we get a lower bound by truncating this series. Using the (known) values for $c\_k$, $k=1,\ldots,71$, we get
$$
E(T) > 4.58607909
$$
Now, you can get an upper bound by bounding the neglected part of the series using $c\_k\leq 4 \cdot 3^{k-1}$. This gives you a very narrow interval containing the right value: if I have made no mistake ;) , we get
$$
4.58607909 < E(T) < 4.58607911
$$
| 12 | https://mathoverflow.net/users/5709 | 23597 | 15,518 |
https://mathoverflow.net/questions/23608 | 13 | Consider the field of fractions $K$
of the quotient algebra $\mathbb{R}[x,y,z,t]/(x^2+y^2+z^2+t^2+1)$,
where $\mathbb{R}$ is the field of real numbers and $x,y,z,t$ are variables.
Clearly $-1$ is a sum of 4 squares in $K$.
How can one prove that $-1$ is not a sum of 2 squares in $K$?
Serre mentions without proof this (probably known or easy) fact
in a letter to Eva Bayer of May 1, 2010,
and I am stuck: I cannot prove it.
| https://mathoverflow.net/users/4149 | Is -1 a sum of 2 squares in a certain field K? | This is a special case of a theorem of A. Pfister. It is well known to quadratic forms specialists. See e.g. Theorem XI.2.6 in T.Y. Lam's *Introduction to Quadratic Forms over Fields*.
I believe the original paper is
>
> Pfister, Albrecht,
> *Zur Darstellung von $-1$ als Summe von Quadraten in einem Körper*. (German)
> J. London Math. Soc. 40 1965 159--165.
>
>
>
In this same paper Pfister defines the "stufe" (which Lam has successfully campaigned to be called the "level") of a non-formally real field, namely the least positive integer $n$ such that $-1$ is a sum of $n$ squares. Among his other achievements, he proves that the level is always a power of $2$ (so that Kevin Buzzard's recollection is correct). It is also worth remarking that his work is an insightful and rapid response to previous work of J.W.S. Cassels.
| 22 | https://mathoverflow.net/users/1149 | 23609 | 15,524 |
https://mathoverflow.net/questions/23607 | 2 | Let X be a smooth projective variety over the complex numbers. Recall that a Cohen-Macaulay curve is a one-dimensional closed subscheme without embedded or isolated points (fat components are allowed).
>
> I want to show that if you fix a curve class β in H2(X), then there is a bounded interval which contains the genus of every CM curve whose homology class is β.
>
>
> Do you know of a reference for this result?
>
>
>
Motivation: I am trying to prove a certain class of sheaves forms a bounded family (namely, the collection of sheaves underlying [stable pairs](http://arxiv.org/abs/0707.2348)). The above result will allow me to reduce to the case where the support is a fixed CM curve, and from there, I know how to finish the proof.
I am aware that Le Potier (who built the moduli space of stable pairs, among others things, in "System Coherents et Structures de Niveau" --- pardon my lack of accents) has proven a much stronger result. However, for my purposes, it would be very helpful to have a proof (hopefully much less technically demanding than Le Potier's) that is no more general than what I need.
EDIT: in hindsight, and in light of the negative answer below, I realize that Le Potier does not claim to prove quite what I claimed he claimed.
| https://mathoverflow.net/users/1231 | Are the arithmetic genera of Cohen-Macaulay curves in a fixed homology class bounded? | I don't think this is true. Take $X = \mathbb P^1 \times \mathbb P^2$. Let $C$ be $\mathbb P^1 \times 0$, and let $C\_1$ be the first infinitesimal neighborhood of $C$. The curve $C\_1$ is the relative spectrum of $\mathcal O\_{\mathbb P^1} \oplus O\_{\mathbb P^1}^{\oplus 2}$, where $\mathcal O\_{\mathbb P^1}^{\oplus 2}$ is a square-zero ideal. Any sheaf $\mathcal O\_{\mathbb P^1}(d)$, with $d \ge 0$, is a quotient of $\mathcal O\_{\mathbb P^1}^{\oplus 2}$. If $C(d)$ denotes the relative spectrum of $\mathcal O\_{\mathbb P^1} \oplus O\_{\mathbb P^1}(d)$, then $C(d)$ is contained in $C\_1$ for all $d \ge 0$; hence it is embedded in $X$ with fundamental class $2[C]$. But the arithmetic genus of $C(d)$ is $-d-1$.
[Added later] For a surface, a Cohen-Macaulay curve is a divisor, and the adjunction formula shows that the arithmetic genus is determined by the cohomology class, so the answer is positive. I believe that the answer is negative for all $X$ of dimension at least three.
| 8 | https://mathoverflow.net/users/4790 | 23613 | 15,528 |
https://mathoverflow.net/questions/23567 | 9 | A projective 3-manifold is a smooth manifold that admits an atlas with values in the real projective 3-space such that all transition maps are restrictions of projective transformations. A Möbius 3-manifold is defined similarly, with the projective space replaced with the standard 3-sphere and projective transformations replaced with Möbius transformations. (Recall that a Möbius transformation of the sphere $S^n$ is a self-diffeomorphism of the sphere that preserves the angles of the standard metric; such transformations form a Lie group isomorphic to $SO\_{n+1,1}(\mathbf{R}))/\pm I$).
Both projective and Möbius manifolds are particular cases of manifolds admitting an $(M,G)$-structure in the sense of W. Thurston.
Every closed (=compact, orientable and without boundary) 2-surface admits both a Möbius structure and a projective one.
I vaguely remember having been to a talk some time ago where the speaker said that (conjecturally?) the situation in dimension 3 is similar. But I don't remember the details at all. So I would like to ask if anyone knows whether either of the statements (each closed 3-manifold admits a Möbius, resp. projective structure) is a theorem, a conjecture or becomes one or the other after eliminating some counter-examples.
A related question: if memory serves, in the same talk it was mentioned that the $PGL\_{n+1}(\mathbf{R})$ and the Möbius group are (conjecturally?) the maximal groups that can act faithfully on an $n$-manifold. I was wondering if anyone knows a reference for this.
| https://mathoverflow.net/users/2349 | Möbius and projective 3-manifolds | The first examples of closed 3-manifolds not admitting a conformally flat (Mobius) structure [were given by Goldma](http://www.ams.org/mathscinet-getitem?mr=701512)n: 3-manifolds modeled on the Nil geometry (this link was given by Macbeth in the comments above). Sol manifolds also don't have a Mobius structure, whereas 3-manifolds modeled on the other six geometries do. [Misha Kapovich](http://www.math.ucdavis.edu/~kapovich/) has many results on conformally flat structures. In his thesis, he shows that certain classes of Haken 3-manifolds have finite-sheeted covers which are (uniformizable) conformally flat, and he shows there is a graph manifold which has no conformally flat structure, but which has a finite-sheeted cover which does. Kulkarni showed that connect sums of conformally flat manifolds are conformally flat. On the other hand, Kapovich's student Hwang showed that for any 3-manifold $M$, there is a 3-manifold $N$ such that $M \# N$ is conformally flat. I don't know of any recent activity on the topic.
As for your question on maximal Lie group actions on 3-manifolds, I've heard this before too, but I don't know a reference. I think you can prove it by analyzing the action of the isotropy group of a point on the jet space at that point.
Addendum: I had a discussion with Cooper about the last question (maximal Lie groups acting faithfully on an $n$-manifold), and we have an idea how to approach it (at least for smooth actions). If a Lie group $G$ acts smoothly and faithfully on a manifold $M$, then one obtains a homomorphism
$\Phi: G \to Diff(M)$. Then we get a map $\phi:g \to Vect(M)$, where $g$ is the Lie algebra of $G$, and $Vect(M)$ is the Lie algebra of $Diff(M)$. So one wants to classify maximal finite-dimensional Lie algebras of $Vect(M)$. First, if $g$ is not semisimple, then it has a non-trivial center $c$, which is generated by a non-zero smooth vector field $V$. One should then be able to take a quotient $M/V$ of $M$ with action on $M/V$ by the Lie algebra $g/c$ and apply an inductive argument. Actually, one should try to do this only locally, since the quotient might not be nice. Then assume $g$ is semisimple. Its Cartan subalgebra gives $R^m$ acting on $R^n$. This gives $m$ commuting vector fields, and in particular gives $m$-dimensional coordinates
at a point, so $m\leq n$. Now, one needs to appeal to the classification of semisimple Lie algebras to finish off the proof (the Cartan subalgebra together with Weyl group determines the Lie algebra), and then apply the inductive argument to deal with the radical of the Lie algebra. I haven't worked out how to do this, but it seems like a plausible approach. I suspect an argument like this may be well-known in the right circles.
| 13 | https://mathoverflow.net/users/1345 | 23615 | 15,529 |
https://mathoverflow.net/questions/23578 | 6 | Graphical representations of intersection of sets as logical combinations are much older than Venn.
Euler and Leibniz are often quoted and the current Wikipedia article also quotes Ramon Llull but I do not really find the illustrations provided in the Wiki Commons for Llull very compelling.
I expect that these kind of ideas can be found in many other places and even older times, perhaps in disguise.
In this context I find the heraldic uses of theological diagrams such as shown [here](http://www.absoluteastronomy.com/topics/Shield_of_the_Trinity) quite fascinating as a kind of medieval fashion statement.
Do you know of older examples of graphical representation of logical and/or set relations, for instance of Chinese, Arabic and Greek origin ?
(ps: at least one of the tags is a joke)
| https://mathoverflow.net/users/5387 | What are the oldest illustrations of "Venn" diagrams? | You may already be familiar with Ruskey and Weston's ["A Survey of Venn Diagrams,"](http://www.combinatorics.org/Surveys/ds5/VennEJC.html) which includes a discussion of Borromean rings. Such rings are similar to the valknut and the triskelion, of which the gankyil is a type. All of these figures are quite old.
Of course these observations don't answer your question about the use of graphical representations of logical and/or set relations in antiquity.
Plato *refers* to diagrams, for example, in his discussion of the double-divided line in Book 6 of *Republic* and in *Meno* when Socrates questions Meno's slave about a problem in geometry -- how to find a square double in area to any given square. I imagine more examples can be found.
An interesting project would be to find examples of "visual" language in the works of Plato, Aristotle, and Euclid.
While writing this post I came across the following two references:
Edwards, Anthony W. F. *Cogwheels of the Mind: The Story of Venn Diagrams.* Baltimore, Maryland: The John Hopkins University Press, (2004).
Kuehni, Rolf. ["On the Source of d’Aguilon’s Arc Color Mixture Diagram."](http://www4.ncsu.edu/~rgkuehni/unpubAndPresentations.html) Unpublished manuscript, 2003.
| 3 | https://mathoverflow.net/users/965 | 23616 | 15,530 |
https://mathoverflow.net/questions/23629 | 20 | This is a question in elementary linear algebra, though I hope it's not so trivial to be closed.
Real symmetric matrices, complex hermitian matrices, unitary matrices, and complex matrices with distinct eigenvalues are diagonalizable, i.e. conjugate to a diagonal matrix.
>
> I'd just like to see an example of a complex symmetric $n\times n$ matrix that is not diagonalizable.
>
>
>
| https://mathoverflow.net/users/4721 | Non-diagonalizable complex symmetric matrix | $$\begin{pmatrix} 1 & i \\ i & -1 \end{pmatrix}.$$
How did I find this? Non-diagonalizable means that there is some Jordan block of size greater than $1$. I decided to hunt for something with Jordan form $\left( \begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix} \right)$. So I want trace and determinant to be zero, but the matrix not to be zero. The diagonal entries made sure the trace vanished, and then the off diagonal entries were forced.
| 38 | https://mathoverflow.net/users/297 | 23631 | 15,540 |
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