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https://mathoverflow.net/questions/40111 | 3 | I have 3 more questions about [maximal words](https://mathoverflow.net/questions/38657/minimal-words-of-length-n) (which are just another way of talking of necklaces).
Let W be a finite word on a two symbol alphabet {0,1}; let us say that W is maximal if it is the last item in the list of all its cyclic permutation (ordered lexicographically).
The number w(n) of maximal words of length n can be expressed with the aid of Eulers' totient function (see
[Minimal words of length n](https://mathoverflow.net/questions/38657/minimal-words-of-length-n/38798#38798)).
**Question 1**: What can be said about the asymptotics of w(n)? I expect that $lim (1/n) \log w(n) = h$ for some positive value h ...
**Question 2:** **The bisection scheme.**
Let us play the following game: start with the two string list [1,0] (which are both maximal) then we put in between the string obtained concatenating them, so we obtain the list [1,10,0], we go on like this (for any two neighbouring strings S,T we put the string ST in between), obtaining in turn the lists [1,110,10,100,0], [1,1110,110,11010,10,10100,100,1000,0] ... und so weiter.
I guess (and almost can prove) that in this way you generate all (and only) **primitive maximal words** of any length (let us say that a maximal word is *primitive* if it is not the repetition of a shorter one).
Has anybody a nice proof of this?
**Question 3:** is there an asymptotic distribution for the lengths of primitive maximal words generated by n runs of the prvious algorithm?
| https://mathoverflow.net/users/7979 | Maximal words (reloaded) | I can answer your first question fully, and the second question only partially.
**Question 1:** Assuming you meant $\log\_2$ in your expression, the answer is $h=1$. That is because $w(n)=\frac{1}{n}\sum\_{d|n}\phi(d)2^{n/d}\geq \frac{1}{n}2^n$ by just considering the first summand, and $w(n)=\frac{1}{n}\sum\_{d|n}\phi(d)2^{n/d}\leq n 2^n$ by upper-bounding the number of summands to be $n$ each at most $n2^n$.
**Question 2:** I'm afraid not all necklaces are generated by your method. Continue two more steps to notice the only two full-period necklaces of length $6$ you produce are $111110$ and $100000$.
| 3 | https://mathoverflow.net/users/9044 | 40124 | 25,684 |
https://mathoverflow.net/questions/40125 | 4 | Is there an ordinal $\alpha$ such that $ZF$ believes that $V\_{\alpha}$ is a model of $ZF$? (If it is problematic to state this since we have to check infinitely many axioms at once, formalize logic in $ZF$. ) If $\alpha > \omega$ is a limit ordinal, then $V\_{\alpha}$ is a model of $ZF - R$, where $R$ stands for the axiom of replacement. The reflection principle tells us that $ZF$ knows that the set $S$ of ordinals $\alpha$ such that $R$ holds relative to $V\_{\alpha}$ is unbounded. So is there some limit ordinal in $S$, thus answering the question? What is the smallest $\alpha$ such that $V\_{\alpha}$ is a model of $ZF$, if it exists? I already know that if $\kappa$ is a strongly inaccessible cardinal, then $V\_{\kappa}$ is a model of ZF, but the existence of such $\kappa$ is independent from $ZF$.
| https://mathoverflow.net/users/2841 | Replacement in von Neumann hierarchy of sets | You cannot prove that there is such an ordinal, but (under a suitable large cardinal assumption) it is consistent that there is such an ordinal.
If you could prove that there was such an ordinal, then you will have proved Con(ZF) in ZF, contrary to the incompleteness theorem.
Another way to see it is: if there were such an ordinal, let $\alpha$ be the least ordinal with $V\_\alpha\models$ZF. Thus, $V\_\alpha$ is a model of ZF having no $\beta$ with $V\_\beta\models$ZF, since the $V\_\beta$ of $V\_\alpha$ is the same as the $V\_\beta$ of $V$.
However, if $\kappa$ is an inaccessible cardinal, then $V\_\kappa\models$ZFC. In fact, there are many smaller $\alpha\lt\kappa$ with $V\_\alpha\models$ZFC, and so the consistency strength of having an $\alpha$ with $V\_\alpha\models$ZFC is strictly lower than an inaccessible, if it is consistent.
Your remark about using the Reflection Theorem to get $\alpha$ with $V\_\alpha$ with Replacement is not quite right. The Replacement Axiom is a *scheme* of axioms, an infinite list of axioms, and the Reflection idea will only produce $\alpha$ with $V\_\alpha$ satisfying any one (or finitely many) of them. But we cannot get a model of the whole scheme this way.
Lastly, it is interesting to note that every nonstandard model $M$ of ZF, having a nonstandard $\omega$, will have a $V\_\alpha$ that is a model of ZF as viewed from outside $M$. The reason is that for any finite collection of the ZF axioms, we may apply the Reflection Theorem as you indicated to get a $V\_\alpha^M$ satisfying them, but then since the $\omega$ of $M$ is nonstandard and $M$ cannot identify its standard cut, it follows by overspill that there must be some nonstandard finite set of ZF axioms in $M$ that $M$ thinks is satisfied in one of its $V\_\alpha^M$. But since this includes all the standard axioms, we have thus obtained a $V\_\alpha^M$ satisfying the true ZF as viewed externally.
| 9 | https://mathoverflow.net/users/1946 | 40126 | 25,685 |
https://mathoverflow.net/questions/40082 | 109 | I'm not teaching calculus right now, but I talk to someone who does, and the question that came up is why emphasize the $h \to 0$ definition of a derivative to calculus students?
Something a teacher might do is ask students to calculate the derivative of a function like $3x^2$ using this definition on an exam, but it makes me wonder what the point of doing something like that is. Once one sees the definition and learns the basic rules, you can basically calculate the derivative of a lot of reasonable functions quickly. I tried to turn that around and ask myself if there are good examples of a function (that calculus students would understand) where there isn't already a well-established rule for taking the derivative. The best I could come up with is a piecewise defined function, but that's no good at all.
More practically, this question came up because when trying to get students to do this, they seemed rather impatient (and maybe angry?) at why they couldn't use the "shortcut" (that they learned from friends or whatever).
So here's an actual question:
What benefit is there in emphasizing (or even introducing) to calculus students the $h \to 0$ definition of a derivative (presuming there is a better way to do this?) and secondly, does anyone out there actually use this definition to calculate a derivative that couldn't be obtained by a known symbolic rule? I'd prefer a function whose definition could be understood by a student studying first-year calculus.
I'm not trying to say that this is bad (or good), I just couldn't come up with any good reasons one way or the other myself.
**EDIT**: I appreciate all of the responses, but I think my question as posed is too vague. I was worried about being too specific, so let me just tell you the context and apologize for misleading the discussion. This is about teaching first-semester calculus to students straight out of high school in the US, most of whom have already taken a calculus course in high school (and didn't do well or retake it for whatever reason). These are mostly students who have no interest in mathematics (the cause for this is a different discussion I guess) and usually are only taking calculus to fulfill some university requirement. So their view of the instructor trying to get them to learn how to calculate derivatives from the definition on an assignment or on an exam is that they are just making them learn some long, arbitrary way of something that they already have better tools for.
I apologize but I don't really accept the answer of "we teach the limit definition because we need a definition and that's how we do mathematics". I know I am being unfair in my paraphrasing, and I am NOT trying to say that we should not teach definitions. I was trying to understand how one answers the students' common question: "Why can't we just do this the easy way?" (and this was an overwhelming response on a recent mini-evaluation given to them). I like the answer of $\exp(-1/x^2)$ for the purpose of this question though.
It's hard to get students to take you seriously when they think that you're only interested in making them jump through hoops. As a more extreme example, I recall that as an undergraduate, some of my friends who took first year calculus (depending on the instructor) were given an oral exam at the end of the semester in which they would have to give a proof of one of 10 preselected theorems from the class. This seemed completely pointless to me and would only further isolate students from being interested in math, so why are things like this done?
Anyway, sorry for wasting a lot of your time with my poorly-phrased question. I know MathOverflow is not a place for discussions, and I don't want this to degenerate into one, so sorry again and I'll accept an answer (though there were many good ones addressing different points).
| https://mathoverflow.net/users/321 | Why do we teach calculus students the derivative as a limit? | This is a good question, given the way calculus is currently taught, which for me says more about the sad state of math education, rather than the material itself. All calculus textbooks and teachers claim that they are trying to teach what calculus is and how to use it. However, in the end most exams test mostly for the students' ability to turn a word problem into a formula and find the symbolic derivative for that formula. So it is not surprising that virtually all students and not a few teachers believe that calculus means symbolic differentiation and integration.
My view is almost exactly the opposite. I would like to see symbolic manipulation banished from, say, the first semester of calculus. Instead, I would like to see the first semester focused purely on what the derivative and definite integral (*not* the indefinite integral) are and what they are useful for. If you're not sure how this is possible without all the rules of differentiation and antidifferentiation, I suggest you take a look at the infamous "Harvard Calculus" textbook by Hughes-Hallett et al. This for me and despite all the furor it created is by far the best modern calculus textbook out there, because it actually tries to teach students calculus as a useful tool rather than a set of mysterious rules that miraculously solve a canned set of problems.
I also dislike introducing the definition of a derivative using standard mathematical terminology such as "limit" and notation such as $h\rightarrow 0$. Another achievement of the Harvard Calculus book was to write a math textbook in plain English. Of course, this led to severe criticism that it was too "warm and fuzzy", but I totally disagree.
Perhaps the most important insight that the Harvard Calculus team had was that the key reason students don't understand calculus is because they don't really know what a function is. Most students believe a function is a formula and nothing more. I now tell my students to forget everything they were ever told about functions and tell them just to remember that a function is a box, where if you feed it an input (in calculus it will be a single number), it will spit out an output (in calculus it will be a single number).
Finally, (I could write on this topic for a long time. If for some reason you want to read me, just google my name with "calculus") I dislike the word "derivative", which provides no hint of what a derivative is. My suggested replacement name is "sensitivity". The derivative measures the sensitivity of a function. In particular, it measures how sensitive the output is to small changes in the input. It is given by the ratio, where the denominator is the change in the input and the numerator is the induced change in the output. With this definition, it is not hard to show students why knowing the derivative can be very useful in many different contexts.
Defining the definite integral is even easier. With these definitions, explaining what the Fundamental Theorem of Calculus is and why you need it is also easy.
Only after I have made sure that students really understand what functions, derivatives, and definite integrals are would I broach the subject of symbolic computation. What everybody should try to remember is that symbolic computation is only one and not necessarily the most important tool in the discipline of calculus, which itself is also merely a useful mathematical tool.
ADDED: What I think most mathematicians overlook is how large a conceptual leap it is to start studying functions (which is really a process) as mathematical objects, rather than just numbers. Until you give this its due respect and take the time to guide your students carefully through this conceptual leap, your students will never really appreciate how powerful calculus really is.
ADDED: I see that the function $\theta\mapsto \sin\theta$ is being mentioned. I would like to point out a simple question that very few calculus students and even teachers can answer correctly: Is the derivative of the sine function, where the angle is measured in degrees, the same as the derivative of the sine function, where the angle is measured in radians. In my department we audition all candidates for teaching calculus and often ask this question. So many people, including some with Ph.D.'s from good schools, couldn't answer this properly that I even tried it on a few really famous mathematicians. Again, the difficulty we all have with this question is for me a sign of how badly we ourselves learn calculus. Note, however, that if you use the definitions of function and derivative I give above, the answer is rather easy.
| 140 | https://mathoverflow.net/users/613 | 40136 | 25,693 |
https://mathoverflow.net/questions/40071 | 8 | Let N be a prime integer. We know that the element $c=(0)-(\infty)$ generates the torsion subgroup of $J\_0(N)$ and it has order Num( (N-1)/12). Now, there is a natural map $\pi^\*:J\_0(N) \rightarrow J\_1(N)$, coming from the covering map $\pi:X\_1(N) \rightarrow X\_0(N)$. My question is what is the image of c under this map? Specifically, is it possible for $\pi^\*(c)=0$?
| https://mathoverflow.net/users/92 | Image of the cuspidal subgroup of J_0(N) in J_1(N) | The fact you mention about $(0) - (\infty)$ generating the torsion subgroup of $J\_0(N)$ is Theorem 1 (Ogg's conjecture) on the first page of Mazur's paper "The Eisenstein Ideal". I recommend you actually read this paper. If you get as far as page 2, you will find a "Theorem 2 (twisted Ogg's conjecture)" which concerns the Shimura subgroup $\Sigma$. The construction of this subgroup *essentially* identifies it with the kernel of $J\_0(N) \rightarrow J\_1(N)$, and Proposition 11.6 of *ibid*. shows that $\Sigma$ is of multiplicative type, so BCnrd's remarks apply to the general case.
| 5 | https://mathoverflow.net/users/nan | 40143 | 25,697 |
https://mathoverflow.net/questions/40120 | 11 | **Question.** Is it true that each infinite hyperbolic group
has a torsion-free subgroup of finite index?
Are there counterexamples, or positive results for some large subclasses of hyperbolic groups?
For example, is the answer positive for orbifold fundamental groups of negatively curved orbifolds? More precisely, I am interested the most in the case of orbifolds with locally $CAT(0)$ metric. I guess it will be hard to construct a counterexample in this category.
**Related question**. Is it known that every nontrivial hyperbolic group has a proper subgroup of finite index?
Just to recall, a definition of hyperbolic group is here <https://en.wikipedia.org/wiki/Hyperbolic_group> .
**Added.** Note, that every hyperbolic group is finitely presented (thanks to Sam Nead).
| https://mathoverflow.net/users/943 | Existence of finite index torsion-free subgroups of hyperbolic groups | This is a well known open problem. The following properties are equivalent
a) Every hyperbolic group is residualy finite
b) Every hyperbolic group has a finite index torsion-free subgroup.
The proof is either here: Olʹshanskiĭ, A. Yu.
On the Bass-Lubotzky question about quotients of hyperbolic groups.
J. Algebra 226 (2000), no. 2, 807--817 or here: Kapovich, Ilya; Wise, Daniel T. The equivalence of some residual properties of word-hyperbolic groups. J. Algebra 223 (2000), no. 2, 562--583 or can be given by exactly the same methods as in these two papers (I do not remember exactly which of these three possibilities hold).
| 21 | https://mathoverflow.net/users/nan | 40149 | 25,700 |
https://mathoverflow.net/questions/40144 | 4 | Suppose $H$ is a Hilbert space, $B(H)$ is the algebra of bounded linear operators on it, $K(H)$ is ideal of compact operators in $B(H)$, $Inv(B(H)/K(H))$ is the topological group of invertible operators in $B(H)/K(H)$, $Inv(B(H)/K(H))\_0$ --- connected component of $id$ in $Inv(B(H)/K(H))$. $ind\colon Inv(B(H)/K(H))\to \mathbb{Z}$ --- Fredholm index.
I want to find a reference for the following fact:
>
> Fact 1. If $H$ is infinite-dimensional and separable then $ind$ is locally-constant and provides an isomorphism between $Inv(B(H)/K(H))/Inv(B(H)/K(H))\_0$ and $\mathbb{Z}$.
>
>
>
In Murphy's [textbook](http://rads.stackoverflow.com/amzn/click/0125113609), where I've read almost all I know about Fredholm index, there is [Atkinson's theorem](http://en.wikipedia.org/wiki/Atkinson%27s_theorem), the fact, that Fredholm index is locally constant and $ind(ab)=ind(a)+ind(b)$.
But in order to check the fact 1, I need also the fact, stated in the head of the question, or, equivalently, the following:
>
> Fact 2. If $H$ is infinite-dimensional and separable then the set $\{a\in B(H)/K(H)\mid ind(a)=0\}$ is connected.
>
>
>
Where can I find the references? Where can I read that?
| https://mathoverflow.net/users/8134 | Set of invertible operators in B(H) is connected. Is it true? Is there a reference? | You can use some spectral theory to show that the set of unitary operators in $B(H)$ is path-connected. Then the (path-)connectedness of the invertibles follows easily from the polar decomposition. See Theorem 5.29 and Corollary 5.30 in Douglas's *Banach Algebra Techniques* (1st edition).
| 4 | https://mathoverflow.net/users/430 | 40150 | 25,701 |
https://mathoverflow.net/questions/37603 | 15 | Introduction
------------
Graphs are not only important combinatorial objects, but also related to many topological/algebraic structures. In this question I am going to talk about various group structures with combinatorial flavor that one can relate to a graph. In what follows all graphs are connected.
The first example that comes to mind is the fundamental group of a graph when viewed as a topological space. This is a free group and the information it carries is the Euler characteristic, or in simpler words the difference between the number of edges and vertices.
The second example is the critical group of a graph (also called the Sandpile group, Picard group and Jacobian group by different authors) which we can denote as $\mathcal{K}(G)$ for a graph $G$. If we let $L(G)$ be the Laplacian of $G=(V,E)$ this group is nothing more than just $$\mathcal{K}(G) := \mathbb{Z}^{V}/L(G)\mathbb{Z}^{V}$$
so maybe another name for it should be "Laplacian cokernel". By [Kirchhoff's theorem](http://en.wikipedia.org/wiki/Kirchhoff%27s_theorem) the order of $\mathcal{K}(G)$ is precisely the number of spanning trees of $G$. A similar situtation occurs when we consider directed graphs instead. The nice thing about this group is that it behaves nicely under quotients. This means that if one has an automorphism $\phi$ of $G$ then $\mathcal{K}(G/\phi)$ is a subgroup of $\mathcal{K}(G)$. This is not hard to prove without using the critical group itself, see my answer [here](https://mathoverflow.net/questions/25312/number-of-spanning-trees-of-a-quotient-graph/25325#25325). The bad thing here is that it is impossible to put a "natural" bijection between $\mathcal{K}(G)$ and the set of spanning trees of $G$, however this is asking for too much.
A last example is that one can do something similar in the case of perfect matchings for planar bipartite graphs, and get [Kasteleyn cokernels](http://arxiv.org/abs/math/0108150). Here one considers the Kasteleyn matrix instead of the Laplacian, and in some cases can give the group structure explicitly, for example G. Kuperberg in that paper shows that the Kasteleyn-Percus cokernel of the [Aztec diamond](http://en.wikipedia.org/wiki/Aztec_diamond) is $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}/2^{n}\mathbb{Z}$.
Question
--------
Can we construct a group whose cardinality counts the number of non-intersecting paths in a directed graph which start and end in specified sources and sinks? Can we say anything about the structure of such a group and have they been studied before? I am particularly interested in subgraphs of $\mathbb{Z}^2$. Also any comments regarding the philosophy of counting objects by passing through a group structure first, are welcome.
Motivation
----------
A positive answer to this question might help in giving a combinatorial proof to [this](https://mathoverflow.net/questions/26440/non-arithmetic-proof-of-the-integrality-of-a-rational-expression) question (see Qiaochu's comment for example). It would be interesting in general to study the subgroup structure of this (hypothetical) group in general.
| https://mathoverflow.net/users/2384 | Is there a group whose cardinality counts non-intersecting paths? | Hi Gjergji. First, when there is a Gessel-Viennot matrix to count non-intersecting lattice paths, there is also a Gessel-Viennot cokernel. This will happen when the graph is planar and when the sources are all "on the left" and the sinks are all "on the right". Otherwise, if you can't find an integer matrix whose determinant counts the families of lattice paths or whatever, then there is no particular reason to believe in a cokernel.
Second, the Gessel-Viennot determinant is not essentially different from the Kasteleyn-Percus determinant. Every planar non-intersecting lattice path problem can be expressed as a planar perfect matching problem for a modified graph, and the Kasteleyn-Percus matrix reduces to the Gessel-Viennot matrix in a predictable way. In particular, the cokernels are the same. This is explained in my paper that you cite.
Third, even the tree group of a planar graph is not all that different from either Gessel-Viennot cokernel or the Kasteleyn cokernel. Again, you can modify a planar, bipartite graph to turn the matchings into trees. The tree group is more general in the sense that it does not require planarity.
---
Actually, although the Gessel-Viennot method is a very nice and very important result, it is something of a social accident that it is (or has at times been) much more popular than the Kasteleyn method in enumerative combinatorics circles. Kasteleyn published in mathematical physics journals, and his work was known but not widely known among combinatorialists until the 1990s. Then, he published a more complicated Pfaffian expression that applies to non-bipartite graphs, and this was only simplified by Percus in the bipartite case. (Although this simplification is easy.) Then, although Kasteleyn-Percus determinants explain and generalize Gessel-Viennot determinants, the matrices are larger and the more condensed Gessel-Viennot form can look more convenient for explicit calculations. Then too, unless you are studying cokernels, you might not need to know that Gessel-Viennot matrices can come from Kasteleyn-Percus matrices.
On the other hand, there are some things that Gessel-Viennot matrices do not easily show you. Three examples: (1) The same Kasteleyn-Percus matrix may have more than one Gessel-Viennot reduction, which will then have the same cokernel. (2) The minors of a Kasteleyn-Percus matrix give you edge probabilities, a fact which has been used to great effect by Rick Kenyon. (3) You may have to discard symmetry to write down the Gessel-Viennot matrix, which can make it more difficult to analyze matchings that are invariant under a group action.
| 10 | https://mathoverflow.net/users/1450 | 40154 | 25,703 |
https://mathoverflow.net/questions/27701 | 7 | This question is about a theorem in the Haag-Kastler axiomatic approach to quantum field theory (QFT), also known as axiomatic or algebraic or local QFT.
PCT stands for parity, charge and time, a "PCT theorem" says roughly that if a quantum field theory describes a universe, then after reversing the parity, charges and the arrow of time in the theory, the resulting theory still describes the same universe.
It is one of the fundamental symmetries of today's theoretical physics.
Various versions of PCT theorems in different QFT frameworks have been around at least since the 1950ties.
My question is: What versions of the PCT theorem exist that are stated and proven using some version of the Haag-Kastler axioms?
There are three papers that I am aware of:
H.J. Borchers, J. Yngvason: [On the PCT--Theorem in the Theory of Local Observables](http://arxiv.org/abs/math-ph/0012020)
H.J. Borchers: “On Revolutionizing of Quantum Field Theory with Tomita’s Modular Theory”
available free for download [here](http://www.esi.ac.at/preprints/ESI-Preprints.html).
Longo, Guido: [An Algebraic Spin Statistics Theorem](http://arxiv.org/abs/funct-an/9406005)
My motivation is twofold:
1. The "conceptional proofs" of physicists tend to be rather simple and general, but are not rigorous. The proofs in the papers above are rigorous, but rather involved and not as general as I expected (which could simply be a misunderstanding or too big expectations on my part). I would like to know if there is a proof that is either simpler (does use less mathematical machinery, e.g. does not use modular theory) or more general (does not need one of the assumptions or weakens one of the assumptions).
2. I would like to know who constructed the first proof of a PCT theorem in the Haag-Kastler approach and when.
BTW: Any information about the twin of the PCT symmetry, the spin-statistics theorem, in the Haag-Kastler approach would be welcome, too.
| https://mathoverflow.net/users/1478 | Proof of PCT theorem for Haag-Kastler nets in QFT | Jens Mund has some papers on spin-statistics and PCT in the case of massive particles in d=2+1 (e.g. [here](http://arxiv.org/abs/0902.4434) and [here](http://arxiv.org/abs/0801.3621)), but as far as I recall he also uses modular theory, so this might not provide a full answer to your question.
| 2 | https://mathoverflow.net/users/9545 | 40166 | 25,711 |
https://mathoverflow.net/questions/37737 | 27 | A neat construction of Bjorn Poonen shows that the Grothendieck ring of varieties (over a field of char. 0) is not a domain: <http://arxiv.org/abs/math/0204306>
Is the Grothendieck ring of varieties reduced? (My guess: the answer is yes, the proof is easy enough that several people have observed this without writing it up anywhere. But I don't know how to show it.)
| https://mathoverflow.net/users/1310 | Is the Grothendieck ring of varieties reduced? | Qing Liu's example probably works, only we don't know if an abelian variety in
positive characteric is determined by its class in
$K\_0(\mathrm{Var}\_k)$. However, we do know that in characteristic zero (this is
what Bjorn Poonen uses in his examples) and the non-cancellation is a purely
arithmetic phenomenon and hence can be realised in characteristic zero.
Hence, we let $\mathcal A$ be a maximal order in a definite (i.e., $\mathcal
A\otimes\mathbb R$ is non-split) quaternion algebra over $\mathbb Q$. There is
an abelian variety $A$ over some field $k$ of characteristic zero with $\mathcal
A=\mathrm{End}(A)$ (Bjorn works hard to get his example defined over $\mathbb
Q$, here I make no such claim). For any (right) f.g. projective (i.e., torsion
free) $\mathcal A$-module $M$ we may define an abelian variety
$M\bigotimes\_{\mathcal A}A$ characterised by $\mathrm{Hom}(M\bigotimes\_{\mathcal
A}A,B)=\mathrm{Hom}\_{\mathcal A}(M,\mathrm{Hom}(A,B))$ for all abelian
varieties (concretely it is constructed by realising $M$ is the kernel of an
idempotent of some $\mathcal A^n$ and then taking the kernel of the same
idempotent acting on $A^n$). In any case we see that $M$ and $N$ are isomorphic
precisely when $M\bigotimes\_{\mathcal A}A$ is isomorphic to
$N\bigotimes\_{\mathcal A}A$.
Now (all the arithmetic results used below can be found in for instance Irving
Reiner: Maximal orders, Academic Press, London-New York), the class group of
$\mathcal A$ is equal to the ray class group of $\mathbb Q$ with respect to the
infinite prime, i.e., the group of fractional ideals of $\mathbb Q$ modulo
ideals with a strictly positive generators. As that is all ideals we find that
the class group is trivial. Furthermore, we have the Eichler stability theorem
which says that projective modules of rank $\geq2$ are determined by their rank and
image in the class group and hence are determined by their rank (the rank
condition comes in in that $\mathrm{M}\_k(\mathcal A)$ is a central simple
algebra which is indefinite at the infinite prime). In particular if $M\_1$ and
$M\_2$ are two rank $1$ modules over $\mathcal A$ and $A\_1$ and $A\_2$ are the
corresponding abelian varieties we get that $A\_1\bigoplus A\_2\cong A\bigoplus A$
as the left (resp. right) hand side is associated to $M\_1\bigoplus M\_2$ (resp.
$\mathcal A^2$). Therefore, to get an example it is enough to give an example of
an $\mathcal A$ for which there exist $M\_1\not\cong M\_2$. The number (or more
easily the mass) of isomorphism classes of ideals can be computed using mass
formulas and tends to infinity with the discriminant of $\mathcal A$. It is
interesting to note that when the discriminant is a prime $p$ we can go
backwards using supersingular elliptic curves: The mass is equal to the mass of
supersingular elliptic curves in characteristic $p$ and the latter mass can be
computed geometrically to be equal to $(p-1)/24$.
| 15 | https://mathoverflow.net/users/4008 | 40174 | 25,716 |
https://mathoverflow.net/questions/40170 | 13 | A $3$-dimensional compact manifold of negative sectional curvature admits (by geometrisation?) a metric of curvature $-1$, and so its fundamental group has subgroups of finite index. I wonder if an analagous question is open already in dimension $4$ -- i.e. it is not known that $M^4$ with negative sectional curvature always have a cover of finite degree? Are there any positive results in this direction?
This a folow up to question that turned up to be open
[Existence of finite index torsion free subgroups of hyperbolic groups](https://mathoverflow.net/questions/40120/existence-of-finite-index-torsion-free-subgroups-of-hyperbolic-groups)
| https://mathoverflow.net/users/943 | Does a compact negatively curved manfiold of dimension 4 admit a cover of finite degree? | This is a well-known open problem. In fact, there are very few tools for studing general negatively curved manifolds. Even in dimension 3 it is unknown (I think) how to prove existence of proper finite index subgroups without using the geometrization. Geometrization implies residual finiteness of f.g. 3-manifold groups, and hence existence of proper finite index subgroups. As for positive results, lattices in semisimple Lie groups are residually finite. There is only one known method of constructing compact negatively curved manifolds that are not homotopy equivalent to locally symmetric ones, namely, branched covers (with examples given by Mostow-Siu, Gromov-Thurston, and Deraux). I do not know the answer to your question for these examples.
| 18 | https://mathoverflow.net/users/1573 | 40181 | 25,718 |
https://mathoverflow.net/questions/40177 | 2 | Let $\mathcal{R}$ be a Markov partition for the cat map. (How) can it be shown that the Lebesgue measure of a rectangle $R\_j \in \mathcal{R}$ satisfies $\mu(R\_j) = \phi^{-n}$ for some $n$, where $\phi = \frac{1+\sqrt{5}}{2}$?
A "physicist's proof" would be based on the *Ansatz* that $\mathcal{R}$ can be constructed extending local stable and unstable manifolds around the origin *à la* Gallavotti, but it's not clear to me how to build a rigorous argument along these lines.
Any references to work informing an answer would be particularly appreciated. Best of all would be a pointer accounting for the relative measures and multiplicities of *all* rectangles.
| https://mathoverflow.net/users/1847 | Measures of rectangles in Markov partitions for the cat map | Lebesgue measure is both the unique SRB measure and the unique measure of maximal entropy for the cat map. Any Markov partition into $p$ rectangles gives a topological (semi-)conjugacy between the cat map and a subshift of finite type on $p$ symbols with some 0-1 transition matrix $A$. This conjugacy carries the measure of maximal entropy for the cat map (Lebesgue measure) into the measure of maximal entropy for the subshift, which is a Markov measure (namely, the Parry measure). In particular, the measures of the rectangles in the partition are the entries in the probability vector that defines the Parry measure. But these are given very explicitly in terms of the transition matrix $A$: if $u$ and $v$ are the left and right eigenvectors for $A$ corresponding to the maximal eigenvalue (which is $e^{h\_\mathrm{top}(f)} = \phi$), then the entries in the probability vector are proportional to $u\_i v\_i$ (one needs to normalise so that they sum to 1).
At this point it becomes linear algebra; given a 0-1 matrix whose maximal eigenvalue is $\phi$, show that the corresponding eigenvectors $u$ and $v$ have the property that $u\_i v\_i / (\sum\_j u\_j v\_j)$ is of the form $\phi^n$. I'm not sure how difficult this is, but that's the direction I'd attack the problem from.
| 2 | https://mathoverflow.net/users/5701 | 40184 | 25,721 |
https://mathoverflow.net/questions/40062 | 34 | I am interested in learning Mirror Symmetry, both from the SYZ and Homological point of view. I am taking a reading course in Mirror Symmetry, which will focus on the SYZ side.
I know basic Complex geometry, Kahler manifolds, Symplectic manifolds in the geometric side and also reading some material for my course on SYZ conjecture. My major concern is Homological side, about which I have little knowledge.
I am seeking a list of good references for SYZ conjecture, Homological Mirror Symmetry, physics of the theory, modern developments and on its relation to other areas of mathematics and some original papers (preferably in Chronological order).
What are your views about the Claire Voisin's book on Mirror Symmetry.
And what is the present status of research in Mirror Symmetry, I mean what type of problems are people working on.
| https://mathoverflow.net/users/9534 | Roadmap for Mirror Symmetry | [Auroux's notes for a course on mirror symmetry at Berkeley](http://math.berkeley.edu/~auroux/277F09/index.html).
They look interesting and they cover a lot of material.
| 19 | https://mathoverflow.net/users/6658 | 40188 | 25,725 |
https://mathoverflow.net/questions/40198 | 0 | Hi
I have the following matrix
A=[a\_11 a\_12 a\_13 1;
a\_21 a\_22 a\_23 1;
.
.
.
a\_n1 a\_n2 a\_n3 1]
I have seen that when some of a\_ij are big for instance in the order of 200 , then condition
number is also big.
I would like to know is it possible to show it theoretically.
Is it possible to find a lower bound for this problem?
Regards,
Reza
| https://mathoverflow.net/users/9557 | condition number | One possible way to proceed would be to get the singular value decomposition of your matrix and then look at the ratio of the largest singular value to the smallest singular value (a.k.a. the 2-norm condition number); largeness of this condition number implies largeness of the condition number with respect to the other norms of interest, and you can probably just manipulate inequalities at that point.
| 1 | https://mathoverflow.net/users/7934 | 40201 | 25,730 |
https://mathoverflow.net/questions/40207 | 6 | Let $\mathcal{M}$ be an infinite model of a first-order language, and for each $n$, let $\mathcal{B}\_n$ be the algebra of definable sets of $n$-tuples from $|\mathcal{M}|$.
1. Given $\{\mathcal{B}\_n\mid \_{n\in\mathbb{N}}\}$ (and, obviously, $|\mathcal{M}|$, is it possible to describe explicitly some $\mathcal{M}'$ whose definable sets are $\{\mathcal{B}\_n\}$? (I think the question makes the most sense assuming that the language itself is unknown, so I'm asking if there's a natural way to invent a language and a model which gives the chosen definable sets. Obviously there is such a model, namely the original one, but I'm open minded about what it would mean to construct $\mathcal{M}'$ "explicitly".)
2. How well do the definable sets "pin down" the model? Need two models with the same definable sets in the same language be elementarily equivalent? Can anything be said about the relationship between two different models of different languages, but with the same definable sets?
3. There are some obvious restrictions on the $\mathcal{B}\_n$: $\mathcal{B}\_n\times\mathcal{B}\_m\subseteq\mathcal{B}\_{n+ m}$, each $\mathcal{B}\_n$ is a Boolean algebra, $\mathcal{B}\_m$ contains all projections from $\mathcal{B}\_{m+n}$. Are there any others?
| https://mathoverflow.net/users/8991 | Reconstructing a model from its definable sets | The answer to the first part of #2 is no. In a language with (just) one binary relation, let the two models be $\omega$ with the relation interpreted as $<$ in one model and as $>$ in the other.
For #1, the canonical choice would be to take all the relations from the $\mathcal{B}\_n$'s as the interpretations of relation symbols. Presumably, you want a smaller language, but I don't see any good criterion for what would be small enough.
For #3, I think the restrictions you list are all you need (provided "Boolean algebra" means subalgebra of the power set of $|\mathcal{M}|^n$ for the appropriate $n$). They seem to be exactly what you need to do quantifier elimination for the language and interpretation I suggested for #1.
| 5 | https://mathoverflow.net/users/6794 | 40210 | 25,734 |
https://mathoverflow.net/questions/40209 | 4 | In Mendelson's Introduction to Mathematical Logic, the proof of Godel's Theorem for S (his axiomatic arithmetic) goes via proving that a sentence that can be interpreted as "This statement has no proof in S" cannot be proved either false or true in S, if S is consistent.
According to the completeness of the Predicate Calculus, any logically valid wf of a theory K is a theorem of K. The statement I interpret as "This statement has no proof in S" cannot be proved either false or true, so I presume that there are models of arithmetic (or rather of axiom system S) in which it is false, and models in which it is true. Is this correct?
A model of arithmetic in which it is true seems sane enough. Are there models of S in which a proof of what is interpreted as "This statement has no proof in S" turns up as some sort of non-standard number, or have I got completely confused? I have a vision in my mind of a non-standard number encoding "1 is not a proof of S. 2 is not a proof of S. 3 is not a proof of S...." or in some other way satisfying the equation that asserts that X is a proof of Y, if not the mathematician posing the equation :-)
| https://mathoverflow.net/users/9562 | How to reconcile Godel's theorem with the completeness of the Predicate Calculus? | A model of arithmetic in which the G"odel sentence "I am unprovable" is false is necessarily a non-standard model. It contains an infinite element which satisfies, in the model, the formula expressing the property of being a proof of the G"odel sentence --- a formula that is not satisfied by any standard natural number (not even in a non-standard model).
| 4 | https://mathoverflow.net/users/6794 | 40213 | 25,736 |
https://mathoverflow.net/questions/40118 | 5 | Recently, during the research, I came across a sum, denoted by $H(n,L)$, involving irreducible characters of the symmetric group,
\begin{equation}
H(n,L)\colon=\sum\_{Y\_{i,j,w}} \frac{\chi^{Y\_{i,j,w}([2^n])} \chi^{Y\_{i,j,w}}(\tau)}{\chi^{Y\_{i,j,w}}([1^{2n}])} s\_{Y\_{i,j,w}}(1,\ldots,1).
\end{equation}
Notations:
Let $S\_{2n}$ denote the symmetric group of all permutations over $2n$ objects. Let $\chi^{Y}$ denote the irreducible character of $S\_{2n}$ labeled by Ferrers diagram (or shape) $Y$.
Let $Y\_{i,j,w}$ denote a special type of Ferrers diagram having one row of $j+w+1$ boxes, one row of $j+1$ boxes, $i-1$ rows of two boxes and $w$ rows of one boxes, where $i\geq 1$, $j\geq 1$ and $i+j+w=n$.
Let $[2^n]$ and $[1^{2n}]$ denote the conjugacy classes of $S\_{2n}$ having the cycle types $[2^n]$ and $[1^{2n}]$, respectively. Set $\tau = (1,\ldots,n) \circ (n+1,\ldots,2n)$, a permutation of cylce type $[n^2]$.
Let $s\_{Y\_{i,j,w}}(x\_1,\ldots,x\_L)$ denote the Schur-polynomial of $Y\_{i,j,w}$ over $L\geq 2n$ indeterminants.
The sum $H(n,L)$ runs over all the Ferrers diagrams $Y\_{i,j,w}$ such that $i\geq 1$, $j\geq 1$ and $i+j+w=n$.
Using the Murnaghan-Nakayama Rule, we obtain $\chi^{Y\_{i,j,w}}(\tau)=2 (-1)^w$. The Schur-polynomial can be experssed as $s\_{Y\_{i,j,w}}(1,\ldots,1)= \prod\_{(p,q)\in Y\_{i,j,w}}\frac{(L-p+q)}{h\_{pq}} $, where $h\_{pq}$ denote the hook length at position $(p,q)$ of $ Y\_{i,j,w}$. Utilizing the hook length formula, we have $\chi^{Y\_{i,j,w}}([1^{2n}]) =\frac{(2n)!}{\prod\_{(p,q)\in Y\_{i,j,w}}h\_{pq} }$. Therefore, we can simplify the sum $H(n,L)$
\begin{equation}
H(n,L)=\frac{2}{(2n)!} \sum\_{Y\_{i,j,w}} (-1)^w \chi^{Y\_{i,j,w}}([2^n])
\prod\_{(p,q)\in Y\_{i,j,w}} (L-p+q)
\end{equation}
We can also use the Murnaghan-Nakayama Rule to compute $\chi^{Y\_{i,j,w}}([2^n])$. But I can't find a formula! We can also show that the character values $\chi^{Y\_{i,j,w}}([2^n])$ have the property: when $n$ is odd, $\chi^{Y\_{i,j,w}}([2^n])=-\chi^{Y\_{j,i,w}}([2^n])$, and when $n$ is even, $\chi^{Y\_{i,j,w}}([2^n])=\chi^{Y\_{j,i,w}}([2^n])$. For specific $n$, the sum $H(n,L)$ is a polynomial of $L$. A few examples are listed below
\begin{equation}
H(2,L) = \frac{1}{6} L^2 (L^2-1)
\end{equation}
\begin{equation}
H(3,L) = \frac{1}{20} L^3 -\frac{1}{20} L^5
\end{equation}
\begin{equation}
H(4,L) = \frac{1}{140} L^2 -\frac{131}{5040} L^4+\frac{47}{2520}L^6+\frac{1}{5040} L^8
\end{equation}
Questions:
I am wondering if there exists a formula for these character values $\chi^{Y\_{i,j,w}}([2^n])$. Are there any other ways to compute this sum $H(n,L)$ smartly? Is there a formula for the sum $H(n,L)$? Any things about these character values $\chi^{Y\_{i,j,w}}([2^n])$ and sums $H(n,L)$ would also be appreciated.
| https://mathoverflow.net/users/6594 | A sum involving irreducible characters of the symmetric group | For any irreducible character $\chi^\lambda$ of $S\_{2n}$, the value
$\chi^\lambda([2^n])$ can be computed as follows. If $\lambda$ has a
nonempty 2-core, then $\chi^\lambda([2^n])=0$. Otherwise $\lambda$ has
a 2-quotient $(\mu,\nu)$, where $\mu$ and $\nu$ are partitions
satisfying $|\mu|+|\nu|=n$. Say $\mu$ is a partition of $k$, so $\nu$
is a partition of $n-k$. Then $\chi^\lambda([2^n])=\pm{n\choose k}f^\mu
f^\nu$, where $f^\rho$ denotes the dimension of the irrep of $S\_n$
indexed by $\rho$ (given explicitly by the hook-length formula). The
sign is $(-1)^{m/2}$, where $\lambda$ has $m$ odd parts. This result
follow from the theory of cores and quotients, e.g., Section 2.7 of
James and Kerber, *The Representation Theory of the Symmetric Group*.
| 12 | https://mathoverflow.net/users/2807 | 40219 | 25,742 |
https://mathoverflow.net/questions/40223 | 0 | I would like to know an example of a compact Riemannian manifold $M$ and a smooth vector field $X$ on $M$ such that the flow $F$ associated to $X$ is defined for all time and for some vector $v\in T\_pM$, the norm $$|dF\_t(v)|\rightarrow \infty$$ as $t\rightarrow\infty$. Thanks.
| https://mathoverflow.net/users/9563 | An example where the norm of the differential of a flow grows unboundedly | For a minimal example, take $M:=\mathbb{S}^1\times \mathbb{S}^1$ where $\mathbb{S}^1:=\mathbb{R}/2\pi \mathbb{Z},$ with the field $X(u,v):=\sin(u)\partial\_v.$ Then $F\\ ^t(u,v)=\big(u,v+\sin(u)t\big)$ and the differential of the flow at time $t$, computed on the vector $\partial\_u$ at $p=(0,0)$ is just $\partial\_u+t\partial\_v.$
| 2 | https://mathoverflow.net/users/6101 | 40228 | 25,747 |
https://mathoverflow.net/questions/40220 | 2 | This is really two questions. First, consider a normal toric variety $X\_\Sigma$. Its homogeneous coordinate ring
$$R=\mathbb C[x\_1,...,x\_{|\Sigma(1)|}]$$
is graded by $A\_{n-1}(X)$. In analogy with projective space, I guess that there is an analogue of the Proj construction: homogeneous ideals of $R$ not contained in the Stanley-Reisner ideal $B(\Sigma)$ of the fan $\Sigma$.
>
> If $X\_\Sigma$ is projective, is $X\_\Sigma = Proj\_{B(\Sigma)}(R)$?
>
>
>
Assuming this is true, I am curious about the case when $X\_\Sigma$ is quasi-projective; $R$ has non-trivial elements in "negative" degree.
>
> If $X\_\Sigma$ is quasi-projective, is there a analogue of Proj with $X\_\Sigma = QProj\_{B(\Sigma)}(R)$?
>
>
>
| https://mathoverflow.net/users/8363 | Is a (quasi)projective toric variety (Q)Proj of its homogeneous coordinate ring? | No variant is necessary, $X\_{\Sigma}$ is $\mathrm{Proj}\_{B(\Sigma)} (R)$. Note: I'm assuming you already understand how this construction works in the projective case, so that I can jump in and start working an example.
Let's work through the example of $\mathbb{P}^2$ with a point deleted. The corresponding fan has three rays, in directions $e\_1= (1,0)$, $e\_2 = (0,1)$ and $e\_3 = (-1, -1)$ and with two dimensional faces $\mathrm{Span}(e\_1, e\_3)$ and $\mathrm{Span}(e\_2, e\_3)$. The ring $R$ is $k[x,y,z]$, with $x$, $y$ and $z$ corresponding to $e\_1$, $e\_2$ and $e\_3$. The grading is that $x$, $y$ and $z$ are in degree $1$. The ideal $B(\Sigma)$ is $\langle x, y \rangle$.
So, the points of $k[x,y,z]$ correspond to those hemogenous primes of $k[x,y,z]$ which do not contain $\langle x,y \rangle$. Sure enough, that's $\mathbb{P}^2$ with a point removed! Note that we don't change the ring or the grading $R$, both of which are determined by the set of rays of the fan. We just change the irrelevant ideal $B(\Sigma)$.
See David Cox's paper [The Homogeneous Coordinate Ring of a Toric Variety](http://arxiv.org/abs/alg-geom/9210008) for the full generalities.
| 1 | https://mathoverflow.net/users/297 | 40242 | 25,754 |
https://mathoverflow.net/questions/40233 | -1 | Hi people. Can you help me realize why this is true? I can tell you that $P\_i$ and $P\_j$ are probabilities, i.e. $0 \leq P\_i, P\_j \leq 1$.
$\displaystyle \sum\_{i=1}^\infty \sum\_{j=1}^\infty ijP\_iP\_j \leq \sum\_{i=1}^\infty \sum\_{j=1}^\infty j^2P\_jP\_i$.
| https://mathoverflow.net/users/9566 | Why does this inequality hold? | As Will Jagy said it is not true in general.
But assume $S=\sum\_{j=1}^\infty j^2 P\_j$ converges, and apparently you are assuming
$\sum\_{i=1}^\infty P\_i = 1$. Then the right side converges to $S$.
You also know that $i^2+j^2\ge 2ij$ (because $(i-j)^2\ge 0$).
Absolute convergence of the right side lets you rearrange it to
$\sum\_j \sum\_i j^2 P\_jP\_i = \sum\_i\sum\_j i^2P\_iP\_j$. So the right side
is $\frac{1}{2}\sum\_i\sum\_j (i^2+j^2)P\_iP\_j$, which is then greater than
or equal to the left.
| 3 | https://mathoverflow.net/users/6998 | 40246 | 25,758 |
https://mathoverflow.net/questions/40249 | 8 | Let (X,d) be a metric space such that for all points p and q in X, there exists an isometry f such that f(p) = q. Does it follow that for all points p and q in X, there exists an isometry f such that f(p) = q and f(q) = p?
This seems like an obvious enough question that I would be surprised if the answer isn't simply a reference, but I haven't found it mentioned anywhere.
| https://mathoverflow.net/users/nan | Are all homogeneous metric spaces bihomogeneous? | The vertices of a [snub cube](http://en.wikipedia.org/wiki/Snub_cube) form a metric space with 24 points that is homogeneous but not bihomogeneous: the edges of the squares have a "direction" associated with them.
Added later: here is an example with just 6 points: take an equilateral triangle with sides of length 1, and take the 6 points on the edges that are distance 1/4 from a vertex.
Added later: There are no examples with less than 6 points; for example, for 5 points there are 10 edges so there are at most 2 possible lengths with 5 edges of each length, which gives essentially only 1 configuration and this is bihomogeneous. Less than 5 points is easy to do case by case.
| 16 | https://mathoverflow.net/users/51 | 40254 | 25,764 |
https://mathoverflow.net/questions/40001 | 1 | If $G$ is a graph with $n$ vertices and $\frac{nk}{2}$ edges, $k\ge -1,$ then $a(G)\ge \frac{n}{k+1}$. Why?
(Here $a(G)$ is the independence number).
| https://mathoverflow.net/users/9523 | Bounds on the independence number of a graph | by turan theorem, that is very simple:
a(G)=w(G')≥n^2/(n^2-2(n(n-1)/2-m))=n^2/(2m+n)
| 0 | https://mathoverflow.net/users/9523 | 40271 | 25,777 |
https://mathoverflow.net/questions/40195 | 7 | In answer to Pete L. Clark's question [Must a ring which admits a Euclidean quadratic form be Euclidean?](https://mathoverflow.net/questions/39510/) on Euclidean quadratic forms, I gave an example in seven variables, repeated below. Pete's Euclidean property is simply that for any point $\vec x \in \mathbf Q^7$ but $\vec x \notin \mathbf Z^7,$ we require that there be at least one $\vec y \in \mathbf Z^7$ such that $$ q(\vec x - \vec y) < 1. $$
[Edit: This is the definition for positive definite integral quadratic forms. --PLC]
I think my answer works (and the easier 6 variable one), based on extensive computer calculations, and Pete has been too polite to express much doubt.
Could someone please try to prove that this example works (and the 6 variable one)? It seems likely that this lies in the field <http://en.wikipedia.org/wiki/Geometry_of_numbers> but who can say?
$$ q( \vec x) = x\_1^2+ x\_1 x\_2 + x\_2^2 + x\_2 x\_3 + x\_3^2 + x\_3 x\_4 + x\_4^2 + x\_4 x\_5 + x\_5^2 + x\_5 x\_6 + x\_6^2 + x\_6 x\_7 + x\_7^2 + x\_7 x\_1. $$ This has the Euclidean property, its worst behavior is either when all $x\_i = \frac{1}{4}$ or when all $x\_i = \frac{3}{4},$ with ``Euclidean minimum'' equal to $\frac{7}{8}.$
Notice that with $\vec x = \left( \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4} \right),$ the integer lattice points $\vec y$ such that $ q( \vec x - \vec y)=\frac{7}{8} $ include $\vec y = \left( 0,0,0,0,0,0,0\right)$ and all seven cyclic permutations (including the identity) of
$\vec y = \left( 0,1,0,1,0,1,0\right),$ another seven for
$\vec y = \left( 1,0,0,0,0,0,0\right),$ another seven for
$\vec y = \left( 1,0,1,0,0,0,0\right),$ finally seven for
$\vec y = \left( 1,0,0,1,0,0,0\right),$ a total of 29 lattice points on the ellipsoid, of 128 in the standard unit 7-cube. The Gram matrix for the form is
$$ Q \; \; = \; \;
\left( \begin{array}{ccccccc}
1 & \frac{1}{2} & 0 & 0 & 0 & 0 & \frac{1}{2}\\\
\frac{1}{2} & 1 & \frac{1}{2} & 0 & 0 & 0 & 0 \\\
0 & \frac{1}{2} & 1 & \frac{1}{2} & 0 & 0 & 0 \\\
0 & 0 & \frac{1}{2} & 1 & \frac{1}{2} & 0 & 0 \\\
0 & 0 & 0 & \frac{1}{2} & 1 & \frac{1}{2} & 0 \\\
0 & 0 & 0 & 0 & \frac{1}{2} & 1 & \frac{1}{2} \\\
\frac{1}{2} & 0 & 0 & 0 & 0 & \frac{1}{2} & 1
\end{array}
\right) , $$
which has determinant $\frac{1}{32}$ and characteristic polynomial
$$ \left( \frac{1}{64} \right) \left(x - 2 \right) \left(8 x^3 - 20 x^2 + 12 x - 1 \right)^2. $$ So the ellipsoids described are not oblate spheroids, there is less symmetry than that.
EDIT. I think it wise to describe what I am completely certain about and what is unclear. What I did is make a cubic grid, where each variable takes on values $\frac{i}{M}$ for $0 \leq i < M.$ So that makes a grid with $M^7$ points. For each point $\vec r$ in the grid, I find the 128 different values of $q(\vec r - \vec y)$ for
$\vec y \in \mathbf Z^7$ and all coordinates of $\vec y$ are either 0 or 1. For that point $\vec r,$ I take the smallest of the 128 values. Now, for every $M$ I have tried, and for every $\vec r$ in the grid, this best value out of 128 has never been larger than $\frac{7}{8}.$ Now, using the fact that for any $\vec x \in \mathbf Q^7$ that is not in the grid, there is some point $\vec r$ such that $ | \vec r -\vec x | \leq \frac{1}{2 M \sqrt 7},$
I get that I can always find a $\vec y \in \mathbf Z^7$ such that
$$ g(x-y) \leq \frac{7}{8} + \frac{1}{7 M} + \frac{1}{14 M^2}, $$ using Cauchy-Schwarz and the maximum eigenvalue of $Q$ being 2. Anyway, this does not show that the Euclidean minimum is really $\frac{7}{8},$ although I believe it is. What it does show is that the Euclidean minimum is less than 1, as soon as $M \geq 2.$
| https://mathoverflow.net/users/3324 | Verifying an example in the Geometry of Numbers and Quadratic Forms | Consider the form
$$ Q(x) = 2q(x) = (x\_1+x\_2)^2 + (x\_2+x\_3)^2 + \ldots + (x\_7+x\_1)^2.$$
You have to show that it has Euclidean minimum $\frac74$ attained at $X\_1 = x\_1+x\_2 = \frac12$, ..., $X\_7 = x\_7 + x\_1 = \frac12$, but unfortunately not over the lattice ${\mathbb Z}^7$, where it would be trivial, but over a lattice of index $2$ defined by the condition $y\_1 + y\_2 + \ldots + y\_7 \equiv 0 \bmod 2$. It is clear that $(\frac12, \ldots, \frac12)$ has Euclidean minimum $\frac 74$, and it remains to show that all other points in the fundamental domain of the lattice have a minimum at most $\frac74$.
This is not difficult to see: assume you have the point $(\frac12 + \delta, \frac12 + \varepsilon, ...)$ for small $\delta, \varepsilon \ge 0$. Subtracting the point $(1,1,0,\ldots, 0)$ you will get a point with coordinates $(-\frac12 + \delta, -\frac12 + \varepsilon, ...)$. Repeating this procedure you will eventually reach a point with
$|X\_2|, \ldots, |X\_7| \le \frac12$. If $|X\_1| \le\frac12$, we are done. If not, you can
make $|X\_1| < \frac12$ at the cost of making another coordinate $> \frac12$ in absolute values. If you think about this for a minute you will see that we can always reach a point of the form
$$ X = \Big(\frac12 + \delta\_1, \frac12 - \delta\_2, . . . , \frac12 - \delta\_7\Big)$$
with $0 \le \delta\_j \le \frac12$ and $\delta\_1 \le \delta\_i$ for all $i > 1$. It remains to show that $q(X) \le \frac74$, which is equivalent to
$$ \delta\_1 - \delta\_2 - \ldots - \delta\_7 + \delta\_1^2 + \ldots + \delta\_7^2 \le 0. $$
Does this inequality hold?
| 5 | https://mathoverflow.net/users/3503 | 40276 | 25,779 |
https://mathoverflow.net/questions/40287 | 22 | My question is simple:
>
> How do the little disks operad and $Gal (\bar {Q}/Q)$ relate?
>
>
>
I realize that a huge amount of heavy-machinery can be brought into an answer to this, but I'm struggling with the basics. All papers I've found just seem to jump into the deep-end or involve musings that are more inspirational than precise; so I'm eager to read what people here say.
| https://mathoverflow.net/users/7867 | Little disks operad and $Gal (\bar {Q}/Q)$ | A short answer would be: $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ acts faithfully on the profinite fundamental groupoïd of the operad of little discs.
If $X$ is an algebraic variety over $\mathbb{Q}$ we have an exact sequence
$$
1 \to\pi\_1(X\otimes \overline{\mathbb{Q}},p) \to
\pi\_1(X,p) \to Gal(\overline{\mathbb{Q}}/\mathbb{Q}) \to 1
$$
Here $\pi\_1(X\otimes \overline{\mathbb{Q}};p)$ is canonically identified with the profinite completion of the usual topological fundamental group $\pi\_1(X(\mathbb{C}),p)$.
If the basepoint is defined over $\mathbb{Q}$, this split and we have an action
$$
Gal(\overline{\mathbb{Q}}/\mathbb{Q}) \to Aut(\widehat{\pi}\_1(X(\mathbb{C}),p)).
$$
The (profinite completion of the) fundamental groupoïds of the $C\_2(n)$ inherit the operad structure. The trick is that all of it can be defined over $\mathbb{Q}$ as $C\_2(n)$ is homotopy equivalent to the configuration space of points on the affine line $F(\mathbb{A}^1\_{\mathbb{Q}},n)(\mathbb{C})$. One has to define rational "tangential base points" and check that the operad structure on the fundamental groupoïds is also defined over $\mathbb{Q}$. The resulting operad is described [here](https://mathoverflow.net/questions/39945/compatibility-of-braids-as-a-simplicial-set-and-as-a-braided-monoidal-category/39981#39981). One can explicitly compute its automorphism group. This is the Grothendieck-Teichmuller group $\widehat{GT}$.
As everything is defined over $\mathbb{Q}$, $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ operates on the whole operad. So we have a morphism
$$
Gal(\overline{\mathbb{Q}}/\mathbb{Q}) \to \widehat{GT}
$$
It follows from a theorem of Belyi that it is injective.
| 26 | https://mathoverflow.net/users/1985 | 40293 | 25,788 |
https://mathoverflow.net/questions/40300 | 2 | Let $\mathcal{V}$ be an outer measure on $X$,
$(A\_\alpha)\_{\alpha\in I}$ be a chain of increasing subsets of $X$.
1. Is it true that $\mathcal{V}(\bigcup\_{\alpha\in I}A\_\alpha)=\sup\_{\alpha\in I}\mathcal{V}(A\_\alpha)$?
2. If this is not true in general, are there classes of spaces $X$ and outer measures $\mathcal{V}$ (for example Hausdorff measures on metric spaces) for which this holds?
| https://mathoverflow.net/users/1272 | Increasing sets Lemma for chains | One can see a counterexample easily for the reals $\mathbb{R}$ if the Continuum Hypothesis holds, for in this case the reals $\mathbb{R}$ are the union of a chain of countable sets. Simply well-order the reals in order type $\omega\_1$ and for countable ordinals $\alpha$ let $X\_\alpha$ be the first $\alpha$ many points in this enumeration. So every $X\_\alpha$ is countable, but the union is all of $\mathbb{R}$.
More generally, avoiding the CH assumption, let $\kappa$ be the cardinality of the smallest non-measure $0$ set $X$ of reals. We can enumerate $X$ in order type $\kappa$, and the initial segments of this enumeration all have measure $0$, but the union is $X$, which is non-measure $0$.
The cardinal $\kappa$ used above is known as the *uniformity number* for Lebesgue measure in the theory of cardinal characteristics; for example, see [this MO answer](https://mathoverflow.net/questions/8972#9027) or [Andreas Blass' handbook article](http://www.math.lsa.umich.edu/~ablass/hbk.pdf).
| 3 | https://mathoverflow.net/users/1946 | 40301 | 25,790 |
https://mathoverflow.net/questions/40298 | 1 | Let, $V$ be a vector space over a field $K.$ Let, $T$ be a function from $V$ to $V$ such that
$T(kX) = kT(X)$ for all $k \in K$ and for all $X \in V$ and also
$T(k + X) = T(X)$ for all $k \in K$ and for all $X \in V$.
If $X = (x\_1, x\_2, \ldots, x\_n)$, then $k + X = (k + x\_1, k\_ + x\_2, \ldots, k + x\_n)$.
I would like to know how to study the behavior of $T$ and its effect on the vector space $V$ be studied.
| https://mathoverflow.net/users/9586 | How to study the behavior of a particular function on a Vector Space. | It seems like the question means to set $X=K^n$. The first condition means that $T$ is homogeneous, and the second that $T(k1+x)=T(x)$ for all $x\in X$ and $k\in K$, where $1=(1,\cdots,1)\in X=K^n$.
As rpotrie says, move to projective space $PK^{n-1}$. This is the set of lines through the origin, or the $K^n$ mod the equivalence relation that $x \sim kx$ for any $k\not=0$. Write the equivalece class of $(x\_1,\cdots,x\_n)$ as $[x\_1,\cdots,x\_n]$. As $T$ is homogeneous, it drops to a map $T:PK^{n-1}\rightarrow K^n$. The second condition is just that $T [x\_1+k,\cdots,x\_n+k] = T[x\_1,\cdots,x\_n]$ for any $k\in K$. This is equivalent to $T[0,x\_2-x\_1,\cdots,x\_n-x\_1] = T[x\_1,\cdots,x\_n]$.
So it seems to me that $T$ is completely determined by some map (which need satisfy no further conditions at all) $PK^{n-2}\rightarrow K^n$.
| 2 | https://mathoverflow.net/users/406 | 40304 | 25,793 |
https://mathoverflow.net/questions/40282 | 11 | Hello!
In "Homological Algebra on a Complete Intersection", Eisenbud proves the following:
Let $A$ be a commutative ring, $M$ be an $A$-module and $F^{\ast}\to M$ an $A$-free resolution. Further, assume that $M$ is annihilated by $I := (x\_1,...,x\_n)$, and that $I$ contains a non zero divisor of $A$. Then there exist maps $s\_{\alpha}: F^{\ast}\to F^{\ast}$, indexed by multiindices $\alpha$ of length $n$, of degree $2|\alpha|-1$, with the following properties:
* $s\_0$ is the differential of $F^{\ast}$.
* $s\_j := s\_{(0,0,...,0,1,0,...,0)}$ is a nullhomotopy for the multiplication by $x\_j$.
* For any $\gamma$ with $|\gamma|\geq 2$ we have
$\sum\limits\_{\alpha+\beta=\gamma} s\_{\alpha} s\_{\beta} = 0$.
Now he asks the following: Assuming $(x\_1,...,x\_n)$ is regular and $F^{\ast}$ is the minimal free resolution of $M$, is is possible to choose the nullhomotopies $s\_j$ in such a way that $s\_j^2=0$ and $s\_i s\_j = -s\_j s\_i$? I.e. can we make $F^{\ast}$ into a differential graded module over the Koszul-Algebra of $(x\_1,...,x\_n)$?
I'm interested in this question and would like to know if progress has been made to answer it. Is it possible to choose the $s\_i$ as above, and, if not, how can the obstruction be described?
**Edit:**
Is it possible to handle all the ways the $s\_j$ can be constructed? Does it have something to do with the $A\_{\infty}$-stuff?
Thank you!
Hanno
| https://mathoverflow.net/users/3108 | Differential graded structures on free resolution? | It is true if the projective dimension of $M$ over $A$ is at most $3$, and counter examples exist when the projective dimension is $4$.
The first counter example was given in Lucho Avramov's [paper](http://www.jstor.org/pss/2374187)
"Obstructions to the existence of a multiplicative structure on minimal resolutions". A simplified example, ($A=k[t\_1,t\_2,t\_3,t\_4], M= A/(t\_1^2, t\_1t\_2, t\_2t\_3, t\_3t\_4, t\_4^2)$, here you can choose the $x\_i$s to be any regular sequence in the annihilator of $M$)
together with discussion of related and more recent results can be found in Section 2 of this [note](http://books.google.com/books?id=C5j4xCl8Q80C&lpg=PA1&ots=F6RUMrczXS&dq=Infinite%20free%20resolutions&pg=PA1#v=onepage&q=Infinite%20free%20resolutions&f=false) (available on Avramov's [website](http://www.math.unl.edu/~lavramov2/papers.html)).
| 12 | https://mathoverflow.net/users/2083 | 40307 | 25,794 |
https://mathoverflow.net/questions/40309 | 12 | Question as in title (Diff = category of smooth manifolds and smooth maps)
I thought I'd convinced myself this is true, so this is just a sanity check.
Also, what about for settings other than smooth manifolds? (like analytic manifolds, complex manifolds, or less differentiable - say, $C^2$ manifolds)
| https://mathoverflow.net/users/4177 | In Diff, are the surjective submersions precisely the local-section-admitting maps? | There are two possible meanings for the sentence "*f* : *M* → *N* admits local sections",
so let's first disambiguate.
**Meaning 1:** For every point of *N*, there exists a neighborhood of that points and a section from that neighborhood back to *M*.
That's what people typically check in order to verify that, say, a map is a $G$-principal bundle.
**Meaning 2:** For every point *m* ∈ *M*, there exists a neighborhood of $f(m)$, and a section *s* from that neighborhood back to *M*, subject to the extra condition that $s(f(m))=m$.
Clearly, you care about the ***second*** meaning of that sentence.
---
It is correct that a map is a submersion (not necessarily surjective!) iff it admits local sections.
If a map has local sections, then the maps on tangent spaces are sujective: that's just obvious.
Conversely, if a map is surjective at the level of tangent spaces, you first pick a local section of the maps of tangent spaces. Then, to finish the argument, you use the fact that
any subspace of the tangent space $T\_mM$ is the tangent space of a submanifold of *M*, and apply the implicit function theorem.
**Note:** if you care about infinite dimensional Banach manifolds, then the existence of a section for the map to tangent spaces needs to be assumed a separate condition. Indeed, it's not enough to assume that the map of tangent spaces is surjective, since it's not true that any surjective map of Banach spaces has a section.
**Note:** For complex varieties, you don't have the implicit function theorem, so it doens't work. Counterexample: the map $z\mapsto z^2$ from ℂ\* to itself. The fix is to pass the the "étale topology"... but that's another story.
| 17 | https://mathoverflow.net/users/5690 | 40312 | 25,798 |
https://mathoverflow.net/questions/40296 | 6 | In his book on set theory, Kunen often emphasizes how important it is to distinguish between statements in the theory and the meta-theory. I have two questions:
a) When we are talking about set theory, isn't this distinction superfluous? For example when we formalize logic in set theory, there exists an enumeration of all formulas and you can make recursive definitions with them. This makes some definitions easier and somehow more natural, I think. Or is it possible that we lose something with this approach, perhaps just the philosophical strength of the statements in the meta-theory? Or is it even possible that we can "prove" wrong statements?
b) As I said, Kunen seems to make a distinction between the natural numbers in the meta-theory and the natural numbers in models of set theory. Now, for example at the beginning of chapter V, there is the following lemma:
>
> Let $\phi(x\_0,...,x\_{n-1})$ be any formula whose free variables are among $x\_0,...,x\_{n-1}$; then
>
>
> $\forall A ( \{s \in A^n : \phi^A(s(0),...,s(n-1))\} \in Df(A,n))$.
>
>
>
Here, $Df(A,n)$ is the set of all definable subsets of $A^n$, which was defined by a recursion involving operations such as intersection, complement etc. Thus this lemma is not a definition, although it could be one when we agree with a). Anyway, my problem is the appearence of this natural number $n$. We fix a formula with $n$ free variables in our meta-theory. But how does set theory "know" which $n$ is meant? Actually I don't even know if it is possible to give a formal definition of the set $\{s \in A^n : \phi^A(s(0),...,s(n-1))\}$. I know what this means, everyone knows it, but how can we define this without an "induction on the structure of $\phi$", which I have learned in lectures, but probably also involves a fusion of theory and meta-theory?
b') Another example is an analoguous theorem for ordinal definable sets:
>
> For each formula $\phi(y\_1,...,y\_n,x)$, we have
> $\forall \alpha\_1 ... \forall \alpha\_n (\forall x : \phi(\alpha\_1,...,\alpha\_n,x) \leftrightarrow x=a) \rightarrow a \in OD$
>
>
>
In the definition of OD, we need some natural number $n$ and a definable set $R \in Df(R(\beta),n+1)$, etc. ... but why do we know this natural number $n$ in the proof?
| https://mathoverflow.net/users/2841 | Is it important to distinguish between meta-theory and theory? | Just to add to Carl's answer:
If $M=(M,E)$ is a model of set theory ($M$ and $E$ sets), for instance one obtained from
the completeness theorem using the assumption that ZFC is consistent, then $M$ typically
is a nonstandard model, with the internal natural numbers actually being "longer" than
our familiar $\mathbb N$. Now, such a model will also have nonstandard formulas.
While every formula in the real world has a translation in $M$, not every object that
$M$ considers to be a formula corresponds to a formula in the real world.
Now logic as a mathematical theory can be developed inside $M$, with formulas and structures being objects in $M$, and a relation definable in $M$ that tells us which structures are models of which formulas.
It could happen, by the second incompleteness theorem, that $M$ is a model of $\neg\text{Con}(\text{ZFC})$. What does this mean? This means that $M$ does not know a structure
that is a model of $M$'s version of ZFC. But it also means that $M$ knows a proof
of the inconsistency of ZFC. Clearly, this proof cannot be translated back into the real world, in other words, it must be a nonstandard proof (nonstandard length, using nonstandard axioms or rules).
These are just some points why we have to separate "mathematics" (in this case stuff concerning objects in $M$) and
"metamathematics" (stuff going on in the real world concerning $M$).
| 8 | https://mathoverflow.net/users/7743 | 40338 | 25,810 |
https://mathoverflow.net/questions/40191 | 7 | I am trying to understand the difference between PCA and FA. Through google research, I have come to understand that PCA accounts for all variance, while FA accounts for only common variance and ignores unique variance.
However, I am having a difficult time wrapping my head around how exactly this occurs. I know PCA rotates the axis used to describe the data in order to eliminate all covariance. Does this step still occur in FA? If not, what differentiates FA from PCA? Thanks in advance.
| https://mathoverflow.net/users/9556 | The difference between Principal Components Analysis (PCA) and Factor Analysis (FA) | The difference between PCA and FA can be thought of in terms of the underlying statistical models (regardless of estimation methods, although these will change depending on the model used).
Consider $n$ iid observations of a $p$ dimensional (column) vector $X$. Suppose that for each $X\_i$, $i \in \lbrace 1, \dots, n\rbrace$, we also had a $k$ dimensional vector $f\_i$, with $k \leq p$. These are our "latent factors". A (linear) factor model assumes that $\mbox{E}(X\_i \mid f\_i) = Bf\_i$, where $B$ is a $p \times k$ "factor loadings" matrix and $\mbox{Cov}(X\_i \mid f\_i) = \Psi$, a diagonal matrix. If we further assume that $\mbox{V}(f\_i) = \mbox{I}\_k$ so that the factors are independent we see that the marginal covariance is $\Sigma \equiv \mbox{Cov}(X\_i) = BB^t + \Psi$.
Roughly, you can think of PCA as making the assumption that $\Psi$ is the zero matrix. In both cases the goal is to find/estimate rotations ($B$) that explain covariance patterns.
If we remove the estimation part of the problem and assume we have $\Sigma$ in hand, the difference is between two ways of decomposing a covariance matrix. We either want a "factor decomposition" $\Sigma = BB^t + \Psi$ or a principle component decomposition $\Sigma = BB^t$.
>
> I think the key really is this: Any
> covariance matrix will admit either
> kind of decomposition, but often the
> rank of $B$ will be substantially
> smaller if we allow the diagonal
> elements of $\Psi$ to be non-zero as
> in the factor decomposition.
>
>
>
Incidentally, finding the factor decomposition for a given covariance that minimizes the rank of $B$ is known as the Frisch problem and is computationally demanding.
PS. I hope this isn't merely a restatement of your remark that "PCA accounts for all variance, while FA accounts for only common variance and ignores unique variance".
| 9 | https://mathoverflow.net/users/8719 | 40352 | 25,816 |
https://mathoverflow.net/questions/40346 | 2 | Let G be an F-algebra group(G=1+J , where J is the jacobson radical of a finite dimensional F-algebra ,where F is a field of prime characteristic)
In a paper of Isaacs ("Characters of groups associated with finite algebras" from 1995) there is a claim of Gutkin with a wrong proof.It says:
Let x be an irreducible character of G ,then $ x=a^G $ ,where a is a linear character of some subgroup H<=G of the form : H=1+U where U is multiplicativly closed F-subspace of J.
This result would be a generalisation of the result of Isaacs paper,but its unclear to me if it has been proven until today.Does someone know if the result is true and can recommand me a paper/link for more information?
| https://mathoverflow.net/users/9514 | A result about Characters of F-algebra groups | I think the result you are referring to is over a finite field $F$ (Gutkin's claim was also stated over local fields, however).
The result over finite fields was proved by Z. Halasi in *On the characters and commutators of finite algebra groups*, J. Algebra 275 (2004), 481-487. Halasi's proof uses some results from the paper by Isaacs in a counting argument, and therefore only proves Gutkin's claim for algebra groups over finite fields.
Recently, M. Boyarchenko has freed Halasi's proof from its dependence on Isaacs's results, and has proved Gutkin's claim also for algebra groups over local fields (see *Representations of unipotent groups over local fields and Gutkin's conjecture*, arXiv:1003.2742v1).
| 3 | https://mathoverflow.net/users/2381 | 40363 | 25,826 |
https://mathoverflow.net/questions/40348 | 16 | I discovered experimentally that a certain finite poset (sorry, I cannot give its definition here) seems to be in fact a (non-distributive, non-graded) lattice. The covering relations are reasonably simple, but it seems not so easy to find out whether one element is smaller than the other, let alone find the meet (or join) of two elements. However, it is (relatively) easy to see that the poset has a minimum and a maximum.
I wonder whether there are standard techniques for proving that a poset is a lattice, that do not need knowledge about how the meet of two elements looks like. (In fact, any example would be very helpful.)
Some more hints:
1. I don't see a way to embed the poset in a larger lattice...
2. the poset is (in general) not self-dual, but the dual poset is itself a member of the set of posets I am looking at.
3. to get an idea, [here](http://service.ifam.uni-hannover.de/%7Erubey/poset.pdf) ([Wayback Machine](https://web.archive.org/web/20150512115725/http://service.ifam.uni-hannover.de/%7Erubey/poset.pdf)) is a picture of one example (produced by sage-combinat and dot2tex)
| https://mathoverflow.net/users/3032 | Proving that a poset is a lattice | I have often found the following lemma of Björner, Eidelman and Ziegler to be useful:
>
> Let $P$ be a bounded poset of finite rank such that, for any $x$ and $y$ in $P$, if $x$ and $y$ both cover an element $z$, then the join $x \vee y$ exists. Then $P$ is a lattice.
>
>
>
See Lemma 2.1 "[Hyperplane arrangements with a lattice of regions](https://mathscinet.ams.org/mathscinet-getitem?mr=1036875)", *Discrete and Computational Geometry*, Volume 5, Number 1, 263--288.
The hypothesis that $P$ is bounded means that $P$ has a minimal and maximal element. In fact, this hypothesis is slightly stronger than necessary: One can show that if $P$ has a minimal element and the other hypotheses hold then $P$ has a maximal element.
| 11 | https://mathoverflow.net/users/297 | 40364 | 25,827 |
https://mathoverflow.net/questions/40351 | 6 | Is there a more interesting name for this graph invariant: edges minus vertices? It seems to have been called 'complexity' in
* Remco van der Hofstad, Joel Spencer, *Counting Connected Graphs Asymptotically*, European Journal of Combinatorics **27** Issue 8 (2006) 1294–1320, doi:[10.1016/j.ejc.2006.05.006](https://doi.org/10.1016/j.ejc.2006.05.006), arXiv:[math/0502579](https://arxiv.org/abs/math/0502579)
and in
* Joel Spencer, *Probabilistic Methods in Combinatorics* In: Chatterji S.D. (eds) Proceedings of the International Congress of Mathematicians, Birkhäuser, Basel (1995), doi:[10.1007/978-3-0348-9078-6\_132](https://doi.org/10.1007/978-3-0348-9078-6_132), [Wayback Machine pdf of article](https://web.archive.org/web/20170811094846/http://www.mathunion.org/ICM/ICM1994.2/Main/icm1994.2.1375.1383.ocr.pdf), [big IMU pdf of whole volume](https://www.mathunion.org/fileadmin/ICM/Proceedings/ICM1994.2/ICM1994.2.ocr.pdf)
The motivation is that we want to talk about a quantity that is preserved under the graph transformation of collapsing two distinct vertices connected by an edge to a single vertex (thereby removing one edge and one vertex, preserving 'edges minus vertices'). So for example if the quantity 'edges minus vertices plus one' is more natural for some reason and has a name, then this would also be helpful. The concept should not be restricted to e.g. planar graphs.
| https://mathoverflow.net/users/9501 | edges minus vertices | Whether you're considering a multigraph (which may have multiple edges and/or loops) or a simple graph, both are CW complexes. For any finite CW complex $G$, the *Euler characteristic* $\chi(G)$ is defined as the alternating sum (#0-cells)-(#1-cells)+(#2-cells)-... (see [Wikipedia](http://en.wikipedia.org/wiki/Euler_characteristic)). Thus for a finite graph, the Euler characteristic is $|V|-|E|$. It's a homotopy invariant, and the operation of collapsing one edge and its vertices to a single vertex is a homotopy equivalence, so any function of $|V|-|E|$ is invariant under this operation.
When the graph is connected, the quantity $|E|-|V|+1$ ($=1-\chi(G)$) is the smallest number of edges that must be removed to yield a graph with no cycles, called the *cyclomatic number* or the *circuit rank* (see [Mathworld](http://mathworld.wolfram.com/CircuitRank.html)). But if the graph is not connected, then "$+1$" must be replaced by "$+k$," where $k$ is the number of components.
| 13 | https://mathoverflow.net/users/6751 | 40375 | 25,837 |
https://mathoverflow.net/questions/40268 | 58 | This is a heuristic question that I think was once asked by Serge Lang. The gaussian: $e^{-x^2}$ appears as the fixed point to the Fourier transform, in the punchline to the central limit theorem, as the solution to the heat equation, in a very nice proof of the Atiyah-Singer index theorem etc. Is this an artifact of the techniques (such as the Fourier Transform) that people like use to deal with certain problems or is this the tip of some deeper platonic iceberg?
| https://mathoverflow.net/users/9569 | Why is the Gaussian so pervasive in mathematics? | Quadratic (or bilinear) forms appear naturally throughout mathematics, for instance via inner product structures, or via dualisation of a linear transformation, or via Taylor expansion around the linearisation of a nonlinear operator. The Laplace-Beltrami operator and similar second-order operators can be viewed as differential quadratic forms, for instance.
A Gaussian is basically the multiplicative or exponentiated version of a quadratic form, so it is quite natural that it comes up in multiplicative contexts, especially on spaces (such as Euclidean space) in which a natural bilinear or quadratic structure is already present.
Perhaps the one minor miracle, though, is that the Fourier transform of a Gaussian is again a Gaussian, although once one realises that the Fourier kernel is also an exponentiated bilinear form, this is not so surprising. But it does amplify the previous paragraph: thanks to Fourier duality, Gaussians not only come up in the context of spatial multiplication, but also frequency multiplication (e.g. convolutions, and hence CLT, or heat kernels).
One can also take an adelic viewpoint. When studying non-archimedean fields such as the p-adics $Q\_p$, compact subgroups such as $Z\_p$ play a pivotal role. On the reals, it seems the natural analogue of these compact subgroups are the Gaussians (cf. Tate's thesis). One can sort of justify the existence and central role of Gaussians on the grounds that the real number system "needs" something like the compact subgroups that its non-archimedean siblings enjoy, though this doesn't fully explain why Gaussians would then be exponentiated quadratic in nature.
| 61 | https://mathoverflow.net/users/766 | 40382 | 25,842 |
https://mathoverflow.net/questions/40383 | 2 | I am interested in a formula which relating two functions over a multiset.
I have a multiset $X$ of sets where each element in $X$ is a set $x \subseteq \{1,2,\ldots,m\}$. Now I have two ``count'' functions
$p\_s = |\{x \in X : s = x\}|$
$\eta\_s = |\{x \in X : s \subseteq x\}|$
One can expand the formula for the marginal count $\eta\_s$ as
$\eta\_s = \sum\_{s \subseteq t,|t|\leq m} p\_t$
I have confirmed for up to $m = 4$ that the following results holds
$p\_s = \sum\_{s \subseteq t,|t|\leq m} (-1)^{|s|-|t|}\eta\_t$
Does the above result hold for arbitrary $m$? This seems like it must be related to the inclusion/exclusion principle (<http://en.wikipedia.org/wiki/Inclusion-exclusion_principle>) but there is a subtle difference, in that the summation is over set which include $s$ as subsets. Perhaps this difference is immaterial, but I don't see the argument just yet. Also, in the general problem that I wish to solve I will have $x \subseteq \{(i,a\_i)\}\_{i \in I}$ where $I$ is all combinations of $\{1,\ldots,m\}$ and $a\_i$ is drawn from a finite set $A:|A|=n$.
| https://mathoverflow.net/users/6908 | Inverse formula for counting marginals | The answer is yes, and this is known as Moebius inversion. See Section E.1, p.286 in [Graphical models, exponential families, and variational inference.](http://www.eecs.berkeley.edu/~wainwrig/Papers/WaiJor08_FTML.pdf)
| 5 | https://mathoverflow.net/users/7655 | 40384 | 25,843 |
https://mathoverflow.net/questions/40329 | 6 | I am a physicist, and I have the following problem. Consider a locally compact group G acting over a measure space $(X, {\cal B}, \mu)$, and assume that $\mu$ is G-invariant. My problem is how to "quotient" the measure $\mu$ for obtaining a measure $\mu/G$ on the quotient space $X/G$, i.e., the space wose elements are the orbits of G. The simple answer consisiting of defining $\mu/G(\Delta):=\mu(\Delta)$ for $\Delta \in X/G$ is not appropriate, as one can see from the following example.
Let $X:=\mathbb{R}^2$ be the configuration space of two one-dimensional particles, let $\mu$ be the Lebesgue measure, and let $G:=\mathbb{R}$ be the group of translations: $a(x, y)=(x + a, y + a)$ for $a \in G$ and $(x, y) \in X$. The orbit of a point $(x, y)$ is the line at $45^°$ passing for $(x, y)$. It is easy to see that the Lebesgue measure of any set of orbits is either $0$ or $\infty$ .
My tentative answer is the following. Let S be a section of the partition of the orbits, i.e., a subset of X composed by an element for every orbit. Let the map $h\_S:G \times X/G \to X$ be defined as follows: $h\_S(g, \xi)=g s\_\xi$, where $s\_\xi$ is the element of the section $S$ belonging to the orbit $\xi$. It is easy to see that $h\_S$ is a bijection between $G \times X/G$ and $X$. It is also easy to see that a measure $\nu\_S$ on $X/G$ exists such that $\mu=h\_S(\alpha \times \nu\_S)$, where $\alpha$ is the Haar measure on G. The measure $\nu\_S$ is the measure I am looking for. The problem is to prove rigorously that it does not depend on the chosen section S.
I guess that this problem has already been addressed. Can anybody give me some reference?
| https://mathoverflow.net/users/9584 | How to define the quotient of a measure which is invariant under group action? | There is a fair amount of work on this. Since measures are roughly the same as cohomology, the standard approach in quantum field theory (one situation where such integrals are needed) roughly boils down to computing equivariant cohomology. The physics buzzword for this is "BRST".
For the case of finite-dimensional manifolds, I think the best mathematical analysis is:
* A. Weinstein, 2009. The volume of a differentiable stack. *Lett. Math. Phys.* 90(1-3) pp.353--371.
In particular, he explains what are the correct type of smooth measures when a Lie group acts on a manifold with compact stabilizers (the group itself need not be compact).
It is important to keep in mind that the search for invariant measures is simply the wrong way to go about defining measures on the quotient in general. In particular, if you consider a group $G$ acting on a space $M$, then if $G$ is *unimodular* (left Haar measure equals right Haar measure), then the smooth measures on the stacky quotient $M/G$ are in bijection with the invariant measures on $M$ (a choice of bijection corresponds to a choice of Haar measure), but if $G$ is not unimodular (e.g. the two-dimensional nonabelian Lie group) then the correct thing to do is pick a measure on $M$ that fails to be invariant in precisely a way that cancels the failure of unimodularity.
More generally, the Weinstein point of view is that you should not think about a space and a group separately — that's too much data — but just think about the *groupoid* that encodes the action. Then there is a well-developed theory of "equivalence" of groupoids, and Weinstein measures push and pull well across equivalences.
A particular example of an equivalence is given by a "full open neighborhood": let $U \subseteq M$ be any open set that intersects every orbit of $M/G$, and restrict your groupoid to it. (Note: $G$ does not act on $U$, but we define a "partial action" in which two points in $U$ are equivalent iff they're equivalent in $M$, and with the same stabilizers.) Then the restricted groupoid is equivalent to the original one.
So, let's take $M = \mathbb R^2$ and $G = \mathbb R$ as in your example. Then we can restrict to the full subset $U = \{(x,y) \text{ s.t. } |x+y|<1$.
Now it is clear that we can pick an invariant Lebesgue measure with lots of finite-sized orbits. We should divide the volume of any region — this is part of the definition, and in general see [ibid.] — by the length of the orbits: each orbit pulls back to some open subset of $G = \mathbb R$, and measure its length there against Haar measure. You do get well-defined groupoid measure doing this. At the end of the day, it's what you want it to be: the measure of a cylinder over a subset of the line $\{x+y = 0\}$ is just the length of the subset (up to your choice of normalizations).
| 4 | https://mathoverflow.net/users/78 | 40387 | 25,846 |
https://mathoverflow.net/questions/40399 | 43 | Let $X$ be a connected CW complex. One can ask to what extent $H\_\ast(X)$ determines $\pi\_1(X)$. For example, it determines its abelianization, because the Hurewicz Theorem implies that [$H\_1(X)$ is isomorphic to the abelianization of $\pi\_1(X)$](http://en.wikipedia.org/wiki/Fundamental_group#Relationship_to_first_homology_group).
I'm thinking about invariants of 2-knots which can be extracted from have to do with the second homology of (covers of) their complements, and I'm therefore very much interested in the answer to the following question:
> What part of the fundamental group is detected by $H\_2(X)$?
In particular, is there an obvious map from $H\_2(X)$ (or from part of it) into $\pi\_1(X)$?
Where in the derived series of $\pi\_1(X)$ would the image of $H\_2(X)$ live?
| https://mathoverflow.net/users/2051 | What part of the fundamental group is captured by the second homology group? | $H\_2(X)$ is all about $\pi\_1(X)$ and $\pi\_2(X)$. If $\pi\_2(X)$ is trivial (as for knot complements) then it is a functor of $\pi\_1(X)$.
Let $H\_n(G)$ be $H\_n(BG)$, the homology of the classifying space ($K(G,1)$). If $X$ is path-connected than there is a surjection $H\_2(X)\to H\_2(\pi\_1(X))$ whose kernel is a quotient of $\pi\_2(X)$, the cokernel of a map from $H\_3(\pi\_1(X))$ to the largest quotient of $\pi\_2(X)$ on which the canonical action of $\pi\_1(X)$ becomes trivial.
This $H\_2(G)$ isn't anything like the next piece of the derived series after $H\_1(G)=G^{ab}$, though. For example, if $G$ is abelian then $H\_2(G)$ is the second exterior power of $H\_1(G)$ (EDIT: so it can be nontrivial even though it knows no more than $H\_1(G)$ does), while if $H\_1(G)$ is trivial $H\_2(G)$ is often nontrivial (EDIT: so, even when it does carry some more information than $H\_1(G)$, it is not necessarily derived-series information).
EDIT: The previous paragraph comes from looking at the integral homology Serre spectral sequence of $X\to K(\pi\_1(X),1)$, where the homotopy fiber is the universal cover $\tilde X$. Since $H\_1\tilde X=0$, the groups $E^\infty\_{p,1}$ are trivial and we get an exact sequence
$$
0\to E^\infty\_{0,2}\to H\_2(X)\to E^\infty\_{2,0}\to 0,
$$
therefore
$$
E^2\_{3,0} \to E^2\_{0,2}\to H\_2(X)\to E^2\_{2,0}\to 0.
$$
Since $H\_2(\tilde X)=\pi\_2(\tilde X)=\pi\_2(X)$, this looks like
$$
H\_3(\pi\_1(X)) \to H\_0(\pi\_1(X);\pi\_2(X))\to H\_2(X)\to H\_2(\pi\_1(X))\to 0.
$$
The place to look for the rest of the derived series would be homology with nontrivial coefficients, for example homology of covering spaces.
| 46 | https://mathoverflow.net/users/6666 | 40405 | 25,857 |
https://mathoverflow.net/questions/40390 | 22 | Let p(n) be the number of partial orders on the set {1,...,n}. From the Online Encyclopedia of Integer Sequences, we find that the known values of p(n) are
{1,1,3,19,219,4231,130023,6129859,431723379,44511042511,6611065248783,1396281677105899,414864951055853499,171850728381587059351,98484324257128207032183,77567171020440688353049939,83480529785490157813844256579,122152541250295322862941281269151,241939392597201176602897820148085023}.
We see that the units digits of these numbers appear to cycle with a period of length four: 1, 3, 9, 9.
Experiments with other moduli indicate that given a prime modulus m, the sequence cycles with a period of length m-1. If the modulus m is a prime power, then the period appears to be of length phi(m), where phi is Euler's phi-function. For any modulus m, the period appears to be of length the least common multiple (LCM) of the constituent period lengths. For example, if m=12, the period appears to be of length LCM(phi(4),phi(3))=LCM(2,2)=2.
I don't know how to prove this conjecture and I don't see any reference to it. If proved, perhaps this result together with an asymptotic estimate for p(n) could be used to find higher values of p(n).
| https://mathoverflow.net/users/9609 | number of partial orders modulo a fixed number | For q prime, enlarge $\{ 1,\cdots,m \}$ to a set of size $n=m+(q-1)$ by replacing $m$ by $q$ clones $m\_1 , m\_2 , \cdots , m\_q$ and consider the $q$-cycle $\sigma=(m\_1\ m\_2\ \cdots \ m\_q)$. It acts on the set of partial orders of the $n$-set and each of its orbits has size 1 or size q. Each orbit of size 1 arises from a unique partial order of the $m$-set by having all $p$ clones behave identically to the original. This proves that $p(m+(q-1)) \equiv p(m) \mod q $ I think I see how to generalize to $q^k$ but I'll have to think about it. The same idea should apply to a wider variety of structures, but which ones?
**later** The argument seems as if it should work for bipartite graphs on n labelled vertices and also connected bipartite graphs *except for powers of 2* The data at OEIS supports this as far as it goes, ignoring the numbers for less than 3 vertices.
<http://www.oeis.org/A047864> <http://www.oeis.org/A001832>
It also works for appropriate restricted classes such as series parallel networks with n labelled vertices and parallel edges allowed. <http://www.oeis.org/A053554>
Here is my argument for why $p(n+\phi(q^2)) \equiv p(n) \mod q^2$. I think it generalizes to $q^k$: Further enlarge the $n$ set above to one of size $m+q^2-1=n+\phi(q^2)=N$ by replacing each clone $m\_i$ by $q$ clones $m\_{i1}, m\_{i2}, \cdots ,m\_{iq}$ and consider the $q^2$ cycle $$\tau=(m\_{11}m\_{21}\cdots m\_{q1}m\_{12}m\_{22}\cdots m\_{q,q})$$ It acts on partial orders of the $N$-set and the action has orbits of size 1, $q$ and $q^2$. The orbits of size less than $q^2$ are in bijective correspondence with the orbits of the same size for the action of $\sigma$ on partial orders of the $n$-set.
| 17 | https://mathoverflow.net/users/8008 | 40421 | 25,872 |
https://mathoverflow.net/questions/40422 | 10 | This is perhaps most naturally phrased as a promise problem. Given numbers $n$ and $s$, where $s$ is the sum of the prime factors of $n$ (distinct or with multiplicity; I imagine both variants will have the same answer), find the factorization of $n$. Can this be done in deterministic polynomial time?
Alternately, and slightly weaker: is FACTORIZATION (any of the standard decision-problem versions, perhaps "does $n$ have a prime factor between $a$ and $b$?") in $\text{P}^\text{sopf}$?
---
I'm essentially trying to prove to myself that sopf($n$) cannot be calculated faster than by factoring $n$, but the problem seems hopeless (unless $\text{FACTORIZATION}\in\text{P}$, in which case it is uninteresting). This is interesting because it seems 'obvious' that there could be no better approach, but I can't think of a way to formalize it that would be true, let alone have a hope to be proved.
Other approaches to this problem would be welcome.
| https://mathoverflow.net/users/6043 | Can a number be factored quickly, given the sum of its prime factors? | I don't know about the promise problem, but my educated hunch is that computing the sum of the prime factors should indeed be roughly as hard as factoring. Here's why: let $N$ be odd and squarefree. Then if we can compute $sopf(N)$, we'll know the parity of the number of prime factors of $N$, which I've [gone on record](https://mathoverflow.net/questions/3820/how-hard-is-it-to-compute-the-number-of-prime-factors-of-a-given-integer/3842#3842) as believing to be hard. (And nobody's contradicted me, so far...)
Terry Tao's answer to that question, by the way, is incredibly useful in thinking about these types of problems. It also indicates that we can sometimes compute the parity even when we can't do anything else, though; I don't know if that applies to number-of-prime-factors. (I am decidedly not a number theorist...)
| 5 | https://mathoverflow.net/users/382 | 40436 | 25,882 |
https://mathoverflow.net/questions/40440 | 2 | Original Question
=================
Consider an infinite tree of constant degree $k$. For such a tree we can consider the total number of nodes at depth $n$, $g(f)$, and the total number of paths from the root, $p(f)$, to be a function of the constant function, $f=k$. We define $G(f)$ to be the resulting infinite tree. Now let us generalize this idea to functions $f(n)$, with the normal convention that the root has depth $n=0$.
Some examples:
$g(1)=1$
$p(1)=n+1$
$G(1)$ is the infinite tree (path) of constant degree 1
$g(2) = 2^n$
$p(2) = 2^{n+1}-1$
$G(2)$ is the infinite complete binary tree
$g(a) = a^n$
$p(a) = \frac{a^{n+1}-1}{(a-1)}$ for a>1
$G(a)$ is the infinite complete tree of constant degree a
$g(f=n) = n!$
$p(f=n) = !n$
$G(f=n)$ is the infinite complete tree of incremental degree
$g(f=n+1)=(n+1)!$
$g(f=2n)=2n!!$
where !! is the double-factorial
$g(f=3n)=3n!!!$
where !!! is the triple-factorial
$g(f=n^2) = n!^2$
$g(f=n^a) = n!^a$
$g(f=an^b) = a^{n}n!^b$ ; Sequences not in Sloan for a>1
$g(f=n^2+n+1)$ = ? ; Related to absolute values for Sloan A130031
$p(f=n^2+n+1)$ = ? ; Sequence: [1, 2, 7, 62, 1107, 31412, 1273917, ... ]
$g(f=2^n)=2^{((n+1)^2-(n+1))/2}$
$p(f=2^n)= ?$ ; Sequence: [1,3,11,75,1099,33867, .... ]
$g(f=a^n)=a^{((n+1)^2-(n+1))/2}$
$g (f=n!)$ = Sloan A000178
$g (f=2n!)$ = ? ; Related to Sloan A156926. Sequence: [1,2,8,96,4608,1105920,....]
$p (f=2n!)$ = ? ; Sequence: [1,3,11,107,4715,....]
$g (f=3n!)$ = ? ; Sequence: [1,3,18,324,23328,8398080,....]
$p (f=3n!)$ = ? ; Sequence: [1,4,22,346, 23674, 8421754,....]
$g(f=a^{a^n})= a^{a^{n+1}-a(3^n + 1)/2}$
My questions are: is this graph construction well known? I would be interested in any references to similar functional transformations on graphs.
Also, could anyone tell me what is the cardinality of the set of all paths in G(f=n)? Clearly it has at least Continuum cardinality. Since the factorial grows faster than $a^n$, yet slower than $2^{2^n}$, I would think it lands in the Continuum. I am not sure, though...
Addendum
========
We can express p and g as functional equations:
$g(f(n))=f(n) g(f(n-1))$
$g(f(0))=1$
$p(f(n))=\sum\_{i=1}^{n} g(f(i))$
If we extend to the complex domain and consider the special case f(z)=z we have the functional equation:
$g(z)= z g(z-1)$
Which has the rather well known solution $g(z) = \Gamma (z+1)$
| https://mathoverflow.net/users/1320 | Infinite graphs as functional operators | You ask for the size of $G(f)$. This tree is always countable, i.e., its set of nodes is countable. My guess is that you are really asking about is the number of branches (maximal chains).
If every node has a node above it that has at least 2 immediate successors, then the number
of branches is $2^{\aleph\_0}$.
If the tree has too few nodes with at least two immediate succesors, then there are only countably many branches. Any number of branches between $\aleph\_0$ and $2^{\aleph\_0}$
is impossible (this is essentially the Cantor-Bendixson theorem).
Here I should add that I view the tree as a partial order. The root is the smallest element, all its neighbors are above the root and incomparable and so on.
This construction of trees comes up in set theory quite a lot, especially in forcing
($a$-Sacks forcing is forcing with subtrees of $G(f=a)$ ($a$ constant) in
which every node has a node above it that has exactly $a$ immediate successors.
The tree of incremental degree give rise to a forcing notion that I called Miller lite forcing in [A dual open coloring axiom, Annals of Pure and Applied Logic 140 (2006), 40-51], which can be found [here](http://www.hausdorff-center.uni-bonn.de/people/geschke/publications).)
But I have never seen your computations.
| 4 | https://mathoverflow.net/users/7743 | 40449 | 25,888 |
https://mathoverflow.net/questions/40161 | 33 | If $X$ is a scheme, the Hilbert scheme of points $X^{[n]}$ parameterizes zero dimensional subschemes of $X$ of degree $n$.
>
>
> >
> > Why do we care about it?
> >
> >
> >
>
>
>
Of course, there are lots of "in subject" reasons, which I summarize by saying that $X^{[n]}$ is maybe the simplest modern moduli space, and as such is an extremely fertile testing ground for ideas in moduli theory. But it is not clear that this would be very convincing to someone who was not already interested in $X^{[n]}$.
The question I am really asking is:
>
>
> >
> > Why would someone who does not study moduli care about $X^{[n]}$?
> >
> >
> >
>
>
>
The main reason I ask is for the sake of having some relevant motivation sections in talks. But an answer to the following version of the question would be extremely valuable as well:
>
>
> >
> > What can someone who knows ~~a lot~~ something\* about $X^{[n]}$ contribute to other areas of algebraic geometry, or mathematics more generally, or even other subjects?
> >
> >
> >
>
>
>
---
\*reworded in light of the answer of Nakajima
| https://mathoverflow.net/users/4707 | Why do we care about the Hilbert scheme of points? |
>
> What can someone who knows a lot about $X^{[n]}$ contribute to other areas of algebraic geometry, or mathematics more generally, or even other subjects?
>
>
>
I know a little about $X^{[n]}$. And I have no contribution to mathematics nor other areas of algebraic geometry. But I find study of Hilbert schemes is very interesting. Isn't it enough to motivate to study Hilbert schemes ?
| 47 | https://mathoverflow.net/users/3837 | 40459 | 25,891 |
https://mathoverflow.net/questions/40427 | 5 | Does anyone know of a way to simplify this sum?
$$S(n)=\sum\_{j=1}^{\rho(n)}\sum\_{k=1}^\infty\frac{\sin[2\pi k n 2^{-j}]-\sin[2\pi k (n-1) 2^{-j}]}{k}$$
where $\rho(n)=[\log\_2(n)]$ (and $[x]$ denotes the greatest integer less than $x$).
Note: This question is a follow-up to a previous question I asked:
[Greatest power of two dividing an integer](https://mathoverflow.net/questions/29828/greatest-power-of-two-dividing-an-integer)
EDIT: After following all the given suggestions, I found that for integer $n$,
$$\frac{S(n)}{\pi}=2^{-\rho(n)}-1+\frac{1}{1+(-1)^n}\sum\_{j=1}^{\rho(n)}\left[\frac{n}{2^j}\right]-\left[\frac{n-1}{2^j}\right].$$
This is pretty much what I started with in my previous post, so if anyone knows of a way to take this sum, please let me know. Anyway, I will leave this result here in case anyone ever comes across $S(n)$ in some other context. Thanks to everyone who helped.
| https://mathoverflow.net/users/7154 | Difficult Infinite Sum | Mathematica computes just the sum on k as
$$
\frac{1}{2} i \left(\log \left(1-e^{-i 2^{1-j} (n-1) \pi }\right)-\log \left(1-e^{i 2^{1-j} (n-1) \pi }\right)-\log \left(1-e^{-i
2^{1-j} n \pi }\right)+\log \left(1-e^{i 2^{1-j} n \pi }\right)\right)
$$
One can then simplify the sum on j of logarithms as the logarithm of a product.
| 3 | https://mathoverflow.net/users/6756 | 40483 | 25,905 |
https://mathoverflow.net/questions/40491 | 1 | Short question
--------------
Can we describe a quasi-coherent module on a scheme by usual modules with respect to an affine cover, which satisfy some compatibility conditions, which can be formulated in the language of commutative algebra (actually tensor products of modules)? Thus, restrictions to non-affine subsets are not allwed.
Example
-------
If the scheme is separated, the answer is of course yes: Choose an affine cover $U\_i$ of our scheme $X$, thus also $U\_i \cap U\_j$ is affine. If $M\_i$ is a $\mathcal{O}(U\_i)$-module such that we have isomorphisms $\phi\_{ij} : M\_i \otimes\_{\mathcal{O}(U\_i)} \mathcal{O}(U\_i \cap U\_j) \cong M\_j \otimes\_{\mathcal{O}(U\_j)} \mathcal{O}(U\_i \cap U\_j)$ of $\mathcal{O}(U\_i \cap U\_j)$-modules, which satisfy the cocycle conditions, namely $\phi\_{ii} = id$ and $\phi\_{ijk} = \phi\_{kji} \phi\_{ikj}$ where $\phi\_{ijk} = \phi\_{ij} \otimes\_{\mathcal{O}(U\_i \cap U\_j)} \mathcal{O}(U\_i \cap U\_j \cap U\_k)$. Then the quasi-coherent modules $\widetilde{M\_i}$ on $U\_i$ can be glued to a quasi-coherent module on $X$.
Long question
-------------
In the general case, there is an affine cover $U\_i \cap U\_j = \cup\_k W\_k$, such that $W\_k$ is basic-open in $U\_i$ and $U\_j$ simulteneously. In this setting we should use isomorphisms $\phi\_{ijk} : M\_i \otimes\_{\mathcal{O}(U\_i)} \mathcal{O}(W\_k) \cong M\_j \otimes\_{\mathcal{O}(U\_j)} \mathcal{O}(W\_k)$. However, in order to formulate the cocycle condition in affine terms, we actually have to cover triple overlaps $U\_i \cap U\_j \cap U\_k$ by affines $R\_n$, such that $R\_n$ is basic-open in $U\_i,U\_j,U\_j$ and then cover the $U\_i \cap U\_j$ with these $R\_n$, where we vary $k$. Thus we assume isomorphisms $\phi\_{ijkn} : M\_i \otimes\_{\mathcal{O}(U\_i)} \mathcal{O}(R\_n) \cong M\_j \otimes\_{\mathcal{O}(U\_j)} \mathcal{O}(R\_n)$ and may formulate the cocycle condition. Now in order to glue these isomorphisms to an isomorphism on $U\_i \cap U\_j$ between $\widetilde{M\_i}$ and $\widetilde{M\_j}$, we also have to ensure that there is some intersection compatibility when we vary $k$. However, this cannot be formulated with the given data.
Before I end up defining refinements of these affine covers by transfinite induction ;-), I thought it would be better to ask here if there is an easy way which I overlook.
| https://mathoverflow.net/users/2841 | Quasi-coherent module given by modules and compatibility conditions in the language of commutative algebra | I think the explicit description that you suggest can be wrapped up as follows.
For every $i,j$, let $C\_{ij}$ be a family of indexes such that
$$U\_i\cap U\_j=\bigcup\_{a\in C\_{ij}} W\_a,$$
with $W\_a$ being basic open in $U\_i$ and in $U\_j$.
We may assume that $C\_{ii}$ is a singleton set and $C\_{ij}=C\_{ji}$.
Let $\phi\_{ij,a}$ be as in your question, $a\in C\_{ij}$. Impose the cocycle condition
$$\phi\_{ij,a}\circ\phi\_{jk,b}=\phi\_{ik,c}$$
on $W\_a\cap W\_b\cap W\_c$ for all $a\in C\_{ij}$, $b\in C\_{jk}$, $c\in C\_{ik}$. Notice that
$W\_a\cap W\_b\cap W\_c$ is a basic open in $W\_a$, $W\_b$, $W\_c$, and also in $U\_i$, $U\_j$, $U\_k$,
so this is an identity in algebraic terms, as required.
A more natural description:
A quasicoherent sheaf on $X$ is given by a $\Gamma(U,O)$-module $M\_U$ for every affine open $U\subset X$ and an isomorphism $\phi\_{U,V}:M\_U|\_V\to M\_V$ whenever $V\subset U$ is a basic
open, such that $\phi\_{U,V}$ satisfies obvious cocycle condition on triples $W\subset V\subset U$.
| 3 | https://mathoverflow.net/users/2653 | 40501 | 25,913 |
https://mathoverflow.net/questions/40334 | 12 | Recently, while playing around with infinite-divisibility, i arrived at the following metric:
$$d(x,y) := \sqrt{\log\left(\frac{x+y}{2\sqrt{xy}}\right)},$$
defined for positive reals $x$ and $y$. Proving that $d$ is a metric is trivial, except for the triangle-inequality. However, we can bypass a direct proof by appealing to Schoenberg's theorem (I. J. Schoenberg. *Metric spaces and positive-definite functions*, TAMS, 1938), from which the metricity follows easily because $-\log(x+y)$ is a conditionally positive-definite kernel.
However, i have been searching for following:
>
> 1. Applications / situations where this metric shows up?
> 2. An elementary proof of $d(x,y)$ being a metric.
>
>
>
**Remarks**
a. A google search on "ratio arithmetic geometric mean" yields some applications of the *ratio* alone;
b. An elementary proof should exist, but my initial attempts have not been that successful, especially as i stubbornly did not want to use differential calculus.
c. Notice that while proving
$$d(x,y) \le d(x,z) + d(y,z),$$ we may assume wlog $x < 1$ and $y > 1$ and $z=1$, as proving the other cases ranges from very-trivial to trivial.
| https://mathoverflow.net/users/8430 | Logarithm of AM/GM ratio: $\sqrt{\log((x+y)/(2\sqrt{xy}))}$ | Take the coordinate transformation from $\mathbb{R}$ to $\mathbb{R}\_+$ by the exponential map. Then $(x+y) / \sqrt{xy} = e^{a-b} + e^{b-a}$ where $x = e^{2a}$ and $y = e^{2b}$. So we re-write
$$ d(e^{2a}, e^{2b}) = \sqrt{\log \cosh (a-b) } $$
So it suffices to consider the function $q(x) = \sqrt{ \log \cosh x}$, which is an even function with a unique minimum at $x = 0$, where $q(0) = 0$. Triangle inequality reduces then to checking that $q$ is sub-additive. If you were willing to use calculus, then this follows from the fact that $x q'(x)$ is negative except at $0$, where the derivative is not well defined.
In any case, a simple lemma for situations like this is
**Lemma** Let $f(x)$ be an increasing function on the positive real line with $f(0) = 0$. Then $f$ is sub-additive (i.e. $f(s+t) \leq f(s) + f(t)$) if $f(x)/x$ is decreasing.
Thus the triangle inequality boils down to checking $\log \cosh x / x^2$ is a decreasing function of (positive) $x$. At this point, I am lost as to how to check this fact without using any calculus at all. (Somewhere you will need to delve into properties of the natural logarithm or hyperbolic cosine, and calculus is the most natural tool there.)
As to where this metric comes up: the function $q(x)^2$ behaves like $x^2/2$ near the origin and $|x| - \log 2$ for large values. This makes it a Huber function\*, which has found many uses in the literature of applied mathematics, especially with image processing (related to fMRI and things like that) and in robust statistics. (My knowledge here is only superficial, so you are probably better off asking a local expert in those fields about the $\log \cosh$ function.)
\*Huber, P. J., 1973, Robust regression: Asymptotics, conjectures, and Monte Carlo: Ann. Statist., 1, 799-821
| 10 | https://mathoverflow.net/users/3948 | 40502 | 25,914 |
https://mathoverflow.net/questions/40453 | 11 | Is there a unique geodesics between any two points in the [NIL (resp. SOL) geometry](http://en.wikipedia.org/wiki/Geometrization_conjecture#Nil_geometry)?
If so, is there a nice way of parametrizing them? For example geodesics in $S^3$ can be parametrized using the embedding in $\mathbb{R}^4$ and $\sin , \cos$ functions. Geodesics in hyperbolic space can be parametrized using the [hyperboloid model](http://en.wikipedia.org/wiki/Hyperbolic_space) and the functions sinh,cosh.
| https://mathoverflow.net/users/3969 | Are there unique geodesics in the NIL and SOL geometry? | The geodesics between points are not unique in both cases. Moreover the following is true: if $M$ is a universal cover of a compact Riemannian manifold whose fundamental group is virtually solvable but not virtually abelian, then there are conjugate points on some geodesics in $M$ and hence geodesics between some points are not unique.
See Croke and Schroeder "[The fundamental group of compact manifolds without conjugate points](https://eudml.org/doc/140046)", Comment. Math. Helv. 61 (1986), no. 1, 161--175, [MR847526](https://mathscinet.ams.org/mathscinet-getitem?mr=847526), for the case when the metric is analytic, and Lebedeva "On the fundamental group of a compact space without conjugate points", [here](ftp://ftp.pdmi.ras.ru/pub/publicat/preprint/2002/05-02.ps.gz), for the general case.
| 17 | https://mathoverflow.net/users/4354 | 40503 | 25,915 |
https://mathoverflow.net/questions/40493 | 4 | Let $X$ be a smooth projective algebraic variety and $D^b(X)$ be the derived category of coherent sheaves on $X$. Denote by $Sym^nX$ the $n$-th symmetric product of $X$. Can we describe the derived category $D^b(Sym^nX)$ in terms of $D^b(X)$. If so, how are they related? Is there any reference?
This question is intrigued by the question [Hilbert schemes of points and exceptional collections](https://mathoverflow.net/questions/34479/hilbert-schemes-of-points-and-exceptional-collections) asked by Cat.
| https://mathoverflow.net/users/2348 | Derived categories of symmetric products | There is a category closely related to $D^b(Sym^n X)$ which can be described. Namely, the $S\_n$-equivariant derived category of coherent sheaves on $X^n$. This category can be considered as a noncommutative resolution of singularities of $D^b(Sym^n X)$ (the latter category is singular when $\dim X > 1$). The description of the $S\_n$-equivariant derived category I have in mind is the following. First, one can consider the $n$-th tensor power'' of $D^b(X)$ --- if one fixes a DG-enhancement for $D^b(X)$ then it is the derived category of DG-modules over the $n$-th tensor power of the underlying DG-category. The symmetric group acts naturally on this category, so one can consider the corresponding equivariant category. That's it.
| 1 | https://mathoverflow.net/users/4428 | 40508 | 25,918 |
https://mathoverflow.net/questions/40530 | 1 | Is it possible ?
| https://mathoverflow.net/users/9645 | Need an example of not finitely generated graded algebra such that its Poincaré series is a rational function. | Rather obviously yes.
Let $A$ be the algebra over the field $K$ generated by elements $a\_1,a\_2,\ldots,$
with $a\_i$ in dimension $i$ and with $a\_ia\_j=0$ for all $i$ and $j$.
This is an incredibly uninteresting example, but since each graded piece
is one-dimensional, its Poincare series is $\sum\_{n=0}^\infty t^n=1/(1-t)$.
| 8 | https://mathoverflow.net/users/4213 | 40537 | 25,936 |
https://mathoverflow.net/questions/40518 | 7 | A subset of a geodesic metric space is called convex if for every two points in the subset one of the geodesics connecting these points lies in the subset. Is it true that every convex subset of a product of two trees with $l\_1$-metric is a median space, that is for every three points A,B,C in the subset there exists a point D in the subset such that D is on (some) geodesics connecting A and B, B and C, A and C?
| https://mathoverflow.net/users/nan | subsets of products of trees | Yes. Let $A=(A\_1,A\_2)$, $B=(B\_1,B\_2)$ and $C=(C\_1,C\_2)$. For the triangle $A\_1B\_1C\_1$ in the first tree, there is a "center" $M\_1$ such that the (unique) geodesics $[A\_1B\_1]$, $[A\_1C\_1]$ and $[B\_1C\_1]$ contain $M\_1$. Similarly, there is a "center" $M\_2$ for the triangle $A\_2B\_2C\_2$ in the second tree. Let us show that $(M\_1,M\_2)$ belongs to any convex set $S$ containing $A$, $B$ and $C$.
Suppose $A\_1\ne M\_1$ and $A\_2\ne M\_2$. Then consider a geodesic $A(t)=(A\_1(t),A\_2(t))$ connecting $A$ to $B$ in $S$. Its coordinate projections are geodesics in the two trees. Let $A\_1(t)$ hit $M\_1$ before $A\_2(t)$ hits $M\_2$ and this first hit happens at $t=t\_0$. Replace $A$ by $A'=A(t\_0)$. It suffices to prove the assertion for $A'BC$ in place of $ABC$ (note that the centers of the new coordinate triangles are the same points $M\_1$ and $M\_2$). What we gained is that now one of $A\_1$ and $A\_2$ coincides with the corresponding center $M\_1$ or $M\_2$.
Repeat this procedure for $B$ and $C$. Now, for each of $A$, $B$ and $C$, one of the coordinate projections is at $M\_1$ or $M\_2$. So at least two of these projections are at the same point $M\_1$ or $M\_2$. Assume w.l.o.g. that $A\_1=B\_1=M\_1$. Consider a geodesic from $A$ to $B$ in $S$. Its first coordinate projection is constant $M\_1$, and the second one is a geodesic from $A\_2$ to $B\_2$ that must go through $M\_2$ eventually.
| 11 | https://mathoverflow.net/users/4354 | 40546 | 25,942 |
https://mathoverflow.net/questions/40241 | 8 | Let $N$ be a prime number. Let $J(N)$ be the jacobian of $X\_\mu(N)$, the moduli space of elliptic curves with $E[N]$ symplectically isomorphic to $Z/NZ \times \mu\_N$. Over complex numbers we get that $J(N)$ is isogeneous to product of bunch of irreducible Abelian varieties. Is there a way of describing these Abelian varieties using $J\_1(M)$ and $J\_0(M)$? Specifically, what can we say about the decomposition of $J(11)$?
Note that $X\_\mu(N)$ is birationally isomorphic as a curve to the fibre product $X\_0(N^2) \times\_{X\_0(N)} X\_1(N)$. (This is because $\Gamma(N)$ is conjugate to $\Gamma\_0(N^2) \cap \Gamma\_1(N)$, and the group generated by $\Gamma\_0(N^2)$ and $\Gamma\_1(N)$ is $\Gamma\_0(N)$.)
Therefore, we have $J\_1(N)$ and $J\_0(N^2)$ are both some of the factors in $J(N)$. In fact, we know that $J(7)$ is three copies of $J\_0(49)$. For $N=11$, the above fibre product to $X\_0(121)$ is an unramified covering. If I were going to make a guess on what $J(11)$ going to decompose as, I would guess that it is five copies of $J\_0^\text{new}(121)$ and six copies of $J\_1(11)$. Is that reasonable? Is there a geometric way of arguing this?
Also, I'm guessing that the question about
[$\operatorname{SL}\_2(F\_N)$ decompoposition of space of cuspforms](https://mathoverflow.net/questions/4763/sl2-z-n-decomposition-of-space-of-cusp-forms-for-gamman) is related to this, and Jared Weienstein's thesis will come into play here, but I'm not sure how.
| https://mathoverflow.net/users/92 | Geometric decomposition of J(11) | The decomposition of $J(11)$ was known (at least over $\mathbf{C}$) to Hecke. It turns out that the Jacobian of the compactification of $\Gamma(11) \backslash \mathfrak{h}$ is isogenous to a product of 26 elliptic curves. All this is very well explained in the following article :
MR0463118 (57 #3079) Ligozat, Gérard . Courbes modulaires de niveau $11$.
(French) Modular functions of one variable, V (Proc. Second Internat. Conf.,
Univ. Bonn, Bonn, 1976), pp. 149--237. Lecture Notes in Math., Vol. 601, Springer, Berlin, 1977.
The idea is to look at the natural representation of the group $\mathrm{PSL}\_2(\mathbf{F}\_p)$ on the space of cusp forms $S\_2(\Gamma(p))$. So, you're right that there is a geometric interpretation.
If I remember well, there are, among the factors of $J(11)$, elliptic curves of conductor $121$ which are $11$-isogenous to itself. These can be seen as rational points of the modular curve $X\_0(11)$ which are not cusps (there are three such points).
EDIT : I remembered somewhat incorrectly. The three non-cuspidal points of $X\_0(11)(\mathbf{Q})$ correspond to the elliptic curves 121B1, 121C1 and 121C2. The subgroups of order $11$ of these curves are described as follows : the elliptic curve 121B1 has CM by $\mathbf{Z}[\frac{1+i\sqrt{11}}{2}]$, so it is $11$-isogenous to itself, whereas 121C1 and 121C2 are $11$-isogenous to each other. Using the notations of Cremona's tables, the Jacobian of the compactification of $\Gamma(11)\backslash \mathfrak{h}$ is then isogenous to $(11A)^{11} \times (121B)^5 \times (121C)^{10}$.
| 9 | https://mathoverflow.net/users/6506 | 40547 | 25,943 |
https://mathoverflow.net/questions/40484 | 5 | Let $X$ be an homogeneous projective variety, written as the quotient $G/P$, where $G$ is a Lie group and $P$ is a parabolic subgroup of it. It seems it is well-known that the monoid of effective curves on $X$, as a submonoid of $H\_2(X, \mathbb{Z})$, is generated by finitely many curves $\beta\_1, \ldots, \beta\_p$. Moreover the $\beta\_i$ are embedded $\mathbb{P}^1$'s in $X$. For instance in the expository notes <http://arxiv.org/abs/alg-geom/9608011>, the authors use this fact (e.g. at the beginning of p. 37).
The question is simply: how does one prove this fact? Is there a reference for a precise proof of it (perhaps in a slightly different generality, or proving the same statement for the effective monoid inside the Chow group, instead of homology)?
| https://mathoverflow.net/users/9391 | Mori cone of homogeneous varieties | One place this is treated is in Brion's notes, <http://arxiv.org/abs/math/0410240>, particularly Section 1.4, and the references in the notes at the end of that section.
A little more precisely, one shows that a divisor (line bundle) is nef iff it is globally generated, and this cone is generated by the Schubert divisors. Using the fact that the Schubert classes form a self-dual basis for $H^\*(X,{\Bbb Z})$ (as well as $A^\*X$) under the intersection pairing, it follows that the "dual classes" to the Schubert divisors generate the cone of curves. (As with many such things in $G/P$ world, the arguments for cohomology and Chow groups are exactly the same.)
By the way, it's an easy consequence of the existence of a dense open $B$-orbit that the pseudo-effective cone is generated by the complement of the open orbit --- that is, again by the Schubert divisors. So for $G/P$, the nef and pseudo-effective cones are the same.
| 3 | https://mathoverflow.net/users/5081 | 40548 | 25,944 |
https://mathoverflow.net/questions/40553 | 4 | I've seen the fact that the loopspace $\Omega K(G,n)$ is homotopy equivalent to $K(G,n-1)$ mentioned in some places, but I have no idea why. Can anyone offer a good explanation? Also, what happens when $n=1$? Does that just mean the loopspace is a discrete space with $|G|$ points?
| https://mathoverflow.net/users/9653 | Loopspace of an Eilenberg Maclane space K(G,n) | In general,the map $P(X,x\_0)\to X$ from the space of based paths in $X$ is a fibration with fiber $\Omega(X,x\_0)$. Since $P(X,x\_0)$ is contractible, by shrinking paths back toward the base point $x\_0$, the homotopy long exact sequence of the fibration shows that $\pi\_k(X,x\_0)\cong \pi\_{k-1}(\Omega(X,x\_0))$. The result follows. And, yes, the loop space on a $K(G,1)$ is discrete, in one-to-one correspondence with $G$.
| 12 | https://mathoverflow.net/users/1822 | 40560 | 25,951 |
https://mathoverflow.net/questions/40555 | 9 | I'm looking for an example in the literature where $\mbox{Pic}^0(X)$, $\mbox{Pic}(X)$, and $NS(X)$ of a projective surface $X$ over a field are calculated. I want them for an example I'm trying to work out, so ideally $X$ would be relatively simple, perhaps a cubic hypersurface in $\mathbb{P}^3$, or something along those lines. I know it's out there, but googling and browsing arXiv and MathSciNet haven't quite panned out.
| https://mathoverflow.net/users/926 | Calculations of Pic^0, Pic, NS of surfaces | Manin's book "Cubic forms" contains the calculations of these groups when $X$ is a smooth projective cubic surface. In particular, $\operatorname{Pic}^0(X)=\{0\}$ and $\operatorname{Pic}(X)=\operatorname{NS}(X)$ is a free commutative group of rank 7.
Another class of examples is provided by products $X=E \times E'$ of two elliptic curves. In particular, if $E=E'$ has no complex multiplication then $NS(E\times E)$ is a free commutative group of rank 3 generated by the classes of $E \times \{0\}$, $\{0\}\times E$ and the diagonal while $\operatorname{Pic}^0(E\times E)=E \times E$. See Mumford's *Abelian Varieties*.
| 16 | https://mathoverflow.net/users/9658 | 40563 | 25,952 |
https://mathoverflow.net/questions/40562 | 11 | A number of topological invariants take the form of functors $\mathscr{T}\to\mathscr{G}$, where $\mathscr{T}$ is the category of all topological spaces and continuous functions, and $\mathscr{G}$ is the category of all groups and homomorphisms. For examples, consider the homology groups $H\_{n}(X)$ or the homotopy groups $\pi\_{n}(X)$. Of course, a problem with these invariants is that they are not fully faithful functors, i.e., $H\_{n}(X)\cong H\_{n}(Y)$ does not imply that $X$ and $Y$ are homeomorphic. The existence of a fully faithful functor $F:\mathscr{T}\to\mathscr{G}$ would imply that $\mathscr{G}$ has a subcategory $F\mathscr{T}$ equivalent to $\mathscr{T}$. This would be both rather disturbing and extremely interesting. First, it would mean that in a sense, all of topology is just a subset of group theory, which would be rather disturbing to topologists, but it would also reveal a fundamental connection between two seemingly disparate disciplines. My question is: is it possible to prove that no such functor exists? In other words, could one exhibit some categorical property that $\mathscr{T}$ posesses that $\mathscr{G}$ does not. This question can naturally be extended to other important categories, like $\mathscr{M}$, the category of all modules, or $\mathscr{R}$, the category of all rings. So in general, given arbitrary categories $\mathscr{C}$, $\mathscr{D}$, is there any natural way of showing that no fully faithful functor $F:\mathscr{C}\to\mathscr{D}$ exists, i.e., are there any nice "categorical invariants?"
EDIT: A couple people pointed out that I really ought to be discussing fully faithful functors, rather than just faithful functors. Also, I have changed the title in accordance with Martin Brandenburg's recommendation.
| https://mathoverflow.net/users/6856 | Proving the impossibility of an embedding of categories | One property that $Top\_{cgwh}$ has (if we let $Top\_{cgwh}$ be the full subcategory of $Top$ consisting of compactly generated, weakly Hausdorff spaces) that $Grp$ doesn't is cartesian closedness: the hom-set $Grp(G,H)$ is not a group. Another property that $Top$ has that $Grp$ doesn't is an initial object distinct from its final object, but these are just off the top of my head.
But thinking of these functors as being intrinsic to $Top$ is a small mistake, because $Top$ is just a presentation of the $(\infty,1)$-category of homotopy types, and these functors are really giving invariants of homotopy types.
This doesn't really answer your question, but I hope it indicates that the sort of functor you are looking for won't be anything like cohomology or homotopy.
---
Edit: what about the composite $Top \to Set \to Grp$ where the first is the 'underlying set' functor and the second is the 'free group' functor? This is faithful, but perhaps not very useful for topological purposes. (Edit again: I just noticed that Tom Goodwillie also pointed out a similar example)
| 3 | https://mathoverflow.net/users/4177 | 40569 | 25,956 |
https://mathoverflow.net/questions/39934 | 26 | I've heard asserted in talks quite a few times that Lusztig's canonical basis for irreducible representations is known to not always have positive structure coefficents for the action of $E\_i$ and $F\_i$. There are good geometric reasons the coefficents have to be positive in simply-laced situations, but no such arguments can work for non-simply laced examples. However, this is quite a bit weaker than knowing the result is false.
>
> Does anyone have a good example or reference for a situation where this positivity fails?
>
>
>
| https://mathoverflow.net/users/66 | When does Lusztig's canonical basis have non-positive structure coefficients? | Hi,
The following formulas are examples of non-positive structure coefficients
for non-symmetric cases which are easily verified by the algorithm presented
in Leclerc's paper "Dual Canonical Bases, Quantum Shuffles, and q-characters"
or quagroup package in GAP4.
Professor Masaki Kashiwara told me that he has known such non-positive
structure coefficient for $G\_2$ since Shigenori Yamane found it in 1994
as treated in his master thesis at Osaka University (written in Japanese).
You can see similar negative coefficients in at least case $A\_{2n}^{(2)}, D\_{n+1}^{(2)}$.
Anyway, conjecture 52 in Leclerc's paper is false
(I already told Professor Leclerc about it).
Shunsuke Tsuchioka
Notation: $G(i\_1,\cdots,i\_n)$ stands for the canonical basis element
corresponds to a crystal element
$b(i\_1,\cdots,i\_n)=\tilde{f}\_{i\_n}b(i\_1,\cdots,i\_{n-1})=\cdots$.
$G\_2$ (1 is the short root) :
$f\_2 G(121112211)
= G(1211122211)
+ [2]G(1111222211)
+ G(2111112221)$
$ + [2]G(1211112221)
+ G(1111122221)
- G(1112211122)
+ [2]G(1122111122)$
$C\_3$ (1,2 are short roots) :
$f\_3 G(23122312)
= [2]G(222333121)
+ [2]G(312222331)
+ [2]G(231222331)$
$ + [2]G(122223331)
+ G(231223312)
+ [2]G(122233312)
- G(223112233)
+ [2]G(231122233)$
$B\_4$ (1,2,3 are long roots) :
$f\_1G(4342341234)
= [2]G(43344423211)
+ [2]G(43423443211)
- G(44233443211)$
$ + [2]G(43423344211)
+ [2]G(43423442311)
+ [2]G(34234442311)$
$ + [2]G(43422334411)
+ G(43423412341)$
| 23 | https://mathoverflow.net/users/9661 | 40577 | 25,960 |
https://mathoverflow.net/questions/40539 | 10 | EDIT: The original question was answered very quickly (and very nicely!) but the answer leads to a pretty obvious subsequent question, which I will now ask. The original question is maintained for motivational purposes below.
I now know that not every sequence of zeros and ones can be realized as the Stiefel-Whitney numbers for some manifold- as I'm sure many of you all already knew. What I don't know, and what I suspect is a more delicate question, is: Which ones are? Is there a relatively easy necessary condition? Any sufficient conditions?
Along similar lines: are there estimates as to the number of cobordism classes in any dimension that are tighter than the number of "possible" Stiefel-Whitney numbers? A tighter bound, as it were.
Thanks!
---
(original question)
Well I just learned a very cool fact over tea: apparently there are finitely many (unoriented) cobordism classes of compact manifolds in any given dimension! The cobordism class is completely determined by the Stiefel-Whitney characteristic numbers (which were explained to me as "the various numbers one gets by cupping characteristic classes of the tangent bundle together and applying them to the fundamental class, all mod 2")... so that's pretty awesome.
While I get over this initial shock, I was wondering if anyone knew the answer to the following: we have an upper bound on the number of cobordism classes by looking at the number of possible Stiefel-Whitney numbers. But is this upper bound realized?
In other words, given a sequence of zeros and ones (the right number of them), can I always construct a manifold that has precisely that sequence of zeros and ones as its Stiefel-Whitney numbers?
| https://mathoverflow.net/users/6936 | Which sets of Stiefel-Whitney characteristic numbers can be realized as coming from a manifold? | The Steenrod operations on mod 2 cohomology imply the vanishing of some characteristic numbers. Specifically, if $p(w\_1,w\_2,\ldots)\in H^k(M^n;Z/2)$ for $k\lt n$ then $0=\langle \sum\_{i+j=n-k}u\_i{\rm Sq}^{j} p, [M^n]\rangle$ where $u\_i$ is the Wu class of $M$ (but take this with a grain of salt, since I'm quoting from memory here). Thom showed that *all* relations between characteristic numbers arise in this way. This allowed him to compute the bordism ring of unoriented manifolds exactly: it is $Z/2[m\_k\vert k$ not of the form $2^j-1]$. This gives you the tight bound you were asking for.
But, as others have already pointed out, there are plenty of good books & papers on this, so you should probably just dive into the library.
| 7 | https://mathoverflow.net/users/8824 | 40578 | 25,961 |
https://mathoverflow.net/questions/40564 | 1 | The representation of SO(6) is $[i,j,k]$;
The representation of SO(10) is $[i,j,k,m,n]$.
Is there any analytical formula to calculate the dimensions of those representations?
For example,
for SO(6):
dim([1,0,0],D3)=6
dim([0,0,1],D3)=4
dim([0,1,0],D3)=4
dim([0,1,1],D3)=15
dim([0,0,2],D3)=10
dim([i,j,k],D3)=?
for SO(10):
dim([1,0,0,0,0],D5)=10
dim([0,0,0,0,1],D5)=16
dim([0,0,0,1,0],D5)=16
dim([0,0,1,0,0],D5)=120
dim([0,1,0,0,0],D5)=45
dim([i,j,k,m,n],D5)=?
| https://mathoverflow.net/users/6577 | How to calculate the dimensions of representations of SO(6) and SO(10)? | Let $[x\_1,\ldots,x\_\ell]$ denote the vector corresponding to the highest weight of
$D\_\ell$. Then the dimension of the representation is given by
$\prod\_{1\leq i < j \leq \ell} ( 1+ \frac{x\_i+\cdots +x\_{j-1}}{j-i} )
\times$
$\prod\_{1\leq i \leq \ell-1} ( 1+ \frac{x\_i+\cdots + x\_{\ell-2}+x\_{\ell}}{\ell-i} )$
$\times
\prod\_{1\leq i < j\leq \ell-1} ( 1+ \frac{x\_i+\cdots +x\_{j-1}+2(x\_j+\cdots + x\_{\ell-2})+x\_{\ell-1}+x\_{\ell}}{2\ell-i-j} )
$
| 6 | https://mathoverflow.net/users/9666 | 40581 | 25,963 |
https://mathoverflow.net/questions/40499 | 8 | The question is simple:
*Let $P$ be an infinite direct product of copies of $\mathbb Z$. Do there exist any nontrivial extensions
$$0 \to \mathbb Z \to E \to P \to 0$$
in the category of commutative groups?*
In other words, I am asking whether the group $\mathrm{Ext}^1(P,\mathbb Z)$ is trivial. The problem here is of course that the group $P$ is not a free group.
Already a funny thing happens with $\mathrm{Hom}(P,\mathbb Z)$. For any finite or infinite index set $I$, the canonical evaluation map
$$\bigoplus\_{i\in I}\mathbb Z \to \mathrm{Hom}\Big(\mathrm{Hom}\Big(\bigoplus\_{i\in I}\mathbb Z,\:\mathbb Z \Big),\:\mathbb Z \Big) \cong \mathrm{Hom}\Big(\prod\_{i\in I}\mathbb Z,\:\mathbb Z \Big)$$
is an isomorphism! That is a nontrivial statement (due to??), whose proof is not a formality at all. Replacing $\mathbb Z$ by, say, $\mathbb Z/p\mathbb Z$, the corresponding statement is wrong for infinite $I$.
| https://mathoverflow.net/users/5952 | Extensions of an infinite product of copies of Z by Z | Here is a complete answer; I think it is more or less what Steve wrote in his comment, except I don't understand the appearance of $\mathbb{R}$ there. If $I$ is the infinite index set, let $L=\mathbb{Z}^{(I)}\subset P$ be the obvious free submodule. Then $\mathrm{Ext}^1(P,\mathbb{Z})=\mathrm{Ext}^1(P/L,\mathbb{Z})$.
EDIT: the last formula is wrong, see Martin's and Steve's comments below.
Now $P/L$ has a big divisible subgroup $D$, whose inverse image in $P$ consists of maps $I\to\mathbb{Z}$ converging to zero in $\widehat{\mathbb{Z}}$ (the profinite completion of $\mathbb{Z}$). (For instance, if $I=\mathbb{N}$ take the sequence $n\mapsto n!$). Since $P/L$ is torsion-free (imediate), $D$ is a nonzero $\mathbb{Q}$-vector space. Since $D$ is divisible it is a direct summand of $P/L$; hence, $P/L$ admits $\mathbb{Q}$ as a direct summand. But it is well known (and easy to see) that $\mathrm{Ext}^1(\mathbb{Q}/\mathbb{Z},\mathbb{Z})\cong\widehat{\mathbb{Z}}$, hence $\mathrm{Ext}^1(\mathbb{Q},\mathbb{Z})=\widehat{\mathbb{Z}}/{\mathbb{Z}}\neq0$.
| 3 | https://mathoverflow.net/users/7666 | 40586 | 25,966 |
https://mathoverflow.net/questions/40593 | 10 | I consider a bounded open set $A$ in ${\mathbb R}^d$. Is the Hausdorff dimension of the boundary of $A$ at least $d-1$ ? I thought I would have found a result on this problem in any textbook about Hausdorff dimension but I failed. As you may guess I have never work with Hausdorff dimension.
| https://mathoverflow.net/users/9668 | Hausdorff dimension of the boundary of an open set in the Euclidean space - lower bound | If a compact set $C$ separates a connected space $X$ of dimension $d$ (with some homogeneity properties, for example, an euclidean space) in two connected components, then it must have topological dimension at least $d-1$ (see for example the book by Hurewicz and Wallman on topological dimension, Theorem IV.4). In general, [Hausdorff dimension](http://en.wikipedia.org/wiki/Hausdorff_dimension) exceeds (or is equal to) topological dimension (see chapter VII.2 of the book by Hurewicz-Wallman mentioned above).
| 13 | https://mathoverflow.net/users/5753 | 40597 | 25,969 |
https://mathoverflow.net/questions/40324 | 13 | Let $p$ be an irregular prime, which means that $p$ divides some Bernoulli number: $p \mid B\_k$ (for some even $k\in[2,p-3]$). This implies that the class number of the field $K$ of $p$-th roots of unity is divisible by $p$. Let $L$ be the field of $p^2$-th roots of unity. What, if anything, is known about the capitulation of ideal classes in $L/K$ ( we say that an ideal class from $K$ capitulates in $L$ if an ideal generating this class becomes principal there)? It is possible to write down criteria in terms of units that are or are not norms from $L$, but this does not seem to help a lot. I am mainly interested in the question whether there is a connection between the index $k$ and the capitulation of the subgroup of order $p$ corresponding to $k$ via Herbrand-Ribet. I am pretty sure that classical algebraic number theorists did not do an awful lot in this direction but I am not familiar with any advances in Iwasawa theory: whether an ideal class capitulates in $L/K$ is encoded in the Hilbert class field, so the structure of the maximal abelian unramified $p$-extension of the cyclotomic Iwasawa extension of $K$ might contain relevant information. Does it?
| https://mathoverflow.net/users/3503 | Capitulation in cyclotomic extensions | Assume $p$ is an irregular prime for which
[Vandiver's conjecture](https://en.wikipedia.org/wiki/Kummer%E2%80%93Vandiver_conjecture) holds, e.g. $p<12'000'000$. This conjecture asserts that $p$ does not divide the $+$-part of the class group.
Then there is no capitulation in the class group from the first layer of the cyclotomic $\mathbb{Z}\_p^{\times}$-tower to any other in this tower. See Proposition 1.2.14 in [Greenberg's book](http://math.washington.edu/~greenber/book.pdf), which says that the capitulation kernel lies in the $+$-part. See also the discussion on page 102 where it is discussed what happens when Vandiver's conjecture does not hold.
Generally capitulations in Iwasawa theory are well studied. The capitulation is linked to the question of whether there are non-trivial finite sub-$\Lambda$-modules in the Iwasawa module $X$, here the projective limit of the $p$-primary parts of the class groups in the tower, or equivalently the Galois group mentioned in the question.
| 8 | https://mathoverflow.net/users/5015 | 40605 | 25,973 |
https://mathoverflow.net/questions/40602 | 2 | Let $S$ be a finite set. Let $R$ be a complex vector space with basis indexed by subsets of $S$. Define a product on $R$ by defining it on the basis elements as $1\_A\cdot 1\_B=1\_{A\Delta B}$, where $A\Delta B$ is the symmetric difference of $A$ and $B$. This gives $R$ the structure of a commutative and associative $C$-algebra.
Is this a well-understood algebra? Does it have a name?
| https://mathoverflow.net/users/9672 | An algebra constructed from symmetric differences | It is the complex group algebra over $((\mathbb{Z}/2)^S,+)$. This may be also described as the tensor product of $S$ copies of $\mathbb{C}[\mathbb{Z}/2] = \mathbb{C} \times \mathbb{C}$.
| 6 | https://mathoverflow.net/users/2841 | 40607 | 25,974 |
https://mathoverflow.net/questions/40481 | 4 | Let $A$ be a non-zero symmetric $n \times n$-matrix with integer entries and suppose that $\det(A) =0$.
>
> **Question:** How long is the shortest non-zero integer vector in the kernel of $A$?
>
>
>
Example: If $A$ has non-zero eigenvalues $\xi\_1,\dots,\xi\_k$ (not necessarily counting with multiplicities), then the polynomial $p(t) = (\xi\_1-t) \cdots (\xi\_k -t)$ has integer coefficients. Hence, $p(A) \in M\_n {\mathbb Z}$ and at the same time $Ap(A)=0$. Clearly,
$$p(A) = {\rm det}'(A) \cdot Q\_{\ker(A)},$$
where $\det'(A) := \xi\_1 \cdots \xi\_k$ and $Q\_{\ker(A)}$ denotes the orthogonal projection onto the kernel of $A$. Hence, we get for the operator norm $\|p(A)\| = |\det'(A)|$. Now, there exists an index $1 \leq i \leq n$, such that $v:=p(A)e\_i \neq 0$ and one gets $\|v\| \leq \det'(A)$. At the same time $v$ has integer entries and lies in the kernel of $A$.
In many examples one can do much better. I would hope someone is able to bound the length in terms of the operator norm of $A$ alone. Is that possible?
The problem can also be phrased in terms of additive combinatorics. Indeed, if one identifies the image of ${\mathbb Z}^n$ under $A$ with ${\mathbb Z}^{k}$ for some $k < n$, then injectivity of $A$ on the set $X=\lbrace-m,\dots,m\rbrace ^n$ implies that $X$ can be embedded in ${\mathbb Z}^k$ preserving the addition whenever it made sense on $X$. This alone seems to require a lot of distortion, i.e. a large operator norm of $A$.
What kind of literature can be recommended?
| https://mathoverflow.net/users/8176 | Integer vectors in the kernel of an integer matrix |
>
> I would hope someone is able to bound the length in terms of the operator norm of alone. Is that possible?
>
>
>
This is not possible. Consider the group ring $\mathbb ZC\_k$ of the cyclic group $C\_k$ of order $k$.
Consider $1-t\in \mathbb ZC\_k$. As an operator on $l^2C\_k$ it has 1-dimensional kernel generated by $v\_k:=1+t+\ldots +t^{k-1}$. Vector $v\_k$ has arbitrary large length, yet the oeprator $1-t$ has a bounded norm.
If you want to get a symmetric operator use the standard trick, i.e. take $(1-t^{-1})(1-t)$ which has the same kernel and only slightly bigger norm.
I also thought about this question in relation to [my recent question](https://mathoverflow.net/questions/40465/regular-languages-of-matrices-and-their-generating-functions). I don't know what application do you have in mind, but maybe what you're looking for is a statement like this (using notation from [my question](https://mathoverflow.net/questions/40465/regular-languages-of-matrices-and-their-generating-functions)): let $\mathcal M$ be a regular family of matrices recognized by an automaton $A$. Suppose there exists $M\in \mathcal M$ such that $\det M=0$. Then there exists a matrix $N\in \mathcal M$ and a vector $v\in \mathbb Z^{\dim N}$ such that $Nv=0$ and the length of $v$ can be bounded by "size of $A$".
"Size of $A$" is some function which depends only on number of letters in the alphabet of $A$ and number of states of $A$. I'm not sure yet the above statement is true but, well, I'm "convinced" it is :-).
The family of operators $1-t$ above gives rise to a regular family of matrices if we allow finite automata which work with a circular tape.
| 4 | https://mathoverflow.net/users/2631 | 40616 | 25,981 |
https://mathoverflow.net/questions/40615 | 3 | Not every real algebraic surface can be endowned a structure of a complex algebraic curve. The only obstruction I know is orientability.
Are there any others?
| https://mathoverflow.net/users/nan | Obstruction for a real algebraic surface to be a complex algebraic curve | If you're not requiring any compatibility criterion between the real and complex structure, then the only obstruction is in fact orientability. Every smooth projective real algebraic surface is a smooth compact real 2-manifold (without boundary). If it's orientable, it must then be a surface of genus $g$ for some $g$. But every surface of genus $g$ admits a complex structure, and every Riemann surface is algebraic.
I don't study real algebraic geometry much, but I'm not aware of a good compatibility condition to impose on your complex structure. If you've got something in mind, let me know.
| 1 | https://mathoverflow.net/users/7399 | 40619 | 25,983 |
https://mathoverflow.net/questions/40604 | 6 | For all that follows, $p$ is a fixed odd prime. In the formulation of the Noncommutative Main Conjecture of Iwasawa theory one uses étale cohomology to define an algebraic object analogous to Iwasawas 'charakteristic ideal' in $\Lambda(G)$ for $G=Gal(k\_\infty/k)$, with $k^{cyc}\subset k\_\infty$ and $\mu=0$:
Let $M\_\Sigma$ be the maximal abelian, pro-$p$, outside of $\Sigma$ unramified extension of $k\_\infty$ for a finite set of primes of $k$, $\Sigma$. Then $X:=Gal(M\_\Sigma/k\_\infty)$ is a $\Lambda(G)$ module. If $G$ has elements of order $p$ we are prevented from seeing this $X$ a relative $K\_0$ group, associated to a denominator set, $S$, of $\Lambda(G)$.
Now étale cohomology enters the picture: We use it to define a complex $C$ of $\Lambda(G)$-modules which is $S$-acyclic and quasi-isomorphic to a bounded complex of finitely generated $\Lambda(G)$-modules. Although it looks quite technical, I will give here for the sake of quick reference the definition of
$C=RHom(R\Gamma\_{ét}(Spec(\mathcal{O\_{k\_\infty}}[\frac{1}{\Sigma}]),\mathbb{Q}\_p/\mathbb{Z}\_p),\mathbb{Q}\_p/\mathbb{Z}\_p)$.
This $C$ is strongly correlated to $X$, namely $H^0(C)=\mathbb{Z}\_p$ and $H^{-1}(C)=X$.
Now my question: An expert in the field told me that this $\mathbb{Z}\_p$ is 'moraly' related to the pole of a zetafunktion. **How is this?**
Is this even related to the Main Conjecture, where evalutations at representations of $G$ and $p$-adic interpolation play the lead role? As far as I understand it, the trivial representation, leading to the zeta function, is left undealt with.
I apologize for my ignorance on this basic question of the field.
| https://mathoverflow.net/users/448 | Characteristic Complexes in Iwasawa theory | The reason why one cannot take the class of $X$ in the relative $K\_0$ when $G$ has $p$-torsion is because $X$ may not have finite resolution by finitely generated projective $\Lambda(G)$-modules. This is necessary even if $G$ is abelian. Hence we take the complex $C$ above. $\mathbb{Z}\_p$ appearing there is indeed interpreted as a pole. One way thinking about this in the commutative Iwasawa theory is as follows- if $G$ is of the form $H \times \mathbb{Z}\_p$, with $H$ a finite abelian group, then for each one dimensional character $\chi$ of $H$ there is a $p$-adic $L$-function, say $L\_p(\chi)$, constructed by Deligne-Ribet, Cassou-Nogues, Barsky. $L\_p(\chi)$ is expected to have a simple pole only when $\chi$ is the trivial character. On the algebraic side one has to make sense of what a characteristic ideal for $X$ means when order of $H$ is divisible by $p$. One can either use the $K$-theory formulation as suggested by Fukaya-Kato or one can do something more directly but only get weaker information. Look at $V:=X \otimes \overline{\mathbb{Q}}\_p$. This is a finite dimensional $\overline{\mathbb{Q}}\_p$ vector space. For each $\chi$ we can consider $V^{(\chi)}$, the $\chi$ isotypic part. Let $f(\chi)$ be the characteristic polynomial of $1-\gamma$ acting on $V$ (Here $\gamma$ is a fixed topological generator of $\mathbb{Z}\_p$). Then for every non-trivial $\chi$ the polynomial $f(\chi)$ generates the same ideal as $L\_p(\chi)$ in the ring $\mathbb{Z}\_p[\chi][[\mathbb{Z}\_p]] = \mathbb{Z}\_p[\chi][[T]]$. However, when $chi$ is the trivial character, we have that $f(\chi)$ and $TL\_p(\chi)$ generate the same ideal. Or $f(\chi)/T$ is essentially same as the $p$-adic $L$-function. So apart from the information about the characteristic element of $X$, the $p$-adic $L$-function has this additional pole (the $T$ in the denominator) corresponding to the module $\mathbb{Z}\_p$ with the trivial action. This is exactly what happens in the noncommutative situation as well. It is not a new phenomenon of the noncommutative theory- it was always there.
The answer to your next question is yes, it is related to the main conjecture. We conjecture that there is an element $\zeta \in K\_1(\Lambda(G)\_S)$ which maps to the class of $C$ in $K\_0(\Lambda(G), \Lambda(G)\_S)$ and which is related to values of $L$-functions at Artin representations of $G$ at odd negative integers i.e. evaluations at even positive Tate twists of Artin representations of $G$. We now have this $\zeta$ and the main conjecture in the noncommutative situation and again one expects $\zeta$ a simple pole when evaluated at the trivial representation of $G$.
| 5 | https://mathoverflow.net/users/2259 | 40626 | 25,987 |
https://mathoverflow.net/questions/40632 | 38 | Given a continuous map $f \colon X \to Y$ of topological spaces, and a sheaf $\mathcal{F}$ on $Y$, the inverse image sheaf $f^{-1}\mathcal{F}$ on $X$ is the sheafification of the presheaf
$$U \mapsto \varinjlim\_{V \supseteq f(U)} \Gamma(V, \mathcal{F}).$$
If $X$ and $Y$ happen to be ringed spaces, $f$ a morphism of ringed spaces, and $\mathcal{F}$ an $\mathcal{O}\_X$-module, one then defines the pullback sheaf $f^\* \mathcal{F}$ on $X$ as
$$f^{-1}\mathcal{F} \otimes\_{f^{-1} \mathcal{O}\_Y} \mathcal{O}\_X.$$
However, I cannot think of any other usage of the inverse image sheaf in algebraic geometry. Moreover, if $X$ and $Y$ are schemes and $\mathcal{F}$ is quasicoherent, there is an alternate way of defining $f^\* \mathcal{F}$. Given $f \colon \mathrm{Spec} B \to \mathrm{Spec} A$, and $\mathcal{F} = \widetilde{M}$, where $M$ is an $A$-module, one defines $f^\* \mathcal{F}$ to be the sheaf associated to the $B$-module $M \otimes\_A B$. To extend this to arbitrary schemes, it is necessary to prove that it is well-defined; but I still think it is easier to work with than the other definition, which involves direct limits and two sheafifications of presheaves (the inverse image, and the tensor product). I have not checked, but I imagine that something similar can be done for formal schemes.
Hence, my question:
>
> What uses, if any, does the inverse image sheaf have in algebraic geometry, other than to define the pullback sheaf?
>
>
>
A closely related question is
>
> In a course on schemes, is there a good reason to define the inverse image sheaf and the pullback sheaf for ringed spaces in general, rather than simply defining the pullback of a quasicoherent sheaf by a morphism of schemes?
>
>
>
To go from the first question to the second question, I suppose one must also address whether there are $\mathcal{O}\_X$-modules significant to algebraic geometers that are not quasicoherent.
**Edit:** I think the question deserves a certain amount of clarification. Several people have given interesting descriptions or explications of the inverse image sheaf. While I appreciate these, they are not the point of my question; I am, specifically, interested to know whether there are constructions or arguments in algebraic geometry that cannot reasonably be done without using the inverse image sheaf. So far, the answer seems to be that such things exist, but are not really within the scope of, say, a one-year first course on schemes. There are other constructions (such as the inverse image ideal sheaf) that do not, strictly speaking, require the inverse image sheaf, but for which it may be more appropriate to use the inverse image sheaf as a matter of taste.
| https://mathoverflow.net/users/5094 | What is the inverse image sheaf necessary for in algebraic geometry? | By some coincidence, I have a student going through this stuff now, and we got to this point this just yesterday.
The definition of $f^{-1}$ is certainly disconcerting at first, but it's not that bad.
You'd like to say
$$f^{-1}\mathcal{F}(U) = \mathcal{F}(f(U))$$
except it doesn't make sense as it stands, unless $f(U)$ is open. So we approximate
by open sets from above. A section on the left is a germ of a section of $\mathcal{F}$ defined in some
open neighbourhood of $f(U)$, where by germ I mean the equivalence class where you identify two sections if
they agree on a smaller neigbourhood.
Even if you're still unhappy with this, the adjointness property tells you that it
is the right thing to look at.
Also, some of us work with non-quasicoherent sheaves (e.g. locally constant sheaves or constructible sheaves), so it's nice to have a general construction.
**Addendum:** In my answer yesterday, I had somehow forgotten to mention
the etale space or sheaf as a bunch of stalks
$$\coprod\_y \mathcal{F}\_y\to Y$$
viewpoint
discussed by Emerton and Martin Brandenburg. Had you started with this "bundle picture",
we would be having this discussion in reverse, because pullback is the natural operation here and pushforward is the thing that seems strange.
| 28 | https://mathoverflow.net/users/4144 | 40639 | 25,998 |
https://mathoverflow.net/questions/40587 | 11 | Let $X$ be a scheme. It is known that $Qcoh(X)$ is cocomplete, co-wellpowered and has a [generating set](https://mathoverflow.net/questions/39941/does-qcohx-admit-a-generating-set). The special adjoint functor theorem tells us that then every(!) cocontinuous functor $Qcoh(X) \to A$ has a right-adjoint. Here $A$ is an arbitrary category (which I always assume to be locally small).
a) Is there a nice description of the right-adjoint to the forgetful functor $Qcoh(X) \to Mod(X)$? Here, you may impose finiteness conditions on $X$. This functor may be called a quasi-coherator.
b) Let $f : X \to Y$ be a morphism of schemes. Then $f^\* : Qcoh(Y) \to Qcoh(X)$ is cocontinuous, since $f^\* : Mod(Y) \to Mod(X)$ is cocontinuous and the forgetful functor preserves and reflects colimits. In particular, there is a right-adjoint $f\_+ : Qcoh(X) \to Qcoh(Y)$. If $f$ is quasi-separated ans quasi-compact, then this is the direct image functor $f\_\*$. Is there a nice description in general? Note that $f\_+$ is the composition $Qcoh(X) \to Mod(X) \to Mod(Y) \to Qcoh(Y)$, where the latter is the quasi-coherator. This is only nice if we have answered a).
c) Since $Mod(X)$ is complete and $Qcoh(X) \to Mod(X)$ has a right adjoint, $Qcoh(X)$ is also complete. Is there a nice description for the products? They are given by taking the quasi-coherator of the product, can we simplify this? I mean, perhaps they turn out to be exact although the products in $Mod(X)$ are not exact?
**Answer** (after reading the article Leo Alonso has cited)
We have the following description of the quasi-coherator: Let $X$ be a concentrated scheme, i.e. quasi-compact and quasi-separated. If $X$ is separated, say $X = \cup U\_i$ with finitely many affines $U\_i$ such that the intersections $U\_i \cap U\_j$ are affine, then the quasi-coherator of a module $M$ on $X$ is the kernel of the obvious map
$\prod\_i (u\_i)\_\* \tilde{M(U\_i)} \to \prod\_{i,j} (u\_{i,j})\_\* \tilde{M(U\_i \cap U\_j)}$,
where $u\_i : U\_i \to X$ and $u\_{ij} : U\_i \cap U\_j \to X$ are the inclusions. If $X$ is just quasi-separated, there is a similar description using the separated case.
The idea is quite simple and can be generalized to every flat ring representation of a finite partial order, which has suprema (for example the dual of the affine subsets of a quasi-compact separated scheme). On an affine part, the quasi-coherator consists of sections of all other affine parts over it, which are compatible in the obvious sense.
If we have no finiteness conditions, the description is basically also valid, but you have to take the quasi-coherators of the products or the direct images, since they don't have to be quasi-coherent. In general there is no nice description. Also in nice special cases, b) and c) have no nice answers (and infinite products are not exact, even in the category of quasi-coherent modules on nice schemes).
| https://mathoverflow.net/users/2841 | Quasi-coherent envelope of a module | A very nice reference for the *coherator* functor together with a nice description of this functor is written down in Thomason and Trobaugh "Higher algebraic $K$-theory of schemes and of derived categories" in The Grothendieck Festschrift, Vol. III, 247--435, Progr. Math., 88, Birkhäuser, Boston, 1990. ([MR11069118](http://ams.rice.edu/mathscinet/search/publdoc.html?pg1=IID&s1=172225&vfpref=html&r=12&mx-pid=1106918)). Look for appendix B.
The original reference goes back to SGA6 (exposé II 3.2, by Illusie). It contains an appendix with counterexamples due to Verdier showing that:
* An affine scheme $\mathrm{Spec}(A)$ together with an injective $A$-module $I$ such that $\widetilde{I}$ is **not** injective as a quasi-coherent sheaf.
* A morphism $f$ between concentrated schemes such that the right derived functors of $f\_\*$ are different when considered from all modules or from quasi-coherent modules
* A concentrated scheme $S$ together with a quasi-coherent sheaf that it is not acyclic for the quasi-coherator.
The word concentrated is a shorthand for quasi-compact and quasi-separated. Under separation (or just semi-separation) hypothesis the last two pathologies do not show up.
| 4 | https://mathoverflow.net/users/6348 | 40641 | 26,000 |
https://mathoverflow.net/questions/40618 | 5 | Let $f\_{=}$ be a function from $\mathbb{R}^{2}$ be defined as follows:
(1) if $x = y$ then $f\_{=}(x,y) = 1$;
(2) $f\_{x,y} = 0$ otherwise.
I would like to have a proof for / a reference to a textbook proof of the following theorem (if it indeed is a theorem):
$f\_{=}$ is uncomputable even if one restricts the domain of $f\_{=}$ to a proper subset of $\mathbb{R}^{2}$, viz. the set of the computable real numbers
Thanks!
| https://mathoverflow.net/users/9679 | Uncomputability of the identity relation on computable real numbers | Suppose that $f\_=$ is computable when restricted to computable real numbers, which means that there exists a Turing machine that, given as input the encoding of two Turing machines $M\_1$ and $M\_2$ that compute the fractional digits of two computable real numbers $r\_1$ and $r\_2$ in $[0,1]$, produces $1$ if $r\_1 = r\_2$ and $0$ otherwise. I will use this assumption to show that the Halting problem is also computable, which is impossible.
Given a Turing machine $M$ and an input $x$ for which we want to know if $M$ on input $x$ halts or not, let $M\_x$ be the Turing machine that acts as follows: given an integer $i$ as input, $M\_x$ starts a simulation of $M$ on input $x$ for up to $i$ steps, and if the simulation does not halt within that number of steps, it outputs $0$ and otherwise it outputs $1$. By definition, $M\_x$ computes the digits of a computable real number (more precisely, it computes the $i$-th digit for every given $i$). Moreover, that real number is $0$ if $M$ on input $x$ does not halt, and the real number $0.0\cdots 011 \cdots = 2^{-k}$ otherwise for some $k \geq 1$. In other words, $M\_x$ computes the real number $0$ if and only if $M$ on input $x$ does not halt. To complete the argument, note that $0$ is a computable real number, so if you could tell whether two computable real numbers are equal you would also be able to tell if $M$ on input $x$ halts or not.
| 14 | https://mathoverflow.net/users/9355 | 40647 | 26,004 |
https://mathoverflow.net/questions/40666 | 19 | All the automorphisms of $SU(2)$ seem to be inner, which would mean that $\mathrm{Out}$ $SU(2)$ is trivial. Is that correct? Is this true in general $SU(n)$? I can't quite see -- any thoughts would be helpful.
| https://mathoverflow.net/users/7671 | What is the outer automorphism group of SU(n)? | $SU(n)$ for $n>2$ has complex fundamental representations. Complex conjugation is an automorphism which exchanges the fundamental representation with its complex conjugate, hence it cannot be an inner automorphism.
---
Upon further reflection (no pun intended), I think that this is all: basically for simply connected simple Lie groups, the outer automorphisms come from the automorphisms of the Dynkin diagram and for $SU(n)$, $n>2$, the only automorphism is reflection along the midpoint of the diagram. This sends the module with highest weight $(1,0,...,0)$ to $(0,...,0,1)$, hence the fundamental representation to its complex conjugate.
| 21 | https://mathoverflow.net/users/394 | 40668 | 26,014 |
https://mathoverflow.net/questions/40653 | 9 | Is there a nice condition on a closed subscheme $Y$ of $X$ such that for every flat family $Z\to Y$, there is a flat family $W\to X$ whose restriction to $Y$ is $Z$? In particular, I'm interested in the case when the closed subscheme is two lines in $\mathbb{P}^2$, or three planes in $\mathbb{P}^3$, or generally $n$ hyperplanes in $\mathbb{P}^n$.
| https://mathoverflow.net/users/1474 | When can one extend a flat family from a subscheme to the whole scheme? | I'm pretty sure the answer is not in general.
Take a Hilbert scheme which is reducible, for example, that of the Hilbert polynomial $3t + 1$ in $\mathbb{P}^3$. This Hilbert scheme has two irreducible components - the one corresponding to twisted cubics and the one corresponding to degree 3 plane curves union a point. The intersection of the these two components corresponds to degree 3 nodal plane curves with an embedded point "pointing out of the plane". It turns out that both components are rational (actually this is pretty easy to see) and even that the intersection is rational.
Anyhow, take a line in the twisted cubic component that meets the other component (see III.9.8.4 in Hartshorne's Algebraic Geometry) at a point $[C]$ and a line through $[C]$ contained in the other component. This gives you a flat family over the union of two intersecting lines in $\mathbb{P}^2$, but you will not be able to "fill it in" because the Hilbert scheme in question is not irreducible.
**EDIT**: I realized that, in this example, the family extended to $\mathbb{P}^2$ doesn't have to be as a family of closed subschemes in $\mathbb{P}^3$. Still, it cannot extend to a flat family in this example - just for a different reason. The general fiber of the extension would be a rational curve, but over a line in $\mathbb{P}^2$ you would have a disconnected family. This is impossible.
| 6 | https://mathoverflow.net/users/397 | 40675 | 26,019 |
https://mathoverflow.net/questions/40672 | 3 | This is a followup to my [previous](https://mathoverflow.net/questions/40621/singular-locus-of-a-homogeneous-polynomial) poorly-worded question.
Consider a finite collection of points $S \subset \mathbb N^n$ lying in a hyperplane $H$. These points define exponents of a collection of monomials in $\mathbb C[x\_1, \cdots, x\_n]$, which may be combined with some choice of coefficients to produce a polynomial that is homogeneous under the $\mathbb C^\*$ action with weights determined by $H$.
>
> Are there conditions on $S$ and/or $H$ such that by choosing generic coefficients, the resulting polynomial $p$ has singular locus at the origin?
>
>
>
I am happy to assume that the sum of the entries of $\alpha \in S$ (the naive degree) is greater than two, if this is easier.
In case it is still not clear, three examples: #1 is no good, but #2,3 are allowed.
1. If I choose $\{(3,0)\}\subset \mathbb N^2$, then for any non-zero coefficient $a$, the polynomial $p=ax^3$ has singularities away from the origin in $\mathbb C^2$.
2. If I choose $\{(3,0), (0,3)\} \subset \mathbb N^2$, then for any choice of non-zero coefficients $a,b$, the polynomial $p = ax^3 + by^3$ will be singular only at the origin.
3. If I choose $\{(6,0,0), (0,3,0), (0,0,2), (1,1,1))\} \subset \mathbb N^3$, then for non-zero $a,b,c,d$, the polynomial $a x^6 + b y^3 + cz^2 + d x y z$ has singularities only at the origin.
Please note that I am asking about singularities of the **affine** hypersurface, not the (weighted) projective hypersurface (hence *conical* singularity).
| https://mathoverflow.net/users/8363 | Constructing affine hypersurfaces with one singularity | I guess that to show that if the $\mathbb{Q}$-span of $S$ is not $\mathbb{Q}^n$ then $H$ is automatically singular is a not too hard exercise. I.e., in this case you construct an affine cone over a weighted projective cone. (Since it is almost midnight here, I cannot be bothered to do this..)
If $S$ spans all of $\mathbb{Q}$ then I am not so sure that you always have a $\mathbb{C}^\*$-action. A necessary and sufficient condition seems the following:
Let $m=|S|$ and let $M$ be the $m\times n$-matrix where the rows are the elements of S. Then the vector consisting of m 1s should be in the column span of M.
If this is the case then $p$ is weighted homogeneous. For such polynomials one has the weighted Euler relation. If $p$ has weighted degree $d$, $w\_i$ is the weight of $x\_i$, and $p\_{x\_i}$ is the derivative of $p$ wrt $x\_i$. Then
$d p=\sum w\_ix\_ip\_{x\_i}$.
Hence to find singularities of $p=0$ you only need to consider the partials of $p$. Since you consider only general coefficients, it suffices to determine if there is a monomial that divides $p\_{x\_i}$.
I.e., for each $i$ let $T\_i\subset S$ of elements such that the $i$-coordinate is non-zero. Let $f\_i$ be the product of coordinates $x\_j$ such that for each element in $T\_i$ the $j$-th coordinate is non-zero (if $j\neq i$) or at least 2 (if $j=i$.).
For a general choice of coefficients the singular locus of $p$ coincides with $V(f\_0,\dots,f\_n)$.
I hope you can work out the details.
| 2 | https://mathoverflow.net/users/8621 | 40677 | 26,020 |
https://mathoverflow.net/questions/40684 | 2 | This may be a naive question, but if I have random variables X and Y and take logs of both, would corr(log X, log Y) be greater than corr(X, Y)? Thank you in advance for your answer.
| https://mathoverflow.net/users/9699 | Does taking logs of two variables increase correlation between the two? | short answer: not necessarily. for example:
let X be a positive random variable with a pdf supported on some non-degenerate interval,
[0, 1] say.
let Y = 1 + X.
then X and Y are perfectly correlated. but U = logX and V = log Y = log (1 + X) = log(1 + e$^U$)
are not linearly related, so their correlation is less than 1.
[X can also be discrete, as long as it assumes at least 3 different values with positive probability.]
this example can be tweaked so that X and Y start out not perfectly correlated, but still
corr(U,V) < corr(X,Y).
| 6 | https://mathoverflow.net/users/8977 | 40691 | 26,027 |
https://mathoverflow.net/questions/40701 | 5 | Here is a question I get from sitting in my Lie algebra class:
Fix a Lie algebra $\mathfrak{h}$, we know there is a unique simply connected Lie group $H$ which serves as the universal cover of other connected Lie groups with the same Lie algebra. Now assuming $H$ is a $n\_1$ sheeted cover of $H\_1$, and a $n\_2$ sheeted cover of $H\_2$, (we are not restricting the Deck groups yet). we know $\phi: \mathfrak{h\_1}\cong \mathfrak{h\_2} (\cong \mathfrak{h})$ but we only cares about the isomorphism between the first two factors. Now as we travel along all $a\in \mathfrak{h\_1}$, $(a, \phi (a))$ will form a Lie subalgebra in $\mathfrak{h\_1}\bigoplus\mathfrak{h\_2}$, denoted by $\mathfrak{h'}$. By the existence theorem of Lie subgroups we have a uniqueLie subgroup $G\subset H\_1\times H\_2$ corresponding to $\mathfrak{h'}$. Note $\mathfrak{h\_1},\mathfrak{h\_2}, \mathfrak{h'}$ are isomorphic Lie subalgebras in $\mathfrak{h\_1}\bigoplus\mathfrak{h\_2}$, however since they sort of "sit in different directions" in the ambient Lie algebra when we form the exponential map we are producing non isomorphic Lie subgroups. Anyhow, Project down from $G$ to $H\_1$ and $H\_2$ induce an iso on the Lie algebra therefore we know $G$ actually covers $H\_i$. That's the set up of the question.
Now if I know $n\_1$ and $n\_2$ are coprime to each other, then automatically $G$ has to be the universal cover $H$. However, if say the maximal common divisor of $n\_1$ and $n\_2$ is 2, and that there are two non isomorphic Lie groups that could cover both $H\_1$ and $H\_2$, namely H and the one doubly covered by H donoted $\tilde{H}$, (here I need some compatible condition on the Deck group, but let assume $\tilde{H}$ covers $H\_i$). So my question is which one is isomorphic to $G$ when I draw the graph? I am worried a little bit that the there may not a canonical way of choosing this isomorphism $\phi$ to make this question make sense, (or is there)? My guess is that if there is a "canonical choice", then the $G$ we get from the graph should be the smaller $\tilde{H}$, while my colleage thinks that by choosing different $\phi$, you can get both. When $H\_1$ and $H\_2$ have a lot of non isomorphic common covers, then by choosing different $\phi$ you can produce all of them as subgroups of $H\_1\times H\_2$? (fixing $H\_1$ and $H\_2$.) I really hope someone ccan shed lights on this, thanks very much!
| https://mathoverflow.net/users/1877 | Which covers of Lie groups will I get | If we fix universal covering maps $H \to H\_1$ and $H \to H\_2$, then $G$ is uniquely defined as the image of the diagonally embedded $H \subset H \times H$ under the covering map to $H\_1 \times H\_2$. If you transform one of the covering maps from $H$ using the deck group, you will get a group isomorphic to $G$.
| 2 | https://mathoverflow.net/users/121 | 40704 | 26,034 |
https://mathoverflow.net/questions/40722 | 7 | Hi Folks,
i'm looking for a reference on the 2-grothendieck construction for a functor $F:\mathcal{I}\to \mathcal{B}\mathrm{icat}$ from a bicategory $\mathcal{I}$ to the tricategory of bicategories. Actually for my purposes it would be sufficient to consider functors going only to $\mathcal{C}\mathrm{at}$.
| https://mathoverflow.net/users/1261 | Reference request: 2-Grothendieck Construction | I. Bakovic, [Grothendieck construction for bicategories](http://www.irb.hr/users/ibakovic/sgc.pdf).
| 9 | https://mathoverflow.net/users/447 | 40727 | 26,047 |
https://mathoverflow.net/questions/40689 | 3 | Let $x, y\in R^n$ and $x, y$ are nonzero, it is well known
$\frac{x^Ty}{\parallel x\parallel\_2\parallel y\parallel\_2}(\parallel x\parallel\_2+\parallel y\parallel\_2)\le \parallel x+y\parallel\_2$. How to extend this to complex vectors? $arcos\frac{x^Ty}{\parallel x\parallel\_2\parallel y\parallel\_2} $
is the angle between $x$ and $y$. What is the appropriate definition of the angle between two complex vectors? I know
$\frac{\mid x^\*y\mid}{\parallel x\parallel\_2\parallel y\parallel\_2}(\parallel x\parallel\_2+\parallel y\parallel\_2)\le \parallel x+y\parallel\_2$
does not hold generally.
| https://mathoverflow.net/users/3818 | What is the angle between two complex vectors? | Let $x,y$ be two nonzero complex vectors, let $\hat x=x/\|x\|$ and $\hat y=y/\|y\|$, and consider the
parabola
$$\phi(t)=\|t\hat x+(1-t)\hat y\|^2=1+2(t^2-t)(1-\Re(\hat x \overline{\hat y})). $$
You easily check that $\phi(t)\ge\phi(1/2)$ for all $t$. This gives the inequality
$$
\|t\hat x+(1-t)\hat y\|\ge \sqrt{\frac{1+\Re(\hat x \overline{\hat y})}2}
$$
for all real $t$. This is stronger than your inequality, which can be obtained by choosing
$$t=\frac{\|x\|}{\|x\|+\|y\|}$$
at the left hand side, and noticing that
$$
\sqrt{\frac{1+\sigma}2}\ge\sigma
$$
for all real $|\sigma|\le1$ at the right hand side. So yes, the correct extension is using $\Re(x \cdot\overline{y})$ instead of $x\cdot y$.
| 4 | https://mathoverflow.net/users/7294 | 40733 | 26,050 |
https://mathoverflow.net/questions/40724 | 2 | I'm trying to fill a woeful gap in my topological knowledge and learn a little knot and link theory (I'll be recording my progress on the nLab, starting with a page on [links](http://ncatlab.org/show/link)). Not wishing to write anything incorrect, I found myself with the following question:
>
> Is the Hopf link a Brunnian link?
>
>
>
According to Wikipedia, a [Brunnian link](http://en.wikipedia.org/wiki/Brunnian_link) is a link with the property that removing any component produces an unlink (of the appropriate size). That's certainly true of the Hopf link! One could outlaw the Hopf link on size grounds and add "of at least 3 components" to the definition of a Brunnian link, but then the family of [rubberband](http://katlas.math.toronto.edu/wiki/%22Rubberband%22_Brunnian_Links) Brunnian links is missing its first member (actually, its second; but the first is a fancy way of drawing the unlink).
This feels a bit like the question "Is 1 prime?" so I suspect that the answer is purely a matter of convention but as I'm not a knot theorist, I don't know what the convention is. So I'm hoping that one Steeped in the Depths of Knot Theory can help me out.
| https://mathoverflow.net/users/45 | Is the Hopf link a Brunnian link? | The Hopf link is normally regarded as Brunnian.
| 5 | https://mathoverflow.net/users/9417 | 40739 | 26,054 |
https://mathoverflow.net/questions/40485 | 20 | I'm trying to understand how to compute a fast Fourier transform over a finite field. This question arose in the analysis of some BCH codes.
Consider the finite field $F$ with $2^n$ elements. It is possible to define a (discrete) Fourier transform on vectors of length $2^n-1$ as follows. Choose a $2^n-1$ root of unity $\omega\in F$. Then given a vector $V=(V\_0,...,V\_{2^n-2})\in F^{2^n-1}$, we can define its Fourier transform $W=(W\_0,...,W\_{2^n-2}) \in F^{2^n-1}$ as: $$W\_i=\sum\_{j=0}^{2^n-2} \omega^{ij} V\_j$$
To find such an $\omega$ we can use any primitive elements of $F$.
Suppose we are given $V$ and we would like to compute efficiently $W$. If we were operating over the complex numbers, we could choose any of a number of fast Fourier transform algorithms. The mixed radix [Cooley-Tukey algorithm](http://en.wikipedia.org/wiki/Cooley-Tukey_FFT_algorithm) translates unchanged in this context, so if $2^n-1$ is the product of small factors (i.e. is smooth), then we are all set.
However, $2^n-1$ may be prime (after all, these numbers include the Mersenne primes) or contain large prime factors. The traditional Cooley-Tukey algorithm is no longer efficient. Over the complex numbers, this does not pose a problem-- there exist fast algorithms (like [Bluestein's algorithm](http://en.wikipedia.org/wiki/Bluestein%27s_FFT_algorithm) and [Rader's algorithm](http://en.wikipedia.org/wiki/Rader%27s_FFT_algorithm)) for handling that case. These algorithms typically involve rephrasing an $N$-point Fourier transform as a convolution, and evaluating the convolution by performing a $2^m$ point FFT, where $2^m\geq 2N-1$.
Unfortunately for us, there is no $2^m$ root of unity in any finite field of characteristic 2. Adjoining such a root to our field produces a much larger ring, and the added complexity of handling these elements appears to cancel the speed-up we would have gotten from the FFT. In (1), Pollard suggests using Bluestein's algorithm, but his argument doesn't seem to apply to fields of characteristic 2 (unless I'm misunderstanding him).
My question is: in the case described above, how do I compute an fast Fourier transform? For my original purpose, I was hoping to find a radix-two algorithm, but at this point I'd be interested in any fast algorithm.
(1) J. M. Pollard. "The Fast Fourier Transform in a Finite Field". Mathematics of Computation, Vol. 25, No. 114 (Apr., 1971), pp. 365-374.
| https://mathoverflow.net/users/8938 | FFTs over finite fields? | There are a few different approaches to this:
1) As Peter Shor mentioned you can use a $3^n$ point transform with Bluestein's algorithm.
2) Even though there are no $2^n$ roots of unity there are substitutes for them (first implicitly discussed by Leonard Carlitz, and then explicitly by David Cantor). Look here for a good account of this <http://www.math.clemson.edu/~sgao/papers/GM10.pdf>
This algorithm is most likely the most efficient one in practice.
3) There's an observation of Shmuel Winograd who noticed that, using the relation between multiplicative and additive characters you can convert an FFT on $N$ points to the calculation of a cyclic convolution on $N-1$ points. And $N-1$ might factor nicely.
| 11 | https://mathoverflow.net/users/2784 | 40741 | 26,056 |
https://mathoverflow.net/questions/40713 | 12 | Let $R$ be a commutative ring, $M$, $N$ $R$-modules, and $f: M\rightarrow N$ a homomorphism. It is known that $f$ is injective (surjective) if and only if $f\_m$ is injective (surjective) for all maximal ideal $m$. But I don't know whether $f$ is split if and only if $f\_m$ is split? Maybe it is true for finitely generated modules over noetherian rings, I suspect.
| https://mathoverflow.net/users/5775 | An elementary lemma of commutative algebra | If you want to avoid the use of Ext-groups, you could prove it like this (which is basically the same proof):
Let $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ be a short exact sequence of $R-$modules with $C$ finitely presented and assume it splits after localisation at every maximal ideal.
Use the natural isomorphism $\operatorname{Hom}\_{R\_m}(C\_m,A\_m)=R\_m\otimes\operatorname{Hom}\_R(C,A)$ (which uses the flatness of localisation and the fact that $C$ is finitely presented) and the assumption to see that the map $\operatorname{Hom}\_R(C,B)\rightarrow\operatorname{Hom}\_R(C,C)$ is surjective since all of its localisations are.
Since the giving a splitting is equivalent to this map being surjective we are done.
P.S.: Of course one could prove "if and only if" in the statement like this.
Counterexample in the general case: Take $R=\prod\_{\mathbb N} \mathbb F\_2$, $I=\sum\_{\mathbb N} \mathbb F\_2$ and $C$ the cokernel of the inclusion.
Now I claim three things:
1.) $R$ has dimension zero
Proof: Every element of $R$ is an idempotent and every prime ideal in $R$ has to contain exactly one of $e$ or $1-e$ for every idempotent $e$ in R. If we had a chain of prime ideals, the larger one would necessarily have to contain $e$ and $1-e$ for one such idempotent and so couldn't exist.
2.) The localisation of $R$ at every maximal ideal is a field.
Proof: It is a zero-dimensional local ring by 1.) and reduced since $R$ doesn't contain nilpotent elements. Thus this localisation is a field.
3.) $I$ is not a direct summand of $R$.
Proof: Direct summands correspond to idempotents in $R$. Since every element in $R$ is idempotent we need to analyse all principal ideals. If an element has only finitely many non-zero entries the ideal created by it has only finitely many elements, thus can't be $I$.
If on the other hand it contains infinitely many non-zero entries, the ideal created by it has uncountably many elements.
Since $I$ contains countably many elements, it can't be a direct summand of $R$.
This establishes the counterexample, since the short exact sequence $0\rightarrow I\rightarrow R\rightarrow C\rightarrow 0$ splits in every localisation at a maximal ideal.
| 8 | https://mathoverflow.net/users/9710 | 40746 | 26,059 |
https://mathoverflow.net/questions/40742 | 10 | It is well known that the symmetric groups have a very nice and explicit representation theory. This is in particular true when one works with the collection of all symmetric groups simultaneously, in which case the ring of virtual representations is the ring of symmetric functions. This has many advantages: it allows one to reduce hard questions in representation theory to formal combinatorial manipulations with symmetric functions; there is a lot of extra structure like inner product, outer product, and plethysm; it leads directly to the general theory of λ-rings, etc...
From the point of view of Coxeter groups there are two other infinite families of groups that seem natural to study from the same point of view, i.e. Bn and Dn. Is there an analogue of the theory of symmetric functions in this situation, i.e. a "nice" (i.e. highly structured and purely combinatorial) description of their rings of virtual representations?
| https://mathoverflow.net/users/1310 | Symmetric functions in type B and type D | A partial answer:
If you want something like the Frobenius map from the ring of characters (with induced product) of $\bigcup\_n S\_n$ to the ring of symmetric functions, then something like this exists for any wreath product $G \wr S\_n$, namely there is a Frobenius map from the ring of characters of $\bigcup\_n G \wr S\_n$ into a tensor product of copies of the ring of symmetric functions, one for each conjugacy class of $G$. And there's an inner product and Schur functions, power sum, etc. This is explained in Appendix B of Macdonald's *Symmetric Functions and Hall Polynomials*.
So this takes care of type B ($G = {\bf Z}/2$) but I'm not sure about type D.
| 7 | https://mathoverflow.net/users/321 | 40753 | 26,063 |
https://mathoverflow.net/questions/40488 | 6 | Consider a collection of points $S \subset \mathbb R^d$. I would like to understand all possible fans $\Sigma$ whose support is the cone over $S$: $|\Sigma| = cone(S)$.
I have heard that the secondary fan does this for me, but I am having trouble parsing the relevant sections of GKZ. I would like to understand this completely, but to begin I would really like to know
>
> How to construct the resulting set of Stanley-Reisner ideals for each possible $\Sigma$.
>
>
>
Can someone explain this and/or give a readable explanation of
>
> How the secondary fan is built and how each cone gives a refinement of $cone(S)$?
>
>
>
| https://mathoverflow.net/users/8363 | Secondary fans and Stanley Reisner ideals | Consider any function $f$ from $S \to \mathbb{R}$. Define a function $\tilde{f}: \mathrm{cone}(S) \to \mathbb{R}$ by
$$\tilde{f}(w) = \max \{ \sum a\_i f(s\_i) : \ \sum a\_i s\_i = w,\ a\_i \geq 0 \}.$$
That is, for every way of writing $w$ as a positive linear combination of the $a$'s, try extending $f$ linearly, and find the largest possible extension.
Then $\tilde{f}$ is a piecewise linear convex function. The domains of linearity for $\tilde{f}$ form a fan, each of whose cones is spanned by a subset of $S$.
Subdivide $\mathbb{R}^S$ into cones, where functions $f$ and $g$ are in the same cone if $\tilde{f}$ and $\tilde{g}$ have the same domains of linearity. This is the secondary fan.
Thus, the cones of the secondary fan are in bijection with fans $F$ whose support is $\mathrm{cone}(S)$, where every face of $F$ is of the form $\mathrm{cone}(T)$ for some $T \subseteq S$, and where there is a convex piecewise linear function whose domains of linearity are the faces of $F$.
---
Several warnings:
(0) You didn't say that you want every cone of your fan to be of the form $\mathrm{cone}(T)$, for $T \subseteq S$, but I assume that you meant to. If not, there will be infinitely many fans meetng your conditions, because you can always insert new vertices.
(1) There can be fans which do not support any convex piecewise linear function. Such fans are called non-coherent or non-regular. They do not correspond to cones of the secondary fan.
(2) As I have described it, the secondary fan lives in $\mathbb{R}^S$. Note that, if $\lambda$ is a linear function on $\mathbb{R}^d$, then $f$ and $f+\lambda$ always lie in the same cone of the secondary fan, so the secondary fan is invariant under this action of $(\mathbb{R}^d)^{\vee}$. Many references quotient by this action.
(3) Many references specialize to the case that $S$ lies in an affine hyperplane, and then draw their pictures in $\mathbb{R}^{d-1}$. I think GKZ does this, which might be part of your difficulty.
(4) You mention Stanley-Reisner ideals. You don't say what your motivation is, but it sounds like you might be computing Grobner degenerations of toric varieties. If so, remember that these can be nonreduced, and the secondary fan will only see the reduced structure. See Sturmfels paper *Gröbner bases of toric varieties* for details.
---
Have you tried reading the Billera, Filliman and Sturmfels paper, *Constructions and complexity of secondary polytopes*? I seem to remember it is very clear, although it also has weakness (3).
| 5 | https://mathoverflow.net/users/297 | 40755 | 26,065 |
https://mathoverflow.net/questions/40699 | 2 | Let $B$ be a Banach space and $f : [0,+\infty)\times B \to B$ be a continuous function which is Lipschitz continuous in the second argument with Lipschitz constant $L$ (which does not depend on the first argument). By Picard-Lindelof, there is a unique function $x : [0,+\infty) \to B$ such that $x(0) = \mathbf{0}$ and for all $t\in [0,+\infty)$, $x'(t) = f(t,x(t))$. Define the family of functions $Euler\_h : [0,+\infty) \to B$ to be the result of linear interpolation between the points obtained by the Euler method on $f$ with initial value $\mathbf{0}$ and step size $h$. Does it follow that $Euler\_h$ converges compactly to the unique solution $x$ as $h$ goes to 0 from the right? If yes, is an explicit rate known?
The results I have found are only for the points provided by the Euler method, and not for the approximating functions obtained from them.
| https://mathoverflow.net/users/nan | Do the Euler method's approximations always approach the true solution? | The Picard-Lindelof Theorem is not quite correctly stated in your question. Recall that it is usually referred to as the LOCAL existence and uniqueness theorem, and it only guarantees a solution on a certain maximal interval [0,T), and for as simple a system as $x'=x^2$ the maximal existence time T is finite. That said, it is correct (with caveats) that the Euler method approximating solutions will converge to this "true" solution at all points of this maximal interval. The detailed story, with all the error estimates, is too complicated to state here, but you can find a careful discussion of this not only for Euler's Method, but also for a number of other methods, in Chapter 5 (Numerical Methods) of the book Differential Equations, Mechanics, and Computation (which I wrote together with my son Bob). There is a website for the book at <http://ode-math.com> where you can download for free more than half the book. In particular clicking here:
<http://ode-math.com/PDF_Files/ChapterFirstPages/First38PagesOFChapter5.pdf>
will download the first 38 pages of chapter 5, where starting on page 144 you will find a careful discussion of the rate of convergence and stability properties etc, for Euler's method, starting from scratch.
| 10 | https://mathoverflow.net/users/7311 | 40763 | 26,070 |
https://mathoverflow.net/questions/40770 | 25 | I've seen that P != LINSPACE (by which I mean SPACE(n)), but that we don't know if one is a subset of the other.
I assume that means that the proof must not involve showing a problem that's in one class but not the other, so how else would you go about proving it?
Thanks!
| https://mathoverflow.net/users/9714 | How do we know that P != LINSPACE without knowing if one is a subset of the other? | Suppose by contradiction that P=SPACE(n). Then there exists an algorithm to simulate an n-space Turing machine in (say) nc time, for some constant c. But this means that there exists an algorithm to simulate an *n2*-space Turing machine in n2c time. Therefore SPACE(n2) is also contained in P. So
P = SPACE(n) = SPACE(n2).
But SPACE(n) = SPACE(n2) contradicts the Space Hierarchy Theorem. QED
(Notice that in this proof, we showed neither that P is not contained in SPACE(n), nor that SPACE(n) is not contained in P! We only showed that *one or the other* must be true, by using the different closure properties of polynomial time and linear space. It's conjectured that P and SPACE(n) are incomparable.)
| 51 | https://mathoverflow.net/users/2575 | 40771 | 26,076 |
https://mathoverflow.net/questions/39976 | 4 | Let $V$ be a finite dimensional vector space over some field (say, $\mathbb C$). Consider the set $\operatorname{GLI}(V)$ of all linear isomorphisms between subspaces of $V$. This is a monoid under natural multiplication (in fact an inverse monoid). Its elements can be represented by triples: two elements of the Grassmannian of $V$ of degree $k\le n$ representing the domain and the range, and a non-singular $k\times k$-matrix representing the map. I am interested in developing a theory of representations of finite inverse monoids (pseudogroups) in $\operatorname{GLI}(V)$. What is the structure of $\operatorname{GLI}(V)$ from the algebraic geometry or geometric topology point of view?
*Edit:* It looks like the question is not completely clear. For comparison, if somebody gives me a group and asks what can I say about it, I would try to decide whether the group is finite or infinite, solvable or not, hyperbolic or not, what is the derived subgroup and the lower central series, is it residually finite and what is the profinite completion, etc. I want a similar analysis of $\operatorname{GLI}$ (but from the algebraic geometry point of view). One of the goals is to study representation varieties of groupoids (=pseudogroups, inverse semigroups). These varieties are complicated even for easy finite groupoids. The starting point would be to understand $\operatorname{GLI}$ itself.
| https://mathoverflow.net/users/nan | General linear inverse monoid | Some small comments.
Let $n=dim(V)$, so I'll think of $V$ as $\mathbb R^n$, then as a space, $GLI(V)$ you could think of as
$$ V\_{n,k} \times\_{O\_k} V\_{n,k} $$
where $V\_{n,k}$ is the Stiefel manifold of orthonormal $k$-frames in the vector space $V$. i.e. this is the space $V\_{n,k}^2$ mod the diagonal action of $O\_k$.
So you could view it as a bundle over $G\_{n,k}^2$ with fiber $O\_k$, or as a bundle over $G\_{n,k}$ with fiber $V\_{n,j}$. $G\_{n,k}$ is the Grassmannian of $k$-dimensional subspaces of $V$.
The map $V\_{n,k} \times\_{O\_k} V\_{n,k}$ to $GLI(V)$ is given by sending a pair $(A,B) \in V\_{n,k} \times V\_{n,k}$ to:
The span of $A$, the span of $B$ and the corresponding linear isometry represented by $B\circ A^{-1}$ where we think of $A$ and $B$ as representing isometric embeddings $\mathbb R^k \to \mathbb R^n$.
So the homotopy-type of this space is at least fairly reasonable as $V\_{n,k}$ is highly connected. I think this bundle likely has a lot of other nice properties lurking near the surface. Is this the kind of thing you're asking about? In particular as a bundle over $G\_{n,k}^2$ you'd have some nice Schubert-cell type constructions. i.e. you could view $V\_{n,k} \times\_{O\_k} V\_{n,k}$ as the "diagonal" $V\_{n,k}$ subspace union "Schubert cells".
| 1 | https://mathoverflow.net/users/1465 | 40781 | 26,082 |
https://mathoverflow.net/questions/40731 | 8 | Hello!
Let $n,m\geq 0$ be integers. If I understand it correctly, there is the following description of the cohomology of the complex Grassmannian $\text{Gr}(m+n;m)$: denote by $\text{Sym}(n,m)$ the subring of the polynomial ring in $n+m$ variables consisting of polynomials which are symmetric in the first $n$ and the last $m$ variables, and by $\text{Sym}(n+m)$ the ring of symmetric polynomials in $n+m$ variables. Then $\text{Sym}(n,m)$ is a free $\text{Sym}(n+m)$ module, and its graded rank (written as a $q$-polynomial) equals the Poincare polynomial of the Grassmannian $\text{Gr}(m+n;m)$.
I'd like to know if this is true for more general flag varieties: Is it true that the graded rank of $\text{Sym}(m\_1,...,m\_k)$ over $\text{Sym}(m\_1+...+m\_k)$ is the Poincare polynomial of the variety of flags $(\{0\}=U\_0,U\_1,...,U\_k=V)$ in $V := {\mathbb C}^{m\_1+...+m\_k}$ such that $\text{dim}(U\_{i}) - \text{dim}(U\_{i-1}) = m\_i$ for $i=1,...,k$? If yes: how to prove it? :-)
On the other hand, there is a presentation of the cohomology of the full flag variety of ${\mathbb C}^N$ as the quotient of ${\mathbb C}[X\_1,...,X\_N]$ by the ideal generated by the symmetric polynomials of positive degree. How do these two descriptions of the cohomology of flag varieties relate?
Thank you!
Hanno
| https://mathoverflow.net/users/3108 | Presentation of the cohomology of generalized flag varieties as graded ranks of rings of symmetric polynomials | Yes, this is all true. The cohomology of the partial flag variety $F(m\_1,\dots,m\_k)$ is the quotient of the polynomials that are symmetric under $S\_{m\_1}\times \cdots \times S\_{m\_k}$ by the positive degree ones which are fully symmetric (a special case of this is the presentation you mention for the full flag variety). Since partially symmetric polynomials are a free module over full symmetric ones, the q-rank of the free module is the same as the q-dimension of the quotient by all positive degree fully symmetric polynomials.
The "reason" this presentation works is that $F(m\_1,\dots,m\_k)$ carries tautological vector bundles $U\_i/U\_{i-1}$ (these are subquotients of the trivial bundles of rank $m\_1+\cdots m\_k$). The Chern classes of these bundles are the elementary symmetric polynomials in the variables corresponding to $S\_{m\_i}$; thus, the fully symmetric polynomials, written in terms of these are the Chern classes of the the trivial bundle of rank $m\_1+\cdots m\_k$ by the Whitney sum formula (which says that if I think of $c\_i(V)$ as elementary symmetric functions in one set of variables, and $c\_j(W)$ as elementary symmetric functions in another, the Chern classes $c\_k(V\oplus W)$ are the elementary symmetric functions in the union of the variables) and the splitting principle (which says that when I have a filtration $ V\_1\subset V\_2\subset \cdots\subset V\_n=V$, then $c\_i(V)=c\_i(V/V\_{n-1}\oplus \cdots \oplus V\_2/V\_1\oplus V\_1)$) are thus 0. Admittedly, you then have to do some actual geometry to see that these Chern classes generate and that these are all the relations, but to me, this is the heart of the argument.
| 4 | https://mathoverflow.net/users/66 | 40782 | 26,083 |
https://mathoverflow.net/questions/40776 | 14 | EQP is the class of problems solvable deterministically using a quantum computer in polynomial time - that seems to me to be a good analogue to P, whereas BQP is the quantum analogue of BPP.
It doesn't seem like much is known about EQP! Just like BPP is not known to be contained in NP, is it known whether EQP \subseteq BQP \subseteq QMA? Is there a corresponding "Derandomization" of BQP into EQP?
What about the relationship of EQP and P?
| https://mathoverflow.net/users/5534 | What's known about the relationship about EQP and BQP? | Hi Henry,
One reason why EQP isn't studied more is that it's not even uniquely defined! In particular, which complexity class you get might depend on the specific quantum gates you assume are available. (For BQP, by contrast, the Solovay-Kitaev Theorem assures us that any universal set of quantum gates can approximate any other universal set to within exponentially small error.)
Still, you *could* forge ahead, fix a particular universal set of gates (say, Hadamard and Toffoli), and study the resulting class EQP.
In that case, it's not hard to construct an oracle relative to which BQP (and even BPP) are not contained in EQP -- for example, just take the MAJORITY function with a promised (1/3,2/3) gap in the Hamming weight. (You can prove that's not in EQP using the polynomial method.) Nor is it hard to construct an oracle relative to which EQP is not contained in BPP (Bernstein and Vazirani's Recursive Fourier Sampling problem, for example).
Outside the oracle world, the main result about EQP I know of is due to Mosca (don't have the reference offhand), who showed that, if you're careful to define EQP with a large enough set of gates, then it contains FACTORING (i.e., Shor's algorithm can be made zero-error). This gives pretty good evidence that EQP (suitably defined) is not contained in BPP.
I would guess that EQP ≠ BQP in the unrelativized world, but I don't have any evidence for that (and of course, it's possible that EQP equals BQP under some definitions but doesn't under others).
| 16 | https://mathoverflow.net/users/2575 | 40787 | 26,087 |
https://mathoverflow.net/questions/40784 | 2 | Constructing trees with the same degree sequences I've got this problem.
Let $G$, $H$ be the trees (simple graphs) with the same degree sequences. Is it true that there always be vertices $q\in V(G)$ and $q′\in V(H)$ such that $(q,p)\in E(G)$ and $(q′,p′)\in E(H)$ for some endvertices $p\in V(G)$ and $p\in V(H)$, and $d(q)=d(q′)$?
$d(q)$ - degree of the vertex $q$.
I haven't found counterexample for trees up to $8$ vertices, and it's seems impossible to me.
Have you references for some results concerned with trees with the same degree sequences?
| https://mathoverflow.net/users/9720 | Trees with the same degree sequences | Consider two trees $G$ and $H$ with 14 vertices. Both will have degree sequence $(2,0,6,6)$ i.e. having two vertices of degree 4. $G$ will have the two 4-vertices connected to 3 leaves each and with a 6 vertex long chain between them. $H$ will have the two 4-vertices connected with a single edge. In addition, each will have three 2 vertex long chains connected to them (one 2-vertex connected to a leaf).
Finally, each leaf of $G$ is connected to a 4 vertex and each leaf of $H$ is connected to a 2 vertex.
A picture would do the trick better.
| 3 | https://mathoverflow.net/users/7412 | 40793 | 26,088 |
https://mathoverflow.net/questions/40789 | 2 | Hi all,
Let $\mathcal{E}$ be an elementary topos with natural number object $N$, and let $+: N \times N \to N$ be the the addition arrow; I expect that the nature of $N$ and $+$ will turn out to be irrelevant to my question, but if so they should at least make its motivation clear. Let $E$ be the pullback of $+$ along itself, with projections $p, q: E \to N \times N$; for example if $\mathcal{E}$ is the topos of sets then $E$ may simply be taken to be the set of quadruples $(n, m, n', m') \in N^4$ such that $n + m' = n' + m$, with $a(n, m, n', m') = (n, m')$, $b(n, m, n', m') = (n', m)$. Let $f\_1, f\_2: E \to N \times N$ be given by
$f\_1 \equiv \left< p\_1 a, p\_2 b \right>$
$f\_2 \equiv \left< p\_1 b, p\_2 a \right>$
(here $p\_1, p\_2: N \times N \to N$ are the projections and $\left< f, g \right>$ denotes the product arrow $X \to N \times N$ of arrows $f, g: X \to N$). For example in the topos of sets again, $f\_1 (n, m, n', m') = (n, m)$ etc.. Let $c: N \times N \to Z$ be the coequaliser of $f\_1$ and $f\_2$, so $Z$ is the integer object in $\mathcal{E}$.
My question is: if $g, h, g', h': X \to N$ are such that $c \left< g, h \right> = c \left< g', h' \right>$, is it always the case that $+ \left< g, h' \right> = + \left< g', h \right>$? Equivalently, is $E$ with the arrows $f\_1$, $f\_2$ the pullback of $c$ along itself?
I've spent a while trying to prove it is but I just keep going round in circles, so any hints will be much appreciated.
| https://mathoverflow.net/users/7842 | Question about equivalence relation defining integers in an elementary topos | I think what you want first is a lemma that $\mathbb{N}$ is a cancellative monoid (which is the case in any topos). Just think about how you would prove your statement in the category of sets using ordinary elements, and I think it will become clear.
The usual construction of the left adjoint to the forgetful functor from abelian groups to abelian monoids involves the observation that, for $m, n, m', n'$ in an abelian monoid $M$, the relation
$$\exists\_{j \in M} m' + n + j = m + n' + j$$
defines an equivalence relation $\langle m, n \rangle \sim \langle m', n' \rangle$ on $M \times M$. (Only transitivity need be checked.) Cancellation means that from this we can infer
$$m' + n = m + n'$$
which is what you want.
We thus need to show
$$\forall\_{j \in \mathbb{N}} ((x + j = y + j) \Rightarrow (x = y))$$
in the natural numbers object. This is done by induction on $j$ (the subobject of such $j$ contains 0 and ... and therefore is all of $\mathbb{N}$).
| 2 | https://mathoverflow.net/users/2926 | 40799 | 26,091 |
https://mathoverflow.net/questions/40790 | 5 | I am trying to locate a modern account of the problem of determining the number of pieces into which a certain geometric set is divided by given subsets. An example of such a problem could be to count the chambers of a real hyperplane arrangements which was solved by Zaslavsky.
I am trying to find an article which has some history of related problems. What I have right now are accounts by Grunbaum written 70's.
| https://mathoverflow.net/users/9724 | history of the topological dissection problem | Maybe look over the Table of Contents of the 1992 book [*Arrangements of Hyperplanes*](http://books.google.com/books?id=IgJrzHYBQp0C&printsec=frontcover&dq=Arrangements+of+hyperplanes++By+Peter+Orlik,+Hiroaki+Terao&source=bl&ots=40U-5qbMJp&sig=TU-vcJxqEJEGfyzmWErZirHjEqs&hl=en&ei=TnWmTKjwN4GKlweSnqQX&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBcQ6AEwAA#v=onepage&q&f=false) by Peter Orlik, Hiroaki Terao to see if what you seek is there.
The paper "[On the number of arrangements of pseudolines](http://portal.acm.org/citation.cfm?id=237232)" by Stefan Felsner uses some very nice analysis to improve the known upper bound at the time (1996), and cites earlier relevant work.
This may be more algorithmic than you care about, but an authoritative
survey as of 2000 was "Arrangements and their applications" by Pankaj Agarwal and Micha Sharir in the
[*Handbook of Computational Geometry*](http://www.elsevier.com/wps/find/bookdescription.cws_home/620809/description#description). If you are more interested in the combinatorics of,
say, pseudosphere arrangements, then
perhaps you should look at oriented matroids, e.g., Jurgen Richter-Gebert, and Günter Ziegler, "Oriented Matroids" in the [*Handbook of Discrete and Computational Geometry*](https://www.crcpress.com/Handbook-of-Discrete-and-Computational-Geometry-Third-Edition/Toth-ORourke-Goodman/p/book/9781498711395), 2004.
Perhaps none of this helps, but it may lead to a sharpening of your question.
| 3 | https://mathoverflow.net/users/6094 | 40804 | 26,092 |
https://mathoverflow.net/questions/39418 | 7 | A classical result of Larson and Sweedler says that a finite dimensional Hopf algebra over a field has invertible antipode. Does this result extend to the setting of Hopf algebras in braided categories? In other words, if a Hopf algebra in a braided category is dualizable, is its antipode necessarily invertible? (Equivalently, if a dualizable bialgebra has invertible fusion operator, is its opfusion operator invertible?) If this result is not true, is there an "easy" example where it fails? Of course, if it fails, it should do so in the free braided category equipped with a Hopf algebra, but it seems difficult to see that things *aren't* invertible using string diagrams.
| https://mathoverflow.net/users/396 | Do dualizable Hopf algebras in braided categories have invertible antipodes? | Yes, they do. See Theorem 4.1 in "Finite Hopf algebra in braided tensor categories" M. Takeuchi, Journal Pure and Applid Algebra 138 (1999) 59-82
| 6 | https://mathoverflow.net/users/6517 | 40806 | 26,094 |
https://mathoverflow.net/questions/40809 | -1 | If you divide 1 by 7, you'll of course have a repeating decimal sequence of `142857`. This of course repeats forever. Are there any scenarios where a quotient seems to repeat forever, but then changes at some point? Is this even possible? I'm not a mathematician (obviously), but rather a programmer. The question came up when I was thinking of a programmatic way to determine whether a division result is repeating.
Example (not real of course):
```
113548718971135487189711354871897113548718974898711
```
P.S. I can't even find an appropriate tag for this question, so feel free to edit my tags.
| https://mathoverflow.net/users/4019 | Is it possible for a large repeating sequence to appear in a non-repeating division quotient? | If you want to check if the result of the division is repeating you can make use of this (from
<http://www.mathlesstraveled.com/?p=134>):
> the decimal representation of any rational number will either terminate, or eventually become periodic. (As a bonus challenge, can you figure out how to tell the difference between rational numbers whose decimal expansion terminates, and those whose expansion repeats?) Note that we also know something else: since there are only (q-1) possible non-zero remainders when dividing by q, the repeating portion of the decimal expansion of a rational number with a denominator of q can be at most (q-1) digits long. It could be shorter, but it can’t be any longer. You can also look at this in reverse; for example, if you see a periodic decimal whose repeating portion is ten digits long, you know that the rational number it represents must have a denominator of at least 11.
So in your example I think you would only need to check up to the 7th digit and for sleepless's example, you would only need to check up the 13th digit.
I'm also a programmer and not a mathematician so I probably have this all wrong :)
| -1 | https://mathoverflow.net/users/9732 | 40812 | 26,097 |
https://mathoverflow.net/questions/40813 | 12 | In the second half of the section "[Operations with Natural Transformations](http://en.wikipedia.org/wiki/Natural_transformation#Operations_with_natural_transformations)" of the wikipedia article on natural transformations, they define the operation taking a natural tranformation $\eta:F\to G$ in the functor category $Cat(C,D)$ and functor $H:D\to E$ which produces a new natural transformation $(H\circ F)\to (H\circ G)$ in $Cat(C,E)$ by applying the functor to every coordinate of the original natural transformation. This operation frequently comes up in the definition of an adjunction.
Does this operation have a name? It's not the "Godement multiplication", but might be related to it.
Is there an agreed-upon symbol for this operation, other than juxtaposition?
*(Minor rant: I think that using juxtaposition for this operation is a horribly cruel thing to do to people learning category theory. Juxtaposition is already used for composition of morphisms -- and therefore for composition of functors-with-functors and naturaltransformations-with-naturaltransformations, since they too are the morphisms in $Cat$ and functor categories, respectively. Recycling juxtaposition yet again for this (non-associative!) operation on functors and natural transformations is just asking for trouble.)*
| https://mathoverflow.net/users/2361 | What is the name for the composition of a functor with a natural transformation? | It's called "whiskering" -- the 1-cells/functors composed on either side of the 2-cell/transformation look like "whiskers". See for example page 24 of [this paper](http://math.ucr.edu/home/baez/2rep.pdf). This terminology is pretty widespread in the categorical community.
| 22 | https://mathoverflow.net/users/2926 | 40814 | 26,098 |
https://mathoverflow.net/questions/40451 | 3 | My question arises from a discussion on an answer given by Maurizio Monge [here](https://mathoverflow.net/questions/40005/generalizing-a-problem-to-make-it-easier/40030#40030).I do not know if there is a known terminology for such matrices. By "sign matrices," I mean square matrices whose entries are in ${-1,+1}$.
For instance,
$\begin{bmatrix}
1 &-1 \\
-1& -1
\end{bmatrix}$ ,
$\begin{bmatrix}
-1&1&1 \\
1&1&-1 \\
-1&-1&-1
\end{bmatrix}$
Clearly, there are $2^{n^2}$ sign matrices of size $n\times n$. So, we start their theory by enumerating them as follows. For a matrix of size $n\times n$ we consider a truth table of $n^2$ arguments and therefore $2^{n^2}$ rows. Each row corresponds to the entries in one matrix$(a\_{11},a\_{12},\dots,a\_{1n},a\_{21},a\_{22},\dots,a\_{nn})$.
Let $M\_{(n,k)}$ be the $n \times n$ sign matrix corresponding to the $k^th$ row of the truth table.
**Question**: Does the following matrix product give the zero matrix for sign matrices of even size?
$\prod\_{k=1}^{2^{n^2}}M\_{(n,k)}$
Thank you. As usual, I will be delighted if you point me to good references on this.
| https://mathoverflow.net/users/5627 | 'Sign matrices'-(-1,+1) square matrices | Much is known about sign nonsingular patterns (sign patterns for which nonsingularity does not depend on the numerical values), if I remember correctly there is a characterization. Less is known about sign patterns which have allow (but do not require) nonsingularity. I suggest looking at the book *Matrices of sign-solvable linear systems* by Brualdi and Shader.
| 5 | https://mathoverflow.net/users/9733 | 40815 | 26,099 |
https://mathoverflow.net/questions/40816 | 10 | I'm trying to write a program with an input of numbers $n$ and $k$ (where $n<10^{1000}$ and $k<10^9$), where I compute fib[n] % k.
What is a good FAST way of computing this?
I realize that the resulting series is periodic, just not sure how to find it efficiently.
| https://mathoverflow.net/users/9734 | fibonacci series mod a number | This is really just an expansion of Gerhard's comment. One has the matrix formula
$$\begin{pmatrix}
1&1\\\
1&0
\end{pmatrix}^n=
\begin{pmatrix}
F\_{n+1}&F\_n\\\
F\_n&F\_{n-1}
\end{pmatrix}
$$
so the problem reduces to computing $A^n$ modulo $k$ where
$$A=\begin{pmatrix}
1&1\\\
1&0
\end{pmatrix}.$$
This can be done by the [repeated squaring](http://en.wikipedia.org/wiki/Exponentiation_by_squaring) method often used in
[modular exponentiation](http://en.wikipedia.org/wiki/Modular_exponentiation). The idea is to compute $A^n$ recursively
either as $(A^m)^2$ or $A(A^m)^2$ according to whether $n=2m$ or $n=2m+1$.
| 15 | https://mathoverflow.net/users/4213 | 40818 | 26,100 |
https://mathoverflow.net/questions/40795 | 4 | Recall that we say that a closed space $F$ of a Banach space $E$ is complemented if there exists a contractive projection $P$ from $E$ onto $F$.
>
> Do you know a charaterization of discrete amenable groups by the existence of a complementation of a closed space $F$ of a Banach space $E$?
>
More precisely, the required charaterization is
For all discrete group $G$, there exists a Banach space $E\_G$ and a closed space $F\_G$ of $E\_G$ such that
$G$ is amenable if and only if $F\_G$ is complemented in $E\_G$.
| https://mathoverflow.net/users/5210 | existence of charaterization of amenable groups by complementation? | Yes: a *discrete* group $G$ is amenable if and only if the reduced group C\*-algebra $C^\*\_r(G)$ is nuclear, see E.C. Lance, On nuclear $C^{\ast} $-algebras.
J. Functional Analysis 12 (1973), 157--176. This is then equivalent to $W^\*(G) = C^\*\_r(G)^{\*\*}$ being an *injective* von Neumann algebra: which by definition means that if $W^\*(G) \subseteq B(H)$ then there is a *contractive* projection from $W^\*(G)$ to $B(H)$.
I'm pretty sure you could look at the group von Neumann algebra $VN(G)$ instead, but I cannot recall the correct reference (but it's all in Runde's book "Lectures on Amenability"). Note that all this only works because $G$ is discrete.
Now, the problem is that you do need ``contractive'' projection here: it's still a conjecture if just having a bounded projection is enough.
Also, I'm sure there are other answers (and perhaps some that are easier: even a streamlined approach to all this uses a lot of operator algebra theory)...
| 2 | https://mathoverflow.net/users/406 | 40820 | 26,101 |
https://mathoverflow.net/questions/37677 | 11 | This question occurred to me while I was reading [Klartag's papers on central limit theorems for convex bodies](http://www.math.tau.ac.il/~klartagb/publications.html).
Given probability measures $\mu$, $\nu$ on (the Borel $\sigma$-field of) $R^d$ with finite first moments, their Wasserstein distance is given by:
$$W\_{R^d}(\mu,\nu) = \sup \mbox{ of }\int\_{R^d}f d\mu-\int\_{R^d}f d\nu\mbox{ over all 1-Lipschitz }f:R^d\to R.$$
(**NB**: there was an error in this formula -- an inf in the place of the sup.)
Given a vector $v\in R^d$, let $\mu\_v$ be the distribution of $X.v$, where $X$ has distribution $\mu$. Define $\nu\_v$ analogously. Note that $\mu\_v$ and $\nu\_v$ are distributions over $R$.
**Question:** is there a constant $C\_d>0$ depending on $d$ only such that: $$W\_{R^d}(\mu,\nu)\leq C\_d\sup\_{v\in R^d, |v|=1}W\_{R}(\mu\_v,\nu\_v)?$$
If so, how does $C\_d$ grow with $d$?
**An illustrative example:** Assume $Z$ is uniform over a $D-1$ dimensional sphere $S^{D-1}$ in $R^D$. Any one-dimensional marginal of $\sqrt{D-1}Z$ is approximatelly Gaussian. Now let $\mu$ be the law of the first $d$ coordinates of $\sqrt{D-1}Z$, and $\nu$ be the standard Gaussian distribution on $R^d$. Can one deduce from the previous statement alone that $\mu$ and $\nu$ are close?
**Another example:** Let $Z$ be a random vector in $R^D$ with mean $0$ and covariance matrix $I\_D$, $D\gg 1$. Old results of Sudakov (quoted [here](https://doi.org/10.1007/s004400050087)) show that "most" one-dimensional marginals of $Z$ are close to $|Z|N$ where $N$ is standard normal and independent from $Z$. A positive answer to the above question would lead to typical results for $d$-dimensional projections of $Z$.
| https://mathoverflow.net/users/8354 | Wasserstein distance in R^d from one dimensional marginals | There is a result which contains an answer to your question in a somewhat different form. Instead of the transportation metric it uses another metric which metrizes the weak topology in the space of measures on $\mathbb R^d$:
$$
\lambda(\mu,\nu) \le \delta \iff \exists\; T\ge 1/\delta : \langle \exp(i(t,\cdot)),\mu-\nu\rangle \le \delta \quad\forall\; |t|\le T \;,
$$
which might still be OK for your purposes. [This article](https://mathscinet.ams.org/mathscinet-getitem?mr=1485529 "Klebanov, L. B.; Rachev, S. T. Proximity of probability measures with common marginals in a finite number of directions.") by Klebanov and Rachev actually contains a stronger result (Theorem 4): it gives an explicit upper estimate for $\lambda(\mu,\nu)$ in terms of the maximal distance $\lambda(\mu\_v,\nu\_v)$ between the projections of $\mu$ and $\nu$ onto a finite (growing) number of directions $v$ in $\mathbb R^d$.
PS In spite of a sufficiently long tradition of misnaming the transportation metric (some people even go as far as calling it after Hutchinson), I would still insist on using the name of Kantorovich (or Monge-Kantorovich, Kantorovich-Rubinshtein), see, for instance, [this](https://encyclopediaofmath.org/index.php?title=Wasserstein_metric) historical article.
| 6 | https://mathoverflow.net/users/8588 | 40824 | 26,104 |
https://mathoverflow.net/questions/40821 | 5 | I have read several times that assuming Con(ZFC), and using compactness it can be proved the existence of a model of ZFC with an ill-founded $\omega$. How is that? Any reference will be welcome.
| https://mathoverflow.net/users/6466 | Existence of an $\omega$-nonstandard model of ZFC from compactness | This is a standard application of the Compactness Theorem, and works basically the same in producing nonstandard models of ZFC as it does for producing nonstandard models of PA or real-closed fields.
Consider the theory $T$, in the language of set theory
augmented with an additional constant symbol $c$,
consisting of all the ZFC axioms, plus the assertions that
$c$ is a natural number, but not equal to $0$, not equal to
$1$, and so on, including for each natural number $n$ the statement $\varphi\_n$ that $c$ is not
equal to $n$. (Note that
all such $n$ are definable in set theory, and so we use the
definition of $n$ in $\varphi\_n$ when asserting that $c$ is
not $n$.)
If there is a model $M$ of ZFC, then every finite subtheory
of $T$ is consistent, since any finite subtheory of $T$
makes only finitely many assertions about $c$, and we may
therefore interpret $c$ as any natural number of $M$ not
mentioned in the subtheory.
Thus, by compactness, $T$ has a model. Any such model of
$T$ will be $\omega$-nonstandard, since the interpretation
of $c$ in the model will be a nonstandard natural number.
There are numerous other ways to produce
$\omega$-nonstandard models of ZFC from existing models,
the most common being ultrapowers.
The conclusion is that if there is any model of ZFC, then
there are nonstandard models of ZFC. The converse of this
is not true, for if there are any transitive models of ZFC,
then there is an $\in$-minimal such model $M$, and being
standard, $M$ will have the same arithmetic as the ambient
universe, thus thinking Con(ZFC), but being minimal, will
have no transitive models of ZFC inside it.
| 14 | https://mathoverflow.net/users/1946 | 40826 | 26,105 |
https://mathoverflow.net/questions/40819 | 6 | I had this question bothering me for a while, but I can't come up with a meaningful answer.
The problem is the following:
Let integers $a\_i,b\_j\in${$1,\ldots,n$} and $K\_1,K\_2\in$ {$1,\ldots,K$}, then how small (as a function of $K$ and $n$), but strictly positive, can the following absolute difference be.
$\biggl|(\sum\_{i=1}^{K\_1} \frac{1}{a\_i})-(\sum\_{j=1}^{K\_2} \frac{1}{b\_j})\biggr|$
As an example for $K\_1=1,$$K\_2=1$ choosing $a\_1=n,$$b\_1=n-1$gives the smallest positive absolute difference, that is $\frac{1}{n(n-1)}$. What could the general case be?
| https://mathoverflow.net/users/2763 | The difference of two sums of unit fractions | A sum of $K$ unit fractions, each of denominator $n\_i \leq n$, can
be rewritten as a fraction with a denominator bounded by
the product of the $n\_i$, i.e. by $n^K$. (A small improvement is possible, with product of distinct integers $\leq n$)
A difference of two such fractions
(which are themselves sums of $K\_1$ and $K\_2$ fractions, respectively),
is a fraction with a demominator bounded by $n^{K\_1+K\_2}$.
(If $K\_1\neq K\_2$ the order of magnitude of the difference is actually
$\frac{1}{n^{\min(K\_1,K\_2)}}$.)
Let us assume that $K\_1=K\_2$.
So, the smallest nonzero difference $d(K,n)$ of sums of
$K$ unit fractions,
with $n\_i \leq n$ is $d(K, n)\leq \frac{1}{n^{2K}}$.
I believe that this order of magnitude, with slightly weaker constants,
can be achieved with a constructive parametrization.
Example: If $K\_1=K\_2=2$, then choose
$\frac{1}{x^2 + 4 x + 1} + \frac{1}{x^2 + 4 x + 3} -
\frac{1}{x^2 + 3 x + 1} - \frac{1}{x^2 + 5 x + 5}=
\frac{2}{(1 + 3 x + x^2) (1 + 4 x + x^2) (3 + 4 x + x^2) (5 + 5 x + x^2)}$.
Now, taking $n=x^2+5x+5$, one has a set of 4 unit fractions,
for which the difference of the sums above is asymptotically
$ \frac{2}{n^4}$. This example is (for $K\_1=K\_2=2$)
possibly the best one can find, (but I did not prove this).
I conjecture that for larger values of
$K$ one can construct similar polynomial examples.
Quite possibly this has applications in questions in diophantine approximation,
exponential sums, large sieve etc.
| 5 | https://mathoverflow.net/users/9737 | 40827 | 26,106 |
https://mathoverflow.net/questions/40835 | 7 | In Jech's Set Theory he defines a $\kappa$-scale as a family of functions $\langle f\_\alpha\colon\omega\to\omega | \alpha < \kappa \rangle$ for which:
1. $f\_\alpha < f\_\beta$ except maybe for a finite set
2. For any $g\colon\omega\to\omega$ there is some $f\_\alpha>g$ (again for all but perhaps a finite set)
(the definition can be found in chapter 10 which deals with measurable cardinals)
Clearly if a $\kappa$-scale does exist then $\kappa \le 2^{\aleph\_0}$ (as there are only that many functions to begin with).
If we look at the quasi-order on $\omega^\omega$ defined by as above, that is $f < g$ if $f(n) < g(n)$ for all but a finite number of $n\in\omega$, then we can immediately say that:
1. There are no maximal elements (for any $f$ we can define $g(n) = f(n) +1$ and clearly $f< g$)
2. For any given $f,g$ there exists an upper-bound for both (namely $\max (f(n), g(n))$)
3. If there exists a $\kappa$-scale then there is a cofinal subset of the quasi-order of cardinality $\le \kappa$ (cofinal in the sense that for every $f$ you can find some $g$ in the cofinal subset for which $f< g$)
My question is, if so, assuming $\kappa$ is the least cardinal number for which a $\kappa$-scale exists, and $A\subseteq \omega^\omega$ some $ < $-cofinal subset. Does it follow that $|A|\ge\kappa$? Does the assumption that $2^{\aleph\_0} = \lambda$ holds some property (e.g. regularity) affects the existence of such $\kappa$ or $A$?
| https://mathoverflow.net/users/7206 | $\kappa$-scales and the continuum | This is a question regarding two famous cardinal characteristics:
* $\mathfrak{b}$ the minimal cardinality of an unbounded family in $(\omega^\omega,{<^\*})$.
* $\mathfrak{d}$ the minimal cardinality of a cofinal family in $(\omega^\omega,{<^\*})$.
It is not difficult to show that a scale exists if and only if $\mathfrak{b} = \mathfrak{d}$, and then the minimal length of a scale is the common value of these two cardinal characteristics (which is also the cofinality of any scale). However, this equality need not hold. In fact, the only inequalities that must hold are
$$\aleph\_1 \leq \mathfrak{b} = cf(\mathfrak{b}) \leq cf(\mathfrak{d}) \leq \mathfrak{d} \leq \mathfrak{c}.$$
Using Hechler's Theorem (see this [answer of mine](https://mathoverflow.net/questions/29624/how-many-orders-of-infinity-are-there/29626#29626)) any pattern of cardinals consistent with the above is possible to attain by a forcing. See Andreas Blass's handbook of set theory chapter (available [here](http://www.math.lsa.umich.edu/~ablass/set.html)) for more on this topic.
| 5 | https://mathoverflow.net/users/2000 | 40838 | 26,111 |
https://mathoverflow.net/questions/40836 | 1 | ### Motivation
I'm studying an approach to axiomatic thermodynamics based on the notion of commutative semigroup $(S,+)$ with a preorder relation $\to$ on $S$. In other words, $S$ is non-empty set, the operation $+: S \times S \to S$ is commutative and associative, the relation $\to$ on $S$ is reflexive and transitive, and in addition there is a compatibility condition:
$$ (\forall a,b,c \in S) \qquad a \to b \iff a + c \to b + c $$
### Question
>
> Is this a standard (or well-known) structure? And if so, is there an accepted term for it?
>
>
>
Naively I thought that it would be a *pre-ordered commutative semigroup*, but I googled and found very little with this name and the little I found suggests that the compatibility condition obeyed by a pre-ordered commutative semigroup is the weaker
$$ (\forall a,b,c \in S) \qquad a \to b \implies a + c \to b + c $$
| https://mathoverflow.net/users/394 | Is this a pre-ordered commutative semigroup? | Your condition certainly was not considered by algebraists studying commutative semigroups. It implies, for example, that if the semigroup has a $0$ (i.e. $0+x=0$), then for every two $a,b$ $a\to b$ and $b\to a$. Your condition makes more sense for semigroups satisfying cancelation law: $a+c=b+c\to a=b$. Then you can embed your semigroup into a group where the two conditions are equivalent.
| 4 | https://mathoverflow.net/users/nan | 40840 | 26,112 |
https://mathoverflow.net/questions/40839 | 1 | Let $\Delta=\{\alpha\_1,\alpha\_2\}$ be the simple root system
of the exceptional Lie group $G\_2$
with $\alpha\_1$ is short and $\alpha\_2$ is long,
so $\lambda\_1=2\alpha\_1+\alpha\_2,\lambda\_2=3\alpha\_1+2\alpha\_2$
are the fundamental dominant weights.
Let $T$ be the maximal torus of $G\_2,$ then $H^\*(BT;Z)$ is a polynomial algebra
on two generators. Can we see $\alpha\_1$ and $\alpha\_2$ as the generators?
What about $\lambda\_1$ and $\lambda\_2$?
| https://mathoverflow.net/users/8152 | What does the weights of Lie group mean? | Let *T* be an arbitrary compact torus.
The second cohomology group of *BT* (with arbitrary coefficients, call that ring *k*) generates the full cohomology freely as an algebra. In other words, if you pick a *k*-basis *x*1, *x*2,... of *H*2(*BT*), then you get an isomorphism of *H*\*(*BT*) with *k*[*x*1, *x*2,...].
Now let's specialise to the case *k*=ℤ. In that case, the second cohomology group of *BT* is canonically isomorphic to the group of characters of *T*, i.e., to the group of homomorphisms from *T* → *S*1. Given a character χ :*T* → *S*1, the corresponding element of *H*2(*BT*) is represented by (the first Chern class of) the complex line bundle *ET* ×T ℂχ → *BT* = *ET*/*T*.
Now back to your question. The elements α1 and α2 form a basis of *BT*, where *T* now refers to the maximal torus of *G*2. So you get an isomorphism *H*\*(*BT*;ℤ) $\xrightarrow{\sim}$ ℤ[α1, α2]. But
λ1 and λ2 also form a basis of *BT*. So you get another isomoprhism *H*\*(*BT*;ℤ) $\xrightarrow{\sim}$ ℤ[λ1, λ2].
---
Let me also answer the question in the title of your question:
By "weight of a Lie group", one means a homomorphism from its maximal torus to *S*1.
| 3 | https://mathoverflow.net/users/5690 | 40843 | 26,113 |
https://mathoverflow.net/questions/40842 | 4 | It is my understanding that Dennis Johnson defined a `relative weight filtration,' of the mapping class group of an oriented surface. My question is what is this filtration, and how does it relate to the lower central series of the mapping class group? In particular, why is the relative weight filtration the "right" filtration as opposed to the lower central series.
| https://mathoverflow.net/users/9417 | What is the relative weight filtration of the mapping class group of a surface? | Let $S$ be a surface (for simplicity, assume that $S$ has exactly one boundary component) and let $Mod(S)$ be its mapping class group. Let's assume that the genus of $S$ is at least $3$. To begin with, $Mod(S)$ is perfect, so its lower central series is not interesting. Define
$\mathcal{I}(S)$ to be the Torelli group, ie the kernel of the action of $Mod(S)$ on $H\_1(S)$. Things work better here. Johnson proved that the intersection of the lower central series for $\mathcal{I}(S)$ is trivial. Let me describe exactly what he did.
For a group $G$, let $\gamma\_k(G)$ be the kth term in the lower central series for $G$, indexed so that $\gamma\_0(G) = G$. Since $S$ has a boundary component, we can stick a basepoint $p$ on that boundary component and get an honest action of $Mod(S)$ on $\Gamma:=\pi\_1(S,p)$ (if $S$ had no boundary, then we would only get an outer action). Johnson defined $\mathcal{I}(S,k)$ to be the kernel of the action of $Mod(S)$ on $\Gamma / \gamma\_k(\Gamma)$. This gives a filtration
$$\mathcal{I}(S) = \mathcal{I}(S,1) \supset \mathcal{I}(S,2) \supset \cdots$$
I believe that this is the filtration you are referring to (it has become known as the Johnson filtration). Johnson proved that $\cap\_{k=1}^{\infty} \mathcal{I}(S,k) = 1$.
It is not hard to show that $\mathcal{I}(S,k) / \mathcal{I}(S,k+1)$ is abelian. One might thus be led to conjecture that $\mathcal{I}(S,k) = \gamma\_{k-1}(\mathcal{I},k)$. Johnson proved that this is false. More specifically, he calculated $\mathcal{I}(S) / \gamma\_1(\mathcal{I}(S))$ and showed that it contains a lot of 2-torsion coming from the Rochlin invariants of homology 3-spheres (the appropriate quotients were originally constructed by Birman and Craggs).
It is true that $\mathcal{I}(S,2)$ is the kernel of the universal torsion-free abelian quotient of $\mathcal{I}(S,1)$. One might thus conjecture that $\mathcal{I}(S,k)$ is the "torsion-free lower central series" of $\mathcal{I}(S)$. This hope was dashed by Morita, who showed that $\mathcal{I}(S,2)$ has a $\mathcal{I}(S)$-invariant $\mathbb{Z}$-quotient coming from the Casson invariant of homology 3-spheres that doesn't vanish on $\mathcal{I}(S,3)$.
We thus have two filtrations of $\mathcal{I}(S)$, the Johnson filtration and the "torsion-free lower central series". I'm not sure which is "better", but it is certainly true that the "torsion-free lower central series" is better understood due to a lot of work by Hain. In particular, he calculated a presentation for the Malcev completion of $\mathcal{I}(S)$, which is the completion of the filtered Lie algebra associated to the "torsion-free lower central series" filtration.
| 10 | https://mathoverflow.net/users/317 | 40847 | 26,115 |
https://mathoverflow.net/questions/40849 | 4 | This should be an easy question, but I don't quite know how to approach it. It may be somewhat related to the concepts mentioned in the context of [this past question](https://mathoverflow.net/questions/32033/), though it was motivated mainly by the college calculus course I am teaching.
Question: Characterize those subsets *S* of the reals for which there exists a function *f* defined on the reals such that the set of points in the reals where *f* is continuous is precisely *S*.
Variant(s): Same as above, but we require *f* to be Borel (inverse images of Borel sets are Borel) or Lebesgue measurable (inverse images of Borel sets are measurable).
What I know:
1. If *S* is open, we can set *f* as the function that is 0 on *S*, 1 on the rationals outside *S* and -1 on the irrationals outside *S*. This is continuous only on *S*.
2. If *S* is closed, we can set *f* as follows. Define $d(x,S)$ as the distance from *x* to *S*. Define $f(x) = 0$ if $d(x,S)$ is rational and $f(x) = d(x,S)$ if $d(x,S)$ is irrational. Since $d(x,S)$ is continuous and is zero precisely on *S*, our *f* works.
3. We can combine the above two tricks to handle *S* locally closed, i.e., the intersection of an open and a closed set.
I suspect that the set of possible *S*es that arise as the sets of continuity form a $\sigma$-algebra. Is this true? If it is (or perhaps through some other method) we could probably show that all Borel sets are of this form.
[ADDED: For instance, there is the famous example of the function that is $1/n$ on all rationals with denominator $n$ in simplified form and $0$ on all irrationals -- that is an example of a function that is continuous only on the irrationals.]
Assuming the axiom of choice, and allowing non-measurable functions, can we get *every* possible subset of the reals?
| https://mathoverflow.net/users/3040 | Possible subsets of reals that equal the set of continuity of a function | The sets which arise as the points of continuity of a real-to-real function are precisely the $G\_{\delta}$ sets.
I think the constructive direction only makes use of Borel sets, so the answer to the variant should be the same.
| 7 | https://mathoverflow.net/users/9068 | 40855 | 26,118 |
https://mathoverflow.net/questions/39941 | 13 | Let $X$ be a scheme (or more generally a ringed space, if it works). Does $Qcoh(X)$, the category of quasi-coherent sheaves on $X$, admit a generating set? This would be useful, because then every cocontinuous functor on $Qcoh(X)$ has a right adjoint (SAFT).
If $X$ is affine, then $\mathcal{O}\_X$ is a generator. I doubt that this is true in general. If $X$ is quasi-separated, perhaps the direct images of the $\mathcal{O}\_U$, $U$ affine, do the job, but the naive proof does not work. If $Qcoh(X)$ does not have a generating set in general, what conditions on $X$ guarantee this?
EDIT: It is true when $X$ is concentrated, i.e. quasi-compact and quasi-separated, in particular when $X$ is noetherian (see Philipp's comment). This is already satisfying. Anyway, are there other (counter)examples?
PS: Note that this question is somehow unnatural with the background of [this question](https://mathoverflow.net/questions/38009/abelian-categories-enriched-over-sheaves); $\underline{Qcoh}(X)$, considered as a stack of abelian categories, always has a "stack-generator", namely $U \mapsto \mathcal{O}\_U$. Nevertheless, I think the question above is interesting.
| https://mathoverflow.net/users/2841 | Does Qcoh(X) admit a generating set? | Gabber's argument also appears in print in
[Enochs and Estrada, "Relative homological algebra in the category of quasi-coherent sheaves," Adv. in Math. 194 (2005) 284--295](http://dx.doi.org/10.1016/j.aim.2004.06.007).
| 9 | https://mathoverflow.net/users/2628 | 40859 | 26,121 |
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