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https://mathoverflow.net/questions/17152 | 26 | Sorry if this is a silly question. I was wondering, under what axioms of set theory is it true that if $\alpha$,$\beta$ are cardinals, and $2^\alpha=2^\beta$, then $\alpha=\beta$? Do people use these conditions to prove interesting results?
This question is prompted from a recent perusing of Johnson's "Topics in the Theory of Group Presentations", where in the first few pages he "proves" free groups of different rank are non-isomorphic:
the number of mappings from a free group of rank $\omega$ to the group $\mathbb{Z}/2\mathbb{Z}$ is $2^\omega$, which would be invariant under isomorphism; and then he assumes the topic of my question: $2^\alpha=2^\beta$ implies $\alpha=\beta$.
But I remember reading something a few years ago about a student of R.L. Moore (I only remember his last names was Jones) "proving" the Moore Space conjecture, and using that $\alpha > \beta$ implied $2^\alpha > 2^\beta$, but that this was incorrect.
Anyway, I realized I don't know anything about when this is true or false, so I thought I'd ask.
| https://mathoverflow.net/users/1446 | When $2^\alpha = 2^\beta$ implies $\alpha=\beta$ ($\alpha,\beta$ cardinals) | François gives the correct affirmative answer. For the negative side, the usual method of proving that the negation of the Continuum Hypothesis is consistent with ZFC is to use the method of forcing to add Aleph2 many Cohen reals, so that 2ω = ω2 in the forcing extension V[G]. In this model V[G], it is also true that 2ω1 = ω2. Thus, this model shows that it is not necessarily true that different-sized sets have different sized power sets. The case of symmetric groups is likely more interesting than free groups, because in this model, the symmetric groups Sω and Sω1 have the same cardinality ω2. Nevertheless, these two groups are not isomorphic, as explained in [this MO question](https://mathoverflow.net/questions/12943).
The general answer about what can be true for the continuum function κ --> 2κ is exactly provided by [Easton's Theorem](http://en.wikipedia.org/wiki/Easton's_theorem). This remarkable theorem states that if you have any class function E, defined on the regular cardinals κ, with the properties that
* κ ≤ λ implies E(κ) ≤ E(λ)
* κ < E(κ)
* κ < Cof(E(κ))
then there is a forcing extension V[G] in which 2κ = E(κ) for all regular cardinal κ. In particular, this shows that the sizes of the power sets (on regular cardinals) are restricted to obey only and exactly the hypotheses listed explicitly above. Each of these properties corresponds to a well-known fact about cardinal exponententiation.
Using Easton's theorem, we can build models of set theory where 2κ = κ++ for all regular κ. The added power of the Woodin/Foreman result mentioned by François is that they also get this for singular cardinals.
The point now is that there are innumerable examples provided by Easton's theorem that satisfy your hypothesis that the continuum function is one-to-one. If one begins with a model of GCH and selects *any* injective Easton function E, then the resulting model of set theory V[G] will have E as it's continuum function κ --> 2κ = E(κ) for regular κ, and the GCH will continue to hold at singular κ, preserving injectivity. So one is quite free to satisfy your hypothesis while having any kind of crazy failures of GCH.
| 21 | https://mathoverflow.net/users/1946 | 17154 | 11,478 |
https://mathoverflow.net/questions/17138 | 21 | I browsed Dirichlets Werke today and was kind of surprised by two remarks that he made on p. 354 (Über die Bestimmung ...) and p. 372 (Sur l'usage ...). In the second paper, he claims (my translation)
*I have applied these principles to a demonstration of the remarkable formula given by Legendre for expressing in an approximate manner how many prime numbers there are below an arbitrary, but very large, limit.*
In a handwritten note on the reprint he sent to Gauss he remarked that $\sum 1/\log n$ (this is Gauss's version of the PNT, at least if you replace the sum by an integral) is a better estimate than Legendre's.
I am a little bit puzzled as to why Dirichlet's claim to have proved the prime number theorem is not discussed anywhere in the literature. Or is it?
| https://mathoverflow.net/users/3503 | Dirichlet and the prime number theorem | Dirichlet's remark from the first paper is extracted and translated on page 98 of The Development of Prime Number Theory by Narkiewicz. So this has not passed completely unnoticed. Narkiewicz remarks that Dirichlet believed that his analytic methods would enable him to prove Legendre's conjecture, and that Dirichlet never returned to the problem.
Dirichlet remained interested in the asymptotic growth laws ("Asymptotische Gesetze") of arithmetic functions for the rest of his life, as seen from his 1849 paper with the estimate
$$
\sum\_{n \leq x}d(n) = x\log(x) + (2\gamma - 1)x + O(x^{1/2}),
$$
and a couple of other estimates, and a letter of 1858 to Kronecker reprinted in Dirichlet's Werke, where he mentions having obtained a substantial improvement of the error term $O(x^{1/2})$ by a new method.
Since Dirichlet demonstrably did not lose interest in such questions, and never returned to the PNT in print, it seems reasonable to believe that he discovered that his real-variable method would not yield the PNT.
| 7 | https://mathoverflow.net/users/3304 | 17163 | 11,483 |
https://mathoverflow.net/questions/17115 | 28 | I am pretty sure that the following statement is true. I would appreciate any references (or a proof if you know one).
Let $f(z)$ be a polynomial in one variable with complex coefficients. Then there is the following dichotomy. Either we can write $f(z)=g(z^k)$ for some other polynomial $g$ and some integer $k>1$, or the restriction of $f(z)$ to the unit circle is a loop with only finitely many self-intersections. (Which means, more concretely, that there are only finitely many pairs $(z,w)$ such that $|z|=1=|w|$, $z\neq w$ and $f(z)=f(w)$.)
EDIT. Here are a couple reasons why I believe the statement is correct.
1) The statement is equivalent to the following assertion. Consider the set of all ratios $z/w$, where $|z|=1=|w|$ and $f(z)=f(w)$ (here we allow $z=w$). If $f$ is a nonconstant polynomial, then this set is finite.
[[ Here is a proof that the latter assertion implies the original statement. Suppose that there are infinitely many pairs $(z,w)$ such that $|z|=1=|w|$, $z\neq w$ and $f(z)=f(w)$. Then some number $c\neq 1$ must occur infinitely often as the corresponding ratio $z/w$. However, this would imply that $f(cz)=f(z)$ (as polynomials). It is easy to check that this forces $c$ to be a root of unity, and if $k$ is the order of $c$, then $f(z)=g(z^k)$ for some polynomial $g(z)$. ]]
Going back to the latter assertion, note that the set of all such ratios is a compact subset of the unit circle, and it is not hard to see that 1 must be an isolated point of this set. So it is plausible that the whole set is discrete (which would mean that it is finite).
2) If I am not mistaken, experiments with polynomials that involve a small number of nonzero monomials (such as 2 or 3) also confirm the original conjecture.
| https://mathoverflow.net/users/4384 | Restriction of a complex polynomial to the unit circle | You're right. Quine proved in "[On the self-intersections of the image of the unit circle under a polynomial mapping](http://www.jstor.org/stable/2039005)" that if the degree is $n$ and $f(z)\neq g(z^k)$ with $k>1$, then the number of points with at least 2 distinct preimage points is at most $(n-1)^2$. An example shows that this is sharp. Here's the [review](http://www.ams.org/mathscinet-getitem?mr=313485) in MR.
After the proof, there is a remark:
>
> As a simple consequence of this theorem we note that a polynomial $p$
> cannot map $|z| < 1$ conformally onto a domain with a slit, for in this case
> $p(e^{i\phi})$ would have an infinite number of vertices.
>
>
>
| 20 | https://mathoverflow.net/users/1119 | 17164 | 11,484 |
https://mathoverflow.net/questions/17166 | 2 | It is well known that the 1-dimensional [heat equation](http://en.wikipedia.org/wiki/Heat_equation) $$\frac{\partial}{\partial t} u(x,t)=a\cdot\frac{\partial^2}{\partial x^2} {u(x,t)}$$ has the fundamental solution $$K(x,t)=\frac{1}{\sqrt{4\pi a t}} \ \exp\left(-\frac{x^2}{4at}\right)$$.
**My question**
I am looking for English or German references for an easy derivation of this particular solution which are comprehensible to undergraduates.
| https://mathoverflow.net/users/1047 | Undergraduate Derivation of Fundamental Solution to Heat Equation | I think the term fundamental solution (at least sometimes) conventionally includes the integral around your $K$. I will assume this. If I recall correctly then the following argument is from "Partial Differential Equations" by Strauss.
A particularly simple solution follows from the self-similarity principle, i.e.
If $u(x,t)$ is a solution then so is $u(cx, a c^2t)$
This suggests looking for a particular solution of the form $K(x,t) = g(p)$, where $p = \frac{x}{\sqrt{4at}}$
Substituting $g$ into the heat equation leads to the differential equation
$$g''+\frac{p}{2}g' = 0 $$
Then the fundamental solution as above follows from solving this.
| 9 | https://mathoverflow.net/users/3623 | 17167 | 11,486 |
https://mathoverflow.net/questions/17180 | 2 | Motivation
----------
The problem I am facing can be considered a variant of the standard set packing problem. However; instead of being given a list of sets, I am given a function $\nu : 2^N \rightarrow \{0,1\}$ and want to find a partitioning $P$ of $N$ that maximizes $g(P) = \sum\_{S \in P} \nu(S)$. This can be shown to require somewhere between $O(2^{|N|})$ and $O(3^{|N|})$ operations.
The above problem can (almost) be reduced to the problem of finding a partition $P\_3$ of $N$ into three sets that maximizes $g(P\_3)$. However there are still $O(3^{|N|})$ partitionings of $N$ into three sets.
Lets say we construct such a 3 partition $(S\_1,S\_2,S\_3)$ as follows: For each element $i \in N$, we add $i$ to the first set with probability $x\_i$, to the second set with probability $y\_i$ and to the third with probability $z\_i$, where $x\_i+y\_i+z\_i = 1$ and $0 \leq x\_i, y\_i, z\_i$.
It can be shown that the expected value of such a probability distribution, $x,y,z$, over the 3 partitions of $N$ is $f(x,y,z) = E[g(P)] = \sum\_{c \subset N} \nu(C)\left[\Pi\_{i \in C} x\_i \Pi\_{i \not \in C} (y\_i+z\_i) + \Pi\_{i \in C} y\_i \Pi\_{i \not \in C} (x\_i+z\_i) + \Pi\_{i \in C} z\_i \Pi\_{i \not \in C} (x\_i+y\_i)\right]$.
I am considering the situation in which we relax the constraint $x\_i+y\_i+z\_i = 1$ to $x\_i+y\_i+z\_i \leq 1$ and then using techniques akin to interior point methods for standard convex programming.
This relaxation clearly does not change maximum of $f(x,y,z)$ and with it in place $f(x,y,z)$ can be shown to be convex over our feasible set $0 \leq x\_i,y\_i,z\_i$ and $x\_i +y\_i+z\_i \leq 1$ for $i \in N$.
---
Given the above, my question is: Is there any general theory for maximizing convex functions over convex compact sets? (Apart from that the maximum must appear on the boundary?)
First time poster, so my apologies if I have tagged this inappropriately.
I know of much work in convex programming (minimizing convex functions over convex sets) but haven't been able to find similar work for maximization.
| https://mathoverflow.net/users/4401 | Algorithmic aspects of maximizing a convex function over a convex set | For one, the KKT conditions still apply. But in general even simple problems can be tough. For instance, maximizing a p-norm over a fairly nice type of convex polytope ("parallelotope") is NP hard -- see Bodlaender et al, "Computational Complexity of Norm Maximization." The general wisdom is: minimizing convex functions is easy; maximizing convex functions (equivalently, minimizing concave functions) is hard, though there are obviously exceptions.
| 4 | https://mathoverflow.net/users/1557 | 17183 | 11,494 |
https://mathoverflow.net/questions/12737 | 2 | Let S and S' be closed [Edit: orientable] surfaces, then it is well known that for S' to cover S it is necessary and sufficient that chi(S)|chi(S'). (Here 'chi' denotes the Euler characteristic).
However, if S and S' are punctured surfaces then the above condition is necessary but no longer sufficient.
>
> Is the question of determining necessary and sufficient criteria for S' to cover S answered in the research or expository literature?
>
>
>
I think that I know such criteria and want to use them a paper that I'm working on but (a) may be deceiving myself and (b) want to know whether I should write up a proof anew or whether there's a suitable reference. Surely the relevant criteria have been rediscovered many times, but I've never seen them discussed in writing.
Edit: Thanks Pete, I should have demanded that my surfaces be orientable
| https://mathoverflow.net/users/683 | Necessary and sufficient criteria for a surface to cover a surface | In a [recent paper](http://arxiv.org/abs/1003.0411) by Calegari, Sun, and Wang, the authors cite
W.S. Massey, Finite covering spaces of 2-manifolds with boundary. Duke Math. J. 41 (1974),
no. 4, 875-887.
which proves that Dmitri's conditions (1) and (2) are sufficient if the S and S' are compact, orientable with non-empty boundary.
(Dmitri's condition (3) is redundant as Massey points out in the second page of his article: writing chi(S') = 2 - 2g(S') - p(S') and chi(S) = 2 - 2g(S) - p(S) Dimitri's condition (1) gives
2 - 2g(S') - p(S') = d(2 - 2g(S) - p(S))
and reducing (mod 2) gives that p(S') and dp(S) have the same parity. But the proof therein is longer than the one that Dimitri for the case that he treats (if one takes the Ore result as given).
| 2 | https://mathoverflow.net/users/683 | 17194 | 11,502 |
https://mathoverflow.net/questions/17197 | 35 | This is a question, or really more like a cloud of questions, I wanted to ask awhile ago based on [this SBS post](http://sbseminar.wordpress.com/2008/09/10/a-curious-speculation/) and [this post I wrote](http://qchu.wordpress.com/2009/06/07/the-catalan-numbers-regular-languages-and-orthogonal-polynomials/) inspired by it, except that Math Overflow didn't exist then.
As the SBS post describes, the Catalan numbers can be obtained as the moments of the trace of a random element of $SU(2)$ with respect to the Haar measure. This is equivalent to the integral identity
$$\int\_{0}^{1} (2 \cos \pi x)^{2k} (2 \sin^2 \pi x) \, dx = C\_k.$$
I can prove this identity "combinatorially" as follows: let $A\_n$ denote the Dynkin diagram with $n$ vertices and $n-1$ undirected edges connecting those vertices in sequence. The adjacency matrix of $A\_n$ has eigenvectors $\mathbf{v}\\_i$ with entries $\mathbf{v}\\_{i,j} = \sin \frac{\pi ij}{n+1}$ with corresponding eigenvalues $2 \cos \frac{\pi i}{n+1}$. If $k \le n-1$, then a straightforward computation shows that the number of closed walks from one end of $A\_n$ to itself of length $2k$ is
$$\frac{1}{n+1} \sum\_{i=1}^{n} \left( 2 \cos \frac{\pi i}{n+1} \right)^{2k} 2 \sin^2 \frac{\pi}{n+1} = C\_k$$
by the combinatorial definition of the Catalan numbers. Taking the limit as $n \to \infty$ gives the integral identity; in other words, the integral identity is in some sense equivalent to the combinatorial definition of the Catalan numbers in terms of closed walks on the "infinite path graph" $A\_{\infty}$. (Is this the correct notation? I mean the infinite path graph with one end.)
Now, closed walks of length $2k$ from one end of $A\_n$ to itself can be put in bijection with **ordered rooted trees** of depth at most $n$ and $k$ non-root vertices. (Recall that the Catalan numbers also count ordered rooted trees of arbitrary depth.) The generating function $P\_n$ of ordered rooted trees of depth at most $n$ satisfies the recursion
$$P\_1(x) = 1, P\_n(x) = \frac{1}{1 - x P\_{n-1}(x)}.$$
This is because an ordered rooted tree of depth $n$ is the same thing as a sequence of ordered rooted trees of depth $n-1$ together with a new root. (Recall that the generating function of the Catalan numbers satisfies $C(x) = \frac{1}{1 - x C(x)}$. In other words, $C(x)$ has a continued fraction representation, and $P\_n$ is its sequence of convergents.) On the other hand, since $P\_n(x)$ counts walks on the graph $A\_n$, it is possible to write down the generating function $P\_n$ explicitly in terms of the characteristic polynomials of the corresponding adjacency matrices, and these polynomials have roots the eigenvalues $2 \cos \frac{\pi i}{n+1}$. This implies that they must be the **Chebyshev polynomials** of the second kind, i.e. the ones satisfying
$$q\_n(2 \cos x) = \frac{\sin (n+1) x}{\sin x}.$$
But the Chebyshev polynomials of the second kind are none other than the characters of the irreducible finite-dimensional representations of $SU(2)$! In particular, they're orthogonal with respect to the Haar measure.
The overarching question I have is: *how does this sequence of computations generalize, and what conceptual framework ties it together?* But I should probably ask more specific sub-questions.
**Question 1:** I remember hearing that the relationship between the Catalan numbers and the Chebyshev polynomials generalizes to some relation between moments, continued fractions, and orthogonal polynomials with respect to some measure. Where can I find a good reference for this?
**Question 2:** I believe that adding another edge and considering the family of cycle graphs gives the sequence ${2k \choose k}$ and the Chebyshev polynomials of the **first** kind, both of which are related to $SO(2)$. According to the SBS post, this is a "type B" phenomenon, whereas the Catalan numbers are "type A." What exactly does this mean? What would happen if I repeated the above computations for other Dynkin diagrams? Do I get continued fractions for the other infinite families?
**Question 3:** Related to the above, in what sense is it natural to relate walks on the Dynkin diagram $A\_n$ to representations of $SU(2)$? This seems to have something to do with [question #16026](https://mathoverflow.net/questions/16026/the-finite-subgroups-of-sl2-c). How do the eigenvectors of the adjacency matrices fit into the picture? I want to think of the eigenvectors as "discrete harmonics"; does this point of view make sense? Does it generalize?
As you can see, I'm very confused, so I would greatly appreciate any clarification.
| https://mathoverflow.net/users/290 | How does this relationship between the Catalan numbers and SU(2) generalize? | The Catalan numbers enumerate (amongst everything else!) the bases of the Temperley-Lieb algebras. These algebras $TL\_n(q)$ are exactly $\operatorname{End}\_{U\_q \mathfrak{su}\_2}(V^{\otimes n})$ where $V$ is the standard representation.
If $q$ is a $2k+4$-th root of unity, the semisimplified representation theory of $U\_q \mathfrak{su}\_2$ becomes '$A\_{k+1}$': that is, it has $k+1$ simples, and the principal graph for tensor product with the standart is $A\_{k+1}$. The algebra $\operatorname{End}\_{U\_q \mathfrak{su}\_2}(V^{\otimes n})$ is now smaller (some of the summands of the tensor power cancel out when you use the quantum Racah rule), and in fact it is enumerated by loops on $A\_{k+1}$ based at the first vertex. You can pick out a subset of the entire Temperley-Lieb that gives a basis: just use your description of paths as trees, and take the dual graph.
Since we're looking at a unitary tensor category, the dimension of the standard object is exactly the Frobenius-Perron eigenvalue (largest real eigenvalue of the adjacency matrix) of the principal graph.
| 28 | https://mathoverflow.net/users/3 | 17200 | 11,505 |
https://mathoverflow.net/questions/17208 | 2 | I believe that what I'm about to describe has a name—I'm almost certain that I've seen this in model theory and term-rewriting systems, possibly having something to do with ‘signature’ or ‘carrier’?—so please feel free to tell me if I'm using terms in bad, or non-standard, ways.
Consider a set $C$ of real constants. (There's no reason that we couldn't consider constants from an arbitrary field, or ring, or … whatever; but let's say ‘real’ for definiteness.) Fix also, for each $n$, a (possibly empty) set $S\_n$ of (partially defined) $n$-place, real-valued functions on $\mathbb R$.
Write $\Sigma = (C, (S\_n)\_n)$, and define the set of *$\Sigma$-terms* in the following (obvious) way. The collection of $\Sigma$-terms is the smallest subset $\mathcal T$ of the set of strings over the alphabet consisting of $C \sqcup \bigsqcup\_n S\_n$, together with 3 distinguished symbols `(`, `)`, and `,`, satisfying:
1. Each element of $C$ is in $\mathcal T$.
2. If $\sigma \in S\_n$, and $w\_1, \ldots, w\_n$ are $\Sigma$-terms, then $\sigma(w\_1, \ldots, w\_n) \in \mathcal T$.
Notice that there is an obvious (partially defined) map from $\Sigma$-terms to $\mathbb R$. Notice also that I am avoiding all questions of representability of elements of $C$; if this is worrisome, we can assume that $C$ is countable, fix an enumeration, and replace all elements of $C$ by their labels.
Now call $\Sigma$ *decidable* if the problem of whether a given $\Sigma$-term represents $0$ is decidable. (To avoid trivialities like $C = \mathbb R\_{\ge0}$, with all operations preserving positivity, let's assume that subtraction lies in $S\_2$.) Are there any results about what ensembles $\Sigma$ are decidable?
For example, it's obvious that, if $C = \mathbb Q$ and the operations allowed are the field operations, then $\Sigma$ is decidable; but what if we enlarge our operations to allow extraction of positive square roots, or enlarge $C$ to the algebraic closure of $\mathbb Q$?
| https://mathoverflow.net/users/2383 | Decidable real arithmetic | There are at least two approaches to what you describe. In symbolic computation people ask questions of the sort you are asking, except that more generally one wants to know how to compute normal or canonical forms of terms, not just decide equality.
Another line of attack comes from computability theory. There is computability and computational complexity over an arbitrary structure (a set with operations and relations). This goes under the names "recursive model theory" and "effective model theory". They study not only decidability of equality, but other general questions about computability and computational complexity of model-theoretic concepts.
I am not really an expert on this so I am hoping someone else can provide the best references, but a good place to start might be the Handbook of Recursive Mathematics (volume 1, Recursive model theory). I do not know what to suggest for symbolic computation.
Also have a look at the Blum-Shub-Smale model of real number computation, see [Complexity and real computation](http://books.google.com/books?id=zxtrVqUP-AwC). It is not what you are asking about, because they *assume* equality as a decidable operation, but it is similar to what you are suggesting in other respects. It is a popular model of real number computation in some circles (computational geometry), and harshly criticized in others (computable analysis).
| 3 | https://mathoverflow.net/users/1176 | 17214 | 11,510 |
https://mathoverflow.net/questions/17195 | 2 | Let $A\_k$ be a sequence of real, rank $r$, $n$ x $m$ matrices such that $A\_k$ converges to a rank $r$ matrix $A$. Let $v\_k, u\_k$ be sequences of vectors such that $u\_k\rightarrow u$ and $A\_k v\_k=u\_k$. I would like to know if it is possible to show that there exist a vector $v$ such that $Av=u$. Appreciate any help.
| https://mathoverflow.net/users/1172 | Sequence of constant rank matrices | I think it is best to settle this problem geometrically, that is if you think of matrices as linear maps from $\mathbb R^m$ to $\mathbb R^n$. The images of these maps are $r$-dimensional linear subspaces of $\mathbb R^n$. Let $X\_k$ denote the image of $A\_k$, then $u\_k\in X\_k$, and you want to prove that the limit vector $u$ belongs to the image of $A$.
Let $X$ denote the set of all possible limits of sequences such that the $k$th element of the sequence belongs to $X\_k$ for every $k$ (for example, $\{u\_k\}$ is one such sequence, hence $u\in X$). Clearly $X$ is a linear subspace of $\mathbb R^n$, and it contains the image of $A$. And the dimension of $X$ is no greater than $r$. Indeed, suppose that $X$ contains $r+1$ linearly independent vectors $w\_1,\dots,w\_{r+1}$. Each $w\_i$ is a limit of a sequence of vectors $u\_k^{(i)}\in X\_k$. For a sufficiently large $k$, the vectors $u\_k^{(1)},\dots,u\_k^{(r+1)}$ are linearly independent because the set of linearly independent $(r+1)$-tuples is open. This contradicts the fact that $\dim X\_k=r$.
Since $X$ is a linear subspace of dimension at most $r$ and it contains the ($r$-dimensional) image of $A$, it must coincide with that image. Since $u\in X$ by definition, it follows that $u$ belongs to the image of $A$, q.e.d.
| 5 | https://mathoverflow.net/users/4354 | 17220 | 11,515 |
https://mathoverflow.net/questions/17221 | 2 | If I have a group $G$ with two two subgroups $H$ and $I$ such that $G/H \approx G/I$, what can I say about the relationship between $H$ and $I$? Are they equal; isomorphic?
Sorry if this question is below the level of the site.
| https://mathoverflow.net/users/4409 | Can a Quotient of a Group by Two Different Subgroups be Isomorphic? | There exists groups $G$ with normal proper non-trivial subgroups $H$ such that $G \cong G/H$. Even finitely presented ones, as the one given by [Higman, Graham. A finitely related group with an isomorphic proper factor group. J. London Math. Soc. 26, (1951). 59--61. [MR0038347](http://www.ams.org/mathscinet-getitem?mr=MR0038347)], namely the group $G$ freely generated by $a$, $b$ and $c$ subject to $$a^{-1}ca=b^{-1}cb=c^2,$$ with $H$ normally generated by adding the relation $$aca^{-1}=bcb^{-1}.$$
| 19 | https://mathoverflow.net/users/1409 | 17222 | 11,516 |
https://mathoverflow.net/questions/17223 | 4 | According to "The Geometry of Four-Manifolds" by Donaldson and Kronheimer, indefinite unimodular forms are classified by their rank, signature and type. This is the Hasse-Minkowski classification of indefinite forms, they say.
However, this seems to be a bit of a folklore theorem, as I cannot find a single citation for it; all of my searches for any permutation of "hasse minkowski indefinite quadratic classification" yield instead a different theorem, namely one about solving the equation $Q(x) = r$ over $\mathbb{Q}$ for a given quadratic form $Q$.
Is it simply the case that the integral classification (which essentially states, for the case of an even form, that it is a sum of $E\_8$ lattices and Hyperbolics) is an easy consequence of this other theorem? I'm not familiar enough with quadratic forms to see how this should be so.
If it isn't an easy consequence, is there a reference for the integral classification that I just haven't found?
| https://mathoverflow.net/users/1703 | What is a reference for the Hasse-Minkowski classification of indefinite forms? | For a survey of the topic check out Milnor and Husemoller, Symmetric Bilinear Forms, Springer, 1973, II.3. They cite Borevich-Shafarevich (Number Theory), O'Meara (Introduction to Quadratic Forms), and Serre (Cours d'Arithmétique).
| 5 | https://mathoverflow.net/users/1822 | 17224 | 11,517 |
https://mathoverflow.net/questions/17212 | 8 | Is there a description of finite groups whose all quotients have trivial center? Is it true that only direct products of non-abelian simple groups have this property?
| https://mathoverflow.net/users/4408 | Finite groups with centerless quotients | The answer to your second question is negative. Take a finite simple group $G$ and its nontrivial irreducible representation $V$ over finite field. Then the semi-direct product of $G$ and $V$ has a unique nontrivial quotient, the group $G$ itself.
| 10 | https://mathoverflow.net/users/4158 | 17225 | 11,518 |
https://mathoverflow.net/questions/17209 | 154 | I assume a number of results have been proven conditionally on the Riemann hypothesis, of course in number theory and maybe in other fields. What are the most relevant you know?
It would also be nice to include consequences of the generalized Riemann hypothesis (but specify which one is assumed).
| https://mathoverflow.net/users/828 | Consequences of the Riemann hypothesis | I gave a talk on this topic a few months ago, so I assembled a list then which could be appreciated by a general mathematical audience. I'll reproduce it here. (Edit: I have added a few more examples to the end of the list, starting at item m, which are meaningful to number theorists but not necessarily to a general audience.)
Let's start with three applications of RH for the Riemann zeta-function only.
a) Sharp estimates on the remainder term in the prime number theorem: $\pi(x) = {\text{Li}}(x) + O(\sqrt{x}\log x)$, where ${\text{Li}}(x)$ is the logarithmic integral (the integral from 2 to $x$ of $1/\log t$).
b) Comparing $\pi(x)$ and ${\text{Li}}(x)$. All the numerical data shows $\pi(x)$ < ${\text{Li}}(x)$, and Gauss thought this was always true, but in 1914 Littlewood used the Riemann hypothesis to show the inequality reverses infinitely often. In 1933, Skewes used RH to show the inequality reverses for some
$x$ below 10^10^10^34. In 1955 Skewes showed without using RH that the inequality reverses for some $x$ below 10^10^10^963. Maybe this was the first example where something was proved first assuming RH and later proved without RH.
c) Gaps between primes. In 1919, Cramer showed RH implies $p\_{k+1} - p\_k = O(\sqrt{p\_k}\log p\_k)$, where $p\_k$ is the $k$th prime. (A conjecture of Legendre is that there's always a prime between $n^2$ and $(n+1)^2$ -- in fact there should be *a lot* of them -- and this would imply $p\_{k+1} - p\_k = O(\sqrt{p\_k})$. This is better than Cramer's result, so it lies deeper than a consequence of RH. Cramer also conjectured that the gap is really $O((\log p\_k)^2)$.)
Now let's move on to applications involving more zeta and $L$-functions than just the Riemann zeta-function. Note that typically we will need to assume GRH for infinitely many such functions to say anything.
d) Chebyshev's conjecture. In 1853, Chebyshev tabulated the primes
which are $1 \bmod 4$ and $3 \bmod 4$ and noticed there are always at least as many $3 \bmod 4$ primes up to $x$ as $1 \bmod 4$ primes. He conjectured this was always true and also gave an analytic sense in which there are more $3 \bmod 4$ primes:
$$
\lim\_{x \rightarrow 1^{-}} \sum\_{p \not= 2} (-1)^{(p+1)/2}x^p = \infty.
$$
Here the sum runs over odd primes $p$. In 1917, Hardy-Littlewood and Landau (independently) showed this second conjecture of Chebyshev's is equivalent to
GRH for the $L$-function of the nontrivial character mod 4. (In 1994, Rubinstein and Sarnak used simplicity and linear independence hypotheses on zeros of $L$-functions to say something about Chebyshev's first conjecture, but as the posted question asked only about consequences of RH and GRH, I leave the matter there and move on.)
e) The Goldbach conjecture (1742). The "even" version says all even integers $n \geq 4$ are a sum of 2 primes, while the "odd" version says all odd integers $n \geq 7$ are a sum of 3 primes. For most mathematicians, the Goldbach conjecture is understood to mean the even version, and obviously the even version implies the odd version. The odd version turns out to be a consequence of GRH. In 1923, assuming all Dirichlet $L$-functions are nonzero in a right half-plane ${\text{Re}}(s) \geq 3/4 - \varepsilon$, where $\varepsilon$ is fixed (independent of the $L$-function), Hardy and Littlewood showed the odd Goldbach conjecture is true for all sufficiently large odd $n$. In 1937, Vinogradov proved the same result without needing GRH as a hypothesis. In 1997, Deshouillers, Effinger, te Riele, and Zinoviev showed GRH implies the odd Goldbach conjecture is true for all odd $n \geq 7$. So GRH completely settles the odd Goldbach conjecture.
Update: This is now an obsolete application of GRH since the odd Goldbach Conjecture was proved by Harald Helfgott in 2013 without needing GRH as a hypothesis.
An account of the current status of Helfgott's work is [here](https://math.stackexchange.com/questions/3164020/the-significance-and-acceptance-of-helfgott-s-proof-of-the-weak-goldbach-conject).
f) Polynomial-time primality tests. By results of Ankeny (1952) and Montgomery (1971), GRH for *all* Dirichlet $L$-functions implies that the first nonmember of every proper subgroup of the unit group $({\mathbf Z}/m{\mathbf Z})^\times$ is $O((\log m)^2)$, where the $O$-constant is independent of $m$. In 1985, Bach showed GRH for all Dirichlet $L$-functions implies the constant in that $O$-estimate can be taken to be 2. That is, GRH for all Dirichlet $L$-functions implies that each proper subgroup of $({\mathbf Z}/m{\mathbf Z})^\times$ is missing some integer from 1 to $2(\log m)^2$. Put differently, GRH for all Dirichlet $L$-functions implies that if a subgroup of $({\mathbf Z}/m{\mathbf Z})^\times$ contains all positive integers below $2(\log m)^2$ then the subgroup is the whole unit group mod $m$. (To understand one way that GRH has an influence on that upper bound, if all the nontrivial zeros of all Dirichlet $L$-functions have ${\text{Re}}(s) \leq 1 - \varepsilon$ then the first nonmember of every proper subgroup of $({\mathbf Z}/m{\mathbf Z})^\times$ is $O((\log m)^{1/\varepsilon})$. Set $\varepsilon = 1/2$ to get the previous result I stated that uses GRH.) In 1976, Gary Miller introduced a deterministic primality test that he could prove runs in polynomial time using GRH for all Dirichlet $L$-functions. (Part of the test involves deciding if a subgroup of units mod $m$ is proper or not.) Shortly afterwards, Solovay and Strassen described a different primality test using Jacobi symbols. GRH for Dirichlet $L$-functions implies their test runs in polyomial time, and the subgroups of units mod $m$ occurring for their test contain $-1$, so the proof that their test runs in polynomial time "only" needs GRH for Dirichlet $L$-functions of even characters. (Solovay and Strassen described their test as a probabilistic test rather than a deterministic test, so they didn't mention GRH one way or the other.)
In 2002 Agrawal, Kayal, and Saxena created a new primality test that they could prove runs in polynomial time without needing GRH. This is a nice example showing how GRH guides mathematicians in the direction of what should be true and then people tried to find a proof of those results by methods not involving GRH.
g) Euclidean rings of integers. In 1973, Weinberger showed that GRH for all Dedekind zeta-functions implies that every ring of algebraic integers with an infinite unit group (so ignoring $\mathbf Z$ and the ring of integers of imaginary quadratic fields) is Euclidean if it has class number 1. As a special case, in concrete terms, if $d$ is a positive integer that is not a perfect square then a ring ${\mathbf Z}[\sqrt{d}]$ that is a unique factorization domain must be Euclidean. The same theorem is true for rings of $S$-integers: an infinite unit group plus class number $1$ plus GRH for zeta-functions of all number fields implies the ring is Euclidean. Progress has been made by Ram Murty and others in the direction of proving class number 1 and infinite unit group implies Euclidean for rings of $S$-integers without needing GRH, and as a striking special case let's consider ${\mathbf Z}[\sqrt{14}]$. It has class number 1 (which must have been known to Gauss in the early 19th century, in the language of quadratic forms) and an infinite unit group, so it should be Euclidean. This particular real quadratic ring was first proved to be Euclidean only in 2004 (by M. Harper). So $\mathbf Z[\sqrt{14}]$ is a ring that was known to have unique factorization for over 100 years before it was proved to be Euclidean.
h) Artin's primitive root conjecture. In 1927, Artin conjectured that each nonzero integer $a$ that is not $-1$ or a perfect square is a generator of $({\mathbf Z}/p{\mathbf Z})^\times$ for infinitely many primes $p$, and in fact for a positive proportion of such $p$.
As a special case, taking $a = 10$, this says for primes $p$ the unit fraction $1/p$ has decimal period $p-1$ for a positive proportion of $p$. (For each prime $p$ other than $2$ and $5$, the decimal period for $1/p$ is a factor of $p-1$, so this special case is saying the largest possible period is realized infinitely often in a precise sense.) In 1967, Hooley showed GRH for zeta-functions of number fields implies Artin's primitive root conjecture. More precisely, Artin's primitive root conjecture for $a$ follows from GRH for the zeta-functions of all the number fields $\mathbf Q(\sqrt[n]{a},\zeta\_n)$ where $n$ runs over the squarefree positive integers. In 1984, R. Murty and Gupta showed Artin's primitive root conjecture is true for infinitely many $a$ without having to use GRH, but their proof couldn't pin down even one specific $a$ for which Artin's primitive root conjecture is true. In 1986, Heath-Brown showed without using GRH that Artin's primitive root conjecture is true for all prime values of $a$ with at most two exceptions (and of course there should not be any exceptions). Without using GRH, no definite $a$ is known for which Artin's conjecture is true.
i) First prime in an arithmetic progression. If $\gcd(a,m) = 1$ then there are infinitely many primes $p \equiv a \bmod m$. When does the first one appear, as a function of $m$? In 1934, Chowla showed GRH implies the first prime $p \equiv a \bmod m$ is $O(m^2(\log m)^2)$. In 1944, Linnik showed without GRH that the bound is $O(m^L)$ for some universal exponent $L$. The latest choice for $L$ (Xylouris, 2009) without using GRH is $L = 5.2$.
j) Gauss' class number problem. Gauss (1801) conjectured in the language of quadratic forms that there are only finitely many imaginary quadratic fields with class number 1. (He actually conjectured more precisely that the 9 known examples are the only ones, but for what I want to say the weaker finiteness statement is simpler.) In 1913, Gronwall showed this is true if the $L$-functions of all imaginary quadratic Dirichlet characters have no zeros in some common strip $1- \varepsilon < {\text{Re}}(s) < 1$. That is weaker than GRH (we only care about $L$-functions of a restricted collection of characters), but it is a condition like GRH for infinitely many $L$-functions. In 1933, Deuring and Mordell showed Gauss' conjecture is true if the ordinary RH (for the Riemann zeta-function) is *false*, and then in 1934 Heilbronn showed Gauss' conjecture is true if GRH is *false* for some Dirichlet $L$-function of an imaginary quadratic character. Since Gronwall proved Gauss' conjecture is true when GRH is true for the Riemann zeta-function and the Dirichlet $L$-functions of all imaginary quadratic Dirichlet characters and Deuring--Mordell--Heilbronn proved Gauss' conjecture is true when GRH is false for at least one of those functions, Gauss' conjecture is true by baby logic. In 1935, Siegel proved Gauss' conjecture is true without using GRH, and in the 1950s and 1960s Baker, Heegner, and Stark gave separate proofs of Gauss' precise "only 9" conjecture without using GRH.
k) Missing values of a quadratic form. Lagrange (1772) showed every positive integer is a sum of four squares. However, not every integer is a sum of three squares: $x^2 + y^2 + z^2$ misses all $n \equiv 7 \bmod 8$. Legendre (1798) showed a positive integer is a sum of three squares iff it is *not* of the form $4^a(8k+7)$. This can be phrased as a local-global problem: $x^2 + y^2 + z^2 = n$ is solvable in integers iff the congruence $x^2 + y^2 + z^2 \equiv n \bmod m$ is solvable for all $m$. More generally, the same local-global
phenomenon applies to the three-variable quadratic form $x^2 + y^2 + cz^2$ for all integers $c$ from 2 to 10 *except* $c = 7$ and $c = 10$. What happens for these two special values? Ramanujan looked at $c = 10$. He found 16 values of $n$ for which there is local solvability (that is, we can solve $x^2 + y^2 + 10z^2 \equiv n \bmod m$ for all $m$) but not global solvability (no integral solution for $x^2 + y^2 + 10z^2 = n$). Two additional values of $n$ were found later, and in 1990 Duke and Schulze-Pillot showed that local solvability implies global solvability except for (ineffectively) finitely many positive integers $n$. In 1997, Ono and Soundararajan showed GRH implies the 18 known exceptions are the only ones.
l) Euler's convenient numbers. Euler called an integer $n \geq 1$ *convenient* if each odd integer greater than 1 that has a unique representation as $x^2 + ny^2$ in positive integers $x$ and $y$, and which moreover has $(x,ny) = 1$, is a prime number. (These numbers were convenient for Euler to use to prove certain numbers that were large in his day, like $67579 = 229^2 + 2\cdot 87^2$, are prime.) Euler found 65 convenient numbers below 10000 (the last one being 1848). In 1934, Chowla showed there are finitely many convenient numbers. In 1973, Weinberger showed there is at most one convenient number not in Euler's list, and that GRH for $L$-functions of all quadratic Dirichlet characters implies that Euler's list of convenient numbers is complete. What he needed from GRH is the lack of real zeros in the interval $(53/54,1)$.
m) Removing a condition in the Brauer-Siegel theorem. In 1947, Brauer proved the Brauer-Siegel theorem for sequences of number fields $K\_n$ such that (i) $[K\_n:\mathbf Q]/\log |{\rm disc}(K\_n)| \to 0$ as $|{\rm disc}(K\_n)| \to \infty$ and (ii) $K\_n$ is Galois over $\mathbf Q$. If the zeta-functions $\zeta\_{K\_n}(s)$ all satisfy GRH (what is actually needed is no real zero in $(1/2,1)$ for those zeta-functions) then we can drop condition (ii). That is, GRH implies the Brauer-Siegel theorem holds for sequences of number fields $K\_n$ fitting condition (i).
n) Lower bounds on root discriminants. In 1975, Odlyzko showed GRH for zeta-functions of number fields implies a lower bound on root discriminants of number fields:
$$
|{\rm disc}(K)|^{1/n} \geq (94.69...)^{r\_1/n}(28.76...)^{2r\_2/n} + o(1)
$$
as $n = [K:\mathbf Q] \to \infty$. (The lower bound was improved later to $136^{r\_1/n}34.5^{2r\_2/n}$ for sufficiently large $n$.) Building on ideas of Stark, he also showed that GRH for zeta-functions of number fields implies there are only finitely many CM number fields with a given class number (a big generalization of the known fact that only finitely many imaginary quadratic fields have any particular class number).
o) Lower bounds on class numbers. In 1990, Louboutin showed GRH for zeta-functions of imaginary quadratic fields (what is really needed is the lack of real zeros in $(1/2,1)$ for these zeta-functions) implies the lower bound $h(\mathbf Q(\sqrt{-d})) \geq (\pi/(3e))\sqrt{d}/\log d$ for imaginary quadratic fields with discirminant $-d$. The point here is the explicit constant factor $\pi/(3e)$. (Hecke had shown such a lower bound with a "computable constant" $c$ in place of $\pi/(3e)$ but he did not compute the constant.) Without GRH, lower bounds for $h(\mathbf Q(\sqrt{-d}))$ are on the order of $\log d$, which is far smaller than $\sqrt{d}/\log d$. For example, Louboutin's GRH-based lower bound shows if $h(\mathbf Q(\sqrt{-d})) \leq 100$ then $d \leq $ 18,916,898. To put this 8-digit upper bound in perspective,
when Watkins determined all imaginary quadratic fields with class number up to 100 in 2004, he used lower bounds on $h(\mathbf Q(\sqrt{-d}))$ that do not depend on GRH and the upper bound of his search space for $d$ was $e^{298368000}$, a number with 129,579,576 digits. Just the exponent in that upper bound on $d$ is greater than the upper bound on $d$ coming from GRH.
p) Proof of André-Oort conjecture. In 2014, Klingler and Yafaev showed GRH for zeta-functions of CM number fields implies the André-Oort conjecture. Daw and Orr gave another proof also using the same version of GRH.
Update: This is now an obsolete application of GRH since the André-Oort conjecture has been proved without GRH. See [here](https://arxiv.org/abs/2109.08788).
q) Class number calculations. In 2015, J. C. Miller showed that GRH implies the class number $h\_p^+$ of the real cyclotomic field $\mathbf Q(\zeta\_p)^{+}$ for prime $p$ is $1$ for all $p$ from 157 to 241 except $h\_{163}^+ = 4$, $h\_{191}^{+} = 11$, and $h\_{229}^+ = 3$.
r) Elliptic curve rank values. In 2019, Klagsbrun, Sherman, and Weigandt used GRH for $L$-functions of elliptic curves and for zeta-functions of number fields to prove that the elliptic curve found by Elkies in 2006 with 28 independent rational points has rank equal to 28.
| 255 | https://mathoverflow.net/users/3272 | 17232 | 11,521 |
https://mathoverflow.net/questions/17202 | 114 | I am interested in the function $$f(N,k)=\sum\_{i=0}^{k} {N \choose i}$$ for fixed $N$ and $0 \leq k \leq N $. Obviously it equals 1 for $k = 0$ and $2^{N}$ for $k = N$, but are there any other notable properties? Any literature references?
In particular, does it have a closed form or notable algorithm for computing it efficiently?
In case you are curious, this function comes up in information theory as the number of bit-strings of length $N$ with Hamming weight less than or equal to $k$.
Edit: I've come across a useful upper bound: $(N+1)^{\underline{k}}$ where the underlined $k$ denotes falling factorial. Combinatorially, this means listing the bits of $N$ which are set (in an arbitrary order) and tacking on a 'done' symbol at the end. Any better bounds?
| https://mathoverflow.net/users/4405 | Sum of 'the first k' binomial coefficients for fixed $N$ | I'm going to give two families of bounds,
one for when $k = N/2 + \alpha \sqrt{N}$ and one for when $k$ is fixed.
The sequence of binomial coefficients ${N \choose 0}, {N \choose 1}, \ldots, {N \choose N}$ is symmetric. So you have
$\sum\_{i=0}^{(N-1)/2} {N \choose i} = {2^N \over 2} = 2^{N-1}$
when $N$ is odd.
(When $N$ is even something similar is true
but you have to correct for whether you include the term ${N \choose N/2}$ or not.
Also, let $f(N,k) = \sum\_{i=0}^k {N \choose i}$.
Then you'll have, for real constant $\alpha$,
$ \lim\_{N \to \infty} {f(N,\lfloor N/2+\alpha \sqrt{N} \rfloor) \over 2^N} = g(\alpha) $
for some function $g$. This is essentially a rewriting of a special case of the central limit theorem. The Hamming weight of a word chosen uniformly at random is a sum of Bernoulli(1/2) random variables.
For fixed $k$ and $N \to \infty$, note that
$$ {{N \choose k} + {N \choose k-1} + {N \choose k-2}+\dots \over {N \choose k}}
= {1 + {k \over N-k+1} + {k(k-1) \over (N-k+1)(N-k+2)} + \cdots} $$
and we can bound the right side from above by the geometric series
$$ {1 + {k \over N-k+1} + \left( {k \over N-k+1} \right)^2 + \cdots} $$
which equals ${N-(k-1) \over N - (2k-1)}$. Therefore we have
$$ f(N,k) \le {N \choose k} {N-(k-1) \over N-(2k-1)}.$$
| 78 | https://mathoverflow.net/users/143 | 17236 | 11,524 |
https://mathoverflow.net/questions/17230 | 10 | Let $\rho : S\_n \rightarrow \text{GL}(n, \mathbb{C})$ be the homomorphism mapping a permutation $g$ to its permutation matrix. Let $\chi(g) = \text{Trace}(\rho(g))$.
What is the value of $\langle \chi, \chi \rangle = \displaystyle \frac{1}{n!} \sum\_{g \in S\_n} \chi(g)^2$ ? Computing this expression for small $n$ yields $2$. Is this always true?
| https://mathoverflow.net/users/4197 | Permutation representation inner product | You are computing the inner product of $\chi$ with itself.
Since $\chi=\mathrm{triv}+\mathrm{std}$ as a $S\_n$-module, with $\mathrm{triv}$ being the trivial reprsentation, and $\mathrm{std}$ its orthogonal complement, which is an irreducible $S\_n$-module, and since the inner product is, well, an inner product and distinct irreducible characters are orthogonal, your $2$ follows from $$\langle\chi,\chi\rangle=\langle\mathrm{triv}+\mathrm{std},\mathrm{triv}+\mathrm{std}\rangle=\langle\mathrm{triv},\mathrm{triv}\rangle+\langle\mathrm{std},\mathrm{std}\rangle=1+1$$.
| 23 | https://mathoverflow.net/users/1409 | 17237 | 11,525 |
https://mathoverflow.net/questions/17231 | 2 | D is a central division algebra over F. We know that we can always find a maximal subfield K inside D such that K/F is separable. I want to know can we always make it Galois?
| https://mathoverflow.net/users/2008 | Maximal subfield inside a central division algebra | I believe you are just asking if every central division algebra over F is a crossed product. This is not true, and the first example was given in:
Amitsur, S. A. "On central division algebras."
Israel J. Math. 12 (1972), 408-420.
[MR 318216](http://www.ams.org/mathscinet-getitem?mr=318216)
[DOI: 10.1007/BF02764632](http://dx.doi.org/10.1007/BF02764632)
A survey of Amitsur's contributions on division algebras which mentions this point in particular is the introduction by Saltman starting on page 109 of:
Amitsur, S. A. *Selected papers of S. A. Amitsur with commentary. Part 2.*
Edited by Avinoam Mann, Amitai Regev, Louis Rowen, David J. Saltman and Lance W. Small.
American Mathematical Society, Providence, RI, 2001. xx+615 pp. ISBN: 0-8218-2925-4
[MR 1866637](http://www.ams.org/mathscinet-getitem?mr=1866637)
[Google](http://books.google.com/books?id=p2uvFqAEhLIC&pg=PA109)
| 6 | https://mathoverflow.net/users/3710 | 17240 | 11,527 |
https://mathoverflow.net/questions/17250 | 20 | Let $ \lambda\_1 \ge \lambda\_2 \ge \dots \ge \lambda\_{2n} $
be the collection of eigenvalues of an adjacency matrix of an undirected graph $G$ on $2n$ vertices. I am looking for any work or references that would consider the middle eigenvalues $\lambda\_n$ and $\lambda\_{n+1}$. In particular, the bounds on $R(G) = \max \left( |\lambda\_n|, |\lambda\_{n+1}| \right)$ are welcome.
For instance, a computer search showed that most connected graphs with maximum valence $3$ have $R(G) \le 1$. The only known exception is the Heawood graph.
The motivation for this question comes from theoretical chemistry, where the difference $\lambda\_n - \lambda\_{n+1}$ in Hueckel theory is called the [HOMO-LUMO gap](https://en.wikipedia.org/wiki/HOMO_and_LUMO).
**Edit:** Note that in the original formula for $R(G)$ the absolute value signs were missing.
| https://mathoverflow.net/users/4400 | The middle eigenvalues of an undirected graph | One thing that you might find useful is the [Cauchy interlacing theorem](https://en.wikipedia.org/wiki/Min-max_theorem#Cauchy_interlacing_theorem).
In response to the comment, presumably Tomaz's interest is in some particular sorts of graphs. It may be the case that such a graph has some well-understood subgraphs. Then you could reduce the question about the size of the middle eigenvalues of the big graph to a question about the size of the "approximately middle" eigenvalues of the smaller graph.
| 7 | https://mathoverflow.net/users/22 | 17254 | 11,536 |
https://mathoverflow.net/questions/17246 | 5 | Suppose a finite group $G$ acts on a standard probability space $(X, \mu)$ by measure-preserving actions (i.e. $\mu(g(A)) = \mu(A)$ for all $g \in G$ and $A \subset X$ measurable). In addition, suppose that for $g \in G$ and $g$ not the identity then $\mu(\{x:g(x) = x\})=0$. I am wondering if it is always possible to find a fundamental domain $D$ of the action of $G$, i.e. is there a measurable $D \subset X$ such that $G$ is the disjoint union (up to measure 0) of the sets $g(D)$?
| https://mathoverflow.net/users/792 | Fundamental domains of measure preserving actions | I believe the answer is yes. You can first show, that once you have a set $A$ of non-zero measure, such that $g(A) = A$ for any $g \in G$, there is some $B \subset A$ of non-zero measure, such that $g(B) \cap h(B) = \emptyset$ for $g \ne h$. If this is shown, then you can prove your statement by Zorn-like argument, showing that the set $A$ of maximal measure, such that $g(A) \cap h(A) = \emptyset$ will be the fundamental domain.
Here is a hint how to show the first claim. Take $g \in G$, $g \ne 1$, and a $G$-invariant set $A$. Since $g(x) \ne x$, there is some $B\_0 \subset A$ of non-zero measure such that $g(B\_0) \ne B\_0$, set $B\_1 = B\_0 \setminus (g(B\_0))$. Then $g(B\_1) \cap B\_1 = \emptyset$. Now do the same thing for other $h \in G$ starting from $B\_1$.
Alternatively, once can pass to a Stone space and use continuity to get this statement, though this requires more machinery.
| 4 | https://mathoverflow.net/users/896 | 17259 | 11,538 |
https://mathoverflow.net/questions/17233 | 3 | Let $G$ be a reductive (or just semisimple) algebraic group over an algebraically closed field $k$ of characteristic $p > 0$, let $T$ be a maximal Torus and let $f:G \rightarrow G$ be an isogeny. Suppose the induced morphism of root data $f^\star: \Psi(G,T) \rightarrow \Psi(G,T)$ is a Frobenius morphism multiplying roots by $q = p^r$ (confer 9.6.3 in Springer's book on algebraic groups). Is it true that $f$ is the geometric Frobenius of an $\mathbb{F}\_{q}$-structure on $G$?
Equivalent formulation: Is the morphism $F:G \rightarrow G$ defined by $F( u\_\alpha(c) ) = u\_\alpha(c^q)$ for all roots $\alpha$ and $F(t) = t^q$ for all $t \in T$ the geometric Frobenius of an $\mathbb{F}\_{q}$-structure on $G$?
I'm pretty sure this is true (perhaps under some additional conditions). I know that there is an easy characterization of Frobenius morphisms in terms of the comorphism of the coordinate rings but I was not able to verify this.
| https://mathoverflow.net/users/717 | If the morphism of root data induced by an isogeny of a reductive group is a Frobenius, is then the isogeny itself a Frobenius? | The answer is "yes", but not in a good way: the descent it arises from is the split form, so this does not encode an interesting $\mathbf{F}\_ q$-structure. More precisely, $f$ arises from the $q$-Frobenius of a split descent down to $\mathbf{F}\_ q$ (and so even to the prime field $\mathbf{F}\_ p$). In particular, this is definitely not the way to encode the information of "interesting" descents of $(G,T)$ to $\mathbf{F}\_ q$, if that may be your ultimate intention.
To see this, let $(G\_0, T\_0)$ be the split form over $\mathbf{F}\_ q$, and $F\_0:(G\_0, T\_0) \rightarrow (G\_0,T\_0)$ the $q$-Frobenius morphism. This induces the self-map of the root datum given by $q$-multiplication on ${\rm{X}}(T\_0)$. Extending scalars to $k$ has no effect on the root datum for split groups, so $(F\_ 0)\_ k$ induces the endomorphism of the root datum of $((G\_ 0)\_ k, (T\_ 0)\_ k)$ given by $q$-multiplication on the root datum. Pick a $k$-isomorphism between the $k$-split pairs $(G,T)$ and $((G\_ 0)\_ k, (T\_ 0)\_ k)$ (as we may, by the
isomorphism Theorem). The resulting isomorphism of root data identifies the maps from $(F\_ 0)\_ k$ and $f$ since $q$-multiplication is functorial with respect to all homomorphisms between abelian groups. Thus, these maps coincide up to the action of some $t \in \overline{T}(k) = (\overline{T}\_ 0)(k)$, where
$\overline{T} := T/Z\_G$ and
$\overline{T}\_ 0 := T\_ 0/Z\_ {G\_ 0}$ are the "adjoint tori''. Using an isomorphism
$\overline{T}\_ 0 \simeq \mathbf{G}\_m^r$, $t$ goes over to some $r$-tuple $(t\_i)$ with $t\_i \in k^{\times}$. Since $k$ is separably closed, we can solve $t\_i = y\_i^{q-1}$ with $y\_i \in k^{\times}$. In other words, $t = f(y)/y$ for some $y \in T\_0(k)$. So if we modify the identification of $(G\_0, T\_0)$ as an $\mathbf{F}\_ q$-descent of $(G,T)$ by composing with the action of $y$ or $y^{-1}$ (depending on direction of maps) then $f = (F\_ 0)\_ k$ via the new identification of $(G\_ 0, T\_ 0)$ as a split $\mathbf{F} \_q$-descent of $(G,T)$.
QED
Remark: A refinement of the argument (using some care if $k$ is not algebraic over
$\mathbf{F}\_ p$) proves that the split descent to $\mathbf{F} \_q$ is the only one which can work, but I won't get into that here. (It comes down to the fact that the solutions to $y^{q-1} = 1$ in $k^{\times}$ lie in $\mathbf{F} \_q$.)
Note that the above works over any separably closed field of characteristic $p > 0$, not necessarily algebraically closed, provided the Isomorphism Theorem is valid over such fields. The Existence, Isomorphism, and even Isogeny Theorems are valid for split connected reductive groups over any field. The paper of Steinberg mentioned in the previous answer is really nice, but unfortunately assumes from the outset that the ground field is algebraically closed. Good news, in case you may care, is that one can deduce the results over any field from that case as a "black box'' by using descent theory. See Appendix A.4 of the book "Pseudo-reductive groups''.
| 5 | https://mathoverflow.net/users/3927 | 17263 | 11,542 |
https://mathoverflow.net/questions/17257 | 11 | What's the current state of Yang–Mills mass gap question, is there any place that does this problem? Especially I want to know if there is any progress (out of that mentioned in the introduction article by Witten and Jaffe). Is it too hard for a mathematician? Thanks!
| https://mathoverflow.net/users/2391 | What's the current state of Yang–Mills mass gap question? | There's some guidelines about open problems in the FAQ that you might want to read, but there was a good article by Faddeev last November that you should know about:
[Mass in Quantum Yang-Mills Theory, by Faddeev](http://arxiv.org/abs/0911.1013)
| 9 | https://mathoverflow.net/users/3623 | 17267 | 11,543 |
https://mathoverflow.net/questions/16966 | 3 | Let $\mathfrak{g}$ be a finite dimensional simple Lie algebra over $\mathbb{C}$.
The polynomial current Lie algebra $\mathfrak{g}[t] = \mathfrak{g} \otimes \mathbb{C} [t]$
has the bracket
$$[xt^r, yt^s] = [x,y] t^{r+s}$$
for $x,y \in \mathfrak{g}$. It is graded with deg$(t) = 1$.
If we set $h=0$ in Drinfeld's first presentation of the Yangian (given in Theorem 12.1.1 of Chari and Pressley's [Guide to Quantum Groups](http://books.google.com/books?id=bn5GNkLfnsAC&lpg=PP1&dq=chari%2520pressley%2520guide%2520to%2520quantum%2520groups&pg=PP1#v=onepage&q=&f=false)) then we get a presentation of $U(\mathfrak{g}[t])$
where the generators are the elements $x \in \mathfrak{g}$ and $J(x) = xt$ of $\mathfrak{g}[t]$ with degree $=0,1$, and the relations all have degree of both sides less than $3$.
---
Specifically we require that all the relation in $\mathfrak{g}$ are satisfied for the elements with degree 0, and
(for all $x,y, x\_i, y\_i, z\_i \in \mathfrak{g}$ and complex numbers $\lambda, \mu$):
$$\lambda xt + \mu yt = (\lambda x + \mu y)t$$
$$[x, yt] = [x,y]t,$$
$$\sum\_i [x\_i, y\_i] = 0 \implies \sum\_i [x\_i t, y\_i t ] = 0$$
$$ \sum\_i [[x\_i, y\_i], z\_i] = 0 \implies \sum\_i [[x\_i t, y\_i t], z\_i t]=0$$
Then assuming that all the relations of degree less than or equal to $3$ hold is enough to get the remaining ones.
The elements $xt^2, xt^3, \ldots$ are defined inductively.
This can be proved by induction, using the Serre presentation of the finite-dimensional Lie algebra and then checking all the required relations in several cases.
But even in the $\mathfrak{sl}\_2$ case the argument is laborious.
>
> Is there a better way of seeing that one needs only relations of degree less than three in order to get the rest?
>
>
>
| https://mathoverflow.net/users/3316 | Why are relations of degree 3 or less enough in a presentation of the polynomial current Lie algebra g[t]? | For a nilpotent Lie algebra L, generators could be described in terms of $H\_1(L) = L/[L,L]$, and relations in terms of $H\_2(L)$. While this is not applicable directly to $\mathfrak g \otimes \mathbb C[t]$, it is close enough: it could be decomposed, for example, as the semidirect sum $\mathfrak g \oplus (\mathfrak g \otimes t\mathbb C[t])$, or, better yet, as $(\mathfrak g\_- \oplus \mathfrak h) \oplus (\mathfrak g\_+ \oplus (\mathfrak g\otimes t\mathbb C[t]))$, where $\mathfrak g = \mathfrak g\_- \oplus \mathfrak h \oplus \mathfrak g+$ is the triangular decomposition. From here, I presume, one may glue the defining relations of the whole algebra from the defining relations of the second summand, and Serre defining relations for $\mathfrak g$. The second summand, $\mathfrak g\otimes t\mathbb C[t]$ or close to it, is, formally, still not nilpotent, but it is an $\mathbb N$-graded algebra with finite-dimensional components, and the isomorphism between the space of defining relations and $H\_2(L)$ still applies. So the whole thing boils down to computation of $H\_2(\mathfrak g\otimes t\mathbb C[t])$. The whole cohomology $H\_\*(\mathfrak g\otimes t\mathbb C[t])$ was computed in the celebrated 1976 paper by Garland and Lepowsky, or, one may use a more direct and pedestrian approach and derive it from the known formulae which describe the second (co)homology of the current Lie algebra $L\otimes A$ in terms of symmetric invariant bilinear forms of $L$, cyclic (co)homology of $A$, etc. I am not sure that, written accurately will all the details, this will provide a shorter way, but it is definitely a different one. The case of $sl(2)$ would be exceptional in a sense ($H\_2(sl(2)\otimes t\mathbb C[t])$ is bigger than in the case of generic $\mathfrak g$).
| 6 | https://mathoverflow.net/users/1223 | 17271 | 11,545 |
https://mathoverflow.net/questions/17205 | 13 | I am supervising an undergraduate for a project in which he's going to talk about the relationship between Galois representations and modular forms. We decided we'd figure out a few examples of weight 1 modular forms and Galois representations and see them matching up. But I realised when working through some examples that computing the conductor of the Galois representation was giving me problems sometimes at small primes.
Here's an explicit question. Set $f=x^4 + 2x^2 - 1$ and let $K$ be the splitting field of $f$ over $\mathbf{Q}$. It's Galois over $\mathbf{Q}$ with group $D\_8$. Let $\rho$ be the irreducible 2-dimensional representation of $D\_8$. What is the conductor of $\rho$? Note that I don't particularly want to know the answer to this particular question, I want to know how to work these things out in general. In fact I think I could perhaps figure out the conductor of $\rho$ by doing calculations on the modular forms side, but I don't want to do that (somehow the point of the project is seeing that calculations done in 2 different ways match up, rather than using known modularity results to do the calculations).
Using pari or magma I see that $K$ is unramified outside 2, and the ideal (2) is an 8th power in the integers of $K$. To compute the conductor of $\rho$ the naive approach is to figure out the higher ramification groups at 2 and then just use the usual formula. But the only computer algebra package I know which will compute higher ramification groups is magma, and if I create the splitting field of $f$ over $\mathbf{Q}\_2$ (computed using pari's "polcompositum" command)
```
Qx<x>:=PolynomialRing(Rationals());
g:=x^8 + 20*x^6 + 146*x^4 + 460*x^2 + 1681;
L := LocalField(pAdicField(2, 50),g);
DecompositionGroup(L);
```
then I get an instant memory overflow (magma wants 2.5 gigs to do this, apparently), and furthermore the other calculations I would have to do if I were to be following up this idea would be things like
```
RamificationGroup(L, 3);
```
which apparently need 11 gigs of ram to run. Ouch. Note also that if I pull the precision of the $p$-adic field down from 50 then magma complains that the precision isn't large enough to do some arithmetic in $L$ that it wants to do.
I think then my question must be: are there any computer algebra resources that will compute higher ramification groups for local fields without needing exorbitant amounts of memory? Or is it a genuinely an "11-gigs" calculation that I want to do?? And perhaps another question is: is there another way of computing the conductor of a (non-abelian finite image) Galois representation without having to compute these higher ramification groups (and without computing any modular forms either)?
| https://mathoverflow.net/users/1384 | Computing (on a computer) higher ramification groups and/or conductors of representations. | You can also compute some higher ramification groups in Sage. At the moment it gives lower numbering, not upper numbering, but here it is anyway:
`sage: Qx.<x> = PolynomialRing(QQ)`
`sage: g=x^8 + 20*x^6 + 146*x^4 + 460*x^2 + 1681`
`sage: L.<a> = NumberField(g)`
`sage: G = L.galois_group()`
`sage: G.ramification_breaks(L.primes_above(2)[0])`
`{1, 3, 5}`
You can also get explicit presentations of G as a permutation group and generators for ramification and decomposition subgroups. The above only takes about half a second on my old laptop -- no 2.5 gigs computations here.
(The point is that it is much easier to do computations over a number field, because everything is exact, rather than over a p-adic field which is represented inexactly.)
| 28 | https://mathoverflow.net/users/2481 | 17272 | 11,546 |
https://mathoverflow.net/questions/17226 | 22 | Recently, the Department of Mathematics at
our University issued a recommendation encouraging its members
to publish their research in non-specialized, mainstream mathematical journals. For
numerical analysts this will make an additional obstacle for their promotions. But even
for discrete mathematicians this recommendation is causing concerns.
For several top mainstream journals I checked with tools offered by MathSciNet what percentage
of discrete mathematical papers they published in recent years. Some statistics indicate that in some journals the number of papers with primary MSC classification, say 05 or 06 decreased significantly in the past 30 years. There are several possible explanations to this fact.
* The quality of research in DM is dropping.
* The majority of research in discrete mathematics is so specialized that it is of no interest for the rest of mathematics
* Some discrete mathematics journals attract even the best work of discrete mathematicians.
* Some top journals may be biased against discrete math.
* Maybe discrete math is no longer part of mainstream mathematics and will, like theoretical computer science, eventually develop into an independent body of research.
But the key issue is whether discrete math is nowadays perceived as mainstream mathematics.
| https://mathoverflow.net/users/4400 | Is discrete mathematics mainstream? | There are two different questions here, one objective and one subjective. I will try to give my view, for what it's worth. Bear with me.
First, you are asking what is the publication history of discrete mathematics? (Even if I suspect you know this much better than I do.) Well, originally there was no such thing as DM. If I understand the history correctly, classical papers like [A logical expansion in mathematics](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-38/issue-8/A-logical-expansion-in-mathematics/bams/1183496087.full) by Hassler Whitney (on coefficients of chromatic polynomials) were viewed as contributions to "mainstream mathematics". What happened is that starting maybe late 60s there was a rapid growth in the number of papers in mathematics in general, with an even greater growth in discrete mathematics. While the overall growth is relatively easy to explain as a consequence of expansion of graduate programs, the latter is more complicated. Some would argue that CS and other applications spurned the growth, while others would argue that this area was neglected for generations and had many easy pickings, inherent in the nature of the field. Yet others would argue that the growth is a consequence of pioneer works by the "founding fathers", such as Paul Erdős, Don Knuth, G.-C. Rota, M.-P. Schützenberger, and W.T. Tutte, which transformed the field. Whatever the reason, the "mainstream mathematics" felt a bit under siege by numerous new papers, and quickly closed ranks. The result was a dozen new leading journals covering various subfields of combinatorics, graph theory, etc., and few dozen minor ones. Compare this with the number of journals dedicated solely to algebraic geometry to see the difference. Thus, psychologically, it is very easy to explain why journals like *Inventiones* even now have relatively few DM papers - if the DM papers move in, the "mainstream papers" often have nowhere else to go. Personally, I think this is all for the best, and totally fair.
Now, your second question is whether DM is a "mainstream mathematics", or what is it? This is much more difficult to answer since just about everyone has their own take. E.g. **miwalin** suggests above that number theory is a part of DM, a once prevalent view, but which is probably contrary to the modern consensus in the field. Still, with the growth of "arithmetic combinatorics", part of number theory is definitely a part of DM. While most people would posit that DM is "combinatorics, graph theory + CS and other applications", what exactly are these is more difficult to decide. The split of *Journal of Combinatorial Theory* into Series A and B happened over this kind of disagreement between Rota and Tutte (still legendary). I suggest [combinatorics wikipedia page](https://en.wikipedia.org/wiki/Combinatorics) for a first approximation of the modern consensus, but when it comes to more concrete questions this becomes a contentious issue sometimes of "practical importance". As an editor of *[Discrete Mathematics](https://en.wikipedia.org/wiki/Discrete_Mathematics_(journal))*, I am routinely forced to decide whether submissions are in scope or not. For example if someone submits a generalization of [R–R identities](https://en.wikipedia.org/wiki/Rogers-Ramanujan_identities) — is that a DM or not? (If you think it is, are you sure you can say what exactly is "discrete" about them?) Or, e.g. is Cauchy theorem a part of DM, or metric geometry, or both? (or neither?) How about "IP = PSPACE" theorem? Is that DM, or logic, or perhaps lies completely outside of mathematics? Anyway, my (obvious) point is that there is no real boundary between the fields. There is a large spectrum of papers in DM which fall somewhere in between "mainstream mathematics" and applications. And that's another reason to have separate "specialized" journals to accommodate these papers, rather than encroach onto journals pre-existing these new subfields. Your department's "encouragement" to use only the "mainstream mathematical journals" for promotion purposes is narrow minded and very unfortunate.
| 35 | https://mathoverflow.net/users/4040 | 17273 | 11,547 |
https://mathoverflow.net/questions/17255 | 11 | ### Informal Statement
In the $n\times n \times n$ grid, we can places rooks (those from chess) such that no two rooks can attack each other. One way to achieve this is to place a rook in position $(i,j,k)$ if and only if $i+j+k=0\mod n$. In general, there are "many" ways to do this.
Each such "attack-free" rook position can be colored with $c$ colors. When we fix an $i$, we can then count the colors in the matrix $(i,.,.)$, and can do similarly for each $j$ and $k$. Call this set of tuples of colors-counts the "color profile". For each color profile, there is either an even or odd number of colored rook positions that achieve it. I want to know the largest $c$ such that all color profiles have an even number of colored rook positions achieving it. In particular, I want to say that $c=\omega(n)$. This question came up in some complexity theory research, but the question seems interesting in its own right.
---
### Formal Statement
Define $[n]$ to be the set ${1,\ldots, n}$, and define $[n]^3=[n]\times[n]\times[n]$. Define a *$c$-coloring* of a set $S\subseteq[n]^3$ to be a function $C:S\to[c]$. We can say that this is a $c$-coloring of $[n]^3$ with the convention that $C(i,j,k)=0$ for $(i,j,k)\notin S$. A $c$-coloring $C$ induces a *color profile* $P$, which is a function from $P:[n]\times[3]\times[c]\to[n]$, via the rules
* $P(i,1,c)$ is the number of $(j,k)\in[n]^2$ such that $C(i,j,k)=c$.
* $P(j,2,c)$ is the number of $(i,k)\in[n]^2$ such that $C(i,j,k)=c$.
* $P(k,3,c)$ is the number of $(i,j)\in[n]^2$ such that $C(i,j,k)=c$.
where we keep in mind the convention above on $(i,j,k)\notin S$.
Call a set $S\subseteq [n]^3$ to be a *rook set*, if
* for all $i,j\in[n]$, there is exactly one $k\in[n]$ such that $(i,j,k)\in S$
* for all $j,k\in[n]$, there is exactly one $i\in[n]$ such that $(i,j,k)\in S$
* for all $i,k\in[n]$, there is exactly one $j\in[n]$ such that $(i,j,k)\in S$
Let a *colored rook set* $C\_S$ correspond the coloring $C$ of a rook set $S$.
Define $N(P)$ to be the number of colored rook sets $C\_S$ that induce the color profile $P$.
The question is:
>
> For each fixed $n$, what is the largest $c$ such that for all color profiles $P$, $N(P)\equiv 0\mod 2$? In particular, is the largest $c$ asymptotically $\omega(n)$?
>
>
>
---
### What I know
It should be clear that this problem can be defined analogously in any dimension, and I'm interested in this more general question. I state it with $d\ge3$ because I can solve the $d=2$ case exactly. In particular
>
> For each $n$, for any $c\le n-1$, and any color profile $P$ on grid $[n]^2$, $N(P)\equiv 0 \mod 2$. For $c\ge n$, there are profiles $P$ where $N(P)\equiv 1\mod 2$.
>
>
>
This can be proven by exhibiting a bijection between colored rook sets (which in d=2 are just permutation matrices). Specifically, using the pigeonhole principle $c\le n-1$ implies that there are two rooks with the same color. If they were at positions $(i,j)$ and $(i',j')$, then we replace them with the rooks of the same color at positions $(i,j')$ and $(i',j)$. (Of course, one needs to make this well-defined to ensure a bijection.)
---
### Possible Methods
I see two possible methods of proof
* generalize the above bijection proof to the 3-dimensional case
+ I don't know how to use the pigeonhole principle to get such an extension, but it seems possible that there is a method to show that some motif exists in any colored rook set, and then argue that we can alter this motif to get the bijection
* define a system of polynomials (over $\mathbb{F}\_2$) such that the solution set corresponds to exactly the colored rook sets inducing a color profile $P$. Then try to apply the Chevalley-Warning theorem.
+ I've tried this, but can't seem to get systems of polynomials where the sum of the total degrees is strictly less than the number of variables, so the C-W theorem does not apply.
+ One can observe the the C-W theorem is an "iff" here: if $N(P)$ is even then there is a multilinear polynomial with degree strictly less than the number of variables, such that the solution set encodes those $C\_S$'s that induce $P$.
* Instead of using "rook sets", one can ask the question for other classes of subsets of $[n]^3$. I'd be happy with establishing $c=\omega(n)$ for any class of subsets (although I'd like to be able to compute at least one example of such a subset efficiently).
>
> Are there other methods for counting modulo two that I missed?
>
>
>
| https://mathoverflow.net/users/4416 | Counting colored rook configurations in the cube - when is it even? | This can be phrased as a problem concerning [Latin squares](http://en.wikipedia.org/wiki/Latin_square). Eg. a "rook set" is equivalent to a Latin square. For example:
```
123 100 010 001
231 <-> 001 100 010
312 010 001 100
```
A colouring of the Latin square is a partition of its entries (corresponding to a coloured rook set).
We can therefore readily construct colour profiles P with c=n2 such that N(P)=1 (that is, by partitioning some Latin square into n2 parts). But we can do much better...
A *defining set* is a partial Latin square with a unique completion. A *critical set* is a minimal defining set. Let scs(n) be the size of a smallest critical set for Latin squares of order n. From a Latin square L containing a critical set of size scs(n), we can choose a partition (i.e. a c-colouring) such that the entries in the critical set are in parts of size 1 and the remaining entries of L are in a single part. This also will give rise to a colour profile P in which N(P)=1. It has been shown that scs(n)≤n2/4 for all n. (see J. Cooper, D. Donovan and J. Seberry, Latin squares and critical sets of minimal size, Australasian J. Combin 4 (1991), 113–120.) Hence we can deduce that for some c<=n2/4+1 we have N(P)≡1 (mod 2). But with a more intelligent choice of the partition, we can do better...
In fact, we can use the constructions in Cooper et al. to construct a colour profile P with c=n for which N(P)=1. I will only be able to prove this by example here (but it should be clear it can be readily generalised): Partition the "back circulant" Latin square of order 6 as follows.
```
123456 100000 020000 003000 000000 000000 000456
234561 000000 200000 030000 000000 000000 004561
345612 <-> 000000 + 000000 + 300000 + 000000 + 000000 + 045612
456123 000000 000000 000000 000000 000000 456123
561234 000000 000000 000000 000004 000000 561230
612345 000000 000000 000000 000040 000005 612300
```
Now observe that the three colour profiles are:
```
100005 111003 111003
020004 011004 011004
003003 001005 001005
000204 000006 000006
000015 000105 000105
000006 000114 000114
```
Together these form P. Given P, we can observe that any Latin square of order 6 with colour profile P must contain the following partial Latin square:
```
123...
23....
3.....
......
.....4
....45
```
Which is a critical set -- and therefore admits a unique completion (that is, to the "back circulant" Latin square of order 6). Therefore, if c=n there exists a color profile P for which N(P)=1, not just N(P)≡1 (mod 2).
**EDIT**: I presented this problem to our research group at Monash and we improved the upper bound to c=1 (which was a bit surprising!). The idea came from the following critical set (call it C) of the back-circulant Latin square, which is related to the one above, but contains more entries than the one above (but this doesn't matter for this problem).
```
12345.
2345..
345...
45....
5.....
......
```
We observed that if every entry in the critical set above is assigned one colour, and the remaining entries another colour, then there is a unique Latin square with that colour profile. That is, we derive the colour profile from in the following way.
```
123456 12345. .....6
234561 2345.. ....61
345612 <-> 345... + ...612
456123 45.... ..6123
561234 5..... .61234
612345 ...... 612345
```
which gives rise to the following colour profile P:
```
15 51 51
24 42 42
33 33 33
42 24 24
51 15 15
06 06 06
```
We can deduce that any Latin square with the colour profile P will, in fact, contain the critical set C. Hence, N(P)=1.
To see how we deduce the critical set from the colour profile, we note that for the first colour, it contains 5 entries in the first row and column, 4 entries in the second row and column, and so on (and the same for columns). This selection of cells can take only one shape -- that is, it is unique. Then placing the symbols 1..5 in it can only be achieved in one way (to preserve the Latin property).
Since this can be generalised for all n≥2, the answer to your question "what is the largest c such that for all colour profiles P, N(P)≡0 (mod 2)" is c=1.
**EDIT 2**: We also discussed at our research meeting the complexity side of the problem.
*Instance*: colour profile P
*Question*: is N(P)≥1?
In fact, there is an easy way to see that this problem is NP-complete, since (a) we can embed instances of the problem of partial Latin square completion in the above problem and (b) a Latin square together with its colouring can be used as a certificate.
The problem of partial Latin square completion was shown to be NP-complete in: C. J., Colbourn, The complexity of completing partial Latin squares, *Discrete Appl. Math.* 8 (1984), no. 1, 25--30.
| 13 | https://mathoverflow.net/users/2264 | 17274 | 11,548 |
https://mathoverflow.net/questions/13928 | 8 | Let $p$ be a prime number, $K$ a finite extension of $\mathbb{Q}\_p$, $\mathfrak{o}$ its ring of integers, $\mathfrak{p}$ the unique maximal ideal of $\mathfrak{o}$, $k=\mathfrak{o}/\mathfrak{p}$ the residue field, and $q=\operatorname{Card} k$.
Recall that a polynomial $\varphi=T^n+c\_{n-1}T^{n-1}+\cdots+c\_1T+c\_0$ ($n>0$) in $K[T]$ is said to be **Eisenstein** if $c\_i\in\mathfrak{p}$ for $i\in[0,n[$ and if $c\_0\notin\mathfrak{p}^2$.
**Question.** When is the extension $L\_\varphi$ defined by $\varphi$ galoisian (resp. abelian, resp. cyclic) over $K$ ?
**Background.** Every Eisenstein polymonial $\varphi$ is irreducible, the extension $L\_\varphi=K[T]/\varphi K[T]$ is totally ramified over $K$, and every root of $\varphi$ in $L\_\varphi$ is a uniformiser of $L\_\varphi$. There is a converse.
If the degree $n$ of $\varphi$ is prime to $p$, then the extension $L\_\varphi|K$ is tamely ramified and can be defined by the polynomial $T^n-\pi$ for some uniformiser $\pi$ of $K$. Thus $L\_\varphi|K$ is galoisian if and only if $n|q-1$, and, when such is the case, $L\_\varphi|K$ is actually cyclic.
**Real question**. Is there a similar criterion, in case $n=p^m$ is a power of $p$, for deciding if $L\_\varphi|K$ is galoisian (resp. abelian, resp. cyclic) ?
| https://mathoverflow.net/users/2821 | When is the extension defined by an Eisenstein polynomial galoisian or abelian or cyclic ? | In the case where the ground field $K$ is $\mathbb{Q}\_p$, some old work of Lbekkouri has recently been published [here](https://doi.org/10.1007/s00013-009-0026-3 "On the construction of normal wildly ramified extensions over Q_p, (p \neq 2). Arch. Math. 93, 331 (2009)"). In particular, for that case, i.e. for finite totally wildly ramified extensions of $\mathbb{Q}\_p$, normality is equivalent to cyclicity. Furthermore:
When $n=p$, this was answered by Ore in the 30's:
the extension is normal if and only if $p^2|c\_j$ for $1\leq j\leq p-2$ and $p^2|(c\_0+c\_{p-1})$.
When $n=p^2$, Lbekkouri gives a list of necessary and sufficent congruence conditions on the coefficients $c\_j$.
More generally for $n=p^m$, he gives some necessary conditions but since the methods require detailed computations with the ramification filtration, it seems unlikely that one could extend the sufficient conditions much beyond the $p^2$ case.
| 3 | https://mathoverflow.net/users/3143 | 17287 | 11,559 |
https://mathoverflow.net/questions/17298 | 6 | Background
----------
The beth function is defined recursively by: $\beth\_0 = \aleph\_0$, $\beth\_{\alpha + 1} = 2^{\beth\_\alpha}$, and $\beth\_\lambda = \bigcup\_{\alpha < \lambda} \beth\_\alpha$. Since the beth function is strictly increasing and continuous, it is guaranteed to have arbitrarily large fixed points by the [fixed-point theorem on normal functions](http://en.wikipedia.org/wiki/Fixed-point_lemma_for_normal_functions).
The cofinality of an ordinal $\alpha$ is the smallest ordinal $\beta$ such that there are unbounded increasing functions $f : \beta \to \alpha$.
| https://mathoverflow.net/users/4087 | Does the beth function have fixed points of arbitrarily large cofinality? | Yes. The definable class $C$ of fixed points $\beth\_\kappa = \kappa$ is closed and unbounded and therefore contains ordinals of all possible cofinalities. Specifically, let $\kappa\_\alpha$ denote the $\alpha$-th element of $C$. Note that $\kappa\_\delta = \sup\_{\alpha<\delta} \kappa\_\alpha$ for every limit ordinal $\delta$. If $\lambda$ is a regular cardinal then $\kappa\_\lambda$ has cofinality $\lambda$.
| 4 | https://mathoverflow.net/users/2000 | 17300 | 11,569 |
https://mathoverflow.net/questions/17295 | 14 | Given a group of order $n$ where $n$ is either a specific number, or a number of a particular form, e.g. square-free, when does $n$ completely determine a particular group property among all groups of that order? Vipul's group theory wiki has several stubs on this topic, and in the language of his wiki, I will call this a $P$-forcing number, where $P$ is a particular group theoretic property.
We already have quite a few easy examples, for instance, orders $pq$, $pqr$, and $p^2q$ force solvability, and $p^2$ forces abelian. Then there are more specific results like 99 is an abelian-forcing number.
I am interested in general, in any results of this flavor beyond what would be considered a common result in a standard graduate-level group theory book.
| https://mathoverflow.net/users/2616 | Results about the order of a group forcing a particular property. | The numbers n such that every group of order n is cyclic, abelian, nilpotent, supersolvable, or solvable are known. Most are described in an easy to read survey:
Pakianathan, Jonathan; Shankar, Krishnan. "Nilpotent numbers."
Amer. Math. Monthly 107 (2000), no. 7, 631-634.
[MR 1786236](http://www.ams.org/mathscinet-getitem?mr=1786236)
[DOI: 10.2307/2589118](http://dx.doi.org/10.2307/2589118)
If you want to go beyond results like this, you may have better luck looking at a slightly more refined version of the order: the isomorphism type of the Sylow subgroups. Sometimes a p-group P has the property that every group G containing it as a Sylow p-subgroup has a normal subgroup Q of order coprime to p such that G is the semi-direct product of P and Q. An easy version of this that does appear in many group theory texts is that if n=4k+2, then in each group G of order n there is a normal subgroup Q of order 2k+1 so that G is the semi-direct product of any of its Sylow 2-subgroups and Q.
Groups all of whose Sylow p-subgroups are cyclic have very nice properties, subsuming those of groups of square-free order. Groups all of whose Sylows are abelian have more flexibility, but are still basically under control.
| 17 | https://mathoverflow.net/users/3710 | 17304 | 11,573 |
https://mathoverflow.net/questions/17305 | 3 | Let $f \in \mathbf{F}\_q[T]$ be irreducible. I know that the ray class field for $\mathrm{Cl}((f)) \cong (\mathbf{F}\_q[T]/(f))^\times$ can be constructed by adjoining torsion points of a Carlitz module. Is there an easy explicit minimal polynomial for a generator of this extension?
| https://mathoverflow.net/users/nan | ray class field of rational function field | The minimal polynomial is $\phi\_f(X)/X$, where $\phi\_g$ (the Carlitz module) is defined by being $\mathbb{F}\_q$-linear in $g$, satisfy
$\phi\_{T^{n+1}} = \phi\_T(\phi\_{T^n})$ and $\phi\_T =X^q+TX$.
It even has the bonus of being an Eisenstein polynomial at $f$.
| 7 | https://mathoverflow.net/users/2290 | 17308 | 11,575 |
https://mathoverflow.net/questions/17268 | -2 | As we can see,there are some conditions given in a proposition.
If there are 2 propositions having approximate conclusions.Usually，we can name one propositions gives stronger conditions than the other.
My question is that whether there is a good system to weigh the conditions given in a proposition.e.g. a condition "uniform convergence"is stronger than a condition"convergence".
Can we give each condition a value to show their sharpness?
| https://mathoverflow.net/users/4419 | how to weigh the conditions given in a proposition | I think that one may have there three kinds of valuations for "strength" of conditions in theorem.
* practical ones: for example some conditions may be easier to check, or even define whilst other may be difficult to check or even define in practical causes. Then You may use fro example "**computational complexity**" of conditions as such kind of valuations for theorems. Conditions which needs more computations or resources or time in order to qualify are less useful. It may be somehow tricky because some conditions may be obvious although complex to check in practice: for example continuity of function is difficult to check in computational way but sometimes is obvious in physics for example. So it should be distinguished if You want build "theory of such valuations" or only measure difficulty in practical cases. In this kinds of valuation You measure practical obstacles connected with some condition. In fact probably "strong" conditions from point of view of pure mathematics, which gives You theorem which may be hard to use in theoretical math because of its requirements, maybe have easier conditions to check computationally, as they will be very specific, and have broad applications in practical computations and physics for example.
* aesthetic one: this is very interesting cause. It is extremely difficult to formalize someone aesthetics. You may see that sometimes aesthetic means to be symmetric but it is somehow non clear what kind of symmetry You mean when we spoke about conditions of theorems. Also it is well known that sometimes very simple axioms for example, may be most restrictive - strong, whilst very complicated one and not so pretty, may be week enough to give very interesting structure for theory for example. So here You will probably do not get any progress...
* last one criterion by size of structure: restrictive in intuitive meaning is as follows: the more restrictive criteria has less positive cases. So it may be measure by size of positive cases space. This is probably the most obvious case, because beautiful and meaningful theorem should have week conditions and strong thesis, so it may concern about big class of objects and say something very certain about them. But if You measure size of structure, You may probably get the same size structures not equally interesting. For example class of continuous functions is very big, but modular functions may be concerned as much more interesting as "single object". So You have to relate "strength of requirements" to "size of set obtained in thesis". Regarding criteria, should You add also criteria defining modular function to the list in this case? You want to compare theorems from different theories or within certain one only?
I suppose that aesthetic conditions are interesting but hard to formalization, whilst practical one are something You may think about in constructive and useful way but probably it is not You are asking for...
| 1 | https://mathoverflow.net/users/3811 | 17311 | 11,578 |
https://mathoverflow.net/questions/17286 | 10 | This question is on "computing" the Grothendieck group of the projective $n$-space with $m$ origins ($m\geq 1$). For any (noetherian) scheme $X$, let $K\_0(X)$ be the Grothendieck group of coherent sheaves on $X$.
Firstly, let me sketch that $K\_0(\mathbf{P}^n) \cong K\_0(\mathbf{P}^{n-1})\oplus K\_0(\mathbf{A}^n)$.
Let $X=\mathbf{P}^n$ be the projective $n$-space.
(I omit writing the base scheme in the subscript. In fact, you can take any noetherian scheme as a base scheme in the following, I think.)
Let $H\cong \mathbf{P}^{n-1}$ be a hyperplane with complement $U\cong \mathbf{A}^n$. By a well-known theorem on Grothendieck groups, we have a short exact sequence of abelian groups $$K\_0(H) \rightarrow K\_0(X) \rightarrow K\_0(U) \rightarrow 0.$$ Now, let $i:H\longrightarrow X$ be the closed immersion. Then the first map in the above sequence is given by the "extension by zero", which in this case is just the K-theoretic push-forward $i\_!$, or even better, just the direct image functor $i\_\ast$. Now, there is a projection map $\pi:X\longrightarrow H$ such that $\pi\circ i = \textrm{id}\_{H}$.
By functoriality of the push-forward, we conclude that $\pi\_! \circ i\_\ast = \pi\_! \circ i\_! = \textrm{id}\_{K\_0(H)}$.
Therefore, we may conclude that $i\_\ast$ is injective and that we have a split exact sequence $$0 \rightarrow K\_0(H) \rightarrow K\_0(X) \rightarrow K\_0(U) \rightarrow 0.$$ Thus, we have that $K\_0(\mathbf{P}^n) \cong K\_0(\mathbf{P}^{n-1})\oplus K\_0(\mathbf{A}^n)$.
**Q1**: Let $\mathbf{P}^{n,m}$ be the projective $n$-space with $m$ origins ($m\geq 1$). For example, $\mathbf{P}^{n,1} = \mathbf{P}^n$. (Again the base scheme can be anything, I think.) Now, is it true that $$K\_0(\mathbf{P}^{n,m}) \cong K\_0(\mathbf{P}^{n-1,m}) \oplus K\_0(\mathbf{A}^n)?$$
**Idea1**: Take a hyperplane $H$ in $\mathbf{P}^{n,m}$. Is it true that $H\cong \mathbf{P}^{n-1,m}$ and that its complement is $\mathbf{A}^n$? Also, even though the schemes are not separated, the closed immersion $i:H\longrightarrow \mathbf{P}^{n,m}$ is proper, right? Also, is the projection $\pi:\mathbf{P}^{n,m}\rightarrow H$ proper? If yes, the above reasoning applies. If no, how can one "fix" the above reasoning? I think that in this case one could still make sense out of $i\_!$ and $\pi\_!$ (even if they are not proper maps.)
**Idea2**: Maybe it is easier to show that $K\_0(\mathbf{P}^{n,m}) \cong K\_0(\mathbf{P}^{n-1})\oplus K\_0(\mathbf{A}^{n,m})$, where $\mathbf{A}^{n,m}$ is the affine $n$-space with $m$ origins. Then one reduces to computing $K\_0(\mathbf{A}^{n,m})$...
**Idea3**: One could also take $m=2$ as a starting case and look at the complement of one of the origins. Then we get a similar exact sequence as above and one could reason from there.
Which of these ideas do not apply and which do?
**Note**: Suppose that the base scheme is a field. Since $K\_0(\mathbf{A}^n) \cong \mathbf{Z}$ and $K\_0(\mathbf{P}^n) \cong \mathbf{Z}^{n+1}$, this would show that $$K\_0(\mathbf{P}^{n,m}) \cong \mathbf{Z}^{n+m}.$$ More generally, if $S$ is the base scheme, $K\_0(\mathbf{P}^{n,m}) \cong K\_0(S)^{n+m}$.
| https://mathoverflow.net/users/4333 | $K_0$ of a non-separated scheme | This is exactly the sort of example where it is relevant which version of K-theory you are employing. (In particular, which theorems you can employ here depends on this!) The issue here is a tad subtle.
To illustrate, let's restrict attention to the case of affine $n$-space $X$ with a doubled origin ($n\geq 2$). (An iteration of the discussion below, combined with your localization argument in Idea2, will let you address your general multiply-origined projective spaces.) For convenience, let me assume that the base is a regular noetherian scheme $S$.
Let me first answer your question as it was asked. Quillen's localization sequence gives fiber sequences of spectra or spaces
$$G(S)\to G(\mathbf{A}\_S^n)\to G(\mathbf{A}\_S^n-\{0\})$$
and
$$G(S)\to G(X)\to G(\mathbf{A}\_S^n)$$
Here I'm using $G$-theory is the $K$-theory spectrum (or space) of coherent sheaves. This has the property that $G(S)\simeq G(\mathbf{A}\_S^n)$. Putting this together, we see that
$$G(X)\simeq G(\mathbf{A}\_S^n)\times G(S)\simeq G(S)\times G(S)$$
This answers your question for $G$-theory; in particular $\pi\_0$ of $G$ is the Grothendieck group you seek, and so we conclude that $G\_0(X)=G\_0(S)\oplus G\_0(S)$. Now use your localization arguments to get that $G\_0(\mathbf{P}^{n,2})$ is $n+2$ copies of $G\_0(S)$.
However: for $K$-theory, the way this computation gets done depends critically on which model of $K$ you use.
(A) If we define $K$-theory as in Thomason-Trobaugh (as the Waldhausen $K$-theory spectrum of perfect complexes), then in this case $K(X)\simeq G(X)$ and $K(S)\simeq G(S)$. (This follows from "Poincaré Duality;" see the end of section 3 of Thomason-Trobaugh.)
(B) If, however, we define $K^{\mathrm{naive}}(X)$ as the Quillen $K$-theory of the category of algebraic vector bundles on $X$, then things look different. [N.B. that this is the name given by Grothendieck, Illusie, et al. in SGA 6. It is not meant to be insulting!] Indeed, the inclusion $\mathbf{A}\_S^n\to X$ induces an equivalence between the catgories of algebraic vector bundles on $X$ and those on $\mathbf{A}\_S^n$, since the origin has codimension at least $2$. So then
$$K^{\mathrm{naive}}(X)\simeq K^{\mathrm{naive}}(\mathbf{A}\_S^n)\simeq K(\mathbf{A}\_S^n)\simeq K(S).$$
The second equivalence here follows from the fact that "naive" $K$-theory agrees with $K$-theory for schemes that admit an ample family of line bundles (see section 3 of Thomason-Trobaugh). The difference between (A) and (B) here reflects the failure of our $X$ to admit such a family.
An example like this is discussed is Thomason-Trobaugh I think at the end of section 8 (wherever they talk about Mayer-Vietoris). [Sorry, this is from memory; I don't have a copy of the Festschrift handy.]
Hope this helps!
| 7 | https://mathoverflow.net/users/3049 | 17316 | 11,581 |
https://mathoverflow.net/questions/17323 | 2 | A paper I'm reading implicitly assumes the statement: Let $K\_0$ be the completion of $\mathbb {Q}\_ p^{un}$. Then any finite extension of $K\_0$ is complete with residue field $\bar {\mathbb {F}} \_p$. So here are a few questions:
1. Why is this true? Is it true in general that if you complete, and then take algebraic closure - then that is complete?
2. Is it true that any finite extension of $K \_0$ comes from the completion of a finite extension of $\mathbb {Q} \_p ^ {un}$?
| https://mathoverflow.net/users/3238 | Fiddling with p-adics | A finite extension of a complete field is complete. This is standard and proved in lots of places e.g. one of the first two chapters of Cassels-Froehlich, or Bosch-Guentzer-Remmert, or lots of other places. The residue field will only go up a finite amount too, so it must have stayed the same!
Your Q1 appears to have two parts: the first is "why is this true", which I think I just answered. The second part is not true: if you complete Q at p you get Q\_p and if you then take alg closure you get Q\_p-bar which isn't complete. What is true is that if you take alg closure and then complete, you're still alg closed. This is proved in BGR too.
Q2 is also true and this will follow from Krasner's lemma, also proved in BGR (and in lots of other places). A finite extension of K\_0 is K\_0(alpha) with alpha a root of a polynomial in K\_0[x]. By Krasner I can perturb the coefficients of this polynomial without changing the extension. So I perturb them until they're in Q\_p^{un} and there's my finite extension of Q\_p^{un}.
| 5 | https://mathoverflow.net/users/1384 | 17326 | 11,589 |
https://mathoverflow.net/questions/17331 | 1 | I have been asked to provide an "approximation at infinity" of an expression that at the end simplifies to $-\frac{b e^{-a t}-a e^{-b t}}{a-b}$, in a course about extreme value theory.
In the course, we saw "approximations" such as $2 t^{-\alpha }-t^{-2 \alpha }$ being approximated to $2 t^{-\alpha }$ whatever the vague word approximation means. In words, this "approximation" states that the distribution tail is dominated by the term $2 t^{-\alpha }$ at infinity. I think that there is no polynomial term $t^{-\alpha }$ which dominates at infinity in the given question, since $e^{-k t}$ decreases faster than any term $t^{-\alpha }$. However, is there any sense in which this "approximation at infinity" can be taken? I tried to take the Taylor series at infinity and Mathematica returns the same expression (unevaluated?) and computing manually, the first order approximation of $e^{-k t}$ is 0.
Any ideas or references about how to compute these sort of approximations in extreme value theory are appreciated.
| https://mathoverflow.net/users/3899 | Extreme value theory | Do you have information on the relative sizes of $a$ and $b$? If we have, say $a>b$, then we can say that as $t \to \infty$, $-\frac{b e^{-a t}-a e^{-b t}}{a-b}=-e^{-b t}(\frac{b e^{-(a-b) t}-a }{a-b}) \sim \frac{a e^{-b t}}{a-b} =\frac{e^{-b t}}{1-\frac{b}{a}}$. It is rather simple, but I don't think there's much more that you can say about it.
| 0 | https://mathoverflow.net/users/4436 | 17336 | 11,597 |
https://mathoverflow.net/questions/17341 | 3 | Let $M$ be a smooth manifold. We can get a Riemannian metric on $M$ by at least two methods: first by partitions of unity and second by the Whitney embedding theorem: we can embed $M$ into a Euclidean space of sufficiently large dimension, and we thus get a Riemannian metric on $M$ by restricting the Euclidean metric on the ambient space.
Can all Riemannian metrics on $M$ be constructed in the second way above?
[Edited for for punctuation, grammar and clarity -- PLC]
| https://mathoverflow.net/users/4437 | Are all Riemannian metrics induced by Euclidean metrics? [Nash Embedding Theorem] | Yes, see e,g, <http://en.wikipedia.org/wiki/Nash_embedding_theorem>
Briefly, every $C^1$--metric on a $C^1$-manifold is induced by a $C^1$-embedding $M^n\to\mathbf{R}^{2n+1}$; every $C^\infty$ metric on a $C^{\infty}$ manifold is induced by $C^\infty$-embedding $M^n\to\mathbf{R}^{n^2+5n+3}$.
| 12 | https://mathoverflow.net/users/2349 | 17343 | 11,603 |
https://mathoverflow.net/questions/17345 | 1 | we all know that if we consider the sheaf of germs of a holomorphic functions defined on a domain in C^n,we have too many beautiful theorems characterizing the geometry of the domain by consider the Cech-cohomology of the sheaf.Then i think that plurisubharmonic functions is in some sense a weaker function than holomorphic functions.So we may get some beautiful theorems as the case of holomorphic case, for example if we can proof that for a domain in C^n,the first Cech-cohomology of the sheaf of germs of plurisubharmonic functions vanishes ,we then can choose any good plurisubharmonic functions as we want. What i want to ask is that have you ever considered such a question ,and i don't know whether this is a good question ? I want to hear some suggestions.
| https://mathoverflow.net/users/4437 | the Cech-cohomology of the sheaf of germs of plurisubharmonic functions defined on a domain in C^n | As Petya has pointed out, plurisubharmonic functions on an open set do not form a group, so when one sheafifies, one gets a sheaf of sets, not groups; it has H^0, but no higher cohomology.
| 4 | https://mathoverflow.net/users/2349 | 17350 | 11,608 |
https://mathoverflow.net/questions/17344 | 20 | In a paper I developed some theory; some of the applications require extensive computations that are not part of the paper. I wrote a Mathematica notebook. I want to publish a PDF and .nb version somewhere to refer to from the paper. arXiv.org seems a good choice, but they won't accept the .nb file. I do not want to put this through peer review. Where to publish it?
| https://mathoverflow.net/users/4438 | Where to publish computer computations | You should upload the notebook along with the sources of the actual article to the arXiv, of course.
The official (but rather hard to find) advice on this from the arXiv is to place your code in a directory called /aux/. (This is [problematic](http://en.wikipedia.org/wiki/LPT#Naming) for windows users.)
You can see an example of this in [my recent paper](http://arxiv.org/abs/0909.4099) on the extended Haagerup subfactor. A footnote in the text of the article explains that the code is available along with the source download. It would also be appropriate to use the comments field in an arxiv submission to explain that source code is available in the /aux/ directory.
| 31 | https://mathoverflow.net/users/3 | 17351 | 11,609 |
https://mathoverflow.net/questions/17325 | 32 | You don't need a metric to define the differential of a function,
and the cotangent bundle carries a canonical one-form.
But you do need a metric to define the gradient, and the
tangent bundle does not have a canonical vector field.
These are not difficult truths, but still... why the preference
toward "co"?
| https://mathoverflow.net/users/1186 | Why is cotangent more canonical than tangent? | If you want to differentiate functions *from* a manifold to (say) the real line R, then you want to use the cotangent bundle on the manifold.
If instead you want to to differentiate functions *to* the manifold from the real line (i.e. parameterised curves), then you want to use the tangent bundle on the manifold.
So the preference comes from whether you want to use the manifold as the domain or as the range of the functions one is differentiating.
| 69 | https://mathoverflow.net/users/766 | 17355 | 11,610 |
https://mathoverflow.net/questions/17358 | 6 | Is finding a cycle of fixed even length in a bipartite graph any easier than finding a cycle of fixed even length in a general graph? This question is related to the question on [Finding a cycle of fixed length](https://mathoverflow.net/questions/16393/finding-a-cycle-of-fixed-length)
**Edit:** There is natural refinement of this question: what happens if the graph has boundend valence, e.g. if the bipartite graph is cubic?
| https://mathoverflow.net/users/4400 | Finding a cycle of fixed length in a bipartite graph | Finding a cycle of length 2k in an arbitrary graph is the same thing as finding a cycle of length 4k in the bipartite graph formed by subdividing every edge. So in general even cycles of fixed length are no easier to find in bipartite graphs than in arbitrary graphs. But it's possible that the lengths that are 2 mod 4 are easier to find than the lengths that are 0 mod 4, in bipartite graphs.
| 11 | https://mathoverflow.net/users/440 | 17359 | 11,612 |
https://mathoverflow.net/questions/17360 | 6 | A friend recently asked me if i had heard anything about a stable Serre Spectral Sequence or one constructed with spectra, has any one else ever heard of this? is there any reason other than historical that we don't think of the Serre SS this way initially?
Any thoughts or references would be great!
| https://mathoverflow.net/users/3901 | Serre spectral sequence with spectra | Since the ordinary Serre spectral sequence is about a fibration of spaces, I don't think you can just talk about a "fibration of spectra" and expect that to be a generalization, since the suspension spectrum functor doesn't preserve fibration sequences. However, there is a version of the Serre spectral sequence involving *parametrized spectra*, which one can think of as a fibration whose base is an ordinary space and whose fibers are spectra. It can be found in section 20.4 of May-Sigurdsson, *Parametrized Homotopy Theory*.
| 10 | https://mathoverflow.net/users/49 | 17361 | 11,613 |
https://mathoverflow.net/questions/17376 | 0 | How can one show that rational functions satisfy a Lipschitz condition *in the chordal metric* on the Riemann sphere?
| https://mathoverflow.net/users/nan | How to prove that rational functions satisfy a Lipschitz condition in the *chordal metric*? | By compactness, you only need to prove the Lipschitz property locally. First prove that Möbius transforms are Lipschitz (easy – they are compositions of translations, multiplications by constants, and inversions). Then, by composing with suitable Möbius transforms, you only need to show that a rational function which maps 0 to 0 is Lipshitz on a neighbourhood of 0. This is trivial, I hope you agree.
| 2 | https://mathoverflow.net/users/802 | 17379 | 11,624 |
https://mathoverflow.net/questions/17371 | 14 | In the undergraduate toplogy course we were given examples of spaces that are $T\_i$ but not $T\_{i+1}$ for $i=0,\ldots,4$. However, no example of a space which is $T\_3$ but not $T\_{3.5}$ was given. Later I was told by a colleague that such examples are rare and difficult to construct.
I know there is an example of such a space, called the Tychonoff corkscrew (or the spiral staircase), in the "*Counterexamples in topology*" book by Steen and Seebach. I've also found the following paper, though at the moment I'm not able to view it: A.B. Raha "*An example of a regular space that is not completely regular*", Proceedings Mathematical Sciences 102 (1992), 49-51.
Are there any other, folklore examples of regular spaces that are not completely regular? Are there any relatively easy ones?
| https://mathoverflow.net/users/2578 | Regular spaces that are not completely regular | These examples seem to be very difficult to construct. The problem is that any local compactness or uniformity will automatically boost your space to a Tychonoff space, and Tychonoff spaces are closed under passing to subspaces or products. Consequently, there's doesn't seem to be a "machine" for producing these kinds of spaces.
The idea of all the counterexamples $X$ is to write down enough open sets of $X$ to make it clear that points can be separated from closed subsets, but to somehow rig things so that any continuous real-valued function on $X$ identifies two distinct points of the space.
The example in Munkres's textbook that Elencwajg mentions is a pretty straightforward one (relatively speaking); it's the same in spirit as Raha's example, which is the easiest I've found. Here it is:
For every even integer $n$, set $T\_n:=\{n\}\times(-1,1)$, and let $X\_1=\bigcup\_{n\textrm{ even}}T\_n$. Now let $(t\_k)\_{k\geq 1}$ be an increasing sequence of positive real numbers converging to $1$.
For every odd integer $n$, set $$T\_n:=\bigcup\_{k\geq 1}\{(x,y)\in\mathbf{R}^2\ |\ (x-n)^2+y^2=t\_k^2\}$$ and let $X\_2=\bigcup\_{n\textrm{ odd}}T\_n$. Now let $$X=\{a,b\}\cup\bigcup\_{n\in\mathbf{Z}}T\_n$$
Topologize $X$ so that:
1. every point of $X\_2$ except the points $(n,t\_k)$ are isolated;
2. a neighborhood of $(n,t\_k)$ consists of all but finitely many elements of $\{(x,y)\in\mathbf{R}^2\ |\ (x-n)^2+y^2=t\_k^2\}$;
3. a neighborhood of a point $(n,y)\in X\_1$ consists of all but a finite number of points of $\{(z,y)\ |\ n-1<z<n+1\}\cap(T\_{n-1}\cup T\_n)$;
4. a neighborhood of $a$ is a set $U\_c$ containing $a$ and all points of $X\_1\cup X\_2$ with $x$-coordinate greater than a number $c$;
5. a neighborhood of $b$ is a set $V\_d$ containing $b$ and all points of $X\_1\cup X\_2$ with $x$-coordinate less than a number $d$.
This is a space that is $T\_3$, but every continuous map $f:X\to\mathbf{R}$ has the property that $f(a)=f(b)$, so it is not $T\_{3\frac{1}{2}}$.
| 15 | https://mathoverflow.net/users/3049 | 17382 | 11,626 |
https://mathoverflow.net/questions/17396 | 24 | To be more precise (but less snappy): is there an example of a group $G$ with finitely many infinite-index subgroups $H\_1,\dots, H\_n$ and elements $k\_1,\dots, k\_n$ such that $G$ is the union of the left cosets $k\_1 H\_1 , ..., k\_n H\_n\ ?$ And what if we relax the requirement that these all be *left* cosets, and ask: can $G$ be the union of finitely many such cosets, some being left cosets, others being right cosets?
If $G$ is [amenable](http://en.wikipedia.org/wiki/Amenable) then this can't happen, since any coset of an infinite-index subgroup must have measure $0$. So this immediately rules out any abelian group $G$.
I've tried playing around with the only non-amenable groups that I'm comfortable with, the free groups on two or more generators. A few months ago I thought I found a simple counterexample in the free group on $\aleph\_0$ generators, but now I've lost my notes and am beginning to doubt I ever had such an example.
(This question was asked to me by a friend who's interested in some kind of application to model theory, but I think it's interesting as a stand-alone puzzle.)
| https://mathoverflow.net/users/93 | Can a group be a finite union of (left) cosets of infinite-index subgroups? | No. This follows from a beautiful theorem of B.H. Neumann:
Let $G$ be a group. If $\{x\_iH\_i\}\_{i=1}^n$ is a covering of $G$ by cosets of proper subgroups, then $n \geq \min\_{i} [G:H\_i]$.
Explicitly, this is Lemma 4.1 in
[http://alpha.math.uga.edu/~pete/Neumann54.pdf](http://alpha.math.uga.edu/%7Epete/Neumann54.pdf)
As Neumann remarks, the identity $gH = (g H g^{-1}) g$ shows that it is no loss of generality to restrict to coverings by left cosets.
| 38 | https://mathoverflow.net/users/1149 | 17398 | 11,636 |
https://mathoverflow.net/questions/17395 | 1 | As the title shows,we know that there is some points the series not approaching to the function.
Now,take the convergence theorem into consideration.As there is some the first break-points，the series is still convergent.And,the Gibbs phenomenon always takes place on the first break-points.
Why does Gibbs phenomenon take place?What does it show the nature of Fourier Series?
| https://mathoverflow.net/users/4419 | What does Gibbs phenomenon shows the nature of Fourier Series | A Fourier series truncated to order $n$ is the best approximation to the given function *in the $L^2$ sense* using trigonometric polynomials of order $n$. As such, small rapid deviations don't matter much. Since there is a limit to how big the derivatives of a trigonometric polynomial of fixed order can be (without the coefficients being big), in order to fit such a polynomial to a discontinuity it pays to overshoot a bit on each side of the discontinuity in order to “gather speed” so you can get from one value to the other fast. When I say it “pays”, i mean to say that you what you lose by not approximating the function too well at the overshoot, you more than gain back by doing the jump faster.
| 11 | https://mathoverflow.net/users/802 | 17399 | 11,637 |
https://mathoverflow.net/questions/17388 | 10 | Suppose A and B are compact connected sets in the XY plane and XZ plane respectively in R^3. Suppose further that the the range of x-values taken by A and B are the same (i.e, projections of A and B onto the x-axis are the same closed interval). Is there always a connected set in R^3 whose projections onto XY and XZ planes are A and B respectively?
| https://mathoverflow.net/users/996 | existence of a connected set with given connected projections. | The answer is yes.
Let $\alpha,\beta:[0,1]\to[0,1]\times \mathbb R$ be two paths;
$\alpha(t)=\left(\alpha\_1(t),\alpha\_2(t)\right)$ and $\beta(t)=\left(\beta\_1(t),\beta\_2(t)\right)$.
Assume that $\alpha\_1(0)=\beta\_1(0)=0$, $\alpha\_1(1)=\beta\_1(1)=1$.
>
> **Claim.** The points $a=\left(0,\alpha\_2(0),\beta\_2(0)\right)$ and $b=\left(1,\alpha\_2(1),\beta\_2(1)\right)$ lie in the same connected coponent of the set $\Sigma\subset \mathbb R^3$ formed by points of the following type $ \left(\alpha\_1(t),\alpha\_2(t),\beta\_2(\tau)\right)$ such that $\alpha\_1(t)=\beta\_1(\tau)$.
>
>
>
**Proof.** Note that for generic smooth choice of $\alpha$ and $\beta$ the set $\Sigma$ is a smooth 1-dimensional manifold which might be not connected, but it has only two boundary points in $a$ and $b$.
Thus, in this case one can connect these points by a curve.
The general case can be done by approximation. $\square$
The rest is easy: one can approximate $A$ and $B$ by path-connected sets $A'$ and $B'$ with the same property.
Thus one can present $A'$ and $B'$ as a union of curves $\alpha$ and $\beta$ with the above property.
Moreover we can assume that the ends of $\alpha$ and $\beta$ are fixed (i.e. $a$ and $b$ are fixed).
For each pair $(\alpha,\beta)$, choose the connected component of $a$ in $\Sigma$ and take the union of all of them.
| 6 | https://mathoverflow.net/users/1441 | 17405 | 11,643 |
https://mathoverflow.net/questions/17411 | 8 | Where's the notion of interpretation (model) originally introduced?
I find it used in Skolem's paper "Logico-combinatorial investigations in the satisfiability or provability of mathematical propositions" (1920).
However, I do not find any explicit reference to semantical ideas in Frege's Begriffsschrift. Where did the formal idea of semantics come up? What's the relation of this to Tarski's theory of truth?
I really have a mess in my head.
| https://mathoverflow.net/users/3872 | Where's the notion of interpretation (model) originally introduced? | **Updated answer:**
As best as I have been able to figure out, the pre-Tarskian notions of "semantics" in mathematical logic grew out of the "algebra of logic" introduced by Boole ("An investigation into the laws of thought," 1854, and some earlier papers) and elaborated by Charles Peirce, Schröder, and others.
It's difficult for me to follow all the arguments in the old papers, but roughly the idea behind Boole's logic was to study logical equations such as "$x + (1 - x) = 1$" or "$x \times (1 - x) = 0.$" Here, + means exclusive or, multiplication is "and," and subtracting $x$ means taking a conjunction with not-$x$. The number 1 should be interpreted as an always-true proposition, 0 as an always-false proposition, and $x$ as a propositional variable (or as Boole might say, a proposition with "indeterminate truth value").
Boole was interested in this analogy between logic an algebra, and here maybe we see the beginnings of the notion of interpretation in logic: we can check the validity of these formulas by considering whether they are true for all possible propositions $x$. (At least, I think this is what Boole meant -- you should track down the Dover reprinting of *The Laws of Thought* if you want to do some more historical investigation.)
Peirce considered the possibility of different "domains of individuals," which could be finite, infinite, or even uncountable. I think it was Peirce who first generalized Boole's calculus of logical propositions to the "calculus of relatives," where a "relative" is the interpretation of some $n$-ary predicate in a domain of individuals. To track down the beginnings of this, I would try Peirce's 1870 "Description of a notation for the logic of relatives," which unfortunately I cannot access right now from where I am.
Peirce, Schröder, and even Löweinheim in his 1915 "On possibilities in the calculus of relatives" continued to use algebraic notation along the lines of Boole, with many $0$'s and $1$'s. Even the "domain of individuals'' was denoted by $1^1$!
One thing in particular that is confusing about reading Löweinheim's paper is that, while he is clearly aware that the domain of individuals $1^1$ could be one of any number of possible collections of things (some finite, others infinite), he seems to insist on talking about "*the* domain of individuals $1^1$" and referring to every possible domain by the same name $1^1$! Obviously, this is confusing if you want to think about comparing two different such domains, and maybe one of Tarski's key contributions here was simply to introduce a notation ''$\mathfrak{A} \models \varphi$'' which explicitly names the universe $\mathfrak{A}$ and suggests comparison with other universes $\mathfrak{B}, \mathfrak{C}, \ldots$.
**My original answer (missing the key point):**
My knee-jerk answer to this question was going to be, "Tarski!" But you seem to already be aware of Tarski's work, so maybe you're looking for something different?
In particular: Alfred Tarski's 1933 article "The concept of truth in formalized languages" (in Polish, unfortunately) seems to be generally regarded as the first place where the concept of "logical satisfaction" (in the modern sense) was first defined.
There was already an "application" of semantic methods in logic by 1940: Gödel's proof of that Con(ZFC) implies Con(ZFC + GCH + AC). (It might be fun to try to find an even earlier application of semantic methods to prove a syntactic result.) Certainly by the 1960's the field of model theory was coming into its own with the work of A. Robinson, Vaught, Morley, and others.
| 3 | https://mathoverflow.net/users/93 | 17419 | 11,651 |
https://mathoverflow.net/questions/17422 | 7 | Let $Q$ be a quadratic form in $n$ variables with integer coefficients. Let us say that $Q$ has the "special property" mod $p$, if the relation $Q(x\_1,...,x\_n)=0$ (mod $p$) implies that $(x\_1,...,x\_n)=(0,...,0)$ (mod $p$). (There must be a name for this property, but I don't know it, which is why I'm calling it "the special property".) Let us say that $Q$ is "special infinitely often" if there are infinitely many primes $p$ such that $Q$ has the special property mod $p$. For example, the one-variable quadratic form $Q(x)=x^2$ is special infinitely often. Another simple example is that $Q(x,y)=x^2+y^2$ is special infinitely often because it is special mod $p$ whenever -1 is a quadratic non-residue mod $p$. In fact, any two-variable quadratic form that is non-degenerate over the rationals is special infinitely often because $Q(x,y)=ax^2+bxy+cy^2$ is special whenever $p$ is odd and the discriminant $b^2-4ac$ is a quadratic non-residue mod $p$.
I've been trying to find a quadratic form in more than two variables that is special infinitely often, but I'm doubtful that such a thing exists. As far as I know, each of the following statements could either be valid or invalid. (Although obviously [c] implies [b] implies [a].)
[a] For every integer quadratic form $Q$ in three or more variables, the set of $p$ such that $Q$ has the special property mod $p$ is finite.
[b] The set of primes $p$ such that there exists an integer quadratic form $Q$ in three or more variables having the special property mod $p$ is finite.
[c] The set of primes $p$ such that there exists an integer quadratic form $Q$ in three or more variables having the special property mod $p$ is empty.
Can anyone help resolve any of these questions?
| https://mathoverflow.net/users/4467 | Quadratic forms that evaluate to zero mod p only when their input is zero. | The Chevalley–Warning theorem (see wikipedia) implies that any quadratic form in at least three variables has a non-trivial solution modulo p.
| 11 | https://mathoverflow.net/users/nan | 17426 | 11,654 |
https://mathoverflow.net/questions/17425 | 17 | Suppose I have a small category $ \mathcal{C} $ which is fibered over some category $\mathcal{I}$ in the categorical sense. That is, there is a functor $\pi : \mathcal{C} \rightarrow \mathcal{I}$ which is a fibration of categories. (One way to say this, I guess, is that $\mathcal{C}$ has a factorization system consisting of vertical arrows, i.e. the ones that $\pi$ sends to an identity arrow in $\mathcal{I}$, and horizontal arrows, which are the ones it does not. But there are many other characterizations.)
Now let $F : \mathcal{C} \rightarrow s\mathcal{S}$ be a diagram of simplicial sets indexed by $\mathcal{C}$. My question concerns the homotopy limit of $F$. Intuition tells me that there should be an equivalence
$$ \varprojlim\_{\mathcal{C}} F \simeq \varprojlim\_{\mathcal{I}} \left (\varprojlim\_{\mathcal{C}\_i} F\_i \right ) $$
where I write $\mathcal{C}\_i = \pi^{-1}(i)$ for any $i \in \mathcal{I}$, $F\_i$ for the restriction of $F$ to $\mathcal{C}\_i$ and $\varprojlim$ for the homotopy limit.
Intuitively this says that when $\mathcal{C}$ is fibered over $\mathcal{I}$, I can find the homotopy limit of a $\mathcal{C}$ diagram of spaces by first forming the homotopy limit of all the fibers, realizing that this collection has a natural $\mathcal{I}$ indexing, and then taking the homotopy limit of the resulting diagram.
Does anyone know of a result like this in the model category literature?
Update: After reading the responses, I was able to find a nice set of exercises [here](http://www-math.mit.edu/~mbehrens/TAGS/Isaacson_exer.pdf) which go through this result in its homotopy colimit version.
| https://mathoverflow.net/users/4466 | Homotopy Limits over Fibered Categories | I can't think of a reference for this. But here is what I would do:
Given any functor $\pi\colon C\to I$ (not necessarily fibered), there's a "homotopy right Kan extension" functor
$$\lim{}^\pi \colon Func(C,sS) \to Func(I,sS),$$
and a weak equivalence $\lim\_C = \lim\_I \lim{}^\pi$. There's a formula to compute $\lim{}^\pi$ in terms of ordinary homotopy limits on comma categories; it looks like
$$(\lim{}^\pi F)(i) = \lim{}\_{i/\pi} F\_i',$$
where $i/\pi$ is the comma category with objects $(c, f:i\to \pi c)$ where $c$ is an object of $C$. The functor $F\_i'$ is the composite of $F$ with the forgetful functor $(i/\pi)\to C$.
I suspect that in your fibered category case, you are able to show that for each $i$ the evident inclusion $C\_i\to (i/\pi)$ is "final" (or, possibly, "cofinal"; I can never remember which is which), and thus that $\lim\_{i/\pi} F' = \lim\_{C\_i} F\_i$.
The "yellow monster" of Bousfield-Kan is a reference for (co)finality condition in context of homotopy (co)limits. It may also discuss homotopy Kan extensions, though I'd have to check.
| 13 | https://mathoverflow.net/users/437 | 17434 | 11,661 |
https://mathoverflow.net/questions/17424 | 1 | I am looking for reference talking about how torsion theory play roles in algebraic geometry. I will be really happy to see some concrete examples. Say, talking about torsion theory in $Coh(P^{1})$.
Thanks in advance
| https://mathoverflow.net/users/1851 | Looking for reference talking about torsion theory on coherent sheaves on projective space | Depending upon how strict you are with your definition of torsion theory a good source of examples is the theory of semi-orthogonal decompositions. A really nice example of this is the appearance of such decompositions which parallel operations in the minimal model program. A good introduction to this is [Kawamata's survey](http://arxiv.org/abs/0804.3150). Of course there are other interesting things one can do with such decompositions in algebraic geometry (e.g. the work of Bondal, Orlov, Kapranov, and many others).
For torsion theories on abelian categories a good example is stability conditions. Here what is interesting is the interplay between torsion theory on hearts and t-structures. The [original paper](http://arxiv.org/abs/math/0212237) is by Bridgeland and in the case of $\mathbb{P}^1$ the stability manifold has been computed by [Okada](http://arxiv.org/abs/math/0411220) (I suggest looking at the journal version, I recall that there were at one point some typos in the arxiv version which were fixed in the published one).
As far as torsion theories on $\mathrm{Coh}(\mathbb{P}^1)$ goes it is a reasonable exercise to actually classify them (I did this at one point but never wrote it up properly). The closest place to this being written down that I know of is in the paper of Gorodentsev, Kuleshov, and Rudakov "t-stabilities and t-structures on triangulated categories" where they classify the minimal t-stabilities on the derived category of coherent sheaves on $\mathbb{P}^1$.
An example of something similar but that is not quite what you asked for is the application of cotorsion theories to relative homological algebra.
**Definition**: Suppose that $\mathcal{A}$ is an abelian category and that $(\mathcal{F},\mathcal{C})$ is a pair of full subcategories. Then we say that $(\mathcal{F},\mathcal{C})$ is a *cotorsion theory* if
$\mathcal{F} = \{F \in \mathcal{A} \; \vert \; \mathrm{Ext}^1(F,\mathcal{C}) = 0\}$ and $\mathcal{C} = \{C \in \mathcal{A} \; \vert \; \mathrm{Ext}^1(\mathcal{F},C) = 0\}$
where the subcategories appearing in the Ext's just signifies that it is true for every object of that subcategory.
There is a notion of a cotorsion theory having enough injectives and projectives and this guarantees for sufficiently good $\mathcal{A}$, say $R$-modules, (by a theorem of Eklof and Trlifaj) that $\mathcal{F}$-covers and $\mathcal{C}$-envelopes exist. In particular this can be used to show that flat covers exist. A good reference for this is Chapter 7 of Relative Homological Algebra by Enochs and Jenda.
The application to algebraic geometry/commutative algebra is using this formalism to build Gorenstein injective/projective/flat covers and envelopes.
| 3 | https://mathoverflow.net/users/310 | 17438 | 11,663 |
https://mathoverflow.net/questions/17437 | 5 | Let $a\_1, ... a\_n$ be real numbers. Consider the operation which replaces these numbers with $|a\_1 - a\_2|, |a\_2 - a\_3|, ... |a\_n - a\_1|$, and iterate. Under the assumption that $a\_i \in \mathbb{Z}$, the iteration is guaranteed to terminate with all of the numbers set to zero if and only if $n$ is a power of two. A friend of mine knows how to prove this, but wants to be able to reference a source where this problem (and/or its generalization to real numbers) is mentioned, and we can't figure out what search terms to use. Can anyone help us out?
(If someone can figure out a better title for this question, that would also be appreciated.)
| https://mathoverflow.net/users/290 | Does this problem have a name? [Ducci Sequences] | This is known as [Ducci's problem](http://en.wikipedia.org/wiki/Ducci_sequence). "Ducci map" or "Ducci sequence" as key words should let you search on most of the articles studying properties of the above map.
| 11 | https://mathoverflow.net/users/2384 | 17439 | 11,664 |
https://mathoverflow.net/questions/17409 | 10 | There are some 'standard' applications of the adjoint functor
theorem (AFT) and the special adjoint functor theorem (SAFT), for
example, the existence of a free $\tau$-algebra (where
$\tau=$(operations,identities)) on a small set by the AFT,
Stone-Cech compactification by the SAFT, and, if I am not mistaken, the proof that the category of $\tau$-algebras is cocomplete (by
using the AFT to establish a left adjoint to the appropriate diagonal
functor).
However, I was not able to find any applications of the *duals* of
the AFT and the SAFT, neither in MacLane, nor in the Joy of
Cats.
The Joy of Cats contains the following intriguing
remark on p. 311:
>
>
> >
> > Since many familiar categories have separators but
> > fail to have coseparators, the dual of the Special Adjoint Functor
> > Theorem is applicable even more often than the theorem itself.
> >
> >
> >
>
>
>
But what *are* the mentioned application of the dual of the SAFT?
So my questions is: **What are the 'standard' applications of the duals
of the AFT and the SAFT?**
Googling for combinations of phrases like ''adjoint functor theorem''
and ''dual'' is not very useful, so I have tried ''dual of the adjoint
functor theorem'' and ''dual of the special adjoint functor theorem.''
This resulted in a total of 7 papers/books, from which I was not able
to get a clear answer to the current question. I have also tried in
[The Wikipedia article on adjoint
functors](http://en.wikipedia.org/wiki/Adjoint_functors), in [nLab's
article on the adjoint functor
theorems](http://ncatlab.org/nlab/show/adjoint+functor+theorem), and
in [some](https://mathoverflow.net/questions/5786/how-do-i-check-if-a-functor-has-a-left-right-adjoint)
[MO](https://mathoverflow.net/questions/6376/why-forgetful-functors-usually-have-left-adjoint/6406#6406)
questions, but without success.
| https://mathoverflow.net/users/2734 | What are the 'standard' applications of the duals of the adjoint functor theorems? | One example is the construction of [geometric morphisms](http://ncatlab.org/nlab/show/geometric+morphism). Any colimit-preserving functor between Grothendieck toposes has a right adjoint, so if it also preserves finite limits, then it is part of a geometric morphism. Of course, in many cases in practice, the right adjoint is also easy to write down explicitly.
| 7 | https://mathoverflow.net/users/49 | 17445 | 11,669 |
https://mathoverflow.net/questions/17440 | 3 | We consider $n\times n$ complex matrices. Let $i\_+(A), i\_-(A), i\_0(A)$ be the number of eigenvalues of $A$ with positive real part, negative real part and pure imaginary. It is well known if two Hermitian matrices $A$ and $B$ are $\*-$congruent, then
$$(i\_+(A), i\_-(A), i\_0(A))=(i\_+(B), i\_-(B), i\_0(B)).\qquad{(1)}$$
If two general matrices $A$ and $B$ are $\*-$congruent, (1) may not hold (can you provide an example?).
Moreover, whether a matrix and its transpose are always $\*-$congruent?
| https://mathoverflow.net/users/3818 | A question on star-congruence. | An answer to the second question: Yes, a square complex matrix is always $\*$-congruent to its transpose, according to a more general result proved by Horn and Sergeichuk in "[Congruences of a square matrix and its transpose](http://arxiv.org/abs/0709.2489)". They prove the result for all fields with involution in characteristic other than 2.
*Added:*
Here's a counterexample for your first question:
$\left( \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right) \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right) \left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right) = \left( \begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix} \right)$.
| 4 | https://mathoverflow.net/users/1119 | 17446 | 11,670 |
https://mathoverflow.net/questions/17452 | 9 | I was wondering about the paper by Bernstein, Gel'fand, and Gel'fand on [Schubert Cells](http://iopscience.iop.org/0036-0279/28/3/R01;jsessionid=E4D2896AE55597FA4F4F63297F63FBC6.c3). This paper is fairly old(and often cited) so I figured someone must have represented this material. In particular, I was wondering if this was treated in an expository paper. More generally, I was wondering if there was a paper that explained the usefulness of the Schubert Calculus for representation theory, and even better one that talked about how Schubert Calculus came into the picture for BBD, again hopefully in an expository way.
Thanks in advance!
| https://mathoverflow.net/users/348 | Expository treatment of Schubert Cells Paper | The Lecture Notes in Mathematics number 1689, "Schubert Varieties and Degeneracy Loci" by Fulton and Pragacz seems to be exactly what you're looking for. I think chapter 6 is particularly relevant.
| 5 | https://mathoverflow.net/users/788 | 17455 | 11,675 |
https://mathoverflow.net/questions/17458 | 3 | I can calculate a likelihood ratio statistic to measure how well my data fits either of two mutually exclusive models:
$\Lambda({\text{data}}) = \frac{ f( \theta\_0 | {\text{data}} )}{f( \theta\_1 | {\text{data}} )} $
Calculating this statistic is straightforward to me, but I'd like some measure of statistical confidence to be carried along with the likelihood ratio. Is there something analogous to a confidence interval for a likelihood ratio? Or is all of the information about the relative confidence of each model?
| https://mathoverflow.net/users/4476 | Confidence intervals for likelihood ratios | Yes, in principle, the distribution function of the likelihood ratios statistic can be computed, since it is a function of the random variables "data" and the model parameters theta\_0 and theta\_1. Suppose, that cumulative distribution function is calculated, the p-value can be computed giving a confidence estimate.
One example where the distribution of a likelihood ratio is known is when the distribution function "f" is Gaussian (For example in the case of Anova). In this case the likelihood ratio is [F-distributed](https://en.wikipedia.org/wiki/F-test) (it is a ratio of two chi-squared random variables) and the p-values can be explicitely computed.
In the case where "f" is not Gaussian , in general there is no explicit expression for the distribution function of the likelihood ratio, however, asymptotic distributions in the case of large samples can be obtained, see for example the following article by: [S.S. Wilks](https://projecteuclid.org/journals/annals-of-mathematical-statistics/volume-9/issue-1/The-Large-Sample-Distribution-of-the-Likelihood-Ratio-for-Testing/10.1214/aoms/1177732360.full), where a chi squared asymptotic distribution of the log-likelihood is obtained.
| 2 | https://mathoverflow.net/users/1059 | 17459 | 11,676 |
https://mathoverflow.net/questions/17258 | 3 | The most appealing statement of the Bessis-Moussa-Villani conjecture is as follows:
>
> **Conjecture:** For all Hermitian positive semidefinite $n\times n$ matrices $A$ and $B$,
> and all positive integer $m$, the polynomial function
> $$t \in \mathbb{R}\mapsto g(t) \equiv tr[(A + t B)^m] =
> \sum\limits\_{
> k=0}^m
> a\_kt^k$$
> has only nonnegative coefficients $a\_k, k=1,\cdots,m$.
>
>
>
Most recent and past research concerns on the quantities $m$ and $n$. What about trying to prove the conjecture in the following way: first show that $a\_m\ge0$, $a\_0\ge0$, $a\_{m-1}\ge0$, $a\_1\ge0$, $a\_{m-2}\ge0$, $a\_2\ge0$. And then go on to show $a\_3\ge 0$ (and so is $a\_{m-3}\ge 0$)...
But it seems difficult to show $a\_3\ge 0$. Can anyone share some idea on this particular coefficient?
**UPDATED**
What is the largest term in $S\_{2m,m}(AB)$ ? More precisely, consider the word in two positive definite letters $A^{k\_1}BA^{k\_2}B\cdots A^{k\_m}B$ , where $(k1,k2,\cdots,k\_m)$ is a pair of nonnegative integer solution of $k\_1+k\_2+\cdots+k\_m=m$ . Is it true that $tr(A^{k\_1}BA^{k\_2}B\cdots A^{k\_m}B)\le tr(A^mB^m)$ ?
| https://mathoverflow.net/users/3818 | What's known about the 3rd coefficient in the BMV conjecture? | Look at <http://arxiv.org/abs/0802.1153>.
Here the fourth coefficient is shown to be non-negative which implies by
Hillar's decent theorem
<http://arxiv.org/abs/math/0507166>
also the third coefficient to be non-negative.
| 2 | https://mathoverflow.net/users/4479 | 17462 | 11,678 |
https://mathoverflow.net/questions/17472 | 3 | Consider $\mathbb{R}^n$ as measurable space with the Borel algebra. If $\mathbb{R}^n$ and $\mathbb{R}^m$ are isomorphic (in the category of measurable spaces, i.e. there are measurable maps in both directions, which are inverse to each other), can we conclude $n=m$? Note that this statement is stronger than the invariance of dimension in topology and I doubt that it is true. Can you give a counterexample?
| https://mathoverflow.net/users/2841 | Dimension of the measurable space $\mathbb{R}^n$ | All uncountable standard Borel spaces are Borel isomorphic (see A. S. Kechris, Classical Descriptive Set Theory, page 90, Theorem 15.6).
| 16 | https://mathoverflow.net/users/3096 | 17475 | 11,684 |
https://mathoverflow.net/questions/2147 | 128 | What are really helpful math resources out there on the web?
Please don't only post a link but a short description of what it does and why it is helpful.
Please only one resource per answer and let the votes decide which are the best!
| https://mathoverflow.net/users/1047 | Most helpful math resources on the web | I occasionally find [mathoverflow.net](https://mathoverflow.net/) rather helpful.
In particular, there's a good list of answers to your specific question [here](https://mathoverflow.net/questions/2147/most-helpful-math-resources-on-the-web).
| 168 | https://mathoverflow.net/users/35575 | 17476 | 11,685 |
https://mathoverflow.net/questions/17466 | 7 | Is there a notion (for schemes or just locally ringed spaces) of cohomology with compact support? I guess there is for algebraic schemes over $\mathbf{C}$, but what about schemes in general? Does anybody have a good reference?
| https://mathoverflow.net/users/4481 | Cohomology with compact support for coherent sheaves on a scheme | Yes there is. Take $S$ a scheme and take $f \colon X \to S$ a compactifiable morphism of schemes. By definition this means that there exists a proper $S$-scheme Y which contains $X$ as an open subscheme. Then, given a compactification and a sheaf on $X$, you may define the cohomology with proper support of this sheaf as the cohomology of the pushforward of your sheaf to $Y$.
I think this works for many different cohomologies, but you need to check that the compact support cohomology does not depend on the chosen compactification. At least for etale sheaves I know this is so.
| 8 | https://mathoverflow.net/users/4398 | 17494 | 11,698 |
https://mathoverflow.net/questions/17486 | 14 |
>
> Is it possible to construct smooth embedded of 2-discs $\Sigma\_1$ and $\Sigma\_2$ in $\mathbb R^3$ with the same boundary curve such that there is no pair of points $p\_1\in \Sigma\_1$ and $p\_2\in \Sigma\_2$ with parallel tangent planes?
>
>
>
**Comments:**
* The question is inspired by [this](https://mathoverflow.net/questions/16335); you will find there a construction of three such discs with no triples of points.
* More formally, *"the same boundary curve"* means that $\Sigma\_1=f\_1(D^2)$ and $\Sigma\_2=f\_2(D^2)$ for some smooth embedding $f\_1$ and $f\_2$ of the unit disc $D^2$
such that $f\_1|\_ {\partial D^2}\equiv f\_2|\_ {\partial D^2}$.
| https://mathoverflow.net/users/1441 | Two discs with no parallel tangent planes | Suppose two such disks Σ1 and Σ2 exist, and pull back TΣ2 to Σ1 by some homeomorphism. Viewed as a subbundle of TR3|Σ1, this plane bundle intersects TΣ1 in a line bundle L over Σ1 since no two tangent planes are parallel. Furthermore, L|∂Σ1 is exactly the bundle of lines tangent to ∂Σ1.
Since a line bundle over a disk is trivial, we can take a nonzero section of L, and thus we get a nonvanishing vector field on the disk Σ1 which is tangent to the boundary at every point of ∂Σ1. But then by doubling we can construct a nonvanishing vector field on S2, and this is impossible.
| 19 | https://mathoverflow.net/users/428 | 17497 | 11,700 |
https://mathoverflow.net/questions/17495 | 17 | Let $\mathbb{H}$ denote the quaternions. Let $G$ be a group, and define a *representation of $G$ on $\mathbb{H}^n$* in the natural way; that is, its a map $\rho:G\rightarrow Hom\_{\mathbb{H}-}(\mathbb{H}^n,\mathbb{H}^n)$ such that $\rho(gg')=\rho(g)\rho(g')$ (where $Hom\_{\mathbb{H}-}$ denotes maps as left $\mathbb{H}$-modules). Representations of algebras and Lie algebras can be defined in a similar way.
Any quaternionic representation of $G$/$R$/$\mathfrak{g}$ can restrict to a complex representation by choosing a $\gamma\in \mathbb{P}^1$ such that $\gamma^2=-1$, and using $\gamma$ to define a map $\mathbb{C}\hookrightarrow \mathbb{H}$. In this way, any quaternionic representation gives a $\mathbb{C}\mathbb{P}^1$-family of complex representations which parametrize the choice of $\gamma$. Furthermore, since every element in $\mathbb{H}$ is in the image of some inclusion $\mathbb{C}\hookrightarrow \mathbb{H}$, this family of complex representations determines the quaternionic representation.
This observation almost seems to imply that there is nothing interesting to say about quaternionic representations that wouldn't come up while studying complex representations. However, this is neglecting the fact that there might be interesting information in *how* the $\mathbb{C}\mathbb{P}^1$-family of complex representations is put together.
For instance, any finite group will have a discrete set of isomorphism classes of complex representations, and so any quaternionic representation will have all complex restrictions isomorphic. However, the quaternion group $\mathbf{Q}:= ( \pm1,\pm i, \pm j,\pm k)$ has an 'interesting' quaternionic representation on $\mathbb{H}$ (more naturally, it is a representation of the opposite group $\mathbf{Q}^{op}$ by right multiplication, but $\mathbf{Q}^{op}\simeq \mathbf{Q}$).
My question broadly is: **What other groups, rings and Lie algebras have quaternionic representations that are interesting (in some non-specific sense)?**
This question came up when I was reading [a paper of Kronheimer's](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=TI&pg6=PC&pg7=ALLF&pg8=ET&review_format=html&s4=kronheimer&s5=&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=28&mx-pid=1083440), where he describes a non-canonical hyper-Kahler structure on a coadjoint orbit of a complex semisimple Lie algebra. At any point in such a coadjoint orbit determines a representation of $\mathfrak{g}$ on the tangent space to the coadjoint orbit, which by the hyper-Kahler structure is naturally a quaternionic vector space. I wondered if this representation could be 'interesting', and then realized I had no real sense of what an interesting quaternionic representation would be.
| https://mathoverflow.net/users/750 | What rings/groups have interesting quaternionic representations? | Recall that if G is a group, k a field, and V\_k an irreducible representation of G over k then End\_G(V\_k) is a division algebra D over k. For example, if $V\_{\mathbb{R}}$ is a real representation then the endomorphism ring is $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$. Such representations are typically called *real*, *complex*, or *quaternionic*.
This usage is complimentary to your definition. Suppose that $V\_{\mathbb{R}}$ is a quaternionic representation, then $\mathbb{H}$ acts on $V\_{\mathbb{R}}$ on the right. Since $\mathbb{H}$ is a division ring, it follows that $V\_{\mathbb{R}} \cong \mathbb{H}^k$. Hence a real representation which is quaternionic is canonically a quaternionic representation in your sense.
Now Frobenius-Schur indicator theory explains how to determine which representations over $\mathbb{C}$ "come from" which kind of real representations. Explicitly, if $V\_{\mathbb{C}}$ is not selfdual then it plus its dual is the complexification of a representation over $\mathbb{R}$ with $D=\mathbb{C}$; if $V\_{\mathbb{C}}$ orthogonal (i.e. has an invariant bilinear form) then it is the complexification of a representation over $\mathbb{R}$ with $D=\mathbb{R}$; finally, if $V\_{\mathbb{C}}$ is symplectic then it plus itself is the complexification of a representation over $\mathbb{R}$ with $D=\mathbb{H}$.
In summary, given any symplectic representation $V\_{\mathbb{C}}$ the representation $V\_{\mathbb{C}} \oplus V\_{\mathbb{C}}$ is the complexification of a representation $W\_\mathbb{R}$ which is itself a module over $\mathbb{H}$. In particular, there is an "interesting" representation over $\mathbb{H}$ whose quaternionic dimension is half the complex dimension of $V\_{\mathbb{C}}$. Your example fits into this scheme, where you start with the 2-dimensional irrep of $\mathbf{Q}$. Another great example involves the symplectic Lie group which can also be realized as a unitary group over the quaternions ([see wikipedia](http://en.wikipedia.org/wiki/Symplectic_group)).
| 12 | https://mathoverflow.net/users/22 | 17500 | 11,702 |
https://mathoverflow.net/questions/17511 | 6 | I know a middle school math teacher looking for some suggestions for hands-on activities to teach the concept of real numbers. I'm new to this site, so this may be a little off topic.
| https://mathoverflow.net/users/4488 | Best way to teach concept of real numbers using a hands-on activity? | What do you mean exactly? Show them that not all numbers are rational? That not all numbers are algebraic? I think that you could explain historically how the Pythagoreans thought all numbers were rational. Then challenge them to find a number which is not rational. You might be surprised with the creativity you get. Maybe they know rational numbers have repeating decimal expansions and so they will invent one which does not repeat. Challenge them to explain why rational numbers have repeating decimal expansions. If you like geometry, you could lead them to the essential step in the geometric argument here:
<http://blog.plover.com/math/sqrt-2-new.html>
that the square root of two is irrational.
| 11 | https://mathoverflow.net/users/1106 | 17513 | 11,712 |
https://mathoverflow.net/questions/17501 | 46 | I was asked the following question by a colleague and was embarrassed not to know the answer.
Let $f(x), g(x) \in \mathbb{Z}[x]$ with no root in common. Let $I = (f(x),g(x))\cap \mathbb{Z}$, that is, the elements of $\mathbb{Z}$ which are linear combinations of $f(x), g(x)$ with coefficients in $\mathbb{Z}[x]$. Then $I$ is clearly an ideal in $\mathbb{Z}$. Let $D>0$ be a generator of this ideal. The question is: what is $D$?
Now, we do have the standard resultant $R$ of $f,g$, which under our hypotheses, is a non-zero integer. We know that $R \in I$ and it's not hard to show that a prime divides $R$ if and only if it divides $D$. I thought $R = \pm D$ but examples show that this is not the case.
| https://mathoverflow.net/users/2290 | The resultant and the ideal generated by two polynomials in $\mathbb{Z}[x]$ | This quantity $D$ is known as the "congruence number" or "reduced resultant" of the polynomials f and g. I first saw this in a preprint by Wiese and Taixes i Ventosa, <http://arxiv.org/abs/0909.2724>. They ascribe the concept to a paper which I don't have a copy of:
M. Pohst. *A note on index divisors*. In *Computational number theory (Debrecen, 1989)*, 173–
182, de Gruyter, Berlin, 1991.
| 22 | https://mathoverflow.net/users/2481 | 17514 | 11,713 |
https://mathoverflow.net/questions/17516 | 15 | (Warning: a student asking)
Let $E$ be an elliptic curve over $\mathbf Q$. Let $P(a,b)$ be a (nontrivial) torsion point on $E$. Is there an easy description of the ring of algebraic integers of $\mathbf Q(a,b)$? I'm curious about the answer for general elliptic curves, but I'm not sure whether such an answer is possible.
(This question is motivated by the nice description of the rings of integers of cyclotomic fields $\mathbf Q(\zeta\_n)$)
| https://mathoverflow.net/users/3777 | The ring of algebraic integers of the number field generated by torsion points on an elliptic curve | [Comment: what follows is not really an answer, but rather a focusing of the question.]
In general, there is not such a nice description even of the number field $\mathbb{Q}(a,b)$ -- typically it will be some non-normal number field whose normal closure has Galois group $\operatorname{GL}\_2(\mathbb{Z}/n\mathbb{Z})$, where $n$ is the order of the torsion point.
In order to maintain the analogy you mention above, you would do well to consider the special case of an elliptic curve with **complex multiplication**, say by the maximal order of an imaginary quadratic field $K = \mathbb{Q}(\sqrt{-N})$, necessarily of class number one since you want the elliptic curve to be defined over $\mathbb{Q}$. In this case, the
field $K(P)$ will be -- up to a multiquadratic extension -- the anticyclotomic part of the $n$-ray class field of $K$.
And now it is a great question exactly what the rings of integers of these very nice number fields are. One might even venture to hope that they will be integrally generated by the x and y coordinates of these torsion points on CM elliptic curves (certainly there are well-known integrality properties for torsion points, although I'm afraid I'm blanking on an exact statement at the moment; I fear there may be some problems at 2...).
I'm looking forward to a real answer to this one!
| 5 | https://mathoverflow.net/users/1149 | 17517 | 11,715 |
https://mathoverflow.net/questions/17523 | 5 | In "continuous" mathematics there are several important notions such as covering space, fibre bundle, Morse theory, simplicial complex, differential equation, real numbers, real projective plane, etc. that have a "discrete" analog: covering graph, graph bundle, discrete Morse theory, abstract simplicial complex, difference equation, finite field, finite projective plane, etc. I would like to know if there are others. But the real question is:
**Are there any important "continuous" mathematical concepts without "discrete" analog and vice versa?**
| https://mathoverflow.net/users/4400 | Are there any important mathematical concepts without discrete analog? | A lot of ideas from topology and analysis don't have obvious discrete analogues to me. At least, the obvious discrete analogues are vacuous.
* Compactness.
* Boundedness.
* Limits.
* The interior of a set.
I think a better question is which ideas have surprisingly interesting discrete analogues, like cohomology or scissors congruence.
| 4 | https://mathoverflow.net/users/2954 | 17537 | 11,726 |
https://mathoverflow.net/questions/17560 | 139 | I'm fascinated by this open problem (if it is indeed still that) and every few years I try to check up on its status. Some background: Let $x$ be a positive real number.
1. If $n^x$ is an integer for every $n \in \mathbb{N}$ then $x$ must be an integer. This is a fun little puzzle.
2. If $2^x$, $3^x$ and $5^x$ are integers then $x$ must be an integer. This requires fairly sophisticated tools, and can be derived from the results in e.g. Lang, Algebraic values of meromorphic functions. II., Topology 5, 1966.
3. Finally, if all you know is that $2^x$ and $3^x$ are integers, then as far as I know it is not known if $x$ is forced to be an integer (unbelievable, isn't it?). Although of course one can never be *certain*, I am quite sure this was still the case as recently as 2003.
So the question is, is that still an open problem, and is there any sort of relevant progress that may provide some hope?
| https://mathoverflow.net/users/25 | If $2^x $and $3^x$ are integers, must $x$ be as well? | Still open, to the best of my knowledge. The $2^x,3^x,5^x$ result follows from the [Six Exponentials Theorem](http://en.wikipedia.org/wiki/Six_exponentials_theorem), q.v., and the $2^x,3^x$ would follow from the Four Exponentials Conjecture, q.v.
| 39 | https://mathoverflow.net/users/3684 | 17564 | 11,743 |
https://mathoverflow.net/questions/17545 | 53 | Because the theory of sheaves is a functorial theory, it has been adopted in algebraic geometry (both using the functor of points approach and the locally ringed space approach) as the "main theory" used to describe geometric data. All sheaf data in the LRS approach can be described by bundles using the éspace étalé construction. It's interesting to notice that the sheafification of a presheaf is the sheaf of sections of the associated éspace étalé.
However, in differential geometry, bundles are for some reason preferred. Is there any reason why this is true? Are there some bundle constructions which don't have a realization as a sheaf? Are there advantages to the bundle approach?
| https://mathoverflow.net/users/1353 | Sheaves and bundles in differential geometry | If $X$ is a manifold, and $E$ is a smooth vector bundle over $X$ (e.g. its tangent bundle),
then $E$ is again a manifold. Thus working with bundles means that one doesn't have to leave
the category of objects (manifolds) under study; one just considers manifold with certain extra structure (the bundle structure). This is a big advantage in the theory; it avoids introducing another class of objects (i.e. sheaves), and allows tools from the theory of manifolds to be applied directly to bundles too.
Here is a longer discussion, along somewhat different lines:
The historical impetus for using sheaves in algebraic geometry comes from the theory of several complex variables, and in that theory sheaves were introduced, along with cohomological techniques, because many important and non-trivial theorems can be stated as saying that certain sheaves are generated by their global sections, or have vanishing higher cohomology. (I am thinkin of Cartan's Theorem A and B, which have as consequences many
earlier theorems in complex analysis.)
If you read Zariski's fantastic report on sheaves in algebraic geometry, from the 50s, you will see a discussion by a master geometer of how sheaves, and especially their cohomology,
can be used as a tool to express, and generalize, earlier theorems in algebraic geometry.
Again, the questions being addressed (e.g. the completeness of the linear systems of hyperplane sections) are about the existence of global sections, and/or vanishing of higher cohomology. (And these two usually go hand in hand; often one establishes existence results about global sections of one sheaf by showing that the higher cohomology of some related sheaf vanishes, and using a long exact cohomology sequence.)
These kinds of questions typically don't arise in differential geometry. All the sheaves
that might be under consideration (i.e. sheaves of sections of smooth bundles) have global sections in abundance, due to the existence of partions of unity and related constructions.
There are difficult existence problems in differential geometry, to be sure: but these are very often problems in ODE or PDE, and cohomological methods are not what is required to solve them (or so it seems, based on current mathematical pratice). One place where a sheaf theoretic perspective can be useful is in the consideration of flat (i.e. curvature zero) Riemannian manifolds; the fact that the horizontal sections of a bundle with flat connection form a local system, which in turn determines the
bundle with connection, is a useful one, which is well-expressed in sheaf theoretic language. But there are also plenty of ways to discuss this result without sheaf-theoretic language, and in any case, it is a fairly small part of differential geometry, since typically the curvature of a metric doesn't vanish, so that sheaf-theoretic methods don't seem to have much to say.
If you like, sheaf-theoretic methods are potentially useful for dealing with problems,
especially linear ones, in which local existence is clear, but the objects are suffiently rigid that there can be global obstructions to patching local solutions.
In differential geomtery, it is often the local questions that are hard: they become difficult non-linear PDEs. The difficulties are not of the "patching local solutions"
kind. There are difficult global questions too, e.g. the one solved by the Nash embedding theorem, but again, these are typically global problems of a very different type to those that are typically solved by sheaf-theoretic methods.
| 112 | https://mathoverflow.net/users/2874 | 17570 | 11,746 |
https://mathoverflow.net/questions/17468 | 23 | Grothendieck, before he disappeared, was working on a manuscript called "Les Derivateurs", which detailed the theory of derivators. Prof. Cisinski has done work with them as he mentioned in this [post](https://mathoverflow.net/questions/17425/homotopy-limits-over-fibered-categories/17463#17463). However, most of his work is in French, and I was wondering if there are any typed up references in english (the nLab has written notes from a seminar, but that's it). I intend to read the references in French later, but could someone explain or give a reference that explains in English the definition of a derivator and the motivation for them?
| https://mathoverflow.net/users/1353 | Derivators (in English) | For a few references in English, there are the papers of Heller, the main one being:
A. Heller, Homotopy theories, Mem. Amer. Math. Soc. 71 (383) (1988)
There is also a paper I wrote with A. Neeman, in which there is a little introduction to derivators in the second half of:
Additivity for derivator K-theory, Adv. Math. 217 (2008), no. 4, 1381-1475
One can see derivators in action in the work of G. Tabuada (he explains Bousfield localization and stabilization in this setting, and compares with the model category point of view):
Higher K-theory via universal invariants, Duke Math. J. 145 (2008), no. 1, 121–206
(availabe as arXiv:0706.2420).
| 20 | https://mathoverflow.net/users/1017 | 17589 | 11,760 |
https://mathoverflow.net/questions/17483 | 5 | Is there a definition of what is a 'free monoid' which does not pre-suppose that the natural numbers has already been defined? The definitions that I have been able to track down all use the natural numbers (since sequences/words need them).
In other words, I am trying to define the *theory* of free monoids (as a signature with sorts, operations and axioms), and I would like to know if I can do this without having the theory of the naturals already defined. For exhibiting/constructing a free monoid, I do expect to need Nat.
Note that my ambient logic is higher-order, so I am fine with a second-order axiomatization. If dependent-types are needed, that would also be acceptable.
| https://mathoverflow.net/users/3993 | Defining 'free monoid' without Nat? | As I pointed out in the comments, the theory of free monoids is somewhat ill defined. It is still unclear what your logic is and what you really want, but you have two basic options which were proposed by Pete Clark and sigfpe. Here are a few additional remarks that may help you sort things out.
*Universal Property à la Pete Clark.* The free monoid functor $F:Set\to Mon$ is left-adjoint to the forgetful functor $G:Mon \to Set$. This adjunction completely determines the isomorphism classes of free monoids. Moreover, many of the essential properties of free monoids follow directly from properties of sets and adjoint functors.
Alternately, the bijection between $Set(A,GB)$ and $Mon(FA,B)$ can be described as Pete Clark suggested, without any reference to adjoint functors. This latter option preferable if your logic can handle the categories $Set$ and $Mon$, but not functors between them.
The obvious advantage of this approach is that this is the true freeness property of free monoids (the fact that word monoids are free is nothing but an interesting accident from this point of view). Other than the fact that $F1$ is a natural number object, there is no mention of natural numbers here. However, I always thought that there was a lot of unnecessary tedium to show that free monoids can be viewed as word monoids in this context.
*Induction Axiom à la sigfpe.* The idea here is to attempt to capture freeness via induction. Here is a variant that fixes the minor problem I pointed out in a comment to sigfpe's answer -- free monoids are structures $(F,A,e,{\cdot})$ where $(F,e,{\cdot})$ is a monoid and $A \subseteq F$ (understood as a unary predicate) is a distinguished set of generators. The axioms to describe this setup are the usual monoid axioms together with the obvious modifications of sigfpe's two axioms
* $A(a) \land A(a') \land x\cdot a = x' \cdot a' \to x = x' \land a = a'$,
* $\forall P\,(P(e) \land \forall x,a\,(P(x) \land A(a) \to P(x\cdot a)) \to \forall x\,P(x))$,
where $P$ varies over all unary predicates $P \subseteq F$. In first-order logic, quantification over such $P$ is not possible and the induction axiom must be replaced by the corresponding induction scheme where $P$ varies over all formulas of the language instead. (In higher-order logics without predicate types, you can usually replace $P(x)$ by $f(x) = g(x)$ for appropriate $f$ and $g$.)
In the standard semantics for second-order logic, these axioms completely describe free monoids. However you must be careful since not all formulations of higher-logic are equal. Much like first-order logic, some of these formulations will implicitly allow for models of the above theory which are not free in the categorical sense. In truth, this is unlikely to be a problem for you but I am compelled to warn you of this possibility.
| 12 | https://mathoverflow.net/users/2000 | 17603 | 11,768 |
https://mathoverflow.net/questions/17614 | 105 | This question is of course inspired by the question [How to solve f(f(x))=cosx](https://mathoverflow.net/questions/17605/how-to-solve-ffx-cosx)
and Joel David Hamkins' answer, which somehow gives a formal trick for solving equations of the form $f(f(x))=g(x)$ on a *bounded* interval. [EDIT: actually he can do rather better than this, solving the equation away from a bounded interval (with positive measure)].
I've always found such questions ("solve $f(f(x))=g(x)$") rather vague because I always suspect that solutions are highly non-unique, but here are two precise questions which presumably are both very well-known:
Q1) Say $g:\mathbf{R}\to\mathbf{R}$ is an arbitrary function. Is there always a function $f:\mathbf{R}\to\mathbf{R}$ such that $f(f(x))=g(x)$ for all $x\in\mathbf{R}$?
Q2) If $g$ is as above but also assumed continuous, is there always a continuous $f$ as above?
The reason I'm asking is that these questions are surely standard, and perhaps even easy, but I feel like I know essentially nothing about them. Apologies in advance if there is a well-known counterexample to everything. Of course Q1 has nothing to do with the real numbers; there is a version of Q1 for every cardinal and it's really a question in combinatorics.
EDIT: Sergei Ivanov has answered both of these questions, and Gabriel Benamy has raised another, which I shall append to this one because I only asked it under an hour ago:
Q3) if $g$ is now a continuous function $\mathbf{C}\to\mathbf{C}$, is there always continuous $f$ with $f(f(x))=g(x)$ for all $x\in\mathbf{C}$?
EDIT: in the comments under his answer Sergei does this one too, and even gives an example of a continuous $g$ for which no $f$, continuous or not, can exist.
Related MO questions: [f(f(x))=exp(x) and other functions just in the middle between linear and exponential](https://mathoverflow.net/questions/4347/ffxexpx-1-and-other-functions-just-in-the-middle-between-linear-and-expo), and [Does the exponential function has a square root](https://mathoverflow.net/questions/12081/does-the-exponential-function-have-a-square-root).
| https://mathoverflow.net/users/1384 | solving $f(f(x))=g(x)$ | Q1: No. Let $g(0)=1, g(1)=0$ and $g(x)=x$ for all $x\in\mathbb R\setminus\{0,1\}$.
Assuming $f\circ f=g$, let $a=f(0)$, then $f(a)=1$ and $f(1)=g(a)=a$ since $a\notin\{0,1\}$.
Then $g(1)=f(f(1))=f(a)=1$, a contradiction.
Q2: No. Let $g(x)=-x$ or, in fact, any decreasing function $\mathbb R\to\mathbb R$. Then $f$ must be injective and hence monotone. Whether $f$ is increasing or decreasing, $f\circ f$ is increasing.
| 121 | https://mathoverflow.net/users/4354 | 17621 | 11,778 |
https://mathoverflow.net/questions/17615 | 12 | As the title says.
In particular, I am interested in the story for a general reductive group $G$, say defined over $\mathbb{Q}$. I can imagine that mod-$\ell$ (algebraic) automorphic representation should correspond to conjugacy classes of homomorphisms from the motivic Galois group into $\widehat{G}(\overline{\mathbb{F}\_\ell})$, but this does not seem like a very workable definition. Is there a more pragmatic approach?
| https://mathoverflow.net/users/1464 | Is there a canonical notion of "mod-l automorphic representation"? | This is in my mind a central open problem.
Here is an explicit example which I believe is still wide open. Serre's conjecture (the Khare-Wintenberger theorem) says that if I have a continuous odd irreducible $2$-dimensional mod $p$ representation of the absolute Galois group of the rationals then it should come from a mod $p$ modular form. We have a perfectly good definition of mod $p$ modular forms (sections of the usual line bundles on mod $p$ modular curves).
But what is the "even" analogue of Serre's conjecture?
One problem is that *in characteristic zero* (i.e., for $2$-dimensional continuous complex representations, which must have finite image) these guys are expected to come from algebraic Maass forms, which have, as far as anyone knows, no algebraic definition: they are genuinely analytic objects and it is a complete miracle that their Hecke eigenvalues are algebraic. So there's a big open problem: give an algebraic definition of an algebraic Maass form which doesn't use analytic functions on the upper half plane which are eigenvectors for the Laplacian with eigenvalue $1/4$. I have no idea how to do this and I don't think that anyone else has either. The problem is that the forms are not holomorphic (so you can't use GAGA and algebraic geometry) and they're not cohomological (so you can't use group cohomology either), and those are in some sense the only tricks we have (other than Langlands transfer, but I don't know any functorial transfer of these guys which (a) loses no information and (b) gives rise to a form which is cohomological or holomorphic). It might be an interesting PhD problem to check this out in fact. Blasius and Ramakrishnan once tried to go up to an im quad field and then to $\mathrm{Sp}\_4$ over $\mathbf{Q}$ but the resulting form isn't holomorphic. One might define an algebraic automorphic rep to be "accessible" if it's cohomological or perhaps limit of discrete series (perhaps these are the ones for which there's a chance of proving they're arithmetic), and then try and find examples of algebraic auto reps which should never transfer to an "accessible" rep on any other group.
But even after that, there's another problem, which is that as far as I know one does not expect a continuous even irreducible mod p representation of this Galois group to lift to a de Rham representation in characteristic zero (and indeed perhaps Frank Calegari might be able to give explicit counterexamples to this, after his recent observation that oddness sometimes follows from other assumptions). So even if one could give an algebraic definition of a Maass form, one wouldn't have enough Maass forms---they would all (at least conjecturally) give rise to Galois representations with images all of whose Jordan-Hölder factors are cyclic or $\mathrm{A}\_5$ (by the classification of finite subgroups of $\mathrm{PGL}\_2(\mathbf{C}))$. So there's another non-trivial sticking point.
In summary---nice question but I don't think that mathematics has a good answer yet (unless you're willing to just cheat and say that an automorphic representation "is" a Galois representation with some properties).
| 18 | https://mathoverflow.net/users/1384 | 17625 | 11,782 |
https://mathoverflow.net/questions/17626 | 0 | Let $G$ be a (complex algebraic) group, $H$ a subgroup, and ${\cal O}(G)$ and ${\cal O}(H)$ the coordinate algebras of complex regular functions of $G$ and $H$ respectively. Can ${\cal O}(H)$ ever embedded as a subalgebra of ${\cal O}(G)$? For example, I'm looking at $U(k-1)$ as a subalgebra of $SU(k)$ (embedded in the bottom right-hand corner with $1$ in the $u^1\_1$ entry and $0$ everywhere else) and can't an embedding of ${\cal O}(U(k-1)$) into ${\cal O}(SU(k)$).
| https://mathoverflow.net/users/4409 | Subgroup Groups and Coordinate Algebra Subalgebras | It may be possible in rare circumstances, but the natural thing is that $\mathcal{O}(H)$ is a quotient of $\mathcal{O}(G)$, not a subalgebra. The quotient map is dual to the inclusion $H\to G$.
I guess in the case that $G$ is a direct product $H \times K$, you get
$$ \mathcal{O}(G) \simeq \mathcal{O}(H) \otimes \mathcal{O}(K), $$
and in this case you have what you want. But in general, the arrow should go the other direction.
| 1 | https://mathoverflow.net/users/703 | 17627 | 11,783 |
https://mathoverflow.net/questions/17610 | 10 | Let $V$ be a finite dimensional highest weight representation of a (semi)-simple Lie algebra. For each $n\ge 0$ take $a\_n$ to be the dimension of the space of invariant tensors in $\otimes^n V$.
In certain cases there is a formula for $a\_n$. For example, for $V$ the two dimensional representation of $sl(2)$ we get $a\_n=0$ if $n$ is odd and for $n$ even we get the ubiquitous Catalan numbers. In general I don't expect a formula but the sequence does satisfy a linear recurrence relation with polynomial coefficients (known as D-finite).
For example, for the seven dimensional representation of $G\_2$ this sequence starts:
1, 0, 1, 1, 4, 10, 35, 120, 455, 1792, 7413, 31780, 140833, 641928, 3000361, 14338702, 69902535, 346939792, 1750071307, 8958993507, 46484716684, 244187539270, 1297395375129, 6965930587924
for more background see <http://www.oeis.org/A059710>
This satisfies the recurrence
$(n+5)(n+6)a\_n=2(n-1)(2n+5)a\_{n-1}+(n-1)(19n+18)a\_{n-2}+ 14(n-1)(n-2)a\_{n-3}$
**Question** How does one find these recurrence relations?
Then I also have a more challenging follow-up question. The space of invariant tensors in $\otimes^n V$ also has an action of the symmetric group $S\_n$ and so a Frobenius character which is a symmetric function of degree $n$.
**Question** How does one calculate these symmetric functions?
I know these can be calculated using plethysms individually. I am hoping for something along the lines of the first question.
**Further remarks** David's answer solves the problem theoretically but I want to make some remarks about the practicalities. This is in case anyone wants to experiment and also because I believe there is a more efficient method.
The $sl(2)$ example can easily be extended. For the $n$-dimensional representation $a\_k$ is the coefficient of $ut^k$ in
$$\frac{u-u^{-1}}{1-t\left(\frac{u^n-u^{-n}}{u-u^{-1}}\right)}$$
For the case $n=3$ see <http://www.oeis.org/A005043> and
<http://www.oeis.org/A099323>
I am not aware of any references for $n\ge 4$. I don't know if these are algebraic.
The limitation of this method is that there is a sum over the Weyl group. This means it is impractical to implement this method for $E\_8$. For the adjoint representation of $E\_8$ the start of the sequence is
1 0 1 1 5 16 79 421 2674 19244 156612 1423028 14320350
(found using LiE)
| https://mathoverflow.net/users/3992 | Can you find linear recurrence relation for dimensions of invariant tensors? | Finding the recurrence (and proving it is correct) can be done by the standard techniques for extracting the diagonal of a rational power series.
Let $\beta\_1$, $\beta\_2$, ..., $\beta\_N$ be the weights of $V$. Let $\rho$ be half the sum of the positive roots and $\Delta = \sum (-1)^{\ell(w)} e^{w(\rho)}$ be the Weyl denominator. Then
$$\sum\_{n=0}^{\infty} t^n \chi \left( V^{\otimes n} \right) = \frac{1}{1- \sum\_{i=1}^N t e^{\beta\_i}}$$
and
$$\sum\_{n=0}^{\infty} t^n \dim \left( V^{\otimes n} \right)^{\mathfrak{g}} = \mbox{Coefficient of}\ e^{\rho}\ \mbox{in} \ \left( \Delta \frac{1}{1- \sum\_{i=1}^N t e^{\beta\_i}} \right).$$
For example, if $\mathfrak{g}=\mathfrak{sl}\_2$ and $V$ is the two dimensional irrep, the right hand side is
$$ \mbox{Coefficient of}\ u \ \mbox{in} \left( \frac{(u-u^{-1})}{1-tu^{-1} - tu} \right)$$
which can be seen without too much trouble to be the generating function for Catalan numbers.
The diagonal of a rational generating function is $D$-finite by a [result of Lipshitz](http://www.ams.org/mathscinet-getitem?mr=929767). The particular recurrence can be found by [Sister Celine's method](http://www.ams.org/mathscinet-getitem?mr=647562) (see theorems 10 and 11). I found these references in Stanley, *Enumerative Combinatorics Vol. II*, solution to exercise 6.61. Stanley warns that there is a gap in Zeilberger's argument, but hopefully his algorithm is right.
| 8 | https://mathoverflow.net/users/297 | 17628 | 11,784 |
https://mathoverflow.net/questions/17638 | 16 | Let $p \equiv 1 \bmod 4$ be a prime number. Define the polynomial
$$ f(x) = \sum\_{a=1}^{p-1} \Big(\frac{a}{p}\Big) x^a. $$
Then $f(x) = x(1-x)^2(1+x)g(x)$ for some polynomial
$g \in {\mathbb Z}[x]$ (this follows from elementary properties
of quadratic residues).
For $p = 5$, we have $g = 1$; for $p = 13$, we find
$$ g(x) = x^8 + 2x^6 + 2x^5 + 3x^4 + 2x^3 + 2x^2 + 1. $$
pari tells me that the Galois group is "2^4 S(4)"
and has order $384 = 16 \cdot 24$.
My questions:
* Have these polynomials been studied anywhere? Since I did not just make them up, I am tempted to believe that they are natural enough to have shown up somewhere else.
* Is $g$ always irreducible? pari says it is for all p < 400.
| https://mathoverflow.net/users/3503 | Irreducibility of polynomials related to quadratic residues | These are known as Fekete polynomials:
<http://en.wikipedia.org/wiki/Fekete_polynomial> .
I don't know of any results on their Galois groups.
| 9 | https://mathoverflow.net/users/4213 | 17642 | 11,789 |
https://mathoverflow.net/questions/17126 | 3 | Let P and Q be simple polytopes such that P = Q ∩ H and let H be a halfspace with normal vector n. Let projn(e) denote the length of the projection of edge e onto vector n.
Consider the set E of edges of Q that cross through H, ie, edges with one endpoint in H and one outside H.
For any e ∈ E, does there always exist a projective transformation φ of P such that for every f ∈ E, projφ(n)(φ(e)) ≥ projφ(n)(φ(f))?
In other words, is there always a projective transformation making an arbitrary edge crossing H maximal with respect to H?
My first approach would be to choose φ to extend the endpoint of e as far as possible, but this does not appear to be sufficient to guarantee that projφ(n)(φ(e)) is maximized without adjusting other edges of Q.
| https://mathoverflow.net/users/2122 | Edge-maximizing projective transformation on polytopes | Let $e$ be the desired edge to be maximized. let point $a$ and $b$ be the endpoint of the edge. Let $a$ lie in $H$, $b$ outside of $H$. Let $e$ pass through the boundry of $H$, $B$ at $c$. Let $Q$ and $H$ lie in a space of dimension $n$ let this space be in a space of dimension $n+1$. Take point $a'$
which lies in this higher dimensional space and has the coordinates same as $a$
except that the coordinate in the extra dimension. Now recall that $B$ was the
boundary of $H$ and has dimension one less than $n$. Add the new coordinate to the $n-1$ dimensional hyperplane that intersects $H$ in $B$ this gives $n$ dimensional space $G$. then project $Q$ onto $G$ through point $a'$. This will map point $a$ to the point at infinity. So if we take the original $n$-dimensional space and roatate it to $G$ in the $n+1$ dimensional space, with the rotation keeping the hyperplane of dimension $n-1$ fixed then the image of $a$ will go to the point at infinity as the angle approaches 90 degrees. Furthermore the the image of $a$'s projection on $n$ will go to infinity and be greater than all the other projections on $n$ of the points of $Q$. Because of this the image of $e$
will maximize all edges of the image of $Q$ that do not contain $a$. For any edge f that passes through a in Q the image of e will maximize f, in fact $ac$ will have the same normal vector and the fact that f is inside q means that $ab$ will have the greater projection.
| 1 | https://mathoverflow.net/users/1098 | 17649 | 11,796 |
https://mathoverflow.net/questions/7664 | 3 | I am interested in the theory of reductive groups which is useful in the theory of automorphic forms. But the trouble boring me so long time is that I don't know the appropriate material for beginners or outsiders who wants to pave in this field and learn more about automorphic forms.
So my question is that what kind of introduction materials, books or papers, fits me as an introduction and for further reading?
Thanks in advance!
| https://mathoverflow.net/users/1930 | How do we study the theory of reductive groups? | Sit at a table with the books of Borel, Humphreys, and Springer. Bounce around between them: if a proof in one makes no sense, it may be clearer in the other. For example, Springer's book develops everything needed about root systems from scratch, and has lots of nice exercises relate to that stuff. On the other hand, Borel is better about systematically allowing general ground fields from early on (so one doesn't have to redo the proofs all over again upon discovering that it is a good idea to allow ground fields like $\mathbf{R}$, $\mathbf{Q}$, $\mathbf{F}\_ p$, and $\mathbf{F} \_p(t)$). Pay attention to the power of inductive arguments with centralizers and normalizers (especially of tori).
Unfortunately, none makes good use of schemes, which clarifies and simplifies many things related to tangent space calculations, quotients, and positive characteristic. (For example, the definition of central isogeny in Borel's book looks a bit funny, and if done via schemes becomes more natural, though ultimately equivalent to what Borel does.) So if some proofs feel unnecessarily complicated, it may be due to lack of adequate technique in algebraic geometry. (Everyone has to choose their own poison!) Waterhouse's book has nothing serious to say about reductive groups, but the theory of finite group schemes that he discusses (including Cartier duality and structure in the infinitesimal case) is very helpful for a deeper understanding isogenies between reductive groups in positive characteristic. The exposes in SGA3 on quotients and Grothendieck topologies (etale, fppf, etc.) are helpful a lot too (some of which is also developed in the book "Neron Models"). Galois cohomology is also useful when working with rational points of quotients.
| 22 | https://mathoverflow.net/users/3927 | 17651 | 11,798 |
https://mathoverflow.net/questions/17634 | 13 | Usually (eg, intro. in M. Rost's 'Cycle modules with coefficients'), for a variety, $X$, over a field one can define the Chow group of p-cycles, $CH\_p (X)$, as
$$CH\_p (X) = coker\; \left[\bigoplus\_{x\in X\_{p+1}} k(x)^\times \rightarrow \bigoplus\_{x\in X\_p} \mathbb{Z}\; \right]$$.
What about for an arithmetic scheme, eg when $X$ is, say, normal, separated, of finite type and flat over $Spec \; \mathbb{Z} $? Does something go wrong with the above definition?
Peter Arndt had posed part of this question already, but it seems without an answer.
| https://mathoverflow.net/users/4235 | Definition of Chow groups over Spec Z | One can define Chow groups over any Noetherian scheme $X$. Let $Z\_iX$ be the free abelian group on the $i$-dimensional
subvarieties (closed integral subschemes) of $X$. For any $i+1$-dimensional
subscheme $W$ of $X$, and a rational function $f$ on $W$, we can
define an element of $Z\_iX$ as follows:
$$ [div(f,W)] = \sum\_{V} ord\_V(f)[V] $$
summing over all codimension one subvarieties $V$ of $W$. Then the $i$-Chow group $CH\_i(X)$ is defined
as the quotient of $Z\_iX$ by the subgroup generated by all elements of the form $[div(f,W)]$. The order function is defined as length of the corresponding local ring, so it does not need any further assumptions on $X$, In fact, $X$ being locally Noetherian is enough. This definition also agrees with the one in the original question.
| 12 | https://mathoverflow.net/users/2083 | 17652 | 11,799 |
https://mathoverflow.net/questions/17653 | 15 | This question is related to this one: [Continued fractions using all natural integers](https://mathoverflow.net/questions/6222/continued-fractions-using-all-natural-integers). Suppose we have the set of natural numbers $N$ with order and we perform permutation on it. So we obtain the same elements with different order. Suppose we describe such permutations by usual notation when (1,2,3,4,5...) means identity permutation. Then **lets say** that permutation denoted by (1,3,2,4,5,6...) ( from the 4th place there is list of natural numbers in usual order) **is finite** because it only mixes numbers 1,2,3 -> 1,3,2 and for remaining elements it is identity permutation. As I find [here](https://mathoverflow.net/questions/1072/definition-of-infinite-permutations) there is definition of such objects, namely a few possibilities as states the [answer](https://mathoverflow.net/questions/1072/definition-of-infinite-permutations/1073#1073) of Qiaochu Yuan.
Questions:
1. Are infinite permutation decomposable into cycles? Transpositions?
2. **Is possible to find such permutation of natural numbers that it cannot be a limit of finite permutations?**
| https://mathoverflow.net/users/3811 | infinite permutations | The first thing to notice is that infinite permutations may have infinite support, that is, they may move infinitely many elements. Therefore, we cannot expect to express them as finite compositions of permutations having only finite support.
But if we allow (well-defined) infinite compositions, then the answer is that every permutation can be expressed as a composition of disjoint cycles and also expressed as a composition of transpositions. So the answer to question 1 is yes, and the answer to question 2 is no.
To see this, suppose that f is a permutation of ω. First, we may divide f into its disjoint orbits, where the orbit of n is defined as all the numbers of the form fk(n) for any integer k. The action of f on each of these orbits commute with each other, because the orbits are disjoint. And the action of f on each such orbit is a cycle (possibly infinite). So f can be represented as a product of disjoint cycles. For the transposition representation, it suffices to represent each such orbit as a suitable product of transpositions. The finite orbits are just finite cycles, which can be expressed as a product of transpositions in the usual way. An infinite orbit looks exactly like a copy of the integers, with the shift map. This can be represented in cycle notation as (... -2 -1 0 1 2 ...). This permutation is equal to the following product of transpositions:
* (... -2 -1 0 1 2 ...) = [(0 -1)(0 -2)(0 -3)...][...(0 3)(0 2)(0 1)]
I claim that every natural number is moved by at most two of these transpositions, and that the resulting product is well-defined. On the right hand side of the equality, I have two infinite products of transpositions. Using the usual order of product of permutations, the right-most factor is first to be applied. Thus, we see that 0 gets sent to 1, and subsequently fixed by all later transpositions. So the product sends 0 to 1. Similarly, 1 gets sent to 0 and then to 2, and then unchanged. Similarly, it is easy to see that every non-negative integer n is sent to 0 and then to n+1 as desired. Now, the right-hand factor fixes all negative integers, which then pass to the left factor, and it is easy to see that again -n is sent to 0 and then to -n+1, as desired. So altogether, this product is operating correctly. An isomorphic version of this idea can be used to represent the action of any infinite orbit, and so every permutation is a suitable well-defined product of transpositions, as desired.
Thus, the answers to the questions in (1) are yes, and the answer to question (2) is no.
| 10 | https://mathoverflow.net/users/1946 | 17655 | 11,801 |
https://mathoverflow.net/questions/17658 | 1 | Every closed immersion is a finite morphism. Can you give an example of quasi-projective varieties $X\subset Y$ such that inclusion $X\hookrightarrow Y$ is not finite? Same with Y projective?
Thanks!
**Edit:** Sorry this question is very simple, I made a mistake asking the question. For a corrected version, check out [this one.](https://mathoverflow.net/questions/17678/example-of-restriction-of-a-finite-morphism-which-is-not-finite)
| https://mathoverflow.net/users/3637 | Example of inclusion which is not a finite morphism | An open immersion is never finite unless it is also a closed immersion (for finite morphisms are proper). So you just need to take a non-empty open subset $X$ which is not a connected component in $Y$.
| 13 | https://mathoverflow.net/users/3485 | 17660 | 11,805 |
https://mathoverflow.net/questions/17594 | 9 | Let $G$ be a reductive group over an (say, algebraically closed) field $k$. Springer (in his book on algebraic groups) calls for a chosen maximal torus $T$ in $G$ a family $(u\_\alpha) \_{\alpha \in \Phi(G,T)}$ of immersions $u \_\alpha:\mathbf{G}\_a \rightarrow G$ such that
(i) $t u\_\alpha(c) t^{-1} = u\_\alpha( \alpha(t) c)$ for all $c \in \mathbf{G}\_a$ and $t \in T$,
(ii) $n\_\alpha := u\_\alpha(1) u\_{-\alpha}(-1) u\_\alpha(1)$ lies in $\mathrm{N}\_G(T) \setminus T$,
(iii) $u\_\alpha(x) u\_{-\alpha}(-x^{-1}) u\_\alpha(x) = \alpha^\vee(x) n\_\alpha$ for all $x \in k^\times$,
a *realization* of (G,T) (or $\Phi(G,T)$) in $G$. We then have $\mathrm{Im}(u\_\alpha) = U\_\alpha$.
In the book of Conrad-Gabber-Prasad on pseudo-reductive groups a *pinning* of $G$ is defined as a tuple $(T,\Phi^+,(\varphi\_\alpha)\_ {\alpha \in \Delta})$ where $T$ is a maximal torus, $\Phi^+$ is a positive system for $\Phi(G,T)$, $\Delta$ is the corresponding basis and $\varphi \_{\alpha}: (\mathrm{SL} \_2, \mathrm{SL} \_2 \cap \mathrm{D} \_2) \rightarrow (G \_\alpha, G \_\alpha \cap T)$ are central isogenies such that $\varphi \_\alpha( \mathrm{diag}(x,x^{-1}) ) = \alpha^\vee(x)$ for all $x \in k^\times$, where $G \_\alpha = \langle U \_\alpha,U \_{-\alpha} \rangle$.
My question is: are these two notions somehow equivalent? If a pinning is given, then by defining
$u\_\alpha(x) = \varphi\_\alpha\begin{pmatrix} 1 & x \\\\ 1 & 0 \end{pmatrix}$, $u\_{-\alpha}(x) = \varphi\_\alpha\begin{pmatrix} 1 & 0 \\\\ x & 1 \end{pmatrix}$
I get closed immersions satisfying the properties above, but unfortunately, as I have $\varphi\_\alpha$ only for $\alpha \in \Delta$ this does not yet define a realization. How can I define the $u\_\alpha$ for $\alpha \notin \Delta \cup -\Delta$? What about the other direction?
Moreover (as C-G-P also mentions) in SGA3, exposé XXIII, there is defined the notion of *épinglages* and Conrad mentions that these carry the same information as the pinnings above. Can somebody make this precise? Moreover in SGA, it is mentioned that an épinglage induces monomorphisms $p\_\alpha: \mathbf{G}\_a \rightarrow G$ for $\alpha \in \Delta \cup -\Delta$. I suspect that these are the morphisms I defined above, but again, can I get a realization from this?
A further problem is the following: For a given realization and a total order on $\Phi(G,T)$ Springer defines *structure constants* which appear in the expression of the commutator $\lbrack u\_\alpha(x), u\_\beta(y) \rbrack $ in terms of $u\_\gamma$ for linearly independent $\alpha, \beta \in \Phi$. Springer shows that for root systems NOT of type $G\_2$ a realization with integral structure constants exist. Demazure also calculates these commutators in SGA3, exposé XXII, for the $p\_\alpha$ mentioned above in case of rank 2 root systems. Here, I was surprised that the structure constants seem to be *independent* of the pinning chosen. Is this now a rank 2 phenomenon that is also true for realizations or does this mean that pinnings/épinglages are more restrictive than realizations?
I hope, somebody can help me here.
| https://mathoverflow.net/users/717 | Realizations and pinnings (épinglages) of reductive groups | OK, here's the deal.
I. First, the setup for the benefit of those who don't have books lying at their side. Let $(G,T)$ be a split connected reductive group over a field $k$, and choose $a \in \Phi(G,T)$ (e.g., a simple positive root relative to a choice of positive system of roots). Let $G\_a$ be the $k$-subgroup generated by the root groups $U\_a$ and $U\_{-a}$. (Recall that $U\_a$ is uniquely characterized as being a nontrivial smooth connected unipotent $k$-subgroup normalized by $T$ and on which $T$ acts through the character $a$.) This is abstractly $k$-isomorphic to ${\rm{SL}}\_ 2$ or ${\rm{PGL}}\_ 2$, and $G\_a \cap T$ is a split maximal torus.
So there is a central isogeny $\phi:{\rm{SL}}\_ 2 \rightarrow G\_a$ (either isomorphism or with kernel $\mu\_2$), and since ${\rm{PGL}}\_ 2(k)$ is the automorphism group of ${\rm{SL}}\_ 2$ and of ${\rm{PGL}}\_ 2$ over $k$ there is precisely this ambiguity in $\phi$ (via precomposition with its action on ${\rm{SL}}\_ 2$). The burning question is: to what extent can we use $T$ and $a$ to nail down $\phi$ uniquely?
The action of $G\_a \cap T$ on $U\_a$ is via the nontrivial character $a$, and among the two $k$-isomorphisms $\mathbf{G}\_ m \simeq G\_a \cap T$ the composition with this character is $t \mapsto t^{\pm 2}$ in the ${\rm{SL}}\_ 2$-case and $t \mapsto t^{\pm 1}$ in the ${\rm{PGL}}\_ 2$-case. Fix the unique such isomorphism making the exponent positive.
Now back to the central isogeny $\phi:{\rm{SL}}\_ 2 \rightarrow G\_a$. By conjugacy of split maximal tori (when they exist!) we can compose with a $G\_a(k)$-conjugation if necessary so that $\phi$ carries the diagonal torus $D$ onto $G\_a \cap T$. Recall that we used $a$ to make a preferred isomorphism of $G\_a \cap T$ with $\mathbf{G}\_ m$. The diagonal torus $D$ also has a preferred identification with $\mathbf{G}\_ m$, namely $t \mapsto {\rm{diag}}(t, 1/t)$. Thus, $\phi: D \rightarrow G\_a \cap T$ is an endomorphism of $\mathbf{G}\_ m$ with degree 1 or 2, so it is $t \mapsto t^{\pm 2}$ (${\rm{PGL}}\_ 2$-case) or $t \mapsto t^{\pm 1}$ (${\rm{SL}}\_ 2$-case). Since the standard Weyl element of ${\rm{SL}}\_ 2$ induces inversion on the diagonal torus, by composing with it if necessary we can arrange that $\phi$ between these tori uses the positive exponent. That is exactly the condition that $\phi$ carries the standard upper triangular unipotent subgroup $U^+$ onto $U\_a$ (rather than onto $U\_{-a}$).
II. So far, so good: we have used just $T$ and the choice of $a \in \Phi(G,T)$ to construct a central isogeny $\phi\_a:{\rm{SL}}\_ 2 \rightarrow G\_a$ carrying $U^+$ onto $U\_a$ and $D$ onto $G\_a \cap T$, with the latter described uniquely in terms of canonically determined identifications with $\mathbf{G}\_ m$ (as $t \mapsto t$ or $t \mapsto t^2$). The remaining ambiguity is precomposition with the action of elements of ${\rm{PGL}}\_ 2(k)$ that restrict to the identity on $D$, which is to say the action of $k$-points of the diagonal torus $\overline{D}$ of ${\rm{PGL}}\_ 2$. Such action restrict to one on $U^+$ that identifies $\overline{D}(k)$ with the $k$-automorphism group of $U^+$ (namely, $k^{\times}$ with its natural identification with ${\rm{Aut}}\_ k(\mathbf{G}\_ a)$).
Summary: to nail down $\phi$ uniquely it is equivalent to specify an isomorphism of $U^+$ with $\mathbf{G}\_ a$. But $\phi$ carries $U^+$ *isomorphically* onto $U\_a$. So it is the same to choose an isomorphism of $U\_a$ with $\mathbf{G}\_ a$. Finally, the Lie functor clearly defines a bijection $${\rm{Isom}}\_ k(\mathbf{G}\_ a, U\_ a) \simeq
{\rm{Isom}}(k, {\rm{Lie}}(U\_ a)).$$
So we have shown that to specify a pinning in the sense of Definition A.4.12 of C-G-P is precisely the same as to specify a pinning in the sense of SGA3 Exp. XXIII, 1.1.
III. Can we improve a pinning to provide unambiguous $\phi\_c$'s for all roots $c$? No, there is a discrepancy of units which cannot be eliminated (or at least not without a tremendous amount of work, the value of which is unclear, especially over $\mathbf{Z}$), and if we insist on working over $k$ and not $\mathbf{Z}$ then there are further problems caused by degeneration of commutation relations among positive root groups in special cases in small nonzero characteristics (e.g., ${\rm{G}}\_ 2$ in characteristic 3 and ${\rm{F}}\_ 4$ in characteristic 2).
As we saw above, to nail down each $\phi\_c$ it is equivalent to do any of the following 3 things: fix an isomorphism $\mathbf{G}\_ a \simeq U\_c$, fix a basis of ${\rm{Lie}}(U\_ c)$, or fix a central isogeny ${\rm{SL}}\_ 2 \rightarrow G\_c$ carrying $D$ onto $G\_c \cap T$ via $t \mapsto t$ or $t \mapsto t^2$ according to the canonical isomorphisms of $\mathbf{G}\_ m$ with these two tori (the case of $G\_c \cap T$ being determined by $c$). This latter viewpoint provides $\phi\_{-c}$ for free once $\phi\_c$ has been defined (compose with conjugation by the standard Weyl element), so the problem is to really define $\phi\_c$ for $c \in \Phi^+$.
Consider the unipotent radical $U$ of the Borel corresponding to $\Phi^+$, so $U$ is directly spanned (in any order) by the $U\_c$'s for positive $c$. If we choose an enumeration of $\Phi^+$ to get an isomorphism $\prod\_c U\_c \simeq U$ of varieties via multiplication, then for simple $c$ we have a preferred isomorphism of $U\_c$ with $\mathbf{G}\_ a$ and one can ask if the isomorphism $\mathbf{G}\_ a \simeq U\_c$ can be determined for the other positive $c$ so that the group law on $U$ takes on an especially simple form. This amounts to working out the commutation relations among the $U\_a$'s for $a \in \Delta$ (when projected in $U\_c$ for various $c$), and by $T$-equivariance such relations will involve monomials in the coordinates of the $U\_a$'s along with some coefficients in $k^{\times}$ (and some coefficients of 0). These are the confusing "structure constants". Chevalley developed a procedure over $\mathbf{Z}$ to make the choices so that the structure constants come out to be positive integers (when nonzero), but there remained some ambiguity of signs since
$${\rm{Aut}}\_ {\mathbf{Z}}(\mathbf{G}\_ a) = \mathbf{Z}^{\times} = \{\pm 1\}.$$
Working entirely over $k$, there are likewise $k^{\times}$-scalings that cannot quite be removed. I am told that Tits developed a way to eliminate all sign confusion, but honestly I don't know a reason why it is worth the heavy effort to do that. For most purposes the pinning as above is entirely sufficient, and this is "justified" by the fact that all ambiguity from $(T/Z\_G)(k)$-action is eliminated in the Isomorphism Theorem when improved to include pinnings (see Theorem A.4.13 in C-G-P).
IV. What about the realizations in the sense of Springer's book? All he's doing is making use of the concrete description of the group law on ${\rm{SL}}\_ 2$ to describe a central isogeny $\phi\_c$ for *every* positive root $c$. (His conditions relate $c$ with $-c$, which amounts to the link between $\phi\_c$ and $\phi\_{-c}$ in the preceding discussion.) As long as he restricts to $\alpha \in \pm \Delta$ then he's just defined a pinning in the above sense. But he goes further to basically do what is described in II but without saying so explicitly. He then has to confront the puzzle of the structure constants. (It is a real puzzle, since in the theory over $\mathbf{Z}$, which logically lies beyond the scope of his book, the structure constants are *not* always $0$ and $\pm 1$; in some rare cases one gets coefficients of $\pm 2$ or $\pm 3$, which implies that if one insists on working over fields and not over $\mathbf{Z}$ then life in characteristic 2 and 3 will look a bit funny in some cases.) The only conceptual way I know of to overcome the puzzle of the structure constants is to work over $\mathbf{Z}$ and to follow either SGA3 or Chevalley in this respect. For the former, one has to really do the whole theory of reductive groups over a base that is not necessarily a field. For the latter, perhaps the (unpublished?) Yale notes of Steinberg are the best reference.
| 17 | https://mathoverflow.net/users/3927 | 17673 | 11,815 |
https://mathoverflow.net/questions/17590 | 2 | Let $\phi:\mathbb{P}^1\to\mathbb{P}^1$ be a rational function of degree $d\geq2$. How can one prove, using the normalized spherical measure, that
$$\int\_{\mathbb{P}^1(\mathbb{C})}|(\phi^n)'(z)|\ d\mu (z) \sim d^n$$ as $n\to\infty$?
| https://mathoverflow.net/users/nan | Help determining the asymptotic behavior of an integral involving rational functions. | If I understand your problem correctly, i.e. that $\mu$ is the [usual metric on the Riemann sphere](http://en.wikipedia.org/wiki/Riemann_sphere#Metric), then you're asking if essentially all orbits are expansive, at least with respect to that metric.
This will almost always be the case. For example, it will be so whenever the Julia set is holomorphically removable. The only case which is doubtful is when the Julia set is the whole of the Riemann sphere and the forward orbit of the critical point(s) are dense in the sphere -- which can happen [there are nice non-Lattes examples of this by M. Rees, I think the paper dates from 1984 or so]. It could still be the case that that is not enough to dampen the integral, if return times to a given neighborhood are too long.
I have a sneaky feeling that this might happen only in the case of an invariant line field -- and a little Googling brings up a [recent paper](http://journals.cambridge.org/action/displayAbstract;jsessionid=1D8D07DE0E8B003AB94A03D517FA2623.tomcat1?fromPage=online&aid=6622112) by Xiaoguang Wang showing that this (essentially) doesn't happen.
[Take the conjectures with a grain of salt, last time I did complex dynamics was 13 years ago!]
| 1 | https://mathoverflow.net/users/3993 | 17676 | 11,818 |
https://mathoverflow.net/questions/17526 | 7 | Hello all, if $a\_1,a\_2, \ldots a\_t$ are $t$ integers $\geq 2$, the set
$G(a\_1,a\_2, \ldots a\_t)=\lbrace N \geq 1 |$ In any sequence of $N$ consecutive
integers there is at least one not divisible by any of $a\_1,a\_2, \ldots a\_t\rbrace$
is nonempty (it contains $a\_1a\_2 \ldots a\_t$) so it has a minimal element
which we denote by $g(a\_1,a\_2, \ldots a\_t)$.
Question 1 : Is there a uniform bound $\gamma (t)$, depending
only on $t$, such that $\gamma (t) \geq g(a\_1,a\_2, \ldots a\_t)$ for any
$a\_1,a\_2, \ldots a\_t$ ? For example, we may take $\gamma(2)=4$.
Question 2 : If $\gamma$ is well-defined,
are any asymptotics known about $\gamma(t)$ ?
| https://mathoverflow.net/users/2389 | Smallest integer not divisible by integers in a finite set | Given an integer $n$, the Jacobsthal function $g(n)$ is the least integer, so that among any $g(n)$ consecutive integers $a,a+1,\dots,a+g(n)-1$ there is at least one that is coprime to $n$. Let $\nu(n)$ count the distinct prime factors of $n$. You can define $$C(r)=\max\_{\nu(n)=r} g(n)$$ and as Jonas Meyer points out in the comments this is precisely $C(t)=\gamma (t)$ (i.e. it is enough to consider when all $a\_i$ are prime).
For the bounds $$\frac{c\_1t (\log t)^2 \log \log \log t}{(\log\log t)^2}\le C(t)\le c\_2 t^{c\_3}$$
see the paper "On the integers relatively prime to n and on a number-theoretic function considered by Jacobsthal"" by Erdos. I don't know if there are better bounds.
| 6 | https://mathoverflow.net/users/2384 | 17682 | 11,820 |
https://mathoverflow.net/questions/17617 | 52 | I'm merely a grad student right now, but I don't think an exploration of the sporadic groups is standard fare for graduate algebra, so I'd like to ask the experts on MO. I did a little reading on them and would like some intuition about some things.
For example, the order of the monster group is over $8\times 10^{53}$, yet it is simple, so it has no
normal subgroups...how? What is so special about the prime factorization of its order? Why is it $2^{46}$ and not $2^{47}$? Why is it not possible to extend it to obtain that additional power of 2 without creating a normal subgroup? Some of the properties seem really arbitrary, and yet must be very fundamental to the algebra of groups.
I don't think I'm the only person curious about this, but I hesitated posting due to my relative inexperience.
| https://mathoverflow.net/users/1876 | Why are the sporadic simple groups HUGE? | The question seems to be made of several smaller questions, so I'm afraid my answer may not seem entirely coherent.
I have to agree with the other posters who say that the sporadic simple groups are not really so large. For example, we humans can write down the full decimal expansions of their orders, where a priori one might think we'd have to resort to crude upper bounds using highly recursive functions. (In contrast, one could say that almost of the groups in the infinite families are too large for their orders to have a computable description that fits in the universe.) Furthermore, as of 2002 we can load matrix representatives of elements into a computer, even for the monster. Noah pointed out that the monster has a smaller order than $A\_{50}$, but I think a more apt comparison is that the monster has a smaller order than even the smallest member of the infinite $E\_8$ family. Of course, one could ask why $E\_8$ has dimension as large as 248...
There was a more explicit question: how is it possible that a group with as many as $8 \times 10^{ 53 }$ elements doesn't have any normal subgroups? I think the answer is that the order of magnitude of a group says very little about its complexity. There are prime numbers very close to the order of the monster, and there are simple cyclic groups of those orders, so you might ask yourself why that fact doesn't seem as conceptually disturbing. Perhaps slightly more challenging is the fact that there aren't any elements of order greater than 119, but again, there is work on the [bounded and restricted Burnside problems](http://en.wikipedia.org/wiki/Burnside%27s_problem) that shows that you can have groups of very small exponent that are extremely complicated.
A second point regarding the large lower bound on order is that there are smaller groups that could be called sporadic, in the sense that they fit into reasonably natural (finite) combinatorial families together with the sporadics, but they aren't designated as sporadic because small-order isomorphisms get in the way. For example, the Mathieu group $M\_{10}$ is the symmetry group of a certain Steiner system, much like the simple Mathieu groups, and it is an index 11 subgroup of $M\_{11}$. While it isn't simple, it contains $A\_6$ as an index 2 subgroup, and no one calls $A\_6$ sporadic. Similarly, we describe the 20 "happy family" sporadic subquotients of the monster, but we forget about the subquotients like $A\_5$, $L\_2(11)$, and so on. Since the order of a nonabelian simple group is bounded below by 60, there isn't much room to maneuver before you get to 7920, a.k.a. "huge" range.
The question about the why the 2-Sylow subgroup has a certain size is rather subtle, and I think a good explanation would require delving into the structure of the classification theorem. A short answer is that centralizers of order 2 elements played a pivotal role in the classification after the Odd Order Theorem, and there was a separation into cases by structural features of centralizers. One of the cases involved a centralizer that ended up having the form $2^{1 + 24} . Co1$, which has a 2-Sylow subgroup of order $2^{46}$ (and naturally acts on a double cover of the Leech lattice). This is the case that corresponds to the monster.
Regarding the prime factorization of the order of the monster, the primes that appear are exactly the supersingular primes, and this falls into the general realm of "monstrous moonshine". I wrote a longer description of the phenomenon in reply to [Ilya's question](https://mathoverflow.net/questions/1269/what-does-supersingular-mean), but the question of a general conceptual explanation is still open.
I'll mention some folklore about the organization of the sporadics. There seems to be a hierarchy given by
* level 0: subquotients of $M\_{24}$ = symmetries of the Golay code
* level 1: subquotients of $Co1$ = symmetries of the Leech lattice, mod $\{ \pm 1 \}$
* level 2: subquotients of the monster = conformal symmetries of the monster vertex algebra
where the groups in each level naturally act on (objects similar to) the exceptional object on the right. I don't know what explanatory significance the sequence [codes, lattices, vertex algebras] has, but there are some level-raising constructions that flesh out the analogy a bit. One interesting consequence of the existence of level 2 is that for some finite groups, the most natural (read: easiest to construct) representations are infinite dimensional, and one can reasonably argue using lattice vertex algebras that this holds for some exceptional families as well. John Duncan has some recent work constructing structured vertex superalgebras whose automorphism groups are sporadic simple groups outside the happy family.
I think one interesting question that has not been suggested by other responses (and may be too open-ended for MO) is why the monster has no small representations. There are no faithful permutation representations of degree less than $9 \times 10^{ 19 }$ and there are no faithful linear representations of dimension less than 196882. Compare this with the cases of the numerically larger groups $A\_{50}$ and $E\_8(\mathbb{F}\_2)$, where we have linear representations of dimension 49 and 248. This is a different sense of hugeness than in the original question, but one that strongly impacts the computational feasibility of attacking many questions.
| 77 | https://mathoverflow.net/users/121 | 17696 | 11,829 |
https://mathoverflow.net/questions/17692 | 9 | Is there any way to define the orientation of an orientable smooth manifold using sheaves (when our smooth manifold is viewed as a locally ringed space) without our definition being overly contrived?
| https://mathoverflow.net/users/1353 | Orientation of a smooth manifold using sheaves | In the study of (finite-dimensional?) paracompact and locally compact (?) spaces there is Verdier's topological duality theorem, expressed in terms of a dualizing complex (which is built up from a sheafification process using duals of compactly-supported cohomologies of open subspaces, or something like that). It is pure topology, having nothing to do with ringed spaces (just like the orientation sheaf!). In the special case of smooth (paracompact) manifolds, this recovers the orientation sheaf up to a shift on each connected component. It is analogous to the fact that the super-abstract dualizing complexes in Grothendieck duality for (quasi-)coherent cohomology collapses to the old friend "top-degree relative differentials" (up to shifting) in the smooth case.
But that's all just fancy mumbo-jumbo which puts the orientation sheaf into a broader perspective (like many duality theories for cohomologies). This does not qualify as a good way to initially "define" the orientation sheaf, much as appealing to Grothendieck duality would be a strange (and even circular, from some viewpoints) way to "define" top-degree relative differentials in the smooth case. To get a real theorem out, we have to put some content in.
It seems more illuminating at a basic level to understand how the orientation sheaf is constructed using punctured neighborhoods along the lines of Emerton's comment or the oriented double cover as in David Roberts' answer, and how one can remove some orientability hypotheses in some classical results on "constant coefficient" cohomology by instead allowing coefficients in the locally constant sheaf given by the orientation sheaf. And likewise to understand why the constant sheaf associated to $\mathbf{Z}(1) = \ker(\exp)$ has $n$th tensor power that serves as an orientation sheaf on a complex manifold of dimension $n$ (and so the *natural* isomorphism $\mathbf{C}(1) = \mathbf{C}$ via multiplication explains the absence of needing to choose orientations for various cohomological calculations on complex manifolds (very relevant if one is to have a hope to translating things into algebraic geometry).
| 9 | https://mathoverflow.net/users/3927 | 17700 | 11,830 |
https://mathoverflow.net/questions/17678 | 2 | Every closed immersion is a finite morphism. Therefore, restriction of a finite morphism to a closed subset is always a finite morphism itself. Can you give an example of quasi-projective varieties $X\subset Y$, $Z$ and a finite morphism $f:Y\to Z$ such that restriction $f:X\to f(X)$ is not finite? Same with Y -- projective?
PS. Sorry the [original version](https://mathoverflow.net/questions/17658/example-of-inclusion-which-is-not-a-finite-morphism) of this question was hilariously stupid.
| https://mathoverflow.net/users/3637 | Example of restriction of a finite morphism which is not finite | Almost the same counterexample works. Take any non-closed (so non-finite) open immersion $U\hookrightarrow Z$. Then the trivial double cover $Z\sqcup Z\to Z$ is finite, but the restriction to $U\sqcup Z\to Z$ is not (but is still surjective).
| 2 | https://mathoverflow.net/users/1 | 17701 | 11,831 |
https://mathoverflow.net/questions/17709 | 4 | This is something I am stuck on (it might well be trivial- in which case this is an embarassing question):
Let V be a dimension r vector space over Fp, the field with p prime elements (I also care about this when V is over Zn with n composite). Let t be a **given** automorphism of V of order q (prime different from p, although I am also interested in more general cases), meaning tq=1. **edit**: It is given that for any non-zero vector v in V, the elements of the set {ti v} together generate V, where i is between 1 and q. This implies that 1-t is also an automorphism of V. (thanks David!)
> **Question**: What is the order of 1-t? When is it q?
The context is representations of commutator subgroups of knot groups onto vector spaces. Here t is induced by the deck transformations of the infinite cyclic cover.
| https://mathoverflow.net/users/2051 | Order of "one minus automorphism" | Perhaps the answer is just that the order is what it is, and that sometimes it's q. Do you have a compelling reason to believe that there is any more structure to your question than that?
Here's an example of the situation. $V$ could be the finite field $F$ with $p^r$ elements, and $q$ could be a prime divisor of $p^r-1$. Now $F^\times$ is cyclic of order $p^r-1$ so will contain elements of order $q$. Let $t$ be such an element. Multiplication by $t$ gives an endomorphism of $V$. The minimal polynomial of $t$ over the prime subfield will have degree dividing $r$, and if $q$ doesn't divide $p^s-1$ for any $s$ less than $r$ then it will have degree equal to $r$. So if $q\geq r$ and $q$ divides $p^r-1$ but not $p^s-1$ for any $s$ less than $r$, and such primes are easy to find (for example take $r$ to be 1---or let $r$ be prime; then for any given $p$ there is almost always a $q$) then the conditions of your question are satisfied (this "degree $r$" business is just making sure the vectors span). Now you're asking what the order of $1-t$ is, and it will be some divisor of $p^r-1$ and it might be $q$ and it might not be.
For example let's choose everyone's favourite prime $p=691$, let $r-1$, let $q=23$ be the biggest divisor of $p-1$, and let's consider the 22 primitive 23rd roots of unity in $\mathbf{Z}/691\mathbf{Z}$. For each of these $t$, let's ask if 1-t is also a 23rd root of 1. The answer is a resounding "sometimes" (namely when $t=20$ or 672).
| 6 | https://mathoverflow.net/users/1384 | 17724 | 11,842 |
https://mathoverflow.net/questions/17722 | 4 | For any $\alpha \in \mathbb{R}$ and a parameter $Q$, we can write $\alpha = a/q + \theta$, for integers $a, q$ with $q \leq Q$, and real $\theta$ with $|\theta|\leq (qQ)^{-1}$, a simple application of the Dirichlet approximation theorem. I'm looking for a similar statement for number fields.
**Setup**: $K$ is a fixed number field of degree $ n $ over $ \mathbb{Q} $, with ring of integers $O\_K$. $\omega\_1, \dots , \omega\_n$ is a fixed $\mathbb{Z}$-basis for $O\_K$.
$\sigma\_i, \dots \sigma\_{n}$ are the distinct embeddings of $K$.
$V$ is the $n$- dimensional commutative $\mathbb{R}$-algebra $K \otimes\_\mathbb{Q} \mathbb{R}$.
We define a distance function $| \cdot |$ on $V$ as follows:
$$|x| = |x\_1 \omega\_1 + \cdots + x\_n \omega\_n| = \max\limits\_{i} | x\_i |.$$ I would just like to point out that I am not necessarily attached to this distance function, if you can say anything sensible using another distance function, then I am interested.
**Precise Statement**: Given $\alpha \in V$ and a parameter $Q$, is it always possible to find $\lambda, \mu \in O\_K$, such that $|\mu| \ll Q$ and $$|\alpha - \dfrac{\lambda}{\mu}| \ll \dfrac{1}{Q |\mu|}?$$
**Equivalent Statement**: can we find $\gamma \in K$ such that $\mathcal{N}=\textrm{N}(\bf{a}\_\gamma) \ll Q^n$, and $$|\alpha - \gamma| \ll \dfrac{1}{Q \mathcal{N}^{1/n}}, $$ where $\bf{a}\_\gamma$ is the denominator ideal of $\gamma$?
Note that it is easy to find an analogous statement to Dirichlet's original theorem, ie $\exists \lambda, \mu \in O\_K$, such that $|\mu| \ll Q$ and $$|\alpha \mu - \lambda| \ll \dfrac{1}{Q},$$ by an application of the pigeonhole principle, but unlike in $\mathbb{R}$, we cannot just divide through by $\mu$ at this point, as the only decent bound (that I know of) for $|\mu^{-1}|$ is $$|\mu^{-1}| \ll \dfrac{|\mu|^{n-1}}{\textrm{N}(\mu)}$$.
Does anyone have a reference for dealing with the fractional form like this? The closest I have managed to find was a generalisation to number fields of the Thue-Siegel-Roth theorem by LeVeque.
| https://mathoverflow.net/users/4426 | Dirichlet Approximation over a Number Field | Schmidt, Wolfgang M.
Diophantine approximation.
Lecture Notes in Mathematics, 785. Springer, Berlin, 1980.
| 3 | https://mathoverflow.net/users/2290 | 17725 | 11,843 |
https://mathoverflow.net/questions/17745 | 9 | Next semester I will teach an elementary statistic course for the first time (which I am actually quite excited about). A brief description can be found [here](https://www.math.ku.edu/cgi/hb/index.cgi?action=coursedetail&number=365&title=Elementary%20Statistics). I am told to expect very little math background from the students. The class size is about 40-50, meeting twice a week , each class lasts about one and half hour. I do have a graduate student helping with grading and recitation.
My first question is about textbook. Please let me know of suitable texts which are:
1) Fun to teach and learn from, especially for students with less background. I would like a book with lot of interesting, engaging, probably more exotic examples ( so less diseases and more, say, gambling! ).
2) Relatively cheap (preferably less than 100 USD). In this economy we don't want to make students pay too much. Most texts I came across seem quite expensive. (Unless if the text is really a treasure, then one can worry less about price.)
The second question is about the best way to teach such a course. My instinct is that a heavy lecture style would not work very well, especially since the class is one and half hour long. *Do you know ways (preferably with references, especially visual references) to teach this materials without too much lecturing?*
Thanks in advance.
(I did not make this community wiki since I want to reward the best answer, and also I am not sure there will be many equally good answers given how much I wanted. Let me know if you disagree, I am open to change on this issue.)
EDIT: The student body will be quite varied. I am told there will be psychology majors and some engineering students, but the majority would prefer examples to proofs.
| https://mathoverflow.net/users/2083 | How to teach introductory statistic course to students with little math background? | A bit of background: a few years ago, I designed such a course, after noticing that many of our social science majors were ending up taking a precalculus course (spending much time learning trig), which was mostly useless for their later study. I created a case-study approach to probability and statistics for students with weaker mathematics backgrounds (i.e., most of my students have been terrified by math in the past).
I wonder how much time the OP has to prepare -- it took me a long time (and a bit of grant support), as a pure mathematician to not only learn the basic probability, but more importantly to change my mindset from pure to applied mathematics. The formulas in basic probability and statistics are nearly trivial for a professional mathematician. The real work is identifying sources of statistical bias, interpreting results correctly, and accepting the fact that no studies are perfect. In statistical mechanics, you have an absurdly large sample size of molecules behaving in a very well-controlled environment. In practical statistics, you have a smaller, usually "dirtier" sample; as a pure mathematician, it's hard to accept this sometimes.
I'd begin preparing yourself by looking at three books -- the classic "How to Lie with Statistics", the new classic "Freakonomics", and Edward Tufte's "Visual display of quantitative information" (and/or his other books). None of these is directly relevant to your course material, but they will give you many ideas for teaching, for caution in the application of statistics, and for good and bad aspects of visual display of statistical information.
Directly regarding your questions: I'm not familiar with textbooks enough to advise you on this one (I wrote my own notes). But I strongly disagree with your assumption that "less diseases and more gambling" will make your class more engaging. Most people don't care about gambling; this is supposed to be a useful class, not training for a poker team. Real statistical studies are extremely interesting, especially given their life-and-death importance. Your students should be able to answer questions like "what is the probability a person has HIV, if their test result is "reactive"? How does the answer differ for populations in the U.S. vs. Mexico vs. South Africa?". Diseases, discrimination, forensic testing, climate extremes, etc.., are important issues to consider.
You might find gambling more interesting than diseases, but a teacher of math for social science students has a responsibility to approach important questions, and not contrived examples. Take it seriously!
There are many activities that you can enjoy with students. You might play a version of the Monty Hall game, for one. There are many activities with coin-tosses (e.g., illustrating the central limit theorem). You can illustrate sample size effects, by randomly sampling students in the course. You can certainly find cases in the media and recent studies, and use them as jumping-off points for discussion: you can even find funny ones in magazines like Cosmo, or on CNN.com so that the students can practice picking apart statistical arguments and deceptive rhetoric. I often make students find an article in popular media (like NYTimes.com) that refers to a study, then track down the original study, and compare the media summary to the published study to analyze how statistics are used and misused. This can make great classroom discussion.
Finally, it might be personal taste, but I would place a heavy emphasis on probability, especially Bayesian probability. Otherwise, the course can become mechanical and reinforce a common malady: students will think of statistics as the process of collecting data, putting data through a set of software/formulas to compute correlation coefficients, p-values, standard deviations, etc.., interpreting these numbers as facts about nature, and being done. The Bayesian approach, I think, requires more thought in setting up a problem, and yields more applicable results. In particular, there are significant Bayesian criticisms of "null-hypothesis statistical testing" that is the centerpiece of many studies; especially the overreliance on p-values is disturbing to me, and you might want to include criticisms of such things.
| 16 | https://mathoverflow.net/users/3545 | 17754 | 11,860 |
https://mathoverflow.net/questions/17753 | 2 | Greetings,
I'm teaching a one-off course (perhaps never to be repeated) in a curriculum that's in transition, and I'm looking for advice on a textbook, or stories from people who have taught similar transitional-curriculum courses would be interesting as well.
The context is that we are creating a 2nd year analysis course which would be our students 1st exposure to analysis. This is to be followed up by a 3rd year analysis course which would be something of a "rigorous multi-variable calculus course". Next year the 3rd year course is going to be offered but the students will not have the 2nd year course as background (in future years the 2nd year analysis course will be a prerequisite).
What they *will* have is a fairly extensive "service calculus" background, consisting of four courses: a more or less standard 2-course 1st year single-variable calculus sequence (text is Edwards and Penney), plus a 2nd year multi-variable calculus course (also Edwards and Penney - this is a standardish calculus in $\mathbb R^n$ for $n \leq 3$ text). They follow that up with a a 3rd year course on multi-variable calculus in $\mathbb R^n$, the text is Folland. This course also covers some material that is traditionally taught in analysis classes, things like uniform convergence, Fourier transforms and such. But very little time is spent fussing about with open subsets of $\mathbb R^n$, they don't get to see bump functions, what a function *is* isn't discussed in much detail, what numbers are isn't dwelled on (not even axiomatically).
So, that's the setup. It can be safely assumed these students are motivated to study analysis, as they're taking this course to transition into our upper-level analysis courses. But I can't do too much too fast. And I don't want to bore them. So things of importance for this course to dwell on are things like what numbers are (at least axiomatically), maybe even a bit of set theory, completeness, fussy details about continuous functions like when extreme values exist, the various formulations of continuity for functions on open subsets of $\mathbb R^n$. Bump functions. And fussy details from multi-variable calculus, such as the inverse and implicit function theorem.
Are there texts out there that are designed for situations like this?
One initial inclination would be to supplement something like Hubbard's calculus text with some additional foundational notes. I suppose there are many standard intro analysis courses but some of these might be oddly paced for this group of students. I presume it's unlikely anyone has written a text for just this situation but who knows?
Thanks in advance for your comments.
| https://mathoverflow.net/users/1465 | Text/structure for an analysis course for students with pre-existing understanding of some applied aspects of analysis | If you want a reference that will not bore them, supplement the main text with "Metric Spaces: Iteration and Application" by Victor Bryant. The book is *short* and it shows in several contexts how the concept of a fixed-point property, via iteration, can be used to solve worthwhile problems. It is very nicely written and Bryant makes a real effort to motivate the material and explain where things are going and why.
| 5 | https://mathoverflow.net/users/3272 | 17768 | 11,867 |
https://mathoverflow.net/questions/17774 | 29 | I have been looking at Serre's conjecture and noticed that there are two conventions in the literature for a p-adic representation $\rho:\mbox{Gal}(\bar{\mathbb Q}/\mathbb Q)\to \mbox{GL}(n,V).$ In some references (eg Serre's book on $\ell$-adic representations), $V$ is a vector space over a finite extension of $\mathbb Q\_p$. However, in more recent papers (eg Buzzard, Diamond, Jarvis) $V$ is a vector space over $\bar{\mathbb Q\_p}$. It is easy to show that the former definition is a special case of the latter, but I suspect, and would like to prove that they are actually the same. That is, I would like to show that the image of any any continuous Galois representation over $\bar{\mathbb{Q}\_p}$ actually lies in a finite extension of $\mathbb Q\_p$.
Is this the case?
I think that a proof should use the fact that $G\_{\mathbb Q}$ is compact and that $\bar{\mathbb Q}\_p$ is the union of finite extensions. I have tried to mimic the proof that $\bar{\mathbb Q}\_p$ is not complete, but have not been able to find an appropriate Cauchy sequence in an arbitrary compact subgroup of GL($n,V$).
(This is my first question, so please feel free to edit if appropriate. Thanks!)
| https://mathoverflow.net/users/3000 | Does the image of a p-adic Galois representation always lie in a finite extension? | A proof of the result you're after is contained at the beginning of section two of a recent paper of Skinner [here](http://www.math.uiuc.edu/documenta/vol-14/10.html). Skinner mentions that references for this fact seem to be rare.
| 17 | https://mathoverflow.net/users/1979 | 17777 | 11,873 |
https://mathoverflow.net/questions/17782 | 15 | I have just realised that a group scheme I've known and loved for years is probably a bit wackier than I'd realised.
In [this question](https://mathoverflow.net/questions/17101/how-do-you-explicitly-compute-the-p-torsion-points-on-a-general-elliptic-curve-in), in Charles Rezk's answer, I erroneously claim that his construction of the space representing Drinfeld $\Gamma\_1(p)$ structures on elliptic curves must be flawed, because the global properties of $Y\_1(p)$ that I know from Katz-Mazur seemed to contradict global properties that his construction appeared to me to have. We took the conversation to email and I also started writing down my thoughts more carefully to check there were no problems with them. I found a problem with them---hence this question.
Let $p$ be prime, let $N\geq4$ be an integer prime to $p$, and consider the fine moduli space $Y\_1(N)$ over an algebraically closed field $k$ of characteristic $p$. The $N$ isn't important, it just saves me having to use the language of stacks. Let $Y^o$ denote the open affine of $Y\_1(N)$ obtained by removing the supersingular points. Over $Y^o$ we have an elliptic curve $E$ (obtained from the universal family over $Y\_1(N)$).
In brief: here's the question. The $p$-torsion $E[p]$ of $E$---it's a group scheme and its identity component is non-reduced. But (regarded as an abstract scheme) does it have a component which is reduced? I think it might! This goes against my intuition.
Now let me go more carefully. Let's consider the scheme $E[p]$ of $p$-torsion points. This is finite flat over $Y^o$ and hence as an an abstract scheme over $k$ it's going to be some sort of 1-dimensional gadget. It also sits in the middle of an exact sequence of group schemes over $Y^o$:
$0\to K\to E[p]\to H\to 0$
with $K=ker(F)$, $F$ the relative Frobenius map (an isogeny of degree $p$). Now at every point in $Y^o$, the fibre of $K$ is isomorphic to $\mu\_p$ and the fibre of $E[p]$ is isomorphic to $\mu\_p\times\mathbf{Z}/p\mathbf{Z}$. In particular all components of all fibres are isomorphic and non-reduced. Now here is where my argument in the thread in the question linked to above must become incorrect. I wanted to furthermore claim that
(a) $K$ (as an abstract curve) is non-reduced, and then
(b) hence (because $K$ is the identity component of $E[p]$ and "all components of a group are isomorphic as sets") all components of $E[p]$ are non-reduced.
I now think that (b) is nonsense. In fact I know (b) is nonsense in the sense that $\mu\_p$ over $\mathbf{Q}$ has only two components and they look rather different when $p$ is odd, but in some sense I feel here that the difference is more striking. In fact I now strongly suspect that $E[p]$ as an abstract scheme has two components, one being $K$ and the other being a *regular* scheme (an Igusa curve) mapping down in an inseparable way onto $Y^o$ (so the component isn't smooth over $Y^o$ but abstractly it's a smooth curve).
If someone wants a proper question, then there is one: am I right? The identity component of $E[p]$ is surely non-reduced---but does $E[p]$ have any regular components? I know how to prove this but it will be a deformation theory argument and I've got to go to bed :-/ If so then I think it's the first example I've seen, or at least internalised, of a group scheme where the behaviour of a non-identity component is in some sense a lot better than the behaviour of the identity component. I say "in some sense" because somehow it's the map down to $k$ that is better-behaved, rather than the map down to $Y^o$. Someone please tell me I'm not talking nonsense ;-)
| https://mathoverflow.net/users/1384 | components of E[p], E universal in char p. | Speaking of "connected components" is a delicate thing since you really mean in a relative sense, and more specifically the etale quotient $H$ can have its open and closed non-identity part with very nontrivial $\pi\_1$-action (so more subtle than on geometric fibers over the base). But even if the $\pi\_1$-action is trivial over whatever base, there are generally no no nontrivial sections through the non-identity part, so you can't do translation arguments, so there's no reason to expect intuition about "homogeneity" to have any relevance. Likewise for any property which isn't local for whatever topology the thing admits local sections. (In this case the fppf topology, for which regularity is not a local property, and ditto for reducedness.)
In this case Katz-Mazur (or better: Kummer!) did all of the deformation theory work. If we pass to the complete local ring at a geometric point of the base curve then (by the Serre-Tate deformation theorem) you're really asking a question about something over the universal deformation ring $R = k[[x]]$ of the $p$-divisible group
$$\mu\_{p^{\infty}} \times (\mathbf{Q}\_ p)/\mathbf{Z}\_ p$$
over which the universal $p$-divisible group $\Gamma$ has finite flat $p$-torsion $G = \Gamma[p]$ with a connected etale sequence that is described explicitly in Katz-Mazur. So there you can stare at the non-identity factors, and if those are regular then you're done. And if not regular somewhere then likewise for the global case over the modular curve.
If you look at (8.7.1.1) in KM (with $i \ne 0$) and then the *proof* of Prop. 8.10.5 there, or instead think about Kummer theory for group schemes, you'll see that there is a unit $q$ unique up to $p$-power unit multiple which "classifies" (up to isomorphism) the extension structure on $G$ (as $p$-torsion extension of $\mathbf{Z}/p \mathbf{Z}$ by $\mu\_p$). We can scale so $q$ is a 1-unit.
Now I claim that $q-1$ has ord equal to 1 in the deformation ring $k[[x]]$. Indeed, otherwise it would say that every first-order deformation of the $p$-divisible group has split $p$-torsion, which we know is nonsense (since we can use the unit $1+x$ to build a deformation violating that).
So then we can change $x$ so $q = 1 + x$ and the equations of the non-identity components are $T^p - (1+x)^i$ for $1 \le i \le p-1$ (from K-M, or thinking on our own). I claim the quotient in each case is a discrete valuation ring. Since $(1+x)^i = 1 + ix + x^2(\dots)$, by change of $x$ it is always the same as $T^p - (1+y)$ over $k[[y]]$, and writing it as $(T-1)^p - y$ since in characteristic $p$ we see it is Eisenstein, so we're done.
Since function fields of the modular curves you had in mind are not perfect, perhaps a more amusing example for you of funny behavior is to give a reduced group scheme over a field which is not smooth, and a non-reduced group scheme whose underlying reduced scheme is not a subgroup scheme (affine groups of finite type, but ground field must be imperfect of course).
| 10 | https://mathoverflow.net/users/3927 | 17785 | 11,877 |
https://mathoverflow.net/questions/17662 | 11 | Prof. Conrad mentioned in a recent [answer](https://mathoverflow.net/questions/7664/how-do-we-study-the-theory-of-reductive-groups/17651#17651) that most of the (introductory?) books on reductive groups do not make use of scheme theory. Do any books using scheme theory actually exist? Further, are there any books that use the functor of points approach? Demazure-Gabriel's second book would have covered general group schemes in this way, but it was never written, and it's not clear whether or not it would have covered reductive groups anyway. There is a lot of material in SGA 3 using more modern machinery to study group schemes, ~~but I'm not aware of any significant treatment of reductive groups in that book~~ (although I haven't read very much of it).
Correction: Prof. Conrad has noted that SGA 3 does contain a significant treatment of reductive groups using modern machinery.
| https://mathoverflow.net/users/1353 | Books on reductive groups using scheme theory | Personally, I find the "classical" books (Borel, Humphreys, Springer) unpleasant to read because they work in the wrong category, namely, that of reduced algebraic group schemes rather than all algebraic group schemes. In that category, the isomorphism theorems in group theory fail, so you never know what is true. For example, the map $H/H\cap N\rightarrow HN/N$ needn't be an isomorphism (take $G=GL\_{p}$, $H=SL\_{p}$, $N=\mathbb{G}\_{m}$ embedded diagonally). Moreover, since the terminology they use goes back to Weil's Foundations, there are strange statements like "the kernel of a homomorphism of algebraic groups defined over $k$ need not be defined over $k$". Also I don't agree with Brian that if you don't know descent theory, EGA, etc. then you don't "know scheme theory well enough to be asking for a scheme-theoretic treatment'.
Which explains why I've been working on a book whose goal is to allow people to learn the theory of algebraic group schemes (including the structure of reductive algebraic group schemes) without first reading the classical books and with only the minimum of prerequisites (for what's currently available, see my website under course notes). In a sense, my aim is to complete what Waterhouse started with his book.
So my answer to the question is, no, there is no such book, but I'm working on it....
| 32 | https://mathoverflow.net/users/930 | 17793 | 11,882 |
https://mathoverflow.net/questions/17758 | 13 |
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> What is an example (as simple as possible, please!) of a closed hyperbolic three-manifold with a right-angled polyhedron as fundamental domain?
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If we allow cusps then the Whitehead link or the Borromean rings are good answers (fundamental domains have not too many sides and the gluings can be understood). If we allow orbifolds then the (4,0) filling on all components of the Borromean rings is a good answer. (This is carefully explained in the [NotKnot](http://www.youtube.com/watch?v=MKwAS5omW_w) video.) I tried to answer the original question, using [SnapPea](http://www.geometrygames.org/SnapPea/) (see also [SnapPy](http://www.math.uic.edu/~t3m/SnapPy/doc/)), to build a cyclic, four-fold, manifold cover of the Borromean orbifold. This was not successful. It is "obvious" that the resulting manifold has a right-angled fundamental domain, but the domain is huge (two octagons, eight hexagons, 16 pentagons) and it is hard (for me) to describe the face pairings or check that I haven't messed up in some way.
More generally, I guess that one could use Andreev's theorem to build as "simple as possible" right-angled polyhedra and then look for low index torsion free subgroups of the resulting reflection groups. However I don't know how large an index we'd have to sacrifice or even if the resulting manifold will have a right-angled domain...
**Edit**: I've accepted bb's answer below, because the first paper cited gives the required construction. However, I didn't understand this until I asked an expert off-line, who puckishly told me that this was equivalent to the four color theorem! Here is the construction:
Suppose that $R$ is a right-angled three-dimensional hyperbolic polyhedron. Note that the edges of $R$ form a cubic graph in $\partial R$. Let $G\_R$ be the subgroup of isometries of $H^3$ generated by reflections in the sides of $R$. Now, by the four-color theorem there is a four coloring of the faces of $R$ (so no two adjacent faces have the same color). This defines surjective homomorphism from $G\_R$ to $(Z/2)^4$. Let $\delta = (1,1,1,1) \in (Z/2)^4$ and let $\Delta$ be the preimage in $G\_R$ of the subgroup generated by $\delta$. Note that $\Delta$ has index eight in $G\_R$ and is torsion-free. Finally, a fundamental domain for $\Delta$ is a obtained by gluing eight copies of $R$ around a vertex.
Other cryptic tidbits I was fed: There is such a hyperbolic manifold in dimension four, coming from the 120-cell. This is the only example known. There are no such examples in dimensions five and higher. See Bowditch and Mess, referring to Vinberg and Nikulin.
**Further edit**: In fact the argument using the four-color theorem can be found in the second paper of Vesnin, as cited by bb.
| https://mathoverflow.net/users/1650 | Closed hyperbolic manifold with right-angled fundamental domain | A. Yu. Vesnin has some articles on these Lobell manifolds. The first one describes how to construct arbitrarily many non-isometric closed hyperbolic manifolds from one right-angled polyhedron.
1. "Three-dimensional hyperbolic manifolds with a common fundamental polyhedron" Math. Notes 49 (1991), no. 5-6, 575--577
2. "Three-dimensional hyperbolic manifolds of Löbell type" Siberian Math. J. 28 (1987), no. 5, 731--733
| 10 | https://mathoverflow.net/users/4325 | 17795 | 11,883 |
https://mathoverflow.net/questions/17792 | 9 | Let $G$ be a group and let $H$ be a subgroup. If $H$ is normal in $G$, then $G/H$ has a group structure. But in general, can there be a groupoid structure on $G/H$(left cosets or right cosets) that generalizes the normal case?
| https://mathoverflow.net/users/4538 | Groupoid structure on G/H? | One answer to your question is that there is always the notion of an "action groupoid", although this does not reproduce the *group* structure on $G/H$ when $H$ is normal.
Let $G$ be a group acting on a set $X$. (There are generalizations when both $X,G$ are groupoids.) Then the **action groupoid** $X//G$ is the groupoid with objects $X$, and morphisms $X\times G$. More precisely, if $x,y \in X$, then $\hom(x,y) = \{ g\in G \text{ s.t. } gx = y\}$. The groupoid axioms are essentially obvious.
For example, let $X = G/H$ be the set of left $H$-cosets. Then the action groupoid is very simple: it is connected (any object is isomorphic to any other), with $\text{aut}(e) = H$, where $e = eH$ is the trivial left $H$-coset. And $\hom(e,g) = gH$, where $g = gH$ is a coset.
So as a discrete groupoid, this action groupoid is equivalent to $\{\text{pt}\}//H$, also sometimes called the "classifying groupoid" $\mathcal B H$, because the geometric realization of the nerve of this groupoid is the usual classifying space of $H$.
In short-hand, we have the following equation:
$$ (G/H) //G \cong 1//H$$
where $\cong$ denotes equivalence of groupoids. (Actually, since I'm talking about left actions, I should probably write $G \backslash \backslash X$ for the action groupoid, and so the equation really should be $G \backslash \backslash (G/H) = H \backslash \backslash 1$, but typing "/" is much faster than typing "\backslash", so I won't use the better notation.)
But you are probably asking a different question. Recall that when $H$ is normal, then $G/H$ has a group structure, which is to say there is a groupoid with one object and whose morphisms are elements of $G/H$. Of course, as you know, if $H$ is not normal, then $G/H$ does not have a natural group structure, because in general $g\_1Hg\_2H$ is not a left $H$-coset.
You can try to do the following. Any set is naturally a groupoid with only trivial morphisms, and then the set $G/H$ is equivalent to the groupoid $G//H$, where $H$ acts on the set $G$ by translation = right multiplication. (This is because $H$ acts on $G$ freely.) But $G$ is actually more than a set: it is a group. So let's think about it as a "groupal groupoid" or "2-group", i.e. a 2-groupoid with only one object; in this case, it will also only have identity 2-morphisms.
Then I guess you should try to form the "action 2-group" or something, by adding 2-morphisms for the translations by $H$. But I think that if you do, you no longer have a groupal groupoid: I think that if $H$ is not normal, then the group multiplication is not a functor from the action groupoid $G//H$.
The other only thing I can think of is to define $K = \text{Norm}\_GH$, the normalizer, and then $K/H$ is a group that embeds in $G/H$, so let the objects of your groupoid be cosets of $K$ and the morphisms given by $K/H$?
So, long story short: in the way that I think you are hoping, no, $G/H$ is not naturally a groupoid.
| 11 | https://mathoverflow.net/users/78 | 17799 | 11,887 |
https://mathoverflow.net/questions/17817 | 7 | It is typical to find a corollary that following theorems, but is it right to use the word corollary for a statement following a conjecture, where the statement is true only if the unproven conjecture is true?
| https://mathoverflow.net/users/4545 | Can a corollary follow a conjecture? | I think it's generally bad form to have a corollary dependent on an earlier conjecture. I recommend one of the following:
**Theorem**: Assuming Conjecture A, properties X, Y and Z are true.
or
**Theorem**: Conjecture A implies X, Y and Z.
Most importantly, it should be crystal clear that the result is dependent on the conjecture.
| 12 | https://mathoverflow.net/users/2264 | 17818 | 11,899 |
https://mathoverflow.net/questions/17811 | 5 | I am confused regarding supermanifolds. Suppose I consider R^(1,2) (1 "bosonic", 2 "fermionic"), This map (x,a,b) -> (x+ab, a,b) (a,b are fermionic) is supposed to be a morphism of this supermanifold. But I thought a morphism should be a continuous map from R->R together with a sheaf map of the sheaf of supercommutative algebra of smooth functions. How is this (x-> x+ab) giving me a continuous map from R->R?
| https://mathoverflow.net/users/3709 | Morphisms of supermanifolds | The ring of functions on your supermanifold is $C^\infty(\mathbb{R}) \otimes \mathbb{C}[a,b]$, where $a$ and $b$ are odd. The even part is then $C^\infty(\mathbb{R}) \oplus C^\infty(\mathbb{R})ab$, where $(ab)^2 = 0$, so there is an even nilpotent direction. You might want to view it as a thickening in a perpendicular direction. The map in question is the identity on the reduced quotient $C^\infty(\mathbb{R})$, and this yields the identity map of manifolds (which is continuous). The $(x \mapsto x+ab)$ can be viewed as an infinitesimal shearing on the even part.
| 4 | https://mathoverflow.net/users/121 | 17823 | 11,901 |
https://mathoverflow.net/questions/17826 | 30 | I hope this question isn't too open-ended for MO --- it's not my favorite type of question, but I do think there could be a good answer. I will happily CW the question if commenters want, but I also want answerers to pick up points for good answers, so...
Let $X,Y$ be smooth manifolds. A smooth map $f: Y \to X$ is a **bundle** if there exists a smooth manifold $F$ and a covering $U\_i$ of $X$ such that for each $U\_i$, there is a diffeomorphism $\phi\_i : F\times U\_i \overset\sim\to f^{-1}(U\_i)$ that intertwines the projections to $U\_i$. This isn't my favorite type of definition, because it demands existence of structure without any uniqueness, but I don't want to define $F,U\_i,\phi\_i$ as part of the data of the bundle, as then I'd have the wrong notion of morphism of bundles.
A definition I'm much happier with is of a **submersion** $f: Y \to X$, which is a smooth map such that for each $y\in Y$, the differential map ${\rm d}f|\_y : {\rm T}\_y Y \to {\rm T}\_{f(y)}X$ is surjective. I'm under the impression that submersions have all sorts of nice properties. For example, preimages of points are embedded submanifolds (maybe preimages of embedded submanifolds are embedded submanifolds?).
So, I know various ways that submersions are nice. Any bundle is in particular a submersion, and the converse is true for proper submersions (a map is **proper** if the preimage of any compact set is compact), but of course in general there are many submersions that are not bundles (take any open subset of $\mathbb R^n$, for example, and project to a coordinate $\mathbb R^m$ with $m\leq n$). But in the work I've done, I haven't ever really needed more from a bundle than that it be a submersion. Then again, I tend to do very local things, thinking about formal neighborhoods of points and the like.
So, I'm wondering for some applications where I really need to use a bundle --- where some important fact is not true for general submersions (or, surjective submersions with connected fibers, say).
| https://mathoverflow.net/users/78 | Why should I prefer bundles to (surjective) submersions? | One would be that a fibre bundle $F \to E \to B$ has a homotopy long exact sequence
$$ \cdots \to \pi\_{n+1} B \to \pi\_n F \to \pi\_n E \to \pi\_n B \to \pi\_{n-1} F \to \cdots $$
This isn't true for a submersion, for one, the fibre in a submersion does not have a consistent homotopy-type as you vary the point in the base space.
| 38 | https://mathoverflow.net/users/1465 | 17829 | 11,905 |
https://mathoverflow.net/questions/17836 | 5 | There are $n$ students in a class, and they must be divided into, say, $k$ groups. Each student ranks the other students in order of preference of working together. Is there a way to generally optimize student happiness (where happiness is based on working with preferred teammates). We could assume for simplicity that happiness is correlated in a simple (say linear) way with preference rank of group members.
When will there be a unique optimal grouping?
What if the happiness is not linearly correlated to preference rank?
| https://mathoverflow.net/users/4241 | How to optimize student happiness in group work? | This is a generalization of the [stable roommate problem](http://en.wikipedia.org/wiki/Stable_roommates_problem) (which is the same thing where $k = n/2$, ie, groups of 2). In general, there exist groups in which under any pair of groups contain members who would both like to switch teams.
From wikipedia:
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> For a minimal counterexample, consider 4 people A, B, C and D where all prefer each other to D, and A prefers B over C, B prefers C over A, and C prefers A over B (so each of A,B,C is the most favorite of someone). In any solution, one of A,B,C must be paired with D and the other 2 with each other, yet D's partner and the one for whom D's partner is most favorite would each prefer to be with each other.
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| 9 | https://mathoverflow.net/users/1610 | 17841 | 11,914 |
https://mathoverflow.net/questions/17803 | 12 | This question is motivated by the recent paper [An invitation to higher gauge theory](http://golem.ph.utexas.edu/category/2010/03/an_invitation_to_higher_gauge.html) by Baez and Huerta, and the 2007 paper [Parallel Transport and Functors](http://arxiv.org/abs/0705.0452) by Schreiber and Waldorf.
Let $M$ be a smooth, finite-dimensional manifold. A **lazy path** in $M$ is a smooth function $\gamma: [0,1]\to M$ such that all derivatives of $\gamma$ vanish at $0,1$. A **homotopy of lazy paths** $\gamma\_0,\gamma\_1: [0,1] \to M$ is a smooth function $\Gamma: [0,1]^2 \to M$ such that all derivatives of $\Gamma(s,t)$ vanish near $t=0,1$, and such that $\Gamma(s,t) = \gamma\_s(t)$ for $s = 0,1$. A homotopy of lazy paths is **lazy** if additionally we have that for each $t$, all the $s$ derivatives of $\Gamma(s,t)$ vanish near $s = 0,1$. A homotopy (of possibly non-lazy paths) is **thin** if $\text{rank}\\, d\Gamma < 2$ everywhere. Note that (non-lazy) thin homotopies include all reparameterizations, and so any (possibly non-lazy) piecewise-smooth path is thinly homotopic to a lazy path. Note also that lazy paths concatenate smoothly, and the concatenation of lazy paths is associative up to thin homotopy. Note also that if $\gamma^{-1}(t) = \gamma(1-t)$, then the concatenation $\gamma^{-1}\gamma$ is thinly homotopic to a constant path. Note also that lazy thin homotopies concatenate, and so define an equivalence relation, and if two paths are thinly homotopic, then they are lazily-thinly homotopic. So define $\mathcal P^1(M)$ to be the set of all lazy thin homotopy equivalence classes of lazy paths in $M$. It is a groupoid with base $M$.
The idea is to consider $\mathcal P^1(M) \rightrightarrows M$ as not just a groupoid but an infinite-dimensional Lie groupoid. I think I understand the smooth structure on $\mathcal P^1(M)$: a **curve** in $\mathcal P^1(M)$ should be precisely a (non-thin) homotopy of lazy thin paths, up to thin homotopy. It's not entirely clear to me that this defines a [smooth structure](http://arxiv.org/abs/0802.2225). But it probably works in some formalism.
But if I really want to think of $\mathcal P^1(M) \rightrightarrows M$ as a Lie groupoid, then I should treat $\mathcal P^1(M)$ not just as a smooth space, but actually as an (infinite-dimensional) manifold, and there are various things to check about the maps (the source and target maps should be surjective submersions, etc.). And it's not clear to me how to write down a smooth manifold structure on $\mathcal P^1(M)$.
Here's what I'd like. Given a point in $\mathcal P^1(M)$, I'd like a neighborhood of it and a "diffeomorphism" between that neighborhood and some (Fréchet, maybe?) vector space, and I'd like it to be clear that the gluings are smooth. I can make a start: it's clear that the space of lazy paths in a finite-dimensional vector space is a vector space, and that thin homotopies respect addition, so that $\mathcal P^1(\mathbb R^n)$ is a vector space. It's not clear to me how to put a topology on it, and it's not clear that I can approximate $\mathcal P^1(M)$ by chopping $M$ into trivializable pieces, take $\mathcal P^1$ of each piece, and try to glue back together — thin homotopies can take a path in one trivializable patch and make it wrap around $M$ in a complicated way, providing it wraps back, for example.
Hence the question:
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> What is the manifold structure on $\mathcal P^1(M)$?
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| https://mathoverflow.net/users/78 | What is the infinite-dimensional-manifold structure on the space of smooth paths mod thin homotopy? | Okay, you asked for it!
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> **Question:** What is the manifold structure on $P^1(M)$?
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> **Answer:** There isn't one.
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**Update:** The biggest failing is actually that the obvious model space is **not** a vector space. The space of paths mod thin homotopy in $\mathbb{R}^n$ does not inherit a well-defined addition from the space of paths in $\mathbb{R}^n$. Full details at the nLab page <http://ncatlab.org/nlab/show/smooth+structure+of+the+path+groupoid>.
(Update added here, rather than at the end, as it's the most direct answer to the specific question; the rest should be viewed as extra for those interested in more than just whether or not this space is a smooth manifold.)
---
It is, as you say, a smooth space. This is formal: whatever category of generalised smooth spaces you like, take the quotient of $P(M)$ by thin homotopies. All the proposed categories of generalised smooth spaces admit quotients, so the quotient exists and is a smooth space. Depending on your choice of category, the description of this smooth space may vary. For example, its Frolicher structure and its diffeological structure are very different.
But it is not "locally linear" in any sense. The basic problem is that, as you say, within an equivalence class you have paths wrapping all the way around the manifold. This destroys any hope of local linearity.
As for the proposed local model, you hit the nail on the head when you say:
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Absolutely! Topologising these spaces can lead to quite strange behaviour. You want a LCTVS structure, else you haven't a hope of even starting, and that can distort the topology from what you expect. For example, if you take piecewise-smooth paths (with no quotient) then the LCTVS topology on that is the $C^0$-topology! Indeed, simply taking so-called "lazy paths" could be fraught with difficulties (I notice that you define "lazy" slightly differently to how I've seen it done before with sitting instances). Is that space a manifold? (I know the answer to this one, but if you don't then you should start with that one as it is a *much* easier question and will hone your skills a little.)
If you really want a manifold, the solution is to go one step further. Rather than quotienting out by thin homotopies, make your "thing" into a 2-structure and put the thin homotopies in at the 2-level. Keep *all* paths at the 1-level. Then each level has a manifold structure and by mapping into a 1-structure you effectively quotient out by the 2-structure but never actually have to consider the quotient itself.
To coin a phrase:
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> Quotients are horrible, it's a shame so many people think otherwise.
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Lastly, that's not to say that there is *no* way of making $P^1(M)$ into a manifold. There may well be. But if there is, it'll be so convoluted and contrived that it won't look anything like the quotient of $P(M)$. A cautionary tale here is the case of *all* paths in a manifold, $C^\infty(\mathbb{R},M)$. That can be made into a manifold, but it has uncountably many components, for example, so looks absolutely horrid.
Okay, not quite lastly. There's lots of details here that have been glossed over. If you are really interested in working out the smooth space structure of this particular space then I (and I suspect Urs and Konrad) would be very interested in seeing it done and helping out. But MO isn't the place for that. Hop on over to the nLab, create a spin-off of <http://ncatlab.org/nlab/show/path+groupoid>, and start working.
### Further Reading
1. *[Constructing smooth manifolds of loop spaces](https://arxiv.org/abs/math/0612096)*, Proc. London Math. Soc. **99** (2009) pp195–216 (doi:[10.1112/plms/pdn058](https://doi.org/10.1112/plms/pdn058), arXiv:[math/0612096](https://arxiv.org/abs/math/0612096)). The point of this is to figure out exactly when the "standard method" (alluded to by Tim) works. The distinction between "loop" and "path" is irrelevant.
2. *[The Smooth Structure of the Space of Piecewise-Smooth Loops](https://arxiv.org/abs/0803.0611)*, Glasgow Mathematical Journal, **59** (2017) pp27-59. (arXiv:[0803.0611](https://arxiv.org/abs/0803.0611), doi:[10.1017/S0017089516000033](https://doi.org/10.1017/S0017089516000033)). Why you should be very, very nervous whenever anyone says "consider piecewise-smooth maps"; and take as a cautionary tale as to the inadvisability of going beyond smooth maps in general.
3. [Work of David Roberts on the nLab](http://ncatlab.org/davidroberts/show/which+smooth+paths+do+I+use). This is where I got the 2-idea that I mentioned above.
4. Other relevant nLab pages: <http://ncatlab.org/nlab/show/generalized+smooth+space>, <http://ncatlab.org/nlab/show/smooth+loop+space> and further.
5. Of course, the magnificent [book by Kriegl and Michor](http://www.ams.org/online_bks/surv53/). (I'm going to create a separate MO account for that book; its role will be to post an answer on relevant questions simply saying "Read Me".)
---
In response to Konrad's comment below, I've started an nlab page to work out the smooth structure of this space. The initial content considers the linear structure of the space of paths in some Euclidean space modded out by thin homotopy. The page is <http://ncatlab.org/nlab/show/smooth+structure+of+the+path+groupoid>.
| 10 | https://mathoverflow.net/users/45 | 17843 | 11,916 |
https://mathoverflow.net/questions/17844 | -1 | I want to do something about ”games of incomplete information“，like "Computer poker program".I know,Albert university(in canada) have do a lot of things to that field,they write a program called: "Polaris"(deal with computer poker) .computer poker is more difficult than computer cheese,because the former don't have enough information.
And I am interested in this field. I think "Monte Carlo Method" perhaps to solve it in some degree.
Now my question is a soft question: recommend me some books that deal with Monte Carlo Method that about above field, or some books that explain Monte Carlo Method in a detail and deep way.
In website, I only see some Science Introduction :)
thanks very much
| https://mathoverflow.net/users/4548 | Monte Carlo method and possible applications to computer poker? | Monte Carlo methods are appropriate for analyzing some systems involving chance, not incomplete information. Monte Carlo methods tell you nothing about how to model a poker strategy.
For general games of incomplete information, you should look up [game theory](http://en.wikipedia.org/wiki/Game_theory) (and not combinatorial game theory), a branch of mathematics which applies well to games of incomplete information such as poker. Some of the earliest work on game theory involved the analysis of model poker games. A common misconception is that bluffing is not mathematical, but this is simply wrong. A book which seems to have been written for mathematicians is ["The Mathematics of Poker"](http://rads.stackoverflow.com/amzn/click/1886070253) by Bill Chen and Jerrod Ankenman. For example, they study many model poker games where players are dealt a uniformly distributed number on [0,1] with restricted betting options, as did [Borel and von Neumann](http://www.math.ucla.edu/~tom/papers/poker1.pdf).
Polaris plays one form of poker, 2-player limit hold'em. This is not the form of poker you see on TV, which is usually multiplayer No Limit hold'em. The 2-player game with fixed bet sizes is still too large combinatorially to solve completely, but half-size problems can be solved (preflop games, and postflop games), and some of the research has been based on trying to glue these half-solutions together. The result, after much effort, has been strong heads-up limit hold'em programs like Polaris which crush casual players, and are only behind the best human players. However, these techniques do not extend easily to No Limit Hold'em, or to multiplayer versions of the game.
Other variants such as tournaments with low blinds or different poker games such as Razz and draw poker (which is rarely played now) are more susceptible to complete or numerical solutions. Here is an approximate [Nash equilibrium calculator](http://www.holdemresources.net/hr/sngs/icmcalculator.html?action=calculate&bb=200&sb=100&ante=0&structure=0.5,0.3,0.2&s1=2000&s2=3000&s3=2500&s4=3500&s5=2500&s6=&s7=&s8=&s9=) for single table tournaments when players are restricted to raising all-in or folding, and at most 3 players can enter the pot, which is a reasonable approximation to a commonly played variant. In practice, exploitive adjustments are important as well.
If you want to understand the current state of poker AI, then I recommend starting by exploring the web page of the [Computer Poker Research Group](http://poker.cs.ualberta.ca/) (University of Alberta) which contains some history and research articles.
| 7 | https://mathoverflow.net/users/2954 | 17851 | 11,921 |
https://mathoverflow.net/questions/17771 | 35 | Fix a prime $p$. What is the smallest integer $n$ so that there is a simplicial complex on $n$ vertices with $p$-torsion in its homology?
For example, when $p=2$, there is a complex with 6 vertices (the minimal triangulation of the real projective plane) with 2-torsion in its homology. I'm pretty sure that it's the smallest possible: with 5 or fewer vertices, there should be no torsion at all. When $p=3$, there is a complex with 9 vertices (a triangulation of the mod 3 Moore space, for instance) with 3-torsion. Is there one with 8 vertices? With $p=5$, there is a complex with 11 vertices, found by randomly testing such complexes on my computer.
We can refine this: fix $p$ and also a positive integer $d$. What's the smallest $n$ so that there is a simplicial complex $K$ on $n$ vertices with $p$-torsion in $H\_d(K)$? Or we can turn it around: for fixed $n$, what kinds of torsion can there be in a simplicial complex on $n$ vertices?
(A [paper by Soulé](http://arxiv.org/abs/math/9812171) ("Perfect forms and the Vandiver conjecture") quotes a result by Gabber which leads to a bound on the size of the torsion for a fixed number $n$ of vertices; however, this bound is far from optimal, at least for small $n$.)
| https://mathoverflow.net/users/4194 | Small simplicial complexes with torsion in their homology? | **UPDATE** This version is substantially improved from the one posted at 8 AM.
I now think I can achieve $\mathbb{Z}/p$ using $O( \log p)$ vertices. I'm not trying to optimize constants at this time.
Let $B$ be a simplicial complex on the vertices $a$, $b$, $c$, $a'$, $b'$, $c'$ and $z\_1$, $z\_2$, ..., $z\_{k-3}$, containing the edges $(a,b)$, $(b,c)$, $(c,a)$, $(a',b')$, $(b',c')$ and $(c',a')$ and such that $H^1(B) \cong \mathbb{Z}$ with generator $(a,b)+(b,c)+(c,a)$ and relation
$$2 {\large (} (a,b)+(b,c)+(c,a) {\large )} \equiv (a',b') + (b',c') + (c',a').$$
I think I can do this with $k=6$ by taking damiano's construction with $p=2$ and adding three simplices to make the hexagon $(h\_1, h\_2, \ldots, h\_6)$ homologous to the triangle $(h\_1, h\_3, h\_5)$.
Let $B^n$ be a simplicial complex with $3+nk$ vertices $a^i$, $b^i$, $c^i$, with $0 \leq i \leq n$, and $z^i\_j$ with $0 \leq i \leq n-1$ and $1 \leq j \leq k-3$. Namely, we build $n$ copies of $B$, the $r$-th copy on the vertices $a^r$, $b^r$, $c^r$, $a^{r+1}$, $b^{r+1}$, $c^{r+1}$ and $z^r\_1$, $z^r\_2$, ..., $z^r\_{k-3}$. Let $\gamma\_i$ be the cycle $(a^i,b^i) + (b^i, c^i) + (c^i, a^i)$.
Then $H^1(B^n) = \mathbb{Z}$ with generator $\gamma\_0$ and relations
$$\gamma\_n \equiv 2 \gamma\_{n-1} \equiv \cdots \equiv 2^n \gamma\_0$$
Let $p = 2^{n\_1} + 2^{n\_2} + \cdots + 2^{n\_s}$.
Glue in an oriented surface $\Sigma$ with boundary $\gamma\_{n\_1} \sqcup \gamma\_{n\_2} \sqcup \cdots \sqcup \gamma\_{n\_s}$, genus $0$, and no internal vertices.
In the resulting space, $\sum \gamma\_{n\_i} \equiv 0$ so $p \gamma\_0 \equiv 0$, and no smaller multiple of $\gamma\_0$ is zero. We have use $3 + k \log\_2 p$ vertices. This is the same order of magnitude as Gabber's bound.
| 16 | https://mathoverflow.net/users/297 | 17852 | 11,922 |
https://mathoverflow.net/questions/17809 | 10 | There was a rather [cute question last week](https://mathoverflow.net/questions/17269/let-g-be-a-graph-such-that-for-all-u-v-v-g-u-no-equal-to-v-n-u-n-v) about graphs where every pair of distinct vertices has an odd number of mutual neighbours.
The question was to show that such a graph must have an odd number of vertices, and it can be accomplished with a nice algebraic graph theory argument.
But let's up the ante a bit: can we actually characterize the graphs with this property?
Here are some examples in the family:
* complete graphs of odd order
* anything obtained by gluing together a bunch of odd complete graphs at a single vertex
* a graph of the form A - B - C where A and C have the "even" version of this property (every pair of vertices have even number common neighbours) B is an odd complete graph, and A is completely joined to B, B completely joined to C.
Is this the lot?
| https://mathoverflow.net/users/1492 | Graphs where every two vertices have odd number of mutual neighbours | Take a Steiner triple system on $v$ points. Let $X$ be the graph with the $v(v-1)/6$ triples
as its vertices, two triples adjacent if the have exactly one point in common. We need
$v\equiv1,3$ modulo 6. Then two adjacent triples have exactly $(v+3)/2$ common neighbours, and two disjoint triples have exactly 9 common neighbours. If we take $v\equiv3,7$ modulo
12 we get examples.
Of course I am just constructing strongly regular graphs with $\lambda$ and $\mu$ odd.
The are strongly regular graphs with this property besides the ones listed, for example generalized quadrangles with $s$ and $t$ even. Further examples appear in Andries Brouwer's on-line tables (<http://www.win.tue.nl/~aeb/graphs/srg/srgtab.html>), or Gordon Royle's
(<http://units.maths.uwa.edu.au/~gordon/remote/srgs/>).
This suggest that a classification might be difficult.
| 4 | https://mathoverflow.net/users/1266 | 17853 | 11,923 |
https://mathoverflow.net/questions/17854 | 1 | I am talking about a relation that is what Wikipedia describes as [left-unique and right-unique](http://en.wikipedia.org/wiki/Relation_%28mathematics%29#Special_types_of_binary_relations). I never heard these terms before, but I have heard of the alternatives (injective and functional). The question is, *which terminology do you recommend*? Should I include short definitions? (The context is a text in the area of formal methods. I'm not sure if this helps.)
These are some trade-offs that I see:
* I think that *left-unique* and *right-unique* are not widely known, but I'm not sure at all.
* *functional* is overloaded
* *injective* sounds too fancy (subjective, of course)
* *left-unique* and *right-unique* are symmetric (good, of course)
**Edit:** It seems the question is unclear. Here are more details. I describe sets *X* and *Y* and then say:
1. now we must find an injective and functional relation between sets *X* and *Y* such that...
2. now we must find a left-unique and right unique relation between sets *X* and *Y*...
Which one do you recommend? What other information would you add? The relation does *not* have to be total. For example, various different ranges correspond to different 'feasible' relations. Technically I should not need to say that the relation does not have to be total, but will many people assume that it has to be total if I don't say it?
| https://mathoverflow.net/users/840 | Do you need to say what left-unique and right-unique means? | Injective and functional are completely standard in this case. This is what you should use. The term "functional" is not overloaded, when you are using it to say that something is a function. Being functional means exactly that the relation is a function.
A relation that is injective and functional is precisely an injective function on its domain. It is a bijection of its domain with its range.
If you don't want to think of the relation as a function, then you can also describe it as a *one-to-one correspondence* of its domain with its range.
(And I don't think any of these terms I suggest would need to be defined, since their meaning is fairly universally known. This would definitely not be true of left-unique and right-unique.)
| 6 | https://mathoverflow.net/users/1946 | 17855 | 11,924 |
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