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https://mathoverflow.net/questions/11077 | 21 | Can someone please explain what the symbol $\stackrel{!}{=}$, consisting of an exclamation mark (!) above an equals sign (=) means? Below is the example I'm trying to decipher:
---
The normalization factor is chosen such that in average, Dynamic Θ Time passes as fast as physical time. In practice it is determiend by the condition that the interval in Dynamic Θ Time corresponding to a 4-year reference period [$T\_0$, $T\_1$] should be of exactly the same length:
$T\_1 - T\_0 \stackrel{!}{=} \int\_{T\_0}^{T\_1} a(t) dt$
| https://mathoverflow.net/users/3062 | What does ! above = mean | In my classes I use it to indicate anxiety. So an equals with ! over it means "we want to show this equality is true". An equals without ! means "I am asserting this is true".
I don't know how universal this convention is, though. I do know I'm not the only person to use this convention.
| 12 | https://mathoverflow.net/users/1465 | 11078 | 7,530 |
https://mathoverflow.net/questions/11073 | 1 | Let {a1,a2,...,an} and {b1,b2,...,bn} be two bases for a vector space E. Fix p, 1 ≤ p ≤n. Is there a permutation σ such that
{a1,a2,...,ap,bσ(p+1),...,bσ(n)} and {bσ(1),bσ(2),...,,bσ(p),ap+1,...,an} are both bases of E?
This question is the last exercise of the first chapter in the book Linear Algebra by Greub. I can prove the case p=n-1.
| https://mathoverflow.net/users/3061 | On permutation of elements of two bases of a vector space (Greub´s book) | It's also Theorem 7.2 in [Prasolov's Problems in Linear Algebra](http://www2.math.su.se/~mleites/books/prasolov-1994-problems.pdf), which gives a proof and attributes it to Green 1973.
| 4 | https://mathoverflow.net/users/2530 | 11080 | 7,532 |
https://mathoverflow.net/questions/10907 | 2 | Let X be a separable reflexive Banach space and f:X\to\mathbb{R} be a
Lipschitz function. Say that a point x in X is a *local supporting point
of* f if there exist x^\* in X^\* and an open neighborhood U of x
such that either x^\* (y-x)\leq f(y)-f(x) for all y in U or
x^\* (y-x)\geq f(y)-f(x) for all y in U.
Question: is true that the set of local supporting points of f is dense in X?
This question is obviously related to differentiability; it might be difficult.
I would be very much interested to know whether it has been asked by others.
| https://mathoverflow.net/users/3038 | Local supporting points of Lipschitz functions | My guess is that you did not formulate question correctly --- in the present form the answer is NO.
One can take strictly saddle $f$ on $\mathbb R^2$, say $f(x,y)=\sqrt{1+x^2}-\sqrt{1+y^2}$.
| 2 | https://mathoverflow.net/users/1441 | 11083 | 7,533 |
https://mathoverflow.net/questions/11069 | 6 | While reading Demazure-Gabriel's construction of $\mathcal{S}ch$ as a full subcategory of $\mathcal{P}sh(CRing^{op})$, I've been trying to translate their exposition into the language of covering sieves and Grothendieck topologies. The requirement that a scheme be a sheaf in the Zariski topology on $CRing^{op}$ is essentially already in the language of sieves, but the requirement that there exists a cover of affines is giving me some trouble.
Question, then:
In general, is there a natural induced Grothendieck topology on $\mathcal{P}sh(\mathcal{C})$ or $\mathcal{S}h(\mathcal{C})$, where $\mathcal{C}$ is a site (not a pre-site!)?
Edit: I removed the other parts of the question regarding t-Schemes and algebraic spaces as functors of points to ask them at some other time.
| https://mathoverflow.net/users/1353 | Induced Grothendieck topology on a presheaf or sheaf category of a site? | The answer to your first question is yes. Suppose $C$ a site.
The category $C^{\sim}$ of sheaves on $C$ simply has the canonical topology (SGA 4, Vol. 1, Exp. II, 2.5). The sheaves here are precisely the representable sheaves. (I'm ignoring questions about universes.)
The category $C^{\wedge}$ of presheaves on $C$ inherits a topology in the following manner (SGA 4, Vol. 1, Exp. II, §5). Declare a morphism $F\to G$ of presheaves on $C$ to be a *covering morphism* if the induced morphism $aF\to aG$ of associated sheaves is an epimorphism. Now say that a collection of morphisms $F\_i\to G$ is a *covering family* if the morphism $\amalg F\_i\to G$ is a covering morphism. This gives $C^{\wedge}$ the finest subcanonical topology such that covering families in $C$ give rise to covering families in $C^{\wedge}$. This topology on $C^{\wedge}$ can be seen as the lift of the topology on $C^{\sim}$ described above along the sheafification functor $a$.
| 8 | https://mathoverflow.net/users/3049 | 11085 | 7,534 |
https://mathoverflow.net/questions/11081 | 19 | There is a conjecture of Kontsevich which states that Hochschild (co)homology of the Fukaya category of a compact symplectic manifold $X$ is the (co)homology of the manifold. (See page 18 of Kontsevich's "Homological algebra of mirror symmetry" paper and page 16 of Costello's paper "TCFTs and CY categories".)
Moreover:
* The Hochschild cohomology of the Fukaya category of $X$ should be the Lagrangian Floer cohomology $HF^\ast(X,X)$ of the diagonal $X \to X \times X$.
* $HF^\ast(X,X)$ is, at least according to Costello's paper, known to coincide with the quantum cohomology of $X$. But I don't know a reference for this?
What is the current status of these conjectures? Are there any cases where any of this is known to be true, or known to be false? Costello's paper states "I really don't know of much evidence" --- perhaps our state of knowledge is better by now?
| https://mathoverflow.net/users/83 | Hochschild (co)homology of Fukaya categories and (quantum) (co)homology | The statement that $HF^{\ast}(X,X)$ is isomorphic to $QH^\ast(X)$ is a version of the Piunikhin-Salamon-Schwarz (PSS) isomorphism (proved, under certain assumptions, in McDuff-Salamon's book "J-holomorphic curves in symplectic topology"). PSS is a canonical ring isomorphism from $QH^{\ast}(X)$ to the Hamiltonian Floer cohomology of $X$, and the latter can be compared straightforwardly to the Lagrangian Floer cohomology of the diagonal.
Now to Hochschild cohomology of the Fukaya category $F(X)$. There's a geometrically-defined map $QH^{\ast}(X) \to HH^{\ast}(F(X))$, due to Seidel in a slightly different setting (see his "Fukaya categories and deformations"), inspired by the slightly vague but prescient remarks of Kontsevich from 1994. One could define this map without too much trouble, say, for monotone manifolds. It's constructed via moduli spaces of pseudo-holomorphic polygons subject to Lagrangian boundary conditions, with an incidence condition of an interior marked point with chosen cycles in $X$. The question is whether this is an isomorphism.
This statement is open, and will probably not be proven true in the near future, for a simple reason: $QH^\*(X)$ is non-trivial, while we have no general construction of Floer-theoretically essential Lagrangians.
There are two positive things I can say. One is that Kontsevich's heuristics, which involve interpreting $HH^{\ast}$ as deformations of the identity functor, now have a natural setting in the quilted Floer theory of Mau-Wehrheim-Woodward (in progress). This says that the Fukaya category $F(X\times X)$ naturally embeds into the $A\_\infty$-category of $A\_\infty$-endofunctors of $F(X)$.
The other is that for Weinstein manifolds (a class of exact symplectic manifolds with contact type boundary), there seems to be an analogous map from the symplectic cohomology $SH^{\ast}(X)$ (a version of Hamiltonian Floer cohomology on the conical completion of $X$) to $HH^{\ast}$ of the wrapped Fukaya category, which involves non-compact Lagrangians. (Edit August 2010: I was careless about homology versus cohomology. I should have said that $HH\_{\ast}$ maps to $SH^{\ast}$.) Proving that this is an isomorphism is more feasible because one may be able to prove that Weinstein manifolds admit Lefschetz fibrations. The Lefschetz thimbles are then objects in the wrapped Fukaya category.
One might then proceed as follows. The thimbles for a Lefschetz fibration should generate the triangulated envelope of the wrapped category (maybe I should split-close here; not sure) - this would be an enhancement of results from Seidel's book. Consequently, one should be able to compute $HH\_{\ast}$ just in terms of $HH\_{\ast}$ for the full subcategory generated by the thimbles. The latter should be related to $SH^{\ast}$ by ideas closely related to those in Seidel's paper "Symplectic homology as Hochschild homology".
What could be simpler?
ADDED: Kevin asks for evidence for or against $QH^{\ast}\to HH^{\ast}$ being an isomorphism. I don't know any evidence contra. Verifying it for a given $X$ would presumably go in two steps: (i) identify generators for the (triangulated envelope of) $F(X)$, and (ii) show that the map from $QH^{\ast}$ to $HH^{\ast}$ for the full subcategory that they generate is an isomorphism. There's been lots of progress on (i), less on (ii), though the case of toric Fanos has been studied by Fukaya-Oh-Ohta-Ono, and in this case mirror symmetry makes predictions for (i) which I expect will soon be proved. In simply connected disc-cotangent bundles, the zero-section generates, and both $HH\_{\ast}$ for the compact Fukaya category and $SH^{\ast}$ are isomorphic to loop-space homology, but I don't think it's known that the resulting isomorphism is Seidel's.
Added August 2010: Abouzaid ([1001.4593](http://arxiv.org/abs/1001.4593)) has made major progress in this area.
| 23 | https://mathoverflow.net/users/2356 | 11098 | 7,544 |
https://mathoverflow.net/questions/11105 | 11 | Is there an easy example of an integral domain and two elements on it which do not have a greatest common divisor? It will have to be a non-UFD, obviously.
"Easy" means that I can explain it to my undergrad students, although I will be happy with any example.
| https://mathoverflow.net/users/3065 | An example of two elements without a greatest common divisor | [Here's](http://en.wikipedia.org/wiki/Greatest_common_divisor) an example stolen blatantly from wikipedia.
Let $R=\mathbb{Z}[\sqrt{-3}]$, let $a=4=2\*2=(1+\sqrt{-3})(1-\sqrt{-3})$ and $b=2(1+\sqrt{-3})$. Now, $2$ and $1+\sqrt{-3}$ are both maximal among divisors, but are not associates, thus, there is not GCD.
| 14 | https://mathoverflow.net/users/622 | 11107 | 7,549 |
https://mathoverflow.net/questions/11106 | 11 | I want to start out by making this clear: I'm NOT looking for the modern proofs and rigorous statements of things.
What I am looking for are references for classical enumerative geometry, back before Hilbert's 15th Problem asked people to actually make it work as rigorous mathematics. Are there good references for the original (flawed!) arguments? I'd prefer perhaps something more recent than the original papers and books (many are hard to find, and even when I can, I tend to be a bit uncomfortable just handling 150 year old books if there's another option.)
More specifically, are there modern expositions of the original arguments by Schubert, Zeuthen and their contemporaries? And if not, are there translations or modern (20th century, say...) reprints of their work available, or are scanned copies available online (I couldn't find much, though I admit my German is awful enough that I might have missed them by not having the right search terms, so I'm hoping for English review papers or the like, though I'll deal with it if I need to.)
| https://mathoverflow.net/users/622 | Classical Enumerative Geometry References | And actually, as a partial answer to my own question, I just stumbled across Schubert's "[Kalkul](http://books.google.com/books?id=1zQAAAAAQAAJ&printsec=frontcover&dq=hermann+schubert&lr=&ei=_dJGS8SYGoaszATUh-XNAg&cd=11#v=onepage&q=&f=false)" on Google Books, and it looks complete, which makes me rather happy, though other portions of the question still stand.
EDIT: A friend of mine has informed me that Zeuthen's "[Lehrbuch](http://books.google.com/books?id=hnVtAAAAMAAJ&printsec=frontcover#v=onepage&q=&f=false)" is also there, and now it's linked.
| 3 | https://mathoverflow.net/users/622 | 11116 | 7,556 |
https://mathoverflow.net/questions/11117 | 15 | Any Eilenberg-MacLane space $K(A,n)$ for abelian $A$ can be given the structure of an $H$-space by lifting the addition on $A$ to a continuous map $K(A\times A,n)=K(A,n)\times K(A,n)\to K(A,n)$.
Does somebody know an explicit way to describe this structure in the cases $K({\mathbb Z}/2{\mathbb Z},1)={\mathbb R}P^\infty$ and $K({\mathbb Z},2)={\mathbb C}P^{\infty}$?
| https://mathoverflow.net/users/3067 | H-space structure on infinite projective spaces | Look at $\mathbb R^\infty\setminus 0$ as the space of non-zero polynomials, which you can multiply. Pass to the quotient to construct the projective space and, from the multiplication, its $H$-space product.
The complex case is quite the same.
**NB:** Jason asks in a comment below if this is the same $H$-space structure that Hanno had in mind.
To check, we can use the fact that Hanno's is characterised by the fact that if $\mu:K(\mathbb Z\_2,1)\times K(\mathbb Z\_2,1)\to K(\mathbb Z\_2,1)$ is his product and $\alpha\in H^1(K(\mathbb Z\_2,1), \mathbb Z\_2)$ is the class represented by the identify map $K(\mathbb Z\_2,1)\to K(\mathbb Z\_2,1)$, then $\mu^\\*(\alpha)=\alpha\times 1+1\times\alpha$.
One should be able to check that this holds for the map given by multiplication of polynomials in a very small skeleton.
| 20 | https://mathoverflow.net/users/1409 | 11121 | 7,558 |
https://mathoverflow.net/questions/11053 | 23 | Let $p$ be an odd prime and $\left( \frac{a}{p} \right)$ the Legendre symbol. The **Gauss sum**
$\displaystyle g\_p(a) = \sum\_{k=0}^{p-1} \left( \frac{k}{p} \right) \zeta^{ak},$
where $\zeta\_p = e^{ \frac{2\pi i}{p} }$, is a periodic function of period $p$ which is sometimes invoked in proofs of quadratic reciprocity. [As it turns out](http://sbseminar.wordpress.com/2008/10/11/the-sign-of-the-gauss-sum/), $g\_p(a) = \left( \frac{a}{p} \right) i^{ \frac{p-1}{2} } \sqrt{p}$, so $g\_p(a)$ is essentially an eigenfunction of the discrete Fourier transform. Now, if $(a, p) = 1$, we can write
$\displaystyle g\_p(a) = \sum\_{k=0}^{p-1} \zeta^{ak^2}$
so Gauss sums are some kind of finite analogue of the normal distribution $e^{-\pi x^2}$, which is itself well-known to be an eigenfunction of the Fourier transform on $\mathbb{R}$. I remember someone claiming to me once that the two are closely related, but I haven't been able to track down a reference. Does anyone know what the precise connection is? Is there a theory of self-dual locally compact abelian groups somewhere out there?
| https://mathoverflow.net/users/290 | What's the relationship between Gauss sums and the normal distribution? | It's true (as the answer below and some of the commenters note) that it's easy to interpret this question in a way that makes it seem trivial and uninteresting. I'm quite sure, however, that pursuing typographical similarity between $e^{x^2}$ and $\zeta^{m^2}$ leads to interesting mathematics, and so here's a more serious attempt at propoganda for some of Ivan Cherednik's work.
Pages 6,7,8 and 9 of Cherednik's [paper](http://arxiv.org/abs/math/0110024) "Double affine Hecke algebras and difference Fourier transforms" explain how to ``interpolate'' between integral formulas relating the Gaussian to the Gamma function and (a certain generalization of) Gauss sums.
More explicitly, he shows that the formula (for many people, it's really just the definition of the Gamma function)
$$\int\_{-\infty}^{\infty} e^{-x^2} x^{2k} dx=\Gamma \left( k+\frac{1}{2} \right)$$
(for $k \in \mathbb{C}$ with real part $>-1/2$) and the Gauss-Selberg sum
$$\sum\_{j=0}^{N-2k} \zeta^{(k-j)^2/4} \frac{1-\zeta^{j+k}}{1-\zeta^k} \prod\_{l=1}^j \frac{1-\zeta^{l+2k-1}}{1-\zeta^l}=\prod\_{j=1}^k (1-\zeta^j)^{-1} \sum\_{m=0}^{2N-1} \zeta^{m^2/4}$$
(where $N$ is a positive integer, $\zeta=e^{2\pi i/N}$ is a prim. $N$th root of $1$, and $k$ is a positive integer at most $N/2$) can both be obtained as limiting cases of the same $q$-series identity. The common generalization of the Gaussian and the function $k \mapsto \zeta^{k^2}$ is the function $x \mapsto q^{x^2}$, and the measures weighting the integral and sum get replaced by Macdonald's measure---essentially the same one that shows up in the constant term conjecture for $A\_1$, and that produces the Macdonald polynomials and kick-started the [DAHA](https://mathoverflow.net/questions/6517/double-affine-hecke-algebras-and-mainstream-mathematics). The Fourier transform is deformed along with everything else to produce the "Cherednik-Fourier" transform.
I don't know how much of the roots of unity story generalizes to higher rank root systems.
Note: In the Gauss-Selberg sum, replacing $k$ by the integer part of $N/2$ and manipulating a little (as in the nice exposition by David Speyer linked to in the question above) gives the usual formula for the Gauss sum.
| 12 | https://mathoverflow.net/users/nan | 11138 | 7,572 |
https://mathoverflow.net/questions/11133 | 5 | Can anyone name a undecidable problem that is genuinely graph-related? (Genuine means: not a standard one in graph's disguise.)
| https://mathoverflow.net/users/2672 | Undecidable graph problems? | From a MathSciNet search:
---
Földes, Stéphane; Steinberg, Richard
A topological space for which graph embeddability is undecidable.
J. Combin. Theory Ser. B 29 (1980), no. 3, 342--344.
From the introduction: ``From Edmonds' permutation theorem and a generalization due to Stahl, it follows that graph embeddability is decidable for all surfaces, orientable as well as nonorientable. We show the existence of a topological space $\hat G$ such that there is no algorithm to decide whether a finite graph is embeddable in $\hat G$. In fact, $\hat G$ will be a path-connected subspace of the real plane.''
---
| 12 | https://mathoverflow.net/users/1149 | 11142 | 7,575 |
https://mathoverflow.net/questions/11130 | 1 | Prove that the [Normal (Gaussian) Distribution](https://en.wikipedia.org/wiki/Normal_distribution) with a given Variance $ {\sigma}^{2} $ maximizes the [Differential Entropy](https://en.wikipedia.org/wiki/Differential_entropy) among all distributions with defined and finite 1st Moment and Variance which equals $ {\sigma}^{2}
$.
| https://mathoverflow.net/users/2285 | Differential Entropy of Random Signal | Cover and Thomas's book is indeed the right place to learn about this.
The statement basically follows by convexity, in the form of Jensen's inequality. Here is the way it is usually presented:
Let $f$ be the probability density of a real random variable. Then the Shannon entropy is given by
$-\int f\log f dx$
You want to prove among all real random variables with finite Shannon entropy and variance equal to $1$, the Shannon entropy is maximized only for Gaussians.
Given two probability densities $f$ and $g$, since $\log$ is a concave function, Jensen's inequality tells us that
$\int f \log (g/f) dx < \log \int f(g/f) dx = \log \int g dx = 0$
Moreover, since $\log$ is strictly concave, equality holds if and only if $g = f$. If you now set $g$ equal to the probability density of a Gaussian with the same variance as $f$ and plug in the explicit formula for $g$, you get what you want.
| 5 | https://mathoverflow.net/users/613 | 11144 | 7,577 |
https://mathoverflow.net/questions/11109 | 7 | I understand the definition of a Killing Form $B$ as $B(X,Y)=Tr(ad(X)ad(Y))$
And when the Lie group is semi-simple the negative of the Killing Form can serve as a Riemannian metric.
{Wonder if thats why some people say that a semi-simple Lie Group has only one metric!}
I would like to know what is the difference between a "Killing Metric" and "Cartan-Killing metric" and the "Killing Form" ?
Or are they just different names for the same concept?
{Like the books by Knapp or Fulton and Harris have nothing called "Cartan-Killing metric"}
| https://mathoverflow.net/users/2678 | A terminology issue with the Killing Form | Let me add a few comments to the answers by Mariano and Theo.
There is a one-to-one correspondence between bi-invariant metrics (of any signature) in a Lie group and ad-invariant nondegenerate symmetric bilinear forms on its Lie algebra.
In a simple Lie algebra every nondegenerate symmetric bilinear form is proportional to the Killing form which you wrote down in your question, hence a simple Lie group has precisely one conformal class of bi-invariant metrics. If (and only if) the group is compact, are these metrics positive-definite. (Some people call this *riemannian*, reserving the word *pseudo-riemannian* (or sometimes also *semi-riemannian*) for indefinite signature metrics. Personally I prefer to use *riemannian* for general metrics.)
One can ask the question: *which Lie groups admit bi-invariant metrics of any signature?* which is the same thing as asking which Lie algebras admint ad-invariant non-degenerate symmetric bilinear forms. Such Lie algebras are called *metric* (or also sometimes *quadratic*, *orthogonal*,...) and although there is no classification except in small index (index 0 = positive-definite, index 1 = lorentzian, etc...) there is a structure theorem proved by Alberto Medina and Philippe Revoy in [this paper](http://www.numdam.org/item?id=ASENS_1985_4_18_3_553_0) (in French). Their theorem says that the class of such Lie algebras is generated by the simple and the one-dimensional Lie algebras under two operations: orthogonal direct sum and *double extension*, a construction explained in that paper.
Double extension always results in indefinite signature, so if you are only interested in the positive-definite case, you get back to the well-known result that every positive-definite metric Lie algebra $\mathfrak{g}$ is isomorphic to the orthogonal direct sum of a compact semisimple Lie algebra and an abelian Lie algebra, or in other words,
$$\mathfrak{g} \cong \mathfrak{s}\_1 \oplus \cdots \oplus \mathfrak{s}\_N \oplus \mathfrak{a}$$
where the $\mathfrak{s}\_i$ are the simple factors and $\mathfrak{a}$ is abelian.
Up to automorphisms, the most general positive-definite inner product on such a Lie algebra is given by choosing for each simple factor $\mathfrak{s}\_i$ a positive multiple $\lambda\_i > 0$ of the Killing form.
These Lie algebras are precisely the Lie algebras of compact Lie groups. Their metricity can also be understood as follows: take any positive-definite inner product on $\mathfrak{g}$ and averageng it over the adjoint representation.
So in summary, although there are metric Lie algebras which are not semisimple (or even reductive), their inner product is always an additional structure, unlike the Killing form which comes for free with the Lie algebra.
As for question concerning the difference between *Killing form* and *Cartan(-Killing) metric* it depends on who says this. In much of the Physics literature people refer to an inner product on a vector space as a "metric". But assuming that this is not the case, then the Killing form is a bilinear form on the Lie algebra, whereas the metric is a metric (in the sense of riemannian geometry) on the Lie group. If $G$ is a Lie group whose Lie algebra is semisimple, then the Killing form on its Lie algebra defines a bi-invariant metric on the Lie group, which I suppose you could call the *Cartan-Killing metric*.
| 15 | https://mathoverflow.net/users/394 | 11152 | 7,581 |
https://mathoverflow.net/questions/11149 | 5 | I could sample a set of m elements from the uniform distribution over a universe $U$ of n >> m elements. Alternately, I could select a random probability distribution $\mathcal{D}$, and sample $m$ elements from $\mathcal{D}$.
EDIT (per Michael Lugo): When I say "select a random probability distribution", I mean select a point uniformly at random from the standard $n$-simplex: {$\{(x\_1,\ldots,x\_n) : x\_i \geq 0, x\_1+\ldots+x\_n = 1\}$}.
Do these two methods lead to the same distribution over my sample? If not, how do they differ? If some event (my sample lies in some set of samples of size m) occurs with probability p using the second method, what can I say about its probability using the first method?
| https://mathoverflow.net/users/3028 | Sample from uniform distribution vs. Sample from random distribution | You're right that things change for m>1; I was thinking sloppily.
Assume $U=\{1,\ldots,n\}$ for concreteness. If $Y\_1,\ldots,Y\_m$ are chosen independently and uniformly from $U$, then for any $k\_1,\ldots,k\_m\in U$, we of course have
$$
\Pr[Y\_1=k\_1,\ldots,Y\_m=k\_m] = \frac{1}{n^m}.
$$
On the other hand, if $x=(x\_1,\ldots,x\_m)$ is chosen uniformly from the standard $n$-simplex and $Y\_1,\ldots,Y\_m$ are then chosen independently according to $x$, then
$$
\Pr[Y\_1=k\_1,\ldots,Y\_m=k\_m] = \mathbb{E}\Pr[Y\_1=k\_1,\ldots,Y\_m=k\_m|x]
= \mathbb{E}\prod\_{i=1}^m x\_{k\_i} = \frac{n!}{(n+r)!}\prod\_{j=1}^n r\_j!,
$$
where $r\_j = \#\{1\le i \le m : k\_i=j\}$ and $r=r\_1 + \cdots r\_n$. This last expectation can be proved most easily from Lemma 1 in [this paper](http://www.jstor.org/stable/2048262).
| 3 | https://mathoverflow.net/users/1044 | 11160 | 7,587 |
https://mathoverflow.net/questions/11164 | 16 | Notation
--------
Let $\mathfrak g$ be a the Lie algebra of an algebraic group $G\subseteq GL(V)$ over a(n algebraically closed) field $k$ (I'm actually thinking $G=GL\_n$, so $\mathfrak g=\mathfrak{gl}\_n$). Then any element $X$ of $\mathfrak g$ can be uniquely written as the sum of a semi-simple (diagonalizable) element $X\_s$ and a nilpotent element $X\_n$ of $\mathfrak g$, where $X\_s$ and $X\_n$ are polynomials in $X$. The nilpotent cone $\mathcal N$ is the subset of nilpotent elements of $\mathfrak g$ (elements $X$ such that $X=X\_n$).
---
People often talk about the nilpotent cone as having the structure of a subvariety of $\mathfrak g$, regarded as an affine space, but usually don't say what the scheme structure really is. To really understand a scheme, I'd like to know what its functor of points is. That is, I don't just want to know what a nilpotent matrix is, I want to know what a family of nilpotent matrices is (i.e. what a map from an arbitrary scheme $T$ to $\mathcal N$ is). Since any scheme is covered by affine schemes, it's enough to understand what an $A$-valued point (a map $\mathrm{Spec}(A)\to \mathcal N$) is for any $k$-algebra $A$. So my question is
>
> What functor should $\mathcal N$ represent?
>
>
>
### A guess
Well, an $A$-point of $\mathfrak g$ is "an element of $\mathfrak g$ with entries in $A$" (again, I'm really thinking $\mathfrak g=\mathfrak{gl}\_n$, so just think "a matrix with entries in $A$"), so I would expect that such an $A$-point happens to be in $\mathcal N$ exactly when the given matrix is nilpotent. That is, $\mathcal N(\mathrm{Spec}(A))=\{X\in \mathfrak{g}(\mathrm{Spec}(A))| X^N=0$ for some $N\}$.
However, this is wrong. That functor isn't even an algebraic space, even for the nilpotent cone of $\mathfrak{gl}\_1$. If it were, the identity map on it would correspond to a nilpotent regular function $f$ (a nilpotent $1\times 1$ matrix), and this would be the *universal* nilpotent regular function; every other nilpotent regular function anywhere else would be a pullback of this one. But whatever the degree of nilpotence of this function (say $f^{17}=0$), there are some nilpotent regular functions which cannot be a pullback of it (something with nilpotence degree bigger than 17). If this version of the nilpotent cone were representable, you can show that the $\mathfrak{gl}\_1$ version would be too.
### Another guess
I think the answer might be that an $A$ point of $\mathcal N$ is a matrix ($A$ point of $\mathfrak g$) so that all the coefficients of the characteristic polynomial vanish. This is a scheme and it has the right field-valued points, but why should this be the nilpotent cone? What is the meaning of having all coefficients of the characteristic polynomial vanish for a matrix with entries in $A$?
| https://mathoverflow.net/users/1 | What does the nilpotent cone represent? | The ideal which defines the nilpotent cone is generated by the homogeneous elements of positive degree in $\mathbb{C}[\mathfrak g]^G$. In the case of $GL\_n$, this ideal is generated by the functions $\mathrm{tr}\;X^k$ for $k$ up to the dimension, and also by the coefficients of the characteristic polynomial. See, for example, [Springer, T. A. Invariant theory. Lecture Notes in Mathematics, Vol. 585. Springer-Verlag, Berlin-New York, 1977. iv+112 pp. [MR0447428](http://www.ams.org/mathscinet-getitem?mr=MR0447428)]
| 14 | https://mathoverflow.net/users/1409 | 11167 | 7,591 |
https://mathoverflow.net/questions/11087 | 5 | In an [opinion piece](http://www.ams.org/notices/201001/rtx100100005p.pdf) which appeared in the AMS Notices of January 2010, John Wermer tells us that he once heard about a seminar given by Grothendieck which was described as "a telegram by Grothendieck to Serre". Is this anecdote recorded somewhere?
| https://mathoverflow.net/users/1409 | A telegram by Grothendieck to Serre | I think I heard Grothendieck's talk at the 1958 ICM described as a "a telegram by Grothendieck to Serre". But this would have been in a conversation (with neither Serre nor Grothendieck). I don't know whether anyone said it in a lecture, much less wrote it down.
MR0130879 (24 #A733) Grothendieck, Alexander The cohomology theory of abstract algebraic varieties. 1960 Proc. Internat. Congress Math. (Edinburgh, 1958) pp. 103--118 Cambridge Univ. Press, New York
Added: I don't know who first described Grothendieck's 1958 ICM talk as a "telegram to Serre", but I probably heard it from Lubin. Since Lubin and the article's author spent their careers at Brown, this fits.
Grothendieck's talk is available [here](http://www.math.jussieu.fr/~leila/grothendieckcircle/CohomologyVarieties.pdf). In it he lays out his plans for schemes and their cohomology. In 1958, not many people would have understood it.
| 15 | https://mathoverflow.net/users/930 | 11172 | 7,595 |
https://mathoverflow.net/questions/11178 | 5 | As far as I know, one way to take a homotopy colimit in a model category is to replace (up to acyclic fibration) all arrows in the diagram with cofibrations, and take the strict colimit of the resulting diagram.
In Top with the model structure given by Serre fibrations, cofibrations, and weak equivalences, if one wants to obtain a homotopy pushout of the diagram $X \leftarrow A \rightarrow Y$, it is "enough" to replace only one of these arrows with a cofibration: that is, there is a natural map (by the universal property of the pushout) $Cyl(X)\cup\_A Cyl(Y) \to Cyl(X) \cup\_A Y$ that is a homotopy equivalence of spaces.
>
> Question 1: What conditions on the model category $\mathcal{C}$ (or objects $X,Y,A$) will guarantee that the natural map $Cyl(X) \cup\_A Cyl(Y) \to Cyl(X) \cup\_A Y$ is a weak equivalence?
>
>
> Question 2: This question is less precise, but if the map above is a weak equivalence, does that mean $Cyl(X) \cup\_A Y$ is a good model for the homotopy pushout?
>
>
>
| https://mathoverflow.net/users/2536 | Homotopy Pushouts via Model Structure in Top | Question 1: The model category $\mathcal{C}$ should be *left proper*, i.e. the pushout of a weak equivalence along a cofibration is again a weak equivalence. (Dually, there is a notion of right proper.) Top is left proper, as is any model category in which every object is cofibrant, such as SSet. There is some information on this notion of properness [at the nlab](http://ncatlab.org/nlab/show/proper+model+category), and I think it's also discussed more thoroughly in Hirschhorn's book *Model Categories and their Localizations* (and probably many other places).
Question 2: Yes. People often say that a square in a model category is a homotopy pushout square if the induced map from the (strict) pushout of a cofibrant replacement (meaning cofibrant objects and maps) of the "initial" three objects to the last object is a weak equivalence, and that is the case here.
| 9 | https://mathoverflow.net/users/126667 | 11180 | 7,601 |
https://mathoverflow.net/questions/11177 | 8 | I have heard that for a locally ringed space $X$ whose topology is second countable and Hausdorff, $X$ is a smooth manifold if and only if it is locally ringed space which is locally isomorphic to the sheaf of differentiable functions of some open sets in $\mathbb{R}^n$.
Question: What about the maps between smooth manifolds? Let $M, N$ smooth manifolds and $\phi: M \rightarrow N$ be a continuous map. Does every locally ringed space morphism $\mathcal{O}\_N \rightarrow \phi\_\* \mathcal{O}\_M$ come from $\phi$?
I'm asking this because I thought it was true - so that I agree that locally ringed spaces are useful generalizations of (whatever) manifolds - until I tried to prove it tonight. In the process it seems to me that this is NOT true. But if this is not true, why is locally ringed space a good generalization of manifolds if the maps don't behave well? (As opposed to rings $A \rightarrow B$ and spectrum maps $SpecB \rightarrow SpecA$?)
Thanks!
| https://mathoverflow.net/users/nan | Smooth maps considered as locally ringed space morphisms? | For a morphism $f: (X,{\mathcal O}\_X)\to (Y,{\mathcal O}\_Y)$ of locally ringed spaces the locality of the maps $f^{\sharp}\_x: {\mathcal O} \_{Y,f(x)}\to {\mathcal O} \_{X,x}$ implies that for each function $\lambda\in{\mathcal O}\_Y(U)$ we have $v(f^{\sharp}\_U(\lambda)) = f^{-1}(v(\lambda))$ (here $v(\lambda)$ denotes the zero set of $\lambda$, i.e. the set of all points $y\in U$ such that $\lambda\_y\in{\mathfrak m}\_y$.
(\*) Therefore the zero set of $f^{\sharp}\_U (\lambda)$ is determined entirely by the underlying continuous map of $f$, and it equals $f^{-1}(v(\lambda))$
From now on, let $X,Y$ be topological spaces and ${\mathcal O} \_X$, ${\mathcal O} \_Y$ denote the sheaves of continuous real-valued functions on $X$ and $Y$, respectively. If you know that $f^{\sharp}$ preserves constant functions, then by translating a function $\lambda\in{\mathcal O}\_Y (U)$ by a real number $\alpha\in{\mathbb R}$, you see by (\*) that the niveau set $v\_\alpha(f^{\sharp}\_U(\lambda)) := \{x\in f^{-1}(U)\ |\ f^{\sharp}\_U(\lambda)(x) = \alpha\}$ equals $f^{-1}(v\_\alpha(\lambda))$, so $f^{\sharp}\_U(\lambda) = \lambda\circ f$ as suspected.
Now $f^{\sharp}$ has to preserve constant functions, because each map $\overline{f^{\sharp}\_x}: {\mathbb R} = {\mathcal O}\_{Y,f(x)}/{\mathfrak m}\_{Y,f(x)}\to{\mathcal O}\_{X,x}/{\mathfrak m}\_{X,x} = {\mathbb R}$ is an automorphism of ${\mathbb R}$, thus equals the identity. (Ok, one could use this to argue directly that $f^{\sharp}\_U(\lambda)(x)=\lambda(f(x))$ - didn't see this when starting to write)
For a field different from ${\mathbb R}$, this argument doesn't work any more, because there may be nontrivial automorphims. For example, if you take ${\mathbb C}$-valued continuous functions, you could set $f^{\sharp}\_U(\lambda) := \text{conj}\circ \lambda\circ f$ for the algebraic component of $f$.
| 9 | https://mathoverflow.net/users/3067 | 11184 | 7,603 |
https://mathoverflow.net/questions/11182 | 7 | Remove the closure of simply connected region from the interior of a simply connected region. Is it true that the resulting domain can be mapped conformally to some annulus?
| https://mathoverflow.net/users/2938 | Riemann mapping for doubly connected regions | The answer is yes. This is a special case of theorem 10 in Ahlfors' *Complex Analysis*, section 5, chapter 6. (Special in that the theorem more generally says that if the complement of the domain has $n$ connected components not reduced to points in the extended plane, then the domain is equivalent to an annulus from which $n-2$ concentric slits have been removed. In your case $n=2$.)
| 7 | https://mathoverflow.net/users/1409 | 11185 | 7,604 |
https://mathoverflow.net/questions/11192 | 34 | Let $\mathcal{X}$ be a real or complex Banach space.
It is a well known fact that $\mathcal{X}$ is a Hilbert space (i.e. the norm comes from an inner product) if the parallelogram identity holds.
Question: Are there other (simple) characterizations for a Banach space to be a Hilbert space?
| https://mathoverflow.net/users/2989 | When is a Banach space a Hilbert space? | From [this article](https://doi.org/10.3103/S0027132208050070) by O. N. Kosukhin:
>
> A real Banach space $(X, \|\cdot\|)$ is a Hilbert space if and only if for any three points $A$, $B$, $C$ of this space not belonging to a line there are three altitudes in the triangle $ABC$ intersecting at one point.
>
>
>
Many other references show when Googling
>
> "is a hilbert space if" banach
>
>
>
| 24 | https://mathoverflow.net/users/1847 | 11194 | 7,610 |
https://mathoverflow.net/questions/11188 | 3 | let $(X\_i)\_{i \in I}$ be an infinite family of sets with $|X\_i| \geq 2$. we define an equivalence relation on $X = \prod\_{i \in I} X\_i$ by $x \sim y \Leftrightarrow \{i : x\_i \neq y\_i\}$ is finite. what is the cardinality of $X/\sim$? we may endow the $X\_i$ with group structures and write this set as $\prod\_{i \in I} X\_i / \oplus\_{i \in I} X\_i$.
it is easy to see that $\min\_i |X\_i| \leq |X/\sim|$ and $|X| \leq |X/\sim| \* \max\_i |X\_i|$. in particular, if $|X\_i|$ is constant, $|X/\sim| = |X|$.
if all the $X\_i$ are finite, it can be shown $|X/\sim|=2^{|I|}$. the same equation holds, if every $X\_i$ satisfies $\aleph\_0 \leq |X\_i| \leq |I|$ (I'll add the proofs if needed).
the general case struggles me.
| https://mathoverflow.net/users/2841 | cardinality of product modulo direct sum | The cardinality of the reduced product is always the same as that of the product, modulo omitting finitely many unusually large $X\_i$'s. (Even when $I$ is finite, in which case all $X\_i$'s should be omitted.)
Arrange the sets $X\_i$ in nondecreasing order of size in a wellordered sequence $(X\_\alpha)\_{\alpha<\tau}$. We may assume that $\tau$ is a limit ordinal. Otherwise, the last coordinate $\tau-1$ (like any single coordinate) contributes nothing to the reduced product. Omit $X\_{\tau-1}$ and repeat as long as necessary.
I will show that then $|X| = |X/{\sim}|$ where $X = \prod\_{\alpha<\tau} X\_\alpha$.
Let $\kappa = \sup\_{\alpha<\tau} |X\_\alpha|$. Since each ${\sim}$-equivalence class has size $|\tau|\cdot\kappa$ (see note) we have
$|X| \leq |X/{\sim}|\cdot|\tau|\cdot\kappa = \max(|X/{\sim}|,|\tau|,\kappa).$
Since $|X| \geq 2^{|\tau|} > |\tau|$, we conclude that either $|X/{\sim}| = |X|$ or $|X/{\sim}| \leq |X| \leq \kappa$.
Since $\tau$ is a limit ordinal, the diagonal embedding $d:\kappa \to \prod\_{\alpha<\tau} |X\_\alpha|$, where
$d\_\alpha(\xi) = \begin{cases} \xi & when\ \xi < |X\_\alpha| \\ 0 & otherwise,\end{cases}$
shows that $\kappa \leq |X/{\sim}|$. So, in the case $|X/{\sim}| \leq |X| \leq \kappa$, we in fact have $|X/{\sim}| = |X| = \kappa$.
Note: The elements ${\sim}$-equivalent to a given $x \in X$ are obtained by selecting finitely many new values from the sets $X\_\alpha-\{x\_\alpha\}$ to replace the corresponding value of the sequence $x$. There are at least $\sum\_{\alpha<\tau} |X\_\alpha-\{x\_\alpha\}|$ and no more than $\left(\sum\_{\alpha<\tau} |X\_\alpha|\right)^{<\omega}$ ways of doing this. Since $\tau$ is infinite and the $X\_\alpha$'s all have two or more elements, these two bounds are equal to $\sum\_{\alpha<\tau} |X\_\alpha| = |\tau|\cdot\kappa$.
| 5 | https://mathoverflow.net/users/2000 | 11204 | 7,616 |
https://mathoverflow.net/questions/11203 | 1 | In my ODE class, we proved that if $\exp(L) = \exp(L')$ then the eigenvalues are congruent mod $2 \pi i$. Here, $L$ and $L'$ are two $n \times n$ matrices. I wanted to know if something more precise was true.
In a way, we should expect that matrix logs are multiple valued since this is the case in $\mathbb{C}$.
$$\log(r e^{i \theta}) = \log r + i\theta + 2 \pi i k$$
with $k \in \mathbb{Z}$. In this way we can construct an branched infinite cover of the complex plane.
We'll define **multiplicity** mod $2 \pi i$ of an eigenvalue $\lambda$ to be the number of eigenvalues congruent to $\lambda$ mod $2 \pi i$ up to multiplicity. If $\exp(L) = \exp(L')$ are the spectra of $L$ and $L'$ the same including multiplicity mod $2 \pi i$?
To put this another way, I could imagine two $5 \times 5$ matrices $\exp(L) = \exp(L')$ where
* the spectrum of $L$ is $(\lambda\_1, \lambda\_1, \lambda\_2, \lambda\_2 + 2 \pi i, \lambda\_2 + 4 \pi i)$
while
* the spectrum of $L'$ is $(\lambda\_1, \lambda\_1, \lambda\_1, \lambda\_2 + 2 \pi i, \lambda\_2 + 4 \pi i)$
Here the multiplicities mod $2 \pi i$ are different. One would be $(2,3)$ while the other would be $(3,2)$. Could $\exp(L) = \exp(L')$ in this case?
| https://mathoverflow.net/users/1358 | Matrix logarithms are not unique | No, they cannot. Note that $\exp(PAP^{-1})=P\exp(A)P^{-1}$, so wlog, both are in Jordan form. Then, we can compute by exponentiating Jordan blocks, and the first will have a two by two block (or two one by one) depending on whether it is diagonal or not $\delta=0,1$, and three $\exp(\lambda\_2)$ eigenvalues. The second has a 3 by 3 block and two $\exp(\lambda\_2)$ eigenvalues. These will exponentiate to new Jordan blocks for the eigenvalus $\exp(\lambda\_1)$, and so the two matrices have different spectra, and so cannot be the same.
| 8 | https://mathoverflow.net/users/622 | 11205 | 7,617 |
https://mathoverflow.net/questions/11208 | 1 | How do you recenter a spherical coordinate system. For example, if the center were at $\left (0, 0, 0 \right )$ and I wanted to move the center of the spherical coordinate system to $\left (\rho\_{1}, \Theta\_{1}, \Phi\_{1} \right )$, then what transformation would I apply to $\left (\rho\_{2}, \Theta\_{2}, \Phi\_{2} \right )$?
In cartesian coordinates, you would simply subtract the two vectors.
| https://mathoverflow.net/users/3084 | Recentering a Spherical Coordinate Sytem | This is going to be unsightly...
The following Mathematica code:
```
Needs["VectorAnalysis`"]
Simplify@ CoordinatesFromCartesian[
CoordinatesToCartesian[{r, theta, phi}, Spherical]
+ CoordinatesToCartesian[{r0, theta0, phi0}, Spherical],
Spherical
]
```
gives the following output (doctored so that it looks nicer):
$$ r' = \sqrt{r^2+2 r\_0 r \left(\sin (\theta ) \sin
\left(\theta \_0\right) \cos \left(\phi -\phi
\_0\right)+\cos (\theta ) \cos \left(\theta
\_0\right)\right)+r\_0^2} $$
$$ \theta' = \cos ^{-1}\left(\frac{r \cos (\theta )+r\_0 \cos
\left(\theta \_0\right)}{\sqrt{r^2+2 r\_0 r
\left(\sin (\theta ) \sin \left(\theta
\_0\right) \cos \left(\phi -\phi
\_0\right)+\cos (\theta ) \cos \left(\theta
\_0\right)\right)+r\_0^2}}\right) $$
$$ \phi' = \tan ^{-1}\left(r \sin (\theta ) \cos (\phi
)+r\_0 \sin \left(\theta \_0\right) \cos
\left(\phi \_0\right),r \sin (\theta ) \sin
(\phi )+r\_0 \sin \left(\theta \_0\right) \sin
\left(\phi \_0\right)\right) $$
In this last line, there is a two-argument variant of arctan, which is explained [here](http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Two-argument_variant_of_arctangent), for example.
| 5 | https://mathoverflow.net/users/1409 | 11214 | 7,621 |
https://mathoverflow.net/questions/11207 | 9 | Here, I'm primarily concerced about zeta functions of hypersurfaces over fields of finite characteristic.
Assume $F\_q$ to be a finite field with q elements. Consider the zeta function of the hypersurface defined by $-y\_0^2+y\_1^2+y\_2^2+y\_3^2=0$ in $\mathbb{P}^3$.
If $-1$ is a square in $F\_q$, the zeta function is
$$Z(u)=\frac{1}{(1-uq^2)(1-uq)^2(1-u)}.$$
It has a pole of order $2$ at $1/q$. If not, it's
$$Z(u)=\frac{1}{(1-uq^2)(1-uq)(1+uq)(1-u)}.$$
It has a pole of order $1$ at $1/q$.
>
> How does orders of poles indicate any geometric information?
>
>
>
| https://mathoverflow.net/users/2902 | How does the order of a pole of a zeta function indicate any geometric information? | For a smooth projective surface, the order of the pole at 1/q is conjectured to be the rank of the Neron-Severi group of the surface. That's a conjecture of Tate and is an analog of the Birch and Swinnerton-Dyer conjecture. Tate has formulated a more general conjecture for higher dimensional varieties too. For the case of quadrics, as in your example, these conjectures are known.
Edit: maybe you don't want a fancy answer. In the first case, the quadric contains lines and in the second, it doesn't.
| 15 | https://mathoverflow.net/users/2290 | 11215 | 7,622 |
https://mathoverflow.net/questions/11209 | 11 | I can (barely) understand the definition of the higher algebraic K-groups a la the plus construction right now (I have some past familiarity with K-theory for C\*-algebras and can recall the rudiments of the situation for vector bundles). So by "simple", I mean to a mathematical layman. If you have a complicated answer, feel free to answer as well, but I probably won't be able to understand much.
| https://mathoverflow.net/users/1847 | Is there a simple relationship between K-theory and Galois theory? | Perhaps the other Bloch-Kato conjecture is more relevant; it relates *Milnor's* higher $K$-groups and Galois cohomology.
The following text is lifted from the expository account [on the arXiv.](http://arxiv.org/abs/math/0311099)
Let $F$ be a field, $n>0$ an integer which is invertible in $F$, $\bar F$ a
separable closure of $F$ and $\Gamma=\operatorname{Gal}(\bar F|F)$. There
is an exact sequence
$$
\{1\}\to
\mathbb{Z}/n\mathbb{Z}(1)\to
{\bar F}^\times\to
{\bar F}^\times\to
\{1\}
$$
of discrete $\Gamma$-modules, where $\mathbb{Z}/n\mathbb{Z}(1)$ is the group of $n$-th roots of $1$ in $\bar F$. The associated long exact cohomology sequence and
Hilbert's theorem 90 furnish an isomorphism $\delta\_1:F^\times/F^{\times n}\to H^1(\Gamma,\mathbb{Z}/n\mathbb{Z}(1))$. Cup product on cohomology
$$
\smile\;:H^r(\Gamma,\mathbb{Z}/n\mathbb{Z}(r))
\times H^s(\Gamma,\mathbb{Z}/n\mathbb{Z}(s))\to
H^{r+s}(\Gamma,\mathbb{Z}/n\mathbb{Z}(r+s))
$$
then provides a bilinear map
$
\delta\_2:F^\times/F^{\times n}\times F^\times/F^{\times n}\to
H^2(\Gamma,\mathbb{Z}/n\mathbb{Z}(2)).
$
**Lemma** (Tate, 1970)
*The map* $\delta\_2(x,y)=\delta\_1(x)\smile\delta\_1(y)$ *is a
symbol on* $F$.
A symbol is a bilinear map $s:F^\times\times F^\times\to A$ to a commutative
group such that $s(x,y)=0$ whenever $x+y=1$ in $F^\times$.
There is a universal symbol $F^\times\times F^\times\to K\_2(F)$, giving rise
to Milnor's theory of higher $K$-groups $K\_r(F)$ for every $r\in\mathbb{N}$,
as explained in Milnor's book.
This symbol also gives rise to a homomorphism
$$
\delta\_r:K\_r(F)/nK\_r(F)\to
H^r(\Gamma,\mathbb{Z}/n\mathbb{Z}(r)).
$$
**Conjecture** (Bloch-Kato, 1986)
*The map*
$\delta\_r$ *is an isomorphism for all fields* $F$, *all integers* $n>0$
(*invertible in* $F$) *and all indices* $r\in\mathbb{N}$.
The main theorem of Merkurjev-Suslin (1982) says that the map
$\delta\_2$ is
always an isomorphism ; Tate had proved this earlier (1976) for global fields.
Bloch-Gabber-Kato prove this conjecture when $F$ is a field of
characteristic $0$ endowed with a henselian discrete valuation of residual
characteristic $p\neq0$ and $n$ is a power of $p$.
Somebody should ask a qustion about the current status of the Bloch-Kato
conjecture and get some experts (such as Weibel) to answer. My impression is that it is now a theorem by the work of Rost and Voevodsky, but that a proof with all the
details is not available in one place.
The Bloch-Kato conjecture makes the remarkable prediction that the graded
algebra $\oplus\_r H^r(\Gamma,\mathbb{Z}/n\mathbb{Z}(r))$ is generated by
elements of degree 1. Galois groups should thus be very special among
profinite groups in this respect.
| 12 | https://mathoverflow.net/users/2821 | 11218 | 7,625 |
https://mathoverflow.net/questions/11219 | 59 | I am interested in learning about Shimura curves. Unlike most of the people who post reference requests however (see this [question](https://mathoverflow.net/questions/1291/a-learning-roadmap-for-algebraic-geometry) for example), my problem is not sorting through an abundance of books but rather dealing with what appears to be an extreme paucity of sources.
Anyway, I'm a graduate student and have spent the last year or so thinking about the arithmetic of orders in quaternion algebras (and more generally in central simple algebras). The study of orders in quaternion algebras seems to play an important role in Shimura curves, and I'd like to study these connections more carefully.
Unfortunately, it has been very difficult for me to find a good place to start. I only really know of two books that explicitly deal with Shimura curves:
* Shimura's *Introduction to the arithmetic theory of automorphic functions*
* Alsina and Bayer's *Quaternion Orders, Quadratic Forms, and Shimura Curves*
Neither book has been particularly helpful however; the first only mentions them briefly in the final section, and the second has much more of a computational focus then I'd like.
**Question 1**: Is there a book along the lines of Silverman's *The Arithmetic of Elliptic Curves* for Shimura curves?
I kind of doubt that such a book exists. Thus I've tried to read the introductory sections of a few papers & theses, but have run into a problem. There seem to be various ways of thinking about a Shimura curve, and it has been the case that every time I look at an article I'm confronted with a different one. For example, this [talk](http://www.uni-math.gwdg.de/tschinkel/cmi/shimura-clay-nopause.pdf) by Voight and this [paper](http://www.jmilne.org/math/articles/2003a.pdf) by Milne. By analogy, it seems to be a lot like trying to learn class field theory by switching between articles with ideal-theoretic statements and articles taking an adelic slant without having a definitive source which tells you that both are describing the same theorems.
My second question is therefore:
**Question 2**: Can anyone suggest a 'roadmap' to Shimura curves? Which theses or papers have especially good expository accounts of the basic properties that one needs in order to understand the literature.
Clearly I need to say something about my background. As I mentioned above, I'm an algebraic number theorist with a particular interest in quaternion algebras. I don't have the best algebraic geometry background in the world, but have read Mumford's *Red Book*, the first few chapters of Hartshorne and Qing Liu's *Algebraic Geometry and Arithmetic Curves*. I've also read Silverman's book *The Arithmetic of Elliptic Curves* and Diamond and Shurman's *A First Course in Modular Forms*.
Thanks.
| https://mathoverflow.net/users/nan | What is a good roadmap for learning Shimura curves? | First of all, Kevin is being quite modest in his comment above: his paper
---
Buzzard, Kevin. Integral models of certain Shimura curves. Duke Math. J. 87 (1997), no. 3, 591--612.
---
contains many basic results on integral models of Shimura curves over totally real fields, and is widely cited by workers in the field: 22 citations on MathSciNet. The most recent is a paper of mine:
---
Clark, Pete L. On the Hasse principle for Shimura curves. Israel J. Math. 171 (2009), 349--365.
[http://alpha.math.uga.edu/~pete/plclarkarxiv7.pdf](http://alpha.math.uga.edu/%7Epete/plclarkarxiv7.pdf)
---
Section 3 of this paper spends 2-3 pages summarizing results on the structure of the canonical integral model of a Shimura curve over $\mathbb{Q}$ (with applications to the existence of local points). From the introduction to this paper:
"This result [something about local points] follows readily enough from a description of their [certain Shimura curves over Q] integral canonical models. Unfortunately I know of no unique, complete reference for this material. I have myself written first (my 2003 Harvard thesis) and second (notes from a 2005 ISM course in Montreal) approximations of such a work, and in so doing I have come to respect the difficulty of this expository problem."
I wrote that about three years ago, and I still feel that way today. Here are the documents:
1. [http://alpha.math.uga.edu/~pete/thesis.pdf](http://alpha.math.uga.edu/%7Epete/thesis.pdf)
is my thesis. "Chapter 0" is an exposition on Shimura curves: it is about 50 pages long.
2. For my (incomplete) lecture notes from 2005, go to
[http://alpha.math.uga.edu/~pete/expositions2012.html](http://alpha.math.uga.edu/%7Epete/expositions2012.html)
and scroll down to "Shimura Curves". There are 12 files there, totalling 106 pages [perhaps I should also compile them into a single file]. On the other hand, the title of the course was Shimura *Varieties*, and although I don't so much as attempt to give the definition of a general Shimura variety, some of the discussion includes other PEL-type Shimura varieties like Hilbert and Siegel moduli space. These notes do not entirely supercede my thesis: each contains some material that the other omits.
When I applied for an NSF grant 3 years ago, I mentioned that if I got the grant, as part of my larger impact I would write a book on Shimura curves. Three years later I have written up some new material (as yet unreleased) but am wishing that I had not said that so directly: I would need at least a full semester off to make real progress (partly, of course, to better understand much of the material).
Let me explain the scope of the problem as follows: there does not even exist a single, reasonably comprehensive reference on the arithmetic geometry of the classical modular curves (i.e., $X\_0(N)$ and such). This would-be bible of modular curves ought to contain most of the material from Shimura's book (260 pages) and the book of Katz and Mazur *Arithmetic Moduli of Elliptic Curves* (514 pages). These two books don't mess around and have little overlap, so you get a lower bound of, say, 700 pages that way.
Conversely, I claim that there is some reasonable topology on the arithmetic geometry of modular curves whose compactification is the theory of Shimura curves. The reason is that in many cases there are several ways to establish a result about modular curves, and "the right one" generalizes to Shimura curves with little trouble. (For example, to define the rational canonical model for classical modular curves, one could use the theory of Fourier expansions at the cusps -- which won't generalize -- or the theory of moduli spaces -- which generalizes immediately. Better yet is to use Shimura's theory of special points, which nowadays you need to know anyway to study Heegner point constructions.) Most of the remainder concerns quaternion arithmetic, which, while technical, is nowadays well understood and worked out.
| 42 | https://mathoverflow.net/users/1149 | 11229 | 7,631 |
https://mathoverflow.net/questions/11226 | 41 | Typically, in the functor of points approach, one constructs the category of algebraic spaces by first constructing the category of locally representable sheaves for the global Zariski topology (Schemes) on $CRing^{op}$. That is, taking the full subcategory of $Psh(CRing^{op})$ which consists of objects $S$ such that $S$ is a sheaf in the global Zariski topology and $S$ has a cover by representables in the induced topology on $Psh(CRing^{op})$. This is the category of schemes. Then, one takes this category and equips it with the etale topology and repeats the construction of locally representable sheaves on this site (Sch with the etale topology) to get the category of algebraic spaces.
*Can we "skip" the category of schemes entirely by putting a different topology on $CRing^{op}$?*
My intuition is that since every scheme can be covered by affines, and every algebraic space can be covered by schemes, we can cut out the middle-man and just define algebraic spaces as locally representable sheaves for the global etale topology on $CRing^{op}$. If this ends up being the case, is there any sort of interesting further generalization before stacks, perhaps taking locally representable sheaves in a flat Zariski-friendly topology like fppf or fpqc?
Some motivation: In algebraic geometry, all of our data comes from commutative rings in a functorial way (intentionally vague). All of the grothendieck topologies with nice notions of descent used in Algebraic geometry can be expressed in terms of commutative rings, e.g., the algebraic and geometric forms of Zariski's Main theorem are equivalent, we can describe etale morphisms in terms of etale ring maps, et cetera. What I'm trying to see is whether or not we can really express all of algebraic geometry as "left-handed commutative algebra + sheaves (including higher sheaves like stacks)". The functor of points approach for schemes validates this intuition in the simplest case, but does it actually generalize further?
The main question is italicized, but feel free to tell me if I've incorrectly characterized something in the motivation or the background.
| https://mathoverflow.net/users/1353 | Commutative rings to algebraic spaces in one jump? | Yes. The category of algebraic spaces is the smallest subcategory of the category of sheaves of sets on Aff, the opposite of the category of rings, under the etale topology which (1) contains Aff, (2) is closed under formation of quotients by etale equivalence relations, and (3) is closed under disjoint unions (indexed by arbitrary sets). An abstract context for such things is written down in "Algebraization of complex analytic varieties and derived categories" by Toen and Vaquie, which is available on the archive. Toen also has notes from a "master course" on stacks on his web page with more information. It might be worth pointing out that their construction of this category also goes by a two-step procedure, although in their case it's a single construction performed iteratively (and which stabilizes after two steps). This is unlike the approach using scheme theory in the literal sense, as locally ringed topological spaces, where the two steps are completely different. After the first step in T-V, you get algebraic spaces with affine diagonal. Also worth pointing out is that their approach is completely sheaf theoretic. The only input you need is a category of local models, a Grothendieck topology, and a class of equivalence relations. You then get algebraic spaces from the triple (Aff, etale, etale). But the general machine (which incidentally I believe is not in its final form) has nothing to do with commutative rings. I think it would be interesting to plug opposites of other algebraic categories into it.
| 43 | https://mathoverflow.net/users/1114 | 11234 | 7,636 |
https://mathoverflow.net/questions/11238 | 17 | Before I ask the question, I need to recall what Bernoulli numbers
$(B\_k)\_{k\in\mathbb{N}}$ are, and what von Staudt and Clausen discovered about
them in 1840. The numbers $B\_k\in\mathbb{Q}$ are the coefficients in the formal power series
$$
{T\over e^T-1}=\sum\_{k\in\mathbb{N}}B\_k{T^k\over k!}
$$
so that $B\_0=1$, $B\_1=-1/2$, and it is easily seen that $B\_k=0$ for $k>1$
odd.
**Theorem** (von Staudt-Clausen, 1840) *Let* $k>0$ *be an even integer, and let* $p$ *run through the primes. Then*
$$
B\_k+\sum\_{p-1|k}{1\over p}\in\mathbb{Z}.
$$
John Coates remarked at a recent workshop that the analogue of this theorem
for a totally real number field $F$ (other than $\mathbb{Q}$) is an open problem;
even a weak analogue would imply Leopoldt's conjecture for $F$. I missed the
opportunity of pressing him for details.
**Question** : *What is the analogous statement over a totally real number field ?*
| https://mathoverflow.net/users/2821 | von Staudt-Clausen over a totally real field | I also can't answer the question, but I'll say some things that could help. One thing von Staudt-Clausen tells you is the denominator of the Bernoulli number $B\_k$: it is precisely, the product of primes p for which $p-1\mid k$ (when $p-1\nmid k$, a result of Kummer says that $B\_k/k$ is p-integral). As Buzzard commented, the Bernoulli numbers should be thought of (at least in this situation) as appearing in special values of *p*-adic *L*-functions, specifically, for *k* a positive integer
$$\zeta\_p(1-k)=(1-p^{k-1})(-B\_k/k),$$
where $\zeta\_p$ is the p-adic Riemann zeta function (see chapter II of Koblitz's "p-adic numbers, p-adic analysis, and zeta-functions", for example).
For a totally real field *F*, a generalization of the *p*-adic Riemann zeta function exists, namely the *p*-adic Dedekind zeta function $\zeta\_{F,p}$ (as proved independently by Deligne–Ribet ([Inv Math 59](http://www.ams.org/mathscinet-getitem?mr=579702)), Cassou-Noguès ([Inv Math 51](http://www.ams.org/mathscinet-getitem?mr=524276)), and Barsky ([1978](http://www.ams.org/mathscinet-getitem?mr=525346))). One link between these and the Leopoldt conjecture is through the *p*-adic analytic class number formula which is the main theorem of Colmez's "Résidue en *s* = 1 des fonctions zêta *p*-adiques" ([Inv Math 91](http://www.ams.org/mathscinet-getitem?mr=922806)):
$$\lim\_{s\rightarrow1}(s-1)\zeta\_{F,p}(s)=\frac{2^{[F:\mathbf{Q}]}R\_phE\_p}{w\sqrt{D}}$$
where *h* is the class number,
$$E\_p=\prod\_{\mathfrak{p}\mid p}\left(1-\mathcal{N}(\mathfrak{p})^{-1}\right)$$ is a product of Euler-like factors, *w* = 2 is the number of roots of unity, *D* is the discriminant and $R\_p$ is the interesting part here: the *p*-adic regulator (as Colmez notes, $\sqrt{D}$ and $R\_p$ both depend on a choice of sign, but their ratio does not).
**Theorem:** The Leopoldt conjecture is equivalent to the non-vanishing of the *p*-adic regulator.
(For this, see, for example, chapter X of Neukirch-Schmidt-Wingberg's "Cohomology of number fields").
A clear consequence of this is that if $\zeta\_{F,p}$ does not have a pole at *s* = 1, then the Leopoldt conjecture is false for (*F*, *p*). Perhaps an understanding of the denominators of values of $\zeta\_{F,p}$ could lead to an understanding of the pole at *s* = 1 of $\zeta\_{F,p}$.
Added (2010/04/09): So here's how you can use von Staudt–Clausen to see that the $p$-adic zeta function (of **Q**) has a pole at *s* = 1. It is clear from your statment of vS–C that it is saying that for $k\equiv0\text{ (mod }p-1)$, $B\_k\equiv -1/p\text{ (mod }\mathbf{Z}\_p)$ (i.e. it is not $p$-integral). Let $k\_i=(p-1)p^i$, the $k\_i$ is $p$-adically converging to 0, so $\zeta\_p(1-k\_i)$ is approaching $\zeta\_p(1)$ (since $\zeta\_p(s)$ is $p$-adically continuous, at least for $s\neq1$). By the aforementioned interpolation property of $\zeta\_p(1-k)$, we have
$$v\_p(\zeta\_p(1-k\_i))=v\_p(B\_{k\_i}/k\_i)=-1-i\rightarrow -\infty$$
hence $1/\zeta\_p(1-k\_i)$ is approaching 0.
| 8 | https://mathoverflow.net/users/1021 | 11242 | 7,641 |
https://mathoverflow.net/questions/11239 | 11 | In another [question](https://mathoverflow.net/questions/11182/riemann-mapping-for-doubly-connected-regions) here on MO, Anweshi asks if any doubly connected region in the complex plane can be conformally mapped to some annulus. The answer to this is yes. But the fact is that two annuli are conformally equivalent iff the ratio of the outer radius and the inner radius is the same for the two. Thus each is conformally equivalent to a unique "standard" annulus
$r < |z| < 1$ with $ 0 < r < 1$. Now my question is the following:
Is there any way to see what the radius of a "standard" annulus conformally equivalent to a doubly connected region is, just by looking at the region? That is without constructing an explicit map.
| https://mathoverflow.net/users/135 | Conformal maps of doubly connected regions to annuli. | By "see" I will assume you mean in a geometric sense. Then your question falls within a standard topic in geometric complex analysis. First some terminology: A doubly connected domain $R$ on the Riemann sphere is called a *ring domain*, and if you map it onto $r < |z| < s$ as a canonical domain, then $\mathrm{mod}(R) = (2\pi)^{-1}\log(s/r)$ is called the *conformal modulus* or just *modulus* of the ring domain. By the way I have defined it, it is nearly trivially a conformal invariant, but it need not be defined this way. There is a geometric theory, the Ahlfors-Beurling theory of extremal length of curve families, within which the modulus of a ring domain can be defined directly and geometrically, without any preliminary conformal mapping onto some canonical domain. Extremal length can be proved to be a conformal invariant, and then one quickly sees that the two definitions coincide. There is an exposition of the theory of extremal length in *Conformal Invariants* by Ahlfors.
It would be unreasonable to expect to "see" the *exact* value of the modulus of a ring domain. The boundary of a ring domain can be extremely complicated geometrically, and every tiny wiggle impacts on the modulus. But extremal length yields inequalities for the modulus from geometric data. I will quote a single, rather striking, result of this kind:
If a ring domain $R$ contains no circle on the Riemann sphere separating its two boundary components, then $\mathrm{mod}(R) \leq 1/4$. The constant $1/4$ is sharp. The result is due to D. A. Herron, X. Y. Liu and D. Minda.
Small modulus means "thin" ring domain; if the modulus is large enough, the ring domain is so "fat" that it has to contain a separating circle.
| 16 | https://mathoverflow.net/users/3304 | 11243 | 7,642 |
https://mathoverflow.net/questions/11248 | 3 |
>
> What can be said about the polynomials $f\in\mathbb Q[x, y]$ which are nonnegative on $\mathbb R\times \mathbb R$?
>
>
>
**Motivation:** this may lead to progress in the question about [polynomial onto map $\mathbb Z\times \mathbb Z\to\mathbb N$](https://mathoverflow.net/questions/9731/polynomial-representing-all-nonnegative-integers), but I post it separately as it's interesting in itself.
**Note:** there are examples of polynomials nonnegative on $\mathbb Z\times \mathbb Z$, but not bounded from below on $\mathbb R\times \mathbb R$, e.g. $(x^2-x)y^2$, so this doesn't apply directly.
| https://mathoverflow.net/users/65 | Nonnegative polynomial in two variables | The following theorem of Artin -- his solution of Hilbert's 17th problem, but in a stronger form than Hilbert himself asked for -- answers the question.
Theorem (Artin, 1927): Let $F$ be a subfield of $\mathbb{R}$ that has a unique ordering, and let $f(t) = f(t\_1,\ldots,t\_n) \in F(t\_1,\ldots,t\_n)$ be a rational function such that
$f(a) \geq 0$ for all $a = (a\_1,\ldots,a\_n) \in F^n$ for which $f$ is defined. Then $f$ is a sum of squares of rational functions with coefficients in $F$.
A proof can be found in Jacobson, *Basic Algebra II*, Section 11.4.
Note that the tempting strengthening -- that if $f$ is a polynomial, it is a sum of squares of polynomials -- is false, as Hilbert himself showed.
| 5 | https://mathoverflow.net/users/1149 | 11256 | 7,650 |
https://mathoverflow.net/questions/11240 | 31 | I'm trying to understand the notion of an *accessible category*. This isn't the first time I've tried to do this; but every time I try to make sense of the definition, I become perturbed by the following issue: *not every small category is accessible*.
You can find a definition of accessible category at [nLab](http://ncatlab.org/nlab/show/accessible+category). In brief a (possibly large) category C is accessible if there's a regular cardinal $\kappa$ such that C has $\kappa$-directed colimits, and that there's a set of $\kappa$-compact objects so that every object C is a $\kappa$-directed colimit of things in this set. Thus, the behavior of the category C is somehow "controlled" by a small subcategory; roughly speaking, all objects of C "look like" filtered colimits of objects in the small subcategory.
Any small category is of course "controlled" by a small subcategory, namely itself, so you'd think that all small categories are accessible. Not quite! The correct statement is:
>
> A small category is accessible if and only if it is idempotent complete
>
>
>
This is proved (I think) in Adamek & Rosicki, *Locally Presentable and Accessible Categories*. There is a also a proof of an $\infty$-category version of this in Lurie's *Higher Topos Theory*.
My question is not about the proof of this claim (which I think I understand), but about the underlying motivation for the notion of accessible category. Basically, I'd like one of two things:
1. Make me understand why it's such a good thing that not every small category is accessible, or
2. Tell me that "accessible category" is not *exactly* the right idea, and that there's a generalization of it which includes all small categories a special case.
(Note: the class of accessible categories is closed under a bunch of constructions, such as taking undercategories, or taking functors from a fixed small category. The generalization of 2 ought to have the same properties.)
| https://mathoverflow.net/users/437 | Why aren't all small categories accessible? | To me, the "obvious" guess at (2) would be a category whose idempotent-splitting-completion (aka "Cauchy completion" or "Karoubi envelope") is accessible. While I don't have an explicit counterexample, I doubt that these have all the same good properties. The two properties you mention are special cases of closure under pseudo-limits, but the pseudo-limit of a Karoubi envelope is not in general the same as the Karoubi envelope of the pseudo-limit. For instance, let $C$ be the "walking split idempotent", containing two objects $x$ and $y$ with $y$ a retract of $x$, let $F,G\colon C\to Set$ send $x$ to a set $S$ and $y$ to a nonempty subset $T\subseteq S$ with a chosen retraction $S\to T$, and let $\alpha,\beta\colon F\to G$ be natural transformations which are equal on $S$ but not on all of $T$. Then the *equifier* of $\alpha$ and $\beta$ consists only of $y$, whereas if $C'\subseteq C$ contains only $x$, then the equifier of $\alpha$ and $\beta$ restricted to $C'$ is empty.
One answer to (1) is to consider some of the other characterizations of accessible categories. For instance, a category is accessible iff:
* it is the category of $\kappa$-flat functors from some small category to Set (for some $\kappa$), or iff
* it is the category of models in Set of a small sketch, or iff
* it is the category of models in Set of some suitable logical theory.
These sorts of categories clearly always have split idempotents.
| 17 | https://mathoverflow.net/users/49 | 11264 | 7,657 |
https://mathoverflow.net/questions/4463 | 6 | It seems that most authors use the phrase "elementary number theory" to mean "number theory that doesn't use complex variable techniques in proofs."
I have two closely related questions.
1. Is my understanding of the usage of "elementary" correct?
2. It appears that advanced techniques from other areas (e.g. algebra) are allowed, just not complex variables. Are there historical reasons for why complex analysis singled out as a tool to avoid?
NB: I'm asking about how "elementary" usually **is** defined and why, not how it **should be** defined.
| https://mathoverflow.net/users/136 | Definition of elementary number theory | Your usage of "elementary" is correct; your definition is the one that most number theorists would use. You don't have to take my word for it however; just consider the first sentence of [Selberg's Elementary Proof of the Prime Number Theorem](http://www.jstor.org/pss/1969455):
*In this paper will be given a new proof of the prime-number theorem, which is elementary in the sense that it uses practically no analysis, except the simplest properties of the logarithm.*
Ironically, of the many known proofs of the prime-number theorem, this *elementary* proof ranks as one of the most complicated.
| 4 | https://mathoverflow.net/users/nan | 11265 | 7,658 |
https://mathoverflow.net/questions/11246 | 2 | Let $a\_n$ be a bounded sequence of positive real numbers. Is it the case that the Dirichlet series $\sum \frac{a\_n}{n^s}$ can be meromorphically continued up to the right of zero, with at the most a pole at $1$?
This question occurred to me while trying to extend the Dedekind zeta function up to zero. If the above is true, then we get the result.
Related question: What if we assume $a\_n$ to be only a bounded sequence of complex numbers?
| https://mathoverflow.net/users/2938 | Continuation up to zero of a Dirichlet series with bounded coefficients | The following paper seems to (among other things) give a detailed construction roughly along the lines of my comment above:
---
Bhowmik, Gautami, Schlage-Puchta, Jan-Christoph
Natural boundaries of Dirichlet series. (English summary)
Funct. Approx. Comment. Math. 37 (2007), part 1, 17--29.
In this paper, the authors prove some conditions for the existence of natural boundaries of Dirichlet series, and give applications to the determination of asymptotic results.
Let $n\_\nu$ be rational integers, assume the series $\sum\frac{n\_\nu}{2^{\epsilon\nu}}$ converges absolutely for every $\epsilon>0$, and let $\mathcal{P}$ be the set of prime numbers $p$ such that $n\_p>0$. Assume that the Riemann $\zeta$-function has infinitely many zeros on the line $\frac{1}{2}+it$, and suppose that $f$ is a function of the form $$ f(s)=\prod\_{\nu\geq1}\zeta\left(\mu\left(s-\frac{1}{2}\right)+\frac{1}{2}\right)^{n\_\nu}. $$ Then $f$ is holomorphic in the half-plane $\Re s>1$ and has a meromorphic continuation to the half-plane $\Re s>\frac{1}{2}$. If, for all $\epsilon>0$, $ \mathcal{P}((1+\epsilon)x)-\mathcal{P}(x)\gg x^{\frac{\sqrt{5}-1}{2}}\log^2x, $ then the line $\Im s =\frac{1}{2}$ is the natural boundary of $f$; more precisely, every point of this line is an accumulation point of zeros of $f$.
As an example on the existence of a natural boundary, $\Omega$-results for Dirichlet series associated to counting functions are obtained. It is proved that if $D(s)=\sum\frac{a(n)}{{n^s}}$ has a natural boundary at $\Re s=\sigma$, then there does not exist an explicit formula of the form $ A(s) := \sum\_{n\leq x}a\_n=\sum\_{\rho}c\_\rho x^\rho+O(x^\sigma), $ where $\rho$ is a zero of the Riemann zeta-function, and hence it is possible to obtain a term $\Omega(x^{\sigma-\epsilon})$ in the asymptotic expression for $A(x)$.
Reviewed by Roma Kačinskaitė
---
In the above review, where $\Im(s) = \frac{1}{2}$ appears, I'm sure $\Re(s) = \frac{1}{2}$ is intended. Also the "assume" is a bit strange, since it is an old, famous theorem of G.H. Hardy that $\zeta(s)$ has infinitely many zeros on the critical line.
| 3 | https://mathoverflow.net/users/1149 | 11266 | 7,659 |
https://mathoverflow.net/questions/11268 | 2 | Let ${\mathcal D}$ be a triangulated category, ${\mathcal C}$ a triangulated subcategory and $Q: {\mathcal D}\to {\mathcal D}/{\mathcal C}$ the corresponding Verdier-localization. Now suppose we have a triangulated functor ${\mathbb F}: {\mathcal D}\to {\mathcal T}$ to some other triangulated category ${\mathcal T}$.
My question is the following: Under which circumstances do we have some kind of "right derived" functor of ${\mathbb F}$ with respect to ${\mathcal C}$? By that I mean a triangulated functor $\textbf{R}{\mathbb F}: {\mathcal D}/{\mathcal C}\to {\mathcal T}$ together with a natural transformation ${\mathbb F}\Rightarrow \textbf{R}{\mathbb F}\circ Q$ which is initial with this property.
Does there exist such a treatment of derived functors in arbitrary triangulated categories?
Thank you.
| https://mathoverflow.net/users/3108 | Derived Functors in arbitrary triangulated categories | Yes, there exists such a treatment by Deligne, see "Cohomologie a supports propres", SGA4, Tome 3, Lect. Notes Math. 305, subsections 1.2.1-1.2.2. Basically, what one needs is that for any object X in D there exists a morphism X→Y in D with a cone in C such that for any morphism Y→Z in D with a cone in C there exists a morphism Z→W in D with a cone in C such that F(Y)→F(W) is an isomorphism. Then one defines RF(X) as F(Y).
| 7 | https://mathoverflow.net/users/2106 | 11272 | 7,663 |
https://mathoverflow.net/questions/10919 | 7 | Two questions:
1. Suppose *a* and *b* are two uncountable cardinals. Consider the symmetric groups on sets of sizes *a* and *b* respectively (the symmetric group on a set is the group of all bijections from the set to itself, under composition). Consider the first-order theories of these as "pure groups" (i.e., just the group structure, no additional information). Are these elementarily equivalent? (does the answer change if we allow one of *a* or *b* to be the countable cardinal?)
2. Suppose $a\_1$, $a\_2$, $b\_1$ and $b\_2$ are uncountable cardinals with $a\_1 < a\_2$ and $b\_1 < b\_2$. Consider the symmetric group on a set of size $a\_2$ and the subgroup of those bijections that have support of size at most $a\_1$. Consider the pure theory of this group-subgroup pair (i.e., the pure theory of the group along with a membership predicate for the subgroup). Similarly, for $b\_1$ and $b\_2$. Are these two pure theories elementarily equivalent? Does the answer change if, instead, we look at the subgroup of those bijections that have support of size strictly less than $a\_1$? Does the answer change if we allow the countable cardinal? (The support of a bijection is the set of elements that are moved).
By the Baer-Schreier-Ulam theorem, the only normal subgroups of symmetric groups on infinite sets are the subgroups comprising bijections with support of size strictly less than *a* for some infinite cardinal *a* or the subgroups comprising bijections with support of size less than or equal to *a* for some infinite cardinal *a*, plus the trivial subgroup and the finitary alternating group. All these are also characteristic subgroups.
If (2) is true, this would give examples of distinct characteristic subgroups of the same group that are elementarily equivalently embedded as subgroups (i.e., there is no first-order sentence true for one subgroup that is not true for the other).
| https://mathoverflow.net/users/3040 | elementary equivalence of infinitary symmetric groups | For each ordinal $\alpha < \omega^\omega$ the symmetric group on $\aleph\_{\alpha}$ is first order definable in the class of all symmetric groups; i.e. there is a sentence in the first order language of groups that is true in $Sym(A)$ iff $|A| = \aleph\_\alpha$. See Mckenzie - On elementary types of symmetric groups, Algebra Universalis 1 (1971), 13--20.
Shelah provides an interesting necessary and sufficient condition for the elementary equivalence of $Sym(\aleph\_\alpha)$ and $Sym(\aleph\_\beta)$ - First order theory of permutation groups, Israel J. Math. 14 (1973), 149–162; corrections: ibid. 15 (1973), 437–44.
| 12 | https://mathoverflow.net/users/2689 | 11276 | 7,665 |
https://mathoverflow.net/questions/11283 | 11 | This question is partly inspired by a problem in Stewart's Calculus: "Find the area of the region enclosed by $y=x^2$ and $x=y^2$."
Suppose $f\colon [0,1]\to [0,1]$ is a convex increasing function that fixes 0 and 1. The graphs $y=f(x)$ and $x=f(y)$ enclose a convex region. To find its area, students will likely integrate the length of vertical sections: $f^{-1}(x)-f(x)$. They are unlikely to ask if two different functions can yield the same integrand. But I'll ask:
>
> Does $f^{-1}-f$ determine $f$? (Among all convex increasing functions that fix 0 and 1).
>
>
>
| https://mathoverflow.net/users/2912 | Determination of a symmetric convex region by parallel sections | Yes.
Assume we have two distinct functions $f$ and $g$ such that $f^{-1}-f\equiv g^{-1}-g$.
Take a sequence $x\_n=f(x\_{n-1})$.
Clearly $f(x\_n)-g(x\_n)=0$ or $(-1)^n[f(x\_n)-g(x\_n)]$ has the same sign for all $n$.
Sinse $\int\_0^1f=\int\_0^1g$, there are two sequences $x\_n$ and $y\_n$ as above such that $f(x\_n)=g(x\_n)$, $f(y\_n)=g(y\_n)$ and say $(-1)^n[f(x)-g(x)]>0$ for any $x\in(x\_n,y\_n)$.
Note that $x\_n,y\_n\to 0$ and $\int\_{x\_n}^{y\_n}|f-g|=const>0$.
It follows that $\limsup\_{x\to0} |f(x)-g(x)|\to\infty$, a contradiction.
| 9 | https://mathoverflow.net/users/1441 | 11288 | 7,672 |
https://mathoverflow.net/questions/11295 | 1 | (I am a very, very new to mathematics, so I apologise in advance for posing a question so basic, but am out of ideas).
In [*Idoneal Numbers and some Generalizations*](http://www.mast.queensu.ca/~kani/papers/idoneal.pdf), pp. 15, Ernst Kani quotes Euler's criterion for idoneal numbers:
>
> An integer n ≥ 1 is idoneal if and only if for every k = 1,..., [√(n∕3)] with (k, n) = 1 we have that n + k2 = p, p2 , 2p or 2s, for some odd prime p and some integer s ≥ 1.
>
However, my interpretation of this criterion leads to false positives: non-idoneal integers being recognised as idoneal. Kani discusses that others were dissatisfied with this formulation, but it is my understanding that these criticisms are mainly theoretical. That is, the formulation above is supposed to correctly identify all known idoneal numbers.
I take k to be every value between 1 and √(n∕3) inclusive that is coprime to n. I assume k should only take integer values, but am unsure of how to round the square root, i.e. nearest, floor, or ceiling.
For every value of k I compute sum = n + k2. If sum is odd and prime, I take this to be the n + k2 = p result. If sum is a perfect square whose square root is prime, I take it to be the n + k2 = p2 result. If sum is even and sum∕2 is prime, I assume n + k2 = 2p. Lastly, if sum is a power of 2 whose exponent is ≥ 1, I assume
n + k2 = 2s.
I require one of the above results for each value of k to regard n as idoneal.
Let n = 36. √(n∕3) = 3.4641016151377544. Therefore, k = 1, 2, or 3. Only 1 is coprime to n, so this is k's sole value. n + k2 = p = 36 + 1 = 37. 37 is both odd and prime. This seems to satisfy the criterion as I understand it, but 36 is not an idoneal number. n = 100 is but one of other false positives.
Any clues on either how to interpret this criterion correctly or a better algorithm (short of brute force) to recognise idoneal numbers?
| https://mathoverflow.net/users/3114 | Interpreting Euler's Criterion for Idoneal Numbers | In the remark below the statement of Euler's criterion in the paper you linked to, notice that Grube, who tried to correct Euler's original "proof" of this criterion, actually only provided a correct version in one direction -
>
> Frei [17], p. 57, points out that Grube only proved one direction of this criterion and
> says, “whether [this criterion] is also sufficient is still an open problem”
>
>
>
which explains why you might get false positives.
| 0 | https://mathoverflow.net/users/1916 | 11299 | 7,676 |
https://mathoverflow.net/questions/10463 | 8 | Is there a simple example of a smooth proper variety $X$ over $K=\mathbb{Q}\_p$ ($p$ prime) such that
--- $X(K)\neq\emptyset$,
--- the $l$-adic étale cohomology $H^i(X\times\_K\bar K,\mathbb{Q}\_l)$ is unramified for every prime $l\neq p$ and every $i\in\mathbb{N}$,
--- the $p$-adic étale cohomology $H^i(X\times\_K\bar K,\mathbb{Q}\_p)$ is crystalline for every $i\in\mathbb{N}$,
and yet,
--- $X$ is not the generic fibre of any smooth proper $\mathbb{Z}\_p$-scheme ?
[My example](https://mathoverflow.net/questions/416/existence-of-smooth-models/10264#10264) of a Châtelet surface with these properties is simple enough, but can one do better ?
| https://mathoverflow.net/users/2821 | A nice variety without a smooth model | As Minhyong suggests, a curve $C$ (with $C(\mathbb{Q}\_p)\neq0$) which has bad reduction but whose jacobian $J$ has good reduction would do the affair. This works because the cohomology of $C$ is essentially the same as that of $J$, and because an abelian $\mathbb{Q}\_p$-variety has good reduction if and only if its $l$-adic étale cohomology is unramified for some (and hence for every) prime $l\neq p$ (Néron-Ogg-Shafarevich) or its $p$-adic étale cohomology is crystalline (Fontaine-Coleman-Iovita).
I asked Qing Liu for explicit examples. He suggested the curve
$$
y^2=(x^3+1)(x^3+ap^6)\qquad (a\in\mathbb{Z}\_p^\times)
$$
when $p\neq2,3$, and $y^2=(x^3+x+1)(x^3+a3^4x+b3^6)$, with $a,b\in\mathbb{Z}\_3^\times$, for $p=3$.
He refers to Proposition 10.3.44 in his book for computing the stable reduction of these $C$, and to Bosch-Lütkebohmert-Raynaud,
*Néron models,* Chapter 9, for showing that $J$ has good reduction.
I "accept" this answer as coming from Minhyong Kim and Qing Liu.
| 3 | https://mathoverflow.net/users/2821 | 11302 | 7,678 |
https://mathoverflow.net/questions/11306 | 44 | From the example $D\_4$, $Q$, we see that the character table of a group doesn't determine the group up to isomorphism. On the other hand, Tannaka duality says that a group $G$ is determined by its representation ring $R(G)$.
What is the additional information contained in $R(G)$ as opposed to the character table?
| https://mathoverflow.net/users/nan | Character table does not determine group Vs Tannaka duality | If I'm not mistaken, the extra information is not contained in the representation ring, you have to look at the category of representations. In particular, you want to look at the representation category equipped with its forgetful functor to vector spaces. Then the group can be recovered as the automorphisms of this functor. Here's a blogpost I wrote which may be helpful: <http://concretenonsense.wordpress.com/2009/05/16/tannaka%E2%80%93krein-duality/>
| 23 | https://mathoverflow.net/users/321 | 11308 | 7,680 |
https://mathoverflow.net/questions/11307 | 3 | What would you call the product of an annulus and $S^1$ (a 'thickened' torus like 3-manifold)?
More generally, is there an archive or list online of names assigned to various (non-standard) manifolds by people? Or a set convention by which to name them?
| https://mathoverflow.net/users/3121 | What do you call the product of a circle and an annulus? | I would call it a thickened torus. I don't know how standard that is, but it is quite normal to speak of thickened manifolds, where one means that manifold times a closed interval.
I have long felt that there should be a mathematical dictionary- not an encyclopaedia, by a dictionary- in order to fix and record standard usage.
| 6 | https://mathoverflow.net/users/2051 | 11315 | 7,684 |
https://mathoverflow.net/questions/11296 | 39 | A set-theoretical reductionist holds that sets are the only abstract objects, and that (e.g.) numbers are identical to sets. (Which sets? A reductionist is a relativist if she is (e.g.) indifferent among von Neumann, Zermelo, etc. ordinals, an absolutist if she makes an argument for a priviledged reduction, such as identifying cardinal numbers with equivalence classes under equipotence). Contrasting views: classical platonism, which holds that (e.g.) numbers exist independently of sets; and nominalism, which holds that there are no abstract particulars.
I'm interested in the relationship between "structuralism" as it is understood by philosophers of science and mathematics and the structuralist methodology in mathematics for which Bourbaki is well known. A small point that I'm hung up on is the place of set theory in Bourbaki structuralism. I'm weighing two readings.
* (1) conventionalism: Bourbaki used set theory as a convenient "foundation", a setting in which models of structures may be freely constructed, but "structure" as understood in later chapters is not essentially dependent on the formal theory of structure developed in Theory of Sets,
* (2) reductionism: sets provide a ground floor ontology for mathematics; mathematicians study structures in the realm of sets.
In favor of conventionalism:
* (a) Leo Corry's arguments in “Nicolas Bourbaki and the Concept of Mathematical Structure” that the formal structures of Theory of Sets are to be distinguished from and play only a marginal role in the subsequent investigation of mathematical structure,
* (b) ordered pairs: definitions reducing pairs to sets like Kuratowski's bring "baggage" (i.e., extra structure) and Bourbaki used primitive ordered pairs in the first edition of Theory of Sets, showing no excess concern for complete reduction,
* (c) statements of Dieudonne to the Romanian Institute indicating chs. 1 and 2 are mostly to satisfy bothersome philosophers (like me I suppose) before getting on to topics of greater interest,
* (d) the discussion of axiomatics and structure in "The Architecture of Mathematics", placing no special emphasis on sets,
* (e) this interpretation serves my selfish philosophical agenda.
In favor of reductionism:
* (a) linear ordering of texts suggests perceived logical dependence on Theory of Sets,
* (b) reductionism makes sense of unity of mathematics,
* (c) 1970 edition includes Kuratowski pairs,
* (d) makes sense of controversies over category theory,
* (e) makes sense of some outsider criticisms (e.g., Mac Lane in "Mathematical Models" that Bourbaki was dogmatic and stifling),
* (f) I fear that in leaning towards conventionalism I'm self-deceiving to serve my selfish philosophical agenda.
Apologies: not sure this is MO appropriate, any answers may be anachronistic, probably no univocality of opinion among Bourbaki members, my views are based on popular expositions, interviews, and secondary literature and not close study of the primary texts.
Discussion related to this question has recently occurred at [n-category cafe](http://golem.ph.utexas.edu/category/2009/11/interview_with_manin.html), occasioned by Manin's recent claim that Bourbaki provided "pragmatic foundations". The conventionalist interpretation, I think, helps make sense of Manin's claim and would show some criticisms levelled toward Bourbakism to misapprehend their intention (if not their impact). I have Borel's "Twenty-Five Years With Bourbaki" which discusses Grothendieck and the controversy over the direction following the first six books. Corry makes the claim that the Theory of Sets approach had limitations in dealing with category theory. I would especially appreciate references or answers that help me better understand these issues in particular, which are accessible to a philosopher with some grad coursework in mathematics and with only a self taught rudimentary understanding of categories.
| https://mathoverflow.net/users/2833 | Were Bourbaki committed to set-theoretical reductionism? | First, most mathematicians don't really care whether all sets are "pure" -- i.e., only contain sets as elements -- or not. The theoretical justification for this is that, assuming the Axiom of Choice, every set can be put in bijection with a pure set -- namely a von Neumann ordinal.
I would describe Bourbaki's approach as "structuralist", meaning that all structure is based on sets (I wouldn't take this as a philosophical position; it's the most familiar and possibly the simplest way to set things up), but it is never fruitful to inquire as to what kind of objects the sets contain. I view this as perhaps the key point of "abstract" mathematics in the sense that the term has been used for past century or so. E.g. an abstract group is a set with a binary law: part of what "abstract" means is that it won't help you to ask whether the elements of the group are numbers, or sets, or people, or what.
I say this without having ever read Bourbaki's volumes on Set Theory, and I claim that this somehow strengthens my position!
Namely, Bourbaki is relentlessly linear in its exposition, across thousands of pages: if you want to read about the completion of a local ring (in Commutative Algebra), you had better know about Cauchy filters on a uniform space (in General Topology). In places I feel that Bourbaki overemphasizes logical dependencies and therefore makes strange expository choices: e.g. they don't want to talk about metric spaces until they have "rigorously defined" the real numbers, and they don't want to do that until they have the theory of completion of a uniform space. This is unduly fastidious: certainly by 1900 people knew any number of ways to rigorously construct the real numbers that did not require 300 pages of preliminaries.
However, I have never in my reading of Bourbaki (I've flipped through about five of their books) been stymied by a reference back to some previous set-theoretic construction. I also learned only late in the day that the "structures" they speak of actually get a formal definition somewhere in the early volumes: again, I didn't know this because whatever "structure-preserving maps" they were talking about were always clear from the context.
Some have argued that Bourbaki's true inclinations were closer to a proto-categorical take on things. (One must remember that Bourbaki began in the 1930's, before category theory existed, and their treatment of mathematics is consciously "conservative": it's not their intention to introduce you to the latest fads.) In particular, apparently among the many unfinished books of Bourbaki lying on the shelf somewhere in Paris is one on Category Theory, written mostly by Grothendieck. The lack of explicit mention of the simplest categorical concepts is one of the things which makes their work look dated to modern eyes.
| 29 | https://mathoverflow.net/users/1149 | 11319 | 7,686 |
https://mathoverflow.net/questions/11322 | 10 | Somewhere in Mumford's GIT, he seems to imply that any linear algebraic group is rational? This seems strange to me. Is it true?
| https://mathoverflow.net/users/nan | any linear algebraic group rational? | A (reduced, irreducible) linear algebraic group over an algebraically closed field is rational -- i.e., birational to projective space. This is a nontrivial result.
Briefest sketch of proof [EDIT: in characteristic 0 only; see the comment below]: use the Levi decomposition to reduce to the case of reductive groups, then use the Bruhat decomposition to handle the reductive case.
This does not hold for geometrically integral linear groups over an arbitrary ground field. For instance, if $k$ is any field which admits a nondegenerate [i.e., degree $4$] biquadratic extension $l = k(\sqrt{a},\sqrt{b})$, then the norm torus associated to $l/k$ is a three-dimensional nonrational algebraic torus. I think this example is in some sense minimal.
See the [Springer Online Reference Works](http://eom.springer.de/R/r077670.htm) for more information, including references.
| 18 | https://mathoverflow.net/users/1149 | 11324 | 7,690 |
https://mathoverflow.net/questions/11297 | 6 | Hironaka's theorem guarantees an existence of resolution of singularities in characteristic 0. If I am not wrong, it also guarantees (or at least some other result does), that if the resolution is a singular point, one can get the "Exceptional Fiber" to be a simple normal crossing divisor. Very likely, if the singular locus is of higher dimension, then too one can get the "Exceptional Fiber" to be a simple normal crossing divisor.
However, if the nature of singularity varies along the singular locus, (perhaps) one cannot expect the dimensions of the fibers at each point to be constant in the given resolution.
What should be the most general result known in this direction? Can one expect, for example, a stratification such that inverse image of each strata, is "like simple normal crossing" (eg smooth irreducible components, as well as all k-fold intersections being smooth)?
| https://mathoverflow.net/users/2870 | Resolution of singularities, nature of | Hironaka in fact says that you can resolve singularities by a sequence of blow ups, and the universal property of blowing up is that the exceptional locus is a Cartier divisor. So in fact, the exceptional locus of the whole thing will be a Cartier divisor. Making sure that the exceptional locus is a snc divisor is called "embedded resolution" and is also known to be true. This is covered by Kollár in his book *[Lectures on Resolution of Singularities](http://books.google.com/books?id=Oygejj1QFhgC&printsec=frontcover&dq=kollar+singularities&ei=yOVJS4e8HZTIzAS0i-z0DQ&cd=1#v=onepage&q=kollar%2520singularities&f=false)*, which I believe is an expanded version of his notes *Resolution of Singularities – Seattle Lecture*, [arXiv:math/0508332](https://arxiv.org/abs/math/0508332), but also pretty much everywhere else that proves resolution of singularities.
| 8 | https://mathoverflow.net/users/622 | 11326 | 7,692 |
https://mathoverflow.net/questions/11327 | 13 | This is a very minor point, but one which had been grating me for a while. I apologize for asking a relatively trivial question, but nevertheless hope that it is suitable for MO since it should have a definite answer.
In Mumford's books, for instance Curves on Surfaces or Red Book, there is thing called "prescheme" which looks like a scheme, and scheme is something else.
But this terminology does not seem to be used elsewhere, and if at all is the case, prescheme seems to be something cruder than scheme.
I will be grateful for clarifications regarding this terminology. "Curves on surfaces" is a nice book, but whenever I pick it up I find myself wondering about this without any avail.
| https://mathoverflow.net/users/2938 | Preschemes and schemes | The prescheme usage is outdated. As indicated in [nLab](https://ncatlab.org/nlab/show/EGA),
>
> our schemes are in EGA called preschemes; EGA’s schemes are what we call separated schemes
>
>
>
| 18 | https://mathoverflow.net/users/nan | 11328 | 7,693 |
https://mathoverflow.net/questions/11301 | 29 | The Mumford conjecture states that for each integer $n$, we have: the map $\mathbb{Q}[x\_1,x\_2,\dots] \to H^\ast(M\_g ; \mathbb{Q})$ sending $x\_i$ to the kappa class $\kappa\_i$, is an isomorphism in degrees less than $n$, for sufficiently large $g$. Here $M\_g$ denotes the moduli of genus $g$ curves, and the degree of $x\_i$ is the degree of the kappa class $\kappa\_i$. This conjecture was proved by Madsen-Weiss a few years ago.
1. What are the heuristic or moral reasons for the conjecture? (EDIT: I am particularly interested in algebraic geometric reasons, if there are any. Though algebraic topologial reasons are very welcome too.) What lead Mumford to formulating the conjecture in the first place?
2. I know very little about the Madsen-Weiss proof, but I know that it mainly uses algebraic topology methods. Are there any approaches to the conjecture which are more algebraic-geometric?
3. Is there any analogous theorem or conjecture regarding the (topological) $K$-theory of $M\_g$? Or the Chow ring of $M\_g$? etc.
| https://mathoverflow.net/users/83 | Mumford conjecture: Heuristic reasons? Generalizations? ... Algebraic geometry approaches? | All current proofs of Mumford's conjecture in fact prove a far stronger result, the "Strong Mumford conjecture", first formulated by Ib Madsen. This says the following (where by "moduli space" in the following we mean a homotopy type classifying concordance classes of surface bundles with perhaps some extra structure): there is a stable moduli space
$$\mathcal{M}\_\infty := \mathrm{colim} \,\, \mathcal{M}\_{g, 1}$$
where $\mathcal{M}\_{g, 1}$ denotes the moduli space of genus $g$ surfaces with a single boundary component, and the colimit is formed by gluing on a torus with a single boundary component using the "pair of pants" product.
There is also a space, usually called $\Omega^\infty MTSO(2)$, which classifies cobordism classes of "formal surface bundles": that is, codimension -2 submersions with an orientation of the (stable) vertical tangent bundle. As a surface bundle is a formal surface bundle, there is a map
$$\alpha: \mathcal{M}\_\infty \to \Omega^\infty MTSO(2).$$
The strong Mumford conjecture says that this is an integral homology equivalence. For the record, there are currently four distinct known proofs, due to:
1. [The stable moduli space of Riemann surfaces: Mumford's conjecture](http://arxiv.org/abs/math/0212321), Madsen and Weiss,
- [The homotopy type of the cobordism category](http://arxiv.org/abs/math/0605249) , Galatius, Madsen, Tillmann and Weiss,
- [Monoids of moduli spaces of manifolds](http://arxiv.org/abs/0905.2855), Galatius and myself,
- [Madsen-Weiss for geometrically minded topologists](http://arxiv.org/abs/0907.4226), Eliashberg, Galatius and Mishachev.
For part 1) of your question, from this point of view (I do not know what Mumford had in mind): the map $\alpha$ can be thought of as comparable to the map which compares holonomic sections to formal sections in the statement of Gromov's $h$-principle for a sheaf. The idea is then that given a "formal surface bundle" one may begin improving it to be more and more like a bundle, but in this process one cannot really control the genus of the fibres one ends up with. This is why it gives the infinite genus moduli space. This is more or less the approach to proving the conjecture that Eliashberg, Galatius and Mishachev take. The other approaches are less direct and use more algebraic topological machinery.
So, for part 3) of your question: the map $\alpha$ is also an equivalence in any other (co)homology theory (by the Atiyah-Hirzebruch spectral sequence, for example). Thus the topological K-theory of $\mathcal{M}\_\infty$ is "known" in the sense that it is the K-theory of the infinite loop space of a well-understood spectrum, which fits into various simple cofibration sequences and so on. On the other hand, it is "not known" in the sense that I don't think anyone knows what $K^0(\mathcal{M}\_\infty)$ is as a group, though I once tried to compute it without success. On the other hand, even knowing this group, it does not necessarily tell you anything about what you are really interested in, $K^0(\mathcal{M}\_g)$, because stability for ordinary homology does not imply stability in non-connective (co)homology theories.
| 26 | https://mathoverflow.net/users/318 | 11339 | 7,698 |
https://mathoverflow.net/questions/11318 | 4 | I read that the Brauer-Manin obstruction $A(\mathbb{A}\_K)^{\mathbf{Br}}$ of an Abelian variety $A$ over a number field $K$ equals (naturally?) its Tate-Shafarevich group $\mathrm{III}(A)$.
Is this true? And if so, where can I find a proof?
| https://mathoverflow.net/users/nan | Brauer-Manin obstruction and Tate-Shafarevich group of an Abelian variety | The quotient of what you called the Brauer-Manin obstruction by the closure of $A(K)$ within it is related to the divisible part of Sha. In particular, if Sha has no divisible part (e.g. if it is finite) then the Brauer-Manin obstruction
is the closure of $A(K)$. See L. Wang, Brauer-Manin obstruction to weak approximation on abelian varieties, Israel J. Math. 94 (1996), 189–200.
Note that these two groups in your question are very different, so they can't be equal. For instance, Sha is torsion, but the Brauer-Manin obstruction usually isn't.
| 4 | https://mathoverflow.net/users/2290 | 11340 | 7,699 |
https://mathoverflow.net/questions/11294 | 3 | The Kendall tau distance was originally defined as a correlation coefficient. It seems clear to me that every metric function $d$ that is bounded by $b$, induces a correlation coefficient. That is:
Let $d$ be a metric. Then $-d(x,y)/b+1$ is a correlation coefficient.
I wonder if the converse is true, too:
Let $c(x,y)$ be a correlation coefficient. Then $-c(x,y) + 1$ is a pseudometric.
Is the latter statement true or false?
| https://mathoverflow.net/users/3116 | Do all correlation coefficients induce a pseudometric? | The statement is false. Consider the Pearson sample correlation coefficient:
c(x,y) = (x-mean(x)).(y-mean(y))/sqrt(|x-mean(x)|^2\*|y-mean(y)|^2)
Here is an example, where the triangle inequality for the distance defined in
the question is not satisfied
x = [0.5847 -0.3048 -0.4431 0.5032 -0.3401],
y = [0.2018 0.4547 -0.2230 0.3350 -0.7685],
z = [0.5226 -0.5159 -0.5439 0.3701 0.1671].
| 3 | https://mathoverflow.net/users/1059 | 11341 | 7,700 |
https://mathoverflow.net/questions/11338 | 3 | Iasked me the question what the interpretation of the irreducibility of a moduli space is for the functor it represents. For proper, there is the valuative criterion and for (formally) smooth, there is the infinitesimal lifting property, but I don't know something similar for irreducible. Can someone shed some light on this?
| https://mathoverflow.net/users/nan | functorial meaning of irreducibility of a moduli space | Your question is a bit vague, but let me try to say something. Being irreducible is a global property so there can be no local characterization like being smooth, regular, etc. If $\mathcal{M}$ is the "space" representing your functor, we say that it is irreducible if it admits a surjective map from an irreducible variety (this is just topological).
If you think about what it means to be irreducible, we could also say that any two closed points of $\mathcal{M}$ can be connected by an irreducible variety. From the functorial point of view, this would mean that any two objects being parametrized live in a family over some irreducible variety. So for example, saying that $M\_g$ (the moduli space of smooth geneus $g$ curves is irreducible, is the same as saying that for any two smooth curves $C\_1, C\_2$ of genus $g$ (say over a field), there is a family $f: S \rightarrow B$ such that $B$ is irreducible, every geometric fiber is a smooth genus $g$ curve, and both $C\_1$ and $C\_2$ are fibers of $f$. In fact, you can take $B$ to be a curve.
| 8 | https://mathoverflow.net/users/397 | 11347 | 7,705 |
https://mathoverflow.net/questions/10569 | 14 | [**Edit:** Question 1 has been moved [elsewhere](https://mathoverflow.net/questions/11316/smooth-proper-schemes-over-z-with-points-everywhere-locally) so that an answer to Question 2 can be *accept*ed.]
**Question 2.** *Is there a number field* $K$, *and a smooth proper scheme* $X\to\operatorname{Spec}(\mathfrak{o})$ *over its ring of integers, such that* $X(K\_v)\neq\emptyset$ *for every place* $v$ *of* $K$, *and yet* $X(K)=\emptyset$ ?
I believe the answer is Yes.
**Remark.** Let $K$ be a real quadratic field, $\mathfrak{o}$ the ring of integers of $K$, and $A$ the quaternion algebra over $K$ which is ramified exactly at the two real places. Then the conic $C$ corresponding to $A$ is a smooth projective $\mathfrak{o}$-scheme such that $C(\mathfrak{o})=\emptyset$ (because $C(K\_v)=\emptyset$ for each of the real places $v$). But if we insist that $C(K\_v)\neq\emptyset$ at these two real places $v$, then $A$ would have to split at these $v$ (in addition to all the finite places), and we would have $C=\mathbb{P}\_{1,\mathfrak{o}}$.
More generally, let $K$ be a number field, $\mathfrak{o}$ its ring of integers, and let $C$ be a smooth proper $\mathfrak{o}$-scheme whose generic fibre $C\_{K}$ is a twisted $K$-form of the projective space of some dimension $n>0$. If $C$ has points everywhere locally, then $C=\mathbb{P}\_{n,\mathfrak{o}}$. This remark shows that $X$ cannot be a twisted form of a projective space.
| https://mathoverflow.net/users/2821 | Smooth proper schemes over rings of integers with points everywhere locally | Chandan asked Vladimir and me for an example of an elliptic curve over a real quadratic field that has everywhere good reduction and non-trivial sha, with an explicit genus $1$ curve representing some element of sha. Here's one we found:
The elliptic curve $y^2+xy+y = x^3+x^2-23x-44$ over $\mathbb Q$ (Cremona's reference 4225m1) has reduction type III at 5 and 13. These become I0\* over $K=\mathbb Q(\sqrt{65})$, and I0\* can be killed by a quadratic twist. Specifically, the original curve can also be written as $y^2 = x^3+5x^2-360x-2800$ over $\mathbb Q$, and its quadratic twist over $K$.
$E: \sqrt{65}Uy^2 = x^3+5x^2-360x-2800$
has everywhere good reduction over $K$; here $U = 8+\sqrt{65}$ is the fundamental unit of $K$ of norm $-1$.
Now 2-descent in Magma says that the 2-Selmer group of $E/K$ is $(\mathbb Z/2\mathbb Z)^4$, of which $(\mathbb Z/2\mathbb Z)^2$ is accounted by torsion. So it has either has rank over K or non-trivial Sha[2], and according to BSD its rank is 0 as L(E/K,1)<>0 (again in Magma). Actually, because $K$ is totally real, I think results like those of Bertolini and Darmon might prove that E has Mordell-Weil rank $0$ over $K$ unconditionally. So it has non-trivial Sha[2]. After some slightly painful minimisation, one of its non-trivial elements corresponds to a homogeneous space
$C: y^2 = (23562U+1462)x^4 + (4960U+240)x^3 + (1124U-291)x^2 + (141U-833)x + (50U-733)$
with $U$ as above. So here is a curve such that $J(C)$ has everywhere good reduction and the Hasse principle fails for C.
Hope this helps!
Tim
| 12 | https://mathoverflow.net/users/3132 | 11356 | 7,710 |
https://mathoverflow.net/questions/11354 | 11 | Let $G$ be a finite group. If $L$ is a finite free $\mathbf{Z}$-module with an action of $G$, say $L$ is *induced* if it's isomorphic as a $G$-module to $Ind\_H^G(\mathbf{Z})$ with $H$ a subgroup of $G$ and $H$ acting trivially on $\mathbf{Z}$. And, for want of better terminology, let's say $L$ is a *sum-of-induceds* if it's isomorphic to a direct sum of induced modules in the sense above. [EDIT: Ben Webster points out that "permutation representation" is a rather better name for this notion! It's just the $\mathbf{Z}$-module coming from the action of $G$ on a finite set.]
The question: Is a finite free $\mathbf{Z}$-module which is an extension of one sum-of-induceds by another, also a sum-of-induceds?
Over $\mathbf{Q}$ this is trivial because every short exact sequence splits. But this is not true over $\mathbf{Z}$. For example there are two actions of $\mathbf{Z}/2\mathbf{Z}$ on $\mathbf{Z}^2$ which become the group ring over $\mathbf{Q}$: one has the non-trivial element acting as $(1,0;0,-1)$ and the other has it acting as $(0,1;1,0)$. The latter is a non-split extension of the trivial 1-d representation by the non-trivial one (and also a non-split extension of the non-trivial one by the trivial one). Note that this non-split extension is induced.
There are mod $p$ extensions that don't split either. For example over $\mathbf{Z}/p\mathbf{Z}$ there is a non-trivial extension of the trivial representation by itself. But this extension does not lift to an extension of the trivial $\mathbf{Z}$-module by itself.
Why am I interested? For those that know what a $z$-extension of a connected reductive group over a number field is, my "real" question is: is a $z$-extension of a $z$-extension still a $z$-extension? I've checked the geometric issues here but the arithmetic one above is the one I haven't resolved. $G$ is a Galois group and the $\mathbf{Z}$-modules are the character groups of the central tori in question. If I've understood things correctly, a $z$-extension of a $z$-extension is a $z$-extension iff the question I ask above has a positive answer.
Note finally that applying the long exact sequence of cohomology, and using the fact that induced representations have no cohomology by Shapiro's Lemma, we see that the extension I'm interested in also has no cohomology (and furthermore its restriction to any subgroup has no cohomology either). Is this enough to show it's induced?
| https://mathoverflow.net/users/1384 | Extension of induced reps over Z: is it a sum of induced reps? | In my interpretation of your description, a sum-of-induceds is just the permutation representation of some (not necessarily connected) G-set. The representation $\mathrm{Hom}(V\_1,V\_2)$ for two permutation representations is itself permutation: it's the permutation action on product of the G-sets.
So, $\mathrm{Ext}^1(V\_1,V\_2)=H^1(G,\mathrm{Hom}(V\_1,V\_2))$ which you claim vanishes. So it sounds to me like every extension between permutation representations splits. [EDIT: I'm no longer nervous] I'm still a little nervous about this cohomology vanishing, but I don't know integral representation theory so well, so it's hard for me to say.
| 12 | https://mathoverflow.net/users/66 | 11357 | 7,711 |
https://mathoverflow.net/questions/11366 | 21 | I ran into the following problem. I have some lengthy paper in which I develop some theory to attack some problem. While I was working on this paper, I got some nice result which might interest a bigger audience and can stand by itself (that is, it is interesting out of context of the big work, and has a fairly short proof).
The problem: should I write one long paper or two papers, one long and the other short? If I write just one paper, the short result will strengthen the paper, and it will be, perhaps, more whole. On the other hand, for someone that is only interesting in the short one, it will be better if it appears in a separated paper.
| https://mathoverflow.net/users/2042 | When to split/merge papers? | According to [Gian-Carlo Rota](http://www.math.vt.edu/people/day/advice/YMN4_25.html), one of the secrets of mathematical success is to publish the same result many times.
| 26 | https://mathoverflow.net/users/297 | 11369 | 7,720 |
https://mathoverflow.net/questions/10468 | 9 | I know that many graph problems can be solved very quickly on graphs of bounded degeneracy or arboricity. (It doesn't matter which one is bounded, since they're at most a factor of 2 apart.)
From Wikipedia's article on the clique problem I learnt that finding cliques of any constant size k takes linear time on graphs of bounded arboricity. That's pretty cool.
I wanted to know more examples of algorithms where the bounded arboricity condition helps. This might even be well-studied enough to have a survey article written on it. Unfortunately, I couldn't find much about my question. Can someone give me examples of such algorithms and references? Are there some commonly used algorithmic techniques that exploit this promise? How can I learn more about these results and the tools they use?
| https://mathoverflow.net/users/1042 | Algorithms on graphs of bounded degeneracy/arboricity | Bounded degeneracy or arboricity just means that the graph is sparse (number of edges is proportional to number of vertices in all subgraphs).
Some ideas that have been used for fast algorithms on these graphs:
* Order the vertices so that each vertex has only d neighbors that are later in the ordering, where d is the degeneracy. Then if one can similarly order the structure one is looking for, there are not too many different choices to try. For instance (though this is not the method of the Chiba & Nishizeki paper that you indirectly refer to) one can find all cliques by trying all subsets of later neighbors of each vertex. This idea also works to color these graphs with at most d+1 colors: just choose colors for vertices one at a time in the opposite of the above ordering. See e.g. [Matula and Beck, JACM 1983](http://dx.doi.org/10.1145/2402.322385).
* Find a low degree vertex, do something to it to reduce the size of the graph while preserving its overall sparsity, and continue. This is how one finds an ordering as above (repeatedly remove the smallest degree vertex) and is also how many planar coloring algorithms work.
* Find a big independent set (or a big independent set of bounded-degree vertices), do something on it, and repeat on the remaining smaller graph. This often leads to linear time algorithms because every graph of bounded degeneracy has an independent set of Ω(n) vertices, so the size of the graph goes down by a constant factor at each repetition and the total time can be bounded by a geometric series. This is a variation of the "low degree vertex" idea that works better in the parallel algorithms setting.
* Observe that there can only be very few vertices with high degree (O(dk) vertices with degree greater than n/k) or else they would have too many edges. So if you are looking for a structure that needs high degree vertices you don't have many choices to try. See e.g. [Alon and Gutner, Algorithmica 2009](http://dx.doi.org/10.1007/s00453-008-9204-0).
| 11 | https://mathoverflow.net/users/440 | 11374 | 7,725 |
https://mathoverflow.net/questions/11377 | 2 | The standard definition is that $a\in\mathbb{Z}$ is a primitive root modulo $p$ if the order of $a$ modulo $p$ is $p-1$.
Let me rephrase, to motivate my generalization: $a\in\mathbb{Z}$ is a primitive root modulo $p$ if the linear recurrence defined by $x\_0=1$, $x\_n=ax\_{n-1}$ for $n\geq1$ has maximum possible period in $\mathbb{Z}/p\mathbb{Z}$.
So, we could define $(a\_1,\ldots,a\_r)\in\mathbb{Z}^r$ to be an $r$-primitive root modulo $p$ when the order $r$ linear recurrence defined by $x\_0=\cdots=x\_{r-2}=0$, $x\_{r-1}=1$, and $x\_n=a\_1x\_{n-1}+\cdots+a\_rx\_{n-r}$ for $n\geq r$ has the maximum possible period (for such sequences) in $\mathbb{Z}/p\mathbb{Z}$.
Is anything known about the generalized version of primitive roots I've described? I chose my initial values $x\_0=\cdots=x\_{r-2}=0$, $x\_{r-1}=1$ in analogy with the Fibonacci numbers, but is there a standard / most general choice? The starting values affect the period a lot, so it's important to be working with the "right" ones, I guess. What is the maximum achievable period for a linear recurrence in $\mathbb{Z}/p\mathbb{Z}$? An obvious upper bound is $p^r-1$, but it's not clear to me that this is the correct number to replace $p-1$ in the standard definition of primitive root. Is anything known in the direction of Artin's conjecture for "$r$-primitive roots", as I've called them?
Other thoughts: because there is no "formula" for the order of $a$ modulo $p$ (this would amount to a formula for the discrete logarithm), there certainly isn't a formula for the period of the linear recurrence defined by $(a\_1,\ldots,a\_r)$ modulo $p$. I tried to come up with one for a while before I realized this, so I just wanted to save anyone else the trouble.
| https://mathoverflow.net/users/1916 | Generalization of primitive roots | If $(a\_1,\ldots,a\_r)$ is an $r$-primitive root in your definition, then the polynomial $x^r-a\_1x^{r-1}-\cdots-a\_r$ is irreducible in $Z/pZ[x]$ and any of
its roots is a generator (i.e. a primitive root) of the multiplicative group of the field of $p^r$ elements and conversely. This maximal period is $p^r-1$. See, e.g., Lidl-Niederreiter Finite Fields.
| 7 | https://mathoverflow.net/users/2290 | 11381 | 7,729 |
https://mathoverflow.net/questions/11335 | 18 | This is a question I have since longer time, but I have absolutely no idea how to proceed on it.
Let $p>3$ be a prime. Prove that $\displaystyle\sum\limits\_{k=1}^{p-1}\frac{1}{k}\binom{2k}{k}\equiv 0\mod p^2$. Here, we work in $\mathbb{Z}\_{\mathbb{Z}\setminus p\mathbb{Z}}$ (that is, $\mathbb{Z}$ localized at all numbers not divisible by $p$).
I know that it is $0\mod p$ (though I can't find the reference at the moment; it was some hard olympiad problem on MathLinks). The $0\mod p^2$ assertion is backed up by computation for all $p<100$. I am sorry if this is trivial or known. I would be delighted to see a combinatorial proof (= finding a binomial identity which reduces to the above when computed $\mod p^2$). Some number-theoretical arguments would be nice, too. However, I fear that if you use analytic number theory, I will not understand a single word.
EDIT: Epic fail at question title fixed.
| https://mathoverflow.net/users/2530 | A binomial sum is divisible by p^2 | Your binomial sum is divisible by $p^2$ as is shown in a recent paper of Sun and Tauraso:
<https://arxiv.org/abs/0805.0563>
Even more remarkably, they compute that it's equivalent to $\frac{8}{9}p^2B\_{p-3} \bmod p^3$, as well as various other congruences for the corresponding alternating sum.
| 18 | https://mathoverflow.net/users/3143 | 11401 | 7,742 |
https://mathoverflow.net/questions/11404 | 22 | I tried to read the definition of it on Rubin's book "Euler systems" but it looks highly technical. Can anyone shed some light on it? In particular, is there some starting examples?
The wiki entry is too short and does not contain much useful information.
And what is the motivation of such objects? In particular, why is it called "Euler"?
And if possible, can someone say something about how it is used in number theory? such as Iwasawa theory?
| https://mathoverflow.net/users/1238 | what is an Euler system and the motivation for it? | My understanding is that they're named "Euler systems" because that "Frobenius acting on T" in the definition (line 4, p. 22 of Rubin's book) is an "Euler factor" as in Euler's product decomposition of the Riemann zeta function.
The two easiest examples of Euler systems are the so-called cyclotomic units (not the roots of unity, but slightly more complicated, but still classical, beasts built out of them) and the elliptic units (built out of torsion points on CM elliptic curves). Not coincidentally, these are related to the only two types of global fields that we know how to explicitly construct abelian extensions of. Then there is Kato's more complicated Euler system of Heegner points [EDIT - This phrase was wrong- please see post below!], and Kolyvagin's Euler system of Stickelberger elements. All these are described in Rubin's book, but if you haven't seen them before, it might help to have more references.
If group cohomology is still new to you, the cyclotomic units are the best Euler systems to start with, since you don't need Galois cohomology to define them. (Norm-coherent units map to corestriction-coherent cohomology classes under the Kummer map, which is why these global units form an Euler system in the sense of Rubin's book). Rubin's appendix in Lang's republished books Cyclotomic Fields I and II is easier reading for this. Rubin's Inventiones paper on the main conjecture for CM elliptic curves also contains a nice introduction to the technique.
The application of Euler systems to number theory is the following: Euler systems allow us to bound Selmer groups of p-adic Galois representations. These generalize the Selmer group attached to an abelian variety, the ideal class group, and other objects of arithmetic interest. Bounding them is good because it allows us to prove Iwasawa Main Conjectures, which link the behavior of Selmer groups to p-adic L-functions and encompass basically every classical arithmetic tool for computing p-parts of special values of L-functions.
Washington's book on cyclotomic fields explains this in the cyclotomic example, and the spirit and flavor of the general theory is the same. Coates and Sujatha's book Cyclotomic Fields and Zeta Values is also an excellent introduction to the Euler system technique in the cyclotomic setting.
| 19 | https://mathoverflow.net/users/1018 | 11405 | 7,743 |
https://mathoverflow.net/questions/11393 | 16 | I'm aware to varying extents of the existence of certain decompositions of the space of $k$-forms on a compact complex or compact Riemannian manifold that split into closed, co-closed, and harmonic forms, and that the space of harmonic forms becomes isomorphic to the de Rham cohomology groups. However, these are defined differently, but there seems to be analogy here in the theorems and to some extent in the proofs; this is also suggested by the common name. Is there one? Or is there a more general way to encapsulate all this?
| https://mathoverflow.net/users/344 | Why the similarity between Hodge theory for compact Riemannian and complex manifolds? | For any Riemannian (in particular, for any Hermitean) manifold there is the decomposition: all forms are the direct sum of the harmonic ones, exact ones and those in the image of the conjugate operator of the de Rham differential. Every de Rham cohomology class is represented by a unique $d$-harmonic form.
For any complex (Hermitean) manifold there is a similar theory for the $\bar\partial$ operator, and we get a similar decomposition for each complex $({\cal E}^{p,\bullet},\bar\partial)$ where ${\cal E}^{p,q}$ stands for complex valued smooth $(p,q)$-forms. See e.g. Chern, Complex manifolds. Every Dolbeault cohomology class is represented by a unique $\bar\partial$-harmonic form.
For general complex hermitean manifolds the above decompositions have nothing to do with one another. However, if the metric is Kaehler, some miracles happen:
1. the Laplacians of $d$ and of $\bar\partial$ coincide (more precisely, one is twice the other), so the spaces of harmonic forms coincide as well.
2. the $(p,q)$-projection of a $d$-harmonic form is again $d$-harmonic. Indeed, a local check shows that the $(p,q)$ projection operator commutes with the $d$-Laplacian. So each of the $(p,q)$ components of a $d$-harmonic form is closed and none is exact (if non-zero). So the $(p,q)$-decomposition of forms gives a $(p,q)$-decomposition of the cohomology classes (this does not exist for general complex manifolds). This is the Hodge decomposition.
3. since the conjugate of a $d$-harmonic form is again $d$-harmonic, the $(p,q)$ and the $(q,p)$ parts of the Hodge decomposition are conjugate to one another.
As a consequence of the above, the Hodge-to-de Rham spectral sequence degenerates in the first term. This implies that the $dd^c$-lemma holds for Kaehler manifolds, and hence, they are formal. See Deligne, Griffiths, Morgan, Sullivan, Real homotopy theory of Kaehler manifolds.
The cohomological Hodge decomposition holds also for arbitrary bimeromorphically Kaehler compact complex manifolds (in particular, for smooth, complete but not necessarily projective complex algebraic varieties). See e.g. Peters, Steenbrink, Mixed Hodge structures, p. 49.
| 14 | https://mathoverflow.net/users/2349 | 11407 | 7,744 |
https://mathoverflow.net/questions/11398 | 3 | Hi,
I know that the answer is no, yet I dont know how to prove it wrong. Finding a counterexample is not a good solution because it is a past written exam question with no calculators allowed. The first counterexample is about 30000... Is there a, simple preferred, solution to this problem? (This is asked in a CS discrete math exam)
The original problem is from Rosen Discrete Math, and as follows:
Prove or disprove that $p\_1p\_2 ... p\_n+1$ is prime for every integer n, where $p\_i$ is the ith smallest prime number.
Thanks in advance.
| https://mathoverflow.net/users/3142 | Is the product of first $n$ prime numbers $+1$ another prime number? | Here's a possible intended solution to show that $30031 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1$ is composite without factoring it. Recall the Fermat primality test: if $a^{n-1} \not \equiv 1 \bmod n$, then $n$ cannot be prime. It turns out that $2^{30030} \equiv 21335 \bmod 30031$, so $30031$ must indeed be composite. There is a well-known algorithm called [binary exponentiation](http://en.wikipedia.org/wiki/Exponentiation_by_squaring) that is reasonably fast to implement by hand and that could conceivably be done on an exam. (I am not totally convinced that this would be faster than trial division until $p = 59$, though. And if you followed Leonid's suggestion in the comments your life would be even easier.)
**Edit:** Here is the solution by trial division. It suffices to check the primes from $17$ up. Note that the fact that we know the prime factorization of $30030$ helps a lot. All of the arithmetic necessary beyond what I wrote down was mental.
For $17$ write $30031 = 30 \cdot 77 \cdot 13 + 1 \equiv -4 \cdot 9 \cdot -4 + 1 \equiv -8 \bmod 17$.
For $19$ write $30031 \equiv -8 \cdot 1 \cdot -6 + 1 \equiv 11 \bmod 19$.
For $23$ write $30031 \equiv 7 \cdot 8 \cdot -10 + 1 \equiv -99 \bmod 23$.
For $29$ write $30031 \equiv 1 \cdot -10 \cdot 13 + 1 \equiv -100 \bmod 29$.
For $31$ write $30031 \equiv 30000 \bmod 31$.
For $37$ write $30031 \equiv -7 \cdot 3 \cdot 13 + 1 \equiv -13 \bmod 37$.
For $41$ write $30031 \equiv -11 \cdot 5 \cdot 13 + 1 \equiv -14 \cdot 13 + 1 \equiv -140 \bmod 41$.
For $43$ write $30031 \equiv -13 \cdot -9 \cdot 13 + 1 \equiv -26 \bmod 43$.
For $47$ write $30031 = 210 \cdot 143 + 1 \equiv 25 \cdot 2 + 1 \equiv 4 \bmod 47$.
For $53$ write $30031 \equiv -2 \cdot -16 + 1 \equiv 33 \bmod 53$.
Finally, for $59$ write $30031 \equiv 33 \cdot 25 + 1 \equiv 11 \cdot 16 + 1 \equiv 0 \bmod 59$.
| 10 | https://mathoverflow.net/users/290 | 11412 | 7,747 |
https://mathoverflow.net/questions/11427 | 11 | Can anyone give me references where I would see a detailed exposition of how to translate gauge field theory as known to physicists into the language of connections. I am looking for a detailed exposition on the mathematical formulation of Yang-Mills field theory. Something which might also give an exposition about Chern-Simons theory and the related whole bag of what get called "topological actions"
I had read a nice long discussion on the geometrical formulation of gauge field theory in a post at Terence Tao's blog namely [this article](http://terrytao.wordpress.com/2008/09/27/what-is-a-gauge/%20) and also probably something on Secret Blogging Seminar (but I can't locate that link)
Along similar lines I had seen a very old book by Atiyah and Hitchin on this.
I would like to know what books/expository papers on this are read by graduate students today when they try entering this field?
Also advanced references on the topic would also be helpful.
| https://mathoverflow.net/users/2678 | Looking for reference on gauge fields as connections. | *[Geometry, Topology, and Physics](http://rads.stackoverflow.com/amzn/click/0750306068)* by Nakahara
*[Classical Theory of Gauge Fields](http://rads.stackoverflow.com/amzn/click/0691059276)* by Rubakov
*[Modern Geometry](http://rads.stackoverflow.com/amzn/click/0387961623)*, Part 2 by Dubrovin, Fomenko, and Novikov
| 5 | https://mathoverflow.net/users/1847 | 11433 | 7,760 |
https://mathoverflow.net/questions/11422 | 28 | If we have a finite dimensional Lie algebra g, then the flag variety of g is a projective scheme.
My question is what is Hirzebruch-Riemann-Roch formula for this projective scheme? Are there any interesting results?
I wonder know whether Riemann-Roch in this setting have some beautiful representation theory explanations.
Thanks
| https://mathoverflow.net/users/3156 | What is the Hirzebruch-Riemann-Roch formula for the flag variety of a Lie algebra? | Riemann-Roch for the flag variety is the Weyl Character formula!
More specifically, let $L$ be an ample line bundle on $G/B$, corresponding to the weight $\lambda$. According to [Borel-Weil-Bott](http://en.wikipedia.org/wiki/Borel%25E2%2580%2593Weil%25E2%2580%2593Bott_theorem), $H^0(G/B,L)$ is $V\_{\lambda}$, the irreducible representation of $G$ with highest weight $\lambda$, and $H^i(G/B,L)=0$ for $i>0$. So the holomorphic Euler characteristic of $L$ is $\mathrm{dim} \ V\_{\lambda}$.
As we will see, computing the holomorphic Euler characteristic of $L$ by Hirzebruch-Riemann-Roch gives the [Weyl character formula](http://en.wikipedia.org/wiki/Weyl_character_formula) for $\mathrm{dim} \ V\_{\lambda}$.
Notation:
----------
$G$ is a simply-connected semi-simple algebraic group, $B$ a Borel and $T$ the maximal torus in $B$. The corresponding Lie algebras are $\mathfrak{g}$, $\mathfrak{b}$, $\mathfrak{t}$. The Weyl group is $W$, the length function on $W$ is $\ell$ and the positive roots are $\Phi^{+}$. It will simplify many signs later to take $B$ to be a **lower Borel**, so the weights of $T$ acting on $\mathfrak{b}$ are $\Phi^{-}$.
We will need notations for the following objects:
$$\rho = (1/2) \sum\_{\alpha \in \Phi^{+}} \alpha.$$
$$\Delta = \prod\_{\alpha \in \Phi^{+}} \alpha.$$
$$\delta = \prod\_{\alpha \in \Phi^{+}} (e^{\alpha/2}-e^{-\alpha/2}).$$
They respectively live in $\mathfrak{t}^\*$, in the polynomial ring $\mathbb{C}[\mathfrak{t}^\*]$ and in the power series ring $\mathbb{C}[[\mathfrak{t}^\*]]$.
Geometry of flag varieties
--------------------------
Every line bundle $L$ on $G/B$ can be made $G$-equivariant in a unique way. Writing $x$ for the point $B/B$, the Borel $B$ acts on the fiber $L\_x$ by some character of $T$. This is a bijection between line bundles on $G/B$ and characters of $T$. Taking chern classes of line bundles gives classes in $H^2(G/B)$. This extends to an isomorphism $\mathfrak{t}^\* \to H^2(G/B, \mathbb{C})$ and a surjection $\mathbb{C}[[\mathfrak{t}^\*]] \to H^\*(G/B, \mathbb{C})$. We will often abuse notation by identifiying a power series in $\mathbb{C}[[\mathfrak{t}^\*]]$ with its image in $H^\*(G/B)$.
We will need to know the Chern roots of the cotangent bundle to $G/B$.
Again writing $x$ for the point $B/B$, the Borel $B$ acts on the tangent space $T\_x(G/B)$ by the adjoint action of $B$ on $\mathfrak{g}/\mathfrak{b}$. As a $T$-representation, $\mathfrak{g}/\mathfrak{b}$ breaks into a sum of one dimensional representations, with characters the positive roots. We can order these summands to give a $B$-equivariant filtration of $\mathfrak{g}/\mathfrak{b}$ whose quotients are the corresponding characters of $B$. Translating this filtration around $G/B$, we get a filtration on the tangent bundle whose associated graded is the direct sum of line bundles indexed by the positive roots. So the Chern roots of the tangent bundle are $\Phi^{+}$. (The signs in this paragraph would be reversed if $B$ were an upper Borel.)
The Weyl group $W$ acts on $\mathfrak{t}^\*$. This extends to an action of $W$ on $H^\*(G/B)$. The easiest way to see this is to use the diffeomorphism between $G/B$ and $K/(K \cap T)$, where $K$ is a maximal compact subgroup of $G$; the Weyl group normalizes $K$ and $T$ so it gives an action on $K/(K \cap T)$.
We need the following formula, valid for any $h \in \mathbb{C}[[\mathfrak{t}^\*]]$:
$$\int h = \ \mbox{constant term of}\left( (\sum\_{w \in W} (-1)^{\ell(w)} w^\*h)/\Delta \right). \quad (\*)$$
Two comments: on the left hand side, we are considering $h \in H^\*(G/B)$ and using the standard notation that $\int$ means "discard all components not in top degree and integrate." On the right hand side, we are working in $\mathbb{C}[[\mathfrak{t}^\*]]$, as $\Delta$ is a zero divisor in $H^\*(G/B)$.
**Sketch of proof of `(*)`:** The action of $w$ is orientation reversing or preserving according to the sign of $\ell(w)$. So $\int h = \int (\sum\_{w \in W} (-1)^{\ell(w)} w^\*h) / |W|$. Since the power series $\sum\_{w \in W} (-1)^{\ell(w)} w^\*h$ is alternating, it is divisible by $\Delta$ and must be of the form $\Delta(k + (\mbox{higher order terms}))$ for some constant $k$. The higher order terms, multiplied by $\Delta$, all vanish in $H^\*(G/B)$, so we have $\int h = k \int \Delta/|W|$. The right hand side of $(\*)$ is just $k$.
By the Chern root computation above, the top chern class of the tangent bundle is $\Delta$. So $\int \Delta$ is the (topological) Euler characteristic of $G/B$. The [Bruhat decomposition](http://en.wikipedia.org/wiki/Bruhat_decomposition) of $G/B$ has one even-dimensional cell for every element of $W$, and no odd cells, so $\int \Delta = |W|$ and we have proved formula $(\*)$.
The computation
---------------
We now have all the ingredients. Consider an ample line bundle $L$ on $G/B$, corresponding to the weight $\lambda$ of $T$. The Chern character is $e^{\lambda}$. HRR tells us that the holomorphic Euler characteristic of $L$ is
$$\int e^{\lambda} \prod\_{\alpha \in \Phi^{+}} \frac{\alpha}{1 - e^{- \alpha}}.$$
Elementary manipulations show that this is
$$\int \frac{ e^{\lambda + \rho} \Delta}{\delta}.$$
Applying $(\*)$, and noticing that $\Delta/\delta$ is fixed by $W$, this is
$$\mbox{Constant term of} \left( \frac{1}{\Delta} \frac{\Delta}{\delta} \sum\_{w \in W} (-1)^{\ell(w)} w^\* e^{\lambda + \rho} \right)= $$
$$\mbox{Constant term of} \left( \frac{\sum\_{w \in W} (-1)^{\ell(w)} e^{w(\lambda + \rho)}}{\delta} \right).$$
Let $s\_{\lambda}$ be the character of the $G$-irrep with highest weight $\lambda$. By the Weyl character formula, the term in parentheses is $s\_{\lambda}$ as an element of $\mathbb{C}[[\mathfrak{t}^\*]]$. More precisely, a character is a function on $G$. Restrict to $T$, and pull back by the exponential to get an analytic function on $\mathfrak{t}$. The power series of this function is the expression in parentheses. Taking the constant term means evaluating this character at the origin, so we get $\dim V\_{\lambda}$, as desired.
| 53 | https://mathoverflow.net/users/297 | 11440 | 7,764 |
https://mathoverflow.net/questions/11091 | 20 | I was sure it is known, but it appears to be an open problem (see the answer of Terry Tao).
>
> Assume for a group $G$ there is a
> polynomial $P$ such that given
> $n\in\mathbb N$ there is set of
> generators $S=S^{-1}$ such that
> $$|S^n|\leqslant P(n)\cdot|S|\ \ (\text{or stronger condition}\ |S^k|\leqslant P(k)\cdot|S|\ \text{for all}\ k\le n) .$$
> Then $G$ is virtually nilpotent (or equivalently it has polynomial growth).
>
>
>
**Comments:**
* Note that typically, $|S|\to \infty$ as $n \to\infty$ (otherwise it follows easily from the original Gromov's theorem).
* If it is known, then it would give a group-theoretical proof that manifolds with almost non-negative Ricci curvature have virtually nilpotent fundamental group (see [Kapovitch--Wilking, "Structure of fundamental groups..."](https://arxiv.org/abs/1105.5955)). This proof would use only one result in diff-geometry: Bishop--Gromov inequality.
**Edit:** The actual answer is "Done now: [arxiv.org/abs/1110.5008](http://arxiv.org/abs/1110.5008)" --- it is a comment to the accepted answer. (Published reference: *E. Breuillard, B. Green, T. Tao. The structure of approximate groups. Publ. Math. Inst. Hautes Études Sci. 116 (2012), 115–221.* [Link under Springer paywall](https://link.springer.com/article/10.1007%2Fs10240-012-0043-9))
| https://mathoverflow.net/users/1441 | А generalization of Gromov's theorem on polynomial growth | My [paper with Shalom](http://terrytao.wordpress.com/2009/10/23/a-finitary-version-of-gromovs-polynomial-growth-theorem/) does settle the question when $S = S\_n$ is known to have size polynomial in n (and maybe is allowed to grow just a little bit faster than this, something like $n^{(\log \log n)^c}$ or so), but I doubt that the result is known yet if S is allowed to be arbitrarily large.
Note that even the bounded case is nontrivial - it's not obvious why having $|S\_n^n| \leq n^{O(1)} |S\_n|$ implies polynomial growth. (There is no reason why growth has to be uniform for fixed cardinality of generators; for instance, I believe it is a major open problem (due to Gromov?) as to whether exponential growth is the same as uniform exponential growth for finitely generated groups.)
If we had a good non-commutative Freiman theorem, then one may possibly be able to settle your question affirmatively (note from the pigeonhole principle that if $|S\_n^n| \leq n^C |S\_n|$ for some large n, then there exists an intermediate $m=m\_n$ between 1 and n such that the set $B := S\_n^m$ has small doubling, in the sense that $|B^2| = O(|B|)$). The best result in this direction for general groups currently is [due to Hrushovski](http://arxiv.org/abs/0909.2190), which does show that sets of small doubling contain some vaguely "virtually nilpotent" structure, but it is not yet enough to give Gromov's theorem, let alone the generalisation you mention above. More is known if one already has some additional structure on the group (e.g. it has a faithful linear representation of bounded dimension, or if it is already known to be virtually solvable).
| 17 | https://mathoverflow.net/users/766 | 11447 | 7,770 |
https://mathoverflow.net/questions/11435 | 12 | Suppose *A* is a set and *S* is a collection of subsets closed under arbitrary unions and intersections. Can we find a collection *F* of functions from *A* to itself such that a subset *B* of *A* is in *S* if and only if $f(B) \subseteq B$ for all $f \in F$ (in other words, is *S* precisely the collection of invariant subsets under a collection of functions)?
P.S.: I don't really know what subject tag to give this, so I'm giving it "combinatorics", which seems the closest, though it is more like a question from lattice theory.
| https://mathoverflow.net/users/3040 | Collection of subsets closed under union and intersection | The answer is **Yes**. Furthermore, such a family can be found of size at most the cardinality of A, even when S is much larger.
The key to the solution is to realize that every such family S arises as the collection of downward-closed sets for a certain partial pre-order on A, which I shall define. (Conversely, every such order also leads to such a family.)
An interesting special case occurs when the family S is linearly ordered by inclusion. For example, one might consider the family of cuts in the rational line, that is, downward-closed subsets of Q. (I had thought briefly at first that this might be a counterexample, but after solving it, I realized a general solution was possible by moving to partial orders.)
Suppose that S is such a collection of subsets of A. Define the induced partial pre-order on A by
* a <= b if whenever B in S and b in B, then also a in B.
It is easy to see that this relation is transitive and reflexive.
I claim, first, that S consists of exactly the subsets of A that are downward closed in this order. It is clear that every set in S is downward closed in this order. Conversely, suppose that X is downward closed with respect to <=. For any b in X, consider the set Xb, which the intersection of all sets in S containing b as an element. This is in S. Also, Xb consists of precisely of the predecessors of b with respect to <=. So Xb subset X. Thus, X is the union of the Xb for b in X. So X is in S.
Next, define fa(b) = a if a <= b, and otherwise fa(b) = b. Let F be the family of all such functions fa for a in A.
Clearly, every B in S is closed under every fa, by the definition of <=. Conversely, suppose that X is closed under all fa. Thus, whenever b is in X and a <= b, then a is in X also. So X is downward closed, and hence by the claim above, X is in S.
Incidently, the sets S are exactly the open sets in the topology on A induced by the lower cones of <=.
| 15 | https://mathoverflow.net/users/1946 | 11451 | 7,772 |
https://mathoverflow.net/questions/11436 | 1 | Suppose we have a monoidal category $(\mathcal{C},\otimes)$. I am interested in the conditions or situations where the following do and do not hold:
1. For any objects *A*, *B* of $\mathcal{C}$, the induced map $\operatorname{End}(A) \times \operatorname{End}(B) \to \operatorname{End}(A \otimes B)$ is injective (by induced map, I mean the map induced by the bifunctoriality).
2. For any objects *A*, *B* of $\mathcal{C}$, the induced map $\operatorname{Aut}(A) \times \operatorname{Aut}(B) \to \operatorname{Aut}(A \otimes B)$ is injective (this is a restriction of the map in (1)).
(1) is stronger than (2). I think I have a proof that in the case that $\otimes$ is a product or coproduct (i.e., we have a Cartesian or co-Cartesian (?) monoidal category) [EDIT: My proof was wrong, as the counterexamples below show], both (1) and (2) hold, i.e., the maps on the two factors are "determined" by the map on the (co)-product. On the other hand, when $\otimes$ is the tensor product of modules over a commutative unital ring, (1) and (2) need not hold, though I think they do hold when the ring is a field.
So my question is: what are more examples where (1) and (2) do hold, what are more examples where they don't, and are there some other conditions/properties which would imply (1) and (2)?
ADDED LATER: It seems that zero objects provide some immediate counterexamples to (1) and (2) even for the category of sets. So I'm modifying (1) and (2) to the following:
1.' For what objects *A*, *B*, is the induced map $\operatorname{End}(A) \times \operatorname{End}(B) \to \operatorname{End}(A \otimes B)$ injective? i.e., How would we characterize such *A* and *B*? The most general possibility seems to be if it is true for all *A* and *B* that are not initial or final objects.
2.' For what objects *A*, *B*, is the induced map $\operatorname{Aut}(A) \times \operatorname{Aut}(B) \to \operatorname{Aut}(A \otimes B)$ injective? i.e., How would we characterize such *A* and *B*? The most general possibility seems to be if it is true for all *A* and *B* that are not initial or final objects.
| https://mathoverflow.net/users/3040 | Monoidal operations on categories where the maps on Aut, End are injective | Your conditions don't seem to obtain very often, unfortunately.
Let's begin with the one-object case (where the one object is the unit, as it must be). This is the same thing as a monoid object in monoids, or equivalently, a commutative monoid $A$, by Eckmann-Hilton. Your requirement (1) appears to be that the multiplication $A\times A\to A$ is injective; requirement (2) is the same, applied to the invertible elements. This says that every element of $A$ can be uniquely factored, so that $A$ is trivial. So in the one-object case, (1) happens exactly once, and (2) is equivalent to the nonexistence of nontrivial invertible elements.
Using this, we deduce that the unit in a monoidal category satisfying either (1) or (2) is either *very rigid* in the sense that it admits no nontrivial endomorphisms or, respectively, *rigid* in the sense that it admits no nontrivial automorphisms.
If the category $\mathcal{C}$ has an initial object $\varnothing$ that is preserved by the monoidal structure (so that $X\otimes\varnothing=\varnothing\otimes X=\varnothing$ for all $X\in\mathcal{C}$), then every object has to (1) very rigid or (2) rigid. This kills off the bulk of the "algebraic" examples (like modules over a ring, etc.)
Also, note that your conditions are symmetric, i.e., (1) or (2) holds for $\mathcal{C}$ iff it holds for $\mathcal{C}^{\mathrm{op}}$. So Reid's example showing that not all cartesian categories satisfy (2) also works to show that not all cocartesian categories satisfy (2). (We contemplate the coproduct in $\mathrm{Set}^{\mathrm{op}}$...)
I don't want to be overly negative; so here's a situation in which I think your conditions do hold: suppose $\mathcal{C}$ a category with all finite coproducts. Suppose the morphism $X\to X\sqcup Y$ is monic for any $X,Y\in\mathcal{C}$. Then $\mathcal{C}$ (equipped with the coproduct) satisfies your condition. (So this works for finite sets with disjoint union, for instance.) Dually, if $\mathcal{C}$ is a category with all finite products such that the morphism $X\times Y\to X$ is epic for any $X,Y\in\mathcal{C}$, then $\mathcal{C}$ (equipped with the product) satisfies your condition. (So this works for the product on nonempty finite sets, for instance.)
| 3 | https://mathoverflow.net/users/3049 | 11452 | 7,773 |
https://mathoverflow.net/questions/11457 | 9 | In their paper *[Computing Systems of Hecke Eigenvalues Associated to Hilbert Modular Forms](http://arxiv.org/abs/0904.3908)*, Greenberg and Voight remark that
...it is a folklore conjecture that if one orders totally real fields by their discriminant, then a (substantial) positive proportion of fields will have strict class number 1.
I've tried searching for more details about this, but haven't found anything.
Is this conjecture based solely on calculations, or are there heuristics which explain why this should be true?
| https://mathoverflow.net/users/nan | Strict Class Numbers of Totally Real Fields | One heuristic is the following: if one imagines that the residue at $s = 1$ of the $\zeta$-function doesn't grow too rapidly, then the value is a combination of the regulator and the class number. I don't know any reason for the regulator not to also grow (there
are a lot of units, after all!), and hence one can imagine that the class number then stays
small.
This is part of a general heuristic that in random number fields there tends to be a trade-off between units and class number, so especially in the totally real case, when there are so many units, the class number should often be 1.
I learnt some of these heuristics from a colleague of mine who regards it more-or-less as an axiom that a random number field has very small class number. I think this view was formed through a mixture of back-of-the-envelope ideas of the type described above, together with a lot of experience computing with random number fields. So the answer to your question might be that it is a mixture of heuristics and computations.
Incidentally, in the real quadratic case, it is compatible with Cohen--Lenstra, but I think it goes back to Gauss. Also, there are generalizations of Cohen--Lenstra to the
higher degree context, and I'm pretty sure that they are compatible with the class group/unit group trade-off heuristic described above.
| 6 | https://mathoverflow.net/users/2874 | 11463 | 7,778 |
https://mathoverflow.net/questions/11450 | 19 | I wonder why the Artin conjecture is so important. The only reason I could figure out is that one could use the holomorphy of Artin L-series and Weil's converse theorem to show modularity of two-dimensional Galois representations.
Are there other reasons?
| https://mathoverflow.net/users/nan | Applications of Artin's holomorphy conjecture | From a modern view-point, its importance stems from the relationship with modularity/automorphy. Namely, a stronger conjecture, due to Langlands, should be true:
if $\rho:G\_K \to GL\_n({\mathbb C})$ is a continuous irreducible representation, for some number field $K$, then there is a cuspidal automorphic representation $\pi$ of $GL\_n({\mathbb A}\_n)$ such that $\rho$ and $\pi$ have the same $L$-function; in short,
$\rho$ is determined by $\pi$.
This is a non-abelian class field theory. In the case $K = {\mathbb Q}$ and $n = 2$,
it says that $2$-dimensional irred. cont. reps. of $G\_{\mathbb Q}$ are classified by either
weight one holomorphic cuspidal eigenforms (if the rep. is odd) or Maass cuspidal eigenforms with $\lambda = 1/4$ (if the rep. is even). The odd case is now a fully proved theorem
(of Deligne--Serre to go from the modular forms to the Galois reps., and of Khare, Wintenberger, and Kisin to go the other way), while the even case is still open. (If the Maass form is dihedral, one can construct a corresponding dihedral Galois rep. --- this is due to Maass himself --- but otherwise that direction is open; if the image of $\rho$ is solvable, then one can construct the corresponding Maass form --- this is due to Langlands and Tunnell.)
As you note, in the two-dimensional case over $\mathbb Q$, Weil's converse theorem shows that Langlands' conjecture is equivalent to the Artin conjecture. (In fact, proving this over any number field $K$ was one of the goals of the famous book of Jacquet--Langlands).
In general, the two are also very closely linked, so that no modern number theorist thinks about one without the other. In general, one working principle of modern number theory is that the only way to establish holomorphy of $L$-functions arising from Galois representations is by simultaneously proving automorphy.
| 26 | https://mathoverflow.net/users/2874 | 11466 | 7,780 |
https://mathoverflow.net/questions/11488 | 26 | Is there a characterization of the class of varieties which can be described as an intersection of quadrics, apart from the taulogical one?
Lots of varieties arise in this way (my favorite examples are the Grassmanianns and Schubert varieties and some toric varieties) and I wonder how far can one go.
| https://mathoverflow.net/users/1409 | Varieties cut by quadrics | In fact the answer *is* in some sense tautological: every projective variety can be realized as a scheme-theoretic intersection of quadrics! See e.g.
D. Mumford, "Varieties defined by quadratic equations", Questions on algebraic varieties, C.I.M.E. Varenna, 1969 , Cremonese (1970) pp. 29–100,
for quantitative refinements of this question.
| 32 | https://mathoverflow.net/users/1149 | 11490 | 7,800 |
https://mathoverflow.net/questions/11480 | 18 | Let $V=L$ denote the axiom of constructibility. Are there any ***interesting*** examples of set theoretic statements which are independent of $ZFC + V=L$? And how do we construct such independence proofs? The (apparent) difficulty is as follows: Let $\phi$ be independent of $ZFC + V=L$. We want models of $ZFC + V=L + \phi$ and $ZFC + V=L + \neg\phi$. An inner model doesn't work for either one of these since the only inner model of $ZFC + V = L$ is $L$ and whatever $ZFC$ can prove to hold in $L$ is a consequence of $ZFC + V=L$. Forcing models are of no use either, since all of them satisfy $V \neq L$.
| https://mathoverflow.net/users/2689 | On statements independent of ZFC + V=L | There are numerous examples of such statements. Let me organize some of them into several categories.
First, there is the hierarchy of [large cardinal](http://en.wikipedia.org/wiki/Large_cardinal) axioms that are relatively consistent with V=L. See the [list of large cardinals](http://en.wikipedia.org/wiki/List_of_large_cardinal_properties). All of the following statements are provably independent of ZFC+V=L, assuming the consistency of the relevant large cardinal axiom.
* There is an inaccessible cardinal.
* There is a Mahlo cardinal.
* There is a weakly compact cardinal.
* There is an indescribable cardinal.
* and so on, for all the large cardinals that happen to be relatively consistent with V=L.
These are all independent of ZFC+V=L, assuming the large cardinal is consistent with ZFC, because if we have such a large cardinal in V, then in each of these cases (and many more), the large cardinal retains its large cardinal property in L, so we get consistency with V=L. Conversely, it is consistent with V=L that there are no large cardinals, since we might chop the universe off at the least inaccessible cardinal.
Second, even for those large cardinal properties that are not consistent with V=L, we can still make the consistency statement, which is an arithmetic statement having the same truth value in V as in L.
* Con(ZFC)
* Con(ZFC+'there is an inaccessible cardinal')
* Con(ZFC+'there is a Mahlo cardinal')
* Con(ZFC+'there is a measurable cardinal')
* Con(ZFC+'there is a supercompact cardinal').
* and so on, for any large cardinal property. Con(ZFC+large cardinal property).
These are all independent of ZFC+V=L, assuming the large cardinal is consistent with ZFC, since on the one hand, if W is a model of ZFC+Con(ZFC+phi), then LW is a model of ZFC+V=L+Con(ZFC+phi), as Con(ZFC+phi) is an arithmetic statement. And on the other hand, by the Incompleteness theorem, there must be models of ZFC+¬Con(ZFC+phi), and the L of such a model will have ZFC+V=L+¬Con(ZFC+phi).
Third, there is an interesting trick related to the theorem of Mathias that Dorais mentioned in his answer. For any statement phi, the assertion that there is a countable well-founded model of ZFC+phi is a Sigma12 statement, and hence absolute between V and L. And the existence of a countable well-founded model of a theory is equivalent by the Lowenheim-Skolem theorem to the existence of a well-founded model of the theory. Thus, the truth of each of the following statements is the same in V as in L.
* There is a well-founded set model of ZFC. This is equivalent to the assertion: there is an ordinal α such that Lα is a model of ZFC.
* There is a well-founded set model of ZFC with ¬CH. (This is also equivalent to the previous statement.)
* There is a well-founded set model of ZFC with Martin's Axiom.
* and so on. For all the statements known to be forceable, you can ask for a well-founded set model of the theory.
* There is a well-founded set model of ZFC with an inaccessible cardinal.
* There is a well-founded set model of ZFC with a measurable cardinal.
* There is a well-founded set model of ZFC with a supercompact cardinal.
* and the same for any large cardinal notion.
These are all independent of ZFC+V=L, since they are independent of ZFC, and their truth is the same in V as in L. I find it quite remarkable that there can be a model of V=L that has a transitive model of ZFC+'there is a supercompact cardinal'. The basic lesson is that the L of a model with enormous large cardinals has very different properties and kinds of objects in it than a model of V=L arising elsewhere. And I believe that this gets to the heart of your question.
Since all these statements are studied very much in set theory, and are very interesting, and are independent of ZFC+V=L, I find them to be positive instances of what was requested.
**However**, how does this relate to Shelah's view in Dorais's excellent answer? He seems there to dismiss the entire class of consistency strength statements as combinatorics in disguise. What does he mean exactly? Since we set theorists are very interested in these statements, I don't think that he means to dismiss them as silly tricks with the Incompleteness theorem. Perhaps he means something like: to the extent that we believe that a large cardinal property LC is consistent, then we don't really want to consider the theory ZFC+V=L, but rather, the theory ZFC+V=L+Con(LC). That is, we aren't so interested in models having the wrong arithmetic theory, so we insist that Con(LC) if we are comitted to that. And none of the examples I have given exhibit independence from *that* corresponding theory.
| 26 | https://mathoverflow.net/users/1946 | 11491 | 7,801 |
https://mathoverflow.net/questions/11495 | 3 | It's sometimes convenient to have different notations for "$A$ is a subset of $B$" depending on what the inclusion map does:
1. If it's non-surjective, $A\subsetneq B$ or $A\subset B$, depending on your religion
2. If it's surjective, $A=B$ :)
3. If the image is a precompact set, $A\Subset B$
Does there exist notation to indicate that the inclusion $A\hookrightarrow B$ is a homotopy equivalence? I'd like to use something similar to 1-3.
| https://mathoverflow.net/users/2912 | Notation for "the inclusion map is a homotopy equivalence" | $A\stackrel{\sim}{\hookrightarrow}B$? Alternatively, using Oberdiek's stackrel.sty you could say something like
```
A \mathrel{\raisebox{2pt}{$\stackrel[\raisebox{1pt}{$\sim$}]{}\subset$}} B
```
and play a little with the raiseboxes so that this aligns more or less correctly (this depends on your final font, and your publisher's typographer is not going to love you for this...)
| 6 | https://mathoverflow.net/users/1409 | 11496 | 7,804 |
https://mathoverflow.net/questions/11502 | 35 | [Une traduction française suit la version anglaise.]
The question is only about elliptic curves $E$ over $\mathbb{Q}$ and concerns only the aspect
(order of vanishing of $L(E,s)$ at $s=1$)$\ =\ $(rank of $E(\mathbb{Q})$).
Let $r$ be the LHS and $d$ the RHS, so that (a special case of ) the Birch and Swinnerton-Dyer Conjecture is
**BSD?.** $r=d$.
By the end of the last millenium, we knew
**Theorem** (1977--2000). *If* $\ r=0,1$, *then* $d=r$ (*and* $\ \operatorname{Sha}(E)$ *is finite*).
Some years ago, I heard that there was some progress in proving $(r>0)\Longrightarrow (d>0)$ under the assumption of the finiteness of $\operatorname{Sha}(E)$. What is the current status of the
**Statement.** *Suppose that* $\operatorname{Sha}(E)$ *is finite. If* $r>1$, *then* $d>0$ ?
---
*L'état actuel de la conjecture de Birch et Swinnerton-Dyer*
On s'interesse uniquement aux courbes abéliennes $A$ sur $\mathbf{Q}$ et à
l'aspect
(ordre d'annulation de $L(A,s)$ en $s=1$)$\ =\ $(rang de $A(\mathbf{Q})$).
Désignons par $r$ le membre de gauche et par $d$ le membre de droite, de sorte
que la conjecture de Birch et Swinnerton-Dyer prédit (en particulier)
**BSD?** $r=d$.
Vers la fin du millénaire dernier, on avait démontré le
**Théorème** (1977--2000). *Si* $r=0,1$, *alors* $d=r$ (*et* $\ \operatorname{Cha}(A)$ *est fini*).
Il y a quelques années, j'avais entendu dire qu'on a fait des progrès
concernant l'implication $(r>0)\Longrightarrow(d>0)$ sous l'hypothèse de la
finitude de Cha$(A)$. Quel est l'état actuel de l'
**Énoncé.** *Supposons que* $\operatorname{Cha}(A)$ *est fini*. *Si* $r>1$, *alors*
$d>0$ ?
| https://mathoverflow.net/users/2821 | The current status of the Birch & Swinnerton-Dyer Conjecture | The parity conjecture is known, i.e. it is known that if the order of vanishing of the $L$-function is even/odd, then the corank of $p$-Selmer is of the same parity (and I think this is known for every $p$ at this point; Nekovar handled the good ordinary or multiplicative case,
and B.D. Kim the good supersingular case. T. and V. Dokchitser then gave a new proof that handled the general case). This would imply that if Sha(E) is finite (even after passing to the $p$-part of Sha for some prime $p$) and the $L$-function has odd order vanishing, then $E({\mathbb Q})$ has positive rank.
There has also been recent work on establishing cases of positive even order vanishing, and trying to prove that the Selmer group has rank at least two. This has been investigated by both Bellaiche--Chenevier and Skinner--Urban. I don't know precise statements though, and I'm not sure if either pair of authors can handle elliptic curves. (In both cases, the arguments involve deforming along an eigenvariety, and there are problems at low weights. So they may only have results for modular forms of weight $k > 2$.)
Incidentally, although it wasn't part of your question, I don't think anything new is known about finiteness of Sha, beyond what you recalled in your question.
| 27 | https://mathoverflow.net/users/2874 | 11507 | 7,811 |
https://mathoverflow.net/questions/11508 | 18 | Given points $u, v\_1, \dots,v\_n \in \mathbb{R}^m$, decide if $u$ is contained in the convex hull of $v\_1, \dots, v\_n$.
This can be done efficiently by linear programming (time polynomial in $n,m$) in the obvious way. I have two questions:
1. Is there a different (or more efficient algorithm) for this? If not, there ought to be a simple reduction from linear programming to the above problem as well. What is it?
2. Are there interesting families of instances for which the problem can be solved significantly faster than by means of linear programming, e.g., nearly linear time in $m+n$.
| https://mathoverflow.net/users/825 | Deciding membership in a convex hull | It isn't just that you can do the problem with linear programming. The reduction to linear programming actually shows that the convex hull problem is equivalent by dualization to the feasible-point problem in linear programming. In other words, you are asking whether there exist coefficients $\alpha\_1,\ldots,\alpha\_n \ge 0$ such that
$$u = \alpha\_1 v\_1 + \alpha\_2 v\_2 + \cdots + \alpha\_n v\_n.$$
The vector of coefficients $\vec{\alpha}$ is a feasible point in a general linear programming problem, and you are asking whether a feasible point exists. Thus the problem can't be any harder or easier than linear programming.
Your second question then asks whether there are special cases of (the feasible point problem of) linear programming that are easier than the general case of linear programming. The answer is yes. For instance, the "quickhull" type algorithm that Steve Huntsman mentions performs well in any specific low dimension. It amounts to an algorithm for linear programming for the case that the coefficients are nonnegative and there are few equalities. You can also expect a faster algorithm (maybe even equivalent to quickhull) at the other extreme, when there are almost as many equalities as variables. And you can always think of new types of structured linear programming problems that are more convenient or faster, e.g. with a sparse matrix.
| 15 | https://mathoverflow.net/users/1450 | 11510 | 7,813 |
https://mathoverflow.net/questions/11464 | 18 | There was a problem in an Olympiad selection test, which went as follows: Consider the set $\{1,2,\dots,3n \}$ and partition it into three sets *A*, *B* and *C* of size *n* each. Then, show that there exist *x*, *y* and *z*, one in each of the three sets, such that *x + y = z*.
This has a tricky-to-get but otherwise straightforward solution, that starts by assuming 1 to be in *A*, finding the smallest *k* not in *A*, assuming that to be in *B*, and then arguing that no two consecutive elements can be present in *C* (for that would give an infinite descent). Finally, cardinality considerations solve the problem.
I managed to prove a corresponding statement for *4n*, namely: for $\{ 1,2,3, \dots, 4n \}$, partitioned into four sets of size *n* each, there exist *x*, *y*, *z*, and *w*, one in each set, such that *x + y = z + w*.
The question here is whether analogues of this hold for all *m*, with $m \ge 3$ and $n \ge 2$. In other words, if $\{ 1,2, \dots, mn \}$ is divided into $m$ sets of size $n$ each, can we always make a choice of one element in each set such that the sum of floor $m/2$ of the elements equals the sum of the remaining ceiling $m/2$ elements ($(m-1)/2$ and $(m + 1)/2$ for $m$ odd, $m/2$ each for $m$ even). Note we need $n \ge 2$ due to parity considerations when $m$ is congruent to $1$ or $2$ modulo $4$.
| https://mathoverflow.net/users/3040 | Balancing problem | [This](http://www.emis.de/journals/INTEGERS/papers/a9int2003/a9int2003.pdf) article is a nice survey of "Rainbow Ramsey theory". In this jargon what you are trying to prove is that the vector $(1,1,\dots,1,-1,-1,\dots,-1)$ is rainbow partition $m$-regular.
The case of $(1,1,-1,-1)$ being rainbow partition 4-regular, was proved in "Rainbow solutions for the sidon equation x+y=z+w", J. Fox, M. Mahdian, and R. Radoicic. They actually proved that as long as each of the four parts of $[n]$ has at least $(n+1)/6$ members then one can always find rainbow solutions to $x+y=z+w$ (i.e. each variable coming from a different partition.)
Though these results were mostly inspired from their monochromatic version (the 3 variable case dates back to Schur, and then R.Rado classified all linear equations that are partition regular), the analogy hasn't proven very faithful. The rainbow Hales-Jewett theorem is false, and so is the rainbow Van der Waerden theorem!
Another thing worth mentioning is that if we color $\mathbb{Z}/p\mathbb{Z}$ in $k$ colors with each color having at least $k$ elements, then the equation $\sum\_{i=1}^k a\_i x\_i\equiv b\pmod{p}$ always has a rainbow solution given that not all $a\_i$ are the same. A proof is in [this](http://www.dpmms.cam.ac.uk/~dc340/RE.pdf) article by D.Conlon.
| 16 | https://mathoverflow.net/users/2384 | 11512 | 7,815 |
https://mathoverflow.net/questions/11537 | 1 | I know that if we have some data representing some wave, for example image line values, we can use fourier transform to get frequency function of that wave. But we have N values at points x=0...N-1 And we get only N frequencies at the output. So I want to analyze the wave everywhere in the range [0, N-1] For example at point u = 1.5. How can I do it?
| https://mathoverflow.net/users/3195 | Calculating fourier transform at any frequency | You need to have some a priori knowledge about your wave between and beyond the sampling points to get a meaningful guess about the full Fourier transform. The $N$ values that you get doing the discrete Fourier transform have anything to do with the continuous Fourier transform only for indices much less than $N$. Note that different assumptions will lead to different answers for large frequences. If you need only relatively low frequences, your signal is compactly supported in time, and your sampling points cover the entire support, you can safely interpolate the values of the discrete FT to guess the continuous one but there is no way to get reliable high precision values for the continuous Fourier transform at frequences comparable to $N$. The $N$-point resolution on $[0,1]$ is just not high enough to catch those frequences without an error.
Also note that you want the frequences not on "on the interval $[0,N-1]$" as you wrote, but rather on the interval $[-M,M]$ where $M\ll N$. For decent signals, the frequences with indices close to $N$ in the discrete $FT$ actually catch the negative low frequences in the continuous FT, not the high ones.
| 7 | https://mathoverflow.net/users/1131 | 11551 | 7,838 |
https://mathoverflow.net/questions/11549 | 8 | In Vitali's Lemma it uses outer measure rather than measure. What are some results that depend on it this theorem applying to sets with only outer measure rather than measurable sets?
**Vitali's Lemma:**
Let $E$ be a set of finite outer measure and $G$ a collection of intervals that cover $E$ in the sense of Vitali. Then given $\varepsilon> 0$ there is a finite disjoint collection of intervals in $G$ such that $m^\*(E - \bigcup\_{n=1}^N I\_n) < \varepsilon$.
I'm trying to learn this theorem and I keep replacing "outer measure" with "measure" and I want a reason to stop doing that.
| https://mathoverflow.net/users/2907 | Why is this generality in Vitali's Lemma useful? | The short answer (to the question in the title) is so that you don't need to worry about whether E is measurable. If you happen to know that E is measurable then you can drop "outer" everywhere.
There may be a longer answer (better addressing the question as stated in the page) involving specific applications where E is in fact nonmeasurable, but I personally don't know of such applications offhand.
| 8 | https://mathoverflow.net/users/1044 | 11552 | 7,839 |
https://mathoverflow.net/questions/11536 | 9 | Theorem 3.2 of the [paper](http://arxiv.org/abs/1001.0056) "Quantum cohomology of the Springer resolution" by Braverman, Maulik and Okounkov relates equivariant quantum cohomology of the cotangent bundle of $G/B$ to the trigonometric Dunkl operators. A naive guess (or maybe [WAG](https://mathoverflow.net/questions/9764/sheaves-and-differential-equations)?) is that replacing cohomology with $K$-theory gives the Dunkl representation of the [DAHA](https://mathoverflow.net/questions/6517/double-affine-hecke-algebras-and-mainstream-mathematics). An explanation (or reference to an explanation) of what (if any) significant obstacles remain for doing the $K$-theoretic analog of BMO will be greatly appreciated! Is it the sort of thing that should be straightforward (though perhaps technical) at this point? I'd prefer an answer that addresses these particular varieties, rather than general foundational problems.
| https://mathoverflow.net/users/nan | Quantum equivariant $K$-theory and DAHA. | My best guess is that either
1. this is true for $\mathbb{CP}^1$, and it's pretty easy to generalize that given what's already in that paper, or
2. this is false for $\mathbb{CP}^1$, and you're hosed.
The bit one has to understand is the map from the 2 point genus 0 moduli space to the Steinberg variety. BMO get away with just noting that the two spaces have the same dimension, so the pushforward of the fundamental class of the moduli space has to be a sum of fundamental classes of components of Steinberg, whose coefficients they work out by deforming to an almost generic situation and doing the calculation for $\mathbb{CP}^1$.
I think by looking at the pushforward of the structure sheaf on the 2-point moduli space, you'll find that quantum correction is some K-class on the Steinberg variety and thus something in the affine Hecke algebra, and I think it should be the sum of SL(2) contributions for each root by the same deformation arguments that BMO use.
---
I just spoke to Davesh Maulik about this, and it seems my intuition has failed me: he claims it is just hard, and the techniques of that paper will not work.
| 2 | https://mathoverflow.net/users/66 | 11556 | 7,842 |
https://mathoverflow.net/questions/11558 | -2 | I want to create Taylor series of a complex function that has complex conjugate in it. Obviously I cannot do a total derivative but derivations over real and imag parts exist.
Bonus question: Can I produce a Taylor series using only derivations over real part?
| https://mathoverflow.net/users/269 | Taylor series of a complex function that is not holomorphic | Another option is $\sum c\_{mn}z^m \bar z^n$, which still keeps track of the complex structure. For instance, harmonic functions will have $c\_{mn}=0$ unless $mn=0$.
| 2 | https://mathoverflow.net/users/2912 | 11568 | 7,850 |
https://mathoverflow.net/questions/11580 | 6 | There are a few algorithms around for finding the minimal bounding rectangle (OBB) containing a given (convex) polygon.
Does anybody know about an algorithm for finding a minimal-area bounding quadrilateral (any quadrilateral, not just rectangles)?
I've been refered to this site from stackoverflow.com ([original post](https://stackoverflow.com/questions/2048024/minimum-area-quadrilateral-algorithm)), since the guys over there did not know the answer to this...
(PS: I'm a programmer and not a mathematician, so I would appreciate especially if you could point me to exisiting implementations if there are any... Thanks a lot!)
| https://mathoverflow.net/users/3204 | Minimum-area bounding quadrilateral algorithm | I think what you want is "Geometric applications of a matrix searching algorithm", Aggarwal et al, Algorithmic 1987, [doi:10.1007/BF01840359](http://dx.doi.org/10.1007/BF01840359). It builds on previous work of Aggarwal, Chang, and Yap (their reference [2]) to show that the minimum area enclosing k-gon of a geometric figure can be found in time O(n^2) whenever k is constant — they explain it very briefly towards the bottom of the 11th page of their paper (page 205 of the journal).
| 6 | https://mathoverflow.net/users/440 | 11584 | 7,862 |
https://mathoverflow.net/questions/11576 | 6 | This may be totally trivial or wrong. I am just posting this because I am sick and tired of trying to understand this myself and I am sure someone out here can just answer it out of his head in 2 minutes.
Let $G$ be a finite group, and $k$ an algebraically closed field whose characteristic is not a divisor of $\left|G\right|$. The Maschke theorem in its standard form states that
$$
\displaystyle k\left[G\right]\cong\bigoplus\_{V\text{ irreducible }k\left[G\right]\text{-module}}\mathrm{End}V
$$
as $k$-algebras. This quickly yields that
$$
\displaystyle k\left[G\right]\cong\bigoplus\_{V\text{ irreducible }k\left[G\right]\text{-module}}\mathrm{End}V
$$
as left $k\left[G\right]$-modules, where the $k\left[G\right]$-module structure on $\mathrm{End}V$ is given by $\left(gF\right)\left(v\right)=gF\left(v\right)$ for every $g\in k\left[G\right]$, $f\in\mathrm{End}V$ and $v\in V$.
Now, I've overheard that there exists a stronger form of this, stating that
$$
\displaystyle k\left[G\right]\cong\bigoplus\_{V\text{ irreducible }k\left[G\right]\text{-module}}\mathrm{End}V
$$
as left $k\left[G\times G\right]$-modules, where the $k\left[G\times G\right]$-module structure on $k\left[G\right]$ is defined by $\left(g,h\right)u=guh^{-1}$ for any $g\in G$, $h\in G$ and $u\in k\left[G\right]$, and the $k\left[G\times G\right]$-module structure on $\mathrm{End}V$ is defined by $\left(g,h\right)F=gFh^{-1}$ for any $g\in G$, $h\in G$ and $F\in \mathrm{End}V$. This would follow from
$$
\displaystyle k\left[G\right]\cong\bigoplus\_{V\text{ irreducible }k\left[G\right]\text{-module}}\mathrm{End}V
$$
as $k$-Hopf algebras, where the Hopf algebra structure on $k\left[G\right]$ is the standard one ($S\left(g\right)=g^{-1}$ for every $g\in G$), and the Hopf algebra structure on $\mathrm{End}V$ is the standard one as well.
[**EDIT:** as the comments explained, there is no such thing as a standard Hopf algebra structure on $\mathrm{End}V$, and so "$k$-Hopf algebras" should be "$k$-algebras".]
Is this correct, and how can this be proven?
| https://mathoverflow.net/users/2530 | Apocryphal Maschke theorem? | The result about bimodules is true, and standard. Here is one way to see it.
By Frobenius reciprocity, $\operatorname{Hom}\_G(V,k[G]) = \operatorname{Hom}\_k(V,k)$, since $k[G]$ is the induction (or coinduction, depending on your terminology) of the trivial representation of the trivial subgroup of $G$ to $G$.
Since Frobenius reciprocity is functorial, one easily sees that this canonical isomorphism
is an isomorphism of right $G$-representations, where the source has a right $G$-action coming from the right $G$-action on $k[G]$, and the target has a right $G$-action coming as the transpose of the left $G$-action on $V$.
Now by Maschke's semisimplicity theorem, we know that
$k[G] = \bigoplus\_V V \otimes\_k \operatorname{Hom}\_G(V,k[G])$, where the sum is over all irreducible left $G$-representations. (Indeed, Maschke shows that this is true
for any left $G$-module in place of $k[G]$.) Again, this is a natural isomorphism, and so respects the right $G$-actions on source and target.
Combined with the preceding computation, we find that
$k[G] = \bigoplus\_V V\otimes\_k \operatorname{Hom}\_k(V,k) = \bigoplus\_V \operatorname{End}(V),$ as both left and right $G$-modules,
as required.
[Edit:] Leonid's remark about $k$ being needing to be big enough in his answer below is correct. Each simple $V$ comes equipped with an associated division algebra of $G$-endomorphisms
$A\_V := \operatorname{End}\_G(V)$. The representation $V$ is absolutely irreducible (i.e stays irred. after passing to any extension field) if and only if $A\_V = k$. When we consider $\operatorname{Hom}\_k(V,W)$ for another left $G$-module $W$, this is naturally an $A\_V$-module, and Maschke's theorem
will say that $W = \bigoplus\_V \operatorname{Hom}\_k(V,W)\otimes\_{A\_V} V$. (I have written the factors in the tensor product in this order because $V$ is naturally a left $A\_V$-module (if we think of endomorphisms acting on the left), and then $\operatorname{Hom}\_k(V,W)$ becomes a right $A\_V$-module.)
So in the case of $W$ being the group algebra, we have
$$k[G] = \bigoplus\_V \operatorname{Hom}\_k(V,k)\otimes\_{A\_V} V$$
(an isomorphism of $G$-bimodules).
If all the $V$ are absolutely irreducible, e.g. if $k$ is algebraically closed,
then all the $A\_V$ just equal $k$, and the preceding direct sum reduces to what I wrote above, and what was written in the question.
| 12 | https://mathoverflow.net/users/2874 | 11590 | 7,867 |
https://mathoverflow.net/questions/11567 | 32 | There are several theorems I know of the form "Let $X$ be a locally ringed space obeying some condition like existence of partitions of unity. Let $E$ be a sheaf of $\mathcal{O}\_X$ modules obeying some nice condition. Then $H^i(X, E)=0$ for $i>0$."
What is the best way to formulate this result? I ask because I'm sure I'll wind up teaching this material one day, and I'd like to get this right.
I asked a similar question over at [nLab](http://ncatlab.org/nlab/show/fine+sheaf). Anyone who really understands this material might want to write something over there. If I come to be such a person, I'll do the writing!
---
Two versions I know:
(1) Suppose that, for any open cover $U\_i$ of $X$, there are functions $f\_i$ and open sets $V\_i$ such that $\sum f\_i=1$ and $\mathrm{Supp}(f\_i) \subseteq U\_i$. Then, for $E$ any sheaf of $\mathcal{O}\_X$ modules, $H^i(X,E)=0$. Unravelling the definition of support, $\mathrm{Supp}(f\_i) \subseteq U\_i$ means that there exist open sets $V\_i$ such that $X = U\_i \cup V\_i$ and $f\_i|\_{V\_i}=0$.
Notice that the existence of partitions of unity is sometimes stated as the weaker condition that $f\_i$ is zero on the closed set $X \setminus U\_i$. If $X$ is [regular](http://en.wikipedia.org/wiki/Regular_space), I believe the existence of partitions of unity in one sense implies the other. However, I care about algebraic geometry, and affine schemes have partitions of unity in the weak sense but not the strong.
(2) Any quasi-coherent sheaf on an affine scheme has no higher sheaf cohomology. (Hartshorne III.3.5 in the noetherian case; he cites EGA III.1.3.1 for the general case.) There is a similar result for the sheaf of analytic functions: see [Cartan's Theorems](http://en.wikipedia.org/wiki/Cartan%27s_theorems_A_and_B).
I have some ideas about how this might generalize to locally ringed spaces other than schemes, but I am holding off because someone probably knows a better answer.
---
It looks like the answer I'm getting is "no one knows a criterion better than fine/soft sheaves." Thanks for all the help. I've written a [blog post](http://sbseminar.wordpress.com/2010/02/02/when-fine-just-aint-enough/) explaining why I think that fine sheaves aren't such a great answer on non-Hausdorff spaces like schemes.
| https://mathoverflow.net/users/297 | What is the right version of "partitions of unity implies vanishing sheaf cohomology" | Although we clearly all have more or less the same answers, here is how I like to organize things.
I) Let $\mathcal F$ be a sheaf of abelian groups on the topological space $X$. It is said to be soft if every section $s \in \Gamma (A,\mathcal F)$ over a closed subset $A\subset X$ can be extended to $X$.
Notice carefully that the definition of $s$ is NOT that it is the restriction to $A$ of some section of $\mathcal F$ on an open neighbourhood of $A$ [but that it is an element $ s\in \prod \limits\_{x\in X} \mathcal F\_x$ satisfying some more or less obvious conditions]
II) Consider the following condition on the [not necessarily locally] ringed space $(X, \mathcal O)$ :
The space $X$ is metrizable and given an inclusion $A\subset U \subset X$ with $A$ closed and $U$ open there exists a global section $s\in \Gamma (X,\mathcal O)$ such that
$s|A=1$ and $ s|X \setminus U=0 \quad \quad (SOFT)$.
We then have the
$\textbf{Theorem }$ : If the ringed space satisfies (SOFT), then every sheaf of $\mathcal O\_ X -Modules$ is soft.
III) A metrizable space endowed with its sheaf of continuous functions satisfies $(SOFT)$.
A metrizable differential manifold endowed with its sheaf of smooth functions satisfies $(SOFT)$.
IV) On a metrizable space every soft sheaf is acyclic
Put together these results yield all standard acyclicity results on functions,vector bundles, distributions,etc.
It is interesting to notice that you use partitions of unity only once: in the proof of III). But never more afterwards; you just check that your sheaves are $\mathcal O -Modules$. I like this approach (which I learned from Grauert-Remmert) more than the usual one, where a proof of acyclicity is given for the sheaf of smooth functions, followed by the ( correct!) assertion that you have to repeat it with minor changes for, say, vector bundles. Moreover fine sheaves needn't even be mentioned if you follow this route.
| 13 | https://mathoverflow.net/users/450 | 11594 | 7,870 |
https://mathoverflow.net/questions/11589 | 6 | Let ${\mathcal C}$ be the class of topological spaces which carry a CW-structure (note that I do not want to fix some particular CW-structure).
Is it true that for a covering map $E\stackrel{f}{\to} B$ with $E\in{\mathcal C}$ we have $B\in{\mathcal C}$, too?
It *is* true that the total space of a covering lies in ${\mathcal C}$ if the base space does, but the reverse implication is not clear to me.
**Edit**
As Algori pointed out, the quotient space is not even Hausdorff in general. What about finite regular coverings, i.e. those which come from a free action of a finite group on the total space? Is it true then that the quotient space carries a CW-structure, too?
I'm interested in that because this would imply that given a free group action of a finite group on a "nice" space like a CW-complex, one can always choose a CW-structure with respect to which $G$ just permutes cells. Then the corresponding cellular complex would be a (possibly nice) complex of ${\mathbb Z}G$-modules (for example, if the space was a sphere, then this procedures can be used to construct a periodic ${\mathbb Z}G$-resolution of the trivial module ${\mathbb Z}$, showing that the group has to have periodic invariants like homology and cohomology; in this particular case, however, things behave well as the quotient space ${\mathbb S}^n/G$ is still a compact manifold).
Thank you.
| https://mathoverflow.net/users/3108 | Properties of the class of topological spaces possessing a CW-structure | Ok, here is a simple example when this fails: let $X$ be $\mathbf{R}^2$ minus the origin and consider the automorphism of $X$ given by $(x,y)\mapsto (2x,y/2)$. This automorphism generates a group $G$. The quotient map $X\to X/G$ is a covering, but $X/G$ is not Hausdorff.
This answers the question as it is stated, but maybe you wanted $B$ to be Hausdorff (in which case I don't know what to do).
| 10 | https://mathoverflow.net/users/2349 | 11602 | 7,874 |
https://mathoverflow.net/questions/10237 | 24 | The [Convolution Theorem](http://mathworld.wolfram.com/ConvolutionTheorem.html%20%22Convolution%20Theorem%22) is often exploited to compute the convolution of two sequences efficiently: take the (discrete) Fourier transform of each sequence, multiply them, and then perform the inverse transform on the result.
The same thing can be done for convolutions in the quotient ring Z/pZ via the analogous [Number Theoretic Transform](http://mathworld.wolfram.com/NumberTheoreticTransform.html).
Does this procedure generalize to other algebraic structures? Arbitrary rings? Semirings? Fast convolutions over the semiring (min,+) would be particularly useful.
I'm led to suspect this because both sorting and FFT can be computed using the same butterfly-like network with simple operations like "min", "max", "+", and "\*" at the nodes of the network, and sorting can be thought of as a kind of convolution.
| https://mathoverflow.net/users/2361 | does the "convolution theorem" apply to weaker algebraic structures? | In general, it is a major open question in discrete algorithms as to which algebraic structures admit fast convolution algorithms and which do not. (To be concrete, I define the *$(\oplus,\otimes)$ convolution* of two $n$-vectors $[x\_0,\ldots,x\_{n-1}]$ and $[y\_0,\ldots,y\_{n-1}]$, to be the vector $[z\_0,\ldots,z\_{n-1}]$ with $$z\_k = (x\_0 \otimes y\_k) \oplus (x\_1 \otimes y\_{k-1}) \oplus \cdots \oplus (x\_k \otimes y\_0).$$ Here, $\otimes$ and $\oplus$ are the multiplication and addition operations of some underlying semiring.)
For any $\otimes$ and $\oplus$, the convolution can be computed trivially in $O(n^2)$ operations. As you note, when $\otimes = \times$, $\oplus = +$, and we work over the integers, this convolution can be done efficiently, in $O(n \log n)$ operations.
But for more complex operations, we do not know efficient algorithms, and we do not know good lower bounds. The best algorithm for $(\min,+)$ convolution is $n^2/2^{\Omega (\sqrt{\log n})}$ operations, due to combining my recent APSP paper
>
> Ryan Williams: Faster all-pairs shortest paths via circuit complexity. STOC 2014: 664-673
>
>
>
and
>
> David Bremner, Timothy M. Chan, Erik D. Demaine, Jeff Erickson, Ferran Hurtado, John Iacono, Stefan Langerman, Perouz Taslakian: Necklaces, Convolutions, and X + Y. ESA 2006: 160-171
>
>
>
A substantially subquadratic algorithm for $(\min,+)$ convolution would (to my knowledge) imply a subcubic algorithm all-pairs shortest paths in general graphs, a longstanding open problem. The above ESA06 reference also gives a $O(n^2 (\log \log n)^2/\log n)$ algorithm for a "(median,+) convolution".
The situation is subtle. It's not clear when convolution over a semiring is easy and when it's hard. For instance, the $(\min,\max)$ convolution *can* be computed in subquadratic time: I believe that $O(n^{3/2} \log n)$ operations suffice. This can be obtained from adapting the $(\min,\max)$ matrix multiplication algorithm in my work with Vassilevska and Yuster on all-pairs bottleneck paths. Basically you reduce the problem to computing $\sqrt{n}$ instances of a $(+,\times)$ ring convolution.
| 24 | https://mathoverflow.net/users/2618 | 11606 | 7,877 |
https://mathoverflow.net/questions/11610 | 10 | A Poisson manifold is a real manifold $M$ along with a Lie bracket $[\cdot,\cdot]$ on $C^\infty(M)$ which is a derivation in each variable. Poisson manifolds are interesting for a few reasons, among them:
1. You can define the notion of an integrable system structure on a Poisson manifold, which allows them to be applied to solving problems in physics with enough symmetry.
2. Every symplectic manifold is automatically Poisson
3. Any Poisson manifold has a foliation by symplectic leaves.
(Reference for all of this: anything on Poisson manifolds, in particular, [Wikipedia](https://en.wikipedia.org/wiki/Poisson_manifold).)
Now, I've seen people seriously (for instance, in Vanhaecke's [book](https://books.google.com/books?id=gOseWkOhgWwC&printsec=frontcover&dq=Vanhaecke&ei=3TFNS9yNF5TIzAT1xIjqCw&cd=1#v=onepage&q=&f=false)) extend this notion to affine varieties over $\mathbb{C}$, where being Poisson means that the structure sheaf is a sheaf of [Poisson Algebras](https://en.wikipedia.org/wiki/Poisson_algebra), specifically, a Poisson algebra is an associative algebra along with a Lie bracket that is a derivation in each variable.
Now, what I'm interested in is how far this can be generalized and still have something where there are interesting (new!) examples. For instance, is "Poisson Scheme $X$ over $S$" a real object of interest? Specifically, I'm wondering if there are any examples where $S$ is not the spectrum of a field of characteristic zero, say $S$ is a finite field, or something positive dimensional, or nonreduced, etc, and if there are examples of this form, what makes them interesting? For instance, one reason that Poisson manifolds are interesting is that they are applicable to physics and, in fact in many cases to problems related to the geometry of the moduli space of vector bundles on a Riemann surface.
| https://mathoverflow.net/users/622 | Examples of Poisson schemes | Of course, thinking of Poisson things in a relative way isn't going to give anything new, since a family of Poisson things is itself Poisson.
But to answer whether people think about these things in an abstract way, the answer is yes. There's now a huge and absolutely beautiful theory of symplectic singularities; here's a survey by Kaledin: *[Geometry and topology of symplectic resolutions](https://arxiv.org/abs/math/0608143)* If you read his papers, you'll see lots of modern algebraic geometry; theorems like the local existence of tilting generators depend on reduction to characteristic $p$.
| 8 | https://mathoverflow.net/users/66 | 11615 | 7,883 |
https://mathoverflow.net/questions/11531 | 7 | ### Preliminaries
Let $n \in \mathbb{N}$ and $v$ be a vertex of a graph $G$. Let the *$n$-neighbourhood of $v$*, $N\_n(v)$, be the induced subgraph of $G$ containing $v$ and all vertices at most $n$ edges away from $v$.
With $\epsilon(v)$ the [eccentricity](http://mathworld.wolfram.com/GraphEccentricity.html) of $v$, $N\_{\epsilon(v)}(v)$ is obviously nothing but the connected component of $G$ containing $v$, so it is natural to restrict $n$ for a given $v$ to the values $0,1, ..., \epsilon(v)$.
Consider for any two vertices $v$, $w$ the greatest $s = s(v,w)$ such that $N\_s(v)$ and $N\_s(w)$ are isomorphic **[added:]** with the isomorphism sending $v$ to $w$ (for short: $N\_s(v) \cong\_{vw} N\_s(w)$). If $s(v,w) = \epsilon(v) = \epsilon(w)$, $v$ and $w$ are conjugate. For non-conjugate vertices $v$, $w$ the number $s= s(v,w)$ reflects the size of the smallest neighbourhood that is needed to distinguish $v$ and $w$, since $N\_{s+1}(v) \ncong\_{vw} N\_{s+1}(w)$ by definition.
Let's call the positive number
$$\sigma(v,w) = \frac{2 \cdot s(v,w)}{\epsilon(v) + \epsilon(w)}$$
the *similarity index* of $v$ and $w$.
$\sigma(v,w) = 0$ indicates that $v$, $w$ have different $1$-neighbourhoods (and are maximally dissimilar), $\sigma(v,w) = 1$ indicates that $v$, $w$ are conjugate (i.e. maximally similar = indistinguishable by their neighbourhoods).
The matrix $\Sigma(G) = \lbrace \sigma(v,w) \rbrace\_{v,w \in V(G)}$ reflects the symmetry of the graph $G$:
* If $\sigma(v,w) = 1$ only if $v = w$, the graph is asymmetric.
* If the $1$'s of the matrix come in square blocks along the diagonal, these blocks indicate the orbits of the graph.
$\Sigma(G)$ is a graph invariant up to matrix equivalence. (Is this the right wording?)
>
> Definition: An $n \times n$-matrix $S$
> is a *similarity matrix* iff there is
> a graph $G$ such that $S = \Sigma(G)$.
>
>
>
### Questions
>
> Is the notion of *$n$-neighbourhood* treated in other contexts, maybe under another name?
>
>
>
---
>
> Is there already research on this concept of *similarity* (or a related one)?
>
>
>
---
>
> How might similarity matrices be
> characterized (sufficient/necessary conditions)? ("A matrix is a similarity matrix if ...") Any idea?
>
>
>
---
>
> How will graphs with the same similarity matrix (plus same eccentricity vector) be related? (They will probably not be don't have to be isomorphic, but maybe something weaker?)
>
>
>
| https://mathoverflow.net/users/2672 | Yet another graph invariant: the similarity matrix | This is nice. What is the motivation for dividing by $\epsilon(u)+\epsilon(v)$?
To partly answer your first question, I suspect the notion of $n$-neighborhood is relatively common, but one place where it is very common, if not a staple, in Finite Model Theory (read also Database Theory). It is a key component of various locality properties, notably Gaifman locality. The quantity $s(u,v)$ is closely related to the complexity of distinguishing $u$ from $v$ by a first-order query in the language of graphs. Specifically, a query of quantifier depth $k$ cannot distinguish between vertices such that $s(u,v) > (3^{k+1}-1)/2$. This has many uses, a typical example is to measure the complexity of distinguishing database entries. A good intro can be found in the first few chapters of Libkin's *Elements of Finite Model Theory*.
| 3 | https://mathoverflow.net/users/2000 | 11619 | 7,887 |
https://mathoverflow.net/questions/11622 | 24 | I'm a little embarrassed to ask this one, but it could help for a class I'm teaching, so here goes:
Let $X$ be a metric space. We all know that $X$ admits a **completion**, which is a complete metric space $\hat{X}$ together with an isometric embedding $\iota: X \hookrightarrow \hat{X}$ with dense image. Moreover, one learns that this completion is essentially unique.
From a modern perspective, one would like to realize the completion as satisfying some universal mapping property: this makes precise the "essentially unique" above and gives some functorial properties. But it seems to me that the completion satisfies two different such properties:
1) It is universal with respect to isometric embeddings into complete metric spaces.
2) It is universal with respect to uniformly continuous maps into complete metric spaces.
Of course 1) is the more obvious one. I gather from some internet research that 2) is supposed to be the "right" choice, and its usefulness is related to the fact that uniformly continuous maps have the extension property (again, I don't quite remember this from my undegraduate days; is it in Rudin's *Principles*, for instance?). However, it seems strange to me that by taking 2), we also get for free that the mapping $\iota$ is an isometric embedding (in particular, from 2) it doesn't even seem completely obvious that it is injective). Certainly one can see this by constructing the completion, but is there a more direct way?
I suspect that this is an instance when the more categorical thinkers have one up on me, and I stand ready to be enlightened.
| https://mathoverflow.net/users/1149 | What is the "right" universal property of the completion of a metric space? | This doesn't quite answer your questions about (2), but one could say the following:
given a metric, there is an underlying uniform structure. We can then form the completion
with respect to this uniform structure, which is universal for uniformly continuous maps into complete uniform spaces. That this map is injective is due to the existence of "enough" complete uniform spaces. (And to show this, one constructs the completion! I am still thinking about alternative, less constructive, approaches, to this.)
But I think now one can relate (2) to (1) via the following lemma:
If $X$ is a uniform space, and $d$ is a metric on $X$ inducing the given uniform structure,
then $d$ extends uniquely to the completion of $X$.
Proof: Something along the lines of: $d$ is uniformly continuous from $X \times X$ to
$\mathbb R$, and so extends.
Thus the universal object for (2) had to also be the universal object for (1).
| 15 | https://mathoverflow.net/users/2874 | 11625 | 7,891 |
https://mathoverflow.net/questions/11614 | 9 | Let's consider a vector bundle $E$ of rank $n$ over a compact manifold $X$. Consider the associated Grassmannian bundle $G$ for some $k < n$, obtained by replacing each fiber $E\_x$ by $Gr(k,E\_x)$.
Let's suppose that there is a full flag of subbunldes $F\_1 \subset F\_2 \dots \subset F\_n \subset E$. I think that in this case we are able to define relative Schubert cycles on G which restrict to usual Schubert cycles on each fiber so that we can apply Leray-Hirsh theorem to deduce that $H^\*(G) = H^\*(X) \otimes H^\*(Gr(k,n))$.
1. Is the reasoning above correct?
2. Can we still compute $H^\*(G)$ in the case when the full flag of subbundles doesn't exist?
EDIT: I meant complex vector bundles and complex Grassmannians. Also the bundle can be assumed holomorphic or algebraic if it makes a difference.
EDIT: Ben in his answer mentions Serre's spectral sequence that can be used in this case. Is there a reason why it will degenerate to leave $H^\*(X) \otimes H^\*(Gr(k, n))$ as a result?
| https://mathoverflow.net/users/2260 | Grassmannian bundle theorem | Re 1.: the reasoning is correct additively, but not multiplicatively (did you check the case $k=1$?).
Re 2.: given any complex bundle $E$ over $X$ of rank $n$, the cohomology of the associated Grassmannian bundle of $k$-planes is $$H^{\bullet}(X,\mathbf{Z})\otimes\mathbf{Z}[c\_1,\ldots,c\_k,c\_1',\ldots,c\_{n-k}']/(1+c\_1+\cdots +c\_k)(1+c\_1'+\cdots +c\_{n-k}')=c(E)$$
where $\deg c\_i=\deg c'\_i=2i$ and $c(E)$ is the total Chern class of $E$.
This a modification of the Grothendieck trick which computes the cohomology of the projectivized vector bundle.
upd: here is a sketch of a proof of the above formula. On the total space $G$ of the Grassmannian vector bundle there is the tautological $k$-plane bundle $S$, which is a subbundle of the pullback $E'$ of $E$ under the bundle projection $p:G\to X$. Let $Q$ be the quotient bundle. We have $c(S)c(Q)=c(E')$. So we get an algebra map from the above algebra to the cohomology of $G$ (taking an element of $H^{\bullet}(X,\mathbf{Z})$ to its pullback, $c\_i$ to $c\_i(S)$ and $c\_i'$ to $c\_i(Q)$) and we have to show that it is an isomorphism.
Surjectivity: let us pick a $\mathbf{Z}$-basis of the cohomology of the Grassmannian, express each element as a polynomial in the Chern classes of the tautological bundle and take the resulting polynomials in $c\_i(S)$'s. When restricted to any fiber these form a $\mathbf{Z}$-basis of the cohomology of the fiber, so by the Leray-Hirsch principle, $H^{\bullet}(G,\mathbf{Z})$ is a free $H^{\bullet}(X,\mathbf{Z})$-module spanned by these classes.
Injectivity: the above algebra is also a free module over $H^{\bullet}(X,\mathbf{Z})$ and the the above map from it to $H^{\bullet}(G,\mathbf{Z})$ takes a basis to a basis.
| 17 | https://mathoverflow.net/users/2349 | 11627 | 7,893 |
https://mathoverflow.net/questions/11628 | 4 | As is well known, one can view $\mathbb{CP}^n$ as a quotient of the unit $(2n + 1)$-sphere in $\mathbb{C}^{n+1}$ under the action of $U(1)$, since every line in $\mathbb{C}^{n+1}$ intersects the unit sphere in a circle.
Moreover, we have $S^{2n + 1} = SU(n+1)/SU(n)$, where $SU(n)$ embeds into the bottom right-hand corner (say).
My question is: Is there an embedding $j$ of $U(1)$ into $SU(n+1)/SU(n)$ that gives the $U(1)$ action as a left (right) multiplication, i.e. such that $A.e^{i \theta} = Aj(e^{i \theta})$, for all $A \in SU(n+1)/SU(n)$?
For $n=1$, it's easy: $SU(2) = S^3$, and we embed $e^{i\theta}$ as
$\left(
\begin{array}{cc}
e^{i \theta} & 0 \\\\
0 & e^{-i \theta}
\end{array}
\right)$.
| https://mathoverflow.net/users/1977 | Complex Projective Space as a $U(1)$ quotient | The U(1) group is the torus in SU(N+1) which commutes with SU(N). In the example
given in the question where SU(N) is chosen as the bottom N-dimensional block. U(1) consists
of the diagonal matrices
diag{exp(N\*i\*theta), exp(-i\*theta), . . . . (N-times) exp(-i\*theta)}
Please observe that the restriction to the bottom N-dimensional block is proportional to the unit matrix thus it commutes with the whole of SU(N), also it belongs to SU(N+1), since its has a unit determinant.
The reason that the U(1) and the SU(N) factors commute is due to a theorem by [A. Borel](http://www.pnas.org/content/40/12/1147.full.pdf+html) which states that the denominator subgroup of homogeneous Kaehlerian spaces must be the centralizer of a torus. In our case the torus is the U(1) subgroup and the certralizer is SU(N)\*U(1)
| 4 | https://mathoverflow.net/users/1059 | 11640 | 7,901 |
https://mathoverflow.net/questions/11641 | 6 | There are subsets of the real line that has infinite counting measure, but Lebegue measure 0, so the Lebegue measure is used for measuring larger sets than the counting measure. My question is: Is there a translation invariant measure m such that for some sets with Lebegue measure 0 the m-measure is infinite and for some sets with infinite counting measure, the m-measure is 0?
I have found one example: m(A)=0 if A is countable, and m(A)=infinite otherwise. So I will require that the measure can take the value 1.
If such a measure exist, can we find a measure between this and the counting measure? and between this and the Lebegue measure? and so on.
| https://mathoverflow.net/users/2097 | Measure between the counting measure and the Lebegue measure | Hausdorff measures of dimensions between 0 and 1 are a continuous spectrum of examples.
| 10 | https://mathoverflow.net/users/2368 | 11642 | 7,902 |
https://mathoverflow.net/questions/11629 | 9 | The basic logic course in school gives the impression that logic has both the syntax and the semantics aspects. Recently, I wonder whether the syntax part still plays an essential role in the current studies. Below are some of my observations, I hope the idea from the community can make them more complete.
Model theory: Even though model theory is stated in the language of logic, it can be viewed as the study of local isomorphism (see Poizat's "A course in Model Theory"). The syntax part is therefore a natural (though might be uncomfortable for some) way to view the theory rather than a necessity.
Recursion theory: The object of study is the notion of computability in different context. If we believe in Church-Turing Thesis, then these concept are independent of the formalism chosen.
Set theory: The intimate relationship between large cardinal and determinacy perhaps can suggest that this is a universal phenomenon. Will this phenomenon disappear if we change the language of mathematics to, for example, category theory?
Proof theory: I know too little to say anything.
If the observation is true, is it justified to demand that Turing degrees, and large cardinals receive the same mathematical status as, for example, prime numbers?
| https://mathoverflow.net/users/2701 | How much of the current logic is about syntax? | I think your observation is a very good one, but this phenomenon is limited to classical logic and does not continue to hold when we move to intuitionistic or substructural logics.
One way of understanding the role of syntax is to take the connectives of logic as explaining what counts as a legitimate proof of that proposition. So a conjunction $A \land B$ can be proven with a proof of $A$ and a proof of $B$, and an implication $A \implies B$ can be can be proven with a proof of $B$, assuming a hypothetical proof of $A$, and so on. Conversely, we also give rules explaining how to use true propositions -- e.g., from $A \land B$, we can re-derive $A$, and we can also rederive $B$.
If you work this out formally, you get Gentzen's system of natural deduction. The natural deduction systems for good logics admit a normal form theorem for proofs. The normalization procedure also gives us an equivalence relation on proofs (two proofs are equivalent if they have the same normal form), and it so happens that for classical logic, all proofs are equivalent. (This is a small lie: we can give more refined accounts of equivalence of classical proofs which don't equate everything, but the right answer here is still not entirely settled....)
The equivalence of proofs means that we can take the view that the meaning of a classical proposition is its truth value -- i.e., its provability -- and so algebraic models of classical logic contain all the information contained in a classical proposition. We don't need the proofs, and so syntax takes a secondary role.
But in intuitionistic and substructural logics like linear logic, not all proofs are equated. This means that we can't take the view that all the relevant information about a proposition is contained in its truth value, and so syntax retains a more important role.
| 7 | https://mathoverflow.net/users/1610 | 11646 | 7,905 |
https://mathoverflow.net/questions/11583 | 1 | The following question is related to "Remark 2.2" in Christophe Cazanave's paper "[Algebraic homotopy classes of algebraic functions](http://arxiv.org/abs/0912.2227v1)". I decided to add the arxiv article-id to the questions title to invite other people who like to study this article to do the same. My hope is that this will lead to a culture of discussing arxiv articles on the overflow.
**Question:**
Let $F\_n$ be the open subscheme of $\mathbb{A}^{2n}=\mathrm{Spec}(k[a\_{0},\ldots,a\_{n-1},b\_{0},\ldots,b\_{n-1}])$ complementary to the hypersurface of equation $res\_{n,n}(X^{n}+a\_{n-1}X^{n-1}+\ldots+a\_{0},b\_{n-1}X^{n-1}+\ldots+b\_{0})$. Let $R$ be a ring. The claim is that an $R$-point of $F\_{n}$ is a pair $(A,B)$ of polynomials of $R[X]$, where $A$ is monic of degree $n$, $B$ is of degree strictly less than $n$ and the scalar $res\_{n,n}(A,B)$ is invertible. How can I see that a morphism $\mathrm{Spec}(R)\rightarrow F\_n$ gives (and is the same as) a pair of polynomials in $R[X]$?
| https://mathoverflow.net/users/2146 | A rational point in the scheme of pointed degree n rational functions [0912.2227] | I'm not so good on the scheme-theoretic language, so let me embed $F\_n$ as the affine variety $\text{res}\\_{n,n}(X^n + ..., b\_{n-1} X^{n-1} + ...) y = 1$ one dimension up. Then a morphism $k[a\_0, ... a\_{n-1}, b\_0, ... b\_{n-1}, y]/(\text{stuff}) \to R$ is precisely (assuming that Cazanava means either $k = \mathbb{Z}$ or $R$ a $k$-algebra) a choice, for each variable $a\_i, b\_i, y$, of an element of $R$ subject to the condition that the resultant times $y$ is equal to $1$, i.e. the resultant is invertible in $R$.
| 3 | https://mathoverflow.net/users/290 | 11656 | 7,911 |
https://mathoverflow.net/questions/11664 | 13 | Using the chern character, it can be shown that there is no complex structure on $S^n$ if $n > 6$: See May's book: if $S^{2n}$ has a complex structure, let $\tau$ be the tangent bundle. $c\_n(\tau) = \chi(S^{2n}) = 2$ must be divisible by $(n-1)!$ by Husemöller, Fibre bundles, chapter 20, Theorem 9.8. So the only case left is $S^4$ and $S^6$.
Is there a complex structure on $S^4$ or $S^6$?
| https://mathoverflow.net/users/nan | complex structure on S^n | It is known that $S^4$ doesn't even have an almost complex structure. (The sphere $S^6$ does have some, but whether any of them is the underlying almost complex structure of a complex manifold is open.) The lack of almost complex structure can be proved a number of ways, one way is by showing that an almost complex, compact, four manifold with $\dim\_{\mathbb{Q}}H^2(X,\mathbb{Q})=0$ has $\chi(X)=0$, but the four sphere doesn't. (It follows from the index theorem, here's a quick reference, first result: [Srinivasacharyulu, "Topology of complex manifolds", Canad. Math. Bull. 1966](https://doi.org/10.4153/CMB-1966-003-9).)
| 27 | https://mathoverflow.net/users/622 | 11667 | 7,918 |
https://mathoverflow.net/questions/11660 | 3 | This question was motivated by the answers in [D-module as quasi coherent sheaves on deRham stack](https://mathoverflow.net/questions/11200/geometry-of-triangulated-category-and-d-modules-theory). What I am interested in is the case of D-module on flag variety of Lie algebra. So,in this case, if we realize the D-module as quasi coherent sheaves on De Rham stack X^DR. Then the stack X^DR is a scheme or not?
Why do I ask this question?
Because from the point of view of noncommutative algebraic geometry, if we realize D-module on flag variety of Lie algebra as a quasi coherent sheaves on **"space"**, then **this space** is actually a noncommutative separated scheme.
So I guess, this DeRham stack will be a scheme in this special case
| https://mathoverflow.net/users/1851 | Is D-module on flag variety of Lie algebra a scheme? | The de Rham space of a scheme is essentially never a scheme or algebraic space (unless I guess you're Spec of an Artin ring, in which case you'll get a discrete set of points). In particular this applies to the flag variety. I'm not sure which perspective of NC AG you're taking, but certainly if you define the field as the study of Grothendieck categories, or pretriangulated dg categories, etc then D-modules are a very nice (nonproper) noncommutative space. (An interesting comment on this is found at the end of Kontsevich's letter [here](http://www.math.uchicago.edu/~mitya/langlands/kontsevichletter.txt). Also from the point of view of "function theory" D-modules on a scheme are great, i.e. all functors are given by integral transforms, sheaves on a (fiber) product are tensor product of categories of sheaves, etc, see [here](http://arxiv.org/abs/0904.1247).) But I don't think this says anything about the de Rham stack in a classical commutative sense..
| 10 | https://mathoverflow.net/users/582 | 11671 | 7,920 |
https://mathoverflow.net/questions/11669 | 13 | Hi,
Currently, I'm taking matrix theory, and our textbook is Strang's Linear Algebra. Besides matrix theory, which all engineers must take, there exists linear algebra I and II for math majors. What is the difference,if any, between matrix theory and linear algebra?
Thanks!
| https://mathoverflow.net/users/3142 | What is the difference between matrix theory and linear algebra? | Let me elaborate a little on what Steve Huntsman is talking about. A matrix is just a list of numbers, and you're allowed to add and multiply matrices by combining those numbers in a certain way. When you talk about matrices, you're allowed to talk about things like the entry in the 3rd row and 4th column, and so forth. In this setting, matrices are useful for representing things like transition probabilities in a Markov chain, where each entry indicates the probability of transitioning from one state to another. You can do lots of interesting numerical things with matrices, and these interesting numerical things are very important because matrices show up a lot in engineering and the sciences.
In linear algebra, however, you instead talk about **linear transformations,** which are **not** (I cannot emphasize this enough) a list of numbers, although sometimes it is convenient to use a particular matrix to write down a linear transformation. The difference between a linear transformation and a matrix is not easy to grasp the first time you see it, and most people would be fine with conflating the two points of view. However, when you're given a linear transformation, you're not allowed to ask for things like the entry in its 3rd row and 4th column because questions like these depend on a **choice of basis.** Instead, you're only allowed to ask for things that don't depend on the basis, such as the rank, the trace, the determinant, or the set of eigenvalues. This point of view may seem unnecessarily restrictive, but it is fundamental to a deeper understanding of pure mathematics.
| 55 | https://mathoverflow.net/users/290 | 11679 | 7,924 |
https://mathoverflow.net/questions/11677 | 21 | Are there infinitely many (linearly independent) cuspidal eigenforms for $\Gamma(1)$ with integer coefficients?
Someone told me that the Hecke algebra is conjectured to act irreducibly on the space of modular forms of level 1, so there would be no eigenforms if $\mathrm{dim} S\_k > 1$, i.e. for $k \geq 12$.
| https://mathoverflow.net/users/nan | modular eigenforms with integral coefficients [Maeda's Conjecture] | I will replace *modular form* by *cuspform* in your question, just to avoid the trivial answer of *yes*, because of the Eisenstein series. Given this, numerical evidence suggests that the eigenforms of weight $k$ are all conjugate (in the sense that their $q$-expansions are all algebraically conjugate under the action of $Gal(\overline{\mathbb Q}/{\mathbb Q})$). If this were the case, then the answer to your question would be *no*. Since this question is open, I'm pretty sure the answer to your question is not known.
There is a lot of evidence for this conjugacy statement, but the only suggestion I know for why it's true is essentially that it's the simplest possibility, given that no reason is known for it to be false (in contrast to the case of say fixing the weight to be 2, but letting the
level grow; in that case elliptic curves give rise to integral eigenforms, and there
are elliptic curve of arbitrarily high conductor).
| 17 | https://mathoverflow.net/users/2874 | 11681 | 7,926 |
https://mathoverflow.net/questions/11674 | 28 | I am not by any means an expert in category theory. Anyway whenever I have studied a concept in category theory I have always had the feeling that most of the subtleties introduced are artificial.
For a few examples:
-one does not usually consider isomorphic, but rather equivalent categories
-universal objects are unique only up to a canonical isomorphism
-the category of categories is really a 2-category, so some natural constructions do not yield functors into categories, but only pseudofunctors
-cleavages of fibered categories do not always split
....
My question is: can skeleta be used to simplify all this stuff? It looks like building everything using skeleta from the beginning would remove a lot of indeterminacies in these constructions. On the other hand it may be the case that this subtleties are really intrinsic, and so using skeleta, which are not canonically determined, would only move the difficulties around.
| https://mathoverflow.net/users/828 | Can skeleta simplify category theory? | The first, and perhaps most important, point is that *hardly any categories that occur in nature are skeletal*. The axiom of choice implies that every category is equivalent to a skeletal one, but such a skeleton is usually artifical and non-canonical. Thus, even if using skeletal categories simplified category theory, it would not mean that the *subtleties* were artifical, but rather that the naturally occurring subtleties could be removed by an artificial construction (the skeleton).
In fact, however, skeletons don't actually simplify much of anything in category theory. It is true, for instance, that any functor between skeletal categories which is part of an equivalence of categories is actually an isomorphism of categories. However, this isn't really useful because, as mentioned above, most interesting categories are not skeletal. So in practice, one would either still have to deal either with equivalences of categories, or be constantly replacing categories by equivalent skeletal ones, which is even more tedious (and you'd still need the *notion* of "equivalence" in order to know what it means to replace a category by an "equivalent" skeletal one).
In all the other examples you mention, skeletal categories don't even simplify things that much. In general, not every pseudofunctor between 2-categories is equivalent to a strict functor, and skeletality won't help you here. Even if the hom-categories of your 2-categories are skeletal, there can still be pseudofunctors that aren't equivalent to strict ones, because the *data* of a pseudofunctor includes coherence isomorphisms that may not be identities. Similarly for cloven and split fibrations. A similar question was raised in the query box [here](http://ncatlab.org/nlab/show/Grothendieck+fibration): important data can be encoded in coherence isomorphisms even when they are automorphisms.
The argument in CWM mentioned by Leonid is another good example of the uselessness of skeletons. Here's one final one that's bitten me in the past. You mention that universal objects are unique only up to (unique specified) isomorphism. So one might think that in a *skeletal* category, universal objects would be unique on the nose. This is actually *false*, because a universal object is not just an object, but an object together with data exhibiting its universal property, and a single object can have a given universal property in more than one way.
For instance, a product of objects A and B is an object P *together with* projections P→A and P→B satisfying a universal property. If Q is another object with projections Q→A and Q→B and the same property, then from the universal properties we obtain a unique specified isomorphism P≅Q. Now if the category is skeletal, then we must have P=Q, but *that doesn't mean the isomorphism P≅Q is the identity*. In fact, if P is a product of A and B with the projections P→A and P→B, then composing these two projections with any automorphism of P produces another product of A and B, which happens to have the same vertex object P but has different projections. So assuming that your category is skeletal doesn't actually make anything any more unique.
| 66 | https://mathoverflow.net/users/49 | 11687 | 7,931 |
https://mathoverflow.net/questions/11688 | 25 | How does one prove that the splitting of primes in a non-abelian extension of number fields is not determined by congruence conditions?
| https://mathoverflow.net/users/nan | Why do congruence conditions not suffice to determine which primes split in non-abelian extensions? | Here's an answer for the special case when the base field is $\mathbb{Q}$. It involves a large bit of class field theory over $\mathbb{Q}$, so I'll be terse.
We start with the lemma which Buzzard mentioned.
**Lemma** - Let $K$, $L$ be finite Galois extensions of $\mathbb{Q}$. Then $K$ is contained in $L$ if and only if $\operatorname{sp}(L)$ is contained in $\operatorname{sp}(K)$ (with at most finitely many exceptions).
The proof of the lemma follows from the Chebotarev Density Theorem.
We now show that if the rational primes splitting in $K$ can be described by congruences, then $K/k$ is abelian.
**Proof**. Assume that the rational primes splitting in $K$ can be described by congruences modulo an integer $a$. This allows us to assume that $\operatorname{Sp}(K)$ contains the ray group $P\_a$. The next step is to show that the rational primes lying in $P\_a$ are precisely the primes of $\operatorname{sp}(\Phi\_a(x))$. By the above lemma, this means that $K$ is contained in a cyclotomic field, hence is abelian.
| 17 | https://mathoverflow.net/users/nan | 11692 | 7,933 |
https://mathoverflow.net/questions/11690 | 13 | Let $R$ be a complete discrete valuation ring and let $K$ be its field of fractions. Suppose $X$ is a smooth rigid-anaytic space over $K$. Often it is convenient to have a model of $X$ whose reduction has singularities which are as mild as possible--a semistable model. This amounts to having an admissible covering of $X$ by open affinoids $X\_i$, each of which has good reduction, such that the reductions of any pair $X\_i$ and $X\_j$ meet transversally, if at all. (See the paper of Bosch/Lütkebohmert for definitions.) Let us assume such a semi-stable model of $X$ exists. Then the étale cohomology of $X$ can be computed from the combinatorics of the covering $X\_i$, together with the étale cohomology of each $X\_i$, via the weight spectral sequence of Rapaport-Zink.
Now suppose I have an open affinoid $Z\subset X$ which happens to have good reduction. My question is: Does there admit a semi-stable model of $X$ for which $Z$ belongs to the covering? Failing this, is there some sense one can make of my intuition that the cohomology of the reduction of $Z$ ought to contribute to the cohomology of $X$?
Feel free to edit/criticize my question to smithereens if you like.
| https://mathoverflow.net/users/271 | Cohomology of rigid-analytic spaces | Here's a first pass at your question; hopefully it will suggest something more definitive.
Let's imagine we were in the simplest case, where $X$ is a disk, with its smooth model
being the formal affine line over $R$, and that $Z$ was the sub-disk of elements of
absolute value less than or equal the absolute value of the uniformizer. Then we can find a semistable model in which $Z$ is one of the covering opens, by blowing up the formal affine line at the origin.
So in this test case, the answer seems to be *yes* .
Now in general, I think that Raynaud (and/or his collaborators or those who followed in his
tradition) will say that the open immersion $Z \rightarrow X$ extends to an open immersion
of formal models. So we can blow up the smooth model of $X$ and the smooth model of $Z$
so that the latter sits inside the former. What I'm not very certain about is how much
you can control the nature of these blow-ups. (Presumably not at all in general, but you're starting in a fairly nice situation.)
Have you tried asking Brian Conrad yet?
| 6 | https://mathoverflow.net/users/2874 | 11696 | 7,937 |
https://mathoverflow.net/questions/11704 | 8 | Let k be a perfect field. By a k-variety, I shall mean a geometrically reduced separated scheme of finite type over k. I think that is enough conditions that the following data determine an affine k-variety:
1. A subset $X(\bar{k})$ of $\bar{k}^n$ which is defined by polynomials
2. A continuous action of $\mathop{\mathrm{Gal}}(\bar{k}/k)$ on $X(\bar{k})$, such that each $\sigma \in \mathop{\mathrm{Gal}}(\bar{k}/k)$ acts as $\sigma \circ f$ where f is a $\bar{k}$-regular map
When I say that these data determine an affine k-variety, I mean that there is a unique affine k-variety X whose $\bar{k}$-points are $X(\bar{k})$ with the correct Galois action.
Given these data, I want to work out the functor of points of X (which I consider to have domain the category of k-algebras). You can do that by following through the proof that these data determine a k-variety: first construct the coordinate ring A of X, as the Galois-fixed points of the ring of regular functions $X(\bar{k}) \to \bar{k}$; then $X(R) = \mathop{\mathrm{Hom}}(A, R)$ for any k-algebra R.
But if L is an algebraic extension of k, then there is a much simpler way of working out the L-points of X: just take the subset of $X(\bar{k})$ fixed by $\mathop{\mathrm{Gal}}(\bar{k}/L)$.
If L is a transcendental extension of k (or even a k-algebra which is not a field), is there a direct way of writing down the L-points of X which does not require going through the coordinate ring (or essentially equivalently, going through defining equations for X)?
| https://mathoverflow.net/users/1046 | Points of a variety defined by Galois descent | Despite what you learn in logic classes, there's truth to the conventional wisdom that you can't prove a negative. That is, it's a lot harder to explain why something necessarily can't work than why it can. (One almost invariably settles for: "it cannot work under the following explicit conditions, plus possibly others that I have left implicit.")
With that proviso, my preliminary answer is **no**. The data of the coordinate ring is of course equivalent to that of the set of polynomials $\{P\_i\}$ in (1). Thus your question sounds to me like asking: is there some way to dispense with condition (1)? Of course not: just because the set is Galois invariant doesn't mean it has any kind of algebraic structure (e.g. take $k = \overline{k}$ and we are merely saying that not just any old subset of affine space defines an affine variety).
Moreover, I don't see any shortcut around actually using the data of (1) and (2) to compute the coordinate ring. This is a very basic Galois descent argument involving Hilbert 90 applied to the ideal of $\overline{k}[x\_1,\ldots,x\_n]$ defined via (1).
| 6 | https://mathoverflow.net/users/1149 | 11706 | 7,943 |
https://mathoverflow.net/questions/11691 | 5 | Let the sequence $u\_1, u\_2, \ldots$ satisfy $u\_{n+1} = u\_n - u\_n^2 + O(u\_n^3)$. Then it can be shown that if $u\_n \to 0$ as $n \to \infty$, then $u\_n = n^{-1} + O(n^{-2} \log n)$. (See N. G. de Bruijn, *Asymptotic methods in analysis*, Section 8.5.)
This can be used to obtain asymptotics for $v\_{n+1} = Av\_n - Bv\_n^2 + O(v\_n^3)$, where $A$ and $B$ are constants. Let $w\_n = A^{-n} v\_n$; this gives
$$ A^{n+1} w\_{n+1} = A^{n+1} w\_n - B A^n w\_n^2 + O(A^n w\_n^3)$$
and so
$$ w\_{n+1} = w\_n - BA^{-1} w\_n^2 + O(w\_n^3). $$
Then let $w\_n = Ax\_n/B$ to get
$$ Ax\_{n+1}/B = Ax\_n/B - B/A \cdot (Ax\_n/B)^2 + O(x\_n^3) $$
and after simplifying
$ x\_{n+1} = x\_n - x\_n^2 + O(x\_n^3)$. This satisfies the initial requirements for $u\_n$ (with some checking of the side condition); then substitute back.
But say I actually know that $u\_{n+1} = P(u\_n)$ for some polynomial $P$, with $P(z) = z - z^2 + a\_3 z^3 + \cdots + a\_d z^d.$ In this case it seems like it should be possible to get more explicit information about $u\_n$. Is there a known algorithm for computing an asymptotic series for $u\_n$ as $n \to \infty$?
| https://mathoverflow.net/users/143 | Asymptotics of iterated polynomials | Have you tried the method of section 8.7, i.e., solving Abel's equation $\phi(P(x))-\phi(x)=1$? Here we expect $\phi(t)=t^{-1}+\sum\_{n=1}^{\infty}c\_n t^n$, and you can find the coefficients of $\phi$ one by one. For example, I took $P(x)=x-x^2+x^3+x^4$ and immediately found $c\_1=-2$, $c\_2=-5/2$, $c\_3=-7/2$, $c\_4=-17/4$... Not a general formula, but you can get as many terms as you want for a given $P$.
| 4 | https://mathoverflow.net/users/2912 | 11707 | 7,944 |
https://mathoverflow.net/questions/11709 | 3 | I'm wondering if there's an algorithm, or a program I can use, to compute integral closures. Specifically, what I have in mind are variants of questions of the sort: what is the integral closure of ℤ[x] in Quot(ℚ[x,y]/fℚ[x,y]), for some specific f(x,y).
| https://mathoverflow.net/users/3238 | Computing Integral Closures | If you are looking for prepackaged software, [Macaulay II](http://www.math.uiuc.edu/Macaulay2/doc/Macaulay2-1.3.1/share/doc/Macaulay2/IntegralClosure/html/index.html) can do the job.
| 2 | https://mathoverflow.net/users/297 | 11712 | 7,948 |
https://mathoverflow.net/questions/11631 | 24 | Obviously, graph invariants are wonderful things, but the usual ones (the Tutte polynomial, the spectrum, whatever) can't always distinguish between nonisomorphic graphs. Actually, I think that even a combination of the two I listed will fail to distinguish between two random trees of the same size with high probability.
Is there a known set of graph invariants that *does* always distinguish between non-isomorphic graphs? To rule out trivial examples, I'll require that the problem of comparing two such invariants is in P (or at the very least, not obviously equivalent to graph isomorphism) -- so, for instance, "the adjacency matrix" is not a good answer. (Computing the invariants is allowed to be hard, though.)
If this is (as I sort of suspect) in fact open, does anyone have any insight on why it should be hard? Such a set of invariants wouldn't require or violate any widely-believed complexity-theoretic conjectures, and actually there are complexity-theoretic reasons to think that something like it exists (specifically, under derandomization, graph isomorphism is in co-NP). It seems like it shouldn't be all that hard...
Edit: Thorny's comment raises a good point. Yes, there is trivially a complete graph invariant, which is defined by associating a unique integer (or polynomial, or labeled graph...) to every isomorphism class of graphs. Since there are a countable number of finite graphs, we can do this, and we have our invariant.
This is logically correct but not very satisfying; it works for distinguishing between finite groups, say, or between finite hypergraphs or whatever. So it doesn't actually tell us anything at all about graph theory. I'm not sure if I can rigorously define the notion of a "satisfying graph invariant," but here's a start: it has to be *natural*, in the sense that the computation/definition doesn't rely on arbitrarily choosing an element of a finite set. This disqualifies Thorny's solution, and I think it disqualifies Mariano's, although I could be wrong.
| https://mathoverflow.net/users/382 | Complete graph invariants? | A complete graph invariant is computationally equivalent to a canonical labeling of a graph. A canonical labeling is by definition an enumeration of the vertices of every finite graph, with the property that if two graphs are isomorphic as unlabeled graphs, then they are still isomorphic as labeled graphs. If you have a black box that gives you a canonical labeling, then obviously that is a complete graph invariant. On the other hand, if you have a complete graph invariant for unlabeled graphs, then you also have one for partially labeled graphs. So given a black box that computes a complete graph invariant, you can assign the label 1 to the vertex that minimizes the invariant, then assign a label 2 to a second vertex than again minimizes the invariant, and so on.
There are algorithms to decide graph isomorphism for certain types of graphs, or for all graphs but with varying performance, and there are algorithms for canonical labeling, again with varying performance. It is understood that graph isomorphism reduces to canonical labeling, but not necessarily vice versa. The distinction between the two problems is discussed [in this classic paper](http://portal.acm.org/citation.cfm?id=808746) by Babai and Luks.
One natural canonical labeling of a graph is the one that is lexicographically first. I think I saw, although I don't remember where, a result that computing this canonical labeling for one of the reasonable lex orderings on labeled graphs is NP-hard. But there could well be a canonical labeling computable in P that doesn't look anything like first lex.
As Douglas says, nauty is a graph computation package that includes a canonical labeling function. It is often very fast, but [not always](http://www.cs.trincoll.edu/~miyazaki/rutgers.ps). Nauty uses a fancy contagious coloring algorithm. For a long time people thought that contagious coloring algorithms might in principle settle the canonical labeling and graph isomorphism problems, but eventually counterexamples were found in [another classic paper](http://www.springerlink.com/index/H7185G8U46247443.pdf) by Cai, Furer, and Immerman. It was not clear at first whether this negative result would apply to nauty, but it seems that it does.
| 28 | https://mathoverflow.net/users/1450 | 11715 | 7,950 |
https://mathoverflow.net/questions/11718 | 1 | Let $R$ be the ring of Dirichlet series with integer coefficients. I'd often wondered about whether $R$ was a UFD; [this post](https://mathoverflow.net/questions/5522/dirichlet-series-with-integer-coefficients-as-a-ufd) cleared that up, because it turns out that $R\simeq\mathbb{Z}[[x\_1,x\_2,\cdots]]$ (the $x\_i$ correspond to primes, apparently, but I'm not sure what the explicit isomorphism is).
My first (slightly mundane) question is: what is the group of units $U(R)$? I know that $f\in R$ is a unit iff $f(0)$ is a unit (which in this case means $f(0)=\pm1$); similarly, $f\in\mathbb{Z}[[x\_1,x\_2,\cdots]]$ is a unit iff $f$'s constant term is $\pm1$, and part D of [this link](http://books.google.com/books?id=ZpDt86aI15kC&lpg=PP1&pg=PA318#v=onepage&q=&f=false) would seem to help a bit (using $\mathbb{Z}[[x\_1,x\_2,\cdots]]\simeq \mathbb{Z}[[x\_2,\cdots]][[x\_1]]$), but I couldn't get very far figuring out what $U(R)$ actually is.
Now, my main question: Can we take arbitrary $n$th roots (and hence, arbitrary fractional powers) of Dirichlet series which are units in $R$? I believe this is equivalent to asking whether the group of units is divisible, but I'm not sure.
A motivating example / special case of my question
--------------------------------------------------
$\mu$ and 1 (where $\mu$ is the Mobius function, $1$ is the constant 1 function) are units in $R$. In fact, $\mu\cdot 1=\epsilon$ (where $\epsilon(0)=1$, $\epsilon(n)=0$ for $n>0$ is the identity). Expressing this using the actual series,
$\displaystyle\left(\sum\_{n=1}^\infty \frac{\mu(n)}{n^s}\right)\left(\sum\_{n=1}^\infty\frac{1}{n^s}\right)=\sum\_{n=1}^\infty \frac{\epsilon(n)}{n^s} = 1$
and hence $\displaystyle\sum\_{n=1}^\infty \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)}$. Indeed, we can find the Dirichlet series for $\zeta(s)^k$ for any $k\in \mathbb{Z}$ by looking at the corresponding element $1^k\in R$ (note that $1^{-k}=\mu^k$). However, I would like to know what Dirichlet sequence corresponds to $\displaystyle\zeta(s)^{\frac{a}{b}}$ for $\frac{a}{b}\in\mathbb{Q}$.
| https://mathoverflow.net/users/1916 | Fractional powers of Dirichlet series? | For your first question: let $R$ be any commutative ring, and let $D(R)$ be the ring of formal Dirichlet series over $R$, i.e., the set of all functions $f: \mathbb{Z}^+ \rightarrow R$ under pointwise addition and convolution product.
Then the unit group of $R$ is precisely the set of formal Dirichlet series $f$ such that
$f(1)$ is a unit in $R$.
As for your second question, it is indeed equivalent to asking whether $U(D(R))$ is $n$-divisible. Here, if we take $R = \mathbb{Z}$ as you asked, the answer is that for all $n \geq 2$, $U(D(\mathbb{Z}))$ is **not** $n$-divisible and that even the Dirichlet series $\zeta(s)$ is not an $n$th power in $D(\mathbb{Z})$.
[Now, for some reason, I switch back to the classical notation, i.e., I replace the arithmetical function $f$ by its "Dirichlet generating series" $\sum\_{n=1}^{\infty} \frac{f(n)}{n^s}$. It would have been simpler not to do this, but too late.]
Let
$f(s) = \sum\_{n=1}^{\infty} \frac{a\_n}{n^s}$ be any formal Dirichlet series, and suppose
that $g(s) = \sum\_{n=1}^{\infty} \frac{b\_n}{n^s}$ be a formal Dirichlet series such that
$g^2 = f$. Thus
$a\_1 + \frac{a\_2}{2^s} + \ldots = (b\_1 + \frac{b\_2}{2^s} + ... )(b\_1 + \frac{b\_2}{2^s} + \ldots)$
$= b\_1^2 + \frac{2 b\_1 b\_2}{2^s} + \frac{2 b\_1 b\_3}{3^s} + \frac{2b\_1 b\_4 + b\_2^2}{4^s} +
\ldots$
(This multiplication is formal, i.e., it is true by definition.)
Thus $b\_1 = \pm \sqrt{a\_1}$. Suppose we take the plus sign, for simplicity. Then for all primes $p$,
$a\_p = 2 b\_1 b\_p$, so
$b\_p = \frac{a\_p}{2 \sqrt{a\_1} }$,
so we need $2 \sqrt{a\_1}$ to divide $a\_p$, so at least we need $a\_p$ to be even for all primes $p$. Further conditions will come from the composite terms.
These same considerations show that if we replaced the coefficient ring $\mathbb{Z}$ by
$\mathbb{Q}$ (or any coefficient field of characteristic $0$), then any formal Dirichlet series with $a\_1 = 1$ is $n$-divisible for all positive integers $n$. In particular, you can write $\zeta(s)^{\frac{1}{n}}$ as a Dirichlet series with $\mathbb{Q}$-coefficients just by applying the above procedure and successively solving for the coefficients. Whether there is a nice formula for these coefficients is a question for a better combinatorialist than I to answer.
**EDIT**: Based on your comments below, I now understand that you are looking for a characterization of $U(D(\mathbb{Z}))$ as an astract abelian group. I believe it is isomorphic to
$\{ \pm 1 \} \times \prod\_{i=1}^{\infty} \mathbb{Z}$. (Or, more transparently, to
the product of $\{ \pm 1\}$ with the product of infinitely many copies of $\prod\_{i=1}^{\infty} \mathbb{Z}$, one for each prime number. But as abstract groups it amounts to the same thing.)
| 3 | https://mathoverflow.net/users/1149 | 11724 | 7,953 |
https://mathoverflow.net/questions/6918 | 10 | I'm looking for natural conditions on $a\_{ij}$ to guarantee that the null space of the $n\times m$ matrix $A=(a\_{ij})$ has a nice basis.
The null space of { {1,-2,1,0,0}, {0,1,-2,1,0}, {0,0,1,-2,1} } is the set vectors $\langle x\_1,x\_2,x\_3,x\_4,x\_5\rangle^T$ with $x\_1,\dots,x\_5$ in arithmetic progression or constant, i.e., there is a degree zero or one polynomial $p(t)$ with $x\_i = p(i)$. The null space of { {3,3,-23,21,-4}, {6,3,-38,36,-7} } consists of points for which there is an at-most-quadratic $p(t)$ with $x\_1=p(1),x\_2=p(2),x\_3=p(3),x\_4=p(4),x\_5=p(6)$, with that last 6 not being a typo.
In particular, I need a basis for the null space of the form $\{\langle 1,1,\dots,1\rangle^T,\langle x\_1,\dots,x\_m\rangle^T, \dots, \langle x\_1^{m-n-1},\dots,x\_m^{m-n-1}\rangle^T\}$, with the $x\_i$ distinct (not necessarily integers).
As another specific example, consider the matrix { {3,-3,1,0,-1}, {20,-16,5,-9,0} }. I happen to know that the null space of this matrix has basis $\langle 1,1,1,1,1\rangle^T, \langle 1,4,7,-1,-2 \rangle^T, \langle 1^2,4^2,7^2 ,(-1)^2,(-2)^2\rangle^T$, but only because I made the matrix that way. Even with a specific matrix such as this, I don't know how to compute such a basis, or to guarantee that one exists or doesn't exist.
Here are the obvious necessary conditions: the rows must be independent; each row must add up to 0; no row can have exactly two nonzero components.
As a specific problem (I've no interest in this as a particular problem, mind you, but it may help the discussion) consider the matrix { {35,-3,-42,10,0}, {15,3,-8,0,-10} }. Does it have such a basis?
For background, I'm looking at constructions of sets $X$ of integers that contain no solutions to a system of linear equations. Such a basis as above means that a solution has x\_i in the image of a polynomial, and I already know how to construct sets that don't have those (arithmetic progressions are a special case).
| https://mathoverflow.net/users/935 | Matrices whose nullspace is nicely shaped | After quite a bit of tinkering, I decided that the example and a more fully realized generalization merited separate answers, not least because my initial answer entered community wiki due to the number of edits I made.
Let $N$ be a null space matrix for $A$, i.e., the columns of $N$ are annihilated by $A$. We want vectors $w,w',\dots,w^{(m-n-2)}$ s.t. $(Nw)^{\ell+1} = Nw^{(\ell)}$. Define $z^{(\ell)}$ by $z^{(\ell)}\_{i(\alpha)} := w^\alpha$, where $i(\alpha)$ is the grlex index of
$\alpha \in$ $ X\_{m-n,\ell+1} \equiv$ {$\beta \in \mathbb{Z}^{m-n}: \sum\_k \beta\_k = \ell + 1$}.
Now let $P^{(\ell)}$ be the $m \times |X\_{m-n,\ell+1}|$ matrix with entries given by
$P^{(\ell)}\_{j,i(\alpha)} :=$ coefficient of $w^\alpha$ in $(\sum\_k N\_{jk}w\_k)^{\ell+1}$.
>
> To obtain this coefficient explicitly,
> note that
>
>
> $(\sum\_k N\_{jk}w\_k)^{\ell+1} =
> > \sum\_{\alpha \in X}
> > \binom{\ell+1}{\alpha}N\_{j,\cdot}^\alpha
> > w^\alpha$
>
>
> whence $P^{(\ell)}\_{j,i(\alpha)}$
> equals
>
>
> $\binom{\ell+1}{\alpha}
> > N\_{j,\cdot}^\alpha$.
>
>
> For example, with $N\_{j,\cdot} =
> > (2,3,5,7)$, $\ell = 2$, and $\alpha =
> > (0,1,1,1)$, so that $i(\alpha) = 6$,
> we have that $P^{(\ell)}\_{j,i(\alpha)}
> > =$
>
>
> $\binom{3}{1,1,1}
> > N\_{j,\cdot}^{(0,1,1,1)} = 3! \cdot 3
> > \cdot 5 \cdot 7 = 630$.
>
>
> Extending this example,
>
>
> $P^{(2)}\_{j,\cdot} =
> > (343,343,735,525,125,441,630,225,189,135,27,294,420,150,252,180,54,84,60,36,8)$.
>
>
>
Then the existence (ignoring distinctness of entries) of $w^{(\ell)}$ s.t. $(Nw)^{\ell+1} = Nw^{(\ell)}$ is equivalent to the existence of a solution to
$(Nw)^{\ell+1} = P^{(\ell)}z^{(\ell)}$.
Note that for all $\ell$ this is really an equation in the components of $w$, viz.
$\left(\sum\_k N\_{jk} w\_k \right)^{\ell+1} = \sum\_{\alpha \in X} \binom{\ell+1}{\alpha} N\_{j,\cdot}^\alpha w^\alpha$
and it should (at least) be amenable to solution in a computer algebra routine.
| 1 | https://mathoverflow.net/users/1847 | 11725 | 7,954 |
https://mathoverflow.net/questions/11723 | 6 | Let X\_0(p) be the modular curve of level p where p is prime. The Jacobian variety J\_0(p) has a natural family of quotients defined over Q with dimensions summing to dim(J\_0(p)), each quotient corresponding to a Galois conjugacy class of normalized Hecke eigenforms on X\_0(p), the dimension of a quotient being equal to the degree of the number field generated by the Fourier coefficients of the corresponding form(s).
Let f(p, d) be the number of such quotients of J\_0(p) of dimension d
>
> Has anyone made a compelling and
> nontrivial conjecture about the size
> of f(p, d) as p ---> infinity?
>
>
>
Same question for the average value of f(p, d) for primes p <= N as N ---> infinity (which *a priori* may amount to the same thing).
| https://mathoverflow.net/users/683 | Distribution of dimensions of factors of the Jacobian of X_0(p) | Let me begin with what was formerly a comment above: the feeling among most experts is probably\* that for each fixed $d$, the number of isogeny factors of $J\_0(p)$ of dimension $d$ should be small compared to the dimension of $J\_0(p)$, i.e. $o\_d(p)$.
In what follows, I will cite some results which point in this direction. My overall response is not as logically coherent as I might have liked: perhaps a true expert will do better.
When $p$ is prime, it follows from a 1975 theorem of Ribet (reference below) that the $\mathbb{Q}$-rational endomorphism algebra of $J\_0(p)$ is the same as the geometric (i.e., $\mathbb{C}$-rational) endomorphism algebra, and that this algebra is a product of formally real fields, each being the subfield of $\mathbb{Q}$ obtained by adjoining the Fourier coefficients of the various weight $2$ cuspforms of level $p$.
Thus the problem can be viewed as a special case of a popular one in pure algebraic geometry: for which genera $g$ do there exist complex algebraic curves of genus $g$ with, e.g., Jacobians isogenous to a product of elliptic curves? (Or, more generally, with endomorphism algebra containing at least $N$ semisimple factors?) If you count codimensions in the Siegel moduli space corresponding to (say) principally polarized abelian varieties with certain nontrivial endomorphism algebras and the Torelli locus (i.e., of Jacobians), then you find that (at least in many special cases) the sum of these codimensions adds up to more than $\frac{g(g+1)}{2}$, the dimension of the Siegel moduli space. Thus unless there is **excess intersection** between these two loci, for sufficiently large $g$ one expects Jacobians not to be highly decomposed. [This is a cue for Prof. JSE to weigh in on the matter.]
For instance, I think it is at least conjectured that for sufficiently large $g$, no $g$-dimensional Jacobian splits completely into a product of elliptic curves. This is known to be true over finite fields, but maybe not over $\mathbb{C}$.
Serre has worked on both the general geometric question and on this special case. In his 1997 JAMS paper, Serre showed that the maximal dimension of a simple isogeny factor of $J\_0(N)$ approaches infinity with $N$. I think the result is quantitative, so if you look there you may get the most information currently known on the question you asked.
---
Ribet, Kenneth A. Endomorphisms of semi-stable abelian varieties over number fields. Ann. Math. (2) 101 (1975), 555--562.
---
Serre, Jean-Pierre
Répartition asymptotique des valeurs propres de l'opérateur de Hecke $T\_p$. (French) [Asymptotic distribution of the eigenvalues of the Hecke operator $T\_p$]
J. Amer. Math. Soc. 10 (1997), no. 1, 75--102.
---
\*: I think it's true, and the small number of people I've spoken to about this explicitly (several years ago) thought it was true. We'll see what others have to say.
| 4 | https://mathoverflow.net/users/1149 | 11727 | 7,956 |
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