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https://mathoverflow.net/questions/6883 | 5 | Suppose we have a quadratic eigenvalue problem $\lambda^2 M + \lambda C + K$. Under what conditions is the following statement true: If $\lambda$ is an eigenvalue, so is $1/\lambda$?
Here, $M$, $C$, and $K$ are square matrices (not necessarily full rank). This is of interest to me since I have such systems for which I know (based on physical arguments) that the eigenvalues must come in reciprocal pairs, but I don't know what this necessarily implies about the matrix properties. From a cursory look through Google, it seems that palindromic QEPs have this property, but I'm wondering if this property is more general.
| https://mathoverflow.net/users/1074 | Under what conditions do eigenvalues of a quadratic eigenvalue problem come in reciprocal pairs? | Consider the linearization
$$
L(\lambda)=\lambda\begin{pmatrix}M&0\cr
0&1
\end{pmatrix}+\begin{pmatrix}C&K\cr
-1&0\end{pmatrix}=\lambda A+B.
$$
The eigenvalues of the original problem coincide with those for the linearization.
Now, the eigenvalues of the linearized problem are the roots of the polynomial $\det(\lambda A+B)$, while their reciprocals are roots of the polynomial $\lambda^{2n}\det(\lambda^{-1}A+B)=\det(\lambda B+A)$. If we require that these coincide with multiplicities, then the two polynomials must be linearly dependent. (In fact, they will be $\pm 1$ times each other.) If $A$ is nondegenerate, this implies that $A^{-1}B$ is similar to its inverse matrix. The latter condition does not give us much immediately, but at least we know that $\det A=\pm\det B$; that is, $\det M=\pm\det K$. We also get that the traces of $A^{-1}B$ and $B^{-1}A$ coincide, which can be rewritten as a condition on $M$, $C$, and $K$ (hopefully).
| 4 | https://mathoverflow.net/users/1704 | 6905 | 4,710 |
https://mathoverflow.net/questions/6912 | 13 | As anyone who follows the arxiv, I notice every now and then "proofs" and "disproofs" of Riemann Hypothesis. I looked on several such articles, and it seemed to me quite nonsense, but I didn't make the effort to find a mistake. My question is whether someone reads these "proofs"?
BTW, I wanted to refer to some of these papers in the arxiv, but it turned out that there are too many of them.
| https://mathoverflow.net/users/2042 | On the proofs (and disproofs) of Riemann Hypothesis | Whenever someone claims a proof (or disproof) of a big conjecture, many people leap to the question of whether the proof is correct. The problem then is it that it takes an enormous amount of work to confirm that a proof is correct. Even a clear mistake in a proof could be reparable. Moreover, attempted proofs have inferences that amount to gaps of different sizes. Even in a naive attempt, it can take a lot of work to decide which gaps are so big that the proof has to be called incomplete.
There is a much simpler standard that experts use in practice: "As I start to read this paper, am I learning from it?" You would expect a proof of a big conjecture to have very interesting lemmas, and otherwise to teach you new things along the way. This is not always obvious either; there have been a few grievous misunderstandings in which initial readers dismissed a great paper. Even so, it's a somewhat reliable standard, and it's the most that authors can expect.
When Perelman posted the first of his three papers on geometrization, experts in differential geometry quickly embraced it as exciting and teachable, before they had even checked half of that paper or seen the other two papers. From the beginning, this was very different from most claimed proofs of the Poincare conjecture, even most of the noble failures. The great ideas in these papers were more important than the fact that they had a lot of gaps (by common standards) and even some inessential mistakes (or so I was told).
I know for a fact that experts sometimes do study weird-looking claims of big results, in the arXiv and elsewhere. They have little incentive to broadcast their attention to it if they think that it's shoddy work, but sometimes they try to be fair. For starters, the math arXiv has moderators, and they often take a look. I think that usually (not quite always), several people have looked long enough to decide that they aren't getting anything out of the paper. But hey, there could always be a diamond in the rough, or even a diamond in the garbage.
| 46 | https://mathoverflow.net/users/1450 | 6916 | 4,718 |
https://mathoverflow.net/questions/6889 | 29 | I've been learning about Dedekind zeta functions and some basic L-functions in my introductory algebraic number theory class, and I've been wondering why some functions are called L-functions and others are called zeta functions. I know that the zeta function is a product of L-functions, so it seems like an L-function is somehow a component of a zeta function (at least in the case of Artin L-functions, they correspond to specific representations). Is this the idea behind the distinction between "zeta function" and "L-function"? How do things generalize to other kinds of zeta- and L-functions?
| https://mathoverflow.net/users/1355 | What is the difference between a zeta function and an L-function? | Let me say first that a Dedekind zeta function is *always* a product of Artin L-functions. It is the structure of the Galois closure which is relevant here. Let me give a nice example which is indicative of the general case. Let $p(x) \in \mathbb{Z}[x]$ be an irreducible cubic, and let $\alpha$ be a root of $p$. Then $K=\mathbb{Q}(\alpha)$ has trivial automorphism group, and its Galois closure (say $L/\mathbb{Q}$) is an S3-extension. The group S3 has three irreducible representations: the trivial representation, the "sign representation" $\chi$ which is also one-dimensional, and an irreducible two-dimensional representation which we will call $\rho$. Then we have the relations $\zeta\_K(s)=\zeta\_{\mathbb{Q}}(s)L(s,\rho)$ and $\zeta\_L(s)=\zeta\_{\mathbb{Q}}(s)L(s,\chi)L(s,\rho)^2$. The proofs of these facts are part of the formalism of Artin L-functions.
Generally, the distinction is really a matter of history. Certain objects were named zeta functions - Hasse-Weil, Dedekind - while Dirichlet chose the letter "L" for the functions he made out of characters. However, one feature is that "zeta" functions tend to have poles, and they often "factor" into L-functions. These vagaries are made more precise in various places, for example Iwaniec-Kowalski Ch. 5 and some survey articles on the "Selberg class" of Dirichlet series.
| 24 | https://mathoverflow.net/users/1464 | 6927 | 4,723 |
https://mathoverflow.net/questions/6928 | 47 | What papers should we read to start? What basic knowledge do we need to understand the question? What is this area really about? And what are people researching on it?
| https://mathoverflow.net/users/2110 | How do we study Iwasawa theory? | Iwasawa theory has its origins in the following counterintuitive insight of Iwasawa: instead of trying to describe the structure of any particular Galois module, it is often easier to describe every Galois module in an infinite tower of fields at once.
The specific example that Iwasawa studied was the $p$-Sylow subgroup of the class group of $K\_n = \mathbb{Q}(\zeta\_{p^n})$. It's naturally a $\mathbb{Z}\_p$-module as well as a $G\_n$ = Gal$(K\_n/K\_1)$-module, but the group ring $\mathbb{Z}\_p[G\_n]$ isn't very nice; it's not a domain, for instance. If we instead look at the inverse limits of the $p$ parts of the class groups of all the fields $K\_n$ at once, as modules over $\mathbb{Z}\_p[G\_n]$, we get a module over the inverse limit $\varprojlim\mathbb{Z}\_p[G\_n]$. This ring is much easier to understand; it's a complete 2-dimensional regular local ring that is (non-canonically) isomorphic to a power series ring, and there is a strong structure theorem for modules over this ring. Using this structure theorem, Iwasawa proved many theorems about the class numbers of cyclotomic fields. For a simple example: $p$ divides the class number of one of the fields $K\_n$ if and only if it divides the class number of all of the fields $K\_n$.
There's an even bigger payoff to the theory: a profound connection with special values of $L$-functions. In the function field case, Weil had interpreted the Hasse-Weil $L$-function as computing the characteristic polynomial of Frobenius acting on the Jacobian of a curve. Iwasawa's idea was that the analogue for number fields should be the "characteristic ideal" of the ring $\varprojlim\mathbb{Z}\_p[G\_n]$ acting on ideal class groups. It turns out this characteristic ideal has a generator that is essentially the same as a $p$-adic $L$-function closely related to the ordinary Dirichlet $L$-functions. This was Iwasawa's "main conjecture" and is now a theorem. It implies the Herbrand-Ribet theorem and essentially every classical result relating cyclotomic fields and zeta values.
There have been many generalizations since but it's safe to call an area "Iwasawa theory" if it studies some Galois representation ranging over an infinite tower of fields and connects it to $p$-adic $L$-functions. The most fruitful Galois modules from the point of view of $L$-functions seem to be Bloch and Kato's generalized Selmer groups; the ideal class group can be interpreted as a Selmer group, and so can the classical Selmer group of an abelian variety. There's a lot of current research in this area.
To start reading, I recommend Washington's book on cyclotomic fields. Chapter 13 is fun and is a good use of some of the main techniques of Iwasawa theory. You don't need anything but the basic background in chapters 1-4 to read sections 1-4 of Chapter 13, which contain the types of theorem I was referring to in the first two paragraphs of this answer. The explicit computations in the first ten chapters also give the link to $p$-adic L-functions. If you know some algebraic number theory, you should be fine to read this book. I also strongly recommend Greenberg's PCMI notes on the Iwasawa theory of elliptic curves, which you can find here:
<http://www.math.washington.edu/~greenber/Park.ps>
If you're comfortable with class field theory, and have read the first few sections of Chapter 13 in Washington, then Coates and Sujatha's recent book, *Cyclotomic Fields and Zeta Values*, is a pleasure to read.
| 62 | https://mathoverflow.net/users/1018 | 6933 | 4,726 |
https://mathoverflow.net/questions/6345 | 5 | How does Hermite normal form (over $Z$) vary in families? I.e. if I have an $n\times m$ matrix $M$ whose entries are integral polynomials in some integral variable $x$, how does the Hermite normal form of the integral matrix $M(p)$ (obtained by setting $x$ equal to $p$) vary as a function of $p$? What about the special case that the entries are (at most) linear in $x$?
The question is a bit open ended so answers could be of several kinds, eg:
(i) how certain integer programming problems associated to $M(p)$ depend on $p$;
(ii) by explaining how the answer can be expressed in a way that generalizes Ehrhart theory;
(iii) by specializing to an important case that is well-understood;
(iv) in some other interesting way.
I would also really appreciate a pointer to any relevant literature.
| https://mathoverflow.net/users/1672 | Hermite normal form in families | Hi "DC". I think that I have worked out that the Hermite normal form is a "trichotomous quasipolynomial" in the variable $p$. If $f:\mathbb{Z} \to \mathbb{Z}$ is a function, then my definition is that $f$ is a trichotomous quasipolynomial if it is a quasipolynomial for $x \gg 0$, possibly a different quasipolynomial for $x \ll 0$, and in between finitely many unrestricted values.
I think that if $R$ is a canonically Euclidean ring — Euclidean with canonically chosen quotients and remainders — then there is a Hermite normal form for matrices over $R$. In particular, I think that $R$ does not have to be a Euclidean domain.
As a first try, let $A$ be the ring of all functions $f:\mathbb{Z} \to \mathbb{Z}$, using pointwise quotients and remainders. Hermite normal form over this ring is a model of computing Hermite normal form for any $\mathbb{Z}$ family of integer matrices. $A$ is sort-of a Euclidean ring, except it isn't Noetherian.
Let $B$ be the subring of $A$ consisting of trichotomous quasipolynomials. ~~Then I believe that $B$ is Noetherian and it is a Euclidean ring. If that is correct, then you obtain a Hermite normal form that is also a trichotomous quasipolynomial.~~
---
It's not correct, at least not in any obvious way. It is easy to check that $B$ is not only a subring, but is also closed under quotients with pointwise remainders. In order to compute how $b(x)$ divides into $a(x)$, you can reduce to the case in which $a(x)$ and $b(x)$ are both polynomials. Then for starters there is a quotient and a remainder in $\mathbb{Q}[x]$:
$$a(x) = q(x)b(x) + r(x).$$
Since $r(x)$ has lower degree than $b(x)$, its values are smaller than those of $b(x)$ when $x \gg 0$, so that part is okay; but it may be negative and $q(x)$ may not be integral. We can fix all that by rounding $q(x)$, which creates quasipolynomial behavior; and by changing it by 1 to make $r(x)$ positive. Also since $r(x)$ might have odd degree, there may be a different quasipolynomial solution when $x \ll 0$.
The part that is either not true or far from obvious is why $B$ is Euclidean. My argument for that fell apart. However, I can still show that the Euclidean algorithm for finitely many elements $a\_1,\ldots,a\_n$ of $B$ terminates in a finite number of steps, and consequently that the Hermite normal form stabilizes in a finite number of steps with trichotomous quasipolynomial entries. The proof is a two-stage induction. The outer stage is the sum of the degrees of $a\_1,\ldots,a\_n$. If $\deg a\_j < \deg a\_k$ for some $j$ and $k$, then dividing $a\_j$ into $a\_k$ reduces the degree of $a\_k$. On the other hand, suppose that the degrees are all equal. Then we can pass to a congruence class for the input $x$ in $\mathbb{Z}$ and apply a linear change of variables so that the leading coefficients are all integers. Then (in the inner induction) the Euclidean algorithm on these polynomials, for $x \gg 0$, amounts to the Euclidean algorithm on their coefficients. It is important, in this inner inductive part, to only change the variable $x$ once; don't worry if the lower-order coefficients are not integers. Eventually a $0$ is produced and the degrees once again decrease.
Alas, this is a very informal writeup, but this time I think that it works.
| 3 | https://mathoverflow.net/users/1450 | 6938 | 4,731 |
https://mathoverflow.net/questions/6950 | 50 | This question might not have a good answer. It was something that occurred to me yesterday when I found myself in a pub, needing to do an explicit calculation with 2-cocycles but with no references handy (!).
Review of group cohomology.
---------------------------
Let $G$ be a group acting (on the left) on an abelian group $M$. Then $H^0(G,M)=M^G=Hom\_{\mathbf{Z}[G]}(\mathbf{Z},M)$ and hence $$H^i(G,M)=Ext^i\_{\mathbf{Z}[G]}(\mathbf{Z},M).$$ Now $Ext$s can be computed using a projective resolution of the first variable, so we're going to get a formula for group cohomology in terms of "cocycles over coboundaries" if we write down a projective resolution of $\mathbf{Z}$ as a $\mathbf{Z}[G]$-module.
There's a very natural projective resolution of $\mathbf{Z}$: let $P\_i=\mathbf{Z}[G^{i+1}]$ for $i\geq0$ (with $G$ acting via left multiplication on $G^{i+1}$) and let
$d:P\_i\to P\_{i-1}$ be the map that so often shows up in this sort of thing: $$d(g\_0,\ldots,g\_i)=\sum\_{0\leq j\leq i}(-1)^j(g\_0,\ldots,\widehat{g\_j},\ldots,g\_i).$$
Check: this is indeed a resolution of $\mathbf{Z}$. So now we have a "formula" for group cohomology.
But it's not the usual formula because I need to do one more trick yet. Currently, the formula looks something like this: the $i$th cohomology group is $G$-equivariant maps $G^{i+1}\to M$ which are killed by $d$ (that is, which satisfy some axiom involving an alternating sum), modulo the image under $d$ of the $G$-equivariant maps $G^i\to M$.
The standard way to proceed from this point.
--------------------------------------------
The "formula" for group cohomology derived above is essentially thought-free (which was why I could get this far in a noisy pub with no sources). But we want something more useful and it was at this point I got stuck. I remembered the nature of the trick: instead of a $G$-equivariant map $G^{i+1}\to M$ we simply "dropped one of the variables", and considered arbitrary maps of sets $G^i\to M$. So we need to give a dictionary between the set-theoretic maps $G^i\to M$ and the $G$-equivariant maps $G^{i+1}\to M$. I could see several choices. For example, given $f:G^{i+1}\to M$ I could define $c:G^i\to M$ by $c(g\_1,g\_2,\ldots,g\_i)=f(1,g\_1,g\_2,\ldots,g\_i)$, or $f(g\_1,g\_2,\ldots,g\_i,1)$, or pretty much anything else of this nature. The point is that given $c$ there's a unique $G$-equivariant $f$ giving rise to it. Which choice of dictionary do we use. *Each one will give a definition of group cohomology as "cocycles" over "coboundaries"*. But which one will give the "usual" definition? Well---in some sense, who cares! But at some point maybe someone somewhere made a decision as to what the convention from moving from $f$ to $c$ was, and now we all stick with it. The standard decision was the rather clunky
$$c(g\_1,g\_2,\ldots,g\_i)=f(1,g\_1,g\_1g\_2,g\_1g\_2g\_3,\ldots,g\_1g\_2g\_3\ldots g\_i).$$
Why is the standard definition ubiquitous?
------------------------------------------
Actually though, I bet that no-one really made that decision. I bet that the notion of a 1-cocycle and a 2-cocycle preceded general homological nonsense, and the dictionary between $f$s and $c$s was worked out so that it agreed with the definitions which were already standard in low degree.
But this got me thinking: if, as it seems to me, there is no "natural" way of moving from $f$s to $c$s, then *why do the 1-cocycles and 2-cocycles that naturally show up in mathematics all satisfy the same axioms??*. Why doesn't someone do a calculation, and end up with $c:G^2\to M$ satisfying some random axiom which happens not to be the "standard" 2-cocycle axiom but which is an axiom which, under a non-standard association of $c$s with $f$s, becomes the canonical cocycle axiom for $f$ that we derived without moving our brains? Does this ever occur in mathematics? I don't think I've ever seen a single example.
In some sense it's even a surprise to me that there is a uniform choice which specialises to the standard choices in degrees 1 and 2.
Examples of cocycles in group theory.
-------------------------------------
1) Imagine you have a 2-dimensional upper-triangular representation of a group $G$, so it sends $g$ to the $2\times 2$ matrix $(\chi\_1(g),c(g);0,\chi\_2(g))$. Here $\chi\_1$ and $\chi\_2$ are group homomorphisms. What is $c$? Well, bash it out and see that $c$ is precisely a 1-cocycle in the sense that everyone means when they're talking about 1-cocycles. So we must have used the dictionary $c(g)=f(1,g)$ when moving between $c$s and $f$s above. Why didn't we use $c(g)=f(g,1)$?
2) Imagine you're trying to construct the boundary map $H^0(C)\to H^1(A)$. Follow your nose. Claim: your nose-following will lead you to the standard cocycle representing the cohomology class. Why did it not lead to a non-standard notion? Why do the notions (1) and (2) agree?
3) Imagine we have a group hom $G\to P/M$ where $M$ is an abelian normal subgroup of the group $P$. Can we lift it to a group hom $G\to P$? Well, let's take a random set-theoretic lifting $L:G\to P$. What is the "error"? A completely natural thing to write down is the map $(g,h)\mapsto L(g)L(h)L(gh)^{-1}$ because this will be $M$-valued. This is a 2-cocycle in the standard sense of the word. Why did the natural thing to do come out to be the standard notion of a 2-cocycle? Aah, you say: there are other natural things that one can try. For example we could have sent $(g,h)$ to $L(gh)^{-1}L(g)L(h)$. For a start this "looks slightly less natural to me" (why does it look less natural? That's somehow the question!). Secondly, this is really just applying a canonical involution to everything: we're inverting on $G$ and inverting on $M$, which is something that we'll always be able to do. It certainly does *not* correspond to the "more natural" dictionary that I confess I tried down the pub, namely $c(g,h)=f(1,g,h)$, which gave much messier answers. Why does my "obvious" choice of dictionary lead me to 20 minutes of wasted calculations? Why is it "wrong"?
The real question.
------------------
Has anyone ever found themselves in a situation where a natural cocycle-like construction is staring them in the face, and they make the construction, and find themselves with a non-standard cocycle? That is, a cocycle which will induce an element of a cohomology group but only after a non-standard dictionary is applied to move from $f$s to $c$s?
Edit: Here is what I hope is a clarification of the question. To turn the cocycles from $G$-equivariant maps $G^{i+1}\to M$ to maps of sets $G^i\to M$ we need to choose a transversal for the action of $G$ on $G^{i+1}$ and identify this with $G^i$. There seems to be one, and only one, way to do this that gives rise to the "standard" definition of n-cocycle that *I (possibly incorrectly) percieve* to be ubiquitous in mathematics. I call this "the clunky way" because the map seems odd to me. Why is it so clunky? And why, whenever a 2-cocycle falls out of the sky, does it always seem to satisfy the axioms induced by this clunky method? Why aren't there people popping up in this thread saying "well here's a completely natural "cocycle" $c(g,h)$ coming from theory X that I study, and it doesn't satisfy the usual cocycle axioms, we have to modify it to be $c'(g,h)=c(g,gh)$ before it does, and this boils down to the fact that in theory X we would have been better off if history had chosen a non-clunky identification?"
| https://mathoverflow.net/users/1384 | Why is the standard definition of cocycle the one that _always_ comes up?? | The difference between f(1,g) and f(g,1) is generally an issue of whether mathematicians give preference to "domains" or "ranges" of maps.
Here is one way that you could think of this. I can write EG for a category whose objects are objects are elements of G, and where each pair of objects has a unique map between them. This category has an action of G on it, and you can ask about G-equivariant functors from this category to another category what has a G-action on it.
To define such a functor on the level of objects it suffices to define F(1), where 1 is the unit; equivariance forces us to define F(g) = g F(1). On the level of morphisms, however, we have to make a choice. The unique morphism g→h becomes a morphism F(g)→F(h), and to make such maps compatible with the G-action it suffices to make one of the following sets of choices:
* We could define maps fh:F(1)→F(h) for all h, and get all the other maps as g fh:F(g)→F(gh). To be a functor, we need this to satisfy the cocycle condition fgh = (g fh) fg.
* We could define maps dh:F(h)→F(1) for all h, and get all the other maps as g dh:F(gh)→F(g). To be a functor, we need this to satisfy the cocycle condition dgh = dg (g dh).
In group cohomology, H1(G,M) classifies splittings in the semidirect product of G with M, and the cocycle condition we get comes from our convention of writing this group as pairs (m,g) (which is in the same order as the exact sequence it fits into) and not (g,m). Similarly for H2(G,M).
I would say that I've hit nonstandard cocycle definitions several times because I've been too lazy to come up with sensible conventions about when I'm thinking about domains and ranges or trying to sweep it under the rug, especially when dealing with Hopf algebroids and cohomological calculations there.
I don't have a good answer for higher cocycle conditions other than saying that writing 2-cochains using f(g,1,h) is somehow more unusual than either of the other 2 choices because it's somehow derived from focusing on the "middle" object in a double composite of maps.
| 18 | https://mathoverflow.net/users/360 | 6952 | 4,740 |
https://mathoverflow.net/questions/6941 | 8 | Article [Exotic $\mathbb{R}^4$](http://en.wikipedia.org/wiki/Exotic_R4) on Wikipedia says that there is at least one maximal smooth structure on $\mathbb{R}^4$, that is such an atlas on $\mathbb{R}^4$ that any other smooth $\mathbb{R}^4$ can be embedded into it. Is the construction of such a maximal exotic $\mathbb{R}^4$ explicit? Can anyone give a reference to the construction? What is a good source (or sources) with examples of exotic smooth structures on $\mathbb{R}^4$? Thanks.
| https://mathoverflow.net/users/896 | Maximal exotic $\mathbb{R}^4$ | The paper by Freedman and Taylor mentioned by Carsten Schultz (above or below) is indeed the place to find the explicit construction.
Very roughly, the idea of the construction is as follows. Recall that a Casson handle is, among other things, a smooth 4-manifold which is homeomorphic (but not diffeomorphic) to the standard open 2-handle. It turns out that a countable collection of diffeomorphism classes of Casson handles suffice for solving a certain 5-dimensional h-cobordism problem. The universal $R^4$, call it $U$, is constructed by gluing together countably many copies of each of the Casson handles in the countable collection. Given an arbitrary smooth 4-manifold homeomorphic to $R^4$, we can construct an embedding (but not a proper embedding) into $U$ using the fact that $U$ contains enough Casson handles to solve any link slice problem we might encounter along the way.
Well, that was kind of vague, but hopefully not inaccurate.
| 12 | https://mathoverflow.net/users/284 | 6965 | 4,751 |
https://mathoverflow.net/questions/6964 | 18 | A question I got asked I while ago:
If $T$ is a triangle in $\mathbb R^2$, is there a function $f:T\to\mathbb R$ such that the integral of $f$ over each straight segment connecting two points in the boundary of $T$ not on the same side is always $1$?
(Of course, you can change $T$ for your favorite convex set... and the problem should really be seen as asking for what sets is the answer affirmative, mostly)
| https://mathoverflow.net/users/1409 | Egalitarian measures | In general, no. For the double integral $\iint\_T f(x,y)\,dx\,dy$ will be the height on any side, as is seen by turning the triangle with one side parallel to an axis and performing the integral. So at least, $T$ has to be equilateral. I don't know the answer in that case.
**Edit:** Wait, wait – the same trick works even if I turn the triangle at any angle, hence all heights (defined as $\sup\_{RT} y-\inf\_{RT} y$ where $R$ is a rotation) must be the same. That is never true for a triangle, and limits the number of convex sets seriously – but there are still non-circles that might satisfy the criterion. Once more, I don't know the answer, but for circles at least, it should in principle be straightforward to check if a radially symmetric function will do. And of course, if there is a solution, there is a radially symmetric one, as can be seen by rotating the solution and taking the average of all its rotated variants.
**Edit2:** For the unit disk (and radially symmetric $f$) the answer should in principle be obtainable by the Abel transform (which is really nothing but the Radon transform on radially symmetric functions). The required Abel transform $\Phi$ should be the characteristic function of the interval $[0,1]$ (we only use positive $x$ due to symmetry), and the inverse Abel transform provides the answer:
$$f(r)=\frac{-1}{\pi}\int\_r^\infty \frac{\Phi'(x)\,dx}{\sqrt{x^2-r^2}}=\frac{1}{\pi\sqrt{1-r^2}}$$
when $0\le r<1$. Being lazy, I checked the answer using Maple, and it seems right.
**Addendum:** Anton Petrunin pointed out in a comment that the above measure is the push-forward of the surface measure on the unit sphere on the unit disk under projection. It is well-known that the surface area of the portion of a disk between two parallel planes depends only on the distance between the planes (and is proportional to said distance), which ties in nicely with the desired property of $f$ on the disk.
| 17 | https://mathoverflow.net/users/802 | 6967 | 4,753 |
https://mathoverflow.net/questions/6980 | 4 | Let $\overline{\rho}\_{\Delta,\ell}$ be the mod-$\ell$ representation associated to Ramanujan's $\Delta$-function. It is well-known that (the semisimplification of) this representation is reducible if, say, $\ell=5$ or $\ell=691$. Is there a general name for primes like this? Serre calls them (in a more general context) "exceptional primes," but the word exceptional always strikes me as vague. "Primes of residual reducibility"?
| https://mathoverflow.net/users/1464 | A name for primes where residual Galois representations are reducible | "Eisenstein" ?
| 4 | https://mathoverflow.net/users/1125 | 6993 | 4,770 |
https://mathoverflow.net/questions/6929 | 0 | Let's say a series of 10 bits is output randomly. Now lets do that 256 times. I'd like to find out what the expected number of streaks of 1s or 0s are for each of the possible sizes 1-10.
For example, the stream 0111001011 has 3 1-bit streaks, 2 2-bit streaks, and 1 3-bit streak. Note that even though the 3-bit streak has two 2-bit streaks in it, those are not counted, as the 3-bit streak removes them from consideration. Note that the sumproduct of the streak counts and the streak lengths = (3 \* 1)+(2 \* 2)+(1 \* 3) = 10 (the number of bits), and that should always work.
So, given 256 of these random 10-bit streams, how many 1-bit, 2-bit, 3-bit, 4-bit... 10-bit streaks can I expect? Fractional values are fine, as I'm sure the # of 10-bit streaks is less than 1. I'd really appreciate knowing HOW this is done, as my integrals don't seem to be accomplishing it.
Thanks!
| https://mathoverflow.net/users/2111 | Chances of streaks in small bit-streams | I'm going to call your "streams" "strings" instead, because "streams" looks too much like "streaks" to me. This becomes a much easier problem if we translate it into an enumeration problem. Since each of the 210 possible bit strings occur with equal probability, it suffices to count the total number of streaks of length k in all these strings (and then as Kristal says just divide by 210 for the expected value per string). Let S(k,n) be the number of k-streaks in all strings of length n. For k = 1 we can make a straightforward recursion:
$S(1,n) = 2S(1,n-1) + 2^{n-1} - 2^{n-2} = 2S(1,n-1) + 2^{n-2}$
since for every n string, we have two copies of every (n-1)-string as a prefix (with their associated 1-streaks), plus half the time we add a new 1-streak at the end (if the bit we add at the end is different from the old last bit). Sometimes, though, we break a 1-streak at the end; this happens if our previous last bit was a 1-stream, so it happens once for each possible n-2 string as a prefix. (Note that this means the recursion requires n > 2!).
This recursion is easy to solve; we have S(1,2) = 4 from which it follows $S(1,n) = (n+2)2^{n-2}$.
Now there's a bijection between (k-1)-streaks in (n-1)-strings and k-streaks in n-strings (just add / remove another bit to the streak), so S(k-1,n-1) = S(k,n). We conclude
$S(k,n) = (n-k+3)2^{n-k-1}$
for k < n, and S(n,n) = 2. For n=10, we have the sequence
3072, 1408, 640, 288, 128, 56, 24, 10, 4, 2
for streaks of length 1, 2, 3, ..., 10 out of the 210 = 1024 possible strings, so dividing through should give the expected number in a random string.
The sequence S(1,n) is not in Sloane, but the total number of streaks in all n-strings
$\sum\_{k=1}^n S(k,n)$
is - this sequence starts 2, 6, 16, 40, 96, 224,... and has general form
$\sum\_{k=1}^n (n-k+3)2^{n-k-1} = (n+1)2^{n-1}$.
This sequence [A057711](http://www.research.att.com/~njas/sequences/A057711) apparently has quite a few combinatorial interpretations.
| 3 | https://mathoverflow.net/users/2121 | 6995 | 4,772 |
https://mathoverflow.net/questions/6925 | 4 | I'm trying to get a better handle on characteristic subgroups, and many nice examples are given with some sort of "natural" definition. For example, it's clear that the center, torsion subgroup, and commutator subgroup of a given group are all characteristic, just because of the way they are defined. How can we formalize this "naturality"? The latter two have functors associated to them, but I'm not entirely certain if that's the reason for the subgroups being characteristic (and there isn't a similar functor for the center of a group).
EDIT: Let me explain why I'm asking this question. It's useful to know how various characteristic subgroups interact with direct products, quotients, and other group constructions. Henry's construction below is perhaps the right formalism, but it isn't clear to me what extra that buys us. On the other hand, knowing that a certain subgroup arises as the image of a functor means that we ought to be able to use categorical considerations to determine some properties of this subgroup. So here are a few related questions:
1. Is it possible to use definable subsets to see how characteristic subgroups act with respect to direct product, quotients, or other group constructions?
2. What properties does the Comm functor on Grp have? What about the Tors functor on Ab? Furthermore, which properties does a functor have to have in order to define a characteristic subgroup? (Other than the obvious property that F(G) is a subgroup of G for all G!)
| https://mathoverflow.net/users/913 | How can we formalize the naturality of certain characteristic subgroups? | The commutator subgroup of a group is given by a functor on the category whose objects are groups and whose morphisms are all homomorphisms. We can say the similar statement for the torsion subgroup of an abelian group, and this is why the subgroups are fully invariant and not just characteristic. The center is not fully invariant (see e.g., the dihedral group of order two times four), and it is only given by a functor on the [core](http://ncatlab.org/nlab/show/core) of Grp, i.e., the category whose objects are groups and whose morphisms are isomorphisms.
If you have an endofunctor on the category of (abelian) groups, whose morphisms are all homomorphisms, such that objects are taken to subobjects, then those subobjects are fully invariant (hence characteristic) subgroups. If you have the endofunctor that is only defined on the core of Grp (or Ab), such that objects are taken to subobjects in Grp (not in the core), then the subobjects are characteristic subgroups. Proof: look at what the functor does to an endomorphism of the source.
| 6 | https://mathoverflow.net/users/121 | 6996 | 4,773 |
https://mathoverflow.net/questions/6984 | 1 | Can you suggest an introduction to structural equation modeling for math majors and mathematicians?
| https://mathoverflow.net/users/812 | Introduction to Structural Equation Modeling | There seem to be a number of topics in science that use mostly standard mathematics, but bury it with a lot of new terminology and non-mathematical ideas. Maybe there is no reasonable way to avoid or even criticize this, since mathematicians also make up new terms all the time. Still, it can be weird and frustrating to have to learn more and more terminology for the same math. "Structural equation modeling" seems to be one of these cases.
My impression is that structural equation modeling is a graph-theoretic generalization of correlations between random variables. Very often in statistics you have a bunch of random variables, hypothesized correlations among some sparse sets of the random variables, and observed correlations among other sparse sets of the random variables. You'd like to fit the hypotheses to to the observations. The fact that the relationships under consideration are sparse makes the question graph-theoretic. The hypotheses chain together in paths through the graph to make predictions that can be compared to the data.
The [publications page of Judea Pearl](http://bayes.cs.ucla.edu/csl_papers.html), who is one of the founders of the field, seems to be relatively sane for a mathematical audience. He is a CS professor. In particular, the thesis of his PhD student Carlos Brito, posted on this page, could be a good introduction.
| 3 | https://mathoverflow.net/users/1450 | 6997 | 4,774 |
https://mathoverflow.net/questions/6990 | 4 | The Fourier transform of periodic function $f$ yields a $l^2$-series of the functions coefficients when represented as countable linear combination of $\sin$ and $\cos$ functions.
* In how far can this be generalized to other countable sets of functions? For example, if we keep our inner product, can we obtain another Schauder basis by an appropiate transform? What can we say about the bases in general?
* Does this generalize to other function spaces, say, periodic functions with one singularity?
* What do these thoughts lead to when considering the continouos FT?
| https://mathoverflow.net/users/2082 | Generalize Fourier transform to other basis than trigonometric function | It is not what you want, but may be worth mentioning. There is a huge branch of abstract harmonic analysis on (abelian) locally compact groups, which generalizes Fourier transformation on reals and circle. The main point about sin and cos (or rather complex exponent $e^{i n x}$) is that it is a character (continuous homomorphism from a group to a circle) and it is not hard to see that those are the only characters of the circle. That what makes Fourier transform so powerful. If you generalize it along the direction which drops characters, you'll probably get a much weaker theory.
| 3 | https://mathoverflow.net/users/896 | 6999 | 4,775 |
https://mathoverflow.net/questions/6922 | 14 | I am looking for references related to the terms "Harish-Chandra pair" and "Harish-Chandra modules", and also to the term "category O". I know what these are, or I think I do (a Harish-Chandra pair is a pair (Lie algebra; subgroup) with the subgroup acting in the Lie algebra, satisfying some natural conditions). The question is about any standard or classical sources I could refer to.
1. What are the standard or classical references for the terms "Harish-Chandra pair" and "Harish-Chandra module"?
2. The same question for algebraic Harish-Chandra pairs and modules (with an algebraic or proalgebraic subgroup).
3. One example of a category of Harish-Chandra modules is the category "O" of representations of, e.g., a simple Lie algebra, integrable to the Borel subgroup. Another version is the category of representations integrable to the maximal unipotent subgroup. What are the standard or classical references for either or both of the above definitions of the category "O"?
4. My understanding is that what was called "Harish-Chandra modules" in the classical representation theory was not the above example 3. at all, but rather the modules over a real Lie algebra integrable to the maximal compact subgroup. What are the standard or classical references for this notion of Harish-Chandra modules?
| https://mathoverflow.net/users/2106 | References for Harish-Chandra pairs and modules, category "O"? | Harish-Chandra pairs and their use in localization is discussed in Beilinson-Bernstein *A proof of Jantzen Conjectures*, available on [Joseph Bernstein's web page](http://www.math.tau.ac.il/~bernstei/Publication_list/Publication_list.html). See in particular sections 1.8 and 3.3.
Other sources:
1. Beilinson-Drinfeld, *Quantization of Hitchin's Integrable System and Hecke Eigensheaves*, sections 7.7-7.9. Available [on line](http://www.math.harvard.edu/~gaitsgde/grad_2009/).
2. Beilinson-Feigin-Mazur *Notes on Conformal Field Theory (Incomplete)*. The discussion is centered around groupoids and the Virasoro algebra. This is on [Barry Mazur's web page](http://www.math.harvard.edu/~mazur/index.html) (link on the side to "older material").
| 4 | https://mathoverflow.net/users/121 | 7001 | 4,776 |
https://mathoverflow.net/questions/6981 | 5 | I have a heavily symmetric regular graph whose automorphisms I know. I remove one subgraph and insert another one in a consistent manner; for example, this could be a Delta-Y transformation (replacing a node with a complete subgraph). I'd like to compute the automorphisms of the new graph using the automorphisms of the old graph.
I have the sense that this problem isn't too difficult, but am not sure how best to approach it and am new to algebraic graph theory.
What are some good references or suggested starting points?
| https://mathoverflow.net/users/2122 | How are graph automorphisms are affected by transformations? | I do not think there are any results that relate the automorphism group of a graph after
subgraph replacement to the group of the original graph.
If the original graph was "heavily symmetric", you would expect the group of the new graph to be smaller - most local operations would destroy vertex transitivity, for example - but the details would depend on the operation.
I have the sense that this problem is extremely difficult. There are very few results
in algebraic graph theory which determine the full automorphism group of a graph.
There are results that describe the automorphism group of a graph product in terms of the
automorphism groups of its factors. (See "Products Graphs" by Imrich and Klavsar, for example.)
| 6 | https://mathoverflow.net/users/1266 | 7005 | 4,778 |
https://mathoverflow.net/questions/6987 | 10 | What is the definition of a singular value over a finite field $\mathcal{F}$ of a matrix ${\bf A}$ in $\mathcal{F}^{m\times n}$? Is there a geometric intuition in the same manner as with the real case where the eigenvalues are the radii of the ellipse $\frac{\|{\bf A}{\bf x}\|^2}{\|{\bf x}\|^2}$?
| https://mathoverflow.net/users/2124 | Singular value decomposition over finite fields? | There is no definition of a singular value of a matrix over a finite field. You could
define it to be a non-zero eigenvalue of $A^TA$, but this does not really work as you might expect.
Over the reals, the eigenvalues of $A^TA$ are non-negative and the smallest singular value
is a measure of how close $A$ is to being a non-invertible matrix. Further, there are
stable algorithms to compute it, whereas we cannot compute the rank of $A$ in a stable fashion.
Over the complex numbers, you would use the eigenvalues of ${\bar A}^TA$ as singular values
instead the eigenvalues of $A^TA$. Over finite fields you might use $\sigma(A^T)A$,
where $\sigma$ is a field automorphism. This is a problem, we have ore choice than we do over the reals and complexes.
There is no difficulty in computing ranks over matrices over finite fields anyway.
Over finite fields, eigenvalues are of limited use. We can get the characteristic polynomial of a matrix, and factor it; the zeros of the factors are the eigenvalues. These eigenvalues
would lie in some extension $E$ of your finite field $F$; if I came along with $E$ and asked
which elements of $E$ were the eigenvalues, you would have a lot of work and your final
answer would depend on exactly how I described $E$. (Even over the reals, eigenvalues
are useful if the matrix is small, or normal, but they are much less use otherwise.
Google 'pseudospectra')
| 15 | https://mathoverflow.net/users/1266 | 7012 | 4,783 |
https://mathoverflow.net/questions/7004 | 5 | I have \$3. I flip a coin. If I get heads, I get \$1. If I get tails, I lose \$1. The game stops when I have \$0 or \$7. What is the probability I get \$7?
I solved this by creating a system of linear equations, where $P\_0 = 0$, $P\_7 = 1$, and $P\_x = 0.5 \cdot P\_{x-1} + 0.5 \cdot P\_{x+1}$. Solving them, I got $P\_3 = 3/7$. Moreover, $P\_x = x/7$. Why does it work out to such a simple fraction?
More generally, it seems that $P\_{x,y} = x/y$, which is the probability that, starting from x dollars, I end up with y dollars. I haven't proven this, but why is this the case?
Finally, what is $P\_{x,y,p}$, where I gain 1 dollar with probability $p$ instead of probability 0.5 .
| https://mathoverflow.net/users/1646 | Intuitive explanation to Probability question | I really like Vigleik's answer, but I'll throw in yet another way to look at your original problem. Px = (Px-1+Px+1)/2 is an example of a (discrete) harmonic function; i.e., a function whose value is the average of the adjacent values. In this case, Px is a harmonic function on a chain graph. For purposes of intuition, we can move from a discrete to a continuous line and think about the criterion for a function of one (real) variable to be harmonic: it is harmonic if and only if its second derivative vanishes; i.e., it's linear. This provides some intuition why your solution just linearly interpolates between 0 and 1.
Your general problem of Px,y,p is no longer harmonic, so it will not have as easy a solution, as you may be discovering. For notational simplicity, I'll write Pn for Pn,y,p (preferring n as the index of a sequence to x). If you write down your new recurrence, you will get equations
Pn = (1-p)Pn-1+pPn+1
subject to P0 = 0, Py = 1. We can work with this, or we can use a trick. Let k = (1-p)/p (so p = 1/(1+k)). Then you can verify that
Pn = kPn-1 + 1
satisfies the original equation (with the additional freedom to scale all Pn by a constant factor - we've broken the homogeneity of our original recurrence). [It actually takes some doing to verify this: consider using this new recurrence to write down Pn-Pn+1. When you solve that out for Pn, you retrieve the original recurrence.]
This is much easier to handle, with solution
$P\_n = \frac{k^n-1}{k-1}$.
This gives P0 = 0 as desired, but you'll need to scale down all solutions so that Py = 1.
| 7 | https://mathoverflow.net/users/2121 | 7015 | 4,786 |
https://mathoverflow.net/questions/6979 | 16 | What is etale descent? I have a vague notion that, for example, given a variety $V$ over a number field $K$, etale descent will produce (sometimes) a variety $V'$ over $\mathbb{Q}$ of the same complex dimension which is isomorphic to $V$ over $K$ and such that $V(K)=V'(\mathbb{Q})$. Is this at all right? How does one do such a thing?
| https://mathoverflow.net/users/1464 | What is etale descent? | Let $L/K$ be a Galois field extension and consider a variety $Y$ over $L$. The theory of (Galois) descent addresses the question whether $Y$ can be defined over $K$.
More precisely, the question is: "does there exist a variety $X$ over $K$ such that $Y = X \times\_{Spec(K)} Spec(L)$".
Now assume such $X$ does exist. In this case $Y$ is endowed with $Gal(L/K)$ action coming from an action on the second factor.
Conversely, if $Y$ has a Galois action compatible with the action on $Spec(L)$, then $Y$ descends to some $X$ defined over $K$. $X$ is actually a quotient of $Y$ by $Gal(L/K)$ (so that the conjugate points glue together to form one point on $X$).
Note that the set of $K$-points of $X$ is the set of Galois fixed points.
Example. $K = \mathbf R$ and $L = \mathbf C$.
For any real variety, the set of complex points admits the action of $
\mathbf Z/2$ by complex conjugation.
Conversely, if a complex variety is endowed with conjugation, it descends to a real variety. This is in fact an exercise in Hartshorne.
Remarks. Theory of descent also classifies all possible $X$'s arising from $Y$. Such $X$'s are called forms of $Y$. They are in 1-1 correspondence with a certain Galois cohomology group.
| 16 | https://mathoverflow.net/users/2132 | 7024 | 4,791 |
https://mathoverflow.net/questions/7037 | 0 | Theorem: (SRL)
For every $\epsilon>0$ and integer $m\geq 1$ there is an $M$ such that every graph $G$, with $|G|\geq m$ has an $\epsilon$-regular partition $V(G)=V\_0\cup\ldots\cup V\_k$ for some $m\leq k\leq M$.
Can someone explain to me why this statement is not trivial? For instance, what stops me choosing $M$ larger than $|G|$ and picking $k=|G|$, so I can split $G$ up into singletons, which is trivally $\epsilon$-regular for any $\epsilon>0$.
| https://mathoverflow.net/users/2011 | Szemeredi's Regularity Lemma | Quantifier error. You have to fix your M before you are given a graph G; whereas your approach would require one to have the graph G at hand, before choosing M.
| 5 | https://mathoverflow.net/users/763 | 7038 | 4,799 |
https://mathoverflow.net/questions/6982 | 5 | Can you suggest a book that has a thorough introduction to Singular Value Decomposition?
| https://mathoverflow.net/users/812 | Thorough Introduction to Singular Value Decomposition | I find Numerical Linear Algebra by N. Trefreten and D.Bau an extremely well-written book. It not only introduces the Singluar value decomposition but explains applications and history.
| 4 | https://mathoverflow.net/users/2011 | 7047 | 4,805 |
https://mathoverflow.net/questions/7046 | 1 | What i understand about strata for the nullcone is this: (from Mumford's "Geometric Invariant Theory" and Hesselink's paper "Desingularizations of Varieties of Nullforms")
**ADDED BY DAVID SPEYER** In this setting, we are studying a reductive group $G$ acting on a vector space $V$. We write $T$ for a maximal torus of $G$ and $v$ for a nonzero vector in $V$.
There is a bilinear form on the set $Hom(G\_m, T) \otimes \mathbb{R}$ that is invariant under the Weyl group, this bilinear form defines a norm on the set of one-parameter subgroups. Consider the set of one-parameter subgroups $\lambda : \mathbb{R} \rightarrow G$ satisfying the condition $\lambda(t).v = 0$. Also this may mean that there is a morphism $f : A^{1} \rightarrow V$ with $f(0)=0, f(t) = \lambda(t)(v)$ for $t \neq 0$; define $m(v, \lambda)$ to be the multiplicity of the fibre $f^{-1} (0)$. Then the strata are defined via the $\Gamma$ which associates to each $v$ a set of one-parameter subgroups of "shortest length", and also satisfying that $m(v, \lambda) \geq 1$; and two vectors are int the same strata if they have the same set of one-parameter subgroups linked with them.
**Questions:**
When $G = GL\_{k}(\mathbb{C})$, how would you define the norm on the space $Hom(G\_m, T) \otimes \mathbb{R}$ standard torus $T$ of diagonal matrices? Is there some classification of all one-parameter multiplicative subgroups of $GL\_{k}(\mathbb{C})$? And I don't understand how $m(v, \lambda)$ is defined - how can the multiplicity of the fibre $f^{-1}(0)$ be anything other than $1$? Is there some exception in the case where $ \lambda(t).v=0$ for all $t$, then what is the multiplicity (is it $0$)? How can the multiplicity be, for instance, $2$?
| https://mathoverflow.net/users/2623 | Strata for the nullcone (from Hesselink's paper) | Example of how the multiplicity can be greater than $1$: Let $\mathbb{G}\_m$ act on $\mathbb{A}^2$ by $t: (x,y) \mapsto (t^2x, t^3 y)$. Let $v$ be any element of $\mathbb{A}^2$ not on the coordinate axes, for example, $(1,1)$. So $\mathbb{G}\_m$ maps to $\mathbb{A}^2$ by $t \mapsto (t^2, t^3)$. This extends to a map $\mathbb{A}^1 \to \mathbb{A}^2$, given by the same formula.
I'm not sure what your favorite definition of multiplicity is. Mine is to consider the map of rings in the opposite directions: $k[x,y] \to k[t]$ by $x \to t^2$, $y \to t^3$. The point $(0,0)$ corresponds to the ideal $\langle x,y \rangle$; this maps to the ideal $\langle t^2, t^3 \rangle = \langle t^2 \rangle$ in $k[t]$. Then, by definition, the multiplicity of $f^{-1}(0)$ is the dimension of $k[t]/\langle t^2 \rangle$, which is $2$.
The geometric point here is that the map $\mathbb{A}^1 \to \mathbb{A}^2$ has vanishing derivative at $0$, so the preimage of zero has multiplicity greater than $1$.
---
You also asked about a Weyl invariant form on $\mathrm{Hom}(\mathbb{G}\\_m, T)$, where $T$ is a maximal torus of $GL\_n$. Every hom from $\mathbb{G}\\_m$ to $T$ is of the form $t \mapsto (t^{w\_1}, t^{w\_2}, \ldots, t^{w\_n})$, where the coordinates on the right hand side are the entries in your diagonal matrix. Let's call this $s(w)$.
The obvious choice of Weyl invariant form is
$$\langle s(v), s(w) \rangle = \sum v\_i w\_i.$$
Technically, I note that $\sum v\_i w\_i + c (\sum v\_i) (\sum w\_i)$ would also be Weyl invariant, for any integer $c$. No one ever makes the choice of using nonzero $c$, but I don't know a general principle which would exclude it.
| 5 | https://mathoverflow.net/users/297 | 7049 | 4,807 |
https://mathoverflow.net/questions/7025 | 22 | Many commutative algebra textbooks establish that every ideal of a ring is contained in a maximal ideal by appealing to Zorn's lemma, which I dislike on grounds of non-constructivity. For Noetherian rings I'm told one can replace Zorn's lemma with countable choice, which is nice, but still not nice enough - I'd like to do without choice entirely.
So under what additional hypotheses on a ring $R$ can we exhibit one of its maximal ideals while staying in ZF? (I'd appreciate both hypotheses on the structure of $R$ and hypotheses on what we're given in addition to $R$ itself, e.g. if $R$ is a finitely generated algebra over a field, an explicit choice of generators.)
Edit: I guess it's also relevant to ask whether there are decidability issues here.
| https://mathoverflow.net/users/290 | When can we prove constructively that a ring with unity has a maximal ideal? | I suspect that the most general reasonable answer is a ring endowed with a constructive replacement for what the axiom of choice would have given you.
How do you show in practice that a ring is Noetherian? Either explicitly or implicitly, you find an ordinal height for its ideals. Once you do that, an ideal of least height is a maximal ideal. This suffices to show fairly directly that any number field ring has a maximal ideal: The norms of elements serve as a Noetherian height.
The Nullstellensatz implies that any finitely generated ring over a field is constructively Noetherian in this sense.
Any Euclidean domain is also constructively Noetherian, I think. A Euclidean norm is an ordinal height, but not at first glance one with the property that $a|b$ implies that $h(a) \le h(b)$ (with equality only when $a$ and $b$ are associates). However, you can make a new Euclidean height $h'(a)$ of $a$, defined as the minimum of $h(b)$ for all non-zero multiples $b$ of $a$. I think that this gives you a Noetherian height.
I'm not sure that a principal ideal domain is by itself a constructive structure, but again, usually there is an argument based on ordinals that it is a PID.
| 7 | https://mathoverflow.net/users/1450 | 7062 | 4,816 |
https://mathoverflow.net/questions/6589 | 5 | There is a well known theorem that says that the functor associating to a perverse sheaf $F$ on $X$ the data $(F|\_U,\phi\_f(F),can:\psi\_f(F) \to \phi\_f(F),var:\phi\_f(F)\to \psi\_f(F)(-1))$ where $U = X \setminus (f=0)$ is an equivalence of categories.
In dimension 1, this gives that a perverse sheaf on $(\mathbb C,0)$ is described by a quiver $\psi\_z(F) = V\_0 \leftrightarrows V\_1 = \phi\_z(F)$ .
There is an analog statement for perverse sheaves $(\mathbb C,0)^n$. My question is how do you define directly the vector spaces in the quiver representation. For example, in dimension 2, I think we should have $V\_{00} = \psi\_x\psi\_y(F) = \psi\_x\psi\_y(F)$, $V\_{10} = \phi\_x\psi\_y(F) = \psi\_y\phi\_x(F)$, $V\_{01} = \psi\_x\phi\_y(F) = \phi\_y\psi\_x(F)$, $V\_{11} = \phi\_x\phi\_y(F) = \phi\_y\phi\_x(F)$ but I'm not sure why these functors should commute. Is the key point here that we have a normal crossing divisor?
Also, what if we want to make things more canonical replacing nearby cycles $\psi\_f(F)$ by the restriction of the Verdier specialisation $\nu\_Z(F)|\_{T\_Z^0X}$ and $\phi\_f(F)$ by the evanescent part of $\nu\_Z(F)$? What would be the precise statement in this case?
| https://mathoverflow.net/users/1985 | Hypercube decomposition of perverse sheaves | In some sense there is no canonical way to extract these vector spaces. Algebraically this is because projective objects usually have automorphisms. Geometrically because these vector spaces are the stalks of local systems on a manifold with no natural base point. But the manifold and the local system are canonical.
Each of the 2^n coordinate subspaces of C^n has a conormal bundle in C^{2n}. This gives an arrangement of 2^n Lagrangian subspaces of the symplectic vector space C^2n, which are in general position (as general as possible for Lagrangian subspaces). The smooth part of this arrangement, the points that lie on exactly one of these Lagrangians, is a disjoint union of 2^n copies of (C-0)^n. The 2^n vector spaces in your hypercube are the stalks of local systems on these.
I am not sure what Verdier specialization is, but I bet it's exactly the construction of these local systems. In general, if Y is a smooth subvariety of X, then it is possible to extract from a perverse sheaf a local system on an open subset of the conormal variety to Y in X, by the following recipe:
1. Take nearby cycles for the deformation-to-the-normal-bundle family. This family has a C^\*-action so taking nearby cycles is more canonical than usual--the monodromy action (var compose can) is trivial.
2. Take the Fourier transform. (As you know, I'm still pretty confused about this, but I think this has to be the topological version, or Fourier-Sato transform. In particular even with the C^\*-action on the family, I don't think the sheaf you get at the end of 1. is C^\*-equivariant.)
3. The result is a perverse sheaf on the conormal bundle to Y. This sheaf is locally constant on an open subset.
At one time S. Gelfand, MacPherson, and Vilonen had a project to understand the natural maps between the fibers of these local systems on different strata, i.e. the analogs of can and var. From a certain perspective the work of Nadler and Zaslow is a (not very concrete) solution to this problem.
In general, whether \psi\_f \psi\_g F = \psi\_g \psi\_f F depends on the blowup behavior of F with respect to the map (f,g) to C^2. For a counterexample, consider F = constant sheaf on C^2, f = x, g = xy. But if F is constructible with respect to a stratification that satisfies Thom's condition a\_{f,g}--that is, the stratification is without blowups/sans eclatement with respect to (f,g)--then you are in good shape. (Although even here there is no canonical isomorphism between the two. They behave like stalks of a local system on an open subset of the base C^2.)
For sheaves without blowups I am not sure about references, but try Sabbah's "Morphismes analytiques stratifies sans eclatement et cycles evanescents," or from the etale point of view Illusie's recent note: <http://www.math.u-psud.fr/~illusie/vanishing1b.pdf>
| 3 | https://mathoverflow.net/users/1048 | 7065 | 4,818 |
https://mathoverflow.net/questions/7071 | 5 | In reading the paper of Green and Tao on arithmetic progressions within the primes, I became very interested in the notion of a k-pseudorandom measure discussed in that paper.
A measure here is a function $\nu:\mathbf{Z}\_N\to\mathbf{R}$ such that $\mathbf{E}\nu=1+o(1)$, and it is k-pseudorandom if it obeys the ($k2^{k-1}$,$3k-1$,$k$) (I think) linear forms condition, which basically asserts that it behaves independently with respect to at most $k2^{k-1}$ independent linear forms in $3k-1$ variables, and if it also obeys the correlation condition, which is a weaker form controlling the linear forms $x+h\_i$.
They show that a relative Szemeredi's theorem applies to functions bounded by a k-pseudorandom measure, and then construct one that (effectively) bounds the primes.
My question is where else these type of functions have been studied, whether their theory has been expanded, and whether other explicit examples have been found and applied in other situations.
| https://mathoverflow.net/users/385 | k-pseudorandom measures | Linear forms condition says that these functions are morally the functions that are close to $1$ in appropriate $U^k$ norm. What I mean is that $U^k$ norms are a special kind of linear forms, and so linear forms condition implies proximity to $1$ in $U^k$, on one hand. On the other hand,if one controls $\nu-1$ in $U^t$ norm for sufficiently large $t=t(k)$, then by Cauchy-Schwarz argument one can control arbitrary linear forms.
[EDIT: The rest of the answer is result of my misunderstanding. See the comments.]
There is an unpublished work of David Conlon and Timothy Gowers on Szemerédi-type results in random sets, in which, if I understood correctly what David explained to me, they show as a special case that the control in an appropriate $U^t$ norm is enough. (In particular the correlation condition is no longer necessary, and was an artifact of the original proof.)
So, the answer to your question is that the theory of these functions is essentially the theory of functions with small $U^k$ norm.
| 5 | https://mathoverflow.net/users/806 | 7077 | 4,827 |
https://mathoverflow.net/questions/7018 | 35 | One of my favorite results in algebraic geometry is a classical result of AX (see <http://terrytao.wordpress.com/2009/03/07/infinite-fields-finite-fields-and-the-ax-grothendieck-theorem/>) I'll recall the version of the theorem that I learned in an undergraduate class in model theory.
An algebraic map $F: \mathbb{C}^{n} \to \mathbb{C}^{n}$ is injective iff it is bijective.
The model theoretic proof of this result is very simple. Without discussing details ( truth in TACF\_0 is the same as truth in TACF\_p for p big enough )one can replace $\mathbb{C}$ by the algebraic closure of $F\_p$ and prove the result there. Although the algebraic proof is also "simple" (Hilbert's Nullstellensatz is what one needs ) I personally think that the model theoretic proof is great by its simplicity and elegance.
So my first question is: Are there more examples like AX's theorem in which the model theoretic proof is much more simple that the ones given by other areas. (I should mention that some algebraist I know consider quantifier elimination for TACF non a model theoretic fact, so for them the proof that I refer above is an algebraic proof)
My second question: One of the most famous model theoretic applications to algebra and number theory is Hrushovski's proof on Mordell-Lang of the function field Mordell-Lang conjecture. I'd like to know what are the research questions that applied model theorists are currently working on, besides continuation to Hrushovski's work . In particular, I'd like to know if there is any model theorist that work in applications to Iwasawa theory.
| https://mathoverflow.net/users/2089 | Model theoretic applications to algebra and number theory(Iwasawa Theory) | It's hard for me to think of an area of algebra that applied model theorists **haven't** touched recently. I have not heard of any logicians working on Iwasawa theory, but it wouldn't surprise me if there are some.
Diophantine geometry: [here is a survey article by Thomas Scanlon](http://math.berkeley.edu/~scanlon/papers/bsl4ap00.pdf) on applications of model theory to geometry, including discussions of Mordell-Lang and the postivie-characteristic Manin-Mumford conjecture.
Number fields: [Bjorn Poonen has shown](http://www-math.mit.edu/~poonen/papers/uniform.pdf) that there is a first-order sentence in the language of rings which is true in all finitely-generated fields of characteristic 0 but false in all fields of positive characteristic. It was conjectured by Pop that any two nonisomorphic finitely-generated fields have different first-order theories.
Polynomial dynamics: [see here](http://math.berkeley.edu/~scanlon/papers/pd.pdf) for a recent preprint by Scanlon and Alice Medvedev. It turns out that first-order theories of algebraically closed difference fields where the automorphism is "generic" are quite nice.
Differential algebra: By some abstract model-theoretic nonsense ("uniqueness of prime models in omega-stable theories"), it follows that any differential field has a "differential closure" (in analogy to algebraic closure) which is unique up to isomorphism over the base field. There are much more advanced applications, e.g. [here](http://math.berkeley.edu/~scanlon/papers/jetB.pdf).
Geometric group theory: Zlil Sela has recently shown that any two finitely-generated nonabelian free groups are elementarily equivalent (i.e. they have the same first-order theory). According to [the wikipedia article,](http://en.wikipedia.org/wiki/Zlil_Sela) this work is related to his solution of the isomorphism problem for torsion-free hyperbolic groups, but I don't understand this enough to say whether this counts as an "application" of model theory.
Exponential fields: Boris Zilber has suggested a model-theoretic approach to attacking Schanuel's Conjecture. His conjecture that the complex numbers form a "pseudo-exponential field" is actually a *strengthening* of Schanuel's Conjecture, but the picture that it suggests is appealing. See [here](http://en.wikipedia.org/wiki/Schanuel%27s_conjecture) for more.
This is in addition to the work on Tannakian formalism, valued fields, and motivic integration that have already been mentioned in other answers, and I haven't even gotten to all the work by the model theorists studying o-minimality. This was just a pseudo-random list I've come up with spontaneously, and no offense is meant to the areas of applied model theory that I've left off of here!
| 34 | https://mathoverflow.net/users/93 | 7092 | 4,836 |
https://mathoverflow.net/questions/7059 | 30 | In short: what does Labesse-Langlands say?
Slightly more precise: what are the cuspidal automorphic representations of $SL\_2(\mathbf{A}\_{\mathbf{Q}})$, together with multiplicities? Let's say that I have a complete list of the cuspidal automorphic representations of $GL\_2/\mathbf{Q}$ and I want to try and deduce what is happening for $SL\_2$. I am looking for "concrete examples of the phenomena that occur".
Now let me show my ignorance more fully. My understanding from trying to read Labesse-Langlands is the following. The local story looks something like this: if $\pi$ is a smooth irreducible admissible representation of $GL(2,\mathbf{Q}{}\_p)$ then its restriction to $SL(2,\mathbf{Q}{}\_p)$ is either irreducible, or splits as a direct sum of 2 non-isomorphic representations, or, occasionally, as a direct sum of 4. One interesting case is the unramified principal series with Satake parameter $X^2+c$ for any $c$; this splits into two pieces (if I've understood correctly) (and furthermore these are the only unramified principal series which do not remain irreducible under restriction). Does precisely one of these pieces have an $SL(2,\mathbf{Z}{}\_p)$-fixed vector? And does the other one have a fixed vector for the other hyperspecial max compact (more precisely, for a hyperspecial in the other conj class)? Have I got this right?
Packets: Local $L$-packets are precisely the J-H factors for $SL(2,\mathbf{Q}{}\_p)$ showing up in an irreducible $GL(2,\mathbf{Q}{}\_p)$-representation. So they have size 1, 2, or 4.
Now globally. a global $L$-packet is a restricted product of local $L$-packets (all but finitely many of the components had better have an invariant vector under our fixed hyperspecial max compact coming from a global integral model). Note that global automorphic $L$-packets might be infinite (because $a\_p$ can be 0 for infinitely many $p$ in the modular form case). My understanding is that it is generally the case that one element of a global $L$-packet is automorphic if and only if all of them are, and in this case, again, generally, each one shows up in the automorphic forms with the same multiplicity. What is this multiplicity? Does it depend?
Finally, my understanding is that the above principle (multiplicities all being equal) fails precisely when $\pi$ is induced from a grossencharacter on a quadratic extension of $\mathbf{Q}$. In this case there seems to be error terms in [LL]. Can someone explain an explicit example where they can say precisely which elements of the packet are automorphic, and what the multiplicities which which the automorphic representations occur in the space of cusp forms?
I find it very tough reading papers of Langlands. My instinct usually would be to press on and try and work out some examples myself (which is no doubt what I'll do anyway), but I thought I'd ask here first to see what happens (I know from experience that there's a non-zero chance that someone will point me to a website containing 10 lectures on Labesse-Langlands...)
Edit: I guess that there's no reason why I shouldn't replace "cuspidal" by "lies in the discrete series" with the above (in the sense that the questions then still seem to make sense, I still understand everything (in some sense) for $GL\_2$ and I still don't know the answers for $SL\_2$)
| https://mathoverflow.net/users/1384 | Overview of automorphic representations for $SL(2)/{\mathbf{Q}}$? | Warning: not an expert, so could be major mistakes in this.
The multiplicity is one for every element in the global packet, in the non-CM case. In the CM case,
half of the packet has multiplicity one and the other half has multiplicity zero.
For the general story, I'd suggest looking at Arthur's conjectures, which at least conjecturally give a very nice picture. In general one should talk of Arthur packets, not Langlands packets. They coincide in your case.
For CM representations on $SL\_2$, the associated global parameter has dihedral image inside $PGL\_2(\mathbb{C})$. Its centralizer $S$ has size $2$. According to Arthur, the obstruction to an element of the global packet being automorphic is valued in the dual of $S$; thus, "one-half" is automorphic. More precisely, the set of irreducible representations in a global $L$-packet is a principal homogeneous space for a certain product of $Z/2Z$s (in the obvious way: one $Z/2Z$ for each $p$ where $a\_p = 0$; for simplicity, suppose that local multiplicity $4$ doesn't occur). The automorphic ones correspond exactly to one of the fibers of the summation map to Z/2Z. Thus, if you take an automorphic representation in this packet, and switch it at one place to the other element of the local packet, it won't be automorphic any more.
Concretely: Start with $\Pi$ an automorphic representation for $GL\_2$, it maps
by restriction to the space of automorphic forms for $SL\_2$. The above remarks
suggest that this restriction map must have a huge kernel when $\Pi$ is CM. But one can almost see this by hand: $\Pi$ is isomorphic
to its twist by a certain quadratic character $\omega$. This, and multiplicity one for $GL\_2$,
means that, for $f \in \Pi$, the function $g \mapsto f(g) \omega(\det g)$ also belongs to $\Pi$. This forces extra vanishing, although I didn't work out the details: For instance, if $\omega$ is everywhere unramified and $f$ the spherical vector, then $f(g) \omega (\det g)$ must be proportional to $f$, so $f$ vanishes whenever $\omega(\det g)$ is not $1$, i.e. various translates of $f$ do vanish when restricted to $SL\_2$.
Remark/warning: If you are concerned with multiplicity one, there is a further totally distinct phenomenon that causes it to fail for $\mathrm{SL}\_n, n \geq 3$: Two non-conjugate homomorphisms of a finite group into $\mathrm{PGL}\_n$ can be conjugate element by element.
There's a paper of Blasius about this. It has nothing to do with packets.
| 11 | https://mathoverflow.net/users/1253 | 7096 | 4,839 |
https://mathoverflow.net/questions/7091 | 13 | Let K/Q be a field, probably not a finite extension. Is it possible for a polynomial to be irreducible over K but have a root in every completion of K? What about all but finitely many completions?
This question is related to the question "Can a non-surjective polynomial map from an infinite field to itself miss only finitely many points?", and should help prove that such a polynomial can not exist for any subfield of the algebraic closure of the rationals.
The idea is that we make the candidate polynomial monic and have algebraic integers for coefficients, then take any maximal ideal in the ring of integers of the candidate field and complete it using the ideal - since the polynomial must have a root in the residue field, it will have a root in the completion. I'm wondering if this forces the polynomial to have a root in the original field - hence the question.
The same question only for function fields is also interesting, in order to prove the above for subfields of the algebraic closure of Fp(t)
| https://mathoverflow.net/users/2024 | Irreducible polynomial over number field with roots in every completion? | For a finite extension $K/\mathbf{Q}$, the answer is no. Suppose that $f$ is an irreducible polynomial with coefficients in $K$ and splitting field $L$. If $G$ is the Galois group of $L/K$, then the polynomial $f$ gives rise to a faithful transitive permutation representation $G \rightarrow S\_d$, where $d$ is the degree of $f$. If $P$ is a prime in $O\_K$ that is unramified in $L$, then $f$ has a root over $O\_{K,P}$ if and only if the corresponding Frobenius element $\sigma\_P \in S\_d$ has a fixed point. On the other hand, a theorem of Jordan says that every transitive subgroup of $S\_d$ contains an element with no fixed points. By the Cebotarev density theorem, it follows that $f$ fails to have a point modulo $P$ for a positive density of primes $P$.
For an infinite extension $K/\mathbf{Q}$, the answer is (often) yes. Let $K$ be the compositum of all cyclotomic extensions. Then $x^5 - x - 1$ is irreducible over $K$, because its splitting field over $\mathbf{Q}$ is $S\_5$. On the other hand, it has a root in every completion (easy exercise).
Finally, your claim that "since a the polynomial must have a root in the residue field, it will have a root in the completion" is false. Hensel's lemma comes with hypotheses.
| 24 | https://mathoverflow.net/users/nan | 7100 | 4,841 |
https://mathoverflow.net/questions/5994 | 16 | If G is a graph then its adjacency matrix has a distinguished Peron-Frobenius eigenvalue x. Consider the field Q(x). I'd like a result that says that if G is a "random graph" then the Galois group of Q(x) is "large" with high probability. (Well, what I really want is just that the Galois group is non-abelian, but I'm guessing that for a typical graph it will be the full symmetric group.)
Here's another version of this question. Take a graph with a marked point and start adding a really long tail coming into that point. This gives a sequence of graphs G\_n. If G is sufficiently complicated (i.e. you're not building the A or D type Dynkin diagrams as the G\_n) is Gal(Q(x\_n)/Q) symmetric for large enough n?
The reason behind these questions is that fusion graphs of fusion categories always have cyclotomic Peron-Frobenius eigenvalue. So results along these lines would say things like "random graphs aren't fusion graphs" or "fusion graphs don't come in infinite families." So the particular details of these questions aren't what's important, really any results on the number theory of the Peron-Frobenius eigenvalue of graphs would be of interest.
| https://mathoverflow.net/users/22 | Number theoretic spectral properties of random graphs | Just in case anyone else is still thinking about this question...
The answer is the following. Either:
1. The eigenvalues of $G\_n$ are all of the form $\zeta + \zeta^{-1}$ for roots of unity $\zeta$, and the graphs $G\_n$ are subgraphs of the Dynkin diagrams
$A\_n$ or $D\_n$.
2. For sufficiently large $n$, the largest eigenvalue $\lambda$ is greater than two, and
$\mathbf{Q}(\lambda^2)$ is not abelian.
The proof is effective, but a little long to post here. The main ingredients are some basic facts about Weil height, some ideas due to Cassels, and an amplification step using Chebyshev polynomials.
---
The [paper](http://arxiv.org/abs/1004.0665) is now on the ArXiv. See the last section for the proof of this result, and the second to last section for a more effective but logically weaker result.
| 19 | https://mathoverflow.net/users/nan | 7102 | 4,843 |
https://mathoverflow.net/questions/7099 | 8 | I know at least two definitions of *correspondence*, and my question might as well be about both of them.
1. Let $X,Y$ be objects in your favorite category. A *correspondence* is a *span*, namely a diagram $X \leftarrow Z \rightarrow Y$.
2. Let $X,Y$ be objects in your favorite category with products. A *correspondence* is a [subobject](http://en.wikipedia.org/wiki/Subobject) of $X\times Y$.
If your category has pull-backs, then there is a good notion of composition of spans. My understanding is that definition 2. does not usually give a good notion of composition. Also, sometimes these definitions should be modified. For example, when the category consists of spaces with structure (symplectic manifolds, for example), often I should replace $Y$ above with its opposite. In particular, the correct notion of any of these should be such that the graph of a function is an example of a correspondence.
I'm generally interested in very concrete categories — the category of smooth manifolds, for example — and I'm hoping for open-ended answers to my open-ended question. Which is: to what extent should I treat correspondences like functions?
For example, if $K$ is a ring object in my category, what conditions make the set of correspondences between $X$ and $K$ into a ring? (Or higher-categorical analogue, since really the span-category is a two-category, etc.)
| https://mathoverflow.net/users/78 | How should I think about correspondences? | You can think of replacing your category by its category of spans as a kind of "linearization" of the category. For instance if you start with the category of finite sets, a correspondence between X and Y is the same thing as a map from the free commutative monoid on X to the free commutative monoid on Y. (Here we have to ignore the automorphisms of correspondences.) More generally, the category of spans is enriched in commutative monoids (assuming your original category has finite coproducts).
If we want to work with the category of all sets, we can use, instead of the free commutative monoid on X, the category SetX of X-tuples of sets. Associated to a function f : X → Y we have the restriction or pullback functor f\* : SetY → SetX and its left adjoint the pushforward functor f! : SetX → SetY which takes the union of all the sets corresponding to elements of X sent by f to a given element of Y. Both these functors preserve colimits, because f\* also has a right adjoint. Given a correspondence f : X ← Z → Y : g we can form the composite g!f\* : SetX → SetY which is also a colimit-preserving functor. In fact, this construction identifies colimit-preserving functors from SetX to SetY with correspondences, as 2-categories. When we restrict to the elements of SetX which are finite at each element of X and decategorify, we recover the homomorphism between the free commutative monoids on X and Y (under a suitable finiteness condition on Z).
When we start with the category of schemes, rather than SetX we should consider the derived category of quasicoherent sheaves on X. This leads to the subject of [geometric function theory](http://ncatlab.org/nlab/show/geometric+function+theory).
| 9 | https://mathoverflow.net/users/126667 | 7105 | 4,845 |
https://mathoverflow.net/questions/7114 | 47 | As compared to classes of graphs embeddable in other surfaces.
Some ways in which they're exceptional:
1. Mac Lane's and Whitney's criteria are algebraic characterizations of planar graphs. (Well, mostly algebraic in the former case.) Before writing this question, I didn't know whether generalizations to graphs embedded in other surfaces existed, but some lucky Google-fu turned up some references -- in particular there seems to be a generalization of Whitney's criterion due to Jack Edmonds for general surfaces, although frustratingly I can't find the paper, and the main reference I found implies that there might be a small problem on the Klein bottle. Anyone know if Edmonds' result is as easy to prove as Whitney's?
2. Kuratowski's classic characterization of planar graphs by forbidden minors. Of course this does generalize to other surfaces, but this [result](http://en.wikipedia.org/wiki/Forbidden_minor) is both incredibly deep and difficult (as opposed to the proof of Kuratowski, which is by no means trivial but is obtainable by a sufficiently dedicated undergraduate -- actually my working it as an exercise is largely what motivated the question) and is in some sense "essentially combinatorial" in that it applies to a wider class of families that aren't inherently topologically defined.
3. In the other direction of difficulty, the four-color theorem. It's apparently not difficult to show (except for the plane) that what turns out to be [the tight upper bound](http://en.wikipedia.org/wiki/Heawood_conjecture) on the chromatic number of a graph embeddable on a surface (other than, for whatever reason, the Klein bottle) is, in fact, an upper bound -- the problem is showing tightness! Whereas it's pretty much trivial to show that $K\_4$ is planar (to be fair, though, tightness is easy to check for surfaces of small genus -- the problem's in the general case), but the four-color theorem requires inhuman amounts of calculation and very different, essentially ad-hoc methods.
I realize that the sphere has genus 0, which makes it unique, and has trivial fundamental group, which ditto, but while I imagine this information is *related* to the exceptionalness of the plane/sphere, it's not really all that satisfying as an answer. So, why is it that methods that work everywhere else tend to fail on the sphere?
Related questions: [reasons-for-the-importance-of-planarity-and-colorability](https://mathoverflow.net/questions/51682/reasons-for-the-importance-of-planarity-and-colorability)
| https://mathoverflow.net/users/382 | Why are planar graphs so exceptional? | (I think that the question of why planar graphs are exceptional is important. It can be asked not only in the context of graphs embeddable on other surfaces. Let me edit and elaborate, also borrowing from the remarks.)
**Duality:** Perhaps duality is the crucial property of planar graphs. There is a theorem asserting that the dual of a graphic matroid M is a graphic matroid if and only if M is the matroid of a planar graph. In this case, the dual of M is the matroid of the dual graph of G. (See [this wikipedia article](http://en.wikipedia.org/wiki/Matroid)).
This means that the circuits of a planar graph are in one to one correspondence with cuts of the dual graph.
One important manifestation of the uniqueness of planar graphs (which I believe is related to duality) is Kasteleyn's formula for the number of perfect matchings and the connection with counting trees.
**Robust geometric descriptions:** Another conceptual difference is that (3-connected or maximal) planar graphs are graphs of convex 3-dimensional polytopes and thus have extra geometric properties that graphs on surfaces do not share.
The geometric definition of planar graphs (unlike various generalizations) is very robust. A graph is planar if it can be drawn in the plane such that the edges do not intersect in their interiors and are represented by Jordan curves; The class of planar graphs is also what we get if we replace "Jordan curves" by "line intervals," or if we replace "no intersection" by "even number of crossings". The Koebe-Andreev-Thurston theorem allows to represent every planar graph by the "touching graph" of nonoverlapping circles. Both (related) representations via convex polytopes and by circle packings, can respect the group of automorphisms of the graph and its dual.
**Simple inductive constructions.** Another exceptional property of the class of planar graphs is that planar graphs can be constructed by simple inductive constructions. (In this respect they are similar to the class of trees, although the inductive constructions are not so simple as for trees.) This fails for most generalizations of planar graphs.
A related important property of planar graphs, maps, and triangulations (with labeled vertices) is that they can be enumerated very nicely. This is Tutte theory. (It has deep extensions to surfaces.)
It is often the case that results about planar graphs extend to other classes. As I mentioned, Tutte theory extends to triangulations of other surfaces. Another example is the fundamental Lipton-Tarjan separator theorem, which extends to all graphs with a forbidden minor.
**The study of planar graphs have led to important graph theoretic concepts** Another reason (of a different nature) why planar graphs are exceptional is that several important graph-theoretic concepts were disvovered by looking at planar graphs (or special planar graphs.) The notion of vertex coloring of graphs came (to the best of my knowledge) from the four color conjecture about planar graphs. Similarly, Hamiltonian paths and cycles were first studied for planar graphs.
---
**Graphs on surfaces and other notions generalizing planarity.** Considering the class of all graphs that can be embedded in a given surface is a natural and important extension of planarity. But, in fact, for various questions, graphs embeddable on surfaces may not be the right generalization of planar graphs.
David Eppstein mentioned another generalization via the colin de Verdier invariant. This describes a hiearachy of graphs where the next class after planar graphs are "linklessly embeddable graphs". Those are graphs that can be embedded in space without having two disjoint cyles geometrically link. As it turned out this is also a very robust notion and it leads to a beautiful class of graphs. (They all have at most 4v-10 edges where v is the number of vertices; The known case of Hadwiger's conjecture for graphs not having K\_6 minor implies that they are all 5 colorable.) Further classes in this hierarchy are still very mysterious. Other extensions of planarity are: 3) (not literally) Graphs not having K\_r as a minor; 4;5) (Both very problematic) As Joe mentioned, graphs of d-polytopes, and also graphs obtained from sphere packings in higher dimensions; 6) (not graphs) r-dimensional simplicial complexes that cannot be embedded in twice the dimension, 7) [A notion that I promoted over the years:] graphs (and skeleta) of d-polytope with vanishing second (toric) g-number, and many more.
**Forbidden minors and coloring.** As for the second and third items in the question. About coloring I am not sure if we should consider 4-coloring planar graphs and coloring graphs on other surfaces as very related phenomena. Regarding forbidden minors. Kuratowski's theorem on surfaces is a special case (and also an important step of the proof) of a much more general result (Wagner's conjecture proved by Robertson and Seymour) about any minor-closed class of graphs. This result can be regarded as extending Kuratowski theorem and also (and perhaps more importantly) extending Kruskal and Nash-Williams theorem on trees. Indeed Kuratoski's theorem is related nicely to the more general picture of topological obstruction to embeddibility. If you would like to propose a different (perhaps topological) understanding of the extension of Kuratowski's theorem for surfaces, then maybe you should start by the well-quasi ordering theorem for trees.
| 34 | https://mathoverflow.net/users/1532 | 7116 | 4,852 |
https://mathoverflow.net/questions/7129 | 3 | I think it's clear that commutative semigroups S that are also bands, i.e. $e^2 = e$ for all e, correspond to finite posets (consider the elements of the semigroups as sets, where the intersection of two sets is their product), satisfying the condition that there is a unique minimal element in this finite poset (to make sure multiplication is well-defined), and also satisfying the condition that there is a unique maximal element in this semigroup if we want it to have an identity. (Unless my reasoning is wrong?)
I was wondering if there is a standard representation theory for such semigroups covered in a paper somewhere. I've thought about it a bit and reduced it to a combinatorial problem, but I'd rather not spend time on it since it is almost certainly a well-known (and easy probably) result.
Is there a more general representation theory for bands available? What about representation theory of rectangular bands? (which I presume is slightly more challenging, unless the problem above grows much more difficult as the poset grows more complicated, which is also a possibility).
| https://mathoverflow.net/users/2623 | Representations of finite commutative band semigroups | Firstly, if you're only interested in finite semigroups, I suggest amending the title of the question ;)
Commutative semigroups in which ever element is idempotent are called *semilattices*, and are a special case of the more general notion of *inverse semigroup*. They have been much studied, although their infinite-dimensional representation theory seems slightly less well mapped out. In any case, the correspondence you describe between semilattices as semigroups and semilattices as certain kinds of poset is indeed well known.
If you're only interested in finite semilattices, then I think the paper to look at is one by Solomon where he gives explicit formulas for the central idempotents that generate the semigroup ring of a given finite (semi)lattice. Unfortunately I've forgotten the exact reference, but I think similar versions or improvements are discussed in
C. Greene, On the M\"obius algebra of a partially ordered set, Adv. in Math. 10 (1973), 177--187.
G. Etienne, On the M\"obius algebra of geometric lattices, Eur. J. Combin. 19 (1998), 921--933.
(Off the top of my head, for representations of more general finite semigroups, see recent work of Benjamin Steinberg, which is where I first became aware of some of the older work.)
I would have thought that the representation theory of rectangular bands should actually be much easier than that of general bands, since every rectangular band can be rewritten as $L \times R$ for index sets $L, R$ and with multiplication defined by $(l\_1,r\_1)\cdot (l\_2,r\_2)=(l\_1,r\_2)$.
| 6 | https://mathoverflow.net/users/763 | 7131 | 4,862 |
https://mathoverflow.net/questions/7126 | 2 | Here's a question I've been thinking about, it's a curiosity that I don't know how to answer. There could be a simple counterexample, or it could be true and I don't know how difficult it would be to prove.
If we fix $m$, is it always possible to find a sufficiently large $n$ satisfying the conditions of the following question: [Note: My original question was to determine whether it is true for arbitrary $m,n$ which was answered below negatively; I have edited it to make the question more interesting].
Define $ \phi: S\_m \rightarrow S\_{m+n}$ is a canonical embedding, and $\phi^{\*} : F[S\_m] \rightarrow F[S\_{m+n}]$ and similarly embeddings $\theta: S\_{n} \rightarrow S\_{m+n}$, and the induced map, such that $\phi(S\_{m}) \times \theta(S\_{n})$ is a direct product.
Given an element $x \in \phi^{\*}(F[S\_m]), x \neq 0$, is it necessary that there exist an element $x' \in F[S\_{m+n}]$ so that the product $xx' \in \theta^{\*}(F[S\_n]), xx' \neq 0$. It seemed true in the cases that I have tried, but they are quite small so I'm not certain if this is true.
Making the assumption $ \text{char} F = 0$ would make it easier I'm sure, but even in this case I can't prove it.
| https://mathoverflow.net/users/2623 | Embedding group algebra $F[S_m X S_n]$ into a group algebra $F[S_{m+n}]$ | **NOTE:** *This answer was given in response to an earlier version of the question above, and so probably seems completely irrelevant if you can't see the old version. (More strikethrough and less deletion in editing, please?)*
Firstly, I'm going to assume you want $x'$ to be nonzero. Secondly, do you want $x$ and $xx'$ to also both be non-zero?
If so, then there is a silly counterexample -- take $n=1$, and take $x$ to be $id - t$ where $t$ is the permutation $1\to 2\to \dots \to m\to 1$. Then $x$ lies in the augmentation ideal of $F[S\_{m+n}]$ and so if $xx'$ lies in $\theta^\*(F[S\_1]) = F$ then it would have to be zero.
I strongly suspect that one could play similar tricks with higher values of $n$.
| 3 | https://mathoverflow.net/users/763 | 7132 | 4,863 |
https://mathoverflow.net/questions/7134 | 7 | Why doesn't the join operation on the category of simplicial sets commute up to unique isomorphism? I mean, aren't products and coproducts commutative up to isomorphism? That leads me to conclude at first glance that the join is commutative, but it's not. Recall, given two simplicial sets $S$ and $S'$, we define the join to be the simplicial set such that for all finite nonempty totally ordered sets $J$, $$(S\star S')(J)=\coprod\_{J=I\cup I'}S(I) \times S'(I')$$
Where $\forall (i \in I \land i' \in I') i < i'$, which implies that $I$ and $I'$ are disjoint.
Now the thing is, clearly my conclusion is stupid, because we use the fact that it doesn't commute to distinguish between over quasi-categories and under quasi-categories. Where did I go wrong?
I hope this is up to the standards of MO, but if it's not, I'll delete the topic.
| https://mathoverflow.net/users/1353 | Joins of simplicial sets | Implicit in the index of the coproduct is that you're writing J as an ordered disjoint union of I and I', where I comes first.
EDIT: Some elaboration.
For a simplicial set $T$, let's write $T\\_n$ for the "n-simplices", i.e. the value of on the ordered set $\{0,1,...,n\}$; these together with the maps between them determine the functor $T$ completely. (Your formula for the join requires the convention that $T$ takes the empty set to a single point.)
Given $S$ and $S'$, let's determine the 0- and 1-simplices of the join.
First, $(S \star S')\\_0$. There are exactly two ways to write $\{0\} = I \cup I'$ in an order preserving way as indexed by the coproduct: either $I'$ is empty and $I$ is everything, or vice versa. Thus $(S \star S')\\_0 = S\\_0 \cup S'\\_0$ accordingly. The zero-simplices of the join are the zero-simplices of the original simplicial set.
Next, the 1-simplicies. Similarly
$$
(S \star S')\\_1 = S(\{0,1\}) \cup (S(\{0\}) \times S'(\{1\})) \cup S'(\{0,1\})= S\\_1 \cup (S\\_0 \times S'\\_0) \cup S'\\_1
$$
There are then 3 types of 1-simplices: the 1-simplices from S, those from S', and for each choice of a point of S and a point of S' there is a new 1-simplex.
The two boundary maps $(S \star S')\\_1 \to (S \star S')\\_0$ are induced by the inclusions of $\{0\}$ and $\{1\}$ into $\{0,1\}$ (the "back" and "front" boundaries respectively). In particular, on the new 1-simplices $S\\_0 \times S'\\_0$ the back boundary is the projection to $S\\_0$ and the front boundary is projection to $S'\\_0$. There is *asymmetry* here because the only ways we're allowed to decompose $\{0,1\}$ in the coproduct have $I$ (the subset corresponding to $S$) first and $I'$ second. None of the "new" edges start at a vertex of $S'$ and end at a vertex of $S$.
| 9 | https://mathoverflow.net/users/360 | 7138 | 4,866 |
https://mathoverflow.net/questions/7112 | 15 | This is an extension of [this question](https://mathoverflow.net/questions/7080/definition-of-the-symmetric-algebra-in-arbitrary-characteristic-for-graded-vector/) about symmetric algebras in positive characteristic. The title is also a bit tongue-in-cheek, as I am sure that there are multiple "correct" answers.
Let $\mathfrak g$ be a Lie algebra over $k$. One can define the universal enveloping algebra $U\mathfrak g$ in terms of the adjunction: $$\text{Hom}\_{\rm LieAlg}(\mathfrak g, A) = \text{Hom}\_{\rm AsAlg}(U\mathfrak g, A)$$ for any associative algebra $A$. Then it's easy enough to check that $U\mathfrak g$ is the quotient of the free tensor algebra generated by $\mathfrak g$ by the ideal generated by elements of the form $xy - yx - [x,y]$. (At least, I'm sure of this when the characteristic is not $2$. I don't have a good grasp in characteristic $2$, though, because I've heard that the correct notion of "Lie algebra" is different.)
But there's another good algebra, which agrees with $U\mathfrak g$ in characteristic $0$. Namely, if $\mathfrak g$ is the Lie algebra of some algebraic group $G$, then I think that the algebra of left-invariant differential operators is some sort of "divided-power" version of $U\mathfrak g$.
So, am I correct that this notions diverge in positive characteristic? If so, does the divided-power algebra have a nice generators-and-relations description? More importantly, which rings are used for what?
| https://mathoverflow.net/users/78 | Which is the correct universal enveloping algebra in positive characteristic? | The notions do indeed diverge in positive characteristic: there is the enveloping algebra, and then (in the case that $\mathfrak g$ is the Lie algebra of an algebraic group G) there is also the hyperalgebra of G, which is the divided-power version you mention. In characteristic 0 these two algebras coincide, but in positive characteristic they differ very much. In particular, the hyperalgebra is not finitely-generated in positive characteristic; see Takeuchi's paper "Generators and Relations for the Hyperalgebras of Reductive Groups" for the reductive case. There is also a good exploration of the hyperalgebra in Haboush's paper "Central Differential Operators of Split Semisimple Groups Over Fields of Positive Characteristic."
One can obtain the hyperalgebra as follows. I don't know in what generality the following construction holds, so let's say that $\mathfrak g$ is the Lie algebra of an algebraic group G defined over $\mathbb Z$. Then there is a $\mathbb Z$-form of the enveloping algebra of G (the Kostant $\mathbb Z$-form) formed by taking divided powers, and upon base change this algebra becomes the hyperalgebra. Alternatively, one can take an appropriate Hopf-algebra dual of the ring of functions on G (cf Jantzen's book "Representations of Algebraic Groups").
As for their uses, in positive characteristic the hyperalgebra of G captures the representation theory of G in the way that the enveloping algebra does in characteristic 0, i.e. the finite-dimensional representations of G are exactly the same as the finite-dimensional representations of the hyperalgebra. This fails completely for the enveloping algebra: instead, the enveloping algebra sees only the representations of the first Frobenius kernel of G.
| 24 | https://mathoverflow.net/users/1528 | 7142 | 4,870 |
https://mathoverflow.net/questions/7133 | 41 | Is there a classification of finite commutative rings available?
If not, what are the best structure theorem that are known at present?
All I know is a result that every finite commutative ring is a direct product of local commutative rings (this is correct, right?) in some paper which computes the size of the general linear group over that ring.
| https://mathoverflow.net/users/2623 | Classification of finite commutative rings | Yes, a finite ring $R$ is a finite direct sum of local finite rings. As a first step, for each prime $p$ there is a subring $R\_p$ of $R$ corresponding to the elements annihilated by the powers of $p$. $\require{enclose} \enclose{horizontalstrike}{R\_p\ \style{font-family:inherit;}{\text{is then an}}\hspace{-7mm}}$
$\enclose{horizontalstrike}{\style{font-family:inherit;}{\text{algebra over}}\ \mathbb{Z}/p.}$ $R\_p$ then resembles an algebra over $\mathbb{Z}/p$ and it could be one, but it can also have a more complicated structure as an abelian $p$-group (see below). This step generalizes to maximal ideals: For each maximal ideal $m$, $R\_m$ is the subring of elements annihilated by $m^n$ for some $n$, and $R$ is the direct sum of these subrings, which are local rings.
It is not difficult to write down a rough partial classification of of local finite rings. If $R$ is local with maximal ideal $m$, it ~~is~~ resembles an algebra over the finite field $F = R/m$; the associated graded ring is such an algebra. If you choose a basis $x\_1,\ldots,x\_n$ for $m/m^2$, then $R$ or its associated graded is a quotient of the polynomial ring $F[\vec{x}]$ in which only finitely many monomials are non-zero. You can make a diagram of these non-zero monomials; they can be any order ideal in the $n$-dimensional orthant. Or, in basis-independent form, $R$ has a length, which is the largest nonvanishing power of $m$, and each $m^k/m^{k+1}$ is some quotient of the $k$th symmetric power of the generating vector space $V = m/m^2$.
After that, the non-zero monomials may be linearly dependent (and never mind that $R$ might be more complicated than its associated graded). Informally, there will be an endless stream of partial results and there will never be a complete classification when the length of the local ring is 3 or more. To see this, suppose that $m^4 = 0$, and suppose that $m^3$ is only one dimension shy of $S^3(V)$. Then the ring is defined by an arbitrary symmetric trilinear form in $V$. These make a "wild" sequence of algebraic varieties, in the same sense that people say that the representation theories of certain rings are wild. For instance, I think (not quite sure) that it is NP-hard to determine when two such trilinear forms are equivalent. NP-hardness is not by itself rigorously equivalent to no classification, but informally the classification is an intractable mess.
If the nonvanishing monomials in $R$ are linearly independent, then it is a toric local ring. Toric local rings are certainly a tractable class of finite rings.
The situation is similar to non-commutative $p$-groups, which are also wild and will never be classified. In both cases, certain classes have a nice structure. It is also interesting to make estimates for how many there are.
**Note:** Corrected per comment.
| 17 | https://mathoverflow.net/users/1450 | 7146 | 4,874 |
https://mathoverflow.net/questions/7153 | 49 | I'm not sure whether this is non-trivial or not, but do there exist simple examples of an affine scheme $X$ having an open **affine** subscheme $U$ which is not principal in $X$? By a principal open of $X = \mathrm{Spec} \ A$, I mean anything of the form $D(f) = \{\mathfrak p \in \mathrm{Spec} \ A : f \notin \mathfrak p\}$, where $f$ is an element of $A$.
| https://mathoverflow.net/users/1107 | Open affine subscheme of affine scheme which is not principal | Let X be an elliptic curve with the identity element O removed. Let U=X-P where P is a point of infinite order. Then U is affine by a Riemann-Roch argument. Now suppose that U=D(f). Then on the entire elliptic curve, the divisor of f must be supported at P and O only. This implies that P is a torsion point, a contradiction.
| 54 | https://mathoverflow.net/users/425 | 7185 | 4,909 |
https://mathoverflow.net/questions/7192 | 10 | There is a simple, intuitive "construction" of twisted K-theory if we are allowed to ignore that many things only hold up to homotopy. We know that maps to $K(Z,2)$ give line bundles on a space and that $K(Z,2)$ forms a group corresponding to the tensor product of line bundles. Line bundles also act as endomorphisms of K-theory given by the tensor product. Thus, there is an action of $K(Z,2)$ on $F$ (where $F$ is the classifying space for $K^0$). $K(Z,2)$ principal bundles are classified by maps to $BK(Z,2) \cong K(Z,3)$, ie, elements of $H^3$. Choosing such a map, we get a principal $K(Z,2)$ bundle, $E$, and we can form the associated bundle $E \times\_{K(Z,2)} F$. Twisted K-theory is then the homotopy classes of sections of this bundle.
The usual constructions of twisted K-theory that I have seen make the above precise by choosing representatives of the relevant objects so that all the needed relations hold on the nose. My question is whether you can avoid doing that. In other words, can you define all the various notions up to homotopy and obtain a definition of twisted K-theory that way?
| https://mathoverflow.net/users/947 | Constructing Twisted K-theory | The answer is yes if you're working on the level of $\infty$-categories (and I'm pretty sure no if you're working on the level of homotopy categories). In other words, in the $\infty$-world there's no problem talking about a principal bundle for K(Z,2)=BBZ on any space, and they're indeed classified by maps (in the $\infty$-category of spaces) to BBBZ,
as are elements of $H^3$. Moreover for any spectrum $E$ there's a well defined group object $GL\_1(E)$, and we have a map $BBZ\to GL\_1(E)$ in the case of E=K-theory. For any principal $GL\_1(E)$ bundle (eg one induced from a BBZ bundle) we have an associated "E line bundle", and its global sections are the twisted K-theory of your space (i.e. a spectrum whose homotopy groups are the usual twisted K-groups).
Or again in short, everything works intuitively as you think without chosing representatives or strictifying if you work in the wonderful world of $\infty$-categories.. for example, the idea of sheaves of spectra (hence twisted cohomology theories) are as simple to work with formally as ordinary sheaves. When it comes to calculating things.. well that's a whole other story.
| 12 | https://mathoverflow.net/users/582 | 7198 | 4,920 |
https://mathoverflow.net/questions/7152 | 10 | Suppose $A \rightarrow B$ is a faithfully flat map of rings. Then the Amitsur complex is exact:
$0 \rightarrow A \rightarrow B \rightarrow B \otimes\_A B \rightarrow \dots$
(the second map is $id \otimes 1 - 1 \otimes id$, and the subsequent maps are alternating sums of the different ways of putting in a 1.)
That this is exact makes sense! Its saying that element of $b\in B$ such that $1\otimes b=b\otimes 1$ comes from $A$. The way to prove it is exact is roughly the following (the details can be found in Milne's online notes on étale cohomology.)
First, one can check that the sequence is exact if the first map has a section. Second, it is exact iff it is exact after a faithfully flat base change. Finally, if we base change by $B$, we have a section to the first map: the map we want to find a section to is $B \rightarrow B\otimes\_A B, b \mapsto b\otimes 1$, and this has a section given by multiplication.
Why do we care? This fact is important in showing schemes are sheaves on the étale site. (Again, this can be found in Milne's notes.)
Okay, so my question is: what is really going on behind the scenes? I know a section is supposed to somehow "solve an equation", but I have no idea what is really being solved here. The proof feels like magic.
(One note: Its probably instructive to look at the special case of $A \rightarrow \prod A\_f$ , a cover by basic open affines. I believe this proof is showing that the sequence is exact simultaneously on each member of the cover, and because exactness is a local property we have that the sequence is exact.)
| https://mathoverflow.net/users/791 | intuition about the "section after base-change" for flat descent and exactness of the Amitsur complex | I think it feels like magic because there's something tautological going on. The story should really culminate with the definition of "faithfully flat" rather than begin with it.
As you suggested, let's consider the case where you cover an affine scheme Spec(A) by finitely many basic open affines Spec(Af). The cover Spec(B)=Spec(∏Af) has the following special property:
>
> (∗) If you want to check exactness of any sequence of A-modules, it's enough to check exactness after restricting to Spec(B) (i.e. tensoring with B over A).
>
>
>
Now if you'd like to show that the Amitsur complex is exact, you know it's enough to find a section B→A. By (∗), we know what it's actually enough to find a section *locally*, and in the case Spec(B)=∐Spec(Af)→Spec(A), it's obvious that there's a section locally. This is all very tautological, but it allows you to prove that schemes are sheaves in the Zariski topology, which maybe doesn't impress you so much.
But we have extra assumptions in this argument. We didn't really need B to be of the form ∏Af, we just needed it to satisfy (∗). So let's make a new definition, saying that B is "faithfully flat" over A if (∗) holds. Now we get the result: "If B is faithfully flat over A, then the Amitsur complex is exact." As a consequence, we can show that schemes are sheaves in the faithfully flat topology on affine schemes, but all we've really done is shown that "schemes are sheaves in the strongest topology for which this proof works," and just defined "faithfully flat" to be that topology. Delightfully, we can prove (with commutative algebra) that lots of different properties of affine schemes and morphisms of affine schemes are local in this topology.
Okay, but we'd really like to prove things about schemes and morphisms of schemes, not about affine schemes and morphisms of affine schemes. I think part of your confusion arises from the desire to understand the etale topology (a good desire), rather than continuing the strategy of making definitions which make results trivial (or at least straight-forward). If you indulge your generality tooth, you'll ask the question,
>
> What is the strongest topology on the category of schemes so that every cover can be understood as a combination of (a) Zariski covers and (b) faithfully flat covers of affine schemes by affine schemes?
>
>
>
The answer is the fpqc topology. Basically by construction, if a property of schemes (resp. morphisms of schemes) is local in the Zariski topology and is local in the faithfully flat topology when you restrict to the category of affine schemes, then it is local in the fpqc topology. Similarly, if a functor is a sheaf in the Zariski topology, and its restriction to the category of affine schemes is a sheaf in the faithfully flat topology, then it's a sheaf in the fpqc topology. In particular, we get that
* Schemes are sheaves in the fpqc topology.
* A whole bunch of properties of morphisms are local in the fpqc toplogy. See Theorem 2.36 of Vistoli's [Notes on Grothendieck topologies, fibered categories and descent theory](http://arxiv.org/abs/math.AG/0412512) for a long list.
This sounds very impressive given that nobody understands what general fpqc morphisms look like (I think), but that's just because we defined the fpqc topology to be whatever it has to be to make the straight-forward proofs work. One thing we do know is that fppf morphisms are fpqc (by [EGA IV](http://www.numdam.org/numdam-bin/item?id=PMIHES_1964__20__5_0), Corollary 1.10.4). In particular, etale morphisms are fpqc.
| 7 | https://mathoverflow.net/users/1 | 7204 | 4,925 |
https://mathoverflow.net/questions/7212 | 7 | Is there a good place to learn about the structure of **moduli stack** of flat $G$-bundles on an algebraic curve?
Of course, we're just studying representations of a group $\pi\_1(X)\to G$ modulo some conjugation (that's why it should be a stack). Since this is very similar to **Galois representations** in number theory, I wonder if there's a reference that also explains the similarities and differences between the two cases.
| https://mathoverflow.net/users/65 | Moduli space of flat bundles | You have to be a bit careful here. Over $\mathbb{C}$ the stack of representations of $\pi\_{1}(X)$ in $G$ and the stack of flat algebraic $G$-bundles on $X$ are isomorphic as complex analytic stacks but are *not* isomorphic as algebraic stacks. In fact the algebraic structure on the stack of flat $SL\_{n}(\mathbb{C})$ bundles on $X$ reconstructs the curve $X$, while the stack of representations of $\pi\_{1}(X)$ into $SL\_{n}(\mathbb{C})$ depends only on the genus of $X$ and not on the particular curve $X$.
As for references I suggest you take a look at Carlos Simpson's papers "Moduli of representations of the fundamental group of a smooth projective variety, I and II" which you can find [here](http://archive.numdam.org/article/PMIHES_1994__79__47_0.pdf) and [here](http://archive.numdam.org/article/PMIHES_1994__80__5_0.pdf).
| 16 | https://mathoverflow.net/users/439 | 7221 | 4,939 |
https://mathoverflow.net/questions/7229 | 15 | I am a combinatorist by training and I am interested in learning about the connections between combinatorics and Schubert varieties. The theory of Schubert varieties seems to be a difficult area to break into if one has not already studied it in graduate school. I don't have a formal course in Algebraic Geometry, but I do know some Commutative Algebra. Does anyone have any recommendations on what Algebraic Geometry one needs to understand (and what books/papers one should read) in order to begin studying Schubert varieties? Algebraic Geometry is such a huge subject that I am trying to figure out what is essential to this particular study and what is not. Thanks, in advance, for any suggestions!
| https://mathoverflow.net/users/2176 | Learning About Schubert Varieties | I would suggest Part III of Fulton's Young Tableaux book (of which you should skip Part I) as the best starting point for learning about Schubert varieties.
One can get very far in this subject with a naive 19th century view of algebraic geometry, especially if one is willing to occasionally accept without proof a few foundational facts (for example the basics of intersection theory). I would suggest that you don't need to learn algebraic geometry in general for now, though if you're serious about working in this area you'll eventually need to get some feel for what kinds of questions algebraic geometers are interested in.
| 17 | https://mathoverflow.net/users/3077 | 7234 | 4,948 |
https://mathoverflow.net/questions/7095 | 22 | There is a fact that I should have learned a long time ago, but never did; I was reminded that I did not know the answer by Qiaochu's excellent series of posts, the most recent of which is [this one](http://qchu.wordpress.com/2009/11/28/the-noetherian-condition-as-compactness/).
Let $X$ be a topological space. I can think of at least four rings of continuous functions to attach to $X$:
1. $C(X)$ is the ring of continuous functions to $\mathbb R$.
2. $C\_b(X)$ is the ring of bounded functions to $\mathbb R$.
3. $C\_0(X)$ is the ring of continuous functions that "vanish at infinity" in the following sense: $f\in C\_0(X)$ iff for every $\epsilon>0$, there is a compact subset $K \subseteq X$ such that $\left|f(x)\right| < \epsilon$ when $x \not\in K$.
4. The ring $C\_c(X)$ of functions with compact support.
Option 2. is not very good. For example, it cannot distinguish between a space and its Stone-Cech compactification. My question is what kinds of spaces are distinguished by the others.
For example, I learned from [this question](https://mathoverflow.net/questions/3871/maximal-ideals-in-the-ring-of-continuous-real-valued-functions-on-r) that MaxSpec of $C(X)$ is the Stone-Cech compactification of $X$. But it seems that $C(X)$ knows a bit more, in that $C(X)$ has residue fields that are not $\mathbb R$ if $X$ is not compact?
On the other hand, my memory from my C\*-algebras class (which was a few years ago and to which I didn't attend well), is that for I think locally-compact Hausdorff spaces $C\_0(X)$ knows precisely the space $X$?
So, MOers, what are the exact statements? And if I believe that the correct notion of "space" is "algebra of observables", are there good arguments for preferring one of these algebras (or one I haven't listed) over the others?
| https://mathoverflow.net/users/78 | Which is the correct ring of functions for a topological space? | I work with infinite dimensional manifolds so am extremely distrustful of anything that requires some sort of compactness condition. Most of the time, it's just too restrictive.
Consider a really nice simple space: the coproduct of a countably infinite number of lines, $\sum\_{\mathbb{N}} \mathbb{R}$ (coproduct taken in the category of locally convex topological vector spaces). This has the property that any compact subset is contained in a finite subspace. However, any neighbourhood of the origin has to be absorbing (the union of the scalar multiples of it is the whole space) so there aren't any non-zero continuous functions with compact support. That defenestrates option 4.
Particularly simple functions on an infinite dimensional vector space are the **cylinder** functions. These are important in measure theory on such spaces. A cylinder function has the property that it factors through a projection to a finite dimensional vector space. Such functions can be continuous and can be bounded, but (apart from the zero function) never vanish at infinity and never have compact support. Thus option 3 joins option 4 in the flowerbed.
As for option 2, I have no particular qualms about it except that it's not stable under partitions-of-unity. Assuming that I have such, then any continuous function can be written as a sum of bounded functions so when doing standard p-of-1 constructions I have to assume that my starting family is uniformly bounded (if that's the right term).
Well, I just remembered one qualm about option 2: if I go up the scale of differentiability then it gets increasingly hard to justify global bounds on the derivatives. I have a memory of John Roe telling me of some result that he'd proved which was to do with bounding all derivatives of a smooth function in some fashion. I don't recall the exact conditions, but the conclusion was that the only functions that satisfied them were trigonometric.
As others have said, if you are really only interested in (locally) compact spaces then the other options have meaning (functionally Hausdorff - points separated by functions - is assumed). But then the title of your question should have been: "Which is the correct ring of functions for a **Locally Compact Hausdorff Space**?".
| 10 | https://mathoverflow.net/users/45 | 7255 | 4,960 |
https://mathoverflow.net/questions/835 | 6 | This is a somewhat vague question which came up MSRI a few days ago: Suppose I have a family of curves over a one dimensional base, given in a computationally explicit way. For example, maybe I have a family F\_t(x,y,z) of homogenous polynomials, whose coefficients are polynomials in t, and which cut out smooth curves for t \neq 0.
Is there an algorithmically practical way to write down the limit of this family in \overline{M}\_g?
| https://mathoverflow.net/users/297 | Algorithms for semistable reduction of families of curves | If your curves are in P^n (specifically in P^2 - as in your example), I think there is something you can do: project your curves from a general P^{n-2} to P^1. This means that you
are now looking for a limit in a Hurwitz scheme. This can be broken into two problems:
* looking for the limit on the underlying M\_{0,n}
* tracing the ramification structure.
Here is an example: find the limit of F+t Q^2 where F is a plane quartic, and Q is a plane quadric.
Project from your favorite random point. You can verify that the limit of the ramification points on the family are
* the eight intersection points of F and Q
* twice on each of ramification points of the projection of p from Q.
From here you can continue in a variety of ways (e.g. you have a pencil of g^1\_4 s on the limit curve which break through a map from the limit curve to a plane conic, which has 8 ramification points)
| 4 | https://mathoverflow.net/users/404 | 7268 | 4,970 |
https://mathoverflow.net/questions/7264 | 3 | Can one formulate those version of Weak Lefschetz that uses tubular neighbourhoods purely in terms of cohomology of (some) algebraic varieties?
Theorem in 5.1 of Part II in Goresky-MacPherson's "Stratified Morse theory" implies (in particular) that:
for a smooth projective P (over the field of complex numbers), X open in P, and a small enough tubular neighbourhood $H\_\delta$ of an arbitrary (!) hyperplane section H of P (in P!) a Weak Lefschetz theorem for $(H\_\delta\cap X,X)$ is valid i.e.:
the map on singular cohomology $H^{i}\_{sing}(X)\to H^{i}\_{sing}(H\_\delta\cap X)$ is an isomorphism for $i<\dim X-1$, and is an injection for $i=\dim X-1$. A caution: $H\_\delta\cap X$ is not (usually) a tubular neighbourhood of $H\cap X$ in $X$.
My question is: could one formulate an analogue of this statement purely in terms of algebraic geometry? I would be completely satisfied with cohomology with $Z/l^n Z$-coefficients i.e. etale cohomology. I only want to replace the cohomology of $H^{i}\_{sing}(H\_\delta\cap X)$ in the statement by something that could be computed without using differential geometry.
My guess: one should probably replace $H\_\delta$ with an etale tubular neighbourhood of $H$ in $P$ (then $H\_\delta\cap X$ will be replaced by the corresponding fibre product); this is 'my conjecture'. Etale tubular neighbourhoods were defined and studied by Cox and Friedlander. Yet though they proved that etale tubular neighbourhoods share several properties with 'ordinary' tubular neighbourhoods, it seems that no comparison statement that would allow to deduce my conjecture from the Goresky-MacPherson's theorem is known. One should probably use nice properties of the comparison of the etale site with the fine one; yet this seems to require a site-theoretic definition of a tubular neighbourhood. Also, etale tubular neighbourhoods seem to be rather 'implicit', so I don't know how to check my conjecture on examples. Certainly, I do not object against proving my conjecture 'directly', yet this seems to be difficult (since Goresky-MacPherson's proof heavily relies upon stratified Morse theory).
Any suggestions would be very welcome!
| https://mathoverflow.net/users/2191 | On algebraic tubular neighbourhoods and Weak Lefschetz | regarding the comparison theorem between (Z/l^n-cohomology of) etale and complex-analytic tubular neighborhoods, my feeling is that it should hold, but won't follow formally from the comparison theorem for varieties.
one can at least get the comparison map, i think, as follows (this also addresses your question about sites): both in the etale and complex-analytic cases, the intersection of an open subset U of X with a tubular neighborhood of a closed subvariety Z in X can be represented by the topos which is the lax 2-fiber product of Sh(Z) and Sh(U) over Sh(X). so the comparison map of topoi comes from ordinary considerations. but to show that it's an isomorphism on cohomology, one probably should argue as in the proof of artin's comparison theorem, locally fibering by curves, etc. hopefully this vague sketch works out, but i definitely haven't considered the details.
| 1 | https://mathoverflow.net/users/2200 | 7296 | 4,993 |
https://mathoverflow.net/questions/7247 | 17 | I recently had the idea that maybe measure spaces could be viewed as sheaves, since they attach things, specifically real numbers, to sets...
But at least as far as I can tell, it doesn't quite work - if $(X,\Sigma,\mu)$ is a measure space and $X$ is also given the topology $\Sigma$, then we do get a **presheaf** $M:O(X)\rightarrow\mathbb{R}\_{\geq0}$, where $\mathbb{R}\_{\geq0}$ is the poset of non-negative real numbers given the structure of a category, and $M(U) = \mu(U)$, and $M(U \subseteq V)$ = the unique map "$\geq$" from $\mu(V)$ to $\mu(U)$, but this is not a sheaf because for an open cover {$U\_i$} of an open set $U$, $\mu(U)$ is in general not equal to sup $\mu(U\_i)$ (and sup is the product in $\mathbb{R}$).
First off, is the above reasoning correct? I'm still quite a beginner in category theory, but I've seen on Wikipedia something about a sheafification functor - does applying that yield anything meaningful/interesting? Does anyone have good references for measure theory from a category theory perspective, or neat examples of how this perspective is helpful?
| https://mathoverflow.net/users/1916 | measure spaces as presheaves? | For a sheaf-theoretical interpretation of measure theory, measure spaces are the wrong objects, you want *measure algebras* and then consider certain Grothendieck topologies on a Boolean algebra.
For measure algebras, check out volume 3 of FRemlin's 5-volume opus dedicated to measure theory. For more sheaf-theoretical stuff, google for Mathew Jackson's phd thesis on measure theory in the context of topos theory.
Hope it helps, regards,
G. Rodrigues
| 17 | https://mathoverflow.net/users/2562 | 7307 | 5,002 |
https://mathoverflow.net/questions/7283 | 37 | In a workshop about the geometry of $\mathbb{F}\_1$ I attended recently, it came up a question related to a mysterious but "not-so-secret-anymore" seminar about... an hypothetical Topological Langlands Correspondence!
I had never heard about this program; I have found this page via Google:
<http://www.math.jhu.edu/~asalch/toplang/>
I only know a bit about the Number-Theoretic Langlands Program, and I still have a hard time trying to understand what is happening in the Geometrical one, so I cannot even start to draw a global picture out of the information dispersed in that site.
So, the questions are: What do you know (or what can you infere from the web) about the Topological Langlands Correspondence? Which is the global picture? What are its analogies with the (original) Langlands Program? Is it doable, or just a "little game" for now? What has been proved until now? What implications would it have?
(Note: It is somewhat difficult to tag this one, feel free to retag it if you have a better understanding of the subject than I have!)
| https://mathoverflow.net/users/1234 | Topological Langlands? | I would cautiously venture that there is not really something we could call a topological Langlands program to outsiders at this point. My objection is to the final word - we don't really know what we're doing. For example, I don't think we even have a conjecture at this point relating representations of something to something else, or have the right idea of L-functions that are supposed to play a role.
The "topological Langlands program" banner is more of an idea that the scattered pieces of number-theoretic content we see in stable homotopy theory should be part of a general framework. There is the appearance of groups whose orders are denominators of Bernoulli numbers in stable homotopy groups of spheres, Andrew Salch's calculations at higher chromatic levels that seem to be related to special values of L-functions, the relationship between K(1)-local orientation theory and measures p-adically interpolating Bernoulli numbers (see [Mark Behrens'](http://math.mit.edu/~mbehrens/) website for some material on this), the surprisingly canonical appearance of realizations of Lubin-Tate formal group laws due to work of Goerss-Hopkins-Miller in homotopy theory, et cetera, et cetera.
Perhaps at this point I'd say that we have a "topological Langlands program" program, whose goal is to figure out why on earth all this arithmetic data is entering homotopy theory and what form the overarching structure takes in a manner similar to the Langlands program itself.
| 35 | https://mathoverflow.net/users/360 | 7312 | 5,006 |
https://mathoverflow.net/questions/7308 | 1 | It is well know that the genus three non orientable surface, N3, has only periodic and reducible auto-homeomorphisms, meanwhile the surface N4 is the first non orientable surface with pseudo Anosov maps. Also, recently profesor B.Szepietowski gave the MCG presentation of N4, from where, I calculated that there are seven periodic mapping classes. The question is: are there more?
Curiously, the torus and N3 show also seven periodic mapping classes each but we would like to understand better why N3 loses pseudo Anosov maps which contrast with the fact that N3 is got from the 2-torus via an one-point blow up...
| https://mathoverflow.net/users/2196 | N_3 and N_4 periodic and pseudo Anosov auto-homeomorphisms | Just to lend some context to the above question: the mapping class group of the two-torus is naturally isomorphic to GL(2, Z). If we restrict to orientation preserving homeomorphism the mapping class group is SL(2, Z). The periodic mapping classes (isotopy classes of homeomorphisms) are exactly those with trace less than two in absolute value. (Hmm, and +/- Id, I guess!) Now we need to count the number of conjugacy classes of periodic elements. There should be a cool algebraic way to do this. (Perhaps it would help to give a purely algebraic proof that the order of torsion is at most 6?)
I think that there is a geometric way to do this: every periodic element occurs as the symmetry of some flat torus (= parallelogram with opposite sides identified). All tori have have the hyperelliptic symmetry, corresponding to rotation by 180 degrees about any point. (These maps lie in the mapping class of the negative identity.) Other symmetries:
Rombic tori have a reflection symmetry as do rectangular tori.
The square torus has a rotation by 90 degrees.
The hexagonal torus has a rotation by 60 degrees.
So I count:
1. the identity, Id
2. the hyperelliptic = -Id = rotation by 180
3. rotation by 90
4. rotation by 60
5. rotation by 120
6. the reflection [[-1,0],[0,1]] (reflection in an axis) and
7. the reflection [[0,1],[1,0]] (exchange axes).
You can prove that the last two are distinct algebraically. Perhaps the lack of 45 degree rotation is a geometric proof.
Now, we could perform similar geometric tricks to obtain symmetries of $N\_4$ and get at least all of the rotations... [Edit: For example, it is possible to build a copy of $\rm{Sym}\_4$ by placing the cross-caps at the vertices of a tetrahedron.]
| 2 | https://mathoverflow.net/users/1650 | 7314 | 5,007 |
https://mathoverflow.net/questions/7252 | 5 | As I understand, when we have a nilpotent cone, or a nullcone of a Lie group representation, what seems to be done in a lot of the literature (e.g. Achar&Henderson-"Orbit closures in the enhanced nilpotent cone") is to compute the intersection cohomology sheaves and find polynomials that determine the dimensions of various stalks.
But what other cohomology theories (that are different to intersection cohomology, I understand sometimes two different cohomology/homology can coincide under special circumstances), would be interesting in nilpotent cones?
Here's a bit about the problem I'm working on, and some theories that I hope (?) might be interesting, can anyone tell me some more that might be interesting? I am very far from being knowledgeable about cohomology, so if some-one could tell me if the following questions are stupid/trivial/ill-defined or not, please tell me.
* I have the orbits, which themselves are usually quasi-affine or quasi-projective varieties, which I could compute cohomology of? (perhaps Cech cohomology?)
* The set of orbits inherits a Zariski topology structure from the Zariski topology structure (that coincides with that inherited from the classical topology), perhaps I can compute some (co)homology of this topological space? In my case the set of orbits is uncountably infinite, but I am not completely sure if it has a triangulation - any theory that doesn't involve triangulations?
* As standard, one computes the orbit closures, and instead of doing intersection cohomology of these singular varieties, compute perhaps some of the lower K-groups?
* perhaps Hochschild cohomology of the coordinate rings of some of the affine coordinate rings of these varieties could be interesting?
| https://mathoverflow.net/users/2623 | What cohomology theories would be interesting for nilpotent cones/nullcones? | I think most questions of interest regarding the nilpotent cone have to do with categories of equivariant sheaves on it - either equivariant perverse sheaves, like IC complexes of orbits, or equivariant coherent sheaves, like structure sheaves of orbit closures. So cohomologies that help elucidate the structure of these categories would be great. For example, equivariant IC of orbit closures fits into this, as does ordinary cohomology of orbit closures. Hochschild homology/cohomology of structure sheaves or K-theory likewise control aspects of the category of coherent sheaves, and you could ask for equivariant analogs.
But I would say history suggests it's best to emphasize two things:
-the full structure of the category (ie what are simples, standards, relations
between them)
-if you have an analog of the Springer resolution for these nullcones, its cohomologies might be even more interesting (or if you'd like, the pushforwards of standard sheaves from there to the nullcone..)
| 2 | https://mathoverflow.net/users/582 | 7319 | 5,008 |
https://mathoverflow.net/questions/7317 | 4 | It turns out that joins of simplicial sets are fairly easy to define, but hard to manage. In lots of cases, we'd like to compute what a join is, does it look like a horn?, a boundary?, etc? and identify it as such, so we can figure out when our morphisms from the join have certain nice properties like being anodyne, having lifting properties, and all of that wonderful stuff.
For example, consider the join, $\Lambda^n\_j \star \Delta^m$. The problem that I currently face is, I can't tell what this thing looks like from the definition.
Consider an even simpler case, $\Delta^n \star \partial \Delta^m$. From the definition, we get a very nasty definition of this join, and I'm having trouble applying it and computing the join in terms of nicer simplicial sets.
I ask this, because on p.62 of Higher Topos Theory by Lurie, for example, he states that for some $0 < j \leq n$ $$\Lambda^n\_j \star \Delta^m \coprod\_{\Lambda^n\_j \star \partial \Delta^m} \Delta^n \star \partial \Delta^m$$
and says that we can identify this with the horn $\Lambda^{n+m+1}\_j$. Unraveling the definitions seems to make it harder to understand, and I just don't see how this result was achieved. However, my aim here is to understand how the computation was actually carried out, since it is completely omitted.
For convenience, here is the definition of the join of $S$ and $S'$ for each object $J \in \Delta$
$$(S\star S')(J)=\coprod\_{J=I\cup I'}S(I) \times S'(I')$$
Where $\forall (i \in I \land i' \in I') i < i'$, which implies that $I$ and $I'$ are disjoint.
**EDIT AFTER ANSWER:** Both Reid and Greg provided good answers to the question, and I only accepted the one that I did because Greg commented more recently. So for anyone reading this at some point in the future, read both answers, as they are both good.
| https://mathoverflow.net/users/1353 | Computation of Joins of Simplicial Sets | Since the join of simplicial sets is associative and $\Delta^m = \Delta^0 \star \cdots \star \Delta^0$ ($m+1$ times), we should start by trying to understand things like $\Lambda^n\_j \star \Delta^0$, a.k.a. the "final" cone on $\Lambda^n\_j$. It's not too hard to see that this is the subcomplex of $\Delta^{n+1}$ consisting of those faces which do not contain the (codimension 2) face $\{0, \ldots, r-1, r+1, \ldots, n\}$. In other words, we are missing the face opposite $r$ and $n+1$, because we were originally missing the face opposite $r$ of $\Delta^n$, as well as the three other faces (including the interior of $\Delta^{n+1}$) it contains. Similarly $\Delta^0 \star \partial \Delta^n$ is the horn $\Lambda^{n+1}\_0$ (we are missing the interior of $\Delta^{1,\ldots,n}$ and the cone on it).
In general all the simplicial sets that come up have the form of the subcomplex of $\Delta^N$ consisting of those faces which do not contain a fixed face $\Delta^S$, $S \subset \{0, \ldots, N\}$. Forming the cone (on either side) on such a space results in another such space with $N$ replaced by $N+1$ and $S$ unchanged (as a subset of the vertices of the original $\Delta^N$, which if we formed a cone on the left, means we increment each index in $S$).
After doing these sorts of computations, I expect that $\Lambda^n\_j \star \Delta^m$ and $\Delta^n \star \partial \Delta^m$ will be two subcomplexes of $\Delta^{n+m+1}$ each characterized by avoiding faces containing a certain face, and that $\Lambda^n\_j \star \partial \Delta^m$ is their intersection and $\Lambda^{n+m+1}\_j$ is their union, from which the claim would follow.
| 5 | https://mathoverflow.net/users/126667 | 7323 | 5,009 |
https://mathoverflow.net/questions/7315 | 4 | So I've been skimming Bowen's 1972 paper "Symbolic Dynamics for Hyperbolic Flows" hoping it would give me some insight into how to build a Markov family for the cat flow (i.e., the Anosov flow obtained by suspension of the cat map with unit height). For the sake of completeness, the cat flow $\phi$ is obtained as follows:
i. Consider the cat map $A$ on the 2-torus and identify points $(Ax,z)$ and $(x,z+1)$ to obtain a 3-manifold $M$
ii. Equip $M$ with a suitable metric (e.g., $ds^2 = \lambda\_+^{2z}dx\_+^2 + \lambda\_-^{2z}dx\_-^2 + dz^2$, where $x\_\pm$ are the expanding and contracting directions of $A$ and $\lambda\_\pm$ are the corresponding eigenvalues.)
iii. Consider the flow generated by the vector field $(0,1)$ on $M$--that's the cat flow.
Unfortunately I'm getting stuck at the first part of Bowen's quasi-constructive proof, which requires finding a suitable set of disks and subsets transverse to the flow. Rather than rehash the particular criteria for a set of disks and subsets used in Bowen's construction, I will relay a simpler but very similar set of criteria, for a *proper family* (which if it meets some auxiliary criteria is also a Markov family):
$\mathcal{T} =$ {$T\_1,\dots,T\_n$} is called a proper family (of size $\alpha$) iff there are differentiable closed disks $D\_j$ transverse to the flow s.t.
1. the $T\_j$ are closed
2. $M = \phi\_{[-\alpha, 0]}\Gamma(\mathcal{T})$, where $\Gamma(\mathcal{T}) = \cup\_j T\_j$
3. $\dim D\_j = \dim M - 1$
4. diam $D\_j < \alpha$
5. $T\_j \subset$ int $D\_j$ and $T\_j = \bar{T\_j^\*}$ where $T\_j^\*$ is the relative interior of $T\_j$ in $D\_j$
6. for $j \ne k$, at least one of the sets $D\_j \cap \phi\_{[0,\alpha]}D\_k$, $D\_k \cap \phi\_{[0,\alpha]}D\_j$ is empty.
I've been stuck on even constructing such disks and subsets (let alone where the subsets are rectangles in the sense of hyperbolic dynamics). Bowen said this sort of thing is easy and proceeded under the assumption that the disks and subsets were already in hand. I haven't found it to be so. The thing that's killing me is 6, otherwise neighborhoods of the Adler-Weiss Markov partition for the cat map would fit the bill and the auxiliary requirements for the proper family to be a Markov family.
I've really been stuck in the mud on this one, could use a push.
| https://mathoverflow.net/users/1847 | Proper families for Anosov flows | Take Adler-Weiss on $0\times\mathbb T^2$, $1/3\times\mathbb T^2$ and $2/3\times\mathbb T^2$. Take neighborhoods of this tripled Adler-Weiss. Then this collection would satisfy all the properties with $\alpha=1/3$.
I am not sure why are you particularly interested in suspension flow, everything is determined by the base Anosov diffeo.
Edit: Indeed, this has to be tinkered a bit. Say Adler Weiss has two rectangles with neighborhoods $D\_1$ and $D\_2$. Then take collection
$0\times D\_1$,
$1/3\times D\_1$,
$2/3\times D\_1$,
$\varepsilon\times D\_2$,
$1/3+\varepsilon\times D\_2$,
$2/3+\varepsilon\times D\_2$.
To ensure the second property take $\alpha=1/3+\varepsilon$.
| 32 | https://mathoverflow.net/users/2029 | 7325 | 5,011 |
https://mathoverflow.net/questions/7322 | 4 | Given an undirected connected graph, our goal is to remove some edges to make the graph disconnected. The constraint is that **each** node of the graph can not lose more than $m$ edges incident to it. I want to find the minimum $m$ for which the goal is achievable. Is there any efficient algorithm to compute this minimum $m$ (and/or which edges to remove)? Or is it NP-complete?
| https://mathoverflow.net/users/1401 | A graph connectivity problem (restated) | It appears to be NP-complete even when m=1: see The Complexity of the Matching-Cut Problem, Maurizio Patrignani and Maurizio Pizzonia, WG 2001, <http://dx.doi.org/10.1007/3-540-45477-2_26>
| 5 | https://mathoverflow.net/users/440 | 7334 | 5,016 |
https://mathoverflow.net/questions/7326 | 12 | Let (X,x) be a pointed projective variety. Then there exists an abelian variety V which is universal for maps of pointed varieties $(X,x) \to (A,e\_A)$, called the [albanese variety](http://en.wikipedia.org/wiki/Albanese_variety). When X is a curve, the variety V is isomorphic to the Jacobian of X (in higher dimensions this fails) which is a principally polarized abelian variety.
**Question**: When is an albanese variety principally polarized? If it is not always principally polarized, can one describe the degree of the polarization in terms of data intrinsic to X?
| https://mathoverflow.net/users/2 | When is an Albanese variety principally polarized? | In general it could happen that the Albanese variety does not admit a principal polarization at all. For instance the Albanese variety of an abelian variety is the Abelian variety itself. So choose $X$ to be some abelian variety that has no principal polarization and you will get an example.
On the other hand it can happen that the Albanese variety is principally polarized. For instance you can take the Albanese of the $n$-th symmetric product of a curve. It is equal to the Jacobian of the curve and so admits a principal polarization. Or if you want to be fancier you can take a hyperplane section in the symmetric product of a curve. It will also have the Jacobian of the curve as its Albanese variety.
Another useful comment is that the Albanese of $X$ is the dual of $Pic^{0}(X)$ and so $Alb(X)$ admits a principal polarization if and only if $Pic^{0}(X)$ does. If you fix an ample line bundle $L$ on an $n$-dimensional complex projective variety $X$, then $L$ induces a natural polarization on $Pic^{0}(X)$: the universal cover of $Pic^{0}(X)$ is naturally identified with $H^{1}(X,O\_{X}) = H^{1,0}(X)$, the integral $(1,1)$ form $c\_{1}(L)$ then induces a Hermitian pairing on $H^{1,0}(X)$ by the formula
$$
h(\alpha,\beta) := -2i \int\_{X} \alpha\wedge \bar{\beta} \wedge c\_{1}(L)^{\wedge (n-1)}.
$$
This $h$ defines a polarization on $Pic^{0}(X)$. The construction of $h$ is purely cohomological and so it is straightforward to check if it defines a principal polarization by computing the divisors of this polarization.
| 14 | https://mathoverflow.net/users/439 | 7336 | 5,017 |
https://mathoverflow.net/questions/7329 | 16 | I recently discovered [The College Mathematics Journal](http://www.maa.org/pubs/cmj.html) and enjoyed reading through some of the articles on fun applications of mathematics. I'd like to send some of the articles to my younger sister, a high school sophomore, but unfortunately most of them require calculus, a subject she hasn't studied yet.
Are there any other journals or websites that publish quality articles on applied math that a high school student taking pre-calculus could understand reasonably well?
| https://mathoverflow.net/users/2205 | Math journal for high school students? | An excellent journal published by the University of New South Wales (my alma mater!) is [Parabola](http://www.parabola.unsw.edu.au/content/about-us), aimed at interested secondary students.
| 14 | https://mathoverflow.net/users/3 | 7355 | 5,034 |
https://mathoverflow.net/questions/7318 | 28 | $\newcommand{\bb}{\mathbb}\DeclareMathOperator{\gal}{Gal}$
Before stating my question I should remark that I know almost nothing about etale cohomology - all that I know, I've gleaned from hearing off hand remarks and reading encyclopedia type articles. So I'm looking for an answer that will have some meaning to an etale cohomology naif. I welcome corrections to any evident misconceptions below.
Let $E/\bb Q$ be an elliptic curve the rational numbers $\bb Q$: then to $E/\bb Q$, for each prime $\ell$, we can associate a representation $\gal(\bar{\bb Q}/\bb Q) \to GL(2n, \bb Z\_\ell)$ coming from the $\ell$-adic Tate module $T\_\ell(E/\bb Q)$ of $E/\bb Q$ (that is, the inverse limit of the system of $\ell^k$ torsion points on $E$ as $k\to \infty$). People say that the etale cohomology group $H^1(E/\bb Q, \bb Z\_\ell)$ is dual to $T\_\ell(E/\bb Q)$ (presumably as a $\bb Z\_\ell$ module) and the action of $\gal(\bar{\bb Q}/\bb Q)$ on $H^1(E/\bb Q, \bb Z\_\ell)$ is is the same as the action induced by the action of $\gal(\bar{\bb Q}/\bb Q)$ induced on $T\_\ell(E/\bb Q)$.
Concerning this coincidence, I could imagine two possible situations:
(a) When one takes the definition of etale cohomology and carefully unpackages it, one sees that the coincidence described is tautological, present by definition.
(b) The definition of etale cohomology (in the case of an elliptic curve variety) and the action of $\gal(\bar{\bb Q}/\bb Q)$ that it carries is conceptually different from that of the dual of the $\ell$-adic Tate module and the action of $\gal(\bar{\bb Q}/\bb Q)$ that it carries. The coincidence is a theorem of some substance.
Is the situation closer to (a) or to (b)?
Aside from the action $\gal(\bar{\bb Q}/\bb Q)$ on $T\_\ell(E/\bb Q)$, are there other instances where one has a similarly "concrete" description of representation of etale cohomology groups of varieties over number fields and the actions of the absolute Galois group on them?
Though I haven't seen this stated explicitly, I imagine that one has the analogy [$\gal(\bar{\bb Q}/\bb Q)$ acts on $T\_\ell(E/\bb Q)$: $\gal(\bar{\bb Q}/\bb Q)$ acts on $H^1(E/\bb Q; \bb Z\_\ell)$]::[$\gal(\bar{\bb Q}/\bb Q)$ acts on $T\_\ell(A/K)$: $\gal(\bar{\bb Q}/\bb Q)$ acts on $H^1(A/K; \bb Z\_\ell)$] where $A$ is an abelian variety of dimension $n$ and $K$ is a number field: in asking the last question I am looking for something more substantively different and/or more general than this.
I've also inferred that if one has a projective *curve* $C/\bb Q$, then $H^1(C/\bb Q; \bb Z\_\ell)$ is the same as $H^1(J/\bb Q; \bb Z\_\ell)$ where $J/\bb Q$ is the Jacobian variety of $C$ and which, by my above inference I assume to be dual to $T\_\ell(J/\bb Q)$, with the Galois actions passing through functorially. If this is the case, I'm looking for something more general or substantially different from this as well.
The underlying question that I have is: where (in concrete terms, not using a reference to etale cohomology as a black box) do Galois representations come from aside from torsion points on abelian varieties?
---
[Edit (12/09/12): A sharper, closely related question is as follows. Let $V/\bb Q$ be a (smooth) projective algebraic variety defined over $\bb Q$, and though it may not be necessary let's take $V/\bb Q$ to have good reduction at $p = 5$. Then $V/\bb Q$ is supposed to have an attached 5-adic Galois representation to it (via etale cohomology) and therefore has an attached (mod 5) Galois representation. If $V$ is an elliptic curve, this Galois representation has a number field $K/\bb Q$ attached to it given by adjoining to $\bb Q$ the coordinates of the 5-torsion points of $V$ under the group law, and one can in fact write down a polynomial over $\bb Q$ with splitting field $K$. The field $K/\bb Q$ is Galois and the representation $\gal(\bar{\bb Q}/\bb Q)\to GL(2, \bb F\_5)$ comes from a representation $\gal(K/\bb Q) \to GL(2, \bb F\_5)$. (I'm aware of the possibility that knowing $K$ does not suffice to recover the representation.)
Now, remove the restriction that $V/\bb Q$ is an elliptic curve, so that $V/\bb Q$ is again an arbitrary smooth projective algebraic variety defined over $\bb Q$. Does the (mod 5) Galois representation attached to $V/\bb Q$ have an associated number field $K/\bb Q$ analogous to the (mod 5) Galois representation attached to an elliptic curve does? If so, where does this number field come from? If $V/\bb Q$ is specified by explicit polynomial equations is it possible to write down a polynomial with splitting field $K/\bb Q$ explicitly? If so, is a detailed computation of this type worked out anywhere?
I'm posting a bounty for a good answer to the questions succeeding the "Edit" heading.
| https://mathoverflow.net/users/683 | Etale cohomology and l-adic Tate modules | IMO, the scenario is closer to your (a). I'll sketch an explanation of the duality between $H^1(E,\mathbf{Z}\_l)$ and the dual to the Tate module. We have $H^1(E,\mathbf{Z}\_l)=\text{Hom}(\pi\_1(E),\mathbf{Z}\_l)$,
where that $\pi\_1$ means etale fundamental group with base point the origin $O$ of $E$. Thus the isomorphism we really want is between $\pi\_1(E)\otimes\mathbf{Z}\_l$ and $T\_\ell(E)$.
What is $\pi\_1(E)$? In the topology world, we'd consider the universal cover $f\colon E'\rightarrow E$ and take $\pi\_1(E)$ to be its group of deck transformations. Then $\pi\_1(E)$ has an obvious action on $f^{-1}(O)$. If $E$ is the complex manifold $\mathbf{C}/L$ for a lattice $L$, this is just the natural isomorphism $\pi\_1(E)\cong L$.
But in the algebraic geometry world, there is no universal cover in the category of varieties, so the notion of universal cover is replaced with the projective system $E\_i\to E$ of etale covers of $E$. Then $\pi\_1(E)$ is the projective limit of the automorphism groups of $E\_i$ over $E$.
One nice thing about $E$ being an elliptic curve is that any etale cover $E'\rightarrow E$ must also be an elliptic curve (once you choose an origin on it, anyway); if $E\rightarrow E'$ is the dual map then the composition $E\rightarrow E'\rightarrow E$ is multiplication by an integer. So it's sufficient to only consider those covers of $E$ which are just multiplication by an integer. Since it's $\pi\_1(E)\otimes\mathbf{Z}\_l$ we're interested in, it's enough to consider the isogenies of $E$ given by multiplication by $l^n$.
What are the deck transformations of the maps $l^n\colon E\rightarrow E$? Up to an automorphism of $E$, they're simply translations by $l^n$-division points. And now we see the relationship to the Tate module: A compatible system of deck transformations of these covers is the exact same thing as a compatible system of $l^n$-division points. Thus we get the desired isomorphism. Naturally, it's Galois compatible!
In the end, we see that torsion points were tucked away in the construction of the etale cohomology groups, so it wasn't exactly a coincidence. Hope this helps.
Re the edit: I believe your best bet is to work locally. First of all, you didn't mention which Galois representation you wanted exactly; let's say you want the representation on $H^i$ of your variety for a given $i$. Let's assume this space has dimension $d$.
Step 1. For each prime $p$ at which your variety $V$ has good reduction, you can compute the local zeta function of $V/\mathbf{F}\_p$ by counting points on $V(\mathbf{F}\_{p^n})$ for $n\geq 0$. In this way you can compute the action of the $p^n$th power Frobenius on $H^i(V\otimes\overline{\mathbf{F}}\_p,\mathbf{F}\_5)$ for various primes $p$.
Step 2. Do this enough so that you can gather up information on the statistics of how often the Frobenius at $p$ lands in each conjugacy class in the group $\text{GL}\_d(\mathbf{F}\_5)$. In this way you could guess the conjugacy class of the image of Galois inside $\text{GL}\_d(\mathbf{F}\_5)$.
Step 3. Now your job is to find a table of number fields $F$ whose splitting field has Galois group equal to the group you found in the previous step. I found a table here: <http://hobbes.la.asu.edu/NFDB/>. You already know which primes ramify in $K$ -- these are at worst the primes of bad reduction of $V$ together with 5 -- and you can distinguish your $F$ from the other number fields by the splitting behavior your found in Step 1. Then $K$ is the splitting field of $F$.
A caveat: Step 1 may well take you a very long time, because unless your variety has some special structure or symmetry to it, counting points on $V$ is Hard.
Another caveat: Step 3 might be impossible if $d$ is large. If $d$ is 2 then perhaps you're ok, because there might be a degree 8 number field $F$ whose splitting field has Galois group $\text{GL}\_2(\mathbf{F}\_5)$. If $d$ is large you might be out of luck here.
You are free not to accept this answer because of the above caveats but I really do think you've asked a hell of a tough question here!
| 37 | https://mathoverflow.net/users/271 | 7359 | 5,036 |
https://mathoverflow.net/questions/7365 | 5 | I know that for $X$ a connected space, $THH(\Sigma^\infty \Omega X) = \Sigma^\infty \Lambda X$, the suspension spectrum of the free loop space of $X$. The computation can be carried out in spaces and then transferred to spectra via $\Sigma^\infty$. What is $TC(\Sigma^\infty \Omega X)$? Can it also be computed from some kind of $TC$ on the level of spaces?
Edit: Tyler answered my question, but I want to ask a followup question: Is it fair to say that $TC(\Omega X)$ in the world of spaces, after $p$-completion, is just $X$, and is there a map (not an equivalence, because we take limits to build $TC$) $\Sigma^\infty X \to $ the thing Tyler wrote down? (Note: all my $\Sigma^\infty$ are $\Sigma^\infty\_+$.)
| https://mathoverflow.net/users/126667 | What is $TC(\Sigma^\infty \Omega X)$? | The TC spectrum, at a prime $p$, of this is the homotopy pullback of a diagram
$S^1 \wedge (\Sigma^\infty\_+ \Lambda X)\_{hS^1} \to \Sigma^\infty\_+ \Lambda X \leftarrow \Sigma^\infty\_+ \Lambda X$
after $p$-completion. Here the left-hand map is the $S^1$-transfer from homotopy orbits back to the spectrum and the right-hand map is the difference between the identity and the "$p$'th power" maps on the loop space.
This is in Bökstedt-Hsiang-Madsen's original paper defining topological cyclic homology, in section 5.
ADDED LATER: This doesn't really work on the space level, because they don't have all the structure necessary. They have the $F$ maps, but not the $R$ ones which only come about from stable considerations. Spaces with a group action really only have one notion of "fixed points," namely the honest fixed points of the group action.
However, the associated equivariant spectrum of $\Lambda X$ is built out of spaces like
$$\Omega^V \Sigma^V \Lambda X = Map(S^V, S^V \wedge \Lambda X\_+)$$
where $V$ ranges over representations of $S^1$. This has two "fixed-point" objects for any cyclic group $C$: there's the fixed points, which is the space
$$Map^C(S^V, S^V \wedge \Lambda X\_+)$$
of equivariant maps. There is also the collection of maps-on-fixed-points
$$Map((S^V)^C, (S^V \wedge \Lambda X\_+)^C)$$
which is called the "geometric" fixed point object, and it accepts a map from the ordinary fixed points. The fact that $(\Lambda X)^C \cong \Lambda X$ implies that you can interpret this as a map $(Q \Lambda X)^C \to (Q \Lambda X)$ where the latter uses an accelerated circle. These maps give rise to the $R$ maps in the definition of $TC$, and they definitely rely on the fact that you're considering the associated spectra.
| 5 | https://mathoverflow.net/users/360 | 7382 | 5,050 |
https://mathoverflow.net/questions/7432 | 14 | Let $(A,m\_A)$ and $(B,m\_B)$ be noetherian local rings and $f:A\rightarrow B$ a local homomorphism. Let $F = B/m\_AB$ be the fiber ring and assume that
$$\mathrm{dim}(B) = \mathrm{dim}(A) + \mathrm{dim}(F).$$
The following Theorem (23.1 in Matsumura's CRT) is really quite a miracle:
Theorem: If $A$ is regular and $B$ is Cohen-Macaulay then $f$ is flat.
I am wondering to what extent this theorem can be generalized. What I have in mind is a statement of the type:
"Theorem": If $A$ is $X$ and $B$ is $Y$ then $f$ is of finite Tor-dimension
(i.e. $\mathrm{Tor}^i\_A(B,A)=0$ for all $i$ sufficiently large).
Here, $X$ and $Y$ are ring-theoretic conditions which should be *strictly weaker* than
"regular" and "CM" respectively. Is the "Theorem" above true just requiring $A$ and $B$ to be normal?
How about both CM? Or maybe CM plus finitely many $(R\_i)$?
Any thoughts/ counterexamples?
| https://mathoverflow.net/users/2215 | Generalizing miracle flatness (Matsumura 23.1) via finite Tor-dimension | The "Theorem" isn't true with both rings just normal, or just CM, or even normal *and* CM. Let $A = k[[x,y,z]]/(xz-y^2) \cong k[[a^2,ab,b^2]]$ and let $B = k[[a,b]]$, with $f$ the natural inclusion. The dimensions add up as they must, since $f$ is module-finite. In this case finite flat dimension is the same as finite projective dimension, but $B$ does not have finite projective dimension over $A$.
I don't expect that any addition of assumptions $(R\_i)$ would help.
| 8 | https://mathoverflow.net/users/460 | 7438 | 5,094 |
https://mathoverflow.net/questions/7439 | 3 | Any non-singular projective variety over $\mathbb{C}$ is easily seen to be a smooth manifold. Presumably the same is not true for algebraic varieties - one would not expect varieties with singular points to have a smooth structure. But do there exist non-singular varieties that are not smooth manifolds?
| https://mathoverflow.net/users/1977 | Algebraic Varieties which are also Manifolds | Every non-singular algebraic variety over $\mathbb C$ is a smooth manifold. See for instance:
<http://en.wikipedia.org/wiki/Manifold>
under "Generalizations of Manifolds".
In fact, Arminius' suggested answer in the comments seems to give a proof of this fact, and I'll attempt to flesh it out a small amount. Every algebraic variety is locally a quasi-affine variety. So we may take an open cover $U\_i$ of the variety, where each $U\_i$ is a closed subset of an open subset of affine n-space. We may then check smoothness at each point of $U\_i$ via the Jacobian criterion. The same procedure illustrates that each $U\_i$ is a complex manifold. Since the gluing maps are algebraic, they are smooth, and hence our non-singular variety is also a smooth manifold.
| 7 | https://mathoverflow.net/users/4 | 7448 | 5,101 |
https://mathoverflow.net/questions/7454 | 13 | The complex structure on a complex manifold pulls back to what's called a CR structure on any real codimension 1 submanifold. The structure induced on a submanifold of higher codimension is a CR structure if a non-degeneracy condition holds. It's possible to describe these structures intrinsically, without reference to an embedding. I don't know anything else.
I'd be happy with whatever kind of answer to the title question, but here are some more specific ones:
1. Does CR stand for Cauchy-Riemann, or what?
2. What kind of local invariants do CR manifolds have? Are there coordinates around every point that look like a real hyperplane in C^n? Or can there be some curvature or something.
3. Can there be continuous families of CR structures on a given manifold? If the manifold is compact can these families (mod diffeomorphism) be infinite-dimensional?
4. I have the impression, just from arxiv postings and seminar titles, of CR geometry being studied more in analysis than in softer geometric fields. Is that accurate, and if so what accounts for it?
| https://mathoverflow.net/users/1048 | What are CR manifolds like? | CR does stand for Cauchy-Riemann.
CR structures on 3 dimensional manifolds arise as the boundaries of complex (or almost-complex) 4 manifolds; if these boundaries are strictly *pseudo-convex* (i.e. convex in "holomorphic directions") the CR structure on the 3-manifold is a contact structure (if the boundary is only (pseudo-)convex or (Levi) flat, the CR structure integrates to a confoliation or a foliation respectively). There can be infinite dimensional families of foliations on a 3-manifold; more generally, whenever the CR structure is "non-generic" or integrable, one has continuous moduli, otherwise (eg in the contact structure case) one has discrete moduli (to be explicit: what has discrete moduli is the contact structure, not the "CR+contact structure".)
| 13 | https://mathoverflow.net/users/1672 | 7455 | 5,106 |
https://mathoverflow.net/questions/7446 | 15 | I am taking a first course on Algebraic Geometry, and I am a little confused at the intuition behind looking at schemes over a fixed scheme. Categorically, I have all the motivation in the world for looking at comma categories, but I would like to make sense of this geometrically.
Here is one piece of geometric motivation I do have: A family of deformations of schemes could be thought of as a morphism $X \rightarrow Y$, where the fibers of the morphism are the schemes which are being deformed, and these are indexed by the scheme Y.
This is all well and good, but I am really interested in Schemes over $Spec(k)$ are thought of as doing geometry "over" $k$. I know that their is a nice "schemeification functor" taking varieties over $k$ to schemes over $k$, but this is somewhat unsatisfying. All that I see algebraically is that $k$ injects into the ring of global sections of the structure sheaf of $X$, but this does not seem like much of a geometric condition...
Any help would be appreciated.
EDIT: The answers I have received so far have been good, but I am looking for something more. I will try to flesh out an example that give the same style of answer that I am asking for:
The notion of a k-valued point: Every point $(a,b)$ of the real variety $x^2 + y^2 - 1 = 0$ has a corresponding evaluation homomorphism $\mathbb{R}[x,y] \rightarrow \mathbb{R}$ given by $x \mapsto a$ and $y \mapsto b$. Since $a^2 + b^2 - 1 = 0$, this homomorphis factors uniquely through $\mathbb{R}[x,y]/(x^2 + y^2 -1)$. So the real valued points of the unit circle are in one to one correspondence with the homomorphisms from $\mathbb{R}[x,y]/(x^2 + y^2 -1)$ to $\mathbb{R}$. Similarly, the complex valued points are in one to one correspondence with the homomorphisms from $\mathbb{R}[x,y]/(x^2 + y^2 -1)$ to $\mathbb{C}$. Actually, points of the unit circle valued in any field $k$ are going to be in one to one correspondence with homomorphisms from $\mathbb{Z}[x,y]/(x^2 +y^2 -1)$ to $k$.
Dualizing everything in sight, we are saying that the $k$- valued points of the scheme $Spec(\mathbb{Z}[x,y]/(x^2 +y^2 -1))$ are just given by homomorphisms from the one point scheme $Spec(k)$ into $Spec(\mathbb{Z}[x,y]/(x^2 +y^2 -1))$.
EDIT 2: Csar's comment comes very close to answering the question for me, and I will try and spell out the ideas in that comment a little better. I wish Csar had left his comment as an answer so I could select it.
So it seems to me that the most basic reason to think about schemes over a field $k$ is this:
I already spelled out above why $k$-valued points of a scheme are important. But a lot of the time, a morphism from $Spec(k)$ to $X$ will point to a generic point of $X$, not a closed point. Different morphisms can all point to the same generic point. For instance the dual of the any injection $\mathbb{Z}[x,y] \rightarrow \mathbb{C}$ (of which there are many), will all "point" to the generic point of $\mathbb{Z}[x,y]$.
On the other hand if we are looking at $\mathbb{Q}[x,y]$, this is a $\mathbb{Q}$ scheme. $Spec(\mathbb{Q})$ is also a $\mathbb{Q}$ scheme via the identity map. A morphism of $\mathbb{Q}$ schemes from $Spec(\mathbb{Q})$ to $\mathbb{Q}[x,y]$ will correspond to a **closed** $\mathbb{Q}$ -valued point! So this is a nice geometric reason for looking at schemes over a fixed field $k$: $k$ morphisms of $Spec(k)$ to $X$ correspond to $k$-valued closed points of $X$.
It will take some work for me to internalize the vast generalization that S-schemes entail, but I think this is a good start. Does everything I said above make sense?
| https://mathoverflow.net/users/1106 | Intuition about schemes over a fixed scheme | This is going to be perhaps vague, but I'l try to write down the idea.
I've been told that in doing scheme theory, most of the time what is studied is not a scheme but a morphism of schemes $f:X\rightarrow S$. The studied of such a maps can be enriched allowing a chance of "base" $S$. Such a scheme $S$ can be crazy, but I'll impose finite properties to the map $f$ to keep everything under control.
Now, as you say, if we have two $S$-schemes $f:X\rightarrow S$ and $Y\rightarrow S$ then we can consider the fiber product of them (which is a categorical product of $X$ and $Y$ over $S$). Such a process can be though of as replacing the base $S$ by the scheme $Y$. In doing so, the new map $f\_Y:X\_Y\rightarrow Y$ (standing for $p\_2:X\times\_{S}Y\rightarrow Y$) may be easier to work with than the map itself $f$. This very construction is generalizing the idea of "extending scalars".
Here is a sort of example of a product described above. Given the point $s\in S$ and its residue field $k(s)$ of the local ring $\mathcal{O}\_s$ at that point. Then it is well known that $X\_s=X\times\_S Spec(k(s))$ has an underlying space the "fiber" $f^{-1}(s)$ (as long as $f$ is "finite"). This space can be considered as an **algebraic variety** over the field $k(s)$. This way $S$ is the scheme parameterizing the varieties $X\_s$ some of which a priori may have different ground fields $k(s)$.
On the other hand, back to the fiber product, in considering a variety $X$ over the field $k$ (scheme of finite type over $k$), we can extend the scalars from $k$ to a field extension $K$. In doing so, we may think of such a variety now over a field extension $K$. Here is where the product comes into play. Vaguely enough, this is saying that if you have an equation and you have solved it over the field $k$ now you might ask yourself if you'd be able to solved it again but over a field extension $K$, here you want to perform an extension of scalars and the ideas written above may help out.
Needless to say that we can consider as well $X\times\_k Spec(\overline{k})$ and due to the fact that field $k$ may not have been algebraically closed, such a fiber product can be handful.
Now, to get intuition, let's take a look at an example: $$\pi:Spec \mathbb{Z}[i]\rightarrow Spec(\mathbb{Z})$$
Let $X=Spec\mathbb{Z}[i]$ and notice that the fiber $$X\_p=X\times\_{\mathbb{Z}}\mathbb{F}\_p=Spec(\mathbb{Z}[i]\otimes\mathbb{F}\_p)$$
Such a fiber is going to have cardinality 2 if $p\equiv 1 mod(4)$ and cardinality 1 if $p\equiv 3 mod(4)$ (this is the fact that $p=a^2+b^2$ where $a,b\in \mathbb{Z}$ if $p\equiv 1mod 4$). These fibers are *$X$ reduce mod $p$*. Notice that at the generic point we have $X\_0=\mathbb{Q}[i]$.
Now the $T-points$ in $X$ where $T=\mathbb{F}\_p[i]$, and all that story give rise to the functor which represents the scheme $X$. If I am not wrong, the info that carries such a functor is nothing but that of the fibers of the map $\pi$. So naively it is like having $X$ fibers wise.
| 5 | https://mathoverflow.net/users/1547 | 7461 | 5,111 |
https://mathoverflow.net/questions/7470 | 23 | There is a nice formula for the area of a triangle on the 2-dimensional sphere;
If the triangle is the intersection of three half spheres, and has angles $\alpha$, $\beta$ and $\gamma$, and we normalize the area of the whole sphere to be $4\pi$ then the area of the triangle is
$$
\alpha + \beta + \gamma - \pi.
$$
The proof is a cute application of inclusion-exclusion of three sets, and involves the fact
that the area we want to calculate appears on both sides of the equation, but with opposite signs.
However, when trying to copy the proof to the three dimensional sphere the parity goes the wrong way and you get 0=0.
Is there a simple formula for the volume of the intersection of four half-spheres of $S^3$ in terms of the 6 angles between the four bounding hyperplanes?
| https://mathoverflow.net/users/2229 | Is there a neat formula for the volume of a tetrahedron on $S^3$? | [On the volume of a hyperbolic and spherical tetrahedron](http://www.f.waseda.jp/murakami/papers/tetrahedronrev3.pdf), by Murakami and Yano. The volume is obtained as a linear combination of dilogarithms and squares of logarithms. The origin of their formula is really interesting: Asymptotics of quantum $6j$ symbols. (These asymptotics have also been studied by many other people: D. Thurston, [Roberts](http://xxx.soton.ac.uk/abs/math-ph/9812013), Woodward, Frohman, Kania-Bartoszynska, etc.)
Note that the 3-dimensional formula has to be much more complicated. The 2-dimensional formula comes from Euler characteristic and Gauss-Bonnet, but the Euler characteristic of the 3-sphere, or any odd-dimensional manifold, vanishes. In fact every characteristic class of a 3-sphere vanishes, because the tangent bundle is trivial. There can't be a purely linear treatment of volumes in isotropic spaces in odd dimensions. In even dimensions, there is always a purely linear extension from lower dimensions using generalized Gauss-Bonnet.
| 27 | https://mathoverflow.net/users/1450 | 7475 | 5,121 |
https://mathoverflow.net/questions/7373 | 11 | Let's suppose we have a Scheme $X$ over the the field $k$, where such a field can be though to be either $\mathbb{C}$ or a finite field $\mathbb{F}\_q$. Then having this in mind,
Where do we find some representative examples where Geometry governs arithmetic? That is to say, examples where the geometry (or topology) of $X$ over $\mathbb{C}$ dictates the arithmetic behavior over $\mathbb{F}\_q$.
Answers along with references would be highly appreciated.
| https://mathoverflow.net/users/1547 | Geometry Vs Arithmetic of schemes | Let's start with the most elementary example: **projective space** $\mathbb P^n$. It's not hard to see that that the number of points on it is always $q^n + q^{n-1} + \dots + q + 1.$
Note that this is because $\mathbb P^n$ can be always decomposed into simpler pieces: $\mathbb A^n \cup \mathbb A^{n-1}\cup\dots\cup \mathbb A^0$. Interestingly, something similar applies to **all $\mathbb F\_q$-varieties**. Specifically, the Lefschetz fixed points formula from topology applied to arithmetics gives the following statement for a variety $X/\mathbb F\_q:$
>
> There exist some algebraic numbers $\alpha\_i$ with $|\alpha\_i| = q^{n\_i/2}$ for some $(n\_i)$ such that the number of points $$\\# X(\mathbb F\_{q^l}) = \sum\_i (-1)^{n\_i}\alpha^l\_i\quad \text{for}\\ l > 0 .$$
>
>
>
Numbers $\alpha\_i$ in fact come from geometry: they are eigenvalues of some operators acting on etale cohomology groups $H\_{et}(X)$. In particular, the numbers $n\_i$ can only occupy an interval between 0 and $\text{dim}\\, X$ and there are as many of them as the dimension of this group.
These groups can directly compared to the case of $\mathbb C$ whenever you construct your variety in a geometric way. To see how, consider the example of curves. Over $\mathbb C$ the cohomology have the form $\mathbb C \oplus \mathbb C^{2g} \oplus \mathbb C\ $ for some $g$ called *genus*; the same holds over $\mathbb F\_q$:
* **projective line** $\mathbb P^1$ has genus 0, so it always has $n+1$ points
* **elliptic curves** $x^2 = y^3 + ay +b$ have genus 1, so they must have exactly $n + 1 + \alpha + \bar\alpha$ points for some $\alpha\in \mathbb C$ with $|\alpha| = \sqrt q.$ This is exactly the **[Hasse bound](http://en.wikipedia.org/wiki/Hasse_bound)** mentioned in another post.
These theorems, which provided an unexpected connecion between topology and arithmetics some half-century ago, were just the beginning of studying varieties over $\mathbb F\_q$ using the geometric intuition that comes from the complex case.
You can read more at any decent introduction to [arithmetic geometry](http://en.wikipedia.org/wiki/Arithmetic_geometry) or [étale cohomology](http://en.wikipedia.org/wiki/Etale_cohomology). There are also some questions here about [motives](https://mathoverflow.net/questions/tagged/motives) which are a somewhat more abstract version of the above picture.
---
As a reply to Ben's comment above about reconstructing the genus if you know $X\_n = \#X(F\_{q^n})$:
* You know with certainty that $1 + q^n - X\_n = \sum \alpha\_i^n\ $ for some algebraic numbers $\alpha\_i, i = 1, 2, \dots $ having property $|\alpha\_i| = \sqrt q.$
* There cannot be two different solutions $(\alpha\_i)$ and $(\beta\_i)$ for a given sequence of $X\_n$ because if $N$ is a number such that both $\alpha\_i = \beta\_i = 0$ for $i>N$ then both $\alpha$ and $\beta$ are uniquely determined from the first $N+1$ terms of the sequence.
* So a given sequence uniquely determines the genus.
I don't know, however, if a constructive algorithm that guarantees to terminate and return genus for a sequence $X\_n$ is possible. The first idea is to loop over natural numbers testing the conjecture that genus is less then $N$, but there seem to be some nuances.
| 5 | https://mathoverflow.net/users/65 | 7501 | 5,138 |
https://mathoverflow.net/questions/7507 | 10 | Recall that a module is called
1. semisimple if every submodule is a direct summand
2. pure semisimple if every pure submodule is a direct summand
There is quite a bit of work on semisimple and pure semisimple modules, of course.
My question is
What is a module called if every submodule is pure?
and
What is known about these modules and where can I read about them?
| https://mathoverflow.net/users/1698 | When is every submodule pure? | Hmmm, I seem to have found the answer to this question. In
[Regular and semisimple modules](https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-65/issue-2/Regular-and-semisimple-modules/pjm/1102866791.full)
by Cheatham and Smith in 1976, they call a module regular if every submodule is pure. Regular is of course an overused word, and maybe other people have called this different things. But the justification makes some sense: if I is a 2-sided ideal of R, then R/I is a regular R-module if and only if R/I is a von Neumann regular ring.
| 8 | https://mathoverflow.net/users/1698 | 7512 | 5,143 |
https://mathoverflow.net/questions/7521 | 0 | What's your usual action online in order to browse math journals? Like check Arxiv or MathSciNet. Any other good link directs you to most updated articles in major math journals. Or the traditional way of browsing periodical section of your library is still a better way to get a glimpse of current development in math.
| https://mathoverflow.net/users/1947 | A website linking to most major math journals? | The AMS Digital Mathematics Registry has a huge list at <http://www.ams.org/dmr/JournalList.html>
| 3 | https://mathoverflow.net/users/532 | 7529 | 5,155 |
https://mathoverflow.net/questions/6749 | 27 | The salamander lemma is a lemma in homological algebra from which a number of theorems quickly drop out, some of the more famous ones include the snake lemma, the five lemma, the sharp 3x3 lemma (generalized nine lemma), etc. However, the only proof I've ever seen of this lemma is by a diagram chase after reducing to R-mod by using mitchell's embedding theorem. Is there an elementary proof of this lemma by universal properties in an abelian category (I don't know if we can weaken the requirements past an abelian category)?
If you haven't heard of the salamander lemma, here's the relevant paper: [Bergman - Diagram chasing in double complexes](http://sbseminar.files.wordpress.com/2007/11/diagramchasingbergman.pdf).
And here's an article on it by our gracious administrator, Anton Geraschenko: [The salamander lemma](http://sbseminar.wordpress.com/2007/11/13/anton-geraschenko-the-salamander-lemma/).
Also, small side question, but does anyone know a good place to find some worked-out diagram-theoretic proofs that don't use Mitchell and prove everything by universal property? It's not that I have anything against doing it that way (it's certainly much faster), but I'd be interested to see some proofs done without it, just working from the axioms and universal properties.
**PLEASE NOTE THE EDIT BELOW**
**EDIT:** Jonathan Wise posted an edit to his [answer](https://mathoverflow.net/a/7531) where he provided a great proof for the original question (doesn't use any hint of elements!). I noticed that he's only gotten four votes for the answer, so I figured I'd just bring it to everyone's attention, since I didn't know that he'd even added this answer until yesterday. The problem is that he put his edit notice in the middle of the text without bolding it, so I missed it entirely (presumably, so did most other people).
| https://mathoverflow.net/users/1353 | A proof of the salamander lemma without Mitchell's embedding theorem? | There's [a proof of the snake lemma without elements](https://math.colorado.edu/%7Ejonathan.wise/papers/snake.pdf) (a non-elementary proof?) on my website.
**Edit:** I added a section about the salamander lemma.
**Much later edit:** As Charles Rezk points out below ([1](https://mathoverflow.net/questions/6749/a-proof-of-the-salamander-lemma-without-mitchells-embedding-theorem#comment423375_7531) [2](https://mathoverflow.net/questions/6749/a-proof-of-the-salamander-lemma-without-mitchells-embedding-theorem#comment423378_7531)), my proof of the salamander lemma is correct only in a special case. I will correct the proof when I find the tex file.
**Much, much later edit:** I found the tex file and corrected the link above (including the corrected proof of the salamander lemma).
What makes working with elements in an abelian category easier than working with objects is that elements of the target of an epimorphism can be lifted to the source. If your abelian category has enough projectives, then a proof with elements can usually be adapted to one without elements by replacing each element by a surjection from a projective object. If you don't have enough projectives, you can still get by without elements. You have to replace the concept of "element" with "epimorphism from something"; then every "element" can still be lifted by passing to a more refined epimorphism.
This is just code for working locally in the topology generated by epimorphisms. (That it is a topology is implied by AB2.) Since there are always enough injective sheaves of abelian groups, this gives an exact embedding of any abelian category in an abelian category with enough injectives (or, if we work with "cotopologies", a category with enough projectives). This permits one to apply the simpler approach outlined above (using projectives) rather than "pro-epimorphisms".
Once one has an embedding in a Grothendieck abelian category (the category of sheaves of abelian groups always is one), it is not much further to a proof of Mitchell's embedding theorem anyway.
| 39 | https://mathoverflow.net/users/32 | 7531 | 5,157 |
https://mathoverflow.net/questions/7532 | 0 | For a power series $f(z) = \sum\_{i=0}^{\infty} a\_i z^i$ with $a\_1$ nonzero, [Lagrange's inversion formula](http://en.wikipedia.org/wiki/Lagrange_inversion_theorem) gives an explicit way to compute the Taylor coefficients of the inverse function.
Is there any analogous formula for Laurent series?
| https://mathoverflow.net/users/83 | Inversion of Laurent series | The Lagrange inversion formula is meant to give you the Taylor series expansion of $f^{-1}$ at the point $f(0)$. If $f$ has a Laurent series instead, then it means that $f(0) = \infty$ and that $f$ is meromorphic. The Taylor series at $\infty$ of $f^{-1}$ then doesn't particularly mean anything unless you change to a different coordinate patch on the Riemann sphere, for instance $\zeta = 1/z$. So you can first switch to the function $1/f$, which has a usual Taylor series, and then use the standard Lagrange inversion formula for $(1/f)^{-1}$.
(If I have understood the question correctly. Maybe this answer is too straightforward to address the real question.)
| 5 | https://mathoverflow.net/users/1450 | 7533 | 5,158 |
https://mathoverflow.net/questions/7535 | 0 | Let A = $\{a\_1,...,a\_n\}$ be a set of numbers. We can assume all elements of A are integers.
Is there any efficient way to partition A into two sets B = $\{b\_1,...,b\_k\}$ and C = $\{c\_1,...,c\_l\}$ such that $|(b\_1...b\_k) - (c\_1...c\_l)|$ is minimal?
Is the problem anything easier if we let A be a set of strictly positive integers? What if we only let prime numbers?
| https://mathoverflow.net/users/1737 | Paritioning a set of numbers A into two sets B,C so that abs(prod(B) - prod(C)) is minimal | I suspect that it's NP-hard even to check whether you can get prod(B) - prod(C) = 0, although there's a problem with the obvious argument that I don't know off the top of my head how to fix.
"Reduction" from subset sum: If you have a set S of integers, replace each integer $k \in S$ with $2^k$. Then this new set can be partitioned into two parts with the same product iff the original set could be partitioned into two parts with the same sum.
The problem is that our new integers are exponentially large compared to the original ones, which means that this isn't actually allowed as a reduction. But I think it's morally correct, since the hardness of subset sum is controlled by the size of the set rather than the lengths of the elements.
| 2 | https://mathoverflow.net/users/382 | 7540 | 5,162 |
https://mathoverflow.net/questions/7539 | 2 | I understand the ordinary Springer correspondence gives a bijection between orbits in the nilpotent cone for the adjoint representation and irreducible representations of the Weyl group, through action of the Weyl group on the top intersection cohomology. (I'm still learning about this, I know a very vague picture).
My question is: does the same method give a correspondence (not injective nor surjective) between orbits in the nullcone of any representation (not the adjoint representation), and representations (not necessarily irreducible) of the Weyl group? Or is the construction so specific that it breaks down entirely when we replace the nilpotent cone by more generally, a nullcone, and we cannot get any Weyl group representations out of it?
| https://mathoverflow.net/users/2623 | Springer corresponding for nullcones other than the standard nilpotent cone | Let me make another guess as to what you are seeing:
The Lie/algebraic group acts on the coordinate ring for the closure of the nilpotent orbit (since it acts on the nilpotent orbit). Take the weight 0 weight space of this coordinate ring; the Weyl group (realized as N(T)/T) acts on it since T acts trivially on this weight space.
In the case of SL\_n, this gives the total cohomology ring of the Springer fiber associated to the conjugate partition (if I remember correctly, and I might not, this result is due to DeConcini and Procesi back in the early 80s). I seem to recall there is a way to recover the grading on the cohomology ring, which allows you to recover the top cohomology (and hence the original Springer correspondence), but cannot remember what it is.
This isomorphism between the weight 0 subring and the cohomology ring is specific to SL\_n.
What I just wrote does make sense for any nullcone, though a priori there is no guarantee that you get a finite dimensional representation of the Weyl group.
| 2 | https://mathoverflow.net/users/3077 | 7551 | 5,170 |
https://mathoverflow.net/questions/7556 | 9 | This is related to the [question](https://mathoverflow.net/questions/6704/how-to-think-about-cm-rings) I asked last time. This sounds a bit too specific, I hope this question is still acceptable on MO.
I am still not quite comfortable with the concept of depth, and there is this exercise in Matsumura's book that goes as follows:
>
> Find an example of a noetherian local ring $A$ and a finite $A$-module $M$ such that $\operatorname{depth}M > \operatorname{depth}A$. Also find $A$, $M$ and $P \in \operatorname{Spec}A$ such that $\operatorname{depth}M\_P > \operatorname{depth}\_P(M)$.
>
>
>
I hope I have found correct examples, but I am still quite lost about why one can find such examples, and what the generic ones are. So if someone can just give me some representative examples I would be grateful.
---
The examples I found myself:
For the first one, it is clear that $A$ must not be Cohen-Macaulay. Then I set $A = \frac {k[x,y,z]}{(xz,yz)}\_{(x,y,z)}$, which is of depth 1, and I consider its quotient by $(z)$, which is $k[x,y]\_{(x,y)}$ and should be of depth 2 (at least $x,y$ is a regular sequence I think).
For the second one, I try to fix $\operatorname{depth}\_P(M) = 0$, which means $P$ should lie in some associated primes of $M$, so I consider $M = \frac {k[x,y,z]}{(x^2,xy,xz)}\_{(x,y)}$, such that $(x,y)$ is not associated prime when localized.
| https://mathoverflow.net/users/nan | Some examples of depth | 1) Start with a regular local ring $R$. Take 2 ideals $I,J$ such that $I$ does not contain $J$, $R/I$ is CM and $\dim R/J <\dim R/I$. Then $A=R/(I\cap J)$ and $M=R/I$ work. In your example, $I=(x)$ and $J=(y,z)$. The reason is that CM means unmixed, so by having components of different dimensions one makes sure A is not CM.
2) Take $(A,m,k)$ to be any CM rings of dimension at least 2. Let $M=A\oplus k$. Then for any non-minimal $P\in Spec(A)-{m}$, $depth M\_P =depth A\_P$, but $depth M=0$.
The common theme: depth is usually the minimum depth of all components, while dimension is the maximal of those.
| 9 | https://mathoverflow.net/users/2083 | 7564 | 5,178 |
https://mathoverflow.net/questions/7561 | 7 | Let $E$ be a spectrum. Then $E \wedge E$ is a $\mathbb{Z}/2$-spectrum in the naivest possible sense, i.e., an object with $\mathbb{Z}/2$-action in the (∞,1)-category of spectra. Can I make it a $\mathbb{Z}/2$-spectrum in the less naive, but still not genuine, sense? (That is, a $\mathbb{Z}/2$-spectrum indexed on the trivial universe.)
I'm thinking of something like the following. I may represent $E$ as an (reduced & continuous) excisive functor from pointed spaces to pointed spaces. Then define
$$G(X) = \mathrm{colim}\_{I \times I} \mathrm{Map}(S^{x\_1} \wedge S^{x\_2}, E(S^{x\_1}) \wedge E(S^{x\_2}) \wedge X)$$
where $I$ is the category of finite sets and inclusions. Hopefully $G$ is a functor from spaces to $\mathbb{Z}/2$-spaces. If I forget about the $\mathbb{Z}/2$-fixed point set, I can think of it as $E \wedge E$ with its $\mathbb{Z}/2$ action. What spectrum does $G(X)^{\mathbb{Z}/2}$ correspond to? Is there a more familiar name for it? **Edit:** I seem to be getting $E \vee (E \wedge E)^{h\mathbb{Z}/2}$, but without much confidence.
[Leftover part of the question: If so, by my question [here](https://mathoverflow.net/questions/3154/1-categorical-description-of-equivariant-homotopy-theory) I can think of the resulting object as a functor from the opposite of the orbit category of $\mathbb{Z}/2$ to spectra. Unpacking this amounts to giving some spectrum $F$ together with a map $F \to (E \wedge E)^{h\mathbb{Z}/2}$. What is $F$?]
| https://mathoverflow.net/users/126667 | Naive Z/2-spectrum structure on E smash E? | Given a spectrum $E$ there is a "standard" lift of $E \wedge E$ to a $\mathbb{Z}/2$-spectrum using the basic technique you describe. One way to describe it as follows.
You can construct the category of genuine $\mathbb{Z}/2$-spectra (indexed on the full universe) via collections of $\mathbb{Z}/2$-spaces $X\\_n$ with equivariant structure maps $\sigma: S^V \wedge X\\_n \to X\\_{n+1}$, where $S^V = S^1 \wedge S^1$ is the 1-point compactification of the regular representation $\mathbb{R} \times \mathbb{R}$ with the "flip" action. Under this description, if $E$ is a spectrum made up of spaces $E\\_n$ and structure maps $S^1 \wedge E\\_n \to E\\_{n+1}$, then you can construct $E \wedge E$ as a genuine $\mathbb{Z}/2$-spectrum with spaces $E\\_n \wedge E\\_n$ and structure maps $S^1 \wedge S^1 \wedge E\\_n \wedge E\\_n$ that simply twist and apply the structure map on each factor. (This, e.g., is one way to pass forward the equivariant structure on $TC$).
This fully genuine spectrum has an underlying spectrum indexed on the trivial subgroup. The $\mathbb{Z}/2$-fixed point object is the homotopy pullback of a diagram
$$
(E \wedge E)^{h\mathbb{Z}/2} \to (E \wedge E)^{t\mathbb{Z}/2} \leftarrow E
$$
where $Z^{t\mathbb{Z}/2}$ is the so-called "Tate spectrum" of $Z$, which is to Tate cohomology as the homotopy fixed point spectrum is to group cohomology.
If $E = \Sigma^\infty W$ for a space $W$ then the map from $E$ to the Tate spectrum lifts (via the diagonal) to a map to the homotopy fixed point spectrum, and so the homotopy pullback will actually be homotopy equivalent to $E \vee (E \wedge E)\_{h\mathbb{Z}/2}$. (This homotopy orbit is the fiber of the map from homotopy fixed points to Tate fixed points.)
For finite spectra the map from $E$ to its Tate spectrum is 2-adic completion; this is one way to state the content of the Segal conjecture that Carlsson proved (at least at the prime 2). Sverre Lunoe-Nielsen extended this result to a number of other spectra like the Brown-Peterson spectra. In these cases the fixed-point object is equivalent after 2-adic completion to the homotopy fixed point object.
All the above plays out the same way for a cyclic group of prime order.
| 4 | https://mathoverflow.net/users/360 | 7579 | 5,188 |
https://mathoverflow.net/questions/7541 | 16 | The following question was posed to me a while ago. No one I know has a given a satisfactory (or even a complete) proof:
Suppose that $M$ is an $n$ x $n$ matrix of non-negative integers. Additionally, suppose that if a coordinate of $M$ is zero, then the sum of the entries in its row and its column is at least $n$.
What is the smallest that the sum of all the entries in $M$ can be?
The conjecture posed to me was that it was $\frac{n^2}{2}$ which is obtained by the diagonal matrix with $\frac{n}{2}$ in all diagonal entries.
[I'm guessing that there should be a "suppose that" in describing M. -- GJK]
| https://mathoverflow.net/users/2043 | Extremal question on matrices | The following looks too simple, so perhaps there's a mistake, but here goes.
Let $m$ be the smallest among all row sums *and* column sums. If $m\geq n/2$, we are done.
Otherwise, $m=cn$ with $c\lt 1/2$. Suppose it is a column which has sum $m$. This column has at least $n-m$ zeroes, and each of the corresponding rows has sum $\geq n-m$. The remaining $m$ rows each has sum $\geq m$.
In total we have a sum of at least $(n-m)^2+m^2 = ((1-c)^2+c^2)n^2$. Finally, note that $(1-c)^2+c^2\gt 1/2$ when $c\lt 1/2$.
So this gives a lower bound of $n^2/2$, and equality requires that any row and any column sums to exactly $n/2$, so the matrix is a sum of $n/2$ permutation matrices by König's Theorem.
Now what about the case when $n$ is odd?
**Edit:** Since the sum is an integer, the lower bound $n^2/2$ actually gives $(n^2+1)/2$, which can be attained by for example taking the direct sum of an $m\times m$ matrix of $1$s with an $(n-m)\times (n-m)$ matrix of $1$, where $m=(n-1)/2$. *When $n$ is odd, this is the only extremal example up to column and row permutations.* Here is a proof.
Let $m$ now, as originally, denote the minimum over all row and column sums. If $m\geq (n+1)/2$, then the total sum is too large: at least $nm\geq n(n+1)/2$. Therefore, $m\leq(n-1)/2$, and the $(n-m)^2+m^2$ lower bound now gives a total sum of at least $(n-n(n-1)/2)^2+((n-1)/2)^2 = (n^2+1)/2$ (using the fact that $(1-c)^2+c^2$ is decreasing when $c\lt 1/2$).
So up to now we have only rederived the lower bound for $n$ odd.
However, if equality now holds, we get that $m=(n-1)/2$, that each row adds up to either $m$ ($m$ times) or $n-m$ ($n-m$ times), and that in a column that adds up to $m$, there are exactly $n-m$ $0$s, so all entries are $0$ or $1$ (and similar statements with columns and rows interchanged). By permuting the rows and columns we may assume that the first $m$ rows [columns] each add up to $m$.There can't be a $0$ in the upper left $m\times m$ submatrix, since then the sum of the row and column containing the $0$ adds up to only $2m\lt n$. We have found a direct sum of an $m\times m$ and an $(n-m)\times(n-m)$ all $1$ matrix.
| 13 | https://mathoverflow.net/users/932 | 7581 | 5,190 |
https://mathoverflow.net/questions/7490 | 7 | Let R be a normal noetherian domain.
What is the difference between a finitely generated reflexive module and a finitely generated projective module?
Can anybody recommend any references or make any suggestions about this?
---
Finitely generated projective modules can be identified with idempotents matrix...
Finitely generated projective modules correspond with vector bundles over topological space...
Are there similar results about reflexives modules?
| https://mathoverflow.net/users/2040 | Differences between reflexives and projectives modules | Well, the answer is well known of course. For a finitely generated module over a commutative normal Noetherian domain TFAE
1. M is reflexive
2. M is torsion-free and equals the intersection of its localizations at the codimension 1 primes
3. M satisfies Serre's condition S2 and its support is Spec R.
4. M is the dual of a f.g. module N
As you say, a finite projective module is the same as a locally free sheaf on Spec R. Similarly, a finite reflexive module is the same as the push forward of a locally free sheaf from an open subset U of Spec R whose complement has codimension $\ge2$.
So for an easy example take a Weil divisor D which is not Cartier, the associated divisorial sheaf (corresponding to an R-module of rank 1) is reflexive, not projective. Your example with a line on a quadratic cone is of this form.
This stuff is standard and used all the time in higher-dimensional algebraic geometry around the Minimal Model Program. For an old reference covering some of this, see e.g. Bourbaki, Chap.7 Algebre commutative, VII.
| 16 | https://mathoverflow.net/users/1784 | 7588 | 5,193 |
https://mathoverflow.net/questions/7585 | 5 | Are any finitely generated reflexive module a second syzygy?
(I´m thinking especially in normal noetherian domains)
More general...
Are any divisorial lattice a second syzygy?
(I´m thinking especially in Krull domains)
| https://mathoverflow.net/users/2040 | Are any finitely generated reflexive module a 2nd syzygy? | Over a normal domain (in fact, you only need Gorenstein in codimension 1, being second syzygy and reflexive are equivalent). This is Theorem 3.6 of Evans-Griffith "Syzygies" book.
| 6 | https://mathoverflow.net/users/2083 | 7592 | 5,197 |
https://mathoverflow.net/questions/7304 | 41 | Is there an example of a complex bundle on $\mathbb CP^n$ or on a Fano variety (defined over complex numbers), that does not admit a holomorphic structure? We require that the Chern classes of the bundle are $(k,k)$ Hodge classes (which is automatic for $\mathbb CP^n$ or Fanos of dimension<4). If by any chance such examples are known, what is the smallest dimension of the variety (or the bundle)?
For $\mathbb CP^1$ it is elementary to see that all bundles are holomorphic. In the book of Okonek and Schneider it is stated, that all complex bundles on $\mathbb CP^2$ and $\mathbb CP^3$ are also holomorphic. But for $\mathbb CP^n$, $n\ge 4$ this is stated as an open problem (as for 1980).
| https://mathoverflow.net/users/943 | Complex vector bundles that are not holomorphic | Here is the answer to the question, kindly explained to me by Burt Totaro.
EDITED. This is an OPEN PROBLEM.
0) Apparently in the case of $\mathbb CP^n$ existence of a complex bundle without holomorphic structure is still an OPEN PROBLEM. Though it is believed that there should be plenty of examples starting from $n\ge 5$, coming from topologically indecomposable rank two bundles, apparently no such bundle was proven to be non-holomorphic as for today.
1) A topologically non-trivial rank 2 complex bundle with $c\_1=0$, $c\_2=0$ was constructed in
Rees, Elmer, Some rank two bundles on ${\rm P}\_{n}\mathbb C$, whose Chern classes vanish. Variétés analytiques compactes (Colloq., Nice, 1977).
It was also claimed in this article that this bundle does not admit a holomorphic structure. But this claim was deduced from an article that contained a gap. So it is yet unknown if this particular bundle has holomorphic structure or not. This is discussed in
M. Schneider. Holomorphic vector bundles on ${\rm P}^n$. Seminaire Bourbaki
1978/79, expose 530.
This is why Okonek and Schneider write in their book p. 137 that this is an open problem.
2) On the positive side it is proven that every complex vector bundle on a smooth projective rational 3-fold has an holomorphic structure.
C. Banica and M. Putinar. On complex vector bundles on projective
threefolds. Invent. Math. 88 (1987), 427-438.
3) If one wants to construct examples of bundles on projective manifolds that are not necessarily Fanos it is possible to use the fact that the integral Hodge conjecture fails. Namely there are elements in $H^{2p}(X,\mathbb Z)$ which are in $H^{p,p}$
but which are not represented by an algebraic cycle. Kollar gave
such examples with $dim(X)=3$. A recent reference, which refers back to earlier results, is:
C. Soule and C. Voisin. Torsion cohomology classes and algebraic cycles
on complex projective manifolds. Adv. Math. 198 (2005), 107-127
4) One reason to expect examples of such bundles in higher dimensions is
Schwarzenberger's conjecture that every rank-2 algebraic vector bundle $E$,
on $\mathbb CP^n$ with $n\ge 5$ is a direct sum of two line bundles. So, for example,
if $c\_1(E)=0$ then $c\_2(E)=-d^2$ for some integer $d$, according to the
conjecture.
| 34 | https://mathoverflow.net/users/943 | 7596 | 5,200 |
https://mathoverflow.net/questions/7586 | 22 | Here $\zeta(s)$ is the usual Riemann zeta function, defined as $\sum\_{n=1}^\infty n^{-s}$ for $\Re(s)>1$.
Let $A\_n=${$s\;:\;\zeta(s)=n$}. The behaviour of $A\_0$ is basically just the Riemann hypothesis; my question concerns $A\_n$ for $n\neq0$.
1) Is determining this just as hard as the Riemann hypothesis?
2) If we know the behaviour of some $A\_n$, does it help in deducing the behaviour of other $A\_m$?
3) For which $n$ is $A\_n$ non-empty?
Question 3 has now been answered for all strictly positive $n$ - it is non-empty, and has points on the real line to the left of $s=1$. For $n=0$, it is known to be non-empty. Any idea for negative $n$? (the same answer won't work, since $\zeta(s)$ is strictly positive on the real line to left of $s=1$.
Big Picard gives it non-empty for all but at most one $n$. How can we remove the 'at most one'?
| https://mathoverflow.net/users/385 | When does the zeta function take on integer values? | Regarding 3), this "Big Picard" stuff is serious overkill.
Think like an undergraduate real analysis student:
The p-series $\zeta(p)$ converges for real $p > 1$, whereas $\zeta(1)$ = sum of the harmonic series = oo.
An easy argument using (e.g.) the integral test shows that
$$lim\_{p \rightarrow \infty} \zeta(p) = 1$$
The function $\zeta(p)$ is continuous in p [the convergence is uniform on right half-planes, hence on compact subsets], so by the intermediate value theorem it takes on every positive integer value $n \ge 2$ at least once -- and, since it is a decreasing function of p, exactly once -- on the real line.
Thus $A\_n$ is nonempty for all $n > 1$.
EDIT: Let me show that zeta(s) takes on all real values infinitely many times on the negative real axis.
For this, note that for all $n > 0$,
$$\zeta(-(2n-1)) = - \frac{B\_{2n}}{(2n)}$$,
where $B\_{2n}$ is the $(2n)$th Bernoulli number. It is known that the $B\_{2n}$'s alternate in sign and grow rapidly in absolute value:
$$|B\_{2n}| \sim 4 \sqrt{\pi n} \left(\frac{n}{(\pi e)^{2n}}\right)$$
The claim follows from this and the Intermediate Value Theorem.
| 29 | https://mathoverflow.net/users/1149 | 7600 | 5,204 |
https://mathoverflow.net/questions/7603 | 21 | From [a post](https://mathoverflow.net/questions/7374/the-jouanolou-trick/7602#7602) to [The Jouanolou trick](https://mathoverflow.net/questions/7374/the-jouanolou-trick/):
>
> Are all **topologically trivial** (contractible) **complex algebraic varieties** necessarily affine? Are there examples of those not birationally equivalent to an affine space?
>
>
>
The examples that come to my mind are similar to a singular $\mathbb P^1$ without a point given by equation $x^2 = y^3$. This particular curve is clearly birationally equivalent to affine line.
Perhaps the "affine" part follows from a comparison between Zariski cohomology and complex cohomology?
| https://mathoverflow.net/users/65 | Topologically contractible algebraic varieties | No. Counterexamples were first constructed by Winkelmann, as quotients of $\mathbb A^5$ by algebraic actions of $\mathbb G\_{\text{a}}$. I learned this from Hanspeter Kraft's very nice article available here:
[Challenging problems on affine $n$-space](http://www.numdam.org/numdam-bin/item?id=SB_1994-1995__37__295_0).
Recently Aravind Asok and Brent Doran have been studying these kinds of examples in the setting of $\mathbb A^1$-homotopy theory, on the arxiv as [On unipotent quotients and some A^1-contractible smooth schemes](http://arxiv.org/abs/math/0703137).
| 26 | https://mathoverflow.net/users/1048 | 7606 | 5,208 |
https://mathoverflow.net/questions/7441 | 7 | Not sure how to tag this one so feel free to edit and add tags.
When I initially started graduate school my choice for an area of study was quite nebulous. I had only figured out enough to know that I wanted to do some work involving a lot of category theory. So when I applied to schools I figured I could find some interesting topic to work on since almost anything involves category theory. Now I'm a bit more mature and I have a much better perspective and a much better idea of what I would like to do. My question then is how do I go about finding an adviser working in an area I would like to specialize in and if this person is not at my current school what are my possible courses of action?
| https://mathoverflow.net/users/nan | Choice of adviser | I strongly recommend finding a good advisor (someone who you get along with, who has compatible understanding of how hands-on the advisor will be, who will keep you funded, who can get you a postdoc, who actually wants a student etc.) over choosing a particular subfield. There's too little correlation between what you learn about a subject as a starting graduate student and what it's like to do research in that field for you to make decisions on that basis. That said, it's reasonable to narrow your search to broad fields (say "I want to do algebra") because that still allows a huge range of subfields to work in. As long as your tastes are reasonably broad (and if they're not you should work on broadening them) you and your advisor should be able to find something that interests you both for you to work on no matter what your advisor's particular speciality.
| 15 | https://mathoverflow.net/users/22 | 7611 | 5,213 |
https://mathoverflow.net/questions/7604 | 7 | As is well known, the line bundles over \**CP*\*$^1$ are indexed by the integers. My question is how are the line bundles over \**CP*\*$^n$, $n > 1$, and \**Gr*\*$(n,k)$ indexed? Moreover, do there exist any other interesting classifications of line bundles over spaces (I remember something about Atiyah and elliptic curves)?
| https://mathoverflow.net/users/1648 | Indexing the line bundles over a Grassmannian. | Algebraic line bundles on a smooth variety $X$ are classified by the Picard group $Pic(X) = H^1(X, \mathbf O\_X^\*)$. This is an exercise in Hartshorne's book, basically every line bundle is mapped to it's gluing cocycle.
The group $Pic(X)$ is also equal to the group $CH^1(X)$ of divisors modulo rational equivalence. The map sends a line bundle to it's first Chern class.
Now for a projective space Picard group is $\mathbf Z$ generated by $\mathbf O(1)$.
Picard group of a Grassmannian is also $\mathbf Z$. I believe that the generator is a pullback of $\mathbf O(1)$ for a Plucker embedding $Gr(n,k) \subset \mathbf P^N$.
One can prove it as follows: both $\mathbf P^n$ and $Gr(n,k)$ are algebraically cellular, meaning that they consist of pieces isomorphic to affine spaces: for $\mathbf P^n$ this is obvious, and for $Gr(n,k)$ the cells are Schubert cells.
For such varieties Chow groups coincide with cohomology groups and are generated by cells (it's like computing cohomology of CW complex without cells of odd dimension).
Since there is only one cell of complex codimension 1 for projective spaces and Grassmannians, we get $Pic(X) = H^2(X, \mathbf Z) = \mathbf Z$.
EDIT:
* Actually, a projective space IS a Grassmannian. :)
* As other people commented, the generator of the $Pic(Gr(n,k))$ is the n'th wedge power of the canonical vector bundle of rank n.
| 18 | https://mathoverflow.net/users/2260 | 7614 | 5,216 |
https://mathoverflow.net/questions/7523 | 21 | This question is closely related to [this one](https://mathoverflow.net/questions/124/is-every-finite-dimensional-lie-algebra-the-lie-algebra-of-an-algebraic-group).
Ado's theorem states that given a finite-dimensional Lie algebra $\mathfrak g$, there exists a faithful representation $\rho\colon\mathfrak g \to \mathfrak{gl}(V)$, with $V$ a finite-dimensional vector space. In the real or complex case one can take the exponent of the image and obtain a (virtual) Lie subgroup $\exp\rho(\mathfrak g)$ in $GL(V)$ having Lie algebra $\rho(\mathfrak g)$. But nothing guarantees that this subgroup will be closed in $GL(V)$.
So the question is: is every finite-dimensional Lie algebra the Lie algebra of some closed linear Lie group? I am primarily interested in the real and complex case, but it might be interesting to ask what happens in the ultrametric case as well.
| https://mathoverflow.net/users/2164 | Is every finite-dimensional Lie algebra the Lie algebra of a closed linear Lie group? | I think that the answer is yes. It looks like you can prove it by relying on a convenient proof of Ado's theorem.
Procesi's book, "Lie groups: an approach through invariants and representations", has the following theorem preceding the proof of Ado's theorem:
**Theorem 2.** Given a Lie algebra $L$ with semismiple part $A$, we can embed it into a new Lie algebra $L'$ with the following properties:
1. $L'$ has the same semismiple part $A$ as $L$.
2. The solvable radical of $L'$ is decomposed as $B' \oplus N'$, where $N'$ is the nilpotent radical of $L'$, $B'$ is an abelian Lie algebra acting by semisimple derivations, and $[A, B'] = 0$.
3. $A \oplus B'$ is a subalgebra and $L' = (A \oplus B') \ltimes N'$.
With all of that, the idea is to first prove the refinement of Ado's theorem for $L'$. We need a particular refinement: Let $\tilde{A}$ be the maximal algebraic semisimple Lie group with Lie algebra $A$, and let $\tilde{B'}$ and $\tilde{N'}$ be the contractible Lie groups with Lie algebras $B'$ and $N'$. If we can find a closed embedding of $(\tilde{A} \times \tilde{B'}) \ltimes \tilde{N'}$ in a matrix group, then it will restrict to a closed embedding of the Lie subgroup of the original $L$.
In the proof of Ado's theorem that follows, the action of $N'$ is nilpotent, so the representation of $\tilde{N'}$ is closed and faithful. The Lie algebra $L'$ has a representation which is trivial on $B'$ and $N'$ and generates $\tilde{A}$. It has another representation which is trivial on $N'$ and $A$ and for which the action of $B'$ is nilpotent. If I have not made a mistake, the direct sum of these three representations is the desired representation of $L'$.
| 9 | https://mathoverflow.net/users/1450 | 7623 | 5,222 |
https://mathoverflow.net/questions/7626 | 13 | This seems like an obvious fact, but I'm not sure what the necessary meaning of "nice" is to get a result like this. I'm wondering if there is a theorem of the form:
For any <1> field extension $K/F$, a map from $\phi:K\rightarrow F$ that satisfies <2> is the field norm (or trace).
where <1> could be something like finite, algebraic, etc., and <2> could be anything (obviously there would be different <2>'s for norm and trace).
| https://mathoverflow.net/users/1916 | Are the field norm and trace the unique "nice" maps between fields? | The field norm and trace exist when $K$ is a finite algebraic extension of $F$. In this case, an element $\alpha \in K$ can be interpreted as an $F$-linear map on $K$ by multiplication. The field norm is just the determinant of $\alpha$ as a linear map, while the trace is the trace of $\alpha$ as a linear map. This yields an evident generalization: Norm and trace are part of a family of nice maps, namely the coefficients of the characteristic polynomial of $\alpha$.
---
Since Zev asks for a uniqueness theorem in the comments, here is one that shows both the merits and limitations of the characteristic polynomial as an answer.
For simplicity let $F$ have characteristic 0. Let $K$ be a field extension of degree $n$ which is generic in the sense that the Galois group is $S\_n$. Then any Galois-invariant polynomial in $\alpha \in K$ and its Galois conjugates, is a symmetric polynomial. The theorem is that the algebra of symmetric polynomials is generated by elementary symmetric polynomials, which are exactly the coefficients of the characteristic polynomial of $\alpha$. (This is using the fact that the eigenvalues of $\alpha$ as a map are itself and its Galois conjugates.) In particular, the trace is the unique linear such map up to a scalar; and any multiplicative polynomial of this type is a power of the norm. You can also describe the norm as the last Galois-invariant polynomial (the one of degree $n$) that provides new information.
But if the Galois group is smaller, then the ring of invariant polynomials in $\alpha$ and its Galois conjugates is larger, and any of these other invariant polynomials is also "nice". These extras are somewhat hidden by the fact that, for any Galois group, the trace is still the only linear example and the norm is still the only multiplicative example.
Well, the original question was open-ended. I think that this answer does fit one interpretation of the question, but maybe it is too standard and maybe there are also other interesting answers.
| 27 | https://mathoverflow.net/users/1450 | 7628 | 5,223 |
https://mathoverflow.net/questions/7624 | 0 | I am puzzled with the following discrete logarithm problem:
Given positive integers `b, c, m` where `(b < m) is True` it is to find a positive integer `e` such that
```
(b**e % m == c) is True
```
where two stars is exponentiation (e.g. in Ruby, Python or ^ in some other languages) and % is modulo operation. Using general math symbols it looks like:($b^e \equiv c (\mod m)$).
What is the most effective algorithm (with the lowest big-O complexity) to solve it ?
Example:
Given b=5; c=8; m=13 this algorithm must find e=7 because 5\*\*7%13 = 8
Thank you in advance!
| https://mathoverflow.net/users/2266 | The Discrete Logarithm problem | The question is not phrased to our taste at mathoverflow, but the user has a point that this particular Wikipedia page is under-developed. As David Speyer suggests, it is a very different problem for very large primes than for small ones. For small primes the simplest algorithms described in Wikipedia are probably the most appropriate. If the question is instead about the theoretical time complexity, see [this review article](http://www.math.leidenuniv.nl/~psh/ANTproc/12oliver.pdf).
| 5 | https://mathoverflow.net/users/1450 | 7629 | 5,224 |
https://mathoverflow.net/questions/7569 | 2 | How would one classify the strata for the standard nilpotent cone for $GL\_{k}(\mathbb{C})$, using the definition from Hesselink's paper "[Desingularizations of Varieties of Nullforms](https://doi.org/10.1007/BF01390087)"? I know that they correspond to partitions / nilpotent orbits etc, but from first principles why aren't two different nilpotent orbits possibly in the same strata - how would you prove that? (preferably using the definition of Hesselink)
I would like to classify the strata for the problem I'm working on, but don't completely understand how to do it for the more basic case (for which the method is probably well-known), I get stuck on the details, so that would be very helpful. Thanks.
| https://mathoverflow.net/users/2623 | Classifying strata for the adjoint representation of GL from first principles | I'm going to give a partial answer here for two reasons: (1) I am lazy and (2) this is starting to feel a little homeworky to me. Obviously, no one would assign this material as homework, but part of reading a math paper is taking the time to work out lots of simple examples and see how the definitions work. I feel like you are pushing the boundaries of how much of this work it is reasonable to ask other people to do. Not a major criticism, certainly not a vote to close the question, but my input.
---
On to the math. I've scanned the first 3 pages of Hesselink's paper. He make the following definitions. G acts on V, v is a point of V and $\star$ a chosen base point of V fixed by G. In your setting, G is $GL\_n$, V is the $n \times n$ matrices where G acts by conjugation, and $\star$ is zero. Hesselink writes Y(G) for what is essentially $\mathrm{Hom}(\mathbb{C}^\*, G)$. More precisely, Hesselink tensors with $\mathbb{Q}$, so that he can talk about maps like $t \mapsto \left( \begin{smallmatrix} t^{1/3} & 0 \\\\ 0 & t^{-2/7} \end{smallmatrix} \right)$. I'll ignore this detail.
For $\lambda \in Y(G)$, Hesselink defines a rational number $m(\lambda)$. We talked about this in [your previous question](https://mathoverflow.net/questions/7046/strata-for-the-nullcone-from-hesselinks-paper). In this setting, where V is an $N$-dimensional vector space, Hesselink gives an explicit formula for m on the bottom of page 142/top of page 143: Diagonalize the action of $\lambda$ as $t \mapsto \mathrm{diag}(t^{m\_1}, \cdots, t^{m\_N})$ and write $v = \sum v\_i e\_i$.. Then $m(\lambda) = \min(m\_i : v\_i \neq 0)$ if this number is nonnegative, and is $- \infty$ if this minimum is negative.
Let's see what this definition means in your setting. We can conjugate any $\lambda$ into diagonal form as $t \mapsto \mathrm{diag}(t^{c\_1}, \cdots, t^{c\_n})$. I've replaced $m\_i$ by $c\_i$ to point out that these $c$'s are not the $m$'s of the previous paragraph. In our notation, the $N$ of the previous paragraph is $n^2$. The vector space $V$ has dimension $n^2$ with basis $e\_{ij}$. The action of $\lambda(t)$ on $e\_{ij}$ is by $t^{c\_i - c\_j}$. (**Exercise!**).
So $m(\lambda) > 0$ if and only if $c\_i \leq c\_j$ implies $v\_{ij} =0$.
We may as well order our basis such that $c\_1 \geq c\_2 \geq \cdots \geq c\_n$.
If $c\_1 > c\_2 > \cdots >c\_n$ then we see that $m(\lambda) > 0$ if and only if $v$ is a strictly upper triangular matrix. When there are some equalities among the $c$'s, you want $v$ to be strictly block upper triangular. For such a $v$, $m(\lambda) = \min(c\_i - c\_j : v\_{ij} \neq 0)$. In particular, notice that there exists a $\lambda$ such that $m(\lambda) > 0$ if and only if $v$ is nilpotent.
Hesselink defines $\Lambda(v)$ to be the locus in $\{ \lambda : m(\lambda) = 1 \}$ where $q(\lambda)$ is minimized, where $q$ is the inner product from your [previous question](https://mathoverflow.net/questions/7046/strata-for-the-nullcone-from-hesselinks-paper). What you want to show is that $\Lambda(v)$ determines the Jordan normal form of $v$.
I must admit that I haven't thought out how to prove this. But I hope this makes things explicit enough that you can attack it.
| 3 | https://mathoverflow.net/users/297 | 7636 | 5,229 |
https://mathoverflow.net/questions/7567 | 3 | Let $R\to S$ be a ring map such that $S$ is projective over $R$ (I am willing to assume $S=R[X\_1,...,X\_n]$). Let $M,N$ be finite $S$-modules. Let $P\in Spec R$ such that $M\_P$ is $R\_P$-flat. Under what condition can one say that $Ext^1\_R(M,N)\_P=0$?
This is trivial if $M$ is finite over $R$, but in general $Ext$ does not commute with localization. I would appreciate any reference on this matter.
| https://mathoverflow.net/users/2083 | When can one localize Ext? | I would suggest having a look at the article ["Compactifying the Picard Scheme"](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6W9F-4CRY60R-1H3&_user=2459438&_rdoc=1&_fmt=&_orig=search&_sort=d&_docanchor=&view=c&_acct=C000057302&_version=1&_urlVersion=0&_userid=2459438&md5=deca64ad5b5501727ed918a10bc1ffd9)
by Altman-Kleiman. They discuss base change issues for Ext. I'm not sure if this will be applicable in your particular context, but it may be a start.
| 5 | https://mathoverflow.net/users/2215 | 7651 | 5,237 |
https://mathoverflow.net/questions/7650 | 36 | This is a follow up to Harrison's question: [why planar graphs are so exceptional](https://mathoverflow.net/questions/7114/why-are-planar-graphs-so-exceptional). I would like to ask about (and collect answers to) various notions, in graph theory and beyond graph theory (topology; algebra) that generalize the notion of planar graphs, and how properties of planar graphs extend in these wider contexts.
| https://mathoverflow.net/users/1532 | Generalizations of Planar Graphs | I guess one possible generalization could be: an $m$-dimensional [stratified space](https://en.wikipedia.org/wiki/Topologically_stratified_space) (i.e. "manifold with singularities") which is embeddable in $2m$-dimensional Euclidean space. Every smooth manifold can be so embedded (by Whitney's theorem), but singularities may force the ambient dimension higher, as witnessed by the simple case $m=1$ in which "stratified space" is just a graph and the embeddability condition is just planarity.
(This was just invented on the spot - I have no idea if this is actually an interesting definition...)
| 17 | https://mathoverflow.net/users/25 | 7652 | 5,238 |
https://mathoverflow.net/questions/7639 | 21 | I'm looking for a reference which has the first statement of the twin prime conjecture. According to wikipedia, nova, and several other quasi-reputable resources it is Euclid who first stated it, but according to Goldston
<http://www.math.sjsu.edu/~goldston/twinprimes.pdf>
it was stated nowhere until de Polignac. I'm hoping to resolve this issue by accessing either primary historical documents, or other reputable secondary sources (Goldston being one such example). I have looked at de Polignac's work, and he does indeed make a conjecture, but have been unable to find anything definitive (besides Goldston's statements) that there was no conjecture earlier. If this is too specific for MO, I'll remove the question. Thank you.
| https://mathoverflow.net/users/2043 | Twin Prime Conjecture Reference | I don't have it to hand right at this moment, but Narkiewicz' *The Development of Prime Number Theory* is excellent on just this kind of question. It is a historiomathematical survey of prime number theory up to 1910, and also has discussions of later developments directly related to work done before 1910. It is historical, with very many references, and mathematical, in that it sketches many old proofs. It even has exercises.
| 9 | https://mathoverflow.net/users/3304 | 7659 | 5,240 |
https://mathoverflow.net/questions/7320 | 35 | We know that presheaves of any category have enough projectives and that sheaves do not, why is this, and how does it effect our thinking?
[This](https://mathoverflow.net/questions/5378/when-are-there-enough-projective-sheaves-on-a-space-x) question was asked(and I found it very helpful) but I was hoping to get a better understanding of why.
I was thinking about the following construction(given during a course);
given an affine cover, we normally study the quasi-coherent sheaves, but in fact we could study the presheaves in the following sense:
Given an affine cover of X,
$Ker\_2\left(\pi\right)\rightrightarrows^{p\_1}\_{p\_2} U\rightarrow X$
then we can define $X\_1:=Cok\left(p\_1,p\_2\right)$, a presheaf, to obtain refinements in presheaves where we have enough projectives and the quasi-coherent sheaves coincide. Specifically, if $X\_1\xrightarrow{\varphi}X$ for a scheme $X$, s.t. $\mathcal{S}\left(\varphi\right)\in Isom$ for $\mathcal{S}(-)$ is the sheaffication functor, then for all affine covers $U\_i\xrightarrow{u\_i}X$ there exists a refinement $V\_{ij}\xrightarrow{u\_{ij}}U\_i$ which factors through $\varphi$.
This hinges on the fact that $V\_{ij}$ is representable and thus projective, a result of the fact that we are working with presheaves. In sheaves, we would lose these refinements. Additionally, these presheaves do not depend on the specific topology(at the cost of gluing).
In this setting, we lose projectives because we are applying the localization functor which is not exact(only right exact). However, I don't really understand this reason, and would like a more general answer.
A related appearance of this loss is in homological algebra. Sheaves do not have enough projectives, so we cannot always get projective resolutions. They do have injective resolutions, and this is related to the use of cohomology of sheaves rather than homology of sheaves. In paticular, in Rotman's Homological Algebra pg 314, he gives a footnote;
>
> In *The Theory of Sheaves,* Swan writes "...if the base space X is not discrete, I know
> of no examples of projective sheaves except the zero sheaf." In Bredon, *Sheaf Theory*:
> on locally connected Hausdorff spaces without isolated points, the only
> projective sheaf is 0
>
>
>
addressing this situation.
>
> In essence, my question is for a
> heuristic or geometric explanation of
> why we lose projectives when we pass
> from presheaves to sheaves.
>
>
>
Thanks in advance!
| https://mathoverflow.net/users/348 | Heuristic explanation of why we lose projectives in sheaves. | This is pretty much Dinakar's answer from a different view point: He says that it is too easy for a sheaf morphism to be an epi, so, since there are so many epis, it is now a stronger requirement that for *every* epi we find a lift - so strong that is not satisfied most of the times. I just want to call attention to the fact that this problem has nothing to do with module sheaves but is about sheaves of sets - and as such has the following nice interpretation:
The condition of being a projective module sheaf can be split in two conditions: That of existence of the lifting map as a *morphism of sheaves of sets* and that of it being a *morphism of module sheaves*.
In the category of sets the first condition is always satisfied; we have the axiom of choice which says that every epimorphism has a section and composing the morphism from our would-be projective with this section produces a lift - set-theoretically. Then one has to establish that one such lift is a module homomorphism.
But in a sheaf category step one can fail. Sheaves (of sets) are objects in the category of sheaves. This category is a topos and can be seen as an intuitionistic set-theoretic universe (in a precise sense: there is a sound and complete topos semantics for intuitionistic logic, see e.g. [this book](http://books.google.com/books?id=SGwwDerbEowC&pg=PA454&dq=moerdijk+maclane&client=firefox-a&hl=de#v=onepage&q=moerdijk%20maclane&f=false)). Now in an intuitionistic universe of sets, the axiom of choice is not valid in general; there might not be a *"set-theoretic"* section of the epimorphism!
| 26 | https://mathoverflow.net/users/733 | 7661 | 5,241 |
https://mathoverflow.net/questions/7656 | 71 | Defining $$\xi(s) := \pi^{-s/2}\ \Gamma\left(\frac{s}{2}\right)\ \zeta(s)$$ yields $\xi(s) = \xi(1 - s)$ (where $\zeta$ is the Riemann Zeta function).
Is there any conceptual explanation - or intuition, even if it cannot be made into a proof - for this? Why of all functions does one have to put the Gamma-function there?
Whoever did this first probably had some reason to try out the Gamma-function. What was it?
(Best case scenario) Is there some uniform way of producing a factor out of a norm on the rationals which yields the other factors for the p-adic norms and the Gamma factor for the absolute value?
| https://mathoverflow.net/users/733 | Why does the Gamma-function complete the Riemann Zeta function? | To the best of my understanding, the answer is yes, and this uniform way consists of some integration over the local field. This is explained in John Tate's dissertation. One starts with a certain smooth rapidly decreasing function, for which one takes the characteristic function of the p-adic integers in the nonarchimedean case and the function $e^{-|x|^2}$ for an archimedean field. This is being multiplied with $|x|^s$ (approximately) and integrated over the Haar measure of the additive group of the field. This produces the $\Gamma$-factor for an archimedean field and $(1-p^{-s})^{-1}$ for a p-adic field.
| 57 | https://mathoverflow.net/users/2106 | 7662 | 5,242 |
https://mathoverflow.net/questions/7645 | 4 | Let $\alpha\in(0,1)$ and $\eta\in\Lambda\_0^\alpha(\mathbb{R})$ be a compactly supported Hölder continuous function of order $\alpha$. I would like to show that, for any $n\in\mathbb{N}$, it is possible to decompose $$\eta=f+g$$
in such a way that $f\in C^n(\mathbb{R})$ and $||f||\_{C^n}=O(R^C)$, and $g\in L^\infty(\mathbb{R})$ with $\|g\|\_{L^\infty}=O(R^{-1})$.
Here $C$ is a universal constant. On the other hand, the real parameter $R$ can be chosen as large as we want (at the expense of increasing $\|f\|\_{C^n}$).
Thank you!
| https://mathoverflow.net/users/2274 | Decomposition of Hölder continuous functions | I have carried out the suggestion in the last paragraph of Yemon Choi's answer. Choose $\phi\in C^\infty(\mathbb{R})$, $\phi\ge0$ and $\int\_{\mathbb{R}}\phi(x)dx=1$, and let $\phi\_R(x)=R\phi(Rx)$. Define
$$ f=\phi\_R\star\eta,\quad g=\eta-f.$$
Then it is easy to see that
$$ \|f\|\\_{C^n}=O(R^{n-\alpha}),\quad \|g\|\\_\infty=O(R^{-\alpha}),$$
but this is not what you are asking for.
My feeling is that the constant $C$ must show some dependence on $n$.
In response to your last comment, let me prove the estimate on $\|f\|\_{C^n}$. We have
$$f^{(n)}=(\phi\_R)^{(n)}\star\eta=R^n(\phi^{(n)})\_R\star\eta.$$
Since $(\phi^{(n)})\_R$ has mean zero, for any $x\in\mathbb{R}$:
$$ |f^{(n)}(x)|\le R^n\int\_{\mathbb{R}}|\phi^{(n)}(y)||\eta(x-\frac{y}{R})-\eta(x)|dy\le HR^{n-\alpha}\int\_{\mathbb{R}}|\phi^{(n)}(y)||y|^\alpha dy,$$
where $H$ is $\eta$'s Hölder constant.
| 3 | https://mathoverflow.net/users/1168 | 7671 | 5,247 |
https://mathoverflow.net/questions/7647 | 4 | Given a polyhedron consists of a list of vertices (`v`), a list of edges (`e`), and a list of surfaces connecting those edges (`s`), how to break the polyhedron into a list of tetrahedron?
I have a convex polyhedron.
| https://mathoverflow.net/users/807 | Break polyhedron into tetrahedron | If I understand your question correctly, you're saying that the given information is the face structure of a 3-dimensional convex polytope, and you would like a subdivision of the polytope into tetrahedra.
Here is one way to proceed. First, subdivide all the faces into triangles. Then pick your favourite vertex $v\_0$. Connect $v\_0$ to each triangle belonging to a face not containing $v\_0$. This subdivides your polytope into tetrahedra.
| 8 | https://mathoverflow.net/users/468 | 7672 | 5,248 |
https://mathoverflow.net/questions/7666 | 21 | Lax functors of bicategories were introduced at the very inception of bicategories, and I'm trying to get a better feel for them. They are the same as ordinary 2-functors, but you only require the existence of a coherence morphism, not an isomorphism. The basic example I'm looking at are when you have a lax functor from the singleton bicategory to a bicategory B. These are just object b in B with a monad T on B.
My Question: If I have an equivalence of bicategories A ~ A', do I a get equivalent bicategories of lax functors Fun(A, B) and Fun(A', B)? If not, is there any relation between these two categories?
[Edit]
So let me be more precise on the terminology I'm using. I want to look at lax functors from A to B. These are more general then strong/pseudo and much more general then just strict functors. For a lax functor we just have a map like this:
$ F(x) F(g) \to F(fg)$
for a strong or pseudo functor this map is an isomorphism, and for strict functor it is an identity. I don't care about strict functors.
I'm guessing that these form a bicategory Fun(A,B), with the 1-morphism being some sort of lax natural transformation, etc, but I don't really know about this. Are there several reasonable possibilities?
When I said equivalence between A and A' what I meant was I had a strong functor F:A --> A' and a strong functor G the other way, and then equivalences (not isomorphisms) FG = 1, and GF = 1. This seems like the most reasonable weak notion of equivalence to me, but maybe I am naive.
I haven't thought about equivalences using lax functors. Would they automatically be strong? What I really want to understand is what sort of functoriality the lax functor bicategories Fun(A, B) have?
| https://mathoverflow.net/users/184 | Lax Functors and Equivalence of Bicategories? | First of all, for any two bicategories A and B, there is a bicategory $Fun\_{x,y}(A,B)$ where x can denote either strong, lax, or oplax functors, and y can denote either strong, lax, or oplax transformations. There's no problem defining and composing lax and oplax transformations between lax or oplax functors, and the lax/oplax-ness doesn't even have to match up. It's also true that two x-functors are equivalent in one of these bicategories iff they're equivalent in any other one. That is, any lax or oplax transformation that is an equivalence is actually strong/pseudo.
Where you run into problems is when you try to compose the functors. You can compose two x-functors and get another x-functor, but in general you can't whisker a y-transformation with an x-functor unless x = strong, no matter what y is, and moreover if y isn't strong, then the interchange law fails. Thus you only get a tricategory with homs $Fun\_{x,y}(A,B)$ if x=y=strong. (In particular, I think this means that there isn't a good notion of "equivalence of bicategories" involving lax functors.)
For a fixed strong functor $F\colon A\to A'$, you can compose and whisker with it to get a functor $Fun\_{x,y}(A',B) \to Fun\_{x,y}(A,B)$ for any x and y. However, the same is not true for transformations $F\to F'$, and the answer to your question is (perhaps surprisingly) no! The two bicategories are not equivalent.
Consider, for instance, A the terminal bicategory (one object, one 1-morphism, one 2-morphism) and A' the free-living isomorphism, considered as a bicategory with only identity 2-cells. The obvious functor $A' \to A$ is an equivalence. However, a lax functor from A to B is a monad in B, and a lax functor from A' to B consists of two monads and a pair of suitably related "bimodules". If some lax functor out of A' is equivalent to one induced by composition from A (remember that "equivalence" doesn't depend on the type of transfomation), then in particular the two monads would be equivalent in B, and hence so would their underlying objects. But any adjunction in B whose unit is an isomorphism gives rise to a lax functor out of A', if we take the monads to be identity 1-morphisms, the bimodules to be the left and right adjoint, and the bimodule structure maps to be the counit and the inverse of the unit. And of course can have adjunctions between inequivalent objects.
By the way, I think your meaning of "equivalence" for bicategories is becoming more standard. In traditional literature this sort of equivalence was called a "biequivalence," because for strict 2-categories there are stricter sorts of equivalence, where you require either the functors to be strict, or the two composites to be isomorphic to identities rather than merely equivalent to them, or both. These stricter notions don't really make much sense for bicategories, though. For instance, in a general bicategory, even identity 1-morphisms are not isomorphisms, so if "equivalence" were to demand that FG be isomorphic to the identity, a general bicategory wouldn't even be equivalent to itself!
| 18 | https://mathoverflow.net/users/49 | 7702 | 5,270 |
https://mathoverflow.net/questions/7689 | 17 | I was in a lecture not long ago given by C. Teleman and at some point he said "Well, since Riemann-Roch is an index problem we can do..."
Then right after that he argued in favour of such a sentence. Could anyone tell me what did he mean exactly?. That is to say, in this case what is elliptic operator like, what is the heuristic idea which such a result relies on? ...and a little bit of more details about it.
As usual references will be appreciated.
**ADD:** Thanks for the comments below, but I think they do not answer the question of title : **Why** is RR an Index problem?. Up to this point, what I can see is that two numbers happened to be the same.
| https://mathoverflow.net/users/1547 | Why is Riemann-Roch an Index Problem? | Here is a sketch of the argument as I learned it in a complex analysis class: For a Riemann surface $X$ and a holomorphic line bundle $L$, we want
$$\text{dim}H^0(X,L)-\text{dim}H^0(X,L\otimes\Lambda^{0,1})=c\_1(L)+\frac{1}{2}\chi(X)$$
You have an operator $\overline{\partial}$ (differentiation with respect to $d\overline{z}$ taking $\Gamma(X,L)$ to $\Gamma(X,L\otimes\Lambda^{0,1})$. Then $H^0(X,L)$ is the kernel of $\overline{\partial}$ and $H^0(X,L\otimes\Lambda^{0,1})$ is the kernel of its adjoint, $\overline{\partial}^+$. Now define $\Delta^+=\overline{\partial}\overline{\partial}^+$ and $\Delta^-=\overline{\partial}^+\overline{\partial}$. Their spectra are the same, except for the kernels, and we get
$$\text{Tr}(e^{-t\Delta^+})-\text{Tr}(e^{-t\Delta^-})=\text{dim}(\text{ker}\Delta^+)-\text{dim}(\text{ker}\Delta^-)$$
We also have that the kernel of $\overline{\partial}$ is the kernel of $\Delta^-$, and the kernel of $\overline{\partial}^+$ is the kernel of $\Delta^+$, so it's enough to get your hands on the left-hand side. Then you write those traces as integrals of heat kernels, take the limit as $t\rightarrow 0^+$, and show that the integrals go to $c\_1(L)+\frac{1}{2}\chi(X)$. And that's possible because we can interpret Chern classes and Euler characteristics of Riemann surfaces as integrals of curvatures of line bundles. Of course, then there's more work to turn $c\_1(L)+\frac{1}{2}\chi(X)$ into it's more familiar form.
| 22 | https://mathoverflow.net/users/88 | 7710 | 5,277 |
https://mathoverflow.net/questions/7668 | 6 | If I remembered correctly. There are some work done by C.M.Ringel,he defined so called Ringel-Hall algebra on abelian category and then show that Ringel-hall algebra is isomorphic to positive part of quantized enveloping algebra. Some others generalized to triangulated(derived category of coherent sheaves on projective spaces,so on so forth)
I wonder know whether there exists some explicit geometric explanation to these works. I think we can consider the quantized enveloping algebra as noncommutative affine scheme. Therefore, it seems these work are doing some sort of reconstruction of schemes in abelian category or derived category.
I wonder whether Bondal-Orlov's work has any relationship(explicit)to these stuff. Therefore, what I am really interested in is what is the real meaning of Ringel-Hall algebra
At last, it is well known that Belinson-Bernstein's theorem established the equivalence of category of U(g)-module and category of D-modules on flag variety of Lie algebra. And some others, say Bezrukavnikov,Frenkel,gaitsgory generalized these results to Kac-Moody algebra.On the other hand, Van den Berg used generalized Ringel Hall algebra to realize quantum group of Kac-Moody algebra. I wonder whether anybody here can say something about these stuff.
All the other comments are welcomed
| https://mathoverflow.net/users/1851 | What is the geometric meaning of reconstruction of quantum group via Ringel Hall algebra | I tend to think of the geometric content of Hall algebra construction as reflecting the categorification of quantum groups as studied by Lusztig, Rouquier, Khovanov-Lauda, etc. You can read [Lusztig's original paper](http://www.ams.org/mathscinet-getitem?mr=1035415), though I feel like a sadist even suggesting that. I wrote [a bit about](http://ncatlab.org/nlab/show/categorification+via+groupoid+schemes#the_hall_algebra_and_lusztigs_categorification_8) this in nLab recently; you might find that useful.
| 5 | https://mathoverflow.net/users/66 | 7724 | 5,288 |
https://mathoverflow.net/questions/7733 | 12 | I seem to remember written or said somewhere that at some point Thurston decided to stop writing down his theorems in order not to repel mathematicians from his field (maybe this is not correct?). I am really curious if now 25-30 years later there is some nice source, book, or notes, where it is possible to learn some basic ideas about the proof of the fact that Haken manifolds admit a hyperbolic structure? Maybe some of his ideas got a more accessible explanation? Of course his beautiful notes <http://www.msri.org/publications/books/gt3m/> exist, but they don't go so far.
| https://mathoverflow.net/users/943 | The work of Thurston | There are several sources for Thurston's hyperbolization theorem, some published, some not.
Off the top of my head:
1) M.Kapovich, Hyperbolic manifolds and discrete groups.
2) J. Hubbard's Teichmuller theory volume II (not yet published)
3) J. Morgan, H. Bass (eds). The Smith conjecture. (English)
Papers presented at the symposium held at Columbia University, New
York, 1979.
Pure and Applied Mathematics, 112.
Academic Press, Inc., Orlando, Fla., 1984. xv+243 pp.
For only the case of manifolds that fibre over S^1
1) J-P. Otal, The hyperbolization theorem for fibred 3-manifolds.
Of course there's also the new non-Thurston proofs using Ricci flow.
Oh, and regarding that anecdote about repelling people from a field -- I've only heard that comment attributed to one mathematician and it was in reference to Thurston's early work on foliations. I don't think that's a widely held belief, but I wasn't alive then so I'm just going on 2nd hand comments.
| 12 | https://mathoverflow.net/users/1465 | 7735 | 5,296 |
https://mathoverflow.net/questions/7677 | 2 | Start with a distribution $\mu$ on [n], and drop m balls into these n+1 slots independently and according to the distribution &mu. That is, we have iid random variables x 1 through x m distributed according to &mu. Assume that m is close to the same size as n.
I call a collection of tuples (a(i), b (i)), with all a(i), b(i) between 1 and m, a partial matching if no number is repeated, and for all i, x b(i) = x a(i) + 1.
My (purposefully vague) question is: Is there, with high probability, a partial matching that includes many of the balls?
Of course, for some distributions this obviously doesn't happen. I would be happy if somebody had a 'standard argument' for this type of question in nice cases, and maybe we could get some understanding of what makes it run (and so in what cases it doesn't).
Some side comments:
- We need that m be not much smaller than n, as otherwise most points will be fairly far apart.
- For nice distributions, such as the uniform or binomial, one can do calculations much like the birthday paradox, and get some semi-plausible answers. I don't know if these are best possible, and would love to hear it if somebody out there has a nice clearly-tight argument for nice distributions. I was thinking of just asking about the uniform distribution, as I guess somebody out there must have a beautiful argument.
| https://mathoverflow.net/users/2282 | Parity, Balls and Boxes | if $m=\alpha n$ and $\mu$ is uniform, it seems like a basic sub-additivity argument shows that $\frac{M\_n}{n}$ converges almost surely to a constant $C\_{\alpha}$, where $M\_n$ is the cardinal of a maximal partial match of $[n]$. To see that, put a Poisson process $P$ on the real line with intensity $\alpha$ and say that there are $P((k;k+1)) = Poisson(\alpha)$ balls in the slot $k$. Then, if $M\_{m,n}$ is the cardinal of a maximal partial match of $\{m,m+1, \ldots,n-1\}$ (with your notations), then $M\_{p,r} \geq M\_{p,q} + M\_{q,r}$ so that a Kingsman-like sub-additive theorem holds and give the conclusion.
Notice also that the number of missed slots is very concentrated around $n e^{-\alpha}$ (concentration of order $\sqrt{n}$) so that I would not be surprised if one could compute this constant $C\_{\alpha}$ .
| 2 | https://mathoverflow.net/users/1590 | 7745 | 5,305 |
https://mathoverflow.net/questions/7746 | 2 | Please, any information on the periodic mapping classes of the genus two orientable surface, $O\_2$, will be greatly thanked. We had been studying the topological structure of 3d surface bundles and reintrepreting them as a circle bundles over orbifolds.
In the <http://web.archive.org/web/20070316045651/http://www.smm.org.mx/SMMP/html/modules/Publicaciones/AM/Cm/35/artExp08.pdf> -work you would like see the cases $O\_1$, among $N\_1$ and $N\_2$, solved. Any feedback on the results and conjectures, some of them obviously false, will bring a lot of happiness :)
| https://mathoverflow.net/users/2196 | Periodic mapping classes of the genus two orientable surface | If you want to enumerate the finite-order automorphisms (up to conjugacy) I suggest the following exercise. The associated 3-manifold is Seifert fibred. So determine how the genus 2 surface is sitting in the Seifert manifold (horizontal incompressible surface).
This will give you a formula relating the various branch points of the monodromy to the Seifert data. Moreover, you should be able to go back-and-forth between the description of the Seifert-fibred space (unnormalized Seifert data, fibred over a genus 0, 1 or 2 surface) and the monodromy of the surface. So the classification of Seifert-fibred spaces basically gives you a dictionary for walking-through the finite-order automorphisms of a mapping class group.
| 4 | https://mathoverflow.net/users/1465 | 7747 | 5,306 |
https://mathoverflow.net/questions/7750 | 12 | Where can I find the most direct and simplest presentation of what geodesics on a (complex) Grassmannian look like? I know how to do it from scratch, but, if I want to provide a reference to, say, a graduate student in EE who doesn't want to deal with any unnecessary abstract mathematical machinery, what should I point him to?
| https://mathoverflow.net/users/613 | Geodesics on a Grassmannian | Grassmanians are symmetric spaces, and symmetric spaces are "geodesic orbit spaces", that is, their geodesics are orbits of their group of isometries. Your Grassmanians, in particular, are of the form $SU(p+q)/SU(p)\times SU(q)$. If $g$ is the Lie algebra of the big group and $h\subseteq g$ the Lie algebra of the subgroup, then there is a $SU(p)\times SU(q)$-invariant complement $p$ to $h$ in $g$. The geodesics are the orbits of the $1$-parameter subgroups of $SU(p+q)$ whose tangent vectors are in $p$.
So to compute the geodesics, you need only find that complement $p$ and compute exponentials...
| 12 | https://mathoverflow.net/users/1409 | 7757 | 5,314 |
https://mathoverflow.net/questions/7751 | 2 | Let $V$ be a vector space over $\mathbb R$, and $a: V\otimes V\to \mathbb R$ a symmetric bilinear pairing. Recall that the *Morse index* of $a$ is the maximal dimension of any subspace $V\_- \subseteq V$ on which $a$ is negative-definite.
If $V$ is infinite-dimensional, it can be very hard to check every negative-definite subspace and compare their dimensions. Much easier is to exhibit a subspace $V\_- \subseteq V$ on which $a$ is negative-definite, which is not contained within any other negative subspace. But how do I know that the dimension of *this* negative subspace is the maximal dimension of *any* negative subspace? The approach I thought worked fails; see [this question](https://mathoverflow.net/questions/7709/splitting-a-space-into-positive-and-negative-parts/).
| https://mathoverflow.net/users/78 | How can I measure the Morse index in infinite dimensions? | Let $V\_-$ be this subspace of $V$ defined in the weak sense that it is negative-definite and not extendible. There is a linear map $b:V \mapsto V\_-^\*$ given by the formula $b(v) = a(v,\cdot)$, where the right side is interpreted as a dual vector on $V\_-$. Since $V\_-$ is finite-dimensional by hypothesis, $\dim V\_- = \dim V\_-^\*$. Let $V\_-^\perp$ be the kernel of $b$. Now let $W$ be some other negative-definite subspace of $V$. I claim that $\dim W \le \dim V\_-$. Otherwise, $b$ has a kernel in $W$ because $W$ does not fit in the target of $b$. But if $w \in W \cap V\_-^\perp$ is non-zero, then $V\_-$ can be extended by $w$, contrary to the hypothesis.
Thus, any finite-dimensional negative-definite subspace which is not extendible has the maximum-dimension property. The proof has been arranged so that you never take the dual of $V$ itself, only of that of $V\_-$.
| 3 | https://mathoverflow.net/users/1450 | 7761 | 5,317 |
https://mathoverflow.net/questions/7715 | 16 | I am starting on a Phd program and am supposed to read Colliot Thelene and Sansuc's article
on R-equivalence for tori. I find it very difficult and although I have some knowledge over schemes , I am completely baffled by this scalar restriction business of having a field extension $K/k$ , a torus over $K$ and "restricting" it to $k$.
I would be very gratefull for a reference or even better by some explanation . I found nothing in my standard books (Hartshorne, Qing Liu, Mumford etc) so I hope this question is appropriate for the site. Thank you.
| https://mathoverflow.net/users/2292 | What is "restriction of scalars" for a torus? | As said, the sought after concept is also known as Weil restriction. In a word, it is the algebraic analogue of the process of viewing an $n$-dimensional complex variety as a $(2n)$-dimensional real variety.
The setup is as follows: let $L/K$ be a finite degree field extension and let $X$ be a scheme over $L$. Then the Weil restriction $W\_{L/K} X$ is the $K$-scheme representing the following functor on the category of K-algebras:
$A\mapsto X(A \otimes\_K L)$.
In particular, one has $W\_{L/K} X(K) = X(L)$.
By abstract nonsense (Yoneda...), if such a scheme exists it is uniquely determined by the above functor. For existence, some hypotheses are necessary, but I believe that it exists whenever $X$ is reduced of finite type.
Now for a more concrete description. Suppose $X = \mathrm{Spec} L[y\_1,...,y\_n]/J$ is an affine scheme. Let $d = [L:K]$ and $a\_1,...,a\_d$ be a $K$-basis of $L$. Then we make the following "substitution":
$$y\_i = a\_1 x\_{i1} + ... + a\_d x\_{id},$$
thus replacing each $y\_i$ by a linear expression in d new variables $x\_{ij}$. Moreover,
suppose $J = \langle g\_1,...,g\_m \rangle$; then we substitute each of the above equations into $g\_k(y\_1,...,y\_n)$ getting a polynomial in the $x$-variables, however still with $L$-coefficients. But now using our fixed basis of $L/K$, we can regard a single polynomial with $L$-coefficients as a vector of $d$ polynomials with $K$ coefficients. Thus we end up with $md$ generating polynomials in the $x$-variables, say generating an ideal $I$ in $K[x\_{ij}]$, and we put $\mathrm Res\_{L/K} X = \mathrm{Spec} K[x\_{ij}]/I$.
A great example to look at is the case $X = G\_m$ (multiplicative group) over $L = \mathbb{C}$ (complex numbers) and $K = \mathbb{R}$. Then $X$ is the spectrum of
$$\mathbb{C}[y\_1,y\_2]/(y\_1 y\_2 - 1);$$
put $y\_i = x\_{i1} + \sqrt{-1} x\_{i2}$ and do the algebra. You can really see that the
corresponding real affine variety is $\mathbb{R}\left[x,y\right]\left[(x^2+y^2)^{-1}\right]$, as it should be: see e.g.
p. 2 of
[http://alpha.math.uga.edu/~pete/SC5-AlgebraicGroups.pdf](http://alpha.math.uga.edu/%7Epete/SC5-AlgebraicGroups.pdf)
for the calculations.
Note the important general property that for a variety $X/L$, the dimension of the Weil restriction from $L$ down to $K$ is $[L:K]$ times the dimension of $X/L$. This is good to keep in mind so as not to confuse it with another possible interpretation of "restriction of scalars", namely composition of the map $X \to \mathrm{Spec} L$ with the map $\mathrm{Spec} L \to Spec K$ to
give a map $X \to \mathrm{Spec} K$. This is a much weirder functor, which preserves the dimension but screws up things like geometric integrality. (When I first heard about "restriction of
scalars", I guessed it was this latter thing and got very confused.)
| 28 | https://mathoverflow.net/users/1149 | 7765 | 5,320 |
https://mathoverflow.net/questions/7709 | 2 | Let $V$ be a vector space over $\mathbb R$. A *symmetric bilinear pairing* on $V$ is a linear map $a: V\otimes V \to \mathbb R$. Because $\mathbb R$ is characteristic not-two, I will freely confuse symmetric bilinear pairings with quadratic forms; if $v\in V$, I will write $av^2$ for $a(v\otimes v)$; and $av\_1v\_2$ for $a(v\_1\otimes v\_2)$. The pairing $a$ is *positive (negative) definite* on $V$ if $av^2$ is strictly positive (negative) whenever $v\neq 0$. The pairing $a$ is *nondegenerate* if the corresponding map $a: V\to V^\\*: v \mapsto a(v\otimes-)$ has trivial kernel.
Two subspaces $V\_1,V\_2 \subseteq V$ are *orthogonal* if $av\_1v\_2 = 0$ for any $v\_1\in V\_1$ and any $v\_2\in V\_2$. If $V\_1\subseteq V$, its *orthogonal complemenet* is the maximal subspace $V\_2\subseteq V$ so that $V\_1$ and $V\_2$ are orthogonal (it always exists, and may intersect $V\_1$). A subspace $V\_+\leq V$ is *positive (negative)* if $a|\\_{V\_+ \otimes V\_+}$ is positive- (negative-) definite. A subspace is *maximally positive (negative)* if it is positive (negative) and not contained in any other positive (negative) subspace. Maximally positive (negative) subspaces exist by Zorn's lemma.
Let $a$ be a nondegenerate symmetric bilinear pairing on $V$. If $V$ is finite-dimensional, then the following are true (e.g. by running Gram-Schmidt):
1. Let $V\_+$ and $V\_-$ be maximally positive and negative subspaces. Then $V = V\_+ + V\_-$. Since $V\_+ \cap V\_- = 0$, this presents $V$ as a direct sum.
2. Let $V\_+$ be a maximally positive subspace. Then its orthogonal complement is maximal negative.
An important corrolarry of 1. is that any two maximally positive (negative) subspaces have the same dimension. Then we can define the *signature* of the pairing $a$ as the pair $(\dim V\_+,\dim V\_-)$. When $a$ is not nondegenerate, the signature is actually a triple; the third term is the dimension of the kernel of the map $a: V\to V^\\*$. ($a$ induces a nondegenerate symmetric pairing on $V / \ker a$.)
**My question is:** are statements 1., 2. true when $V$ is infinite-dimensional? If not, what (topologico-analytical, say) conditions on $V$ assure that they are? (Or at least that 1. is; I don't really care about 2.)
| https://mathoverflow.net/users/78 | Splitting a space into positive and negative parts | Here's another counter-example, taken from *Loop Groups* (p 128).
Consider the space of continuous functions on the circle and define
$$
f(\theta) = \sum\_{k \gt 1} \frac{\sin k \theta}{k \log k}
$$
The positive part of this function is
$$
f\_+(\theta) = \frac{1}{2 i} \sum\_{k \gt 1} \frac{e^{i k \theta}}{k \log k}
$$
which is unbounded near $\theta = 0$.
Let's move on to the other part of your question: when can we ensure that the splitting exists? What you are asking for is that $V$ be the direct sum of two subspaces, $V\_-$ and $V\_+$. Of course, if you start with two abstract vector spaces, $V\_-$ and $V\_+$, both of which admit symmetric, positive definite bilinear forms, say $b\_-$ and $b\_+$ respectively, then you can construct an example by taking $V = V\_- \oplus V\_+$ and taking $-b\_- + b\_+$. This shows that you can get this situation to work with quite awful spaces, but the point is that all the awfulness of $V$ divides nicely into awfulness of $V\_-$ plus awfulness of $V\_+$.
Presumably, though, you are more interested in the case where you start with $V$ and the quadratic form. Maybe this quadratic form can be fairly arbitrary (perhaps varies in some space of quadratic forms). In this situation, you would want conditions on $V$ that guarantee that the splitting occurs without too much fuss.
Let's examine the question from the other end: suppose that $V = V\_- \oplus V\_+$. Then by changing the sign of the form on $V\_-$, we obtain a positive-definite symmetric bilinear form on $V$. This usually goes by the name of an **inner product** as we're over $\mathbb{R}$. So the problem reduces to finding complements of subspaces in inner product spaces. To guarantee this, you want completeness. Then your bilinear form is related to the original inner product by the operator $2P\_+ - I$ where $P\_+$ is the orthogonal projection on to $V\_+$.
So what you want is to be working with a Hilbert space and the space of self-adjoint square-roots of the identity.
As I said, this isn't an "if and only if". But it is a simple condition that quite often holds. It can be further relaxed since it's enough that the inner product induced by the bilinear form and the original inner product be merely equivalent rather than equal, but I'll leave those details as an exercise.
**Edit:** From your other question related to this it seems as though you are particularly interested in the case where one of the factors is finite dimensional. In that case, the splitting always holds.
| 1 | https://mathoverflow.net/users/45 | 7768 | 5,323 |
https://mathoverflow.net/questions/7687 | 40 | **Background**
For definiteness (even though this is a categorical question!) let's agree that a *vector space* is a finite-dimensional real vector space and that an *associative algebra* is a finite-dimensional real unital associative algebra.
Let $V$ be a vector space with a nondegenerate symmetric bilinear form $B$ and let $Q(x) = B(x,x)$ be the associated quadratic form. Let's call the pair $(V,Q)$ a **quadratic vector space**.
Let $A$ be an associative algebra and let's say that a linear map $\phi:V \to A$ is **Clifford** if
$$\phi(x)^2 = - Q(x) 1\_A,$$
where $1\_A$ is the unit in $A$.
One way to define the Clifford algebra associated to $(V,Q)$ is to say that it is universal for Clifford maps from $(V,Q)$. Categorically, one defines a category whose objects are pairs $(\phi,A)$ consisting of an associative algebra $A$ and a Clifford map $\phi: V \to A$ and whose arrows
$$h:(\phi,A)\to (\phi',A')$$
are morphisms $h: A \to A'$ of associative algebras such that the obvious triangle commutes:
$$h \circ \phi = \phi'.$$
Then the **Clifford algebra of $(V,Q)$** is the universal initial object in this category. In other words, it is a pair $(i,Cl(V,Q))$ where $Cl(V,Q)$ is an associative algebra and $i:V \to Cl(V,Q)$ is a Clifford map, such that for every Clifford map $\phi:V \to A$, there is a unique morphism
$$\Phi: Cl(V,Q) \to A$$
extending $\phi$; that is, such that $\Phi \circ i = \phi$.
(This is the usual definition one can find, say, in the [nLab](http://ncatlab.org/nlab/show/Clifford+algebra).)
**Question**
I would like to view the construction of the Clifford algebra as a functor from the category of quadratic vector spaces to the category of associative algebras. The universal property says that if $(V,Q)$ is a quadratic vector space and $A$ is an associative algebra, then there is a bijection of hom-sets
$$\mathrm{hom}\_{\mathbf{Assoc}}(Cl(V,Q), A) \cong \mathrm{cl-hom}(V,A)$$
where the left-hand side are the associative algebra morphisms and the right-hand side are the Clifford morphisms.
My question is whether I can view $Cl$ as an adjoint functor in some way. In other words, is there some category $\mathbf{C}$ such that the right-side is
$$\mathrm{hom}\_{\mathbf{C}}((V,Q), F(A))$$
for some functor $F$ from associative algebras to $\mathbf{C}$. Naively I'd say $\mathbf{C}$ ought to be the category of quadratic vector spaces, but I cannot think of a suitable $F$.
I apologise if this question is a little vague. I'm not a very categorical person, but I'm preparing some notes for a graduate course on spin geometry next semester and the question arose in my mind.
| https://mathoverflow.net/users/394 | Clifford algebra as an adjunction? | Disqualifier: this isn't a complete answer.
There's a basic "chalk and cheese" problem here. The "categories" that you are comparing are of two different types, although they do seem similar on the surface. On the one hand you have an honest algebraic category: that of associative algebras. But the other category (which, admittedly, is not precisely defined) is "vector spaces plus quadratic forms". This is **not** algebraic (over Set). There's no "free vector space with a non-degenerate quadratic form" and there'll (probably) be lots of other things that don't quite work in the way one would expect for algebraic categories. For example, as you require non-degeneracy, all morphisms have to be injective linear maps which severely limits them. You could add degenerate quadratic forms (which means, as AGR hints, that you regard exterior algebras as a sort of degenerate Clifford algebra - not a bad idea, though!) but this still doesn't get algebraicity: the problem is that the quadratic form goes *out* of the vector space, not into it, so isn't an "operation".
However, you may get some mileage if you work with **pointed** objects. I'm not sure of my terminology here, but I mean that we have a category $\mathcal{C}$ and some distinguished object $C\_0$ and consider the category $(C,\eta,\epsilon)$ where $\eta : C\_0 \to C$, $\epsilon : C \to C\_0$ are such that $\epsilon \eta = I\_{C\_0}$. In Set, we take $C\_0$ as a one-point set. In an algebraic category, we take $C\_0$ as the free thing on one object. Then the corresponding pointed algebraic category is algebraic over the category of pointed sets (I think!).
The point (ha ha) of this is that in the category of pointed associative algebras one does have a "trace" map: $\operatorname{tr} : A \to \mathbb{R}$ given by $(a,b) \mapsto \epsilon(a \cdot b)$. Thus one should work in the category of pointed associative $\mathbb{Z}/2$-graded algebras whose trace map is graded symmetric.
In the category of pointed vector spaces, one can similarly define quadratic forms as operations. You need a binary operation $b : |V| \times |V| \to |V|$ (only these products are of *pointed* sets) and the identity $\eta \epsilon b = b$ to ensure that $b$ really lands up in the $\mathbb{R}$-component of $V$ (plus symmetry).
Whilst adding the pointed condition is non-trivial for algebras, it is effectively trivial for vector spaces since there's an obvious functor from vector spaces to pointed vector spaces, $V \mapsto V \oplus \mathbb{R}$ that is an equivalence of categories.
Assuming that all the $\imath$s can be crossed and all the $l$s dotted, the functor that you want is now the forgetful functor from pointed associative algebras to pointed quadratic vector spaces.
| 14 | https://mathoverflow.net/users/45 | 7769 | 5,324 |
https://mathoverflow.net/questions/7508 | 22 | Ok, it's time for me to ask my first question on MO.
Consider the affine curve $Y+Y^q=X^{q+1}$ over the finite field $\mathbf{F}\_q$. It's interesting because it has the largest number of points over $\mathbf{F}\_{q^2}$ possible relative to its genus, which is $q(q-1)/2$. In other words, this curve realizes the Weil bound over $\mathbf{F}\_{q^2}$. This seems to be well-known in the literature.
Now consider the following hypersurface $\mathcal{X}$ over the finite field $\mathbf{F}\_q$:
$$Z+Z^q+Z^{q^2}=\det\left(\begin{matrix}
0 & X & Y \\ Y^q & 1 & X^q \\ X^{q^2} & 0 & 1\end{matrix}\right)$$
Empirical observation seems to point to the following: The compactly supported cohomology of $\mathcal{X}$ is only nonzero in degrees 2 and 4. In degree 2, the dimension of $H^2(\mathcal{X})$ is $q^2-1$ and the $q^3$-power frobenius acts as the scalar $q^3$. Which is all a fancy way of saying that for all $n$,
$$\#\mathcal{X}(\mathbf{F}\_{q^{3n}})=q^{6n}+q^{3n}(q^2-1).$$
Thus $\mathcal{X}$ has the largest number of $\mathbf{F}\_{q^3}$-points among any hypersurface with the same compactly supported Betti numbers.
Can anyone help me prove the above formula? (I can do $n=1$ alright...) Bonus points if you can also compute the automorphism group of $\mathcal{X}$. Many more bonus points if you can formulate the generalization to hypersurfaces of higher dimension!
The above hypersurface arises in the study of the bad reduction of Shimura varieties, if anyone cares to know.
EDIT: Admittedly, this is a narrow problem about a very particular surface. Therefore I'm going to accept an answer to the following question: Is there an algorithm to compute the zeta function of a hypersurface of this sort, that's quicker than counting points?
(I've already noticed that it's enough to recur over X and Y, and to test for each pair that the expression on the right lies in the image of the linear map defined by the expression on the left. But I can't think of anything faster than this.)
| https://mathoverflow.net/users/271 | A hypersurface with many points | Here is a complete solution to the main question when $n$ and $q$ are both odd, and a partial solution for the other parities. The partial solution includes a reduction to the case $n=2$.
Let $\text{Tr}\_k$ denote the trace map from $\mathbb{F}\_{q^n}$ to $\mathbb{F}\_{q^k}$, assuming that $k|n$.
The equation is
$$z+z^q + z^{q^2} = x^{q^2+q+1} - xy^q - x^{q^2}y.$$
First, make the change of variables $y \leftarrow -yx^{q^+1},$ so that the equation becomes
$$z+z^q+z^{q^2} = x^{q^2+q+1}(y^q+y+1).$$
Second, when $q$ is odd, we can clarify matters a little with the change of variables $y \leftarrow y - \frac12$ to get rid of the constant. The image of the $q$-linear map $z \mapsto z + z^q + z^{q^2}$, acting on $\mathbb{F}\_{q^{3n}}$, consists of those elements $Z'$ such that $\text{Tr}\_3(z') = \text{Tr}\_1(z')$. Whenever such a $z'$ is reached by the right side, there are $q^2$ solutions for $Z$. On the right side, $y \mapsto y' = y^q+y$ is an $q$-linear isomorphism when $n$ is odd, while when $n$ is even its image is the locus of $\text{Tr}\_2(y') = \text{Tr}\_1(y')$. Meanwhile $y' \mapsto x^{q^2+q+1}y'$ is a $q$-linear isomorphism unless $x=0$.
Thus when $n$ and $q$ are both odd, there are $(q^{3n}-q^2)(q^{3n}-1)$ solutions with $x,y \ne 0$. There are $q^2(2q^{3n}-1)$ more solutions when one of them is zero.
When $q$ is odd and $n$ is even, then in principle different $x' = x^{q^2 + q + 1}$ could behave differently in the equation $z' = x'y'$. For a fixed $x'$, the set of possible $x'y'$ is a certain $q$-linear hyperplane with codimension $1$, while the set of possible $z'$ is a certain $q$-linear hyperplane with codimension $2$. In the special case that $\{x'y'\}$ contains $\{z'\}$, then there are $q^{3n+1}$ solutions for that value of $x'$. For generic non-zero values of $x'$, there are $q^{3n}$ solutions. Dualize the hyperplanes with respect to $\text{Tr}\_1$. The dual of $\{y'\}$ is the line $L$ of trace 0 elements in $\mathbb{F}\_{q^2}$, while the dual of $\{z'\}$ is the plane $P$ of trace 0 elements in $\mathbb{F}\_{q^3}$. Meanwhile $\{x'\}$ is the subgroup $G$ of $\mathbb{F}\_{q^{3n}}$ of index $q^2+q+1$. A special value of $x'$ in this subgroup is one such that $x'L \subset P$. A priori I am not sure that it never happens. What I can say is that if $x'$ is special, then it must lie in $\mathbb{F}\_{q^6}$ because both $L$ and $P$ do. So you can reduce the counting problem to the case that $3n = 6$ or $n = 2$.
If $q$ is even, then the equation is
$$z' = x'(y'+1),$$
where as before $z' = z+z^q+z^{q^2}$ and $y' = y + y^q$. In this case $y'$ is any element with zero trace, and the dual line $L$ is just $\mathbb{F}\_q$ itself. I have not worked out exactly how it looks, but I suppose that it reduces to the case $n=1$ for similar reasons as above.
Afterthought: I don't feel like changing all of the equations, but I'm wondering now whether there a dual change of variables to put the right side in the form $y''(x^{q^2}+x)$. I think that the map $x \mapsto x^{q^2}+x$ is always non-singular when $q$ is odd.
---
A remark about where the trace conditions come from. If $a$ is an irreducible element of $\mathbb{F}\_{q^n}$, then the map $x \mapsto x^q$ is a cyclic permutation matrix in the basis of conjugates of $a$. A map such as $z \mapsto z+z^q+z^{q^2}$ is then a sum of disjoint permutation matrices and it easy to compute its image and cokernel.
---
Some remarks about Jared's second, more general question: C.f. the answer to [this other mathoverflow question](https://mathoverflow.net/questions/6279/counting-points-on-varieties-of-low-codimension/6295) about counting points on varieties. For fixed $q$, the equation of a hypersurface is equivalent to a general Boolean expression, and there may not be much that you can do other than count one by one. There are several strategies that work in the presence of special structure: You can use zeta function information, if you have it, to extrapolate to large values of $q$. You can count the points on a variety if you happen to know that it's linear, or maybe quadratic, or the coset space of a group. And you can use standard combinatorial counting tricks, which in algebraic geometry form amount to looking at fibrations, blowups, inclusion-exclusion for constructible sets, etc.
This particular variety decomposes a lot because it can be made jointly linear in $Y$ and $Z$, and $X$ only enters in a multiplicative form.
| 8 | https://mathoverflow.net/users/1450 | 7781 | 5,331 |
https://mathoverflow.net/questions/7776 | 8 | **Question** Is there a nice universal property which captures the notion of "collection of all epimorphisms out of a given object". Of course I will have to consider two epimorphisms $X \rightarrow Y$ the same if they are isomorphic over $X$. The answer to the dual question is yes, at least in a topos: The power object $P(X)=\Omega^X$ , where $\Omega$ is the subobject classifier can be thought of as "the collection of all subobjects of X". The universal property is just the property for exponentials.
**Background** (Not strictly necessary for the question): I have been reading Sheaves in Geometry and Logic by Mac Lane and Moerdijk. Their definition of an elementary topos is this: A category with pullbacks, a terminal object (i.e. all finite limits), a subobject classifier, and a power object for every object. They construct all other exponential objects from these axioms. The construction they use is to basically consider the "collection" of all graphs of morphisms. This is just the standard construction in set theory suped up to toposes.
This construction agrees with the set theoretic convention that a function should be regarded as a set of ordered pairs, i.e. if $f:A \rightarrow B$, then the set theorist will define $f$ as the image of the map $A \rightarrowtail A \times B$ induced by the $1\_A$ and $f$ (this may be the most convoluted sentence I have ever written). Why not define functions dually? There is also a map $A+B \twoheadrightarrow B$ induced by $1\_B$ and $f$. Then we could define $f$ as the partition of $A$ induced by this epimorphism, which seems like a perfectly nice way to define functions.
I was wondering if this construction could be used to construct exponential objects if I was given finite colimits and some kind of epimorphism classifier, or collection of epimorphisms out of a given object.
**Comment** if it turns out that there is no really nice answer to this question, do you think that has bearing on the fact that the formula for the number of subsets of a set is easy ($2^{|X|}$) but the formula for the number of partitions of a set is relatively hard (<http://en.wikipedia.org/wiki/Partition_of_a_set>)?
| https://mathoverflow.net/users/1106 | Universal property for collection of epimorphisms | I think the natural meaning of "collection of all epimorphisms out of $X$" or "epimorphism classifier" in a category $\mathbf{S}$ would be: an object $E$, an object $Y\to E$ of $\mathbf{S}/E$, and an epimorphism $p\colon E\times X \twoheadrightarrow Y$ in $\mathbf{S}/E$, such that for any object $U$ and any epimorphism $q\colon U\times X\twoheadrightarrow Z$ in $\mathbf{S}/U$, there exists a unique morphism $f\colon U\to E$ such that $(f\times 1)^\*q$ is isomorphic to $p$ under $E\times X$ in $\mathbf{S}/E$. In other words, a representing object for the presheaf on $\mathbf{S}$ which sends an object $U$ to the set of (isomorphism classes of) epimorphisms out of $U\times X$.
In a topos, such epimorphism classifiers can be constructed from power objects. Every epimorphism $X\twoheadrightarrow Z$ in a topos is the quotient of its kernel pair, which is an internal equivalence relation on $X$, i.e. a particular element of $P(X\times X)$, and every internal equivalence relation has a quotient. Therefore, the subobject of $P(X\times X)$ which internally "consists of all equivalence relations" can be shown to be an epimorphism classifier in the above sense.
| 7 | https://mathoverflow.net/users/49 | 7789 | 5,335 |
Subsets and Splits