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https://mathoverflow.net/questions/513 | 7 | Suppose that $M$ is a finitely generated module over $A=k[X\_1,\ldots,X\_n]$ of Krull dimension $m$ with $k$ an infinite field. Then one version of Noether normalisation says there is an $m$-dimensional $k$-subspace $W$ of the $k$-vector space spanned by $X\_1,\ldots,X\_n$ such that $M$ is finitely generated over $\operatorname{Sym}(W)$ considered as a subring of $A$.
As is surely well-known, in fact one can show that the set of $m$-dimensional $k$-vector spaces $W$ that work is open in the appropriate Grassmannian. My question is where is there a reference for this fact in the literature?
| https://mathoverflow.net/users/345 | Generic Noether normalisation | Ok, I might be missing something (I often am) but I believe that [this](http://books.google.com/books?id=7ua4WsmpDbMC&pg=PA453&lpg=PA453&dq=noether+normalization+works+generically&source=bl&ots=Rb24oAphIe&sig=fXu2RhCrSf77Dyi0IYHg-mmXMMM&hl=en&ei=ODveSo_2KsbV8AbsiJFn&sa=X&oi=book_result&ct=result&resnum=4&ved=0CBgQ6AEwAw#v=onepage&q=&f=false) does it. Scroll up to page 452, line 3.9. The book is Effective Methods in Algebraic Geometry by Rossi and Spangher.
| 4 | https://mathoverflow.net/users/622 | 1515 | 952 |
https://mathoverflow.net/questions/1510 | 3 | It's hard to prove a number is transcendental (non-algebraic) yet there are some wonderful examples amongst them like π,e and Liouville's number. What's so special about them?
Are most numbers transcendental?
| https://mathoverflow.net/users/836 | What's so special about transcendental numbers? | The set of real numbers is uncountable, but the set of algebraic numbers is countable, so "most" real numbers are transcendental in a very strong sense of "most". This is actually a capsule description of Cantor's proof of the existence of transcendental numbers; just note that an uncountable set cannot be empty. Looking at the interval [0,1], the set of algebraic numbers there have (Lebesgue) measure zero, so a number picked "at random" (uniform distribution) from that interval is transcendental with probability 1. Transcendental numbers are a dime a dozen - but to prove that *particular* real numbers are transcendental is either hard or just *too* hard. It is known that $ e^\pi $ is transcendental, but as to $\pi^e$, nobody knows. Transcendental numbers are studied for their own sake. Important mathematicians found them interesting, so they must be important :) Euler was the first to consider the possibility that there might be real numbers that were not algebraic, Liouville constructed the first ones, Hermite proved e transcendental, and so forth. The closely connected subject of linear forms in logarithms has applications in more mainstream number theory, especially to Diophantine equations.
| 16 | https://mathoverflow.net/users/3304 | 1517 | 954 |
https://mathoverflow.net/questions/1522 | 2 | A submodule of a free module need not be free (for instance, in the free Z[X]-module Z[X] the submodule generated by 2 and X is not free). But over a principal ideal domain, submodules of free modules are free.
I was wondering about the center of a free (as a module) algebra. Is it always free? or are there weaker conditions on the ground ring which guarantee it?
| https://mathoverflow.net/users/336 | Is the center of a free (as a module) algebra free? | Choose a ring R of characteristic not 2 which does not satisfy the condition "every projective is free" (e.g. R is not local). Pick a nonfree projective R-module M and make M into a commutative R-algebra in some way. Pick a complement N such that M + N (direct sum) is free and make N into an anticommutative R-algebra via a nondegnereate skew-symmetric form N x N -> R such that MN = NM = 0. I think this should yield an example, though I'm not sure on all the details.
| 4 | https://mathoverflow.net/users/344 | 1531 | 963 |
https://mathoverflow.net/questions/1380 | 21 | Suppose $X$ and $Y$ are Banach spaces whose dual spaces are isometrically isomorphic. It is certainly true that $X$ and $Y$ need not be isometrically isomorphic, but must it be true that there is a continuous (not necessarily isometric) isomorphism of $X$ onto $Y$?
| https://mathoverflow.net/users/792 | Isomorphisms of Banach Spaces | Indeed, $\ell\_1$ provides a strong counterexample. As noted by Matt, the spaces C(X), where X is countable and compact, provide nonisomorphic Banach spaces whose duals are isomorphic to $\ell\_1$. If X is countable and compact (and Hausdorff, of course!) then X is homeomorphic to a closed ordinal interval [0, *a*] (equipped with its natural order topology) for some countable ordinal a; this result is known as the Mazurkiewicz-Sierpinski theorem. About 50 years ago Bessaga and Pelczynski showed that if *a* and *b* are infinite countable ordinals and *a* < *b*, then C([0, *a*]) is isomorphic to C([0, *b*]) if and only if *b* < *a*^\*w\*, where *w* denotes the first infinite ordinal. Thus C([0, *w*]) is in fact isomorphic to C([0, *w*^2]), contrary to Gerald's assertion above. Moreover, the least infinite ordinal *b* such that C([0, *b*]) is not isomorphic to C([0, *w*]) is *b* = *w^w*. Combining the Mazurkiewicz-Sierpinski result with the Bessaga=Pelczynski result, one has that each space C(X), where X is countable and compact, is isomorphic to C([0, *w*^(*w*^\*a\*)]) for a *unique* countable ordinal *a*. In particular, there are uncountably many nonisomorphic Banach spaces whose duals are isometrically isomorphic to $\ell\_1$, namely the spaces C([0, *w*^(*w*^\*a\*)]), where *a* varies over the set of countable ordinals.
$\ell\_1$ has other preduals too. For example, a 1972 paper from the Israel Journal of Mathematics by Yoav Benyamini and Joram Lindenstrauss exhibits a Banach space E whose dual is isometrically isomorphic to $\ell\_1$, but E is not isomorphic to any space of the form C(X).
Although Gerald's assertion that the Cantor-Bendixson rank distinguishes between spaces of continuous functions on countable compact spaces is not correct, there is a rank/index that does distinguish between the isomorphism classes of these spaces, namely the *Szlenk index*. In fact the Szlenk index may be used to prove the 'only if' direction of the Bessaga-Pelczysnki result.
I strongly recommend Chapter 2 of the book *Biorthogonal Systems in Banach Spaces* by Hajek et al if you want to read about the Szlenk index and a proof of the Bessaga-Pelczynski result. That book also contains a sketch proof of the Mazurkiewicz-Sierpinski theorem; for a full account of that theorem I recommend Section 8 of Semadeni's classic book *Banach spaces of continuous functions*.
If you want a different example, I think the James-tree space *JT* would do. *JT* is separable but its dual is (at least I think so - it is worth checking!) isometrically isomorphic to the dual of the direct sum $\*JT \oplus H\*$, where *H* is Hilbert space of dimension equal to the cardinality of the continuum. In particular, $\*JT \oplus H\*$ cannot be isomorphic to *JT* because they have different density characters, but their duals are isometrically isomorphic. If you want to check the details of this example consult Chapter 13 of the Albiac and Kalton book recommended above by Matthew.
| 17 | https://mathoverflow.net/users/848 | 1538 | 966 |
https://mathoverflow.net/questions/1493 | 9 | I am looking for software that can find a global minimum of a polynomial function over a polyhedral domain (given by, say, some linear inequalities) in $\mathbb R^n$. The number of variables, $n$, is not more than a dozen. I know it can be done in theory (by Tarski's elimination of quantifiers in real closed fields), and I know that the time complexity is awful. However, if there is a decent implementation that can handle a dozen variables with a clean interface, it would be great. I have tried builtin implementations in Mathematica and Maple, and they do not appear terminate on 4-5 variable instances.
If the software can produce some kind of concise "certificate" of its answer, it would be even better, but I am not sure how such a certificate should look like even in theory.
Edit: Convergence to the optimum is nice, but what I am really looking for is ability to answer questions of the form "Is minimum equal to 5?" where 5 is what I believe on a priori grounds to be the answer to optimization to be (in particular, it is a rational number). That also explains why I want a certificate/proof of the inequality, or a counterexample if it is false.
| https://mathoverflow.net/users/806 | Software for rigorous optimization of real polynomials | I used [QEPCAD](https://www.usna.edu/CS/qepcadweb/B/QEPCAD.html) once for this sort of problem, with reasonable success, although I think my problem was a bit smaller than yours.
| 6 | https://mathoverflow.net/users/126667 | 1539 | 967 |
https://mathoverflow.net/questions/1523 | 8 | The Uniformization Theorem states that the universal cover of a Riemann surface is biholomorphic to the extended complex plane, the complex plane or the open unit disk. Each of these three is a domain in the extended complex plane. In particular, then, the universal cover of a domain in the extended complex plane is biholomorphic to a domain in the extended complex plane. This leads to an analogous question in higher dimensions: Is the universal cover of a domain in complex projective space biholomorphic to a domain in complex projective space? More precisely, I am asking for a counterexample. Many results in one complex variable break in several complex variables, and the Uniformization Theorem is fairly delicate, so it seems reasonable to expect it to break. Perhaps there is a counterexample that one can see just by topology?
| https://mathoverflow.net/users/3304 | Universal covers of domains in complex projective space | Consider a tubular neighborhood of three generic lines on P^2.
The fundamental group is Z. The universal covering will contain
an infinite chain of P^1's, and in particular two
disjoint P^1's. Thus it cannot be a domain in P^2.
| 10 | https://mathoverflow.net/users/605 | 1544 | 970 |
https://mathoverflow.net/questions/1546 | 10 | As a follow up to me other [question](https://mathoverflow.net/questions/1527/pushforwards-of-line-bundles-and-stability), what can be said about unstable vector bundles? I know this is rather open ended, but what sorts of horrible things does having a subbundle of strictly greater slope imply?
| https://mathoverflow.net/users/622 | Unstable Vector Bundles | Well, I don't know about horrible. There's a lot you can say that's good! I'll start rambling and see where I end up.
I'm going to pretend you said principal GL(n)-bundle instead of rank n vector bundle. Same thing, really, since we have the standard representation.
The collection Bun(n,C) of all principal GL(n) bundles P on a smooth curve C is a very nice geometric object: it's an Artin stack. It's not connected; the different components are labelled by topological data, like the Chern class. The tangent "space" (complex, really) to Bun(n,C) at a point P is naturally the derived global sections RGamma(C,ad(P)), where ad(P) is the associated bundle with fiber the adjoint representation of GL(n). The zero-th cohomology gives the infinitesimal automorphisms and the 1st cohomology gives the deformations. So the stabilizer group of any point V in Bun(n,C) is finite-dimensional, and the dimension of the stack is n(g-1) (by Riemann-Roch). Bun(n,C) is smooth, and unobstructed, thanks to the vanishing of H^2(C,ad(P)).
Bun(n,C) has a very nice stratification, too. It's an increasing union of quotient stacks [A/G] of projective varieties by finite-dimensional groups. Roughly, A is the stack of pairs (P,t), where t is a trivialization of P in an infinitesimal neighborhood of some point in C. Make the neighborhood large enough, i.e., r-th order, and you can kill off all the automorphisms of P. Unfortunately, except for n=1, there is no uniform bound on r that works for all bundles. So, Bun(n,C) isn't a finite type quotient stack.
You can also realize Bun(n,C) (homotopically) as the infinite type quotient stack of U(n)- connections modulo complexified gauge transformations. That's what Atiyah & Bott do in their paper "The Yang-Mills Equations on Riemann Surfaces". (They also have a nice discussion of slope-stability and the stratification.)
The top component of the stratification (those bundles where the stabilizer group is as small as possible) is the stack of (semi-)stable vector bundles. If you take the coarse moduli space of this substack, you get the usual moduli space of stable bundles.
In summary: If you drop the stability conditions, you get a lot more geometry with a similar flavor, and without the random bits of weirdness that crop up in the theory of moduli spaces. (e.g., the stack always carries a universal bundle, you don't need the rank and the chern class to be coprime.)
OK, I'll stop evangelizing now.
| 7 | https://mathoverflow.net/users/35508 | 1561 | 976 |
https://mathoverflow.net/questions/1564 | 2 | For a collection of points in $\mathbb{R}^n$, is there a statistic that I can compute which will estimate the number of clusters with some level of confidence?
| https://mathoverflow.net/users/812 | Estimating the number of clusters | This is an age-old question, which actually does not have (I think even cannot have) a definite answer, because first you need to define what you mean by a cluster and so on. A famous saying in this regard is that "cluster is in the eye of a beholder". It is easy to construct examples where somebody could see one cluster, but somebody else more than one.
This being said, the MDL (minimum description length) principle would lead you
to devise (IMHO) a clustering cost function in a most principled way, which by
optimizing you could the find the cluster assignments and number of clusters
simultaneously. For multinomial data you can see following:
P.Kontkanen, P.Myllymäki, W.Buntine, J.Rissanen, H.Tirri, *An MDL Framework
for Data Clustering. In Advances in Minimum Description Length: Theory and
Applications*, edited by P. Grünwald, I.J. Myung and M. Pitt. The MIT Press,
2005.
The intuitively-appealing idea behind MDL clustering is that by clustering you
create a model of the data. So the assumption is that a very good model is one
that lets you compress the data well.
Anyway MDL might not be easy to apply, if you are looking for a practical way to detect the number of clusters. BIC ([Bayesian information criteria](http://en.wikipedia.org/wiki/Bayesian_information_criterion)) and the F-ratio have
proven to work OK in practice.
| 4 | https://mathoverflow.net/users/861 | 1571 | 984 |
https://mathoverflow.net/questions/1470 | 4 | Poisson brackets play the very important roles in Symplectic geometry and Dynamical system. I'm interested in some conserved quantities of Poisson brackets.
Let's say we are working on T^n x R^n (T^n is the torus in R^n, T^n = R^n/Z^n). Assume that I have the Hamiltonian H: T^n x R^n \mapsto R, where H=H(x,p) and H is smooth for simplicity.
Every function here is T^n-periodic in x.
For K=K(x,p) then we define the Poisson bracket
{H,K} = D\_p H.D\_x K - D\_x H.D\_p K.
The question is, in general setting or in some particular case, can you find some conserved quantities S such that {H,S}=0?
One easy and important example is for S= f(H) for f is smooth then we have {H,f(H)}=0.
But can you have some other conserved quantities S?
**Edit:** here's one specific example: If H(x,p)=H(p)+V(x\_1+x\_2+...+x\_n) for V: R \mapsto R is T-periodic then this kind of Hamiltonian has some other conserved quantities S=p\_i - p\_j for any i \ne j.
Any deeper intuitions or physical examples?
| https://mathoverflow.net/users/823 | What are some conserved quantities of Poisson brackets? | My impression is that for most choices of H this is a hard question. In general, the largest possible cardinality of a set of independent Poisson commuting functions on a 2n-dimensional symplectic manifold (where by independent I mean that their differentials are linearly independent at each point in their domain--in particular H and f(H) are not independent) is n. Sketch proof: Given k independent Poisson commuting functions, their [Hamiltonian vector fields](http://en.wikipedia.org/wiki/Hamiltonian_vector_field) span a k-dimensional *isotropic* subspace (i.e., the symplectic form dxdp vanishes on the subspace), and it's an easy linear algebra exercise to show that the largest possible dimension of an isotropic subspace is n.
If there is an independent set of n functions, including H, which mutually Poisson commute, then the Hamiltonian system associated to H is called [integrable](http://en.wikipedia.org/wiki/Integrable_system), and can be solved exactly using action-angle coordinates--see Arnold's book *Mathematical Methods of Classical Mechanics*. The example you gave is integrable, but in general this is a fairly-rarely satisfied condition.
There are some obstructions to H having independent functions that Poisson commute with it. For example, Poincare noticed that if the Hamiltonian vector field of H has a periodic orbit, so if the orbit has period T the time-T map F of the flow has some fixed point y, then looking at the derivative of F at y gives the following restriction: if there are k Poisson-commuting functions (including H) which are independent along the orbit through y, then the derivative of F at y has to have the eigenvalue 1 with multiplicity at least 2k. The generic situation is that 1 occurs as an eigenvalue with multiplicity just 2, so in this sense the typical Hamiltonian doesn't have any other functions that Poisson commute with it along its periodic orbits.
Another point that occurs to me is that if almost every level set of H contains a dense orbit for the Hamiltonian flow of H, then since any K such that {K,H}=0 is constant along the Hamiltonian flow of H it would follow that any such K would be constant on almost every level set of H. But then K would be constant on every level set of H just by continuity, and this is only possible if K has the form f(H) that you mentioned. However I'm not sure if the assumption that almost every level set of H has a dense orbit is realistic--in particular I can't think of any examples where this holds on TnxRn.
One positive statement that can be made is the following. If you look at the proof of [Darboux's theorem](http://en.wikipedia.org/wiki/Darboux%27s_theorem) that's given in Arnold's above mentioned book, you'll see that it proves the following slightly stronger statement: if y is any point at which dH is nonvanishing, then there are local coordinates (Hi,Ki) around in terms of which the symplectic form is given by \sum dHidKi *and H1=H.* So (H=H1,H2,...,Hn) is an n-tuple of Poisson commuting functions, just defined near the given point y. So locally there are certainly functions that Poisson commute with H--they just can't generally be extended globally.
| 4 | https://mathoverflow.net/users/424 | 1575 | 988 |
https://mathoverflow.net/questions/1565 | 13 | Let XX be a Deligne-Mumford stack and let XX \to X be a coarse moduli space. Suppose that X is smooth. Is XX smooth? If not, what is an example? What if XX is of finite type over C (the complex numbers)? What are conditions we can put on XX to make this true?
| https://mathoverflow.net/users/2 | Can a singular Deligne-Mumford stack have a smooth coarse space? | The answer is yes, a singular DM stack can have a smooth coarse space. Let U=Spec(k[x,y]/(xy)) be the union of the axes in **A**2, and consider the action of G=**Z**/2 given by switching the axes: x→y and y→x. Then take XX to be the stack quotient [U/G]. This is a singular Deligne-Mumford stack (since it has an etale cover by something singular), but its coarse space is **A**1, which is smooth.
| 17 | https://mathoverflow.net/users/1 | 1584 | 994 |
https://mathoverflow.net/questions/1480 | 2 | I recall being told -- at tea, once upon a time -- that there exist models of the real numbers which have no unmeasurable sets. This seems a bit bizarre; since any two models of the reals are isomorphic, you'd expect any two models to have the same collection of subsets. Can anyone tell me exactly what the story here is? Have I misremembered something? Is this some subtlety involving how strong a choice axiom you use to define your set theory?
| https://mathoverflow.net/users/35508 | Models of the reals which have no unmeasurable sets | As John Goodrick is asking in a few places, you have to be careful in stating what you mean by "a model of the reals". If you're going to talk about sets of reals, then you need to have variables ranging over reals, and also variables ranging over sets of reals. You also of course want symbols in your language for the field operations and ordering, and possibly more.
**Three Options**
One way to do this is to use the language of second-order analysis, which is bi-interpretable with the language of third-order number theory. (It's straightforward to translate between real numbers and sets of natural numbers, and then between sets of real numbers and sets of sets of naturals.)
Another way to do this is to use ZF, which talks about the reals and sets of reals, but also many many other things. (Far more than any mathematician who's not a logician (or perhaps category theorist?) ever uses.)
There's also an intermediate strategy, which is basically what Russell and Whitehead did in Principia Mathematica, where you have some variables ranging over objects at the bottom (which might be real numbers, or anything else), and then variables ranging over sets of objects, and then variables ranging over sets of sets of objects, and so on to arbitrarily high levels. This is still far weaker than ZF, because you don't get sets that mix levels, and you also can't make sense of infinitely high levels.
**First-order and Higher-order logic**
If you take the first or third option, then you have two more choices, which correspond to what David Speyer was saying. You can require that variables that range over sets of things range over "honest subsets" of the collection of things they're supposed to be sets of.
Or you can interpret the set variables in a "whacked model". (The technical term is a "Henkin model".) On this interpretation, the "sets" are just further objects in your domain, and "membership" is just interpreted as some arbitrary relation between the objects of one type and the objects of the "set" type, and you interpret all your axioms in first-order logic.
The difference is that the honest interpretation uses second-order logic, while the Henkin interpretation just uses first-order logic. Second-order logic (and higher-order logic) is nice in that it lets you prove all sorts of uniqueness results - there is a unique model of honest second order Peano arithmetic, and if you require honest set-hood then this means there will be unique models at the third order level and higher, giving you one result that you remember. But first-order logic is nice because there's actually a proof system - that is, there is a set of rules for manipulating sentences such that any sentence true in every first-order model can actually be reached by doing these manipulations. That is, Gödel's Completeness Theorem applies. However, his Incompleteness Theorems also apply - thus, there are lots of models of first-order Peano arithmetic, and then there are even more Henkin models of "second-order" Peano arithmetic, and far far more Henkin models of "third-order" Peano arithmetic, which is the theory you're interested in.
Unfortunately, I don't know what these Henkin models look like. It all depends on what set existence axioms you use. There's a lot of discussion of this stuff for "second-order" Peano arithmetic in Steven Simpson's book *Subsystems of Second-Order Arithmetic*, which is the canonical text of the field known as reverse mathematics. However, none of that talks about arbitrary sets of reals, which is what you're interested in.
**Solovay's results**
The other result you mention, which is cited in one of the other answers here, takes the other option from above. That is, we do everything in ZF and see what different models of ZF are like. (Note that I don't say ZFC - of course if you have choice, then you have non-measurable sets of reals.)
Every model of ZF has a set it calls ω, which is the set it thinks of as "the natural numbers". Set theorists then talk about the powerset of this set as "the real numbers" - you might prefer to think of this set as "the Cantor set", and some other object in the model of ZF as its "real numbers", but there will be some nice translation between the Cantor set and your set, that gives the relevant topological and measure-theoretic properties.
Of course, since we're just talking about *models* of ZF, none of this is going to be the real real numbers. After all, since ZF is a first-order theory, the Löwenheim-Skolem theorem guarantees that it has a countable model. This model thinks that its "real numbers" are uncountable, but that's just because the model doesn't know what uncountable *really* means. (This is called Skolem's Paradox - http://en.wikipedia.org/wiki/Skolem%27s\_paradox>wikipedia, http://plato.stanford.edu/entries/paradox-skolem/>Stanford Encyclopedia of Philosophy.)
What Solovay showed is that if you start with a countable model of ZFC that has an inaccessible cardinal (assuming that inaccessibles are consistent, then there is such a model, and we have almost as much reason to believe that inaccessibles are consistent as we do to believe that ZFC is consistent) then you can use Cohen's method of forcing to construct a different (countable) model of ZF where there are no unmeasurable sets of "reals".
Of course, the first result you stated (that any two models of the reals are isomorphic) holds within any model of set theory, assuming you're talking about "honest" second-order models (that is, models of reals that are "honest" with respect to the notion of "subset" that you get from the ambient model of ZF). But the notion of "honest" second-order model doesn't even translate when you move from one model of set theory to another. So Solovay's model of ZF has the property that every "honest" model of second-order analysis (or third-order number theory) has no non-measurable sets, while any model of ZFC has the property that every "honest" model of second-order analysis (or third-order number theory) does have non-measurable sets.
That's how your two results are consistent.
| 22 | https://mathoverflow.net/users/445 | 1587 | 996 |
https://mathoverflow.net/questions/1501 | 6 | **Background:** Properness is a much more robust notion than projectiveness. For example, properness descends along arbitrary fpqc covers (see, for example, Vistoli's [Notes on Grothendieck topologies, fibered categories and descent theory](http://arxiv.org/abs/math/0412512), Proposition 2.36). This is far from true for projectiveness. In fact, projectiveness doesn't even descend along Zariski covers! The standard example of a proper non-projective morphism is locally projective over the base. (pictured in an appendix of [Hartshorne](http://books.google.com/books?id=3rtX9t-nnvwC&printsec=frontcover&dq=hartshorne+algebraic+geometry&ei=xDHeSvL_E4GEkgSSzqCVAQ#v=onepage&q=&f=false); page 443, unfortunately not on Google books)
---
So how delicate is projectivity? Suppose X is a scheme over a field k, and suppose K is a field extension of k such that XK is projective over K. Does it follow that X is projective over k?
The obvious thing to do (I think) is to pick a very ample line bundle L on XK and try to descend it to X. If K is a finite extension of k, then the *norm* of L will be an ample line bundle on X (if I haven't misunderstood [EGA II](http://www.numdam.org/numdam-bin/item?id=PMIHES_1961__8__5_0), section 6.5 and Corollary 6.6.2). But could it be that there is an *infinite* extension K of k such that XK is projective over K, but X is not projective over k?
| https://mathoverflow.net/users/1 | Does projectiveness descend along field extensions? | For finite extensions, this is stated explicitly as Corollary 6.6.5 in EGA II and it is also stated there that the result is true for arbitrary extensions. One may reduce the general case to the finite case as follows:
First, we may asume that K is finitely generated since any projective scheme is defined by finitely many equations. This allows us to find a finitely generated integral domain A over k with quotient K and a projective scheme Y over Spec(A) with generic fibre equal isomorphic to X\_K. Let Y' = X x Spec(A) and p the projection from Y' to Spec(A). By construction the generic fibre of the two schemes above over Spec(A) are isomorphic. By properness it easily follows that there is an open set U of Spec(A) such that the two schemes become isomorphic over U. U has rational points over a finite extensions of k so we reduce to the finite extension case.
| 4 | https://mathoverflow.net/users/519 | 1591 | 997 |
https://mathoverflow.net/questions/1047 | 11 | The size of a finite skeletal category C [in the sense of Leinster](http://arxiv.org/abs/math.CT/0610260) is defined as follows: Label the objects of C by integers 1,2,...,n and let aij be the number of morphisms from i to j (for i and j between 1 and n). The **size (or Euler characteristic) of C** is defined as the sum of the entries of the inverse of the nxn matrix A=(aij), if the inverse exists.
Let Fq be a finite field with q elements. For every natural number i, there is up to isomorphism exactly one Fq-vector space Vi of dimension i. The number of linear maps from Vi to Vj is equal to qij. We ignore the zero dimensional vector space V0. Consider the infinite matrix
Q=(qij)
where rows and and columns are indexed by positive integers 1,2,3,... From now on let us treat q as a formal parameter, don't care about convergence issues, and set v=q-1.
**Is there a notion of an inverse of Q?** (The entries will probably be formal power series in v.) **If the answer is yes, what is a closed form for the sum of the entries of the inverse (as a formal power series in v), i.e. the size of the category of finite dimensional Fq-vector spaces?**
At least every truncation Qn of Q to an upper left nxn corner has an inverse for every positive integer n, since Qn is a Vandermonde matrix. What is the limit of the sum of the entries of Qn-1 as n goes to infinity? I believe the answer is a power series in v. Is there an explicit form?
How can you interpret the answer? Is it the Euler characteristic of some moduli space? Is it equal or related to a sum of 1/Gl(Vi)? Does something interesting happen at q=1?
| https://mathoverflow.net/users/296 | What is the size of the category of finite dimensional F_q vector spaces? | Following the observations made in the comments one can compute the sum of the entries of Qn-1. It turns out that Kevin Costello's formula is true for every n.
Let (a1, a2, ..., an) be the the transpose of the kth column vector of Qn-1. (Of course, this vector depends on k, but we omit the index k.) Qiaochu Yuan suggested to consider the polynomial
A(x) = a1x + a2x2 + ... + anxn.
The degree of A is n, therefore A is determined by values at n+1 points. But we know that A(0)=0 and that
A(qi) = deltaik for i = 1, 2, ... , n.
By Lagrange interpolation, A(x) is equal to
x(x-q)(x-q2)...(x-qk-1)(x-qk+1)...(x-qn) / qk(qk-q)(qk-q2)...(qk-qk-1)(qk-qk+1)...(qk-qn).
The sum a1+a2+...+an is equal to A(1). Let us work with quantized integers. We use the notation [k] = (1-qk)/(1-q) = 1+q+...+qk-1. (Note that people from quantum groups sometimes use a different convention.) Furthermore, let [n choose k] be the quantized binomial coefficient. Then, A(1) is equal to
(-1)k-1 [n choose k] qk(k-1)/2-kn.
We sum A(1) over all k. A variant of the [quantum binomial theorem](http://en.wikipedia.org/wiki/Gaussian_binomial) gives that the sum of the entries of Qn-1 is equal to 1 - (1-1/q)(1-1/q2)...(1-1/qn).
| 5 | https://mathoverflow.net/users/296 | 1609 | 1,008 |
https://mathoverflow.net/questions/1614 | 15 | Are there any general results on when a closed subscheme X of a quasi-projective smooth scheme M can be written as the zero-set of a section of a vector bundle E on M?
To put it in a diagram: When is X the fiber product of M -> E <- M , where one arrow is the zero section and the other arrow is the section I'm looking for.
If this is not possible, can X be written as a degeneracy locus?
| https://mathoverflow.net/users/473 | When is a scheme a zero-set of a section of a vector bundle? | As for the first question, the class of X has to be the product of the Chern roots of the bundle, so in the Chow ring, it is the class of a complete intersection.
As for the second question, you would have to find classes that will solve the class of X in the Thom-Porteus formula, see Fulton's intersection theory 14.4
| 14 | https://mathoverflow.net/users/404 | 1615 | 1,010 |
https://mathoverflow.net/questions/1607 | -1 | Hello,
I am studying random variables.
Question is this:
if rv X & a function g is known, what is the pdf of random variable Y = g(x)?
in the textbook answer is explained as follows.
P[y ≤ Y ≤ y + dy] = P[x ≤ X ≤ x + dx]
F\_y(y + dy) - F\_y(y) / dy dx = F\_x(x + dx) - F\_x(x) / dx dy
why is left side of dx & right side of dy exists in above equation?
| https://mathoverflow.net/users/877 | about Function of Random variables | What you're looking at is known as "the transformation theorem" and is just an integral change of variables written in probability notation.
Suppose g is an increasing function and Y = g(X). Then
```
F_Y(y) = P( g(X) < y ) = P( X < g^{-1}(y) ) = F_X( g^{-1}(y) )
```
To obtain the PDF, differentiate both sides of the equation above:
```
f_Y(y) = f_X( g^{-1}(y) ) D_y ( g^{-1}(y) )
```
where D\_y means derivative with respect to y. Now if g were a decreasing function we'd have
```
F_Y(y) = P( g(X) < y ) = P( X > g^{-1}(y) ) = 1 - F_X( g^{-1}(y) )
```
and
```
f_Y(y) = f_X( g^{-1}(y) ) | D_y ( g^{-1}(y) ) |.
```
In the last line we would have -D\_y. Since g is a decreasing function, it's derivative is negative and so the absolute values take care of the negative sign.
| 4 | https://mathoverflow.net/users/136 | 1616 | 1,011 |
https://mathoverflow.net/questions/1624 | 28 | Can you prove that 8 is the largest cube in the Fibonacci sequence?
| https://mathoverflow.net/users/887 | Is 8 the largest cube in the Fibonacci sequence? | For a much more accessible treatment, and history of the result, see [Andrejic (2006) "On Fibonacci powers"](http://www.doiserbia.nb.rs/img/doi/0353-8893/2006/0353-88930617038A.pdf)
| 21 | https://mathoverflow.net/users/261 | 1629 | 1,019 |
https://mathoverflow.net/questions/1628 | 21 | I repeatedly heard that K(F\_1) is the sphere spectrum. Does anyone know about the proof and what that means?
| https://mathoverflow.net/users/451 | K(F_1) = sphere spectrum? | I understand that this is because GLn(F1) is supposed to be Sigman, the symmetric group on n letters. Thus K(F1) = K(finite sets) which is the sphere spectrum by the Barratt-Priddy-Quillen-Segal theorem.
But I have no idea why GLn(F1) should be Sigman...
| 16 | https://mathoverflow.net/users/318 | 1630 | 1,020 |
https://mathoverflow.net/questions/1256 | 9 | I've been reading the [wonderful slides by Terry Tao](http://terrytao.files.wordpress.com/2009/07/primes1.pdf) and thought about this question.
Primes appear to be quite random, and the formal statement should be that there are some characteristics of primes that are indistinguishable by any algorithm from the a sequence of random numbers. I think an easy example should be the distribution of the first digit of the prime number (bounded by C where C goes to infinity), which is basically known, so it's possible to say that this distribution is the same as that of some random sequence.
Are there any formal statements of this kind?
| https://mathoverflow.net/users/65 | Primes are pseudorandom? | There is no general statement, but there is a general philosophy.
The general idea in mathematics is that things do not happen for no reason. For example, almost every mathematician would be willing to bet that alpha=e^e+pi^sqrt(2) is irrational, as the 'generic' number is irrational, and a genuine reason is needed for a number to be non-generic. Of course, with current technology there is hardly any hope of *proving* that alpha is irrational, and we must do with proofs of irrationality of numbers such as e and pi, that have more structure that can be exploited in the proofs. Of course, it does not mean that e or pi are indistinguishable from a generic number. For example, the continued fraction expansion of e exhibits a very regular pattern.
Similarly, it is hard to imagine how say a sequence a\_n=floor(n^sqrt(2))+p\_n, where p\_n denotes n'th prime, can behave in substantially different way from a random sequence. Again, there is hardly a hope of proving that. The primes themselves enjoy more noticeable structure than {a\_n} however, making it much easier to prove things about them. Of course, primes are not a generic sequence. For example, there is only one even prime.
With this principle in mind one can easily make a myriad of conjectures expressing the idea that 'primes should behave like a generic sequence unless there is an obvious reason that they do not'. Most of these conjectures will be true, but only a few will be provable with current ideas.
The value of proving such conjectures is that since they involve an object so simply defined as the primes, they are likely to involve general mathematical techniques that are useful elsewhere. Like transcendence proofs that gave rise to many ideas in function interpolation, and algebraic number theory, the proofs of conjectures about pseudorandomness of primes led to much progress. For example, proving law of large numbers for primes (which is generally known as the prime number theorem) stimulated the development of the order of entire complex-analytic function. Dirichlet's theorem on uniform distribution mod q led to the introduction of L-functions that are now useful far beyond the original application.
| 6 | https://mathoverflow.net/users/806 | 1637 | 1,023 |
https://mathoverflow.net/questions/1621 | 15 | I was taught to think of generalized cohomology theories as the homotopy category of (symmetric) spectra. But is there also a category of 'invariants', that is, some category of contravariant functors from a suitable category of topological spaces to a suitable category of algebraic objects, which has a model category structure such that the homotopy category gives the category of generalized cohomology theories, without referring to spectra?
If such a thing exists, why do people prefer to use spectra?
| https://mathoverflow.net/users/798 | Are generalized cohomology theories a homotopy category of some category of invariants? | Here is a short argument why we don't expect generalized cohomology theories to behave so well.
In the stable homotopy category, there is a generalized homology/cohomology theory represented by the sphere spectrum $S$, so that $S\_\*(X)$ are the stable homotopy groups of $X$. It has a multiplication-by-2 self-map $f$ and we can use the triangulated structure to find an exact triangle $S \stackrel{f}{\to} S \to M \to S[1]$ where $S[1]$ is the suspension (equivalently, the shift) of $S$. We think of $M$ as "the sphere mod 2", and it is called the mod-2 Moore spectrum.
In most derived categories coming from algebra, such an exact triangle would have the property that the multiplication-by-2 map on $M$ was zero. However, we know that this is not the case here; the multiplication-by-2 map is not zero, but the multiplication-by-4 map is.
One of the problems with using a "functor" language to get at generalized cohomology theories is that given a natural transformation $E \to F$ of generalized cohomology theories, it is not clear what the associated cofiber should be in order to produce a triangulated structure. Spectra have a natural triangulated structure and they rectify this problem. (A map between spectra also includes "phantom" data that isn't easily detected by the natural transformation between associated generalized cohomology theories.)
| 13 | https://mathoverflow.net/users/360 | 1642 | 1,026 |
https://mathoverflow.net/questions/1610 | 16 | It is known that the binomial coefficient $2n \choose n$ is equal to number of shortest lattice paths from $(0,0)$ to $(n,n)$. The Catalan number $\frac{1}{n+1} {2n\choose n}$is equal to the number of shortest lattice paths that never go above the diagonal. Here, the diagonal may be viewed as a path from $(0,0)$ to $(n,n)$.
Is there a formula for the number of pairs $(P\_{1},P\_{2})$ where each $P\_{i}$ is a shortest lattice path from $(0,0)$ to $(n,n)$ such that $P\_{1}$ never goes above $P\_{2}\ ?$ Here, "$P\_{1}$ never goes above $P\_{2}$" means that $P\_{1}$ lies inside or on the boundary of the region determined by $P\_{2}$, the $x$-axis, and the line $x=n$.
| https://mathoverflow.net/users/296 | Pairs of shortest paths | The answer is (2n)! (2n+1)! / (n)!^2 (n+1)!^2 .
You can get this by the Gessel-Viennot method suggested above. One difficulty is that GV wants to count paths which don't touch at all, even at vertices, while you just want to count paths that don't cross. To solve this, take your lower path and slide it south-east. You are now looking for two paths, one from (0,0) to (n,n) and one from (-1,1) to (n-1,n+1), that don't touch at all.
The GV method gives the determinant
(2n choose n) (2n choose n+1)
(2n choose n-1) (2n choose n)
Expanding this and simplifying gives the above, if I didn't make any mistakes.
| 16 | https://mathoverflow.net/users/297 | 1650 | 1,032 |
https://mathoverflow.net/questions/1036 | 20 | Do I remember a remark in "Sketch of a program" or "Letter to Faltings" correctly, that acc. to Grothendieck [anabelian geometry](http://www.math.okayama-u.ac.jp%2F~h-naka%2Fzoo%2Fpeacock%2FNTM.ps "anabelian survey") should not only enable finiteness proofs, but a proof of FLT too? If yes, how?
Edit: [In this](http://www.ams.org/notices/201009/rtx100901106p.pdf "AMS-Notices-pdf") transcript, Illusie makes a remark that Grothendieck looked for a connection between "FLT" and "higher stacks". BTW, [here](http://madameboisvert.blogspot.com/2009/04/discovering-paca.html "link") a note on (acc. to Illusie) Grothendieck's [favored landscape](http://www.cabris.fr/#/galerie/sample-album-1/point-de-vue-9/ "link").
| https://mathoverflow.net/users/451 | "Fermat's last theorem" and anabelian geometry? | See the papers of Minhyong Kim. For example, begin by looking at the MR review 2181717 of his paper Invent. Math. 161 (2005), no. 3, 629--656.
| 13 | https://mathoverflow.net/users/930 | 1659 | 1,039 |
https://mathoverflow.net/questions/1600 | 4 | Let (X,x) be a pointed space. There is an action of π1(X,x) on πn(X,x) -- determined by considering πn(X,x)=πn-1(ΩxX,**x**), where ΩxX denotes the space of loops in X based at x, and **x** denotes the constant loop -- given simply by conjugation. We can speak unambiguously of πn(X), the *free* (i.e., not necessarily basepoint-preserving) homotopy group exactly when this action is trivial.
On an algebraic level I'm fine with this, but I'm having trouble envisioning how a homotopy class might be conjugated to a different homotopy class in this way. Besides my admittedly small collection of toy examples, my issue could also be that I'm mainly thinking about π1, in which case it might (???) be that the action *is* trivial. (I seem to recall that before learning about general homotopy theory, I heard a statement along the lines of "for path-connected spaces, you may as well ignore basepoints". Certainly the groups are all isomorphic, but I'm not certain whether there is a unique natural isomorphism.)
Also, are there (necessary and/or sufficient) conditions for when the π1 action on πn will or won't be trivial, and does this depend on n?
| https://mathoverflow.net/users/303 | free homotopy groups -- when do they exist? | For the last part of your question: given a group π1 which acts on an abelian group πn, there is always as space X with these homotopy groups with this action, and you can manufacture one using Eilenberg-MacLane spaces. You can make the group π1 act on the Eilenberg-MacLane space K(πn,n) in such a way that realizes the action on πn, and then use this to build a space as a fibration X-->K(π1,1) with fiber K(πn,n). So the only general limitation is in how the group π1 can act on the group πn.
There is one condition on a space X which implies the action is trivial: if X is a loop space (i.e., X is homotopy equivalent to ΩY for some Y), then the action of π1(X) on πn(X) is always trivial; the idea is that ΩX=Ω2Y, in which loop composition is commutative up to homotopy.
| 5 | https://mathoverflow.net/users/437 | 1660 | 1,040 |
https://mathoverflow.net/questions/1647 | 1 | I have several temporal signals of different dimensions, for example the motion of a point throughout time which would be of dimension 3, and the value of a temperature sensor, of dimension 1.
I would like to find out if these signals are correlated. Is there a measure of correlation that works for signals that do not have the same dimensionality?
| https://mathoverflow.net/users/180 | Correlation measure between signals of different dimensions? | Yes, you can use multiple regression, with temperature as the dependent variable and the three space dimensions as the independent variables.
| 2 | https://mathoverflow.net/users/619 | 1663 | 1,042 |
https://mathoverflow.net/questions/1662 | 5 | The Walsh-Hadamard transform is very fast to compute.
Can it be used to compute the convolution of two functions as it can be done with Fourier transform ?
| https://mathoverflow.net/users/903 | Can Walsh-Hadmard transform be used for convolution ? | Not in the sense I think you mean it. First of all, the Walsh-Hadamard transform **is** a Fourier transform - but on the group (Z/2Z)^n instead of on the group Z/NZ. That means you can use it to compute convolutions with respect to the space of functions (Z/2Z)^n -> C. Unfortunately, unlike the case with Z/NZ you can't use this to approximate a compactly supported convolution on Z, at least not directly.
| 5 | https://mathoverflow.net/users/290 | 1678 | 1,051 |
https://mathoverflow.net/questions/1592 | 8 | While the general problem of detecting a Hamiltonian path or cycle on an undirected grid graph is known to be NP-complete, are there interesting special cases where efficient polynomial time algorithms exist for enumerating all such paths/cycles? Perhaps for certain kinds of k-ary n-cube graphs? I hope this question isn't too open-ended...
Update - Is the problem of iterating Hamiltonian path/circuits known to be NP-complete for the N-cube?
| https://mathoverflow.net/users/774 | Special cases for efficient enumeration of Hamiltonian paths on grid graphs? | There are certainly special graphs that are always Hamiltonian (if every vertex of a graph of n vertices has degree at least n/2, say) and these have efficient algorithms associated with them.
For instance, [this paper](http://www.math.cmu.edu/~af1p/Texfiles/5out.pdf) proves the graph of a random 5-outregular digraph is Hamiltonian and there is an algorithm that finds a Hamiltonian cycle in polynomial time.
| 5 | https://mathoverflow.net/users/441 | 1687 | 1,057 |
https://mathoverflow.net/questions/1590 | 62 | Once in a while I run into literature that invokes vanishing cycle machinery with a cryptic sentence like, "this follows from a standard vanishing cycle argument." Is there a good way to look at vanishing cycles, nearby cycles, and specialization? I have a decent idea how some of it works for studying the cohomology of a one-parameter flat family of degenerating complex manifolds (see below), but the general sheaf picture still gives me a headache. Any recognition principles (e.g., "this looks like a place where I can use a vanishing cycle argument") would be most welcome.
Say I have a family of complex manifolds where the fibers are smooth over a punctured unit disc, and have some mild singularities over zero (a priori the singularities could be arbitrarily bad, but say we blow up until we have simple normal crossings). The cohomology of the fibers forms a vector bundle on the punctured disc, and it comes equipped with some extra structure, such as a pure Hodge filtration and a Gauss-Manin connection that identifies nearby fibers. When we attempt to extend the vector bundle over the whole disc, the extra structures degenerate - the Hodge structure becomes "mixed", and the connection acquires logarithmic singularities. These structures aren't immediately relevant to this question, but they seem to be interesting.
As far as I can tell, vanishing cycles and nearby cycles arise when we try to relate the cohomology of smooth fibers Xt with that of the special fiber X0. Each smooth fiber Xt has an inclusion map to the total space X, and X is homotopy equivalent to X0 by a fiberwise retraction. The composition yields a map from Xt to X0, and the pushforward of a sheaf on Xt along this map yields the nearby cycles sheaf. When I start with the constant sheaf on Xt this yields a sheaf on X0 that computes cohomology of Xt for some abstract nonsense reason. So far, I'm okay, but it seems that choosing t is not canonical enough, so one replaces Xt with the homotopy equivalent universal fiber Xoo over the universal cover of the punctured disc (the upper half plane), and defines nearby cycles by some crazy pullback-pushforward-pullback sequence. Specialization and vanishing cycles seem to be similar - I think there is a nice geometric picture somewhere, but the proliferation of upper and lower stars makes me sad. Is there a good way to see through that thicket?
| https://mathoverflow.net/users/121 | Is there a good way to think of vanishing cycles and nearby cycles? | In general I don't think there's anything easy about nearby and vanishing cycles. However, I tend to find it enlightening to just consider their topology. Namely, if $f:X \to \Bbb C$ is a function on a complex algebraic (or analytic) variety, then the stalk cohomology of the nearby cycles functor applied to some complex of sheaves $F$ at a point $x \in f^{-1}(0)$ is simply the the cohomology (with respect to $F$) of the Milnor fiber of $f$ at $x$. Given the utility of Milnor fibers in studying the topology of singular spaces, I think this is good motivation for the nearby cycles functor: in some sense it bundles the information of the all the Milnor fibers together into a single sheaf.
Now you can also use lots of neat results about Milnor fibers to actually say things about these stalk cohomology groups and their monodromy. For example, if f has an isolated singularity, then the singular Milnor fiber is homotopic to a bouquet of n-spheres, where n is given by the so-called Milnor number. The monodromy can often be calculated with the help of the Thom-Sebastiani theorem, which says that if $f = g + h$ (where $g,h$ are functions on distinct subvarieties of $X$ such that their sum makes sense as a function on $X$) then the monodromy of the Milnor fiber with respect to $f$ is the tensor product of the monodromies with respect to $g$ and $h$.
If we're also interested in vanishing cycles, then we can give a similarly topological description of the stalk cohomology groups. Letting $i$ denote the inclusion of $f^{-1}(0)$ into $X$, we have the exact triangle with the specialization map $i^\*F$ to the nearby cycles of $F$ and the canonical map from the nearby cycles of $F$ to the vanishing cycles of $F$ (although perhaps it shouldn't be called canonical because it's the cone of the specialization map). Anyway, looking at the associated long exact sequence provides a topological description of the vanishing cycles (in particular if $f$ is smooth then the vanishing cycles is zero and we get an isomorphism between $i^\*F$ and the nearby cycles of $f$).
I wish I could say more about the kind of recognition principle you're looking for, but the only place that I've encountered nearby cycles is in Springer theory, where the nearby cycles of the adjoint quotient map applied to the constant sheaf is isomorphic to the pushforward of the constant sheaf under the Springer resolution (and hence the monodromy action determines the action of the Weyl group on the cohomology of Springer fibers). Perhaps in this case the thing to notice is that the nilpotent cone is singular and also the fiber over zero of the adjoint quotient map, which makes it seem like a potentially good candidate for nearby cycles arguments?
| 22 | https://mathoverflow.net/users/916 | 1691 | 1,061 |
https://mathoverflow.net/questions/1675 | 13 | This is a follow-up to [this post](https://mathoverflow.net/questions/1039/explicit-direct-summands-in-the-decomposition-theorem) on the Decomposition Theorem. Hopefully, this will also invite some discussion about the theorem and perverse sheaves in general.
My question is how does one use the Decomposition Theorem in practice? Is there any way to pin down the subvarieties and local systems that appear in the decomposition. For example, how do you compute intesection homology complexes using this theorem? Does anyone have a link to a source with worked out examples?
Another related question: What is the deep part of the theorem? Is it the fact that the pushforward of a perverse sheaf is isomorphic to its perverse hypercohomology? Is it the fact that these pieces are semisimple? Or are these both hard statements? And what is so special about algebraic varieties?
| https://mathoverflow.net/users/788 | How to do Computations Using the Decomposition Theorem for Perverse Sheaves | To supplement Ben's answer, basically every aspect of the decomposition theorem is hard.
To give you a simple example of something which is implied by the decomposition theorem but is far from trivial is the following statement: given a proper smooth map of smooth varieties f : X -> Y the direct image of the constant sheaf splits as a direct sum of local systems. Note that this implies (but is stronger than) the degeneration of the Leray-Serre spectral sequence for the fibration. This answers to some extent your question "what is so special about algebraic varieties" because Leray-Serre just doesn't degenerate in general.
I think the situation has been cleared up considerably by the work of de Cataldo and Migliorini which (IMHO) is the first genuinely geometric proof of the decomposition theorem.
One might think of the "smooth map" case above as the "easiest case" (and indeed it does have an easier proof). However de Cataldo and Migliorini point out that in fact the "easiest case" is the case of a semi-small map, for which the decomposition theorem can be deduced from the non-degeneracy of certain bilinear forms. In a difficult work, they deduce the general case by reducing to this case by induction on the "defect of semi-smallness" (how far away a map is from being semi-small) and by taking hyperplane sections to reduce this defect.
An excellent informal survey about the decomposition theorem, with lots of wonderful examples can be found in *[The decomposition theorem, perverse sheaves and the topology of algebraic maps](http://arxiv.org/abs/0712.0349)* by de Cataldo and Migliorini.
Note that there are really three statements in the decomposition theorem, all of which are hard:
1. the direct image is the sum of its perverse cohomology groups;
2. each perverse cohomology is a direct sum of IC extensions of a local system;
3. each local system is semi-simple.
As is often the case in mathematics, a nice way to learn why the decomposition theorem is hard is to go to situations when it fails. This occurs when one takes perverse sheaves with coefficients in positive characteristic (or even Z). Daniel Juteau, Carl Mautner and I have written a survey called "Perverse sheaves and modular representation theory" which contains lots of examples of the failure of the decomposition theorem. (Note that all of 1), 2) and 3) above can fail!)
| 13 | https://mathoverflow.net/users/919 | 1694 | 1,064 |
https://mathoverflow.net/questions/1664 | 7 | Compact connected simply-connected Lie groups have so much structure that you can calculate their cohomology from their Lie algebras using Lie algebra cohomology (certain Ext-groups) and similarly their homology from their Lie algebras using Lie algebra homology (certain Tor-groups).
Is there similar theorem that gives the (co)homology of the loop space of a Lie group in terms of its Lie algebra?
| https://mathoverflow.net/users/798 | Can one calculate the (co)homology of the loopspace of a Lie group from its Lie algebra? | Yes.
At least, rationally.
The result that you want starts on p68 of "Loop Groups" by Pressley and Segal. There, they prove surjectivity of H\*(L𝔤;ℝ) → H\*(LG;ℝ). The basic idea of the argument is as follows: for reasonably simple reasons, the cohomology of LG is easily obtainable from that of G. This yields specific formulae for generators of the de Rham cohomology of LG. Although these forms are not themselves left invariant, they are cohomologous to rational multiples of left invariant forms, and thus come from the cohomology of the Lie algebra, L𝔤. The proof of surjectivity is thus not hard.
The proof that it is an isomorphism is a little trickier and they defer that to section 14.6 (p299). That this is quite some way through the book is a good indication of how much you need to absorb to understand it.
Amazingly, parts of Loop Groups (including pages 68, 69, and 299, but not 300) are available on Google books. It says that it is a "Limited preview" but whether or not that is just limited in pages or limited in time, I do not know. However, Loop Groups is a great book for anyone interested in Lie Groups and infinite dimensional "stuff".
(Incidentally, this is all for G simply connected.)
| 2 | https://mathoverflow.net/users/45 | 1697 | 1,066 |
https://mathoverflow.net/questions/1620 | 9 | A little bit of background: A graph G is, of course, a set of vertices V(G) and a multiset of edges, which are unordered pairs of (not necessarily distinct) vertices. We say that two vertices v\_1, v\_2 are adjacent if {v\_1, v\_2} is an edge. A directed graph, or digraph is essentially the same thing, except that the edges are now ordered pairs. Digraphs have a rather nice categorical interpretation.
A *graph homomorphism* is exactly what it should be, categorically, if you think of graphs as "sets with an adjacency structure." It's a map f: V(G) \rightarrow V(H) is a homomorphism if v\_1, v\_2 \in V(G) adjacent implies f(v\_1), f(v\_2) adjacent. The notion of homomorphism for directed graphs is essentially the same; if there's an edge from v\_1 to v\_2, then there's an edge from f(v\_1) to f(v\_2). Taking these definitions as morphisms, we can define categories of graphs and of digraphs.
Oftentimes in graph theory (particularly when linear algebra methods come into play), it's easier to work with a digraph than a graph, and so we usually orient the edges arbitrarily. But this isn't really natural...
There's a forgetful functor from the category of digraphs to the category of graphs. Does this functor have an adjoint? (I forget which is left and which is right.) Now that I think about it, is the obvious thing (replace each edge by a pair of directed edges, one in each direction) an adjoint, and if so is there a way to fudge the categories so that simple graphs are taken to simple digraphs?
| https://mathoverflow.net/users/382 | Is there a free digraph associated to a graph? | I like to use the following definitions, which give a nonstandard definition of undirected graph but produce particularly nice categories.
>
> A directed graph is a pair of sets V and E together with two maps s, t : E -> V. Call the category of these DirGraph.
>
>
> An undirected graph is a pair of sets V and E together with two maps s, t : E -> V plus a map r : E -> E such that r^2 = 1, sr = t and tr = s. Call the category of these UndirGraph.
>
>
>
The interpretation of directed graphs should be obvious. For an undirected graph, an edge is an orbit under r of an element of E. Note that there are two kinds of loops--orbits of size 1 or 2.
We can represent the categories of directed graphs and undirected graphs as the categories of presheaves on two categories Dir and Undir respectively (each has two objects corresponding to V and E and a small number of morphisms). There's an inclusion functor i : Dir -> Undir which induces a restriction functor UndirGraph -> DirGraph; this is your doubling functor. It has adjoints on both sides (left and right Kan extension). The left adjoint is your "forgetful" functor DirGraph -> UndirGraph. The right adjoint is another functor DirGraph -> UndirGraph which roughly speaking sends a directed graph G to the undirected graph G' with the same vertices where an edge between v and w in G' is a pair of an edges in G, one from v to w and one from w to v.
So with these definitions, not only does the "forgetful" functor have a right adjoint, but its right adjoint also has a right adjoint.
| 8 | https://mathoverflow.net/users/126667 | 1703 | 1,068 |
https://mathoverflow.net/questions/1684 | 62 | The exterior algebra of a vector space V seems to appear all over the place, such as in
* the definition of the cross product and determinant,
* the description of the Grassmannian as a variety,
* the description of irreducible representations of GL(V),
* the definition of differential forms in differential geometry,
* the description of fermions in supersymmetry.
What unifying principle lies behind these appearances of the exterior algebra? (I should mention that what I'm really interested in here is the geometric meaning of the Gessel-Viennot lemma and, by association, of the principle of inclusion-exclusion.)
| https://mathoverflow.net/users/290 | Why is the exterior algebra so ubiquitous? | Just to use a buzzword that Greg didn't, the exterior algebra is the symmetric algebra of a purely odd supervector space. So, it isn't "better than a symmetric algebra," it is a symmetric algebra.
The reason this happens is that super vector spaces aren't just Z/2 graded vector spaces, they also have a slightly different tensor category structure (the flip map on the tensor product of two odd vector spaces is -1 times the usual flip map, and the usual flip map for all other pure vector spaces). If you look at all the formulas from homological algebra, for things like how to take the tensor product of two complexes, they always have a bunch of weird signs showing up; these always can be thought of as coming from the fact that you should take the tensor product on graded vector spaces inherited from super vector spaces, not the boring one.
Of course, this just raises the question of why supervector spaces show up so much. Greg had about as good an answer as I could give for that.
| 38 | https://mathoverflow.net/users/66 | 1705 | 1,069 |
https://mathoverflow.net/questions/1467 | 17 | One makes precise the vague notion of "curve with a fractional point removed" (see for instance [these slides](http://www-math.mit.edu/~poonen/slides/campana_s.pdf)) using stacks -- one should really consider Deligne-Mumford stacks whose coarse spaces are curves, and the "fractional points" correspond to the residual gerbes at the stacky points.
One example: let a,b,c > 1 be coprime integers and let S be the affine surface given by the equation x^a + y^b + z^c = 0. Then there is a weighted Gm action on S (t sends (x,y,z) to (t^bc x, t^ac y, t^ab z) and one can check that the stack quotient [S-{0}/Gm] has coarse space P^1 and, that since the action is free away from xyz = 0 but has stabilizers at those 3 points, one gets a stacky curve with three non-trivial residual gerbes.
>
> **Question**: Are two 1-dimensional DM stacks with isomorphic coarse spaces
> and residual gerbes themselves isomorphic?
>
>
>
I have an idea for how to prove this when the coarse space is P^1, but in general don't know what I expect the answer to be. Also, one may have to restrict to the case when the coarse space is a smooth curve.
This might be analogous to the statement that the `angle' of a node of a rational nodal curve with one node doesn't affect the isomorphism class of the curve.
Also, the recent papers of Abromivich, Olsson, and Vistoli (on stacky GW theory) may be relevant.
| https://mathoverflow.net/users/2 | Are curves with `fractional points' uniquely determined by their residual gerbes? | Here's an example of two non-isomorphic Deligne-Mumford stacks whose coarse spaces are **A**1, and the only non-trivial residual gerbe in each case is B(**Z**/2) at the origin.
First, take the **Z**/2 action on **A**1 given by reflection around 0, given by x→-x. The stack quotient [**A**1/(**Z**/2)] has coarse space **A**1 and there's a B(**Z**/2) gerbe at the origin. Note that this stack is smooth since it has an etale cover by something smooth.
On the other hand, you can take the stack I defined in [this answer](https://mathoverflow.net/questions/1565/can-a-singular-deligne-mumford-stack-have-a-smooth-coarse-space/1584#1584): the stack quotient of the coordinate axes in **A**2 by the **Z**/2 action which switches the two axes. The coarse space is again **A**1 and the stack has a B(**Z**/2) gerbe at the origin. Note that this stack is *not* smooth since it has an etale cover by something singular, so it cannot be isomorphic to the previous stack.
An example where both stacks are smooth
---------------------------------------
The first stack will be the same [**A**1/(**Z**/2)] I used above.
Let G be the affine line with a doubled origin, regarded as a group over **A**1 (most of the fibers are trivial groups, but the fiber over the origin is **Z**/2). This G has the trivial action on **A**1. Consider the quotient stack [**A**1/G]=B**A**1 G. This is a DM stack whose coarse space is **A**1, and there's a B(**Z**/2) gerbe at the origin. Note that this stack is smooth since it has an etale cover by a smooth scheme (namely, **A**1). Note that this stack has non-separated diagonal (since the pullback G→**A**1 is non-separated), but the diagonal of [**A**1/(**Z**/2)] is separated, so the two stacks are non-isomorphic.
| 23 | https://mathoverflow.net/users/1 | 1715 | 1,074 |
https://mathoverflow.net/questions/1722 | 76 | I often use the internet to find resources for learning new mathematics and due to an explosion in online activity, there is always plenty to find. Many of these turn out to be somewhat unreadable because of writing quality, organization or presentation.
I recently found out that "The Elements of Statistical Learning' by Hastie, Tibshirani and Friedman was available free online: <http://www-stat.stanford.edu/~tibs/ElemStatLearn/> . It is a really well written book at a high technical level. Moreover, this is the second edition which means the book has already gone through quite a few levels of editing.
I was quite amazed to see a resource like this available free online.
Now, my question is, are there more resources like this? Are there free mathematics books that have it all: well-written, well-illustrated, properly typeset and so on?
Now, on the one hand, I have been saying 'book' but I am sure that good mathematical writing online is not limited to just books. On the other hand, I definitely don't mean the typical journal article. It's hard to come up with good criteria on this score, but I am talking about writing that is reasonably lengthy, addresses several topics and whose purpose is essentially pedagogical.
If so, I'd love to hear about them. Please suggest just one resource per comment so we can vote them up and provide a link!
| https://mathoverflow.net/users/812 | Free, high quality mathematical writing online? | [John Baez's stuff](http://www.math.ucr.edu/home/baez/) is a fantastic resource for learning about - well, whatever John Baez is interested in, but fortunately that's a lot of interesting stuff. Scroll down for a link to TWF as well as his expository articles.
| 30 | https://mathoverflow.net/users/290 | 1723 | 1,079 |
https://mathoverflow.net/questions/1721 | 16 | We all know how to take integer tensor powers of line bundles. I claim that one should be able to also take fractional or even complex powers of line bundles. These might not be line bundles, but they have some geometric life. They have Chern classes, and one can twist differential operators by them. How should I think about these? What do they have to do with gerbes?
| https://mathoverflow.net/users/66 | What do gerbes and complex powers of line bundles have to do with each other? | Complex powers of line bundles are classes in $H^{1,1}$, or equivalently sheaves of twisted differential operators (TDO) (let's work in the complex topology). This maps to $H^2$ with $\mathbb{C}$ coefficients, or modding out by $\mathbb{Z}$-cohomology, to $H^2$ with $\mathbb{C}^\times$ coefficients. The latter classifies $\mathbb{C}^\times$ gerbes, ie gerbes with a flat connection (usual gerbes can be described by $H^2(X,\mathcal{O}^\times)$). Note that honest line bundles give the trivial gerbe.
In fact the category of modules over a TDO only depends on the TDO up to tensoring with line bundles --- ie it only depends on the underlying gerbe, and can be described as ordinary $\mathcal{D}$-modules on the gerbe. Or if you prefer, regular holonomic modules over a TDO are the same as perverse sheaves on the underlying gerbe. This is explained eg in the encyclopedic Chapter 7 of Beilinson-Drinfeld's Quantization of Hitchin Hamiltonians document, or I think also in a paper of Kashiwara eg in the 3-volume Asterisque on singularities and rep theory (and maybe even his recent $\mathcal{D}$-modules book).
B&D talk in terms of crystalline $\mathcal{O}^\times$ gerbes rather than $\mathbb{C}^\times$ gerbes but the story is the same.
| 7 | https://mathoverflow.net/users/582 | 1737 | 1,091 |
https://mathoverflow.net/questions/1743 | 14 | Or at least it's order of magnitude.
I've only ever heard it described as "huge", and a google search turned up nothing.
Also, given that the Strassen algorithm has a significantly greater constant than Gaussian Elimination, and that Coppersmith-Winograd is greater still, are there any indications of what constant an O(n^2) matrix multiplication algorithm might have?
| https://mathoverflow.net/users/942 | What is the constant of the Coppersmith-Winograd matrix multiplication algorithm | In your second question, I think you mean "naive matrix multiplication", not "Gaussian elimination".
Henry Cohn et al had [a cute paper](http://arxiv.org/pdf/math.GR/0307321.pdf) that relates fast matrix multiply algorithms to certain groups. It doesn't do much for answering your question (unless you want to go and prove the conjectured results =), but it's a fun read.
Also, to back up **harrison**, I don't think that anyone really believes that there's an $O(n^2)$ algorithm. A fair number of people believe that there is likely to be an algorithm which is $O(n^{2+\epsilon})$ for any $\epsilon > 0$. An $O(n^2 \log n)$ algorithm would fit the bill.
**edit:** You can get a back-of-the-envelope feeling for a lower bound on the exponent of Coppersmith-Winograd based on the fact that people don't use it, even for n on the order of 10,000; naive matrix multiplication requires $2n^3 + O(n^2)$ flops, and Coppersmith-Winograd requires $Cn^{2.376} + O(n^2)$. Setting the expressions equal and solving for $C$ gives that the two algorithms would have equal performance for n = 10,000 (ignoring memory access patterns, implementation efficiency, and all sorts of other things) if the constant were about 627. In reality, it's likely much larger.
| 8 | https://mathoverflow.net/users/598 | 1749 | 1,101 |
https://mathoverflow.net/questions/430 | 26 | Homological algebra for abelian groups is a standard tool in many fields of mathematics. How much carries over to the setting of commutative monoids (with unit)? It seems like there is a notion of short exact sequence. Can we use this to define ext groups which classify extensions? What works and what doesn't work and why?
| https://mathoverflow.net/users/184 | Homological algebra for commutative monoids? | Your question can be understood as how to do Homological Algebra over the Field with one Element.
Deitmar, in <http://arxiv.org/abs/math/0608179> , section 6, gives an example of what can go wrong if you try to do sheaf cohomology directly via resolutions...
You might also want to look at his <http://arxiv.org/abs/math/0605429> ; in order to construct K-theory of monoids he sets up an analogue of the Q-construction. The Hom-sets in the resulting category are sort of Exts, maybe something to start with...
Durov, in <http://arxiv.org/abs/0704.2030> , follows the simplicial approach for commutative monads, of which commutative monoids are a special case
| 7 | https://mathoverflow.net/users/733 | 1752 | 1,102 |
https://mathoverflow.net/questions/1672 | 9 | Hello.
I am studying stochastic process.
here,
I don't know what is difference of
"the process is homogeneous"
and
"the process is stationary"
I feel confusing. It seems to similar to me.
| https://mathoverflow.net/users/877 | What is the difference between a homogeneous stochastic process and a stationary one? | A process is (strictly) stationary if any sequence of n consecutive points has the same distribution as any other sequence of n consecutive points. There are weaker definitions, for example weak stationarity is based only on the first two moments.
A (discrete valued) process is homogeneous if the transition probability between two given state values at any two times depends only on the difference between those times.
However, some references uses "homogeneous" rather loosely and confuse the two concepts.
| 6 | https://mathoverflow.net/users/261 | 1761 | 1,107 |
https://mathoverflow.net/questions/1720 | 32 | For an algebraic variety X over an algebraically closed field, does there always exist a finite set of (closed) points on X such that the only automorphism of X fixing each of the points is the identity map? If Aut(X) is finite, the answer is obviously yes (so yes for varieties of logarithmic general type in characteristic zero by Iitaka, Algebraic Geometry, 11.12, p340). For abelian varieties, one can take the set of points of order 3 [added: not so, only for polarized abelian varieties]. For P^1 one can take 3 points. Beyond that, I have no idea.
The reason I ask is that, for such varieties, descent theory becomes very easy (see Chapter 16 of the notes on algebraic geometry on my website).
| https://mathoverflow.net/users/930 | Can algebraic varieties be rigidified by finite sets of points? | I get that the answer is "no" for an abelian variety over the algebraic closure of Fp with complex multiplication by a ring with a unit of infinite order. Since you say you have already thought through the abelian variety case, I wonder whether I am missing something.
More generally, let X be any variety over the algebraic closure of Fp with an automorphism f of infinite order. A concrete example is to take X an abelian variety with CM by a number ring that contains units other than roots of unity. Any finite collection of closed points of X will lie in X(Fq) for some q=p^n. Since X(Fq) is finite, some power of f will act trivially on X(Fq). Thus, any finite set of closed points is fixed by some power of f.
As I understand the applications to descent theory, this is still uninteresting. For that purpose, we really only need to kill all automorphisms of finite order, right?
| 25 | https://mathoverflow.net/users/297 | 1768 | 1,112 |
https://mathoverflow.net/questions/1726 | 15 | So, physicists like to attach a mysterious extra cohomology class in H^2(X;C^\*) to a Kahler (or hyperkahler) manifold called a "B-field." The only concrete thing I've seen this B-field do is change the Fukaya category/A-branes: when you have a B-field, you shouldn't take flat vector bundles on a Lagrangian subvariety, but rather ones whose curvature is the B-field. How should I think about this gadget?
| https://mathoverflow.net/users/66 | How should I think about B-fields? | Let me add a few words of explanation to Aaron's comment. Perturbative string theory is (at least at the level of caricature) concerned with describing small corrections to classical gravitational physics on the spacetime X. So, to do perturbative string theory on X, you need to choose a "background" metric on X. You might need to choose other fields as well, but we can assume for now that those are all set to zero.
Having chosen a metric, you can talk about strings moving in X. In the limit where the string length goes to zero, a single string will look like a particle. What sort of particle it looks like will depend on how it's vibrating inside X. In particular, a closed string has a set of vibrational states which a) appear massless in this limit, and b) fill out a representation R of the Lorentz group. Specifically, R is the representation induced from the tensor square V (x) V, where V is the standard representation of the little group that fixes some light-like vector. You can decompose V into a sum of traceless symmetric square, trace, and antisymmetric traceless square. The states in the first summand are states of the graviton, representing tiny quantum excitations of the metric in X. The states in the last summand, the antisymmetric representation, are tiny excitations of the B-field, which we set equal to zero. (The states in the trace representation are quanta of the "dilaton" field.)
So, we didn't give the B-field any respect when we started, but it turns out to part of the definition of a string background. And once you know about the B-field, it's easy to include it in the action for the sigma model to X: Add to your action the term i<[S],f\*B>, where [S] is the fundamental class of the Riemann surface, and f: S -> X is the function embedding your string's worldsheet into X. Edit: Forgot a factor of i=root(-1), which is necessary to make the action real.
And I forgot to mention that Aaron's H is dB.
| 5 | https://mathoverflow.net/users/35508 | 1772 | 1,115 |
https://mathoverflow.net/questions/1765 | 9 | I was looking at a [paper](http://www.google.com/url?sa=t&source=web&ct=res&cd=1&ved=0CA4QFjAA&url=http%3A%2F%2Fwww.mathematik.hu-berlin.de%2F~farkas%2Fsdg-2.pdf&ei=3KTfSoXmFo-0sgPoydHhCA&usg=AFQjCNHJWSHKXWGpaxtDsY6CxCJEDGYIBg&sig2=0Yq2ygZ4ikWs5tBgAakIeA) of Farkas and the following confusing point came up.
Let $\mathscr{M}\_g$ be the moduli stack of smooth genus $g$ curves and let $\pi: \mathscr{C} \to \mathscr{M}\_g$ be the universal curve. Let $\mathscr{F}$ be $\Omega^1\_\pi \otimes \Omega^1\_\pi$, where $\Omega^1\_\pi$ is the sheaf of relative differentials of $\pi$. Then the pushforward $\pi\_\* \mathscr{F}$ is isomorphic $\Omega^1\_{\mathscr{M}\_g}$.
Why is this true? Farkas says this follows from Kodaira-Spencer theory. I googled for a while and asked a few students, but couldn't figure this out.
| https://mathoverflow.net/users/2 | Kodaira-Spencer Theory and moduli of curves | By standard deformation theory (see e.g., Hartshorne III Ex 4.10, but there are probably better references), the tangent sheaf of $\mathscr{M}\_g$ is $R^1\pi\_{\ast}(\mathscr{C}, T\_{\mathscr{C}/\mathscr{M}\_g})$, which is Serre dual to $\pi\_{\ast}\mathscr{F}$. The tangent sheaf is dual to what you wanted.
| 7 | https://mathoverflow.net/users/121 | 1779 | 1,120 |
https://mathoverflow.net/questions/1781 | 8 | The following question came up in my research. I suspect that it has a slick answer,
but I can't seem to find it.
Fix an integer n>=2 and a prime p. Define X(n) to be the set of primitive
vectors in the Z-module Z^n and Y(n,p) to be the set of "lines" in the vector space (Z/pZ)^n (ie the spans of non-zero vectors). There is a natural surjective map f:X(n)-->Y(n,p) ("reduce mod p and take the span").
Question : Does there exist a map g:Y(n,p)-->X(n) with the following two properties.
1. f(g(L))=L for all L in Y(n,p).
2. If {L\_1,...,L\_n} \subset Y(n,p) spans the vector space (Z/pZ)^n, then {g(L\_1),...,g(L\_n)} is a basis for the Z-module Z^n.
Of course, I expect that the answer is no except in certain simple situations (for instance,
it is yes for n=p=2), but I can't seem to find a proof.
EDIT : Oops! I phrased the question incorrectly. Above is a corrected version.
| https://mathoverflow.net/users/317 | Lifting bases for (Z/pZ)^n to Z^n | Here's a unified argument based on my comments to Scott's post that doesn't use quadratic reciprocity in any form. Suppose n=2 and p >= 5, and lift each line of slope i in Y(2,p) to a point (ai+pbi, iai+pci).
Since each pair of lifts should give a basis of Z2 and thus a matrix with determinant \pm 1, taking each pair from among i=1,2,k+2 (with 1 <= k <= p-3) gives us conditions
a1a2 = \pm 1 (mod p)
k\*a2ak+2 = \pm 1 (mod p)
(k+1)\*a1ak+2 = \pm 1 (mod p).
Combining the first two gives ka22\*a1ak+2 = \pm 1, or a22 = \pm(1+1/k) (mod p).
But for k=1 this gives us a22 = \pm 2, and for k=2 we get a22 = \pm (1 + (p+1)/2) = \pm (p+3)/2, so either (p+3)/2 = 2 (mod p) or (p+3)/2 = -2 (mod p). These imply p=1 and p=7, respectively, so already the only possible solution is p=7. But if p=7 then k=3 gives a22 = \pm 6, which is not \pm 2 (mod 7), so that doesn't work either. Thus a lift with n=2 can only possibly exist if p is 2 or 3.
| 4 | https://mathoverflow.net/users/428 | 1787 | 1,125 |
https://mathoverflow.net/questions/769 | 23 | Let $q$ be a power of a prime. It's well-known that the function $B(n, q) = \frac{1}{n} \sum\_{d | n} \mu \left( \frac{n}{d} \right) q^d$ counts both the number of irreducible polynomials of degree $n$ over $\mathbb{F}\_q$ and the number of [Lyndon words](http://en.wikipedia.org/wiki/Lyndon_word) of length $n$ over an alphabet of size $q$. Does there exist an explicit bijection between the two sets?
| https://mathoverflow.net/users/290 | Exhibit an explicit bijection between irreducible polynomials over finite fields and Lyndon words. | In Reutenauer's "Free Lie Algebras", section 7.6.2:
A direct bijection between primitive necklaces of length $n$ over $F$ and the set of irreducible polynomials of degree $n$ in $F[x]$ may be described as follows: let $K$ be the field with $q^n$ elements; it is a vector space of dimension $n$ over $F$, so there exists in $K$ an element $\theta$; such that the set $\{\theta, \theta^q, ..., \theta^{q^{n-1}}\}$ is a linear basis of $K$ over $F$.
With each word $w = a\_0\cdots a\_{n-1}$ of length $n$ on the alphabet $F$, associate the element $\beta$ of $K$ given by $\beta = a\_0\theta + a\_1\theta^q + \cdots + a\_{n-1} \theta^{q^{n-1}}$. It is easily shown that to conjugate words $w, w'$ correspond conjugate elements $\beta, \beta'$ in the field extension $K/F$, and that $w \mapsto \beta$ is a bijection. Hence, to a primitive conjugation class corresponds a conjugation class of cardinality $n$ in $K$; to the latter corresponds a unique irreducible polynomial of degree $n$ in $F[x]$. This gives the desired bijection.
| 31 | https://mathoverflow.net/users/961 | 1800 | 1,134 |
https://mathoverflow.net/questions/1809 | 14 | Given a homomorphism f:G→H between smooth algebraic groups, we get an induced homomorphism of algebraic stacks Bf:BG→BH, given by sending a G-torsor P over a scheme X to the H-torsor PxGH, whose (scheme-theoric) points are {(p,h)|p∈P,h∈H}/∼, where (pg,h)∼(p,f(g)h).
Is every morphism of algebraic stacks BG→BH of the form Bf? If not, what is an example of a morphism not of this form?
| https://mathoverflow.net/users/1 | Does every morphism BG-->BH come from a homomorphism G-->H? | Depends on the base scheme and the topology being used. For example if you're working over a field k in the etale or the flat topology, and take the group G to be trivial, you're asking if H^1(k,H) is trivial, which is obviously false in general. This is, in a sense, the only obstruction: for any base scheme S, giving a map from BG to any stack Y (in stacks/S) is the same as specifying a point y of Y(S), and a homomorphism G -> Aut\_S(y). In particular, if BH(S) is connected (i.e., if H^1(S,H) = \*) then the answer to your question is positive.
| 8 | https://mathoverflow.net/users/986 | 1819 | 1,147 |
https://mathoverflow.net/questions/1788 | 62 |
>
> Precisely, if an R-module M *has* a finite presentation, and Rk → M is some unrelated **surjection** (k finite), is the kernel necessarily also finitely generated?
>
>
>
Basically I want to believe I can choose generators for M however I please, and still get a finite presentation. I have reasons from algebraic geometry to believe this, but it seems like a very basic result, so I would like to understand it directly in terms of the commutative algebra, which I just can't seem to figure out...
(Here R is an arbitrary commutative ring, with no other hypotheses.)
**Edit**: All maps here are maps of R-modules. Also, the reason this is not the same as "does finite presentation imply coherent?" is that I am only asking for finite type kernels of **surjections** Rk → M. That the hypotheses assume surjectivity is a common misreading of the general definition of "coherent".
If the answer to the above is "yes", then coherent will mean "finite type, and all finite type submodules are finite presentation"
| https://mathoverflow.net/users/84526 | Does "finitely presented" mean "always finitely presented"? (Answered: Yes!) | $\require{begingroup} \begingroup$
$\def\coker{\operatorname{Coker}}$
$\def\im{\operatorname{Im}}$
Suppose that we have a short exact sequence $0 \to K \to R^m \to M \to 0$ with $K$ finitely generated over $R$ and that $0 \to K' \to R^n \to M \to 0$ is another short exact sequence. Your question is: is $K'$ necessarily finitely generated?
The answer is yes and we can see this as follows:
First, we argue for the existence of a commutative diagram
$$
\require{AMScd}
\begin{CD}
0 @>>> K @>>> R^m @>>> M @>>> 0 \\
@. @VV{\tilde{f}}V @VV{f}V @| \\
0 @>>> K' @>>> R^n @>>> M @>>> 0 \\
\end{CD}
$$
Using the fact that free modules are projective we can lift the identity map $M = M$ to an $f\colon R^m\to R^n$ which makes the right hand square commute. Restricting $f$ to a map $\tilde{f}\colon K → K'$ fills in the last square and so we have the diagram as claimed.
Now using Snake's lemma we find that there is an isomorphism $\coker{\tilde{f}} \cong \coker{f}$. Thus We have a short exact sequence;
$$
0\to \im{\tilde{f}}\to K'\to \coker{f}\to 0.
$$
Since $K'$ is squeezed between two finitely generated $R$ modules, it follows (by a [well-known-fact](https://math.stackexchange.com/questions/234753/finitely-generated-modules-in-exact-sequence/3490591#3490591)) that $K'$ is itself finitely generated.
$\endgroup$
| 71 | https://mathoverflow.net/users/493 | 1824 | 1,150 |
https://mathoverflow.net/questions/1818 | 4 | As someone whose knowledge of cohomology is patchy and picked up on a need-to-know basis, and whose algebraic geometry is even worse, I wondered if someone could help with this question. (I ran into it a while back while trying to answer some questions about the *Fourier* algebras of compact Lie groups.) The solution could just be to point me to the right bit of the right literature.
In more detail: using R to denotes the reals, regard the special orthogonal group SO(n,R) as an affine variety in Mn(R)=Rn^2; and let A be the coordinate ring of this variety, regarded as an R-algebra (I think this is the correct terminology; I just mean the algebra of R-valued functions on the group that is generated by the coordinate functions). I would like to know the largest *d* for which Hdalt(A,A) is non-zero, where Hd is Hochschild homology of A with coefficients in itself, and "alt" is the part generated by alternating a.k.a. antisymmetric cycles.
The vague idea I had was that said *d* should be equal to the dimension of the tangent space (as defined via maximal ideals in A) at a point of the original variety, and hence to the dimension of SO(n,R) as a real manifold. I've tried to do some hunting in the literature, but much of what I come across is for general affine/projective varieties, and so is overly general and not exactly self-contained.
Is there a general picture telling us that the Hochschild homology of A can be calculated in terms of the Lie algebra cohomology of *so*(n,R)? If so, does this work for compact forms of other semisimple Lie algebras?
**Edit:** it's been pointed out that, if we know A is a smooth ring in the sense of commutative algebra, then one can wheel in the Hochschild-Kostant-Rosenberg theorem. I had avoided mentioning this, because although I'm sure A should be smooth in this sense, one of my problems is trying to find an explicit reference to this putative fact. It also feels (since the variety SO(n,R) has such good global symmetry) as if there should be a proof which doesn't rely on proving smoothness of A.
**Final edit (for now):** Having gone and done some reading in a few textbooks, it look s like the simplest - if not the most elementary solution is to "observe" or cite the fact that SO(n,R) is a smooth variety over R, use localization to reduce the computation to Hochschild homology of the local ring at a point, and then use HKR or something similar. This still feels like overkill to me, but at least it looks to give a well-defined proof. (If you can see a better route, please feel free to add a new comment!)
| https://mathoverflow.net/users/763 | What is the homology of the real coordinate ring of SO(n,R)? Other compact matrix groups? | I am very far from an expert on the subject, but I think the Hochschild homology of the coordinate ring should be the algebraic de Rham complex of your variety SO(n, R)--*not* the cohomology of the complex, just the groups in the complex (with 0 differential if you like). This is the Hochschild-Kostant-Rosenberg theorem and should just require smoothness, not the fact that you have a Lie group.
I don't know what HH^{alt} is, but maybe you can figure it out from this description. Unless I am very confused about something, HH\_k is nonzero exactly for 0 <= k <= the dimension of SO(n, R), so your guess about d sounds plausible.
| 4 | https://mathoverflow.net/users/126667 | 1834 | 1,156 |
https://mathoverflow.net/questions/452 | 7 | $f:\mathbf{R}^d\to \mathbf{R}\_{\ge 0}$ is *log-concave* if $\log(f)$ is concave (and the domain of $\log(f)$ is convex).
Theorem: For all $\sigma$ on the sphere $\Bbb S^{d-1}$ and $r\in \mathbf{R}$,
$$
g\_\sigma(r) := \int\limits\_{\sigma\cdot x=r}f(x)\,\mathrm{d}S(x)
$$
is a log-concave function of $r$. (Note: $g$, as a function of $\sigma$ and $r$, is the Radon transform of $f$.)
Question: does this characterize log-concavity? That is, if $g\_\sigma(r)$ is log-concave as a function of $r$ for all $\sigma$, is $f$ log-concave?
| https://mathoverflow.net/users/302 | Characterizing the Radon transforms of log-concave functions | If I understand the question correctly, I think the answer is no.
Start with the following : if $f$ is the indicator function of the unit ball, then the function $g\_\sigma(r)$ is strictly log-concave close to 0 (this function does not depend on $\theta$).
Now, let $h$ be the indicator function of the ball of radius $r<1$. Then $f-\epsilon h$ is never log-concave for any $\epsilon>0$, and its Radon transform (which again is independent of $\theta$) remains log-concave if epsilon is small enough.
(this is especially easy to see in dimension 2, in which case the Radon transforms of both $f$ and $h$ are second-degree polynomials on their support)
| 5 | https://mathoverflow.net/users/908 | 1838 | 1,160 |
https://mathoverflow.net/questions/1634 | 22 | I am not too certain what these two properties mean geometrically. It sounds very vaguely to me that finite type corresponds to some sort of "finite dimensionality", while finite corresponds to "ramified cover". Is there any way to make this precise? Or can anyone elaborate on the geometric meaning of it?
| https://mathoverflow.net/users/nan | Finite type/finite morphism | I definitely agree with Peter's general intuitive description.
In response to some of the subsequent comments, here are some implications to keep in mind:
**Finite** ==> **finite fibres** (1971 EGA I 6.11.1) and **projective** (EGA II 6.1.11), hence **proper** (EGA II 5.5.3), but *not conversely*, contrary to popular belief ;)
**Proper** + **locally finite presentation** + **finite fibres** ==> **finite** (EGA IV (part 3) 8.11.1)
When reading about these, you'll need to know that "quasi-finite" means "finite type with finite fibres." Also be warned that in EGA (II.5.5.2) projective means $X$ is a closed subscheme of a "finite type projective bundle" $\mathbb{P}\_Y(\mathcal{E})$, which gives a nice description via relative Proj, whereas "Hartshorne-projective" more restrictively means that $X$ is closed subscheme of "projective n-space" $\mathbb{P}^n\_Y$.
When the target (or "base" scheme) is locally Noetherian, like pretty much anything that comes up in "geometry", a proper morphism is automatically of locally finite presentation, so in that case we do have
**finite** <==> **proper** + **finite fibres**
Regarding "locally finite type", its does *not* imply finite dimensionality of the fibres; rather, it's about finite dimensionality of small neighborhoods of the source of the map. For example, you can cover a scheme by some super-duper-uncountably-infinite disjoint union of copies of itself that is LFT but not FT, since it has gigantic fibres.
| 22 | https://mathoverflow.net/users/84526 | 1839 | 1,161 |
https://mathoverflow.net/questions/1827 | 18 | So here's my problem: I have no intuition for how a "generic" module over a commutative ring should behave. (I think I should never have been told "modules are like vector spaces.") The only examples I'm really comfortable with are
* vector spaces,
* finitely generated modules over a PID, and
* modules over a group algebra.
But when I try to apply these examples to understanding something like Nakayama's lemma I don't have any intuition to bring to the table. So, what other examples of modules should I keep in mind so that
* I'm not fooled by my intuition about vector spaces, and
* I can concretely understand what something like Nakayama's lemma means, at least in an important special case?
| https://mathoverflow.net/users/290 | What representative examples of modules should I keep in mind? | Yes, there is a big class of modules that have an intuition different from the abstract algebra, namely the ones that come from an **algebraic geometry**. If $R$ is a (say, Noetherian) commutative ring, then you consider a scheme $\mathrm{Spec}\, R$ and (finitely generated) modules over $R$ correspond to coherent sheaves on $\mathrm{Spec}\, R$.
For example, a **skyscraper sheaf** on point $p$ corresponds to the module $R/m\_p$, where $ m\_p$ is a maximal ideal that defines point. More interestingly, a **locally trivial bundle** over $\mathrm{Spec}\, R$ corresponds to a projective module (this is exactly the statement locally free = projective).
The objects that seem unnatural on algebraic side, like support of a module, in fact have a clear geometric meaning — **support** of a sheaf is where the sheaf sits, that's it. And this is the geometric setting that explains Nakayama's lemma: the lemma says you can consider things locally for a coherent sheaf (see [wikipedia](http://en.wikipedia.org/wiki/Coherent_sheaf#Definition)).
| 14 | https://mathoverflow.net/users/65 | 1845 | 1,167 |
https://mathoverflow.net/questions/1805 | 6 | Let g be a finite dimensional Lie algebra over k, and let U be its universal enveloping Lie algebra. Is there a left module M of U which is projective but not free? That is, is the Quillen-Suslin theorem still true for enveloping algebras?
Quillen-Suslin says this there are no non-free projectives for S(g), the associated graded algebra of U. Thus, if the associated graded module of a projective is projective, then it is free (and so the original module was also free). Therefore, this question is equivalent to the question "Is the associated graded module of a projective U-module always projective?"
My guess is no, because the Weyl algebra has non-free projectives, even though it's associated graded algebra is a polynomial algebra. However, the tricks I know that work for the Weyl algebra don't work for Lie algebras. I would love a simple example of a non-free projective U-module.
| https://mathoverflow.net/users/750 | Non-free projective modules for a Universal Enveloping Algebra? | In [this paper](http://www.numdam.org/numdam-bin/item?id=CM_1985__54_1_63_0) Stafford shows that whenever g is a finite-dimensional non-abelian Lie algebra the enveloping algebra has non-free but stably free (and therefore projective) right ideals. He also shows how to construct them.
| 5 | https://mathoverflow.net/users/345 | 1860 | 1,179 |
https://mathoverflow.net/questions/1828 | 3 | Suppose w^(2n)=1 (w is a complex number).
For which n (if any) \sqrt(w) \in Q(w) ?
| https://mathoverflow.net/users/966 | Is an nth root of unity a square? | The key point is to understand the field Q(w) for w a primitive kth root of unity. Call this field Qk. In particular, you want to know that Q4n \neq Q2n.
The key fact here is that the field extension Qk/Q has degree phi(k), where phi(k) is the Euler phi function, and phi(4k) \neq phi(2k). For a proof that Qk/Q has degree phi(k), see the early parts of any book on cyclotomic fields. This is probably also done in many Galois theory books but I don't know which ones.
| 3 | https://mathoverflow.net/users/297 | 1862 | 1,180 |
https://mathoverflow.net/questions/1812 | 20 | While many of us have had the experience of learning mathematics informally by osmosis or more formally in classes, there are times when we have to sit down and systematically learn, without the benefit of a class, large amounts of mathematics. For instance, there might be a technique that we need from a field we are not familiar with.
When you find yourself in such a situation, what are your best tricks for teaching yourself new mathematics?
| https://mathoverflow.net/users/812 | Learning new mathematics | Many have remarked that they first understood a subject only when they first taught it.
| 14 | https://mathoverflow.net/users/454 | 1871 | 1,185 |
https://mathoverflow.net/questions/1878 | 5 | Functions on an algebraic subvariety X of A^n are the same as functions on A^n restricted to X. So the statement that functions on X extend to all of A^n follows by the definition. My question is: does the analogous statement hold for C^n and closed complex submanifolds (maybe even closed analytic subvarieties), and if so, how is this proved?
| https://mathoverflow.net/users/788 | Extending Functions on Closed Submanifolds of C^n | Yes, this is true. It follows from "Cartan's Theorem B" which says that H^1 of any coherent analytic sheaf on a closed submanifold of C^n is 0; the same result is also true for analytic subspaces. Look up any book on several complex variables for a proof. (It is quite possible that there is a more elementary proof.)
(One uses the theorem as follow: Let X be the submanifold or analytic space and consider the exact sequence of sheaves on C^n
0 --> I --> O\_{C^n} --> O\_X --> 0
where I is the ideal sheaf of X. The vanishing of H^1(C^n,I) implies that the map H^0(C^,O\_{C^n}) to H^0(X,O\_X) is surjective, which is what you want.)
| 4 | https://mathoverflow.net/users/519 | 1883 | 1,192 |
https://mathoverflow.net/questions/1822 | 9 | The *norm* of a graph is the largest eigenvalue of the adjacency matrix. I'll write ||`G`|| for the norm of `G`.
Now, fix some graph `G` with a chosen vertex `*`, and consider the family of graphs `G_k` obtained by adding a chain of `k` edges to `*`.
For many such examples, the sequence `{`||`G_k`||`^2}_k` appears to be never cyclotomic; I'd like some ideas as to how I might try to prove such statements for particular graphs `G`.
I know how to show individual algebraic integers aren't cyclotomic -- modulo any prime not dividing the discriminant, the minimal polynomial of a cyclotomic integer must factor into factors with uniform degree. This approach seems very hard to make work for a family of numbers, although I'm aware of the work of [Asaeda-Yasuda](http://arxiv.org/abs/0711.4144) in which they did this for the graph
```
o-o-o
/
*-o-o-o
\
o-o-o
```
(with the exception of k=4, where the norm-square is in fact cyclotomic). If anyone has ideas about how one should attack such a question, or examples of similar problems, please let me know!
Finally -- the application here is to subfactors; [Etingof-Nikshych-Ostrik](http://arxiv.org/abs/math.QA/0203060) proved that the index of a subfactor must be a cyclotomic integer, and the index is just the norm square of the principal graph. When we look for possible new examples of subfactors, we tend to get results constraining the principal graph to lie in such a sequence `{G_k}`.
| https://mathoverflow.net/users/3 | How can I prove that a sequence of squares of graph norms is never cyclotomic? | This is a vague thought: is there some simple recurrence for the characteristic polynomials of the charctertistic polynomials of the corresponding matrices. For example, if you look at the A\_n chains, the polynomials are the Chebyshev polynomials, whose roots are cyclotomic, and which obey a simple resursion.
Even if you had a recursion, I do not know how to show that the roots are not cyclotomic, but the problem feels more tractable.
| 3 | https://mathoverflow.net/users/297 | 1885 | 1,194 |
https://mathoverflow.net/questions/1876 | 9 | There are a lot of results in textbooks concerned with canonical forms of matrices under certain complex groups of transformations, e.g. GL(n|C), O(n|C),...
Could anybody give me references where the canonical forms of real matrices under the action of SO(p,q|R) were found. Of most interest is the canonical form of antisymmetric matrices, i.e. that of the adjoint representation. Other related results, e.g. on canonical forms under SU(p,q), SP(p,q) are also appreciated. Thanks!
| https://mathoverflow.net/users/985 | Orbits of real groups, canonical forms of matrices | In Chapter 2 of [the PhD thesis of Charles Boubel](http://cat.inist.fr/?aModele=afficheN&cpsidt=204208) (in French, though), you can find the normal forms of a symmetric or antisymmetric bilinear form under the automorphism group of a nondegenerate bilinear form, again either symmetric or skewsymmetric. In particular you can find there the canonical forms for elements of the Lie algebra of SO(p,q|R) under the adjoint action.
A summary of this work can be found (in German, though) in the Diplom thesis of Thomas Neukirchner *Normalform eines schief- oder selbstadjungierten Endomorphismus auf
einem pseudo-euklidischen oder symplektischen Vektorraum über R oder C — eine Arbeit
von Charles Boubel,* which is a Humboldt University preprint, and also for some special signatures in a paper of Felipe Leitner's *Imaginary Killing spinors in lorentzian geometry* (math.DG/0302024) and also in a paper of mine and Joan Simón's *Supersymmetric Kaluza-Klein reductions of AdS backgrounds* (hep-th/0401206). As in Leitner's paper, we were mostly concerned with the case q=2, though.
I hope this helps. I have PDFs of Boubel's thesis, which I can make available if you want.
| 10 | https://mathoverflow.net/users/394 | 1891 | 1,197 |
https://mathoverflow.net/questions/1858 | 9 | Suppose that X and Y are finite sets and that f : X → Y is an arbitrary map. Let PB denote the pullback of f with itself (in the category of sets) as displayed by the commutative diagram
PB → X
↓ ↓
X → Y
[Terence Tao](http://terrytao.wordpress.com/2007/04/01/open-question-triangle-and-diamond-densities-in-large-dense-graphs/#comment-518) observes in one the comments on his weblog that the product of |PB| and |Y| is always greater than or equal to |X|2. (This is an application of the Cauchy-Schwarz inequality.) This fact may be rephrased as follows: If we ignore in the above diagram all arrows and replace the sets by their cardinalities we obtain a 2x2 matrix with a non-negative determinant.
The question is whether this is a general phenomenon. Suppose that n is a positive integer and that X1, X2, ... ,Xn are finite sets; furthermore we are given maps f1 : X1 → X2, f2 : X2 → X3, ... , fn-1 : Xn-1 → Xn. We construct a pullback diagram of size nxn. The diagram for n=4 is shown below.
PB → PB → PB → X1
↓ ↓ ↓ ↓
PB → PB → PB → X2
↓ ↓ ↓ ↓
PB → PB → PB → X3
↓ ↓ ↓ ↓
X1 → X2 → X3 → X4
Here, the maps between the Xi in the last row and column are the corresponding fi and the PBs denote the induced pullbacks. (Of course, although they are denoted by the same symbol, different PBs are different objects.) The PBs can be constructed recursively. First, take the pullback of X3 → X4 ← X3; it comes with maps X3 ← PB → X3. Having constructed this, take the pullback of X2 → X3 ← PB and so forth.
**Ignore all arrows and replace sets by their cardinalities. Is the determinant of the resulting nxn matrix always non-negative?**
| https://mathoverflow.net/users/296 | Determinant of a pullback diagram | Write Xn = {x1, ..., xk}. For each 1 ≤ i ≤ k let wi be the vector whose jth component is the cardinality of the inverse image of xj in Xi. Then your matrix is the sum w1w1T + ... + wkwkT, a sum of positive semidefinite matrices, so it is positive semidefinite and in particular has nonnegative determininant.
| 10 | https://mathoverflow.net/users/126667 | 1892 | 1,198 |
https://mathoverflow.net/questions/1894 | 5 | Given groups $G\_1, G\_2, G\_3$ and injections $A\_1 \to G\_1$ and $A\_1
\to G\_2$ , from $A\_2 \to G\_2$ and $A\_2 \to G\_3$, let $G\_1 \*\_{A\_1} \*G\_2 \*\_{A\_2} G\_3$ be the amalgam formed these groups and maps.
Then is it true that $G\_1 \*\_{A\_1} \*G\_2 \*\_{A\_2} G\_3$ is the same as (G\_1 \*\_{A\_1} G\_2 ) \*\_{A\_2} G\_3. If yes, how do we see this?
| https://mathoverflow.net/users/996 | is amalgamation of groups associative | Amalgamation of groups is a categorical construction known as a "pushout": <http://en.wikipedia.org/wiki/Pushout_(category_theory)>
By general category theory, pushouts are associative up to unique isomorphism, i.e., the two things you wrote are isomorphic in a unique way (subject to commuting with the inclusions from A\_i, etc.)
| 7 | https://mathoverflow.net/users/321 | 1898 | 1,202 |
https://mathoverflow.net/questions/1887 | 0 | I have a graph with Edge `E` and Vertex `V`, I can find the spanning tree using [Kruskal algorithm](http://www-b2.is.tokushima-u.ac.jp/~ikeda/suuri/kruskal/Kruskal.shtml), now I want to find all the cycle bases that are created by utilitizing that spanning tree and the edges that are not on the tree, any algorithm that allows me to do that, besides brute force search?
I can, of course, starts from one vertex of the non-spanning tree edge, gets all the edges, explore all of them, retracts if I find dead end, until I come back to the other vertex of the edge. But this is a bit, err... brutal. Any other ideas?
| https://mathoverflow.net/users/807 | Given a Spanning Tree and an Edge Not on the Spanning Tree, How to Form a Cycle Base? | I think you're likely to get better answers if you post this question on Stack Overflow rather than here.
But anyways, if your graph doesn't have weights on edges, you don't need Kruskal's algorithm to find a spanning tree; you can just use DFS. As you compute the tree, store for each vertex its parent and the distance to the root. Then when you encounter a non-tree edge, you can find the nearest common ancestor of its two endpoints efficiently.
| 3 | https://mathoverflow.net/users/126667 | 1901 | 1,205 |
https://mathoverflow.net/questions/1893 | 6 | I've been reading some "introduction to categories" type materials and have been impressed with the all-encompassing nature, but the skeptic in me wonders: is there any mathematical object that categories *can't* describe?
To be quite specific, I'd be interested any of these:
a.) Objects that can be described by categories that have properties that can't.
b.) Category equivalents of set-theoretic type limits, like how "the set of all sets" causes problems.
c.) Some type of mathematics so pathological it foils, say, associativity. It doesn't need to be a mathematics that's useful in any sense, just one designed specifically to be impossible to describe with categories.
| https://mathoverflow.net/users/441 | What can't be described by categories? | I don't quite understand your question, but if you're asking whether category theorists should worry about set-theoretic problems the answer seems to be "sometimes". I'm not an expert in this area, but it seems that people tend to avoid universal constructions like limits over large diagrams, and in other cases, people assume the existence of strongly inaccessible cardinals. This seems to avoid standard contradictions, but I must confess that I've never checked such arguments.
I don't know many references for this question. Lurie discusses some constructions in section 5.4 of [Higher topos theory](http://arxiv.org/abs/math.CT/0608040).
| 6 | https://mathoverflow.net/users/121 | 1903 | 1,207 |
https://mathoverflow.net/questions/1908 | 2 | Recall that a ring homomorphism A->B is geometrically regular if for all primes p of A, the fiber of B over p is geometrically regular over k(p). A Grothendieck ring (or, G-ring) is one for which A\_p->A\_p\* is regular for all primes p. These are the maps from the local rings of A to their completions.
If A is an order in a number field, is A a G-ring? Equivalently (in this special case), is A excellent?
I've heard it said that 'all' rings that appear in algebraic geometry are excellent. Since an order A corresponds to a singular curve, I guess I expect A to be excellent as well.
| https://mathoverflow.net/users/100 | Are non-maximal orders in number fields Grothendieck rings? | Yes: if R is excellent, so is any finite type R-algebra (apply this to Z and A).
| 5 | https://mathoverflow.net/users/986 | 1911 | 1,213 |
https://mathoverflow.net/questions/1931 | 27 | Grothendieck's approach to algebraic geometry in particular tells us to treat all rings as rings of functions
on some sort of space. This can also be applied outside of scheme theory (e.g., Gelfand-Neumark theorem
says that the category of measurable spaces is contravariantly equivalent to the category
of commutative von Neumann algebras). Even though we do not have a complete geometric description
for the noncommutative case, we can still use geometric intuition from the commutative case effectively.
A generalization of this idea is given by Grothendieck's relative point of view, which says that a morphism of rings f: A → B
should be regarded geometrically as a bundle of spaces with the total space Spec B fibered over the base space Spec A
and all notions defined for individual spaces should be generalized to such bundles fiberwise.
For example, for von Neumann algebras we have operator valued weights, relative L^p-spaces etc.,
which generalize the usual notions of weight, noncommutative L^p-space etc.
In noncommutative geometry this point of view is further generalized to bimodules.
A morphism f: A → B can be interpreted as an A-B-bimodule B, with the right action of B given by the multiplication
and the left action of A given by f.
Geometrically, an A-B-bimodule is like a vector bundle over the product of Spec A and Spec B.
If a bimodule comes from a morphism f, then it looks like a trivial line bundle with the support being
equal to the graph of f.
In particular, the identity morphism corresponds to the trivial line bundle over the diagonal.
For the case of commutative von Neumann algebras all of the above can be made fully rigorous
using an appropriate monoidal category of von Neumann algebras.
This bimodule point of view is extremely fruitful in noncommutative geometry (think of Jones' index,
Connes' correspondences etc.)
However, I have never seen bimodules in other branches of geometry (scheme theory,
smooth manifolds, holomorphic manifolds, topology etc.) used to the same extent as they are used
in noncommutative geometry.
Can anybody state some interesting theorems (or theories) involving bimodules in such a setting?
Or just give some references to interesting papers?
Or if the above sentences refer to the empty set, provide an explanation of this fact?
| https://mathoverflow.net/users/402 | Bimodules in geometry | In "commutative geometry," I think bimodules tend to be a little concealed. People are more likely to talk about "correspondences" which are the space version of bimodules: A correspondence between spaces X and Y is a space Z with maps to X and Y.
When you think in this language, there are lots of examples you're missing. For example, the right notion of a morphism between two symplectic manifolds is a Lagrangian subvariety of their product, or even a manifold mapping to their product with Lagrangian image (maybe not embedded). See, for example, Wehrheim and Woodward's *[Functoriality for Lagrangian correspondences in Floer homology](https://arxiv.org/abs/0708.2851)*.
Similarly, correspondences are incredibly important in geometric representation theory. See, for example, the work of Nakajima on quiver varieties.
The theory of stacks also is at least partially founded on taking correspondences seriously as objects, and in particular being able to quotients by any (flat) correspondence.
This same philosophy also underlies [groupoidification](http://ncatlab.org/nlab/show/groupoidification) as studied by the Baez school (they tend to use the word "span" instead of "correspondence" but it's the same thing).
| 20 | https://mathoverflow.net/users/66 | 1943 | 1,235 |
https://mathoverflow.net/questions/1949 | 1 | This is not a homework question.
Just wondering if there is a general formula for the gaussian curvature at point (x,y,f(x,y)) in terms of x, y, and f(x,y).
I didn't see any thing like that on the [wikipedia article](http://en.wikipedia.org/wiki/Gaussian_curvature), but maybe it's hidden behind all these letters/symbols which are not actually described.
By the way, I'm studying computer science. For us, a general formula is one that works most of the time. When it doesn't work, we call this an exception, and it requires special handling :). So here of course, the function is assumed everywhere indefinitely continuous and differentiable and bounded inside any box that would fit the RAM - whatever that means - and also in R3, inside a standard classical euclidean space...
Thank you.
| https://mathoverflow.net/users/1007 | Gaussian curvature of a z=f(x,y) function | There are a few explicit formulas at <http://mathworld.wolfram.com/GaussianCurvature.html>.
| 4 | https://mathoverflow.net/users/303 | 1952 | 1,239 |
https://mathoverflow.net/questions/1916 | 6 | In a complex analysis course I have been given the following definition:
Let $X$ be a Riemann surface, denote by $H(1,0)$ the space of all $(1,0)$-holomorphic forms on $X$ and consider the quotient vector space (over $\mathbb{C}$) of $H(1,0)$ by $$\{f \in H(1,0) \mid f = d(\phi) \text{ for some } \phi \in C(X)\}.$$ The dimension of this vector space is called the genus of the surface.
Does anyone know of any good book that deals with this?
Thank you.
| https://mathoverflow.net/users/997 | Space of $(1,0)$-holomorphic forms on a Riemann surface | There is the introductory graduate-level text Riemann Surfaces by Otto Forster which approaches the subject from just the angle suggested by the definition you were given. If you read French there is the book Quelques Aspects des Surfaces de Riemann by Eric Reyssat, a gentle introduction with a broad outlook. Rather more demanding is Compact Riemann Surfaces by Raghavan Narasimhan, a modern treatment that is not overly long but covers considerable ground. Actually, there are many good books on Riemann surfaces, not all from an algebraic geometry viewpoint. If you can get your hands on them, the wonderful Columbia University notes of Lipman Bers show Riemann surfaces from a complex analysis/PDE/differential geometry angle that you should not miss. They date from 1957, so inevitably some things are not there (I don't recommend them for the specific purpose you had in mind). If your taste is towards analysis, there is also Compact Riemann Surfaces by Jürgen Jost.
I think Forster's book is my best response to your question. Or perhaps even more useful if you are in a hurry; Chapter 9 of the second edition of Complex Analysis in One Variable by Narasimhan may be all you need!
| 9 | https://mathoverflow.net/users/3304 | 1953 | 1,240 |
https://mathoverflow.net/questions/1832 | 16 | My understanding is that an analogy along the following lines is (roughly) true:
"The Alexander polynomial is to knot Floer homology is to gl(1|1)
as the Jones polynomial is to Khovanov homology is to sl(2)
as a-lot-of-other-specializations-of-HOMFLY are to Khovanov-Rozansky homology are to sl(n)."
1) To what extent is it possible to add another line that starts with the (unspecialized) HOMFLY polynomial? I think there is a triply-graded complex that I can put here (and that maybe this is what I should be calling Khovanov-Rozansky homology? or at least is also due to them?), but is there an analogous object to put in place of the Lie (super-)algebras appearing above?
2) Why is gl(1|1) here? That seems weird.
| https://mathoverflow.net/users/361 | HOMFLY and homology; also superalgebras | In terms of just the knot polynomials, there's a simple explanation for what's going on that makes $\mathfrak{gl}(1|1)$ seem totally in place:
The knot polynomial attached to the defining representation of $\mathfrak{gl}(m|n)$ only depends on m-n (the dimension of that representation in the categorical sense); you just get the specialization of HOMFLY at $t=q^{m-n}$.
Furthermore, nothing much interesting happens at negative values, since they're basically the same as positive ones. So, our current techniques, which work for $\mathfrak{gl}(n)$ knot homology, can't get at dimension 0, which can be minimally described as $\mathfrak{gl}(1|1)$.
| 10 | https://mathoverflow.net/users/66 | 1970 | 1,252 |
https://mathoverflow.net/questions/1972 | 17 | Can someone explain, explicitly, how to, given a reductive complex algebraic group construct the Langlands dual group? I know it is a group with the cocharacters of G as its characters, but how does one go about writing down what group it is?
| https://mathoverflow.net/users/622 | Langlands Dual Groups | You can construct the dual group in a combinatorial manner as follows: Reductive groups are classified by their root datum. There is an obvious duality on the set of all root data, and the dual group is the reductive group with the dual root datum.
You can see [Wikipedia](http://en.wikipedia.org/wiki/Root_datum) for the notion of the root datum.
| 14 | https://mathoverflow.net/users/425 | 1991 | 1,267 |
https://mathoverflow.net/questions/697 | 2 | Hi,
I was wondering if anyone could point me to any sources regarding the convergence of iterated affine transformation, i.e. sequences where {a\_n} is a set of affine transforms and the sequence:
a\_n (a\_{n\_1}(...a\_1)...) converges to something as n->infinity. (Preferably another affine transform).
I know my question is bit vague, most of what I've been able to dredge up so far seemed to be based on probability, but I'm not necessarily looking at probabilistic results.
Thanks.
EDIT (Second edit): I had put in a restriction here - but on second thought I was wrong about the restriction so the question still stands as is. Sorry.
| https://mathoverflow.net/users/16 | Convergence of Affine Transformations | Hi,
could you perhaps specify what kind of space your transformations are acting on? Before you do that, let me try to still share some things...
If it's a Riemannian (sub)manifold, e.g., Euclidean plane, then your problem fits well within the framework of dynamical systems, in particular "discrete-time" dynamical systems as your transformations are indexed by a countable set. Depending on the transformations, you might end up having an attracting set, and the limiting operator would amount to a kind of "dynamical projection" of the entire space to that attracting set. If transformations are all equal, i.e., a\_k = a\_l, for all k,l, then you have an "affine, time-invariant (autonomous) system". If not, you have an "affine, time-varying (non-autonomous) system".
Most of dynamical systems will not be phrased exactly as you presented it, rather, the sequence of transformations will be generated as a solution of a difference/differential equation, especially at the introductory level. The language you are using is more common in ergodic theory, which I think is the appropriate setting for types of questions you are asking. It deals with limiting processes for (semi)groups of operators, however, the operators in question are often considered to be linear (they are composition operators on spaces of functions/distributions). Perhaps more advanced texts do generalize to non-linear operators. Additionally, reapplication of affine transformation (if it's the same one), translates to a reapplication of a linear operator + a series generated by reapplication of linear operator to the offset vector, in a handwavy way. :) I therefore believe there is hope for your problem within the context of ergodic theory. As an intro text, perhaps you'd want to look into Silva: Invitation to ergodic theory (recently published by AMS). For a more advanced treatment, you'd have to look for something more advanced, e.g., Petersen to start with, but perhaps going to Cornfeld, Fomin, Sinai (I have still to even parse through that).
Hope this helped.
| 1 | https://mathoverflow.net/users/992 | 1998 | 1,270 |
https://mathoverflow.net/questions/2004 | 8 | Let $K$ be a number field and suppose $K$ contains no $p$-power roots of unity. Let $\mathcal{P}$ be a prime of $K$ *above the rational prime $p$*. Can someone prove or disprove the assertion that the local field $K\_{\mathcal{P}}$ will contain no $p$-power roots of unity?
| https://mathoverflow.net/users/1018 | p-power roots of unity in local fields | Looks to me like this is false. Let $K = \mathbb{Q}(z)/(z^p-1-p^2)$. This is an extension of degree $p$, so it is disjoint from the p-th cycloctomic field, and hence does not contain a $p$-th root of $1$. Thus, it also can not contain a $p^k$-th root of 1.
Now, let's see how $z^p - 1 - p^2$ factors in Qp. There is already one p-th root of $1+p^2$ in Qp; call this root a. (To see this, note that the power series $(1+x)^{1/p} = 1+(1/p) x + \binom{1/p}{2} x^2 + ...$ converges for $x=p^2$.)
Let $\mathcal{P}$ be a prime of $K$ corresponding to a factor of $z^p-1-p^2$ other than z-a. (In fact, $( z^p-1-p^2)/(z-a)$ is irreducible over $\mathbb{Q}\_p$, but I don't need that.) So $K\_\mathcal{P}$ contains a root b of $z^p-1-p^2$ other than $a$. But then $b/a$ is in $K\_\mathcal{P}$ and is a $p$-th root of 1.
This might be true if you ask $K/\mathbb{Q}$ to be Galois, but I would bet against it.
| 8 | https://mathoverflow.net/users/297 | 2011 | 1,277 |
https://mathoverflow.net/questions/2014 | 86 | There is a standard problem in elementary probability that goes as follows. Consider a stick of length 1. Pick two points uniformly at random on the stick, and break the stick at those points. What is the probability that the three segments obtained in this way form a triangle?
Of course this is the probability that no one of the short sticks is longer than 1/2. This probability turns out to be 1/4. See, for example, problem 5 in [these homework solutions](http://www.isds.duke.edu/courses/Fall05/sta104/hw/hw08sol.pdf) ([Wayback Machine](http://web.archive.org/web/20100619170143/http://www.isds.duke.edu/courses/Fall05/sta104/hw/hw08sol.pdf)).
It feels like there should be a nice symmetry-based argument for this answer, but I can't figure it out. I remember seeing once a solution to this problem where the two endpoints of the interval were joined to form a circle, but I can't reconstruct it. Can anybody help?
| https://mathoverflow.net/users/143 | If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4. Is there a nice proof of this? | Here's what seems like the sort of argument you're looking for (based off of a trick Wendel used to compute the probability the convex hull of a set of random points on a sphere contains the center of the sphere, which is really the same question in disguise):
Connect the endpoints of the stick into a circle. We now imagine we're cutting at three points instead of two. We can form a triangle if none of the resulting pieces is at least 1/2, i.e. if no semicircle contains all three of our cut points.
Now imagine our cut as being formed in two stages. In the first stage, we choose three pairs of antipodal points on the circle. In the second, we choose one point from each pair to cut at. The sets of three points lying in a semicircle (the nontriangles) correspond exactly to the sets of three consecutive points out of our six chosen points. This means that 6 out of the possible 8 selections in the second stage lead to a non-triangle, regardless of the pairs of points chosen in the first stage.
| 117 | https://mathoverflow.net/users/405 | 2016 | 1,280 |
https://mathoverflow.net/questions/1967 | 19 | I was talking this morning to a colleague who thinks about combinatorial Hopf algebras. He mentioned several rings, which are of interest in combinatorics, for which he didn't know whether a Hopf structure existed. I was able to rule out several by the following result:
If A is a finitely generated commutative algebra over a field of characteristic 0, and A has a Hopf structure, then A is a regular ring.
So, two questions:
(1) The only reference I know for this is Tate's article on group schemes in "Modular Forms and Fermat's Last Theorem." Does anyone know a version which targeted towards a reader who likes algebra better than geometry? (So, for example, "Hopf algebra" is a friendlier term than "group scheme".)
(2) Are there any useful generalizations that take out "commutative" or "finitely generated"?
| https://mathoverflow.net/users/297 | Hopf algebra reference | Oort has an elementary proof that group schemes in char. 0 are reduced -- see MR0206005.
| 8 | https://mathoverflow.net/users/986 | 2023 | 1,285 |
https://mathoverflow.net/questions/1912 | 7 | Sorry for a loaded question.
I'm not an expert on those things, but I do know that a fibration gives rise to the representations of pointed fundamental group of the base on the cohomology of the fiber.
What are the properties of this map for different classes of fibrations? I think it's known what the image of this map can be. And the local properties are governed, at least in the complex case, by what type the manifold is.
And, most importantly, there is something about uppertriangularity. What exactly is that?
| https://mathoverflow.net/users/65 | Properties of monodromy of a fibration? | A **small clarification** on bhargav's answer: in algebraic geometry we only have quasi-unipotency of the *local* monodromy in one-parameter families (which is what bhargav is talking about); or in multi-parameter families but only near a normal crossing point of the discriminant. Global monodromies are reductive and local monodromies near bad points of the discriminant can be more general.
**For concreteness** look at a projective morphism $f : X \to B$, where $X$, $B$ are smooth complex projective varieties. Let $D \subset B$ be the discriminant divisor of $f$, i.e. the divisor where the differential of $f$ is not surjective. The global monodromy of the smooth fibration $f : X - f^{-1}(D) \to B - D$ is always reductive by a theorem of Borel. That is: the Zariski closure of the monodromy in the linear automorphisms of the cohomology of the marked fiber is a complex reductive group. If we take a small analytic ball $U \subset B$ centered at some point of $D$, and if we know that $D\cap U$ is a normal crossings divisor in $U$, then the monodromy of the local fibration $f : f^{-1}(U-D) \to U-D$ is quasi-unipotent as bhargav explained. Note that the normal crossings condition implies that the fundamental group of $U - D$ is abelian, so the quasi-unipotency condition makes sense here.
If however $U\cap D$ does not have normal crossings, then $\pi\_{1}(U-D)$ need not be abelian and the monodromy of $f : f^{-1}(U-D) \to U-D$ need not be quasi-unipotent. An easy example is to look at a generic projective plane $\mathbb{P}^{2}$ in the $9$ dimensional projective space of cubic curves in $\mathbb{P}^{2}$. This plane parametrizes a family of cubics which degenerates along a discriminant curve $D \subset \mathbb{P}^{2}$ and under the genericity assumption $D$ has only nodes and cusps. The cuspidal points of $D$ correspond to cuspidal cubics, and near a cusp of $D$ the local fundamental group of $U - D$ is the amalgamated product of $\mathbb{Z}/4$ and $\mathbb{Z}/6$ over $\mathbb{Z}/2$ and so is isomorphic to $SL\_{2}(\mathbb{Z})$ the local monodromy representation near the cusp, i.e. the representation of $\pi\_{1}(U-D,u\_{0})$ to the linear automorphisms of the first integral cohomology of the cubic corresponding to $u\_{0} \in U - D$, is an inclusion, i.e. has image $SL\_{2}(\mathbb{Z})$. In particular it is not quasi-unipotent.
| 16 | https://mathoverflow.net/users/439 | 2025 | 1,286 |
https://mathoverflow.net/questions/2022 | 46 | I never really understood the definition of the conductor of an elliptic curve.
What I understand is that for an elliptic curve E over ℚ, End(E) is going to be (isomorphic to) ℤ or an order in a imaginary quadratic field ℚ(√(-d)), and that this order is uniquely determined by an integer f, the conductor, so that End(E) ≅ ℤ + f Oℚ(√(-d)) (where O just means ring of integers).
However I feel that this is not very convenient; this definition does not say anything about elliptic curves without complex multiplication.
The other definition I have come across gives the conductor as the product of primes at which the elliptic curve does not have good reduction:
N = ∏ pfp
where fp = 0 if E has good reduction at p, fp = 1 if the reduction is multiplicative, fp = 2 if it is additive and p ≠ 2 or 3, and fp = 2 + δ if p = 2 or 3, where δ is some (seemingly complicated) measure of how bad the reduction is.
I've never been able to make much sense of the second definition, nor have I seen any relation with the first. How did the idea initially appear, and why is this particular definition more useful (or "natural") than other similar definitions?
| https://mathoverflow.net/users/362 | Definition and meaning of the conductor of an elliptic curve | The conductor of the curve and the conductor of the order in the endomorphism ring are not equal in the CM case; it's just unfortunate terminology. For example, y^2 = x^3 - x has complex multiplication by the maximal order Z[i] (conductor = 1) of Q(i), but it certainly doesn't have everywhere good reduction.
The conductor N defined in the rather clunky way, prime by prime, is useful for organizing the information that's packed into the L-function of the elliptic curve. More specifically, it shows up in the functional equation that relates the L-function in the right half-plane to its values in the left half-plane. (Which is conjectural unless E is modular-- including all curves defined over Q-- or E has complex multiplication.) The conceptual reason the funny business shows up at the primes 2 and 3 is that the L-function is a product of local L-functions counting points on reductions, and this counting is harder to do mod 2 or mod 3. This is all sketched in sections 15 and 16 of appendix C of Silverman's first book on elliptic curves and spelled out in his second book.
| 24 | https://mathoverflow.net/users/1018 | 2026 | 1,287 |
https://mathoverflow.net/questions/689 | 11 | Could someone give an overview, or just some examples, of "finiteness conditions" for simplicial sheaves/presheaves and/or simplicial schemes? Any answer or comment about this would be interesting, but I am interested in particular in the following two things:
1) I once heard Toen say something about this, and that one can express some kind of condition on simplicial sheaves/presheaves/schemes in terms of convergence of some power series, or something along these lines. What kind of power series is this, and what are the definitions/statements?
2) As explained for example in Deligne's classical papers on Hodge theory, a cohomology theory for varieties (defined as hypercohomology of a complex of sheaves, say) can be extended to simplicial varieties using a spectral sequence. I suspect (but am not sure) that this works well in some sense only under some kind of condition on the simplicial variety, but what would such a condition be?
Some background: I am interested primarily in simplicial sheaves/presheaves on a site coming from algebraic geometry, which would typically be some category of schemes equipped with the Zariski/Nisnevich/etale/flat topology. A simplicial sheaf/presheaf should be thought of a "generalized space", some of the most important examples being the "motivic spaces" in A1-homotopy theory, and stacks in the sense of Toen.
| https://mathoverflow.net/users/349 | Finiteness conditions on simplicial sheaves/presheaves | I don't know what Toën was talking about, but I suspect that it was about finiteness conditions for Artin stacks: the problem is that the usual finiteness conditions we look at for schemes (like the notion of constructibility for l-adic sheaves) do not extend to stacks in a straightforward way, which gives some trouble if one wants to count points
(i.e. to define things like Euler characteristics). Some notions of finiteness are developed to define Grothendieck rings of Artin stacks (e.g. in Toën's paper arXiv:0509098 and in Ekedahl's paper arXiv:0903.3143), which can be realized by our favourite cohomologies (l-adic, Hodge, etc), but the link with a good notion of finiteness for categories of coefficients over Artin stacks (l-adic sheaves, variation of mixed Hodge structures) does not seem to be fully understood yet, at least conceptually (and by myself).
As for finiteness conditions for sheaves (in some homotopical context), the kind of properties we might want to look at are of the following shape.
Consider a variety of you favourite kind X, and a derived category D(X) of sheaves over
some site S associated to X (e.g. open, or étale, or smooth subvarieties over X etc.).
For instance, D(X) might be the homotopy category of the model category of simplicial sheaves, or the derived category of sheaves of R-modules.
Important finiteness properties can be expressed by saying that for any U in the site S,
we have
(1) hocolimᵢ RΓ(U,Fᵢ)= RΓ(U,hocolimᵢ Fᵢ)
where {Fᵢ} is a filtered diagram of coefficients. If you are in such a context, then
you can look at the compact objects in D(X), i.e. the objects A of D(X) such that
(2) hocolimᵢ RHom(A,Fᵢ)= RHom(A,hocolimᵢ Fᵢ)
for any filtered diagram {Fᵢ}. In good situations, condition (1) will imply that the category of compact objects will coincide with constructible objects (i.e. the smallest subcategory of D(X) stable under finite homotopy colimits (finite meaning: indexed by finite posets) which contains the representable objects).
Sufficient conditions to get (1) are the following:
a) For simplicial sheaves (as well as sheaves of spectra or R-modules...), a sufficient condition is that the topology on S is defined by a cd-structure in the sense of Voevodsky (see arXiv:0805.4578). These include the Zariski topology, the Nisnevich topology, as well as the cdh topology (the latter being generated by Nisnevich coverings as well as by blow-ups in a suitable sense), at least if we work with noetherian schemes of finite dimension. Note also that topologies associated to cd structures define what Morel and Voevodsky call a site of finite type (in the language of Lurie, this means that, for such sites, the notion of descent is the same as the notion of hyperdescent: descent for infinity-stacks over S can be tested only using truncated hypercoverings; this is the issue discussed by David Ben Zvi above).
In practice, the existence of a cd structure allows you to express (hyper)descent using only Mayer-Vietoris-like long exact sequences (the case of Zariski topology was discovered in the 70's by Brown and Gersten, and they used it to prove Zariski descent for algebraic K-theory).
b) For complexes of sheaves of R-modules, a sufficient set of conditions are
i) the site S is coherent (in the sense of SGA4).
ii) any object of the site S is of finite cohomological dimension (with coefficients in R).
The idea to prove (1) under assumption b) is that one proves it first when all the Fᵢ's are concentrated in degree 0 (this is done in SGA4 under assumption b)i)).
This implies the result when the Fᵢ's are uniformly bounded. Then, one uses the fact, that, under condition b)ii), the Leray spectral sequence converges strongly, even for unbounded complexes (this done at the begining of the paper of Suslin and Voevodsky "Bloch-Kato conjecture and motivic cohomology with finite coefficients").
This works for instance for étale sheaves of R-modules, where R=Z/n, with n prime to the residual characteristics. Note moreover that, in the derived category of R-modules, the compact objects (i.e. the complexes A satisfying (2)) are precisely the perfect complexes.
The fact that the six Grothendieck operations preserves constructibility can then be translated into the finiteness of cohomology groups (note however that the notion of constructiblity is more complex then this in general: if we work with l-adic sheaves
(with Ekedahl's construction, for instance), then the notion of constructiblity does not agree with compactness anymore). However, condition (1) is preserved after taking the Verdier quotient of D(X) by any thick subcategory T obtained as the smallest thick subcategory which is closed under small sums and which contains a given small set of compact objects of D(X) (this is Thomason's theorem). This is how such nice properties survive in the context of homotopy theory of schemes for instance.
Note also that, in a stable (triangulated) context, condition (2) for A implies that we have the same property, but without requiring the diagrams {Fᵢ} to be filtering.
For your second question, the extension of a cohomology theory to simplicial varieties is automatic (whenever the cohomology is given by a complex of presheaves), at least if we have enough room to take homotopy limits, which is usually the case (and not difficult to force if necessary). The only trouble is that you might lose the finiteness conditions, unless you prove that your favorite simplicial object A satisfies (2). The fact that Hironaka's resolution of singularities gives the good construction (i.e. gives nice objects for open and/or singular varieties) can be expained by finiteness properties related to descent by blow-ups (i.e. cdh descent), but the arguments needed for this use strongly that we work in a stable context (I don't know any argument like this for simplicial sheaves). The fuzzy idea is that if a cohomology theory satisfies Nisnevich descent and homotopy invariance, then it satisfies cdh descent (there is a nice very general proof of this in Voevodsky's paper arXiv:0805.4576 (thm 4.2, where you will see we need to be able to desuspend)); then, thanks to Hironaka, locally for the cdh topology, any scheme is the complement of a strict normal crossing divisor in a projective and smooth variety. As cdh topology has nice finiteness properties (namely a)), and as
any k-scheme of finite type is coherent in the cdh topos, this explains, roughly, why we get nice extensions of our cohomology theories (as far as you had a good knowledge of smooth and projective varieties). If we work with rational coefficients, the same principle applies for schemes over an excellent noetherian scheme S of dimension lesser or equal to 2, using de Jong's results instead of Hironaka's, and replacing the cdh topology by the h topology (the latter being obtained from the cdh topology by adding finite surjective morphisms): it is then sufficient to have a good control of proper regular S-schemes.
| 16 | https://mathoverflow.net/users/1017 | 2027 | 1,288 |
https://mathoverflow.net/questions/807 | 22 | As is well known, the universal covering space of the punctured complex plane is the complex plane itself, and the cover is given by the exponential map.
In a sense, this shows that the logarithm has the worst monodromy possible, given that it has only one singularity in the complex plane. Hence we can easily visualise the covering map as given by the Riemann surface corresponding to *log* (given by analytic continuation, say).
Seeing how fundamental the exponential and logarithm are, I was wondering how come I don't know of anything about the case when two points are removed from the complex plane.
**My main question is as follows:** how can I find a function whose monodromy corresponds to the universal cover of the twice punctured complex plane (say ℂ∖{0,1}), in the same way as the monodromy of *log* corresponds to the universal cover of the punctured plane.
For example, one might want to try *f*(\*z\*) = *log*(\*z\*) + *log*(\*z\*-1) but the corresponding Riemann surface is easily seen to have an abelian group of deck transformations, when it should be F2.
The most help so far has been looking about the Riemann-Hilbert problem; it is possible to write down a linear ordinary differential equation of order 2 that has the required monodromy group.
Only trouble is that this does not show how to explicitly do it: I started with a faithful representation of the fundamental group (of the twice punctured complex plane) in GL(2,ℂ) (in fact corresponding matrices in SL(2,ℤ) are easy to produce), but the calculations quickly got out of hand.
My number one hope would be something involving the hypergeometric function 2F1 seeing as this solves in general second order linear differential equations with 3 regular singular points (for 2F1 the singular points are 0, 1 and ∞, but we can move this with Möbius transformations), but I was really hoping for something much more explicit, especially seeing as a lot of parameters seem to not produce the correct monodromy. Especially knowing that even though the differential equation has the correct monodromy, the solutions might not.
I'd be happy to hear about any information anyone has relating to analytical descriptions of this universal cover, I was quite surprised to see how little there is written about it.
Bonus points for anything that also works for more points removed, but seeing how complicated this seems to be for only two removed points, I'm not hoping much (knowing that starting with 3 singular points (+∞), many complicated phenomena appear).
| https://mathoverflow.net/users/362 | Describing the universal covering map for the twice punctured complex plane | Others have already given a satisfactory qualitative description as a modular function under a suitable congruence group. Since the quotient in question is necessarily genus zero, there are explicit formulas for such functions.
I was mistaken in my comment to Tyler's answer. The function I provided there is invariant under a larger group than the one we want, and yields the universal cover for the complex plane with one and a half punctures.
The Dedekind eta product eta(z/2)^8/eta(2z)^8 is not only invariant under Gamma(2) (= F2), but it maps H/Gamma(2) bijectively to the twice punctured plane. You will have to post-compose with a suitable affine transformation to move the punctures to zero and one. An alternative description is: eta(z)^24/(eta(z/2)^8 eta(2z)^16).
This function arises in monstrous moonshine: eta(z)^8/eta(4z)^8 + 8 is the graded character of an element of order four, in conjugacy class 4C in the monster, acting on the monster vertex algebra (a graded vector space with some extra structure). It is invariant under Gamma0(4), which is what you get by conjugating Gamma(2) under the z -> 2z map.
Other elements of the monster yield functions that act as universal covers for planes with specific puncture placement and orbifold behavior. For the general case of more than 2 punctures, you have to use more geometric methods, due to nontrivial moduli. I think you idea of using hypergeometric functions is on the right track. I think Yoshida's book, *Hypergeometric Functions, My Love* has a few more cases worked out.
| 9 | https://mathoverflow.net/users/121 | 2034 | 1,294 |
https://mathoverflow.net/questions/2039 | 4 | k is a perfect field. X and Y are two regular varieties over k. Does their fiber product over k remain to be regular?
Note: When k is algebraically closed it's true by Jacobian criterion. When k is not perfect there's counter-example.
| https://mathoverflow.net/users/2008 | Does the fiber product of two regular varieties over perfect field remain regular? | The answer is yes. Indeed, over a perfect field the notions of smooth and regular coincide so it follows from the fact that base change and composition preserve smoothness.
| 6 | https://mathoverflow.net/users/310 | 2042 | 1,299 |
https://mathoverflow.net/questions/1977 | 84 | This is a somewhat long discussion so please bear with me. There is a theorem that I have always been curious about from an intuitive standpoint and that has been glossed over in most textbooks I have read. Quoting [Wikipedia](http://en.wikipedia.org/wiki/Level_set#Level_sets_versus_the_gradient), the theorem is:
>
> The gradient of a function at a point is perpendicular to the level set of $f$ at that point.
>
>
>
I understand the Wikipedia article's proof, which is the standard way of looking at things, but I see the proof as somewhat magical. It gives a symbolic reason for why the theorem is true without giving much geometric intuition.
The gradient gives the direction of largest increase so it sort of makes sense that a curve that is perpendicular would be constant. Alas, this seems to be backwards reasoning. Having already noticed that the gradient is the direction of greatest increase, we can deduce that going in a direction perpendicular to it would be the slowest increase. But we can't really reason that this slowest increase is zero nor can we argue that going in a direction perpendicular to a constant direction would give us a direction of greatest increase.
I would also appreciate some connection of this intuition to [Lagrange multipliers](http://en.wikipedia.org/wiki/Lagrange_multipliers) which is another somewhat magical theorem for me. I understand it because the algebra works out but what's going on geometrically?
Finally, what does this say intuitively about the generalization where we are looking to: maximize $f(x,y)$ where $g(x,y) > c$.
I have always struggled to find the correct internal model that would encapsulate these ideas.
| https://mathoverflow.net/users/812 | Why is the gradient normal? | The gradient of a function is normal to the level sets because it is **defined** that way. The gradient of a function is **not** the natural derivative. When you have a function $f$, defined on some Euclidean space (more generally, a Riemannian manifold) then its derivative at a point, say $x$, is a function $d\_xf(v)$ on tangent vectors. The intuitive way to think of it is that $d\_xf(v)$ answers the question:
>
> If I move infinitesimally in the direction $v$, what happens to $f$?
>
>
>
So $d\_xf(v)$ is not itself a tangent vector. However, as we have an inner product lying around, we can convert it into a tangent vector which we call $\nabla f$. This represents the question:
>
> What tangent vector $u$ at $x$ best represents $d\_xf(v)$?
>
>
>
What we mean by "best represents" is that $u$ should satisfy the condition:
>
> $\langle u,v\rangle = d\_xf(v)$ for all tangent vectors $v$.
>
>
>
Now we look at the level set of $f$ through $x$. If $v$ is a tangent vector at $x$ which is tangent to the level set then $d\_xf(v) = 0$ since $f$ doesn't change if we go (infinitesimally) in the direction of $v$. Hence our vector $\nabla f$ (aka $u$ in the question) must satisfy $\langle\nabla f, v\rangle = 0$. That is, $\nabla f$ is normal to the set of tangent vectors at $x$ which are tangent to the level set.
For a generic $x$ and a generic $f$ (i.e. most of the time), the set of tangent vectors at $x$ which are tangent to the level set of $f$ at $x$ is codimension $1$ so this specifies $\nabla f$ up to a scalar multiple. The scalar multiple can be found by looking at a tangent vector $v$ such that $f$ does change in the $v$-direction. If no such $v$ exists, then $\nabla f = 0$, of course.
| 112 | https://mathoverflow.net/users/45 | 2049 | 1,303 |
https://mathoverflow.net/questions/1886 | 22 | Suppose we have an infinite matrix A = (aij) (i, j positive integers). What is the "right" definition of determinant of such a matrix? (Or does such a notion even exist?) Of course, I don't necessarily expect every such matrix to have a determinant -- presumably there are questions of convergence -- but what should the quantity be? The problem I have is that there are several ways of looking at the determinant of a finite square matrix, and it's not clear to me what the "essence" of the determinant is.
| https://mathoverflow.net/users/913 | Infinite matrices and the concept of "determinant" | There is a class of linear operators that have a determinant. They are, for some strange reason, known as "operators with a determinant".
For Banach spaces, the essential details go along these lines. Fix a Banach space, X, and consider the **finite rank** linear operators. That means that T: X → X is such that Im(T) is finite dimensional. Such operators have a well-defined trace, tr(T). Using this trace we can define a norm on the subspace of finite-rank operators. If our operator were diagonalisable, we would define it as the sum of the absolute values of the eigenvalues (of which only finitely many are non-zero, of course). This norm is finer than the operator norm. We then take the closure in the space of all operators of the space of finite-rank operators with respect to this trace norm. These operators are called **trace class** operators. For such, there is a well-defined notion of a trace.
(Incidentally, these operators form a two-sided ideal in the space of all operators and are actually the dual of the space of all operators via the pairing (S,T) → tr(ST).)
Now trace and determinant are very closely linked via the forumula etr T = det eT. This means that we can use our trace class operators to define a new class of "operators with a determinant". The key property should be that the exponential of a trace class operator should have a determinant. This suggests looking at the family of operators which differ from the identity by a trace class operator. Within this, we can look at the group of units, that is invertible operators.
So an "operator with a determinant" is an invertible operator that differs from the identity by one of trace class.
For more details, I recommend the book "Trace ideals and their applications" by Barry Simon (MR541149) and the article "On the homotopy type of certain groups of operators" by Richard Palais (MR0175130).
But defining the determinant of an arbitrary operator is, of course, impossible. One can always figure out a renormalisation for a *particular* operator but there just ain't gonna be a system that works for everything: obviously det(I) = 1 but then det(2I) = ?
(I should also say that I picked Banach spaces for ease of exposition. One can generalise this to locally convex topological spaces, but that involves handling nuclear materials so caution is advised.)
| 31 | https://mathoverflow.net/users/45 | 2052 | 1,306 |
https://mathoverflow.net/questions/2046 | 14 | I felt like following up on [Kate's question](https://mathoverflow.net/questions/544/why-are-subfactors-interesting). There were some good motivational answers there.
Given a pair of [factors](http://en.wikipedia.org/wiki/Von_Neumann_algebra#Factors) M < N, there is a standard way to construct a 2-category whose objects are M and N, whose morphism categories are the categories of bimodules, and whose composition is described by some kind of Connes product. If I restrict to the endomorphism category of M, I get a monoidal category structure, but I don't know how to say anything about it. Here's a barrage of questions:
1. When people talk about fusion categories coming from subfactors, are they referring to the endomorphism category of one of the factors?
2. How are the endomorphism categories of M and N related? Are they equivalent? Are they Koszul dual?
3. Does the Jones index say something concrete about the category, like Frobenius-Perron dimension? (How does one compute Jones index, anyway?)
4. How do people go about constructing exotic subfactors? Do they just arise in nature? I'm totally okay with pointers to references here.
5. (bonus) I should get a braided tensor structure from a net of factors on a circle. Is this the center of the fusion category, and is it in the literature?
**Edit:** Based on the (fantastically illuminating) responses, it seems that my bonus question doesn't make sense, because the M-M bimodule fusion category depends on the choice of N in an essential way. Maybe the phrase "conformal defect" should be used somewhere. If I come up with a suitable replacement, I'll present it as a separate question.
| https://mathoverflow.net/users/121 | How do I describe a fusion category given a subfactor? | Here are some partial answers:
1- Usually the fusion category is the category of bifinite correspondences, i.e. Hilbert spaces with actions of $N$ and $M$ whose module dimensions are finite. Jones has a result saying that a bifinite correspondence is irreducible if and only if the algebraic module of bounded vectors is irreducible (on his website, two subfactors and the algebraic decomposition...). This means the fusion category of bifinite correspondences should be equivalent to the fusion category of algebraic bimodules (you probably need bifinite in the sense of Lueck, but this is very technical). The former category is generated by the $N-M$ correspondence $L^(M)$, and the later category is generated by $M$ as an $N-M$ bimodule (generated in the sense of taking tensor products and decomposing into irreducibles). In fact, Morrison, Peters, and Snyder use the algebraic category in their recent paper on extended Haagerup (arXiv:0909.4099v1).
2- This isn't what you're asking, but $L^2(M)$ as an $N-M$ bimodule is a Morita equivalence from $N$-Hilbert modules to $M\_1$-Hilbert modules where $M\_1$ is the basic construction of $N\subset M$. I just think it's an interesting point to bring up.
4- One of the best ways of constructing subfactors is via planar algebras. Given a suitable fusion category, one can construct a planar algebra. Typical examples of these nice fusion categories are the fusion categories arising from the representation theory of a finite group or a quantum group. This gives rise to a family of subfactors. In fact, since (for finite groups) there are only finitely many irreducible representations, we have that this planar algebra will be finite depth (see arXiv:0808.0764, section 4.1), and the subfactor constructed from this planar algebra will be finite depth as well. When someone says "exotic subfactor," they mean a finite index, finite depth subfactor that doesn't appear in the well known families coming from these fusion categories. To date, the best way of constructing these subfactors is to stumble upon a finite bipartite graph which doesn't appear as a fusion graph determine if it can be a principal graph for a subfactor. This has inspired a program to classify all principal graphs which can occur (see the extended Haagerup paper for a synopsis of this as well).
Tie in to 3- Two exotic subfactors, namely the Haagerup and extended Haagerup subfactor, have been constructed by finiding a subfactor planar algebra with the appropriate principal graph inside the graph planar algebra of the bipartite graph (this technique was first explored in detail in Peters' thesis). These subfactors have index equal to the square of the norm of the graph, which is the Perron-Frobenius eigenvalue. In fact, if a finite index subfactor is extremal (irreducible implies extremal), then the norm squared of the principal graph is always the Jones index. (One typically computes Jones index by computing the von Neumann dimension of the $N$-Hilbert module $L^2(M)$.)
5- I know that Kawahigashi et al. (see arXiv:0811.4128) have found a net of type $III\_1$-factors corresponding to intervals on the circle. I would recommend starting there.
| 8 | https://mathoverflow.net/users/351 | 2057 | 1,310 |
https://mathoverflow.net/questions/2038 | 8 | Suppose $k$ is an algebraically closed field, and $X$, $Y$ are two normal varieties over $k$. Is the product $X \times Y$ necessarily still normal?
| https://mathoverflow.net/users/2008 | Does the fiber product of two normal varieties remain normal? | The answer is yes.
In general one can define a normal morphism of schemes $f:X \rightarrow Y$ to be a flat morphism such that for every $y \in Y$ the fibre over $y$ is geometrically normal.
Then we have the following theorem on normality and base change (see EGA Ch 2 IV 6.14.1)
Let $g: Y' \rightarrow Y$ be a normal morphism of locally noetherian schemes. Then for every normal $Y$-scheme $X$ the fibre product $X \times\_Y Y'$ is normal.
Over an algebraically closed field flatness and geometric normality reduce to just being normal so the result follows.
| 9 | https://mathoverflow.net/users/310 | 2058 | 1,311 |
https://mathoverflow.net/questions/724 | 16 | Pretty much exactly what it says on the tin. Let G be a connected graph; then the Tutte polynomial T\_G(x,y) carries a lot of information about G. However, it obviously doesn't encode everything about the graph, since there are examples of non-isomorphic graphs with the same Tutte polynomial.
My question is, what information exactly does the Tutte polynomial encapsulate? I'm aware of a few answers to this question, but I don't find any of them particularly satisfying. For instance, T\_G(x,y) can be characterized as "the universal Tutte-Grothendieck invariant," but the definition of Tutte-Grothendieck invariants is just as unintuitive as the definition of the Tutte polynomial (because it's essentially the same definition!) One can also define the coefficients as counting certain spanning trees of G, but this doesn't make apparent the fact that the Tutte polynomial specializes to the chromatic polynomial, or the notion that it carries most of the information one can obtain via linear algebra methods.
So is there a nice way of thinking about what data about G the Tutte polynomial encodes?
ETA: Okay, here's a very rough conjecture. Suppose that there's some "computationally simple" (i.e., testing membership is in NP) class of graphs such that there are two connected graphs G, H with the same Tutte polynomial, and G is in the class and H is not. Then there are spanning trees S, T of G, H respectively, such that S is in the class and T is not.
This would mean, in a sense that I can't make entirely rigorous, that the information about a graph G not encoded in the Tutte polynomial is just information about the structure of spanning trees of G. (Update: As Kevin Costello points out in a comment, this idea appears to be severely limited by the existence of certain pairs of co-Tutte graphs. In particular, we would need to count spanning trees with multiplicity for it to have even a chance of being true.)
As stated, the above conjecture is false for trivial reasons. But is there a way of making it true, perhaps by requiring the property to be, in some sense, natural? Is there a broad notion of "graph properties" for which it is true? Can we at least state a conjecture along these lines which does seem to be true?
| https://mathoverflow.net/users/382 | What is the Tutte polynomial encoding? | No-one so far has mentioned matroids. The Tutte polynomial encodes some of the information from the cycle matroid of the graph. Two graphs with the same cycle matroid (and number of vertices) have the same Tutte polynomials. So if a graph property is not determined by the cycle matroid (and the number of vertices) then it can't be obtained from the Tutte polynomial.
| 20 | https://mathoverflow.net/users/1028 | 2061 | 1,314 |
https://mathoverflow.net/questions/2065 | 2 | I have a problem in computing (i.e. classify) a factor group.
For example The group Z\*Z\*Z/<(3,6,9)> is isomorphic to Z\_3\*Z\*Z. I can show this by contructing a homomorphism f
f(a,b,c) = ( a mod 3 , 2\*a - b, 3\*a - c )
and then show that Ker(f) = <(3,6,9)>.
It is not hard to see that Im(f) = Z\_3\*Z\*Z.
But how would I compute e.g. Z\*Z/<(9,12)> ?
I guess I could create a function
f(a,b) = ( a mod 9, 4\*a - 3\*b ).
then Ker(f) = <(9,12)>, but what is the image?
| https://mathoverflow.net/users/818 | Computing a Factor Group | Since 4 and 3 are coprime, you can obtain every integer as 4a-3b for some a and b, and thus the image is isomorphic to Z/(9) x Z.
In general, for each factor you get the quotient of Z by the ideal generated by the gcd of the coefficients in your expression.
**EDIT** Sorry for the confusion, wrote too quickly, hope this clarifies better:
Imagine that you are working on R^3 (real vectors). If you take *any* nonzero vector v and quotient our R^3/(v) you always get something that is isomorphic to R^2, right?
Well, sort of the same thing is true for Z, but now you need to care about gcd's; take your vector v=(9,12); you cannot extend it to a basis of Z^2 because it has a nontrivial gcd, so write it as 3(3,4). Now, take a vector extending (3,4) to a basis of Z^2, for instance (0,1) (1,1). Now, *every* element v in Z^2 can be written in a unique way as v = a(3,4) + b(1,1); if you quotient out by (3,4), you'd only have the 'b' term, getting a free part of rank one. but you have the 3 multiplying , so the image ox v under the quotient is (a mod 3, b), and thus you get as a quotient Z/(3) x Z. In general, if you take Z^n/(w) the result will be Z^(n-1) x Z/(gcd(w)).
For the case in which you take quotient by the submodule spanned by more than one vector, ref the answer by Armin and the reference to Smith form.
| 0 | https://mathoverflow.net/users/914 | 2067 | 1,319 |
https://mathoverflow.net/questions/1939 | 8 | Let R be a commutative ring, and $A$ an $R$-algebra (possibly non-commutative). Then $A$ is separable if it is finitely generated (f.g.) projective as an $(A \otimes\_R A^{\mathrm{op}})$-algebra. Suppose further that $A$ is f. g. projective as an $R$-module. Does this imply that $A$ is a (symmetric) Frobenius algebra?
There are lots of equivalent definitions of a Frobenius algebra. One (assuming $A$ is a f.g. projective R-module) is that there exists an $R$-linear map $\mathrm{tr}: A\to R$, such that $b(x,y) := \mathrm{tr}(ab)$ is a non-degenerate.
I know that the answer is yes when $R$ is a field. What about other rings?
I am not an expert on algebras, but this question is related to understanding obstructions for extended TQFTs, and so I am very interested in knowing anything I can about it.
| https://mathoverflow.net/users/184 | Separable and finitely generated projective but not Frobenius? | Theorem 4.2 of [On separable algebras over a commutative ring](https://projecteuclid.org/journals/osaka-journal-of-mathematics/volume-4/issue-2/On-separable-algebras-over-a-commutative-ring/ojm/1200691953.full) says that the answer is always yes.
| 3 | https://mathoverflow.net/users/345 | 2070 | 1,321 |
https://mathoverflow.net/questions/2071 | 11 | It is remarked in Shafarevich's Basic Algebraic Geometry 1 that Rees and Nagata constructed examples of quasiprojective varieties such that the ring of regular functions is not finitely generated, but I cannot find the source he is referring to. Can anyone give such examples here? Does that mean we can't really say anything about the ring of regular functions of a quasi-projective variety?
| https://mathoverflow.net/users/nan | Non-finitely generated ring of regular functions | It's a theorem that a quasi-projective variety is affine if and only if it is Stein (we're working over C, say) and its ring of functions is finitely generated. So find a Stein manifold that isn't affine, and that will do it.
And, after a bit of looking, it appears that Vakil may have rediscovered the Rees and Nagata example, [here](http://math.stanford.edu/~vakil/files/nonfg.pdf).
| 8 | https://mathoverflow.net/users/622 | 2073 | 1,323 |
https://mathoverflow.net/questions/1937 | 8 | Let pi: \bar{Mg,1} \to \bar{M\_g} be natural projection of compactified moduli stacks of curves and let omega be the relative dualizing sheaf. Then the Hodge class \lambda of \bar{M\_g} is the first chern class of the pushforward \pi\_\*(ω). Among other things the hodge class, together with the boundary divisors, freely generates the Picard group of \bar{M\_g}.
>
> **Question**: Why is lambda big and nef?
>
>
>
| https://mathoverflow.net/users/2 | Why is the Hodge class of \bar{M_g} big and nef? | Some multiple of lambda is defined on the coarse moduli space and this is the pullback of an ample bundle on \bar{A\_g}, the Satake-Baily-Borel compactification of A\_g. Since \bar{M\_g} maps birationally onto its image in \bar{A\_g}, it follows that lambda is nef and big, in fact also semi-ample (some multiple is base point free) on the coarse moduli space.
(The map to \bar{A\_g } contracts the boundary divisor corresponding to irreducible nodal curves so lambda is not ample.)
| 5 | https://mathoverflow.net/users/519 | 2075 | 1,325 |
https://mathoverflow.net/questions/2040 | 39 | People who talk about things like modular forms and zeta functions put a lot of emphasis on the existence and form of functional equations, but I've never seen them used as anything other than a technical tool. Is there a conceptual reason we want these functional equations around? Have I just not seen enough of the theory?
| https://mathoverflow.net/users/290 | Why are functional equations important? | There are many reasons that functional equations are important. Some background: To most varieties/schemes occurring in arithmetic geometry, you can associate a zeta function/L-function. There are two main ways of constructing these, either from l-adic cohomology, or from counting solutions in various finite fields. Usually the word L-function is used for functions associated to varieties over number fields, constructed from l-adic cohomology, and the word zeta function is used for functions associated to schemes over Z or some other ring of integers, or over a finite field. However, there is some confusion about the terminology, and there is also some overlap between the two, since there is a close relation between the L-function of a variety and the zeta function of an integral model for the variety.
Over finite fields, the functional equation is part of the famous Weil conjectures, proved by Deligne. One reason that the functional equation is cool is that it reflects Poincare duality in the etale cohomology of the variety, so it is in some sense a deep geometric statement. For background on the Weil conjectures, see for example Freitag and Kiehl: Etale cohomology and the Weil conjectures, and the survey of Mazur: Eigenvalues of Frobenius acting on algebraic varieties over finite fields, in some conference proceedings.
For the definition of L-functions and zeta functions and lots of other background, see Manin and Panchishkin: Introduction to modern number theory, chapter 6. There are at least two deep reasons for being interested in the functional equations for these functions. Firstly, the existence of a functional equation seems to always be directly related to the L-function coming from an automorphic representation, and the idea that "every L-function from algebraic geometry (aka motivic L-function) also comes from an automorphic representation" is in some sense the number-theoretic incarnation of the global Langlands program. See for example Bump et al: An introduction to the Langlands program. The most famous cases where this has been proved is (1) Tate's thesis, which treats Hecke L-functions, and where the corresponding automorphic representation is one-dimensional, i.e. a character on the ideles, and (2) the work by Wiles and others related to Fermat's last theorem, where they show that the L-function associated to an elliptic curve over Q also comes from a modular form, and hence satisfies the expected functional equation. See the book of Diamond and Shurman on modular forms.
The other deep reason for thinking about the functional equation is that some optimistic people dream of an "arithmetic cohomology theory", which would allow us to mimick the proof of the Weil conjectures, but for zeta functions over Z or L-functions over Q. Then the functional equation should be related to Poincare duality for this cohomology. All this is related to the Riemann hypothesis, noncommutative geometry, and the field with one element. See for example Deninger's *Motivic L-functions and regularized determinants* ([pdf](https://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/deninger3.pdf)), his more recent survey *[Arithmetic Geometry and Analysis on Foliated Spaces](https://arxiv.org/abs/math/0505354)*, some [slides of Paugam](http://people.math.jussieu.fr/%7Efpaugam/documents/transparents-expose-equa-fonc-baltimore.pdf)link broken on the functional equation, and also the later chapters of Manin-Panchishkin. Some of the key names if you want to find more references: Deninger, Connes, Consani, Marcolli; most of them have lots of stuff on their webpages and on the arXiv.
| 28 | https://mathoverflow.net/users/349 | 2085 | 1,331 |
https://mathoverflow.net/questions/2083 | 26 | When is a Stein manifold a complex affine variety? I had thought that there was a theorem saying that a variety which is Stein and has finitely generated ring of regular functions implies affine, but in the comments to my answer [here](https://mathoverflow.net/questions/2071/non-finitely-generated-ring-of-regular-functions), Serre's counterexample was brought up. I'm guessing that the answer is that the ring of regular functions must be nontrivial somehow, like it must separate points, but I'm curious about what the exact condition is.
| https://mathoverflow.net/users/622 | Stein Manifolds and Affine Varieties | Charlie, it is funny answering this way but here it is.
The criterion you are thinking about is a criterion that is relative to an embedding. It says that if $X$ is a quasi-affine complex normal variety, whose associated analytic space $X^{an}$ is Stein, then $X$ is affine if (and only if) the algebra $\Gamma(X,\mathcal{O}\_{X})$ is finitely generated. This is a theorem of Neeman.
You can reformulate the requirement of $X$ being quasi-affine as a separation of points property: for any point $x \in X$ consider the subset $S\_{x} \subset X$ defined as the set of all points $y \in X$ such that all regular functions on $X$ have equal values at $x$ and $y$. Then by an old theorem of Goodman and Hartshorne $X$ is quasi-affine if $S\_{x}$ is finite for all $x$. So you can say that $X$ is affine if it satisfies: 1) $X^{an}$ is Stein; 2) $S\_{x}$ is finite for all $x \in X$; 3) $\Gamma(X,\mathcal{O}\_{X})$ is finitely generated.
| 26 | https://mathoverflow.net/users/439 | 2087 | 1,333 |
https://mathoverflow.net/questions/2054 | 9 | I've been asking myself this question all the time. Let's say you are given a large set of time series data. Your task is to find out patterns that are meaningful or that you can use for future trend prediction.
The issue now is, how do you know for sure that the patterns you extract are valid, in the sense that they don't suffer from data snooping bias or a case of "torture-the-data-until-it-confesses"?
I can always test my hypothesis as new data comes in, but even if it can predict all the trends in the past, that doesn't mean that it will continue to do so in the future. No?
| https://mathoverflow.net/users/807 | Data Mining-- how do you know whether the pattern you extract is valid? | A good way to do this is called 'cross-validation'. The data can be divided into three disjoint sets: the training set, the test set and the validation set.
Different models are developed using the training set. A reasonable way to do this is to take different subsets of points from the training set uniform randomly (for instance by generating 100 subsets of size 90 from a training set of 100 points) and to fit your model as you normally would. This will give you a set of models with varying ability to predict. Pick the model that gives the best prediction for the test set. (Taking random subsets has the benefit of eliminating outliers.) Now, having done all this, the training set and the test set have been 'tainted' by the fact that you used them to build your model.
Therefore, finally, the model should be evaluated on the validation set. This set gives a more honest estimate of how well the model generalizes to a new set of data.
There is some sophistication involved in picking the right relative sizes of the sets and in getting more out of less data. For instance, you can look up k-fold cross-validation.
I also support looking at the Elements of Statistical Learning. The second edition is available free online! <http://www-stat.stanford.edu/~tibs/ElemStatLearn/>
| 13 | https://mathoverflow.net/users/812 | 2096 | 1,340 |
https://mathoverflow.net/questions/959 | 11 | David Corfield asked the following questions yesterday: Is the
$n$-dimensional Fourier transform of $\exp(-\|x\|)$ always non-negative,
where $\|\cdot\|$ is the Euclidean norm on $\Bbb R^n$? What is its support?
I want to ask a more general question: what happens when $\|\cdot\|$ is the
$p$-norm, for arbitrary $p\in [1, 2]$?
David's question is here: [Is the Fourier transform of $\exp(-\|x\|)$ non-negative?](https://mathoverflow.net/questions/723/is-the-fourier-transform-of-exp-x-non-negative)
| https://mathoverflow.net/users/586 | Fourier transform of $\exp(-\|x\|_p)$: more general question | Okay, I think I do have an answer now. I'm borrowing arguments from the proof of Lemma 2.27 in the book "Fourier Analysis in Convex Geometry" by A. Koldobsky (apparently not available online at all). That lemma states that the Fourier transform of the function (on $\Bbb R$) $\exp(-|x|^p)$ is positive everywhere for $p \in (0,2]$.
The central tool is a theorem of Berstein, which in particular implies that if $s$ is in $(0,1]$ then $\exp(-z^s)$ is the Laplace transform of some finite positive measure $\mu$ on $[0,\infty)$; that is,
$$
\exp(-z^s) = \int \exp(-uz) d\mu(u).
$$
Applying this with $s=1/p$ and $z=\|x\|\_p^p$ yields
$$
\exp(-\|x\|\_p) = \int \exp(-u \|x\|\_p^p) d\mu(u).
$$
Now calculate the Fourier transform on $\Bbb R^n$ of this. Using Fubini you get an integral wrt $\mu$ of a product of Fourier transforms of $\exp(-|x|^p)$, and you can now apply the one-dimensional lemma. (The one-dimensional lemma is proved by using the same theorem of Bernstein to reduce to the case $p=2$.)
| 11 | https://mathoverflow.net/users/1044 | 2106 | 1,349 |
https://mathoverflow.net/questions/665 | 1 | Recall that for k a field, a finite dimensional k-algebra A is called *symmetric* if it is isomorphic to its dual as a bimodule of itself. Which is to say, there's a trace map t:A -> k such that t(ab)=t(ba) and for any nonzero a, there is a b such that t(ab) is not 0.
The most popular examples of symmetric algebras are the cohomology rings of compact manifolds. In this case, the trace is integration of top forms.
Various algebras arising in representation theory are symmetric and this has some nice consequences for their categories of representations (for example, the Serre functor is trivial).
Khovanov and Lauda have defined some algebras by explicit generators and relations (see, for example [0909.1810](https://arxiv.org/abs/0909.1810), the cyclotomic KLR algebras, which I strongly believe are symmetric, but after quite a bit of trying, have had no luck with. Anyone else care to try their hand?
| https://mathoverflow.net/users/66 | Are cyclotomic Khovanov-Lauda-Rouquier algebras symmetric? | This is now proven both as Theorem 1.7 in my paper "[Knot invariants and higher representation theory I](https://arxiv.org/abs/1001.2020)" and by Kang and Kashiwara in "[Categorification of Highest Weight Modules via Khovanov-Lauda-Rouquier Algebras](https://arxiv.org/abs/1102.4677)."
| 2 | https://mathoverflow.net/users/66 | 2109 | 1,351 |
https://mathoverflow.net/questions/2045 | 1 | Can you help me understand the class of problems solvable by a nondetermimistic Turing machine with an oracle for SAT running in polynomial time?
| https://mathoverflow.net/users/1027 | How can one characterize NP^SAT? | Surely this class, being $\text{NP}^\text{NP}$, is by definition equal to $\Sigma\_2^p$. In particular, if the Polynomial Hierarchy (PH) does not collapse, then it does not contain $\Pi\_2^p$.
| 5 | https://mathoverflow.net/users/1046 | 2112 | 1,353 |
https://mathoverflow.net/questions/2118 | 7 | The rationals are clearly dense in the real number system, i.e. for every pair a < b of real numbers there exists a rational number p/q s.t. a < p/q < b. I conjecture the same to be true with p and q both primes. Any idea of how one could prove it? It should depend on some strong result on the distribution of prime numbers.
| https://mathoverflow.net/users/1049 | Density of a subset of the reals | Yes. Take q sufficiently big and fixed (in terms of a and b). Then the question is, is there some prime p between qa and qb? Use the prime number theorem to estimate pi(qb) - pi(qa) > 0, where q is chosen to be big enough so that the main term is bigger than the error terms. QED.
| 6 | https://mathoverflow.net/users/1050 | 2120 | 1,358 |
https://mathoverflow.net/questions/2077 | 17 | Let $SX$ be the suspension of CW complex. What are some results available to determine the homotopy groups of $SX$?
| https://mathoverflow.net/users/1034 | How to determine the homotopy groups of the suspension of a space? | This, in general, an incredibly difficult problem. Even we just want to compute the *rational* homotopy groups of the suspension of $X$ and $X$ is simply connected, where we can do everything using rational homotopy theory to reduce things to commutative DGAs, this starts to involve things like free Lie algebras and the like.
Locally at the prime 2, there is actually a famous long exact sequence when $X$ is a sphere called the EHP long exact sequence. It relates the suspension homomorphism, the Hopf map, and a "whitehead product" map. This gives rise to the [EHP spectral sequence](http://en.wikipedia.org/wiki/EHP_spectral_sequence) that, funnily enough, starts with the 2-local homotopy groups of *odd*-dimensional spheres and computes the 2-local homotopy groups of spheres. Miller and Ravenel have a paper titled "Mark Mahowald's work on the homotopy groups of spheres" that covers some of this material in detail.
Another approach is to say: The "stable" homotopy groups of $X$ are a first-order approximation using the Freudenthal suspension theorem that Andrea mentioned. There is then a "quadratic" correction term that you can try to use to get an approximation of the homotopy groups that is correct out to roughly three times the connectivity, and so on. These lead into the subject of Goodwillie calculus.
For a classifying space $K(G,1)$, neither of these approaches work very well, because the higher homotopy groups are going to depend pretty intricately on your group itself. For instance, $\pi\_3$ of the suspension of the classifying space of a *free* group is the set of symmetric elements in $G\_{\text{ab}}\otimes G\_{\text{ab}}$ where $G\_{\text{ab}}$ is the abelianization of $G$, and for a general group it lives in an exact sequence between something involving such symmetric elements and the second group homology of $G$. I don't know a closed form for it but maybe someone else knows better.
**EDIT**: Let me at least be precise, there's an exact sequence
$$\pi \_4 (\Sigma BG)\rightarrow H\_3 G \rightarrow (G\_{\text{ab}}\otimes G\_{\text{ab}})^{\mathbb Z/2} \rightarrow \pi\_3 (\Sigma BG)\rightarrow H\_2 G\rightarrow 0$$
Note that for $R$ a ring, an element of $(G\_{\text{ab}}\otimes G\_{\text{ab}})^{\mathbb Z/2}$ gives rise to an $R$-valued symmetric bilinear pairing on $\mathsf{Hom}(G\_{\text{ab}},R)$.
**EDIT FOR THE FINAL TIME**: sorry for the multiple revisions, switching back and forth between homology and cohomology gave me errors. the exact sequence above should be correct now.
| 29 | https://mathoverflow.net/users/360 | 2121 | 1,359 |
https://mathoverflow.net/questions/1951 | 19 | One could try to apply the Eilenberg-Moore spectral sequence to the pullback diagram • → X ← •, obtaining a spectral sequence TorH•(X, R)(R, R) => H•(ΩX, R), but could there be differentials or extension problems which differ for different spaces X with the same cohomology ring?
| https://mathoverflow.net/users/126667 | Does the cohomology ring of a simply-connected space X determine the cohomology groups of ΩX? | Tyler's comment to my earlier answer seems to give a solution; he suggests comparing the space $T=(S^3\vee S^3)\cup\\_{[x,[x,y]]} e^8$ with a wedge $S^3\vee S^3\vee S^8$. It's probably easier to think about homology with the Pontryagin product. Homology of loops on on the wedge will be a tensor algebra on classes in 2,2,7 (since it's loops of a suspension). The homology of loops on Tyler-space $T$ should differ in dimension 6: the homology class [x,[x,y]] will be 0 (where x,y are now the homology generators in dimension 2), "killed" by the new attaching map. So H\_6 (and thus H^6) of the two spaces have different rank.
To make this explicit, we have $S^7 \xrightarrow{f} X \rightarrow T$, where X is the wedge of two 3-spheres. The restriction of $\Omega f: \Omega S^7 \to \Omega X$ is a map $S^6 \to \Omega X$ adjoint to f, and on homology this hits the homology class corresponding to the [x,[x,y]]. The result follows because $\Omega S^7 \to \Omega X \to \Omega T$ is null homotopic. (I'm basically using the Hilton-Milnor theorem to understand $\Omega X$.)
| 14 | https://mathoverflow.net/users/437 | 2122 | 1,360 |
https://mathoverflow.net/questions/383 | 70 | In undergraduate differential equations it's usual to deal with the Laplace transform to reduce the differential equation problem to an algebraic problem.
The Laplace transform of a function $f(t)$, for $t \geq 0$ is defined by $\int\_{0}^{\infty} f(t) e^{-st} dt$.
How to avoid looking at this definition as "magical"? How to somehow discover it from more basic definitions?
| https://mathoverflow.net/users/273 | Motivating the Laplace transform definition | What is also very interesting is that the Laplace transform is nothing else but the continuous version of power series - see this insightful video lecture from MIT:
<https://ocw.mit.edu/courses/18-03-differential-equations-spring-2010/resources/lecture-19-introduction-to-the-laplace-transform/>
| 53 | https://mathoverflow.net/users/1047 | 2141 | 1,369 |
https://mathoverflow.net/questions/2107 | 4 | Let $B$ be a curve (integral but not necessarily smooth) and let $\pi: C --> B$ be a family of curves such that each fiber is a rational curve with $g$ many elliptic tails attached.
Let $\omega$ be the relative dualizing sheaf.
>
> **Question**: Why is the pushforward $\pi\_\* \omega$ trivial (as a vector bundle)?
>
>
>
| https://mathoverflow.net/users/2 | Question about a family of semistable curves | I don't think this is quite right. Here is the right statement: let $E\_1, ..., E\_g$ be the tails, with maps $q\_i: E\_i \longrightarrow B$. Then
$\pi\_\* \omega\_{C/B} = \bigoplus (q\_i)\_\* \omega\_{E\_i/B}$.
So, if your tails don't vary with B, this bundle is trivial.
Explanation: $\omega\_{C/B}$ can be described explicitly: a section of $\omega\_{C/B}$ is a one-form on each component of $C$, with simple poles at the nodes of $C$, so that at every node the residues of the form on the two components match.
Now, on a curve of genus $1$, a one-form with only a single simple pole, must in fact have no poles. So the sections of $\omega\_{C/B}$, restricted to the $E\_i$, are sections of $\omega\_{E\_i/B}$. Moreover, the sections of $\omega\_{C/B}$ restricted to the rational components are one-forms with no poles, and are hence $0$. So to give a section of $\omega\_{C/B}$ is simply to give a section of $\omega\_{E\_i/B}$ on each $E\_i$. QED.
| 8 | https://mathoverflow.net/users/297 | 2145 | 1,370 |
https://mathoverflow.net/questions/2128 | 18 | Let L be the lattice of Young diagrams ordered by inclusion and let Ln denote the nth rank, i.e. the Young diagrams of size n. Say that lambda > mu if lambda covers mu, i.e. mu can be obtained from lambda by removing one box and let C[L] be the free vector space on L. The operators
U lambda = summu > lambda mu
D lambda = sumlambda > mu mu
are a decategorification of the induction and restriction operators on the symmetric groups, and (as observed by Stanley and generalized in the theory of differential posets) they have the property that DU - UD = I; in other words, Young's lattice along with U, D form a representation of the Weyl algebra.
Is this a manifestation of a more general phenomenon? What's the relationship between differential operators and the representation theory of the symmetric group?
Edit: Maybe I should ask a more precise question, based on the comment below. As I understand it, in the language of Coxeter groups the symmetric groups are "type A," so the Weyl algebra can be thought of as being associated to type A phenomena. What's the analogue of the Weyl algebra for types B, C, etc.?
| https://mathoverflow.net/users/290 | Young's lattice and the Weyl algebra | **EDIT** (3/16/11): When I first read this question, I thought "hmmm, Weyl algebra? Really? I feel like I never hear people say they're going to categorify the Weyl algebra, but it looks like that's what the question is about..." Now I understand what's going on. Not to knock the OP, but there's a much bigger structure here he left out. If you have any $S\_n$ representation $M$, you get a functor $$\operatorname{Ind}\_{S\_m\times S\_n}^{S\_{m+n}}-\otimes M: S\_m-\operatorname{rep}\to S\_{m+n}-\operatorname{rep}$$ and these functors all have adjoints which I won't bother writing down. All of these together categorify a Heisenberg algebra, which is what Khovanov proves in the paper linked below (though cruder versions of these results on the level the OP was talking about are much older, at least as far back as [Leclerc and Thibon](http://arxiv.org/abs/q-alg/9602025)).
---
There is a much more general story here, though one my brain is not very up to explaining it this afternoon, and unfortunately, I don't know of anywhere it's summarized well for beginners.
So, how you you prove the restriction rule you mentioned above? You note that the restriction of a $S\_n$ rep to an $S\_{n-1}$ rep has an action of the [Jucys-Murphy](http://en.wikipedia.org/wiki/Jucys%E2%80%93Murphy_element) element $X\_n$ which commutes with $S\_{n-1}$. The different $S\_{n-1}$ representations are the different eigenspaces of the J-M element.
So, one can think of "restrict and take the m-eigenspace" as a functor $E\_m$; this defines a direct sum decomposition of the functor of restriction.
Of course, this functor has an adjoint: I think the best way to think about this is as $F\\_m=(k[S\\_n]/(X\\_n-m))
\otimes\\_{k[S\\_{n-1}]} V$.
These functors `E_m,F_m` satsify the relations of the Serre relations for $\mathfrak{sl}(\infty)$. Over characteristic 0, these are all different, and you can think of this as an $\mathfrak{sl}(\infty)$. If instead, you take representations over characteristic p, then `E_m=E_{m+p}` so you can think of them as being in a circle, an affine Dynkin diagram, so one gets an action of $\widehat{\mathfrak{sl}}(p)$.
Similar categorifications of other representations can deconstructed in general by looking at representations of complex reflection groups given by the wreath product of the symmetric group with a cyclic group. So, Sammy, you shouldn't rescale, you should celebrate that you found a representation with a different highest weight (also, if you really care, you should go talk to Jon Brundan or Sasha Kleshchev; they are some of the world's experts on this stuff).
**EDIT:** Khovanov has actually just posted [a paper](http://arxiv.org/abs/1009.3295) which I think might be relevant to your question.
| 12 | https://mathoverflow.net/users/66 | 2152 | 1,374 |
https://mathoverflow.net/questions/2150 | 17 | Are filtered colimits exact in all abelian categories?
In Set, filtered colimits commute with finite limits. The proof carries over to categories sufficiently like Set (i.e. where you can chase elements round diagrams), in particular A-Mod where A is a commutative ring. This implies that filtered colimits are exact in A-Mod.
I am aware of a vague principle that things that are true in A-Mod are true for all abelian categories, but I have never seen a precise statement of this principle so I am not sure if it applies in this case.
| https://mathoverflow.net/users/1046 | Exactness of filtered colimits | Here's a dumb counterexample. If C is an abelian category, so is Cop. In Cop, filtered colimits are filtered limits in C. And, of course, there are many examples of abelian categories (such as abelian groups) where filtered limits aren't exact.
Of course, your question is really: when is an abelian category C sufficiently close to Set, so that we can ratchet up the fact that filtered colimits are exact in Set to a proof for C.
Any category of sheaves of abelian groups on a space (or on a Grothendieck topos) will have exact filtered colimits, for instance.
| 21 | https://mathoverflow.net/users/437 | 2158 | 1,379 |
https://mathoverflow.net/questions/2084 | 10 | The volume in the orthogonal group is measured by the Haar measure, which is the up to scaling unique measure that is invariant under the group operation. I consider the usual metric that is induced by the spectral norm $|M| = \max |Mx|$ where $x$ ranges over all vectors of length 1 and the vector norm is the Euclidean one. A $\delta$-ball is the set of all orthogonal matrices that have distance less or equal $\delta$ to a fixed matrix $M$. Because of the invariance of the Haar measure, for a fixed $\delta$, all $\delta$-balls have the same volume.
| https://mathoverflow.net/users/1038 | What is the volume of a $\delta$-ball in the orthogonal group $O(n)$? Is there a (simple) lower bound? | The volume of the delta-ball of the special orthogonal group can be computed exactly by applying the Weyl integration formula: (Without loss of generality, we assume that the delta-ball is around the unit group element).
a. One notices (Again due to the invariance under the Haar measure) that the characteristic function of the delta ball is a class function. Thus upon the application of the Weyl integration formula we are left only with the radial part on the eigenvalues which is a $\lfloor N/2\rfloor$-dimensional integral for $\mathrm{SO}(N)$. Here, the radial integral is described explicitely.
b. The eigenvalues of an orthogonal matrix of dimension $N=2m+1$ are $1$ and $m$ pairs $\exp(i \phi\_ m)$ and $\exp(-i \phi\_ m)$, $0\leq\phi\_ 1 \leq\ldots\leq\phi\_m \leq\pi$. In the case of even dimensions, the unit eigenvalue is absent.
c. The delta-ball condition on the eigenvalues becomes:
$$
|\exp(i\phi\_k)-1|\leq\delta ,
$$
which implies:
$$\phi\_k\leq 2 \arcsin\sqrt{\delta/2}.$$
d. Applying the Weyl integration formula, we obtain for the odd case $\mathrm{SO}(2m+1)$:
$$
\mathrm{Vol}(\delta\mathrm{-ball}) = \frac{2^{m^2}}{\pi^m m!} \int\_{\phi\_1\leq\ldots\leq\phi\_m \leq 2 \arcsin\sqrt{\delta/2}}
\prod\_{1\leq j < k \leq m} (\cos\phi\_k-\cos\phi\_j)^2 \prod\_{l=1}^m \sin^2(\phi\_l/2) d\phi\_1 \cdots d\phi\_m.
$$
e. For the even dimensional case, the only changes are $2^{m^2}$ is replaced $2^{(m-1)^2}$ and the sine terms are absent.
| 6 | https://mathoverflow.net/users/1059 | 2168 | 1,388 |
https://mathoverflow.net/questions/2132 | 8 | Trying to understand answer to [this question](https://mathoverflow.net/questions/1988/regulators-of-number-fields-and-elliptic-curves).
What is the **(Beilinson) higher regulator** of a number field?
| https://mathoverflow.net/users/65 | What is the Beilinson regulator? | Here is an attempt to answer, but I hope that someone else can give a better explanation.
As Rob H. pointed out in his answer to the previous question, the survey of Nekovar is very nice, and it is also available online [here](http://people.math.jussieu.fr/~nekovar/pu/mot.pdf).
About your question: The Beilinson regulator can be defined for number fields but also for varieties over number fields. It is a map from motivic cohomology (or algebraic K-theory) with rational coefficients, to Deligne-Beilinson cohomology, and can be thought of as a kind of Chern character.
The precise definition of the regulator is quite nontrivial, and there are several equivalent ways of defining it. Philosophically, it should be a map between certain Ext groups, induced from a "Hodge realization" of motives. Sorry for not explaining this well - it belongs to your other question about the yoga of motives.
Forgetting about philosophy, there are several ways of actually constructing the regulator. One approach is to use the general framework of characteristic classes developed by Gillet. Nekovar has a "direct" construction in his paper. For another construction in terms of explicit complexes, see the recent thesis of Elisenda Feliu, available on [her webpage](http://www.math.ku.dk/~efeliu/research.html). Another reference is the Bourbaki talk by Soule, available [here](http://archive.numdam.org/ARCHIVE/SB/SB_1984-1985__27_/SB_1984-1985__27__237_0/SB_1984-1985__27__237_0.pdf).
If you are only interested in number fields, there is another construction of the regulator, named after Borel. An excellent online reference for this, and its relation to the Beilinson regulator, is the book of Burgos, available [here](http://atlas.mat.ub.es/personals/burgos/files/brbr.pdf). The regulator generalizes the Dirichlet regulator which is covered in most introductory books on algebraic number theory. For a computational approach to Borel's regulator, see recent papers on the arXiv by Choo, Mannan, Sánchez-Garcia and Snaith.
| 10 | https://mathoverflow.net/users/349 | 2187 | 1,400 |
https://mathoverflow.net/questions/1652 | 19 | I try to keep a list of standard ring examples in my head to test commutative algebra conjectures against. I would therefore like to have an example of a ring which is normal but not Cohen-Macaulay. I've found a few in the past, but they were too messy to easily remember and use as test cases. Suggestions?
| https://mathoverflow.net/users/297 | Simple example of a ring which is normal but not CM | Another family of examples is given by the homogeneous coordinate rings of
irregular surfaces (ie 2-dimensional $X$ such that $H^1({\mathcal O}\_X) \neq 0$);
these surfaces cannot be embedded in any way so that their homogeneous coordinate rings
become Cohen-Macaulay. Elliptic scrolls (such as the one in the previous answer)
and Abelian surfaces in P4, made from the sections of the Horrocks-Mumford bundle, are such examples.
The point is that sufficiently positive, complete embeddings of any smooth variety (or somewhat more generally) will have normal homogeneous coordinate rings, and they will be Cohen-Macaulay iff the intermediate cohomology of the variety vanishes. All the examples above fall into this category. It's an interesting general question to ask how positive is "sufficiently positive".
| 22 | https://mathoverflow.net/users/5771 | 2194 | 1,406 |
https://mathoverflow.net/questions/2093 | 9 | I'm interested in graph families which are sparse, and by sparse I mean the number of edges is linear in the number of vertices. |E| = O(|V|).
Besides non-trivial minor-closed families of graphs (these turn out to be sparse), I don't know any other families. Can anyone suggest any interesting graph families (which are not minor-closed) which are sparse?
Please don't suggest the trivial family ("the family of sparse graphs").
EDIT: Thanks to the first few people who replied, I realized that bounded degree graphs (max degree < k) also form an interesting and large class of sparse graphs. So perhaps I'll refine my question to exclude those too. Any interesting sparse graph families where the max degree isn't bounded? For example the family of star graphs is sparse and not bounded degree. (But they're minor-closed.)
| https://mathoverflow.net/users/1042 | Interesting families of sparse graphs? | 1) Graphs of degree at most d. There are a myriad results known about them. Among them there are regular graphs (graphs in which every vertex is of degree exactly d).
2) Graphs of bounded degeneracy. These are the graphs in which every subgraph has a vertex of degree at most d. For example, stars have unbounded degree, but degeneracy 1. These share some common properties with graphs of bounded degree. For example, there are several results in Ramsey theory that have been extended from graphs of bounded degree to graphs of bounded degeneracy.
3) Random graphs of edge density c/n for constant c. Strictly speaking these are not a 'family of graphs', but since for properties of interest a random graph usually either satisfies it with high probability or fails it with high probability, it is fair to add them. There are also various other models of random graph that yield sparse graphs.
4) On a lower level, there are particular subfamilies of the above, such as stars, grids, etc.
| 10 | https://mathoverflow.net/users/806 | 2195 | 1,407 |
https://mathoverflow.net/questions/2127 | 5 | The aim of transforming the Black-Scholes PDE is of course to find a form where an relatively easy solution exists. Most of the steps seem to be straightforward - please use this reference:
<https://planetmath.org/AnalyticSolutionOfBlackScholesPDE>
...all but one, actually the last one where a convection-diffusion equation is being transformed into the basic diffusion equation.
[In the article you find it here: "Under the new coordinate system (z), we have the relations amongst vector fields ... leading to the following transformation of equation..."]
The u\_x-term vanishes by some magic coordinate transformation. When you look at the derivatives they even seems wrong to me because they state that tau=s and then derive delta/delta tau = delta/delta s + some extra term (delta/delta y \* -(r-1/2 sigma^2).
I simply don't get it: first how it works and second how they find that kind of transformation.
| https://mathoverflow.net/users/1047 | Transformation of the Black-Scholes PDE into the diffusion equation - shift of coordinate system | I'm afraid the Planetmath page put my browser into an infinite reload loop, so I can't help you with the formalism there.
I would recommend instead looking at the change of variables in [the Wikipedia article](http://en.wikipedia.org/wiki/Black-Scholes#Derivation). The last time I checked it, it seemed to work.
**Edit:** Okay, I have a formulation that works. I'll write s for sigma, so the equation is initially:
Vt + (1/2)s2S2VSS = rV - rSVS.
Since S follows a lognormal random walk (in particular the stochastic diff eq governing S involves a logarithmic derivative), it is natural to change to x = log S, or S = ex, so the log price x follows normal Brownian motion. This yields the equation:
Vt + (1/2)s2(Vxx - Vx) = r(V - Vx).
Black-Scholes is a final-value problem, i.e., we know the value of the option at time T, and diffusion works backwards. It is therefore natural to negate the time variable (and multiply by a suitable scalar to make things neater). tau = (1/2)s2(T-t). Then we get:
(1/2)s2(Vxx - Vx - Vtau) = r(V - Vx).
Finally we rescale the value function to remove exponential growth effects. u = eax + b(tau)V for undetermined coefficients a and b. We can substitute, multiply the equation by 2eax+b(tau)/s2, and we get:
uxx + (something)ux + (something else)u = utau.
(something) is a degree one polynomial in a and is independent of b. (something else) is a degree one polynomial in b, so we can choose a and b to kill those terms. This yields the diffusion equation.
Hope that helps.
| 5 | https://mathoverflow.net/users/121 | 2199 | 1,410 |
https://mathoverflow.net/questions/2146 | 73 | There are some things about geometry that show why a motivic viewpoint is deep and important. A good indication is that Grothendieck and others had to invent some important and new algebraico-geometric conjectures *just to formulate the definition of motives*.
There are things that I know about motives on some level, e.g. I know what t[he Grothendieck ring of varieties](https://mathoverflow.net/questions/319/spectrum-of-the-grothendieck-ring-of-varieties) is or, roughy, what are the ingredients of the definition of motives.
But, how would you explain the **Grothendieck's yoga of motives**? What is it referring to?
| https://mathoverflow.net/users/65 | What's the "Yoga of Motives"? | So this is a crazy question, but I will try to give at least a partial answer. This [question about the Beilinson regulator](https://mathoverflow.net/questions/2132/what-is-the-beilinson-regulator/2187#2187) is also relevant, and this is also an attempt to reply to the comments of Ilya there. I apologize for simplifying and glossing over some details, see the references for the full story.
First of all, some references: A leisurely but still far from content-free exposition by Kahn on the yoga of motives is available [here](http://people.math.jussieu.fr/~kahn/preprints/lecons.pdf) (in French). For Grothendieck's idea of pure motives, see Scholl: Classical motives, available on [his webpage](http://www.dpmms.cam.ac.uk/~ajs1005/) in zipped dvi format. For mixed motives, see this [survey article of Levine](http://www.math.neu.edu/%7Elevine/publ/MixMotKHB.pdf). There is also lots of stuff in the Motives volumes, edited by Jannsen, Kleiman and Serre, here is the [Google Books page](http://books.google.com/books?id=v2CuklFFV5IC&printsec=frontcover&source=gbs_navlinks_s#v=onepage&q=&f=false). Finally, I would strongly recommend the book by André: Introduction aux motifs - this is has lots of background and "yoga", as well as precise statements about what is known and what one conjectures.
**Pure motives**
The standard way of explaining what motives are is to say that they form a "universal cohomology theory". To make this a bit more precise, let's start with pure motives. We fix a base field, and consider the category of smooth projective varieties, and various cohomology functors on this category. The precise notion of cohomology functor in this context is given by the axioms for a Weil cohomology theory, see this [blog post of mine](http://homotopical.wordpress.com/2009/03/18/weil-cohomology/) for more details.
There are (at least) three key points to mention here: one is that a Weil cohomologies are "geometric" theories, as opposed to "absolute". For example, when considering etale cohomology, we are considering the functor given by base changing the variety to the absolute closure of the ground field, and then taking sheaf cohomology with respect to the constant sheaf Z/l for some prime l, in the etale topology. The "absolute" theory here would be the same, but without base changing in the beginning. In the classical literature, and in number theory, the geometric version is the most important, partly because it carries an action of the Galois group of the base field, and hence gives rise to Galois representations. On the other hand, the absolute version is important for example in the work of Rost and Voevodsky on the Bloch-Kato conjecture, and in comparison theorems with motivic cohomology. Similarly, it seems like cohomology theories in general come in geometric/absolute pairs.
The second key point to mention is that the Weil cohomology groups come with "extra structure", such as Galois action or Hodge structure. For example, l-adic cohomology takes values in the category of l-adic vector spaces with Galois action, and Betti cohomology takes values in a suitable category of Hodge structures. A nice reference for some of this is Deligne: Hodge I, in the ICM 1970 volume.
The third key point is that Weil cohomology theories are always "ordinary" in some sense, i.e. in some framework of oriented cohomology theories they would correspond to the additive formal group law (see Lurie: [Survey on elliptic cohomology](http://www-math.mit.edu/~lurie/papers/survey.pdf)). If we allowed more general (oriented) cohomology theories, the universal cohomology would not be pure motives, but algebraic cobordism.
Now all these cohomology theories are functors on the category of smooth projective varieties, and the idea is that they should all factor through the category of pure motives, and that the category of pure motives should be universal with this property. We know how to construct the category of pure motives, but there is a choice involved, namely choosing an equivalence relation on algebraic cycles, see the article by Scholl above for more details. For many purposes, the most natural choice is rational equivalence, and the resulting notion of pure motives is usually called Chow motives. For a precise statement about the universal property of Chow motives, see André, page 36: roughly (omitting some details), any sensible monoidal contravariant functor on the category of smooth projective varieties, with values in a rigid tensor category, factors uniquely over the category of Chow motives.
Now to the point of realizations raised by Ilya in the question about regulators. Given a category of pure motives with a universal property as above, there must be functors from the category of motives to the category of (pure) Hodge structures, to the category of Q\_l vector spaces with Galois action, etc, simply because of the universal property. These functors are called realization functors.
**Mixed motives**
It seems like all the cohomology functors one typically considers can be defined not only for smooth projective varieties, but also for more general varieties. The right notion of cohomology here seems to be axiomatized by some version of the Bloch-Ogus axioms. One could again hope for a category which has a similar universal property as above, but now with respect to all varieties. This category would be the category of mixed motives, and in the standard conjectural framework, one hopes that it should be an abelian category. It is not clear whether this category exists or not, but see Levine's survey above for a discussion of some attempts to construct it, by Nori and others. If we had such a category, a suitable universal property would imply that there are realization functors again, now to "mixed" categories, for example mixed Hodge structures. The realization functors would induce maps on Ext groups, and a suitable such map would be the Beilinson regulator, from some Ext groups in the category of mixed motives (i.e. motivic cohomology groups) to the some Ext groups which can be identified with Deligne-Beilinson cohomology.
We do not have the abelian category of mixed motives, but we have an excellent candidate for its derived category: this is Voevodsky's triangulated categories of motives. They are also presented very well in the survey of Levine. A really nice recent development is the work of Déglise and Cisinski, in which they construct these triangulated categories over very general base schemes (I think Voevodsky's original work was mainly focused on fields, at least he only proved nice properties over fields).
To end by reconnecting to the Beilinson conjectures, there is extremely recent work of Jakob Scholbach (submitted PhD thesis, maybe on the arXiv soon) which seems to indicate that the Beilinson conjectures should really be formulated in the setting of the Déglise-Cisinski category of motives over Z, rather than the classical setting of motives over Q.
The yoga of motives involves far more than what I have mentioned so far, for example things related to periods and special values of L-functions, the standard conjectures, and the idea of motivic (and maybe even "cosmic") Galois groups, but all this could maybe be the topic for another question, some other day :-)
| 58 | https://mathoverflow.net/users/349 | 2206 | 1,414 |