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https://mathoverflow.net/questions/2214 | 12 | Hello, the Green-Tao theorem says infinitely many k-term Arithmetic Progressions exist for any integer k.
My question is: can we actually partition the primes into 3-term APs only (or is there a simple reason why it cannot be expected) ? And if it were possible, then what would it mean for the set of primes ?
For example fix a large integer (say M=10,000) then take the primes (except 2): 3, 5, 7, 11, 13... and remove the longest possible AP with common difference less than M as you go along. It provides the partition:
3 5 7 -- 11 17 23 29 -- 13 37 61 -- 19 31 43 -- 41 47 53 59 -- 67 73 79 -- 71 89 107 -- 83 131 179 227 -- 97 103 109 -- 101 137 173 -- ... Numerical data for the first 10,000 primes with that M shows that the average length of APs so defined is 3.2 (which is thus quite close to 3, so at least numerically there exists partitions where 3-terms APs cover a large fraction of the primes, hence the question).
| https://mathoverflow.net/users/469 | Covering the primes by 3-term APs ? | Using the greedy algorithm, this would follow if for any fixed prime q, there exist infinitely prime "pairs" of the form p and 2p-q. This follows from standard (difficult) conjectures if q is odd (for example, the case q = -1 corresponds to "Sophie Germaine Primes"). On the other hand, it would be an implication of such a partition that for each odd prime q, there either:
(i) exists at least one prime pair (p,2p-q),
(ii) 2q is the sum of two primes,
(iii) exists at least one prime pair (p,q-2p).
Almost all proofs showing that there *exist* primes of a certain form also prove that there are *infinitely* many such primes. Thus, I suspect, one could not prove this result without also proving the difficult conjectures alluded to above.
| 8 | https://mathoverflow.net/users/nan | 2219 | 1,419 |
https://mathoverflow.net/questions/2215 | 12 | I've read numerous *introductions* to finite fields, but I feel like my intuition about them is fairly lacking. Considering that finite fields are the the most "inert" objects in algebraic geometry, I think I could use a serious surge of perspective.
What I would like to read now is a comprehensive overview that tells me "everything I need to know" about how finite fields and their algebraic closures work, algebraically. I don't mind working out the proofs on my own if they are terse or absent; I'm just looking for quality and quantity of results. Hopefully some intense reading will help steep out some of my insecurities about characteristic p.
Can anyone recommend a single source for such an overview?
Thanks!
| https://mathoverflow.net/users/84526 | A comprehensive overview of finite fields | [Finite Fields](http://books.google.com/books?id=xqMqxQTFUkMC) by R. Lidl and H Niederreiter (CUP). Probably as comprehensive as it gets.
The ams review calls it the ``the Bible of finite fields''. You can find it (the review)[here](http://www.ams.org/bull/1999-36-01/S0273-0979-99-00768-5/S0273-0979-99-00768-5.pdf).
| 10 | https://mathoverflow.net/users/262 | 2223 | 1,423 |
https://mathoverflow.net/questions/2212 | 6 | I ran into this obstacle in a harmonic analysis problem; I know epsilon about coloring problems.
Is it possible to finitely color Z^2 so that the points (x,a) and (a,y) are differently colored for every x, y and a in the integers (excepting, of course, the trivial cases x=y=a)?
| https://mathoverflow.net/users/1071 | Can I finitely color Z^2 such that (x,a) and (a,y) are different for every x,y,a? | No, this isn't possible. First note that the question is the same if we replace Z with any infinite subset of Z. Now, suppose it were possible to color Z2 with k colors, where k is minimum. Pick any row r. We can find an infinite set S of integers not containing r such that for x in S, all the (x, r) have the same color c. No point in SxS can have color c, so SxS is colored with k-1 colors. But this contradicts the minimality of k.
| 11 | https://mathoverflow.net/users/1013 | 2226 | 1,426 |
https://mathoverflow.net/questions/2218 | 11 | What can you say about the complexity class $\text{P}^{\text{NP}}$, i.e. decision problems solvable by a polytime TM with an oracle for SAT? This class is also known as $\Delta\_2^p$.
Obviously $\text{P}^{\text{NP}}$ is in $\text{PH}$ somewhere between $\text{NP} \cup \text{coNP}$, and $\Sigma\_2^p \cap \Pi\_2^p$. What else is known about that complexity class?
| https://mathoverflow.net/users/1027 | Characterize P^NP (a.k.a. Delta_2^p) | The standard complete problem for the "function version" of P^NP is to find the lexicographically last satisfying assignment of a given boolean formula. To be more finicky, a complete language for P^NP is
{ n-variable boolean formulas phi : phi is satisfiable and phi's lexicographically last satisfying assignment has x\_n = 1 }.
This is Krentel's Theorem.
Under a standard complexity assumption one can derandomize the more powerful class BPP^NP down to P^NP. With the power of BPP^NP, one can compute the number of satisfying assignments to a circuit to within a 1 +/- 1/poly(n) factor [Sipser / Stockmeyer / Valiant-Vazirani] and exactly learn unknown polynomial-size circuits [Bshouty-Cleve-Gavalda-Kannan-Tamon].
| 23 | https://mathoverflow.net/users/658 | 2240 | 1,437 |
https://mathoverflow.net/questions/2200 | 26 | What is a simplest example of a manifold $M^{2n}$ that admits two symplectic structures with isotopic almost complex structures, and such that Gromov-Witten invariants of these symplectic structures are different?
(unfortunately I don't know any example...) If we don't impose the condition that almost complex structures are isotopic, such examples exist in dimension 6.
**Added**
**Refined question.** Is there a manifold $M^{2n}$ with two symplectic forms $\omega\_1$, $\omega\_2$,
such that the cohomology classes of $\omega\_1$ and $\omega\_2$ are the same and
the corresponding almost complex structures are homotopic,
but at the same time the Gromov-Witten invariants are different?
| https://mathoverflow.net/users/943 | Manifolds distinguished by Gromov-Witten invariants? | The examples in Ruan's paper "Symplectic topology on algebraic 3-folds" (JDG 1994) seem to qualify: take any two algebraic surfaces V and W which are homeomorphic but such that V is minimal and W isn't. These are nondiffeomorphic, but VxS2 and WxS2 are diffeomorphic, and Ruan gives lots of examples (starting with V equal to the Barlow surface and W equal to the 8-point blowup of CP2) where the diffeomorphism can be arranged to intertwine the first Chern classes, whence by a theorem of Wall the almost complex structures are isotopic. However, the distinction between the GW invariants between V and W (which holds because V is minimal and W isn't) survives to VxS2 and WxS2, so VxS2 and WxS2 aren't symplectic deformation equivalent.
| 21 | https://mathoverflow.net/users/424 | 2242 | 1,438 |
https://mathoverflow.net/questions/2245 | 3 | We call $S(u)$ the space complexity of the vEB tree holding elements in the range $0$ to $u-1$, and suppose without loss of generality that $u$ is of the form $2^{2^k}$.
It's easy to get the recurrence $S(u^2) = (1+u) S(u) + \Theta(u)$. (In Wikipedia's [article](https://en.wikipedia.org/w/index.php?title=Van_Emde_Boas_tree&oldid=314311691) the last term is $O(1)$, but it's wrong because we must count the space for the array.)
Van Emde Boas (and others) gave in [1] the trivial bound $S(u) = O(u \log \log u)$, and later in [2] he found a clever way to combine the data structure with another one in order to reach space complexity $O(u)$, while maintaining the $O(\log \log u)$ time bounds.
But, modern references present the original data structure and state without proof that the space complexity is $O(u)$. For instance, the very recent 3rd edition of "Introduction to algorithms" by Cormen et al. ([ZBL1187.68679](https://zbmath.org/?q=an:1187.68679)) leaves it as an exercise.
I tried with some friends to [dis]prove the $O(u)$ bound without luck.
1. *van Emde Boas, P.; Kaas, R.; Zijlstra, E.*, [**Design and implementation of an efficient priority queue**](https://doi.org/10.1007/BF01683268), Math. Syst. Theory 10(1976), 99β127 (1977). [ZBL0363.60104](https://zbmath.org/?q=an:0363.60104).
2. *van Emde Boas, P.*, [**Preserving order in a forest in less than logarithmic time and linear space**](https://doi.org/10.1016/0020-0190(77)90031-X), Inf. Process. Lett. 6, 80β82 (1977). [ZBL0364.68053](https://zbmath.org/?q=an:0364.68053).
| https://mathoverflow.net/users/961 | Determining the space complexity of van Emde Boas trees | The recurrence S(u2) = (1+u) S(u) + Ξ(u) can be shown to be O(u) by the following method: First assume that the constant in the Ξ(u) is at most 1 and that S(4) is at most 1, by dividing through as necessary.
Then we can prove S(u) < u - 2 by induction. The base case S(4) holds by the above assumption. The inductive case is
S(u^2) < (1+u) (u-2) + u = u2 - 2,
as desired.
Incidentally, I think one can avoid the Ξ(u) term and have Ξ(1) instead by not actually storing the array of pointers to substructures (as in the exposition currently on Wikipedia) but instead having implicit substructures all stored in one big array. Of course, some work would need to be done to show that you can keep track of all the necessary information. Either way, the solution above works.
| 4 | https://mathoverflow.net/users/1079 | 2251 | 1,444 |
https://mathoverflow.net/questions/2173 | 17 | I've been reading about D-modules, and have seen a proof that D\_X, the ring of differential operators on a variety, is "almost commutative", that is, that its associated graded ring is commutative. Now, I know that with the Ore conditions, we can localize almost commutative rings, and so we get a legitimate sheaf D to do geometry with. But how far does the analogy go? What theorems are true for commutative rings but can't be modified reasonably to be true for (nice, say, left and right noetherian or somesuch) almost commutative rings?
| https://mathoverflow.net/users/622 | How much theory works out for "almost commutative" rings? | Don't get too excited about the theory of algebraic geometry for almost commutative algebras. A ring can be almost commutative and still have some very weird behavior. The Weyl algebras (the differential operators on affine n-space) are a great example, since they are almost commutative, and yet:
* They are simple rings. Therefore, either they have no 'closed subschemes', or the notion of closed subscheme must correspond to something different than a quotient.
* The have global dimension n, even though their associated graded algebra has global dimension 2n. So, global dimension can jump up, even along flat deformations.
* There exist non-free projective modules of the nth Weyl algebra (in fact, stably-free modules!). Thus, intuitively, Spec(D) should have non-trivial line bundles, even though it is 'almost' affine 2n-space.
Just having a ring of quotients isn't actually that strong a condition on a ring. For instance, Goldie's theorem says that any right Noetherian domain has a ring of quotients, and that is a pretty broad class of rings.
Also, what sheaf are you thinking of D as giving you? You have all these Ore localizations, and so you can try to build something like a scheme out of this. However, you start to run into some problems, because closed subspaces will no longer correspond to quotient rings. In commutative algebraic geometry, we take advantage of the miracle that the kernel of a quotient map is the intersection of a finite number of primary ideals, each of which correspond to a prime ideal and hence a localization. In noncommutative rings, there is no such connection between two-sided ideals and Ore sets.
Here's something that might work better (or maybe this is what you are talking about in the first place). If you have a positively filtered algebra A whose associated graded algebra is commutative, then A\_0 is commutative, and so you can try to think of A as a sheaf of algebras on Spec(A\_0). The almost commutativity requirement here assures us that any multiplicative set in A\_0 is Ore in A, and so we do get a genuine sheaf of algebras on Spec(A\_0). For D\_X, this gives the sheaf of differential operators on X. Other algebras that work very similarly are the enveloping algebras of Lie algebroids, and also rings of twisted differential operators.
| 23 | https://mathoverflow.net/users/750 | 2253 | 1,446 |
https://mathoverflow.net/questions/2270 | 16 | I have a basic working knowledge of category thoery since I do research in programming languages and typed lambda-calculus. Indeed, I have refereed many papers in my area based on category theory.
But, in doing this refereeing, and in reading many important categorial papers in my area, I simply find the terminology and presentation style extremely opaque compared to style that I prefer which instead emphasizes logic inference rules and extensions of the lambda calculus.
Given the (extended) Curry-Howard Isomorphism between programming languages, logics, and categories, it's clear that I can understand the concepts I need by mapping category theory into the other two. But, am I missing something in the process, or are the papers I'm refereeing just making things more difficult than they need to be?
| https://mathoverflow.net/users/1086 | Why do I find Category Theory mostly just a way to make simple things difficult? | Although to you, category theory is merely an inefficient framework for data about logic and programming languages, to mathematicians working in areas like algebraic geometry and algebraic topology, categories are truly essential. For us, some of our most basic notions make no sense and look extremely awkward (in fact, some of them are from pre-category days, and no one really knew what we wanted intuitively until after we started using categories) until you phrase them in terms of categories and universal properties of objects and morphisms (and 2-morphisms, etc). Additionally, it helps us tell what various types of mathematical objects have in common, and how they relate via functors and natural transformations.
As far as recasting algebraic topology and algebraic geometry in terms of lambda calculus, I'd rather like to see that if anyone can manage to, say, give an alternative definition of stack or gerbe or model structure which are more intuitive without using categories and groupoids and the like.
| 21 | https://mathoverflow.net/users/622 | 2284 | 1,463 |
https://mathoverflow.net/questions/2281 | 15 | Hello, apparently finite groups which are n-transitive with n>5 are only the permutation groups Sn or the alternating groups An+2, see e.g. page 226 this book by Isaacs <http://books.google.fr/books?id=pCLhYaMUg8IC&pg=PA226>
Is this characterization useful at all? For instance, are there famous proofs (maybe in a geometric context), which use it to, say, show that a certain group is in fact some An ?
| https://mathoverflow.net/users/469 | Use of n-transitivity in finite group theory | This fact is used in a nice way by [Dunfield and Thurston](http://arxiv.org/abs/math/0502567) to show that, for any finite simple group Q, the number of Q-covers of a "random" 3-manifold in their sense follows a Poisson distribution. (The multiple transitivity appears in Thm 7.4.)
Also: I don't have a reference for this in mind, but I've seen Nick Katz give talks where he uses a "linear" version of this, showing that an algebraic subgroup of GL(V) (typically a monodromy group) is "as big as you expect", using irreducibility of tensor powers of V.
**EDIT:** In response to Noah's query, I looked up a reference; the relevant theorem is due to Michael Larsen and is often called "Larsen's Alternative." You can read about it in section 1 of [this paper of Katz](http://muse.jhu.edu/journals/american_journal_of_mathematics/info/docs/hida_pdfs/hida18.pdf).
For the specific problem of distinguishing a subgroup from a group by means of moments, [the 2005 paper of Guralnick and Tiep](http://www.ams.org/ert/2005-009-05/S1088-4165-05-00192-5/home.html) is relevant.
| 16 | https://mathoverflow.net/users/431 | 2285 | 1,464 |
https://mathoverflow.net/questions/2264 | 8 | A rig is an algebraic object with multiplication and addition, such that multiplication distributes over addition and addition is commutative. However, instead of requiring that the set forms an abelian group under addition, we require only that it forms an abelian monoid. A commutative rig is a rig in which multiplication induces an abelian monoid.
A distributive category is a small category with finite products and coproducts, which I'll denote \* and + respectively, such that the canonical morphism X \* Y + X \* Z \rightarrow X \* (Y+Z) is an isomorphism. Essentially, taking products distributes over taking coproducts.
There's an apparent formal similarity between the definitions, and in fact you can get a commutative rig out of a distributive category C by taking the objects of the rig to be isomorphism classes of objects in the category (equivalently, considering the skeletal category) and letting coproducts correspond to addition and products to multiplication. For instance, if you start with the category of finite sets, you can skeletonize to obtain a commutative rig isomorphic to the rig of natural numbers.
So the question is, does every commutative rig arise this way from some distributive category? I suspect the answer is "no," and I can even think of some likely counterexamples (the nonnegative real numbers, Z), but I don't see an obvious way to prove that they are counterexamples.
Supposing that my hunch is correct, is there a nice way to classify the rigs that do arise in this manner? Is there a way to classify the commutative rings that arise when you add negatives to these rigs (analogously to the Grothendieck group)?
| https://mathoverflow.net/users/382 | Which commutative rigs arise from a distributive category? | No commutative rig with a nontrivial solution to x + y = 0 (in particular, no nonzero ring) can arise from a distributive category. I believe Schanuel observes this near the beginning of one of his papers on Euler characteristics.
The proof is simple. Suppose C is a distributive category and R is its associated commutative rig. Suppose x + y = 0 for some x, y in R. Note that 0 in R is the isomorphism class of the initial object **0** of C, since the latter is obviously an additive unit. Then there are objects X and Y in C and an isomorphism X β Y -> **0**. In other words, for every object Z of C, β’ = Hom(**0**, Z) = Hom(X β Y, Z) = Hom(X, Z) Γ Hom(Y, Z), and this can only happen when Hom(X, Z) = Hom(Y, Z) = β’, so X and Y are also initial objects and thus x = y = 0.
Of course, group completing the rig you get from a category is one way around this. Alternatively you can work in some kind of homotopical setting, e.g., finite CW complexes, and replace the condition about coproducts with one about homotopy pushouts. Either way, you have to define the ring in such a way that its elements are not isomorphism classes of objects, but in the latter case you can at least find representatives in the category for every ring element.
| 12 | https://mathoverflow.net/users/126667 | 2291 | 1,468 |
https://mathoverflow.net/questions/2302 | 6 | So, I've been running in both stacky circles and logarithmic circles and I've been wondering: is there a definition of log stack that is "useful"? I can imagine two such definitions:
1) A log stack is a stack along with an effective divisor (could be useful for studying moduli of smooth curves, then)
2) A log stack is a stack over the category of log schemes
Is either one in standard use? Are they equivalent?
| https://mathoverflow.net/users/622 | What are Log Stacks | This question was answered in Martin Olsson's thesis ( <http://math.berkeley.edu/~molsson/thesis.ps> ). He gives sufficient conditions for a fibered category on the category of log schemes to arise from "a stack with a log structure", which I think is more or less what Charles intended by Option 1.
| 6 | https://mathoverflow.net/users/35508 | 2307 | 1,479 |
https://mathoverflow.net/questions/2293 | 12 | I've recently started to look at elliptic curves and have three basic questions:
1. Is it correct to say that elliptic curves $E$ in the projective plane are in **bijective** correspondence with lattices $L$ in the complex plane via $E$ <--> $C/L$.
2. If so, is there an explicit expression of the lattice generators in terms of the equation defining the curve? Or, at least, is there a simple example of a curve and its corresponding lattice?
3. Since every elliptic curve is a Lie group, it must have a corresponding Lie algebra. Is there an explicit expression of the Lie algebra in terms of the equation or lattice? Or, again, a simple example of a curve and its Lie algebra (or, even better, an example of a curve, its lattice, and its Lie algebra).
| https://mathoverflow.net/users/1095 | Elliptic Curves, Lattices, Lie Algebras | What you want in terms of the relation between lattices and elliptic curves over C is proposition I.4.4 of Silverman's Advanced topics in the arithmetic of elliptic curves. Additionally, to go from a lattice to the equation of the elliptic curve (explicitly), you use Eisenstein series as in Corollary I.4.3 of that book. To go from the elliptic curve to the lattice is described in the proof of proposition I.4.4: you look at the homology of the curve and compute period integrals (incidentally, this is how you go from an abelian variety over C to a complex torus).
| 10 | https://mathoverflow.net/users/1021 | 2311 | 1,483 |
https://mathoverflow.net/questions/2185 | 35 | I've read about model categories from an Appendix to one of Lurie's papers.
What are the examples of **model categories**? What should be my intuition about them?
E.g. I understand the typical examples come from taking homotopy of something β but are all model categories homotopy categories?
| https://mathoverflow.net/users/65 | How to think about model categories? | Model categories are 1-categorical presentations of (β,1)-categories, which you can just think of as categories enriched in topological spaces, such as the category of spaces itself. (Actually, there are conditions on (β,1)-categories that come from model categories--most importantly they must have all homotopy limits and colimits.) They're particularly convenient for computations, as with any kind of presentation.
| 17 | https://mathoverflow.net/users/126667 | 2317 | 1,488 |
https://mathoverflow.net/questions/2300 | 106 | I've heard of this many times, but I don't know anything about it.
What I do know is that it is supposed to solve the problem of the fact that the final object in the category of schemes is one-dimensional, namely $\mathop{\text{Spec}}\mathbb Z$.
So, what is the field with one element? And, what are typical geometric objects that descend to $\mathbb F\_1$?
| https://mathoverflow.net/users/100 | What is the field with one element? | As other have mentioned, F\_1 does not exist of a field. Tits conjectured the existence of a "field of characteristic one" F\_1 for which one would have the equality G(F\_1) = W, where G is any Chevalley group scheme and W its corresponding Weyl group.
Later on Manin suggested that the "absolute point" proposed in Deninger's program to prove the Riemann Hypothesis might be thought of as "Spec F\_1", thus stating the problem of developing an algebraic geometry (and eventually a theory of motives) over it.
There are several non-equivalent approaches to F\_1 geometry, but a common punchline might be "doing F\_1 geometry is finding out the least possible amount of information about an object that still allows to speak about its geometrical properties". A "folkloric" introduction can be found in the paper by Cohn [Projective geometry over F\_1 and the Gaussian binomial coefficients](https://arxiv.org/abs/math/0407093).
It seems that all approaches so far contain a common intersection, consisting on toric varieties which are equivalent to schemes modeled after monoids. In the case of a toric variety, the "descent data" that gives you the F\_1 geometry is the fan structure, that can be reinterpreted as a diagram of monoids (cf. some works by Kato). What else are F\_1 varieties beyond toric is something that depends a lot on your approach, ranging from Kato-Deitmar (for which toric is all there is) to Durov and Haran's categorical constructions which contains very large families of examples. A somehow in-the-middle viewpoint is Soule's (and its refinement by Connes-Consani) which in the finite type case is not restricted only to toric varieties but to something slightly more general (varieties that can be chopped in pieces that are split tori). No approach is yet conclusive, so the definitions and families of examples are likely to change as the theory develops.
Last month Oliver Lorscheid and myself presented an state-of-the-art overview of most of the different approaches to F\_1 geometry: [Mapping F\_1-land: An overview of geometries over the field with one element](https://arxiv.org/abs/0909.0069) (sorry for the self promotion).
| 63 | https://mathoverflow.net/users/914 | 2327 | 1,497 |
https://mathoverflow.net/questions/2269 | 12 | Manin [stressed](http://www.ihes.fr/document?id=1704&id_attribute=48 "Manin talk") that every projective scheme should have a quantum-cohomology structure. I'd like to know more about that. And since the varieties considered in texts about monodromy resp. vanishing cycles which I have read are projective, I am curious about the behaviour of quantum cohomology under monodromy.
Edit: A coming [seminar](http://www.mpim-bonn.mpg.de/Events/This+Year+and+Prospect/Seminar+Quantum+Motives/ "link"): "Quantum motives: realizations, detection, applications", incl. a lecture "Quantum motives: review (of) the classical idea of how to linearize algebraic geometry with an eye to utilizing it in the quantum setup." and a minicourse "Geometric Langlands and quantum motives: a link".
Edit: The slides on Manin's talk on the concept of classical motives and it's relation to quantum cohomology are [here](http://pdfcast.org/pdf/manin-s-2009-ihes-talk-quantum-motives "some free pdf-sharing site").
Edit: Manin's and Smirnov's [interesting computations and thoughts](http://arxiv.org/abs/1107.4915 "link") on e.g. a "membrane quantum cohomology" (**+** russian **videos** of a talk in june 2011 on the program "... elucidation of this "self-referentiality" of quantum cohomology has just begun. The talk tries to outline the contours of this huge program and the first steps of it.": [part 1](http://narod.ru/disk/17105482001/Globus-Manin-p1.mp4.html "download-site"), [part 2](http://narod.ru/disk/17109473001/Globus-Manin-p2.mp4.html "download-site")). BTW, has anyone the text by Kapranov which is mentioned in the paper above and in [Hacking's introduction](http://arxiv.org/abs/math/0509567 "arxiv")?
| https://mathoverflow.net/users/451 | ubiquitous quantum cohomology | I'm not sure what you mean by "every projective scheme should have a quantum cohomology structure". In the talk abstract that you link to, it does not say "projective scheme" but "smooth projective variety". I don't know whether the theory generalizes to non-smooth things or to things that are not varieties.
Quantum cohomology is a deformation of ordinary cohomology (or Chow ring if you like) of a smooth projective variety (or compact symplectic manifold). This structure comes from (genus 0) Gromov-Witten invariants. GW invariants are constructed using the ordinary cohomology of your variety/manifold and the ordinary cohomology of moduli spaces of stable maps and stable curves. I mostly work over $\mathbb{C}$ so I don't know too much about what I'm about to say, but if you're not working over $\mathbb{C}$, then ordinary cohomology doesn't make sense, but instead you can still work with things like $\ell$-adic cohomology or crystalline cohomology. This is what "motives" refers to. I guess Manin is saying that just as you can do cohomological (and Chow) Gromov-Witten invariants and quantum cohomology, you can also do the analogous things for motives. I suppose the resulting things would be called "motivic Gromov-Witten invariants" and "quantum motives".
I'm not sure whether it makes sense to ask about the behavior of quantum cohomology under monodromy. As I understand it, monodromy refers to using a connection (Gauss-Manin connection) to parallel transport (co)homology classes. You can view quantum cohomology as being simply ordinary cohomology except with coefficients in a Novikov ring and with a deformed cup product. Viewed as such, the monodromy of quantum cohomology should be the same as the monodromy of ordinary cohomology, because "quantum cohomology classes" are no different from ordinary cohomology classes.
| 13 | https://mathoverflow.net/users/83 | 2345 | 1,510 |
https://mathoverflow.net/questions/2333 | 8 | Toeplitz operators provide a natural language with which to do geometric quantization. I don't want to really understand them, and I don't need them in full generality. I'm looking for some references that will provide formulas with which to compute products of Toeplitz operators, and specifically formulas for the asymptotics of such products as Planck's constant *h* β 0. I will give some background (also so the experts can correct any errors I might have), and then sketch the type of calculation I would like to perform.
Background
----------
A *quantization* of a commutative Poisson algebra *A* (a commutative algebra *A* is *Poisson* if it comes equipped with a bilinear bracket {,}: *A* β *A* β *A* that is a derivation in each coordinate (i.e. Leibniz rule) and a Lie bracket (i.e. antisymmetry and Jacobi)) is a smooth bundle of algebras over the real line (or some open subinterval thereof), with certain properties. Think of the bundle as a family *Ah* of (noncommutative) algebras, where *h* is my real variable. The conditions are that *A*0 = *A*; that we have given linear isomorphisms Ο*h* : *A*0 β *Ah*; and that lim*h*β0 Ο*h*-1( *h*-1 [ Ο*h*(a) , Ο*h*(b) ] ) = {a,b}, where [,] is the commutator bracket and {,} is the Poisson bracket. Quantizations are not determined by their Poisson algebras: for example, take any smooth map **R** β **R** that sends 0 to 0 and has derivative 1 at 0, and pull your bundle of algebras back along this map.
(A typical example of a quantization is the universal enveloping algebra of a finite-dimensional Lie algebra. Indeed, if *G* is a finite-dimensional Lie algebra (sorry, I don't know how to make fraktur letters here), then the symmetric algebra S\_G\_ is the algebra of polynomials on the dual space *G*\*, and inherits a Poisson bracket by {f,g}(p) = <p,[df(p),dg(p)]>, where <,> is the pairing *G*\* β *G* β **R**, and since *G*\* is a vector space, I can canonically identify T\*p*G*\* = *G*; [,] is the Lie bracket on *G*. Anyway, let *G**h* be the Lie algebra *G* with the rescaled bracket [a,b]*h* = *h*[a,b]. Then the universal enveloping algebra U\_G\_*h* is a quantization of the Poisson algebra S\_G\_.)
((Another parenthetical: most people actually do quantizations of **C**, and then ask that complex conjugation deform well. They also sometimes decorate their formulas with \_i\_s. This has to do with the ability to find good representations noncommutative algebras in terms of bounded operators on Hilbert spaces. I'll skip such parts of the definition.))
Quantizations were originally invented to describe quantum mechanics on **R**n, and this is the situation I'm trying to understand. Let *A* be an algebra of functions on the cotangent bundle T\***R**n = **R**2n. If we take the algebra of polynomials on **R**2n, it is generated by {p1,..., pn, q1,..., qn}, and the Poisson bracket is defined by {pi,qj} = Ξ΄ij, the Kronecker delta. Here are two quantizations:
* The QP quantization. For *h* nonzero, let *Ah* be the algebra of differential operators on the polynomial ring **R**[q1,..., qn]. Construct the maps Ο*h* by sending qiβ*A* to (multiplication by) q in *Ah*, and send piβ*A* to the partial derivative operator Ξ΄i. For a more complicated monomial, first write it with all qs to the left and all ps to the right, and then apply the above maps letter-by-letter.
* The Z Z-bar quantization. Complexify *A*, and change variables so that zj = qj + ipj and wj = qj - ipj. Write every monomial with the zs to the left and the ws to the right, and let the monomials act on the polynomial algebra **C**[z1,..., zn] analogous to in the QP case. I believe that you can reincorporate by real structure by recording the action of "complex conjugation".
So far, I've been playing with the QP quantization. This has a natural extension to the algebra of functions on T\***R**n that are polynomial in the p variables but smooth in the q variables (i.e. Cβ**R**n β **R**[p1,..., pn]).
What I've been told is that Toeplitz operators quantize (at least) the algebra of analytic functions on T\***R**n, (actually, they work much more generally to quantize symplectic manifolds, but I don't need them to), and that the quantization corresponds to the Z Z-bar quantization above.
The calculation I'd like to do
------------------------------
Starting with a Lagrangian on **R**n, it's relatively straightforward to write down the formal Feynman path integral (a power series in *h*), and I know some things about the derivatives of this formal power series with respect to the physical variables. This series is a putative solution to Schroedinger's equation. When the Lagrangian is quadratic in velocity, I can compute the Hamiltonian explicitly, and sure enough Schroedinger's equation is satisfied. When the Lagrangian is not quadratic, the Hamiltonian is still well-defined, but I don't have explicit enough formulas, and in general it is not polynomial in momentum. I do have various differential equations satisfied by the Hamiltonian, and I'm hoping that with these and the right formulas for the asymptotics of products of Toeplitz operators, I can check the Schroedinger equation (asymptotically) without knowing precisely what the Schroedinger operator is.
| https://mathoverflow.net/users/78 | Where can I learn about (the asymptotics of) Toeplitz operators? | You probably need some conditions to restrict the kind of Schroedinger operators you want to work with. Have a look at the work of Charles and Vu-Ngoc on Toeplitz operators and the semiclassical limit, in particular theorem 1 of this <http://people.math.jussieu.fr/~charles/Articles/BerToep.pdf> and section 1.3 of this <http://people.math.jussieu.fr/~charles/Articles/Half1.pdf> and for spectral asymptotics in more concrete examples (a non-degenerate potential well) see this <http://hal.archives-ouvertes.fr/docs/00/06/73/18/PDF/birkhoff.pdf>
| 2 | https://mathoverflow.net/users/469 | 2365 | 1,527 |
https://mathoverflow.net/questions/2380 | 5 | Suppose I have a regular n-gon. I want to draw some noncrossing diagonals to subdivide it into smaller polygons. In how many ways can I do this? The vertices are unlabeled, so I don't distinguish between rotations or reflections of a given subdivision.
A triangle has 1 subdivision (do nothing!); a square has 2, a pentagon has 3, and a hexagon has at least 9 -- I'm not certain that I haven't missed any.
In fact, what I would really like is not just a count, but an algorithm for generating such subdivisions. There are obvious algorithms that generate some subdivisions multiple times, but what I'd really like is an algorithm that only generates distinct subdivisions, and that generates all of them.
| https://mathoverflow.net/users/913 | Number of subdivisions of an n-gon | I think this could be [A001004](http://www.research.att.com/~njas/sequences/A001004) in Sloane's Encyclopedia. It's hard to be sure without checking the references given there; the sequence is defined as "Number of symmetric dissections of a polygon.", which may or may not be what you mean. (In particular, the OEIS claims the next term is 20 and I'm afraid to try to check that.)
| 4 | https://mathoverflow.net/users/143 | 2384 | 1,542 |
https://mathoverflow.net/questions/2372 | 30 | Suppose I put an ant in a tiny racecar on every face of a soccer ball. Each ant then drives around the edges of her face counterclockwise. The goal is to prove that two of the ants will eventually collide (provided they aren't allowed to stand still or go arbitrarily slow).
My brother told me about this result, but I can't quite seem to prove it. Instead of a soccer ball, we should be able to use any connected graph on a sphere (provided that there are no vertices of valence 1). We may as well assume there are no vertices of valence 2 either, since you can always just fuse the two edges.
I (and some people I've talked to) have come up with a number of observations and approaches:
* Notice that if two ants are ever on the same edge, then they will crash, so the problem is discrete. You can just keep track of which edge each ant is on, and let the ants move one at a time. Then the goal is to show that there is no way for the ants to move without crashing unless some ant only moves a finite number of times.
* You can assume all the faces are triangles. If there is a face with more than three edges, then you can triangulate it and make the ants on the triangles move in such a way that it looks exactly the same "from the outside". If there is a 1-gon, it's easy to show the ants will crash. If there is a 2-gon, it's easy to show that you can turn it into an edge without changing whether or not there is a crash.
* One approach is to induct on the number of faces. If there is a counterexample, I feel like you should either be able to fuse two adjacent faces or shrink one face to a point to get a smaller counterexample, but I can't get either of these approaches to work.
* If you have a counterexample on a graph, I think you get a counterexample on the dual graph. Have the dual ant be on the next edge along which a (non-dual) ant will pass through the given vertex.
* It feels like there might be a very slick solution using the hairy ball theorem.
| https://mathoverflow.net/users/1 | The ants-on-a-ball problem | This is known as Klyachko's Car Crash Theorem. It was proved in order to prove a theorem about finitely presented groups. In fact, the result allows the ants to move at arbitrary nonzero speeds so long as they make infinitely many loops around their 2-cell. The conclusion is that there's either a collision between ants in the interior of an edge, or else there is a `complete collision', which means that there's a collision at a vertex of all ants from adjacent edges.
EDIT: Oh, it's actually important that there are **two** complete collisions, which is somewhat harder to prove than one (a collision in the middle of an edge is also a complete collision.)
You can read an expository article by Colin Rourke [here](http://msp.warwick.ac.uk/~cpr/ftp/klyachko.ps).
| 26 | https://mathoverflow.net/users/1109 | 2387 | 1,544 |
https://mathoverflow.net/questions/2211 | 6 | So, there's a construction of Reshetikhin and Turaev which extracts knot invariants from ribbon monoidal categories, which are (usually) the representation category a Hopf algebra with a choice of ribbon element.
How do these knot invariants change if I pick a different ribbon element in the same Hopf algebra? In particular, will something strange happen with 3-manifold invariants?
| https://mathoverflow.net/users/66 | How do quantum knot invariants change when I pick a funny ribbon element? | Did you look at prop 5.21 in the [paper with Peter](http://arxiv.org/PS_cache/arxiv/pdf/0810/0810.0084v2.pdf)? I think that should answer your question.
There are two slightly different questions you could ask. First how does the framing-dependent invariant change. Here it is just (\pm 1)^#L where # is the number of components. Second how does the framing-corrected invariant change? Here it's (\pm 1)^#L (\pm 1)^writhe. In both cases the \pm 1 just measures whether you've changed the FS indicator of your rep V.
If you want to think about things labelled with components labelled by more than one irrep it'll get yuckier.
| 2 | https://mathoverflow.net/users/22 | 2394 | 1,549 |
https://mathoverflow.net/questions/2329 | 1 | Am I right in assuming that one cannot define an antipode for $M\_q(n)$ the bi-algebra of $nXn$ quantum matrices? If so, does anyone know a proof?
| https://mathoverflow.net/users/1095 | Antipode for quantum matrices. | I think a proof could be provided like this.
Claim: Let $H$ be a bi-algebra. Then an antipode S on H, if it exists, is unique.
Proof: If H is a bialgebra, then the set of linear maps from H to H inherits an algebra structure given by f\*g(x)=f(x\_1)g(x\_2), where \Delta(x)=x\_1\ot x\_2 is Sweedler's notation. This is an associative algebra structure, as can be easily checked. Moreover the axioms of the antipode assert that S is a left and right inverse to the identity map id:H-->H.
Now a general fact about associative algebras is that inverses are unique when the exist:
if xy=id and zx=id, y=(zx)y = z(xy)=z. //
Now suppose that Mat\_q(n) has an antipode, call it S'. Then, this extends to a map of its localization O\_q(GL\_n), so long as S'(det\_q)\neq 0[GAP - see below]. This is guaranteed, though, because S' was assumed to be bijective in order to be an antipode on Mat\_q(n). Okay, so now you have two antipodes on O\_q(GL\_n), but the previous claim implies they are equal. This implies that S' which you started with originally required inverting the quantum determinant in order to be defined in the first place, which is a contradiction, since quantum determinant is not invertible in Mat\_q(n). //
By the way, Mat\_q(n) is a bialgebra in two ways. First of all, it is a bialgebra coming from a coproduct coming viewing matrices as a monoid under composition. It is also, however, a bialgebra with respect to a different coproduct, coming from regarding matrices as a group under addition. The latter notion also quantizes, and so Mat\_q(n) becomes what is called a "braided bialgebra". This means it's not a bialgebra in the usual sense, but it is a bialgebra in the braided tensor category of U\_q(gl\_n)-modules, which just means that when you write down the axiom for compatibility of product and coproduct, you realize that there are some tensor flips that need to be replaced with braiding. If you regard it in this sense, then you would expect it to have a (braided) antipode, and indeed it does.
All of this can be read, for instance in Majid's "Foundations of Quantum Groups", or also Klymik and SchΓΌdgen's "Quantum Groups and Their Representation Theory".
| 1 | https://mathoverflow.net/users/1040 | 2400 | 1,553 |
https://mathoverflow.net/questions/2019 | 14 | Let M be a smooth manifold. The double tangent bundle, TTM,can be viewed as a fibre bundle over TM in two ways, with the projection maps given by T\_ΟM (i.e. the derivative of the projection from TM to M) and Ο\_TM (i.e. the standard projection onto TM). There is also a canonical involution K:TTM->TTM, which basically flips the inner two coordinates. It turns out to be a diffeomorphism and a natural transformation from T^2 to itself. In fact, TΟ\_M and Ο\_TM are related through composition with K.
My question is, what happens if you keep taking higher and higher tangent bundles? Evidently, you should have more ways to write them as fibre bundles over the lesser tangent bundle. Intuitively, to me at least, it feels that there are k ways to view (T^k)M as a fibre bundle over (T^(k-1))M: inductively, there is the derivative of all the previous projection maps, and the canonical way. Are there any others?
Is there always a diffeomorphism (or whatever the suitable notion is here) that will take one projection map to the other as in the case of the canonical involution in k=2? If not, what goes wrong, and does it have any significance?
| https://mathoverflow.net/users/855 | In how many ways can an iterated tangent bundle (T^k)M be viewed as a fibre bundle over (T^(k-1))M? | If we use the notation $(TM, p\_M, M)$ for the tangent bundle of any manifold $M$, then you are right to think that $T^{\ k}M$ has $k$ natural vector bundle structures over $T^{\ k-1}M$ and so on down to $M$, making a diagram which is a $k$-dimensional cube. Such a structure is a $k$-fold vector bundle (See articles by Kirill Mackenzie) and $T^{\ k} M$ is a particularly symmetrical one. An easy way of writing the $k$ bundle maps would be
$$T^{\ a}(p\_{T^{\ b} M}) $$
for $a+b+1 = k$, where this means that we are taking the $a$-th derivative of the tangent bundle projection $T\ (T^{\ b}M)\to T^{\ b}M$, yielding a map now from $T^{\ k}$ to $T^{\ k-1}$.
To see the symmetries of $T^{\ k} M$ it is more convenient to describe the functor in a direct way rather than as a $k$-fold composition -- just as you might think of tangent vectors as infinitesimal curves, you can think of points in $T^{\ k} M$ as infinitesimal maps of a unit $k$-cube into $M$. The restrictions to the $k$ faces of the cube (going through the origin) gives your $k$ maps.
From that point of view, it is clear that you can permute the $k$ coordinate axes and get another map of a cube into $M$, so that the functor $T^{\ k} M$ has a $S\_k$ group of natural-automorphisms.
Incidentally, to make the above into a definition of $T^{\ k} $, you could do the following:
Consider the fat point $fp$, which you should think of as a space whose smooth functions form the ring $\mathbb{R}[x]/(x^2)$. Then the tangent bundle $TM$ can be thought of as the space of maps from the fat point to $M$, i.e. $TM=C^\infty(fp,M)$. Such maps, by the way, are just algebra homomorphisms from the algebra $C^\infty(M,R)$ to $\mathbb{R}[x]/(x^2)$. You can check that such a map has two components $f\_0 + f\_1 x$, and that $f\_0$ is a homomorphism to $\mathbb{R}$ defining a maximal ideal (i.e. a point $p$ in $M$) and $f\_1$ defines a derivation (i.e. a vector at $p$).
In precisely the same way you can consider a cubical fat $k$-point $kfp$ with functions
$$\mathbb{R}[x\_1,...,x\_k]/(x\_1^2,...,x\_k^2)$$
and then define $T^{\ k} M = C^\infty(kfp,M)$. Then you can see the symmetries as automorphisms of the above algebra, and the $k$ maps you ask about as homomorphisms $C^\infty(kfp)\to C^\infty\big((k-1)fp\big).$
There are probably more subtle things to be said about these higher iterated bundles but I hope the above is at least correct.
| 15 | https://mathoverflow.net/users/1116 | 2413 | 1,563 |
https://mathoverflow.net/questions/2414 | 29 | Now, among algebraic geometers, at least, it is well known that there is an equivalence between locally free $\mathcal{O}\_X$-modules of rank $n$ and vector bundles of rank $n$. So, equivalently, principal $\mathrm{GL}(n,\mathbb{C})$-bundles are given by locally free sheaves of rank $n$.
So...what about other groups? I guess that $\mathrm{SL}(n,\mathbb{C})$ bundles are then locally free sheaves of rank $n$ with top exterior power trivial, but can we phrase everything in terms of the properties of a sheaf and a group?
My guess is that in this context, if we can do it, we'll end up with something that's not quite locally free sheaves of rank n for $\mathrm{GL}(n,\mathbb{C})$, but which will be equivalent.
***Note***: I'm aware that we could just say something like "the sheaf of local sections of a $G$-bundle" but I'm looking for something intrinsic, a set of properties of the sheaf without reference to the geometric bundle, which can be reconstructed from the sheaf description.
| https://mathoverflow.net/users/622 | Sheaf description of $G$-bundles | $\newcommand{\O}{\mathcal{O}}$
$\newcommand{\F}{\mathcal{F}}$
The way you get a locally free sheaf of rank $n$ from a $GL(n)$-torsor $P$ is by twisting the trivial rank $n$ bundle $\O^n$ (which has a natural $GL(n)$-action) by the torsor. Explicitly, the locally free sheaf is $\F=\O^n\times^{GL(n)}P$, whose (scheme-theoretic) points are $(v,p)$, where $v$ is a point of the trivial bundle and $p$ is a point of $P$, subject to the relation $(v\cdot g,p)\sim (v,g\cdot p)$. Conversely, given a locally free sheaf $\F$ of rank $n$, the sheaf $Isom(\O^n,\F)$ is a $GL(n)$-torsor, and this procedure is inverse to the $P\mapsto \O^n\times^{GL(n)}P$ procedure above. (Note: I'm identifying spaces over the base $X$ with their sheaves of sections, both for regarding $Isom(\O^n,\F)$ as a torsor and for regarding $\O^n\times^{GL\_n}P$ as a locally free sheaf.)
Similarly, if you have a group $G$ **and a representation** $V$, then you can associate to any $G$-torsor $P$ a locally free sheaf of rank $\dim(V)$, namely $V\times^G P$. But I don't know of a characterization of which locally free sheaves of rank $\dim(V)$ arise in this way.
Operations with the locally free sheaf (like taking top exterior power, or any other operation which is basically defined fiberwise and shown to glue) correspond to doing that operation with the representation $V$, so I think you're right that in the case of $SL(n)$ you get exactly those locally free sheaves whose top exterior power is trivial (since $SL(n)$ has no non-trivial $1$-dimensional representations).
| 17 | https://mathoverflow.net/users/1 | 2425 | 1,572 |
https://mathoverflow.net/questions/1043 | 4 | Hi all,
The short-time fourier transform decomposes a signal window into a sin/cosine series.
How would one approximate a signal in the same way, but using a set of arbitrary basis functions instead of sin/cos? These arbitrary basis functions are likely in my case to be very small discrete chunks of a 1-dimensional non-periodic waveform.
Is this something wavelets are used for?
Please excuse my tag, 'signal-analysis' does not exist and I can not create it.
| https://mathoverflow.net/users/663 | Decomposing a 1-d signal into arbitary basis functions | Wavelets are generally used for nonperiodic signals. They're often used in earthquake detection and things like that. There are many books on the subject, a quick look for "Wavelets" in amazon.com should reveal many.
The Haar Wavelet and Daubchies Wavelet might be good choices. Haar may be better if you don't need a smooth decomposition, the wavelet decomposition their is much easier.
Google scholar (scholar.google.com) may be a good place to look, just look up "Wavelet decomposition" and your particular topic.
You can make your own wavelets per your particular needs, but it's not particularly easy, they need to fulfill certain conditions
This may be useful
<https://doi.org/10.1007/BF01257191>
Generalized multi-resolution analyses and a construction procedure for all wavelet sets in R^n
If you can get this down to 1-d you may have your answer.
| 4 | https://mathoverflow.net/users/429 | 2426 | 1,573 |
https://mathoverflow.net/questions/2402 | 16 | Abel's identity for the dilogarithm (see the wikipedia page about polylogarithms)
plays a role in web geometry as it is one of the abelian relations of the
first example of exceptional web (Bol's 5-web) to appear in the literature.
I have heard it is important in other domains (cohomology of SL(3,C), algebraic K-theory, motives ).
I would like to learn more about it.
I am asking for:
1. Insights on why Abel's identity is relevant in this or that field;
2. References where it plays a role.
---
**Edit.**
I have just learned from this [blog](http://lamington.wordpress.com/) about [Bridgeman's orthospectrum identity](http://lamington.wordpress.com/2009/10/24/bridgemans-orthospectrum-identity/).
Those interest in the question above might want to take a look at it.
| https://mathoverflow.net/users/605 | Abel's equation for the dilog | One basic answer is given by hyperbolic geometry.
Ideal tetrahedra in hyperbolic 3-space $\mathbb{H}^3$ are equivalent (under the action of the automorphism group $PGL\_2(C))$ to tetrahedra with vertices $\{0,1,\infty,z\}$, and their volume is given by $D(z)$, where $D(z)$ is the Bloch-Wigner dilogarithm, which is a slightly modified version of the dilogarithm. This amounts to writing down the hyperbolic metric and evaluating an integral, which turns out to be (very close to) $Li\_2(z)$ (although it is real valued for complex $z$).
The tetrahedron $\{0,1,\infty,z\}$ is equivalent under $PSL\_2(\mathbb C)$ to $\{0,1,\infty,1/(1-z)\}$ and $\{0,1,\infty,1-1/z\}$, and so we get formulae:
$$D(z) = D(1/(1-z)) = D(1 - 1/z).$$
The tetrahedron $\{0,1,\infty,z\}$ is also equivalent to $\{0,1,\infty,1/z\}$, except with an odd permutation of the vertices, and thus: $D(z) = - D(1/z).$
Finally, choose a random point $y$ in the boundary $\mathbb P^1(\mathbb C)$ of $\mathbb H^3$. If we take the tetrahedron $\{0,1,\infty,y\}$, we can break it off into $\{0,1,\infty,x\}$ and three other tetrahedra (just like in Euclidean space). Transforming the coordinates of the other three tetrahedra into the standard form gives the 5-term relation:
$$D(x) - D(y) + D\left(\dfrac yx\right) - D\left(\dfrac {1-x^{-1}}{1-y^{-1}}\right) + D\left(\dfrac {1-x}{1-y}\right) = 0,$$
which gives a proof of Abel's equation.
Let's think some more about a closed hyperbolic 3-manifold $M$. By definition, $M = \mathbb H^3/\Gamma$ for a lattice $\Gamma$ in $PSL\_2(\mathbb{C})$. Since $\mathbb{H}^3$ is contractible, $M$ is a $K(\pi,1)$ space, and so there is a canonical isomorphism $H\_\*(M, \mathbb{Z}) = H\_\*(\Gamma, \mathbb{Z})$, comparing simplicial homology with the group homology of $\Gamma$. Now $M$ has a fundamental class $[M]$ in $H\_3(M, \mathbb{Z})$, which gives an element in $H\_3(\Gamma, \mathbb{Z})$ and hence also a class in $H\_3(PSL\_2(\mathbb{C}), \mathbb{Z})$.
On the other hand, $[M]$ can be decomposed ("triangulated") into ideal tetrehedra with parameters $z\_i$. The set of parameters $[z\_i]$ is not unique, however, the only real "move" is the subdivision of tetrahedra, and so associated to $M$ we get an element of the group generated by $[z\_i]$ for $z\_i$ in $P^1(\mathbb{C})$ and with relations exactly of the form satisfied by $D$ above. This is essentially the definition of the Bloch group. $D$ is a function this group, and this decomposition gives a map from $H\_3(PSL\_2(\mathbb{C}), \mathbb{Z})$ to the Bloch group.
Note that it is not obvious that the $z\_i$ can be taken inside some field $\mathbb{F}$, this is a consequence of Mostow Rigidity. It turns out that if we take the Bloch group $B(\mathbb{F})$ generated by elements of $\mathbb{F}$, this is, by work of Suslin, essentially equal to $K\_3(\mathbb{F})$.
To summarize, the connection between the identity, the cohomology of $PSL\_2(\mathbb{C})$, and the Bloch group is well understood, see some papers by Walter Neumann. For the connection between the Bloch group $B(\mathbb{F})$ and $K\_3(\mathbb{F})$, see papers of Suslin. The connection with motives is more speculative, but here you should look at some papers of Goncharov.
(There are some generalizations/connections to higher regulators for K-groups, but this is a very nice example to understand, being both somewhat accessible yet still very interesting.)
| 13 | https://mathoverflow.net/users/nan | 2427 | 1,574 |
https://mathoverflow.net/questions/2433 | 12 | If we think of the l1 distance as a grid-distance between points, then we can think of l2 distance as what we get when we "shortcut" the grid by going "inside" a cell.
Making the grid finer doesn't change the l1 distance, so there's no obvious sense in which the l2 distance can be seen as a limiting version of the l1.
So here's my question: is there any way to take an l1-like distance and extract an l2 like distance from it (possibly as a limiting case). I'm asking because i have a distance defined in a discrete space that has an l1-like behaviour, and I'd like to generalize it as the discrete space gets finer and finer, but I want to end up with a distance that goes "directly" through the space like l2.
apologies if this is way too vague.
| https://mathoverflow.net/users/972 | Making an l_2 distance out of l_1 distance | L\_2 can be viewed as the limit of a sequence of metrics whose metric balls are (in the plane) regular polygons with 2n sides, where n goes to infinity. In higher dimensions one doesn't have enough regular polytopes to make a limit out of them, but it works just as well to use irregular ones. This technique is sometimes useful in computational geometry, because the polygonal metrics allow for faster range searching data structures than the Euclidean one.
If you want a distance in some sort of discrete graph (the way that L\_1 can be viewed as distance in a regular grid) but that approaches planar L\_2 in the limit, try using the pinwheel tiling. If you measure the shortest path along tile edges between any two points on the edges of any two tiles, and then repeatedly subdivide to produce a finer pinwheel tiling, the shortest path distance will approach the Euclidean distance in the limit.
As for Kore min's suggestion of looking into metric embedding: any metric can be embedded into L\_infinity with no distortion at all. And in the plane, L\_1 is the same as L\_infinity (though they are quite different in higher dimensions) so it might have something to do with what you're talking about. See e.g. [this Wikipedia article](http://en.wikipedia.org/wiki/Tight_span).
| 5 | https://mathoverflow.net/users/440 | 2436 | 1,581 |
https://mathoverflow.net/questions/2283 | 7 | Consider the tome of Bruhat and Tits: Groupes rΓ©ductifs sur un corps local : I. DonnΓ©es radicielles valuΓ©es. Publications MathΓ©matiques de l'IHΓS, 41 (1972), p. 5-251. (available on [NUMDAM](http://www.numdam.org/numdam-bin/fitem?id=PMIHES_1972__41__5_0)). I am interested primarily in the statements of Propositions 4.4.3 and 4.4.4 (and maybe also 7.3.1).
In the swathe of notation and technical conditions present, I find it hard to read exactly what the precise statements of these two propositions are. My question is, can anyone give
(a) a precise version of the statements of 4.4.3 and 4.4.4?
(b) if the request in (a) is too much, some sort of simplified version that is easy to state/comprehend and hopefully still reasonably general.
or
(c) an alternative reference covering at least the case of a split reductive group?
(this question is related to, but more general than [Dinakar's question](https://mathoverflow.net/questions/2229/iwasawa-decomposition))
| https://mathoverflow.net/users/425 | Iwasawa and Cartan Decompositions. | I think the key point is the proposition 4.4.2, where "good" subgroups are caracterised geometrically as stabilisers of special subgroups (ie, stabilisers of a point o such that the Weyl group W is the semidirect product of its translations and of the stabiliser of o in W).
Then the group G is the product of B (the stabiliser of a class of sector, a minimal parabolic group for an algebraic group) and the group K. Moreover, the group B itself is the product of B^0 (which is the union of pointwise fixators of sectors) and the group of translations (acting on some apartment containing o and a sector in this class).
The Cartan decomposition is, as usual, the decomposition of an element g in kvk', where k and k' are element of K and v is an element which sends o to a vertice of the sector starting at o in the class defined by B.
The proposition 4.4.4 is meant to explain the relation between the two decompositions (ie, when you know the translation part in the Iwasawa decomposition, can you deduce it in the Cartan decomposition ?)
If you know how to attach a building to a reductive group, then the book "Buildings" by Abramenko and Brown is a good reference (see chap. 11), much easier to read. They treat every building, but construct only the affine building associated to SL(n). Another reference is the small book of Macdonald, "Spherical functions on a group of p-adic type" (chapter II, Theorem 2.6.11)
| 6 | https://mathoverflow.net/users/915 | 2442 | 1,585 |
https://mathoverflow.net/questions/2459 | 1 | Varieties decompose uniquely into finitely many irreducibles, and each variety is generated by only finitely polynomials. These two finiteness properties make varieties seemingly "manageable" objects, and leads me to the question:
Can a computer, given a variety (finite set of polynomials) produce a list of its irreducible components (finite set of finite sets of polynomials) in finite time?
| https://mathoverflow.net/users/416 | Does automatic decomposition of varieties into irreducibles exist? | This is just about primary decomposition! There are several CAS which can do that, for example Singular.
| 2 | https://mathoverflow.net/users/717 | 2463 | 1,601 |
https://mathoverflow.net/questions/2461 | 10 | This is perhaps too broad or vague (or silly) a question, but here it is anyway: why should I care about constructing line bundles *on* a moduli space? This comes up all of the time, but I seem to be missing the (probably obvious) motivation.
Ideally, it would be nice to attach to this question a particular moduli space (vector bundles on a curve, instantons, etc.), but I think I will leave the task of finding an efficient yet instructive example to someone with more knowledge.
Thanks.
| https://mathoverflow.net/users/1124 | Line bundles on moduli spaces | The reasons for caring which occur to me seem to fall into two broad categories.
First, line bundles are useful for identifying interesting substacks of your moduli stack. You usually first encounter this idea in GIT theory, but the philosophy applies more generally. See, for example, Teleman's papers on the stack of G-bundles on a curve, and Jarod Alper's paper on "good" moduli spaces ( <http://math.columbia.edu/~jarod/good_moduli_spaces.pdf> ).
Second, sections of line bundles are a good substitute for/generalization of functions. For example, on the moduli stack of curves, the only holomorphic functions are constant, but there are plenty of meromorphic functions, which are naturally thought of as sections of line bundles. This point of view shows up, for example, in the theory of modular forms (where it clarifies their transformations under SL(2,z)) and in the Beilinson-Bernstein approach to representation theory (where you use "twisted D-modules", which act on sections of a line bundle on the flag variety rather than on functions, to get representations with non-trivial central character).
Is that roughly what you were wanting?
| 11 | https://mathoverflow.net/users/35508 | 2466 | 1,603 |
https://mathoverflow.net/questions/2464 | 6 | I would like to know what Drinfeld's [scanned manuscript](http://math.uchicago.edu/~drinfeld/bestdream.pdf "low res, high res on his website") "Best Dream" is about: the title makes me curious.
It's in Russian.
| https://mathoverflow.net/users/451 | What is Drinfeld's manuscript "Best Dream" (in Russian!) about? | From the first line it appears to about **D-modules on stacks**.
The next topic is an attempt to construct a "duality" (called F) for derived category of (quasicoherent) D-modules on `Bun_G` (the space/stack of G-bundles). The problem is to find a "reverse Langlands transform" (?) by defining a suitable scalar product on those.
In the middle a single question appears, "What is the Eisenstein fuctor?" (indeed, what is it?) Next, there's a question about Hecke functors.
Part two starts with the self-explanatory
```
D (Bun_G) =?= O(LocSys)
```
The Eisenstein functor for `GL(2)` appears in the discussion, with the standard definition. Some constructions relevant to the formula above appear. Pages 59β60 then are in English. Then the derived categories and Eisentein functor continue.
Page 76 contains some specific questions ("I don't understand!") again along the topics above which continue until the end of paper.
| 7 | https://mathoverflow.net/users/65 | 2498 | 1,627 |
https://mathoverflow.net/questions/2484 | 2 | I just saw this paper recently which mentioned that the optimization on a Grassmanian Manifold can be used to get an achieve an best approximation of a multilinear rank of a tensor (in the sense of a multidimensional matrix, also called a hypermatrix). Does anyone happen to know what Grassmanians have to do with tensors and their analysis? And where could I learn more about Grassmanian manifolds (enough to understand the use here, more is of course welcome =)
The paper is here for those interested
[Paper](http://www.stanford.edu/~lekheng/jobs/qn.pdf)
(I hope this isn't too specific)
| https://mathoverflow.net/users/429 | Algebraic Geometry in an applied setting? | You might also want to check out [this](http://www.math.tamu.edu/~jml/rankvsbrank.pdf) paper by Landsberg and Teitler on getting bounds for the Waring rank (and border rank) of symmetric tensors using geometry.
The point here is that the projective bundle on a vector space captures the geometry of its symmetric tensors so one can reformulate questions about say the rank into questions about the geometry of secant varieties on the projective bundle of interest. It turns out one can prove some new bounds using some very classical algebraic geometry.
| 4 | https://mathoverflow.net/users/310 | 2508 | 1,633 |
https://mathoverflow.net/questions/183 | 18 | The short answer to my question may be a pointer to the right text. I will give all the background I know, and then ask my questions in list form.
Let *A* be an operator (on an infinite-dimensional vector space). You might as well assume that its spectrum is all real and positive. In fact, I only care when the spectrum is discrete and grows polynomially, but I hear that this stuff works more generally.
In general, *A* is not trace-class (the sum of the eigenvalues converges) or determinant-class (the product of the eigenvalues converges) β if the *n*th eigenvalue grows as *n**p* for some *p*>0, then it won't be. But there is a procedure to try to define a "trace" and "determinant" of *A* nevertheless.
Let us hope that for large enough *s*, the operators *A-s* (=exp(-*s* log *A*), and log *A* makes sense if the spectrum of *A* is positive) are trace-class. If so, then we can define ΞΆ*A*(*s*) = tr(*A-s*); it is analytic for Re(*s*) large enough. Let's hope that it has a single-valued meromorphic continuation and that this function (which I will also call ΞΆ*A*(*s*)) is smooth near *s*=0 and *s*=-1. All these hopes hold when the eigenvalues of *A* grow polynomially, whence ΞΆ*A*(*s*) can be compared to the Riemann zeta function.
Then we can immediately define the "regularized trace" TR *A* = ΞΆ*A*(0) and the "regularized determinant" DET *A* = exp(-ΞΆ*A*'(0)), where by ΞΆ*A*'(*s*) I mean the derivative of ΞΆ*A*(*s*) with respect to *s*. (If the eigenvalues Ξ»*n* are discrete, then ΞΆ*A*(*s*) = Ξ£ Ξ»*n*-*s*, and so one would have TR *A* = Ξ£ Ξ»*n* and DET *A* = Ξ£ (log Ξ»*n*) Ξ»*n*-*s* |*s*=0, if they converged.) If *A* is trace- (determinant-) class, then TR *A* = tr *A* (DET *A* = det *A*).
So, here are my questions:
1. Is it true that exp TR *A* = DET exp *A*?
2. Let *A*(*t*) be a smooth family of operators (*t* is a real variable). Is it true that d/d*t* [ log DET *A*(*t*) ] = TR( A-1 d*A*/d*t* )? (I can prove this when A-1d*A*/d*t* is trace-class.)
3. Is DET multiplicative, so that DET(*AB*) = DET *A* DET *B*? (I can prove this using 1. and 2., or using the part of 2. that I can prove if *B* is determinant-class.)
4. Is TR cyclic, i.e. TR(*AB*) = TR(*BA*)?
5. Is TR linear, i.e. TR(*A* + *B*) = TR *A* + TR *B*?
None of these are even obvious to me when *A* and *B* (or d*A*/d*t*) are simultaneously diagonalizable (except of course cyclicity), but of course in general they won't commute.
| https://mathoverflow.net/users/78 | Zeta-function regularization of determinants and traces | I will answer some of my questions in the negative.
**3.**
First consider the case of rescaling an operator *A* by some (positive) number λ. Then ΢λ*A*(s) = λ-s΢*A*(s), and so TR λ*A* = λ TR *A*. This is all well and good. How does the determinant behave? Define the "perceived dimension" DIM *A* to be logλ[ (DET λ*A*)/(DET *A*) ]. Then it's easy to see that DIM *A* = ΢*A*(0). What this means is that DET λ*A* = λ΢*A*(0) DET *A*.
This is all well and good if the perceived dimension of a vector space does not depend on *A*. Unfortunately, it does. For example, the Hurwitz zeta functions ΞΆ(s,ΞΌ) = Ξ£0β(n+ΞΌ)-s (-ΞΌ not in **N**) naturally arise as the zeta functions of differential operators β e.g. as the operator *x*(d/d*x*) + ΞΌ on the space of (nice) functions on **R**. One can look up the values of this function, e.g. in Elizalde, et al. In particular, ΞΆ(0,ΞΌ) = 1/2 - ΞΌ. Thus, let *A* and *B* be two such operators, with ΞΆ*A* = ΞΆ(s,Ξ±) and ΞΆ*B* = ΞΆ(s,Ξ²). For generic Ξ± and Ξ², and provided *A* and *B* commute (e.g. for the suggested differential operators), then DET *AB* exists. But if DET were multiplicative, then:
DET(Ξ»*AB*) = DET(Ξ»A) DET(*B*) = Ξ»1/2 - Ξ± DET A DET B
but a similar calculation would yield Ξ»1/2 - Ξ² DET A DET B.
This proves that DET is not multiplicative.
**1.**
My negative answer to 1. is not quite as satisfying, but it's OK. Consider an operator *A* (e.g. *x*(d/d*x*)+1) with eigenvalues 1,2,..., and so zeta function the Reimann function ΞΆ(s). Then TR *A* = ΞΆ(-1) = -1/12. On the other hand, exp *A* has eigenvalues e, e2, etc., and so zeta function ΞΆexp *A*(s) = Ξ£ e-ns = e-s/(1 - e-s) = 1/(es-1). This has a pole at s=0, and so DET exp *A* = limsβ0 es/(es-1)2 = β. So question 1. is hopeless in the sense that *A* might be zeta-function regularizable but exp *A* not. I don't have a counterexample when all the zeta functions give finite values.
**5.**
As in my answer to 3. above, I will continue to consider the Hurwitz function ΞΆ(*s,a*) = Ξ£*n*=0β (*n*+\_a\_)-*s*, which is the zeta function corresponding, for example, to the operator *x*(d/d*x*)+*a*, and we consider the case when *a* is not a nonpositive integer. One can look up various special values of (the analytic continuation) of the Hurwitz function, e.g. ΞΆ(-*m,a*) = -*B**m*+1(*a*)/(*m*+1), where *Br* is the \_r\_th Bernoulli polynomial.
In particular,
TR(*x*(d/d*x*)+*a*) = -ΞΆ(-1,*a*)/2 = -*a*2/2 + *a*/2 - 1/12
since, for example (from [Wikipedia](http://en.wikipedia.org/wiki/Bernoulli_polynomials)):
*B*2(*a*) = Ξ£*n*=02 1/(n+1) Ξ£*k*=0 *n* (-1)*k* {n \choose k} (*a*+\_k\_)2 = *a*2 - *a* + 1/6
Thus, consider the operator 2\_x\_(d/d*x*)+*a*+\_b\_. On the one hand:
TR(*x*(d/d*x*)+*a*) + TR(*x*(d/d*x*)+*b*) = -(*a*2+*b*2)/2 + (*a*+\_b\_)/2 - 1/6
On the other hand, TR is "linear" when it comes to multiplication by positive reals, and so:
TR(2\_x\_(d/d*x*)+*a*+\_b\_) = 2 TR(*x*(d/d*x*) + (*a*+\_b\_)/2) = -(*a*2+2\_ab\_+*b*2)/4 + (*a*+\_b\_)/2 - 1/6
In particular, we have TR(*x*(d/d*x*)+*a*) + TR(*x*(d/d*x*)+*b*) = TR( *x*(d/d*x*)+*a* + *x*(d/d*x*)+*b* ) if and only if *a*=\_b\_; otherwise 2\_ab\_ < *a*2+*b*2 is a strict inequality.
So the zeta-function regularized trace TR is not linear.
**0./2.**
My last comment is not so much to break number 2. above, but to suggest that it is limited in scope. In particular, for an operator *A* on an infinite-dimensional vector space, it is impossible for *A-s* to be trace-class for *s* in an open neighborhood of 0, and so if the zeta-function regularized DET makes sense, then det doesn't. (I.e. it's hopeless to say that det *A* = DET *A*.) Indeed, if the series converges for *s*=0, then it must be a finite sum.
Similarly, it is impossible for *A* to be trace class and also for *A-s* to be trace class for large *s*. If *A* is trace class, then its eigenvalues have finite sum, and in particular cluster near 0 (by the "divergence test" from freshman calculus). But then the eigenvalues of *A*-*s* tend to β for positive *s*. I.e. it's hopeless to say that tr *A* = TR *A*.
My proof for 2. says the following. Suppose that d*A*/d*t* *A*-1 is trace class, and suppose that DET *A* makes sense as above. Then
d/d*t* [ DET *A* ] = (DET *A*)(tr d*A*/d*t* *A*-1)
I have no idea what happens, or even how to attack the problem, when d*A*/d*t* *A*-1 has a zeta-function-regularized trace.
| 7 | https://mathoverflow.net/users/78 | 2519 | 1,640 |
https://mathoverflow.net/questions/2518 | 5 | What's the background I need to know to understand the conjectural
```
D (Bun_G) =?= O(LocSys)
```
from [this question](https://mathoverflow.net/questions/2464/what-is-drinfelds-manuscript-best-dream-in-russian-about). I know the LHS is about the derived category of D-modules on the space (stack?) of some (stable?) bundles for some preselected group G. What's the RHS? I've seen the physics articles that say something about his topic, but how would you explain it?
I know this goes under the name of Geometric Langlands. Why?
| https://mathoverflow.net/users/65 | Sheaves on Bun_G | Don't need stable. It's the stack of all G bundles on a curve. The right hand side is the derived category of local systems, which are vector bundles with flat connection (G-bundles, for the Langlands dual of G in this equation). As for background, that depends on how deep an understanding you want.
For a lot of the positive characteristic stuff, you want to look at stuff by [Frenkel](https://arxiv.org/search/advanced?advanced=&terms-0-operator=AND&terms-0-term=Frenkel&terms-0-field=author&terms-1-operator=AND&terms-1-term=Langlands&terms-1-field=all&classification-mathematics=y&classification-physics_archives=all&classification-include_cross_list=include&date-filter_by=all_dates&date-year=&date-from_date=&date-to_date=&date-date_type=submitted_date&abstracts=show&size=50&order=-announced_date_first) and [Gaitsgory](https://arxiv.org/search/advanced?advanced=&terms-0-operator=AND&terms-0-term=gaitsgory&terms-0-field=author&terms-1-operator=AND&terms-1-term=Langlands&terms-1-field=all&classification-mathematics=y&classification-physics_archives=all&classification-include_cross_list=include&date-filter_by=all_dates&date-year=&date-from_date=&date-to_date=&date-date_type=submitted_date&abstracts=show&size=50&order=-announced_date_first), is my understanding.
In the complex case, there's a bit of other stuff. I've got to, of course, recommend the work of my [advisor](https://arxiv.org/search/advanced?advanced=&terms-0-operator=AND&terms-0-term=Donagi&terms-0-field=author&terms-1-operator=AND&terms-1-term=Langlands&terms-1-field=all&classification-mathematics=y&classification-physics_archives=all&classification-include_cross_list=include&date-filter_by=all_dates&date-year=&date-from_date=&date-to_date=&date-date_type=submitted_date&abstracts=show&size=50&order=-announced_date_first), as well as the work of [Witten](https://arxiv.org/search/advanced?advanced=&terms-0-operator=AND&terms-0-term=witten&terms-0-field=author&terms-1-operator=AND&terms-1-term=Langlands&terms-1-field=all&classification-mathematics=y&classification-physics_archives=all&classification-include_cross_list=include&date-filter_by=all_dates&date-year=&date-from_date=&date-to_date=&date-date_type=submitted_date&abstracts=show&size=50&order=-announced_date_first) which is closely related. To really get this, you need to have a bit of an understanding of the Hitchin system on Higgs bundles (though not by that name, they're used extensively in the paper *[Spectral curves and the generalised theta divisor](http://math.unice.fr/%7Ebeauvill/pubs/bnr.pdf)*, Crelle 1989).
As for why it's called Geometric Langlands, that'd be because it's, essentially, a geometric formulation related to the classical Langlands conjecture in number theory, which I know fairly little about, except that it involves automorphic forms. I recommend asking your local number theorist for some help, though I do believe that it is explained in the book *[Introduction to the Langlands Program](https://doi.org/10.1007/978-0-8176-8226-2)* by quite a few authors, which includes Gaitsgory explaining roughly what Geometric Langlands is and how it all fits together.
Hope that helps.
| 3 | https://mathoverflow.net/users/622 | 2527 | 1,644 |
https://mathoverflow.net/questions/2522 | 4 | What os the meaning of `a reverse Langlands transform` to which Drinfeld [seems to refer](https://mathoverflow.net/questions/2464/what-is-drinfelds-manuscript-best-dream-in-russian-about)?
| https://mathoverflow.net/users/65 | Reverse Langlands transform | My guess is that, traditionally, the Geometric Langlands program seems to be looking for a functor from D\_{coh}(Loc,O)\to D\_{coh}(^L Bun,D), that is, from the derived category on the space of local systems to the derived category of D-modules on the space of bundles for the Langlands dual. So, by reverse Langlands transformation, he might be asking for a functor in the other direction, to try to establish the equivalence.
| 1 | https://mathoverflow.net/users/622 | 2528 | 1,645 |
https://mathoverflow.net/questions/2517 | 2 | Consider a system of the form: dx/dt = f(x,y) , dy/dt=g(x,y), with the property that the associated ODE dy/dx = g(x,y)/f(x,y) has a unique solution to IVP y(0)=0.
Also, f(x,y) is smooth every *except* the point (0,0), at which it has an infinite discontinuity, and g(x,y) is continuous everywhere. Does it follow that there is a solution to the system which tends to (0,0)?
This problem may be not well-formulated, but it seems like there may be a topological argument for the existence of such a solution to the system.
| https://mathoverflow.net/users/1142 | ODE system question | I may be missing some subtle point here, but it seems to me that if you let Y(x) be your presumed solution to Y'(x)=g(x,Y)/f(x,Y) and then let x(t) solve dx/dt=f(x,Y(x)) and put y(t)=Y(x(t)), you have your answer. The only way this could fail to tend to (0,0) for some value of t is if f(x,Y(x))=0 for arbitrarily small x, in which case I would question the validity of the assumed solution Y to begin with.
(Maybe I should have made this a comment rather than an answer, since the problem as I have understood it does not seem all that interesting. If you agree, feel more than free to not award any points to it, though I'd appreciate the absence of negative points.)
| 2 | https://mathoverflow.net/users/802 | 2531 | 1,646 |
https://mathoverflow.net/questions/2525 | 15 | I know of three ways to define the dimension of a finitely-generated commutative algebra A over a field F:
* The Gelfand-Kirillov (GK) dimension, based on the growth of the Hilbert function.
* The Krull dimension, based on chains of prime ideals.
* The transcendence degree of the fraction field of A over F.
According to Artin, GK dimension is the most robust notion because it applies to certain noncommutative algebras. And in the noncommutative setting one can't form fraction fields, so the transcendence degree is out of the question (right?).
But Krull dimension has the advantage that it applies to arbitrary rings. As far as I can tell, the definition of GK dimension only applies to algebras. So: as a matter of pedagogy, which notion of dimension is most appropriate for what applications? Which is easiest to prove things about when?
| https://mathoverflow.net/users/290 | Different definitions of the dimension of an algebra | In a non-commutative ring, you need to be careful with what you even mean by a prime ideal, and usually there are very few two-sided ideals you might call prime. Oh, and even in the cases when there is a nice ring of fractions, it won't be a field, and so transcedence degree is still bad.
My personal favorite notion of dimension is 'global dimension', the maximum projective dimension of any module of the ring. This concept exists for any ring, and in fact for any abelian category (though, if there aren't enough projectives, you need to play with the definition). The only problem is that it can often be infinity, even for relatively mild rings, like C[x]/x^2. It still makes for a pretty good theory of 'smooth dimension', however.
From a conceptual perspective, Krull dimension seems best suited for geometric perspectives, since it is measuring chains of irreducible closed subsets. The easiest times to work with Krull dimension is when you are in a Cohen-Macaulay ring, and then Krull dimension is equivalent to depth, which is easier to prove things about, since you only need to produce a maximal regular sequence.
| 10 | https://mathoverflow.net/users/750 | 2532 | 1,647 |
https://mathoverflow.net/questions/2503 | 15 | Given a curve C, and a reductive group G, there is a moduli stack Loc\_G(C), the stack of G-local systems. I keep reading that there's a substack of "opers" but am having trouble locating a definition. So what's an oper, and how should I think about them?
| https://mathoverflow.net/users/622 | What is an Oper? | Look at <http://arxiv.org/abs/math/0501398> (*Opers*, Beilinson and Drinfeld, 1993/2005)
| 7 | https://mathoverflow.net/users/439 | 2535 | 1,649 |
https://mathoverflow.net/questions/2520 | 22 | Is it possible give an example of (or explain) how the Voevodsky et al.'s homotopy theory of schemes computes higher Chow groups?
| https://mathoverflow.net/users/65 | Homotopy theory of schemes examples | To keep things simple, let us assume we work over a perfect field.
The easiest part of motivic cohomology which we can get is the Picard group (i.e. the Chow group in degree 1). This works essentially like in topology: in the (model) category of simplicial Nisnevich sheaves (over smooth k-schemes), the classifying space of the multiplicative group $\mathbb G\_m=\mathbb A^1-\{0\}$ has the $\mathbb A^1$-homotopy type of the infinite dimensional projective space. Moreover, as the Picard group is homotopy invariant for regular schemes (semi-normal is even enough), the fact that $H^1(X,\mathbb G\_m)=\text{Pic}(X)$ reads as $[X,B\mathbb G\_m]=\text{Pic}(X)=CH^1(X)$, where $[?,?]$ stands for the $\text{Hom}$ in the $\mathbb A^1$-homotopy category of $k$-schemes, denoted by $H(k)$.
In general, we denote by $K(\mathbb Z(n),2n)$ the $n$-th motivic Eilenberg-MacLane space, i.e. the object of $H(k)$ which represents the $n$-th Chow group in $H(k)$: for any smooth $k$-scheme $X$, one has
$$[X,\Omega^i K(\mathbb Z(n),2n)]=H^{2n-i}(X,\mathbb Z(n))$$
(where $\Omega^i$ stands for the $i$-th loop space functor). For $i=0$, we just get the usual Chow groups:
$$H^{2n}(X, \mathbb Z (n)))\simeq CH^n(X) .$$
Then, there are several models for $K(\mathbb Z(n),2n)$, one of the smallest being constructed as follows. What I explained above is that $K(\mathbb Z(1),2)$ is the infinite projective space.
$K(\mathbb Z(0),0)$ is simply the constant sheaf $\mathbb Z$. For higher $n$, here is a construction (this is Voevodsky's).
Given a $k$-scheme $X$, denote by $L(X)$ the presheaf with transfers associated to X, that is the presheaf of abellian groups whose sections over a smooth $k$-scheme $V$ are the finite correspondences from $V$ to $X$ (i.e. the finite linear combinations of cycles
$\sum n\_iZ\_i$ in $V \times X$ such that $Z\_i$ is finite and surjective over $V$). This is a presheaf, where the pullbacks are defined using the pullbacks of cycles (the condition that the $Z\_i$; are finite and surjective over a smooth (hence normal) scheme $V$ makes that this is well defined without working up to rational equivalences, and as we consider only pullbacks along maps $U \to V$ with $U$ and $V$ smooth (hence regular) ensures that the multiplicities which will appear from these pullbacks will always be integers). The presheaf $L(X)$ is a sheaf for the Nisnevich topology. This construction is functorial in $X$ (I will need this functoriality only for closed immersions).
Let $X$ (resp. $Y$) be the cartesian product of $n$ (resp. $n-1$) copies of the projective line.
The point at infinity gives a family of $n$ maps $u\_i : Y \to X$. Then a model of $K(\mathbb Z(n),2n)$ in $H(k)$ is the sheaf of sets obtained as the quotient (in the category of Nisnech sheaves of abelian groups) of $L(X)$ by the subsheaf generated by the images of the maps $L(u\_i):L(Y)\to L(X)$.
If you want a more conceptual definition, there is also the direction of algebraic cobordism (but for this, you need to understand the stable homotopy of $\mathbb P^1$-spectra, but maybe it is enough at first to think of $\mathbb P^1$-spectra simply as the cohomology theories
allowed in homotopy theory of schemes): the idea is that there is an algebraic cobordism, which is represented by a $\mathbb P^1$-spectrum $MGL$ (the analog of of the spectrum $MU$ which represents complex cobordism in algebraic topology). The idea is that $MGL$ is the universal oriented cohomology theory (for short, this means that, if a cohomology theory $E$ satisfies the projective bundle formula, then the choice of an orientation, i.e. of a generating class in the second cohomology group of $\mathbb P^1$ with coefficients in $E(1)$, is the same as a map of ring spectra $MGL \to E$. The idea is that, as in algebraic topology, formal groups laws classify oriented cohomology theories ($MGL$ corresponding to the initial formal group law). The cohomology theory which corresponds to the multiplicative formal group law is $KGL$, the $\mathbb P^1$-spectrum which represents algebraic K-theory, while the cohomology theory corresponding to the additive formal group law is motivic cohomology (this latter characterization has been announced by F. Morel and M. Hopkins if $k$ is of characteristic zero, but is not published yet, and it is known for any field $k$ if we work with rational coefficients (this is a result of Spitzweck, Nauman, Ostvaer)).
| 29 | https://mathoverflow.net/users/1017 | 2546 | 1,658 |
https://mathoverflow.net/questions/2566 | 8 | Given a group G, we can define Gop to be a new group whose underlying set is the same as that of G but with the new multiplication g.h = hg, i.e. multiply as if you were in G but reverse the order.
Is there anything interesting to say about this construction?
When is a group isomorphic to its opposite group?
One sees the "opposite ring" of a non-commutative ring showing up in a crucial way in number theory in the study of Brauer groups, so this is not necessarily an artificial question. This is just a question that has been poking around my brain for a little while. . . it seems like a potentially neat little problem.
| https://mathoverflow.net/users/493 | Silly Question about "opposite groups" | Every group is isomorphic to its opposite group, the map $g \mapsto g^{-1}$ being an isomorphism.
In ring theory this is not the case, and the opposite ring is an important construction. For instance, if $A$ is a central simple algebra over a field (or an Azumaya algebra over a ring), then the classes of $A$ and $A^{\operatorname{op}}$ are inverse in the Brauer group.
I don't know of anything similarly interesting to say about opposite groups.
| 23 | https://mathoverflow.net/users/1149 | 2568 | 1,673 |
https://mathoverflow.net/questions/2557 | 36 | In "Quantum field theory and the Jones polynomial" (Comm. Math. Phys. 1989 vol. 121 (3) pp. 351-399), Witten writes:
>
> A representation *Ri* of a group *G* should be seen as a quantum object. This representation should be obtained by quantizing a classical theory. The Borel-Weil-Bott theorem gives a canonical way to exhibit for every representation *R* of a compact group *G* a problem in classical physics, with *G* symmetry, such that the quantization of this classical problem gives back *R* as the quantum Hilbert space. One introduces the "flag manifold" *G/T*, with *T* being a maximal torus in *G*, and for each representation *R* one introduces a symplectic structure Ο*R* on *G/T*, such that the quantization of the classical phase space *G/T*, with the symplectic structure Ο*R*, gives back the representation *R*. **Many aspects of representation theory find natural explanations by thus regarding representations of groups as quantum objects that are obtained by quantization of classical physics.** [page 372; emphasis added]
>
>
>
I'm fascinated by this idea β I haven't seen it before, but it seems natural, in that classical objects should not be linear, whereas quantum objects should be. I'm most interested in the last sentence: what examples can y'all come up with of representation-theoretic facts that can be "explained" by "physics" on *G/T*? (Besides, of course, Witten's application in the paper I quoted from.)
More generally, I've read the Wikipedia discussion of the Borel-Weil-Bott theorem, and done some random googling, but I haven't found an elementary description of the symplectic structure Witten refers to. Anyone want to pedantically spell out Witten's comment, please?
| https://mathoverflow.net/users/78 | Examples of applications of the Borel-Weil-Bott theorem? | Look at the orbits of `G` on `g*` the dual of the Lie algebra. These 'coadjoint orbits' have a canonical symplectic structure. Each of these orbits intersects the positive Weyl chamber exactly once; consider those intersecting it at a 'positive weight' (i.e. the the elements of `g*` that lift to characters on `T`, the maximal torus). The positive weights exactly classify the irreducible representations. At the same time, we can build a line bundle over the corresponding coadjoint orbit so that the symplectic form realises the Chern class. Sections of this bundle are automatically a representation of G, and the Borel-Weil-Bott theorem, in one of its friendlier guises, says that this is the representation you expect -- the one with the highest weight we started with.
Generically, a coadjoint orbit is just G/T. When it isn't, it's a deeper quotient, but you can pull back the canonical symplectic structure to G/T, and still do everything there. This is the symplectic structure ΟR Witten is referring to.
| 15 | https://mathoverflow.net/users/3 | 2570 | 1,675 |
https://mathoverflow.net/questions/2015 | 12 | If not, are there any interesting subcategories that can be concertized? If I am not mistaken, the category of reduced finite type varieties over the complex numbers would be an example, where the forgetful functor to sets would be given by looking at the underlying map of points.
| https://mathoverflow.net/users/788 | Can the Category of Schemes be Concretized? | The category of schemes is not small-concrete.
Let $S$ be a generating set. Let $U$ be the set of all rings $A \neq 0$ such that $\mathrm{Spec}(A)$ is an open subscheme of a scheme in $S$. Let $X$ be a set whose cardinality is larger than any element of $U$, for example, $2^{\bigsqcup\_{A \in U} A}$. Let $K$ be the field $\mathbb{Q}(t\_x)\_{x \in X}$, where $t\_x$ are a collection of algebraically independent generators indexed by $X$. So $|K|$ is larger than $|A|$ for any $A \in U$. Since ring maps from a field to a nontrivial ring are always injective, $\mathrm{Hom}(\mathrm{Spec}(A),\mathrm{Spec}(K))=\emptyset$ for every $A \in U$, and therefore $\mathrm{Hom}(s,\mathrm{Spec}(K))=\emptyset$ for every $s \in S$.
There is only one map from the empty set to itself. But $\mathrm{Spec}(K)$ has nontrivial isomorphisms, coming from permuting the generators. So
$\mathrm{Hom}(\mathrm{Spec}(K),\mathrm{Spec}(K)) \longrightarrow \mathrm{Hom}\_{\mathrm{Set}^{S^\mathrm{op}}}( (\mathrm{Spec}(K))(-), (\mathrm{Spec}(K))(-))$
is not injective.
| 22 | https://mathoverflow.net/users/297 | 2595 | 1,695 |
https://mathoverflow.net/questions/2597 | 8 | In the answer to [this](https://mathoverflow.net/questions/2071/non-finitely-generated-ring-of-regular-functions/) question we saw that [there exists](http://math.stanford.edu/~vakil/files/nonfg.pdf) a nonsingular quasi-projective threefold over a field with non-finitely generated global sections.
I was talking about this previous question today and the following question came up - given any countably generated noetherian k-algebra R which is an integral domain and whose field of fractions has finite transcendence degree over k, where k is a field does there exist some quasi-projective variety X (by variety I mean an integral separated scheme of finite type over k) such that the ring of global sections of X is R?
It is possible one needs more hypotheses to make this work - if this is false I think it would be interesting to know the class of algebras which can occur.
| https://mathoverflow.net/users/310 | Can any countably generated k-algebra occur as the ring of global sections of some variety? | No. Take k[t] and invert countably many relatively prime polynomials. This obeys all of your adjectives (localization preserves noetherianness, the others are obvious.)
However, a ring of global sections must be a subring of some finitely generated k-algebra. (I pointed this out in the last discussion.) Hence, its unit group must be a subgroup of the group of units of a finitely generated k-algebra. If A is any finitely generated k-algebra, then Units(A)/Units(k) is a finitely generated abelian group, and thus can't contain the countably generated group of the above example.
I can't find a reference for the fact about Units(A)/Units(k) at the moment, Tevelev describes this as well known in
* *Compactifications of subvarieties of tori*, American Journal of Mathematics **129** no. 4 (2007) pp. 1087β1104, doi:[10.1353/ajm.2007.0029](https://doi.org/10.1353/ajm.2007.0029), arXiv:[math/0412329](https://arxiv.org/abs/math/0412329)
| 10 | https://mathoverflow.net/users/297 | 2601 | 1,697 |
https://mathoverflow.net/questions/2409 | 11 | Let Xi be a sequence of identically-distributed random variables with finite-range dependence (i.e. there exists I such that if |i-i'| β₯ I, then Xi and Xi' are independent), and a finite moment-generating function (i.e. EerXi < β for all r β R).
It's not too hard to show that Xi satisfies a strong law of large numbers, and I've got a proof written. However, I'm sure that this is a standard theorem in the probability literature, and I'd rather just cite it in the paper I'm writing. Do you have a good reference for this result?
Here are two follow-up generalizations: what if Xi instead has only a finite moment condition? Or what if Xi has exponential correlation decay (i.e. EXiXi' β€ Ce-c|i-i'| for some positive c, C)?
| https://mathoverflow.net/users/238 | Strong Law of Large Numbers for weakly dependent random variables | This question sounds like an exercise: Split the sequence into I sequences of iid random variables. Apply the classical SLLN to each sequence. Recombine.
Tom:
Of course it is true with exponential decay of the corellation function, but it is not easy.
The essential difficulty is that one wants to reduce the SLLN to an exponentially growing subsequence of N's. In the classical case, this is done by a martingale inequality. (Prob of a supremum of a martingale is dominated by the probability at end of the martingale.)
Once one moves away from an implicit martingale structure, then tricks have to be employed---of which the most obvious is that if the sequence of random variables is bounded, then obviously you can reduce to an exponentially growing subsequence. This point is much of the content of the paper of Lyons cited already.
Not sure that this would appear in a text book however. My sense is that these considerations are well-known.
| 7 | https://mathoverflow.net/users/1158 | 2605 | 1,700 |
https://mathoverflow.net/questions/2548 | 13 | For a variety V, its [Albenese variety](http://en.wikipedia.org/wiki/Albanese_variety) Alb(V) is a variety with a map V β Alb(V) which factors uniquely into any map from V to an abelian variety. Can we say something similar for an arbitrary scheme? When do we know there exists an "Albanese" scheme Alb(X)? That is,
>
> Under what conditions on a scheme X does there exist a morphism X β Alb(X) which factors uniquely into any map from from X to an abelian scheme?
>
>
>
| https://mathoverflow.net/users/84526 | "Albanese" schemes: When does an "initial abelian scheme" exist under a given scheme? | The construction of an Albanese scheme and an Albanese map for proper and geometrically irreducible schemes over a perfect field goes back to the work of Chevalley, to [this talk](http://archive.numdam.org/article/SCC_1958-1959__4__A10_0.pdf) of Serre, and to Grothendieck, e.g. [Theorem 3.3. in FGA Exp.VI](http://www.math.jussieu.fr/~leila/grothendieckcircle/FGA.pdf) . The idea is always to look at the dual of an appropriate component of the Picard scheme. One just has to pile enough conditions on the setup to ensure that the Picard functor is representable.
One thing that should be said here is that the Albanese scheme is not in general an abelian scheme but only a torsor over a semi-abelian scheme. Also if we drop the properness condition or consider our original scheme to be defined over a base scheme things become more interesting. In full generality it is possible to define an Albanese 1-motive over the base scheme (it is a complex of sheaves of abelian groups of small amplitude with typically representable cohomology) which has the desired universal property. There are various techniques for construction of this derived version of the Albanese scheme. Some use characteristic zero, resolution of singularities, and Nagata compactifications, and some use simplicial scheme resolutions. There are many cool works in this direction, e.g. the paper of [Barbieri-Viale and Srinivas](http://arxiv.org/pdf/math/9906165v1) , and the more recent papers of Niranjan Ramachandran and [Ayoub and Barbieri-Viale](http://arxiv.org/pdf/math/0607738v2) . The appendix of Mochizuki's paper mentioned in Lars' post is also excellent.
| 16 | https://mathoverflow.net/users/439 | 2606 | 1,701 |
https://mathoverflow.net/questions/69 | 14 | For $n$ an integer greater than $2$, Can one always get a complete theory over a finite language with exactly $n$ models (up to isomorphism)?
Thereβs a theorem that says that $2$ is impossible.
My understanding is this should be doable in a finite language, but I donβt know how.
If you switch to a countable language, then you can do it as follows. To get $3$ models, take the theory of unbounded dense linear orderings together with a sequence of increasing constants $\langle c\_i: i < \omega\rangle$. Then the $c\_i$βs can either have no upper bound, an upper bound but no sup, or have a sup. This gives exactly $3$ models. To get a number bigger than $3$, we include a way to color all elements, and require that each color is unbounded and dense. (The $c\_i$βs can be whatever color you like.) Then, we get one model for each color of the sup plus the two sup-less models.
| https://mathoverflow.net/users/27 | Complete theory with exactly n countable models? | You can refine Ehrenfeuchtβs example getting rid of the constants.
Here is what John Baldwin suggested:
Consider the theory in the language $L=\{\le\}$, saying
* $\le$ is a total preorder (transitive, total [hence reflexive], not necessarily anti-symmetric) without least or last element. (Notice that the binary relation defined by $x\le y \land y\le x$ is an equivalence relation. Call it $E$.)
* For each $n$, $E$ has exactly one class of size $n$. Call it $C\_n$.
* $C\_i\le C\_j$ (for $i\le j$) setwise.
* $E$-classes are densely ordered: for any two points there is a point $\le$-between them and not $E$-equivalent to any of them.
Check that this theory is complete.
Note that each finite equivalence class in this new theory plays the role of one of the constants in the classical example, so you get three countable models the same way.
| 17 | https://mathoverflow.net/users/1152 | 2612 | 1,704 |
https://mathoverflow.net/questions/2615 | 31 | Hello, I'm wondering if there is a standard reference discussing the least number of charts in an atlas of a given manifold required to describe it.
E.g. a circle requires at least two charts, and so on (I couldn't manage to get anything relevant neither on wikipedia nor on google, so I guess I'm lacking the correct terminology).
*Edit*: in the case of an open covering of a topological space by n+1 contractible sets (in that space) then n is called the Lusternik-Schnirelman Category of the space, see Andy Putman's answer. The following book seems to be the standard reference <http://books.google.fr/books?id=vMREfNN-L4gC&pg=PP1>
Great, now I'm still interested by the initial question: does anybody know of another theory without this contractibility assumption (hoping that it allows more freedom)? e.g. would it lead to different numbers say for genus-g surfaces?
*Final edit*: yes different numbers for genus-g surfaces (see answers below), but not sure there is a theory without contractibility. Right, really lots of interesting literature on the LS category nevertheless, hence the accepted answer. For example there are estimates for non-simply connected compact simple Lie groups like PU(n) and SO(n) in Topology and its Applications, Volume 150, Issues 1-3, 14 May 2005, Pages 111-123.
| https://mathoverflow.net/users/469 | Least number of charts to describe a given manifold | It's not quite the same thing, but a related object is the LyusternikβSchnirelmann category of a topological space. See
<http://en.wikipedia.org/wiki/Lyusternik-Schnirelmann_category>
| 15 | https://mathoverflow.net/users/317 | 2616 | 1,707 |
https://mathoverflow.net/questions/2305 | 4 | Be done the triangle ABC, it is known the method to finding the point Q that minimises the sum QA+QB+QC among all points Q in the plane (The Fermat point).
I want a hint for solving this problem using the construction of the tangent to ellipse.
(Hadamard, Lesson de Geometrie Elementaire, II, problem no. 745).
| https://mathoverflow.net/users/1093 | How to find the Fermat Point using the construction of the tangent to ellipse? | I have the vague idea that Hadamard is referring to the construction where you erect equilateral triangles BCA', CAB' and ABC' on the sides of the triangle, as described [here](http://www.cut-the-knot.org/Generalization/fermat_point.shtml). The Fermat point is the intersection of the cevians AA', BB' and CC'. It can also be constructed using the various angles of 60 resp. 120 degrees.
In the construction of the tangent from a point P to an ellipse with foci F and F' (in the book you cite), they consider an additional point f. The correspondence should be
F β C,
F' β A,
P β B,
f β P'.
The general philosophy behind both, I think, is to convert a sum of segments into a single segment.
| 1 | https://mathoverflow.net/users/296 | 2625 | 1,715 |
https://mathoverflow.net/questions/1497 | 9 | If (C,tensor,1) is a symmetric monoidal category and f:A-->B is a morphism of PROPs (or monoidal cats = colored PROPs), one gets a forgetful functor f^\*:B-Alg(C)-->A-Alg(C) (where B-Alg(C)=tensor-preserving functors from B to C) defined by precomposing with f.
Does anyone conditions on A,B,C under which this functor has a left or a right adjoint?
(e.g. if C has the monoidal structure coming from products, it has a left adjoint, is there more to say?)
| https://mathoverflow.net/users/733 | When do PROP-morphisms induce adjunctions? | Paul-André Melliès has quite an interesting paper on this topic:
<http://hal.archives-ouvertes.fr/docs/00/33/93/31/PDF/free-models.pdf>
...but phrased in the more general terms of T-algebras of a pseudomonad. The idea is that a pseudomonad on a 2-category (especially Cat), let you put algebraic structures on categories the same way monads let you put them on objects of a category, like sets. This is motivated by the need to put PROPs, PROBs, PROs, Lawvere theories, etc. all under one roof.
He begins by talking about how a T-algebra homomorphism (a monoidal functor in the case where the T-algebras are monoidal categories) j : A -> B induces a forgetful functor U\_j from Models(B,C) to Models(A,C) in the way you mentioned. Looking for left adjoint to U\_j amounts to looking for a way to push some functor backwards along j in a suitably natural way. As Tom already mentioned, this is the left Kan extension. This process is functorial, and usually written Lan\_j : [A,C] -> [B,C]. Furthermore, Lan\_j -| U\_j.
But if we were done there, all PROPs would have free algebras, which we know is not true in general (cf. bialgebras). The hard part is proving the Lan\_j is a *T-algebraic* left Kan-extension. In the case of Lawvere theories, this is easy, because the product structure guarantees all natural transformations of cartesian functors are cartesian, but in the monoidal case, this stuff all needs to be checked.
This is where the story starts to get more complicated. It seems quite tricky to come up with suitably weak conditions under which Lan\_j is T-algebraic. Mellies phrases these in terms of distributers (aka profunctors, modules, depending on who you ask and what country you are in :-P). If functors are like functions, this are a bit like relations. The nice thing about them is they always come in adjoint pairs f\_\* and f^\* for any functor f.
So, thm 1 in the paper is (roughly) this. If j and j^\* are T-algebraic in the suitable 2-categories, C is (T-algebraically) complete and co-complete, and for any model f : A -> C, f\_\* o j^\* factors through the up-star of the Yoneda embedding y : C -> Psh(C), then U\_j has a left adjoint computed as Lan\_j that is indeed the free functor.
This is quite heavy-duty (pro-arrow equipment, ends, etc.), but it seems to get the job done. It would be nice to see more concrete/specific examples of this.
| 3 | https://mathoverflow.net/users/800 | 2631 | 1,719 |
https://mathoverflow.net/questions/2610 | 1 | This question may be too detailed but perhaps somebody knows the answer: Neukirch proofs in his algebraic number theory book in Chapter IV, Proposition 6.2, that his class field axiom implies that the Tate cohomology groups $H^n(G(L|K),U\_L)$ for $n=0,-1$ vanish for finite unramified extensions $L|K$, where $U\_L$ is the group of units. He mentions in the proof that every element $a \in A\_L$ can be written as $a = \epsilon \pi\_K^m$, where $\epsilon \in U\_L$ and $\pi\_K$ is a prime element in $A\_K$. Why does this work?
I absolutely understand this argument when the image of the valuation just lies in $\mathbb{Z}$! But how does this work for a valuation whose image is $\hat{\mathbb{Z}}$? Unless $A$ is not a profinite module, I don't know what $\pi\_K^m$ is for some general $m \in \hat{\mathbb{Z}}$. Unfortunately, this must work in this generality for global class field theory.
| https://mathoverflow.net/users/717 | Neukirch's class field axiom and cohomology of units for unramified extension | You don't need to make sense out of $\pi\_K^m$ for a general $m$ in $\hat{\mathbb{Z}}$. All you really need to know for his argument is that $v\_K(A\_K) = v\_L(A\_L)$ as subgroups of $\hat{\mathbb{Z}}$. I didn't think this through but I think it should be pretty easy to establish from the fact that $\pi\_K$ is prime for both valuations.
All he really uses is that the Galois group fixes $\pi\_K$.
| 2 | https://mathoverflow.net/users/493 | 2632 | 1,720 |
https://mathoverflow.net/questions/2640 | 9 | I know that the Weyl groups of affine Lie algebras don't have a longest element, but are there any good substitutes for w\_0. In particular, is there any good substitute for a reduced decomposition of the longest element?
| https://mathoverflow.net/users/788 | Longest Element of an Affine Weyl Group | If you look at section 4 of Thomas Lam and Pavlo Pylyavskyy's [recent preprint](https://arxiv.org/abs/0906.0610), they study infinite reduced words in W, modulo braid and commutation relations. This is a good substitute for a reduced word for the long word in the context of total positivity, as they explain. I think it should also be useful in the context of canonical bases.
| 9 | https://mathoverflow.net/users/297 | 2642 | 1,728 |
https://mathoverflow.net/questions/2521 | 6 | Consider a random walk on a two-dimensional surface with circular reflecting boundary conditions (say, of radius 'R'). Here, for a fixed-size area, one finds a larger fraction of the probability density (for the position of the walker) near the midpoint of the circle than near its contour.
Given this example, my question is - for a discrete/continuous random walk in a two-dimensional (or higher dimensional) space, now with arbitrary reflecting boundary conditions, how 'well' can one restrict/focus the mass of the probability density function to the smallest possible area relative to the total surface area available to the walker?
In other words, how effectively can one construct a 'trap' (I'm using this term very loosely) for such a walker, given random initial conditions?
(I obviously welcome any help to ask this question in a more appropriate manner.)
| https://mathoverflow.net/users/774 | 'Focusing' the mass of the Probability Density Function for a Random Walk | Hmm, you're asking for concentration for heat kernels. Over long periods of time, these kernels are dominated by the low-energy eigenfunctions, so basically one needs to construct domains which have concentrated low-energy eigenfunctions.
Generally one expects in fact that heat kernels become smoother and disperse over time (parabolic regularity). For instance, all the L^p norms of heat kernels are non-increasing in time, so it's going to be harder and harder to concentrate into a small domain as time goes by. There is a substantial theory on controlling heat kernels (using tools such as the Poincare inequality, maximum principle, integration by parts, etc.) though it isn't quite my field; one may have to ask a parabolic PDE person.
| 6 | https://mathoverflow.net/users/766 | 2648 | 1,734 |
https://mathoverflow.net/questions/2117 | 4 | When it comes to solving the heat diffusion equation u\_t=u\_xx the two most important solutions are
a) a combination (sum) of sin-terms to resemble the function of the initial condition (that is essentially a fourier series)
b) a convolution-integral of the function of the initial condition with the Gauss-curve
In most books you only find a) or b).
My question: How does it come that you get to such different solutions? Why is it that some books end up with a) and others with b)? What is the essential difference of deriving a) or b) in the end?
| https://mathoverflow.net/users/1047 | Solutions to the diffusion equation | The two solutions solve different problems for the same equation.
The Fourier series solution solves the heat equation u\_t=u\_xx on a bounded interval [a,b] with an initial condition at t=0 of the form u(x,0)=f(x), a <=x<=b, and boundary conditions at both ends of the interval. These conditions can be of different types, leading to different series expansions. The most general (homogeneous) conditions are of the form
\alpha u(a)+\beta u\_x(a)=\alpha u(b)+\beta u\_x(b)=0, \alpha^2+\beta^2!=0.
If \beta=0 they are called Dirichlet conditions; if \alpha=0 Neumann conditions, if both \alpha and \beta are non zero, Robin conditions. Conditions can also be mixed: of one type on one end, of another type on the other end.
The convolution solutions solves the pure initial value problem, or Cauchy problem, on the whole real line, with initial value u(x,0)=f(x), x\in R.
| 6 | https://mathoverflow.net/users/1168 | 2652 | 1,737 |
https://mathoverflow.net/questions/2653 | 9 | I heard this from Haskell Rosenthal many years ago.
If V is a complex vector space, say the **opposite** of V is the complex vector space with the same elements, the same operations except switch scalar multiplication to scalar multiplication by the complex conjugate scalar. Of course this definition applies in particular to a complex Banach space. Question: Is every complex Banach space isomorphic to its opposite? (An isomorphism is a complex-linear homeomorphism.)
[ prompted by question [Silly question about opposite groups](https://mathoverflow.net/questions/2566/silly-question-about-opposite-groups) ]
| https://mathoverflow.net/users/454 | opposite Banach space | Does [this paper of Kalton](http://arxiv.org/abs/math/9402207) do the trick? (disclaimer: I haven't read through the details)
| 9 | https://mathoverflow.net/users/763 | 2667 | 1,747 |
https://mathoverflow.net/questions/2659 | 3 | Is there a lower bound on the volume of a delta-ball in the orthogonal group O(n) of the type
f(n) \* delta^{n(n-1)/2}? For which f(n)? How can it be proven?
n(n-1)/2 is the number of degrees of freedom in the orthogonal group.
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/1170 | Lower bound on the volume of a delta-ball in the orthogonal group O(n) of the type f(n)*delta^{n(n-1)/2} | Well, there is certainly a bound of that form. Here is a crude proof:
There is an exponential map from skew-symmetric matrices to O(n). This is differentiable with nonsingular Jacobian near the origin. The preimage of the delta ball, if I understand your notations, is the log(1+delta) ball. So the volume of this preimage is c(n)\*log(1+delta)^{n(n-1)/2} where c(n) is the volume of the unit ball. Now just multiply by some lower bound for the Jacobian near 0.
By the way, do you want to insist on using the sup norm? A lot of these formulas are nicer for the L^2 norm. For example, I think the preimage in the skew symmetric matrices is an ordinary ball, so its volume would be well known.
| 2 | https://mathoverflow.net/users/297 | 2669 | 1,749 |
https://mathoverflow.net/questions/2671 | 4 | Hello,
homotopic maps induce isomorphic pullback bundles, and so isomorphism classes of vector bundles over X correspond to homotopy classes of maps from the grassmannian to X. But I think, in the case of X = S^1 (the 1-sphere), the isomorphism classes of 1-bundles correspond to (the generators of) Ο\_\_\_{1}(S^1), since there is the trivial bundle, the Moebius bundle, and that's it. So my question is: am I right with this, and if yes: what needs to happen that Ο\_n(X) = {homotopy classes of maps:grassmannian -> X}?
| https://mathoverflow.net/users/1057 | (how) are vector bundles and homotopy groups related? | The map goes the other way: vector bundles over X correspond to homotopy classes of maps from X into a grassmannian.
Let BO(n) be the grassmannian of n-plane bundles in R-infinity. Then, if you want to know about n-dimensional real vector bundles over Sk you are led to study the homotopy classes of maps from Sk to BO(n), or in other words Οk BO(n).
In particular, the fact that there are exactly two 1-bundles over S1 comes from a calculation Ο1 BO(1)=Z/2.
| 9 | https://mathoverflow.net/users/910 | 2674 | 1,751 |
https://mathoverflow.net/questions/2672 | 21 | I made the following claim over at the [Secret Blogging Seminar](http://sbseminar.wordpress.com/2009/10/26/concrete-categories/), and now I'm not sure it's true:
Let $f: X \to Y$ and $g: X \to Y$ be two maps between finite CW complexes. If f and g induce the same map on $\pi\_k$, for all k, then f and g are homotopic.
Was I telling the truth?
EDIT: Since I didn't say anything about basepoints, I probably should have said that f and g induce the same map
$[S^k, X] \to [S^k, Y]$.
This will also deal better with the situation where X and Y are disconnected. I'd be interested in knowing a result like this either with pointed maps or nonpointed maps. (Although, of course, if you work with pointed maps you have to take X and Y connected, because $[S^k, -]$ can't see anything beyond the number of components in that case.)
| https://mathoverflow.net/users/297 | Whitehead for maps | This is not true. Consider, for example, a degree 1 map from a torus $S^1 \times S^1$ to $S^2$ (concretely, realize the torus as a square with identifications, and then collapse the boundary of the square to a point). This map is trivial on all homotopy groups (since for any $n>0, \pi\_n$ is 0 for either the domain or the codomain), but it is not homotopically trivial because it is nonzero on $H\_2$.
If you want to demand that the spaces be simply connected, you can get a counterexample by considering cohomology operations: the cup square, for example, gives a map from $K(\mathbb{Z},n)$ to $K(\mathbb{Z},2n)$ which is nontrivial, but for the same reason as the previous example it must be 0 on homotopy groups. This example is not finite-dimensional, but it's probably possible to find one that is--I just don't know how because I don't know how to show a map is trivial on homotopy groups if the spaces have infinitely many nontrivial homotopy groups whose values are unknown, which is the case for most finite-dimensional examples.
| 39 | https://mathoverflow.net/users/75 | 2678 | 1,753 |
https://mathoverflow.net/questions/2677 | 7 | If *G* is a group, its **abelianization** is the abelian group *A* and the map *G* β *A* such that any map *G* β *B* with *B* abelian factors through *A*. Abelianization is a functor, and in general a very lossy operation. The map *G* β *A* is always a surjection/quotient, because we can construct *A* by dividing *G* by the minimal normal subgroup that contains all conjugations *ghg-1h-1* for *g,h*β*G*.
If *V* is a finite-dimensional (super)vector space over a field *K*, then the abelianization of GL(*V*) is isomorphic to the multiplicative group *K*\* of non-zero numbers in *K*. Indeed, the determinant exhibits the desired isomorphism.
Here are two questions I'm curious about:
1. What can be said about the abelianizations of other (finite-dimensional) Lie groups?
2. If *V* is an infinite-dimensional vector space, what can be said about the abelianization of GL(*V*)? Most infinite-dimensional vector spaces have some analytic structure, e.g. topological vector spaces, and so it's reasonable to ask that the operators in GL(*V*) should preserve that structure; you are welcome to take your favorite type of infinite-dimensional vector space and your favorite type of GL(*V*), if you want.
| https://mathoverflow.net/users/78 | Abelianization of Lie groups | I don't have anything to say about specific examples, but here are some general remarks. A way to construct the abelianization of any compact group is to consider its image under the product of all its 1-dimensional unitary representations. This is because a compact abelian group is characterized by its set of characters by Pontrjagin duality. More generally, you can construct the double Pontrjagin dual of a locally compact group to get its locally compact abelianization as a subgroup of a space of maps to U(1) with the compact-open topology.
| 4 | https://mathoverflow.net/users/75 | 2679 | 1,754 |
https://mathoverflow.net/questions/2682 | 1 | This is inspired by [*The Whitehead for maps*](https://mathoverflow.net/questions/2672/whitehead-for-maps) question.
Consider two maps `f, g: X\to Y` which happen to induce the same maps (of discrete spaces) `[Z, X] \to [Z, Y]` for every Z. Does this mean `f` and `g` are homotopic?
And what would be the lessons from the answer to this question? I feel like there's something interesting about the way we should ask it.
| https://mathoverflow.net/users/65 | Something like Yoneda's lemma | Yes, this is a special case of Yoneda. Let Z=X and consider the identity map in [X,X]; the hypothesis says that f1=f and g1=g are then equal as elements of [X,Y].
| 2 | https://mathoverflow.net/users/75 | 2684 | 1,756 |
https://mathoverflow.net/questions/2696 | 14 | I'm a little bit hesitant to ask this here, so please notice the tag. My hope is that someone will have a more satisfying answer than what I've heard before...
A long time ago I read (perhaps 'browsed' is a better word) Wolfram's "A New Kind of Science". There are many many references to the "Rule 30" CA - <http://mathworld.wolfram.com/Rule30.html>. However, no intuitive reasoning for it's random/pseudorandom behavior is provided prior to a digression to the importance of the discovery. I was recently reminded of this when I heard that Mathematica (a program I use quite frequently) uses certain outputs from this CA as its random number generator.
So my question is - beyond 'numerical phenomenology' is there an intuitive understanding why Rule 30 should behave in this random/pseudorandom manner?
| https://mathoverflow.net/users/774 | An intuitive reason why the "Rule 30" CA is random/pseudorandom? | If you look at the [results of elementary cellular automata](http://mathworld.wolfram.com/ElementaryCellularAutomaton.html) from mathworld, most of them seem to have some kind of symmetry. (I don't want to make "symmetry" formal here; what I mean is that you get nice patterns of some sort.)
But I suspect that in general, the results of cellular automata are psuedorandom. (I haven't looked at this too closely.)
So if I had to guess, I would say that the answer is just that most CAs on small neighborhoods are "nice", and most CAs on large neighborhoods aren't. Rule 30 is not sporadic, but it's the first example of a family that eventually predominates.
| 5 | https://mathoverflow.net/users/143 | 2705 | 1,768 |
https://mathoverflow.net/questions/2681 | 9 | This is related to [Theo's question](https://mathoverflow.net/questions/2677/abelianization-of-lie-groups) about the abelianizations of finite dimensionsal Lie groups.
I am interested in a specific (infinite-dimensional) case of the above question. Let H be an infinite-dimensional Hilbert space and GL(H) represent the bounded two-sided invertible operators on H. Is there a nice description of the commutator subgroup G (the group generated be elements of the form ABA^{-1}B^{-1}) and the abelianization GL(H)/G?
| https://mathoverflow.net/users/792 | Abelianization of GL(H) | The solution to Problem 240 in Halmos is: "On an infinite-dimensional Hilbert space, the commutator subgroup of the full linear group is the full linear group itself."
It is mentioned, although the details are not given, that every invertible operator is the product of two(!) commutators. Reference: A. Brown and C. Pearcy, *Multiplicative commutators of operators*, Can. J. Math. **18** (1966) 737-749.
| 10 | https://mathoverflow.net/users/430 | 2709 | 1,771 |
https://mathoverflow.net/questions/2692 | 12 | I was thinking about quivers recently, and the following idea came to me.
Let ei,j denote the matrix unit in Mn for 1 β€ i,j β€ n. Let Ξ denote the complete quiver on vertices {1, β¦, n}: one directed edge Ei,j for each ordered pair (i, j), including self-loops Ei,i.
Mn(k) is then the quotient of the path algebra PΞ by a (rather large) ideal generated by "2-faces" of the simplex: Ei,jEk,l = Ξ΄j,kEi,l.
In this language, for example, the Borel of upper triangular matrices corresponds to the ordered simplex inside Ξ.
* Is this correspondence ***interesting***?
* Can we transport Lie theoretic ideas about gln(k) to the quiver language? Should we?
* What happens if we quotient by a smaller ideal? Say, only reduce paths of length at least 3 (Ei,jEk,lEp,q = Ξ΄j,kΞ΄l,pEi,q).
My apologies in advance for these questions being vague.
| https://mathoverflow.net/users/813 | Matrices into path algebras | There's a slightly different equivalence that is also useful. Consider the quiver with n elements, and an arrow `E_i` from `i` to `i+1` and another `F_i` from `i+1` to `i` for all `i`. The relations are then that `E_i F_i = e_i` and `F_i E_i = e_{i+1}`, where `e_j` is the `j`-th simple idempotent. This gets the same path algebra with fewer arrows and relations, but it has even less symmetry than your presentation.
A first answer to your question is that this perspective can often be **useful**. The reason I say this is because this perspective allows you to realize a lot of other quivers as subalgebras of matrices, and vice versa (for instance, the Borel subalgebra as the path algebra of a subquiver). It's not an extremely useful proving technique, but it can be a good way to produce a lot of quivers, especially when first learning about them.
Is it interesting? That's another question entirely. It's unfortunate that it picks out a basis in a necessary way, and so the `GL_n` action on `M_n` doesn't seem natural. I think the fact the the presentation I mention above is close to what is called a 'double quiver' is somewhat interesting. Especially if you like to think of a semisimple Lie algebra as something like the tangent bundle to the space of Borel subalgebras. Precisely, I mean that BB localization relates certain modules of g to D-modules on the space of Borel subalgebras, and so it is interesting to think of `M_n` as a deformation of the tangent bundle to U\_n, the upper triangular matrices.
| 4 | https://mathoverflow.net/users/750 | 2716 | 1,775 |
https://mathoverflow.net/questions/2713 | 3 | I believe there was an old conjecture that there's **always a prime number between `N` and `2N`**.
What's the history and how is this proven is the easiest/elementary/deepest ways?
| https://mathoverflow.net/users/65 | Bertrand's postulate | Proven. This is called Bertrand's postulate. Here is [Erdos' elementary proof](http://www.nd.edu/~dgalvin1/pdf/bertrand.pdf); the original proof is due to Chebyshev.
| 13 | https://mathoverflow.net/users/143 | 2717 | 1,776 |
https://mathoverflow.net/questions/2708 | 15 | Is there known to be an $x$ such that for all positive integers $N$ there exists some $n>N$ such that $p\_{n+1}-p\_n \leq x$, where $p\_n$ is the $n$th prime? Or, in other words, is it known that limit as $n$ goes infinity of $p\_{n+1}-p\_n$ is not infinity? If such an $x$ is known to exist, what is the current best known $x$? (Showing $x=2$ would imply the Twin Prime Conjecture, of course.)
| https://mathoverflow.net/users/597 | Is there a known bound on prime gaps? | (**Edit**: things have happened since the original post, changing the short answer to yes. See for example <http://arxiv.org/abs/1410.8400> for the status in 2014 where $x \leq 600$ unconditionally. GRP **End Edit**)
The short answer is no, though if one assumes the Elliot-Halberstam conjecture then one can take x=16. See
<http://arxiv.org/abs/math/0605696>
for a comprehensive survey of the best known results (both conditional and unconditional).
There is also the Wikipedia article at
<http://en.wikipedia.org/wiki/Prime_gap>
although this is less comprehensive.
| 27 | https://mathoverflow.net/users/766 | 2718 | 1,777 |
https://mathoverflow.net/questions/2710 | 6 | Let (A,m) be a complete local Noetherian ring and let X and Y be two schemes of finite type over A (and flat over A). Let Xn and Yn be the reductions of X and Y mod mn+1.
>
> **Question**: Suppose there is a compatible system of isomorphisms between Xn and Yn (for all n). Does there necessarily exist an isomorphism between X and Y over A?
>
>
>
In other words, suppose the formal schemes \hat{X} and \hat{Y} are isomorphic; are X and Y isomorphic?
*Remark*: The answer is `no' if we drop flatness (you can just stick an extra component over the generic fiber) or finite type (A[t] vs. A{t} = the completion of A[t]).
| https://mathoverflow.net/users/2 | Can isomorphisms of schemes be constructed on formal neighborhoods? | No. Let A be k[[t]]. Let X be A^1 \setminus {-1,0,1} and Y be A^1 \setminus {1,t,-1}. In explicit equations, X = Spec k[[t]][x, y]/y(x-1)x(x+1)-1 and Y = Spec k[[t]][x, y]/y(x-1)(x-t)(x+1)-1.
Over k[[t]]/t^{n+1}, the reductions of X and Y are isomorphic because all infinitesimal deformations of a smooth affine scheme are trivial. (See corollary 4.7 in [Hartshorne's notes on deformation theory](http://math.berkeley.edu/~robin/math274root.pdf).)
However, X and Y are not isomorphic because the two fibers over the general point are not: For any field K, if we have P\_K^1 \setminus {a,b,c,d} for {a,b,c,d} \in P^1(K), then the cross ratio of a,b,c and d is a well defined element of K. In particular, this applies when K=k((t)).
| 7 | https://mathoverflow.net/users/297 | 2720 | 1,779 |
https://mathoverflow.net/questions/2704 | 32 | I'm taking introductory courses in both Riemann surfaces and algebraic geometry this term. I was surprised to hear that any compact Riemann surface is a projective variety. Apparently deeper links exist.
What is, in basic terms, the relationship between Riemann surfaces and algebraic geometry?
| https://mathoverflow.net/users/416 | Links between Riemann surfaces and algebraic geometry | For simplicity, I'll just talk about varieties that are sitting in projective space or affine space. In algebraic geometry, you study varieties over a base field k. For our purposes, "over" just means that the variety is cut out by polynomials (affine) or homogeneous polynomials (projective) whose coefficients are in k.
Suppose that k is the complex numbers, C. Then affine spaces and projective spaces come with the complex topology, in addition to the Zariski topology that you'd normally give one. Then one can naturally give the points of a variety over C a topology inherited from the subspace topology. A little extra work (with the inverse function theorem and other analytic arguments) shows you that, if the variety is nonsingular, you have a nonsingular complex manifold. This shouldn't be too surprising. Morally, "algebraic varieties" are cut out of affine and projective spaces by polynomials, "manifolds" are cut out of other manifolds by smooth functions, and polynomials over C are smooth, and that's all that's going on.
In general, the converse is false: there are many complex manifolds that don't come from nonsingular algebraic varieties in this manner.
But in dimension 1, a miracle happens, and the converse is true: all compact dimension 1 complex manifolds are analytically isomorphic to the complex points of a nonsingular projective dimension-1 variety, endowed with the complex topology instead of the Zariski topology. "Riemann surfaces" are just another name for compact dimension 1 (dimension 2 over R) complex manifolds, and "curves" are just another name for projective dimension 1 varieties over any field, hence the theorem you described.
As for why Riemann surfaces are algebraic, Narasimhan's book explicitly constructs the polynomial that cuts out a Riemann surface, if you are curious.
| 38 | https://mathoverflow.net/users/1018 | 2722 | 1,781 |
https://mathoverflow.net/questions/2724 | 7 | Is it known that for every epsilon there is `N_0` such that all intervals of the form [N, (1+\epsilon)\*N], where N > `N_0`, contain prime numbers?
| https://mathoverflow.net/users/65 | Strong Bertrand postulate | This follows from the Prime Number Theorem. Let Ο(n) be the number of primes less than n. Then Ο(n) ~ n/log(n); it follows Ο((1+Ξ΅)n)-Ο(n) -> β as n -> β.
| 8 | https://mathoverflow.net/users/143 | 2725 | 1,783 |
https://mathoverflow.net/questions/2748 | 26 | This is somewhat related to [Greg's question](https://mathoverflow.net/questions/2551/why-do-groups-and-abelian-groups-feel-so-different) about groups and abelian groups. Suppose you met someone who was well-acquainted with groups, but who was unwilling to accept rings as a meaningful object of study. How would you describe rings to them in a natural way given that they like talking about groups?
(Admittedly this is not really the question the title asks.)
| https://mathoverflow.net/users/290 | What is the "right" definition of a ring? | Well rings are naturally the objects which act on abelian groups - indeed composition always endows the endomorphisms of an abelian group with the structure of a ring. So if one is interested in the endomorphisms of groups one is actually interested in rings.
One can make this analogy more precise especially if one picks a particular ring and looks at the forgetful functor to abelian groups from its module category. This analogy can then be used again for instance to motivate the definition of plethory which are the natural objects which act on rings.
To address Eric's comment about commutative rings there is an analogue in this case of something which was mentioned in the discussion of groups versus abelian groups. Indeed one can obtain commutative rings by considering the identity in an additive symmetric monoidal category. In this case the endomorphisms of the tensor unit are endowed with an abelian group structure via the augmentation over abelian groups and the Eckmann-Hilton argument applied to tensoring endomorphisms and composing endomorphims forces the composition to be abelian. So from this point of view commutative rings are the gadgets which naturally act on the hom-sets of additive symmetric monoidal categories.
Since I mentioned this one can take this slightly further. If one considers such a category together with an autoequivalence (for instance if we take a tensor triangulated category) then one can consider the graded endomorphism ring of the identity. This naturally gives rise to an integer graded ring which is commutative up to some unit which squares to the identity and which has a natural action on the category.
| 28 | https://mathoverflow.net/users/310 | 2751 | 1,803 |
https://mathoverflow.net/questions/2660 | 4 | Suppose I uniformly sample matrices X from the Gaussian Unitary Ensemble (GUE) with variance \sigma^2. Consider the Ky-Fan d norm, i.e. the sum of the singular values, of X. Let's call this Z=||X||\_1. What is the distribution of Z as a function of the dimension d and the variance \sigma^2? Really, all I want are estimates of the mean and a good tail bound, so maybe also the second moment.
| https://mathoverflow.net/users/1171 | Distribution of 1-norm for Gaussian Unitary Ensemble | Let's normalise the variance of the entries to be $1$. Then GUE asymptotically obeys the semicircular law, i.e., the eigenvalues (which equal the singular values, as GUE is Hermitian), after dividing by $\sqrt{n}$, are distributed according to the law $\frac{1}{2 \pi} (4 - x^2)^{1/2}\_+ dx$. So the Schatten $1$-norm (Ky Fan norm) should asymptotically equal $\sqrt{n}$ times n times the integral
\begin{equation\*}
\int \frac{|x|}{2 \pi} (4 - x^2)^{1/2}\_+ dx,
\end{equation\*}
which Wolfram alpha tells me is $8/3 \pi$. So the answer is $n^{3/2} ( 8 / 3 \pi + o(1) )$ with probability $1-o(1)$. Using an explicit convergence rate for GUE, one can probably replace the $o(1)$ with $O(n^{-c})$ for some explicit constant $c>0$.
Getting the variance may be within current technology - it's some integral of two-point correlations of GUE, which are known - but somewhat tedious. Higher moments should also (in principle) be computable. My guess is that the limiting distribution will be asymptotically gaussian, but I might be wrong about this (the central limit theorem doesn't apply directly because the eigenvalues are correlated with each other).
| 6 | https://mathoverflow.net/users/766 | 2760 | 1,808 |
https://mathoverflow.net/questions/2755 | 74 | As Akhil had great success with his [question](https://mathoverflow.net/questions/1291/a-learning-roadmap-for-algebraic-geometry), I'm going to ask one in a similar vein. So representation theory has kind of an intimidating feel to it for an outsider. Say someone is familiar with algebraic geometry enough to care about things like G-bundles, and wants to talk about vector bundles with structure group G, and so needs to know representation theory, but wants to do it as geometrically as possible.
So, in addition to the algebraic geometry, lets assume some familiarity with representations of finite groups (particularly symmetric groups) going forward. What path should be taken to learn some serious representation theory?
| https://mathoverflow.net/users/622 | A learning roadmap for Representation Theory | I second the suggestion of Fulton and Harris. It's a funny book, and definitely you want to keep going after you finish it, but it's a good introduction to the basic ideas.
You specifically might be happier reading a book on algebraic groups.
While I third the suggestion of Ginzburg and Chriss, I wouldn't call it a "second course." Maybe if what you really wanted to do was serious, Russian-style geometric representation theory, but otherwise you might want to try something a little less focused, like Knapp's "Lie Groups Beyond an Introduction."
If you want Langlandsy stuff, then Ginzburg and Chriss is actually a bit of a tangent; good enrichment, but not directly what you want, since it skips over all the good stuff with D-modules. Look in the background reading for the graduate student seminar we're having in Boston this year: <http://www.math.harvard.edu/~gaitsgde/grad_2009/>
| 25 | https://mathoverflow.net/users/66 | 2763 | 1,811 |
https://mathoverflow.net/questions/2779 | 22 | The projective curve $3x^3+4y^3+5z^3=0$ is often cited as an example (given by Selmer) of a **failure of the [Hasse Principle](http://en.wikipedia.org/wiki/Hasse_principle)**: the equation has solutions in any completion of the rationals $\mathbb Q$, but not in $\mathbb Q$ itself.
I don't think I've ever seen a proof of the latter claim β is someone able to provide an outline? What are the necessary tools?
| https://mathoverflow.net/users/25 | Proof of no rational point on Selmer's Curve $3x^3+4y^3+5z^3=0$ | My friend has written an introduction to algebraic number theory before, which contains a short proof of this statement, but I didn't check its validity.
**Edit:** updated the link of the document, <http://www.2shared.com/document/2d6M7kNU/Introduction_to_Algebraic_Numb.html>
p. 41 of the document, or p. 45 of the PDF.
| 7 | https://mathoverflow.net/users/nan | 2785 | 1,827 |
https://mathoverflow.net/questions/2734 | 2 | Brown defines the classifying space of a crossed complex in the following way.
Given a filtration X\* of a space X, define the fundamental crossed complex by:
C\_0 = X\_0, C\_1=\pi(X\_1,X\_0) (the fundamental groupoid), C\_n = the family of groups \pi(X\_n,X\_n-1,p) for all p in X\_0.
Now let Ξ^n be the cell complex of the standard n-simplex, with its skeletal filtration. The crossed complex \pi(Ο^n) is then written \pi[n]. The nerve NC of a crossed complex C is defined to be the simplicial set given in dimension n by (NC)n = Crs(\pi[n],C), where Crs(-,-) is the internal hom in the category of crossed complexes.
The think I don't understand is that each of the n-simplices are contractible, so why wouldn't the fundamental crossed complex associated to the filtration of Ξ^n be trivial?
| https://mathoverflow.net/users/343 | Classifying space of a crossed complex | It is indeed (weakly equivalent to) the trivial one, in the [model structure on crossed complexes](http://ncatlab.org/nlab/show/folk+model+structure).
Your question seems to indicate that you think this is a problem. But it is not: each cell of NC is contractible (as it is for each cell of a space!) but there may still be simplicial maps
partial Delta^n --> NC
from the simplicial set that is the boundary of the n-simplex into NC that represent non-trivial elements of [simplicial homotopy group](http://ncatlab.org/nlab/show/simplicial+homotopy+group)s of NC.
To see what is going on, in may be helpful to realize that the situation with crossed complexes is the analogue in the context of [*strict* oo-groupoids](http://ncatlab.org/nlab/show/strict+omega-groupoid) of the more familiar situation with [Kan complex](http://ncatlab.org/nlab/show/Kan+complex)es.
A Kan complex is a model for a weak oo-groupoid. A crossed complex is a strict oo-groupoid (a more restrictive notion).
To any topological space X is associated its singular simplicial Kan complex S(X) = Hom(Delta^bullet,X)-- which we may think of as the [fundamental weak oo-groupoid](http://ncatlab.org/nlab/show/fundamental+infinity-groupoid) of X. This is the weak oo-groupoid version of Ronnie Brown's fundamental crossed complex.
Indeed, the nerve of Brown's fundamental crossed complex gives a Kan complex that approximates the full S(X).
| 2 | https://mathoverflow.net/users/381 | 2787 | 1,829 |
https://mathoverflow.net/questions/2596 | 18 | Illusie [mentions](http://www.math.uchicago.edu/~mitya/langlands/Illusie.wav) tape recordings of Grothendieck explaining his trace formula and more. Are they or similar recordings online? I guess, even if (what I doubt) everything he thought about that is somewhere in print, it would give an interesting insight in his way of thinking.
| https://mathoverflow.net/users/451 | Are there any recordings of Grothendieck online? | Illusie told me last year that these tapes of his meetings with Grothendieck are somewhere in his basement where they're very hard to find, so they're certainly not online.
The recording of Illusie's reminisces of Grothendieck to which both posts link has been transcribed [here](http://math.uchicago.edu/~tp/main.dvi). Drinfeld and Illusie will be releasing a more polished version at some point next year.
| 7 | https://mathoverflow.net/users/307 | 2788 | 1,830 |
https://mathoverflow.net/questions/1814 | 11 | It took me some effort to work out Gerashenko's nice simple example [Can a singular Deligne-Mumford stack have a smooth coarse space?](https://mathoverflow.net/questions/1565/can-a-singular-deligne-mumford-stack-have-a-smooth-coarse-space/1584#1584) of a DM stack non-equisingular with its coarse moduli space, which means I must improve my understanding of coarse moduli spaces.
What are your favourite examples of coarse moduli spaces? One per answer, please, so we can rank them.
| https://mathoverflow.net/users/307 | What are some examples of coarse moduli spaces? | Thank you all for your very kind answers! It was silly of mine to suggest we rate the examples, since it's unclear if any examples are better than others. To atone for my mistake, let me offer a summary of the proposed examples in the form of a table.
stack : coarse moduli space
---------------------------
$\mathcal{M}\_1$ : affine line $\mathbf{A}^1$;
line bundles of degree $0$ on a smooth curve : $\mathrm{Pic}^0$;
$[\mathrm{pt}/G]$ : $\mathrm{pt}$.
Good moduli spaces
------------------
One thing that I learned from [Alper, Good moduli spaces](http://math.columbia.edu/~jarod/good_moduli_spaces.pdf) is that an object technically better than the coarse moduli space is a *good moduli space*, a replacement notion which commutes with arbitrary, not just flat, base change, and exists more generally. Explicitly, a good moduli space of an Artin stack $X$ is a morphism $f$ to an algebraic space such that 0) it's quasi-compact a) pushforward along $f$ is exact on quasi-coherent sheaves, and b) the pullback morphism $f\_\*\mathcal{O}\_Y\rightarrow \mathcal{O}\_X$ is an isomorphism.
For example, given a linear algebraic group $G$ acting an an affine scheme of ring A over a field, the morphism
$$
[\mathrm{Spec}(A)/G] \rightarrow \mathrm{Spec}(A^G)
$$
is a good moduli space (ibid., Example 8.3), hence over $\mathbf{Q}$,
$$
[\mathbf{A}^1/\mathrm{Spec}(k)]\rightarrow \mathrm{Spec}(k)
$$
is a good moduli space (ibid., Example 8.1 and Example 12.4 (1)).
And when is the old notion of a coarse moduli space a special case of the new notion of good moduli space, you ask? Well, if the stack in question is *tame* (ibid. Example 8.1), i.e., if it has inertia stack finite over it and if the morphism from it to the coarse moduli space is exact on quasicoherent sheaves. (The latter condition is automatic in characteristic $0$.)
| 2 | https://mathoverflow.net/users/307 | 2792 | 1,833 |
https://mathoverflow.net/questions/2607 | 63 | Does anyone have an answer to the question "What does the cotangent complex measure?"
Algebraic intuitions (like "homology measures how far a sequence is from being exact") are as welcome as geometric ones (like "homology detects holes"), as are intuitions which do not exactly answer the above question.
In particular: Do the degrees have a meaning? E.g. if an ideal $I$ in a ring $A$ is generated by a regular sequence, the cotangent complex of the quotient map $A\twoheadrightarrow A/I$ is $(I/I^2)[-1]$. Why does it live in degree 1?
| https://mathoverflow.net/users/733 | Intuition about the cotangent complex? | One thing the cotangent complex measures is what kind of deformations a scheme has. The precise statements are in Remark 5.30 and Theorem 5.31 in Illusie's article in "FGA explained". Here's the short simplified version in the absolute case:
If you have a scheme $X$ over $k$, a first order deformation is a space $\mathcal{X}$ over $k[\epsilon]/(\epsilon)^2$ whose fiber over the only point of $k[\epsilon]/(\epsilon)^2$ is $X$ again. You can imagine $k[\epsilon]/(\epsilon)^2$ as a point with an infinitesimal arrow attached to it and $\mathcal{X}$ as an infinitesimal thickening of $X$. The cotangent complex gives you precise information on how many such thickenings there are: The set of such thickenings is isomorphic $\mathop{Ext}^1(L\_X, \epsilon^2)$.
Now let's assume that we have chosen one such infinitesimal thickening $\mathcal{X}$ over $k[\epsilon]/(\epsilon)^2$. It is not always true that you can go on and make this thickening into a thickening to the next order. Whether or not you can do this is measured precisely by the cotangent complex: There is a map that takes as input your chosen thickening $\mathcal{X}$ and spits out an element in $\mathop{Ext}^2(L\_X, \epsilon^3)$. If the element in the Ext group is zero you can go on to the next level. If it is not zero, it's game over and your stuck.
| 29 | https://mathoverflow.net/users/473 | 2796 | 1,835 |
https://mathoverflow.net/questions/2765 | 6 | SGA 7, tome 1, exp. IX, contains in its introduction and in section 13.4 remarks about ideas and conjectures of Deligne on a βthΓ©orie de NΓ©ron pour motifs de poids quelconqueβ. Would someone please give an epitome of that theory or references? (This is Thomas Riepe's request <http://sbseminar.wordpress.com/requests/#comment-3717> at the secret blogging seminar, mildly edited.)
| https://mathoverflow.net/users/307 | NΓ©ron theory for motives of arbitrary weight | Deligne commented last year:
""Neron model" is perhaps misleading. It is only the case of unipotent (rather than quasi-unipotent) local monodromy I want to consider. The questions I had in mind were :
-What can one say about a motive about the field of fraction of a discrete valuation ring ; what objects over the residue field can one get (in some/all cohomology theory, or perhaps even motivically ?).
-model : for polarized variations of Hodge or admissible variations of mixed Hodge structures on the punctured disk, nilpotent orbits theorems. This is also the story of tangential base points in my fundamental group of P^1 minus 3 points. One should get motives over the punctured Zariski tangent space. There is work of Ayoub to get motives in a sense better than "system of realizations". How the more precise SL(2) orbit theorem fits in is unclear.
-model (function fields,l-adic) : the relation between monodromy weight filtration and weights (in my Weil II).
-model : Raynaud analytic description of degenerating abelian varieties as, in a rigid analytic category, a quotient by a lattice of an extension of an abelian variety by a torus. The resulting abelian variety over the (complete) field of fractions is an analytic, not algebraic, construct. Modulo any power of the maximal ideal of the discrete valuation ring, it however makes algebraic sense. What to hope for for more general motives is unclear. I wonder about it in [71] of my list of publications.
The classical story of Neron models is about points and extending them to some model. Points make motivic sense (as extensions in a mixed motives category), but I did not have them in mind."
| 4 | https://mathoverflow.net/users/451 | 2799 | 1,837 |
https://mathoverflow.net/questions/2795 | 76 | Last year I attended a first course in the representation theory of finite groups, where everything was over C. I was struck, and somewhat puzzled, by the inexplicable perfection of characters as a tool for studying representations of a group; they classify modules up to isomorphism, the characters of irreducible modules form an orthonormal basis for the space of class functions, and other facts of that sort.
To me, characters seem an arbitrary object of study (why take the trace and not any other coefficient of the characteristic polynomial?), and I've never seen any intuition given for their development other than "we try this, and it works great". The proof of the orthogonality relations is elementary but, to me, casts no light; I do get the nice behaviour of characters with respect to direct sums and tensor products, and I understand its desirability, but that's clearly not all that's going on here. So, my question: is there a high-level reason why group characters are as magic as they are, or is it all just coincidence? Or am I simply unable to see how well-motivated the proof of orthogonality is?
| https://mathoverflow.net/users/1202 | Why are characters so well-behaved? | Orthogonality makes sense without character theory. There's an inner product on the space of representations given by $\dim \operatorname {Hom}(V, W)$. By Schur's lemma the irreps are an orthonormal basis. This is "character orthogonality" but without the characters.
How to recover the usual version from this conceptual version? Notice $$\dim \operatorname{Hom}(V,W) = \dim \operatorname{Hom}(V \otimes W^\*, 1)$$ where $1$ is the trivial representation. So in order to make the theory more concrete you want to know how to pick off the trivial part of a representation. This is just given by the image of the projection $\frac1{|G|} \sum\_{g\in G} g$.
The dimension of a space is the same as the trace of the projection onto that space, so
$$
\def\H{\rule{0pt}{1.5ex}H}
\dim \operatorname{Hom}(V \otimes W^\*, 1) = \operatorname{tr}\left(\frac1{|G|} \sum\_{g\in G} {\large \rho}\_{\small V \otimes W^\*}(g)\right)
= \frac1{|G|} \sum\_{g\in G} {\large\chi}\_{V}(g)\ {\large\chi}\_{W}\left(g^{-1}\right)
\\
$$ using the properties of trace under tensor product and duals.
| 70 | https://mathoverflow.net/users/22 | 2808 | 1,843 |
https://mathoverflow.net/questions/2809 | 47 | It is well known that the operations of differentiation and integration are reduced to multiplication and division after being transformed by an integral transform (like e.g. Fourier or Laplace Transforms).
My question: Is there any intuition why this is so? It can be proved, ok - but can somebody please explain the big picture (please not too technical - I might need another intuition to understand that one then, too ;-)
| https://mathoverflow.net/users/1047 | Intuition for Integral Transforms | It might help you to think about a discrete model: consider complex valued functions on $Z/n$. The discrete Fourier transform takes $f(k)$ to
$g(j) :=\sum\_{k=1}^n f(k) \zeta^{jk}$ where $\zeta=e^{2 \pi i/n}$. It is pretty easy to see that, if we change $f(k)$ to $f(k+1)$, we change $g(j)$ to $g(j)\*\zeta^j$.
Similarly, changing $f(k)$ to $f(k+1)-f(k)$ changes $g(j)$ to $g(j)\*(\zeta^j-1)$. So, in this discrete model, taking a difference becomes multiplication by $(\zeta^j-1)$. In a similar way, in the continuous setting, taking a derivative becomes multiplication by $x$.
| 23 | https://mathoverflow.net/users/297 | 2823 | 1,855 |
https://mathoverflow.net/questions/2848 | 6 | Suppose you have N symbols (e.g. "1","2",...,"N" or "a","b",...,"$") and a string of these symbols (say, the first trillion digits of pi). Then does there exist a prime number whose N-ary representation contains that string of digits?
| https://mathoverflow.net/users/1132 | Prime numbers and strings of symbols | Yes. This follows from the strong version of [Bertrand's postulate](https://mathoverflow.net/questions/2724/strong-bertrand-postulate). For example, to see that there is a prime which contains the digits 314159, use the fact that there is a prime between 314159\*10^N and
314159\*10^N\*(1.000001) for N sufficiently large.
| 9 | https://mathoverflow.net/users/297 | 2850 | 1,875 |
https://mathoverflow.net/questions/2665 | 5 | The following is a result I feel like I've seen some form of before, but can't figure out how to prove or find a reference for. Suppose you have a map p:E \to B, with B paracompact, and suppose that every point in B has a neighborhood U such that there is a map p^{-1}(U) \to U \times F over U which is a fiber homotopy equivalence. Does it follow that p is a Hurewicz fibration? The converse, if B is locally contractible, is standard: a Hurewicz fibration is locally equivalent to a product.
| https://mathoverflow.net/users/75 | Is a map that is locally fiberwise equivalent to a product a Hurewicz fibration? | The answer is no; Allen Hatcher sent me the following:
An example where this fails is the projection of the letter L onto its horizontal base, which I'll call B. The deformation retraction of L onto B is a fiberwise homotopy equivalence. The homotopy lifting property fails: Map a point to the left endpoint of B, then lift this to a point of L - B and take a homotopy that moves the left endpoint of B to the right endpoint.
| 4 | https://mathoverflow.net/users/75 | 2855 | 1,878 |
https://mathoverflow.net/questions/2857 | 6 | Can you build a probabilistic scheme for colouring each edge (independently of all other edges) of the complete graph G on the positive integers such that the probability that G contains an infinite monochromatic complete subgraph is neither 0 nor 1?
| https://mathoverflow.net/users/416 | Non trivial colouring of the edges of an infinite complete graph | It seems like this would go against the [Kolmogorov 0-1 law.](http://en.wikipedia.org/wiki/Kolmogorov%27s_zero-one_law). If we let Xi denote the coloring of all of the edges from i to integers larger than i, wouldn't the existence of an infinite monochromatic subgraph be a tail event?
| 9 | https://mathoverflow.net/users/405 | 2859 | 1,880 |
https://mathoverflow.net/questions/2776 | 6 | If β# exists then why is cof(ΞΈL(β)) = Ο? Also I have the same question for the L(VΞ»+1) generalization (if it's actually a different proof; I presume it isn't), i.e. if ΞΈ is defined as the sup of the surjections in L(VΞ»+1) of VΞ»+1 onto an ordinal, then if VΞ»+1# exists why is cof(ΞΈL(VΞ»+1)) = Ο?
| https://mathoverflow.net/users/1178 | Cofinality of Theta if sharps exist | This is because the pieces of the sharp singularize Theta. Let s\_n be the sequence of the first n cardinals above continuum and let a\_n be the nth cardinal above continuum. Then the theory of reals with a parameter s\_n in L\_{a\_n+1}(R) is a set of reals A\_n. They are Wadge cofinal in Theta, another words the sequence is not in L(R) but each A\_n is and that is why you get a singularization.
| 5 | https://mathoverflow.net/users/20584 | 2862 | 1,882 |
https://mathoverflow.net/questions/2861 | 14 | Given a high precision real number, how should I go about guessing an algebraic integer that it's close to?
Of course, this is extremely poorly defined -- every real number is close to a rational number, of course! But I'd like to keep both the coefficients and the degree relatively small. Obviously we can make tradeoffs between how much we dislike large coefficents and how much we dislike large degrees.
But that aside, I don't even know how you'd start. Any ideas?
**Background**: this comes out of the project that prompted Noah's [recent question](https://mathoverflow.net/questions/2396/solving-polynomial-equations-when-you-know-in-which-number-field-the-solutions-li) on solving large systems of quadratics. We're trying to find subfactors inside their graph planar algebras. We've found that solving quadratics numerically, approximating the solutions by algebraic integers, and then checking that these are actually exact solutions is very effective. So far, we've been making use of Mathematica's "[RootApproximant](http://reference.wolfram.com/mathematica/ref/RootApproximant.html)" function which does exactly what I ask here, but it's an impenetrable black box, like everything out of Wolfram.
| https://mathoverflow.net/users/3 | How should I approximate real numbers by algebraic ones? | The Lenstra-Lenstra-Lovasz lattice basis reduction algorithm is what you need. Suppose that your real number is a and you want a quadratic equation with as small coefficients as possible, of which a is nearly a root. Then calculate 1,a,a^2 (to some precision), find a nontrivial integer relation between them, and use the LLL algorithm to find a much better one from the first one. Exactly this example is discussed in the Wikipedia entry on the LLL algorithm, applied to the Golden Section number. And there is a big literature on the algorithm and its many applications. (For higher degree, calculate 1,a,a^2,...,a^n).
| 29 | https://mathoverflow.net/users/3304 | 2864 | 1,884 |
https://mathoverflow.net/questions/2142 | 6 | I've read an interesting article, [math.NT/0409456](http://arxiv.org/abs/math.NT/0409456) where you're just trying to solve a simple problem:
>
> For a given (finite) set of primes S find all solutions to an equation `a + b = c` with the condition that all prime divisiors of integers a, b, c must be in S.
>
>
>
and this problem turns out to be very geometric. It turns out (and I tell that in comments below) you're actually dealing with sections of certain projective morphism of schemes `R --> Spec ZZ \ S`. The article then proves that the number of solutions to the equation is finite by proving that the number of these sections is finite.
Is there kind of general theory or other methods to prove things about sections of these maps? What is the intuition used here? Would there be a way to count these solutions?
| https://mathoverflow.net/users/65 | Solving "a, b, a+b have given divisors" problem | This is bound to be pretty hard! Knowing the prime divisors of a,b,c implies knowing the radical rad(abc) (the product of the distinct primes dividing abc), and then knowing the solutions a,b,c of a + b = c in positive integers, you would know the largest c with a,b,c coprime (pairwise or not, doesn't matter). So you would be in a position to settle the ABC conjecture c \leq C(epsilon)rad(abc)^{1 + \epsilon} if you could handle all those S-unit equations. The ABC conjecture is *the* outstanding problem in Diophantine analysis.
| 1 | https://mathoverflow.net/users/3304 | 2878 | 1,895 |
https://mathoverflow.net/questions/2876 | 12 | Let X ---> Y be a finite surjective morphism of smooth, projective, connected varieties over a finite field F\_q. Can one describe the zeta function Z(X, t) in terms of the zeta-function Z(Y,t) of Y? Can one say anything at all about how they are related? What if we assume the morphism is finite etale?
| https://mathoverflow.net/users/81 | Behaviour of Zeta-function under Finite Morphism | I don't think you can say much of anything of consequence unless you have better control over things. As an example, consider the kth power map A^1 -> A^1. This is finite, and surjective and even etale if you throw out 0, and has degree k, but the zeta functions are the same.
Another way of rerephrasing Ilya's comment is that it's not just varieties at have zeta functions; all mixed sheaves have them. The zeta function of X is the zeta function of the pushforward of the constant sheaf on X to Y. Now it may be that you can say something useful about this sheaf (for example, if you have a cover, it is a local system, and you can think about L-functions of pi\_1 representations) but in general, that sheaf could be pretty horrible.
| 2 | https://mathoverflow.net/users/66 | 2884 | 1,899 |
https://mathoverflow.net/questions/2890 | 18 | This question is related to [this question](https://mathoverflow.net/questions/1346/representablity-of-cohomology-ring) from Dinakar, which I found interesting but don't yet have the background to understand at that level.
Unless I'm mistaken, the rough statement is that $H^n(X;G)$ (the $n$-dimensional cohomology of $X$ with coefficients in $G$) should somehow correspond to (free?) homotopy classes of maps $X \to K(G,n)$. I want to understand this better, in relatively elementary terms. Here are some questions which (I hope) will point me in the right direction.
1. What category are we working in? My guess is that $X$ should just be a topological space, the cohomology is singular cohomology, and our maps $X \to K(G,n)$ just need to be continuous.
2. Does this carry over if we give $X$ a smooth structure, take de Rham cohomology, and require our maps $X \to K(G,n)$ to be smooth?
3. How does addition in $H^n(X;G)$ carry over?
4. How does the ring structure on $H^\*(X;G)$ carry over? (This has probably been adequately answered to Dinakar already.)
| https://mathoverflow.net/users/303 | Cohomology and Eilenberg-MacLane spaces | 1. We are working in the **homotopy** category of topological spaces where morphisms are homotopy classes of continuous maps. More accurately, we tend to work in the based category where each object has a distinguished base point and everything is required to preserve that base point. The non-based category can be embedded in the based category by the simple addition of a disjoint base point, so we often pass back and forth between the two without worrying too much about it. The cohomology theory itself is slightly more interesting. For CW-complexes, it doesn't matter which one you choose as they are all the same. However, outside the subcategory of CW-complexes then the different theories can vary (as was mentioned in another question). So what we do is the following: using Big Theorems we construct a topological space, which we call $K(G,n)$, which represents the $n$th cohomology group with coefficients in $G$ for **CW-complexes**. So whenever $X$ is a CW-complex, we have a natural isomorphism of functors $\tilde{H}^n(X;G) \cong [X, K(G,n)]$, where the right-hand side is homotopy classes of based maps. For arbitrary topological spaces, we then **define** cohomology as $[X, K(G,n)]$. If this happens to agree with, say, singular cohomology then we're very pleased, but we don't require it.
2. Depends what you mean by "smooth structure". Certainly in the broadest sense, you will get different answers if you insist on everything being smooth. But for smooth manifolds, continuous maps are homotopic to smooth maps (and continuous homotopies to smooth homotopies) so the homotopy category of smooth manifolds and smooth maps is equivalent to the homotopy category of smooth manifolds and continuous maps. However, you need to be careful with the $K(G,n)$s as they will, in general, not be finite dimensional smooth manifolds. However, lots of things aren't finite dimensional smooth manifolds but still behave nicely with regard to smooth structures so this shouldn't be seen as quite the drawback that the other answerants have indicated.
3. Addition in $\tilde{H}^n(X;G)$ translates into the fact that $K(G,n)$ is an $H$-space. The suspension isomorphism, $\tilde{H}^n(X;G) \cong \tilde{H}^{n+1}(\Sigma X, G)$ implies the **stronger** condition that $K(G,n)$ is the loop space of $K(G,n+1)$ and so the $H$-space structure comes from the Pontrijagin product on a (based!) loop space. But the basic theorem on representability of cohomology merely provides $K(G,n)$ with the structure of an $H$-space.
4. As for the ring structure, that translates into certain maps $K(G,n)\wedge K(G,m) \to K(G,n+m)$. I don't know of a good way to "see" these for ordinary cohomology, mainly because I don't know of any good geometric models for the spaces $K(G,n)$ except for low degrees. One simple case where it can be seen is in rational cohomology. Rational cohomology (made 2-periodic) is isomorphic to rational $K$-theory and there the product corresponds to the tensor product of vector bundles.
(It should be said, in light of the first point, that $K$-theory should only be thought of as being built out of vector bundles for compact CW-complexes. For all other spaces, $K$-theory **is** homotopy classes of maps to $\mathbb{Z} \times BU$.)
| 17 | https://mathoverflow.net/users/45 | 2898 | 1,907 |
https://mathoverflow.net/questions/2900 | 17 | This came up in the [question about Eilenberg-MacLane spaces](https://mathoverflow.net/questions/2890/cohomology-and-eilenberg-maclane-spaces). Given the definition of `K(G, n)`, it's easy to prove that there is a map `K(G,n) x K(G,n) --> K(G,n)` that endows cohomology with an additive structure.
>
> **Question:** what's the most geometric way to show the existence of maps `K(G,n) x K(G,m) --> K(G,n+m)` that endow cohomology with multiplicative structure?
>
>
>
| https://mathoverflow.net/users/65 | How to get product on cohomology using the K(G, n)? | If you form the smash product $X = K(A,p) \wedge K(B,q)$ of two Eilenberg-MacLane spaces, then the resulting space is $(p+q-1)$-connected, and the first non-trivial homotopy group in dimension $p+q$ is $A \otimes B$.
To see this "geometrically", I would model the EM spaces as CW-complexes, whose first non-basepoint cells are in dimensions p and q respectively. Then in the smash product $X$, the bottom dimensional cells are products of the bottom dimensional cells of the EM spaces; this gives the connectivity result, and by looking at the attaching maps of the $(p+q+1)$-dimensional cells in $X$, you can compute $\pi\_{p+q}$. Now you can use obstruction theory to produce a map $X \to K(A\otimes B, p+q)$.
| 21 | https://mathoverflow.net/users/437 | 2906 | 1,910 |
https://mathoverflow.net/questions/2904 | 27 | Inspired by [this question](https://mathoverflow.net/questions/2857/non-trivial-colouring-of-the-edges-of-an-infinite-complete-graph), I was curious about a comment in [this article](http://en.wikipedia.org/wiki/Kolmogorov%27s_zero-one_law):
>
> In many situations, it can be easy to
> apply Kolmogorov's zero-one law to
> show that some event has probability 0
> or 1, but surprisingly hard to
> determine which of these two extreme
> values is the correct one.
>
>
>
Could someone provide an example?
| https://mathoverflow.net/users/441 | Examples where Kolmogorov's zero-one law gives probability 0 or 1 but hard to determine which? | There's a set of good examples from percolation theory:
<http://en.wikipedia.org/wiki/Percolation_theory>
If you create a "random network" with a certain probability p of edges between nodes (see article above for precise definitions) then there is an infinite cluster with probability either zero or one. But for a given value of p it can be nontrivial to determine which.
| 26 | https://mathoverflow.net/users/1227 | 2911 | 1,915 |
https://mathoverflow.net/questions/2913 | 9 | There was a previous post on the correspondence between Riemann surfaces and algebraic geometry. I want to ask a related but more detailed question.
BACKGROUND:
Engelbrekt gave an overview of how you start with a compact Riemann surfaces and map them into projective space
[Links between Riemann surfaces and algebraic geometry](https://mathoverflow.net/questions/2704/links-between-riemann-surfaces-and-algebraic-geometry/2849#2849)
In the case of a genus 1 surface X there's a very explicit construction. Namely X can be realized as β/L for a lattice L β
β€xβ€. From here the Weierstrass p function and its derivative can be constructed
<http://en.wikipedia.org/wiki/Weierstrass%27s_elliptic_functions>
and these give you a map β/L --> β^2 via z |--> [p(z), p'(z), 1] which realizes X as a degree three curve in β^2
QUESTION:
Say now X is a compact Riemann surface of genus g > 1. As has been pointed out below I should restrict to say g = 1/2(d-1)(d-2) where d>3, because otherwise there is no hope to realize X as a nonsingular curve in β^2.
Is there
1) a complex manifold Y that is a covering space of X such that X β
Y/G where G is the covering group of Y over X
2) holomorphic functions fβ, fβ, fβ from Y/G to ββͺβ
such that z |--> [fβ,(z), fβ(z), fβ(z)] realizes X as a projective variety of dimension 1 in β^2?
I'm told a good choice for Y would be the hyperbolic plane because then the 4g-gon representation of a genus g surface tiles the plane.
| https://mathoverflow.net/users/7 | Analogues of the Weierstrass p function for higher genus compact Riemann surfaces | I think Hunter and Greg's answers make it hard to see the forest for the trees. Let X be a compact Riem. surface of genus >= g. Let Y be the universal cover of X equipped with the complex structure pulled back from X. As a complex manifold, Y is isomorphic to the upper half plane, and the deck transformations form a subgroup Gamma of PSL\_2(R). There will be characters chi of Gamma for which there are nonzero functions f on Y such that f(gz) = chi(g) f(z). For chi ample enough (not defined here), we will be able to choose functions (f\_1, f\_2, f\_3) such that z --> (f\_1(z) : f\_2(z) : f\_3(z)) gives an immersion X --> P^2. All of this works in any genus.
The technical issue is that this map is an immersion, not an injection, meaning that the image can pass through itself. One can either decide to live with this, or work with maps to P^3 instead.
Most books that I have seen don't lift all the way to the universal cover of X. Instead, they take the covering of X which corresponds to the commutator subgroup of pi\_1(X). This can be motivated in a particular nice way in terms of the Jacobian. This is a complex manifold with the topological structure of a 2g dimensional torus. There is a map X --> J, so that the map pi\_1(X) --> pi\_1(J) is precisely the map from pi\_1(X) to its abelianization. People then work with the universal cover of J, and the preimage of X inside it. This has three advantages: the universal cover of J is C^g, not the upper half plane; the group of Deck transformations is Z^{2g}, not the fundamental group of a surface, and the action on C^g is by traslations, not Mobius transformations. The functions which transform by characters, in this setting, are called Theta functions\*, and they are given by explicit Fourier series.
\*This is a slight lie. Theta functions come from a certain central extension of the group of Deck transformations. It is certain ratios of Theta functions that will transform by characters as sketched above. The P function itself, for example, is a ratio of four Theta functions. In the higher genus case, in my limited reading, I haven't seen names for these ratios, only for the Theta functions.
| 9 | https://mathoverflow.net/users/297 | 2923 | 1,922 |
https://mathoverflow.net/questions/2875 | 19 | I've heard that irreducible unitary representations of noncompact forms of simple Lie groups, the first example of such a group `G` being `SL(2, R)`, can be completely described and that there is a discrete and continuous part of the spectrum of `L^2(G)`.
1. How are those representations described?
2. Do all unitary representations come from `L^2(G)`?
3. How are those related to representation of compact `SO(3, R)`?
4. What happens in the flat limit between `SL(2, R)` and `SO(3, R)`?
Also, is it possible to answer the questions above simultaneously for all Lie groups, not just `SL(2, R)`?
| https://mathoverflow.net/users/65 | Unitary representations of SL(2, R) | I strongly recommend you read the article "Representations of semisimple Lie groups" by Knapp and Trapa in the park city/ias proceedings "Representation theory of Lie groups". It's a very nice introduction to the problem of describing the "unitary dual" (which is what you are asking about) which focusses on SL(2,R). For example, page 9 says "the irreducible unitary representations that appear in L^2(G) do not nearly exhaust the unitary dual" for general semisimple Lie groups (thus answering you question 2). For more info, you can check out knapp's book "Representation theory of semisimple groups: an overview based on examples". For example, sections II.4 and II.5 describe the unitary duals of SL(2,C) and SL(2,R) respectively. The unitary duals of GL(n,C) and GL(n,R) were described by Vogan. Some other unitary duals are known, but in general, I don't think anything else is known. One approach is via Langlands' parametrization of irreducible admissible representations of reductive groups. This result is known for all groups and unitary representations are admissible, so the problem would be to identify which admissible representations are unitary (the knapp-trapa article talks about this). As for 3), every irreducible unitary representation of a compact group is finite-dimensional, so you don't get any of the infinite dimensional representations you get for SL(2,R). I don't know what you mean by 4).
For a full answer to 1) you can check out [link text](http://en.wikipedia.org/wiki/Representation_theory_of_SL2%28R%29#Unitary_representations).
| 19 | https://mathoverflow.net/users/1021 | 2940 | 1,936 |
https://mathoverflow.net/questions/2944 | 23 | Let $\{a\_n\}$ be a sequence of complex numbers indexed by the positive integers. Does there always exist an analytic function $f$ such that $f(n) = \{a\_n\}$ for $n=1,2,...$? If not, are there any simple necessary or sufficient conditions for the existence of such $f$? This analytic function should be defined on some connected domain in the complex plane containing the positive integers.
To make this concrete, consider Ackermann's function, which is defined recursively: first define the sequence of functions $A\_k$, $k=1,2,...$, as
$A\_1(n) = 2n$,
$A\_k(1) = 2$,
$A\_k(n) = A\_{k-1}(A\_k(n-1))$,
and then define Ackermann's function as the diagonal $A(n) = A\_n(n)$ for $n \geq 1$. Does there exist an analytic function $f$ such that $f(n) = A(n)$ for $n = 1,2,...$?
Actually, the individual functions $A\_k$ are interesting as well. $A\_1(n) = 2n$, as given above; $A\_2(n) = 2^n$, and $A\_3(n) = 2^{2^{\ldots^{2^2}}}$ (with $n$ twos in the expression). Obviously, $A\_1$ and $A\_2$ have analytic extensions. According to [Wikipedia](http://en.wikipedia.org/wiki/Ackermann_function) (which uses a slightly different definition and notation), analytic extensions of $A\_3$ or $A\_k$ for any other $k$ aren't known, but from the language, it isn't clear whether the existence of an extension is itself in question, or whether one simply hasn't been found yet. Also, it doesn't say anything about the diagonal $A(n)$ (unless I missed it).
There are many other obvious sequences that don't seem to have obvious analytic extensions, like the prime-counting function (just to name one!). As far as my knowledge is concerned, this seems more like the rule than the exception. My knowledge here is admittedly very limited, though, so anything at all that you can share will probably teach me something.
| https://mathoverflow.net/users/302 | Which sequences can be extended to analytic functions? (e. g., Ackermann's function) | It's a standard theorem in complex analysis that if $z\_n$ is a sequence that goes to infinity, there is an entire function taking any prescribed values at the $z\_n$. [There is a function $f$ vanishing to order 1 at each $z\_n$](http://en.wikipedia.org/wiki/Weierstrass_factorization#Existence_of_entire_function_with_specified_zeroes) (for $z\_n=n$, you could take $f(z)=\sin \pi z$), and then consider $\sum\_n a\_nf(z)/(f'(z\_n)(z-z\_n))$. This may not converge, but you can tweak it by multiplying each term by something that is 1 at $z\_n$ (eg, $\exp(c\_n(z-z\_n))$ for $c\_n$ chosen appropriately) to make it converge.
(I don't know off the top of my head how to choose the $c\_n$; this is copied from Exercise 1 on page 197 of Ahlfors's Complex Analysis.)
EDIT: It's easy to show that such $c\_n$ exist. If you write $b\_n=a\_n/(z\_n f'(z\_n))$, then for any fixed $z$, the terms of the sum will be approximately $b\_n \exp(c\_n z\_n)$ for $n$ large. You can obviously pick $c\_n$ so that this converges.
| 26 | https://mathoverflow.net/users/75 | 2947 | 1,941 |
https://mathoverflow.net/questions/2905 | 25 | Sometimes people say that the Fukaya category is "not yet defined" in general.
What is meant by such a statement? (If it simplifies things, let's just stick with Fukaya categories of compact symplectic manifolds, not Fukaya-Seidel categories or wrapped Fukaya categories or whatever else might be out there)
What should a "correct" definition of the Fukaya category satisfy? (--- aside from, perhaps, "it makes homological mirror symmetry true")
What are some of the things which make defining the Fukaya category difficult?
What are some cases in which we "know" that we have the "correct" Fukaya category?
| https://mathoverflow.net/users/83 | Is the Fukaya category "defined"? | From what I understand, the most fundamental issue obstructing the definition of the Fukaya category in general is the fact that the boundaries of the relevant moduli spaces typically have codimension-one pieces arising from the bubbling off of pseudoholomorphic discs. In situations where there are no bubbles (for instance, there isn't any bubbling when the Lagrangians are exact, by Stokes' theorem and the fact that the symplectic form would have to integrate positively over the bubble), the A-infinity relations are proven by interpreting the various terms as arising from boundary components of these moduli spaces--hence bubbling contributes additional terms which potentially mess the A-infinity relations up.
Fukaya-Oh-Ohta-Ono (at least in the version that I looked at a few years ago) develop an obstruction theory which assigns a sequence of classes in the homology of a given Lagrangian L (arising by evaluating boundaries of pseudoholomorphic discs) such that the obstruction classes of L all have to vanish in order for L to fit into the Fukaya category. And if the obstruction classes do vanish, one needs to choose a "bounding chain" b for L with the isomorphism type of (L,b) potentially depending on b.
There also some issues with defining the Fukaya category having to do with transversality of the moduli spaces, but my impression is that these issues are technical and can/have been addressed--and that the need to address them is part of what accounts for the length of the [current version](http://www.ams.org/bookstore?fn=20&arg1=amsipseries&ikey=AMSIP-46) of FOOO.
To learn more, you could do worse than spending some time at [Fukaya's webpage](http://www.math.kyoto-u.ac.jp/~fukaya/fukaya.html). In particular, the introduction to the old version of FOOO, which you'll find there, lays out some of the properties that the Fukaya category has and/or is supposed to have.
| 19 | https://mathoverflow.net/users/424 | 2948 | 1,942 |
https://mathoverflow.net/questions/2903 | 9 | I've heard that there are different theories providing knot invariants in form of homologies. My understanding is that if you embed knot in a special way into a space, there is a special homology theory called Floer homology.
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> **Question:** what's the definition and properties of a Floer homology of a knot? How is it related to other knot homology theories?
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| https://mathoverflow.net/users/65 | What is Floer homology of a knot? | I can say something about this for Heegaard Floer homology. Given a 3-manifold Y, you can take a Heegaard splitting, i.e. a decomposition of Y into two genus g handlebodies joined along their boundary. This can be represented by drawing g disjoint curves a1,...,ag and g disjoint curves b1,...,bg on a surface S of genus g; then you attach 1-handles along the ai and 2-handles along the bi, and fill in what's left of the boundary with 0-handles and 3-handles to get Y.
The products Ta=a1x...xag and Tb=b1x...xbg are Lagrangian tori in the symmetric product Symg(S), which has a complex structure induced from S, and applying typical constructions from Lagrangian Floer homology gives you a chain complex CF(Y) whose generators are points in the intersection of these tori and whose differential counts certain holomorphic disks in Symg(S). Miraculously, its homology HF(Y) turns out to be independent of every choice you made along the way. We can also pick a basepoint z in the surface S and identify a hypersurface {z}xSymg-1(S) in Symg(S), and we can count the number nz(u) of times these disks cross that hypersurface: if we only count disks where nz(u)=0, for example, we get the hat version of HF, and otherwise we get more complicated versions.
Given two points z and w on the surface S of any Heegaard splitting we can construct a knot in Y: draw one curve in S-{ai} and another in S-{bi} connecting z and w, and push these slightly into the corresponding handlebodies. In fact, for any knot K in Y there is a Heegaard splitting such that we can construct K in this fashion. But now this extra basepoint w gives a filtration on CF(Y); in the simplest form, if we only count holomorphic disks u with nz(u)=nw(u)=0 we get the invariant $\widehat{HFK}(Y,K)$, and otherwise we get other versions. The fact that this comes from a filtration also gives us a spectral sequence HFK(Y,K) => HF(Y).
This was constructed independently by Ozsvath-Szabo and Rasmussen, and it satisfies several interesting properties. Just to name a few:
* for knots K in S3 it has a bigrading (a,m), and the Euler characteristic $\sum\_m (-1)^m HFK\_m(S^3,K,a)$ is the Alexander polynomial of K;
* there's a skein exact sequence relating HFK for K and various resolutions at a fixed crossing;
* the filtered chain homotopy type of CFK tells you about the Heegaard Floer homology of various surgeries on K;
* the highest a for which HFK\*(S3,K,a) is nonzero is the Seifert genus of the knot;
* If Y-K is irreducible and K is nullhomologous, then HFK(Y,K,g(K)) = Z if and only if K is fibered (proved by Ghiggini for genus 1 and Ni in general, and later by Juhasz as well).
For knots in S3 it is also known how to compute HFK(K) combinatorially: see papers by Manolescu-Ozsvath-Sarkar and Manolescu-Ozsvath-Szabo-Thurston.
The relation to other knot homology theories isn't all that well understood, but there are some results comparing it to Khovanov homology. For example, given a knot K in S3:
* Just as Lee's spectral sequence for Khovanov homology gave a concordance invariant s(K), the spectral sequence from HFK(K) to HF(S3) gives a concordance invariant tau(K), and both of these provide lower bounds on the slice genus of K. (Hedden and Ording showed that these invariants are not equal.)
* There's a spectral sequence from the Khovanov homology of the mirror of K to HF of the branched double cover of K.
* For quasi-alternating knots, both Khovanov homology and HFK are determined entirely by the Jones and Alexander polynomials, respectively, as well as the signature; this can be proven using skein exact sequences for both (Manolescu-Ozsvath).
Anyway, that was long enough that I've probably made several mistakes above and still not been anywhere near rigorous. There's a nice overview that's now several years old (and thus probably missing some of the things I said above) on Zoltan Szabo's website, [<http://www.math.princeton.edu/~szabo/clay.pdf>](http://www.math.princeton.edu/~szabo/clay.pdf), if you want more details.
| 16 | https://mathoverflow.net/users/428 | 2955 | 1,948 |
https://mathoverflow.net/questions/2945 | 5 | If f is a weight 2 newform on $\Gamma\_1(N)$ then there exists an abelian variety Af whose endomorphism algebra is isomorphic to the field generated by the coefficients of f.
I've seen this proven in Shimura's book, but was wondering if anyone knows of a different reference (perhaps one that is a bit more readable...).
Thanks.
| https://mathoverflow.net/users/nan | Modular forms reference | Have a look at Section 6.6 of Diamond and Shurman, *A First Course in Modular Forms*:
As an aside, the theorem states a bit more than you have said: for instance, when the field of Fourier coefficients is $\mathbb{Q}$, you are just asserting the existence of an elliptic curve $E\_{/\mathbb{Q}}$ with $\operatorname{End}\_{\mathbb{Q}}(E) \otimes\_{\mathbb{Z}} \mathbb{Q} = \mathbb{Q}$: every elliptic curve over $\mathbb{Q}$ has this property. You want to require an equality of L-series between the abelian variety and the modular form.
| 4 | https://mathoverflow.net/users/1149 | 2962 | 1,954 |
https://mathoverflow.net/questions/2971 | 13 | I am reading some papers which involve D-modules on a Lie algebra g, which are supported on the nilpotent cone n. They are equivariant for the action of G. (In particular, I consider g=sl\_n).
It was explained to me (the statement, not the proof) that the category of such D-modules is semisimple, and that the simple objects are given by (the intermediate extension) of the constant sheaf sheaf of functions on each g-orbit on n, so they are in bijection with Jordan decompositions with all zeroes, so just partitions of n.
I'm not strong with D-modules (I'm learning!). My question is this: Since g is affine, D(g) is an associative algebra, namely the Weyl algebra on the vector space g. How can I describe the D(g)-module M corresponding to partition \lambda explicitly as a module over D(g)?
| https://mathoverflow.net/users/1040 | D-modules supported on the nilpotent cone | Watch out that there are more simple objects than it looks like at first glance, even for sl\_n. Although the orbits are parameterized by partitions, they can carry nontrivial local systems whose intermediate extensions to the nilpotent cone will be new simple D-modules.
I have heard that the general problem of writing down intermediate extensions explicitly, say by generators and relations, is weirdly difficult. Kari Vilonen did this for isolated singularities in his thesis. For the nilpotent cone I wonder how well Ben's suggestion works: is it a simple matter to pick out the isotypic components of this pushforward D-module, using the bare fact (geometric magic) that it's an intermediate extension of a D-module you know how to decompose?
| 6 | https://mathoverflow.net/users/1048 | 2992 | 1,973 |
https://mathoverflow.net/questions/2991 | 3 | The group scheme G\_a here is the one-dimensional additive group.
| https://mathoverflow.net/users/425 | Over which schemes can there exist non-trivial G_a bundles? | Principal Ga-bundles on a scheme X, in any of the Zariski, etale, or flat topologies, are classified by the coherent cohomology group H^1(X,OX). For a smooth complex projective variety, this is the antiholomorphic component of the de Rham group H^1(X,C), which is a topological invariant. So (in this smooth Kahler setting) the existence of nontrivial Ga-bundles depends only on the topological type of X.
I omitted some underscores for typesetting reasons.
| 5 | https://mathoverflow.net/users/1048 | 2995 | 1,975 |
https://mathoverflow.net/questions/3008 | 15 | What are some number theoretic sequences that you know of that occur as (or are closely related to) the sequence of Fourier coefficients of some sort of automorphic function/form or the sequence of Hecke eigenvalues attached to a Hecke eigenform?
I know many such sequences, but am always looking for more.
Some examples
(1) The sequence a(n) deriving from the traces a(p) of the Frobenius elements in a Galois representation (Langlands reciprocity conjecture)
(2) Number of representations of a natural number as a sum of k squares (theta functions)
(3) The sum of powers of divisor functions (Eisenstein series)
(4) The central critical values of L-functions attached to all quadratic twists of a Hecke eigenform (Kohnen, Waldspurger)
(5) Intersection numbers of certain subvarieties of Hilbert modular surfaces (Hirzebruch-Zagier)
I'll end with a question that is ill-posed but nevertheless very interesting (at least to me personally): why do so many familiar and yet diverse sequences appear in this fashion? Note that many of them have a history of study that precedes the recognition that they are essentially coefficients of automorphic functions/forms.
| https://mathoverflow.net/users/683 | Number theoretic sequences and Hecke eigenvalues | Characters of rational vertex operator algebras tend to yield modular functions. This is due to the space of torus partition functions in a chiral conformal field theory being a complex moduli invariant. The standard example is the monster vertex algebra, whose character is j-744. Other examples come from lattice CFTs (presumably describing a bosonic string propagating in a torus), and have the form of a theta function divided by a power of eta. The characters are never Hecke eigenforms, because of the pole at infinity, but traces arising from higher-weight vectors may be. In some cases, the vertex algebra structure is supposed to arise from geometry of a target space, so this phenomenon may be related to Hirzebruch-Zagier (number 5).
Characters of highest weight representations of affine Kac-Moody algebras yield modular forms. One can reasonably argue that this is a special case of the previous paragraph, since (I think) they come from Wess-Zumino-Witten. The Weyl-Kac character formula for such representations is one way to get Macdonald identities, and the smallest case (trivial rep of affine sl2) yields the Jacobi triple product.
| 6 | https://mathoverflow.net/users/121 | 3012 | 1,987 |
https://mathoverflow.net/questions/2638 | 4 | I am looking for some good references on singularity theory. I'm interested in singularity theory in the context of mirror symmetry, so this means I'm interested in things like Picard-Lefschetz theory, oscillating integrals, Frobenius manifolds (Saito theory). I have looked at the book by Arnold, Gusein-Zade, Varchenko, and I have looked at Seidel's book, but I am wondering if there are any other good, readable, relatively introductory, not-overly-heavy-on-notation books out there. Thanks!
| https://mathoverflow.net/users/83 | Singularity theory references | Some of the topics you are interested in are covered (very nicely) in C.Sabbah, Isomonodromic deformations and Frobenius manifolds. An introduction. A more specialised book is C.Hertling, Frobenius manifolds and moduli spaces for singularities.
| 2 | https://mathoverflow.net/users/1220 | 3028 | 1,998 |
https://mathoverflow.net/questions/3007 | 8 | If I'm given a division algebra D with Z(D)=F, then how can I view Dx as an algebraic group defined over F? I'd like to see first how Dx can be given the structure of a variety defined over F, and then to see how the group law on Dx is defined over F.
| https://mathoverflow.net/users/493 | Division Algebras as Algebraic Groups | Choose an F-basis of D. The multiplication is described by certain quadratic functions, with respect to this basis; D\* is given by the nonvanishing of a polynomial function (the norm).
So the multiplication can be understood as defining an algebraic group structure on the complement of a hypersurface in an affine space.
| 11 | https://mathoverflow.net/users/513 | 3035 | 2,003 |
https://mathoverflow.net/questions/2969 | 9 | In Signal Processing books, a fundamental theorem is that linear time invariant systems can be represented as a convolution with a distribution. Could you give a mathematically rigorous statement of this theorem, or refer a book that includes it?
Edit: For example, would the following be a correct statement?
"Let S' be the space of tempered distributions. If L is a linear operator on S' that commutes with translations, then there exists a distribution h in S' such that Lf = f\*h for all f in S'"
| https://mathoverflow.net/users/1229 | What is a rigorous statement for "linear time-invariant systems can be represented as convolutions"? | I think the result you are looking for is the following:
Let T be a linear continuous and translation invariant operator mapping S into S' (rather than S' into S').
Then there exists a distribution K s.t. Tf = f\*K, for every f in S.
The continuity of T is referred to the usual Frechet topology on S and the weak dual topology on S' (you want f -> to be a continuous linear functional on S for every g in S).
You can find a proof (by Sobolev embedding) on Introduction to Fourier analysis on Euclidean spaces (E. Stein).
From this you can prove analogous results for L^p spaces by embedding (translation invariantly) S in L^p and L^q in S'.
To rephrase everything in the language of multipliers it suffices to remember that the Fourier transform F is a topological isomorphism of S' and that F(f\*K) = F(f)F(K) whenever K is a tempered distribution and f is a Schwartz function.
Than the operator T is a multiplier operator F(Tf) = m F(f) for m = F(K).
| 5 | https://mathoverflow.net/users/1049 | 3054 | 2,016 |
https://mathoverflow.net/questions/3024 | 8 | Can someone verify this for me.. or tell me what reference shows me this... is this true:
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> Let $k$ be a field. Then a field extension $K$ of $k$ is separable over $k$ iff for any field extension $L \supseteq k$ the Jacobson radical of the tensor product $K\otimes\_k L$ is trivial.
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I got this idea by looking at some definitions of separable algebras (which is not my field of research.. but somehow this definition got me intruiged). Anyone knows if this is true and why so? or maybe a reference or two about it?
| https://mathoverflow.net/users/1245 | Characterisation for separable extension of a field | You should take a look at Theorem 3.4 (p 85) of Farb and Dennis' book *Noncommutative Algebra*. The statement is:
Let $L/k$ be a finite extension of fields. Then $K\otimes \_k L$ is semisimple for every field $K\supseteq k$ if and only if $L/k$ is a separable extension.
That the tensor product is semisimple implies that its Jacobson radical vanishes. Conversely, any artinian ring with trivial Jacobson radical is semisimple. Therefore your equivalent formulation of separability is true.
| 6 | https://mathoverflow.net/users/nan | 3062 | 2,021 |
https://mathoverflow.net/questions/2985 | 19 | I would like to understand the relationship between the derived category definition of a right derived functor $Rf$ (which involves an initial natural transformation $n: Qf \rightarrow (Rf)Q$, where $Q$ is the map to the derived category) and the "universal delta functor" definition given in Hartshorne III.1.
I already know that $R^if(A) = H^i(Rf(A))$. What I want to know most is:
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I guess it can be thought of as a natural map from a injective resolution of $f(A)$ to $f$(an injective resolution of $A$), but I'm not sure what is the significance of this... Does anyone know a good reference explaining such things?
| https://mathoverflow.net/users/84526 | Derived functors vs universal delta functors | I haven't checked all the details, but I think the story could go like this. (I have to apologize: it's a bit long.)
(1) Let $F:\mathsf A\rightarrow \mathsf B$ be an additive left exact functor between two abelian categories. Take an injective resolution of an object $A$ in $\mathsf A$:
$$0\rightarrow A \rightarrow I^0 \rightarrow I^1 \rightarrow \cdots $$
Let us call $i: A \rightarrow I^0$ the first morphism. Apply $F$ to this exact sequence:
$$0\rightarrow FA \rightarrow FI^0 \rightarrow FI^1 \rightarrow \cdots $$
Now, the *total* right derived functor of $F$ applied to $A$ (thought as a complex concentrated in degree zero) is the complex
$$\mathbb RF(A) = [ FI^0 \rightarrow FI^1 \rightarrow FI^2 \rightarrow \cdots ]$$
and the *classical* right derived functors of $F$ are its cohomology:
$R^nF(A) = H^n(\mathbb RF(A)) = H^n(FI^)$.
These ${R^nF}\_n$ are a universal cohomological delta-functor and we have a natural transformation of functors
$$qF \Rightarrow (\mathbb RF)q$$
which is essentially
$$Fi: FA \rightarrow \mathbb RF(A)$$
(here we have extended $F$ degree-wise to the category of complexes, and this is the degree zero of the natural transformation, because $\mathbb RF(A)^0 = FI^0$ ).
(2) Now, let $T^n : \mathsf A \rightarrow \mathsf B$ be a cohomological delta-functor and $f^0 : F \Rightarrow T^0$ a natural transformation. We have to extend this $f^0$ to a unique morphism of delta-functors ${ f^n : R^nF \Rightarrow T^n }$.
To do this, observe that, in general, given two right-derivable functors between two, say, model categories $$F,G: \mathsf C \rightarrow \mathsf D$$, and a natural transformation between them $t: F \Rightarrow G $, we have a natural transformation between the total right derived functors $\mathbb Rt : \mathbb RF \Rightarrow \mathbb RG$ because of the universal property of the derived functors:
Indeed, if $f : qF \Rightarrow (\mathbb RF)q$ and $g : qG \Rightarrow (\mathbb RG)q$
are the universal morphisms of the derived functors, then we have a natural transformation
$$gt : F \Rightarrow (\mathbb R G)q$$
and, so, because of the universal property of derived functors, a unique natural transformation $\mathbb R t : \mathbb R F \rightarrow \mathbb R G$ such that $(\mathbb R t)qf = g$.
(3) So, take our $f^0 : F \Rightarrow T^0$ , extend it to a natural transformation between the degree-wise induced functors between complexes. Passing to the derived functors, we obtain
$$\mathbb R f^0 : \mathbb R F \Rightarrow \mathbb R T^0.$$
Taking cohomology, for each $n$, we get
$$H^n(\mathbb R f^0) : H^n (\mathbb R F) \Rightarrow H^n (\mathbb R T^0).$$
But these are the classical right derived functors, so we have natural transformations
$$R^nf : R^n F \Rightarrow R^nT^0$$
and because the classical right derived functors are universal delta-functors, we have unique natural transformations
$$i^n : R^nT^0 \Rightarrow T^n$$
which extend the identity
$$i^0 : R^0T^0 = T^0.$$
The composition
$$i^n \circ R^f : R^F \Rightarrow T^n$$
is, I think, the required morphisms of delta-functors that we need.
| 10 | https://mathoverflow.net/users/1246 | 3065 | 2,023 |
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