parent_url
stringlengths 37
41
| parent_score
stringlengths 1
3
| parent_body
stringlengths 19
30.2k
| parent_user
stringlengths 32
37
| parent_title
stringlengths 15
248
| body
stringlengths 8
29.9k
| score
stringlengths 1
3
| user
stringlengths 32
37
| answer_id
stringlengths 2
6
| __index_level_0__
int64 1
182k
|
|---|---|---|---|---|---|---|---|---|---|
https://mathoverflow.net/questions/3764 | 28 | Let f be a complex-valued function of one real variable, continuous and compactly supported. Can it have a Fourier transform that is not Lebesgue integrable?
| https://mathoverflow.net/users/1386 | Does there exist a continuous function of compact support with Fourier transform outside L^1? | Here's a sketch of what I think is an example of the sort you want. Consider a trapezoidal function Tδ, supported on [-1,1], which is 1 on [-1+δ , 1-δ ] and is defined on the remaining intervals by interpolation in the obvious way. Then as δ tends to zero, the Fourier transform of Tδ is going to tend to infinity in the L1(R)-norm -- I can't remember the details of the proof, but since Tδ is a linear combination of Fourier transforms of Féjer kernels one can probably do a fairly direct computation.
Of course, the supremum norm of each Tδ is always 1. So the idea is to now stack scaled copies of these together, so as to obtain a function on [-1,1] which will be continuous (by uniform convergence) but whose Fourier transform is not integrable because its the limit of things with increasing L1-norm.
To be a little more precise: suppose that for each n we can find δ(n) such that Tδ(n) has a Fourier transform with L1-norm equal to n2 3n.
Put Sm = Σj=1m m-2 Tδ(m) and note that the sequence (Sm) converges uniformly to a continuous function S which is supported on [-1,1]. The Fourier transform of S certainly makes sense as an L2 function. On the other hand, the L1-norm of the Fourier transform of Sm is bounded below by
3m - (3m-1 + ... + 3 + 1) ~ 3m /2
which suggests that the Fourier transform of S ought to have infinite L1-norm -- at the moment lack of sleep prevents me from remembering how to finish this off.
Alternatively, one could argue as follows. Consider the Banach space C of all continuous functions on [-1,1] which vanish at the endpoints, equipped with the supremum norm. If the Fourier transform mapped C into L1, then by an application of the closed graph theorem it would have to do so continuously, and hence boundedly. That means there would exists a constant M >0, such that the Fourier transform of every norm-one function in C has L1-norm at most M. But the functions Tδ show this is impossible.
| 24 | https://mathoverflow.net/users/763 | 3790 | 2,524 |
https://mathoverflow.net/questions/3716 | 25 | What are the possible automorphism groups of a principally polarized abelian variety $(A,\lambda)$ of dimension $g,$ say an abelian surface ($g=2$) over the complex numbers or algebraic closure of a finite field? The fact that the moduli stack $A\_g$ is of finite diagonal (over the integers) implies that the automorphism groups are all finite, but do we know more? Like the size.
When $g=1$ this is given in Silverman I, p.103.
**Edit:** Let me make the question more specific. Let $(A,\lambda)$ be an $\mathbb F\_q$-point of $A\_g$ (i.e. an abelian variety $A$ over $\mathbb F\_q$ of dimension $g$ and a principal polarization $\lambda$). We want to consider its automorphism group (over $\mathbb F\_q$).
Let $\pi:A\_{g,N}\to A\_g$ be the natural projection, where $A\_{g,N}$ is the moduli stack of p.p.a.v. of dimension $g$ with a level $N$ structure (a symplectic isomorphism $H^1(A,Z/N)\to(Z/N)^{2g}$). We always assume $q$ is prime to $N.$ Note that $\pi$ is a $G$-torsor, for $G=GSp(2g,Z/N),$ so it gives a surjective homomorphism $\pi\_1(A\_g)\to G.$ The sheaf $\pi\_\*\mathbb Q\_l$ on $A\_g$ is lisse (even locally constant), corresponding to the representation of $\pi\_1(A\_g)$ obtained from the regular representation $\mathbb Q\_l[G]$ of $G$ and the projection $\pi\_1(A\_g)\to G.$ For any $\mathbb F\_q$-point $x$ of $A\_g,$ the local trace $\text{Tr}(Frob\_x,(\pi\_\*\mathbb Q\_l)\_{\overline{x}})$ is either $|G|$ or 0, depending on $Frob\_x\in\pi\_1(A\_g)$ is mapped to 1 in $G$ or not.
We have isomorphisms $H^i\_c(A\_{g,N},\mathbb Q\_l)=H^i\_c(A\_g,\pi\_\*\mathbb Q\_l).$ By Lefschetz trace formula, applied to both $\mathbb Q\_l$ on $A\_{g,N}$ and $\pi\_\*\mathbb Q\_l$ on $A\_g,$ we have
$$|A\_{g,N}(\mathbb F\_q)|=|G|\sum\_{x\in S} 1/\#Aut(A\_x,\lambda\_x),$$
where $S$ is the subset of $[A\_g(\mathbb F\_q)]$ consisting of points $x$ such that all $N$-torsion points of the abelian variety $A\_x$ are rational over $\mathbb F\_q$ (i.e. $|A\_x[N](\mathbb F\_q)|=N^{2g}$), and $(A\_x,\lambda\_x)$ is the pair corresponding to $x.$ This equation gives some constraints (one for each $N$) that $|Aut(A,\lambda)|$ must satisfy. In particular, when $g=N=2$ and $q=3,$ we have $|A\_{2,2}(\mathbb F\_3)|=10$ and $|G|=720$ (in this case $G$ is the symmetric group $S\_6$), and this becomes a puzzle of solving
$$
1/72 = \sum 1/n\_i,
$$
and the $n\_i$'s satisfy some additional conditions. Any idea on how to solve it? I'm considering the contributions of the two parts in $A\_2,$ one for Jacobians of smooth genus 2 curves and one for Jacobians of stable singular ones $E\_1\times E\_2$. Any suggestion is appreciated.
**Edit**: Maybe it's easier to solve it over $\mathbb F\_5,$ since the (orders of the) automorphism groups of smooth genus 2 curves over finite fields of characteristic 5 is known.
| https://mathoverflow.net/users/370 | What are the automorphism groups of (principally polarized) abelian varieties? | The standard proof of finiteness goes as follows: the polarization defines a positive involution \* on the endomorphism algebra, and so the automorphisms are the elements of End(A) tensor the reals that are in End(A) and satisfy a\*a=1. Thus the set of automorphisms is the intersection of a discrete set and a compact set.
Let phi(n) be the degree of the field you get by adjoining an nth root of 1 to Q. Then certainly there exist abelian varieties of dimension g on which the nth roots of 1 act if phi(n) divides 2g (for any CM-field E there is an abelian variety with complex multiplication by E). However, unlike the elliptic curve case, there are other possibilities.
| 15 | https://mathoverflow.net/users/930 | 3792 | 2,526 |
https://mathoverflow.net/questions/3656 | 78 | Singular homology is usually defined via singular simplices, but Serre in his thesis uses singular cubes, which he claims are better adapted to the study of fibre spaces. This young man (25 years old at the time) seemed to know what he was talking about and has had a not too unsuccessful career since.
So my (quite connected) questions are
1) Why do so few books use this approach ( I only know Massey's) ?
2) What are the pro's and con's of both approaches ?
3) Does it matter ? After all the homology groups obtained are the same.
| https://mathoverflow.net/users/450 | Cubical vs. simplicial singular homology | Others have mentioned the advantages to cubical sets and so I don't want to say much on those; I'll just mention some facts about the other direction. The main disadvantage that comes solely from the point of view of homology is that you have this irritating normalization procedure: you have to take the cubical chains on X and take the quotient by degenerate cubes. (You can do the same for simplicial chains and you get the same answer as the original.) The cubical theory gets niceties that others have stated.
The main advantages of simplices are mostly apparent when you move a little further along into homotopy theory. The homotopy theory of simplicial sets is in some senses simply easier than the cubical analogue. For example, the cartesian product of two simplicial sets has as geometric realization the product of the geometric realizations of its factors. On the other hand, the standard "cubical interval" I, which realizes to [0,1], has a self-product I^2, whose geometric realization has fundamental group ℤ. Simplices in some ways play more nicely with degeneracies than cubes do.
Simplices are also more closely tied to categories via the simplicial nerve functor. For example, there are the simplicial constructions of classifying spaces coming from categories, and these give you nice constructions of group cohomology. (Perhaps I just don't know the cubical analogues.)
There is also the ubiquitous "bar construction" which is unreasonably useful in algebraic topology, and which comes most naturally from a simplicial construction. We use this to resolve modules, show equivalences between algebras over different operads, and more applications that are almost too numerous to name. For example, May used it quite heavily in his proofs that spaces X which are algebras over an E∞-operad have infinitely many deloopings BX, B2X, ...
There has been a lot of work done on cubical sets, however, but I'm not the best authority on it. Try [here](http://ncatlab.org/nlab/show/cubical+set) for a starting point.
| 40 | https://mathoverflow.net/users/360 | 3795 | 2,529 |
https://mathoverflow.net/questions/3798 | 2 | It is easy to see that the totally ordered group $\mathbb{Z}$ (the integers) with the natural order has no non-trivial automorphisms. Is this also true for $\mathbb{Z}^n$ with the lexicographical order?
| https://mathoverflow.net/users/717 | Automorphisms of the totally ordered group $\mathbb{Z}{^n}$ with lexicographical order | Here's a counterexample: on $\mathbb{Z}^2$, $f(x,y)=(x,y+x)$.
More generally, the order-preserving automorphisms of $\mathbb{Z}^n$ are exactly the upper triangular matrices with 1s on the diagonal (this should be easy to see by combining Charles's argument with my example in the case $n=2$, and then the generalization to arbitrary $n$ isn't too hard).
| 5 | https://mathoverflow.net/users/75 | 3800 | 2,531 |
https://mathoverflow.net/questions/3797 | 3 | Why are $n$-fold complete segal spaces or $(\infty, n)$-categories (which I'm unsure of how to distinguish from omega-categories) important for $n \geq 3$? Why are they "badly behaved" for $n \geq 3$? (Lurie refers to them this way in his thesis).
Also, I'm particularly interested to connections between $n$-fold complete segal spaces with regards to a question asked recently about "same" proofs. Is a 2-fold complete segal space sufficient in this particular arena?
(Please tell me if this question is ill-posed, I'm just currently learning category theory.)
| https://mathoverflow.net/users/429 | omega-categories and n-fold complete segal spaces | $n$-fold complete Segal spaces are one *model* for $(\infty,n)$-categories; there are other models. More precisely, they are supposed to be a model for *weak* $(\infty,n)$-categories.
The distinction that I think you are asking about is between *weak* and *strict*. Strict $n$-categories can be easily defined by a recursive definition: a strict $n$-category is just a category enriched over strict $(n-1)$-categories. A strict 1-category is just a plain-old category. Though easy to define, strict $n$-categories don't seem to capture the things people want an $n$-category to capture.
One such feature is that strict $n$-categories don't satisfy the "homotopy hypothesis", which says that an $n$-groupoid ($=n$-category in which all morphisms are in some sense invertible) should model homotopy $n$-types ($=$ spaces whose homotopy groups vanish above dimension $n$).
In fact, this failure only occurs for $n \geq 3$; I believe this is the type of bad behavior Lurie refers to. Another failure of strict $n$-categories happens when you try to talk about higher monoidal structures.
If you haven't already, take a look at the papers by Baez-Dolan on arxiv, which discuss a lot of these issues.
| 10 | https://mathoverflow.net/users/437 | 3805 | 2,534 |
https://mathoverflow.net/questions/735 | 6 | Let $w(x,y)$ be a group word in $x$ and $y$.
Let $x$ and $y$ now vary in $\operatorname{SL}\_n(K)$, where $K$ is a field. (Assume, if you wish, that $K$ is an algebraically complete field of characteristic bigger than a constant.)
I would like to know for which words $w$ the map
$$y \mapsto w(x,y)$$
isn't surjective (or even dominant — that is, "almost surjective") for $x$ generic.
It is clear, for example, that the map is surjective for $w(x,y)=xy$, and that it isn't
surjective for $w(x,y)= y x y^{-1}$, or for $w(x,y) = y x^n y^{-1}$, $n$ an integer:
all elements of the image of $y \mapsto y x^n y^{-1}$ lie in the same conjugacy class.
A moment's thought (thanks, Philipp!) shows that $w(x,y) = x y x^n y^{-1}$
isn't surjective either: its image is just $x\* \mathrm{im}(y\mapsto y x^n y^{-1})$,
and, as we just said, $y\mapsto y x^n y^{-1}$ isn't surjective.
I would like to know if the only words $w$ for which the map
isn't surjective for $x$ generic are the $w$'s of the form
$$w(x,y) = x^a v(x,y) x^b (v(x,y))^{-1} x^c,$$
where $v$ is some word and $a,b,c$ are some integers. (This seems to me a sensible guess, though I would actually be quite glad if it weren't true.)
---
\*Edit:\* The discussion has now moved to this subsequent question:
[Surjective maps given by words, redux](https://mathoverflow.net/questions/2082/surjective-maps-given-by-words-redux)
| https://mathoverflow.net/users/398 | When is a map given by a word surjective? | The discussion has now moved to
[Surjective maps given by words, redux](https://mathoverflow.net/questions/2082/surjective-maps-given-by-words-redux)
| 0 | https://mathoverflow.net/users/398 | 3807 | 2,536 |
https://mathoverflow.net/questions/3819 | 71 | Complex analytic functions show rigid behavior while real-valued smooth functions are flexible. Why is this the case?
| https://mathoverflow.net/users/1354 | Why do functions in complex analysis behave so well? (as opposed to functions in real analysis) | Well, real-valued analytic functions are just as rigid as their complex-valued counterparts. The true question is why complex smooth (or complex differentiable) functions are automatically complex analytic, whilst real smooth (or real differentiable) functions need not be real analytic.
As Qiaochu says, one answer is elliptic regularity: complex differentiable functions obey a non-trivial equation (the Cauchy-Riemann equation) which implies a integral representation (the Cauchy integral formula) which then implies analyticity (Taylor expansion of the Cauchy kernel); the ellipticity of the Cauchy-Riemann equation is what gives the analyticity of its fundamental solution, the Cauchy kernel. Real differentiable functions obey no such equation.
Another approach is via Cauchy's theorem. In both the real and complex setting, differentiability implies that the integral over a closed (or more precisely, exact) contour is zero. But in the real case this conclusion has trivial content because all closed contours are degenerate in one (topological) dimension. In the complex case we have non-trivial closed contours, and this makes all the difference.
EDIT: Actually, the above two answers are basically equivalent; the latter is basically the integral form of the former (Morera's theorem). Also, to be truly nitpicky, "differentiable" should be "continuously differentiable" in the above discussion.
| 103 | https://mathoverflow.net/users/766 | 3830 | 2,549 |
https://mathoverflow.net/questions/3841 | 13 | Taking tori in symmetric products and "miraculously" proving that the Floer homology is independent of choices always seemed, well, miraculous. Some time ago Max Lipyanski explained to me the origins of this construction from gauge theory on surfaces, a la Atiyah-Floer conjecture, which I have then forgotten. What is the origin of Heegard Floer?
| https://mathoverflow.net/users/375 | The "miracle" of Heegard Floer. | I think the crude answer is that there is (or maybe just should be) an extended 4 dimensional TQFT that assigns the Fukaya category of a symmetric product to a surface, and the usual Heegard-Floer Lagrangian to a 3 manifold. So, the usual definition of Heegard-Floer is the gluing formula for a Heegard splitting, and invariance is no miracle at all.
| 9 | https://mathoverflow.net/users/66 | 3843 | 2,559 |
https://mathoverflow.net/questions/3849 | 2 | Suppose G is a Lie group (or algebraic group) acting on a manifold (or scheme) X, and H⊆G is a subgroup. Let x,y∈X be points such that x is in the closure of the orbit H⋅y (but not in H⋅y itself). Then obviously x is in the closure of G⋅y, but can it happen that x is actually in the orbit G⋅y (not just in the closure)?
**Background:** I got stuck on this point when trying to understand the very last line of the proof of Theorem 0.3.1 of Kapranov's [Chow quotients of Grassmannian I](http://arxiv.org/abs/alg-geom/9210002), which states that every irreducible component of an algebraic cycle corresponding to a point in the Chow quotient X//G is the closure of a single G-orbit. In this case, H⊆G is a torus and X is a smooth projective variety.
| https://mathoverflow.net/users/1 | Do subgroups respect the orbit-closure relation? | Sure, that can happen. G = PGL\_2, H is the torus, X is the Riemann sphere, x is the north pole, y is some point other than the two poles.
I've read that Kapranov paper recently, I'll see if I can find something more useful to say.
| 3 | https://mathoverflow.net/users/297 | 3853 | 2,564 |
https://mathoverflow.net/questions/3857 | 5 | Some of my friends and I were trying to discover a universal mapping property that characterizes the integers $\mathbb{Z}$ in the category of groups without referring to the underlying sets (So it is a no no to say it is the free group on a one element set). One of the big uses of $\mathbb{Z}$ is that it is a separator, i.e. for any two distinct pair of parallel arrows $f,g:A \rightarrow B$, there is at least one morphism $x:\mathbb{Z} \rightarrow A$ such that $f\circ x \neq g\circ x$. Unfortunately, any free group satisfies this property. I have two questions:
What are the separators in the category of groups (I think they will be just the free groups, but I have not proved this yet). Given this I think I can write down a universal property for Z which stays inside the category of groups.
Whether or not the above claim is correct, does anyone have a UMP that does the job?
| https://mathoverflow.net/users/1106 | Separators in the Category of Groups | According to your definition, the separators will be exactly those groups $G$ with a surjection to $\mathbb{Z}$. One direction: if $f(x)\neq g(x)$, then take the composition $G \twoheadrightarrow \mathbb{Z} \rightarrow A$ where the latter map sends the generator of $\mathbb{Z}$ to $x$. For the other direction, to distinguish the maps $f,g: \mathbb{Z} \rightarrow \mathbb{Z}$ defined by $f(x)=x$ and $g(x)=2x$, your separator must surject to $\mathbb{Z}$.
This is a huge class of groups which has no particularly nice description beyond the definition, as far as I know.
| 6 | https://mathoverflow.net/users/250 | 3861 | 2,569 |
https://mathoverflow.net/questions/3858 | 48 | Can someone indicate me a good introductory text on geometric group theory?
| https://mathoverflow.net/users/1049 | Introductory text on geometric group theory? | de la Harpe's book is quite nice and has an amazing bibliography, but it doesn't really prove any deep theorems (though it certainly discusses them!). Some other sources.
1) Bridson and Haefliger's book "Metric Spaces of Non-Positive Curvature". Very easy to read and covers a lot of ground.
2) Ghys and de la Harpe's book on hyperbolic groups. Another classic, but in French. If you look around the web, you can find English translations.
3) Cannon's survey "Geometric Group Theory" in the Handbook of Geometric Topology is very nice.
4) Bowditch's survey "A course on geometric group theory" is also very nice.
5) Bridson has written two beautiful surveys entitled "Non-Positive Curvature in Group Theory" and "The Geometry of the Word Problem". The latter was one of the first things I read in any depth.
6) Geoghegan's "Topological Methods in Group Theory" is very nice, with a more topological approach.
7) Mike Davis's "The Geometry and Topology of Coxeter Groups" is a bit specific, but covers a lot of important material in a nice way.
8) John Meier's book "Groups, Graphs and Trees: An Introduction to the Geometry of Infinite Groups" is well-written and pretty gentle.
| 47 | https://mathoverflow.net/users/317 | 3868 | 2,572 |
https://mathoverflow.net/questions/3859 | 3 | I am wondering how *B-fields*, which are basic objects in Generalized Geometry, relate to the B-fields of [Ben's question and the answers to it](https://mathoverflow.net/questions/1726/how-should-i-think-about-b-fields).
In Generalized Geometry, the B-field is a (1,1)-form, and when it is closed it preserves the generalized complex structure. Furthermore, the Dolbeault cohomology class of the B-field acts on the equivalence class of a "generalized holomorphic bundle" (the natural generalization of holomorphic bundles to generalized complex manifolds). In what ways are these B-fields connected to the physics B-fields discussed in Ben's post?
| https://mathoverflow.net/users/1177 | Prevalence of B-fields | The short answer is that both B-fields are the same object!
The way the B-field comes to us from string theory it doesn't come alone, but comes together (among other fields) with the Riemannian metric. Due to the way both originate in the string, they are interrelated by what is called [T-duality](http://ncatlab.org/nlab/show/T-duality) that mixes them and shows that both fields have to be regarded as two aspects of one and the same unified entity.
Physicists had understood various aspects of how these two fields unify, when Nigel Hitchin came along and noticed that there is a nice and useful mathemtical formalization of what is going on. This is the origin of [generalized complex geometry](http://ncatlab.org/nlab/show/generalized+complex+geometry).
But some aspects of the picture are still mising. For instance it is well-know that in full beauty [the B-field is a gerbe with connection](http://ncatlab.org/nlab/show/Kalb-Ramond+field) . Last I checked, this is represented in generalized complex geometry only *rationally* , meaning that the integral degree 3 class of this gerbe is seen only in its image in deRham cohomology.
This has to do with the fact that generalized complex geometry is really a theory of [Couran algebroid](http://ncatlab.org/nlab/show/Courant+algebroid)s (certain [Lie 2-algebroids](http://ncatlab.org/nlab/show/Lie+infinity-algebroid)) and it is only their integration that knows about the full Lie 2-groupoids that yield the gerbe.
(I think I know the full story, but it is not written up yet.)
| 3 | https://mathoverflow.net/users/381 | 3872 | 2,574 |
https://mathoverflow.net/questions/3847 | 11 | The Cartan structure equations for a connection and various associated 1-forms can be checked in a straightforward algebraic manner. But is there a geometric or global significance to the equations- can one visualize the proof?
| https://mathoverflow.net/users/344 | What is the geometric significance of Cartan's structure equations? | There is a nice grand story behind all this. I don't know if you like thinking that way, but things do clarify when one looks at it from a more general perspective of [oo-Lie algebroid valued differential forms](http://ncatlab.org/schreiber/show/%E2%88%9E-Lie+algebroid+valued+differential+forms) with [curvature](http://ncatlab.org/schreiber/show/curvature+of+%E2%88%9E-Lie+algebroid+valued+differential+forms).
I am still working on these entries, trying to expose some of my work with Jim Stasheff and Hisham Sati. If you bug me with questions, there is a good chance that I'll improve the exposition taylor-made for your needs.
| 6 | https://mathoverflow.net/users/381 | 3874 | 2,576 |
https://mathoverflow.net/questions/3871 | 25 | For a compact space $K$, the maximal ideals in the ring $C(K)$ of continuous real-valued functions on $K$ are easily identified with the points of $K$ (a point defines the maximal ideal of functions vanishing at that point).
Now take $K=\mathbb{R}$. Is there a useful characterization of the set of maximal ideals of $C(\mathbb{R})$, the ring of continuous functions on $\mathbb{R}$? Note that I'm not imposing any boundedness conditions at infinity (if one does, I think the answer has to do with the Stone–Čech compactification of $\mathbb{R}$ — but I can't say I'm totally clear on that part either). Is this ring too large to allow a reasonable description of its maximal ideals?
| https://mathoverflow.net/users/25 | Maximal ideals in the ring of continuous real-valued functions on ℝ | Peter Johnstone's book [Stone Spaces](https://books.google.com/books?id=66Njdgsk3ukC&lpg=PR7&pg=PA144#v=onepage&q=&f=false) (p. 144) proves that for any X, maximal ideals in $C(X)$ are the same as maximal ideals in $C\_b(X)$ (bounded functions), i.e. the Stone-Cech compactification $\beta X$. Indeed, if I is a maximal ideal, let Z(I) be the set of all zero sets of elements of I; this is a filter on the lattice of all closed sets that are zero sets of functions. Then I is contained in J(Z(I)), the set of functions whose zero sets are in Z(I), so by maximality they are equal. But also, by maximality, Z(I) must be a maximal filter on the lattice of zero sets, and we get a bijection between maximal filters of zero sets and maximal ideals in $C(X)$. Now the exact same discussion applies to $C\_b(X)$ to give a bijection between maximal filters of zero sets and maximal ideals of $C(X)$ (since the possible zero sets of bounded functions are the same as the possible zero sets of all functions). But the maximal ideals of $C\_b(X)$ are just $\beta X$.
The difference between $C\_b(X)$ and $C(X)$ is that for $C\_b(X)$, the residue fields for all of these maximal ideals are just $C$, while for $C(X)$ you can get more exotic things. Indeed, if a maximal ideal in $C(X)$ has residue field $C$, then every function on X must automatically extend continuously to the corresponding point of $\beta X$. This can actually happen for noncompact X, e.g. the ordinal $\omega\_1$.
Section IV.3 of Johnstone's book has a pretty thorough discussion of this stuff if you want more details.
| 32 | https://mathoverflow.net/users/75 | 3876 | 2,577 |
https://mathoverflow.net/questions/3863 | 2 | Some functions are not represented by their power series even when they are continuous and have all the necessary derivatives. What's the best characterization of these functions? Explanations at any level are welcome.
| https://mathoverflow.net/users/812 | What functions are not represented by their power series? | A smooth function is characterized as being analytic if its derivatives
on any closed interval have a certain growth rate. See "Alternate Characterizations"
in the
[Wikipedia article on analytic functions.](http://en.wikipedia.org/wiki/Analytic_function)
| 6 | https://mathoverflow.net/users/1345 | 3881 | 2,581 |
https://mathoverflow.net/questions/2883 | 6 | D. Orlov proved that any equivalence of bounded derived categories *F:Db(X) -> Db(Y)* is a Fourier-Mukai transform, when *X* and *Y* are smooth projective varieties. Is there any example of such equivalence, which is not a Fourier-Mukai transform (it is not an integral transform)?
| https://mathoverflow.net/users/1220 | Equivalence of derived categories which is not Fourier-Mukai | Schlichting gave an example of two categories of singularities which are derived equivalent but whose K-groups are not isomorphic. Dugger and Shipley (arXiv:0710.3070) expanded on this example and noted that it gives two dga's which are derived equivalent but not by an integral transform.
Otherwise, Lunts and Orlov's [results](https://arxiv.org/abs/0908.4187) on uniqueness of enhancements give a large class of triangulated categories for which one might lift exact functors to dg-functors and apply Toen's result.
| 9 | https://mathoverflow.net/users/1404 | 3891 | 2,589 |
https://mathoverflow.net/questions/3877 | 1 | Maybe I'm being dense here, but can someone give me a subset of the set of all languages which is uncountable and the subset is easy to describe? (Some natural subset -- not like "take the set of all languages and remove a few.")
For instance, I thought of the set of recursive or recursively enumerable languages, but these are countable. Perhaps some set in the arithmetic hierarchy?
This is probably a very easy question, and I'm just being silly.
| https://mathoverflow.net/users/1042 | A subset of all languages which is uncountable? | One way to take the question is to ask for a set of languages, for instance a complexity class, which is uncountable "for a good reason". In other words, that people study the class for some substantially different reason, and it's clearly convenient for it to be uncountable.
Probably the most common example is the class P/poly. This can be defined as polynomial-time computations with polynomial-length advice strings. (An advice string is any extra information that depends on the length of the input, but not on the specific value of the input.) By a famous structure theorem, it is also computations performed by polynomial-sized circuits on n input bits, without the requirement that the circuits can be built quickly by a Turing machine. This is clearly not a countable set of langauges, because anything recursive or non-recursive can be done with the input length.
On the other hand, it is a very useful class and construction. A stronger version of the P vs NP problem, inspired by circuit formulations of P vs NP, conjectures that NP is not contained in P/poly. And it is a theorem that BPP (randomized polynomial time) is contained in P/poly.
| 5 | https://mathoverflow.net/users/1450 | 3903 | 2,596 |
https://mathoverflow.net/questions/3913 | 2 | Let g,h be positive integers. Let E be an elliptic curve, C be a genus h curve, and D be a genus g-h-1 curve. Let c,d,e be points on (resp.) C,D, and E.
Let f:CC --> E-e be the family whose fiber over a point e' is the curve obtained by glueing C to E together at the points c and e and D to E at the points d and e'.
>
> Question: what is the pushforward f\* ωf
> ?
>
>
>
It should be trivial, and David Speyer's answer to my question [here](https://mathoverflow.net/questions/2107/question-about-a-family-of-semistable-curves/2145#2145) should answer this question, but I (and a few others I asked earlier) couldn't get it to work.
| https://mathoverflow.net/users/2 | Triviality of the Hodge bundle for a special family of semistable curves | Under (the extension of) Torrelli this curves maps to one point in Ag. On the other hand the hodge class on Mg minus D0 is a pullback (under the extension of Torelli) of the hodge class on Ag.
| 2 | https://mathoverflow.net/users/404 | 3914 | 2,602 |
https://mathoverflow.net/questions/3911 | 13 | SGA 3, expose 12, remark 1.6 says that one can easily construct a group scheme over a discrete valuation ring with generic fiber Gm and special fiber Ga.
What is such an example?
| https://mathoverflow.net/users/2 | Constructing a degeneration (as a group scheme) of G_m to G_a | R[x,y]/(y^2-x^3-\pi x^2) gives the coordinate ring of a bad cubic curve, where \pi is a uniformizer in R. Remove the origin (which is the one singular point), and projectivize the curve by adding a point at infinity, so your group has an identity.
The generic fiber (treating \pi as a unit) is then the smooth part of a nodal cubic, yielding **G**m, and the special fiber (setting \pi to zero) is the smooth part of a cuspdal cubic, yielding **G**a.
| 15 | https://mathoverflow.net/users/121 | 3918 | 2,606 |
https://mathoverflow.net/questions/3864 | 6 | I know the interior point method works both for Linear Programming (LP) and semidefinite programming (SDP). My question is, can the other popular method for solving LP, namely the simplex method, be extended to SDP? If not, what is the barrier?
| https://mathoverflow.net/users/1401 | Simplex method for SDP? | *[the first part of this answer is similar to Dinakar Muthiah's]*
When optimizing a linear function on a convex set, it can always be assumed that the optimal solution lies on an extreme point of the feasible region (if there are several optimal solutions, at least one is an extreme point).
In the case of linear programming, these extreme points are vertices of a polyhedron, with the nice property that
* there are a finite number of vertices
* every vertex admits a simple algebraic description (this is the notion of *basis*, which is essentially a list of active inequalities)
However, for semidefinite programming, the feasible region, altough convex, typically admits an infinite number of extreme points, for which there is no clear equivalent to the concept of basis.
Note that simplex-type methods (also called active-set methods) can be generalized to quadratic programming (minimization of a convex quadratic function over a polyhedron). On the other hand, I am not aware of any such generalization for quadratically constrained quadratic programming (QCQP, i.e. quadratic programming with convex quadratic constraints) or second-order cone programming (a slight extension of QCQP), two problem classes whose instances can be posed as semidefinite programming problems.
| 3 | https://mathoverflow.net/users/1184 | 3923 | 2,609 |
https://mathoverflow.net/questions/3912 | 5 | Let X,Y to be mappings from the sample space Ω to R and suppose Y is measurable with respect to σ(X), the smallest σ-field that makes X measurable.
Does it follow that there exists some Borel-measurable function f: R → R such that
Y=f(X)?
| https://mathoverflow.net/users/1407 | question on sigma-fields | The answer is yes. The proof is quite standard.
**1.** If Y=1\_A, where A is in \simga(X), then by the definition of \sigma(X) there exists a Borel set B such that A = X^{-1}(B) and therefore
Y(\omega) = 1\_A(\omega) = 1\_B(X(\omega)) = f(X(\omega)),
where we put f:= 1\_B (of course f is now a Borel function).
**2.** If now Y is a simple r.v. i.e. it can be written in form Y = \sum\_{i=1}^n c\_i 1\_{A\_i}, where A\_i are sets in \sigma(X), then using the previous point we can find Borel functions f\_i such that 1\_{A\_i} = f\_i(X) and obviously in this case f = \sum\_{i=1}^n c\_i f\_i.
**3.** Finally, any r.v. Y measurable w.r. to \sigma(X) can be approximated by a sequence of simple r.v. Y\_n measurable w.r. to \sigma(X) i.e. Y\_n -> Y almost surely. By the previous point there exist f\_n such that Y\_n = f\_n(X). Now we can define
f(x) = \lim\_n f\_n(x) if the limit exists and put f(x)=0 otherwise. It is easy to check that f is a Borel function (basically it is a limit of Borel functions), and that Y = f(X).
| 6 | https://mathoverflow.net/users/1302 | 3925 | 2,611 |
https://mathoverflow.net/questions/3920 | 20 | When constructing proofs using [natural deduction](http://en.wikipedia.org/wiki/Natural_deduction) what does it mean to say that an assumption or premise is *discharged*? In what circumstances would I want to, or need to, use such a mechanism?
The reason I'm asking this question is that many texts on logic use this term as understood by the reader and don't take the time to adequately explain the technical sense in which they are using it.
| https://mathoverflow.net/users/107 | What does it mean to 'discharge assumptions or premises'? | As I understand it, to discharging a premise or assumption is the opposite of introducing it: you absorb it (for example) into the antecedent of an implication --- this means that it is no longer an assumption. A trivial example:
P 1. Assume P
\_\_
P 2. From 1
\_\_
P->P 3. Discharging 1
Thus I have concluded that P->P without any assumptions (iow |- P->P). If we didn't discharge the assumption, we would have P|-P
| 8 | https://mathoverflow.net/users/1222 | 3930 | 2,613 |
https://mathoverflow.net/questions/3280 | 15 | A Leibniz algebra L may be thought of as a noncommutative generalisation of a Lie algebra. One drops the requirement that the bracket be alternating and substitutes the Jacobi identity for the Leibniz identity
$$ [x,[y,z]] = [[x,y],z] + [y,[x,z]] $$ for all $x,y,z \in L$
**Remark**. What is being defined above is a **left** Leibniz algebra. There is also the notion of a right Leibniz algebra where the Leibniz identity now says that it is the right multiplication $[-,x]$ which is a derivation, instead of the left multiplication $[x,-]$ as in the equation above.
Since the bracket is no longer alternating, left- and right-multiplications are no longer related simply by a sign as in the case of Lie algebras, and this means that representations in general admit two actions of $L$: one on the left and one on the right, satisfying some identities which are explained, for example, in a paper of Loday (who seems to have introduced the concept) and Pirashvili (*Math. Ann.* **296** (1993) pp. 139–158). In that paper they also define the universal enveloping algebra $U(L)$ of a Leibniz algebra $L$ and show that the there is a categorical equivalence between representations of $L$ and left modules over $U(L)$. (Right modules of $U(L)$ correspond to the notion of corepresentation.)
Also notice that in that paper they work with right Leibniz algebras, so everything there is the mirror image to what I'm saying here. One difference with the case of a Lie algebra is that $U(L)$ is a quotient of the tensor algebra of $L\oplus L$, to take into account the two actions of $L$ on a representation.
My question is **whether there is a Hopf algebra structure on $U(L)$.**
My interest in this question is that in some recent work on the deformation theory of n-Leibniz algebras, I studied cohomology with values in a representation $M $of a Leibniz algebra L and also with values on $End(M)$. The action of $L$ on $End(M)$ follows from the formalism and one can check that it is indeed a representation, but it does not follow in any obvious way from the action of $L$ on $M$. In Lie theory, we are used to the fact that if $M$ is a (finite-dimensional) representation of a Lie algebra $G$, then we have an isomorphism $End(M) \cong M \otimes M^\*$ as representations of $G$, where to determine the action of $G$ on $M \otimes M^\*$ we use the Hopf algebra structure on $U(G)$. Hence my question.
**EDIT**: I am adding more details about $U(L)$, as requested in the comment below by Theo Johnson-Freyd.
To motivate it, let us first define a *representation* $M$ of a (left) Leibniz algebra $L$ to be a vector space admitting two actions of $L$:
$$
(x,m) \mapsto [x,m] \textrm{ and } (m,x) \mapsto [m,x], \forall m \in M \textrm{ and } x \in L
$$
satisfying three identities, which are obtained from the Leibniz identity above by replacing $x,y,z$ in turn by $m$; that is,
$$
[m,[x,y]] = [[m,x],y] + [x,[m,y]] \\
[x,[m,y]] = [[x,m],y] + [m,[x,y]] \\
[x,[y,m]] = [[x,y],m] + [y,[x,m]]
$$
To define $U(L)$ we start with the tensor algebra $T(L\oplus L)$ of $L \oplus L$. Let $l\_x = (x,0)$ and $r\_x = (0,x) \in L \oplus L$. Then $U(L)$ is the quotient of $T(L+L)$ by the two-sided ideal generated by the following elements (which can be read off from the conditions defining a representation):
$$
r\_{[x,y]} - r\_y r\_x - l\_x r\_y \\
l\_x r\_y - r\_y l\_x - r\_{[x,y]} \\
l\_x l\_y - l\_y l\_x - l\_{[x,y]}
$$
for all $x, y \in L$, and where I have omitted the $\otimes$'s.
Notice that adding the first two, we can substitute one of them by the simpler
$$
r\_y (l\_x + r\_x) = 0
$$
I don't know what the coalgebra structure is, though. That's part of the original question.
| https://mathoverflow.net/users/394 | Hopf algebra structure on the universal enveloping algebra of a Leibniz algebra? | Do you know the paper [of Loday and Pirashvili](http://www-irma.u-strasbg.fr/~loday/PAPERS/98LodayPira%28linmaps%29.pdf)? They discuss what, in their opinion, should replace the notion of a Hopf algebra in Leibniz setting, "Hopf algebras in the category of linear maps".
| 10 | https://mathoverflow.net/users/1306 | 3934 | 2,617 |
https://mathoverflow.net/questions/3927 | 19 | I really should know the answer to this, but I don't, so I'll ask here.
A *Q-curve* is an elliptic curve E over Q-bar which is isogenous to all its Galois conjugates. A Q-curve is *modular* if it's isogenous (over Q-bar) to some factor of the Jacobian of X\_1(N) for some N>=1 (here X\_1(N) is the compact modular curve over Q-bar).
Has current machinery proved the well-known conjecture that all Q-curves are modular yet?
Remark: I know there are many partial results. What I'm trying to establish is whether things like Khare-Wintenberger plus best-known modularity lifting theorems are strong enough to give the full conjecture yet, or whether we're still waiting.
| https://mathoverflow.net/users/1384 | Are Q-curves now known to be modular? | Yes, this is a consequence of Serre's conjecture. The canonical reference is probably Corollary 6.2 of Ribet's paper on Q-curves:
<http://math.berkeley.edu/~ribet/Articles/korea.pdf>
| 18 | https://mathoverflow.net/users/nan | 3936 | 2,619 |
https://mathoverflow.net/questions/3928 | 2 | Let X be a continuous stochastic process. I know that (t>s)
P(|X(t) - X(s)|>δ) < |t-s|/δ
Is it possible to say anything (e.g. estimate the decay of the tail) about
Y=sup\_{s \in [0,1]} |X(s)|?
I suspect that the answer is no (it would be an easy question if we have |t-s|^{1+a}, for any a>0).
I wonder what are **"standard" methods of estimating the suprema of continuous stochastic processes**.
So far in my problem I tried to use the Garsia-Rodemich-Rumsey inequality.
| https://mathoverflow.net/users/1302 | Suprema of stochastic processes | The answer is, indeed, "No" because there is an unbounded with probability $1$ stochastic process that satisfies the given inequality, namely, $X(t)=0.1\log|t-w|$ where $w$ is equidistributed on $[0,1]$. Truncating it at high level $L$, we get a continuous process such that $E|X(t)|$ is uniformly bounded but the supremum is identically $L$. Taking a suitable mixture of such truncations, we see that the tails may decay arbitrarily slowly.
There is essentially only one universal method to gets bounds for the supremum from the bounds for the increments, which is to consider $\delta\_ k$-nets with diminishing $\delta\_ k>0$ and bound the supremum by the convergent series of suprema taken over finite sets (differences between points from 2 successive nets). Clever choice of the nets may be crucial for the success but not in this case. Whatever you can get from the standard dyadic nets here is the best you can say.
I'm also tempted to ask whether you, indeed, need the estimate for the supremum (which cannot be made without extra assumptions) rather than for some $L^p$ norm.
| 5 | https://mathoverflow.net/users/1131 | 3940 | 2,622 |
https://mathoverflow.net/questions/2888 | 15 | It's not hard to count the number of permutations in a given conjugacy class of Sn. In particular, the number of permutations in Sn whose cycle decomposition has ci i-cycles is n!/(Πi=1n ci!ici). It's also not too hard to see that this is maximized for the conjugacy class that leaves one element fixed and permutes the others in an (n-1)-cycle, and that this is *strictly* the maximum when n ≥ 3.
What I'm looking for is an "injective proof" of this fact. Namely, a set of injections from the other conjugacy classes into the set of (n-1)-cycles. Ideally you should be able to define a single nice function from Sn into these cycles, which is injective but not surjective (for n ≥ 3) when restricted to every other conjugacy class.
| https://mathoverflow.net/users/1060 | Injective proof about sizes of conjugacy classes in S_n | For any cycle decomposition, we can uniquely order the cycles from smallest length to largest length, breaking ties between cycles of the same length in some fixed arbitrary way (say by maximal elements). Let us do this for concreteness.
Suppose there was an (injective) way to "join" and A-cycle and a B-cycle together to form an A+B cycle whenever |A| and |B| are > 1. Then, given any cycle decomposition as above, one can start bubbling the two largest cycles together to eventually form a single cycle of some length m.
If m = n - 1, we are done.
If m = n, write S = (1............) and then omit the last term.
If m = 1 the problem is very easy.
If n - 1 > m > 2, there is an injective map from m-cycles to elements whose cycle decomposition is a product of an m-cycle with a 2-cycle. For concreteness, one can add the 2-cycle with the two lowest missing entries. Now bubble again to form an m+2 cycle.
If m = 2, and n is at least 5, bubble with a 3-cycle.
If m = 2 and n = 4 (the last case), form whatever bijection you like between the two sets of 6 elements.
The key point is therefore to find a way to bubble an A-cycle and a B-cycle when |A|,|B| > 1.
We do this as follows.
Amongst the entries of A and B, there is a unique smallest integer, call it X.
Case 1. If X lies in A, Let Z denote the *largest* element of B.
Then one can (uniquely) write A = (X.....) and B = (.....Y,Z), where Y < Z. Then consider the A+B cycle obtained by concatenating A and B in this form, i.e. (X,......,Y,
Z).
Case 2. If X lies in B, let Z denote the *smallest* element of A.
Then one can (uniquely) write B = (X....) and A = (.....,Y,Z), where now Y > Z.
Then consider the A+B cycle (X,.....,Y,Z).
Given an A+B cycle, one can uniquely write it in the form (X,....,Y,Z), where X is the smallest entry. Then one can break it up into an A-cycle and B-cycle depending on whether Y < Z or Y > Z.
---
Since this was apparently a little confusing, suppose that the cycle lengths of S are
a`_`1 <= a`_`2 <= a`_`3 <= ..... <= a`_`r.
Here I omit the 1-cycle lengths, so a`_`1 > 1, and sum a`_`r = m for some m possibly less than n. Then the cycle lengths of the steps in the algorithm will have lengths:
(a`_`1, ...., a`_`(r-1),a`_`r),
(a`_`1, ...., a`_`(r-2),a`_`(r-1) + a`_`r),
(a`_`1, ...., a`_`(r-3),a`_`(r-2) + a`_`(r-1) + a`_`r),
....
(a`_`1 + a`_`\_2 + ... + a`_`r) = (m)
(2,m)
(m+2)
(2,m+2)
(m+4)
...
(n-1 or n, depending on m mod 2),
then n-1.
(if m = 2 and n is at least 5, then instead it should go
(2) --> (2,3) --> (5) --> (2,5) --> (7) --> (2,7) --> (9) ...,etc.
| 5 | https://mathoverflow.net/users/nan | 3952 | 2,631 |
https://mathoverflow.net/questions/3951 | 52 | I always have trouble memorizing theorems. Does anybody have any good tips?
| https://mathoverflow.net/users/812 | Memorizing theorems | As far as possible, you should turn yourself into the kind of person who does not have to remember the theorem in question. To get to that stage, the best way I know is simply to attempt to prove the theorem yourself. If you've tried sufficiently hard at that and got stuck, then have a quick look at the proof -- just enough to find out what the point is that you are missing. That should give you an Aha! feeling that will make the step far easier to remember in the future than if you had just passively read it.
| 145 | https://mathoverflow.net/users/1459 | 3957 | 2,636 |
https://mathoverflow.net/questions/3971 | 12 | Let $E/\mathbf{Q}\_p$ be an elliptic curve having split multiplicative reduction. Then Tate uniformization gives a surjective homomorphism of $p$-adic analytic groups $G\_m \to E$, with infinite cyclic kernel. Is there an analogue of this fact for $E$ having nonsplit multiplicative reduction, perhaps replacing Gm with a nonsplit torus? E.g., can one uniformize $E$ over the quadratic extension where the reduction splits, and then somehow descend?
(My intuition was as follows. Take $E/\mathbf{Q}\_p$ with nonsplit multiplicative reduction, and let $K/\mathbf{Q}\_p$ be quadratic so that $E$ becomes split semistable over $K$, and let $E'$ be the $K$-twist of $E$ (which has split multiplicative reduction). Then one has a short exact sequence
$$0 \to Z \to \mathbf{G}\_m \to E' \to 0$$
(where $Z$ is the constant analytic group of integers). Extending scalars to $K$ then applying Weil restriction of scalars, we get
$$0 \to X \to T \to A \to 0$$
where $X$ is an etale-locally-constant analytic group, $T$ is a torus, and $A$ is an abelian variety, each of rank $2$ in the appropriate sense. The latter short exact sequence contains the former short exact sequence as a sub (direct factor?); the quotient sequence should be something like
$0 \to Z' \to \mathbf{G}\_m' \to E \to 0$,
where ' still denotes twisting by $K/\mathbf{Q}\_p$. Since $Z'$ has trivial $\mathbf{Q}\_p$-points, then, one should have something like $\mathbf{G}\_m'(\mathbf{Q}\_p) = E(\mathbf{Q}\_p)$, modulo any descent used in forming the quotient.
Does this sound sensical?
If anyone has access to Google Wave and wants to discuss, I've set up a wave here:
<https://wave.google.com/wave/#restored:wave:googlewave.com!w%252BQCn6fZTuZ>
| https://mathoverflow.net/users/367 | Tate uniformization of nonsplit semistable elliptic curves | A form of this is contained in Silverman, second book, Chapter V, Corollary 5.4. I guess
that the image of Gm' in E (at the level of Q\_p-points) may have index 2.
| 4 | https://mathoverflow.net/users/1253 | 3985 | 2,657 |
https://mathoverflow.net/questions/2082 | 8 | I asked some time ago:
Let $w(X,Y)$ be a word in $X$ and $Y$ (i.e., an element in the free group on $X$ and $Y$). Let the variables $x$ and $y$ now range among elements of $SL\_n(K)$, $K$ an algebraically closed field.
For which $w$ is it the case that, for $x$ generic, the image of the map given by
$y \rightarrow w(x,y)$
is not Zariski-dense?
It seems that the map has Zariski-dense image for most w one can try. At the same time, as I said back when I first asked the question, the image of the map $y \rightarrow w(x,y)$ is not Zariski-dense for $w(x,y) = y x y^{-1}:$ the image of the map is contained in the conjugacy class of $x$.
By the same reasoning, the image of the map $y \rightarrow w(x,y)$ is not Zariski-dense for $x$ generic when $w$ is of the form
$w(x,y) = v(x, u(x,y)^{-1} x u(x,y)),\ (\*)$
where $v$ and $u$ are some words. The question can then be made precise: are all examples of this form? That is: is it the case that, for all words $w(x,y)$ not of the special form $(\*)$, the image of the map $y\rightarrow w(x,y)$ is Zariski-dense for $x$ generic?
I would be extremely interested in the correct answer, even in the case $n=3$. I suspect that the answer is "yes", at least for $n=2$. At the same time, it would be rather nice if it were "no".
---
[The most obvious approach may be to take derivatives at the origin. However, while this often proves that a map is surjective, it does not prove that a map is not surjective -
and in this problem it leaves too many candidates of possible words w for which the map $y\rightarrow w(x,y)$ might not be surjective but probably really is.
For those who have asked about the characteristic: if the characteristic is finite, you can assume it's much greater than the sum of the absolute values of the exponents of the word we are considering. In other words, let us not consider things like $y\rightarrow y^p, p = char(K)$.
| https://mathoverflow.net/users/398 | Surjective maps given by words, redux | You need some conditions on K to make sense of the question. For instance, y ⟼ y2 is not surjective when K = ℝ and n = 2, because you cannot reach [[-1,0],[0,-2]]. Of course, you might have meant surjectivity in the sense of algebraic groups rather than in the sense of set-theoretic groups. But I think that that just amounts to asking for K to be algebraically closed. There is another problem when K has characteristic p. In this case y ⟼ yp is not surjective; its image is only the diagonalizable matrices.
The conjecture as stated seems plausible when K = ℂ. Or if K has positive characteristic and is algebraically closed, you could perhaps ask for a Zariski-dense image.
Updates:
Re Tom's comment: I think that Harald's question is clear enough, even though I agree that one has to work a little to see what he means. He must mean that w is an element of the free group in the letters x and y, of course. The second objection is also not essential. The question can be phrased as: For each w, what are the Zariski topology properties of the set of all x for which y ⟼ w(x,y) is surjective? I believe that Chevalley's theorem on constructible images tells you that the set of such x is constructible, so it either contains a Zariski open or it is disjoint from a Zariski open.
Re Harald's comment concerning if-and-only-if conditions. You have probably thought of this, but here goes anyway. If your derivative is non-singular for any z, then the w-map is Zariski dense. Of course, there are interesting maps in algebraic geometry that are Zariski dense but not surjective. You seem to suggest that that cannot happen here, but I do not know how to prove it.
| 3 | https://mathoverflow.net/users/1450 | 3991 | 2,661 |
https://mathoverflow.net/questions/3988 | 3 | Let's say I want to prove that a closed subgroup of GL(n,R) or GL(n,C) is a Lie group, with an atlas given by exponential of matrices (restricted to an appropriate subalgebra of gl(n)), without using any manifold or Lie theory. Can you provide the necessary argument? Maybe it's trivial, but I can't see it at the moment.
| https://mathoverflow.net/users/1049 | Closed subgroups of GL(n) | Well, the crux of the matter is to cook up the linear subspace of $gl(n)$ that will eventually be the subalgebra.
If $H$ is the subgroup: set $h=\{X\in gl(n): (\forall t \in \mathbb{R}) \, exp(tX) \in H \}$.
You need to show two not completely obvious things:
1. $h$ is a linear subspace
2. $exp(h)$ is a nbhd of $1$ in $H$ (give $H$ the induced topology from $GL\_n(\mathbb{R})$).
Adams has ingenious short arguments for these on pages 17-19 of his [Lectures on Lie groups](http://books.google.co.uk/books?id=TC7d3ZcqjfsC&lpg=PP1&dq=adams%20lectures%20lie%20groups&pg=PA17#v=onepage&q=&f=false) which require nothing but a little real analysis.
I remark that all this uses no manifold or Lie theory beyond the necessary definitions and the argument works completely without change for closed subgroups of any Lie group.
| 8 | https://mathoverflow.net/users/1143 | 3996 | 2,665 |
https://mathoverflow.net/questions/3997 | 15 | I've been learning game theory on my own and was just curious how it connected with previous things I've learned. So are there any interesting connections between Game Theory and Algebraic Topology? or possibly other branches of topology?
| https://mathoverflow.net/users/695 | Are there any interesting connections between Game Theory and Algebraic Topology? | One example is in the concept of a Nash equilibrium, whose existence can be proved using various (topological) fixed point theorems. (Google "nash equilibrium proof" for a wide variety of examples... the main topological machinery that comes up is the Kakutani fixed point theorem or the plain-vanilla Brouwer theorem.)
Fixed point theorems come up in many other, similar settings; you see them a lot in certain kinds of mathematical economics. The overall intuition is that given a current set of known player strategies, people will generally want to change to better strategies. This defines a function on the "space of strategies". A fixed point of this function is interesting because it indicates an "equlibrium" of strategies that are in a sense optimal. Often topology is the best way to prove a fixed point exists.
| 19 | https://mathoverflow.net/users/1227 | 3999 | 2,667 |
https://mathoverflow.net/questions/3426 | 6 | This is a question that has haunted me for some time. In the domain of time series you always talk about trends and mean reversion. But at least to me these concepts are either defined axiomaticly within the data generating process (like the drift component in a geometric Brownian motion) or feel more like a heuristic approach (that everybody knows what you are talking about).
**My question:** When you have an empirical time series how can you define trending- and mean-reverting behaviour there? How do you detect it and test for it within some confidence interval.
That this is *not* a trivial question as it might seem at first shows the fact that in the financial markets many people adhere to different camps like "trend following" and "trading systems with moving averages" where there is much controversy going on - much of what could be avoided if you had a better understanding of what you are talking about here. (as a side-note: it of course also touches on the question of efficiency in markets)
**Note:** There is one paper I found on this: <http://hal.inria.fr/docs/00/35/28/34/PDF/FES-Finance.pdf>
From the abstract: "We are settling a longstanding quarrel in quantitative finance by proving the existence of trends in financial time series thanks to a theorem due to P. Cartier and Y. Perrin, which is expressed in the language of nonstandard analysis [...] Those trends, which might coexist with some altered random walk paradigm and efficient market hypothesis, seem nevertheless difficult to reconcile with the celebrated Black-Scholes model. They are estimated via recent techniques stemming from control and signal theory. Several quite convincing computer simulations on the forecast of various financial quantities are depicted. We conclude by discussing the role of probability theory."
Unfortunately I am no expert in non-standard-analysis and cannot fully appreciate the paper
| https://mathoverflow.net/users/1047 | Rigorous definition, detection and test for trending vs. mean-reverting behaviour of stochastic processes | In the abstract, you're looking to see when the derivative of the previsible part of a martingale (a la the Doob-Meyer decomposition) is nonzero. Finding trends with perfect accuracy amounts to performing the D-M decomposition, but of course nobody can do this in practice. Although there are a lot of ad hoc approaches to approximate trend detection, probably the best rigorous tools for this sort of thing are change of measure and change of time techniques along with nonparametric statistical tests.
For point processes a good rigorous technique is the time rescaling theorem along with the Kolmogorov-Smirnov test or a standard Cramer large-deviation inequality. An application to determine abrupt rate changes for inhomogeneous Poisson processes is sketched at
<http://blog.eqnets.com/2009/07/28/why-poissonian-traffic-models-matter-more-now-than-ever-part-4/>
and example code and numerics are provided in the sequel that is linked in the comments section there.
| 2 | https://mathoverflow.net/users/1847 | 4003 | 2,670 |
https://mathoverflow.net/questions/4001 | 2 | How can I compute the SRB measure for the cat map? Also any pointers to references for obtaining Markov partitions and recurrence times would be lovely. Thanks
| https://mathoverflow.net/users/1847 | The Arnold cat map | Unless I'm misinterpreting the question, the SRB measure is just Lebesgue measure... the cat map is hyperbolic, preserves area, and is topologically transitive. See Theorem 3.10 and the following remarks here:
<http://books.google.com/books?id=uu-qeVBvQNEC&pg=PA141#v=onepage&q=&f=false>
| 4 | https://mathoverflow.net/users/1227 | 4004 | 2,671 |
https://mathoverflow.net/questions/3845 | 1 | Let's use the notation of `[A=>B]` for `Hom(A, B)`. Take a 1-dimensional algebraic torus `G``m` and higher-dimensional torus `T` and let's live in the category of commutative algebraic groups over `k`.
Out of four expressions like `[G``m``=> [G``m``=>T]]` etc. half give back `T` `(*)`, others the *dual torus* `T``V`, in the sense that `X``*``(T) :=` [`G``m` `=> T] = [T``V` `=> G``m``] =: X``*``(T``V``)`.
The equality `(*)` can be proven by using the following formula with `B = G``m`
```
(**) A \otimes [B=>B] ==== [[A=>B] => B].
```
>
> **Question:** Is there another example of commutative algebraic group **or a similar generalized object** `B`, for which the identity `(**)` is true or **true in some generalized sense**?
>
>
>
One thing I specifically have in mind is that if we could write `[X => Y] = X``*` `\otimes Y` whenever `X` and `Y` are groups, as if they were vector spaces, the formula would hold for all `A` and `B`. So, what's a category that is related to algebraic groups but which posesses this property?
| https://mathoverflow.net/users/65 | How to make commutative algebraic groups strongly dualizable? | Since Ilya asked I'll write out in a bit more detail the case of finitely generated abelian groups. There is not (as far as I know) any reasonable notion of strong duality on the abelian category of finitely generated abelian groups in the sense you ask for. Indeed, Z is the tensor unit and any torsion group gets killed by Hom(-,Z). But if one passes to D^b(Z-mod) one does get a rigid tensor category. Setting D(-) = RHom(-,Z) one gets that D(A)\otimes^L B \iso RHom(A,B), so in particular D^2 \iso Id and with the suitably derived tensor and internal hom D^b(Z-mod) is a rigid tensor category. It even turns out that this structure is enough to canonically recover Spec Z as a locally ringed space. As Scott pointed out in his comment one does need finiteness conditions for this to work, the full unbounded derived category of all abelian groups is only a closed tensor category.
In fact this story is true for D^{perf}(X) for any quasi-compact quasi-separated scheme X. The finiteness conditions being necessary is also quite general - in good cases for compactly generated triangulated categories with a tensor product (which respects the triangulation and coproducts) one always gets an internal hom and often the compacts form a rigid tensor category (one can give conditions under which this is guaranteed).
For a category of actual algebraic groups with this property I'd recommend having a look at the category of finite flat commutative group schemes over some base S. Cartier duality gives an honest duality of this category with itself and maybe at least for suitably good S one might get further good properties?
| 1 | https://mathoverflow.net/users/310 | 4010 | 2,675 |
https://mathoverflow.net/questions/4011 | 33 | Because I have heard the phrase "totally ordered abelian group", I imagine there should be non-abelian ones. By this I mean a group with a [total ordering](https://en.wikipedia.org/wiki/Total_order) (not to be confused with a well-ordering) which is "bi-translation invariant": a < b should imply cad < cbd.
Does anyone know any examples?
Totally ordered *abelian* groups are easy to come up with: any direct product of subgroups of the reals, with the lexicographic ordering, will do. Knowing some non-abelian ones would help reveal what aspects of totally ordered abelian groups really depend on them being abelian...
**Edit:** Via Andy Putman's answer below, I found this great summary of results about ordered and bi-ordered groups (i.e. groups with bi-translation invariant orderings) on Dale Rolfsen's site:
[Lecture notes on Ordered Groups and Topology](http://www.math.ubc.ca/~rolfsen/papers/luminynotes/lum.pdf)
He shows numerous examples of non-abelian bi-orderable groups, including a bi-ordering (bi-translation invariant ordering) on the free group with two generators. As well, he mentions, due to Rhemtulla, that a left-orderable group is abelian iff every left-ordering is a bi-ordering, which I think really highlights the relationship between ordering and abelianity.
| https://mathoverflow.net/users/84526 | What's a non-abelian totally ordered group? | This concept is usually called biorderability (there is also left- and right-orderability). There are many examples, such as free groups and surface groups. Most spectacularly, the pure braid groups are biorderable, while the full braid groups are left orderable but not biorderable. The left ordering on the braid groups is usually attributed to Dehornoy, though it was discovered even earlier by Thurston (but not published).
Dale Rolfsen has several nice surveys of material related to this on his webpage [here](http://www.math.ubc.ca/~rolfsen/reprints.html). In particular, there is the complete text of a nice book called "Why are braids orderable?" that he wrote with Patrick Dehornoy, Ivan Dynnikov, and Bert Wiest. I believe that a new and much expanded edition of this book was just published.
EDIT 1 : I just found the website for the much-expanded version of Rolfsen et al's book [here](https://bookstore.ams.org/surv-148).
EDIT 2 : Thurston's construction of a left-ordering on the braid groups (which, of course, uses hyperbolic geometry) is very beautiful. It is explained very nicely in the first few pages of the paper "Orderings of mapping class groups after Thurston" by Short and Wiest, which is available on the arXiv [here](https://arxiv.org/abs/math/9907104). The intro sections of this paper also contain a brief but enlightening account of the general theory of group orderings.
Also, I have not read it, but there is a book entitled "Orderable Groups" by Rehmtulla and Mura. However, it is from 1977 and will thus omit a lot of recent work.
| 33 | https://mathoverflow.net/users/317 | 4015 | 2,677 |
https://mathoverflow.net/questions/4028 | 8 | Many special functions including the gamma function have a [duplication formula](http://en.wikipedia.org/wiki/Duplication_formula) of some sorts. In the case of the gamma function it reads:
>
> Gamma(2z) = Gamma(z) Gamma(z+1/2) 22z-1/Gamma(1/2)
>
>
>
On the other hand, there is no algebraic relation between Gamma(2z) and Gamma(z) by themselves, meaning that is there is no nonzero polynomial f(x,y) such that f(Gamma(2z),Gamma(z))=0 for all complex z. I can prove this by chasing poles and their order.
However, I'd be interested in a (simple) argument which shows that the following similar statement is true (which I believe it is):
>
> There is no (nonzero) polynomial f(x,y) such that f((2n)!, n!)=0 for all integers n≥0.
>
>
>
Any ideas? Thank you!
| https://mathoverflow.net/users/359 | No simple duplication formula for factorials? | It's equivalent to show that there is no polynomial relationship f({2n choose n}, n!) = 0. On the other hand, we know that {2n choose n} ~ 4^n/sqrt{n} asymptotically and n! grows much faster.
Terence Tao once remarked that if a sufficiently simple duplication formula were known for the factorial then Wilson's theorem would give an efficient primality test. (Edit: see the other answer. I may be misremembering the stronger remarks that Dick Lipton made.)
| 10 | https://mathoverflow.net/users/290 | 4032 | 2,687 |
https://mathoverflow.net/questions/4023 | 38 | Any suggestions on a good text to use for teaching an introductory Real Analysis course? Specifically what have you found to be useful about the approach taken in specific texts?
| https://mathoverflow.net/users/1462 | Text for an introductory Real Analysis course. | Stephen Abbott, [*Understanding Analysis*](http://books.google.com/books?id=7t1ZhUAc5yMC)
Strongly recommended to students who are ony getting to grips with abstraction in mathematics. Find a review [here](http://www.maa.org/press/maa-reviews/understanding-analysis-0).
| 32 | https://mathoverflow.net/users/532 | 4034 | 2,689 |
https://mathoverflow.net/questions/2826 | 20 | I have long been curious about equivalent forms of the Riemann hypothesis for automorphic L-functions.
In the case of the ordinary Riemann hypothesis, one gets a very good error term for the prime number theorem, one has the formulation involving the Mobius mu function which is a result to the effect of the parity of prime factors in a square free number having a distribution related to that of flips of an unbiased coin, and one also has the reformulation in terms of Farey fractions.
I know that for L-functions attached to Dirichlet characters, one gets a very good error term for the prime number theorem for primes in arithmetic progressions. Presumably if one focuses on Dedekind zeta functions and Hecke L-series one gets a very strong effective Chebotarev density theorem or something like that.
But for L-functions attached to Hecke eigenforms for GL(2), or more abstract things like symmetric n-th power L-functions attached to automorphic forms or automorphic representations, it seems quite unclear to me what the significance of the Riemann hypothesis for these L-functions is. I think that I remember something about a zero free region to the left of the boundary of the critical strip being related to the Sato-Tate conjecture, so I have a vague impression that one might be able to get a good bound on the speed of convergence to the Sato-Tate distribution as an equivalent to the Riemann hypothesis for some of these L-functions.
What are some interesting equivalents to the Riemann hypothesis for automorphic L-functions that you know? I'm particularly interested in statements that have qualitative interpretations.
P.S. I've blurred the distinction between an equivalent of the Riemann hypothesis for a single L-function and equivalents to the Riemann hypothesis for a specified family of L-functions. I am interested in both things
P.P.S. I am more interested in equivalents than in consequences of the Riemann hypotheses for these L-functions in so far as equivalents "capture the essence" of the statement in question to a greater extent than consequences do. Still, I would would welcome references to interesting *consequences* of the Riemann hypothesis for automorphic L-functions, again, especially those with qualitative interpretations.
| https://mathoverflow.net/users/683 | Equivalent forms of the Grand Riemann Hypothesis | Well, suppose pi is a cuspidal automorphic representation of GL(n)/Q. This has the structure of a tensor product, indexed by primes p, of representations pi\_p of the groups GLn(Qp). The Satake isomorphism tells us that at almost all primes, each pip is determined by a conjugacy class A(p) in GLn(C). In this language, the Riemann hypothesis for the L-function associated to pi says that the partial sums of tr(A(p)) over p < X show "as much cancellation as possible," and are of size sqrt(X). But if n>1, we are dealing with very complicated objects, and the local components of these automorphic representations vary in some incomprehensible way...
You are right, there are certainly special cases. If we knew GRH for L-functions associated to Artin representations then the Cebotarev density theorem would follow with an optimal error term. Likewise, GRH for all the symmetric powers of a fixed elliptic curve E implies (and is in fact equivalent to; see Mazur's BAMS article for a reference) the Sato-Tate conjecture for E with an optimal error term. But in general, reformulations like this simply don't exist.
There are many interesting consequences of GRH for various families of automorphic L-functions. I recommend Iwaniec and Kowalski's book (Chapter 5), the paper "Low-lying zeros of families of L-functions" by Iwaniec-Luo-Sarnak, and Sarnak's article at <http://www.claymath.org/millennium/Riemann_Hypothesis/Sarnak_RH.pdf>
| 13 | https://mathoverflow.net/users/1464 | 4037 | 2,692 |
https://mathoverflow.net/questions/3235 | 8 | Let $F$ be a free semigroup (say, $2$-generated) which is embedded in a group $G$, and suppose that $G$ (as a group) is generated by $F$. The most simple such situation would be when $G$ is a free group, but there are lot of groups, besides free ones, which could occur in this situation (for example, $G$ could be solvable). Is it possible that $G$ contains a subgroup isomorphic to $\mathbb{Z} \times \mathbb{Z}$ (the direct product of two copies of the infinite cyclic group $\mathbb{Z}$)?
**Update**: thanks to all the people for a very interesting discussion. Sorry that I cannot award a few "accepted answers", so the only one goes to Greg who supplied the most lucid and explicit example.
| https://mathoverflow.net/users/1223 | Subgroups of a group generated by a free semigroup | I think that Henry's solution doesn't work either. In that example, I get at2a = ta3t = a6b3t2. But I think that this construction can be fixed.
Consider the group of affine linear maps f(x) = αx+β over the reals ℝ. Let a act by multiplication by α, where α is transcendental, let b act by adding 1, let F be the semigroup that they generate, and let G be the group that they generate. F is free because two distinct words in a and b, if they have the same degree in a, have in them two distinct polynomials in α with non-negative integer coefficients. (It is interesting that α must be transcendental for this to work.) Then G contains a ℤ x ℤ, generated by adding 1 and adding α.
| 11 | https://mathoverflow.net/users/1450 | 4041 | 2,696 |
https://mathoverflow.net/questions/4054 | 2 | Suppose $f: M\to N$ is a smooth map between two smooth manifolds, with $M$ compact and connected, and suppose there is a dense subset of $f(M)$ where each fiber is connected, then each fiber of $f$ is connected.
If it helps, you can just consider the case where the set of regular values is dense in $f(M)$ and the fiber of each regular value is connected, and you want to prove every fiber of $f$ is connected.
| https://mathoverflow.net/users/1468 | Can connectedness of fibers of a smooth map be checked on a dense set? | Perhaps I've misunderstood the question, but it looks like it's false.
Let M={(x,y)∈ℝ²|(x,y)≠(0,0)}, N=ℝ, and define f(x,y)=x. This is a smooth map of smooth manifolds, with the fibers over ℝ-{0} connected, but the fiber over 0 disconnected.
**Edit:** Wayne has added the hypothesis that M is compact. I think the statement is true under this hypothesis. Here's a sketch proof. Suppose f-1(x) is disconnected, then I'd like to prove that there is an open neighborhood of x where the fibers are disconnected. Since manifolds are normal, there are two non-empty disjoint open sets U and V in M covering f-1(x). Now prove a generalization of the hotdog lemma, which will say that there is an open neighborhood W of x such that U∪V covers f-1(W). Since U and V are disjoint, this will show that the fibers over points of W are disconnected. To prove the generalized hotdog lemma. use the fact that smooth maps locally "look like products", choose a cover of f-1(x) by "box shaped" open sets contained in U∪V. You can choose a finite number of these by compactness of f-1(x) (it's a closed subset of a compact space), and take W to be the intersection of all of their images in N.
**More Edit:** The above proof doesn't work (see comments below and [Richard Kent's post](https://mathoverflow.net/questions/4054/can-connectedness-of-fibers-of-a-smooth-map-be-checked-on-a-dense-set/4189#4189)). Apparently, I'm confused about the meaning of [Ehresmann's\_theorem](http://en.wikipedia.org/wiki/Ehresmann%27s_theorem), because it looks to me like the map f:S2⊂ℝ3→ℝ given by f(x,y,z)=z is smooth, but it doesn't look like a trivial fibration around the poles. The algebro-geometric analogue says that a smooth morphism X→Y always factors as X→**A**nY→Y, where the first map is etale. But an algebraic geometer would say that the map S2→ℝ is not smooth.
| 1 | https://mathoverflow.net/users/1 | 4056 | 2,707 |
https://mathoverflow.net/questions/234 | 8 | A $k$-component link defines a map $T^k \rightarrow \operatorname{Conf}\_k S^3$. Does the homotopy type of this map capture the Milnor invariants?
Some special cases:
* $k=2$, no, it's null homologous, but you can look instead at the map $T^2 \rightarrow \operatorname{Conf}\_2 R^3$, which captures linking number.
* $k=3$, [Melvin et al.](https://arxiv.org/abs/0901.1612) proved it does.
| https://mathoverflow.net/users/3 | A $k$-component link defines a map $T^k\rightarrow \operatorname{Conf}_k S^3$. Does the homotopy type capture Milnor's invariants? | It's been a while since I've thought about this but I think Koschorke answered much of your question back in 1997 "A generalization of Milnor's mu-invariants to higher-dimensional link maps" Topology 36 (1997), no 2. 301--324. Scanning through the paper I see he recovers many of the mu invariants but not all. He lists it as an open question (6.3) if the homotopy class of the map T^k --> C\_k R^3 is a complete link homotopy invariant of the link.
Related:
Brian Munson put these Koschorke "linking maps" into the context of the Goodwillie calculus in a recent arXiv paper. I've wondered for a while if you could use these types of maps to create a direct construction of the Cohen-Wu correspondence between the homotopy groups of S^2 and their corresponding simplicial quotient object made from the brunnian braid groups.
| 8 | https://mathoverflow.net/users/1465 | 4058 | 2,708 |
https://mathoverflow.net/questions/4064 | 9 | Given a toric variety, is it easy to see if a crepant resolution exists? If so, how can it be explicitly constructed?
| https://mathoverflow.net/users/28 | Crepant resolutions of toric varieties | Whether or not a resolution is crepant only depends on the hypersurfaces in the exceptional locus -- to speak casually, it depends on which hypersurfaces you add. In the toric case, resolution corresponds to subdividing the fan, and the new hypersurfaces correspond to the new rays you add. In particular, if I can resolve without adding any new rays, that resolution will certainly be crepant.
What is going to happen is that there will be some combinatorial rule picking out a finite collection of rays. This will be fairly explicit and simple. You will then have to determine whether there is a subdivision of the original fan to a smooth fan, adding in only these permitted rays. That is a difficult combinatorial question, although it is finite in principal.
Also, whether or not a ray is permitted will depend solely on the cone of the original fan that it is in. (Geometrically, the crepancy of a divisor can be computed locally on the variety being resolved.) So it is enough to, for a given polyhedral cone, give a rule for which rays in that cone are permitted.
Now I'm going to disappoint you by not remembering that rule. The only case I remember is where sigma is a simplicial cone, with generators v\_1, v\_2, ..., v\_n, and there is some monomial w in the dual lattice such that <w, v\_1>= <w, v\_2> = ... = <w, v\_n> = 1. Then a ray, with minimal lattice vector v, corresponds to a crepant divisor if and only if <w, v>=1. This condition is actually fairly natural; it is the case of a quotient singularity of the form C^n/G where G is a finite abelian subgroup of SL\_n.
Everything I know about this, I learned during the 2006 Michigan [working seminar](http://www.math.lsa.umich.edu/seminars/alggeo/topicsAG/Winter06.shtml#124) on the McKay correspondence. You might find the references there to be useful.
UPDATE: I thought a bit harder, and I can state the above rule slightly more generally: The cone sigma doesn't have to be simplicial. It is enough that there is some monomial in the dual lattice which pairs with the generators this way. If the cone is not simplicial, then this gives more linear equations for w than we have variables, but they might happen to be solvable anyway. This is precisely the situation that the singularity is Gorenstein.
| 8 | https://mathoverflow.net/users/297 | 4085 | 2,725 |
https://mathoverflow.net/questions/4086 | 12 | Given an infinite group which is finitely generated, is there a proper maximal normal subgroup?
| https://mathoverflow.net/users/996 | Does every finitely generated group have a maximal normal subgroup? | If you mean nontrivial maximal normal subgroup (not 1 or the whole group), then the answer is no.
Higman constructed a finitely generated infinite group $G$ with no subgroups of finite index. You then get a finitely generated group with no nontrivial normal subgroups by taking the quotient by a maximal normal subgroup.
[Higman's group](https://en.wikipedia.org/wiki/Higman_group) $G$ is $\langle a,b,c,d | a^{-1} b a = b^2, b^{-1}cb = c^2, c^{-1}dc=d^2, d^{-1}ad=a^2 \rangle$
See Higman, Graham. [A finitely generated infinite simple group](https://dx.doi.org/10.1112/jlms/s1-26.1.61). J. London Math. Soc. 26, (1951). 61--64.
Edit:
If you mean does it have a proper maximal normal subgroup, then the answer is yes:
Finitely generated groups have a (possibly trivial) maximal normal subgroup. Higman's reference for this is B.H. Neumann, "[Some remarks on infinite groups](http://jlms.oxfordjournals.org/cgi/reprint/s1-12/2/120) ", Journal London Math. Soc, 12 (1937), 120-127.
| 16 | https://mathoverflow.net/users/1335 | 4091 | 2,730 |
https://mathoverflow.net/questions/4083 | 21 | I'm currently taking a course in Hodge theory ... and I wonder if all the splittings in $\{i,-i\}$ Eigenvalue pairs come from the Galois group action (of the extension $\mathbb{R}\rightarrow\mathbb{C}$) - it seems to me like that (and I couldn't find such a statement in my textbook).
Is this true? If yes, is this a good way to think of Hodge decomposition or does one need more data than just the Galois group? If not, what is my misconception?
I thought (if my assumption is true), this would be a way to generalize to other algebraic field extensions.. are there analogues of Hodge theory for any algebraic field extension? Does it involve the Galois group?
If this question isn't "researchy" enough, just close it ... I will come back asking questions in a year then :-)
| https://mathoverflow.net/users/956 | Is Hodge theory somehow connected with a Galois group action Gal(C/R)? | You are correct: there is a connection to the Galois theory of $\mathbb{C}/\mathbb{R}$ here.
To give a Hodge structure on a real vector space $V$ -- i.e., a direct sum decomposition of its complexification into $(p,q)$ subspaces such that $H^{q,p}$ is the complex conjugate of $H^{p,q}$ -- is equivalent to giving an action of $G = \operatorname{Res}\_{\mathbb{C}/\mathbb{R}} \mathbb{C}^{\times}$ on $V$. Here
$\operatorname{Res}\_{\mathbb{C}/\mathbb{R}} \mathbb{C}^{\times}$ means the "restriction of scalars" from $\mathbb{C}/\mathbb{R}$ of the complex multiplicative group $\mathbb{C}^{\times}$. In plainer terms, it means that we view $\mathbb{C}^{\times}$ not as a one-dimensional complex algebraic group, but as a 2-dimensional real algebraic group, a "nonsplit torus". Then the fact that we have a homomorphism of real groups
$$G \to \operatorname{GL}(V)$$
implies an extra condition on the complexified representation
$\mathbb{C}^{\times} \to \operatorname{GL}(V \otimes \mathbb{C})$: namely that the space $V^{p,q}$ on which $z$ in $\mathbb{C}^{\times}$ acts
as $z^{p} \overline{z}^q$ is the complex conjugate of the space $V^{q,p}$.
A brief (but accurate!) discussion of this can be found at
<http://en.wikipedia.org/wiki/Hodge_structure#Hodge_structures>
| 17 | https://mathoverflow.net/users/1149 | 4092 | 2,731 |
https://mathoverflow.net/questions/4075 | 53 | I'm not sure if the questions make sense:
Conc. primes as knots and Spec Z as 3-manifold - fits that to the Poincare conjecture? Topologists view 3-manifolds as Kirby-equivalence classes of framed links. How would that be with Spec Z? Then, topologists have things like virtual 3-manifolds, has that analogies in arithmetics?
Edit: [New MFO report](http://www.math.nagoya-u.ac.jp/~furusho/paper/MFO2012.pdf "pdf"): "At the moment the topic of most active interaction between topologists and number theorists are quantum invariants of 3-manifolds and their asymptotics. This year’s meeting showed significant progress in the field."
Edit: "What is the analogy of quantum invariants in arithmetic topology?", "If a prime number is a knot, what is a crossing?" asks [this old report](http://www.mfo.de/programme/schedule/2003/43c/Report46_2003.pdf "pdf").
An other such question:
Minhyong Kim [stresses](https://web-beta.archive.org/web/20111016105345/http://londonnumbertheory.wordpress.com/2009/11/04/optimal-proofs/ "London NT seminar blog")
the special complexity of number theory: "To our present day understanding, number fields display exactly the kind of order ‘at the edge of chaos’ that arithmeticians find so tantalizing, and which might have repulsed Grothendieck." Probably a feeling of such a special complexity makes one initially interested in NT. Knot theory is an other case inducing a similar impression. Could both cases be connected by the analogy above? How could a precise description of such special complexity look like and would it cover both cases? Taking that analogy, I'm inclined to answer [Minhyong's question](https://web-beta.archive.org/web/20111016115127/http://londonnumbertheory.wordpress.com/2009/12/18/within-the-mess/ "London NT seminar blog")
with the contrast between low-dimensional (= messy) and high-dimensional (= harmonized) geometry. Then I wonder, if "harmonizing by increasing dimensions"-analogies in number theory or the Langlands program exist.
Minhyong hints in a mail to "the study of moduli spaces of bundles over rings of integers and over three manifolds as possible common ground between the two situations". A google search produces an old article by Rapoport "Analogien zwischen den Modulräumen von Vektorbündeln und von Flaggen" (Analogies between moduli spaces of vector bundles and flags) (p. 24 [here](httP://dml.math.uni-bielefeld.de/JB_DMV/JB_DMV_099_4.pdf "DMV Jahresbericht 99"), [MR](http://ams.mathematik.uni-bielefeld.de/mathscinet-getitem?mr=99e:14010 "MathSciNet review")). There, Rapoport describes the cohomology of such analogous moduli spaces, inspired by a similarity of vector bundles on Riemann surfaces and filtered isocrystals from p-adic cohomologies, "beautifull areas of mathematics connected by entirely mysterious analogies". ([book](http://www.cambridge.org/gb/knowledge/isbn/item4027614/?site_locale=en_GB "book") by R., Orlik, Dat) As interesting as that sounds, I wonder if google's hint relates to the initial theme. What do you think about it? (And has the mystery Rapoport describes now been elucidated?)
Edit:
[Lectures by Atiyah](http://arxiv.org/abs/1009.4827 "link") discussing the above analogies and induced questions of "quantum Weil conj.s" etc.
[This interesting essay by Gromov](http://www.ihes.fr/~gromov/PDF/ergobrain.pdf "pdf") discusses the topic of "interestung structures" in a very general way. Acc. to him, "interesting structures" exist never in isolation, but only as "examples of structurally organized classes of structured objects", Z only because of e.g. algebraic integers as "surrounding" similar structures. That would fit to the guesses above, but not why numbers were perceived as esp. fascinating as early as greek antiquity, when the "surrounding structures" Gromov mentions were unknown. Perhaps Mochizuki has with his ["inter-universal geometry"](https://mathoverflow.net/questions/852/what-is-inter-universal-geometry "MO") a kind of substitute in mind?
Edit: [Hidekazu Furusho](http://de.arxiv.org/abs/1211.5469 "arxiv"): "Lots of analogies between algebraic number theory and 3-dimensional topology are suggested in arithmetic topology, however, as far as we know, no direct relationship seems to be known. Our attempt of this and subsequent papers is to give a direct one particularly between Galois groups and knots."
| https://mathoverflow.net/users/451 | Questions about analogy between Spec Z and 3-manifolds | The analogy doesn't quite give a number theoretic version of the Poincare conjecture. See Sikora, "Analogies between group actions on 3-manifolds and number fields"
(arXiv:0107210): the author states the Poincare conjecture as "S3 is the only closed 3-manifold with no unbranched covers." The analogous statement in number theory is that Q is the only number field with no unramified extensions, and indeed he points out that there are a few known counterexamples, such as the imaginary quadratic fields with class number 1.
The paper also has a nice but short summary of the so-called "MKR dictionary" relating 3-manifolds to number fields in section 2. Morishita's expository article on the subject, arXiv:0904.3399, has more to say about what knot complements, meridians and longitudes, knot groups, etc. are, but I don't think there's an explanation of what knot surgery would be and so I'm not sure how Kirby calculus fits into the picture.
Edit: An [article](http://arxiv.org/abs/math/0602064) by B. Morin on Sikora's dictionary (and how it relates to Lichtenbaum's cohomology, p. 28): "he has given proofs of his results which are very different in the arithmetic and in the topological case. In this paper, we show how to provide a unified approach to the results in the two cases. For this we introduce an equivariant cohomology which satisfies a localization theorem. In particular, we obtain a satisfactory explanation for the coincidences between Sikora's formulas which leads us to clarify and to extend the dictionary of arithmetic topology."
| 30 | https://mathoverflow.net/users/428 | 4094 | 2,733 |
https://mathoverflow.net/questions/3888 | 17 | Background reading: [John Stembridge's webpage](http://www.math.lsa.umich.edu/~jrs/other.html).
The idea is that when you want to prove a theorem for all root systems, sometimes it suffices to prove the result for the simply laced case, and then use the concept of folding by a diagram automorphism to deduce the general case.
I have never seen an example of this in practice. So my question is: What are some (good) examples illustrating this technique?
| https://mathoverflow.net/users/425 | Folding by Automorphisms | One very important use of this technique is the relation between Lie algebras / quantum groups and quiver varieties. I first saw something about this in Lusztig's book, Introduction to Quantum Groups; but see also [this arXiv paper](http://arxiv.org/abs/math.QA/0406073) Alistair Savage. Quiver varieties are important for categorifying many structures related to a simple Lie algebra (its representation theory, its enveloping algebra, etc.). Categorification is a long story that leads to all kinds of interesting things, and it is a sequel to the long story of quantum groups themselves. But even if you're not learning about either one for their own sake, Lusztig already needed it to prove properties of his canonical bases of representations of simple Lie algebras.
A Dynkin-type quiver is an orientation of a Dynkin diagram. A quiver representation is a collection of maps between vector spaces in the pattern of the diagram. A quiver variety is then a variety of (certain of) these representations, for fixed choices of the vector spaces. The point is that you can only define a quiver for a simply laced Dynkin diagram. You need the folding automorphism to obtain quiver varieties or information from quivers in general in the multiply laced case.
| 14 | https://mathoverflow.net/users/1450 | 4102 | 2,741 |
https://mathoverflow.net/questions/4062 | 17 | Let $G$ and $H$ be affine algebraic groups over a scheme $S$ of characteristic 0 and let $\textbf{Hom}\_{S,gp}(G,H)$ be the functor $T \mapsto \text{Hom}\\_{T,gp}(G,H)$
>
> **Theorem** (SGA 3, expose XXIV, 7.3.1(a)): Suppose G is reductive. Then
> $\textbf{Hom}\_{S,gp}(G,H)$ is representable by a scheme.
>
>
>
Can this fail if $G$ is not reductive? I worked out a few example with $G = \mathbb{G}\_a$, but they were representable.
| https://mathoverflow.net/users/2 | Can Hom_gp(G,H) fail to be representable for affine algebraic groups? | $\operatorname{Hom}(\mathbb{G}\_a, \mathbb{G}\_m)$ is not representable.
Let $R$ be a $\mathbb{Q}$-algebra. I claim that $\operatorname{Hom}(\mathbb{G}\_a, \mathbb{G}\_m)(\operatorname{Spec} R)$ is {Nilpotent elements of $R$}. Intuitively, all homs are of the form $x\mapsto e^{nx}$ with $n$ nilpotent.
More precisely, the schemes underlying $\mathbb{G}\_a$ and $\mathbb{G}\_m$ are
$\operatorname{Spec} R[x]$ and $\operatorname{Spec} R[y, y^{-1}]$ respectively. Any hom of schemes is of the form $y \mapsto \sum f\_i x^i $for some $f\_i$ in $R$. The condition that this be a hom of groups says that
$\sum f\_k (x\_1+x\_2)^k = (\sum f\_i x\_1^i)(\sum f\_j x\_2^j)$. Expanding this, $f\_{i+j}/(i+j)! = f\_i/i! f\_j/j!$. So every hom is of the form $f\_i = n^i/i!$, and n must be nilpotent so that the sum will be finite.
Now, let's see that this isn't representable. For any positive integer $k$, let $R\_k = C[t]/t^k$. The map $x \mapsto e^{tx}$ is in $\operatorname{Hom}(\mathbb{G}\_a, \mathbb{G}\_m)(\operatorname{Spec} R\_k)$ for every $k$. However, if $R$ is the inverse limit of the $R\_k$, there is no corresponding map in $\operatorname{Hom}(\mathbb{G}\_a, \mathbb{G}\_m)(\operatorname{Spec} R)$. So the functor is not representable.
| 26 | https://mathoverflow.net/users/297 | 4105 | 2,742 |
https://mathoverflow.net/questions/4104 | 4 | By the GAGA principle we know that a holomorphic vector bundle E->X is analitically isomorphic to an algebraic one, say F->X, and by definition F is locally trivial in the Zariski topology. But since the isomorphism between E and F is analytic, I fail to see if this implies that E is Zariski locally trivial too.
I hope the answer is not "trivially yes" for some stupid reason, but I cannot guarantee that.
| https://mathoverflow.net/users/828 | Is a holomorphic vector bundle on a projective variety locally trivial in the Zariski topology? | You get (analytic) trivializations of E over Zariski-open sets just by composing a trivialization of E with the isomorphism between E and F. Of course, you do not get algebraic trivializations, but for this you would need an algebraic structure on E in the first place.
| 7 | https://mathoverflow.net/users/1476 | 4110 | 2,747 |
https://mathoverflow.net/questions/4117 | 18 | Any topological group $G$ has a classifying space, whose loopspace is a (homotopy) group which is homotopy equivalent to $G$ in a way that preserves the group structure. More generally, if $G$ is an $A\_\infty$-group (a space with a binary operation which satisfies the group axioms up to coherent homotopy), it similarly can be delooped to a classifying space.
Now suppose you have a [cogroup](https://mathoverflow.net/questions/3740/cogroup-objects) in the category of pointed spaces. If it is actually literally a cogroup, it's not hard to show it must be a point. However, up to (coherent) homotopy, the suspension of any space is a cogroup. Is the converse true? Can you desuspend any $A\_\infty$-cogroup? Are there any examples of homotopy cogroups (possible not $A\_\infty$) which are not suspensions? More generally, are there any criteria that you can use to prove that a space does not have the homotopy type of a suspension? The only one I know is that all cup products must vanish, but this also holds automatically for a homotopy cogroup (indeed, for any "co-H-space").
| https://mathoverflow.net/users/75 | When can you desuspend a homotopy cogroup? | Whilst the n-lab page on [co-H-spaces](http://ncatlab.org/nlab/show/co-H-space) could be described as a little meagre, it does nonetheless contain a reference to a paper in the Handbook of Algebraic Topology. Various parts of this book are in the "preview" in google books, in particular page 1153 which contains the magical sentence:
>
> We can now construct cell complexes $S^n \cup\_\alpha e^m$ which are co-H-spaces but not suspensions.
>
>
>
This follows a theorem which classifies when such spaces are co-H-spaces and suspensions.
The specific article is the aptly named "Co-H-spaces" by Martin Arkowitz, and is chapter 23 of Handbook of Algebraic Topology, edited by Ioan James. I would recommend this as a first place to look to find out more about these spaces.
Incidentally, I don't believe your remark about delooping. I'm almost certain that there are H-spaces that can't be delooped so either there are H-spaces that aren't $A\_\infty$-groups or your claim is incorrect. Unfortunately, I'm not currently surrounded by my usual stock of algebraic topology textbooks so can't look this up (and you can do an internet search as well as I can). Can you supply a reference for the delooping claim?
| 9 | https://mathoverflow.net/users/45 | 4128 | 2,758 |
https://mathoverflow.net/questions/4132 | 2 | I know that this is on the boundaries of what's allowed, but hopefully someone'll answer before it gets closed!
What (periodic) function has Fourier series the harmonic series? I really want the even (cosine) terms to be the harmonic series and no odd terms.
Edit: so that the record is perfectly clear, what I wanted was a function with Fourier series
$$
\sum\_{n \ge 1} \frac{1}{n} \cos(n \pi t)
$$
| https://mathoverflow.net/users/45 | What function has fourier series the harmonic series? | It's a standard series computation to show that
$$
\sum\_{n \ge 1} \frac{x^n}{n} = \log \frac{1}{1 - x}
$$
Now substitute $x = e^{i t}$ and take the real part.
(As an aside, the reason I write the identity this way is that this is the version which is combinatorially significant.)
| 11 | https://mathoverflow.net/users/290 | 4133 | 2,762 |
https://mathoverflow.net/questions/4138 | 17 | Let $X$ be a complex analytic space. It is a 'well known fact' that the categories of local systems on $X$ (i.e. locally constant sheaves with stalk $C^n$), and of (holomorphic) vector bundles on $X$ with flat connection, are equivalent. I've been looking for a proof of this, but every reference I can find merely says something like 'this is well known' without further argument. Does anyone know of a proof?
| https://mathoverflow.net/users/1481 | Why are local systems on a complex analytic space equivalent to vector bundles with flat connection? | The important point of the proof is that either of these objects can be locally trivialized with transition functions on each double overlap given by a constant element of GL(n). So, given a local system, you just build the vector bundle with flat connection that has the same transition functions, and vice versa.
**EDIT:** Brian Conrad points out below that while this is a fairly complete sketch in the smooth case, it requires more work in the singular case.
| 12 | https://mathoverflow.net/users/66 | 4140 | 2,766 |
https://mathoverflow.net/questions/4125 | 35 | Rings of functions on a nonsingular algebraic curve (which, over $\mathbb{C}$, are holomorphic functions on a compact Riemann surface) and rings of integers in number fields are both examples of Dedekind domains, and I've been trying to understand the classical analogy between the two. As I understand it, $\operatorname{Spec} \mathcal{O}\_k$ should be thought of as the curve / Riemann surface itself. Is there a good notion of integration in this setting? Is there any hope of recovering an analogue of the Cauchy integral formula?
(If I am misunderstanding the point of the analogy or stretching it too far, please let me know.)
| https://mathoverflow.net/users/290 | If Spec Z is like a Riemann surface, what's the analogue of integration along a contour? | For a variety $X$ over a finite field, I guess one can take $\ell$-adic sheaves to replace differential forms. Then the local integral around a closed point $x$ (like integral over a little loop around that point) is the trace of the local Frobenius $\operatorname{Frob}\_x$ on the stalk of sheaf, the so-called naive local term. Note that $\operatorname{Frob}\_x$ can be regarded as an element (or conjugacy class) in $\pi\_1(X)$, "a loop around $x$". The global integral would be the global trace map
$$
H^{2d}\_c(X,\mathbf{Q}\_{\ell})\to\mathbf{Q}\_{\ell}(-d),
$$
and the Tate twist is responsible for the Hodge structure in Betti cohomology (or the $(2\pi i)^d$ one has to divide by). The Lefschetz trace formula might be the analog of the residue theorem in complex analysis on Riemann surfaces.
For the case of number fields, each closed point $v$ in $\operatorname{Spec} O\_k$ still defines a "loop" $\operatorname{Frob}\_v$ in $\pi\_1(\operatorname{Spec} k)$ (let's allow ramified covers. One can take the image of $\operatorname{Frob}\_v$ under $\pi\_1(\operatorname{Spec} k)\to\pi\_1(\operatorname{Spec} O\_k)$, but the target group doesn't seem to be big enough). For global integral, there's the Artin-Verdier trace map $H^3(Spec\ O\_k,\mathbb G\_m)\to\mathbb{Q/Z}$ and a "Poincaré duality" in this setting, but I don't know if there is a trace formula. The fact that 3 is odd always makes me excited and confused.
So basically I think of trace maps (both local and global) as counterpart of integrals. Correct me if I was wrong.
| 34 | https://mathoverflow.net/users/370 | 4141 | 2,767 |
https://mathoverflow.net/questions/3312 | 16 | For $f: X → Y$ a morphism of schemes, does anybody know conditions for the existence of an adjunction $(f\_!,f^!)$ between the module-categories (*not the quasicoherent*), where $f\_!$ is direct image with proper support and $f^!$ is its *right* adjoint? Can this ever happen at all on the level of the module categories, or only on the derived level?
I gave an answer to [this](https://mathoverflow.net/questions/2877/when-does-the-sheaf-direct-image-functor-f-have-a-right-adjoint/2892#2892) question which is useless without knowing the above...
| https://mathoverflow.net/users/733 | When does direct image with proper support have a right adjoint? | I assume the question holds in contexts where we can glue open immersions and proper morphisms to produce $f\_!$ for $f$ separated of finite type. In particular, we shall have $f\_!=f\_\ast$ for $f$ proper.
Non derived setting ---
If $f:X\rightarrow Y$ is proper, one can ask if $f\_\ast$ has a right adjoint.
Note that, if $Y=\mathrm{Spec}(k)$ for an algebraically closed field, then $f\_\ast$ is essentially the global section functor on $X$ (if we work with reasonnable topologies like Zariski, Nisnevich, or étale). This suggests that $f\_\ast$ does not have a right adjoint in general (otherwise, it would be exact, and this would say that proper schemes have not any interesting cohomology).
If $f$ is quasi-compact and quasi-separed, then $f\_\ast$ preserves filtered colimits (this is the case if $f$ is proper). Hence the obstruction for $f\_\ast$ to have a right adjoint is only its left exactness. There are still cases where $f\_\ast$ has a right adjoint at the level of sheaves: when f is a closed immersion (for the Zariski topology), and when f is finite (for the Nisnevich topology and for the étale topology): the right exactness of $f\_\ast$ is proved by contemplating the behaviour of $f\_\ast$ on stalks of $Y$: for the Zariski topology, one uses that the points are the local rings, and that any quotient of a local ring is local; similarly, for the Nisnevich (resp. étale) topology, one uses the fact that the points are the henselian (resp. strictly henselian) rings, and that any finite extension of an henselian ring is still henselian. So in conclusion, it seems that we might expect $f^!$ to exists when $f$ is an immersion (for the Zariski topology), or when $f$ is quasi-finite (for the Nisnevich or étale topology).
Derived setting --- As, for an open immersion $j, j\_!$ is still the extension by zero, it always has a right adjoint $j^\ast$. The problem is then again to get an adjoint of $f\_!=f\_\ast$ for $f$ proper. The problem is similar to the one in the non derived case, except that the obstruction is smaller: as $f\_\ast$ is an exact functor between well generated triangulated categories (in the sense of Neeman; for this we have to work with unbouded complexes), $f\_\ast$ has a right adjoint if and only if it preserves small direct sums (this is an instance of the Brown representability theorem). There is a nice sufficient condition for this, which I will recall. Remember that an object $X$ in a triangulated category is compact if, for any integer $n$, the functor $\mathsf{Hom}(X[n],-)$ preserves small direct sums. A triangulated category $T$ is compactly generated if it admits small sums, and if there exists a small generating family $G$ in $T$ which consists of compact objects (i.e. all the element of $G$ are compact in $T$, and for an object $M$ in $T$, one has $M=0$ iff $\mathsf{Hom}(X[n],M)=0$ for any integer $n$ and any $X$ in $G$). A compactly generated triangulated category is a basic example of a well generated triangulated category. In particular, if $F:T\rightarrow T'$ is functor between compactly generated triangulated categories, then it has a right adjoint iff it preserves small direct sums. The good news are that, for such an $F$, a sufficient condition for this is the following: assume that $F$ has a left adjoint $L:T'\rightarrow T$ and that there exists a small family $G'$ of compact generators of $T'$ such that $L$ sends the elements of $G'$ to compact objects of $T$. Then $F$ preserves small direct sums (hence, has a right adjoint).
If we come back to our functor $f\_\ast$ (with $f$ proper), a sufficient condition for $f^!$ to exist is then that our derived categories of sheaves are compactly generated and that the representable sheaves form a family of compact generators: as $f^\ast$ obviously preserves representables, these assumptions give the existence of $f^!$. As for the conditions under which representables are compact or not (which is a finiteness assumption on $X$ and $Y$ themselves), some sufficient conditions have already been suggested [here](https://mathoverflow.net/questions/689/finiteness-conditions-on-simplicial-sheaves-presheaves/2027#2027) (the fact that representables form a generating family is obviously always true).
| 18 | https://mathoverflow.net/users/1017 | 4152 | 2,773 |
https://mathoverflow.net/questions/3701 | 14 | $\DeclareMathOperator\SL{SL}\DeclareMathOperator\Sp{Sp}$Suppose that $F/Q$ is a number field.
Using automorphic forms, Borel computed the ($R$-) stable cohomology of $\SL\_n(O\_F)$, and as a result, computed $K\_i(O\_F)\otimes Q$. Nowadays, using work of Suslin, Voevodksy, Rost, etc, one has an almost "complete" understanding of the actual integral groups $K\_i(Z)$, say, modulo Vandiver's conjecture. This does not *directly* give the stable cohomology of $\SL\_n$, because one set of groups is computing homotopy and the other is computing cohomology, but let us not worry about that distinction for the moment.
Borel also computed the ($R$-) stable cohomology of $\Sp\_{2n}(O\_F)$.
My question, loosely speaking, is whether one can describe the stable integral cohomology of $\Sp\_{2n}(Z)$ in as detailed away as the algebraic K-groups of $Z$ "describe" the integral cohomology of $\SL\_n(Z)$.
---
Let me summarize some of what I have found out (following up some of the answers below), mostly though emails from experts.
For affine objects, which certainly includes $Z$, K-theory is about the monoidal category
$P(A)$ of projective finitely generated A-modules, and
Hermitian K-theory is about the monoidal category
$P(A)\_h$ of objects in $P(A)$ equipped with a non-degenerate
symmetric (or skew-symmetric) form. One of the issues with computing or working with such a theory over $Z$ is that irritating issues arise in characteristic $2$, as one might expect with quadratic forms present. It seems that one might be in good shape to understand the groups $K^h\_i(Z[1/2])$. For usual K-theory, there is an excision formula relating $K\_i(Z)$ to $K\_i(Z[1/2])$ and $K\_i(F\_2)$. The latter group is "easy" (or at least was computed by Quillen).
Of interest to me in $K\_i(Z)$ are the Soulé classes. The next thing I will be thinking about is whether Soule classes can give rise to elements in the stable cohomology of $\Sp\_2n(Z)$.
Andy said some very interesting things, but I will probably be awarding the bounty to Oscar, since the paper he linked to was more directly relevant to what I was trying to find out. (Sorry Andy...but it looks like you have lots of reputation anyway!)
| https://mathoverflow.net/users/nan | Stable homology of arithmetic groups | The algebraic K-groups of Z are to the homology of SLn(Z) as the **Hermitian K-groups** of Z are to the homology of Sp2g(Z). There is a paper by Berrick and Karoubi [here](https://faculty.math.illinois.edu/K-theory/0649/BerrickK.pdf "Hermitian K-theory of the integers") in which they discuss, and make some calculations of, the Hermitian K-theory of Z and Z[1/2].
| 4 | https://mathoverflow.net/users/318 | 4158 | 2,775 |
https://mathoverflow.net/questions/4167 | 2 | Say I'm working in the space of linear transformations from $\mathbb R^n$ to $\mathbb R^n$ and I've picked a basis so I can identify with any operator a component matrix in $\mathbb R^{n\times n}$. Transposing an operator swaps components $(i,j)$ and $(j,i)$. In this setting, the transpose operation is itself a linear map from $\mathbb R^{n\times n}$ to $\mathbb R^{n\times n}$.
What does the transpose operator's component representation look like for, say, $n=2$ or $n=3$? Not what does it do when applied, but in what space does it live and what are its actual components?
How does one speak about applying this representation to a matrix? Say $T$ is the transpose representation and I'm applying it to a matrix $A$ to get $TA=A^{T}$. What is precise way to discuss this linear algebra operation in the context of the component representations of both tensors?
Big bonus points for the two lines of Mathematica or Sage to demonstrate the mechanics.
| https://mathoverflow.net/users/1195 | What are the components of a transpose operator from $\mathbb R^{n\times n}$ to $\mathbb R^{n\times n}$? | A linear map from R^{n\*n} to itself can be written as a matrix where its rows and columns are both indexed by pairs (i,j) where i,j are in {1,2,...,n}. The transpose operation is given by the matrix which has a 1 in position ((i,j),(j,i)) and a zero elsewhere. For the 2x2 case we can order the basis on R^(2\*2} as (1,1), (1,2), (2,1), (2,2) and write the matrix as
```
(1 0 0 0)
(0 0 1 0)
(0 1 0 0)
(0 0 0 1)
```
So, we get
```
(a b) -> (a c)
(c d) (b d)
```
because
```
(a) (1 0 0 0) (a)
(c) = (0 0 1 0) (b)
(b) (0 1 0 0) (c)
(d) (0 0 0 1) (d)
```
| 5 | https://mathoverflow.net/users/1233 | 4171 | 2,783 |
https://mathoverflow.net/questions/4156 | 4 | Some branches of math seem to have reasoning which is more global. There is a lot of efficiency in the proofs because the reasoning transfers easily between proofs. For other branches of math, a lot of truths seem to be more local. The proofs tend to have lots of sub-cases and exceptions. There are fewer general principles. Does anybody know why branches of math vary like this? Can you place different branches of math on this scale from being dominated by more ad hoc to being dominated by less ad hoc proofs?
| https://mathoverflow.net/users/812 | Why do branches of math vary in proof styles and what category are different branches in? | The only thing this reminds me of is [Tim Gowers's nice article](http://www.dpmms.cam.ac.uk/~wtg10/2cultures.ps) on the two cultures of mathematics, in which he compares and contrasts "geometry" (very broadly defined) and combinatorics.
The categories in the article don't exactly match up with the categories in the question, but perhaps there's some point of contact.
| 5 | https://mathoverflow.net/users/1463 | 4178 | 2,787 |
https://mathoverflow.net/questions/4183 | 14 | I just saw a post like this one, but particularly for statistical mechanics, I thought I'd ask the question in general.
Where does a mathematically trained person go to learn mathematical physics? By that I mean, what books or manuscripts are demanding in the area of mathematical maturity but not particularly demanding in the area of physics knowledge (physics maturity I guess, idk if they use that word in physics?). I myself am particularly interested in computational fluid dynamics and other kinds of computational physics, but I want to keep this general to help as many people as possible. Also, if someone knows a good book for mathematicians to help with one of the biggest difficulties I've found "Physics INTUITION" that'd be helpful.
As usual, one answer per post so votes can be tabulated well.
| https://mathoverflow.net/users/429 | Mathematical Physics? (Particularly computational) | A classic reference is Courant and Hilbert ([volume 1](http://rads.stackoverflow.com/amzn/click/0471504475), [volume 2](http://rads.stackoverflow.com/amzn/click/0471504394)).
| 2 | https://mathoverflow.net/users/136 | 4191 | 2,793 |
https://mathoverflow.net/questions/4103 | 2 | Where can I learn more about shear matrices?
The Wikipedia article is not enough, and sadly it does not have any references.
I understand they are linear transformations. Do they form a group? How do they look like for n-dimensional vectors? How many independent shears can I do in n-dimensions? Are they related to the orthogonal group SO(D), perhaps contained?
Google takes me to this other [website](http://cse.taylor.edu/~btoll/s99/424/res/mtu/Notes/geometry/geo-tran.htm). It is almost enough, but again no references.
I am trying to do some physics with this, so I do not think you will tolerate my explanation. ;-)
| https://mathoverflow.net/users/1475 | Shear transformations | It might be helpful to note that, in two dimensions, a shear transformation is exactly one whose Jordan canonical form is $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ or $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ if you include the identity as a shear transformation. These are exactly the transformations whose eigenvalues are all $1$, so-called unipotent matrices. The unipotent $2 \times 2$ matrices do not form a group: $\begin{pmatrix}1&1\\0&1\end{pmatrix}\times\begin{pmatrix}1&0\\1&1\end{pmatrix}$ is not unipotent.
| 1 | https://mathoverflow.net/users/126667 | 4193 | 2,795 |
https://mathoverflow.net/questions/4187 | 3 | If $A$ is a C\*-algebra and $n$ is a normal element of $A$, then we have: (By Gelfand duality for example.)
$\operatorname{spec}( |N| ) = | \operatorname{spec}(N) | := \left\{ | \lambda | ; \lambda \in \operatorname{spec}(N) \right\}$
where we define: $|n|:=(n^\*n)^{1/2}$. My question is, does the converse also hold?
That is if $a\in A$ and for $r>0$:
$\left\{ r e^{it} : t \in [0, 2\pi[ \right\} \cap \operatorname{spec}(a)$ is not empty if and only if $r\in \operatorname{spec}(|a|)$
implies that $a$ is normal. (Possibly some exceptions made for the zero-element) Or bluntly speaking if the mapping $a$ to $a^\*a$ does not create any "new" (or removes any "old") elements in the spectrum then $a$ is normal.
For example if $e$ is an idempotent in $A$, then $e$ is a projection if and only if $||e||=1$. Hence if $e$ is a non-projection idempotent we have $\left\{0,1\right\} = \operatorname{spec}(e) \subsetneq \operatorname{spec}(|e|)$, since $||e||>1$ and by the spectral radius $\operatorname{spec}(e^\*e)$ contains an element strictly bigger that one.
Clearly if p is a projection, then p=|p|.
| https://mathoverflow.net/users/1488 | Normal operators and it's spectrum in C*-algebras | You can take a normal operator $N$ with spectrum filling the unit disk and take its tensor product with any operator $T$ of norm less than $1$ thus effectively hiding the non-normal component's contribution to the spectrum in both $A$ and $|A|$.
| 5 | https://mathoverflow.net/users/1131 | 4202 | 2,804 |
https://mathoverflow.net/questions/4137 | 5 | Suppose M is a von Neumann algebra.
Consider a monoidal category of bimodules over M.
Here a bimodule is a Hilbert space with two normal representations of M.
The monoidal structure is given by Connes' fusion.
Alternatively we can take right W\*-correspondences (right Hilbert W\*-modules over M with a normal left action of M)
together with the completed algebraic tensor product.
What can we say about one-parameter “semigroups” of such bimodules?
More precisely, consider a family E of bimodules parametrized by a real number t > 0
such that E\_s ⊗ E\_t = E\_{s+t}. Equality here denotes isomorphism.
Does it have an “infinitesimal generator”?
What is the “exponent” then?
Do we need any “continuity” conditions to guarantee good properties of such “semigroups”?
What kind of additional restrictions we can obtain on M provided that such a family exists?
It appears to me that such objects are known as “continuous tensor product systems”.
Any references on this matter will be appreciated.
| https://mathoverflow.net/users/402 | One-parameter semigroups of bimodules | Such structures have been investigated at depth (welcome to the club!). Let me try to answer some of your questions.
1) First, let me suggest that you look at Bill Arveson's book on this subject, [Noncommutative Dynamics and E-semigroups](http://www.springer.com/math/analysis/book/978-0-387-00151-7). Arveson is considered the pioneer of modern work on these structures, and you will enjoy his book.
2) An "infinitesemal generator" can mean various things, but in some sense, remarkably, it doesn't always exist. Arveson's book contains some treatment of this issue, known as the existence of type III examples, which are due to R. T. Powers and B. Tsirelson (there is also recet work of M. Izumi).
3) Do we need continuity conditions? Measurability conditions suffice. The condition is that the bundle {E\_t}\_{t>0} be isomorphic, as a measurable bundle of Hilbert spaces, to the trivial bundle (0,infinity) X H\_0, with H\_0 some fixed Hilbert space, plus compatibility of the measurable structure with addition, multiplication, etc.
4) Extensive research has been carried out also in the case where the E\_t are Hilbert bimodules (C^\*-correspondences). Search the ArXiv for works of Michael Skeide, or Paul Muhly and Baruch Solel.
5) I should also mention: these product systems arise naturally in the study of, give rise to, and are in a one-to-one correspondence with semigroups of \*-endomorphisms on von Neumann algebras.
| 6 | https://mathoverflow.net/users/1193 | 4211 | 2,813 |
https://mathoverflow.net/questions/4119 | 3 | Lists of stellations of polyhedrons are given particular rules like in the book [*The Fifty Nine Icosahedra*](http://en.wikipedia.org/wiki/The_fifty_nine_icosahedra) which follows "Miller's Rules".
There seems to be no "correct" ruleset to use, so [more stellations are still being discovered](http://www.steelpillow.com/polyhedra/icosa/lost/lost.html) using alternate rules.
However, I have read no mention of the stellations being infinite; unlike cumulations there does seem to be some finite restriction to the total. Is this true? If not, which ruleset(s) will produce an infinite number of stellations?
| https://mathoverflow.net/users/441 | Are there infinite sets of stellations of polyhedra? | Let me quote from "In search of the lost icosahedra" the paper I mentioned above:
"Stellations of a polyhedron are obtained by extending some of its edges or faces until they intersect at a distance from the original polyhedron. One way of studying stellations is to consider the planes in which the faces of the polyhedron lie, that is, its face planes. The face planes of the regular icosahedron intersect eachother (see Appendix) to dissect space into numerous regions, of which 473 are finite cells. These cells come in just 12 shapes which form layers around the original icosahedron, itself the innermost cell. The set of cells of a given shape comprises part or all of a layer, with icosahedral symmetry. The various stellations can be obtained by selecting different combinations of these cell sets. Because there are 12 types of cell and we are not interested in the 'empty' combination, there are 212 - 1 = 4,095 possible combinations."
So in this case or any other case we will be limited to a finite number of cells. Even if we ignored the 12 types and considered all types of cells there would still be 2^473-1
types of combinations. In general there will be a finite number of regions formed by the face planes of a polyhedron and this will result in a limitation on the number combinations to a finite although possibly very large number.
| 2 | https://mathoverflow.net/users/1098 | 4213 | 2,815 |
https://mathoverflow.net/questions/1666 | 19 | This question is pretty technical, but there are some very smart people here.
Fix a quiver Q, WITH oriented cycles. Let k[[Q]] be the completed path algebra. (Like the path algebra, but we allow formal infinite sums.) My question is about what behavior I should expect when I equip Q with a generic choice of potential.
A potential, denoted S, is a formal sum of (nontrivial) closed cycles in k[[Q]]. I'm interested in the case where the coefficients of that sum are chosen generically. Let J(S) denote the Jacobian ideal of S. This is a two sided ideal which is, roughly speaking, generated by the derivatives of S. (See Derksen-Weyman-Zelevinsky, *[Quivers with potentials and their representations I: Mutations](https://arxiv.org/abs/0704.0649)*, for the precise definitions.) [DWZ] show that the space of deformations of (Q,S) is k[[Q]]/(J(S) + [,]) where [,] is the vector space (NOT usually an ideal) spanned by commutators. They define (Q,S) to be rigid if this vector space is spanned by the empty cycle.
Heuristic arguments seem to suggest that a generic potential is always rigid. But example 8.6 in [DWZ] shows that this is false. Can someone give me a better heuristic or, better yet, a theorem?
EDITED to fix errors involving cycles of length 0.
| https://mathoverflow.net/users/297 | When should I expect a quiver with potential to be rigid? | As I understand it, the quotient C(Q) = k[[Q]]/[,] is the (completed) vector space of formal linear combinations of oriented circuits in Q. Define a *detour* in Q to be an edge e and an oriented path p with the same source and sink as e. For every detour (e,p), there is a detour operator D(e,p). Given a circuit s, D(e,p)(s) is a sum of terms for each occurrence of e in s; each term is obtained by replacing that occurrence of e by p. Then the image I(S) of J(S) appears to be the span of all D(e,p)(S).
If I have all of that straight, then I can't think of a significant condition to guarantee that there is a rigid potential S. However, I can think of a significant condition to guarantee that there isn't one. Suppose for simplicity that the quiver Q has no parallel edges. Consider further just the shortest non-trivial circuits in C(Q); they span C(Q)g, where g is the oriented girth of Q. For these circuits, the only detour operators D(e,p) that matter are the trivial ones with p = e. In other words, you have no choice but to replace an edge e with itself. But often C(Q)g is bigger than the space of edges, i.e., Q could have more shortest cycles than it has edges. In this case, just for dimension reasons, I(S) can't contain C(Q)g.
What bothers me is that if this is correct, then Derksen, Weyman, and Zelevinsky worked a little harder than necessary to make their Example 8.6.
---
Now that I read on a bit more of the paper, the authors give some constructions of quivers with rigid potentials. For instance, they establish that the existence of a rigid potential is a mutation-invariant property of quivers. They also provide an example of a rigid quiver which does not mutate to an acyclic quiver. You can go a little further: In studying the cyclic part of the path algebra of a quiver, you can restrict attention to its strongly connected components. So you can start with any collection of strongly connected, rigid quivers, then connect them any way you like acyclically, and then mutate that.
It sounds like the status quo of the question is that there are several constructions of quivers that do not have rigid potentials, and several constructions of quivers that have rigid potentials, but that finding a good characterization of the dichotomy is an open problem.
Also, clearly the quivers that do not have rigid potentials are recursively enumerable. (I.e., there is an algorithm to confirm that a quiver does not have a rigid potential, but it might not terminate if it does have one.) Maybe a good remaining question is whether quivers with rigid potentials are also recursively enumerable, i.e., that the dichotomy is recursive.
---
One way to confirm that a quiver has a rigid potential is to find a J(S) that contains all paths of some length. I(S) then automatically contains all loops of that length or greater. Moreover, it does not matter whether S includes any terms beyond the cutoff length. If a rigid potential always has this property, then determining whether there is one is recursive.
| 6 | https://mathoverflow.net/users/1450 | 4215 | 2,816 |
https://mathoverflow.net/questions/4190 | 5 | Consider the following question:
Input: Two graphs G1 and G2
Question: Is the cycle matroid M(G1) isomorphic to the cycle matroid M(G2)
What is the complexity of this question?
It is well known that the two cycle matroids are isomorphic if and only if the graphs are "2-isomorphic" which means that there is a sequence of "Whitney flips" (where a graph is disconnected at a 2-vertex cutset and then reconnected with one of the pieces flipped) from one to an isomorph of the other.
This suggests that the complexity should be the same as graph isomorphism, but I cannot find a reference to this.
| https://mathoverflow.net/users/1492 | Complexity of determining if two graphs have same cycle matroid? | The following paper seems to show that this problem is polynomial equivalent to graph isomorphism (see section 5):
<http://arxiv.org/abs/0811.3859>
| 4 | https://mathoverflow.net/users/1028 | 4230 | 2,824 |
https://mathoverflow.net/questions/4224 | 81 | Is there a relationship between the eigenvalues of individual matrices and the eigenvalues of their sum? What about the special case when the matrices are Hermitian and positive definite?
I am investigating this with regard to finding the normalized graph cut under general convex constraints. Any pointers will be very helpful.
| https://mathoverflow.net/users/1501 | Eigenvalues of matrix sums | The problem of describing the possible eigenvalues of the sum of two hermitian matrices in terms of the spectra of the summands leads into deep waters. The most complete description was conjectured by Horn, and has now been proved by work of Knutson and Tao (and others?) - for a good discussion, see the [Notices AMS article](http://www.ams.org/notices/200102/) by those two authors
Depending on what you want, there should be simpler results giving estimates on the eigenvalues of the sum. A book like Bhatia's Matrix Analysis might have some helpful material.
| 72 | https://mathoverflow.net/users/763 | 4238 | 2,830 |
https://mathoverflow.net/questions/4243 | 5 | This should be a trivial question for people who know Gödel's 1st incompleteness theorem. I quote the statement the theorem from wikipedia: "Any effectively generated theory capable of expressing elementary arithmetic cannot be both consistent and complete. In particular, for any consistent, effectively generated formal theory that proves certain basic arithmetic truths, there is an arithmetical statement that is true, but not provable in the theory."
My question is: what is the meaning of 'true' in the last sentence? Let me elaborate: the only (introductory) proof of the theorem that I know starts with a specific model of the theory and constructs a sentence which is true in that model, but not provable from the theory.
So does `true arithmetical statement' in the statement of the theorem mean`true in the (implicitly) given model', or true in EVERY model?
| https://mathoverflow.net/users/1508 | Godel's 1st incompleteness theorem - clarification. | It means true in the usual model. For 1st order logic we have Godel's Completeness theorem, which guarantees that if something is true in every model, then it is actually provable in the theory.
| 6 | https://mathoverflow.net/users/828 | 4245 | 2,836 |
https://mathoverflow.net/questions/4260 | 17 | I've heard from multiple sources now that one's CV should include grants you've applied for, even if you didn't receive them or won't find out if you've received them until after your CV goes out. I haven't had much luck finding this in other people's CVs, though. I'd like to have confirmation of this from someone who's been on a hiring committee before. If this is the case, any advice on how to present this information? (example appreciated as always).
General advice about CVs is appreciated too ...
| https://mathoverflow.net/users/699 | Curriculum vitae: including grants you've applied for, not received (or not yet received). | You have a certain amount of leeway as to what kind of material you wish to include in your CV. I have seen things on CVs that were of little or no interest [edit: here I actually meant "of interest to me as a potential hirer", not personal interest]: high school honors, nonmathematical awards, etc. It doesn't make me think less of the person.
I would say that it's within reason to include a grant that you have applied for but not yet heard back from. You are inviting the reader to compute an expected value, which will presumably turn out to be positive. This cannot be said for grants that you have definitely not received, and I wouldn't normally recommend that someone put that on their CV. Still, I could imagine cases where that might be appropriate: for instance, at many research universities, given that you don't have a grant, the information that you have at least applied for one every time you had the opportunity might be looked upon favorably: ask a trusted mentor about this.
| 13 | https://mathoverflow.net/users/1149 | 4265 | 2,846 |
https://mathoverflow.net/questions/4254 | 2 | If X is an algebraic scheme, K\_0(X) has a filtration by taking the subgroups generated by coherent sheaves whose support as at most dimension k. The associated graded groups are the quotients, and there exists a natural map from the k-th Chow group A\_k(X) to the k-th graded part Gr\_k K\_0(X), just by mapping [V] to [O\_V]. This is example 15.1.5 in Fulton's book.
This even becomes an isomorphism after tensoring with Q! That's Corollary 18.3.2 in Fulton's book.
All of the definitions surely make sense for DM-stacks. We've got Chow-groups and K\_0 can be graded in the same way. I'm not sure whether the natural map above actually passes to rational equivalence, but I'll just assume this to be true.
Here's the question: Is the map still an isomorphism?
| https://mathoverflow.net/users/473 | k-th Chow Group and k-th graded part of K_0 ismorphic for DM-stacks? | Maybe you might have a look at Toën's paper:
"On motives for Deligne-Mumford stacks", IMRN No. 17 (2000), 909-928.
He defines Chow groups of a Deligne-Mumford stack X *with coefficients in the characters* of X, and proves that the corresponding graded ring of Chow groups is isomorphic to K\_0 (see the Remark following def. 3.3 of loc. cit.).
| 3 | https://mathoverflow.net/users/1017 | 4270 | 2,848 |
https://mathoverflow.net/questions/4269 | -3 | Is there a proof for no proof ?
| https://mathoverflow.net/users/1519 | what is logic without a proof system | If I understand your question you are asking can something be proved as unprovable? If so I'd suggest that Gödel's incompleteness theorems for a starting place.
>
> "Gödel's incompleteness theorems state
> that any effectively generated formal
> theory in which all arithmetic truths
> can be proved is inconsistent; hence,
> any such consistent formal theory that
> can prove some arithmetic truths can't
> prove all arithmetic truths."
>
>
>
http://en.wikipedia.org/wiki/Gödel's\_incompleteness\_theorems
| 0 | https://mathoverflow.net/users/1214 | 4271 | 2,849 |
https://mathoverflow.net/questions/4235 | 71 | I'm wondering what the relation of category theory to programming language theory is.
I've been reading some books on category theory and topos theory, but if someone happens to know what the connections and could tell me it'd be very useful, as that would give me reason to continue this endeavor strongly, and know where to look.
Motivation: I'm currently researching undergraduate/graduate mathematics education, specifically teaching programming to mathematics grads/undergrads. I'm toying with the idea that if I play to mathematicians strengths I can better instruct them on programming and they will be better programmers, and what they learn will be useful to them. I'm in the process (very early stages) of writing a textbook on the subject.
| https://mathoverflow.net/users/429 | Relating category theory to programming language theory | The most immediately obvious relation to category theory is that we have a category consisting of types as objects and functions as arrows. We have identity functions and can compose functions with the usual axioms holding (with various caveats). That's just the starting point.
One place where it starts getting deeper is when you consider polymorphic functions. A polymorphic function is essentially a family of functions, parameterised by types. Or categorically, a family of arrows, parameterised by objects. This is similar to what a natural transformation is. By introducing some reasonable restrictions we find that a large class of polymorphic functions are in fact natural transformations and lots of category theory now applies. The standard examples to give here are the free theorems, see Philip Wadler's 1989 article [*Theorems for free!*](https://doi.org/10.1145/99370.99404), in FPCA '89: Proceedings of the fourth international conference on Functional programming languages and computer architecture.
Category theory also meshes nicely with the notion of an 'interface' in programming. Category theory encourages us not to look at what an object is made of, but how it interacts with other objects, and itself. By separating an interface from an implementation a programmer doesn't need to know anything about the implementation. Similarly category theory encourages us to think about objects up to isomorphism - it doesn't precisely proclaim which sets our groups comprise, it just matters what the operations on our groups are. Category theory precisely captures this notion of interface.
There is also a beautiful relationship between pure typed lambda calculus and cartesian closed categories (CCC). Any expression in the lambda calculus can be interpreted as the composition of the standard functions that come with a CCC: like the projection onto the factors of a product, or the evaluation of a function. So lambda expressions can be interpreted as applying to any CCC. In other words, lambda calculus is an internal language for CCCs. This is explicated in Lambek and Scott. This means for instance that the theory of CCCs is deeply embedded in Haskell, because Haskell is essentially pure typed lambda calculus with a bunch of extensions.
Another example is the way structural recursion over recursive datatypes can be nicely described in terms of initial objects in categories of F-algebras. You can find some details in Philip Wadler's note [*Recursive types for free!*](https://homepages.inf.ed.ac.uk/wadler/papers/free-rectypes/free-rectypes.txt) (dated originally to 1990, labelled "DRAFT DRAFT DRAFT DRAFT DRAFT").
And one last example: dualising (in the categorical sense) definitions turns out to be very useful in the programming languages world. For example, in the previous paragraph I mentioned structural recursion. Dualising this gives the notions of F-coalgebras and guarded recursion and leads to a nice way to work with 'infinite' data types such as streams. Working with streams is tricky because how do you guard against inadvertently trying to walk the entire length of a stream causing an infinite loop? The appropriate dual of structural recursion leads to a powerful way to deal with streams that is guaranteed to be well behaved. Bart Jacobs, for example, has many nice papers in this area.
| 90 | https://mathoverflow.net/users/1233 | 4274 | 2,852 |
https://mathoverflow.net/questions/4180 | 45 | So, lots of people work on the Geometric Langlands Conjecture, and there have been a few questions around here on it (admittedly, several of them mine). So here's another one, tagged community wiki because there isn't really a "right" answer: what does GLC imply? Lots of big conjectures have well known consequences (Riemann Hypothesis and distribution of primes) but what about GLC? Are there any nice things that are known to follow from this equivalence of derived categories?
EDIT: The Geometric Langlands Conjecture says the following: Let $C$ be an algebraic curve (any field, though I think the formulation I know is only good in characteristic 0), $G$ a reductive algebraic group, $^L G$ its Langlands dual (the characters of G are cocharacters of $^L G$, if I recall correctly). Then there's a natural equivalence of categories from the derived category of coherent sheaves on the stack of $G$-local systems to the derived category of coherent $\mathcal{D}$-modules on the moduli stack of principal $^L G$-bundles, such that the structure sheaf of a point is sent to a Hecke Eigensheaf (and I'm not going to sit down and define that on top of the rest here...the idea is that $G$-local systems on the curve are equivalent to eigensheaves for some collection of operators, but actually making it precise and having a hope of being true gets technical)
Edit 2: [This paper](http://arxiv.org/pdf/math.AG/0012255.pdf) states one version of the conjecture (for $GL(n)$ only) as 1.3, after defining the Hecke operators.
| https://mathoverflow.net/users/622 | Consequences of Geometric Langlands | OK, this is a very broad question so I'll be telegraphic.
There is a sequence of increasingly detailed conjectures going by the name GL -- it's really a "program" (harmonic analysis of $\mathcal{D}$-modules on moduli of bundles) rather than a conjecture -- and only the first of this sequence has been proved (and only for $GL\_n$), but I don't want to get into this.
There are several kinds of reasons you might want to study geometric Langlands:
1. direct consequences. One application is Gaitsgory's proof of de Jong's conjecture (arXiv:math/0402184). If you prove the ramified geometric Langlands for $GL\_n$, you will recover L. Lafforgue's results (Langlands for function fields), which have lots of consequences (enumerated eg I think in his Fields medal description), which I won't enumerate. (well really you'd need to prove them "well" to get the motivic consequences..)
In fact you'll recover much more (like independence of $l$ results). To me though this is the least convincing motivation..
2. Original motivation: by understanding the function field version of Langlands you can hope to learn a lot about the Langlands program, working in a much easier setting where you have a chance to go much further. In particular the GLP (the version over $\mathbb{C}$) has a LOT more structure than the Langlands program -- ie things are MUCH nicer, there are much stronger and cleaner results you can hope to prove, and hope to use this to gain insight into underlying patterns.
By far the greatest example of this is Ngo's proof of the Fundamental Lemma --- he doesn't use GLP per se, but rather the geometry of the Hitchin system, which is one of the key geometric ingredients discovered through the GLP. To me this already makes the whole endeavor worthwhile..
1. Relations with physics. Once you're over $\mathbb{C}$, you (by which I mean Beilinson-Drinfeld and Kapustin-Witten) discover lots of deep relations with (at least seemingly) different problems in physics.
a. The first is the theory of integrable systems -- many classical integrable systems fit into the Hitchin system framework, and geometric Langlands gives you a very powerful tool to study the corresponding quantum integrable systems. In fact you (namely BD) can motivate the entire GLP as a way to fully solve a collection of quantum integrable systems. This has has lots of applications in the subject (eg see Frenkel's reviews on the Gaudin system, papers on Calogero-Moser systems etc).
b. The second is conformal field theory (again BD) --- they develop CFT (conformal, not class, field theory!) very far towards the goal of understanding GLP, leading to deep insights in both directions (and a strategy now by Gaitsgory-Lurie to solve the strongest form of GLP).
c. The third is four-dimensional gauge theory (KW). To me the best way to motivate geometric Langlands is as an aspect of electric-magnetic duality in 4d SUSY gauge theory. This ties
in GLP to many of the hottest current topics in string theory/gauge theory (including Dijkgraaf-Vafa theory, wall crossing/Donaldson-Thomas theory, study of M5 branes, yadda yadda yadda)...
1. Finally GLP is deeply tied to a host of questions in representation theory, of loop algebras, quantum groups, algebraic groups over finite fields etc. The amazing work of Bezrukavnikov proving a host of fundamental conjectures of Lusztig is based on GLP ideas (and can be thought of as part of the local GLP). (my personal research program with Nadler is to use the same ideas to understand reps of real semisimple Lie groups). This kind of motivation is secretly behind much of the work of BD --- the starting point for all of it is the Beilinson-Bernstein description of reps as $\mathcal{D}$-modules.
There's more but this is already turning into a blog post so I should stop.
| 61 | https://mathoverflow.net/users/582 | 4275 | 2,853 |
https://mathoverflow.net/questions/4276 | 24 | It has been many years since I first read Categories for the Working Mathematician, but I still have a question about one of the first exercises. Question 5 in section 1.3 asks you to find two different functors $\mathsf{T}: \mathsf{Groups} \to \mathsf{Groups}$ with object function $\mathsf{T}(\mathsf{G}) = \mathsf{G}$ for every group $\mathsf{G}$. I have played with this for a long time, and none of the obvious choices end up working. Was this a mistake on Mac Lane's part, or am I just missing something very obvious?
If it turns out there are no "obvious" choices, does anyone have an idea of how to prove that there are **not** two such functors?
| https://mathoverflow.net/users/1106 | Two functors from Grp to Grp? | This is an "[evil](http://ncatlab.org/nlab/show/evil)" question, which deserves an evil answer. Pick your favorite pair of an object G0 of Groups and a nontrivial automorphism φ of G0. Define the functor T : Groups → Groups by T(G) = G, T(f) = [φ ∘] f [∘ φ-1] where we compose with φ if the target of f is G0 and compose with φ-1 if the domain of f is G0. This T is clearly different from the identity functor (it acts by φ on Hom(Z, G0) = the underlying set of G0).
This answer isn't very satisfying though because T is naturally isomorphic to the identity. I don't know whether one can find an example where T is not naturally isomorphic to the identity.
| 33 | https://mathoverflow.net/users/126667 | 4278 | 2,854 |
https://mathoverflow.net/questions/4268 | 10 | Suppose $f$ is a weight $2$ level $N$ cusp form. When can we realize the mod-$\ell$ representation of $f$ in a form of weight $2$ and level $Np^3$, where $p$ is some prime not dividing $N$? I assume that, if a simple criterion exists at all, it is a condition on the mod-$\ell$ representation of $f$ restricted to inertia at $p$, but I'm not sure what it would say...
| https://mathoverflow.net/users/1464 | Level raising by prime powers | Presumably you want the form (let me call it g) of level Np^3 to be new at p, otherwise it's trivial.
Let me also assume ell isn't p.
If the form g is new at p, and has level Gamma0(p^3) at p, then the ell-adic representation attached to g will have conductor p^3. But this is a bit of a problem, because the conductor of the mod ell representation can't be that much lower than the conductor of the ell-adic representation. Indeed a theorem of Carayol and, independently, Livne, says that the p-conductor of the mod ell representation will be at least p if the p-conductor of the ell-adic representation is p^3 (the exponent can drop by at most 2). So if you're looking for Gamma\_0(p^3) then you're in trouble. This is just a local calculation and isn't too deep.
Diamond and Taylor, in their second paper on the subject, give a list of the conductors of the newforms that can give rise to a given irreducible modular mod ell representation. You can see that Gamma0(p^3) is too much from the main theorem there. Of course the work in that theorem is realising everything that is possible, not ruling out everything that isn't.
| 10 | https://mathoverflow.net/users/1384 | 4282 | 2,856 |
https://mathoverflow.net/questions/1168 | 4 | This question is related to [my earlier, even more open-ended question](https://mathoverflow.net/questions/406/how-is-tropicalization-like-taking-the-classical-limit) on tropilcalization. I will give some background and ask my question at the end.
On **R**, consider the family of commutative, associative operations ⊕*h*, indexed by positive *h*, given by *x* ⊕*h* *y* = -*h* ln( exp(-*x*/*h*) + exp(-*y*/*h*) ). For *h*>0, the semigroup (**R**,⊕*h*) is isomorphic to the normal additive groupsemi (**R**>0,+). But as *h* → 0, for fixed *x* and *y* we have the limit *x* ⊕*h* *y* → min(*x*,*y*). This defines the *tropical addition*, and it's conventional to include the additive unit ∞ = -*h* ln(0).
There is a continuous/integral version of the observation that in the limit, + (in the guise ⊕*h*) becomes max. Indeed, let *f* : **R***n* → **R** be a continuous function bounded below, and assume that *f* grows to +∞ in all directions, fast enough so that for any *h*>0, the integral ∫**R***n* exp(-*f*(*x*)/*h*) *dx* converges (or anyway for *h* small enough; if it converges for any *h* then it does for all smaller *h*, and to converge for small *h* requires only very mild growth rates; as |*x*|ε for ε>0 is certainly good enough). Then asymptotically as *h* → 0, the integral is supported at the (or, rather, in a formal neighborhood of the) globally-minimal values of *f*. To make the correspondence explicit, note that ∫**R***n* exp(-*f*(*x*)/*h*) *dx* is (exp of -*h*-1 times) the "⊕*h* integral" of *f*, whereas the "⊕0 integral" of a function is its global minimum value.
There is another fact about asymptotic integrals, related by "Wick rotation", which is what the physicists call it any time you switch a variable from pure-real to pure-imaginary. As above, let *f* : **R***n* → **R** continuous and growing reasonably quickly to infinity, but this time for real non-zero *h* consider the integral ∫**R***n* exp(-*f*(*x*)/(*ih*)) *dx*, where *i* = √-1. The integral never converges absolutely (and so does not exist in the sense of Lebesgue), but it converges conditionally as a Riemann integral, e.g. if *f* is differentiable and given mild conditions on the growth of the norm of the derivative. (If *f* grows at least as fast as |*x*|1+ε, we're fine, I think.) In any case, let's assume that the integral converges conditionally for small enough (real, non-zero) *h*. Then the method of stationary phase shows that asymptotically, the integral is supported at (formal neighborhoods of) critical points of *f*.
My question is this: Is there a version of "tropical arithmetic" like the operation ⊕*h* defined above but related to the Wick-rotated integral? The most naive approach, replacing *h* by *ih* and so considering *x* ⊕*ih* *y* = -*ih* ln( exp(-*x*/*ih*) + exp(-*y*/*ih*) ), is not defined because of the problem of picking a branch of the logarithm. But perhaps this problem can be fixed for small *h*, or by approximating each pure-imaginary *ih* by *ih*+ε for some very small positive ε? Put another way: what is the operation on numbers that corresponds to {critical points} in the same way that min(*x*,*y*) corresponds to {global minimum}?
| https://mathoverflow.net/users/78 | "Wick rotation" of tropical geometry | There has been very little activity on this question, so I'm going to take it off the unanswered list. In particular, in [a related question](https://mathoverflow.net/questions/4228/), kilimanjaro linked to [this paper](http://arxiv.org/abs/math.GM/0507014), which answers some of my questions and includes many references.
| 1 | https://mathoverflow.net/users/78 | 4284 | 2,858 |
https://mathoverflow.net/questions/4259 | 8 | The following topic came up in conversation with my office-mate Lionel: Let $p$ be a fixed prime, $c$ a fixed positive real parameter and $n$ a large number. Consider a random $(0,1)$ matrix with entries in $Z/p$, where the probability of a $0$ is $1-\frac{c}{n}$ and that of a $1$ is $\frac{c}{n}$. As $n\rightarrow \infty$, what is the probability that this matrix is singular? UPDATE: As moonface points out, this probability is $1$. Furthermore, his argument points out that we should expect the corank to be something like $a\*n$. So I'll ask more generally what we can say about the behavior of the rank as $n\rightarrow\infty$.
Actually, in our motivating example, the matrix is symmetric and the distribution on the diagonal is different than in the rest of the matrix. I left these details out, but please mention it if this point would strongly effect your answer.
| https://mathoverflow.net/users/297 | Singularity of sparse random matrices | A few observations:
-As n tends to infinity, the function corank/n is highly concentrated for each n -- for example, we can think of exposing the matrix minor by minor (looking at the upper left kxk matrix for k increasing towards n). Since changing what happens at each level of exposure can only affect the rank of the matrix by at most 2, it follows from Azuma's inequality that the rank is concentrated in an interval of width about Sqrt(n).
-A recent preprint of Bordenave and Lelarge ([The rank of diluted random graphs](http://arxiv.org/abs/0907.4244)) may be of relevance here.
There they consider a model which includes the adjacency matrices of Erdos-Renyi graphs with edge probability c/n and show that the ratio rank/n converges to an explicit function (which also turns out to be the size of the maximum matching in the graph).
The differences between their work and your model are
(1) They consider matrices over the reals instead of over a finite field.
and
(2) They make assumptions about the random graphs converging locally to a tree. Their results definitely apply if the diagonal entries are 0, and from interlacing their results should also hold if your distribution doesn't let too many of the diagonal entries become non-zero.
However, I would expect a drastically different corank if a positive proportion of the diagonal entries are non-zero. The main contribution to the corank in these graphs come from non-expanding sets of rows (sets of k rows which have nonzero entries in fewer than k columns). By adjusting the diagonal entries, I can remove THAT source of dependency, though at the same time I'll possibly be creating others. (EDIT: For example, if I set all of the diagonal entries equal to 1, I've created a new constant\*n sources of dependency corresponding to the isolated edges in the graph.
Going back to assumption (1), my guess is that it would make only a small difference in the rank of the matrix. My intuition for this is twofold.
(1) If we consider non-symmetric dense matrices instead of sparse ones (e.g. by requiring the probability any entry takes on any one value to be bounded away from 1), then the probability the matrix has rank n-k is exponentially small in k
(2) If c is sufficiently small, then (as observed by [Bauer and Golinelli](http://arxiv.org/abs/cond-mat/0102011)) we can get a good estimate on the rank of the random matrix by repeatedly following the following procedure: Locate a vertex with exactly one neighbor in the graph (a row and column with exactly one non-zero entry), and remove both that vertex and its neighbor from the graph. Doing so will reduce the rank of the matrix by exactly 2 regardless of the field we are working over. For c less than e, we can keep on following this procedure until o(n) non-isolated vertices remain, implying that the field we are working over will only affect the rank of the matrix by o(n). For c greater than e, I see no reason to expect any different behavior.
Note that the argument in (2) here is, again, heavily dependent on the distribution of the diagonal entries and breaks down if the entries are nonzero. Still, I wouldn't expect it to matter too much.
| 7 | https://mathoverflow.net/users/405 | 4290 | 2,864 |
https://mathoverflow.net/questions/4216 | 21 | From Wiener's tauberian theorem we know that linear combinations of translates of f \in L^1(R) are dense in L^1(R) if and only if the Fourier transform of f never vanishes. It is also known that linear combinations of translates of f \in L^2(R) are dense in L^2 if and only if the Fourier transform of f is nonzero almost everywhere. Is there a characterization (in terms of the Fourier transform) of functions in L^p(R) with the property that linear combinations of its translates are dense in L^p?
If the answer is no can it be shown that no reasonable measure of the size of the zero set of the Fourier transform of f will suffice to give such a characterization?
| https://mathoverflow.net/users/630 | Is there an L^p tauberian theorem? | Actually this is a well known question. N. Lev and A. Olevskii have shown the following theorem:
**Theorem (Lev, Olevskii)** Given any 1 < p < 2 one can find two vectors in $l^1(Z)$, such that one is cyclic in $l^p(Z)$ and the other is not, but their Fourier transforms have an identical set of zeros.
The same result follows for $L^p(R)$.
Look [here](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6X1B-4SK0C3K-2&_user=10&_rdoc=1&_fmt=&_orig=search&_sort=d&_docanchor=&view=c&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=71a092e5dc0badfd237266b26a087a44) for example or on arxiv under Olevskii or Lev. This means more or less that for $p\neq 1,2$, there can be no characterization of $L^p$ generators in terms of the zero set of the Fourier transform. Hope this helps.
PS: Maybe I should add that I have the impression that this is a big open problem so you shouldn't expect an 'easy' answer. It is not clear in what terms one should seek for such a characterization. I would contact Nir Lev for more information (you can look for his e-mail on his web site).
| 17 | https://mathoverflow.net/users/881 | 4291 | 2,865 |
https://mathoverflow.net/questions/4153 | 2 | Suppose we have a Hamiltonian action of a torus $T = T^m = R^m/Z^m$ on a compact, connected symplectic manifold $M$. According to the convexity theorem, we know every fiber of the momentum map $\mu: M\to R^m$ is connected. My question here is about the proof.
We assume the Hamiltonian action is effective without loss of generality, i.e., only the zero point of $T$ fixes $M$. I already know that the set of regular values of $\mu$ is dense in $\mu(M)$, also the set of points $\eta$ in $\mu(M)$ with $(\eta\_1, \dotsc , \eta\_{m-1})$ a regular value for the reduced momentum map $(\mu\_1, ..., \mu\_{m-1})$ is dense in $\mu(M)$. I also know that the fiber of $\eta$ is connected whenever $(\eta\_1, ..., \eta\_{m-1})$ a regular value for the reduced momentum map. "Since the set of such points is dense in $\mu(M)$ it follows by continuity that the fiber of $\eta$ is connected for every regular value $\eta$." (in the book by McDuff and Salamon), and this is my first question. My second question is that then we can imply that every fiber of $\mu$ is connected?
Notice that here the Hamiltonian action must play a special role, as the following is not true:
Suppose $f: M \to N$ is a smooth map between two smooth manifolds, with $M$ compact and connected, and suppose there is a dense subset of $f(M)$ where each fiber is connected, then each fiber of $f$ is connected.
Example: consider a natural smooth surjection from $S^1$ to the figure eight. The fiber over the nodes of the figure eight has two points but every other fiber is a single point.
If anyone knows how to prove the connectedness part of the Convexity Theorem, could you please show us? Thank you very much!
| https://mathoverflow.net/users/1468 | Convexity Theorem of Hamiltonian actions - the connectedness part | I think the key point is that given a Hamiltonian torus action, the components of the moment map are Morse-Bott functions which have *even dimensional* critical submanifolds all with *even index*.
For example, suppose you have a Hamiltonian circle action on X. If p is fixed by the action, the circle acts on the tangent space TpX at p and so TpX decomposes into eigenspaces each of which is necessarily even-dimensional. The spaces where the action is trivial are tangent to the fixed locus, those with negative weight are where the hessian is negative definite and those of positive weight are where the hessian is positive definite. It follows that each component of the critical locus has even dimension and even index.
To see why this implies that the fibres are connected, at least in the case of a circle action, one applies Mores-Bott theory. When t passes a critical level the level set mu-1(t) changes by a certain type of surgery. The only surgeries which can alter the connectedness of the level-set are those of index or coindex 1, but we have just seen that this never happens for a Hamiltonian circle action. So all the fibres are connected.
To pass from a circle action to a torus action one can use induction. If you get stuck, searching for things like "Morse-Bott even index torus" should help you find the proof on Google somewhere. Alternatively, if you are feeling more traditional, you could look at Atiyah's original proof in your library: "Covnexity and commuting Hamiltonians" in Bulletin of the LMS 1982 14(1).
| 2 | https://mathoverflow.net/users/380 | 4303 | 2,875 |
https://mathoverflow.net/questions/3887 | 4 | In what sense does Lusztig's quantum Frobenius, defined on a quantum enveloping algebra, generalise the classical Frobenius mapping on a variety over a finite field?
| https://mathoverflow.net/users/1095 | Quantum Frobenius | There is one sense in which I'd say that Lusztig's Frobenius morphism is a generalization of the Frobenius morphism on a variety: In [this](http://arxiv.org/abs/math/0005246) paper, Kumar and Littelmann show that Lusztig's quantum Frobenius morphism induces a Frobenius morphism on a quantized analog of the multicone over a flag variety (which they call a "lift of the Frobenius morphism to characteristic 0"). Upon specialization and base change this morphism becomes the standard Frobenius morphism on the flag variety.
| 2 | https://mathoverflow.net/users/1528 | 4319 | 2,887 |
https://mathoverflow.net/questions/4318 | 3 | Sorry if this question is too simple.
I once read, on a number theory textbook - forget the title, in one of the problems list that all Pythagorean triplets when multiplied are divisible by 60.
I proved that using the generating functions (is this the correct name? I got the name from my Discrete Mathematics textbook):
\begin{align}
a &= p^2 - q^2 \\\
b &= 2pq \\\
c &= p^2 + q^2.
\end{align}
I proved it by proving all possible parities of $p$ and $q$. It's tedious because I have to prove some cases are not possible (like $a$, $b$, and $c$ can't be all even or odd).
My questions are:
1. Who and how someone came up with the generating functions?
2. If you don't know the generating functions or don't want to prove it like I did, is there any other way to prove it? Geometrically? Using Calculus? I mean there're many ways to prove Pythagorean theorem using Geometry, Number Theory, etc.
| https://mathoverflow.net/users/1538 | Proof of "if $a^2 + b^2 = c^2$ then $abc$ is divisible by 60" | 1. Generating functions is not really the right name. I would say "parameterization."
2. These formulas were known to the Babylonians. The simple proof goes as follows: assume WLOG that a, b, c are relatively prime. Since a, b, c cannot all have the same parity, WLOG b, c have different parity. Then a^2 = (c + b)(c - b) where the factors on the RHS are odd and relatively prime (use the Euclidean algorithm), so they must both be squares, say p^2 and q^2 (use unique prime factorization.) Then c = p^2 + q^2, b = p^2 - q^2, and this gives a = 2pq.
3. It's equivalent to showing that abc is divisible by 3, 4, 5. This is straightforward if you know that squares are congruent to 0, 1 mod 3, congruent to 0, 1 mod 4, and congruent to 0, 1, 4 mod 5 because 1 + 1 != 1 mod 3 or mod 4 and 1 + 1, 1 + 4, and 4 + 4 are not equal to 1 or 4 mod 5. This implies that two squares which are not divisible by 3, 4, 5 cannot sum to a square which is not divisible by 3, 4, 5.
| 14 | https://mathoverflow.net/users/290 | 4320 | 2,888 |
https://mathoverflow.net/questions/4310 | 5 | Let's say we have a sequence $T(n)$ with the corresponding generating function
$$A(t) = \sum\_{n = 0}^\infty T(n) t^n$$
Is there some relationship between the two functions $A(t)$ and $A(t^2)$? And for that matter is there some generalization for any integer power or $t$?
**Edit:** I'm actually trying to solve for the generating function $A(t)$ in the equation
$$A(t) + (1+t)A(t^2) = t/(1-t^2)$$
this is what inspired my question. My intuition suggested to me that I should look for some kind of relationship between $A(t^2)$ and $A(t)$, hence the vagueness of my question.
| https://mathoverflow.net/users/1447 | Generating-functions: is there a relationship between a generating function and the corresponding squared generating function | Alright, so on the one side, you have this:
$$A(t)+(1+t)A(t^{2})=\sum\_{n=0}^{\infty}T(n)t^{n}+\sum\_{n=0}^{\infty}T(n)t^{2n}+\sum\_{n=0}^{\infty}T(n)t^{2n+1}$$
On the other side, you have:
$$\frac{t}{1-t^{2}}=\sum\_{n=0}^{\infty}t^{2n+1}$$
Equating the coefficients of $x^{2k}$, you have the relation: $T(2k)+T(k)=0$.
Equating the coefficients of $x^{2k+1}$, you have the relation: $T(2k+1)+T(k)=1$.
Now you can start computing the coefficients: $T(0)=0$, $T(1)=1$, $T(2)=-1$, $T(3)=0$, etc.
sigfpe correctly identified [the sequence](http://oeis.org/A065359 "the sequence"). You can even see these recurrences mentioned in the formula section.
| 10 | https://mathoverflow.net/users/855 | 4328 | 2,894 |
https://mathoverflow.net/questions/4335 | 5 | At the risk of asking an uninformed question...
Imagine an ant on a compact two-dimensional surface embedded in 3-space. The ant is placed at a point on the surface with random orientation. Once placed and oriented, the ant will walk along a local geodesic path.
Are there examples of such 2D surfaces where we are guaranteed the ant will never return to some starting position?
(Thanks to Henry Wilton for requesting clarification as per his comment below.)
| https://mathoverflow.net/users/774 | A walk on a compact 2D surface embedded in 3-space that never returns home | Silly answer: any plane in 3-space will work (so the answer is "yes").
You probably want a compact surface. These always admit geodesic cycles (so the answer in this case is "no").
| 6 | https://mathoverflow.net/users/121 | 4339 | 2,899 |
https://mathoverflow.net/questions/2828 | 11 | Let $\mathfrak{s} = \mathfrak{s}\_0 \oplus \mathfrak{s}\_1$ be a real Lie superalgebra. (The ground field does not matter much, but at least one formula will not work as written if the characteristic is 2 or 3.) Recall that this means that there is a bilinear 2-graded bracket $[-,-]$ with three components
(a) $\mathfrak{s}\_0 \times \mathfrak{s}\_0 \to \mathfrak{s}\_0$ (skewsymmetric)
(b) $\mathfrak{s}\_0 \times \mathfrak{s}\_1 \to \mathfrak{s}\_1$
(c) $\mathfrak{s}\_1 \times \mathfrak{s}\_1 \to \mathfrak{s}\_0$ (symmetric)
satisfying the Jacobi identity, which splits into 4 components, which can be paraphrased as
(1) $\mathfrak{s}\_0$ is a Lie algebra under (a)
(2) $\mathfrak{s}\_1$ is an $\mathfrak{s}\_0$-module under (b)
(3) the map in (c) is $\mathfrak{s}\_0$-equivariant
(4) $[[x,x],x] = 0$ for all $x \in \mathfrak{s}\_1$
The fact that the first three components can be written using words, whereas the fourth is easiest via a formula, suggests that they should perhaps be treated differently.
Indeed, over time I have come across many examples of superalgebras where the first three components of the Jacobi identity are satisfied but not the fourth. I'd like to call them **3/4-Lie superalgebras**. I would like to know how far can this notion be pushed and in particular how much of the theory of Lie superalgebras still works in the 3/4 case.
To motivate this seemingly random question, let me end by pointing out one generic example where they arise. There are others, but they are lengthier to describe.
Let $\mathfrak{g}$ be a *metric Lie algebra*; that is, a Lie algebra with an ad-invariant inner product $(-,-)$ and let $V$ be a *symplectic $\mathfrak{g}$-module*; that is, one possessing a $\mathfrak{g}$-invariant symplectic form $\langle-,-\rangle$. Now let $\mathfrak{s} = \mathfrak{g} \oplus V$. Then maps (a) and (b) are obvious: given by the Lie bracket on $\mathfrak{g}$ and the action of $\mathfrak{g}$ on $V$, respectively. Map (c) is the transpose of map (b) using the inner products of both $\mathfrak{g}$ and $V$; in other words, if $x,y \in V$ then $[x,y] \in \mathfrak{g}$ is defined by
$$([x,y],a) = \langle a\cdot x,y\rangle$$
for all $a \in \mathfrak{g}$.
Then it is easy to see that $[x,y] = [y,x]$ and that (1)-(3) are satisfied, whereas in general (4) is **not** satisfied and instead defines a subclass of symplectic $\mathfrak{g}$-modules.
| https://mathoverflow.net/users/394 | 3/4-Lie superalgebras: how much of a theory can one develop? | At least in the semisimple case, it doesn't seem like there can be a theory of 3/4-Lie superalgebras other than a fairly predictable set of examples within Cartan-Weyl theory. I don't know how to do all of the relevant calculations, but it is easy to see roughly how they would go.
Suppose that the even part $s\_0$ is a semisimple Lie algebra $g$. Then $s\_1$ is can be a self-dual representation $V$ of $g$. Unless possibly $g$ is $E\_6$, I don't think that $V$ can be anything other than a self-dual representation. So then the form is a symmetric element of $\mathrm{Inv}(g \otimes V \otimes V)$, which is a vector space whose dimension can be computed. The dimension is at most the rank of $g$ when $V$ is irreducible. Indeed it is usually the rank of $g$, when the highest weight of $V$ is far away from the walls of the Weyl chamber. The dimension of the symmetric part is smaller, but it is often non-zero. (This is the part that I don't know how to compute off the top of my head.)
So inevitably there will be a classification of these 3/4 Lie algebras using these branching rules. Besides the classification, the only theory that springs to mind is representations of these 3/4 Lie algebras. A representation $W$ would first off be a representation of $g$ (and I suppose a super vector space?). Then there would be a $g$-invariant map $V \otimes W \to W$ which represents the action of $V$. I don't see a rationale for imposing restrictions on this map other than $g$-invariance, for otherwise $g \oplus V$ would not be the adjoint representation of itself. So once again, there is some Cartan-Weyl classification that says what $W$ can be, and I'm not sure what more you could say.
| 5 | https://mathoverflow.net/users/1450 | 4344 | 2,902 |
https://mathoverflow.net/questions/4331 | 19 | * Is the wedge product of two harmonic forms on a compact Riemannian manifold harmonic? I'm looking for a counter-example that the textbooks say exists.
* I would like to see a counter example that is on a complex manifold, Ricci-flat (or Einstein) manifold or both, if it is at all possible.
* In general, I'm trying to understand the interaction between the wedge product, Hodge star and the Laplacian on forms and it's eigen-vectors, references will be much appreciated.
| https://mathoverflow.net/users/1544 | Is the wedge product of two harmonic forms harmonic? | It is easy to construct examples on Riemann surfaces of genus $>1$. Take any surface like this. Let $A$ and $B$ be two harmonic $1$-forms, that are not proportional. Then $A \wedge B$ is non-zero, but it vanishes at some point, since both $A$ and $B$ have zeros. At the same time a harmonic $2$-form on a Riemann surface is constant.
Explicit examples of $1$-forms on Riemann surfaces can be obtained as real parts of holomorphic $1$-forms.
Note of course that the above example is complex, and Einstein just take the standard metric of curvature $-1$. If you want an example on a Ricci flat manifold you should take a $K3$ surface. It is complex and admits a Ricci flat metric. Now, its second cohomology has dimension $22$. Now it should be possible to find two anti-self-dual two-forms whose wedge product vanishes at one point on $K3$ but is not identically zero. This is because the dimension of the space of self dual forms is 19 which is big enought to get vanishing at one point
| 32 | https://mathoverflow.net/users/943 | 4355 | 2,906 |
https://mathoverflow.net/questions/4374 | 0 | Define a sequence $(C\_n)$ by $C\_0=1$, $C\_1=1$, and $C\_{n+1} = \sum\_{r=0}^n C\_r C\_{n-r}$ for $n\geq 2$. What is the simplest explicit formula for $C\_n$?
| https://mathoverflow.net/users/1553 | What is the simplest non-recursive formulation for the following recursive function? | You want [Catalan number](http://en.wikipedia.org/wiki/Catalan_number)
| 5 | https://mathoverflow.net/users/nan | 4375 | 2,919 |
https://mathoverflow.net/questions/4381 | 2 | I know how to pronounce Dijkstra's name correctly (hear it here: <http://en.wikipedia.org/wiki/Edsger_W._Dijkstra>).
But I'd like to know how people usually say his name. I've heard it in many different ways throughout my career, and since I'm teaching a course on graphs and Dijkstra's algorithm, I don't want to teach the real pronunciation since nobody seems to use it. I want to teach the most common pronunciation (although I shall mention the real one).
I appreciate your reply. Thank you.
| https://mathoverflow.net/users/1172 | Pronunciation: Dijkstra | I've always heard it basically the same way as Wikipedia, except with an American accent. Basically "dike' struh", with the accent on the first syllable as indicated, and struh is the same as in Strunk and strum.
| 11 | https://mathoverflow.net/users/1450 | 4385 | 2,925 |
https://mathoverflow.net/questions/4379 | 15 | Is/should there be a theory of finite solvable extensions over a given base field? Could it be based on/use class field theory? Assume the base field isn't a local field.
| https://mathoverflow.net/users/1555 | Solvable class field theory | As FC says, since solvable extensions are built up out of abelian extensions, class field theory is certainly relevant and helpful in understanding the structure of solvable extensions. On the other hand, to do this in a systematic way requires understanding class field theory of each number field in a tower "all at once". The picture that one gets in this way seems quite blurry compared to the classical goal of class field theory: to describe and parameterize the finite abelian extensions L of a field K in terms of data constructed from K itself. In the case of a number field, this description is in terms of groups of (generalized) ideal classes, or alternately in terms of quotients of the idele class group. I'm pretty sure there's no description like this for solvable extensions of
any number field.
What I can offer is a bunch of remarks:
1) Sometimes one has a good understanding of the entire absolute Galois group of a field K, in which case one gets a good understanding of its maximal (pro-)solvable quotient. Of course this happens if the absolute Galois group is abelian.
2) Despite the OP's desire to exclude local fields, this is one of the success stories: the full absolute Galois group of a $p$-adic field is a topologically finitely presented prosolvable group with explicitly known generators and relations.
3) On the other hand, we seem very far away from an explicit description of the maximal solvable extension of Q. For instance, in the paper
MR1924570 (2003h:11135)
Anderson, Greg W.(1-MN-SM)
Kronecker-Weber plus epsilon. (English summary)
Duke Math. J. 114 (2002), no. 3, 439--475.
the author determines the Galois group of the extension of Q^{ab} which is obtained by taking the compositum of all quadratic extensions K/Q^{ab} such that K/Q is Galois. Last week I heard a talk by Amanda Beeson of Williams College, who is working hard to extend Anderson's result to imaginary quadratic fields.
4) This question seems to be mostly orthogonal to the "standard" conjectural generalizations of class field theory, namely the Langlands Conjectures, which concern
finite dimensional complex representations of the absolute Galois group.
5) A lot of people are interested in points on algebraic varieties over the maximal solvable extension Q^{solv} of Q. The field arithmeticians in particular have a folklore conjecture that Q^{solv} is Pseudo Algebraically Closed (PAC), which means that every absolutely irreducible variety over that field has a rational point. This would have applications to things like the Inverse Galois Problem and the Fontaine-Mazur Conjecture (if that is still open!). Whether an explicit description of Q^{solv}/Q would be so helpful in these endeavors seems debatable. I have a paper on abelian points on algebraic varieties, in which the input from classfield theory is minimal.
The two papers on solvable points that I know of (and very much admire) are:
MR2057289 (2005f:14044) Pál, Ambrus Solvable points on projective algebraic curves. Canad. J. Math. 56 (2004), no. 3, 612--637.
MR2412044 (2009m:11092) Çiperiani, Mirela; Wiles, Andrew Solvable points on genus one curves. Duke Math. J. 142 (2008), no. 3, 381--464.
| 13 | https://mathoverflow.net/users/1149 | 4386 | 2,926 |
https://mathoverflow.net/questions/4329 | 73 | During a talk I was at today, the speaker mentioned that if you truncate the Taylor series for $e^x - 1$, you'll get lots of roots with nonzero real part, even though the full Taylor series only has pure imaginary roots.
If you plot the roots of truncations of $e^x - 1$ (or check out the ready-made plots in [this *Mathematica* notebook](http://www.ma.utexas.edu/users/afenyes/share/truncated-exponential.nb), now available as a [PDF](http://www.ma.utexas.edu/users/afenyes/share/truncated-exponential.pdf)) you can see lots of cool features. I'd like to know where they come from! I know there's a vast literature on polynomials, but I'm a total beginner, and I don't know where to start.
Here are a few specific questions:
1. The roots of a high-degree truncation seem to fall into two categories: roots that lie very close to the imaginary axis, and roots that lie on a C-shaped curve. (Another interpretation is that all of the roots lie on a curve, which has a very sharp kink near the imaginary axis.) Can you write down an equation for the curve?
2. If you put the roots of a lot of consecutive truncations together on the same plot, you'll see definite "stripes" to the right of the imaginary axis. Once a stripe appears, each higher-degree truncation sticks another root onto the end, making the stripe grow outward. Can you write down equations for the stripes?
3. If $k$ is odd, the truncation of degree $k$ has no nonzero real roots. If *k* is even, the truncation of degree $k$ has one nonzero real root. The location of this root depends *almost* linearly on $k$. Why is the dependence so close to linear? Does it get more linear as $k$ increases, or less?
4. Can roots be given identities that persist across time? That is, as $k$ increases, can you point to a sequence of roots and say, "those are all the same individual, which was born at $k$ = so-forth, is following such-and-such trajectory, and will grow up to become the root ($2\pi i\cdot$ whatever) of $e^x - 1$"?
| https://mathoverflow.net/users/1096 | Roots of truncations of $ e^x - 1$ | I finally got around to googling this a bit, and I immediately came up with <http://www.mai.liu.se/~halun/complex/taylor/> which describes the same phenomenon for the exponential function itself. Briefly, if Pn is the Taylor polynomial of ex, then the zeroes of Pn(nx) pile up on the curve |ze1-z|=1, |z|≤1. They credit this discovery to Szegő (1924). See also the paper
*On the Zeros of the Taylor Polynomials Associated with the Exponential Function* by Brian Conrey and Amit Ghosh (The American Mathematical Monthly, **95**, No. 6 (1988), pp. 528-533), <http://www.jstor.org/stable/2322757> if you have JSTOR access.
| 31 | https://mathoverflow.net/users/802 | 4413 | 2,949 |
https://mathoverflow.net/questions/4347 | 72 | The question is about the function f(x) so that f(f(x))=exp (x)-1.
The question is open ended and it was discussed quite recently in the comment thread in Aaronson's blog here <http://scottaaronson.com/blog/?p=263>
The growth rate of the function f (as x goes to infinity) is larger than linear (linear means O(x)), polynomial (meaning exp (O(log x))), quasi-polynomial (meaning exp(exp O(log log x))) quasi-quasi-polynomial etc. On the other hand the function f is subexponential (even in the CS sense f(x)=exp (o(x))), subsubexponential (f(x)=exp exp (o(log x))) subsubsub exponential and so on.
What can be said about f(x) and about other functions with such an intermediate growth behavior? Can such an intermediate growth behavior be represented by analytic functions? Is this function f(x) or other functions with such an intermediate growth relevant to any interesting mathematics? (It appears that quite a few interesting mathematicians and other scientists thought about this function/growth-rate.)
Related MO questions:
* [solving $f(f(x))=g(x)$](https://mathoverflow.net/questions/17614/solving-ffxgx)
* [How to solve $f(f(x)) = \cos(x)$?](https://mathoverflow.net/questions/17605/how-to-solve-ffx-cosx)
* [Does the exponential function has a square root](https://mathoverflow.net/questions/12081/does-the-exponential-function-have-a-square-root)
* [Closed form functions with half-exponential growth](https://mathoverflow.net/questions/45477/closed-form-functions-with-half-exponential-growth)
* [$f\circ f=g$ revisited](https://mathoverflow.net/questions/18882/f-circ-fg-revisited)
* [The non-convergence of f(f(x))=exp(x)-1 and labeled rooted trees](https://mathoverflow.net/questions/44740/the-non-convergence-of-ffx-expx-1-and-labeled-rooted-trees)
* [The functional equation $f(f(x))=x+f(x)^2$](https://mathoverflow.net/questions/49724/the-functional-equation-ffx-xfx2)
* [Rational functions with a common iterate](https://mathoverflow.net/questions/115113/rational-functions-with-a-common-iterate)
* [Smoothness in Ecalle's method for fractional iterates](https://mathoverflow.net/questions/179736/smoothness-in-ecalles-method-for-fractional-iterates).
| https://mathoverflow.net/users/1532 | f(f(x))=exp(x)-1 and other functions "just in the middle" between linear and exponential | Let me see if I can summarize the conversation so far. If we want $f(f(z)) = e^z+z-1$, then there will be a solution, analytic in a neighborhood of the real axis. See either [fedja's](https://mathoverflow.net/questions/4347/ffxexpx-1-and-other-functions-just-in-the-middle-between-linear-and-expo/4398#4398) Banach space argument, or my [sketchier iteration argument](http://scottaaronson.com/blog/?p=263#comment-13954). The previous report of numerical counter-examples were in error; they came from computing $(k! f\_k)^{1/k}$ instead of $f\_k^{1/k}$. We do not know whether this function is entire. If it is, then there must be some place on the circle of radius $R$ where it is larger than $e^R$. (See [fedja's comment](https://mathoverflow.net/questions/4347/ffxexpx-1-and-other-functions-just-in-the-middle-between-linear-and-expo/4353#4353) here.)
If we want $f(f(z)) = e^z-1$, there is no solution, even in an $\epsilon$-ball around $0$. According to mathscinet, this is proved in a paper of
*Irvine Noel Baker*, [**Zusammensetzungen ganzer Funktionen**](http://www.ams.org/mathscinet-getitem?mr=97532), *Math. Z.* **69** (1958), 121--163. However, there are two
half-iterates (or associated Fatou coordinates $\alpha(e^z - 1) = \alpha(z) + 1$) that are holomorphic with very large domains. One is holomorphic on the complex
numbers without the ray $\left[ 0,\infty \right)$ along the positive real axis, the other is holomorphic on the complex numbers
without the ray $\left(- \infty,0\right]$ along the negative real axis. And both have the formal power series of the half-iterate $f(z)$ as asymptotic
series at 0.
If we want $f(f(z))=e^z$, there are analytic solutions in a neighborhood of the real line, but they are known not to be entire.
I'll make this answer community wiki. What else have I left out of my summary?
Here is [a related MO question](https://mathoverflow.net/questions/12081/does-the-exponential-function-have-a-square-root). The answers to the new question contain further interesting information. Let me mention here a link with many references on ["iterative roots and fractional iterations"](http://reglos.de/lars/ffx.html) one particular link on the iterative square root of exp (x) is [here](http://mathpages.com/home/kmath507.htm).
The following two links mentioned in the old blog discussion may be helpful
* <http://www.math.niu.edu/~rusin/known-math/97/sqrt.exp> (outdated link)
* <http://www.math.niu.edu/~rusin/known-math/99/sqrt_exp> (outdated link)
* <http://web.archive.org/web/20140521065943/http://www.math.niu.edu/~rusin/known-math/97/sqrt.exp>
* <http://web.archive.org/web/20140521065943/http://www.math.niu.edu/~rusin/known-math/99/sqrt_exp>
| 31 | https://mathoverflow.net/users/297 | 4423 | 2,956 |
https://mathoverflow.net/questions/4422 | 6 | **Background**: When proving that the group of $k$-isogenies $\mathrm{Hom}\_k(A,B)$ between two abelian varieties is finitely generated, one first shows that the Tate map $$\mathbb{Z}\_\ell\otimes\_{\mathbb{Z}} M \to \mathrm{Hom}\_{\mathbb{Z}\_\ell}(T\_\ell A,T\_\ell B)$$ is injective. Since each Tate module is free of finite rank over $\mathbb{Z}\_\ell$, it follows that the localization $M\_\ell$ is $\mathbb{Z}\_\ell$-finite. One then uses a little trick to deduce the $\mathbb{Z}$-finiteness of $M$ itself. (See Silverman I, for example.)
The above proof needs only a single prime $\ell$,
but disregarding issues of the characteristic of the field (which are apparently surmountable) we actually have an injective Tate map at *every* prime. Thus...
**Question**: Can the $\mathbb{Z}$-finiteness of $M$ be deduced directly from the $\mathbb{Z}\_\ell$-finiteness of $M\_\ell$ for all primes $\ell$?
One can consider this a question about general *torsion-free* abelian groups $M$. A non-counterexample to keep in mind is $M=\mathbb{Z}[1/p]$,
for which $M\_\ell$ is $\mathbb{Z}\_\ell$-finite for all $\ell\neq p$.
(A google search shows that there is actually quite a body of literature on torsion-free abelian groups, so perhaps the answer to this question is well-known, but I'm not sure where to look...)
| https://mathoverflow.net/users/412 | Is a torsion-free abelian group finitely generated, if all of its localizations at primes $p$ are finitely generated over $\mathbb{Z}_p$? | I don't think so. Consider the $\mathbb{Z}$-module $M$ be the additive subgroup of the rationals consisting of rationals with square-free denominator.
| 21 | https://mathoverflow.net/users/1384 | 4425 | 2,957 |
https://mathoverflow.net/questions/4361 | 17 | Are there any general results on the (integral) cohomology of manifolds that are fibrations over the circle? Any literature references much appreciated.
| https://mathoverflow.net/users/1552 | Cohomology of fibrations over the circle | The above Mayer-Vietoris argument gives the cohomology of a fiber bundle over a circle in a concrete fashion. For sake of "mathematical culture", I thought I'd mention what happens for fiber bundles over a higher dimensional sphere (this is also a good excuse for me to test drive the new latex support).
For fiber bundles $F \hookrightarrow E \rightarrow S^n$ with $F$ connected and $n \geq 2$, the Serre spectral sequence degenerates in a very simple fashion into what is known as the Wang exact sequence. Namely, we have a long exact sequence of the form
$$\cdots \rightarrow H^k(E) \rightarrow H^k(F) \rightarrow H^{k-n+1}(F) \rightarrow H^{k+1}(E) \rightarrow \cdots$$
The proof of this is completely analogous to the proof of the better known Gysin exact sequence, which tells you what happens for fiber bundles whose fibers are spheres.
A reference for this material is McCleary's "User's guide to spectral sequences", page 145.
| 19 | https://mathoverflow.net/users/317 | 4428 | 2,959 |
https://mathoverflow.net/questions/4427 | 17 | A $\mathbb{Z}/2\mathbb{Z}$-graded algebra is said to supercommute if $xy = (-1)^{|x| |y|} yx$; in other words, odd elements anticommute. Why is this the "right" definition of supercommutativity? (Put another way, why is this the natural tensor product structure on super vector spaces?) Answers from both a categorical or physical point of view would be great.
| https://mathoverflow.net/users/290 | What is the conceptual significance of supercommutativity? | The categorical answer is that (in characteristic zero) this is the only way that you can make a suitable symmetric tensor category, other than by using group representations. There is a Tannakian theorem of Deligne to this effect in the algebraic setting.
One of the physical answers is equivalent to the categorical answer. "Parastatistics" is the topic of self-consistent linear actions of the symmetric group on identical quantum-mechanical particles. The parastatistics theorem in physics (or theorems or conjectures; the level of rigor of the real point is not entirely clear) is a lot like Deligne's theorem. It says that parastatistical particles come in two kinds, parafermions and parabosons, and that they can all be modeled as fermions and bosons together with internal state spaces which are group representations.
Bosons and fermions may not look exactly the same as commutative or supercommutative algebras. But they are the same topic, because (if you apply second quantization in reverse) the values of their fields commute or anticommute.
For particles in 2D, the correct group action is the braid group, not the symmetric group. So in this case, the parastatistics theorem does not hold and you can have "anyons". Then the allowed statistics is given by a unitary ribbon tensor category. However, since the category in question is no longer symmetric, there is no clear way to define commutativity; at least, nothing that's clearly important.
Note also that isn't just that the principle of available symmetric tensor categories comes from category theory and is needed in physics. It's also needed in topology. The most traditional supercommutativity in mathematics is cohomology.
---
To answer Qiaochu's question below, there's nothing wrong with using the standard switching map $v \otimes w \mapsto w \otimes v$ to define commutativity. It shows up all the time. The point is that the signed switching map $v \otimes w \mapsto (-1)^{|v||w|}w \otimes v$ is another valid and inequivalent symmetric monoidal structure. (The symmetric tensor structure is interpreted as what it means to permute factors of a product, of course.) There is nothing to prevent the signed switching map from arising among topological invariants or in physics, so it does arise. The structure theorems say that all "suitable" choices for the switching map are essentially these two, possibly disguised by a restriction to tensors that are invariant under a group action.
For both good and bad reasons, I was deliberately vague about what it means for the symmetric tensor category to be suitable, in the sense that it will satisfy a structure theorem. You want some extra axioms and properties to hold, some of them related to existence of duals and traces. One version of the structure theorem, due to Deligne, is reviewed in the paper *[The classification of finite-dimensional triangular Hopf algebras over an algebraically closed field of characteristic 0](http://arxiv.org/abs/math/0202258)* by Etingof and Gelaki ([published version](https://doi.org/10.17323/1609-4514-2003-3-1-37-43)). (The theorem cited as [De2] is the relevant one.)
| 21 | https://mathoverflow.net/users/1450 | 4430 | 2,961 |
https://mathoverflow.net/questions/4395 | 12 | Some physicists have told me that if you think about an extended n-dimensional TQFT $F$, then the decategorification is given by $F'(X)=F(X\times S^1)$, which I believe they call "compactification on a circle." Is there any way to make this statement precise?
| https://mathoverflow.net/users/66 | What do decategorification and "compactification on a circle" have to do with each other? | In a general extended TQFT Z, the assignment $Z(X \times S^1)$ is the "dimension" of Z(X), in the following sense. Write the circle as an incoming arc followed by an outgoing arc. The incoming arc is a morphism (coevaluation) from the unit (Z(empty set)) to Z(X) tensor its dual $Z(X^{op})=Z(X)^\*$, followed by a morphism (evaluation) back to the unit.
In particular we learn Z(X) HAS a dual (is dualizable), and these are the two canonical maps that come in the definition of being a dual. The composition is an endomorphism of the unit Z(empty), which is very generally called the dimension of Z(X), or the Hochschild homology of Z(X).
If Z(X) is a vector space, End Z(empty) = numbers and this is the usual dimension. If Z(X) is a category (or an algebra, or a 2-category, or....), this is what is usually known as its Hochschild homology. In particular Hochschild homology is where characters (or traces) of objects in Z(X) live, if it's a category. In simple situations this will be the same as the K-theory of Z(X) (in great generality there's a map from K-theory to Hochschild homology), if you want to compare this to another version of decategorification, which is taking K-groups.
| 11 | https://mathoverflow.net/users/582 | 4446 | 2,970 |
https://mathoverflow.net/questions/4113 | 11 | Consider the transcendental extension Q(t) of the field of rationals.
To Q(t) adjoin the root of the polynomial x^5+t^5=1. The resulting
field Q(t)[x] is a radical extension of Q(t). Is it true that the
only solutions to the equation X^5+Y^5=1 in the field Q(t)[x] are
(0,1), (1,0), (t,x), (x,t), (1/t,-x/t) and (-x/t, 1/t)?
Comment: Using the ABC theorem one can prove that the Fermat
curve X^n+Y^n=1 does not have a non-trivial solution in Q(t) for n>2.
In particular in Q(t) the equation X^5+Y^5=1 does not have
non-trivial solutions.
| https://mathoverflow.net/users/1169 | a question on function fields | First, replace Q by the complex numbers C.
Write A = C[x,y]/(x^n+y^n-1). Then the field you write down, call it K, is the fraction field of A, which is the function field of the Fermat Curve.
Finding a solution (X,Y) to x^n + y^n = 1 is equivalent to finding a map A --> K, where one sends x to X and y to Y. Any map to a field factors though A/P for some prime ideal P of A.
Since (x^n+y^n-1) is irreducible, the only primes in A are either (0) or maximal ideals m with A/m = C.
If A/P = C, then the map A --> K factors through C. Maps from A to C correspond to complex points on the Fermat curve.
If P = 0, then A --> K extends to a non-trivial map from K --> K. Giving a map between function fields is equivalent to giving a map of the corresponding smooth projective curves (in the opposite direction). Thus, the question becomes: what are the non-trivial maps from the Fermat curve to itself? For symmetry reasons, write the Fermat curve as x^n + y^n + z^n = 0 (over C).
Assume that n > 3. Since the genus of the Fermat curve is > 1, the Reimann-Hurwitz formula tells us that any non-trivial map must be an automorphism. On the other hand, the isomorphism group of the Fermat curve is the semi-direct product of (Z/nZ)^2 by S`_`3. (Explicitly, this corresponds to multiplying (x,y,z) by n-th roots of unity, and permuting the entries.)
In terms of the affine coordinates the S\_3 action is generated by (x,y)->(y,x) and (x,y)->(1/y,-x/y).
To summarize, the K points are given by the (finitely many) automorphisms, and the C-points of the curve. You have noted all the automorphisms over Q, I would note that you are also invoking the non-trivial fact that the Fermat curve has no non-trivial points over Q, which (even for n = 5) that is harder than everything else in this answer (and is especially harder for general n > 2!). (The "ABC theorem" for polynomials only tells you that there are no *non-constant* solutions in C[t]. In general, a solution in C[t] will correspond to a map from P^1-->X, which can be non-trivial only if X has genus 0.)
Note that this argument used basically nothing about the Fermat curve itself --- any curve of genus > 1 has only finitely many automorphisms.
If g = 1, then there are some non-trivial maps from a curve of genus g to itself, but Riemann-Hurwitz tells us that they are all unramfied. X/C in this case is an elliptic curve, so there will be infinitely many non-trivial solutions in K, corresponding to the rational functions giving the multiplication by [n] map (or more if X has CM). In the Fermat case, this means that for n = 3 there will be many more solutions.
If g = 0, there are buckets of maps from P^1 to itself, as one knows in the Fermat case when n = 2 or 1.
| 12 | https://mathoverflow.net/users/nan | 4451 | 2,972 |
https://mathoverflow.net/questions/3929 | 8 | The Newlander-Nirenberg theorem states that an almost complex structure is integrable if and only if the Nijenhuis tensor vanishes.
I heard that this statement is not true in infinite dimensions, since for example the Loop space of a Riemannian 3-manifold is counterexample. (In fact, I think NN fails for Fréchet manifolds in general(?))
So my question is:
Is the Newlander Nirenberg theorem valid for Banach- or Hilbertmanifolds? If not, is it possible to weaken the statement (or some conditions) such that it remains true for some class of infinite dimensional manifolds?
EDIT: Added the tag "open-problem", since NN for Hilbertmanifolds seems to be an open problem.
| https://mathoverflow.net/users/675 | Infinite dimensional Newlander-Nirenberg theorem | As far as I understand, in a paper of Petyi [On the ∂-equation in a Banach space. Bull. Soc. Math. France 128 (2000), no. 3, 391–406.] it is shown that the NN theorem does not hold for Banach manifolds in general. However, as you may know, the NN theorem has an easy proof when the almost complex structure is assumed to be real analytic. In this case, the NN theorem is an easy consequence of the Frobenius theorem, which is true for Banach manifolds (reference? Smale?). There is a paper by Daniel Beltita [http://arxiv.org/abs/math/0407395] which confirms that NN is true in this real-analytic case. It would be interesting to see exactly where the NN proof breaks down for Banach manifolds, and exactly what one should assume to allow it to go through. But I am not aware of any such detailed work.
| 6 | https://mathoverflow.net/users/1116 | 4501 | 3,007 |
https://mathoverflow.net/questions/4519 | 3 | Are the sites similar to mathoverflow in other sciences related to mathematics? statistics, computer science, physics, economics, etc?
Let me explain what I mean by "similar": those are sites devoted to posing questions and answers,in these areas. I do not insist on the precise format of "mathoverflow" (reputation points, badges, etc.). But I am looking for general multi-participants forums like this one, so scientific blogs do not qualify.
| https://mathoverflow.net/users/1532 | Something like mathoverflow in other sciences | Here is a brief list of science-related sites that run on the same platform as Math Overflow:
* Science
+ [Science Stack](http://sciencestack.com/)
+ [asksci.com](http://asksci.com/)
* Physics
+ [physics.stackexchange.com](https://physics.stackexchange.com/)
+ [PhysicsOverflow](http://physicsoverflow.org/)
* Electronics
+ [Electronics Exchange](https://electronics.stackexchange.com/)
* Programming
+ [Stack Overflow](https://stackoverflow.com/)
A complete list of these sites is available [here](https://meta.stackexchange.com/questions/4/list-of-stackexchange-sites/5#5).
The one thing that I personally feel is missing is a lower-level mathematics site (a Math Underflow, if you will). I understand the desire to keep Math Overflow relevant and interesting for professional mathematicians, but there is currently no equivalent site for amateurs and students.
**Update:** (Sept 2010)A math site for [university level mathematics exists now](https://math.stackexchange.com/).
There are also sites for [theoretical computer science](https://cstheory.stackexchange.com/) and for [statistics](https://stats.stackexchange.com/).
(April 2011) There is a [TEX Q/A site](https://tex.stackexchange.com/)!
(September 2011) [Area 51: science](http://area51.stackexchange.com/categories/7/science) contains new proposed stackexchange sites related to science. We can mention especially proposals for [Theoretical physics (and mathematical physics)](http://area51.stackexchange.com/proposals/23848/theoretical-physics) (closed), economics (closed), research economics, game theory, [computational sciences](https://scicomp.stackexchange.com/), [philosophy](https://philosophy.stackexchange.com/) (the site is running), [numerical models and simulations](http://area51.stackexchange.com/proposals/1907/numerical-modeling-and-simulation), Mathematica, mathematics in german.
| 10 | https://mathoverflow.net/users/19 | 4524 | 3,027 |
https://mathoverflow.net/questions/4504 | 15 | Let's fix a finite field F and consider abelian varieties of dimension g over F. Can we say how many isogeny classes there are? Is it even clear that there's more than one isogeny class? For g=1, and some fixed F with characteristic not 2 or 3, we could probably just write down all the Weierstrass equations and count isogeny classes by brute force, but is there a cleaner way to do it?
| https://mathoverflow.net/users/88 | Can we count isogeny classes of abelian varieties? | Let q be the order of your finite field. Then the category of abelian varieties over $\mathbb{F}\_q$ up to isogeny is semisimple - any object is isogenous to a product of simple ones in an essentially unique way, so this reduces your question to one about simple objects.
For simple abelian varieties over $\mathbb{F}\_q$, there is the Tate-Honda classification which states that the isogeny classes are in bijective correspondence with Weil $q$-integers (algebraic numbers that have absolute value $q^{1/2}$ in all complex embeddings) up to Galois conjugacy.
I learned this from Milne's "[Points on Shimura varieties mod $p$](http://www.jmilne.org/math/articles/1979a.pdf)" (one of the articles from the Corvallis proceedings that Kevin mentioned), which has a nice and fairly elementary discussion in section 5.
| 10 | https://mathoverflow.net/users/360 | 4525 | 3,028 |
https://mathoverflow.net/questions/4527 | 1 | Does this even make sense what I translated into english?
PS. I am probably gonna delete this question eventually
| https://mathoverflow.net/users/462 | Is it that only with normal matrices, the transition matrix to its [del: inherent] [ins: own] basis is unitary? | I am not sure I understand the question, but if you are talking about linear algebra over the complex numbers, then it is true that normal matrices (those which commute with their hermitian adjoint) are precisely those which can be diagonalised by a unitary transformation.
(This is proven in Herstein's *Topics in algebra*: §10 of the 1964 edition.)
| 2 | https://mathoverflow.net/users/394 | 4528 | 3,029 |
https://mathoverflow.net/questions/1922 | 46 | One can build a projective plane from $\Bbb R^n$, $\Bbb C^n$ and $\Bbb H^n$ and is then tempted to do the same for octonions. This leads to the construction of a projective plane known as $\Bbb OP^2$, the Cayley projective plane.
What are the references for the properties of the Cayley projective plane? In particular, I would like to know its (co)homology and homotopy groups.
Also, what geometric intuition works when working with this object? Does the intuition from real projective space transfer well or does the non-associativity make a large difference? For example, I would like to know why one could have known that there is no $\Bbb OP^3$.
| https://mathoverflow.net/users/798 | What is the Cayley projective plane? | As I recall, the Cayley projective plane is painful to build, but it is a 2-cell complex, with an 8-cell and a 16-cell. The cohomology is Z[x]/(x^3) where x has degree 8, as you would expect. Its homotopy is unapproachable, because it is just two spheres stuck together, so you would pretty much have to know the homotopy groups of the spheres to know it. The attaching map of the 16-cell is a map of Hopf invariant one, from S^15 to S^8, the last such element.
I think the real reason that the Cayley projective plane exists is because any subalgebra of the octonions that is generated by 2 elements is associative. That is just enough associativity to construct the projective plane, but not enough to construct projective 3-space. And this is why you should not expect there to be a projective plane for the sedonions (the 16-dimensional algebra that is to the octonions what the octonions are to the quaternions), because every time you do the doubling construction you lose more, and in particular it is no longer true that every subalgebra of the sedonions that is generated by 2 elements is associative.
Mark
| 40 | https://mathoverflow.net/users/1698 | 4532 | 3,033 |
https://mathoverflow.net/questions/4434 | 22 | Let g(z) be an entire function of a complex variable z. Does there exist an entire function f(z) such that f(z+1)-f(z)=g(z)? As I learned several years back, the answer to this [is apparently 'yes'](http://www.math.niu.edu/~rusin/known-math/01_incoming/coho_func), but I have not felt satisfied with the proof because it goes beyond my expertise.
I tried to find f using the power series expansion of g, for that works when g is a polynomial. But the results of partial inversions kept diverging. Representing g as an integral via Cauchy's formula, and doing inversion inside the integration led to similar problems. Perhaps I am overly optimistic, but a question this elementary should have equally elementary solution. Is there such a solution? If not, is there a reason to expect that no simple and elementary solution should exist?
| https://mathoverflow.net/users/806 | Elementary solutions to f(z+1)-f(z)=g(z) in entire functions | It took me some time to find a solution that satisfies both requirements:
a) If should be based on the power series expansion
b) It should use no tools heavier than contour integration.
So, let $g(z)=\sum a\_ k z^k$. We know that $a\_ k$ decay faster than any geometric progression. We want analytic functions $F\_ k(z)$ such that $F\_ k(z+1)-F\_ k(z)=z^k$ and $F\_ k(z)$ grows not faster than a geometric progression as a function of $k$ in every disk. Then $f=\sum a\_ k F\_ k$ is what we want. So, just choose any odd multiples $r\_ k\in(k,k+2\pi)$ of $\pi$ and put
$$
F\_ k(z)=\frac{k!}{2\pi i}\oint\_ {|z|=r\_ k}\frac {e^{z\zeta}}{e^\zeta-1}\frac{d\zeta}{\zeta^{k+1}}
$$
The key is that $|e^\zeta-1|\ge \frac 12$ on the circle and $r\_ k^k\ge k!$, so $|F\_ k(z)|\le 2e^{|z|(k+2\pi)}$ on the plane.
| 32 | https://mathoverflow.net/users/1131 | 4542 | 3,041 |
https://mathoverflow.net/questions/4547 | 36 | There is a definition of Iwahori-Hecke algebras for Coxeter groups in terms of generators and relations and there is a definition of Hecke algebras involving functions on locally compact groups. Are these two concepts somehow related? I think I read somewhere that the Hecke algebra with functions includes Iwahori-Hecke algebras, is that correct? Is there a good motivation for introducing and studying either type of algebras? How much is known about their representations?
| https://mathoverflow.net/users/717 | Definitions of Hecke algebras | A Hecke algebra describes the most reasonable way to convolve functions or measures on a homogeneous space. Suppose that you have seen the definition of convolution of functions on a vector space, or on a discrete group --- the latter is just the group algebra of the group or some completion. Then how could you reasonably define convolution on a sphere? There is no rotationally invariant way to convolve a general $f$ with a general $g$. However, if $f$ is symmetric around a reference point, say the north pole, then you can define the convolution $f \* g$, even if $g$ is arbitrary.
This is the basic idea of the Hecke algebra. The $(n-1)$-sphere is the homogeneous space $SO(n)/SO(n-1)$. A function $g$ on the sphere is a function on left cosets. A function $f$ on the sphere which is symmetric about a reference point is a function on double cosets. If $H \subseteq G$ is any pair of compact groups, if $f$ is any continuous function on $H\backslash G/H$, and if $g$ is any continuous function on $G/H$, then their product in the continuous group algebra is well-defined on $G/H$. The functions on double cosets make an algebra, the Hecke algebra, and the functions on left cosets are a bimodule of the Hecke algebra and the parental group $G$.
It is important for the same reasons that any other kind of convolution is important.
A particular case studied by ~~Hecke~~ Iwahori and others from before quantum algebra was the finite group $GL(n,q)$ and the upper triangular subgroup $B$. This is "the" Hecke algebra; it turns out that it is one algebra with a parameter $q$. Or as Ben says, this generalizes to the Iwahori-Hecke algebra of an algebraic group $G$ with a Borel subgroup $B$.
---
The other place that the Hecke algebra arises is as an interesting deformation of the symmetric group, or rather as a deformation of its group algebra. It has a parameter $q$ and you obtain the symmetric group when $q=1$. As I said, it is also the Hecke algebra of $GL(n,q)/GL(n,q)^+$, where $B = GL(n,q)^+$ is the Borel subgroup of upper-triangular matrices (all of them, not just the unipotent ones). There is a second motivation for the Hecke algebra that I should have mentioned: It immediately gives you a representation of the braid group, and this representation reasonably quickly leads to the Jones polynomial and even the HOMFLY polynomial.
When the Jones and HOMFLY were first discovered, it was simply a remark that the braid group representation was through the same Hecke algebra as the convolutional Hecke algebra for $GL(n,q)/B$ (or equivalently $SL(n,q)/B$). Even so, it's a really good question to confirm this "coincidence", as Arminius asks in the comment. Particularly because it is now a fundamental and useful relation and not a coincidence at all. As Ben explains in his blog post, the first model of the Hecke algebra is important for the categorification of the second model.
The coset space of $GL(n,q)/B$ consists of flags in $\mathbb{F}\_q^n$, and you can see these more easily using projective geometry. When $n=2$, there is an identity double coset 1 and another double coset $T$. A flag is just a point in $\mathbb{P}^1$, and the action of $T$ is to replace the point by the formal sum of the other $q$ points. Thus you immediately get $T^2 = (q-1)T + q.$ When $n=3$, a flag is a point and a line containing it in $\mathbb{P}^2$. The two smallest double cosets other than the identity are $T\_1$ and $T\_2$. $T\_1$ acts by moving the point in the line; $T\_2$ acts by moving the line containing the point. A little geometry then gives you that $T\_1T\_2T\_1$ and $T\_2T\_1T\_2$ both yield one copy of the largest double coset and nothing else. Thus they are equal; this is the braid relation of the Hecke algebra. When $n \ge 3$, the Hecke algebra is given by these same local relations, which must still hold.
| 65 | https://mathoverflow.net/users/1450 | 4555 | 3,052 |
https://mathoverflow.net/questions/4561 | 38 | Y.I. Manin mentions in a [recent interview](http://golem.ph.utexas.edu/category/2009/11/interview_with_manin.html#c028992 "link to n-category discussion")
the need for a “codification of efficient new intuitive tools, such as … the “brave new algebra” of homotopy theorists”. This makes me puzzle, because I thought that is codified in e.g. Lurie’s articles. But I read only his survey on elliptic cohomology and some standard articles on symmetric spectra. Taking the quoted remark as indicator for me having missed to notice something, I’d like to read what others think about that, esp. what the intuition on “brave new algebra” is.
Edit:
In view of [Rognes' transfer](https://arxiv.org/abs/math/0502183 "arxiv") of Galois theory into the context of "brave new rings" and his [conference](http://folk.uio.no/rognes/yff/alexandra.html "website") last year, I wonder if themes discussed in [Kato's article](http://www.springerlink.com/index/p644163275773165.pdf "article") (e.g. reciprocity laws) have "brave new variants".
Edit: I found Greenlees' introductions ([1](https://arxiv.org/abs/math/0609452 "arxiv"), [2](https://arxiv.org/abs/math/0609453 "arxiv")) and Vogt's ["Introduction to Algebra over Brave New Rings"](http://hdl.handle.net/10338.dmlcz/701626) for getting an idea of the topological background very helpful.
| https://mathoverflow.net/users/451 | What is the "intuition" behind "brave new algebra"? | I hardly know how to begin to reduce this subject to some kind of intuitive ideas, but here are some thoughts:
\* Adding homotopy to algebra allows for generalizations of familiar algebraic notions. For instance, a *topological commutative ring* is a commutative ring object in the category of spaces; it has addition and multiplication maps which satisfy the usual axioms such as associativity and commutativity. But instead, one might instead merely require that associativity and commutativity hold "up to all possible homotopies" (and we'll think of the homotopies as part of the structure). (It is hard to give the flavor of this if you haven't seen a definition of this sort.) This gives one possible definition of a "brave new commutative ring".
\* What is really being generalized is not algebraic objects, but *derived categories* of algebraic objects. So if you have a brave new ring R, you don't really want to study the category of R-modules; rather, the proper object of study is the derived category of R-modules. If your ring R is an ordinary (cowardly old?) ring, then the derived category of R-modules is equivalent to the classical derived category of R.
If you want to generalize some classical algebraic notion to the new setting, you usually first have to figure out how to describe it in terms of derived notions; this can be pretty non-trivial in some cases, if not impossible. (For instance, I don't think there's any good notion of a subring of a brave new ring.)
\* As for Manin's remarks: the codification of these things has being an ongoing process for at least 40 years. It seems we've only now reached the point where these ideas are escaping homotopy theory and into the broad stream of mathematics. It will probably take a little while longer before things are so well codified that brave new rings get introduced in the grade school algebra curriculum, so the process certainly isn't over yet!
| 42 | https://mathoverflow.net/users/437 | 4565 | 3,058 |
Subsets and Splits