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/44057 | 6 | Divide the unit circle into three arcs, and let $z$ be a point in the open unit disk. Is there a simple formula for the probability that Brownian motion started at $z$ will hit one particular arc rather than the other two when it first hits the unit circle? It would also be nice to have a way of representing $z$ in terms of the three probabilities (assuming that the mapping from points to probability triples is invertible); indeed, this would give a natural way to assign "conformal coordinates" to points in the interior of any simply-connected domain (given a choice of three points on the boundary). Is there literature on this?
In a related vein, given a triangle, one might ask for a formula that Brownian motion started at some point $z$ inside the triangle will hit one particular side of the triangle rather than the other two when it first hits the boundary of the triangle. This is equivalent to the preceding question (via the conformal map between the triangle and the disk that sends the vertices to the three marked points) but one picture might lead to nicer formulas than the other.
| https://mathoverflow.net/users/3621 | Coordinatizing the disk via Brownian motion | I think zhoraster's answer is on the right track, but it can be said in a simple geometric way:
You can think of the interior of the disk as the hyperbolic plane, using the Poincaré disk model. Brownian paths are the same in any conformally equivalent metric. It follows (by symmetry) that the hitting probability for paths starting at a point $x$ is proportional to the angle subtended by the three arcs in the hyperbolic plane.
For the same problem in the upper half plane, there's an even simpler description: The hyperbolic angle subtended by an arc on the real line from a point in upper half space is exactly twice the Euclidean angle. Roughly speaking: If you're in a big stuffy room with an open door in the middle of one wall, the amount of fresh air you get is proportional to the visual width of the door.
| 13 | https://mathoverflow.net/users/9062 | 44130 | 28,029 |
https://mathoverflow.net/questions/44109 | 65 | In most basic courses on general topology, one studies mainly Hausdorff spaces and finds that they fit quite well with our geometric intuition and generally, things work "as they should" (sequences/nets have unique limits, compact sets are closed, etc.). Most topological spaces encountered in undergraduate studies are indeed Hausdorff, often even normed or metrizable. However, at some point one finds that non-Hausdorff spaces do come up in practice, e.g. the Zariski topology in algebraic geometry, the Fell topology in representation theory, the hull-kernel topology in the theory of C\*-algebras, etc.
My question is: how should one think about (and work with) these topologies? I find it very difficult to think of such topological spaces as geometric objects, due to the lack of the intuitive Hausdorff axiom (and its natural consequences). With Hausdorff spaces, I often have some clear, geometric picture in my head of what I'm trying to prove and this picture gives good intuition to the problem at hand. With non-Hausdorff spaces, this geometric picture is not always helpful and in fact relying on it may lead to false results. This makes it difficult (for me, at least) to work with such topologies.
As this question is somewhat ambiguous, I guess I should make it a community wiki.
**EDIT**: Thanks for the replies! I got many good answers. It is unfortunate that I can accept just one.
| https://mathoverflow.net/users/7392 | How should one think about non-Hausdorff topologies? | For a variety of reasons, it's often useful to develop an intuition for *finite* topological spaces. Since the only Hausdorff finite spaces are discrete, one will have to deal with the non-Hausdorff case almost all the time.
The fact of the matter is that the category of finite spaces is equivalent to the category of finite preorders, i.e., finite sets equipped with a reflexive transitive relation. In terms of a picture, draw an arrow $x \to y$ between points $x$ and $y$ whenever $x$ belongs to the closure of $y$ (or the closure of $x$ is contained in the closure of $y$). This defines a reflexive transitive relation.
Two points $x$, $y$ have the same open neighborhoods if and only if $x \to y$ and $y \to x$. It follows that the topology is $T\_0$ (the topology can distinguish points) if and only if the preorder is a poset, where antisymmetry of $\to$ is satisfied.
The closure of a point $y$ is the down-set {$x: x \to y$}, and a set is closed iff it is downward closed in the preorder. In the finite case, I believe it is true that every closed *irreducible* set (one that isn't the union of two proper closed subsets) is the closure of a point = principal ideal; if the point is unique, the space is called *sober*. Sober spaces are the kinds of spaces that arise as underlying topological spaces of schemes, and it seems to be true that a finite space is sober iff it is $T\_0$.
| 53 | https://mathoverflow.net/users/2926 | 44135 | 28,033 |
https://mathoverflow.net/questions/44136 | 0 | I have three pairs of points in 3D space. These may or may not be coplanar. I want to find a point such that it is equidistant from each pair of points. I know that may or may not be possible depending on the positions of the points. What I want is the best average point, which I can take safely as the centre and draw a sphere from there whose radius is the maximum distance of this point from any of the six points, then I want all the points to remain inside the sphere.
| https://mathoverflow.net/users/10429 | What is the average center of six points in space | Distance from a pair of points instead of from each one of a pair of points does not seem to be well defined. However, you could simply take the centroid of all six points (add the coordinates and divide by 6 in each axis), then compute the distance to each of the six points, and use the maximum for the radius of the sphere. It is not clear how you want us to use the fact that the points come in pairs instead of six single points.
| 0 | https://mathoverflow.net/users/7408 | 44138 | 28,034 |
https://mathoverflow.net/questions/44120 | 1 | Consider the Lie algebra $sl\_2$
with the standard basis $(e,f,h),$ where
\begin{equation\*}\label{sl2}
[h,e]=2\,e, [h,f]=-2\,f,[e,f]=h.
\end{equation\*}
Let $V$ be finite-dimensional $sl\_2$-module and let we know that element $e$ acts on $V$ as linear operator with a matrix $E$ (in some fixed basis).
**Question**. How one may find the matrix $F,H$ which correspond to elements $f$ and $h.?$
It is posible to recover $F,H$ from $E?$
| https://mathoverflow.net/users/9645 | Representation of Lie algebra sl_2. | Edited in light of clarifications made by OP:
Given a nilpotent matrix $E$ acting on a finite dimensional vector space $V$, it is always possible to extend it to a representation of $sl\_2$ in such a way that it represents $e$. The extension is almost never unique: conjugating the representing matrices $F$ and $H$ by anything in the centralizer of $E$ gives a new extension.
The existence statement is the Jacobson-Morozov lemma (part of whose standard proof is reproduced in another answer) applied to the semisimple Lie algebra $sl(V)$. See Proposition 2 of section 2 of paragraph 11 of Bourbaki's "Lie Groups and Lie Algebras", Chapter VIII (see the Corollary following the Proposition for the extent to which uniqueness is true: basically, up to conjugacy).
On the other hand, if you have some additional rigid structure, there might be a unique extension. For instance, if you know a contravariant form and have an orthonormal basis at your disposal then $F$ is the transpose of $E$ (written in terms of the given orthonormal basis) and $H$ is determined by $H=[E,F]$.
| 3 | https://mathoverflow.net/users/8552 | 44143 | 28,038 |
https://mathoverflow.net/questions/44142 | 1 | Given two torsion free coherent sheaves $M$ and $N$ wit $rk(M)=rk(N)=r$ on an smooth projective surface $S$, by definition $det(M):=\Lambda^r(M)^{\\*\\*}$.
Is the following criterion correct?
$M\cong N$ $\Leftrightarrow$ $M \hookrightarrow N$ and $c\_i(M)=c\_i(N)$ for $i=0,1,2$
One only has to look at "$\Leftarrow$":
So we have $0\rightarrow M\rightarrow N\rightarrow Q \rightarrow0$. Because of $c\_i(M)=c\_i(N)$ we see that $c\_1(Q)=0$ and $c\_2(Q)=0$. Since $M$ and $N$ have the same rank $codim(supp(Q))\geq 1$. We also have an induced map $det(M)\hookrightarrow det(N)$ of line bundles, i.e. $det(N)\cong det(M)\otimes O\_S(D)$ for some effective divisor $D$. Now $det(Q)\cong det(N)\otimes det(M)^{-1}\cong O\_S(D)$.
So by definition $c\_1(Q)=c\_1(det(Q))=D$, but $c\_1(Q)=0$, so $D$ is effective and 0, i.e. $Q$ has no support in codimension 1, so $codim(supp(Q))\geq 2$.
So $Q$ is an Artinian sheaf, and for those one has $c\_2(Q)=-dim(H^0(S,Q))$. Since $c\_2(Q)=0$ we have $H^0(S,Q)=0$, but $H^0(S,Q)=\bigoplus\limits\_{s\in supp(Q)} Q\_s$. So $Q\_s=0$ for all $s\in supp(Q)$, i.e. $Q=0$. So we have $M\cong N$.
| https://mathoverflow.net/users/3233 | Equality of chern classes and isomorphism | That's correct. A slightly shorter argument is: if $\mathcal{Q}$ has support in codimension $d$, then its Chern character $\mathrm{ch}\_d(\mathcal{Q})$ is non-zero and effective. So a sheaf is trivial if and onlfy if $\mathrm{ch} = 0$, which is true if and only if the rank and the Chern classes vanish. In particular, the result holds for any dimension.
| 5 | https://mathoverflow.net/users/7437 | 44144 | 28,039 |
https://mathoverflow.net/questions/29970 | 53 | Grothendieck famously objected to the term "perverse sheaf" in *Récoltes et Semailles*, writing "What an idea to give such a name to a mathematical thing! Or to any other thing or living being, except in sternness towards a person—for it is evident that of all the ‘things’ in the universe, we humans are the only ones to whom this term could ever apply.” (Link [here](https://www.ams.org/notices/200410/fea-grothendieck-part2.pdf), in an excellent article "*Comme Appelé du Néant*: The life of Alexandre Grothendieck", part 2, by Allyn Jackon.) But a google search for '"perverse sheaf" etymology' gives only nine hits, none of which seem informative.
>
> What is the etymology of the term "perverse sheaf"?
>
>
>
| https://mathoverflow.net/users/6950 | What is the etymology of the term "perverse sheaf"? | When MacPherson and I first started thinking about intersection homology, we realized that there was a number that measured the "badness" of a cycle with respect to a stratum. This number had the property that when you (transversally) intersected two cycles, their
"badness" would add. The best situation occurs for cocycles, in which case that number was zero, and the intersection of two cocycles was again a cocycle. The worst situation was for ordinary homology, in which case that number could be as large as the codimension of the stratum. In that case, the intersection of two cycles could even fail to be a cycle. After a while it became clear that we needed a name for this number and we tried "degeneracy", "gap", etc., but nothing seemed to fit. It seemed that the bad cycles were being "obstinate", but "obstinateness" did not sound reasonable. Finally we said, "let's just call it the perversity, and we'll find a better word later". We tried again later, with no success. (We did not realize that in some languages the word is obscene.) When we first went to talk with Dennis Sullivan and John Morgan about these ideas, we were calling the resulting groups "perverse homology", but Sullivan suggested the alternative, "intersection homology", which seemed fine with us. This was 1974-75. Later, when it was discovered that, for any perversity, there is an abelian category of sheaves, whose simple objects are the intersection cohomology sheaves (with that perversity) of closures of strata, Deligne coined the term "faisceaux pervers".
| 145 | https://mathoverflow.net/users/10431 | 44149 | 28,044 |
https://mathoverflow.net/questions/44047 | 11 | Under which conditions localizing an additive category by some class S of morphisms yields and additive category? It seems easy to define certain addition on morphisms if we fix their representatives as zig-zags (i.e. compositions of 'old' morphisms with inverses of morphisms in S; here I use the fact that 'my' S is closed with respect to direct sums of morphisms), but I am not sure at all that this addition will not depend on the choice on representatives. Is there any reasonable condition that will ensure this? I definitely do not want to restrict myself to abelian or triangulated categories.
It seems that in the situations I am interested in, any morphism is a composition of the embedding of a direct summand, an inverse of a morphism from S, and an 'old' morphism (i.e. it is 'almost a fraction'). The Ore conditions are not fulfilled (in general, probably); yet some weakening of them could hold.
I would be deeply grateful for any associations here!
My examples are:
For an additive (pseudo-abelian) category B consider some full triangulated (thick) subcategory D of $K^b(B)$; then my S for B is the set of morphisms in B that yield objects of D (if considered as complexes of length 1).
In particular, S is always closed with respect to compositions and direct sums of morphisms.
In fact, I am interested in all aspects of this setup!
| https://mathoverflow.net/users/2191 | Localizing an arbitrary additive category | This is a very elementary problem. To solve it, it is better not to try to understand the localization explicitely, but to work only with the universal property of the localization (there is no need for any calculus of zig-zags of any kind). You should also think of finite sums not as something defined on each family of objects, but as a left adjoint to the diagonal functor $C\to C^n$ for each $n\geq 0$.
Let us look at a more general situation first. Let $C$ and $C'$ be categories, and $S$ and $S'$ be a class of maps in $C$ and $C'$ respectively, which contains all the identities. Then, it is an easy exercise to check that the canonical functor
$$(S\times S')^{-1}(C\times C')\to S^{-1}C\times {S'}^{-1}C'$$
is an equivalence of categories (or, if you prefer, an isomorphism, depending on whether you prefer to consider the localized category $S^{-1}C$ as the solution of a universal problem in the $2$-category of categories, or in the $1$-category of categories, respectively). Hint: just check that the two categories have the same universal property.
Another elementary exercise is that, given any adjunction
$$L:C\rightleftarrows D:R$$
if $S$ (resp. $T$) is a class of maps in $C$ (resp. in $D$), such that $L(S)\subset T$ and $R(T)\subset S$, then we get a canonical adjunction
$$L:S^{-1}C\rightleftarrows T^{-1}D:R$$
It follows rather immediately from this that, if $C$ admits finite sums (resp. finite products), and if the class $S$ is closed under finite sums (resp. finite products), then the localized category $S^{-1}C$ admits finite sums (resp. finite products), and the canonical functor $\gamma: C\to S^{-1}C$ commutes with them.
It is now obvious that, if $C$ is an additive category, and if $S$ is a class of maps which contains the identities and which is closed under finite sums (hence also under finite products), then the category $S^{-1}C$ is additive, and the canonical functor $\gamma$ is additive. Indeed, an additive category is nothing but a category with finite sums as well as finite products, such that, the initial and terminal object coincide, such that $X\amalg Y\simeq X\times Y$ for any objects $X$ and $Y$, and such that any object has the structure of an internal group object (which is necessarily unique). As the functor $\gamma$ preserves finite products, it preserves group objects, and, as $\gamma$ is essentially surjective, any object of $S^{-1}C$ has a canonical structure of group object...
| 20 | https://mathoverflow.net/users/1017 | 44155 | 28,048 |
https://mathoverflow.net/questions/44075 | 2 | In Nakajima's [Geometric construction of algebras](http://www.kurims.kyoto-u.ac.jp/~nakajima/TeX/HongKong/hongkong.pdf)(pages 3-7), he uses the subalgebra of the convolution algebra of $Gr(k^N)\times Gr(k^N)$ invariant under $GL\_N$ action to construct $U\_q(sl\_2)$. To do this, he finds imposes relations to form some special operators, which he shows are isomorphic to the generators of $U\_q(sl\_2)$. In particular, he uses characteristic functions on certain subsets of $k^N$ with dimensionality conditions. These conditions smell like Schubert conditions.
>
> Can we see these as the Schubert conditions in some way?
>
>
>
I have compared them, but I don't see the link.
EDIT: Read the comments below for a related paper.
| https://mathoverflow.net/users/348 | Can we see the geometric realization of $U_q(sl_2)$'s relations as Schubert Conditions? | Well, it depends what you mean; there's certainly a close link.
Any $GL\_N$ invariant closed subset of the product of two Grassmannians can be described as a family of Schubert varieties. Consider the projection to the first factor. This must be surjective, since $GL\_n$ acts transitively on the Grassmannian. The fiber over a subspace $V\subset \mathbb{C}^N$ must be a closed subset of the other Grassmannian which is preserved by the subgroup stabilizing $V$; that is, it must be a Schubert variety of the form $\{U\subset \mathbb{C}^N | \dim(U)=m, \dim(U\cap V)\geq h\}$.
The same argument can easily be applied to any product of two partial flag varieties with the same conclusion (unfortunately, it completely breaks down for more factors).
| 1 | https://mathoverflow.net/users/66 | 44156 | 28,049 |
https://mathoverflow.net/questions/44154 | 4 | Hi,
I'm trying to understand the group of cycles (modulo numerical equivalence) contracted by a flopping contraction $f$.
More precisely, I'm in the setup of Definition 2.12 of [this paper by Yukinobu Toda](http://arxiv.org/abs/0909.5129).
Let $f: X \to Y$ be a flopping contraction: $X$ is a smooth and projective CY3, f is birational, $Y$ is Gorenstein, $f$ is isomorphic in codimension one, $dim\_\mathbb{R} N^1(X/Y)\_\mathbb{R}=1$.
Where $N^1(X/Y)$ is the group of divisors of $X$ modulo numerical equivalence over $Y$ (viz. $D\_1 \equiv D\_2$ iff $D\_1.C=D\_2.C$ for all curves $C$ contracted by $f$).
(a side question is: what's the correct way to define "isomorphic in codimension d"?)
Denote, $N\_1(X/Y)$ the group of 1-cycles contracted by $f$, modulo numerical equivalence.
>
> What is $N\_1(X/Y)$? (without tonsuring with Q or R)
>
>
>
In the paper cited above, it is written that the exceptional locus of $f$ is a tree of projective lines $C\_1 \cup \ldots \cup C\_m$
>
> Is $C\_i \equiv C\_j$?
>
>
>
In the end I'm really hoping that $N\_1(X/Y) = \mathbb{Z}$. If this is not the case, then I'm also interested in what happens after tensoring with $\mathbb{Q}$.
Thanks.
| https://mathoverflow.net/users/3701 | What is the Exceptional Locus of a flopping contraction between threefolds? | 1) $f$ is *isomorphic in codimension $d$* if it is an isomorphism near any codimension $d$ point in either $X$ or $Y$. Equivalently, there exists closed subsets $Z\subseteq X$ and $W\subseteq Y$ such that ${\rm codim}\_XZ\geq d+1$, ${\rm codim}\_YW\geq d+1$, and $f:X\setminus Z\overset{\simeq}{\longrightarrow} Y\setminus W$ is an isomorphism.
2) By the Theorem of the Base of Néron–Severi, if $f$ is proper of finite type, then $N\_1(X/Y)\_{\mathbb Q}$ and $N^1(X/Y)\_{\mathbb Q}$ are finite-dimensional vector spaces of the same dimension. This is actually more than you need, because even without the finite type assumption it is true that the intersection pairing $N\_1(X/Y)\_{\mathbb Q}\times N^1(X/Y)\_{\mathbb Q}\to {\mathbb Q}$ is non-degenerate.
| 5 | https://mathoverflow.net/users/10076 | 44161 | 28,053 |
https://mathoverflow.net/questions/43968 | 0 | Hi,
I have a problem
>
> (1) where I need to compute the ratio of probabilities of hitting and stopping at a positive vertical barrier x vs hitting and stopping at a negative horizontal barrier y after starting from (0,0).
>
>
>
I feel that by symmetry, the answer to this would be the same as
>
> (2) The probability of hitting -y vs hitting +x, horizontal lines in 2d grid,
>
>
>
which looks like being same as
>
> (3) The probability of hitting -y vs x on a real line.
>
>
>
Can someone please tell me if my 1->2 assumption or 2->3 assumption is wrong. In which case, could someone please tell me how to proceed with the solution to 1.
However, if my assumption is right, can someone tell me how to proceed to prove it. Also, what would be a way to solve the case when both x and y are positive.
I shall be grateful for a response/hint/link.
Thanks.
EDIT: I really wish and hope that someone would answer it. Is there some way to transfer credit from superuser.com/stackoverflow.com here and place a bounty on it? I shall be really grateful if someone could give me an answer. Thanks.
| https://mathoverflow.net/users/10401 | Random walk question on 2D grid, probability of vertical line vs horizontal line hit | Problems 2 and 3 are equivalent by ignoring horizontal steps.
Problems 1 and 2 are not equivalent. I believe the probabilities are known, but I don't know them. However, if the probability of hitting the vertical barrier were really $y/(x+y)$ as in problem 2, then it would not be a martingale. So, whatever symmetry argument you wanted was incorrect. Specifically, consider the function $y/(x+y)$ on the lattice points $(x,y)$ in the first quadrant. The value at $(1,2)$ is $2/3$ but the average of the values at its neighbors is $11/16$.
For a Brownian motion, the angle from the origin is a martingale so you can use problem 3 to determine the answer to problem 1, $\frac{\arctan(y/x)}{\pi/2}$.
| 1 | https://mathoverflow.net/users/2954 | 44170 | 28,057 |
https://mathoverflow.net/questions/44165 | 1 | In my research, I work with certain finitely presented quotients of Coxeter groups. These are the automorphism groups of abstract polytopes, which are combinatorial generalizations of "usual" polytopes. (Essentially, an abstract polytope is an incidence complex.) Now, in this context, there is a useful combinatorial operation that has a nice effect on the automorphism groups. In fact, it's easy to generalize the operation on groups, so I'm curious whether any work has been done with this.
Let $G = \langle X \mid R \rangle$ and $H = \langle X \mid S \rangle$ be finitely presented groups. (I'm not sure that the finiteness of the presentation is essential, but let's assume it for now.) In other words, we have that $G = F(X) / \overline{R}$ and $H = F(X) / \overline{S}$, where $F(X)$ is the free group on $X$ and $\overline{R}$ is the normal closure of $R$ in $F(X)$. Then if $K$ naturally covers $G$ and $H$ (that is, if the identity map on $X$ extends to (surjective) homomorphisms from $K$ to $G$ and $K$ to $H$), we have that $K$ covers the group $F(X) / (\overline{R} \cap \overline{S})$. Similarly, if $G$ and $H$ naturally cover $K$, then the group $F(X) / (\overline{R} \overline{S})$ with presentation $\langle X \mid R \cup S \rangle$ naturally covers $K$ as well.
Therefore, the group $F(X) / (\overline{R} \cap \overline{S})$ is the minimal natural cover of $G$ and $H$, and $F(X) / (\overline{R} \overline{S})$ is the maximal natural quotient of $G$ and $H$. The first group is the fibre product of $G$ and $H$ over the second group.
These seem like such natural operations that I would guess they have been studied before, but I am having trouble finding anything. Any references would be greatly appreciated.
EDIT: Let me expand a little bit. As Mark Sapir points out, I'm essentially looking at the lattice of normal subgroups of F(X). I've looked a little at the general theory of subgroup lattices, but I'm really only interested in normal subgroups of F(X) or other finitely presented groups. Also, I find it difficult to work with the normal subgroups of F(X) directly, whereas it's not too hard to work with the quotients by these normal subgroups via presentations. So I'm hoping to find something that's somewhat more specific than just subgroup lattices; ideally, something that works with presentations.
Here are some examples of the type of questions I'm interested in:
1. Is there a simple way to write down a presentation for $F(X) / (\overline{R} \cap \overline{S})$ without changing the generators?
2. Given G and H, what are some conditions on the relations under which $\overline{R} \overline{S} = F(X)$? (This obviously won't be all-inclusive, but some instructive examples would be nice.)
| https://mathoverflow.net/users/913 | Reference request: lattice operations on the class of finitely presented groups | What you are studying is the lattice of normal subgroups of the free group $F(X)$. The normal subgroup lattices of groups have been studied a lot. For example, this lattice is complete and modular. See the references [here.](http://en.wikipedia.org/wiki/Lattice_of_subgroups)
**Update.** About your newer, more concrete, questions. Even if $R$ and $S$ are finite, the normal subgroup $\bar R\cap \bar S$ may not be finitely generated. If the membership problem for $\bar R$ and $\bar S$ is decidable, then in principle one can decide the membership problem for the intersection, so you will find a generating set of the intersection - the whole intersection. But I am sure that the problem whether the intersection is finitely generated is undecidable (although I did not think about a proof).
The question whether $\bar R\bar S=F\_k$ is the triviality problem which is undecidable (by a theorem of Adian-Rabin) even when $R$ is empty. For example, take any presentation of the trivial group, call a subset of it $R$ and the complement $S$. You get the equality $\bar R\bar S=F\_k$. Another way: take a presentation of a finite group of exponent $n$, say, $S\_3$ has exponent $6$ and presentation $\langle a,b \mid a^2=b^3=1, aba=b^{-1}\rangle$, and add relations that "kill" the generators, say $a^5=1, b^7=1$. The first set of relations is $R$, the second is $S$.
| 5 | https://mathoverflow.net/users/nan | 44189 | 28,070 |
https://mathoverflow.net/questions/35270 | 2 | If A is a commutative algebra and B is an X- algebra, then the tensor product $A \otimes B$ is an X-algebra (so for example, $Com \otimes Lie$ is a Lie algebra). This is seen using the language of operads. Let $Com$ be the commutative operad. Since $Com(n)$ is a one dimensional vector space for every $n$, tensoring $Com$ with an operad $O$ doesn't change the operad $O$.
Does a similar thing hold true for a $C\_\infty$ algebra? That is, if $A$ is a $C\_\infty$ algebra, is $A \otimes B$ an $X\_\infty$ algebra?
I'm still trying to familiarize myself with the language of operads, and perhaps the question can be made more precise in that language, where the infinity version of an operad is cofibrant resolution of the operad.
| https://mathoverflow.net/users/8358 | Extending a property of commutative algebras to C infinity algebras | We can see that this is false by looking at a degenerate example. Consider any operad $P$ so that $P(n)=0$ for $n>1$. If you tensor such an operad with $C\_\infty$ so that $(P\otimes C\_\infty)(n)=P(n)\otimes C\_\infty(n)$ you get $P\otimes C\_\infty=P$, since $C\_\infty(1)$ is one dimensional and $P(n)=0$ for $n>1$. But $P$ is not the same thing as $P\_\infty$ in general.
For a specific example, let $P(1)=k[x]/x^2$. Then $P\_\infty(1)$ can be modeled by $(k[x\_1,x\_2,\cdots,],dx\_k=\sum\_{i+j=k}x\_ix\_j)$.
| 6 | https://mathoverflow.net/users/3075 | 44191 | 28,072 |
https://mathoverflow.net/questions/43638 | 9 | Let $\lambda$ denote a partition of size $n$. Let
$$d\_{\lambda}= \text{number of distinct parts of } \lambda $$
$$o\_{\lambda}= \text{number of odd parts of } \lambda $$
$$f\_{\lambda}= \text{number of standard Young tableau of shape } \lambda $$
Given an involution $\pi \in S\_{n}$, whose insertion tableau has shape $\lambda$, it is well known (via the Robinson-Schensted correspondence, and neatly outlined in Sagan's book on the Symmetric Group) that :
$$ o\_{\lambda^{t}}= \text{number of fixed points in the involution } \pi $$
$$ \sum\_{\lambda \vdash n} f\_{\lambda}= \text{number of involutions in } S\_{n} $$
In the aforementioned formulae, $\lambda^{t}$ refers to the conjugate of the partition $\lambda$.
Now, some computations I have carried out for Kronecker products of two irreducible characters of $S\_{n}$ revealed the following identity in a special case:
$$\sum\_{\lambda \vdash n}d\_{\lambda}f\_{\lambda}=\sum\_{\lambda \vdash n}o\_{\lambda}f\_{\lambda}$$
Note that the right hand side actually counts the total number of fixed points in all involutions in $S\_{n}$. I did manage to prove the above result in general, but I am hoping someone could guide me to a proof which is bijective, i.e say uses the RS correspondence to establish the left hand side equals the the total number of fixed points in all involutions in $S\_{n}$.
Also, I'd like it if I could be directed to where this and/or similar sums appeared.(as an exercise in a book, or in some paper).
Thanks!
Edit: I had a look at Sagan, which I did not have handy last night and made a minor change in saying the number of fixed points in an involution $\pi \in S\_{n}$ is the number of odd columns in the insertion tableau of $\pi$.
Edit(10/27):
I thought I should put down the idea that I had. But since I am not sure if this should count as an answer, I am putting it in the body of the question.
Note that
$$\sum\_{\lambda \vdash n}d\_{\lambda}f\_{\lambda}=\sum\_{\lambda \vdash n+1}f\_{\lambda}-\sum\_{\lambda \vdash n}f\_{\lambda}$$
So all that remains to be shown is the nice fact that the total number of fixed points in all the involutions of $S\_{n}$ is the difference between the number of involutions in $S\_{n+1}$ and the number of involutions in $S\_{n}$.
| https://mathoverflow.net/users/10335 | A Distinct parts/Odd parts identity for standard Young tableaux | "all that remains to be shown is the nice fact that the total number of fixed points in all the involutions of $S\_n$ is the difference between the number of involutions in $S\_{n+1}$ and the number of involutions in $S\_n$."
And this is straightforward: every involution in $S\_n$ can be extended to an involution in $S\_{n + 1}$ either by replacing a fixed point $i$ with a cycle $(i, n + 1)$ or by adding $n + 1$ as a new fixed point.
(It is unclear to me whether you had already seen that this fact has such a simple bijective proof; it's also not clear to me whether this satisfies your desire for a completely bijective proof.)
| 2 | https://mathoverflow.net/users/4658 | 44204 | 28,078 |
https://mathoverflow.net/questions/44203 | 6 | Initial caveat: the following question could probably be answered by Google, MathSciNet or my library, if I could find the right search terms or book... but I've not had any luck today, so I hope someone can point me to a reference.
(The question is related to some of my older questions concerning characters of finite groups. All representations/characters are over the complex field.)
I am trying to estimate a certain invariant associated to finite groups, and recently thought that a useful toy example to play with would be
$$ G = \left\{
\left( \begin{matrix} a & b \\ 0 & 1 \end{matrix} \right) \;\mid\;
a,b\in {\mathbb Z}/p^n{\mathbb Z}, p \nmid a
\right\} $$
where $p$ is a prime and $n\geq 1$. I guess this might be called the "affine group" of the ring $R={\mathbb Z}/p^n{\mathbb Z}$?
Now the invariant can be calculated pretty easily (modulo tedious sums) once we know the character table of $G$, but this means more than knowing the degrees of the irreducible characters; I need to know the values they take on the various conjugacy classes inside $G$.
For $n=1$ this does not take long to do directly and can also be found as an example in various introductory-level textbooks on representation theory. However, for $n=2$ the best I could find was a section in a paper of several authors, where they just work out the character table by hand after first finding the characters via induction ($G$ is a semi-direct product arising from the action of the group of units in $R$ on the additive group of $R$). Now since I want to continue to higher $n$, I seem to be faced with three options:
1) Slog through the computation myself (which is probably good for my mathematical soul, but takes time & brainpower I need to spend on other things)
2) Learn how to ask a computer to do this (see previous parenthetical remark)
3) Find a reference which just gives the table.
So before embarking on 1), I thought I'd ask here. Most sources I could find from a crude skim online and in my library only discussed linear groups over finite fields; but I'm hoping that the construction here is sufficiently natural that it might have been treated already and written up somewhere.
| https://mathoverflow.net/users/763 | Character table for the affine group of Z/p^nZ | The groups you are interested in are sometimes called false Tate extensions in number theorists' jargon. They are Galois groups of the Galois closures of extensions of $\mathbb{Q}$ obtained by adjoining the $p^n$-th roots of a $p$-th power free element. The irreducible representations are very explicitly described in [Vladimir Dokchitser's PhD thesis](http://www.dpmms.cam.ac.uk/~vd209/phdweb.ps), beginning of chapter 3.
Alternatively, the characters, together with their values, are easy to compute using a procedure described in Serre's Linear representation book, part II, Section 8.2. He explains how to obtain all the irreducible characters of any group that is a semi-direct product when the normal subgroup is abelian. Note that the character of an induced representation is easy to compute in terms of the original character and the coset-action.
And of course one more remark: for any given $p$ and $n$, MAGMA will just give you the character table.
Edit: Since you seem unsure, how to ask a computer for the character table, here is MAGMA code as an example. You will easily adopt it to any other package that handles character tables:
```
p:=3; n:=2;
Z:=Integers();
gl:=GL(2,quo<Z|p^n*Z>);
pr:=PrimitiveRoot(p^n);
G:=sub<gl|gl![[1,1],[0,1]],gl![[pr,0],[0,1]]>;
CharacterTable(G);
```
| 6 | https://mathoverflow.net/users/35416 | 44205 | 28,079 |
https://mathoverflow.net/questions/44196 | 2 | Suppose that $(\Omega,\mathcal{A},\mu)$ is a $\sigma$-finite measure space of infinite measure and $T:\Omega\to\Omega$ a measure-preserving transformation with measurable inverse. Let be $\Omega\_k\in \mathcal{A}$ an increasing sequence such that $\Omega\_k\uparrow\Omega$ and
$\mu(\Omega\_k)<+\infty$ for all $k\in\mathbb{N}$.
**Question 1:** Given a set $A\in\mathcal{A}$, such that $\mu(A)>0$, is it true that the set
$$E\_k=\{\omega\in A; T^n(w)\notin A\ \forall n\in\mathbb{N}\ \text{and}\ T^{n\_j}(w)\in \Omega\_k \ \text{for some infinite sequence}\ (n\_j(\omega)) \}$$
has zero measure ?
**Question 2:** If $T$ is not invertible is $\mu(E\_k)=0$, in general ?
| https://mathoverflow.net/users/2386 | Poincare Recurrence Theorem on Infinite Measure Space | If I remember my infinite ergodic theory correctly, any measure-preserving transformation $T$ of a $\sigma$-finite measure space $(\Omega,\mathcal{A},\mu)$ leads to a decomposition of $\Omega$ into a dissipative part $\Omega\_d$ and a conservative part $\Omega\_c$. (This notation is probably non-standard.) The dissipative part is the union of the wandering sets for $T$, where a set $E$ is wandering if it is disjoint from all its preimages $T^{-n}(E)$, $n\geq 0$. The conservative part is thee complement of the dissipative part.
On the dissipative part, everything "goes to infinity" in some sense (eventually leaves $\Omega\_k$, for instance), while on the conservative part, the Poincare recurrence theorem holds. (Conservative measures are exactly those $\sigma$-finite measures for which the Poincare recurrence theorem still works.) I believe this establishes the dichotomy you want.
This may even work for transformations that don't preserve $\mu$, but only preserve the collection of null sets. I'd check a reference on that, though -- details of all of this are in Aaronson's book *An Introduction to Infinite Ergodic Theory*.
| 5 | https://mathoverflow.net/users/5701 | 44214 | 28,083 |
https://mathoverflow.net/questions/44219 | 25 | I am reading some work on Mirror Symmetry from Physics perspective,the physicists seem to use some aspects of BRST quantization and BRST cohomology. What is BRST Quantization and BRST cohomology, in realm of Mathematics. More precisely,
What are the BRST complexes, how do we get this cohomology, what is the relation (if any) of this cohomology theory to say de Rham or Cech cohomology. Where do the mathematicians use this theory. What is it in context of $N =2 $ Superconformal field Theory, most relevant for Mirror Symmetry.
I appreciate both physics and mathematical ideas.
| https://mathoverflow.net/users/9534 | BRST cohomology | There is no way I will be able to answer all of your questions, so instead I will focus on just a tiny part, and try at least to explain "BRST integrals". Much of what I say is probably well-known, but I am also in the process of writing up some conversations on this and related topics with Dan Berwick Evans, and if there's anything here that's new, then it is joint work with him :) For many basic ideas and definitions, I am also indebted to Rajan Mehta, as well as to other people here on MO and in person.
Recall that a *(Lie) algebroid* consists of the data of the following data:
* A smooth manifold $X$
* A vector bundle $A \to X$
* A Lie bracket $[,]: \Gamma(A)^{\wedge 2} \to \Gamma(A)$
* A map of vector bundles $\rho : A \to {\rm T}X$
The Lie bracket need not be $\mathcal C^\infty(X)$-linear — rather, for $a,b \in \Gamma(A)$ and $f\in \mathcal C^\infty(X)$, we require that $[a,fb] = \rho(a)[f] \cdot b + f\cdot [a,b]$, where $\rho(a)[f]$ is the action of the vector field $\rho(a) \in \Gamma({\rm T}X)$ on $f$.
There are two main examples of Lie algebroids:
1. Any integrable subbundle of the tangent bundle ${\rm T}X$ is an algebroid
2. If $\mathfrak g$ is a Lie algebra acting on a manifold $X$, then the action equips the trivial vector bundle $\mathfrak g \times X$ with an algebroid structure.
Recall also that a *graded vector space* is a (real) vector space $V = \bigoplus\_{n\in \mathbb Z} [n]V\_n$, where $V\_n$ is a classical vector space and the functor $[n]$ means "put it in degree $n$". The category of graded vector spaces is equivalent to the category of $U(1)$-modules.(**Edit:** As Scott Carnahan points out in the comments, *complex* graded vector spaces are equivalent to $U(1)$-modules, but *real* graded vector spaces are not. To a physicist, the idea is that a graded vector space has a "charge" or "energy" or "number" or whatever you want to call it operator, which is quantized to only take integer values, and this is often called a "$U(1)$ gauge group".)
This category has a symmetric tensor product, which is the same tensor product as in the category of $U(1)$-modules, but with the Koszul sign rule: if $a\in [m]V\_m$ and $b\in [n]W\_n$, then the canonical isomorphism $[m]V\_m \otimes [n]W\_n \to [n]V\_n \otimes [m]W\_n$ is the one that takes $a\otimes b$ to $(-1)^{mn}b\otimes a$. Given a (finite-dimensional) graded vector space $V$, the algebra of *polynomial functions on $V$* is the symmetric algebra (with respect to the Koszul rule) on its dual. The algebra of *smooth functions on $V$* is the completion of this algebra with respect to the natural Frechet topology: $\mathcal C^{\infty}(V) = \widehat{{\rm S}V^\*}$. (In particular, the generator of $\mathcal C^{\infty}([n]\mathbb R)$ is in degree $-n$.) By partitions of unity, this is a sheaf over $[0]V\_0$.
Finally, recall that a *graded manifold* is a manifold $X$ and a sheaf of algebras over $X$ that looks locally like the sheaf of smooth functions on a graded vector space. Every graded manifold is "affine": to present the whole sheaf, you just have to present the algebra of all smooth functions. The category of graded manifolds behaves much like the category of manifolds. The "shift" functors extend to functors of vector bundles: if $A \to X$ is a vector bundle of (graded) manifolds, then $[n]\_XA \to X$ is the vector bundle with the same base, but all fibers shifted by $n$.
A *Q-manifold* is a graded manifold $X$ along with a square-zero degree-one vector field, i.e. a derivation $Q$ of the algebra $\mathcal C^\infty(X)$ which shifts all homogeneous elements up one degree, and with $Q\circ Q = \frac12 [Q,Q] = 0$. (The "commutator" $[Q,Q]$ is taken with respect to the Koszul rule.) Equivalently, a Q-manifold is something so that $(\mathcal C^\infty(X),Q)$ is a dgca.
Any algebroid gives rise to a Q-manifold, by some form of *Koszul duality*: if $(A\to X,[,],\rho)$ is a (classical) algebroid, then $[-1]\_XA$is naturally a Q-manifold. Let $x^i$ be coordinates on $X$ and $a^\mu$ fiber coordinates on $[-1]\_XA$ with corresponding sections $a\_\mu(x) \in \Gamma(A)$, and adopt the Einstein summation convention. Suppose that the Lie bracket is given by the structure constants $[a\_\mu(x),a\_\nu(x)] = E\_{\mu,\nu}^\lambda(x) a\_\lambda(x)$, and that $\rho(a)(x) = \rho^i\_\mu(x) a^\mu \frac{\partial}{\partial x^i}$. Then:
$$ Q = \rho^i\_\mu(x) a^\mu \frac{\partial}{\partial x^i} + \frac12 E\_{\mu,\nu}^\lambda(x) a^\mu a^\nu \frac{\partial}{\partial a^\lambda}$$
Every Q-manifold $X$ for which $\mathcal C^\infty(X)$ is generated by elements of degrees $0$ and $1$ is of this form in a canonical way.
More generally, and I won't explain this in detail, any $\infty$-algebroid gives a Q-manifold. An *$\infty$-algebroid* consists of a chain complex $0 \to A\_n \to \cdots \to A\_1 \to 0$ of vector bundles over $X$, along with various "brackets" of different degrees satisfying compatibility conditions, and also with an "action" $\rho : A\_1 \to {\rm T}X$. Then $\bigoplus\_X [-n]A\_n$ is a Q-manifold in a natural way. A *Lie-Reinhardt pair* is a slightly more general thing than an algebroid — the only difference is that we do not require $A \to X$ to be a vector bundle, but only that $\Gamma(A)$ be a (quasicoherent?) sheaf over $\mathcal C^\infty(X)$. Then the theory of Lie-Reinhardt pairs is sufficiently geometric that there's a good "infinitized" version of it, and $\infty$-algebroids are the nice version (in the way that vector bundles are nicer than sheaves). The upshot is that $\infty$-LR-pair structures move well across quasiisomorphisms of chain complexes of sheaves. In particular, if $X$ is an *algebraic* manifold and $\mathcal D \subseteq \Gamma({\rm T}X)$ an algebraic integrable subsheaf, then by Hilbert-Syzygy it has a finite resolution in vector bundles, which gives a Q-manifold.
But, anyway, let me continue to talk about just plain algebroids $A\to X$, and their associated Q-manifolds $[-1]A$. Then $\mathcal C^\infty([-1]A)$ is dgca, as I said, and mathematicians will recognize it as the "Chevalley-Eilenberg complex of $A$", or "the complex that computes Lie algebroid cohomology". In the two examples above, this algebra is:
1. For $A = {\rm T}X$, $\mathcal C^\infty([-1]{\rm T}X)$ is nothing but the de Rham complex
2. For "action algebroids" $A = \mathfrak g \times X$, then $\mathcal C^\infty([-1]A)$ is precisely the Chevalley-Eilenberg complex of $\mathfrak g$ with coefficients in $\mathcal C^\infty(X)$.
From a mathematician's point of view, the *BRST construction* is nothing more nor less than a different construction that computes the same cohomology. In particular:
There is a "forgetful" functor from Q-manifolds to graded manifolds, and its right adjoint is precisely $X \mapsto [-1]{\rm T}X$. Let $M$ be a Q-manifold (e.g. $M = [-1]A$ for an algebroid $A$), and $M \to X$ a map of graded manifolds, and let $B \to X$ be a submersion of graded manifolds. By the adjunction, we get a diagram:
$$ \begin{matrix}
&& M \\
&& \downarrow \\
[-1]{\rm T}B & \rightarrow & [-1]{\rm T}X
\end{matrix} $$
The bottom arrow is a submersion since $B \to X$ is. The pullback of this diagram in the category of Q-manifolds is the "Q-manifold pullback of $A$ along $B\to X$". If $M = [-1]A$ for an algebroid $A\to X$, and if $B\to X$ are classical manifolds, then the pullback is naturally an algebroid over $B$.
Let $B \to X$ be a vector bundle now, which is certainly a submersion. The *Euler vector field* is the (degree-zero) vector field on $B$ that generates the $\mathbb R^\times$ action (so in particular it points in the fibers of $B$). It gives rise to a contraction of chain complexes at the level of $[-1]{\rm T}$: the map $B \to X$ determines a map $\mathcal C^\infty([-1]{\rm T}B) \leftarrow \mathcal C^\infty([-1]{\rm T}X)$, which is actually a quasiisomorphism. (I.e.: the de Rham cohomology of the total space of a vector bundle is the same as the de Rham cohomology of the base.) Moreover, if $D$ is the pullback of a Q-manifold $M$ along a vector bundle $B \to X$, then $D,M$ are also quasi-iso.
Thus, if you are interested in the cohomology of some Q-manifold $M$, you are free to pull it back. (Any graded manifold $M$ has a classical-manifolds "base" $X$, which is the vanishing locus of the functions in nonzero degrees, and there always exists a map $M \to X$, although it is not usually canonical. But often $M$ comes as a vector bundle over $X$ — in the algebroid and $\infty$-algebroid cases, for example, it does.)
In applications, we have $M = [-1]A$ for an algebroid $A \to X$, and usually $A = \mathfrak g \times X$. When $\mathfrak g = \operatorname{Lie}(G)$ for $G$ a simply-connected connected compact Lie group, the cohomology of $M$ computes the "equivariant cohomology of $G$ acting on $M$", so this is definitely something people care about.
So much for geometry: physicists want to compute integrals. A good notion of *measure with compact support* on a (graded) manifold $X$ is a continuous linear functional $\mathcal C^\infty(X) \to 0$. Temporarily, let $X$ be a (graded) vector space, and $\langle,\rangle$ symmetric (in the Koszul sense) pairing on $X$, represented by the matrix $\langle x,y\rangle = x^T p x$. Then we would like to have, at least for "smooth" measures $\mu$, a "Gauss" formula: $$\left( \int\_{x\in X} \exp( \frac12 \langle x,x \rangle) \mu(x)\right)^2 \to \frac{\#}{\det p} $$
as the support of $\mu$ expands,
where $\#$ depends on the dimension of $X$ (and the normalization of the measure), and $\det$ is the "super" determinant. When $p$ is nondegenerate, we can achieve this, but we cannot have a good theory of continuity: in $[0]\mathbb R^2 \oplus [-1]\mathbb R^2$, we have:
$$ \det \left( \begin{array}{cc|cc} \alpha &&& \\ & \alpha && \\ \hline &&& \beta \\ && -\beta & \end{array}\right) = \frac{\alpha^2}{\beta^2} $$
and so there are many paths of symmetric (in the Koszul sense) matrices that tend to $0$, say, with different determinants.
In particular, the action on $\mathbb R^2$ by itself by translation, which is represented by the Q-manifold with underlying graded manifold $X = [0]\mathbb R^2 \oplus [-1]\mathbb R^2$, should have volume $1$, as it should be the same as the quotient of $\mathbb R^2$ by its translation, but we cannot compute $\int\_X 1 = \int\_X \exp(0) = \frac 0 0$.
Well, we can if we remember that Q-manifolds put restrictions on which functions are "physical". Namely, really only the "Q-closed" functions — those $f$ with $Q[f] = 0$ — correspond to "physical observables". So a better definition of a "measure" is a continuous linear functional from the algebra of *closed* functions to $\mathbb R$. Moreover, we really should insist that the measure by invariant under the action of $Q$; this condition is the same as the condition that the continuous linear functional vanishes on *exact* functions. So: the space of *Q-measures* on a Q-manifold $M$ is precisely the continuous linear dual to the *cohomology* of $\mathbb C^\infty(M)$ with respect to the Q-structure.
So, what's the point? Often, you want to do an oscillating integral of the form $\int \exp(\frac i \hbar s)$. If $s$ has a nondegenerate critical point, you're in business: you apply the method of stationary phase, and the Feynman/Dyson diagrammatics, and you can really compute things. But if the critical point of $s$ is degenerate, you're stuck. Well, almost: you should try to change $s$. Suppose that you're on a Q-manifold $X$, that $s\in \mathcal C^\infty(X)$ is closed, and that $t \in \mathcal C^\infty(X)$ is exact, and that you're working with respect to a Q-measure. Then you can check for yourself (hint: do the case when $t$ is infinitesimal) that $\int\exp(\frac i \hbar s) = \int \exp( \frac i \hbar (s+t))$. So: maybe you can find a $t$ so that $s+t$ has a nondegenerate critical point?
In general, a Q-manifold does not have enough exact functions for this to be viable. But the point is that measures only see cohomology, so you are free to move to a quasi-isomorphic manifold. (Functions pull back but measures push forward, so you really do want the quasi-iso property.) Then you often can win.
In applications, it goes like this. Let $A \to X$ be an algebroid, $s\in \mathbb C^\infty(X)$ invariant under the $A$-action, and suppose that it has an isolated critical orbit. Let $M = [-1]A$, $B = [1]A^\*$, and form the pullback as above. Pick any degree-$(-1)$ function $\tau$ on $B$, which is the same as picking a section of $A$, and let $t = Q[\tau]$ be your exact function. One reason I picked this $B$ is that the degree-zero measures on the pullback (the ones that assign non-zero values to degree-zero functions) can be represented in the form ${\rm d}x^1 \cdots$, and so we can do computations, whereas the cohomologous measures on $M$ cannot be. Anyway, we wanted $s+t$ to have a nondegenerate critical point. This happens (in the generic situation) when the vanishing locus of $\tau$ intersects the critical orbit of $s$ transversally.
If you write out what the total integral of $\exp(\frac i \hbar (s+t))$ is over the pullback, you can identify the components (up to Fourier and some arguments at a "physical" level of rigor) with an integral over $X$ of $\exp(\frac i \hbar s)$ against a delta distribution supported on the zero locus of $\tau$ (and there's the correct Jacobi term thrown in). This delta-distribution integral is what Faddeev and Popov originally did — the BRST argument that I've sketched came later, and I think that "algebroid" language is pretty new.
| 24 | https://mathoverflow.net/users/78 | 44224 | 28,088 |
https://mathoverflow.net/questions/44208 | 88 | Ultrafinitism is (I believe) a philosophy of mathematics that is not only constructive, but does not admit the existence of arbitrarily large natural numbers. According to [Wikipedia](https://en.wikipedia.org/wiki/Ultrafinitism), it has been primarily studied by Alexander Esenin-Volpin. On his [opinions page](https://sites.math.rutgers.edu/%7Ezeilberg/OPINIONS.html), Doron Zeilberger has often expressed similar opinions.
Wikipedia also says that Troelstra said in 1988 that there were no satisfactory foundations for ultrafinitism. Is this still true? Even if so, are there any aspects of ultrafinitism that you can get your hands on coming from a purely classical perspective?
Edit: Neel Krishnaswami in his answer gave a link to a paper by Vladimir Sazonov ([On Feasible Numbers](https://doi.org/10.1007/3-540-60178-3_78)) that seems to go a ways towards giving a formal foundation to ultrafinitism.
First, Sazonov references a result of Parikh's which says that Peano Arithmetic can be consistently extended with a set variable $F$ and axioms $0\in F$, $1\in F$, $F$ is closed under $+$ and $\times$, and $N\notin F$, where $N$ is an exponential tower of $2^{1000}$ twos.
Then, he gives his own theory, wherein there is no cut rule and proofs that are too long are disallowed, and shows that the axiom $\forall x\ \log \log x < 10$ is consistent.
| https://mathoverflow.net/users/1574 | Is there any formal foundation to ultrafinitism? |
>
> Wikipedia also says that Troelstra said in 1988 that there were no satisfactory foundations for ultrafinitism. Is this still true? Even if so, are there any aspects of ultrafinitism that you can get your hands on coming from a purely classical perspective?
>
>
>
There are no foundations for ultrafinitism as satisfactory for it as (say) intuitionistic logic is for constructivism. The reason is that the question of what logic is appropriate for ultrafinitism is still an open one, for not one but several different reasons.
First, from a traditional perspective -- whether classical or intuitionistic -- classical logic is the appropriate logic for finite collections (but not K-finite). The idea is that a finite collection is surveyable: we can enumerate and look at each element of any finite collection in finite time. (For example, the elementary topos of finite sets is Boolean.) However, this is not faithful to the ultra-intuitionist idea that a sufficiently large set is *impractical* to survey.
So it shouldn't be surprising that more-or-less ultrafinitist logics arise from complexity theory, which identifies "practical" with "polynomial time". I know two strands of work on this. The first is Buss's work on $S^1\_2$, which is a weakening of Peano arithmetic with a weaker induction principle:
$$A(0) \land (\forall x.\;A(x/2) \Rightarrow A(x)) \Rightarrow \forall x.\;A(x)$$
Then any proof of a forall-exists statement has to be realized by a polynomial time computable function. There is a line of work on bounded set theories, which I am not very familiar with, based on Buss's logic.
The second is a descendant of Bellantoni and Cook's work on programming languages for polynomial time, and Girard's work on linear logic. The Curry-Howard correspondence takes functional languages, and maps them to logical systems, with types going to propositions, terms going to proofs, and evaluation going to proof normalization. So the complexity of a functional program corresponds in some sense to the practicality of cut-elimination for a logic.
IIRC, Girard subsequently showed that for a suitable version of affine logic, cut-elimination can be shown to take polynomial time. Similarly, you can build set theories on top of affine logic. For example, Kazushige Terui has since described a set theory, [Light Affine Set Theory](https://doi.org/10.1023/B:STUD.0000034183.33333.6f), whose ambient logic is linear logic, and in which the provably total functions are exactly the polytime functions. (Note that this means that for Peano numerals, multiplication is total but exponentiation is not --- so Peano and binary numerals are not isomorphic!)
The reason these proof-theoretic questions arise, is that part of the reason that the ultra-intuitionist conception of the numerals makes sense, is precisely because they deny large *proofs*. If you deny that large integers exist, then a proof that they exist, which is larger than the biggest number you accept, doesn't count! I enjoyed Vladimir Sazonov's paper ["On Feasible Numbers"](https://doi.org/10.1007/3-540-60178-3_78), which explicitly studies the connection.
I should add that I am not a specialist in this area, and what I've written is just the fruits of my interest in the subject -- I have almost certainly overlooked important work, for which I apologize.
| 65 | https://mathoverflow.net/users/1610 | 44232 | 28,092 |
https://mathoverflow.net/questions/44182 | 6 | Let $f:X\longrightarrow Y$ be a finite separable morphism of smooth projective integral curves over an algebraically closed field.
Then we have a linear equivalence of Weil divisors on $X$: $$ K\_X=f^\ast K\_Y + R.$$ Here $$R=\sum \textrm{length} (\Omega\_{X/Y})\_p [p]$$ is the ramification divisor on $X$. This is the Riemann-Hurwitz theorem.
We have a short exact sequence $$ 0 \longrightarrow \textrm{Pic}^0(X) \longrightarrow \textrm{Pic}(X) \longrightarrow \mathbf{Z} \longrightarrow 0,$$ where $\textrm{Pic}(X)\longrightarrow \mathbf{Z}$ is the degree map.
We know what the Riemann-Hurwitz theorem tells us on the degree part of $\textrm{Pic}(X)$. It gives us the topological data $g(X)$ in terms of the degree of $R$, the genus of $Y$ and the degree of $f$.
But what does it tell us on $\textrm{Pic}^0(X)$?
| https://mathoverflow.net/users/4333 | What does the Riemann-Hurwitz formula tell us on the Picard variety | The answer is quite classical when $f \colon X \to Y$ is an unramified double cover.
In this case Riemann - Hurwitz formula gives
$g(X)-1 = 2g(Y)-2$.
Consider the following three natural maps:
$f^\* \colon J(Y) \to J(X)$,
$Nm \colon \textrm{Pic}^0(X) \to \textrm{Pic}^0(Y), \quad Nm(\sum a\_ip\_i):= \sum a\_if(p\_i)$
$\tau \colon J(X) \to J(X)$,
where $f^\*$ is induced by the pull-back of $0$-cycles, $Nm$ is the norm map and $\tau$ is the involution induced by the double cover $f$.
Then
1. $\textrm{Ker} \; f^\*=\langle L \rangle$, where $L$ is a point of order $2$ in $J(Y)$;
2. the connected component of $Nm^{-1}(0)$ containing the identity coincides with the image of $I-\tau$. It is an Abelian subvariety of $\textrm{Pic}^0(X)$ of dimension $g(Y)-1$, that is denoted by $\textrm{Prym}(X, \tau)$.
Moreover, under the identification of $\textrm{Pic}^0(X)$ with $J(X)$, the principal polarization of $J(X)$ restricts to twice a principal polarization on $\textrm{Prym}(X, \tau)$.
The geometry of Prym varieties is very rich. In particular, Riemann-Hurwitz identity
$K\_X =f^\*K\_Y$
induces subtle relations between the Theta divisor $\Theta$ of $X$ and the Theta divisor $\widetilde{\Theta}$ of $\textrm{Prym}(X, \tau)$.
You can look at [Arbarello-Cornalba-Griffiths-Harris, Geometry of algebraic curves, Appendix C] or at [Birkenhake-Lange, Complex Abelian Varieties, Chapter 12] for further details.
In the general case, it is possible to define the so-called *generalized Prym varieties*, at least where $f \colon X \to Y$ is a tame Galois branched cover. Look for instance at the paper of MERINDOL
"Varietés de Prym d'un revetement galoisien [Prym varieties of a Galois covering]"
Journal Reine Angew. Math. 461 (1995), 49-61.
| 9 | https://mathoverflow.net/users/7460 | 44235 | 28,094 |
https://mathoverflow.net/questions/44187 | 11 | I am not a logician or set theorist, so hopefully this makes sense. Let $T$ be a theory which is expressive enough to make statements like "Statement $A$ has a proof in $T$"; for example, $T$ might be capable of expressing elementary arithmetic and proving certain basic arithmetic facts.
Let $\Pi^0(T)$ consist of those statements $A$ in $T$ such that either $A$ or $\neg A$ admits a proof in $T$. In general, for $i\in \mathbb{N}$ let $\Pi^i(T)$ consist of those statements $A$ such that the statement "Con(T) implies that there does not exist a proof of $A$, or a proof of $\neg A$, in $T$" is in $\Pi^{i-1}(T)$. Intuitively, $\Pi^1$ should consist of those statements $A$ for which one can prove in $T$ that neither $A$ nor $\neg A$ admits a proof in $T$, $\Pi^2$ should consist of those statements $A$ such that one can prove in $T$ that one cannot tell whether or not they admit a proof, but where one knows this fact, and so on.
Godel's first incompleteness theorem implies that $\Pi^0(T)$ does not contain every sentence of $T$; in particular, the proof amounts to constructing a sentence in $\Pi^1(T)$.
Now I'm sure that if this hierarchy is not nonsense for some reason that I'm missing, then it must be well-studied. So here goes:
>
> 1. If this set-up is studied, what is it called?
>
>
>
>
> 2. In standard theories, e.g. ZFC, is every sentence contained in $\bigcup\_{i\in \mathbb{N}} \Pi^i(T)$? It seems to me that any ZFC proof that this is not the case must be non-constructive, and so would be pretty interesting.
>
>
>
>
> 3. Are there any standard theories $T$ of finite (known) depth in this hierarchy, i.e. every sentence is contained in $\Pi^i(T)$ for some fixed $i>0$? If so, can this be proven in $T$?
>
>
>
>
> 4. One can also define this hierarchy in a relative setting; e.g. one can take two theories $T\subset S$ and ask about $S$-proofs of statements about the existence of proofs in $T$. Is this studied? Are answers to the above questions known in these cases?
>
>
>
Motivation: I recently watched [this talk](http://video.ias.edu/voevodsky-80th) by Voevodsky, and ever since I've been wondering about how much we actually know about what we can prove. One might hope that even if we can't prove every "true" statement, then we can at least always prove that a statement does not admit a proof in our theory, if that is indeed the case. That seems unlikely, and amounts to the claim that all statements in our theory are in $\Pi^1(T)$, but I think that it is the best situation one can hope for (at least naively) given Godel's theorems. So, is the situation hopeless? Give it to me straight, doc.
| https://mathoverflow.net/users/6950 | What is the depth of the "provability hierarchy"? | I believe the concept you are looking for is that of "iterated consistency extension." A very nice treatment is given by Torkel Franzén in his book [Inexhaustibility: a non-exhaustive treatment](http://books.google.com/books?id=6gJa6bjf374C&lpg=PP1&ots=5jv2Db6sxZ&dq=inexhaustibility&pg=PA185#v=onepage&q&f=false). See also this [related question](https://mathoverflow.net/questions/12865/using-consistency-to-create-new-axioms-in-set-theory) and this [blog article](http://xorshammer.com/2009/03/23/what-happens-when-you-iterate-godels-theorem/) by Mike O'Connor.
| 11 | https://mathoverflow.net/users/2000 | 44240 | 28,096 |
https://mathoverflow.net/questions/44242 | 2 | Let $p(n)$ denote count of lattices on finite set $G$, $|G|=n$ (without isomorphism). It's know closed formula for $p(n)$?
It's clear, that $1 \leq p(n)$ and also that $p(n-1) \leq p(n)$ for $n \geq 2$. My other estimates are $p(n) \leq 2^{\frac{(n-1)(n-2)}{2}}$ (also $p(n) \leq 2^{\frac{(n-1)}{2}}$) and $p(n-1) < p(n)$ for $n \geq 4$. Better lower bound for $p(n)$ is $\min(1,n - 2) \leq p(n)$
If there are not closed formula for $p(n)$, what we are able say about that function?
Thanks for help. (Sorry for my bad English)
| https://mathoverflow.net/users/9067 | Count of lattices on finite set | 1, 1, 1, 1, 2, 5, 15, 53, 222, 1078, 5994, 37622, 262776, 2018305, 16873364, 152233518, 1471613387, 15150569446, 165269824761, ...
There is a lot of information in [The On-Line Encyclopedia of Integer Sequences.](http://oeis.org/A006966)
| 2 | https://mathoverflow.net/users/4600 | 44247 | 28,099 |
https://mathoverflow.net/questions/44243 | 6 | One defines the $H^n(G,M)$ where $M$ is a $\mathbb{Z}[G]$ module as $Ext^n\_{\mathbb{Z}[G]}(\mathbb{Z},M)$ where $\mathbb{Z}$ is viewed as a trivial $\mathbb{Z}[G]$-module.
Is this part of a general pattern for how to define cohomology for non-abelian categories?
In groups we see that we switch to the abelian category of $\mathbb{Z}[G]$-modules. If this idea is indeed extended for general non-abelian categories, what abelian category do we switch to?
| https://mathoverflow.net/users/5309 | Group cohomology and cohomology in non-abelian categories | Yes, this generalizes to Hochschild cohomology and André-Quillen cohomology.
Given a category $\mathcal{C}$ with finite limits and an object $X$, you can form the category of Beck modules $\operatorname{Ab}(\mathcal{C} / X)$, abelian group objects in the slice category over $X$. When $\mathcal{C}$ is the category of groups, a Beck module over $G$ is precisely a split extension of $G$ with abelian kernel, so Beck modules can be identified with $\mathbb{Z}[G]$-modules.
In nice cases, the category $\operatorname{Ab}(\mathcal{C} / X)$ is abelian, and the forgetful functor $\operatorname{Ab}(\mathcal{C} / X) \to \mathcal{C} / X$ has a left adjoint, called abelianization $\operatorname{Ab}\_X$. The Hochschild cohomology of a Beck module $M$ is defined to be $HH^i(X; M) = \operatorname{Ext}^i(\operatorname{Ab}\_X X, M)$. In the groups case, $\operatorname{Ab}\_G G$ is the augmentation ideal, so Hochschild cohomology is just group cohomology shifted by 1.
Hochschild cohomology itself is only an approximation to André-Quillen cohomology, which is somehow a more homotopically correct notion (I don't know enough about any of this to explain what this means, exactly). In the case of groups, these two cohomology theories coincide. I suggest having a look at [Martin Frankland's thesis](http://www.math.uiuc.edu/~franklan/Frankland_Thesis_20100513.pdf), which gives a good overview of these things. It works out explicit examples (groups, abelian groups, associative algebras, commutative algebras) very nicely in Appendix A.
| 7 | https://mathoverflow.net/users/396 | 44251 | 28,102 |
https://mathoverflow.net/questions/44211 | 5 | For which $n\in \mathbb{N}$, can we find (reps. find explicitly) $n+1$ integers $0 < k\_1 < k\_2 <\cdots < k\_n < q<2^{2n}$
such that
$$\prod\_{i=1}^{n} \sin\left(\frac{k\_i \pi}{q} \right) =\frac{1}{2^n} $$
P.S.: $n=2$ is obvious answer, $n=6 $ is less obvious but for instance we have $k\_1 = 1$, $k\_2 = 67$, $k\_3 = 69$, $k\_4 = 73$, $k\_5 = 81$, $k\_6 = 97$, and $q=130$.
| https://mathoverflow.net/users/3958 | Product of sine | Consider the identity (as quoted by drvitek):
$$\prod\_{k=1}^{n} \sin \left(\frac{(2k-1) \pi}{2n}\right) = \frac{2}{2^n},$$
This is completely correct but doesn't quite answer the question because the RHS is
not $1/2^n$ $\text{---}$ this is an issue related to the fact that $\zeta - \zeta^{-1}$ is not
a unit if $\zeta$ is a root of unity of prime power order.
Replace $n$ by $3n$, and take the ratio of the corresponding
products. Then one finds that
$$\prod\_{(k,6) = 1}^{k < 6m} \sin \left(\frac{k \pi}{6m}\right)=
\sin \left(\frac{\pi}{6m}\right)
\sin \left(\frac{5 \pi}{6m}\right)
\sin \left(\frac{7 \pi}{6m}\right) \cdots
\sin \left(\frac{(6m-1) \pi}{6m}\right) = \frac{1}{2^{2m}}.$$
This provides the identity you request for $n = 2m$, since
$q = 6m < 2^{4m} = 2^{2n}$ is true for all $m \ge 1$.
There are other "obvious" identities that can be written down, but they tend to have length
$\phi(r)$ for some integer $r$, and $\phi(r)$ is always even (if $r > 2$).
For odd $n$, note the "exotic" identity:
$$\sin \left(\frac{2 \pi}{42}\right)
\sin \left(\frac{15 \pi}{42}\right)
\sin \left(\frac{16 \pi}{42}\right) = \frac{1}{8}.$$
Since $42 < 64$, this is an identity of the required form for $n = 3$.
On the other hand, none of the rational numbers
$1/21$, $5/14$, $8/21$ can be written in the form $k/6m$
where $(k,6) = 1$.
Hence
$$\sin \left(\frac{2 \pi}{42}\right)
\sin \left(\frac{15 \pi}{42}\right)
\sin \left(\frac{16 \pi}{42}\right)
\prod\_{(k,6) = 1}^{k < 6m} \sin \left(\frac{k \pi}{6m}\right) = \frac{1}{2^{2m+3}},$$
when written under the common denominator $q = \mathrm{lcm}(42,6m)$, consists of distinct fractional multiples $k\_i/q$ of $\pi$ with $0 < k\_i < q$, and is thus
an identity of the required form for $n = 2m + 3$, after checking that
$$q = \mathrm{lcm}(42,6m) \le 42m \le 2^{4m+6} = 2^{2n}.$$
Thus the answer to your question is that such an identity holds for all $n > 1$.
(It trivially does not hold for $n = 1$.)
The first few identities constructed in this way are:
$$\sin \left(\frac{\pi}{6}\right) \sin \left(\frac{5 \pi}{6}\right) =
\frac{1}{4},$$
$$\sin \left(\frac{2 \pi}{42}\right)
\sin \left(\frac{15 \pi}{42}\right)
\sin \left(\frac{16 \pi}{42}\right) = \frac{1}{8},$$
$$\sin \left(\frac{\pi}{12}\right) \sin \left(\frac{5 \pi}{12}\right)
\sin \left(\frac{7 \pi}{12}\right) \sin \left(\frac{11 \pi}{12}\right) =
\frac{1}{16},$$
$$\sin \left(\frac{2 \pi}{42}\right) \sin \left(\frac{7 \pi}{42}\right)
\sin \left(\frac{15 \pi}{42}\right)
\sin \left(\frac{16 \pi}{42}\right)
\sin \left(\frac{35 \pi}{42}\right) = \frac{1}{32},$$
&. &.
| 11 | https://mathoverflow.net/users/nan | 44253 | 28,103 |
https://mathoverflow.net/questions/44252 | 4 | Suppose $f(z)$ is a function analytic in the strip $|Re(z)|\leq a$. Is the fourier transform $\hat{f}(w)=o(e^{-a|w|})$?
It seems plausible but I can't seem to prove it either.
There is similar result called the Paley-Wiener Theorem that states $e^{a|w|}\hat{f}(w)\in L\_2(\mathbb{R})$, but I don't think that helps.
Thanks in advance.
| https://mathoverflow.net/users/2011 | Decay of the Fourier transform | This is not true without additional integrability conditions on $f(\cdot+iy)$.
$\hat{f}(w)=o(e^{-a|w|})$ implies that $e^{b|w|}\hat{f}(w)\in L\_2(\mathbb{R})$ for all $b < a$. The latter inclusion holds if and only if $f(z)$ is analytic in the strip $|\Im(z)|< a$ *and* $$ \sup\limits\_{|y|\leq b}\|f(\cdot+iy)\|\_{L^2(\mathbb R)}<\infty$$
for all $b< a$ (see *Fourier Analysis, Self-Adjointness* by Reed and Simon (Methods of Modern Mathematical Physics, Vol. 2, Theorem IX.13)).
| 7 | https://mathoverflow.net/users/5371 | 44256 | 28,106 |
https://mathoverflow.net/questions/7258 | 12 | q-Catalan numbers are defined recurrently as C0=1, $C\_{N+1}=\sum\_{k=0}^N q^k C\_k C\_{N-k}$.
What can be said about the asymptotics of Cn when `0<q<1`?
P.S. In the case q>1 it is known that as n goes to infinity, $q^{-{n\choose 2}}C\_n(q)$ tends to the partition function $\prod\_{i=1}^\infty\frac1{1-q^{-i}}$. However, this doesn't help in the case `0<q<1`.
| https://mathoverflow.net/users/979 | Asymptotics of q-Catalan numbers | Re Leonid's comment on a previous answer.
If the ratios $C\_{n+1}/C\_n$ converge, their limit $c(q)$ is such that $C(q,q/c(q))=c(q)$.
Equivalently, $1/c(q)$ is the radius of convergence of the series $z\mapsto C(q,z)$.
Or, writing $C(q,\cdot)$ as the ratio of two $q$-hypergeometric functions, one can show that $F(q,1/c(q))=0$, where
$$
F(q,z)=\sum\_{n\ge0}(-1)^nq^{n^2-n}z^n/(q)\_n.
$$
This implies that $c(q)$ is the sum of a series in $q$ with integer coefficients, whose signs seem to be alternating starting with the coefficient of $q$. The first terms are
$$
c(q)=1+q+q^3-q^4+2q^5-3q^6+6q^7-12q^8+25q^9-52q^{10}+111q^{11}+\ldots
$$
The function $q\mapsto c(q)$ is nondecreasing on $q\ge0$, obvious values are $c(0)=1$ and $c(1)=4$, and as a holomorphic function, $c(\cdot)$ might have a pole inside the unit disk at about $q\approx-.4$.
But apart from that...
| 3 | https://mathoverflow.net/users/4661 | 44260 | 28,108 |
https://mathoverflow.net/questions/44125 | 40 | Hi,everyone. I am looking for an introductory textbook on moduli theory,about the background on algebraic geometry,I have read Hartshorne chapter1~4. could you please show some good books or roadmap for studying moduli theory which emphasis to arithmetic aspect?
how about Mumford's GIT? is it an introductory textbook?
Thank you very much!
| https://mathoverflow.net/users/5274 | What is a good introductory text for moduli theory? | Ian Morrison wrote up some nice lectures in the book Lectures on Riemann surfaces,World Scientific publishers, Proceedings of the college of Riemann surfaces in 1987, at the ICTM in Trieste. They were intended as an informal introduction to the two detailed treatments mentioned below by Mumford (l'Enseignement) and Gieseker (Tata).
There are notes on Ravi Vakil's web page for his course on deformations and moduli:
<http://math.stanford.edu/~vakil/727/index.html>
There is a nice treatment of the chow coordinates of a projective variety in chapter 1 of the book Basic algebraic geometry by Shafarevich. This is very elementary and readable.
There is a good discussion of the existence of the Hilbert scheme in Mumford's book Lectures on curves on an algebraic surface, Annals of math studies #59. Sophisticated, but we were able to use it in a seminar long ago, and got some good insight from it.
Mumford (notes by Morrison) first wrote up the case of stable curves in Stability of projective Varieties, in l'Enseignement mathematique, 1977, based on an idea of Gieseker.
Then Gieseker himself presented his version at the Tata Institute in Bombay (TIFR), and wrote it up in their series of lectures on mathematics and physics, #69, 1982.
The original presentation of the concept of stable curves, due to Alan Mayer and David Mumford, is in talks by Mayer and Mumford at the Woods Hole conference 1964, available on James Milne's web site at Michigan, or that of roy smith (mathwonk) at University of Georgia.
As I recall, even the detailed works by Mumford, (GIT, Enseignement), always include some introductory examples and motivation that anyone can read, so one should not shy away from the actual definitive works completely. In regard to the fine recommendations above, Mukai's is actually a textbook as requested, and not a monograph like most of my recommendations here, but that of course makes it longer.
For beginners, I would observe that the Chow approach is to characterize a projective variety by all lines meeting it, thus getting a subset of the Grassmannian of lines, while the Hilbert approach is to describe a variety by the set of all hypersurfaces of fixed large degree containing it, thus getting a subspace of the vector space of those polynomials, another Grassmannaian. Then to characterize abstract varieties, one first chooses some natural projective embedding, say by a multiple of the canonical class, then considers the corresponding Hilbert or Chow scheme, and tries to collapse together all different embeddings of the same variety, in GIT by taking a quotient by a group action. This then leads to singularities at orbits which are smaller than usual, i.e. at points with non trivial isotropy coming from automorphisms of the variety. These isotropy groups are included in the data of a moduli "stack", but were always considered informative even earlier.
Since the subject is huge, it helps also to know which aspect is of interest. A moduli space is usually the set of isomorphism classes of objects of a given type. Hence, aside from foundational subtleties, it “exists” as a set. Then the problem is to give it more structure and to prove it has some nice properties. In algebraic geometry one often tries to give it structure as an algebraic space, scheme, or quasi projective variety, perhaps progressively in that order. So the first job would be to define a natural structure as abstract topological space or even abstract scheme. Next one wants to capture this structure by some “moduli”.
Classically, “moduli” are numbers that distinguish non isomorphic objects, i.e. numerical invariants such as projective coordinates in a field, so this translates into the stage of giving a structure of quasiprojective variety. This requires finding embedding functions, or sections of line bundles which are constant on equivalence classes. If the equivalence classes are orbits of a group action, one seeks functions constant on orbits, i.e. “invariants”, and this is the subject of “invariant theory".
Since algebraic projective mappings are continuous, their level sets are closed, so the geometric invariant theory problem arises of which orbits are closed. This leads into various concepts of “stability” of objects under a given action, and also, since closure is a relative notion, of determining certain unstable subsets to exclude so that the remaining orbits become closed. This is the subject studied by Mumford in which he adapted ideas of Hilbert.
Finally, one wants to find a good geometric compactification of the given moduli space, since the set of isomorphism classes of a given type is seldom compact. The method of parametrizing moduli spaces by subsets of Hilbert schemes, yields a natural compactification, since Hilbert schemes are projective, but since all isomorphism classes of the original type were already present before compactifying, it is unclear what geometric objects the new points added correspond to. This leads to the challenge of identifying the Hilbert scheme compactification with a more abstract compactification which adds in degenerate versions of the original geometric objects. These abstract objects are called perhaps “moduli stable” objects of the orginal kind, and one must show this abstract compact space can be identified with some version of the Hilbert scheme projective compactified one.
The concept of (moduli) stable curves was introduced by Mayer and Mumford, and the next job was to show they give a good abstract separated compactifiction of M(g). This is presumably the content of the paper of Deligne and Mumford. Then the proof they in fact give a natural projective compactification in the Hilbert scheme GIT sense is apparently accomplished in the references of Mumford and Gieseker.
Aside from these global aspects of moduli there are local questions, such as what is the dimension of a (component of a) moduli space, or what is its tangent space? These are the concern of “deformation theory”, or the local variations of structure of a given object. Here also one distinguishes deformations of the original objects, usually non singular varieties or manifolds, as in the works of Kodaira, from deformations of the degenerate objects included at the boundary of the compactification, i.e. deformations of singularities. For the latter there is a nice Tata lecture note by M. Artin, and a recent book by Greuel, Lossen, and Shustin. All sources rely fundamentally on the unpublished 1964 PhD thesis of M. Schlessinger at Harvard.
After all these foundations are settled, it remains to compute invariant properties of the resulting moduli spaces, their singularities, canonical class, Kodaira dimension, Picard group, cohomology, chow ring, rational curves in them,….. For M(g)bar this is still going in progress.
However it seems to me most answers, especially mine, are oriented to geometric questions as opposed to the requested arithmetic ones. Should one suggest some works say by Faltings and Chai?
| 21 | https://mathoverflow.net/users/9449 | 44267 | 28,112 |
https://mathoverflow.net/questions/44269 | 25 | Let $G$ be a group, $G'=[G, G]$.
>
> "Note that it is not necessarily true that the commutator subgroup
> $G'$ of $G$ consists entirely of
> commutators $[x, y], x, y \in G$ (see [107] for some finite group examples)."
>
>
>
Quoted from <http://www.math.ucdavis.edu/~kapovich/EPR/ggt.pdf> page 8.
Anybody can provide the examples? I can't find the book [107].
| https://mathoverflow.net/users/3922 | Commutator subgroup does not consist only of commutators? | The problem is whether the commutator subgroup may contain elements that are not commutators.
One example are the free groups. For instance, in the free group of rank $4$, freely generated by $x$, $y$, $z$, and $w$, the element $[x,y][z,w]$ of the commutator subgroup cannot be written in the form $[a,b]$ for some $a,b$ in the group.
The smallest finite examples are groups of order 96; there's two of them, nonisomorphic to each other. (This was a result in Robert Guralnick's thesis). See [this Math Stack exchange question](https://math.stackexchange.com/questions/7811/derived-subgroups-and-commutators) for a description of these groups, and some references.
| 33 | https://mathoverflow.net/users/3959 | 44276 | 28,117 |
https://mathoverflow.net/questions/44268 | 19 | Throughout "curve" means smooth projective curve over an algebraically closed field.
**Motivation and Background**
I read somewhere that Atiyah has classified vector bundles on elliptic curves. My understanding is that the story is roughly: every vector bundles breaks up as a direct some of indecomposable vector bundles. The indecomposable vector bundles are further divided by their degree and rank. The set $Ind(d,r)$ of isomorphism classes of indecomposable vector bundles of a fixed degree $d$ and rank $r$ has the structure of variety isomorphic to the Jacobian of the curve; namely the curve itself. In fact $Ind(d,r) \cong Ind(0, gcd(d,r) )$ so its enough to consider the degree $0$ vector bundles.
Now for $V,V'$ vector bundles on any curve $C$ its true that $\deg V \otimes V' = \deg V\cdot rk(V') + \deg V'\cdot rk(V)$. So in the case of an elliptic curve the set $Ind(0,r)$ is a torsor for $Pic^0(C)$. Additionally, there is a unique isomorphism class $V\_r \in Ind(0,r)$ characterized by the fact that $h^0(V\_r) \ne 0$ (in fact $h^0(V\_r)=1$).
**The Question**
Since curves of low genus are usually "simple enough" to easily describe explicitly, I would like to see how explicit I can be with a description of the $V\_r$, or at least $V\_2$. So my question is simply, given an explicit elliptic curve in $\mathbb{P}^2$, say $zy^2 - x(x-z)(x+z)$can you reasonably show how to construct $V\_2$; i.e. give cocyles for it $\phi\_{ij} \colon U\_{ij} \to GL\_2(k)$, or produce a graded module for it?
**Initial Thoughts**
$V\_2$ should correspond to the unique nontrivial extension in $Ext^1(\mathcal{O}, \mathcal{O}) \cong H^1(\mathcal{O}) \cong k$.On a curve $C$, $0 \to \mathcal{O}\_C \to K(C) \to K(C)/\mathcal{O}\_C \to 0$ is a flasque resolution of the structure sheaf and an element in $H^1(\mathcal{O}\_C)$ corresponds to a map $\alpha \colon \mathcal{O}\_C \to K(C)/\mathcal{O}\_C$. Then the desired extension should be the pullback of $0 \to \mathcal{O}\_C \to K(C) \to K(C)/\mathcal{O}\_C \to 0$ via $\alpha$.
The trouble with this is that I can't seem to pin down what $\alpha$ is. I now that, via Serre duality, it corresponds to a global section of $H^0(\omega\_C)$ which I can explicitly describe as a differential on the curve. Using the example I mentioned above, on the affine patch where $z \ne 0$, it is: $\frac{dx}{2y} = -\frac{dy}{3x^2-1}$ but I guess I don't understand Serre duality well enough to determine $\alpha$ from this. Also I'm not even sure if this will lead to a reasonable way of getting at the cocylces of $V\_2$. And this is so close to "just using the definitions" I have to imagine there might be a better way to go about this...
| https://mathoverflow.net/users/7 | How do you describe vector bundles on elliptic curves? | A description via cocycle usually is not convenient to work with. From my point of view, a description as an extension is much more useful. But if you want something else, I would advice the following. Since your curve $C$ is given as a double covering of $P^1$, that is as a relative spectrum of the sheaf of algebras $A = O \oplus O(-2)$ on $P^1$ (with $O$ summand being generated by the unit and with multiplication $O(-2)\otimes O(-2) \to O$ given by the ramification divisor $D$), the category of coherent sheaves on $C$ is identified with the category of sheaves on $P^1$ with a structure of a module over this algebra.
Spelling this out, a sheaf on $C$ is a sheaf $F$ on $P^1$ with a morphism $\phi:F(-2) \to F$ such that the composition $\phi\circ\phi(-2):F(-4) \to F(-2) \to F$ coincides with the morphism given by $D$. The structure sheaf $O\_C$ corresponds to $A$. Hence the extension of $O\_C$ with $O\_C$ corresponds to an extension of $A$ with $A$, that is we have an exact sequence
$$
0 \to O \oplus O(-2) \to F \to O \oplus O(-2) \to 0.
$$
Since $Ext^1\_A(A,A) = Ext^1\_O(O,A) = H^1(A) = H^1(O \oplus O(-2))$, we see that the nontrivial extension corresponds to the extension of $O$ in the right term by $O(-2)$ in the left term. Thus $F = O \oplus O(-1)^2 \oplus O(-2)$ and the above sequence can be rewritten as
$$
0 \to O \oplus O(-2) \to O \oplus O(-1) \oplus O(-1) \oplus O(-2) \to O \oplus O(-2) \to 0
$$
The maps are given by matrices $\left[\begin{smallmatrix}1 & 0\cr 0 & x\cr 0 & y\cr 0 & 0 \end{smallmatrix}\right]$ and $\left[\begin{smallmatrix} 0 & y & -x & 0 \cr 0 & 0 & 0 & 1 \end{smallmatrix}\right]$. The map $\phi:F(-2) \to F$ providing $F$ with a structure of an $A$-module is given by the matrix $\left[\begin{smallmatrix} 0 & g & h & -f \cr x & xy & -x^2 & h \cr y & y^2 & -xy & -g \cr 0 & y & -x & 0 \end{smallmatrix}\right]$. Here $f$ is the polynomial in $x$ and $y$ of degree $4$ such that $div(f) = D$, and $g,h$ are polynomials such that $f = gx +hy$.
| 32 | https://mathoverflow.net/users/4428 | 44279 | 28,118 |
https://mathoverflow.net/questions/44265 | 4 | I would like to check a statement about Schauder bases in $C([0,1])$ to be sure that I don't lie to my students on Monday. The statement(s) that I would like to check are:
1. The family of monomials $\{1,t,t^2,t^3,\dots\}$ is a **topological basis** but not a **Schauder basis** in $C([0,1])$ because there's not a unique choice of coefficients converging to a given continuous function. (An example I thought of: there's a polynomial approximation of $|t|$ on $[-1,1]$ using only even polynomials, but then that's a polynomial approximation of $t$ on $[0,1]$ with zero $t$ coefficient.)
2. Ditto for trigonometric polynomials.
The reason I ask is because when searching for this on the internet, I came across a statement claiming that trigonometric polynomials weren't a Schauder basis because in general (there's that phrase again!) Fourier series don't converge uniformly for continuous functions. That seems to me like a load of dingo's kidneys (not the convergence statement, but the deduction from it) but - and here's the clincher - the statement was made by someone whose answers on MO I've found to be generally reliable. (To be clear, the statement wasn't made on MO and was somewhere fairly obscure and I'm not going to "name and shame" because I don't want to embarrass that person - if I'm right - or myself - if I'm wrong.)
| https://mathoverflow.net/users/45 | Question about Schauder bases in C([0,1]). | I would say, monomials are not a Schauder basis for $C[0,1]$ because functions that admits a representation are analytic functions on the unit disc. Actually, $f(t)=\sum\_k a\_k t^k$ immediately implies differentiability at $t=0,$ which is enough to conclude. So, it's more a matter of non-existence than non-unicity.
With trigonometric polynomials it is a bit more delicate. But the reason is again non-existence of the uniform convergent expansion for some continuous function.
Of course, if $f$ admits a representation as uniform limit of a series $\sum\_k c\_k e^{ikt}$ then the series is its Fourier series, so the claim that $\{ e^{ikt} \}\_{k\in\mathbb{Z} }$ is not a Schauder basis for the continuous $2\pi$-periodic functions is equivalent to the statement that a Fourier series of a continuous $2\pi$-periodic function may fail to converge uniformly. (I do not have an example handy; one can also show it indirectly as a consequence of Uniform Boundedness principle if I remember well).
[edit] For instance, Wheeden & Zygmund's book *Measure and Integral* (p 227) has a nice example of a continuous function $f$ whose Fourier series is unbounded at $0$. The idea is defining $f$ as a uniform limit of a normally convergent series $\sum\_{j=1}^{\infty} Q\_j(t)$ which is formally made out of an unbounded trigonometric series after a rearrangement and a parenthesization. In other words, each $Q\_j$ is a linear combination of $e^{ikt}$ with $k\in I\_j,$ and the $I\_j$'s are pairwise disjoint finite subsets of $\mathbb{Z}$). So computing the Fourier series of $f,$ one finds again the bad series.
| 11 | https://mathoverflow.net/users/6101 | 44280 | 28,119 |
https://mathoverflow.net/questions/44278 | 17 | The Artin braid groups $B\_n$ and the symmetric groups $S\_n$ are closely related by the maps $1 \to P\_n \to B\_n \to S\_n \to 1$. The infinite symmetric group has interesting interactions with homotopy theory, due to a result of Barratt-Priddy(-Segal(-Quillen(-others))) that "identifies" the sphere spectrum $QS^0$ with $S\_{\infty}$.
>
> In light of the short exact sequence above, are there any Barrat-Priddy-*esque* results on the infinite braid group $B\_{\infty}?$ Is there even a loop space structure on $B\_{\infty}?$
>
>
> Ditto for the pure braids $P\_{\infty}$, the kernal of $B\_{\infty} \to S\_{\infty}$.
>
>
>
| https://mathoverflow.net/users/7867 | Loop spaces and infinite braids | Yes, $B \beta\_\infty$ is homology equivalent to $\Omega^2\_0 S^2$, the zero component of the double loop space of $S^2$. The map $B\_\infty \to S\_\infty$ induces the obvious stablisation map $\Omega^2\_0 S^2 \to Q\_0S^0$.
| 18 | https://mathoverflow.net/users/318 | 44281 | 28,120 |
https://mathoverflow.net/questions/44244 | 153 | E.T. Bell called Fermat the Prince of Amateurs. One hundred years ago Ramanujan amazed the mathematical world. In between were many important amateurs and mathematicians off the beaten path, but what about the last one hundred years? Is it still possible for an amateur to make a significant contribution to mathematics? Can anyone cite examples of important works done by amateur mathematicians in the last one hundred years?
For a definition of amateur:
>
> I think that to make the term "amateur" meaningful, it should mean someone who has had no formal instruction in mathematics past undergraduate school and does not maintain any sort of professional connection with mathematicians in the research world. – Harry Gindi
>
>
>
| https://mathoverflow.net/users/nan | What recent discoveries have amateur mathematicians made? | About ten years ago Ahcène Lamari and Nicholas Buchdahl independently proved that all compact complex surfaces with even first Betti number are Kahler. This was known since 1983, but earlier proofs made use of the classification of surfaces to reduce to hard case-by-case verification.
At the time, Lamari was a teacher at a high school in Paris. Apparently he announced his result by crashing a conference in Paris and going up to Siu (who had proved the last case in the earlier proof in 1983) with a copy of his proof. Lamari's proof was published in the Annales de l'Institut Fourier in 1999 (*Courants kählériens et surfaces compactes*, Annales de l'institut Fourier, **49** no. 1 (1999), p. 263-285, doi:[10.5802/aif.1673](https://doi.org/10.5802/aif.1673)), next to Buchdahl's (*On compact Kähler surfaces*, Annales de l'institut Fourier, **49** no. 1 (1999), p. 287-302, doi: [10.5802/aif.1674](https://doi.org/10.5802/aif.1674))
| 125 | https://mathoverflow.net/users/4054 | 44283 | 28,122 |
https://mathoverflow.net/questions/44275 | 2 | Let $V$ be a real algebraic variety and let ${\cal O}(V)$ denote its algebra of regular functions. If we put a group structure on $V$ (not necessarily an algebraic group structure) it will induce a Hopf algebra structure on ${\cal O}(V)$ in the usual manner. My question is, is there a bijective correspondence between the possible group structures on $V$ and the possible Hopf algebra structures on ${\cal O}(V)$?
| https://mathoverflow.net/users/1867 | Hopf algebra and group structure correspondence for algebraic varieties | If $V$ is an affine algebraic variety over any field $k$, then there is a bijection between the algebraic group structures on $V$ and the Hopf algebra structures on $O(V)$. The reason is that the category of affine varieties over $k$ is the just the contravariant category (the category with arrows reversed) of algebras over $k$. So a group law $m:V \times V \to V$ becomes a coproduct $m:O(V) \to O(V) \otimes O(V)$, and this correspondence is a bijection. (Because also, the tensor product of algebras corresponds to the Cartesian product of varieties.)
If you are interested in group structures on $V$ that are not necessarily algebraic group structures, then there isn't necessarily a good relation to $O(V)$ either.
| 7 | https://mathoverflow.net/users/1450 | 44284 | 28,123 |
https://mathoverflow.net/questions/44289 | 4 | This is exercise 20.5 out of Jech:
>
> Let $\lambda \geq \kappa$ and let $U$ be a normal measure on $P\_{\kappa}(\lambda)$. The ultraproduct $\mathrm{Ult} \_U \{ (V \_{\lambda \_x},\in) : x \in P \_{\kappa}(\lambda) \}$ is isomorphic to $(V \_{\lambda}, \in)$
>
>
>
Here $\lambda \_x$ simply denotes the order type of $x$. The function $x \mapsto \lambda \_x$ represents $\lambda$ in the ultrapower of $V$ by $U$. Unless I'm mistaken, the ultraproduct mentioned will be $V^M \_{\lambda} = V \_{\lambda} \cap M$ where $M$ is the ultrapower of $V$ by $U$. I don't see why this would be $V \_{\lambda}$ itself, since I don't see why $V \_{\lambda} \subset M$. Clearly $H \_{\lambda ^+} \subset M$ since $M$ is closed under $\lambda$ sequences, but I sort of doubt that $V \_{\lambda} \subset M$ -- I figure if $\lambda$-supercompactness implied $\lambda$-strongness, I would've seen that mentioned somewhere.
So did I make a mistake in computing the ultraproduct, or is there a mistake in the exercise, or does $\lambda$-supercompactness imply $\lambda$-strongness for some reason I'm not seeing?
| https://mathoverflow.net/users/7521 | Strong Cardinals and Supercompact Cardinals | In general, $\lambda$-supercompactness, if consistent, does
not imply $\lambda$-strongness. One can see this by
observing that the smallest cardinal $\kappa$ that is
$\kappa^+$-supercompact is never $\kappa^+$-strong, and in
fact, cannot be even $(\kappa+3)$-strong. The reason is
that $\kappa^+$-supercompactness is witnessed by a measure
on $P\_\kappa\kappa^+$, which amounts essentially to (is
coded by) a subset of $P(\kappa^+)$, and hence is witnessed inside $V\_{\kappa+3}$. Thus, if $j:V\to M$ were any embedding with
critical point $\kappa$, by minimality it follows that
$\kappa$ is not $\kappa^+$-supercompact in $M$, and hence
$M$ cannot have the true $V\_{\kappa+3}$. Thus, $\kappa$ is
not $(\kappa+3)$-strong and thus definitely not $\kappa^+$-strong.
I haven't looked at the context of the exercise, but perhaps he is merely asking you to make the observation that you did in fact make, that the ultrapower will give you $V\_\lambda^M$?
For some kinds of $\lambda$, it does follow that $\lambda$-supercompactness implies $\lambda$-strongness. For example, if $\lambda$ is a beth-fixed point, then every $\lambda$-supercompactness embedding is also $\lambda$-strong and even $(\lambda+1)$-strong, since in this case $|V\_\lambda|=\lambda$. But if $\lambda$ is not a beth-fixed point, then a version of my argument above will still apply: if $\lambda$ is not a beth-fixed point and $j:V\to M$ is a Mitchell minimal $\lambda$-supercompactness embedding for $\kappa$, then $\kappa$ is not $\lambda$-supercompact in $M$, and this is witnessed inside $V\_\lambda$ by the assumption on $\lambda$, and so $j$ cannot be a $\lambda$-strongness embedding.
| 8 | https://mathoverflow.net/users/1946 | 44291 | 28,127 |
https://mathoverflow.net/questions/44234 | 29 | Hillel Furstenberg conjectured that the only $2$-$3$-invariant probability measure on the circle without atoms is the Lebesgue measure. More precisely:
>
> **Question:** (Furstenberg) Let $\mu$ be a continuous probability measure on the circle such that
> $$\int\_{S^1} f(z) d\mu = \int\_{S^1} f(z^2) d\mu = \int\_{S^1} f(z^3) d\mu, \quad \forall f \in C(S^1).$$ Is $\mu$ the Lebesgue measure?
>
>
>
The assumption implies that the Fourier coefficients $\hat\mu(n)$ satisfy $$\hat\mu(n) = \hat \mu(2^k3^ln), \quad \forall k,l \in \mathbb N, n \in \mathbb Z.$$ Furstenberg's question is known to have an affirmative answer if one makes additional assumptions on the entropy of the measure.
The basic strategy is usually to show that a non-vanishing (non-trivial) Fourier coefficient implies the existence of an atom. The standard tool to construct atoms in a measure on the circle is Wiener's Lemma, which says that as soon as there exists $\delta>0$ such that the set
$\lbrace n \in \mathbb Z \mid |\hat\mu(n)| \geq \delta \rbrace$ has positive density in $\mathbb Z$, $\mu$ has an atom. More precisely, the following identity holds:
$$\sum\_{x \in S^1} \mu(\lbrace x \rbrace)^2 = \lim\_{n \to \infty} \frac1{2n+1} \sum\_{k=-n}^n |\hat \mu(k)|^2.$$
Clearly, $$|\lbrace 2^k3^l \mid k,l \in \mathbb N \rbrace \cap [-n,n] | \sim (\log n)^2 $$ so that Wiener's Lemma can not be applied directly. My question is basically, whether this problem can be overcome for any other subsemigroup of $\mathbb N$.
>
> **Question:** Is there any subsemigroup $S \subset \mathbb N$ of zero density known, such that every $S$-invariant continuous probability measure on $S^1$ is the Lebesgue measure? What about the subsemigroup generated by $2,3$ and $5$?
>
>
>
| https://mathoverflow.net/users/8176 | Furstenberg's Conjecture on 2-3-invariant continuous probability measures on the circle | Manfred Einsiedler and Alexander Fish have a paper (arxiv.org/abs/0804.3586) showing that a multiplicative subsemigroup of $\mathbb N$ which is not too sparse satisfies the desired measure classification. The semigroups they consider are still somewhat large, and in particular not contained in finitely generated semigroups.
One can check that the set of perfect squares $E$ satisfies measure classification as follows: since $\frac{1}{N}\sum\_{n=1}^N \exp(2\pi i n^2 \alpha)\to 0$ for every irrational $\alpha,$ one can conclude as in Wiener's lemma that an atomless measure $\mu$ on $S^1$ satisfies $\lim\_{N\to \infty} \frac{1}{N}\sum\_{n=1}^N|\widehat{\mu}(n^2)|=0.$ As in the original post, it follows that an atomless $E$-invariant measure is Lebesgue.
| 21 | https://mathoverflow.net/users/10457 | 44297 | 28,131 |
https://mathoverflow.net/questions/44303 | 23 | I am trying to define an embedding whose range includes classes. Is there a coherent way of assigning "cardinality" to proper classes?
| https://mathoverflow.net/users/nan | Cardinality of classes | Erin, there is no need to do this. I do not know of any practical reasons for doing it. And, of course, "cardinality" has to be properly interpreted to make some sense of the word.
In extensions of set theory where classes are allowed (not just formally as in ZFC, but as actual objects as in MK or GB), sometimes it is suggested to add an axiom (due to Von Neumann, I believe) stating that any two classes are in bijection with one another. Under this axiom, the "cardinality" of a proper class would be ORD, the class of all ordinals. (By the way, by class forcing, given any proper class, one can add a bijection between the class and ORD without adding sets, so this assumption bears no implications for set theory proper.)
Without assuming Von Neumann's axiom, or the axiom of choice, I know of no sensible way of making sense of this notion, as now we could have some proper classes that are "thinner" than others, or even incomparable. Of course, we could study models where this happens (for example, work in ZF, assume there is a strong inaccessible $\kappa$, and consider $V\_\kappa$ as the universe of sets, and $Def(V\_\kappa)$ in Gödel's sense (or even $V\_{\kappa+1}$) as the collection of classes).
| 21 | https://mathoverflow.net/users/6085 | 44305 | 28,137 |
https://mathoverflow.net/questions/44299 | 1 | Let's $C \subset X$ be a smooth curve inside a three dimensional variety with split normal bundle $N\_C^X= \nu\_1 \oplus \nu\_2$. What is a locally free resolution of $\iota\_{\*}\mathcal{O}\_{C}$ ?
| https://mathoverflow.net/users/5259 | How do we write a locally free resolution for... | In general, there is no straight way to write such a resolution (note by the way, that a resolution is in no way unique!). However, in some cases there is a distinguished resolution. For example, if $C$ is the zero locus of a global section of a rank 2 vector bundle $E$ then there is a Koszul resolution
$$
0 \to \det E^\* \to E^\* \to O\_X \to \iota\_\*O\_C \to 0.
$$
Another case, is when $C$ is the degeneration locus of a morphism $\phi:E \to F$ of vector bundles such that $rank(E) = rank(F) - 1$. Then there is a resolution
$$
0 \to E\otimes L \to F\otimes L \to O\_X \to \iota\_\*O\_C \to 0,
$$
where $L =\det E\otimes (\det F)^{-1}$.
| 7 | https://mathoverflow.net/users/4428 | 44306 | 28,138 |
https://mathoverflow.net/questions/44309 | 7 | Problem
-------
I'm looking for an upper bound for the number $k(G)$ of a finite group $G$, defined as follow:
>
> Let $\mathcal{F}\_k$ be the family of subsets of $G$ with size $k$, and we
> define $k(G)$ be the minimum $k$ such that every subset $X \in \mathcal F\_k$
> contains a non-empty sum-full set $S$, which is a set satisfies
> $$ S \subseteq S+S := \{ x+y \mid x,y \in S \}. $$
>
>
>
Note that the inequality $k(G) \leq |G|$ holds trivially since there is only one
subset in $\mathcal F\_{|G|}$ which is $G$ itself, and $G$ is a semigroup indeed.
Are there any papers or references about this number $k(G)$? Does it have a name? I'm interesting in particularly upper bounds of $k(G)$, but any related results are fine.
---
Motivation
----------
The **restricted Davenport number** $\hat{D}(G)$ of a group $G$, is defined as the smallest number $d$ such that given a subset $A \in \mathcal F\_d$, there exists a **zero-sum** non-empty subset $S \subseteq A$, that is,
$$ \sum\_{x \in S} x = 0, $$
where $0$ is the identity in $G$.
In the paper "On a conjecture of Erdos and Heilbronn", Szemeredi has proved:
$$\hat{D}(G) = O(\sqrt{|G|}). $$
Hamidoune and Zemor set a precise bound $\sqrt{2}$ on the constant of the big-O notation.
I'm trying to provide a link between $\hat{D}(G)$ and the number $k(G)$; it seems to me that the size of sum-full sets in $G$ may related to the zero-sum problem. I'll provide the justification in another post, which is highly related.
| https://mathoverflow.net/users/4248 | Upper bound for size of subsets of a finite group that contains a sum-full set | First of all the $k(G)$ cannot be smaller than the size of any proper subgroup of $G$, because if $H$ is a proper subgroup, $gH$ is a coset, $g\not\in H$, then $gHgH$ does not intersect $gH$ (if $ghgh'=gh''$, then $g\in H$). For example, if $G$ is Abelian, $|G|$ is not prime (i.e. $G$ is not cyclic of prime order), then $k(G)>\sqrt{|G|}$: look at the size of a maximal proper subgroup.
| 4 | https://mathoverflow.net/users/nan | 44325 | 28,149 |
https://mathoverflow.net/questions/44332 | 1 | is there an English translation of the book by Guy Barles, "Solutions de viscosite des equations de Hamiltion-Jacobi"? Springer-Verlag
| https://mathoverflow.net/users/5896 | is there an English translation of the book by Guy Barles? | It's extremely unlikely that this 1994 Springer book has been formally translated into English, since neither Springer's site nor MathSciNet lists a translation. In fact it's uncommon for most French mathematics books to be translated "officially" into English, a subject which has come up in many other posts on MO.
By various historical accidents, the most common languages used internationally in mathematics are English, French, German, Russian (with English becoming most standard as has happened in areas like trade and aviation). Naturally there are some research papers and even books in those languages which get translated unofficially into other languages and circulate privately. But French is usually considered manageable by advanced students who already know English. Translating advanced books into any language is seldom an attractive commercial venture, even for big publishers like Springer.
| 2 | https://mathoverflow.net/users/4231 | 44336 | 28,157 |
https://mathoverflow.net/questions/44323 | 3 | I'm reading the book about moduli spaces by Huybrechts and Lehn, and i'm stuck understanding a proof, it is Theorem 6.1.8.:
Given a K3-surface $X$ and a 2-dimensional space $M$, coherent and torsion free sheafs $F$ on $X$ and $E$ on $M\times X$. We have projections $p,q$ from $M\times X$ to $M$ and $X$ resp.
They claim the class $a:=ch(p\_{\\*}\mathcal{H}om(q^{\\*}F,E))$ as an element in $H^{\\*}(M,\mathbb{Q})$, where $p\_{\\*}\mathcal{H}om(q^{\\*}F,E)=\sum\limits\_{i=0}^2 (-1)^i \mathcal{E}xt^i\_p(q^{\\*}F,E)$, only depends on the classes of $ch(q^{\\*}F)$ and $ch(E)$ as elements in $H^{\\*}(M\times X,\mathbb{Q})$, where $\mathcal{E}xt\_p^i(q^{\\*}F,E)=R^i(p\_{\\*}\mathcal{H}om(q^{\\*}F,E))$.
So using Grothendieck-Riemann-Roch as suggested shows:
$ch(\sum\limits\_{i=0}^2 (-1)^i \mathcal{E}xt^i\_p(q^{\\*}F,E))td(M)=p\_{\\*}(ch(\mathcal{H}om(q^{\\*}F,E)td(M\times X))$
Here i am stuck. Why does this show that $a$ only depends on $ch(E)$ and $ch(q^{\\*}F)$. I think one has to show that $ch(\mathcal{H}om(q^{\\*}F,E))$ only depends on this classes, but i can't see why.
| https://mathoverflow.net/users/3233 | Chern character of Hom-sheaves | If you apply ${\mathcal H}om(-,E)$ to a resolution of a sheaf $G$, you obtain a complex, the cohomology of which are ${\mathcal E}xt^i(G,E)$, hence by additivity of the Chern character, the alternating sum of Chern characters of the terms of the complex equals the alternating sum of Chern characters of the Ext sheaves.
| 2 | https://mathoverflow.net/users/4428 | 44338 | 28,158 |
https://mathoverflow.net/questions/44358 | 9 | Is there a natural reason for defining the compact-open topology on the set of continuous functions between two locally compact spaces. For example "to make ... functions continuous". Or in another way of asking this, is there an adjoint functor of the functor, say F, which assigns the topological space $F(X,Y):=Hom\_C(X,Y)$ (with the compact-open topology on it) to the couple X,Y.
| https://mathoverflow.net/users/10469 | compact-open topology | In regard to your question I recommend *[Topologies on spaces of continuous functions,](http://www.cs.bham.ac.uk/~mhe/papers/newyork.pdf) Topology Proceedings, volume 26, number 2, pp. 545-564, 2001-2002* by Martin Escardo and Reinhold Heckmann.
| 9 | https://mathoverflow.net/users/1176 | 44365 | 28,179 |
https://mathoverflow.net/questions/44249 | 2 | I essentially understand (I think) how this ought to be done. Algebras in a monoidal 2-category $\mathcal{C}$, on the level of 0-cells and 1-cells, should appear as algebras in the 1-category truncation of $\mathcal{C}$. To lift these 1-level algebras we must of course weaken the usual diagrams and then describe (a zoology of possibly 3-dimensional) diagrams required of these weakening factors. How does one determine which diagrams are the "right" ones?
| https://mathoverflow.net/users/8157 | How to explicitly describe algebras in a monoidal 2-category? | In a category $\mathscr{C}$ by finite limits you can have inside it all algebraic classical structures, for example about “Monoids” there is a category called “Monoid theory” $Mnd$ (see pioneristic works of Lawvrere) and the category of $\mathscr{C}$- monoids is the category of finite limits preserving $F: Mnd\to \mathscr{C}$, the some for groups, exist the "Groups theory" category “Gr” ecc. (for all one-sort theories the support of any algebric thery category is maked from the dual of $\mathbb{N}$ where any $n\in N$ is the sum of $n$-copies of 1, adding other more morphisms).
How generalizing this to a monoidal category $(\mathscr{C}, \otimes , I)$ ?, simply considering for axample “Mnd” with the canonical monidal structure gived by finite products $\times $ and final object $1$ and considering monoidal functors (strict or no) instead. There is a truble, for the Monoidal or Algebras theory the things go well. This generalization as a big obstrucion, but for example for groups there are the first troubles: for the “inverse propriety" axiom we need to use the finality of $1$ but the monoidal identity $I$ isnt a final object (neither exixt a similar monoidal surrogate).
I know that Luca Mauri wrote a Thesi about “Algeraic Theory “ in a Monoidal category.
But you ask abot the supplemetary diagrams about coherence of these laxifications of “Monoidal, algebra ecc. “ objects.
A pseudo/lax-monoid (or pseudo/lax-algebras ecc.) in a monoidal category $\mathscr{C}$ (see it as a bicategory, or more genrally as a tricategory i.e. a braided category) is just a pseudo/lax-functors from $Mod$ to $\mathscr{C}$, and you can find (John W. Gray, Formal category theory: adjointness for 2-categories. Gordon, Power, Street. Coherence for tricategories) in supplement to the structural definition of what a pseudo/lax-funtor is (maked it as maps) also the axioms about the coherence thet these pseudo/lax-funtor (these maps) must satisfing, and these are the coherence axioms you need.
| 0 | https://mathoverflow.net/users/6262 | 44367 | 28,181 |
https://mathoverflow.net/questions/43326 | 1 | Are there known sharp upper bounds (in terms of $k$ or $\omega(k)$, the number of distinct prime divisors of $k$) for sums of the form $\sum\_{p \mid k} \frac{1}{p+1}$ for $k > 1$ subject to the constraint $\sum\_{p \mid k} \frac{1}{p+1} < 1$? (This would be a special case of the general result of O. Izhboldin and L. Kurliandchik referenced in the comment.)
A related question: Suppose $(p\_i)$ is a set of $n$ consecutive primes which minimizes $1 - \sum\_{i = 1}^n \frac{1}{p\_i+1} > 0$ for a given $n > 1$. Are there known bounds for $1 - \sum\_{i = 1}^{n} \frac{1}{p\_i + 1}$ from below in terms of $n$, e.g., $n^{-\delta n}$ for some fixed $\delta > 0$?
Thanks!
| https://mathoverflow.net/users/10280 | Sharp upper bounds for sums of the form $\sum_{p \mid k} \frac{1}{p+1}$ | If we are allowed to consider somewhat weaker bounds, both answers depend only on the number of unitary divisors of $k$, which is $2^{\omega(k)}$. By the quoted result of O. Izhboldin and L. Kurliandchik (see Fedor's and Myerson's comments), for any set of $n$ positive integers {$a\_{1}, \dots, a\_{n}$} such that $\sum\_{i = 1}^{n} \frac{1}{a\_{i}} <1$, we have
\begin{eqnarray}
\sum\_{i = 1}^{n} \frac{1}{a\_{i}} \leq \sum\_{i = 1}^{n} \frac{1}{d\_{i}} = \frac{d\_{n+1} - 2}{d\_{n+1} -1} < 1,
\end{eqnarray}
where $d\_{i}$ is the $i^{\text{th}}$-Euler number, which satisfies the quadratic recurrence $d\_{i} = d\_{1} \cdots d\_{i-1} + 1$ with $d\_{1} = 2$. The first few terms of the sequence are 2, 3, 7, 43, 1807.... (A000058). It is relatively straightforward to show that the aforementioned recurrence is equivalent to $d\_{i} = d\_{i-1}(d\_{i-1}-1) + 1$, and it is known from the recurrence that $d\_{i} = \lfloor \theta^{2^{i}} + \frac{1}{2} \rfloor$, where $\theta \approx 1.2640$.... Thus,
\begin{eqnarray}
\sum\_{p \mid k} \frac{1}{p + 1} \leq \sum\_{i = 1}^{\omega(k)} \frac{1}{d\_{i}} = \frac{d\_{\omega(k) + 1} - 2}{d\_{\omega(k)+1} - 1} = \frac{\lfloor \theta^{2^{\omega(k) + 1}} + \frac{1}{2} \rfloor - 2}{\lfloor \theta^{2^{\omega(k) + 1}} + \frac{1}{2} \rfloor - 1}.
\end{eqnarray}
One can also show that $d\_{i} - 1 = \lfloor \vartheta^{2^{i-1}} - \tfrac{1}{2} \rfloor$, where where $\vartheta \approx 1.5979$...., so we have
\begin{eqnarray}
1 - \sum\_{p \mid k} \frac{1}{p + 1} \geq 1 - \frac{d\_{\omega(k)+1} - 2}{d\_{\omega(k)+1} - 1} = \frac{1}{d\_{\omega(k)+1} - 1} = \lfloor \vartheta^{2^{\omega(k)}} - \tfrac{1}{2} \rfloor^{-1} > 0.
\end{eqnarray}
**Remark** E. Deutsche points out that the sequence {$d\_{i} - 1$} (A007018) counts the number of ordered rooted trees with out-going degree up to 2 with all leaves at top level.
| 1 | https://mathoverflow.net/users/10280 | 44401 | 28,208 |
https://mathoverflow.net/questions/44397 | 11 | The standard examples of complete but not model-complete theories seem to be:
- Dense linear orders with endpoints.
- The full theory $\mathrm{Th}(\mathcal{M})$ of $\mathcal{M}$, where $\mathcal{M} = (\mathbb{N}, >)$ is the structure of natural numbers equipped with the relation $>$ (and nothing else, i.e. no addition etc).
Can anyone explain or give a reference to show why any of these two theories are not model-complete, or give another example altogether of a complete but not model complete theory (with explanation)?
| https://mathoverflow.net/users/362 | What is a good example of a complete but not model-complete theory, and why? | For the second example, let $M$ be the natural numbers and let $N$ be the
integers greater than or equal to -1. Then $M$ is a substructure of $N$
but $M\models$ ``0 is the least element", while this is false in $N$. Thus the
theory is not model complete.
The first example is similar, let $M$ be $[0,1]$ and let $N$ be $[-1,1]$. Again $M\models$ ``0 is the least element" but the extension $N$ does not.
One equivalent of model completeness is that every formula is equivalent to an existential formula. So theories like true arithmetic, the theory of the natural numbers in the language
{$+,\cdot,0,1$}, are far from model complete.
| 16 | https://mathoverflow.net/users/5849 | 44405 | 28,212 |
https://mathoverflow.net/questions/27990 | 8 | My question is related to this one: [Computing the Galois group of a polynomial](https://mathoverflow.net/questions/22923/computing-the-galois-group-of-a-polynomial).
I was wondering if there is a faster algorithm just to compute the order of the group rather than the group itself.
Also, has anybody compared the performance of GAP and Magma in computing Galois groups? I just heard Magma is very good at it.
I asked this question because I encounter every so often new bug with Magma's implementation and I wanted to see if I can implement something similar. But at this time I'm just interested in the exponent at the first place. This is the last annoying error that I get for basically any deg 5 poly that has Gal group $S\_5$.
```
k := FiniteField(2);
kx<x> := RationalFunctionField(k);
kxbyb<y> := PolynomialRing(kx);
MinP := y^5 + y + x^2 + x;
print GaloisGroup(MinP);
```
The result is:
```
Runtime error: too much looping
```
Which I don't understand what it means (Magma Ver 2.16-8).
To be more clear, my ultimate goal is to check a lot of polynomials and throw out those with $S\_n$ Gal group and focus on those which are not such. As you see even an upper bound over the exponent is enough for me.
| https://mathoverflow.net/users/6776 | Computing only the order of Galois group (not the group itself). | I am actually one of the authors of the Galois package in Magma. Firstly, the "too much looping" error does not happen anymore (for this example at least) in the current Magma version (2.16-13). Secondly, the way Sn/An is recognized in general is through the use of
factorisation as suggested. More precisely, the polynomial is factored modulo several primes and the resulting factors (well their degrees) are noted. Those give possible cycle
types of the Galois group. If cycle types of certain patterns happen, we know the group
is An/Sn. Those types are very frequent, hence this is trivial. However, then we hit a problem. In order to distinguish An and Sn usually one looks at the discriminant of the polynomial with the idea that the group is An (or in general contained in An) iff the discriminant is a square. This unfortunately breaks down in characteristic 2 and the currently employed test is slow. (And caused the "too much looping" message).
Unfortunately, we don't have an interface like IsAnOrSn(f) which would be sufficient here.
In general, looking at cycle types or even at types and their frequency, will not determine the group nor the group size. All one gets from here are a lower bounds. However
for small degrees (and 5 is small) this would work.
The connection between Kash and Magma here is difficult: Magma used to rely on Kash for the Galois groups, but the algorithm was limited to degree <= 23. This is the PhD of
Katharina Geissler, her thesis can be found on the Kash page in Berln. The current Magma implementation of Galois groups is independent and does not share any code with Kash.
| 9 | https://mathoverflow.net/users/10478 | 44414 | 28,217 |
https://mathoverflow.net/questions/44428 | 5 | In Bourbaki's Commutative Algebra we have the following theorem:
II.5.2 Let $A$ ba a ring and $P$ an $A$-module. TFAE:
(i) $P$ is a f.g. projective module\
(ii) $P$ is a finitely presented module and, for every maximal ideal $m$ of $A$, $P\_m$ is a free $A\_m$-module.\
(iii) $P$ is a f.g. module, for all $p \in$ Spec($A$), the $\_p$-module $P\_p$ is free and, if we denote its rank by $r\_p$, the function $p \mapsto r\_p$ is locally constant in the topological space Spec($A$).
There is more to this theorem but I have stated what I need for my question. Namely, unless I am missing something, for (iii) implies (i) don't we need the extra condition that $A$ is a reduced ring? I see their proof and cannot find the problem but Eisenbud suggests that there exists a counterexample in problem 20.13 in Commutative Algebra with a View Toward Algebraic Geometry. I must be missing something!
| https://mathoverflow.net/users/10483 | On Bourbaki's characterization of projectives... | Dear fishibones, an important point is that at the beginning of Chapter II, Bourbaki states that all the rings he will consider are commutative. This implies that if a free module has finite dimension, this dimension is unambiguously defined. This property is called the Invariant Basis Number property (IBN) and may fail for non-commutative rings (Counterexamples can be found in, say, Lam's Lectures on Modules and Rings, Springer GTM 189.) This unambiguously defined dimension of a finitely generated free module is what Bourbaki calls its *rank*.
Let me emphasize that Bourbaki only uses "rank" in the above sense i.e. for finitely generated free modules. If $M$ is such a module of rank $r$ over the ring $A$, then for any prime ideal $\frak {p}$$ \in A$ the modules
$M\_{\frak p}$ over $A\_{\frak {p}}$ and $M\_{\frak {p}} \otimes\_{A\_{\frak p}} \kappa (p)$ over $\kappa (p)$
also are free of rank $r$. With this definition and use of rank, Bourbaki's Théorème 1 that you mention is perfectly correct (bien sûr!)
There is no contradiction with Eisenbud's exercise: he does *not* assume that his module $M$ has all its localizations $M\_{\frak {p}}$ free over $R\_{\frak {p}}$, whereas Bourbaki does.
Eisenbud only assumes that the dimension of the $\kappa ({\frak p})$- vector space $M\_{\frak {p}} \otimes\_{R\_{\frak p}} \kappa (p)$ is locally constant, while Bourbaki does not even mention this dimension when $M\_{\frak {p}}$ is not $R\_{\frak {p}}$-free.
| 4 | https://mathoverflow.net/users/450 | 44434 | 28,229 |
https://mathoverflow.net/questions/44202 | 3 | In the language of modules, it suffices to restrict our view to *positive-primitive* formulas - that is to say, formulas with one existential quantifier and no negation.
And I mean *existentially closed* in the sense that any witness to a particular positive-primitive formula over the module is already in the module itself.
My intuition is that I should look along the lines of injective modules. Can anyone help by offering direction or perhaps an example / verification?
| https://mathoverflow.net/users/9015 | Example or classification of existentially closed modules | Do primitive positive formulas only allow conjunctions, not disjunctions? If so, then injective modules are existentially closed, but are a strictly stronger concept in that the equivalent for infinite conjunctions are allowed.
The concept you're after is that of [algebraically compact](http://en.wikipedia.org/wiki/Algebraically_compact_module) module (also known as pure inective module). Here's a link to a paper by [Prest](http://eprints.ma.man.ac.uk/1148/01/covered/MIMS_ep2008_83.pdf) that talks about some their properties.
[Baer's criterion](http://en.wikipedia.org/wiki/Injective_module#Baer.27s_criterion) asserts existential closure under infinite conjunctions. Let
$$
(\exists m) r\_1 m = m\_1 \wedge r\_2 m = m\_2 \wedge \cdots
$$
be a particular formula, where $R$ is your coefficient ring, $M$ a module over $R$, with $r\_i \in R$ and $m\_i \in M$. Then consider the ideal $I$ generated by $r\_i$ in $R$. There the map that sends $r\_i$ to $m\_i$ is a homomorphism from $I$ to $M$ if and only if that formula is consistent. (There's a homomorphism if and only if the formula does not imply $s = s'$ for two distinct elements $s, s' \in R$.)
| 2 | https://mathoverflow.net/users/3711 | 44442 | 28,235 |
https://mathoverflow.net/questions/44443 | 7 | Is there any known result about the necessary and sufficient conditions for the existence of zeros for a function $f(x)=\sum\_{n=1}^{N} a\_n e^{b\_n x}$, where $a\_n,b\_n \in \mathbb{R}\, \forall n=1,2,\cdots,N$, $a\_1,a\_N >0$, $b\_1 < b\_2 < \cdots < b\_N $ and $x \in \mathbb{R}$?
It is known (see "Problem and Theorems in Analysis II" by Polya and Szego) that using a generalization of Descartes' rule of signs it possible to say that, named with $Z$ the number of changes of sign in the sequence of the $a\_n$ and with $Z\_0$ the number of zeros of $f(x)$, $Z-Z\_0 \geq 0$ is an even integer.
The number $Z-Z\_0$ should be even since for $x \rightarrow -\infty$, the dominant therm of $f(x)$ is $a\_1 e^{b\_1 x}>0$ and for $x \rightarrow +\infty$ the dominant term is $a\_N e^{b\_N x}>0$.
This gives an upper limit for the number of zeros, but there is any way to say "$f(x)$ *should have at least* $M$ *zeros*", with $0 < M \leq Z$?
Thanks in advance,
Nico
| https://mathoverflow.net/users/6162 | Zeros of a combination of exponentials | Note that we can assume wlog that $b\_n\geq 0.$ In the case they are rationals, writing $b\_n=p\_n/q$, with $p\_n\in\mathbb{N},\\ $ $q\in\mathbb{N}\_+,\\ $ and $t:=e^{x/q},\\ $ puts everything into the case of positive roots of a real polinomial, with not more, nor less generality. The book by Pólya and Szegő has a section on the location and number of positive roots of a polynomial; in any case, whatever you can say for it can clearly be translated for your exponential equation. Then, the case of real $b\_n$ can certainly be treated by approximation.
| 3 | https://mathoverflow.net/users/6101 | 44447 | 28,237 |
https://mathoverflow.net/questions/44408 | 13 | Does the support of a Borel probability measure always have [full measure](https://en.wikipedia.org/wiki/Almost_everywhere) in a metric space?
I know this is true for separable metric spaces, and locally compact metric spaces. Is it true in general?
| https://mathoverflow.net/users/10476 | Measure of the support of a Borel probability on a metric space | Following Pietro's lead, let me observe that if there is a
[measurable cardinal](http://en.wikipedia.org/wiki/Measurable_cardinal), then there is a counterexample.
Suppose that $\kappa$ is a measurable cardinal. Then there
is a $\kappa$-additive 2-valued measure $\mu$, measuring
all subsets of $\kappa$, giving them either measure $0$ or $1$, giving measure $1$ to the whole space and giving measure $0$ to any set
of size less than $\kappa$ (among others). If we give $\kappa$ the
discrete topology, then every set is closed (and hence
Borel), and the support is empty.
| 11 | https://mathoverflow.net/users/1946 | 44448 | 28,238 |
https://mathoverflow.net/questions/44452 | -2 | I know the following result is true in the case of strong convergence. But I don't know whether it is true in the case of weak convergence also.
Let $p>1$. Suppose that each $x\_n$ is a non negative sequence such that $\|x\_n\|\_p=1$ and $\stackrel{w}{x\_{n}\rightarrow x}$ in $\ell^p$. Is it true then that $\stackrel{w}{x\_n^p\rightarrow x^p}$ in $\ell^1$.
| https://mathoverflow.net/users/7699 | weak convergence | No. Consider the unit vector basis of $\ell\_2$.
| 5 | https://mathoverflow.net/users/2554 | 44456 | 28,242 |
https://mathoverflow.net/questions/44468 | 23 | In Andrew Gleason's interview for More Mathematical People, there is the following exchange concerning Gleason's work on Hilbert's fifth problem on whether every locally Euclidean topological group is a Lie group (page 92).
>
> **MP:** Is there some "human" story you can tell us about the breakthrough when it came?
>
>
> **Gleason:** Yes, there's a really remarkable story about that. Sometime -- I can't tell you the exact date but let's say around 1949 -- I was doing other things too, and one of the things that I found very interesting and very curious and which I really felt I should try to understand better was a very famous theorem to the effect that a monotonic function is almost everywhere differentiable. It's a rather remarkable and very difficult theorem -- it's not easy to prove. A very very hard theorem of analysis and a really surprising theorem. Well, at the time I was sort of speculating about this theorem, but it wasn't for at least two years that I suddenly realized that *that* would solve the problem I was dealing with! Knowing that, in connection with some other stuff I had been working on, really put the whole thing together. It was a realization that although this theorem had been on my mind for maybe two years, I had never recognized that it was crucial to the arguments that I was trying to work through in the Hilbert problem. I hadn't realized it. Then suddenly it just came to me.
>
>
> **MP:** It just came to you?
>
>
> **Gleason:** That's right. It just came to me that I could use this technique, this theorem, in connection with these curves in Hilbert space that I was dealing with -- and get the answer! ...
>
>
>
I've never studied Hilbert's Fifth Problem or its solutions, but I've always been curious what Gleason meant by this connection. Can anyone shed some light on this?
| https://mathoverflow.net/users/2926 | Monotone functions are differentiable a.e. and Hilbert's Fifth Problem: what's the connection? | Well, I cannot say for certain, but I did know Gleason well (he was my thesis advisor, and we wrote a paper together after that) and I have written an essay about Gleason's work on the Fifth Problem (in the Gleason Memorial article in the AMS Notices --- <http://www.ams.org/notices/200910/rtx091001236p.pdf> ) and based on that I think I can make a reasonable guess about what he had in mind. Recall that what Gleason actually proved was that a locally compact group without arbitrarily small subgroups is a Lie group (then Montomery and Zippin proved that a locally Euclidean group did not have small subgroups). A key idea in Gleason's proof was the construction of a unique one-parameter subgroup through any point sufficiently close to the identity (i.e., essentially, constructing the exponential map) and this in turn depended on showing the existence of a unique square root for elements near the identity (see his paper "Square roots in locally Euclidean groups"). I believe that it is the step going from square roots to one-parameter subgroups that used ideas from the monotonic implies differentiable a.e. theorem. The "these curves" that he mentions in that More Mathematical People article, can only be the one-parameter subgroups. For more details, see my Notices article above, particularly the section called "Following in Gleason’s Footsteps".
| 23 | https://mathoverflow.net/users/7311 | 44473 | 28,249 |
https://mathoverflow.net/questions/44438 | 2 | I believe to have found a typo in Griffiths & Harris.
In the chapter on surfaces, section Rational Surfaces 1, I am trying to read the result that a holomorphic vector bundle over $\mathbb{P}^1\_{\mathbb{C}}$ is a sum of invertible bundles.
What is the exact sequence that shows up at the start of his argument? Mine only has 2 terms and involves the fibres of E and H, which doesn't really make sense to me.
Any help would be appreciated.
| https://mathoverflow.net/users/8867 | Help with Griffiths & Harris, Surfaces | Let $H$ be the divisor corresponding to the point $x\in \mathbb{P}^1$. Tensoring the exact sequence
$$
0\to O\_{\mathbb{P}^1}(-H)\to O\_{\mathbb{P}^1}\to O\_x\to 0.
$$ with $E\otimes H^k$, gives
$$
0\to O\_{\mathbb{P}^1}(E\otimes H^{k-1})\to O\_{\mathbb{P}^1}(E\otimes H^{k})\to E\_x\otimes H\_x^{k}\to 0.
$$Here GH writes $E\_x\otimes H\_x^{k}$ for $E\otimes H^k\otimes O\_x$, or what is the same, the fiber of $E\otimes O(H^{k})$ over $x$.
| 5 | https://mathoverflow.net/users/3996 | 44479 | 28,254 |
https://mathoverflow.net/questions/44474 | 6 | (1) Let $M$ be a complex manifold of real dimension $2n$, and denote the line bundle of complex $(N,0)$-forms by $\Omega^{(N,0)}(M)$. When $M = CP^N$, the line bundles are indexed by the integers, and so, $\Omega^{(N,0)}(CP^N)$ must correspond to a integer. What is this integer? In the $N=1$ case, the corresponding integer is $2$. This suggests, that in general, $\Omega^{(N,0)}(CP^N)$ corresponds to $2N$. Is this true?
(2) For the anti-canonical spin$^c$ structure of $CP^N$, the spinor bundle is isomorphic to
$$
S := (\Omega^{(0,0)}(CP^N) + \cdots +
\Omega^{(0,N)}(CP^N)) \otimes S\_N,
$$
where $S\_k$ is the square root of $\Omega^{(N,0)}(CP^N)$ (square root wrt tensoring as multiplication). What does the square root mean when when line bundle integer is odd? In the $N=2$ case, this is seen to reduce to $\cal{E}\_{-1} \otimes \cal{E}\_1$, where $\cal{E}\_p$ is the line bundle corresponding to the integer $p$. Is this the $Z2$ grading on the spinor bundle? If so, what does this look like in higher dimensions?
(3) Finally, for a given spin connection $\nabla$, to define a Dirac operator we would need a Clifford representation, ie a map
$$
C:(\Omega^{(1,0)} \oplus \Omega^{(0,1)}) \otimes S \to S.
$$
For $N=2$, this should be given by a map
$$
C: (\Omega^{(1,0)} \oplus \Omega^{(0,1)}) \otimes (\cal{E}\_{-1} \oplus \cal{E}\_1) \to \cal{E}\_1 \oplus \cal{E}\_1.
$$
What is this rep? What does it look like for higher order $N$?
Note: the first subindex in the second $\cal{E} \oplus \cal{E}$ above should be $-1$, I'm having tex problems when I try to write it as such though.
| https://mathoverflow.net/users/1648 | Complex Projective Space Spin and Dirac: Part II | ad 1.): No, the answer is that $\Omega^{1,0} CP^N$ corresponds to $N+1$. Proof sketch:
As complex vector bundles, $TCP^N \oplus C= C^{N+1} \otimes L$, $L$ the tautological line bundle (the one with holomorphic sections). Thus $L^{\otimes (N+1)} = \Lambda^{N+1} (TCP^N \oplus C) = \Lambda^N TCP^N$. Dualizing gives the answer. For a more polished reasoning and more details, consult Griffith-Harris.
ad 2.) I am not sure whether your interpretation of $S\_k$ is correct; Friedrich writes (bottom of page 77) that $S\_0$ is the
determinant line bundle and $S\_N$ is trivial.
Here is how I would figure out the answer, switching perspective on $spin^c$ structures slightly (I do not have a handy reference for the
following, I followed hints I found in Weyl's "The classical groups").
Let $V \to X$ be an oriented vector bundle of rank $2n$. A spin$^c$ structure is a bundle of complex $Cl(V)$-modules that is
(fibrewise) irreducible. Given a line bundle $L$ on $X$, you can tensor a $CL(V)$-module with $L$ and obtain a new module. This gives an action of the group
of line bundles on the set of spin^c-structures.
Note that since $2n$ is even, there is a unique irreducible representation of $Cl(R^{2n})$. Moreover, $Cl\_{\mathbb{C}}(R^{2n})$ is
the matrix algebra $\mathbb{C}(2^n)$, so the irrep has to be $2^n$-dimensional. However, this module is determined only up to isomorphism, which is responsible
for the fact that not any vector bundle is spin^c.
Now assume that a complex structure has been given to $V$. Then
the exterior algebra $\Lambda\_{\bC} V$, together with the Clifford action defined by the formula $c(v)=v \wedge - \iota\_v$ is
a Clifford module. It has to be irreducible for dimension reasons!
This shows that any complex vector bundle is spin^c and gives an explicit desciption of the spin^c structures. Note that the complex
structure was used here essentially, to write the exterior algebra $\lambda\_{\bR} V$ (which is a Clifford module, but not irreducible) as
a tensor product
of $\Lambda\_{\mathbb{C}}(V)$ with its conjugate.
The result is pretty close to what you wrote and identical if you stick to Friedrichs notation.
Now the clifford algebra is graded
$Cl(V)\_{\mathbb{C}}=Cl\_{\mathbb{C}}(V)^0 \_{\mathbb{C}}(V)^1$ and if you trace back the definition of the action, you see that $Cl\_{\mathbb{C}}(V)^0$
preserves the even/odd-decomposition of the exterior algebra. This gives the grading of the spinor bundle, more or less as you suspected.
As I said before, spin^c structures can be twisted by line bundles (tensor products). The only line bundle available on a complex manifold is
the determinant bundle of the tangent bundle (and its powers). By choosing different powers, you can switch between the anticanonical/canonical spin^c structures.
| 3 | https://mathoverflow.net/users/9928 | 44484 | 28,259 |
https://mathoverflow.net/questions/44463 | 5 | A lax limit is [defined](http://ncatlab.org/nlab/show/2-limit) to be a 2-limit, except that the cone is only required to commute up to specified transformations, not up to isomorphism. In particular, the limit is defined up to isomorphism, and on examples such as product where there are no 2-cells in the cone, lax limits and 2-limits coincide.
I'm working with an example, whose properties are similar to smash product on partial orders with bottom. This gives rise to a lax version of a product, where the equalities $\langle f,g\rangle;p=f$ and $\langle f,g\rangle;q=g$ are replaced by 2-cells, and the uniqueness property becomes a universality property: for any other candidate $h$, we have a unique 2-cell $h\Rightarrow\langle f,g\rangle$.
Generalizing, it seems that there should be an "even laxer" notion of limit, where the adjunction used to define the limit is specified itself by an adjunction rather than by an equivalence of categories.
Does such a structure exist in the literature?
| https://mathoverflow.net/users/10491 | Lax universality for lax limits | Adjunctions 'up to adjointness' have been considered before. Marta Bunge (*Coherent extensions and relational algebras*, Trans. AMS 197, 1974) called them 'lax adjunctions', John Gray (*Formal category theory: Adjointness for 2-categories*, LNM 391, 1974) called them 'quasi-adjunctions' (of some sort) and Barry Jay (*Local adjunctions*, JPAA 53, 1988) called them 'local adjunctions'. The formulations differ somewhat, but what they have in common is a family of adjunctions
$$ D(F A, X) \leftrightarrows C(A, G X) $$
between hom categories. Another way to generalize, of course, is to turn the triangle equalities into coherent [modifications](http://ncatlab.org/nlab/show/modification) (3-cells), and these two should be Yonedily equivalent.
Often the functors F and G here are allowed to be lax, which complicates things because they can't in general be whiskered onto lax transformations. The unit and counit are often only lax too. There is a Yoneda lemma for lax transformations, written down [here](http://ncatlab.org/nlab/show/lax+natural+transformation) (by me, so any mistakes are my fault), but I for one haven't yet managed to massage it into a form that would be useful for these local/lax adjunctions.
| 2 | https://mathoverflow.net/users/4262 | 44487 | 28,261 |
https://mathoverflow.net/questions/43961 | 8 | Let $T$ be a finite symmetric set generating a Zariski dense subset of an algebraic group $G$ (specifically, $PSL\_2(\mathbb{C})$ or its subgroups). Is there an $\alpha>0$ such that the set $T^{\leq n}$ of words of length at most $n$ is not in any codimension-1 subvariety of degree $n^{\alpha}$?
"Escape from subvariety" arguments seem to prove similar results that are not polynomial.
| https://mathoverflow.net/users/408 | Degree of balls in finitely-generated subgroups of SL_2(C) | In the case of $\text{PSL}(2,\mathbb{C})$, the set of words of length $O(n^3)$ is not contained in a subvariety of degree $n$. A polynomial of degree $n$ is a linear combination of matrix entries of the irreps of highest weight $\le n$. Let $A\_n$ be the direct sum of the corresponding matrix algebras; its dimension is a sum of consecutive squares which is then $O(n^3)$. The generating set $S$ yields a set of operators in $A\_n$, which then yields an algebra filtration of $A\_n$. Since $S$ generates a Zariski dense subgroup, some term of this algebra filtration is eventually all of $A\_n$. On the other hand, the filtration is generated by the term of degree 1, i.e., one can write
$$A\_n^{(k+1)} = A\_n^{(k)}A\_n^{(1)},$$
taking all linear combinations of all products of pairs on the right side. So the dimensions of the terms of the filtration have to keep going up by at least one until the filtration terminates, which then gives you the $O(n^3)$ bound.
There is a similar argument for any linear algebraic group, except that the algebra $A\_n$ is different for each one.
This is a really interesting question. Although this argument does give you the bound that you wanted, I have the feeling that the bound is not optimal.
| 4 | https://mathoverflow.net/users/1450 | 44488 | 28,262 |
https://mathoverflow.net/questions/44207 | 5 | I am looking for the asymptotic growth of product of consecutive primes. Is there anything that is known about this growth?
| https://mathoverflow.net/users/10035 | Asymptotics of Product of consecutive primes | Denote by $$\Pi(x)=\prod\_{p\leqslant x}p,$$ thus
$$\log\Pi(x)=\sum\_{p\leqslant x}\log p:=\theta(x)\sim x,$$ which is known as the Prime Number Theorem. You may find further information in <http://en.wikipedia.org/wiki/Prime_number_theorem>
| 7 | https://mathoverflow.net/users/9944 | 44498 | 28,269 |
https://mathoverflow.net/questions/42629 | 27 | Fix a dimension $n\geqslant 2$. Let $S= \{M\_1,\ldots, M\_k\}$ be a finite set of smooth compact $n$-manifold with boundary. Let us say that a smooth closed $n$-manifold is *generated* by $S$ if it may be obtained by gluing some copies of elements in $S$ via some arbitrary diffeomorphisms of their boundaries.
For instance:
* Every closed orientable surface is generated by a set of two objects: a disc and a pair-of-pants $P$,
* Waldhausen's graph manifolds are the 3-manifolds generated by $D^2\times S^1$ and $P\times S^1$,
* The 3-manifolds having Heegaard genus $g$ are those generated by the handlebody of genus $g$ alone,
* The exotic $n$-spheres with $n\geqslant 5$ are the manifolds generated by $D^n$ alone.
A natural question is the following:
>
> Fix $n\geqslant 3$. Is there a finite set of compact smooth $n$-manifolds which generate all closed smooth $n$-manifolds?
>
>
>
I expect the answer to be ''no'', although I don't see an immediate proof. In particular, I expect some negative answers to both of these questions:
>
> Is there a finite set of compact 3-manifolds which generate all hyperbolic 3-manifolds?
>
>
>
and
>
> Is there a finite set of compact 4-manifolds which generate all simply connected 4-manifolds?
>
>
>
| https://mathoverflow.net/users/6205 | Can all n-manifolds be obtained by gluing finitely many blocks? | Thanks to Ian Agol for pointing out this question and a related one on levels of Morse functions - [/30567/](https://mathoverflow.net/questions/30567/). In both case, for smooth manifold of dim $> 3$, as expected, there is no finite list of blocks ( or regular level components.) The idea is that one may define the "width" of a group, by representing the group G as the fundamental group of some complex K and then slicing K into "levels". The game to to arrange the slices so that the image of $\pi\_1$ of each component of each level set maps a subgroup of small rank under the inclusion into $\pi\_1(K)$. width( G) is defined as a Minmax over all slicings of all complexes K with $\pi\_1 K = G$ of the rank of these image subgroups. I wrote a few pages to show that width( $\mathbb{Z}^k $) $= k-1$. The only slightly technical ingredient is Lusternick-Schnirelmann category. This answers negatively these finiteness questions since there are $d$ manifolds with $\pi\_1 =\mathbb{Z}^k$ all $k$, as long as $d>3$. As soon as the notes are teXed, I can post them on the arxiv or math overflow.
| 23 | https://mathoverflow.net/users/1643 | 44500 | 28,270 |
https://mathoverflow.net/questions/44183 | 26 | Let $(X,d)$ be a metric space, and let $C\_u(X)$ be the Banach space of bounded *uniformly* continuous functions on $X$ (with the uniform norm). How can I characterize its dual space $C\_u(X)^\*$?
I would guess it can be described as some space of measures. I would even be interested in the case $X=\mathbb{R}$.
Obviously if $X$ is compact this is just the signed (or complex) Radon measures on the Borel $\sigma$-algebra of $X$. If $d$ is a discrete metric then we have all finitely additive measures on $X$. But more generally I do not know.
**Edit:** If $C\_b(X)$ is the Banach space of all bounded continuous functions on $X$, we of course have $C\_u(X) \subset C\_b(X)$ as a closed subspace, and we know that $C\_b(X)^\*$ can be identified with the space of finite, regular, finitely additive signed Borel measures on $X$. Certainly each such measure gives us a continuous linear functional on $C\_u(X)$, so we have a map $C\_b(X)^\* \to C\_u(X)^\*$ which is just the restriction map, but it is not injective. Conversely, by Hahn-Banach each bounded linear functional on $C\_u(X)$ extends to a bounded linear functional on $C\_b(X)$, but not in a canonical way.
Also, it is clear that in general $C\_u(X)^\*$ contains more than just the countably additive measures, since e.g. if $X=\mathbb{R}$ it contains some Banach limits. So we have all the countably additive finite Radon measures, but not all the finitely additive finite regular measures. This is why I would imagine that $C\_u(X)^\*$ consists of all finitely additive measures satisfying some condition that is more than "regular" but less than "countably additive". But I have no idea what it might be.
As mentioned in comments, I am happy to know about nontrivial special cases: $X$ locally compact, $X$ complete and separable, etc.
| https://mathoverflow.net/users/4832 | Dual of bounded uniformly continuous functions | $C\_u(\mathbb R)^\*$ is essentially the space of complex measures on $\beta \mathbb Z\coprod (\beta\mathbb Z\times(0,1)).$ Here $\beta \mathbb Z$ is the Stone-Čech compactification of $\mathbb Z,$ and the $\coprod$ denotes disjoint union.
One can identify $C\_u(\mathbb R)$ with $C\_0(\beta \mathbb Z \coprod (\beta \mathbb Z\times (0,1)))$ in the following way: for $f\in C\_u(\mathbb R),$ and write $f=g+h$, where $g(n)=0$ for all $n\in \mathbb Z$ and $h$ is continuous and linear on each interval $[n,n+1].$ We will identify $g$ with a function $\tilde g:\beta \mathbb Z\times [0,1]\to \mathbb C$ in the following way: since $f:\mathbb R\to \mathbb C$ is uniformly continuous, the functions $g|\_{[n,n+1]}, n\in \mathbb Z$ form an equicontinuous family, considered as functions $g\_n\in C([0,1]).$ By Arzelà-Ascoli, the set $\{g\_n:n\in \mathbb Z\}$ is precompact in the uniform topology. By the universal property of $\beta \mathbb Z$, there is a unique continuous function $\varphi: \beta \mathbb Z\to C([0,1])$ such that $\varphi(n)=g\_n$ for $n\in \mathbb Z.$ Now the function $\tilde g(x,y):=\varphi(x)(y)$ is a continuous function from $\beta \mathbb Z\times [0,1]$ to $\mathbb C$; the joint continuity is obtained by again applying equicontinuity of the family $\{\varphi(x):x\in \beta\mathbb Z\}.$
We have identified $f$ with a pair $(\tilde g, h),$ where $\tilde g: \beta \mathbb Z\times [0,1]\to \mathbb C$, $\tilde g(x,0)=\tilde g(x,1)=0$ for all $x\in \beta \mathbb Z,$ and $h:\mathbb R\to \mathbb Z$ is determined by the sequence of values $h(n),n\in \mathbb Z.$ It's easy to check that every such pair $(\tilde g, h)$ uniquely determines a function $f\in C\_u(\mathbb R)$ by $f(n+x)=\tilde g(n,x)+h(n+x), x\in [0,1), n\in \mathbb Z.$
If we use the norm $|(\tilde g, h)|:=|\tilde g|+|h|$ (these are sup norms), the identification $f\leftrightarrow (\tilde g,h)$ is an identification of $C\_u(\mathbb R)$ with $C\_0(\beta \mathbb Z \coprod (\beta \mathbb Z\times (0,1))),$ as Banach spaces.
So it appears that finding a non-obvious element of $C\_u(\mathbb R)^\*$ is more or less equivalent to finding a non-obvious element of $C\_b(\mathbb Z)^\*,$ as Greg predicted.
| 18 | https://mathoverflow.net/users/10500 | 44508 | 28,273 |
https://mathoverflow.net/questions/44516 | 9 | What is an example of a functor $F : \mathbb{C}\text{-Sch.} \to \text{Sets}$ with the property that
the restriction of $F$ to locally Noetherian $\mathbb{C}$-schemes can be represented by a locally Noetherian $\mathbb{C}$-scheme, but that scheme does not represent $F$.
I'd be particularly nice to see a "real-world" example (though this might be stretching the notion
of "real-world").
| https://mathoverflow.net/users/5337 | Non-representable functor, representable on locally Noetherian schemes? | Define $F(X) = {\rm{Hom}}\_{\mathbf{C}}(X,{\rm{Spec}}(R/tR))$ where $R$ is the valuation ring of an algebraic closure of $\mathbf{C}((t))$. Note that every element of the maximal ideal of $R/tR$ is nilpotent yet also an $N$th power for arbitrarily large $N$. For any noetherian $\mathbf{C}$-algebra $A$, every $\mathbf{C}$-algebra map $R/tR \rightarrow A$ carries
the maximal ideal into the nilradical of $A$. But the nilradical of $A$ has all elements
with vanishing $n$th power for some uniform $n$ (depending on $A$) since $A$ is noetherian, so in fact the maximal ideal of $R/tR$ is killed by any such map. In other words, the restriction of $F$ to the full subcategory of locally noetherian objects is represented by ${\rm{Spec}}(\mathbf{C})$.
This is a perfectly "real world" example, since valuation rings of algebraic closures of complete discretely-valued fields come up all the time in number theory and rigid-analytic geometry.
| 18 | https://mathoverflow.net/users/3927 | 44518 | 28,278 |
https://mathoverflow.net/questions/44507 | 3 | A profinite group is said to be projective if its cohomological dimension is $\leq 1$. Is this related to some other notion of "projective"? How so?
| https://mathoverflow.net/users/5309 | What's "projective" about "projective pro-finite groups"? | A profinite group $P$ is projective if and only if any continuous group homomorphism from it to a profinite quotient group $G/H$ lifts to a continuous group homomorphism to the profinite group $G$.
| 4 | https://mathoverflow.net/users/2106 | 44531 | 28,282 |
https://mathoverflow.net/questions/44533 | 4 | Lets consider this method of finding inverse function:
$$f^{-1}(x) = \sum\_{k=0}^\infty A\_k(x) \frac{(x-f(x))^k}{k!}$$
where coefficients $A\_k(x)$ recursively defined as
$$\begin{cases} A\_0(x)=x \\ A\_{n+1}(x)=\frac{A\_n'(x)}{f'(x)}\end{cases}$$
It is evident that for some classes of functions starting from some point $A\_k(x)$ becomes zero and thus the inverse function can be expressed in closed form.
For example, the expression has limited number of terms for any function of the following form:
$$f(x)=a \sqrt[n]{x+b}+c$$
where $n\in \mathbb{N}$. It is also evident that there are other classes of functions for which the series have limited number of terms.
So my question is for which other classes of functions this series give inverse function is closed form?
| https://mathoverflow.net/users/10059 | For which classes of functions this inverse function formula gives a closed form expression? | The answer is not that difficult. Assume that $A\_{n+1}\equiv0$. This means $A\_n={\rm cst}$, that is $A\_{n-1}'=cf'$. Integrating, $A\_{n-1}=cf+d$, that is $A\_{n-2}'=(cf+d)f'$. Integrating again, $A\_{n-2}=\frac{c}{2}f^2+df+e$. And so on. By induction, we find that $A\_{n-k}=P\_k(f)$, where $P\_\ell$ is a polynomial of degree $\ell$. When $k=n$, we find the necessary and sufficient condition that $x=P\_n(f)$.
In conclusion, the functions $f$ for which the series in the formula has finitely many terms are the roots of equations $P(f)=x$, where $P$ is some polynomial.
>
> Edit. This answer suggests that the formula is nothing but Taylor expansion for $f^{-1}$, which turns out to be finite if and only if $f^{-1}$ is a polynomial. This also suggests that the series converges in a non-trivial interval whenever $f^{-1}$ is analytic.
>
>
>
| 6 | https://mathoverflow.net/users/8799 | 44539 | 28,286 |
https://mathoverflow.net/questions/44541 | 2 | Given a point $x$ in a topological space $X$. I was wondering, whether one can always find a [local basis](http://en.wikipedia.org/wiki/Neighbourhood_system) at $x$, which is [totally ordered](http://en.wikipedia.org/wiki/Totally_ordered_set) (a chain) under inclusion. For example this is true for spaces, which have countable local basis: If $\{U\_i|i\in\mathbb{N}\}$ is a local basis at $X$, then $\{\bigcap\_{i=1}^nU\_i|n\in \mathbb{N}\}$ is totally ordered.
So my question is: Assume, there is a totally ordered local basis at $x\in X$. Does this already imply, that there is a countable local basis at $x\in X$?
The rest of this text contains the motivating example. So if you are motivated enough, you can stop reading.
For example the space $\prod\_{i\in I} \{0;1\}$ does not have a totally ordered local basis at $\{0\}^I$ for uncountable $I$. Assume $P$ is such a basis.
Let for any open neighborhood $U$ of $\{0\}^I$
$N\_U:=\{i\in I| 1\_i\notin U\}$
where $1\_i$ denotes the element of $\prod\_{i\in I} \{0;1\}$ with zeros everywhere except for the $i$-th entry. Note that $N\_U$ is always finite by definition of the product topology and that for $U\subset U'$ we get $N\_U\supset N\_{U'}$ Further note that $I=\bigcup\_{U\in P}N\_U$.
Now construct an injection $f:\mathbb{N}\rightarrow I$ in the following way
Choose any $U\_0\in P$ and enumerate all elements of $N\_0$ in your favourite way. Then choose a $U\_1\in P$ with $N\_{U\_0}\subsetneq N\_{U\_1}$. Then enumerate all elements in $N\_{U\_1}\setminus N\_{U\_0}$. Continue this way to get an injection $\mathbb{N}\rightarrow I$.
As $I$ is uncountable, this map can't be surjective. Choose $i\in I\setminus Im(f)$ and any basis element $U\in P$ with $i\in N\_U$. So for any $n\in \mathbb{N}$, we get
$i\notin N\_{U\_n} \Rightarrow U\_n\nsubseteq U\Rightarrow U\subseteq U\_n\Rightarrow N\_U \supset N\_{U\_n}$.
The second last implication holds, as the base is totally ordered. This holds for any $n$, so $N\_U \supset \bigcup\_nN\_{U\_n}$. But the left side is finite and the right side is (by constuction) infinite. Contradiction! So the cannot be a totally ordered local base.
| https://mathoverflow.net/users/3969 | chains and countability | The space $\omega\_1+1$ under the order topology, where $\omega\_1$ refers to the first uncountable ordinal, has a
linearly ordered local basis at the point $\omega\_1$ (and indeed at every point),
consisting of the intervals $(\alpha,\infty)$, but there is
no countable local basis at that point, because every
countable set of ordinals below $\omega\_1$ is bounded below
$\omega\_1$.
An essentially similar example arises from the long line, by placing a point at the top. This new point will have a linearly ordered local basis consisting of intervals, but no countable local basis.
| 3 | https://mathoverflow.net/users/1946 | 44546 | 28,291 |
https://mathoverflow.net/questions/41143 | 4 | From the Incompleteness theorems, if ZF is consistent, one knows there are models of ZF satisfying ¬Con(ZF). These models must be non-standard (in the sense of being models whose ordinals are not well-ordered), and so must be the proof of an inconsistency from the axioms of ZF in them.
Now, ¬Con(ZF) is a very special kind of arithmetical statement. My question is, are there other kinds of statements consistent with ZF known to require the existence of non-standard models to hold?
If there are any, how does one recognize them?
| https://mathoverflow.net/users/6466 | Statements that require the existence of non-standard models to hold | Here's another example. By a "computable well ordering" I will mean an index for a well-founded computable (total) linear order on $\omega$. Because ZFC is an effective theory, there must be some computable well ordering $\zeta$ that ZFC does not prove is a well ordering. This is because:
* The set of indices of computable well orderings is strictly $\Pi^1\_1$ but, because ZFC is an effective theory, the set of computable linear orderings that ZFC proves are well founded is $\Sigma^0\_1$. So it cannot be true that a computable linear ordering is well founded if and only if ZFC proves it is well founded.
* ZFC is a true theory, so if ZFC proves that a computable linear order $L$ is well founded then $L$ really is well founded.
Because ZFC does not prove that $\zeta$ is a well ordering, there is some model of ZFC in which $\zeta$ is either not a total linear ordering, or $\zeta$ is not well founded. Either of these can only happen in a non-well-founded model of ZFC, since by definition $\zeta$ really is a well ordering.
Although this type of example is related to the type from the question, the proof method sketched here does not make use of consistency statements, so I feel this counts as an essentially different example of the same underlying phenomenon.
| 4 | https://mathoverflow.net/users/5442 | 44556 | 28,298 |
https://mathoverflow.net/questions/44563 | 6 | Non-quasi-coherent sheaves of $\mathcal O\_X$ modules on a scheme seem like a wild concept to me; are they actually used for something?
| https://mathoverflow.net/users/8363 | When are non-quasi-coherent sheaves used? | One can think the adeles on a curve (or higher adeles on other spaces) as a sheaf of $\mathcal O$-algebras. That is, consider the sheaf $B(U)=\prod\_{x\in U}\mathcal O\_x$, where $\mathcal O\_x$ is the completion of $\mathcal O$. Then the sheaf $A=B\otimes K$, where $K$ is the sheaf of rational functions, has the adeles as global sections. There is a short exact sequence $\mathcal O\to K\times B\to A$. One can tensor a quasicoherent sheaf with this to obtain a resolution to compute cohomology. Indeed, Weil introduced the adeles (after the earlier ideles) specifically to prove Riemann-Roch. I'm not sure when this was reinterpreted in terms of sheaves, which were only introduced later.
| 5 | https://mathoverflow.net/users/4639 | 44584 | 28,307 |
https://mathoverflow.net/questions/44569 | 5 | This question is with regards to terminology. I am writing a journal paper that describes the software I develop at my workplace call [Navigator](http://ophid.utoronto.ca/navigator/). The software is used for visualization of networks/graphs. At work, we commonly interchange the use of the words 'networks' and 'graphs' to refer to the same concept. But for writing the journal paper, I would like to know if they can be safely interchanged, or if they refer to slightly different concepts?
The software is used for modeling arbitrary entities as nodes and arbitrary connections between them as edges between the nodes. The graphs are not directed. No flow is implied, but some idea of flow can be added by attaching numerical/textual attributes to the edges.
| https://mathoverflow.net/users/10517 | Can 'network' and 'graph' be used interchangably? | If someone says "graph" without further specification I consider it to mean an unweighted undirected graph without loops or multiple edges. I consider "network" to be a more general term. Especially if the audience of the article is more on the applied side, I would go with network.
But really in your situation I would poke through some other NAViGaTOR papers, particularly Igor Jurisica's, and see what he uses. I bet it will be "network" almost all the time.
| 2 | https://mathoverflow.net/users/4580 | 44586 | 28,308 |
https://mathoverflow.net/questions/44581 | 7 | Leonardo of Pisa is best known as Fibonacci; various stories found in books and on the web claim that the name Fibonacci was invented by Edouard Lucas or Guillaume Libri in the 19th century, and that it means "son of Bonacci" (Leonardo's father was apparently called
Guglielmo Bonaccio). Heinz
[Lueneburg](http://mathforum.org/kb/message.jspa?messageID=1176612&tstart=0) found out that the name Fibonacci had been used already by John Leslie in 1820.
Are there any **facts** known as to who, why and when invented the name Fibonacci?
Or wasn't it invented at all?
| https://mathoverflow.net/users/3503 | Fibonacci = Leonardo Pisano? | From [*The Fabulous Fibonacci Numbers*](http://books.google.co.uk/books?id=-O4ZAQAAIAAJ&q=The+Fabulous+Fibonacci+Numbers&dq=The+Fabulous+Fibonacci+Numbers&hl=en&ei=jETQTIS3LY2-4garhLmwBg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CC8Q6AEwAA) by A.S. Posamentier and I. Lehmann (Prometheus Books, New York (2007), pp. 17-18):
>
> Leonardo Pisano - or Leonardo of Pisa, Fibonacci - his name as recorded in history, is derived from the Latin "filius Bonacci," or a son of Bonacci, but it may have been more likely derived
> from "de filius Bonacci," or family of Bonacci. He was born to Guglielmo (William) Bonacci and his wife in the port city of Pisa, Italy, around 1175, shortly after the start of construction of a famous bell tower, the Leaning Tower of Pisa.
>
>
>
The authors also indicate in a footnote that
>
> it is unclear who first used the name Fibonacci; however, it seems to be attributed to Giovanni Gabriello Grimaldi (1757-1837) at around 1790 or to Pietro Cossali (1748-1815).
>
>
>
| 8 | https://mathoverflow.net/users/5371 | 44587 | 28,309 |
https://mathoverflow.net/questions/44512 | 26 | A philosopher asked me an interesting math question today! We know that there are sets S of integers which don't have a "natural" or "naive" density -- that is, the quantity (1/n)|S intersect [1..n]| doesn't approach a limit. The "analytic" or "Dirichlet" density exists whenever the naive density does, and is equal to it -- but sometimes is well-defined even when the naive density is not, for instance when S is the "Benford set" of integers whose first digit is 1. (See [this related MO question](https://mathoverflow.net/questions/9143/analytic-density-of-the-set-of-primes-starting-with-1).)
Anyway, the philosopher asked: why stop at Dirichlet density? Is there a sequence of probability measures p\_1, p\_2, p\_3, .... on the natural numbers such that
* Whenever S has a Dirichlet density, the limit of p\_i(S) exists and is equal to the Dirichlet density, but
* there are subsets of the natural numbers such that p\_i(S) approaches a limit but S has no Dirichlet density?
(Possibly clarifying addition: to obtain naive density, one can take p\_i to be the measure assigning probability 1/i to each integer in [1..i] and 0 to integers greater than i. To get Dirichlet density, take p\_i to be the measure with p\_i(n) = 1/(n^{1+1/i}zeta(1+1/i)). In either case, the corresponding density of S is the limit in i of p\_i(S), whenever this limit exists. So what I have in mind is densities which can be thought of as limits of probability measures, though perhaps there are reasons to have yet more general entities in mind?)
If so, are there any examples which are interesting or which are used for anything in practice?
| https://mathoverflow.net/users/431 | Is there any sense in which Dirichlet density is "optimal?" | The best way to impress a philosopher is to tell him/her about ultrafilters. A (non-principal) filter on $\mathbb N$ is a set of infinite subsets of $\mathbb N$ closed under intersections and taking super-sets. A maximal filter (under inclusion) is called an ultrafilter. There are plenty of those but nobody saw them since their existence depends on the axiom of choice. For every filter $\omega$ one can define the concept of convergence of sequences of real numbers: a sequence $b\_n$ converges to $b$ if for every $\epsilon$ the set of $i$'s such that $|b-b\_i|\le \epsilon$ is in $\omega$. If $\omega$ is an ultrafilter, then every bounded sequence of real numbers has unique $\omega$-limit. It is not true if $\omega$ is not an ultrafilter. The smallest interesting filter (called Fréchet filter) consists of all sets with finite complements. The limit corresponding to that filter is the ordinary limit studied in Calculus. You can start from the Fréchet filter and add sets to it to produce bigger and bigger filters. Each filter gives you a Dirichlet-like density. If $\omega$ is an ultrafilter, then all sets will have density (between 0 and 1). Otherwise there will always be sets without density assigned.
| 15 | https://mathoverflow.net/users/nan | 44590 | 28,310 |
https://mathoverflow.net/questions/44577 | 8 | Let $E$, $F$ be two complex elliptic curves, and $A=E \times F$. Let us denote by
$\pi\_E \colon A \to E, \quad \pi\_F \colon A \to F$
the natural projections. For all $p \in F$ let us write $E\_p$ instead of $\pi\_F^\*(p)$.
Now let us fix $p \in F$ and consider the unique indecomposable rank $2$ vector bundle $\mathcal{F}$ on $A$ defined by the extension
$0 \to \mathcal{O}\_A \to \mathcal{F} \to \mathcal{O}\_A(-E\_p) \to 0$
My main question is the following:
>
> What is the restriction of $\mathcal{F}$ to a fibre of type $E\_x$, for $x \in F$ general?
>
Related to this, another question is
>
> What are $\pi\_{F\*} \mathcal{F}$ and $R^1 \pi\_{F\*} \mathcal{F}$?
>
Observe that the restriction of $\mathcal{F}$ to $E\_x$ is given by an extension
$0 \to \mathcal{O}\_{E\_x} \to \mathcal{F}|\_{E\_x} \to \mathcal{O}\_{E\_x} \to 0$,
so by Atiyah's classification of vector bundles over an elliptic curve we have that $\mathcal{F}|\_{E\_x}$ is either $\mathcal{O}\_{E\_x} \oplus \mathcal{O}\_{E\_x}$ or the unique non-trivial extension of $\mathcal{O}\_{E\_x}$ with itself.
But I cannot decide what happens generically.
**Motivation**. I met this problem when I started the investigation of some triple covers $f \colon X \to A$ such that
$f\_\*\mathcal{O}\_X=\mathcal{O}\_A \oplus \mathcal{F}^{\vee}$.
It is worth noticing that
the vector bundle $\mathcal{F}$ is the easiest example of indecomposable vector bundle on $A$ which is *not simple* (in fact, $\textrm{End}(\mathcal{F})=\mathbb{C} \oplus \mathbb{C}$).
| https://mathoverflow.net/users/7460 | Rank 2 vector bundle on a product of elliptic curves | It's the nontrivial extension. Here is why:
Let $\pi=\pi\_F$ and consider
$$
0\to \pi\_\*\mathscr O\_A \to \pi\_\*\mathscr F \to \pi\_\*\mathscr O\_A(-E\_p)\to R^1\pi\_\*\mathscr O\_A \to \dots
$$
Now
1) $\pi\_\*\mathscr O\_A\simeq \mathscr O\_F$ and $\pi\_\*\mathscr O\_A(-E\_p)\simeq\mathscr O\_F(-p)$
2) $R^1\pi\_\*\mathscr O\_A\simeq R^1\pi\_\*\omega\_A$ is torsion-free by Kollár's theorem (I guess it is also isomorphic to $\omega\_F\simeq \mathscr O\_F$ so you don't even need Kollár here), so
3) if $\pi\_\*\mathscr F\to \pi\_\*\mathscr O\_A(-E\_p)$ is not surjective, then it is zero.
4) $\pi\_\*\mathscr F$ is torsion free and hence locally free.
Suppose that the restriction of your sequence to any fiber is a trivial extension. Then $\pi\_\*\mathscr F$ has to have rank $2$ and hence by 1) and 3) the map $\pi\_\*\mathscr F\to \pi\_\*\mathscr O\_A(-E\_p)\simeq \mathscr O\_F(-p)$ is surjective.
Next, take the short exact sequence
$$
0\to \mathscr O\_F\to \pi\_\*\mathscr F \to \mathscr O\_F(-p) \to 0.
$$
This is necessarily the trivial extension, but also by construction it is the push-forward of your original sequence.
Then the pull-back of this sequence maps functorially to the original sequence and you get an isomorphism on the sides, so you get an isomorphism in the middle as well. That implies that $\mathcal F$ is a trivial extension, which is an obvious contradiction.
Now as a side result we get that $\mathscr O\_F\simeq \pi\_\*\mathscr O\_A\simeq \pi\_\*\mathscr F$ from the beginning of the long exact sequence and
$$\begin{multline}
0\to \pi\_\*\mathscr O\_A(-E\_p)\simeq\mathscr O\_F(-p)\to R^1\pi\_\*\mathscr O\_A\simeq\mathscr O\_F\to R^1\pi\_\*\mathscr F \to \\ \to R^1\pi\_\*\mathscr O\_A(-E\_p)\simeq \mathscr O\_F(-p)\otimes R^1\pi\_\*\omega\_A\simeq \mathscr O\_F(-p)\to 0
\end{multline}$$
so $\pi\_\*\mathscr F\simeq \mathscr O\_F$ and $R^1\pi\_\*\mathscr F\simeq \mathscr O\_p\oplus \mathscr O\_F(-p)$.
**EDIT:** added the last exact sequence and corrected conclusion according to Francesco's comment.
| 4 | https://mathoverflow.net/users/10076 | 44591 | 28,311 |
https://mathoverflow.net/questions/44417 | 2 | Let $\mathbb Z^2$ denote the two-dimensional integer lattice with norm of $i=(i\_1,i\_2)$ given by $\|i\|=|i\_1|+|i\_2|$.
For each $x\in\mathbb Z^2$, we assign a uniform random variable, $\sigma\_x$ taking values on the set $\{-1,1\}$.
Fix $\omega\in\{-1,0,1\}^{\mathbb{Z}^2}$ and $\beta>0$. For each finite $\Lambda\subset\mathbb{Z}^d$, define a probability measure on the sigma algebra generated by the cylinder sets of $\{-1,1\}^{\mathbb{Z}^2}$, such that for each
$\sigma\in\{-1,1\}^{\mathbb{Z}^2}$ the probability of this configuration is given by
$$
\mu\_{\Lambda}^{\beta,\omega}(\sigma)=
\left\{
\begin{array}{rl}
\frac{\exp(-\beta H\_{\Lambda}^{\omega}(\sigma))}{Z\_{\Lambda}^{\omega}},&\text{if}\ \ \sigma\_i=\omega\_i\ \forall i\in\Lambda^c;\\ \\ \\
0,& \text{otherwise},
\end{array}
\right.
$$
where
$$
H\_{\Lambda}^{\omega}(\sigma)=-\sum\_{i,j\in\Lambda}J\_{ij}\sigma\_i\sigma\_j-\sum\_{i\in\Lambda, j\in\Lambda^c}J\_{ij}\sigma\_i\omega\_j
$$
with $J\_{ij}\equiv J(\|i-j\|)\geq 0$ and $J\_{ij}=0$ if $\|i-j\|\geq R$, for some positive $R$ and $Z\_{\Lambda}^{\omega}$ is a normalizing constant so that
$\mu\_{\Lambda}^{\beta,\omega}$ is a probability measure.
**Question 1:** If $\Lambda\_n\uparrow\mathbb{Z}^2$ and $\omega\_i=0$ for all $i\in\mathbb{Z}^2$, sounds reasonable that any accumulation point of the sequence $\mu\_{\Lambda\_n}^{\beta,\omega}$, in the weak\* topology, is translation invariant. Is this true for any finite $R$ ?
**Question 2:** Suppose $R$ finite and bigger than one, keeping the setting of Question 1 but $\omega\_i=1$ (or $\omega\_i=-1$) for all $i\in\mathbb{Z}^2$ is the weak\* limit
$$w-\lim\_{n\to\infty} \mu\_{\Lambda\_n}^{\beta,\omega}$$
translational invariant ?
**Question 3:** For finite $R$ bigger than one is it true the Aizenman-Higuchi Theorem
$$w-\lim\_{n\to\infty} \mu\_{\Lambda\_n}^{\beta,\omega}\in [\mu^{\beta,+},\mu^{\beta,-}] ?$$
| https://mathoverflow.net/users/2386 | On generalisation of Aizenman-Higuchi Theorem | Concerning questions 1 and 2, if I understand it correctly (i.e., you're simply looking at the finite-range ferromagnetic Ising model with free, resp + or -, b.c.), then the sequences actually converge and the limits are translation invariant. This follows from monotonicity in the volume (GKS for free, GKS of FKG for + or -).
(In all these cases, just fix a local function and take two different increasing sequences of boxes. Show using correlation inequalities that the expectation in both cases necessarily converge to the same limit by sandwiching those in one sequence by those in the other. You can find a version of this classical argument in my lecture notes (in french), see my homepage.)
Concerning 3, the answer is certainly yes, but no proof is known (both the original proofs and the improved one we devised recently with Loren Coquille rely in an essential way on the n.n. nature of the interaction). (There might be some results at very large beta, of course.)
| 3 | https://mathoverflow.net/users/5709 | 44599 | 28,317 |
https://mathoverflow.net/questions/44596 | 5 | The definition of a (geometric) vector bundle over a scheme $X$ can be rewritten as follows in terms of 'not-so-geometrical algebra' if $X=Spec R$ is affine and if I am not missing something.
A *vector bundle* of rank $n$ over $R$ is an $R$-algebra $A$ such that
* for every $p\in Spec R$ there is a isomorphism (belonging to the data)
$$
\phi\_p:k(p)[X\_1,...,X\_n]\xrightarrow{\cong} A\otimes\_R k(p)
$$
where $k(p)$ is the residue field $R\_p/m\_p$ and
* there are elements $\{a\_i\}\_{i\in I}$ of $R$ such that the $D(a\_i)=\{ I\in Spec R\mid a\notin I \}$ cover $Spec R$ and for every $i\in I$ there is an $R\_a$-algebra isomorphism
$$
A\otimes\_R R\_{a\_i}\xrightarrow{\cong}R[X\_1,\ldots,X\_n]\otimes\_RR\_{a\_i}
$$
which induces for every $p\in Spec R$ with $a\_i\notin p$ a $k(p)$-**linear** $k(p)$-algebra isomorphism
$$
A\otimes\_R k(p)~\xleftarrow{\phi\_p}k(p)[X\_1,...,X\_n]\to k(p)[X\_1,...,X\_n]\cong R[X\_1,...,X\_n]\otimes\_R k(p).
$$
The (isomorphism classes) of such vector bundles over $R$ should correspond to (isomorphism classes) of finitely gererated projective modules over the ring $R$.
How can this correspondence be seen?
| https://mathoverflow.net/users/2625 | The correspondence between affine vector bundles and f.g. projective modules | Given any $R$-module $M$, there is a scheme which corresponds to the 'total space' of $M$, given by
$$ Tot(M):=Spec( Sym\_R(M\*))$$
where $M\*$ is the dual module $Hom\_R(M,R)$ and $Sym\_RM\*$ is the symmetric algebra of $M^\*$ over $R$. If $M$ happens to a free rank $n$ $R$-module, then $Sym\_RM\simeq R[X\_1,...,X\_n]$. The scheme $Tot(M)$ has a natural map to $Spec(R)$, which is dual to the obvious inclusion
$$R\rightarrow Sym\_RM$$
If you start with a locally free, finite rank $R$-module $P$, and then consider its total space $Tot(P)$, the corresponding scheme is a vector bundle by your definition. This follows from considering open sets on which $P$ is free, and considering the restriction of $Tot(P)$ over those open sets. Since restriction to an open set is the same as tensoring over the localization, and localization commutes with forming symmetric algebras, the locally freeness becomes your second condition. The first condition is also straightforward.
Then, observe that *every* vector bundle by your definition arises this way. To see this, follow Mike's comment. Associate to a vector bundle $V$ its sheaf of sections over $Spec(R)$, which is an $R$-module in a natural way. It will be free over the open cover $D(a\_i)$, with constant rank $n$.
Edit: As pointed out by roger, the total space construction should use the dual of $M$. As a side note, this means that it is the same if you replace $M$ with $M^{\*\*}$, and so it is not interesting to apply this construction to non-reflexive modules.
| 4 | https://mathoverflow.net/users/750 | 44628 | 28,337 |
https://mathoverflow.net/questions/44630 | 0 | I expect the following relation to be vanishing. But it seems not that obvious.
$\Gamma\_{ab}^{\lambda}t^at^b \Gamma\_{\lambda c(d)}t^c=0$
where $t^a$ are even ghosts, "$ab$" are indices for matrix element, and $\lambda$ denote different Gamma matrices. The Einstein summation convention is used above, i.e. we will sum over all indices except $d$.
I checked for both 3D and 4D Clifford algebra. The relation above seems to be right. But not sure whether it is generally true.
Does the following equation also vanishes?
$\Gamma\_{\lambda a b}t^a t^b C^{\lambda} C^{\alpha}C^{\beta}=0$
where $C^{\lambda}$ are odd ghosts, i.e. $C^{\alpha}C^{\beta}=-C^{\beta}C^{\alpha}$.
The left hand side of the equation above is supposed to be something in $\wedge^2 V$, where $V=\{ C^{\lambda}|\lambda=1,2,\cdots,D \}$. $D$ is the dimension of the space.
| https://mathoverflow.net/users/6577 | Clifford Algebra and Gamma matrices: is this relation generally true for any dimension? | It's not clear to me what you mean by "even ghosts". Do you mean perhaps that $t^a t^b = t^b t^a$?
If so, then you will find that the identity is only valid in 3, 4, 6 and 10 dimensions and with lorentzian signature. Indeed, this identity is essentially the condition for the vanishing of a fermionic trilinear which appears in the supersymmetric variation of Yang-Mills coupled minimally to an adjoint fermion, which in turn is the obstruction to the existence of "pure" supersymmetric Yang-Mills.
It is no accident that those dimensions are 2 plus the dimensions of the real division algebras: $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$ and $\mathbb{O}$. In fact, the identities are well-known identities for these algebras. In particular, when the dust clears, the ten-dimensional identities are the celebrated Moufang identities.
Of course, if I got the definition of the even ghosts wrong, then what I say above is probably wrong.
| 4 | https://mathoverflow.net/users/394 | 44633 | 28,339 |
https://mathoverflow.net/questions/20688 | 14 | I know that there are analytic functions whose composition with itself is the exponential function, the so-called functional square root of the exponential function, with the additional property that it is real on the real line.
Is a similar property possible for a holomorphic function that interpolates the tower function?
Tower function on the positive integers is defined recursively by $f(n+1) = e^{f(n)}, f(1) = 1$.
| https://mathoverflow.net/users/4923 | What's a natural candidate for an analytic function that interpolates the tower function? | The function you want grows too fast to be interpolated by usual method, but there exists an [iterative solution with Cauchy integrals by Dmitry Kouznetsov and Henryk Trappmann](http://www.ams.org/journals/mcom/2010-79-271/S0025-5718-10-02342-2/)
If you relax the condition so to find a solution for $f(x+1)=a^{f(x)}$ such that $$a \le e^{1/e} $$ then there are multiple expressions for your function:
$$f(x)=\sum\_{m=0}^{\infty} \binom xm \sum\_{k=0}^m \binom mk (-1)^{m-k}\exp\_a^{[k]}(1)$$
$$f(x)=\lim\_{n\to\infty}\binom xn\sum\_{k=0}^n\frac{x-n}{x-k}\binom nk(-1)^{n-k}\exp\_a^{[k]}(1)$$
$$f(x)=\lim\_{n\to\infty}\frac{\sum\_{k=0}^{2n} \frac{(-1)^k \exp\_a^{[k]}(1)}{(x-k)k!(2n-k)!}}{\sum\_{k=0}^{2n} \frac{(-1)^k }{(x-k) k!(2n-k)!}}$$
$$f(x)=\lim\_{n\to\infty} \log\_a^{[n]}\left(\left(1-\left(\ln \left(\frac{W(-\ln a)}{-\ln a}\right)\right)^x\right)\frac{W(-\ln a)}{-\ln a}+\ln \left(\frac{W(-\ln a)}{-\ln a}\right)\exp\_a^{[n]}(1)\right)$$
Always here the number in square brackets designates n-th iteration and $W(x)$ is the Lambert's function.
There is also an expression for inverse function:
$$ f^{[-1]}(x)=\lim\_{n\to\infty} \frac{\ln \left(\frac{\frac{W(-\ln a )}{\ln a}+\exp\_a^{[n]}(x)}{\frac{W(-\ln a)}{\ln a}+\exp\_a^{[n]}(1)}\right)}{\ln \ln \left(\frac{W(-\ln a)}{-\ln a}\right)}$$
| 12 | https://mathoverflow.net/users/10059 | 44646 | 28,344 |
https://mathoverflow.net/questions/44514 | 5 | Consider the following situation:
Let $M$ and $M'$ be two closed manifolds and suppose $f:M\to \mathbb{R}$ and $f':M'\to \mathbb{R}$ are smooth morse functions on $M$ and $M'$ respectively. We say the pair $(M,f)$ and $(M',f')$ are *equivalent* if there is a smooth diffeomorphism $\phi:M\to M'$ so that $f'\circ \phi=f$ and consider equivalence classes $[(M,f)]$.
My questions are:
1) Has this been studied at all? and if so
2) Is there any sort classification result for (oriented) surfaces $M$?
What I'm looking for in 2) is that any equivalence class $[(M^2, f)]$ contains some geometrically nice example...for instance an immersed surface in $\mathbb{R}^3$ with the morse function given by restriction of the $z$ coordinate function. I'm mostly interested in examples where $f$ has as few critical points as the topology of $M$ allows.
I apologize if this is trivial as it came up while I was playing around with an idea for a different problem and (like most aspects of topology) is sadly not something I'm very familiar with.
If it helps, the example I have in my head is to take a genus-2 surface in $\mathbb{R}^3$ that looks like a figure 8 and morse function given by restricting the $z$ coordinate and contrast it with the genus-2 surface in $\mathbb{R}^3$ that looks like $\infty$ (i.e. the first one on its side) with Morse function given by restricting the $z$ coordinate. These two pairs shouldn't be equivalent as the number of components of level curves of the first morse function is at most 2 while the second has level curves of the morse function with 3 components (I apologize for the crummy graphics). This is in spite of both morse functions having the same number of critical points (six). I was wondering how many other examples of morse functions there were with 6 critical points that weren't equivalent to these two.
**Edit:**
I felt I should add...heuristically what I am interested in is to what extent can one use a Morse function to say whether the handles (in a surface) are "next to each other" or "on top of one another".
| https://mathoverflow.net/users/26801 | Pair consisting of a compact manifold and Morse function | You might have a look at Hatcher and Thurston's paper "[A presentation for the mapping class group of a closed orientable surface](http://www.ams.org/mathscinet-getitem?mr=579573)". They use Morse functions on the surface to facilitate the derivation of their presentation. On page 223 and following, they discuss how to form a graph from a Morse function on a surface, which is the leaf space of the level sets of the Morse function. This graph, together with its map to $\mathbb{R}$, uniquely determines the Morse function on the surface (up to graph isomorphism preserving the function). In particular, one need only know the values of the Morse function at the vertices of the graph (corresponding to the critical points of the Morse function on the surface) since it is monotonic on each of the edges. The space of such functions will therefore be parameterized by a subset of a vector space, satisfying certain inequalities. As they indicate on p. 224, the surface with the Morse function may be recovered from the graph (with its Morse function) by embedding it in $\mathbb{R}^3$ and taking the boundary of a regular neighborhood (after a small perturbation).
| 3 | https://mathoverflow.net/users/1345 | 44648 | 28,345 |
https://mathoverflow.net/questions/44661 | 3 | Assume that $(P,\le)$ is a notion of forcing. There are several ways to define what it means for $P$ being proper and I would like to know: What is the complexity (in terms of the Levy-Hierarchy) of the statement 'P is proper'?
| https://mathoverflow.net/users/4753 | Complexity of the statement 'P is proper' | Properness is observable in any sufficiently large $V\_\alpha$, and therefore has complexity $\Sigma\_2$. In oktan's answer, it suffices to consider sufficiently large $\lambda$, rather than all $\lambda$. I think this is proved in some of the standard accounts of proper forcing.
| 5 | https://mathoverflow.net/users/1946 | 44667 | 28,358 |
https://mathoverflow.net/questions/44666 | 0 | A simple question from someone new to the field:
In a metric space, the [Hausdorff dimension](http://en.wikipedia.org/wiki/Hausdorff_dimension) of a subset is defined by covering the subset with $\epsilon$-balls and looking at how the number of required balls grows as a power or $\epsilon$ in the limit $\epsilon \to 0$.
My question is this: If my space is $\mathbb{R}^n$ and the metric $d\_p$ is the metric induced by the $p$-norm, $p \in (1, \infty]$, does the Hausdorff dimension of a subset $A \subseteq \mathbb{R}^n$ depend on the choice of $p$?
(My initial thought was that every $d\_p$-ball has the same $d\_q$-dimension for all $q$ (namely $n$), so that I believe all the dimensions should coincide, but I'm sure I'm overlooking something.)
Thanks!
| https://mathoverflow.net/users/10535 | Does the Hausdorff dimension depend on the L^p-norm? | Let $B\_p$ denote the 1-ball with centre 0 with respect to the
$l^p$ norm. For any $p$ and $q$ there is a number $N$ such that $B\_p$
is covered by $N$ translates of $B\_q$. Then any $\epsilon$-ball in
the $l^p$ norm is covered by $N$ $\epsilon$-balls in the $l^q$ norm.
Thus within a constant factor, the number of $\epsilon$-balls
required to cover a set in the $l^p$ and $l^q$ norms is the same.
This constant factor won't affect the asymptotic power in the
number of $\epsilon$-balls required to cover a given set,
so the Hausdorff dimension in both cases is the same.
| 3 | https://mathoverflow.net/users/4213 | 44668 | 28,359 |
https://mathoverflow.net/questions/44673 | 18 | Four Color Theorem is equivalent to the statement: "Every cubic planar bridgeless graphs is 3-edge colorable". There is computer assisted proof given by Appel and Haken. Dick Lipton in of his [beautiful blogs](http://rjlipton.wordpress.com/2009/04/24/the-four-color-theorem/) posed the following open problem:
>
> Are there non-computer based proofs of the Four Color Theorem?
>
>
>
Surprisingly, While I was reading this paper,
Anshelevich and Karagiozova, [Terminal backup, 3D matching, and covering cubic graphs](http://portal.acm.org/citation.cfm?id=1250790.1250849), the authors state that Cahit proved that "every 2-connected cubic planar graph is edge-3-colorable" which is equivalent to the Four Color Theorem (I. Cahit, Spiral Chains: The Proofs of Tait's and Tutte's Three-Edge-Coloring Conjectures. arXiv preprint, math CO/0507127 v1, July 6, 2005).
>
> Does Cahit's proof resolve the open problem in Lipton's blog by providing non-computer based proof for the Four Color Theorem?
>
>
>
Cross posted on math.stackexchange.com as [Human checkable proof of the Four Color Theorem?](https://math.stackexchange.com/questions/8752/human-checkable-proof-of-the-four-color-theorem)
| https://mathoverflow.net/users/8784 | Human checkable proof of the Four Color Theorem? | This is too long for a comment, so I am placing it here.
In this [article of the Notices of the AMS](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.141.714&rep=rep1&type=pdf), Gonthier describes a full formal proof of the four-color theorem, which makes explicit every logical step of the proof.
Although this formal proof has been checked by the [Coq proof system](http://coq.inria.fr/), it would seem to be a category error to view this proof as a computer-based proof of the same kind as Appel and Haken's. The situation with Gonthier's proof is that we essentially have a full written text constituting a verified formal proof of the four-color theorem in first order logic.
And that is a state of certainty that most theorems in mathematics, including many of the classical results, have not yet attained.
| 30 | https://mathoverflow.net/users/1946 | 44681 | 28,364 |
https://mathoverflow.net/questions/44677 | 7 | I'm working on Markov Decision Process and I have not found yet an example of MDP that has a stochastic (non deterministic) optimal policy. Is there MDPs that have a stochastic optimal policy or is it shown that an optimal policy is always deterministic ?
If a stochastic policy exist, is it shown that some algorithms (like Q-Learning) converge to this policy ?
| https://mathoverflow.net/users/10537 | Is there MDPs (Markow Decision Process) which have a non deterministic optimal policy ? | If there is an optimal policy, there is a deterministic optimal policy. Here is a sketch of the argument:
Start with an optimal policy within the class of deterministic optimal policies. By the one-deviation-principle, you only have to check whether you can gain by randomizing after a certain history of the process. If you randomize over two actions that do not lead to the same payoff, you could gain by putting more weight on the action with a higher payoff. So all actions will give you the same payoff and you might as well choose a deterministic one. Now a standard result says that in MDPs, such a history dependent strategy can not improve on all Markovian strategies. Therefore, an optimal, deterministic Markovian strategy exist.
| 4 | https://mathoverflow.net/users/35357 | 44685 | 28,367 |
https://mathoverflow.net/questions/44445 | 6 | Why it is true that, over an algebraically closed field, any abelian variety is isogenous to a principally polarized abelian variety?
| https://mathoverflow.net/users/5329 | About isogenies of abelian varieties | This is to fill some prerequisites to BCnrd's comment-answer.
First of all, there are several definitions of a polarization on an abelian variety, and the most "coordinate-free" one is that it is a homomorphism $\lambda:A\to A^t = Pic^0(A)$ given by some (non-unique) ample divisor $D$, so that $\lambda(a) = \mathcal O\_A( T^\*\_a D - D)$, where $T\_a:A\to A$ is the translation by $a\in A$. A polarization is principal if $\lambda$ is an isomorphism.
Then the basic steps, all requiring proof, are:
1. Every abelian variety (over a field) has a polarization.
2. $\lambda$ is surjective, and $K(\lambda)=\ker\lambda$ is a finite group subscheme of $A$ of length $d^2$. The degree of $\lambda$ is defined to be $d$. Thus, a polarization is principal iff $d=1$.
3. $K(\lambda)$ comes with a perfect skew symmetric Weil pairing $b: K(\lambda)\times K(\lambda) \to \mathbb G\_m$ with the values in the multiplicative group. "Perfect pairing" means that it defines an iso from $K(\lambda)$ to its Cartier dual.
4. If $H\subset K(\lambda)$ is a subgroup which is isotropic w.r.t. this pairing (i.e. $b$ restricted to $H\times H$ is trivial), then $\lambda$ descends to a polarization on $B=A/H$ of degree $d/|H|$.
So then it is enough to find a nontrivial isotropic subgroup $H$, replace $A$ by $A/H$, and continue by induction, until you reach $d=1$.
In char 0, this is trivial: just pick any cyclic subgroup in $K(\lambda)$. Since $b$ is skew-symmetric, it is automatically isotropic.
In char p, you need to get into the classification of finite group schemes, and learn the difference between $\mathbb Z/p\mathbb Z$, $\mu\_p$, and $\alpha\_p$. Once you learn that (*Introduction to affine group schemes* by Waterhouse is a good source), then you are perhaps ready to read BCnrd's comment-answer.
| 9 | https://mathoverflow.net/users/1784 | 44686 | 28,368 |
https://mathoverflow.net/questions/44576 | 7 | Let A be a matrix with entries in Z/nZ. (n is not assumed to be prime.) Then the size of the row span is the size of the column span. All computations are mod n, so both these numbers are finite.
I believe you can prove this using Smith normal form: both the size of the row span and the size of the column span will be the same after passing to the Smith normal form (lift the matrix entries to Z to compute the Smith normal form). When A is in Smith normal form, these quantities can be easily computed.
I would like to find a reference for this fact, and would be grateful to anyone who could provide one.
| https://mathoverflow.net/users/5399 | "Linear algebra" over Z/nZ - reference please! | Your idea of using Smith normal form leads directly to a solution: But you need to verify that for every matrix $M$ with entries in $\mathbb{Z}/n\mathbb{Z}$ there are invertible matrices $A$ and $B$ with entries in $\mathbb{Z}/n\mathbb{Z}$ such that $AMB$ is in Smith normal form. It is essential that $A$ and $B$ be invertible as matrices over $\mathbb{Z}/n\mathbb{Z}$, otherwise the row and column spaces of $AMB$ won't necessarily be the same size as those of $M$.
A ring with the property that every matrix is equivalent to one in Smith normal form is called an elementary divisor ring. In the book "Matrices over Commutative Rings" by William Brown (Marcel Dekker 1993) it is shown (Theorem 15.8 and 15.9) that every principal ideal ring is an elementary divisor ring. It is simple to check that the rings $\mathbb{Z}/n\mathbb{Z}$ are principal ideal rings. So there's the reference you asked for. If you can't find the book by Brown the article by Kaplansky mentioned in the comments has the same material in a more general setting.
| 4 | https://mathoverflow.net/users/5229 | 44691 | 28,372 |
https://mathoverflow.net/questions/44692 | 25 | If ${L}$ is a line bundle over a complex manifold, what does the square root line bundle $L^{\frac{1}{2}}$ mean?
After some google, I got to know that there are certain conditions for the existence of square root line bundle. In particular,I've following questions :
1. What is the square root of a line bundle, what are the conditions for its existence.
2. More importantly, how to think of the square root line bundle. Intuitively, It seems that a square root bundle $L^{\frac{1}{2}}$ is a line bundle s.t. the tensor bundle obtained by taking the tensor product of $L^{\frac{1}{2}}$ with itself gives the line bundle $L$.(correct if I am wrong).
I am particularly interested in square root of the Canonical Line bundle over a Riemann Sphere and the relation of square root bundle to Spinors in QFT.
Please provide some references to look at.
| https://mathoverflow.net/users/9534 | What is a square root of a line bundle? | If $L$ is any line bundle over a compex manifold $X$, a square root of $L$ is a line bundle $M$ such that $M^{\otimes2}=L$. So your guess in part (2) is correct.
This square root (if it exists) is not unique in general, and two of them will differ by a $2$-torsion line bundle, that is a line bundle $\eta$ such that $\eta^{\otimes 2}$ is trivial.
In particular, if $\textrm{Pic}(X)$ is torsion free, then there is at most one square root.
In some cases no square root exists. Some general results are:
1. A line bundle of degree $0$ has always at least one square root. This because $\textrm{Pic}^0(X)$ is a complex torus, hence a divisible group (in fact, there are roots of any order).
2. A line bundle over a Riemann surface of genus $g$ has a square root if and only if it has even degree. The number of different square roots equals in this case $2^{2g}$, the number of $2$-torsion points in $\textrm{Pic}^0(X) \cong \textrm{Jac}(X)$.
3. If $L$ is effective, that is $H^0(X, L) \neq 0$, and $Z \subset X$ is the zero locus of a
holomorphic section of $L$, then the existence of a square root of $L$ is equivalent to the existence of a double cover $Y \to X$ branched over $Z$. In particular, non-trivial square roots of the trivial bundle correspond to non-trivial *unramified* double covers of $X$.
The square root of the canonical bundle of the Riemann Sphere $S$ is unique, since $\textrm{Pic}(S)=\mathbb{Z}$, and it is isomorphic to $\mathcal{O}(-1)$, the dual of the hyperplane bundle (the unique line bundle of degree $1$, whose transition function is $z \to 1/z$).
A readable introduction to spinor bundles is provided in the book of MOORE "Lectures on Seiberg-Witten invariants".
| 37 | https://mathoverflow.net/users/7460 | 44700 | 28,379 |
https://mathoverflow.net/questions/44657 | 5 | What does the principal L-functions on GL(n), $n \geq 3, n \in \mathbb{Z}$, look like?
Where can I find materials about principal L-functions on GL(n)?
| https://mathoverflow.net/users/1930 | Principal L-functions on GL(n) | There are several ways to attack standard L-functions (I prefer "standard" over "principal" because it is associated to the standard representation of GL(n) on an n-dimensional space).
I'm going to assume that by "look like" you mean the formula for the local factors as a function of the local data. At primes where the local representation is spherical (which happens for all but finitely many primes), the local factor is
$$1\over(1-\alpha\_1q\_v^{-s})\ldots(1-\alpha\_nq\_v^{-s})$$
where the $\alpha\_i$ are the Satake parameters for the local representation, which are essentially of the form $q\_v^{-s\_i}$, where the $s\_i$ are the exponents of the character of the unramified principal series into which the (spherical) local representation injects (by the Borel-Matsumoto theorem), up to some normalization issues. I'm not going to touch non-spherical or archimedean primes.
On to the different methods (in what follows, all cusp forms are assumed to generate irreducible representations):
1) Godement-Jacquet: In the spirit of Tate's thesis, take a cusp form f on G=GL(n) (and f' in the dual representation) and a Schwartz-Bruhat function $\Phi$ on M(n) and integrate
$$\int\_{Z\_{\bf A}G\_k\backslash G\_{\bf A}} \langle g\cdot f,f'\rangle \Phi(g)|det(g)|^s dg$$
where the brackets are the inner product pairing. This converges for Re(s) sufficiently large.
The uniqueness of the inner product makes this factor over primes into local integrals of similar shape. You then prove that the local integrals give the correct factor for supercuspidal representations, and then you prove an induction formula to show that the integral gives the correct factor for all representations parabolically induced from supercuspidal. (It may be possible to calculate the factors directly for unramified principal series, but I don't remember.) And then you have to deal with the archimedean primes. This proves that the integral produces the L-function for judicious choice of f, f', $\Phi$.
To prove analytic continuation and functional equation, you invoke a version of Poisson summation for M(n) and proceed as in Tate's thesis. (You end up with lots of weird terms that you show are zero by invoking cuspidality of f.)
2) "Doubling method" Rankin-Selberg integrals: Take a cusp form f on G=PGL(n), a cusp form f' in the dual representation, and a specific Eisenstein series on PGL(n$^2$) and integrate
$$\int\_{{C\_{\bf A}G\_k\times G\_k\backslash G\_{\bf A}\times G\_{\bf A}}}E((g\_1,g\_2))f(g\_1)f'(g\_2) dg\_1dg\_2$$
where C is the center of PGL(n$^2$). The embedding of $G\times G$ in PGL(n$^2$) is defined by the natural bi-regular action of $G\times G$ on M(n).
The Eisenstein series gives the integral a meromorphic continuation and functional equation. Unwind it and out pops the Godement-Jacquet integral.
When dealing with Eisenstein series, you usually have to do some extra work to get that there are only finitely many poles. However, if I recall correctly (and I'm not sure that I do), the Eisenstein series here is the mirabolic one, which you can prove has nice properties (essentially using the Godement-Jacquet arguments).
Added: To extend to GL(n), you have to add in an extra term to deal with the fact that the center of $G\times G$ is not compact after you mod out by the center of GL(n$^2$), so the naive integral would diverge for trivial reasons. (And you can't just mod out by more in the integral because the Eisenstein series is not constant on the center of $G\times G$, even though it has trivial central character)
3) Cogdell's Eulerian integrals: Take f on GL(n) and f' on GL(n'), with n>n'. Integrate
$$\int\_{Z\_{\bf A}GL\_{n'}(k)\backslash GL\_{n'}({\bf A})}Pf(g)f'(g)|det(g)|^{s-1/2}dg$$
where P is a projection operation on f. P is brilliantly designed to let this integral unwind completely into what is essentially a Mellin transform of Whittaker functions (which then factors over primes by the uniqueness of Whittaker models). More work shows that this is the L-function of the product.
Note that the decay of the cusp forms implies that the integral is entire. A little observation gives the functional equation.
4) Langlands-Shahidi method: (edited) The constant term of an Eisenstein series on GL(n+n') attached to cusp forms on GL(n) and GL(n') contains the L-function for the product (in your case take n'=1). In fact, it is
$${L\_S(s,f\otimes f')\over L\_S(1+s,f\otimes f')}\cdot A\_S(s,f\otimes f')$$
where $A\_S$ represents bad-prime factors, which generally you have to work to control (when it is even possible). The meromorphic continuation of the Eisenstein series is equivalent to that of the constant term, so you get that this quotient of L-functions is meromorphic.
When n'=1, you have the mirabolic Eisenstein series, so you know that the only poles in the constant term come from the zeroes of $L(1+s,f\otimes f')$, giving you, up to the $A\_S$, the analytic continuation of the L-function you care about.
I never studied the method, so I hope everything I said was correct.
References:
1) Godement-Jacquet (Zeta functions of simple algebras), Jacquet has two survey articles (in PSPM 26 and 33)
2) Gelbart, Piatetski-Shapiro, and Rallis (Explicit construction of automorphic L-functions)
3) Cogdell's website has his expository articles on his integral representation
4) Shahidi has a new book out on the method, and he has expository articles here and there.
| 18 | https://mathoverflow.net/users/6753 | 44701 | 28,380 |
https://mathoverflow.net/questions/44703 | 1 | Is the following a standard problem in combinatorics? Where can I find reference for it?
Consider $n$ particles in a circle, $k$ white and $n-k$ black, otherwise indistinguishable so that the number of dispositions is $n!/(n-k)!k!$. Different dispositions will have a different number of white/black/white/black... clusters. How many dispositions $d(N,K,C)$ have C clusters?
Example. $n=4, k=2$ I get:
$d(4,2,2) = 4$
$d(4,2,4) = 2.$
| https://mathoverflow.net/users/3441 | clusters of coloured particles | We will let $k\_1=k$ and $k\_2=n-k$, and $\tilde{c}=c/2$. I believe the answer is then
$\frac{k\_1 + k\_2}{\tilde{c}} {k\_1-1 \choose \tilde{c}-1} {k\_2-1 \choose \tilde{c}-1}$.
I have no idea whether there is a reference for this anywhere.
Here's how it works. First we solve (nearly) the same problem on a line. There are $k\_1$ white balls and $k\_2$ black balls, and we want to count how many ways there are of arranging the balls so you have $c$ clusters, starting with a white ball on the left and ending with a black ball on the right. What we do is to make $c/2$ clusters of white balls and $c/2$ clusters of balls balls, and interleave them. To make $\tilde{c} = c/2$ clusters of white balls, we start with $k\_1$ white balls, and put in $\tilde{c}-1$ dividing lines in any of the $k\_1-1$ spots between two balls. We can do the same for the black balls. This means that we have
${k\_1-1 \choose \tilde{c}-1} {k\_2-1 \choose \tilde{c}-1}$.
ways of arranging these balls into $c$ clusters on a line (with a white on the left and a black on the right). Now, we join them into a circle. There are $n=k\_1+k\_2$ positions on the circle where we could place the left endpoint of the line. But once we have a circle with $c$ clusters, there are $\tilde{c}=c/2$ ways we could have gotten to this disposition of balls in the circle; we can go back to a line by cutting the circle between any (black,white) pair of balls and there are $c/2$ of them. We thus obtain the above answer.
| 4 | https://mathoverflow.net/users/2294 | 44711 | 28,385 |
https://mathoverflow.net/questions/44680 | 6 | This question is slightly related to a popular one with the same title (see [here](https://mathoverflow.net/questions/27345)).
Let $k$ be a field with characteristic zero. It is known (see [Exercise 310](http://umpa.ens-lyon.fr/~serre/DPF/exobis.pdf)) that a matrix $A\in M\_n(k)$ is nilpotent if and only if it is a commutator of its own: there exists a $B$ such that $A=AB-BA$. Of course, $B$ is not unique.
>
> Consider the complex case ($k=\mathbb C$). Endow $M\_n(\mathbb C)$ with your beloved norm, preferably either the operator norm $\|\cdot\|\_2$ or the Schur--Frobenius--Hilbert--Schmidt norm $\|\cdot\|\_F$. If $A$ is nilpotent, what is the smallest value of $\|B\|$, where $B$ is a factor in $A=AB-BA$ ? What is the smallest constant $\mu(n)$ such that for every $n\times n$ nilpotent $A$, there exists such a $B$ with $\|B\|\le\mu(n)$ ? Actually, is there such a finite $\mu(n)$ ?
>
>
>
**Edit**. When ${\rm rk}A=1$, that is $A=xy^\*$ with $y^\*x=0$, one can always take $B$ such that $\|B\|\_2=\frac12$ or $\|B\|\_F=\sqrt2/2$. Just take $B$ diagonal in a unitary basis $\{\frac{x}{\|x\|\_2},\frac{y}{\|y\|\_2},\ldots\}$, with eigenvalues $-\frac12,\frac12,0,\ldots,0$.
| https://mathoverflow.net/users/8799 | Norm of commutators (bis) | Consider the matrix $$\left\[ \begin{array}{ccc} 0 & 1 & k \\\ 0 & 0 & 1\\\ 0 & 0 & 0\end{array}\right\].$$ Then the norm of $B$ depends on $k$ (just solve the system of linear equations $AB-BA=A$). So the answer to your question seems to be "no".
| 6 | https://mathoverflow.net/users/nan | 44715 | 28,388 |
https://mathoverflow.net/questions/44705 | 55 | It seems that in most theorems outside of set theory where the size of some set is used in the proof, there are three possibilities: either the set is finite, countably infinite, or uncountably infinite. Are there any well known results within say, algebra or analysis that require some given set to be of cardinality strictly greater than $2^{\aleph\_{0}}$? Perhaps in a similar vein, are any objects encountered that must have size larger than $2^{\aleph\_{0}}$ in order for certain properties to hold?
| https://mathoverflow.net/users/6856 | Cardinalities larger than the continuum in areas besides set theory | The Zariski tangent space at any point of a positive dimensional $C^1$-manifold $X$ has dimension $2^{2^{\aleph\_0}}= 2^{\frak c}$.
Let me explain in the case when $X=\mathbb R$.
Consider the ring $C^1\_0$ of germs of $C^1$- functions at $0\in \mathbb R$ and its maximal ideal $\frak m $ of germs of functions vanishing at zero. The cotangent space at zero of $\mathbb R $ is $Cot\_0=\frak m /\frak m ^2$ and the Zariski tangent space is $T\_0=(Cot\_0)^{\ast}$ (dual $\mathbb R$-vector space).
Now the germs of the functions $x^\alpha $ are linearly independent modulo $\frak m ^2$ for $\; \alpha\in(1,2)$ . Hence $\dim\_{\mathbb R} (Cot\_0)=\frak c$ and so indeed the Zariski tangent space at zero of $\mathbb R$ is $\dim\_{\mathbb R} (T\_0)=2^{\frak c}$.
It is noteworthy that many textbooks erroneously claim that for an $n$-dimensional manifold of class $C^1$ the Zariski tangent space defined above has dimension $n$. Or they make some equivalent mistake like claiming that the vector space of derivations of $C^1\_0$ has dimension $n$ . An example of such an error is on page 42 in Claire Voisin's (excellent!) book *Hodge Theory And Complex Algebraic Geometry I* published by Cambridge University Press.
To end on a positive note, the phenomenon I am describing only raises
its ugly head for $C^k$-manifolds with $k<\infty$. For $n$-dimensional $C^\infty$-manifolds the Zariski tangent space at any point has dimension $n$, as it should. The heart of the matter is that a $C^\infty$ function $f$ ,
on $\mathbb R$ say, which vanishes at zero can be written $f=xg$ for some function $g$
which is also of class $C^\infty$, whereas $g$ would only be of class $C^{k-1}$ if $f$ were of class $C^k$.
| 68 | https://mathoverflow.net/users/450 | 44733 | 28,400 |
https://mathoverflow.net/questions/44652 | 1 | Recall:
Let $FU\_\bullet:Cat\to Cat\_\Delta$ be the bar construction assigned to the comonad $FU$ determined by free-forgetful adjunction $F:Quiv\rightleftarrows Cat:U$. The restriction of $FU\_\bullet$ to the full subcategory $\Delta$ (which is isomorphic to the category of finite nonempty ordinals) naturally determines a colimit-preserving functor $\mathfrak{C}:Set\_\Delta=Set^{\Delta^{op}}\to Cat\_\Delta$. The right adjoint of this functor is called $\mathfrak{N}$, the homotopy-coherent nerve.
Identify $Cat$ (not by $FU\_\bullet$) with the full subcategory of $Cat\_\Delta$ spanned by those simplicially enriched categories with discrete hom-spaces.
Also, recall the definition of the right cone $X^\triangleright$ on a simplicial set $X$ is the join $X\star \Delta^0$. This determines an obvious natural map $X\to X^\triangleright$.
Let $X\to \Delta^1=\mathfrak{N}([1])$ be an object of $(Set\_\Delta\downarrow \Delta^1)$, and let $\epsilon:\mathfrak{C}(\Delta^1)=\mathfrak{C}(\mathfrak{N}([1])\to [1]$ be the counit (here $[1]$ is the category determined by the ordinal number $2$ (two objects, one nonidentity arrow). Form the pushout $M$ of the span $\mathfrak{C}(X^\triangleright) \leftarrow \mathfrak{C}(X)\to \mathfrak{C}(\Delta^1)\to [1]$ (the two arrows in the same direction are replaced by their composite, so this is $M=\mathfrak{C}(X^\triangleright)\coprod\_{\mathfrak{C}(X)} [1]$).
This determines a functor $St\_\epsilon X:[1]\to Set\_\Delta$ defined as $i\mapsto M(i,p)$ where $p$ is the image of the cone point of $\mathfrak{C}(X^\triangleright).$
Question:
The book I'm reading asserts that $St\_\epsilon X(0)$ can be identified with $St\_\*(X\times\_{\Delta^1} \Delta^0)$ (where $\Delta^0\to \Delta^1$ is the map $\mathfrak{N}(\lambda)$ where $\lambda:[0]\to [1]$ is the functor classifying the object $0$ of $[1]$) where $St\_\*S$ is simply defined to be the analogous construction when $\epsilon$ is replaced with the identity $[0]=\mathfrak{C}(\Delta^0)\to [0]$. (Note that here we can identify functors $[0]\to Set\_\Delta$ with simplicial sets themselves, and suggestively, that under this identification, $St\_\epsilon X(0)=\lambda^\*St\_\epsilon X$).
Why is this true? It's not so clear to me.
| https://mathoverflow.net/users/1353 | Reducing straightening over an interval to straightening over a point | (Community Wiki Answer)
Urs Schreiber answered this question over at the [nForum](http://www.math.ntnu.no/~stacey/Mathforge/nForum/comments.php?DiscussionID=2055&page=1#Item_4), so I'll reproduce his answer here (since I don't think he's going to add it himself)
---
Since $\mathfrak{C}$ is left adjoint we can essentially compute the pushout before applying $\mathfrak{C}$. Let me call the analog of $M$ obtained this way $P$
$$
\array{
X &\to& X^{\triangleright}
\\
\downarrow && \downarrow
\\
\Delta[1] &\to& P
}
$$
We have a canonical map $P \to \Delta[1]^{\triangleright}$ induced from the commutativity of
$$
\array{
X &\to& X^{\triangleright}
\\
\downarrow && \downarrow
\\
\Delta[1] &\to& \Delta[1]^{\triangleright}
}
\,.
$$
For evaluating $P(0,p)$ we just need the fiber over $\{0\}^{\triangleright}$, hence the pullback of the diagram
$$
\array{
&& P
\\
&& \downarrow
\\
\{0\}^{\triangleright} &\hookrightarrow& \Delta[1]^{\triangleright}
}
\,.
$$
Now, since colimits commute with pullbacks in $sSet$, this pullback is the pushout of the corresponding pullbacks of $X$, and $X^{\triangleright}$. But that pullback of $X$ is $X \times\_{\Delta[1]} \Delta[0]$. Because you can compute it as this consecutive pullback:
$$
\array{
X \times\_{\Delta[1]} \{0\} &\to& X
\\
\downarrow && \downarrow
\\
\{0\} &\to& \Delta[1]
\\
\downarrow && \downarrow
\\
\{0\}^{\triangleright} &\to & P
}
$$
| 0 | https://mathoverflow.net/users/1353 | 44763 | 28,416 |
https://mathoverflow.net/questions/44722 | 1 | In a discussion today on the Shafarevich-Tate group of an elliptic curve, the following structure and question came up. I will abuse many notations and be very vague about some things, but am very open to suggestions for clarification. Not to mention that some, if not all, of the following is incorrect.
A key ingredient in the rich structure of number fields is Hilbert 90, asserting that the first galois cohomology of the multiplicative group is trivial. Elliptic curves, on the other hand, have no analogy (as far as I know). The discussion was about looking for a natural object to inject an elliptic curve into, that has trivial first galois cohomology.
Given an elliptic curve, an interesting looking (well, maybe not) object is the direct limit of Weil restriction of the curve, going over all the galois extensions of the base field and the morphisms are the natural inclusions. As a $G\_K$ module the Weil restriction is the induced module $Ind\_{G\_L}^{G\_K} E$, so by Shapiro's lemma the cohomology is isomorphic to $H^1(G\_L,E)$. Applying the direct limit, we indeed see that the first cohomology of the direct limit of the Weil restriction is trivial (since $G\_L$ keeps getting smaller).
And so many questions are raised regarding this direct limit. The first one is in the title, and is my question.
>
> Is the direct limit of Weil restrictions, going over all galois extensions of the base field, of an elliptic curve a scheme?
>
>
>
| https://mathoverflow.net/users/2024 | Is the direct limit of Weil restriction of an elliptic curve a scheme? | I agree with Adam Smith that the question seems a bit misguided, but let me show anyway that the answer is negative away from certain silly cases. Well, first to make a more well-posed question, one first has to adjust the definition of the functor so that it is at least a Zariski-sheaf (ideally without changing the "value" on affines). So let's first do that. Let $E$ be an elliptic curve over a field $k$, and for the directed system of finite extensions $k\_i/k$ inside of a fixed separable closure $k\_s/k$ let $E\_i = {\rm{Res}}\_{k\_i/k}(E\_{k\_i})$ denote the corresponding abelian varieties arising by Weil restriction. Define the functor $G = \injlim E\_i$ on the category of $k$-schemes. This is an fpqc sheaf on the category of affine $k$-schemes, since we can explicitly compute it: for any $k$-algebra $R$, $G(R) = \injlim E(R\_{k\_i}) = E(R\_{k\_s})$, the final equality using that the functor of $E$ commutes with the formation of direct limits in $k$-algebras (thanks to Grothendieck's awe-inspiring necessary and sufficient functorial characterization of locally finitely presented morphisms of schemes, from EGA IV$\_3$, 8.14, which we will use in a more impressive converse direction shortly).
By inspection, the functor $G$ satisfies the fpqc sheaf axioms on affines. Thus, if we Zariski-sheafify $G$ on the category of $k$-schemes then the resulting functor $G'$ has the same value on affines and so is easily seen to be an fpqc sheaf. So the "right" question is whether $G'$ is representable. I will now prove the "expected" negative answer whenever $k$ is not separably closed or real-closed, which is to say (by the Artin-Schreier theorem) that $k\_s/k$ has infinite degree. (Obviously need to rule out the cases when $[k\_s:k]$ is finite!)
Suppose $G'$ is represented by a $k$-scheme (which we will also denote by $G'$), so by Grothendieck's functorial criterion we see that $G'$ is locally of finite type over $k$. Thus, the identity component ${G'}^0$ makes sense and it *finite type* (as for any locally finite type group scheme over a field; see SGA3, VI$\_{\rm{A}}$, 2.4). As such, it contains the connected $E\_i$ as closed $k$-subgroup schemes since any monomorphic homomorphism between finite type group schemes over a field is a closed immersion (proved in SGA3, VI$\_{\rm{B}}$, 1.4.2, for example). But $E\_i$ has dimension $[k\_i:k]$, which grows without bound since $[k\_s:k]$ is infinite by hypothesis. Hence, ${G'}^0$ contains closed subschemes of arbitrarily large dimension, an absurdity since ${G'}^0$ is finite type over a field. QED
| 7 | https://mathoverflow.net/users/3927 | 44776 | 28,421 |
https://mathoverflow.net/questions/44774 | 37 | Let $L: C^\infty(\mathbb{R}) \to C^\infty(\mathbb{R})$ be a linear operator which satisfies:
$L(1) = 0$
$L(x) = 1$
$L(f \cdot g) = f \cdot L(g) + g \cdot L(f)$
Is $L$ necessarily the derivative? Maybe if I throw in some kind of continuity assumption on $L$? If it helps you can throw the "chain rule" into the list of properties.
I can see that $L$ must send any polynomial function to it's derivative. I want to say "just approximate any function by polynomials, and pass to a limit", but I see two complications: First $\mathbb{R}$ is not compact, so such an approximation scheme is not likely to fly. Maybe convolution with smooth cutoff functions could help me here. Even if I could rig up something I am concerned that if polynomials $p\_n$ converge to $f$, I may not have $p\_n'$ converging to $f'$. My Analysis skills are really not too hot so I would like some help.
I am interested in this question because it is a slight variant of a characterization given here:
[Why do we teach calculus students the derivative as a limit?](https://mathoverflow.net/questions/40082/why-do-we-teach-calculus-students-the-derivative-as-a-limit/40147#40147)
I am not sure whether or not those properties characterize the derivative, and they are closely related to mine.
If these properties do not characterize the derivative operator, I would like to see another operator which satisfies these properties. Can you really write one down or do you need the axiom of choice? I feel that any counterexample would have to be very weird.
| https://mathoverflow.net/users/1106 | Do these properties characterize differentiation? | Yeah, these force it to be ordinary differentiation. We have to show that for each fixed $x\_0 \in \mathbb{R}$, the composite
$$C^\infty(\mathbb{R}) \stackrel{L}{\to} C^\infty(\mathbb{R}) \stackrel{ev\_{x\_0}}{\to} \mathbb{R}$$
is just the derivative at $x\_0$. For each $f \in C^\infty(\mathbb{R})$, there is a $C^\infty$ function $g$ such that
$$f(x) = f(x\_0) + f'(x\_0)(x - x\_0) + (x - x\_0)^2g(x)$$
and so $(ev\_{x\_0} L)(f) = ev\_{x\_0}(f'(x\_0) + 2(x - x\_0)g(x) + (x - x\_0)^2 L(g)(x))$ by the properties you listed. Of course evaluation at $x\_0$ kills the last two terms and one is left with $f'(x\_0)$, as desired.
| 56 | https://mathoverflow.net/users/2926 | 44778 | 28,422 |
https://mathoverflow.net/questions/44777 | 2 | I derive this question while trying to prove the monotonicity of a differentiable vector function $f(x)$ that maps from $X\subset R^n$ to $R^n$ (Here function $f(x)$ is called monotone if $(x-y)'(f(x)-f(y))\geq 0$, $\forall x,y\in X$). The domain $X$ only consists of vectors $x$ such that $1'x=0$, here $1$ is the vector of all ones.
Using the mean-value theorem, we have that $f(x)$ is locally monotone at $x$ (namely $(y-x)'(f(y)-f(x))\geq 0$, $\forall y\in X$) if its Jacobian matrix evaluated at $x$, which we label as $A$, satisfies the following condition:
$$v'Av\geq 0,\quad \forall v \text{ such that } 1'v=0.$$
This is a weaker condition than positive semidefiniteness. However, while there are a number of ways to characterize positive semidefinite matrices, for example, see [this Wikipedia page](http://en.wikipedia.org/wiki/Positive-semidefinite_matrix#Characterizations), how can I characterize the above defined matrices?
| https://mathoverflow.net/users/7595 | How to characterize real square matrices A, such that v'Av >= 0, for all real vectors v with 1'v=0 (1 is the vector of all ones)? | If $A$ is symmetric, then the matrices that you mention are called:
**Conditionally positive definite** (CPD) --- these are intimately related to the venerable *infinitely divisible matrices*
There is a vast amount of literature on these matrices, some useful pointers can already be found in R. Bhatia's wonderful book: *Positive definite matrices*
There are some basic algorithmic approaches to check whether a matrix is CPD or not (e.g., Ref. 3 below)
A simple characterization is given by the following. Let $A$ be an $n \times n$ Hermitian matrix, and let $B$ be the $(n-1) \times (n-1)$ matrix with entries
$$b\_{ij} = a\_{ij} + a\_{i+1,j+1} - a\_{i,j+1} - a\_{i+1,j}$$
Then $A$ is CPD *if and only if* $B$ is positive-definite.
**References**
1. R. Bhatia. *Positive definite matrices* (Chapter 5)
2. R. B. Bapat and T. E. S. Raghavan. Nonnegative matrices and applications (Chapter 4)
3. Kh. D. Ikramov and N. V. Savel'eva. *Conditionally positive definite matrices*, J. Mathematical Sciences, Vo. 98, No. 1, 2000.
4. R. A. Horn. The theory of infinitely divisible matrices and kernels (e.g. here : <http://www.ams.org/journals/tran/1969-136-00/S0002-9947-1969-0264736-5/S0002-9947-1969-0264736-5.pdf>)
| 8 | https://mathoverflow.net/users/8430 | 44789 | 28,430 |
https://mathoverflow.net/questions/44735 | 3 | By a Classification of Dickson everysubgroup of PSL(2,p) has index at least p+1
is there an easy proof with out this classification??
What can be said about the minimal index of subgroups PSL(r,q)??
There is a classification of subgroups of PSL(3,p) by Bloom , even for this
list i could not calculate all the indexes.
Any references are welcome
| https://mathoverflow.net/users/10551 | minimal index of proper subgroups of PSL(r,q) | The first published proof that the index of a subgroup of PSL$(2,p)$ is at least $p+1$ for primes $p \ge 13$ is in:
C. Jordan, "Note sur les equations modulaires", C.R. Acad. Sci. Paris 66 (1868), 308-312,
a long time before the classification!
The minimal indexes of subgroups of classical simple groups are determined (again pre-classification) in
B.N. Cooperstein, "Minimal degree for a permutation representation of a classical group", Israel J. Math. 30 (1978), 213-225.
There are apparently a couple of mistakes in Cooperstein's paper, but they concern $U\_n(2)$ and orthogonal groups over $F\_3$.
In particular, the minimal index of PSL$(n,q)$ is $(q^n-1)/(q-1)$ except for $(n,q)$ = (2,5), (2,7), (2,9), (2,11), or (4,2).
| 5 | https://mathoverflow.net/users/35840 | 44794 | 28,433 |
https://mathoverflow.net/questions/44738 | -1 | Is finite dual of an algebra morphism a morphism of coalgebras? Does taking finite dual preserves exactness of an exact sequence of algebra morphisms? When is this possible?
| https://mathoverflow.net/users/9492 | Finite dual of an algebra morphism. | Gee, I have not a clue what is up, doc! I have never seen an exact sequence of algebra morphisms because usual linear kernels are not subalgebras. Having said that, algebra morphisms may conceivably have kernels but you need to expand on that
BTW, the answer to the first question is yes! All it is kinda saying that $^0$ is a functor. Say $f:A->B$ is an algebra morphism. The equality $\sum\_{(\beta)} f^0(\beta\_1)\otimes f^0(\beta\_2)= \sum\_{(f^0(\beta))} f^0(\beta)\_1\otimes f^0(\beta)\_2$ can be checked on any pair of elements $a,x\in A$ where it becomes $\beta (f(a)f(x))= \beta (f(ax))$.
| 1 | https://mathoverflow.net/users/5301 | 44795 | 28,434 |
https://mathoverflow.net/questions/44582 | 11 |
>
> **Quick version of the question**. Let $(X, \mu)$ be a probability measure space and let $Z$, the group of integers, act on $X$ in a measure preserving way. How can I decompose $X$ into ergodic componenets? More precisely, can $X$ be equivariantly decomposed into a countable union of subspaces $U\_i$, each of which is isomorphic to a product $A\_i\times B\_i$, such that action on $U\_i$ is a product of an ergodic action on $A\_i$ and the trivial action on $B\_i$?
>
>
>
One can ask the same question also for groups other than integers.
---
My motivation
=============
I'm currently learning basics of ergodic theory. More precisely, I'm interested in the notion of *cost*. Let me recall it for group actions: Let a countable discrete group $G$ act on a probability measure space $(X,\mu)$ in a free and probability measure preserving (pmp) manner. Call the action $\rho$. Let $\mathcal R(\rho)$ be the equivalence realtion on $X$ given by $\rho$ (i.e. two points of $X$ are equivalent iff there's a group element which sends one point to the other). Let $F=(U\_i,g\_i)\_{i=1}^\infty$ be a countable family of pairs, where each $U\_i$ is a measurable set, and each $g\_i$ is an element of $G$. Let $\mathcal R(F)$ be the equivalence relation on $X$ generated by the relation $x \sim y$ iff for some $i$ we have $x\in U\_i$ and $\rho(g\_i)(x)=y$. Define
$$
cost(F) = \sum \mu(U\_i),
$$
and let cost of the action $\rho$ be the infimum of numbers $cost(F)$ over all families $F$ such that $\mathcal R(F) = \mathcal R(\rho)$, perhaps after restricting both relations to subsets of measure $1$.
>
> **Theorem.** Let $\rho$ be a free pmp action of $\mathbb Z \times H$, where $H$ is any countable group. Then $cost(\rho)=1$
>
>
>
Suppose first that restriction of the action $\rho$ to $\mathbb Z$ is ergodic. Fix $\varepsilon$. Then for the family $F$ choose pairs $(X, t), (A\_1, h\_1), (A\_2,h\_2) \ldots $, where $t$ is a generator of $\mathbb Z$, $h\_i$ is an enumeration of elements of $H$, and $A\_i$ is any set such that $\mu(A\_i)= \frac{\varepsilon}{2^i}$.
Clearly $cost(F) \le 1 + \varepsilon$, so it's enough to see that $\mathcal R(F) = \mathcal R(\rho)$. Take a point $x$ of $X$ and fix $h\_i\in H$. We're gonna show that, with probability $1$, $x$ is in relation with $\rho(h\_i)(x)$. By the ergodic theorem, since we assume action of $\mathbb Z$ is ergodic, with probability $1$ for some $j$ we have $\rho(t^j)(x)\in A\_i$, so we have $x \sim \rho(t^j)(x) \sim \rho(h\_it^j)(x) \sim \rho(h\_i)(x)$.
When I heard the argument it wasn't even mentioned that we assume that restricion to $\mathbb Z$ is ergodic. Intuitively it's clear what to do - choose $A\_i$ more cleverly, "perpendicular to ergodic components of $\mathbb Z$".
>
> **Question**. Which theorem from ergodic theory allows to make this choice of $A\_i$ "perpendicular to ergodic components" precise?
>
>
>
| https://mathoverflow.net/users/2631 | Decomposition of a dynamical system into ergodic componenents | Answer to the quick version. Yes it is true as soon as $(X,\mu)$ is a Lebesgue space. Beware that the transformation on the product $A\_i\times B\_i$ is not necessarily a true product, but instead it is a skew-product of the form $(x,y)\mapsto (T\_x(y),y)$. This follows from the ergodic decomposition theorem, together with the classification of measurable partitions.
Recall that if T is an invertible measurable transformation acting on a Lebesgue space X, then there is a measurable partition $C\_i$ (which may have uncountably many elements) and
probability measures $\mu\_i$ on $C\_i$ such that all $C\_i$ are invariant by T, T is ergodic w.r.t $\mu\_i$ and $\mu$ is obtained by integrating the $\mu\_i$.
$$\mu(A) = \int\_X \mu\_i(A) d\mu$$
There are two kinds of ergodic components $C\_i$. The one of positive measure, there are at most countably many such components. Let us remove these components from $X$, together with the periodic points, which are easily dealt with. Rohlin structure theorem on measurable partitions (1947) now says that
there is a isomorphism between $([0,1]\times [0,1], \lambda)$ and $(X,\mu)$ such that the pullback of the measurable partition $(C\_i)$ is the decomposition into horizontal lines $([0,1]\times \{i\})\_{i\in [0,1]}$. A reference is the book of Parry, "entropy generators in ergodic theory".
Here is how the ergodic decomposition is often used. If it happens that a result is true for an ergodic transform, then it is true for an arbitrary transform in restriction to its ergodic components, and you (may) recover the result on the whole space $X$ just by using the integral formula given above.
A reference for the ergodic decomposition for countable groups action is Glasner, "ergodic theory via joinings" th. 3.22.
Finally the result you are alluding in your last question is a section theorem. Given a measure preserving transform between two Lebesgue spaces X and Y, there is a section from Y to X, up to null sets, and some warning is in order here because this is not true in the Borel category. I think this is again due to Rohlin, and it can be deduced from its structure theorem for measurable partitions. Have a look at the book of Parry, but really this is overkill.
**EDIT**: following the comments of R.W., here is a counterexample to having a true product, instead of just a skew-product. On $[0,1]\times [0,1]$ take
$(x,y)\mapsto (x+y\ \ mod\ \ 1,y)$, together with Lebesgue measure. The restrictions to the fibers $[0,1]\times \{y\}$ are ergodic for a.e. y, and give uncountably many different isomorphic systems, as can be shown by looking at their spectra.
| 4 | https://mathoverflow.net/users/6129 | 44806 | 28,440 |
https://mathoverflow.net/questions/44801 | 34 | So I was having tea with a colleague immensely more talented than myself and we were discussing his teaching algebraic number theory. He told me that he had given a few examples of abelian and solvable extensions unramified everywhere for his students to play with and that he had find this easy to construct with class field theory in the back of his head. But then he asked me if I knew how to construct an extension of number fields with Galois group $A\_{5}$ and unramified everywhere. All I could say at the time (and now) is:
1. There are Hilbert modular forms unramified everywhere.
2. There are Hilbert modular forms whose residual $G\_{{F}\_{v}}$-representation mod $p$ is trivial for all $v|p$.
3. There are Hilbert modular forms whose residual $G\_{F}$-representation mod $p$ has image $A\_{5}$ inside $\operatorname{GL}\_{2}(\mathbb F\_{p})$.
Suppose there is a Hilbert modular form satisfying all three conditions. Then the Galois extension through which its residual $G\_{F}$-representation factors would have Galois group $A\_{5}$ and would be unramified everywhere.
Can this be made to work?
Regardless of the validity of this circle of idea, can you construct an extension of number fields unramified everywhere and with Galois group $A\_{5}$?
| https://mathoverflow.net/users/2284 | $A_5$-extension of number fields unramified everywhere | If you take the splitting field of $x^5+ax+b$ and consider it as an extension of its quadratic subfield, then it will be unramified with Galois group contained in $A\_5$ whenever $4a$ and $5b$ are relatively prime. This is a result of [Yamamoto](http://www.ams.org/mathscinet-getitem?mr=266898). For almost all $a$ and $b$ (specifically, on the complement of a [thin set](http://en.wikipedia.org/wiki/Thin_set_%2528Serre%2529)), the group is $A\_5$.
You might also enjoy this [preprint](http://math.mit.edu/~kedlaya/papers/unramified.ps.gz) of Kedlaya, which I found very readable. A note on Kedlaya's webpage, dated May 2003, says that he will not be publishing this because it has been superseded by a recent result of Ellenberg and Venkatesh. I assume he is referring to [this paper](http://arxiv.org/abs/math.NT/0309153), but I can't figure out why that one supersedes his.
| 24 | https://mathoverflow.net/users/297 | 44812 | 28,444 |
https://mathoverflow.net/questions/44804 | 5 | Consider the following system of equations:
$$
\sum\_{i=1}^{2n}a\_i=0
$$
$$
\sum\_{i=1}^{2n}\frac{1}{a\_i}=0
$$
Where for each $i$ $a\_i$ is an odd integer and the $a\_i$ are not necessarly distinct. A solution $(a\_1,\dots,a\_{2n})$ is trivial
if (after some permutation of the coefficients) for each $i$ we have
$$a\_i=-a\_{n+i}$$.
I know that if $n>2$ there exist non trivial solutions. My questions are:
* What is the minimum number of variables for which there exist non trivial solutions ?
* Can you exhibit a minimal solution or at least a solution you think could be minimal ?
| https://mathoverflow.net/users/5001 | A Diophantine problem related to egyptian fractions | Well, -1,3,3,5,5,-15 comes to mind. This is 2n variables for n=3. But you said you know there are non-trivial solutions for n>2. Did you mean "for every n>2" ? If so, what are you asking? Also, why not consider the case of an odd number of integers (some of which would be even)?
| 4 | https://mathoverflow.net/users/8008 | 44820 | 28,447 |
https://mathoverflow.net/questions/44684 | 3 | Consider the following counting problem (or the associated decision problem): Given two positive integers encoded in binary, compute their greatest common divisor (gcd). What is the smallest complexity class this problem is contained in? Can you provide a reference?
In this question I am not primarily interested in asymptotic bounds on the running time, but rather in complexity classes. Is the problem in $AC$? In $AC^1$? Can it be proven not to lie in $AC^0$? What are other complexity classes inside $P$ that are of relevance here?
| https://mathoverflow.net/users/10539 | complexity of greatest common divisor (gcd) | I cross-posted this question on [stackexchange](https://cstheory.stackexchange.com/questions/2708/complexity-of-greatest-common-divisor-gcd%20%22here%22) and John Watrous posted an answer. The gist was that it is not known whether gcd is in NC or P-complete. See, e.g., "J. Sorenson. Two fast GCD algorithms. Journal of Algorithms, 1994."
| 2 | https://mathoverflow.net/users/10539 | 44824 | 28,449 |
https://mathoverflow.net/questions/44823 | 3 | Let $f:E\to F$ be a morphism of vector bundles on an irreducible algebraic variety $X$. Does anybody know any results about the irreducibility or smoothness of the degeneracy locus of $f$? I know only the Connectedness Theorem due to Fulton.
| https://mathoverflow.net/users/33841 | Irreducibility\smoothness of the degeneracy locus | The degeneracy locus can be reducible, and even non-reduced. For instance, take $X= \mathbb{P}^2$ and consider a morphism
$\mathcal{O}(-1)^2 \stackrel{f} \to \mathcal{O}^2$.
$f$ is given by a $2 \times 2$ matrix of linear forms, so its degeneracy locus is a conic. For a general choice of $f$ this conic will be smooth, but for special choice of the matrix it can become singular or even a double line.
The best result I am aware of can be found in Ottaviani's book "Varieta' proiettive di codimensione piccola" [projective varieties of small codimension, unfortunately I do not think an english translation is available].
Set
$D\_k(f):=\{x \in X \; | \; \textrm{rank}(f\_x) \leq k \}$
Then we have the following
**Theorem (of Bertini's type)**
Set $\textrm{rank}(E)=m$, $\textrm{rank}(F)=n$. Assume that $E^{\*} \otimes F$ is globally generated. Then for the generic morphism $f \colon E \to F$, the locus $D\_k(f)$ is either empty or it has the expected codimension $(m-k)(n-k)$, and the singular locus of $D\_k(f)$ is contained in $D\_{k-1}(f)$.
In particular, if
$\dim X < (m-k+1)(n-k+1)$
then $D\_k(f)$ is smooth for a general choice of $f$.
| 9 | https://mathoverflow.net/users/7460 | 44830 | 28,452 |
https://mathoverflow.net/questions/44831 | 2 | Given two smooth projective schemes $X$ and $Y$ over some algebraically closed fields $k$, one has the product $X\times Y$ with the projections $\pi\_X$ and $\pi\_Y$.
Now i have a coherent sheaf $M$ on $X$ and a coherent sheaf $N$ on $Y$, and i have locally free resolutions $M\_{\\*}\rightarrow M$ of length m and $N\_{\\*}\rightarrow N$ of length n for cohomological computations. But i want to work with $\pi\_X^{\\*}M$ and $\pi\_Y^{\\*}N$ on $X\times Y$. The easiest way would be if i could pullback the resolutions too, so i don't have to get new resolutions on the product. But as the pullback is just right exact in general, one needs flatness of the projections for this.
So i tried to see if the projectiosn are flat. But for example $\pi\_X$ is given by the base change of $Y\rightarrow Spec(k)$ by $X\rightarrow Spec(k)$. Since $Y\rightarrow Spec(k)$ is flat, and flatness is preserved under base change one has that $\pi\_X$ is flat. Could it be that easy?
| https://mathoverflow.net/users/3233 | Flatness of the canonical projections | Yes. If $f: X\to S$ is a flat morphism and $S'\to S$ is an arbitrary morphism, then the projection $f\_{S'}: X\times\_S S'\to S'$ is flat. A reference is EGA IV.2.1.4.
| 2 | https://mathoverflow.net/users/8680 | 44836 | 28,454 |
https://mathoverflow.net/questions/44798 | 8 | Assume the topological group $\mathbb{R}$ acts properly on a space $X$. Does then the projection map $p:X\rightarrow \mathbb{R}\backslash X$ have local sections ?
(for every $\mathbb{R}x\in \mathbb{R}\backslash X$, there is a open neighbourhood $U \subset \mathbb{R}\backslash X$) and a section of $p|\_{p^{-1}(U)}:p^{-1}(U)\rightarrow U$).
Are there any nice conditions for $X$, that imply the existence of local sections ?
| https://mathoverflow.net/users/3969 | local structure of free $\mathbb{R}$ actions | Such a theorem is proved for completely regular spaces $X$ in my article:
On the Existence of Slices for Actions of Non-Compact Lie Groups, Richard S. Palais, The Annals of Mathematics, Second Series, Vol. 73, No. 2 (Mar., 1961), pp. 295-323 .
Actually, that paper considers more general groups than just $\mathbb{R}$ (any locally compact group) and for actions a little more general than proper (what I call Cartan G-Spaces). The paper is available from JSTOR. (Note that what I prove under these circumstances is the existence of a slice. But since for a proper action the isotropy group at any point is compact, for the case of $\mathbb{R}$ this means that all isotropy groups are trivial, so the action is automatically free, and a slice is a local section.)
| 10 | https://mathoverflow.net/users/7311 | 44842 | 28,456 |
https://mathoverflow.net/questions/44845 | 8 | Let $X$ and $Y$ be projective varieties. I am assuming that there is some construction of a "moduli space" parametrizing the morphisms $X \to Y$. For instance, if one identifies such morphisms with their graphs in $X \times Y$, one can at least hope that these graphs would correspond to a locally closed subscheme of the Chow or Hilbert schemes for $X \times Y$, although this may not be the best way to construct such a space. In the case of $X = Y = \mathbb{P}^n$, this space should probably be a disjoint union of $PGL\_n$ with $\mathbb{P}^n$ [**Edit:**This is not correct; I know a theorem that the image must be either a point or all $\mathbb{P}^n$, but grossly misapplied it by neglecting maps of degree > 1]. (I am requiring the varieties to be projective because, for instance, if $Y$ is the affine line, then the morphisms $X$ to $Y$ would naturally correspond to the global sections of $X$, which for general $X$ is too big to fit in one nice scheme, being infinite-dimensional. Although I suppose it might work if we confined ourselves to e.g. morphisms of a fixed degree, and we are already doing something of this sort in looking at Chow or Hilbert schemes, so perhaps my concern here is needless.)
Is there a standard space that represents these morphisms, and if so, what is known about it? Is there a good reference for this?
| https://mathoverflow.net/users/5094 | What is known about the "moduli space of morphisms" $X \to Y$? | It is called the Hom scheme, and I think it's defined in Grothendieck's [Bourbaki 221 paper](http://www.numdam.org/item?id=SB_1960-1961__6__249_0), where he constructs the Hilbert and Quot schemes. For any $S$-schemes $X$ and $Y$, $\underline{Hom}\_S(X,Y)$ assigns to any $S$-scheme $T$ the set $Hom\_T(X\_T,Y\_T)$. Osserman has a quick overview of the constructions, which you can find with Google. Olsson's paper "Hom stacks and restriction of scalars" is one possible reference that covers a more general setting.
| 9 | https://mathoverflow.net/users/121 | 44847 | 28,458 |
https://mathoverflow.net/questions/44833 | 5 | Let $S$ be a semigroup. If $S$ is abelian, then it follows that the semigroup algebra $k[S]$ is finitely generated if and only if $S$ is.
What if we relax the condition on $k[S]$, so that $k[S]$ is only noetherian. Does it in this case follow that $S$ is finitely generated?
| https://mathoverflow.net/users/3996 | If $k[S]$ is noetherian, is S finitely generated? | It is an open problem (or was, last time I checked!) whether the noetherianity of $k[S]$ implies finite generation of $S$, when $S$ is not abelian.
This is discussed in chapter 5 of *Noetherian semigroup algebras* by Eric Jespers and Jan Okniński, along with various cases where we know that $S$ is finitely generated. They prove, for example, that this is so if $k[S]$ satisfies a polynomial identity, and this gives the case in which $S$ is abelian as a corollary.
| 9 | https://mathoverflow.net/users/1409 | 44870 | 28,469 |
https://mathoverflow.net/questions/44859 | 25 | It seems like when we assume "niceness" in homotopy theory we assume that $X$ has the homotopy type of a CW complex, and in fiber bundle theory we assume that $X$ is paracompact. How do these two interact? Is any space with the homotopy type of a CW complex paracompact? (In particular, is $I^I$ paracompact?)
(CW complexes are always paracompact and Hausdorff. According to Milnor (<http://www.jstor.org/stable/1993204>) a paracompact space that is "equi locally convex" will have the homotopy type of a CW complex. Also according to that paper, if $X$ has the homotopy type of a CW complex and $K$ is actually a finite complex then $X^K$ has the homotopy type of a CW complex.)
| https://mathoverflow.net/users/1874 | CW complexes and paracompactness | $I^I$ is paracompact. It is a theorem of [O'Meara](http://www.jstor.org/pss/2037695) that for $X$ a separable metric space, and $Y$ a metric space, then $Y^X$ - with the compact-open topology - is paracompact. $I$ is certainly a separable metric space, so the result holds.
As to your first question, I doubt that every space of the homotopy type of a CW-complex is paracompact (but this is intuition only). Something like $\mathbb{R}^{\aleph\_2}$ or a similarly large-dimensional, non-metrizable topological vector space might do the trick, as it is contractible, hence the homotopy type of a CW-complex.
Edit: For any uncountable index set $J$, consider $\mathbb{R}^J$ in the product topology. It isn't normal, so isn't paracompact.
| 10 | https://mathoverflow.net/users/4177 | 44872 | 28,470 |
https://mathoverflow.net/questions/44827 | 2 | I know my question is very imprecise. I am trying to understand Tate-Farrell cohomology of the infinite Lie group $S^1$ (say, with coefficients in $\mathbb C$). I would expect that the answer is something like the space of Laurent polynomials $\mathbb C[t^{-1},t]$. Is there any geometric intuition for this? What would be the meaning of multiplication by $t$? What is the meaning of the completion $\mathbb C((t))$?
| https://mathoverflow.net/users/6772 | Tate-Farrell cohomology of a circle | **Comment:** Farrell-Tate cohomology as defined in Brown's book "Cohomology of Groups"
requires the group to be of finite virtual cohomological dimension (i.e. the group has a finite index subgroup which has a finite projective resolution). But $S^1$ doesn't have finite virtual cohomological dimension because it has finite subgroups of arbitrary order.
There is a generalization of Farrell-Tate cohomology for arbitrary groups due to Benson-Carlson/Mislin, that is usually called "complete cohomoloy" or "complete Tate cohomology". I don't know if that cohomology has been computed for $S^1$ (most computations are done for discrete groups).
| 3 | https://mathoverflow.net/users/10194 | 44875 | 28,473 |
https://mathoverflow.net/questions/44877 | 12 | Suppose we are talking about graphs with $n$ labeled vertices. Which graphs are more common: connected or disconnected?
| https://mathoverflow.net/users/979 | Are there more connected or disconnected graphs on $n$ vertices? | Connectedness wins, since the complement of any disconnected graph is connected.
EDIT: Perhaps you'd like a proof of this. Let G be a disconnected graph, G' its complement. If v and u are in different components of G, then certainly they're connected by an edge in G'. And if they're in the same component of G, then there's some w in another component (since G was disconnected), so v-w-u is a path in G'.
| 63 | https://mathoverflow.net/users/1060 | 44889 | 28,481 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.