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/20423 | 4 | Given a Ferrers board of shape $(b\_1,\ldots,b\_m)$, we define $r\_k$ as number of ways to place $k$ non-attacking rooks (as in Chess). In section 2.4 of Stanley's Enumerative Combinatorics (vol. 1) it's shown the identity:
$$\sum\_k r\_k (x)\_{m-k} = \prod\_i (x+s\_i)$$
where $s\_i = b\_i-i+1$, but I don't know if I can invert this formula or make an efficient algorithm to compute the $r\_k$'s.
If this isn't possible, I would be satisfied if I can compute them efficiently in the following shapes:
$(2,2,4,4,\ldots,2n-2,2n-2,2n)$
$(2,2,4,4,\ldots,2n,2n)$
$(1,1,3,3,\ldots,2n-1,2n-1,2n+1)$
| https://mathoverflow.net/users/961 | How to compute the rook polynomial of a Ferrers board? | You can retrieve the coefficients of a polynomial written in the falling factorial basis by computing finite differences, as follows.
Let $f : \mathbb{Z} \to \mathbb{Z}$ be a function, and let $\Delta f(n) = f(n+1) - f(n)$. Let $\Delta^{r+1} f = \Delta(\Delta^r f)$.
**Lemma 1:** $\displaystyle \Delta {n \choose k} = {n \choose k-1}$.
**Corollary:** If $\displaystyle f(n) = \sum\_{i=0}^d a\_i {n \choose i}$, then $\Delta^i f(0) = a\_i$.
**Lemma 2:** $\displaystyle \Delta^i f(0) = \sum\_{j=0}^{i} (-1)^{i-j} {i \choose j} f(j)$.
Note the similarity to how the Taylor coefficients of a polynomial in the usual basis are extracted, and note that $\displaystyle {n \choose k} = (n)\_k k!$. For the sake of having a final answer, this gives
$$r\_k = \frac{1}{(m-k)!} \sum\_{j=0}^{m-k} (-1)^{m-k-j} {m-k \choose j} \prod\_i (j + s\_i).$$
| 6 | https://mathoverflow.net/users/290 | 20464 | 13,593 |
https://mathoverflow.net/questions/20444 | 11 | Let $K$ be a field and $n \geq 1$. Then the set of isomorphism classes of vector bundles over $\mathbb{P}^n\_K$ is a semiring (i.e. almost a ring, but no additive inverses are possible). By introducing additive inverses and quotienting out exact sequences, we get the $K$-theory of $\mathbb{P}^n\_K$, which is known to be $\mathbb{Z}^{n+1}$. But is it also possible to compute exactly the semiring?
For $n=1$, there is a result by Dedekind-Weber (1892) which proves that the semiring is $\mathbb{N}[x,x^{-1}]$, where $x=\mathcal{O}(1)$ ([related topic](https://mathoverflow.net/questions/16434/using-linear-algebra-to-classify-vector-bundles-over-p1)). Some months ago, I was told that the structure is far more complicated for $n>1$. Can anybody elaborate this or even give a presentation of the semiring?
If necessary, you may assume $K = \mathbb{C}$.
| https://mathoverflow.net/users/2841 | Semiring of algebraic vector bundles on projective space | This semiring carries an enourmous amount of information about vector bundles on $\mathbb{P}^n$, including stuff we don't yet know. For example, you can read from it whether there are indecomposable vector bundles of any given rank; and for small rank we know very little about it (this is discussed, for example, in C. Okonek, M. Schneider, H. Spindler, "Vector bundles on complex projective spaces" , Birkhäuser (1987); I don't have access to the book here, and can't give you a more precise reference). I doubt you can can even get close to what you want.
| 10 | https://mathoverflow.net/users/4790 | 20470 | 13,597 |
https://mathoverflow.net/questions/20442 | 13 | A Serre fibration has the homotopy lifting property with respect to the maps $[0,1]^n \times \{0\} \to [0,1]^{n+1}$. A Dold fibration $E \to B$ has the weak covering homotopy property: lifts with respect to maps $Y\times \{0\} \to Y \times [0,1]$ such that the lift agrees with the map $Y \to E$ up to a vertical homotopy (see the [nLab page for more details](http://ncatlab.org/nlab/show/Dold+fibration). All Hurewicz fibrations are Dold fibrations, but not conversely, and not all Dold fibrations are Serre fibrations. I'm sure I read that not all Serre fibrations are Dold fibrations, but I don't have a counterexample.
My request is thus: an example of a Serre fibration that is not a Dold fibration.
Edit: I have found that a slight variant on this question was asked by Ronnie Brown in Proc. Camb. Phil.Soc. in October 1966, under the caveat that the base is path-connected and the base and the fibre have the homotopy type of a CW complex.
| https://mathoverflow.net/users/4177 | Request: A Serre fibration that is not a Dold fibration | (answering my own question - who would have thought?)
There is a paper by G Allaud (Arch. Math 1968) which describes a counterexample as sought by the question. Let $E$ be the subspace of the plane $\mathbb{R}^2$ consisting of the non-negative integer points $(n,0)$ on the $x$-axis together with $(0,1)$ and a line connecting it each point on the $x$-axis. Let $B$ be the subspace of the plane consisting of the origin and the points $(1/n,0)$ on the $x$-axis for positive $n$ together with $(0,1)$ and a line connecting it to each point $(1/n,0)$ and $(0,0)$. The map $E \to B$ is given by sending $(0,0)$ to itself, $(n,0)$ to $(1/n,0)$ and the obvious map on the line segments. This is then (according to Allaud) a Serre fibration which is not a Dold fibration.
| 9 | https://mathoverflow.net/users/4177 | 20479 | 13,602 |
https://mathoverflow.net/questions/20478 | 3 | What is known about the matrix factorization categories of singularities of type ADE? Any references on this would be greatly appreciated.
Background: For ADE singularities, see for example [this](http://www.mathematik.uni-kl.de/~zca/Reports_on_ca/29/paper_html/node10.html). For matrix factorizations, see for example [this](https://mathoverflow.net/questions/9733/matrix-factorizations-and-physics/).
| https://mathoverflow.net/users/83 | Matrix factorization categories for ADE singularities | See: "Matrix Factorizations and Representations of Quivers II: type ADE case" (math/0511155) by Kajiura, Saito, and Takahashi for a recent account.
Older references include:
"Construction geometrique de la correspondance de McKay" Gonzalez-Sprinberg,and Verdier (1983)
Y. Yoshino, Cohen-Macaulay modules over Cohen-Macaulay rings (1990)
| 2 | https://mathoverflow.net/users/874 | 20481 | 13,603 |
https://mathoverflow.net/questions/20471 | 41 | Why is it that every nontrivial word in a free group (it's easy to reduce to the case of, say, two generators) has a nontrivial image in some finite group? Equivalently, why is the natural map from a group to its profinite completion injective if the group is free?
Apparently, this follows from a result of Malcev's that finitely generated matrix groups over an arbitrary commutative ring are residually finite, but is there a more easily accessible proof if we only want the result for free groups?
| https://mathoverflow.net/users/1474 | Why are free groups residually finite? | Here is a direct proof for free groups.
Let $x\_1,\dots,x\_m$ be the generators of our group. Consider a word $x\_{i\_n}^{e\_n}\dots x\_{i\_2}^{e\_2}x\_{i\_1}^{e\_1}$ where $e\_i\in\{\pm 1\}$ and there are no cancellations (that is, $e\_k=e\_{k+1}$ if $i\_k=i\_{k+1}$).
I'm going to map this word to a nontrivial element of $S\_{n+1}$, the group of permutations of $M:=\{1,\dots,n+1\}$. It suffices to construct permutations $f\_1,\dots,f\_m\in S\_{n+1}$ such that $f\_{i\_n}^{e\_n}\dots f\_{i\_2}^{e\_2}f\_{i\_1}^{e\_1}\ne id\_M$. For each $k=1,\dots,n$, assign $f\_{i\_k}(k)=k+1$ if $e\_k=1$, or $f\_{i\_k}(k+1)=k$ if $e\_k=-1$. This gives us injective maps $f\_1,\dots,f\_m$ defined on subsets of $M$. Assign yet unassigned values of $f\_i$'s arbitrarily (the only requirement is that they are bijections). The resulting permutations satisfy $f\_{i\_n}^{e\_n}\dots f\_{i\_2}^{e\_2}f\_{i\_1}^{e\_1}(1)=n+1$.
**Edit**: As pointed out by Steve D in comments, this proof can be found in a book by Daniel E. Cohen, "Combinatorial group theory: a topological approach" (1989). The book can be found on the net if you are determined; the proof in on page 7 and in Proposition 5 on page 11.
---
Edit [DZ]: I have a hard time reading multiple subscripts, so here is an example of Sergei Ivanov's construction.
Take the word $cca^{-1}bc^{-1}a$. This has length $6$, so we will find a homomorphism to $S\_7$ whose image of this word is nontrivial because it sends $1$ to $7$. We'll choose values of permutations so that the $k$th suffix sends $1$ to $k+1$:
Suffix 1: $a$
$\alpha=\bigg(\begin{array}{} 1&2 &3 &4& 5& 6& 7 \\\ 2&?&?&?&?&?&? \end{array} \bigg)$
Suffix 2: $c^{-1}a$
$\gamma=\bigg(\begin{array}{} 1&2 &3 &4& 5& 6& 7 \\\ ?&?&2&?&?&?&? \end{array} \bigg)$
Suffix 3: $bc^{-1}a$
$\beta=\bigg(\begin{array}{} 1&2 &3 &4& 5& 6& 7 \\\ ?&?&4&?&?&?&? \end{array} \bigg)$
Suffix 4: $a^{-1}bc^{-1}a$
$\alpha=\bigg(\begin{array}{} 1&2 &3 &4& 5& 6& 7 \\\ 2&?&?&?&4&?&? \end{array} \bigg)$
Suffix 5: $ca^{-1}bc^{-1}a$
$\gamma=\bigg(\begin{array}{} 1&2 &3 &4& 5& 6& 7 \\\ ?&?&2&?&6&?&? \end{array} \bigg)$
Suffix 6: $cca^{-1}bc^{-1}a$
$\gamma=\bigg(\begin{array}{} 1&2 &3 &4& 5& 6& 7 \\\ ?&?&2&?&6&7&? \end{array} \bigg)$
These conditions on $\alpha, \beta,$ and $\gamma$ don't conflict, they can be extended to permutations, and then $\gamma\gamma\alpha^{-1}\beta\gamma^{-1}\alpha(1) = 7$.
| 42 | https://mathoverflow.net/users/4354 | 20485 | 13,606 |
https://mathoverflow.net/questions/20493 | 209 | Hi,
given a connection on the tangent space of a manifold, one can define its torsion:
$$T(X,Y):=\triangledown\_X Y - \triangledown\_Y X - [X,Y]$$
What is the geometric picture behind this definition—what does torsion measure intuitively?
| https://mathoverflow.net/users/2837 | What is torsion in differential geometry intuitively? | The torsion is a notoriously slippery concept. Personally I think the best way to understand it is to generalize past the place people first learn about torsion, which is usually in the context of Riemannian manifolds. Then you can see that the torsion can be understood as a sort of obstruction to integrability. Let me explain a little bit first.
The torsion really makes sense in the context of general [G-structures](http://en.wikipedia.org/wiki/G-structure). Here $G \subseteq GL\_n(\mathbb{R}) = GL(V)$ is some fixed Lie group. Typical examples are $G = O(n)$ and $G = GL\_n(\mathbb{C})$. We'll see that these will correspond to Riemannian metrics and complex structures respectively. Now given this data, we have an exact sequence of vector spaces,
$$0 \to K \to \mathfrak{g} \otimes V^\ast \stackrel{\sigma}{\to} V \otimes\wedge^2 V^\ast \to C \to 0 $$
Here $\sigma$ is the inclusion $\mathfrak{g} \subseteq V \otimes V^\ast$ together with anti-symmetrization. K and C are the kernel and cokernel of $\sigma$.
If we are given a manifold with $G$-structure, we then get four associated bundles, which fit into an exact sequence:
$$ 0 \to \rho\_1P \to ad(P) \otimes T^\*M \to \rho\_3P \to \rho\_4P \to 0$$
Now the difference of two connections which are both compatible with the G-structure is a tensor which is a section of the second space $\rho\_2P = ad(P) \otimes T^\*M$. This means that we can write any connection as $$\nabla + A$$
where $A$ is a section of $\rho\_2(P)$.
Now the torsion of any G-compatible connection is a section of this third space. Suppose that we have two compatible connections. Then their torsions are sections of this third space. However since we can write the connections as $\nabla$ and $\nabla + A$, the torsion differ by $\sigma(A)$. Thus they have the same image in the fourth space $\rho\_4(P)$.
The section of this fourth space is the *intrinsic torsion* of the G-structure. It measures the failure of our ability to find a torsion free connection. If this obstruction vanishes, then the torsion free connections form a torsor over sections of the smaller bundle $\rho\_1P$. Now some examples:
1. $G = O(n)$. This is the case of a Riemannian structure. In this case $\sigma$ is an isomorphism so that the there is always a unique torsion free connection. The Levi-Civita connection.
2. $G = GL\_m(\mathbb{C})$. This is the case of a complex structure. More precisely a $GL\_m(\mathbb{C})$-structure is the same as an almost complex structure. In this case the intrinsic torsion can be identified with the Nijenhuis tensor. So it vanishes precisely when the almost-complex structure is integrable (i.e. a ordinary complex structure).
3. $G = Sp(n)$. Having an $Sp(n)$-structure on a manifold for which the intrinsic torsion vanishes is equivalent to having a symplectic manifold.
From these examples you can see that the vanishing of torsion can be viewed as a sort of integrability condition. In these latter two cases the space of torsion free connections consists of more then a single point. There are many such connections. That's one reason why we don't see them popping up more often.
| 137 | https://mathoverflow.net/users/184 | 20506 | 13,621 |
https://mathoverflow.net/questions/20503 | 3 | Here is an exercise from Serre's "local fiels" when he starts to do cohomology: Let G act on an abelian group A, f be an inhomogenous n cochain, i.e. $f\in C^n(G,A).$ Define an operator T on f, $Tf(g\_1,g\_2,\cdots,g\_n)=g\_1g\_2\ldots g\_n f(g\_n^{-1},g\_{n-1}^{-1},\ldots,g\_1^{-1})$. It is clear that $T^2f=f$. It is also not too hard to show $T(df)=(-1)^{n+1}d(Tf)$. Thus f is a cocycle iff Tf is, and f is a coboundary iff Tf is. When n=1, it is straightforward to see -f is cohomologous to Tf.
Then the exercise wants us to show when n= 0,3 mod 4, f is cohomologous to Tf,
while when n=1,2 mod 4, Tf is cohomologous to -f.
Any idea will be appreciated.
| https://mathoverflow.net/users/1877 | An exercise in group cohomology | Fix the signs as Wilberd suggests in the comments, check that you get a natural automorphism of the complex computing cohomology, see that it induces in fact an automorphism of the universal $\delta$-functor $H^\bullet(G,\mathord-)$, and see what it does in degree zero.
| 4 | https://mathoverflow.net/users/1409 | 20509 | 13,623 |
https://mathoverflow.net/questions/20512 | 2 | Let A be a borelian set with postivie measure. I was asking myself if it is possible to find an open set $B\subseteq A$ such that $B$ is an open set minus a set of null measure...
| https://mathoverflow.net/users/4928 | If a Borelian set has positive measure, does it contain a non empty open set (minus a measure null set)? | The Cantor set of positive measure is nowhere dense set. So it is an example.
| 11 | https://mathoverflow.net/users/2823 | 20513 | 13,625 |
https://mathoverflow.net/questions/20516 | 12 | I'm having trouble distinguishing the various sorts of tori.
One definition of torus is the algebraic torus. Groups like $SU(2,\mathbb{C})$ and $SU(3,\mathbb{C})$ have important subgroups that are topologically a circle and a torus, and I guess those were some of the most important Lie groups so the name torus stuck. Groups like $SL(2,\mathbb{C})$ and $SL(n+1,\mathbb{C})$ have a similar important subgroup isomorphic to $\mathbb{C}^\ast$ and $(\mathbb{C}^\ast)^n$, so the name torus gets applied to them too. In general, one calls the multiplicative group of an arbitrary field a torus in many situations, sometimes denoting the entire lot of them as $\mathbb{G}\_m$.
Another definition of a topological torus is a direct product of circles. A standard way to construct various flat geometries on a torus is to take $\mathbb{R}^n$ and quotient out by a discrete rank $n$ lattice $\Lambda$, for instance $\mathbb{R}/\mathbb{Z}$ or $\mathbb{C}/\mathbb{Z}[i]$. A complex torus is defined analogously as $\mathbb{C}^n/\Lambda$ where $\Lambda$ is a rank $2n$ lattice (since $\mathbb{C}^n$ has real rank $2n$).
One reads in various places that every abelian variety is a complex torus, but not every complex torus is an abelian variety. The notation $\mathbb{C}^n/\Lambda$ is usually nearby.
>
> Is the multiplicative group of the field, $\mathbb{G}\_m$ or $\mathbb{C}^\ast$, an abelian variety?
>
>
>
In other words, is an algebraic torus over the complexes a complex torus?
>
> Is an abelian variety isomorphic as a group to $\mathbb{C}^n/\Lambda$, or just topologically?
>
>
>
My dim memory of elliptic curves was that they were finitely generated abelian groups, but since they are uncountable that doesn't make any sense. Presumably I am thinking of their rational points. However, $\mathbb{C}^n/\Lambda$ is always an abelian group, so I don't see what the fuss is about deciding when it is an abelian variety. It seems likely to me the group operations are different.
| https://mathoverflow.net/users/3710 | Complex torus, C^n/Λ versus (C*)^n | The difference between an Abelian variety and $\mathbb{C}^n/\Lambda$ is that an abelian variety is *polarized*; that is, it comes with an ample line bundle, which yields an embedding into $\mathbb{P}^m$ for some $m$.
That is, Abelian varieties are projective algebraic, whereas complex tori (in the sense of $\mathbb{C}^n/\Lambda$) are not necessarily.
The fact that we also call $\mathbb{C}^\*$ a torus is, to the best of my knowledge, unrelated. It is not an Abelian variety.
| 13 | https://mathoverflow.net/users/1703 | 20519 | 13,627 |
https://mathoverflow.net/questions/20511 | 16 | In chapter 13 in [his notes on 3-manifolds](http://www.msri.org/communications/books/gt3m/PDF/13.pdf), Thurston defines the orbifold fundamental group to be the group of deck transformations of the universal cover of the orbifold. He also makes a statement "Later we shall interpret $\pi\_1(O)$ in terms of loops on $O$, but this interpretation doesn't seem to appear in his notes.
My question is, well, what is this interpretation, precisely?
Here are my thoughts so far:
The example I'm currently interested in is the 1-D orbifold S1/ℤ2. Its universal cover is ℝ. Deck transformations are generated by a translation by 2π, which I'll call T and a reflection about the origin R. There's relations $R^2=1$ and $TR=RT^{-1}$. If I haven't messed up, this is the same as the fundamental group of the Klein bottle as well (if someone can explain how to construct the Klein bottle from S1/ℤ2, I would greatly appreciate it as well!). How can I relate paths on S1/ℤ2 to loops in the Klein bottle? Oops, I mixed up something in my head. I'll have another question on this in the future, perhaps.
I think my main trouble is making all of these observations precise, so a good reference with standard terminology / theorems (with lots of examples like the one I've been thinking about) would be appreciated as well.
| https://mathoverflow.net/users/353 | Orbifold fundamental group in terms of loops? | Most of the standard intro sources on orbifolds discuss their fundamental groups in terms of coverings. One exception is Ratcliffe's book "Foundations of Hyperbolic Manifolds", chapter 13 of which contains a discussion of the fundamental group of an orbifold defined via loops.
| 12 | https://mathoverflow.net/users/317 | 20521 | 13,629 |
https://mathoverflow.net/questions/20507 | 10 | Let $S$ be a subset of $\mathbb{R}^n$ defined by a system $\theta$ of polynomial inequalities with integer coefficients. Let $S+\mathbb{Z}^n$ be all points of the form $s+z$ with $s \in S$ and $z \in \mathbb{Z}^n$. Is there a method known for determining, given $\theta$, whether $S+\mathbb{Z}^n=\mathbb{R}^n$? Is this problem known to be effectively undecidable?
| https://mathoverflow.net/users/5229 | A Decision Problem Concerning Diophantine Inequalities | It is undecidable. If you could solve this, you could also solve [Hilbert's 10th problem](http://en.wikipedia.org/wiki/Hilbert%2527s_tenth_problem).
Suppose we have an algorithm solving your problem for all $n$. Given a polynomial $p\in\mathbb[x\_1,\dots,x\_n]$, let's decide whether it has integer solutions. If $p$ is constant, this is trivial. Otherwise we can find $z\_0\in\mathbb Z^n$ such that $|p(x)|>1$ for all $x\in z\_0+[0,1]^n$. Let's work with a polynomial $f(x)=p(x+z\_0)$ rather than $p$. It satisfies $|f(x)|>1$ for all $x\in[0,1]^n$.
Let $g(x)=x\_1(x\_1-1)x\_2(x\_2-1)\dots x\_n(x\_n-1)$. Apply our algorithm to the inequality
$$
r(x):=(f(x)^2-1)\cdot g(x)<0 .
$$
If it says that $S+\mathbb Z^n=\mathbb R^n$, then we know that $S$ contains a point from $\mathbb Z^n$, and this point must a root of $f$. If it says that $S+\mathbb Z^n\ne\mathbb R^n$, then we know that there is $c\in\mathbb R^n$ such that $r(c+z)\ge 0$ for all $z\in\mathbb Z^n$.
This $c$ must belong to $\mathbb Z^n$. Indeed, suppose that e.g. $c\_1\notin\mathbb Z$. We may assume that all coordinates of $c$ are positive and $0<c\_1<1$. Substitute $z=(0,z\_2,\dots,z\_n)$ where $z\_2,\dots,z\_n$ are arbitrary positive integers and conclude that $|f(c+z)|\le1$ for all such $z$. It follows that $f$ is constant on the hyperplane $\{x\_1=c\_1\}$, and the modulus of this constant no greater than 1. This contradicts
the fact that $|f|>1$ on $[0,1]^n$.
Thus we know that $c\in\mathbb Z^n$, or, equivalently, that $r(z)\ge 0$ for all $z\in\mathbb Z^n$. This means that $f$ does not have integer roots except possibly at points where one of the coordinates is 0 or 1. Thus we reduced the problem to the case of $n-1$ variables and can solve it by induction.
| 9 | https://mathoverflow.net/users/4354 | 20522 | 13,630 |
https://mathoverflow.net/questions/20453 | 20 | What is the relationship between $G\_\infty$ (homotopy Gerstenhaber) and $B\_\infty$ algebras?
In Getzler & Jones "Operads, homotopy algebra, and iterated integrals for double loop spaces" (a paper I don't well understand) a $B\_\infty$ algebra is defined to be a graded vector space $V$ together with a dg-bialgebra structure on $BV = \oplus\_{i \geq 0} (V[1])^{\otimes i}$, that is a square-zero, degree one coderivation $\delta$ of the canonical coalgebra structure (stopping here, we have defined an $A\_\infty$ algebra) and an associative multiplication $m:BV \otimes BV \to BV$ that is a morphism of coalgebras and such that $\delta$ is a derivation of $m$.
A $G\_\infty$ algebra is more complicated. The $G\_\infty$ operad is a dg-operad whose underlying graded operad is free and such that its cohomology is the operad controlling Gerstenhaber algebras. I believe that the operad of chains on the little 2-discs operad is a model for the $G\_\infty$ operad. Yes?
It is now known (the famous Deligne conjecture) that the Hochschild cochain complex of an associative algebra carries the structure of a $G\_\infty$ algebra. It also carries the structure of a $B\_\infty$ algebra. Some articles discuss the $G\_\infty$ structure while others discuss the $B\_\infty$ structure. So I wonder: How are these structures related in this case? In general?
| https://mathoverflow.net/users/132 | Are G_infinity algebras B_infinity? Vice versa? | There is a nice summary of the relationship between B infinity and G infinity in the first chapter of the book "Operads in Algebra, Topology and Physics" by Markl, Stasheff and Schnider. The short answer is G infinity is the minimal model for the homology of the little disks operad (the G operad). B infinity is an operad of operations on the Hochschild complex. Many of the proofs of Deligne's conjecture involve constructing a map between these two operads.
| 9 | https://mathoverflow.net/users/4960 | 20530 | 13,635 |
https://mathoverflow.net/questions/20531 | 4 | Consider the usual language and axioms of ZF. Now add constants $x\_1, x\_2, \dots$ to the language together with the axioms $x\_2\in x\_1, x\_3\in x\_2, \dots$ to form a new theory. Then by the compactness theorem, since every finite subset of the axioms has a model, the new theory has a model. But doesn't the set {$x\_1, x\_2, \dots$} have no $\in$-minimal element, contradicting the axiom of foundation?
I'm thinking that maybe {$x\_1, x\_2, \dots$} is not necessarily a set in the model, but isn't it by replacement? Maybe not, since we don't necessarily have a copy of $\mathbb{N}$ in our model... Could someone clarify this please?
| https://mathoverflow.net/users/416 | Contradiction to axiom of foundation | You have already answered your question yourself: the set $\{x\_1, x\_2, \ldots\}$ cannot exist in the model, since this would violate the Foundation Axiom. You cannot get this set from Replacement, since there is no definable function having this set as its range, required to invoke the Replacement Axiom.
What you have done is exactly to prove that if ZF is consistent, then there are non-wellfounded models of set theory. Your proof is one way to justify nonstandard analysis.
Note that the ordinals of any model of your theory will have a non-wellfounded collection of ordinals, since the Levy rank of the sets x\_n will be decreasing.
| 11 | https://mathoverflow.net/users/1946 | 20532 | 13,636 |
https://mathoverflow.net/questions/20538 | 9 | Let $L$ be a lattice in $\mathbb{C}$ with two fundamental periods, so that $\mathbb{C}/L$ is topologically a torus. Let $p:\mathbb{C}/L \mapsto \mathbb{R}^3$ be an embedding ($C^1$, say). Call $p$ conformal if pulling back the standard metric on $\mathbb{R}^3$ along $p$ yields a metric in the equivalence class of metrics on $\mathbb{C}/L$ (i.e. a multiple of the identity matrix).
>
> Is there an explicit formula for such a p in the case of L an oblique lattice?
>
>
>
---
Background
----------
The existence of such $C^1$ embeddings is implied by the Nash embedding theorem (fix a metric on $\mathbb{C}/L$, pick any short embedding, apply Nash iteration to make it isometric and hence conformal).
For orthogonal lattices, the solution is simple: Parametrise the standard torus of radii $r\_1$, $r\_2$ in the usual way. Make the ansatz $\pi(\theta, \phi) = (f(\theta), h(\phi))$, pull back the standard metric on $\mathbb{C}/L$ and solve the resulting system of ODEs. This relates $r\_1/r\_2$ to the ratio of the magnitudes of the periods. This shows that no standard torus can be the image of $p$ in the original question (oblique lattice), although that is geometrically clear anyway.
| https://mathoverflow.net/users/5181 | conformally embedding complex tori into R^3 | You should have a look in Pinkall's Hopf Tori paper. You take the preimage of a curve in $S^2$ under the Hopf fibration. The lattice of the torus and hence the conformal class is then given by the generators $1\in C$ and $L+i/2 A$ (if I remember right), where $L$ is the length and $A$ is the enclosed area of the curve.
| 11 | https://mathoverflow.net/users/4572 | 20540 | 13,640 |
https://mathoverflow.net/questions/20539 | 10 | The conjugacy problem for a free group $F\_n$ on $n$ letters has an easy solution. Each element of $F\_n$ is conjugate to a unique and easily computable "cyclically reduced element" (this means that if you arrange the word around a circle, then there are no cancellations), so two elements of $F\_n$ are conjugate if and only if they have the same cyclically reduced conjugates.
I've been trying unsuccessfully to generalize this to solve the following problem. Let $\{x\_1,\ldots,x\_k\}$ and $\{y\_1,\ldots,y\_{k'}\}$ be two finite sets of elements of $F\_n$. Let $G\_x$ and $G\_y$ be the subgroups of $F\_n$ generated by the $x\_i$ and the $y\_i$, respectively. Is there an algorithm to decide if $G\_x$ and $G\_y$ are conjugate? Does anyone know how to do this? Thank you very much!
| https://mathoverflow.net/users/4682 | Decidability of conjugacy problem for finitely generated subgroups of free groups | There is an algorithm to do this. I would have thought that it was classical, but in any case an algorithm is given in: I. Kapovich and A. Myasnikov "Stallings foldings and the subgroup structure of free groups", J. Algebra 248 (2002), no 2, pp. 608-668. In the online version I found [here](http://www.math.uiuc.edu/~kapovich/PAPERS/gr.pdf), it is Corollary 7.8 on page 18.
| 7 | https://mathoverflow.net/users/1109 | 20544 | 13,643 |
https://mathoverflow.net/questions/20551 | 49 | Does anyone know of a good place to find already-done BibTeX entries for standard books in advanced math? Or is this impossible because the citation should include items specific to your copy? (I am seeing the latter as potentially problematic because the only date I can find in my copy of Hartshorne is 2006, whereas the citations I can find all put the publication date at 1977.)
| https://mathoverflow.net/users/5094 | Sources for BibTeX entries | I recommend using the AMS website MREF, located [here](https://mathscinet.ams.org/mref).
EDIT : Another remark about your question. Don't worry too much about getting things like the printing date for a book correct (it changes every time they make a new printing run). Just make sure that you have the author, title, and edition in some standard format (and for papers, the journal, date, and page number). Every journal will reformat things into their house style and verify that your bibliographic entries are correct. Make those copy-editors work for their money!
| 41 | https://mathoverflow.net/users/317 | 20552 | 13,648 |
https://mathoverflow.net/questions/20548 | 6 | For example, how to solve the equation $\sum^{p-1}\_{i}x\_{i}^{2}=0$ in $F\_{p}$? This is not a homework problem. I think it should have a definite answer, so not an open problem. I just don't know how to solve it.
| https://mathoverflow.net/users/5175 | How to solve Diophantine equations in $F_{p}$? | There is a *deterministic* polynomial-time algorithm for finding solutions to diagonal equations of degree less than or equal to the number of variables over finite fields. See [Christiaan van de Woestijne's thesis](http://www.opt.math.tugraz.at/~cvdwoest/).
(A solution of your example equation can be found much more simply, however: try small integers, not necessarily distinct... . And for quadratic forms, the other solutions can be found by drawing lines through the point and intersecting with the quadric hypersurface: there will either be one more intersection point, or a whole line of points.)
| 14 | https://mathoverflow.net/users/2757 | 20553 | 13,649 |
https://mathoverflow.net/questions/20534 | 8 | Related to the question [link text](https://mathoverflow.net/questions/18636/number-of-invertible-0-1-real-matrices) I was asking myself some time ago the following. Can one precisely describe the invertible n\times n matrices with{0, 1} entries? For example, is anything special about the graph asociated to this matrix?
| https://mathoverflow.net/users/5179 | Characterizing invertible matrices with {0,1} entries | You are unlikely to find a characterization which does not result from simple facts in linear algebra. I am unaware of any characterizations which make interesting statements about graphs.
You may want to choose the ring over which the matrices belong. For example, the same matrix may be invertible over the reals, but if it has even determinant, then it is not invertible over the field of two elements. You can say that, given a matrix A, the parallelipiped associated with the rows (or columns) of A is nontrivial (has nonzero volume in R^n) iff the matrix is invertible over the reals, but this is a simple consequence of a geometric interpretation of determinant; it doesn't give anything new.
Also, the eigenvalues of the adjacency matrix of a directed graph are all nonzero precisely when said matrix is invertible over the reals; big hairy tautological deal, as I am just saying a matrix is invertible when its determinant is nonzero.
Consider the ring above fixed, and look at the {0,1}-matrices over that ring which are invertible. This set includes some lower triangular matrices, some upper triangular matrices, some "comb matrices" where you take an invertible matrix and alternately add
an extra row and column, picking one of them to be mostly zeros and the other mostly 1's, while making the diagonal all 1's, and alternating between rows and columns. In addition to these patterns, you have some block matrices, incidence geometries, certain combinatorial designs, and so on, all belonging to the class of invertible {0,1}- matrices, and looking pretty woolly as a set. The attendant directed graphs will be a similarly woolly-looking set of graphs.
Given the above, it may be possible to describe the class of graphs in an interesting way. For example, if you build the matrices by augmentation, consider the corresponding operation for adding a vertex and certain edges to a graph. You may be able to prove facts about the set of graphs so constructed, especially as a set of representatives of isomorphism classes of graphs. I just don't think the result will look pretty, appealing, or useful without a major shift in perspective.
Gerhard "Ask Me About System Design" Paseman, 2010.04.06
| 6 | https://mathoverflow.net/users/3402 | 20559 | 13,654 |
https://mathoverflow.net/questions/20529 | 3 | Edit: Rewritten with motivation, and hopefully more clarity.
I'm building a site for a card game called [dominion](http://www.boardgamegeek.com/boardgame/36218/dominion). In it, people build 'decks' of 10 distinct cards from a set of (currently) approximately 80. People (will) upload their decks to the site I'm working on, where other users will rate them for quality.
What I would like to do is create a random deck generator that generates decks that are 'similar' to use-created decks. For example, if cards A and B occur frequently in isolation, but infrequently together, the decks generated should share this same property. The state required has to be relatively limited in order for me to be able to do this online.
My tentative idea is to do the following:
1. Compute the sum of all ratings for all decks (call it S)
2. Compute the sum of all ratings for decks that contain a given card (call it S(A))
3. Compute the sum of all ratings for decks that contain any pair of cards (call it S(A∩B))
4. Compute the 'weighted' conditional probability P(A|B) = S(A∩B) / S(B)
Then, to generate a random deck, follow a procedure like the following:
1. Initialize a probability distribution P0(x) such that P0(x) = S(x) / S.
2. Select the first card, c, using the probability distribution P0
3. Compute the updated probability distribution P1(x), such that P1(x) = n P0(x) P(x|c), where n is a normalizing factor such that the integral of the distribution is 1.
4. Repeat from step 2 for the next card.
The problem is, I have no idea if this is valid, or if not, what should be modified to make it so. Based on what I've read, this seems like an application of bayes' theorem, but again I have no idea if I'm getting it wrong.
| https://mathoverflow.net/users/4379 | Random generation of subsets using conditional probabilities | You may do better with an approach that mimics the likely characteristics, and then selects cards that meet the characteristics, and then resolves conflicts. Here is a possible approach:
Consider the gross characteristics of such a deck: number and distribution of costs, number of +n Buys +n Cards +n Actions, number of duration cards, number of attack cards.
Now start the build by choosing a cost distribution, say 2 2's, 3 3's, 2 4's, and 3 5's.
Choose 20 cards at random with cost distribution mirroring the target distribution. Now try a subset of 10 appropriate cards. Check their stats against the others, e.g. number of +1 Buys. If all the stats match up, then check to see how many pairs of cards are disallowed. By whatever means, determine which cards out of the ten chosen do not represent a good fit to a random desired deck, and replace those cards with appropriate choices from the remainder of the 20 cards. If possible, let the stats dictate the
replacement subset. Now evaluate the modified deck, and see how many of the stats are
out of whack. Chances are good that you will converge to an acceptable deck within a
few trials.
If you implement this and find contrarily that chances are bad on converging to a good deck, then try resolving conflicts using a subset of 30 cards instead of a subset of 20 cards. I believe that finding a good set of characteristics will give you a way of
generating many good random decks, rather than just considering how often individual cards and card pairs occur or do not occur in favored decks.
Gerhard "Ask Me About System Design" Paseman, 2010.04.06
| 0 | https://mathoverflow.net/users/3402 | 20569 | 13,661 |
https://mathoverflow.net/questions/20568 | 12 | It is well known that there are strong links between Set Theory and Topology/Real Analysis.
For instance, the study of Suslin's Problem turns out to be a set theoretic problem, even though it started in topology: namely, whether $\mathbb{R}$ is the only complete dense unbounded linearly ordered set that satisfies the c.c.c.
Another instance is when we see that what's behind extending Lebesgue Measure is really the theory of large cardinals, with the introduction of measurable cardinals. Also another example of a real analysis problem that ends up in Set Theory is whether every set of reals is measurable. So the links are clear between Set Theory and Topology/Real analysis.
My question is this: are there links, as strong as the ones I roughly described in the last paragraph, between Set Theory and Abstract Algebra? The only example I know of is the Set Theoretic solution to the famous Whitehead Problem by Shelah (namely that if $V=L$ then every Whitehead group is free and if MA+$\neg$CH then there is a Whitehead group which is not free).
Can we hope to discover more of these type of links between Set Theory and Abstract Algebra? In contrast, Model Theory seems to be strongly grounded in Abstract algebra. I have seen that Shelah has some papers about uncountable free Abelian groups but he seems to be the only one investigating some areas of Abstract Algebra with the help of Set Theory. So again is there hope for links?
| https://mathoverflow.net/users/3859 | Is there a ground between Set Theory and Group Theory/Algebra? | Descriptive set theory also has something to say about algebra ... For example, the Higman-Neumann-Neumann Embedding Theorem
states that any countable group G can be embedded into a
2-generator group K. In the standard proof of this classical
theorem, the construction of the group K involves an
enumeration of a set of generators of the group G; and it is
clear that the isomorphism type of K usually depends upon
both the generating set and the particular enumeration
that is used. So it is natural to ask whether there is a more uniform construction with the property that the
isomorphism type of K only depends upon the isomorphism
type of G. As if ...
Assume the existence of a Ramsey cardinal and
suppose that G |----> F(G) is a Borel map from the space of countable groups to the space of finitely generated groups such that G embeds into F(G). Then there exists
an uncountable set of pairwise isomorphic groups G such that
the f.g. groups F(G) are pairwise incomparable with respect
to relative constructibility; ie while G, H are isomorphic,
F(G) doesn't even lie in the "set-theoretic universe generated
by F(H)."
| 14 | https://mathoverflow.net/users/4706 | 20574 | 13,665 |
https://mathoverflow.net/questions/19907 | 5 | Is there much known about the theory of lax and colax monads on a bicategory? Here, I really mean lax or colax, not weak. I'm aware of some literature about weak monads. I'm interested in distributive laws of lax and colax monads and their relationships with algebras. I've managed to prove some things, but, I don't want to reinvent the wheel.
| https://mathoverflow.net/users/4528 | Lax and Colax Monads | This isn't technically an answer, but depending on your examples, you might want to think about lax/colax monads on (pseudo) double categories instead. Part of the problem with lax monads on bicategories is that there is no tricategory of bicategories and lax functors, whereas there is a 2-category of pseudo double categories and lax functors, so that all of the "formal theory of monads" can be applied directly to lax monads on double categories. Most of the lax functors and lax monads that I've seen on bicategories have "actually" lived on double categories, except that people tend to forget about the extra direction of arrows and think only about the bicategory.
| 2 | https://mathoverflow.net/users/49 | 20575 | 13,666 |
https://mathoverflow.net/questions/19011 | 2 | Assume I have a set of weighted samples, where each samples has a corresponding weight between 0 and 1. I'd like to estimate the parameters of a gaussian mixture distribution that is biased towards the samples with higher weight. In the usual non-weighted case gaussian mixture estimation is done via the EM algorithm. Does anyone know how to modify the algorithm to account for the weights?
If not, can some one give me a hint on how to incorporate the weights in the initial formula of the maximum-log-likelihood formulation of the problem?
Thanks!
| https://mathoverflow.net/users/4806 | Estimate gaussian (mixture) density from a set of weighted samples | The usual EM algorithm can be modified for weighted inputs. Following along the [Wikipedia presentation](http://en.wikipedia.org/wiki/Mixture_model#Expectation_maximization_.28EM.29), you would use these formulas instead:
$a\_i = \frac{\sum\_{j=1}^N w\_j y\_{i,j}}{\sum\_{j=1}^{N}w\_j}$
and
$\mu\_{i} = \frac{\sum\_{j} w\_jy\_{i,j}x\_{j}}{\sum\_{j} w\_jy\_{i,j}}$
where $w\_j \ge 0$ are the weights of the data points.
| 2 | https://mathoverflow.net/users/634 | 20579 | 13,670 |
https://mathoverflow.net/questions/20580 | 9 | This is a follow-up question to this [coend computation](https://mathoverflow.net/questions/20445/coend-computation). Here's an attempt at a slightly simpler computation:
>
> $\int^{a \in A} \mbox{hom}\_A(a,a)$
>
>
>
This should be similar to the trace operator. In attempting to follow the derivation
>
> $\begin{array}{l}\mbox{Set}(\int^{b \in B}\mbox{hom}(a, b) \times F(b), S)\\ \cong \int\_b \mbox{Set}(\mbox{hom}(a,b) \times F(b), S)\\ \cong \int\_b \mbox{Set}(\mbox{hom}(a,b), \mbox{Set}(F(b), S)) \\ \cong \mbox{Nat}(\mbox{hom}(a,-), \mbox{Set}(F(-), S) \\ \cong \mbox{Set}(F(a), S),\end{array}$
>
>
>
I get
>
> $\begin{array}{l}\mbox{Set}(\int^{a \in A} \mbox{hom}\_A(a,a), S) \\ \cong \int\_{a \in A} \mbox{Set}(\mbox{hom}\_A(a,a), S) \\ \cong \mbox{Nat}(\mbox{hom}\_A(-,-), S)\end{array}$
>
>
>
So here I guess we have the set of natural transformations from the hom functor to the constant functor $S$. For any first parameter $a$, we have the set of natural transformations from hom$(a,-)$ to $S(a,-)$, which by Yoneda's lemma is isomorphic to $S(a,a) = S$. So I think it goes
>
> $\begin{array}{l}\cong \displaystyle \prod\_a \mbox{Nat}(\mbox{hom}\_A(a,-), S(a,-)) \\ \cong \prod\_a S\\ \cong S^{Ob(A)} \\ \cong \mbox{Set}(\mbox{Ob}(A), S).\end{array}$
>
>
>
So $\int^{a \in A} \mbox{hom}\_A(a,a) \cong \mbox{Ob}(A).$ Is that right?
| https://mathoverflow.net/users/756 | Coend computation continued | I agree with Reid's answer, but I want to add a bit more.
Putting Reid's calculation into a more general setting, if $A$ is *any* category then
$$
\int^{a \in A} \mathrm{hom}\_A (a, a)
= (\mathrm{endomorphisms\ in\ } A)/\sim
$$
where $\sim$ is the (rather nontrivial) equivalence relation generated by $gh \sim hg$ whenever $g$ and $h$ are arrows for which these composites are defined.
You can see confirmation there that your instinct about traces was right. If we wanted to define a 'trace map' on the endomorphisms in $A$, it should presumably satisfy $\mathrm{tr}(gh) = \mathrm{tr}(hg)$, i.e. it should factor through $\int^a \mathrm{hom}(a, a)$.
In fact, Simon Willerton has done work on 2-traces in which exactly this coend appears. See for instance [these slides](http://www.simonwillerton.staff.shef.ac.uk/ftp/TwoTracesBeamerTalk.pdf), especially the last one.
You can see in that slide something about the dual formula, the *end*
$$
\int\_{a \in A} \mathrm{hom}\_A(a, a).
$$
By the "fundamental fact" I mentioned [before](https://mathoverflow.net/questions/20445/coend-computation), this is the set
$$
{}[A, A](\mathrm{id}, \mathrm{id}) = \mathrm{Nat}(\mathrm{id}, \mathrm{id})
$$
of natural transformations from the identity functor on $A$ to itself. Evidently this is a monoid, and it's known as the **centre** of $A$. For example, when $G$ is a group construed as a one-object category, it's the centre in the sense of group theory. So your set might, I suppose, be called the **co-centre** of $A$.
| 8 | https://mathoverflow.net/users/586 | 20592 | 13,678 |
https://mathoverflow.net/questions/20549 | 8 | Given a character $\chi$ and $k$ odd how can one compute a basis for the space of modular forms $M\_\frac{k}{2}(\Gamma\_0(4),\chi)$. By compute a basis I mean, finding the beginning of the Fourier expansions. I am looking for computer programs, which can do that for me.
I have heard of the package SAGE, which seems to do the job for integral weight modular forms. There is even the function <http://www.sagemath.org/doc/reference/sage/modular/modform/half_integral.html> but the examples all have q-expansions starting with q, so I guess this is not really a basis for the space of all modular forms but only cusp forms.
MAGMA does not seem to include this functionality, either.
So, are there any packages which can do this? Since I have not found a package, I have some doubts that there is really an algorithm working in general. If there is no algorithm known to handle this, what methods are available in order to compute a basis "by hand"?
Thanks.
| https://mathoverflow.net/users/3757 | Basis for modular forms of half-integral weight | **Edit:** Here's a rather silly method that should work if SAGE is just giving you cusp forms: $\Gamma\_0(4)$ has a single normalized cusp form of weight 6, given by $\eta(2\tau)^{12} = q - 12q^3 + 54q^5 - \dots$, so take your basis of cusp forms of weight $k/2 + 6$, and divide each element by this form to get a basis of the space of modular forms of weight $k/2$.
**Edit in response to Buzzard:** Thanks for pointing out that I should make this argument. Here is a proof that the cusp form has minimal vanishing at all cusps. $\Gamma\_0(4)$ is conjugate to $\Gamma(2)$ by $\tau \mapsto 2\tau$, so it suffices to check that $\Delta(\tau)$, the square of $\eta(\tau)^{12}$, vanishes to twice the minimum order at each cusp of $\Gamma(2)$. The quotient $\Gamma(1)/\Gamma(2) \cong S\_3$ acts transitively on the cusps of $X(2)$ with stabilizers of order 2, so the quotient map to $X(1)$ has ramification degree 2 at each cusp. $\Delta(\tau)$ is invariant under the weight 12 action of $\Gamma(1)$, and $\Delta(\tau)$ has minimal vanishing at infinity on $X(1)$.
**Old answer:** If you have a cusp form of weight $k/2$ for $\Gamma\_0(4)$ (e.g., given to you by SAGE), you can multiply it by the modular function $\frac{\eta(\tau)^8}{\eta(4\tau)^8} = q^{-1} - 8 + 20q - 62q^3 + 216q^5 - \dots$ to get a modular form of the same weight, that is nonvanishing at one of the three cusps and vanishing at the other two. If you want a form that is nonzero at one of the other cusps, you can multiply by $\frac{\eta(4\tau)^8}{\eta(\tau)^8}$ (has a pole at zero) or by $\frac{\eta(\tau)^{16}\eta(4\tau)^8}{\eta(2\tau)^{24}}$ (pole at $1/2$). [Constant term $-8$ added Sept. 23, in response to an email correction from Michael Somos.]
| 7 | https://mathoverflow.net/users/121 | 20595 | 13,679 |
https://mathoverflow.net/questions/20590 | 11 | Here's the setup: you have a first-order theory T, in a countable language L for simplicity. Let k be a cardinal and suppose T is k-categorical. This means that, for any two models
M,N |= T
of cardinality k, there is an isomorphism f : M --> N.
Supposing all this happens inside of ZFC, let's say I change the underlying model of ZFC, e.g by restricting to the constructible sets, or by forcing new sets in. I would like to understand what happens to the k-categoricity of T.
I'll assume the set theory doesn't change so drastically that we lose L or T. Then, a priori, a bunch of things may happen:
(i) We may lose all isomorphisms between a pair of models M,N of cardinality k;
(ii) Some models that used to be of cardinality k may no longer have bijections with k;
(iii) k may become a different cardinal, meaning new cardinals may appear below it, or others may disappear by the introduction of new bijections;
(iv) some models M, or k itself, may disappear as sets, leading to a new set being seen as "the new k".
Overall, nearly every aspect of the phrase "T is k-categorical" may be affected. How likely is it to still be true? Do some among (i)-(iv) not matter, or is there some cancellation of effects? (Say, maybe all isomorphisms M-->N disappear, but so do all bijections between N and k?)
| https://mathoverflow.net/users/4367 | How does categoricity interact with the underlying set theory? | Categoricity is absolute.
By the Ryll-Nardzewski theorem, for a countable language, $\aleph\_0$-categoricity of a complete theory $T$ is equivalent to $T$ proving for each natural number $n$ that there are only finitely many inequivalent formulas in $n$ variables. This property is evidently arithmetic and, thus, absolute.
Likewise, again in a countable language, it follows from the Baldwin-Lachlan theorem that a theory is categorical in some (hence, by Morley's theorem, all) uncountable cardinality just in case every model is prime and minimal over a strongly minimal set. Moreover, the strongly minimal formula may be taken to be defined over the prime model and the primality and minimality of every model over this strongly minimal formula is something which will be witnessed by an explicit analysis, hence, something arithmetic and absolute.
For uncountable languages, the situation is a little more complicated, but again categoricity is equivalent to an absolute property. Shelah shows that either the theory is totally transcendental and Morley's analysis in the case of countable languages applies, or the theory is strictly superstable though unidimensional.
| 22 | https://mathoverflow.net/users/5147 | 20601 | 13,681 |
https://mathoverflow.net/questions/20609 | 19 | When studying representation theory, special functions or various other topics one is very likely to encounter the following identity at some point:
$$\det \left(\frac{1}{x \_i+y \_j}\right) \_{1\le i,j \le n}=\frac{\prod \_{1\le i < j\le n} (x \_j-x \_i)(y \_j-y \_i)}{\prod \_{i,j=1}^n (x \_i+y \_j)}$$
This goes under the name of Cauchy's determinant identity and has various generalizations and analogous statements. There is also a lot of different proofs using either analysis or algebra. In my case I have always seen it introduced (or motivated) as an identity that plays an important role in combinatorics, but I realized that I haven't really seen this identity in a combinatorial context before. In this question I'm asking for a combinatorial interpretation of the above identity. A bonus to someone who can give such an interpretation to Borchardt's variation:
$$\det \left(\frac{1}{(x \_i+y \_j)^2}\right) \_{1\le i,j \le n}=\frac{\prod \_{1\le i < j\le n} (x \_j-x \_i)(y \_j-y \_i)}{\prod \_{i,j=1}^n (x \_i+y \_j)} \cdot \text{per}\left(\frac{1}{x \_i +y \_j}\right) \_{1 \le i,j \le n}$$
(This seems a little too ambitious though, and I would be happy to accept an answer of just the first question)
| https://mathoverflow.net/users/2384 | What role does Cauchy's determinant identity play in combinatorics? | See also pp. 397--398 of *Enumerative Combinatorics*, vol. 2. Cauchy's determinant is given in a slightly different but equivalent form. It is explained there that the evaluation of the determinant is equivalent to the fundamental identity $\prod(1-x\_iy\_j)^{-1} =\sum\_\lambda s\_\lambda(x)s\_\lambda(y)$ in the theory of symmetric functions.
| 16 | https://mathoverflow.net/users/2807 | 20639 | 13,706 |
https://mathoverflow.net/questions/20645 | 3 | Let $E$ and $F$ be two abelian varieties of dimension 1 over $\mathbb{C}$. Let $f : E \to F$ be a surjective homomorphism of abelian varieties ($f(0) = 0$). If $\ker (f) \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, does this imply that $E$ and $F$ are isomorphic?
| https://mathoverflow.net/users/5197 | isogeny of elliptic curves | Yes. In zero characteristic the image of an isogeny of elliptic curves
is determined up to isomorphism by its kernel. Your isogeny has the same kernel
as the doubling map from $E$ to itself.
| 12 | https://mathoverflow.net/users/4213 | 20647 | 13,710 |
https://mathoverflow.net/questions/20608 | 1 | I have the following setup:
$X, Y$ are topological spaces (if it helps, they can both be $T\_1$ and normal. They can even be countably paracompact. They *can't* be assumed paracompact). $V$ is a normed space (it can be Banach if you like). $f : X \to Y$ is a perfect surjection.
I have continuous and bounded $g : X \to V$ and given $\epsilon > 0$ would like to find continuous $h : Y \to V$ such that $d(h(x), g(f^{-1}(x))) < \epsilon$
Is there some sort of selection theorem that will let me do this? I've used the Michael selection theorem to good effect elsewhere, but it doesn't apply here due to the lack of convexity of the target sets (even if they were convex the hypotheses don't apply due to potential non-paracompactness of Y, but one might be able to work something out using countable paracompactness and compactness of the targets).
| https://mathoverflow.net/users/4959 | Approximate selection theorems for factoring through perfect maps | Consider $X=Y=S^1$. Let $f:X\to Y$ be a 2-fold covering and $g:X\to\mathbb R^2$ the standard embedding (whose image is a unit circle). Assume $\epsilon<1$, then there is no map $h$ with the desired property.
Indeed, if $d(h(x),g(f^{-1}(x)))<\epsilon$, then there is a unique $y\in f^{-1}(x)$ such that $d(h(x),g(y))<\epsilon$. Obviously $y=:u(x)$ depends continuously on $x$. Thus we obtain a continuous map $u:Y\to X$ such that $f\circ u=id\_Y$. Such a map does not exist because $f$ (and hence any map of the form $f\circ u$) induces a non-surjective homomorphism of fundamental groups.
| 4 | https://mathoverflow.net/users/4354 | 20653 | 13,714 |
https://mathoverflow.net/questions/20604 | 21 | The discovery (or invention) of negatives, which happened several centuries ago by the Chinese, Indians and Arabs, has of course be of fundamental importance to mathematics.
From then on, it seems that mathematicians have always striven to "put the negatives" into whatever algebraic structure they came across, in analogy with the usual "numerical" structure, $\mathbb{Z}$.
But perhaps there are cases in which the notion of a semiring seems more natural than the notion of a ring (I will be very very sloppy!):
1) The Cardinals. They have a natural structure of semiring, and the usual construction that allows to pass from $\mathbb{N}$ to $\mathbb{Z}$ cannot be performed in this case without great loss of information.
2) Vector bundles over a space; and notice that in the infinite rank case the Grothendieck *ring* is trivial just because negatives are allowed.
3) Tropical geometry.
4) The notion of semiring, as opposed to that of a ring, seems to be the most natural for "categorification", in two separate senses: (i) For example, the set of isomorphism classes of objects in a category with direct sums and tensor products (e.g. finitely-generated projective modules over a commutative ring) is naturally a semiring. When one constructs the Grothendieck ring of a category, one usually adds formal negatives, but this can be a very lossy operation, as in the case of vector bundles. (ii) A category with finite biproducts (products and coproducts, and a natural isomorphism between these) is automatically enriched over commutative monoids, but not automatically enriched over abelian groups. As such, it's naturally a "many object semiring", but not a "many object ring".
Do you have any examples of contexts in which semirings (which are not rings) arise naturally in mathematics?
| https://mathoverflow.net/users/4721 | Are rings really more fundamental objects than semi-rings? | Of course the real question is whether abelian groups are really more fundamental objects than commutative monoids. In a sense, the answer is obviously no: the definition of commutative monoid is simpler and admits alternative descriptions such as the one I give [here](https://mathoverflow.net/questions/2551/why-do-groups-and-abelian-groups-feel-so-different/9302#9302). The latter description can be adapted to other settings, such as to the 2-category of locally presentable categories, which shares many formal properties with the category of commutative monoids (such as being closed symmetric monoidal, having a zero object, having biproducts). As such I would claim that any locally presentable closed symmetric monoidal category is itself a categorified version of a semiring, not in the sense you describe, but in that it is an algebra object in a closed symmetric monoidal category, so we may talk of modules over it, etc.
However, it is undeniable that there is a large qualitative difference between the theories of abelian groups and commutative monoids. Observe that an abelian group is just a commutative monoid which is a module over $\mathbb{Z}$ (more precisely a commutative monoid has either a unique structure of $\mathbb{Z}$-module, if it has additive inverses, and no structure of $\mathbb{Z}$-module otherwise). The situation is analogous to the (smaller) difference between abelian groups and $\mathbb{Q}$-vector spaces. I do not know of a characterization of $\mathbb{Z}$ as a commutative monoid that can be transported to other settings. It seems that there is something deep about the fact that $\mathbb{Z}$-modules are so much nicer than commutative monoids, which often is taken for granted.
| 10 | https://mathoverflow.net/users/126667 | 20654 | 13,715 |
https://mathoverflow.net/questions/20651 | 9 | Hello everyone;
i'm looking for a motivation for equivariant sheaves (see <http://ncatlab.org/nlab/show/equivariant+sheaf>) ~ **Why are we interested in them?**
More explicitely: Can I think of G-equivariant sheaves on a space X as a quotient of the category of sheaves (by some action? in a more general sense) by G?
| https://mathoverflow.net/users/1261 | Motivation for equivariant sheaves? | If you know that the sections of a vector bundle form a standard example of a sheaf, then the corresponding example of a $G$-equivariant sheaf on a space $X$ with $G$-action is a vector bundle $V$ over $X$ with a $G$-action compatible with the projection (i.e. making the projection $G$-equivariant, i.e. intertwining the actions). Such actions on vector bundles over homogeneous spaces were considered in representation theory by Borel, Weil, Bott and Kostant ("homogeneous vector bundles"; and later many generalizations to sheaves by Beilinson-Bernstein, Schmid, Miličić etc.). David Mumford introduced $G$-equivariant structures on sheaves under the name $G$-**linearization** for the purposes of geometric invariant theory.
While for a function on a $G$-space the appropriate notion is the $G$-invariance, for sheaves the invariance is useful only up to a coherent isomorphism, what spelled out yields the definition of the $G$-equivariant sheaf. This is an example of a categorification (recall that functions form a set and sheaves form a category). Using the Yoneda embedding one can indeed consider the $G$-equivariant objects as objects in some fibered category of objects on $X$ with an action on each hom-space (see the lectures by Vistoli).
While for a function to be invariant is a property, for a sheaf the $G$-equivariance entails the additional coherence data, so it is a structure.
Category of $G$-equivariant sheaves on $X$ is not a quotient of the category of usual sheaves on $X$, but rather equivalent to the category of sheaves on the geometric quotient $G/X$, in the case when the action of $G$ on $X$ is principal; or in general if we replace the geometric quotient by the appropriate stack $[G/X]$.
| 16 | https://mathoverflow.net/users/35833 | 20662 | 13,721 |
https://mathoverflow.net/questions/20634 | 8 | From the [Peter–Weyl theorem in Wikipedia](https://en.wikipedia.org/wiki/Peter%E2%80%93Weyl_theorem), this theorem applies for compact group. I wonder whether there is a non-compact version for this theorem.
I suspect it because the proof of the Peter–Weyl theorem heavily depends on the compactness of Lie group. It is related to the spectral decomposition of compact operators.
*Thanks Mariano pointing out the Peter–Weyl theorem does not hold for non-compact group. But I really wants to know is: is there any Peter–Weyl analogue decomposition for non-compact group, say decompose to integral representations but not finite dimensional representations?*
Another related questions is about the definition of quantized flag variety. In the work of [Lunts and Rosenberg on localization for quantum group](https://doi.org/10.1007/s000290050044 "Lunts, V., Rosenberg, A. Localization for quantum groups. Sel. math., New ser. 5, 123 (1999)"), they tried to establish the quantum analogue of Beilinson–Bernstein localization theorem. They defined the quantized flag variety in the framework of noncommutative algebraic geometry. They used the Peter–Weyl philosophy for quantum group to define the coordinate ring of quantized base affine space as the direct sum of all simple $U\_{q}(g)$-modules with highest weight $\lambda$(positive).(Denoted by $R\_{+}$)
Then one can define category of quasi coherent sheaves on "quantized flag variety" as proj-category of graded $R\_{+}$.
What I want to ask is there any other way to define quantized flag variety? In the classical case, It is well known that flag variety can be define as $G/B$, say $G$ is general linear group and $B$ is Borel subgroup. Is there any analogue for quantum case? Is there a definition like $G\_{q}$ as "quantum linear group" and $B\_{q}$ as quantum analogue of Borel subgroup?
However, I suspected, because the quantum flag variety is essentially not a **real space**.
| https://mathoverflow.net/users/1851 | Is there analogue of Peter–Weyl theorem for non-compact or quantum group |
>
> What I want to ask is there any other
> way to define quantized flag variety?
> In the classical case, it is well
> known that flag variety can be defined
> as $G/B$, say $G$ is general linear group
> and B is Borel subgroup. Is there any
> analogue for quantum case? Is there a
> definition like $G\_q$ as "quantum linear
> group" and $B\_q$ as quantum analogue of
> Borel subgroup?
>
>
>
The Borel subgroup of the quantized function algebra $G\_q$ is of course certain quotient Hopf algebra $p:\mathcal{O}(G\_q)\to \mathcal{O}(B\_q)$ which then canonically coacts on $G\_q$ from both sides (from the left by $(p\otimes id)\circ\Delta\_{G\_q}:\mathcal{O}(G\_q)\to \mathcal{O}(B\_q)\otimes \mathcal{O}(G\_q)$. In the case of $SL\_q(n)$ you get the lower Borel by quotienting by the Hopf ideal generate by the entries of the matrxi $T$ of the generators standing above the diagonal.
Of course, if you define $Qcoh(G\_q)$ as the category ${}\_{\mathcal{O}(G\_q)}\mathcal{M}$ of left modules over $\mathcal{O}(G\_q)$ then the category of relative left-right [Hopf modules](https://ncatlab.org/nlab/show/Hopf+module) ${}\_{\mathcal{O}(G\_q)}\mathcal{M}^{\mathcal{O}(B\_q)}$ is precisely the category of quasicoherent sheaves over the quantum flag variety $G\_q/B\_q$. If you look at several of my earlier articles they together show that in the case of $SL\_q(n)$ at least, there is a collection of localizations of the above category of relative Hopf modules which are affine, which gives it a structure of a noncommutative scheme with a canonical atlas whose cardinality is the cardinality of the Weyl group (the generalization to other parabolics is easy). Moreover the forgetful functor from the category of Hopf modules to ${}\_{\mathcal{O}(G\_q)}\mathcal{M}$ is the inverse image part of the geometric morphism which is a Zariski locally trivial $B\_q$-fibration; the local triviality, which is obtained with the help of the quantum Gauss decomposition, boils down to the Schneider's equivalence for the extension of the algebras of localized coinvariants into the corresponding localization of $G\_q$. The sketch of the whole program is in
* *Localizations for construction of quantum coset spaces*, in "Noncommutative geometry and Quantum groups", W.Pusz, P.M. Hajac, eds. Banach Center Publications vol.61, pp. 265--298, Warszawa 2003, [math.QA/0301090](https://arxiv.org/abs/math.QA/0301090).
For the Peter-Weyl look at the original works of S. Woronowicz or
* A. Klimyk, K. Schmüdgen, *Quantum groups and their representations*, Springer 1997.
| 5 | https://mathoverflow.net/users/35833 | 20668 | 13,724 |
https://mathoverflow.net/questions/20613 | 12 | Suppose I have a Banach space $V$ and a set $A \subseteq V$ such that for all $\epsilon > 0$ there exists $v$ such that $A \subseteq \overline{B}(v, r + \epsilon)$. Does there exist $c$ such that $A \subseteq \overline{B}(c, r)$?
The answer is clearly yes for finite dimensional normed spaces: Define $T\_\epsilon = \bigcap\_{a \in A} \overline{B}(a, r + \epsilon)$. The $T\_\epsilon$ form a chain of closed sets and for $\epsilon > 0$ are non-empty, so have the finite intersection property. Thus when $V$ is finite dimensional they have non-empty intersection, and any element of the intersection works as $c$.
For more general Banach spaces I feel like you should be able to choose a cauchy sequence $x\_n$ such that $x\_n \in T\_{\epsilon\_n}$ with $\epsilon\_n \to 0$, but I can't seem to make it work.
Note that an arbitrary choice of $x\_n \in T\_{\epsilon\_n}$ can't be guaranteed to be Cauchy: If $V$ is $l^\infty$ and $A = \{ x : x\_0 = 0, ||x|| \leq 1 \}$ then diam$(T\_\epsilon) \geq 2$ because you can choose $c\_0$ arbitrarily in $[-1, 1]$
Note also that the assumption of $V$ a Banach space is essential: If $V$ is not Banach and $c$ is an element of the completion which is not in $V$ then $A = \overline{B}(c, 1) \cap V$ has no center.
| https://mathoverflow.net/users/4959 | Radii and centers in Banach spaces | I believe that the property does not hold for all Banach spaces, but my counterexample is a little involved. If you've the patience then follow me through...
Let $V=\bigoplus\_{n=1}^\infty \ell^n\_2$ where $\ell^p\_2$ is $\mathbb{R}^2$ with
norm $\lVert\cdot\rVert\_p$ (Note: $n$ is taking the role of $p$). For $i\geq1$
and $j\in\{0,1\}$ we have $e\_{i,j}$, the $j^{th}$ standard basis vector of
$\ell^i\_2$ in $V$.
Give $V$ the norm $\lVert v\rVert=\sup\_n\lVert v\_n\rVert\_n$.
Let $W=\{v\in V:\lVert v\_n\rVert\_n\to 0\}$. I assert that $W$ is a Banach space.
Certainly every $e\_{i,j}\in W$.
Let $A=\{e\_{k,0}+e\_{k,1}, e\_{k,0}-e\_{k,1}:k\geq 1\}$.
Fact: Let $r(A)$ be the infimum of radii of balls containing $A$. Then $r(A)\leq1$
Proof:
Let $c\_N=\sum\_{i=1}^n e\_{i,0}$. We wish to compute the distance of each point
of $A$ from $c\_N$.
For $k\leq N$ we have
$\lVert c\_N-e\_{k,0}-e\_{k,1}\rVert$
$=\lVert\sum\_{i=1\ (i\not=k)}^Ne\_{i,0}-e\_{k,1}\rVert$
$=\sup\{\lVert e\_{i,0}\rVert\_i:i\leq N,i\not=k\}\cup\{\lVert-e\_{k,1}\rVert\_k\}$
$=1$
and similarly for $\lVert c\_N-e\_{k,0}+e\_{k,1}\rVert$.
For $k>N$ we have
$\lVert c\_N-e\_{k,0}-e\_{k,1}\rVert$
$=\max(\lVert c\_N\rVert,\lVert e\_{k,0}+e\_{k,1}\rVert\_k)$
$= \max(1,(1+1)^\frac{1}{k})$
$= 2^\frac{1}{k}$
$\leq 2^\frac{1}{N}$
and similarly for $\lVert c\_N-e\_{k,0}+e\_{k,1}\rVert$.
Thus $A\subseteq \overline{B}(c\_N,2^\frac{1}{N})$ and so $r(A)\leq2^\frac{1}{N}$. Letting
$N\to\infty$ we have $r(A)\leq 1$.
QED
Fact: $A$ is not contained in a ball of radius $1$.
Proof:
Suppose $A\subseteq \overline{B}(c,1)$. Then in particular for every $n$ we have
$\lVert c-e\_{n,0}-e\_{n,1}\rVert\leq 1$ and thus $\lVert c\_n-e\_{n,0}-e\_{n,1}\rVert\_n\leq 1$. Similarly $\lVert c\_n-e\_{n,0}+e\_{n,1}\rVert\_n\leq 1$.
Simple consideration of $\ell^n\_2$ shows that this implies $c\_n=e\_{n,0}$.
Thus $\lVert c\_n\rVert=1\not\to0$ and $c\not\in W$, contradicting the assumption.
QED
| 7 | https://mathoverflow.net/users/5213 | 20672 | 13,727 |
https://mathoverflow.net/questions/20650 | 12 | Let $X$ be a Riemannian manifold. If $X$ is simply connected, irreducible, and not a symmetric space then we know that the possible holonomy groups of the metric on $X$ are:
1) $O(n)$ General Riemannian manifolds
2) $SO(n)$ Orientable manifolds
3) $U(n)$ Kahler manifolds
4) $Sp(n)Sp(1)$ Quaternionic Kahler manifolds
5) $SU(n)$ Calabi-Yau manifolds
6) $Sp(n)$ Hyperkahler manifolds
7) $G\_{2}$ (if $X$ has dimension 7) $G\_{2}$ manifolds
8) $Spin(7)$ (if X has dimension 8) $Spin(7)$ manifolds
Of these, cases 5-8 are important in string theory. The reduced holonomy implies that these manifolds have vanishing Ricci curvature and hence are automatically solutions to Einstein's equations of general relativity.
However, physics does not take place on a Riemannian manifold, but rather on a Lorentzian one. Thus my question is: what is known about special holonomy manifolds for metrics of general signature? (I am most interested in the (1,n) case!) In particular I would like to know if there is a classification of allowed holonomy groups and further if there are interesting examples that I should be aware of.
| https://mathoverflow.net/users/5124 | Special Holonomy Groups for Lorentzian Manifolds | I think it would be more accurate to say that the real reason why
Calabi-Yau, hyperkähler, $G\_2$ and $\mathrm{Spin}(7)$ manifolds are of
interest in string theory is not their Ricci-flatness, but the fact
that they admit parallel spinor fields. Of course, in
positive-definite signature, existence of parallel spinor fields
implies Ricci-flatness, but the converse is still open for compact
riemannian manifolds, as discussed in [this
question](http://mathoverflow.net/questions/16818/are-there-ricci-flat-riemannian-manifolds-with-generic-holonomy), and known to fail for noncompact manifolds as pointed
out in an answer to that question.
The similar question for lorentzian manifolds has a bit of history.
First of all, the holonomy principle states that a spin manifold
admits parallel spinor fields if and only if (the spin lift of) its
holonomy group is contained in the stabilizer subgroup of a nonzero
spinor. Some low-dimensional (i.e., $\leq 11$, the cases relevant to
string and M-theories) investigations (by [Robert Bryant](http://arXiv.org/abs/math.DG/0004073) and [myself](http://arXiv.org/abs/hep-th/9904124),
independently) suggested that these subgroups are either of two types:
subgroups $G < \mathrm{Spin}(n) < \mathrm{Spin}(1,n)$, whence $G$ is
the ones corresponding to the cases 5-8 in the question, or else $G =
H \ltimes \mathbb{R}^{n-1}$, where $H < \mathrm{Spin}(n-1)$ is one of
the groups in cases 5-8 in the question. Thomas Leistner showed that
this persisted in the general case and, as Igor pointed out in his
answer, arrived at a classification of possible lorentzian holonomy
groups. Anton Galaev then constructed metrics with all the possible
holonomy groups, showing that they all arise. Their work is reviewed
in [their
paper](http://dx.doi.org/10.4171/051-1/2) ([MR2436228](http://www.ams.org/mathscinet-getitem?mr=2436228)).
The basic difficulty in the indefinite-signature case is that the de
Rham decomposition theorem is modified. Recall that the de Rham
decomposition theorem states that if $(M,g)$ is a complete, connected
and simply connected positive-definite riemannian manifold and if the
holonomy group acts reducibly, then the manifold is a riemannian
product, whence it is enough to restrict to irreducible holonomy
representations. This is by no means a trivial problem, but is
tractable.
In contrast, in the indefinite signature situation, there is a
modification of this theorem due to Wu, which says that it is not
enough for the holonomy representation to be reducible, it has to be
*nondegenerately* reducible. This means that it is fully reducible
and the direct sums in the decomposition are orthogonal with respect
to the metric. This means that it is therefore not enough to restrict
oneself to irreducible holonomy representations. For example,
Bérard-Bergery and Ikemakhen proved that the only lorentzian holonomy
group acting irreducibly is $\mathrm{SO}\_0(1,n)$ itself: namely, the
generic holonomy group.
It should be pointed out that in indefinite signature, the
integrability condition for the existence of parallel spinor fields is
not Ricci-flatness. Instead, it's that the image of the Ricci
operator $S: TM \to TM$, defined by $g(S(X),Y) = r(X,Y)$, with $r$ the
Ricci curvature, be isotropic. Hence if one is interested in
supersymmetric solutions of supergravity theories (without fluxes) one
is interested in Ricci-flat lorentzian manifolds (of the relevant
dimension) admitting parallel spinor fields. It is now not enough to
reduce the holonomy to the isotropy of a spinor, but the
Ricci-flatness equation must be imposed additionally.
| 16 | https://mathoverflow.net/users/394 | 20673 | 13,728 |
https://mathoverflow.net/questions/20666 | 8 | The following question came up in the course on Quantum Groups here at UC Berkeley. (If you care, I have been TeXing [uneditted lecture notes](http://math.berkeley.edu/~theojf/QuantumGroups10.pdf).)
Let $X,Y$ be (infinite-dimensional) [Hopf algebras](http://en.wikipedia.org/wiki/Hopf_algebra) over a ground field $\mathbb F$. A linear map $\langle,\rangle : X\otimes Y \to \mathbb F$ is a **bialgebra pairing** if $\langle x,y\_1y\_2 \rangle = \langle \Delta x,y\_1\otimes y\_2\rangle$ and $\langle x\_1x\_2,y\rangle = \langle x\_1\otimes x\_2,\Delta y\rangle$ for all $x,x\_1,x\_2 \in X$ and $y,y\_1,y\_2 \in Y$. (You must pick a convention of how to define the pairing $\langle,\rangle : X^{\otimes 2} \otimes Y^{\otimes 2} \to \mathbb F$.) And we also demand that $\langle 1,- \rangle = \epsilon\_Y$ and $\langle -,1\rangle = \epsilon\_X$, but this might follow from the previous conditions. (See edit.)
A bialgebra pairing is **Hopf** if it also respects the antipode: $\langle S(x),y \rangle = \langle x,S(y)\rangle$. A pairing $\langle,\rangle : X\otimes Y \to \mathbb F$ is **nondegenerate** if each of the the induced maps $X \to Y^\*$ and $Y \to X^\*$ has trivial kernel.
>
> **Question:** Is a (nondegenerate) bialgebra pairing of Hopf algebras necessarily Hopf? (Does it depend on whether the pairing is nondegenerate?)
>
>
>
My intuition is that regardless of the nondegeneracy, the answer is "Yes": my motivation is that a bialgebra homomorphism between Hopf algebras automatically respects the antipode. But we were unable to make this into a proof in the infinite-dimensional case.
**Edit:** If $\langle,\rangle: X\otimes Y \to \mathbb F$ is nondegenerate, then it is true that as soon as it satisfies $\langle x,y\_1y\_2 \rangle = \langle \Delta x,y\_1\otimes y\_2\rangle$ and $\langle x\_1x\_2,y\rangle = \langle x\_1\otimes x\_2,\Delta y\rangle$, so that the induced maps $X \to Y^\*$ and $Y \to X^\*$ are (possibly non-unital) algebra homomorphisms, then it also satisfies $\langle 1,- \rangle = \epsilon\_Y$ and $\langle -,1\rangle = \epsilon\_X$, so that the algebra homomorphism are actually unital. But I think that this does require that the pairing be nondegenerate. At least, I don't see how to prove it without the nondegeneracy assumption. So probably the nondegeneracy is required for the statement about antipodes as well.
| https://mathoverflow.net/users/78 | Is a bialgebra pairing of Hopf algebras automatically a Hopf pairing? | Theo,
I think one can argue like this in the case of a non-degenerate pairing. I didn't check everything here carefully, so don't believe it unless you confirm it yourself.
One has from the pairing an inclusion of $i:X\hookrightarrow Y^\*$. One has two maps on $i(X)$, $S\_X$, and $S\_Y^\*|\_{i(X)}$. Both of these satisfy the axioms of an antipode on $i(X)$ (which we can check by pairing with elements of $Y$), so they must agree, as desired, since being a Hopf algebra is a property, not a structure.
I'll think about the degenerate situation. My guess is that it's not true, but I don't know.
| 3 | https://mathoverflow.net/users/1040 | 20678 | 13,731 |
https://mathoverflow.net/questions/20675 | 16 | Katzarkov-Kontsevich-Pantev define a *smooth* dg ($\mathbb{C}$-)algebra $A$ to be a dg algebra which is a perfect $A \otimes A^{op}$-module. They say that an $A$-module $M$ is *perfect* if the functor $Hom(M,-)$ from ($A$-modules) to ($\mathbb{C}$-modules) preserves small homotopy colimits. (Definition 2.23 of [KKP](http://arXiv.org/abs/0806.0107)) \*
Kontsevich-Soibelman define *smooth* dg (or $A\_\infty$) algebras in the same way; but they define *perfect* $A$-modules to be ones which are quasi-isomorphic to a direct summand of an extension of a sequence of modules each of which is quasi-isomorphic to a shift of $A$. (Definition 8.1.1 of [KS](http://arXiv.org/abs/math/0606241v2)) \*\*
**First question:** I have trouble wrapping my head around what it means for a functor to preserve small homotopy colimits. I don't understand the definition from Kontsevich-Soibelman, either. What are the "moral" meanings of these conditions?
**Second question:** Presumably these two conditions are equivalent. How do you prove this?
**Third question:** If $A$ is a commutative $(\mathbb{C}$-)algebra, then presumably smoothness of $A$ as a dg algebra is equivalent to smoothness of $\operatorname{Spec} A$ as a ($\mathbb{C}$-)scheme. (Maybe you need $A$ to be finite type?) How do you prove this? Furthermore, Example 8.1.4 of KS gives some more examples of dg algebras that are allegedly smooth (for instance free algebras $k\langle x\_1, \dots, x\_n \rangle$). Again, I don't know how to prove that these are in fact smooth, and Kontsevich-Soibelman don't seem to provide proofs; or maybe I overlooked something.
---
\*The answers below indicate that this is the standard definition of *compact* (also known as *small*) module.
\*\*The answers below indicate that this is the standard definition of *perfect* module.
| https://mathoverflow.net/users/83 | Smooth dg algebras (and perfect dg modules and compact dg modules) | The condition of Hom(M,-) being a continuous functor, i.e. preserving (small homotopy) colimits is equivalent (in the present stable setting) to the maybe more concrete condition of preserving arbitrary direct sums. The issue is not finite direct sums, that's automatic (since the derived Hom is an exact functor, it automatically commutes with FINITE colimits).
This is better known as a compact object of the dg category (see n-lab for lots of discussion).
This is a strong finiteness condition on a module. Its importance in algebraic geometry was realized by Thomason (in a dream form of his friend Trobaugh), who proved some amazing fundamental results about the behavior (and abundance) of compact/perfect objects on schemes.
One can think of Hom(M,-) as the functional on the category defined by M (Hom being a kind of inner product), and this is saying the functional is "continuous"..
There's a discussion of the different common notions of finiteness for modules and their properties (including the foundational results of Thomason and Neeman) in Section 3.1 of [this paper](http://arxiv.org/abs/0805.0157) (sorry for the self-referencing -- none of this is in the least original, but it's a convenient discussion with plentiful references). This includes an explanation of why compactness is the same as being in the thick subcategory category generated by the free module, which is your definition of perfect from K-S (ie built out of the free module by taking sums, cones, summands), and also the same as being a dualizable object with respect to tensor product in case your category is the derived category of quasicoherent sheaves (ie this is a "commutative" notion not applicable in the NC setting you're discussing).
By the way the definition from K-S is the standard definition of perfection, that from K-K-P is the standard definition of compactness.. there are settings where the two notions don't agree (for example for sheaves on the classifying space of a finite group in a modular characteristic, or on BS^1) [hence our terminology of "perfect stack" in the paper with Francis and Nadler, which is a stack where these notions for sheaves agree and these nice finite objects generate].
The fact that perfection of the diagonal (ie A as an A-bimodule) is equivalent to smoothness in the case of a scheme is a reformulation of the homological criterion of Serre for smoothness of a point in terms of finite Tor amplitude of the skyscraper at the point - ie we're saying all points are smooth all at once.
| 27 | https://mathoverflow.net/users/582 | 20679 | 13,732 |
https://mathoverflow.net/questions/20615 | 5 | I have a lemma about antichains that I think should be already known, but I can't find it anywhere. I am looking for a reference to this result that I can use in my paper, so that I don't have to include the proof.
Let $\mathcal{F}$ be an antichain on finite universe $U$, such that there are no $m$ distinct subsets $S\_1, S\_2, \ldots, S\_m \in \mathcal{F}$ such that $|S\_1 \cap S\_2 \ldots \cap S\_m| \geq n$. Then $|\mathcal{F}| \leq 2m|U|^n$.
(The bound can actually be made a little bit sharper, but this is sufficient for my purposes.)
Here is a proof of why this claim is true. We count the number of sets in $\mathcal{F}$ in two steps.
1) Since $\mathcal{F}$ is an antichain, all sets in it are distinct. Hence there are less than $|U|^k$ sets in $F$ of size $k$. Hence the number of sets in $\mathcal{F}$ of size at most $n-1$ is less than $\sum \_{k=0}^{n-1} |U|^k = (U^n - 1)/(U - 1) \leq U^n$.
2) Now we count the number of sets with at least $n$ elements. For each set in $\mathcal{F}$ with at least $n$ elements, select $n$ of its elements arbitrarily and group the sets according to the chosen $n$ elements. If some group contains at least $m$ subsets of $\mathcal{F}$, then the $n$ elements that define the group are in the common intersection of these $m$ subsets, and this violates our assumption. Hence every group has less than $m$ subsets in it. Since there are less than $|U|^n$ different groups, and each group has less than $m$ subsets in it, there are less than $m|U|^n$ subsets in $\mathcal{F}$ of size at least $n$.
Adding the numbers from (1) and (2) yields that the total number of subsets in $\mathcal{F}$ is bounded by $2m|U|^n$.
Can anyone tell me if this is already known and under which name? A reference would be greatly appreciated. Thanks!
| https://mathoverflow.net/users/5200 | Maximum size of antichain if no m subsets have a common intersection of size n | Note that the interesting case is if $n \leq U/2$. Otherwise, your bound is worse than the bound $\binom{U}{\lfloor U/2 \rfloor}$ given by [Sperner's Theorem](http://en.wikipedia.org/wiki/Sperner_family), for the size of *any* antichain on the universe $U$. So assuming that $n \leq U/2$, we can improve the bound in (1) from $U^n$ to $\binom{U}{n-1}$, by the [LYM inequality](http://en.wikipedia.org/wiki/Lubell%E2%80%93Yamamoto%E2%80%93Meshalkin_inequality). And, as Robin noted, the bound in (2) can be improved to $(m-1) \binom{U}{n}$. Adding, (1) and (2) gives the bound $\binom{U}{n-1}+(m-1) \binom{U}{n}$.
| 1 | https://mathoverflow.net/users/2233 | 20680 | 13,733 |
https://mathoverflow.net/questions/20667 | 8 | The Original Fitch Cheney puzzle goes like this:
>
> A Volunteer from the crowd chooses any
> five cards at random from a deck, and
> hands them to you so that nobody else
> can see them. You glance at them
> briefly, and hand one card bakc, which
> the volunteer then places face down on
> the table to one side. You quickly
> place the remaining four cards face up
> on the table, in a row from left to
> right. Your confederate, who has not
> been privy to any of the proceedings
> so far, arrives on the scene, inspects
> the faces of the four cards, and
> promptly names the hidden fifth card.
>
>
>
The solution to this is:
>
> One of the 4 suits must be represented 2 times in your set of 5 cards due to the pigeon hole principle. Consider the 13 cards of that suit A=1,... K=13 to be arranged in a clockwise circle. We can see that the cards are at most 6 away from each other, meaning counting clockwise one of them lies at most 6 vertices past the other. Use the "higher" one as the hidden card and place the "lower" one as the first card face up on the table.
>
>
> Now, using an established value for all 52 cards in the deck, the remaining 3 cards can be placed in 6 orders, Low-Middle-High=1, LHM=2, MLH=3, MHL=4, HLM=5, and HML=6, and give each one of these a value of +1,..., +6 from the first card.
>
>
> For example, Jc, 2c, 3h, 4d, 2s are handed to you. Choosing the lower club, (2-Jack(11) Mod 13 is 4, so the Jack is the lower one and the 2 is +4 from the jack. Hide the 2c and place the Jc in the first spot. Then using MHL, our +4 value, we arrange the remaining 3 cards with the 3 first, than the 4, now the 2. This implies that the hidden card is a club, and it is +4 from the Jc in the first spot, so we know it is the 2c.
>
>
>
The 124 card solution is discussed in the January 2001 issue of Emissary or Michael Kleber’s article in The Mathematical Intelligencer, Winter 2002 (which I don't have access to).
It can be shown that with 5 cards there is a strategy to do the trick on a deck of size up to 124 cards (n!+(n-1)).
My question is this: With the audience choosing *n* cards out of a deck of size (n!+(n-1)) and one card being hidden, how many unique strategies could the magician and the assistant use?
| https://mathoverflow.net/users/2655 | Generalization of Finch Cheney's 5 Card Trick | Kleber's paper will certainly point you in the right direction if you can find it. (I have a printed out copy, and I don't remember where on the web I got it. Sorry.)
In it, he suggests thinking about a strategy as a pairing up of the $(n!+n-1)\\_{n-1}$ messages with the ${{n!+n-1}\choose n} = (n!+n-1)\\_{n-1}$ hands the audience can give you. Of course, you can only pair a message with one of the n! hands that contain its cards, and you can only pair a hand with one of the n! messages you can make with it. So this is just like finding a perfect matching on an n!-regular bipartite graph, and by Hall's Marriage Theorem this is possible.
The bipartite graph you draw here is quite symmetric: aside from being n!-regular, it's also vertex transitive, so any of the n! edges you could choose from a particular vertex v are part of some perfect matching (since one of them is). This is where Kleber's claim that there are "at least n!" strategies come from.
We can get a much better lower bound if [this paper](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6WHT-45SRMS1-9&_user=520880&_coverDate=01%2F31%2F1998&_rdoc=1&_fmt=high&_orig=search&_sort=d&_docanchor=&view=c&_acct=C000023460&_version=1&_urlVersion=0&_userid=520880&md5=aa3f297a22e5087962ec4c451b0cc68b) is to be believed. Here it says there are at least $\left(\frac{(n!-1)^{n!-1}}{n!^{n!-2}}\right)^{{n!+n-1}\choose n}$ strategies, which for n=2 isn't very enlightening (it says there's at least one, when we know there are really two), but for n>2 puts our previous estimate to shame: with a three-card hand (and thus an eight-card deck), we get at least 2.54x10^21 strategies, a number so fantastically larger than our previous estimate of 6 that I'm still a little bit in shock. (It is, at least, many orders of magnitude lower than the naive upper bound of 6^56, where for each of the 3-subsets of 8 we choose one of the possible messages it could send without worrying about overlap with other hands.)
Anyway, I haven't read the linked paper, but the abstract suggests we might not get a much better lower bound than this. To improve, one might look for results on vertex transitive k-regular bipartite graphs, but I haven't found any.
| 6 | https://mathoverflow.net/users/1060 | 20735 | 13,763 |
https://mathoverflow.net/questions/20692 | 4 | Let $G= (\mathbb{Z} \bigoplus \mathbb{Z}) \star (\mathbb{Z} \bigoplus \mathbb{Z})$, where $\star$ denotes the free product, let F be the commutator subgroup of G, it is free by a theorem of Kurosh. Find a proper normal subgroup of F (other than the trivial one) such that it is of infinite index.
| https://mathoverflow.net/users/5218 | Finite index normal subgroups of a free group. | The commutator subgroup $F' = [F:F]$ of $F$. It is normal. $F$ is not abelian, so $F'$ is nontrivial. The quotient $F/F'$ is a free abelian group of infinite rank, so $[F:F']$ is infinite.
| 2 | https://mathoverflow.net/users/1149 | 20743 | 13,768 |
https://mathoverflow.net/questions/20750 | 4 | I remember to have seen a big list in the EGA of properties $(P)$ such that:
if $f : X \to Y$ has $(P)$ then, $f\_{(S')} : X\_{(S')}\to Y\_{(S')}$ has $(P)$, where $f\_{(S')}$ is the morphism $f$ after a base change $S'\to S$., etc. but I can't find it now...
Does anyone know where I can find such a list ?
(I am interested by $(P)$ = "to be a closed map")
| https://mathoverflow.net/users/2330 | Properties stable under base change in algebraic geometry | One list I've seen is in Appendix C from a course 'Rational Points on Varieties' taught by Bjorn Poonen. Here's the link:
<http://math.mit.edu/~poonen/papers/Qpoints.pdf>
Also, the appendix to the book of Gortz and Wedhorn 'Algebraic Geometry 1: Schemes With Examples and Exercises' is a great reference.
| 15 | https://mathoverflow.net/users/386 | 20751 | 13,771 |
https://mathoverflow.net/questions/20646 | 4 | Are there any known reversible pairing functions $f: \mathbb N \times \mathbb N \to \mathbb N$ that can be computed in constant time (FAC⁰)?
| https://mathoverflow.net/users/2644 | Are there any pairing functions computable in constant time (AC⁰) | Interleaving the binary encodings of the two numbers a and b seems to be the best solution:
For example the encoding of
a = 20d = 10100b
b = 5d = 101b
We interleave the bits starting with the least significant bits (we pad shorter numbers with 0's so they are the same length).
The resulting paired number is 0100110010b = 306d
This pairing function can be computed and reversed by a constant depth (depth 1?) circuit and so is in FAC0.
See:
- <http://mathworld.wolfram.com/PairingFunction.html>
- Pigeon, P. Contributions à la compression de données. Ph.D. thesis. Montreal, Université de Montréal, 2001. (page 115)
| 5 | https://mathoverflow.net/users/2644 | 20755 | 13,775 |
https://mathoverflow.net/questions/20702 | 9 | I would like to know some reference (articles, books...) about any kind of moduli spaces of any of the following objects:
* vector bundles
* torsion-free sheaves
* principal bundles
* parabolic bundles
over *singular* algebraic curves (reducible or not), in any of the following frameworks:
* algebraic geometry (in characteristic zero and in positive characteristic)
* holomorphic geometry
* integrable systems
* gauge theory
* differential geometry
* topology
* ...anything you like...
I would be particularly glad to have some reference about torsion-free sheaves in the algebro-geometric setting.
Thanks
---
**Edit:** I should emphasize that my reference request is about some structures over **singular** curves. The freedom I expect in a typical answer should be on the structure (e.g. bundles, torsion-free sheaves,...) and on the viewpoint (e.g. pure algebraic geometry, trascendental methods, ...), but the base curve must be singular (for the answer not to be offtopic).
| https://mathoverflow.net/users/4721 | Reference request: Moduli spaces of bundles over singular curves | Some of the many (semi)standard references are below (with no claims to completeness or representativeness, if that's a word -- just the first references that came to mind). My feeling is the subject is still very much in its infancy however, for example one would like to know the standard package of nonabelian Hodge theory results for singular curves (geometry of Higgs bundles and local systems, Hitchin fibration, its self-duality etc) and there are partial results but no complete picture as far as I know.
Caporaso, Lucia A compactification of the universal Picard variety over the moduli space of stable curves. J. Amer. Math. Soc. 7 (1994), no. 3, 589--660.
Pandharipande, Rahul A compactification over $\overline {M}\_g$ of the universal moduli space of slope-semistable vector bundles. J. Amer. Math. Soc. 9 (1996), no. 2, 425--471.
Seshadri, C. S. Moduli spaces of torsion free sheaves on nodal curves and generalisations. I. Moduli spaces and vector bundles, 484--505, London Math. Soc. Lecture Note Ser., 359, Cambridge Univ. Press, Cambridge, 2009.
(and earlier papers of his)
arXiv:1001.3868 Title: Autoduality of compactified Jacobians for curves with plane singularities
Authors: D.Arinkin
--see this reference for refs to the vast literature by Altman-Kleiman and Esteves-Kleiman on compactified Jacobians
Kausz, Ivan A Gieseker type degeneration of moduli stacks of vector bundles on curves. Trans. Amer. Math. Soc. 357 (2005), no. 12, 4897--4955 (electronic).
Schmitt, Alexander H. W. Singular principal $G$-bundles on nodal curves. J. Eur. Math. Soc. (JEMS) 7 (2005), no. 2, 215--251.
(and earlier papers of his)
| 9 | https://mathoverflow.net/users/582 | 20758 | 13,778 |
https://mathoverflow.net/questions/20712 | 24 | Is there a map $f\colon X \to Y$ of closed, connected, smooth and orientable $n$-dimensional manifolds such that the degree of $f$ is 0 but $f$ is not **homotopic** to a non-surjective map?
**Added**: The motivation is: There is a "mild version" of the Nearby Langrangian conjecture stating: any exact Lagrangian manifold $X \to T^\*Y$ has non-zero degree when composed with the projection $T^\*Y \to Y$. It is known that the map is always surjective. I am looking at a **possible** inbetween stating that the map cannot be homotoped to a non-surjective map.
| https://mathoverflow.net/users/4500 | Do "surjective" degree zero maps exist? | It is a theorem of H. Hopf that a map between connected, closed, orientable n-manifolds of degree 0 is homotopic to a map that misses a point, when n > 2. See D. B. A. Epstein, The degree of a map. Proc. London Math. Soc. (3) 16 1966 369--383, for a "modern" discussion including the analogous situation in the non-orientable case. The same result holds for n = 2, but is more difficult and is due to Kneser. See Richard Skora, The degree of a map between surfaces.
Math. Ann. 276 (1987), no. 3, 415--423, for a thorough discussion of the non-orientable case in dimension 2.
| 37 | https://mathoverflow.net/users/1822 | 20759 | 13,779 |
https://mathoverflow.net/questions/20698 | 6 | I've come up with the following optimization problem in my research. Is this a known problem in graph-theory and/or combinatorial optimization? If not, which of the known problems are the most similar to it?
Let's have a graph $G=(V,E)$ with real positive or negative weights assigned to its edges: $w: E \rightarrow \Re$. The problem is to remove a set of edges ($E\_1$) from $G$ such that the sum of the remaining edges ($E\_2$) is minimized, and no vertex in G has degree less than 2 (i.e. no leaf vertices should exist).
| https://mathoverflow.net/users/5223 | Degree constrained edge partitioning | I'm not sure about published references to this specific problem, but I'm pretty sure it can be solved in polynomial time via a reduction to minimum weight perfect matching, as follows.
Replace each vertex of degree $d$ by a complete bipartite graph $K\_{d-2,d}$ where each of the original edges incident to the vertex becomes incident to one of the vertices on the $d$-vertex side of the bipartition. Set all edge weights in this complete bipartite graph to zero. Next, add a complete graph (with zero edge weights) connecting all of the vertices in the whole graph that are on the $(d-2)$-vertex sides of their bipartitions.
For each vertex $v$ in the original graph, a perfect matching in this modified graph has to include at least two of the original graph's edges incident to $v$, because at least two of the vertices on the $d$-vertex side of the complete bipartite graph that replaces $v$ are not matched within that complete bipartite graph. Because all the other edges have cost zero, the cost of the perfect matching is the same as the cost of the solution it leads to in the original graph.
On the other hand, whenever one has a subgraph of the original graph that includes at least two edges at each vertex, it can be completed to a perfect matching at no extra cost by matching the remaining vertices on the $d$-vertex sides of their bipartition to vertices on the $(d-2)$-vertex sides, and then using the complete graph edges to complete the matching among any remaining unused vertices on the $(d-2)$-vertex sides.
Therefore, the cost of the minimum weight perfect matching in this graph is the same as the cost of the optimal solution to your problem.
Added later: something like this appears to be in "An efficient reduction technique for degree-constrained subgraph and bidirected network flow problems", Hal Gabow, STOC 1983, [doi:10.1145/800061.808776](http://doi.acm.org/10.1145/800061.808776)
| 3 | https://mathoverflow.net/users/440 | 20763 | 13,783 |
https://mathoverflow.net/questions/20773 | 8 | The above title is in fact a special case of what I want to ask.
Certainly we have a well defined notion of Galois extension for $ \mathbb{Q}\_p $. The intersections of these extensions to the ring of integer of the absolute algebraic closure of $\mathbb{Q}\_p$ give us a notion of Galois extensions for $\mathbb{Z}\_p $. ( I know that there is a notion of Galois extension for commutative rings, and I believe that it should give us this. Am I correct?)
Let's go further. Let $A\_K$ be the ring of integer in a finite Galois extension $K$ of $ \mathbb{Q}\_p$. Let $e$ be the ramification degree of $K$ over $\mathbb{Q}\_p$. The injection of $ \mathbb{Z}\_p$ into $A\_K$ will induce an injection of $ \mathbb{Z} / p^n $ into $ A\_K / \mathfrak{p}^{en} $. In this picture, there seems to be some desire to say that $ A\_K / \mathfrak{p}^{en} $ is the correct notion Galois of extension of $ \mathbb{Z} / p^n $. But there are problems; taking this notion of Galois extension, if $K$ is has ramification degree $e >1$, the corresponding extension $ A\_K /p^e $ is not a field (it is not even an integral domain).
Question 1: Is there any notion of Galois extensions corresponding to what I desire?
Question 2: Can a class field theory (i.e a nice description of absolute abelian Galois extension) of $ \mathbb{Z}/p^n$ be developed in this context? Is there any relationship between this and the local class field theory of $\mathbb{Q}\_p$ ( which is the same as that of $\mathbb{Z}\_p $)?
| https://mathoverflow.net/users/2701 | Is there a notion of Galois extension for Z / p^2? | Perhaps not directly answering your questions but something along those lines is Deligne's theory of truncated valuation rings, given in [Les corps locaux de caractéristique $p$, limites de corps locaux de caractéristique 0](http://www.ams.org/mathscinet-getitem?mr=771673).
A truncated valuation ring is an Artin local ring with principal maximal ideal and finite residue field - by Cohen's structure theorem, these are precisely the quotients of rings of integers in local fields by a power of the maximal ideal. Deligne sets up a category using these truncated valuation rings and provides definitions of extensions aswell as a ramification theory for them.
He goes on to show an equivalence between the category of "at most $e$-ramified" separable extensions of a local field $K$ and the category of "at most $e$-ramified" extensions of the length $e$ truncation of the ring of integers of $K$.
The main point of all this is that the behaviour of objects defined over discrete valuation rings is often determined by their reduction modulo a power of the maximal ideal i.e. on truncated data. It also ties in with Krasner's idea (hence the title of Deligne's paper) that local fields of characteristic $p$ are limits of local fields of characteristic 0 as the absolute ramification index tends to infinity.
| 4 | https://mathoverflow.net/users/3143 | 20783 | 13,792 |
https://mathoverflow.net/questions/20782 | 50 | ### Background
Recall that, given two commutative rings $A$ and $B$, the set of morphisms of rings $A\to B$ is in bijection with the set of morphisms of schemes $\mathrm{Spec}(B)\to\mathrm{Spec}(A)$. Furthermore, we know that Spec$(A)$ has a base of open affine subsets, the "basic" or "principal" open affines $D(f)$ for all $f\in A$. Furthermore, $D(f)\cong \mathrm{Spec}(A\_f)$ as schemes, and the inclusion $D(f)\hookrightarrow\mathrm{Spec}(A)$ corresponds to the localization map $A\to A\_f$.
But answers to a [recent MathOverflow question](https://mathoverflow.net/questions/7153/open-affine-subscheme-of-affine-scheme-which-is-not-principal) show that open affine subschemes of affine schemes can arise in other ways.
---
### Question
In order to try to make sense of the situation above, I'd like to know the following.
>
> Given a commutative ring $A$, is there a "ring-theoretic" characterization of the ring homomorphisms $A\to B$ that realize $\mathrm{Spec}(B)$ as an open affine subscheme of $\mathrm{Spec}(A)$ (more precisely, those morphisms such that the induced map $\mathrm{Spec}(B)\to\mathrm{Spec}(A)$ is an open immersion)?
>
>
>
Of course, "ring-theoretic" is a bit vague. Let's certainly avoid any tautological characterizations. I would prefer if an answer didn't make any reference to the Zariski topology (for instance, the morphisms $A\to A\_f$ make perfect sense without the Zariski topology), but I'm not sure whether that's reasonable.
---
**Update:** I received two great answers, thank you both! I chose the one that was closer to the kind of condition that I had in mind. But Dan Petersen's answer was also very interesting and unexpected.
| https://mathoverflow.net/users/778 | Ring-theoretic characterization of open affines? | **Theorem 1**: Let $R$ be an integral domain with field of fractions $K$, and $R \to A$ a homomorphism. Then $Spec(A) \to Spec(R)$ is an open immersion if and only if $A=0$ or $R \to K$ factors through $R \to A$ (i.e. $A$ is birational over $R$) and $A$ is flat and of finite type over $R$.
Proof: Assume $Spec(A) \to Spec(R)$ is an open immersion and $A \neq 0$. It is known that open immersions are flat and of finite type. Thus the same is true vor $R \to A$. Now $R \to K$ is injective, thus also $A \to A \otimes\_R K$. In particular, $A \otimes\_R K \neq 0$. Open immersions are stable under base change, so that $Spec(A \otimes\_R K) \to Spec(K)$ is an open immersion. But since $Spec(K)$ has only one element and $Spec(A \otimes\_R K)$ is non-empty, it has to be an isomorphism, i.e. $K \to A \otimes\_R K$ is an isomorphism. Now $R \to A \to A \otimes\_R K \cong K$ is the desired factorization.
Of course, the converse is not as trivial. It is proven in the paper
>
> Susumu Oda, On finitely generated birational flat extensions of integral domains
> Annales mathématiques Blaise Pascal, 11 no. 1 (2004), p. 35-40
>
>
>
It is available [online](http://ambp.cedram.org/item?id=AMBP_2004__11_1_35_0). In the section "Added in Proof." you can find some theorems concerning the general case without integral domains. In particular, it is remarked that in E.G.A. it is shown that
**Theorem 2**: $Spec(A) \to Spec(R)$ is an open immersion if and only if $R \to A$ is flat, of finite presentation and an epimorphism in the category of rings.
More generally, in EGA IV, 17.9.1 it is proven that a morphism of schemes is an open immersion if and only if it is flat, a (categorical) monomorphism and locally of finite presentation.
There are several descriptions of epimorphisms of rings (they don't have to be surjective), see [this](https://mathoverflow.net/questions/109/what-do-epimorphisms-of-commutative-rings-look-like) MO-question.
| 40 | https://mathoverflow.net/users/2841 | 20792 | 13,797 |
https://mathoverflow.net/questions/20789 | 10 | Suppose we want to approximate a real-valued random variable $X$ by a discrete random variable $Z$ with finitely many atoms. Suppose all moments of $X$ is finite. We want to match the moments of $X$ up to the $m^{\rm th}$ order:
(1) $\mathbb{E}[X^k] = \mathbb{E}[Z^k]$ for $k = 1, \ldots m$.
Here is a positive result, which is a simple consequence of convex analysis ([Caratheodory's theorem](http://en.wikipedia.org/wiki/Carath%25C3%25A9odory%2527s_theorem_%2528convex_hull%2529)): there exists $Z$ with at most $m+1$ atoms such that (1) holds.
Here are my questions:
1) Is there a **converse** result about this? Say $X$ has an absolutely continuous distribution supported on $\mathbb{R}$ (e.g. Gaussian). When $m$ is large, given that $Z$ has only $m$ atoms, can we conclude that we cannot approximate all $2m$ moments of $X$ well, i.e., can we lower bound the error
$\max\_{1 \leq k \leq 2m}|\mathbb{E}[X^k] - \mathbb{E}[Z^k]|$? My intuition is the following: for a Gaussian $X$, $\mathbb{E}[X^k]$ grows like $k^{\frac{k}{2}}$ superexponentially. When we find a $Z$ who matches all moments of $X$ up to $m$, it cannot catch up with higher-order moments $X$; if $Z$ matches all moments from $m+1$ up to $2m$, then its low-order moments will be quite different from $X$.
2) Is there an **efficient algorithm** to compute the location and weights of the approximating discrete distribution? Does there exist a table to record these for approximating common distribution (e.g. Gaussian) for each fixed $m$? It could be very handy...
3) I heard from folklore that when (1) holds, the total variation distance between their distributions can be upper bounded by, say, $e^{-m}$ or $1/m!$. Of course, this won't be true for a discrete $Z$. But let's say $X$ and $Z$ both has smooth and bounded density on $\mathbb{R}$. Could this be true? Now two characteristic functions matches at $0$ up to $m^{\rm th}$ derivatives. They should be pretty close?
| https://mathoverflow.net/users/3736 | approximate a probability distribution by moment matching | For (1) and (2) just forget about probability and recall everything you ever learned about orthogonal polynomials and the Gauss quadrature formulae.
3) is false as stated: there are plenty of Schwartz functions orthogonal to all polynomials, so you can have all moments coincide and still have a large distance (in any sense). *Something like that* may be true but I cannot think of any good formulation right away.
| 5 | https://mathoverflow.net/users/1131 | 20793 | 13,798 |
https://mathoverflow.net/questions/20788 | 2 | Mathematically, I know what a semigroup is: It is a set S along with an *associative* binary operation $\* : S \times S \rightarrow S$. So far, so good.
From a computational perspective, one can represent a semigroup as the tuple $\left< S,\* \right>$, or my preference, as a record { S: type; $\* : S \times S \rightarrow S$} which is dependently typed. So, for example, {S = $\mathbb{N}$; $\* = + $} is a semigroup [assuming I have the naturals with addition already built].
Well, actually, it's not -- who says that I defined $+$ properly? Maybe I made a mistake and the $+$ that I used for $\mathbb{N}$ isn't associative. So I sure shouldn't be able to build {S = $\mathbb{N}$; $\* = + $} and have that have 'type' semigroup.
So is the proof of associativity part of the *type* 'semigroup' (as it is in Coq for example), or is it part of the *input* to the 'constructor' for the semigroup type ? [The constructor is allowed to then forget the proof, at least for computational purposes].
One wrinkle here is that while the proof of associativity doesn't seem to say much, for the *term algebra* over the semigroup, it does induce two 'associator' functions which can perform the rewrite of $\lceil a \* (b \* c) \rceil \leftrightarrow \lceil (a \* b) \* c\rceil$ (where I use the $\lceil\ \cdot\ \rceil$ brackets to denote working over *terms* since semantically $a\*(b\*c)$ and $(a\*b)\*c$ denote the same thing so there is nothing to do). That associator is really quite useful [and much more so when you go to things like rings and fields, where the induced rewrites on the term algebra give rise to what can rightly be called a 'simplifier']. So the proof is quite useful in that sense.
The question boils down to: where does the 'proof' that the binary operation is associative really belong in the theory of a *single* semigroup, where by 'theory' here I mean both the semantic theory and the induced equational theory of the term algebra? Once I have established that $\*$ is associative, can I really throw away the proof as it is not going to be used again?
(I would really want to also ask why is it that in classical mathematics, proofs are crucial, but somehow not so important that definitions of standard objects omit them. Asking that would likely be ruled as off-limits for MO as being too 'philosophical'...)
| https://mathoverflow.net/users/3993 | What is a semigroup or, what do I do with that associativity proof? | The answer to your real question is that all proofs are typically equated in set theories (more accurately, in topoi). That is, the interpretation of a proposition is as an element of the lattice of truth values, which in classical logic can take on either the value true or false.So, this means that if you give a semigroup as a dependent record (indexed over some family of small sets $U$, to avoid thinking about size issues):
$$(S \in U, \* \in S \to S \to S, assoc : \forall x,y,z \in S.\; (x \* y) \* z = x \* (y \* z))$$
then for any two elements of this set, the $assoc$ component corresponding to the associativity proof is always the value true. This means the set of semigroups is trivially isomorphic to the subset
$$\{ (S \in U, \* \in S \to S \to S) \;|\; \forall x,y,z \in S.\; (x \* y) \* z = x \* (y \* z)) \}$$
This holds even in intuitionistic topoi, since the key point is that in this setting the meaning of propositions are elements of a lattice of truth values, and the truth of a proposition is its equality to the lattice element true.
(As an aside, this highlights the fact that a theory may have an intuitionistic entailment relation, but not have a constructive reading of propositions such as a realizability or props-as-types reading. This happens a lot whenever you work relative to some open-endedly extensible gadget -- the property that propositions about your gadget should continue to be valid under extensions basically ends up defining a Kripke structure.)
| 3 | https://mathoverflow.net/users/1610 | 20799 | 13,801 |
https://mathoverflow.net/questions/20764 | 13 | This question is inspired by a statement of Atiyah's in "Geometry and Physics of Knots" on page 24 (chapter 3 - Non-abelian moduli spaces).
Here he says that for a Riemann surface $\Sigma$ the first cohomology $H^1(\Sigma,U(1))$, where $U(1)$ is just complex numbers of norm 1, parametrizes homomorphisms $$\pi\_{1}(\Sigma)\to U(1).$$
This is fine by me, after all by Brown Representability we know $$H^1(\Sigma,G)\cong [\Sigma,BG]=[B\pi\_{1}\Sigma,BG]$$ since we know that Riemann surfaces are $K(\pi\_{1},1)$s. We then use the handy fact from Hatcher Prop. 1B.9 (pg 90) that shows that...
>
> For $X$ a connected CW complex and $Y$ a $K(G,1)$ every homomorphism $\pi\_1(X,x\_0)\to\pi\_1(Y,y\_0)$ is induced by a map $(X,x\_0)\to(Y,y\_0)$ that is unique up to homotopy fixing $x\_0$.
>
>
>
So group homomorphisms $\pi\_{1}(\Sigma)\to U(1)$ correspond on the nose with first cohomology of the $\Sigma$ with coefficients in $U(1)$.
EDIT: To be accurate the Hatcher result shows we have an injection of group homomorphisms into homotopy classes of maps. Does anyone know if every homotopy class of maps is realized by a group homomorphism?
What bothers me is what comes next. Now replace $U(1)$ with $G$ - any compact simply connected Lie group, take $G=SU(n)$ for example. Now Atiyah claims that $H^1(\Sigma,G)$ parametrizes **conjugacy classes** of homomorphisms $\pi\_{1}(\Sigma)\to G$.
Now if the fundamental group or the Lie group were abelian this would be the same statement, but higher genus Riemann surfaces (genus greater than 1) have non-abelian fundamental groups and Atiyah is looking specifically at non-abelian $G$. Also, it seems that the statement of Brown Representability requires abelian coefficient groups, so I am stuck.
Does anyone know a clean way of proving Atiyah's claim?
EDIT2: I renamed the question to draw in the "right" people. I think the answer has to do with the fact that principal G-bundles over a Riemann surface are determined by maps of $\pi\_1(\Sigma)\to G$ (can someone explain why?). This is related to local systems and/or flat connections, which I don't understand well. Thanks!
| https://mathoverflow.net/users/1622 | Representations of \pi_1, G-bundles, Classifying Spaces | Since it is not always clear what $H^1(X;G)$ means for a non-abelian group, Atiyah might have meant this loosely, perhaps using Cech 1-cocycles to construct flat $G$ bundles, with homologous cycles giving isomorphic bundles. The moduli space of flat bundles is homeomorphic (and real analytically isomorphic) to the space of conjugacy classes of $G$ representations via the holonomy. I suspect this is what Ben is hinting at in his answer above.
But another interpretation in terms of homotopy classes of maps to $BG$ is as follows. If you give $G$ the discrete topology, then $BG$ is a pointed space with a fixed choice of isomorphism $\pi\_1(BG)\cong G$. So
$Hom(\pi\_1(X),G)$ is in bijective correspondence with the *pointed* homotopy classes of maps $[X,BG]\_0$. The *free* homotopy classes $[X,BG]$ are then in bijective correspondence with the conjugacy classes $Hom(\pi\_1(X),G)/conj$, since the action of $G=\pi\_1(BG)$ on the pointed homotopy classes corresponds to conjugation, with quotient the free homotopy classes. So if you define $H^1(X;G)=[X,BG]$ (unbased), you get Atiyah's statement.
In the case of Atiyah's book, though, more important that the notation for the space of conjugacy classes of reps is the fact that at a representation $r:\pi\_1(\Sigma)\to G$, the (usual, twisted by the adjoint rep) cohomology $H^1(\Sigma; g\_r)$ ($g$ the lie algebra of $G$)is the Zariski tangent space to the variety of conjugacy classes of representations at $r$, i.e. to the moduli space of flat $G$ connections on $\Sigma$. The cup product
$H^1(\Sigma; g\_r)\times H^1(\Sigma; g\_r)\to R$ determines a symplectic form on this variety, and a holomorphic structure on $\Sigma$ induces a complex structure on this variety, which reflects itself in the Zariski tangent space as an almost complex structure $J:H^1(\Sigma; g\_r)\to H^1(\Sigma; g\_r)$ which coincides with the Hodge $\*$ operator on harmonic forms.
Incidentally, all principal $SU(2)$ bundles over a Riemann surface are topologically trivializable.
| 6 | https://mathoverflow.net/users/3874 | 20804 | 13,804 |
https://mathoverflow.net/questions/20802 | 7 | Let $f: X\to Y$ be a finite (surjective) morphism between two algebraic varieties. I know when $X$ and $Y$ are non-singular and $\dim Y =1$, $f$ is flat. But in general, is it true that $f$ is a flat morphism?
| https://mathoverflow.net/users/2348 | Finite morphisms between algebraic varieties are flat? | If $X$ and $Y$ are both regular, then this is true. In fact, it's true more generally if $Y$ is regular and $X$ is Cohen-Macaulay (Eisenbud, Commutative Algebra, Corollary 18.17). In general it's certainly false.
| 15 | https://mathoverflow.net/users/1594 | 20806 | 13,806 |
https://mathoverflow.net/questions/20225 | 13 | I have read more than once that the Littlewood-Paley (LP) projections of a function (i.e. decomposing a function into parts with frequency localization in different octaves) behave in some sense like iid random variables.
I am also aware of some facts (like inequalities for square functions vs. Khinchine Inequality) which "look similar".
Is there any precise way of stating this similarity?
And why do we have this similarity?
Can we somehow interpret the LP projections as something like independent random variables?
A related question concerns systems of functions of the form
$$
f\_k(\cdot):=f(n\_k \cdot )\quad {k\geq 1} ,
$$
with $(n\_k)\_{k\geq 1}$ a lacunary sequence. Also in this case (under suitable assumptions) the functions $f\_k$ behave like iid random variables in the sense that they satisfy the Central Limit Theorem and the LIL.
| https://mathoverflow.net/users/5112 | Why do Littlewood-Paley projections behave like iid random variables | It is much better to replace 'iid random variables' above by 'martingale differences.'
The usual Littlewood-Paley square function is closely related to the Haar square function.
And the Haar square function is exactly a martingale square function, namely a sum of
squares of martingale differences.
One can pass back and forth, from martingale to continuous analogs. A striking method to
do this was found by Stefanie Petermichl, when she found a simple way to obtain the
Hilbert transform from a modification of a martingale multiplier.
| 7 | https://mathoverflow.net/users/1158 | 20808 | 13,807 |
https://mathoverflow.net/questions/20791 | 12 | Let $A, B, C$ and $D$ be abelian varieties (over $\mathbb{C}$) such that $A \times B \cong C \times D$, and $A \cong C$. From the irreducibility of abelian varieties, we can say that $B$ and $D$ are isogeneous. But do we actually have $B \cong D$?
| https://mathoverflow.net/users/5197 | isomorphism of abelian varieties | This is false even for elliptic curves over $\mathbb{C}$. This was proved by T. Shioda in "Some remarks on abelian varieties" J. Fac. Sci. Univ. Tokyo Sect. IA Math. 24 (1977), no. 1, 11-21, <http://repository.dl.itc.u-tokyo.ac.jp/dspace/bitstream/2261/6164/1/jfs240102.pdf>.
| 20 | https://mathoverflow.net/users/4790 | 20811 | 13,810 |
https://mathoverflow.net/questions/20784 | 4 | Hi everyone,
Let $X$ be a smooth projective variety over a field $k$ and let $L$ be a line bundle on $X$. I'm reading the article *Heights for line bundles on arithmetic varieties* and there one speaks of $\textrm{Pic}^L(X)$. What is that? And above that, if $L$ and $K$ are algebraically equivalent line bundles, why is $\textrm{Pic}^L(X) = \textrm{Pic}^K(X)$? (Here algebraic equivalence boils down to $L = K \mod \textrm{Pic}^0(X)$.) Maybe it is even better to just ask why $\textrm{Pic}^L(X) = \textrm{Pic}^K(X)$ when $L$ and $K$ are isomorphic.
I'm guessing that taking $L=\mathcal{O}\_X$ will give the connected component of the Picard scheme. So I ask, is $\textrm{Pic}^L(X)$ also an abelian variety?
| https://mathoverflow.net/users/4333 | For a line bundle L on a smooth projective variety X, what is meant by Pic^L(X) | Let me address the last part of your question.
Let $X$ be a smooth, projective variety over an arbitrary ground field $k$.
I want to write $Pic^{[L]}(X)$ instead of $Pic^L(X)$ -- i.e., to make explicit that the variety depends only on the Neron-Severi class of $L$ -- for reasons which will become clear shortly.
Suppose first that $L$ is algebraically equivalent to $0$. Then $Pic^{[L]}(X) = Pic^0(X)$, so certainly it is an abelian variety.
Next suppose that $L$ is a $k$-rational line bundle on $X$. Then $Pic^{[L]}(X)$ is not literally an abelian variety, because it is a nonidentity coset of a group rather than a group itself. However, it is canonically isomorphic to the abelian variety $Pic^0(X)$ just by mapping a line bundle $M$ to $M - L$. So it might as well be an abelian variety, really.
Finally, supose that $L$ is not itself $k$-rational but that its Neron-Severi class $L$ is rational -- i.e., $L$ is given by a line bundle over the algebraic closure which is algebraically equivalent to each of its Galois conjugates. Then $Pic^{[L]}(X)$ is a well-defined principal homogenous space of the Picard variety $Pic^0(X)$ but need not have any $k$-rational points. For instance, suppose that $X$ is a curve. Then the Galois action on the Neron-Severi group is trivial, so taking $L/\overline{k}$ to be any degree $n$ line bundle, we get $Pic^{[L]}(X) = Pic^n(X) = Alb^n(X)$, a torsor whose $k$-rational points parameterize $k$-rational divisor classes of degree $n$. (Note that here when I write $Pic^0(X)$ I am talking about the Picard **variety** rather than the degree $0$ part of the Picard group. More careful notation would be $\underline{\operatorname{Pic}}^0(X)$.)
In particular, if $X$ is a genus one curve, then there is a canonical isomorphism $X \cong Pic^1(X)$, so $Pic^1(X)$ can be endowed with the structure of an abelian variety iff $X$ has a $k$-rational point.
Some further material along these lines can be found in Section 4 of
[http://alpha.math.uga.edu/~pete/wc2.pdf](http://alpha.math.uga.edu/%7Epete/wc2.pdf)
| 4 | https://mathoverflow.net/users/1149 | 20813 | 13,812 |
https://mathoverflow.net/questions/20777 | 11 | A Hermitian symmetric space is a connected complex manifold with a hermitian metric on which the group of holomorphic isometries acts transitively, and which satisfies the following extra condition: each point (equivalently, some point) is an isolated fixed point of some (holomorphic, isometric) involution of the space. There is a beautiful classification theory of Hermitian symmetric spaces, and this is the starting point for the study of Shimura varieties.
The first four conditions (connected, complex, hermitian, homogeneous), which define a Hermitian homogeneous space, all seem natural enough to me. My question: why the need for the extra condition regarding the involution?
This is deliberately vague because I'm not quite sure what form the most satisfying answer will take. But here are a couple of more specific versions of the question. (1) What sort of spaces are we excluding by imposing this condition, and why? (2) Are there other conditions which might seem more natural to me which are equivalent (or equivalent some of the time)? For instance if you were to tell me that a Hermitian homogeneous space that's negatively curved and simply-connected always has such an involution, I would be completely satisfied.
| https://mathoverflow.net/users/379 | Hermitian symmetric spaces vs Hermitian homogeneous spaces | Here is a geometric answer to (2), or more precisely a slight and classical reinterpretation. First, note that Hermitian symmetric spaces fit into the more general concept of riemannian symmetric space; this can helps you find references.
Denote your space by $X$ (assumed to be Riemannian, or Hermitian if you prefer) and take a point $p$. Consider the geodesic symmetry around $p$: it maps a point $q$ to the point $\sigma\_p(q)$ so that $\sigma\_p(q), p, q$ lie on a constant speed geodesic, at times $-1,0,1$ respectively. This map is at least defined locally. Then $X$ is symmetric if and only if for all $p\in X$, $\sigma\_p$ is globally defined and an isometry (ask it to be also holomorphic if your are in the Hermitian setting, but I guess it will automatically be so since it is conjugate to $-\mathrm{Id}$ by the exponential map). Of course, $\sigma\_p$ is your involution.
This condition automatically implies that the isometry group of $X$ acts transitively (because any $q$ is mapped to any $q'$ by the map $\sigma\_p$ where $p$ is a midpoint of $[q,q']$).
It also gives you an involution $\theta$ on the Lie algebra $\mathfrak{g}$ of the isometry group of $X$. Now remark that $\theta$ is a linear endomorphism, and $\theta^2=1$ implies that
$\mathfrak{g}$ decomposes into two components, the eigenspace associated to the eigenvalue $1$ of $\theta$ and the one associated to $-1$. They are usually denoted by $\mathfrak{h}$ and $\mathfrak{p}$; the latter identifies with the tangent space to $X$ at $p$. This is called Cartan decomposition and is fundamental to the study of these spaces.
The most common reference is Helgason's *Differential geometry, Lie groups, and symmetric spaces* but I find it quite difficult to read. In the nonpositively curved case, I found Eberlein's *Geometry of nonpositively curved manifolds* useful.
Concerning the motivation, it seems that the interest in symmetric spaces is that they provide examples between constant curvature spaces, which are very constrained, and homogeneous spaces that are very numerous.
Last, a lead concerning (1). You shall find in Berger's *A panoramic view of Riemannian geometry*, section 15.8.1 page 719 a formula enabling one to compute the curvature of a homogeneous space. I guess it can be used to show the existence of many such spaces, even in the Hermitian world, that are negatively curved but not symmetric.
| 5 | https://mathoverflow.net/users/4961 | 20819 | 13,814 |
https://mathoverflow.net/questions/20740 | 227 | The title really is the question, but allow me to explain.
I am a pure mathematician working outside of probability theory, but the concepts and techniques of probability theory (in the sense of Kolmogorov, i.e., probability measures) are appealing and potentially useful to me. It seems to me that, perhaps more than most other areas of mathematics, there are many, many nice introductory (as well as not so introductory) texts on this subject.
However, I haven't found any that are written from what it is arguably the dominant school of thought of contemporary mainstream mathematics, i.e., from a structuralist (think Bourbaki) sensibility. E.g., when I started writing notes on the texts I was reading, I soon found that I was asking questions and setting things up in a somewhat different way. Here are some basic questions I couldn't stop from asking myself:
[0) Define a Borel space to be a set $X$ equipped with a $\sigma$-algebra of subsets of $X$. This is already not universally done (explicitly) in standard texts, but from a structuralist approach one should gain some understanding of such spaces before one considers the richer structure of a probability space.]
1. What is the category of Borel spaces, i.e., what are the morphisms? Does it have products, coproducts, initial/final objects, etc? As a significant example here I found the notion of the product Borel space -- which is exactly what you think if you know about the product topology -- but seemed underemphasized in the standard treatments.
2. What is the category of probability spaces, or is this not a fruitful concept (and why?)? For instance, a subspace of a probability space is, apparently, not a probability space: is that a problem? Is the right notion of morphism of probability spaces a measure-preserving function?
3. What are the functorial properties of probability measures? E.g., what are basic results on pushing them forward, pulling them back, passing to products and quotients, etc. Here again I will mention that product of an arbitrary family of probability spaces -- which is a very useful-looking concept! -- seems not to be treated in most texts. Not that it's hard to do: see e.g.
[http://alpha.math.uga.edu/~pete/saeki.pdf](http://alpha.math.uga.edu/%7Epete/saeki.pdf)
I am not a category theorist, and my taste for how much categorical language to use is probably towards the middle of the spectrum: that is, I like to use a very small categorical vocabulary (morphisms, functors, products, coproducts, etc.) as often as seems relevant (which is very often!). It would be a somewhat different question to develop a truly categorical take on probability theory. There is definitely some nice mathematics here, e.g. I recall an arxiv article (unfortunately I cannot put my hands on it at this moment) which discussed independence of events in terms of tensor categories in a very persuasive way. So answers which are more explicitly categorical are also welcome, although I wish to be clear that I'm not asking for a categorification of probability theory *per se* (at least, not so far as I am aware!).
| https://mathoverflow.net/users/1149 | Is there an introduction to probability theory from a structuralist/categorical perspective? | $\def\Spec{\mathop{\rm Spec}}
\def\R{{\bf R}}
\def\Ep{{\rm E}^+}
\def\L{{\rm L}}
\def\EpL{\Ep\L}$
One can argue that an object of the right category of spaces in measure theory is not a set equipped with a σ-algebra of measurable sets,
but rather a set $S$ equipped with a σ-algebra $M$ of measurable sets and a σ-ideal $N$ of *negligible sets*, i.e., sets of measure 0.
The reason for this is that you can hardly state any theorem of measure theory or probability theory without referring to sets of measure 0.
However, objects of this category contain less data than the usual *measured* spaces, because they are not equipped with a measure.
Therefore I prefer to call them *enhanced measurable spaces*, since they are measurable spaces *enhanced* with a σ-ideal of negligible sets.
A morphism of enhanced measurable spaces $(S,M,N)→(T,P,Q)$ is a map $S\to T$ such that
the preimage of every element of $P$ is a union of an element of $M$ and a subset of an element of $N$
and the preimage of every element of $Q$ is a subset of an element of $N$.
Irving Segal proved in “[Equivalences of measure spaces](https://doi.org/10.2307/2372178)” (see also Kelley's “[Decomposition and representation theorems in measure theory](https://doi.org/10.1007/bf02052840)”)
that for an enhanced measurable space $(S,M,N)$ that admits a faithful measure (meaning $μ(A)=0$ if and only if $A∈N$) the following properties are equivalent.
1. The Boolean algebra $M/N$ of equivalence classes of measurable sets is complete;
2. The space of equivalence classes of all bounded (or unbounded) real-valued functions on $S$ modulo equality almost everywhere is Dedekind-complete;
3. The Radon-Nikodym theorem is true for $(S,M,N)$;
4. The Riesz representation theorem is true for $(S,M,N)$ (the dual of $\L^1$ is isomorphic to $\L^∞$);
5. Equivalence classes of bounded functions on $S$ form a von Neumann algebra (alias W\*-algebra).
An enhanced measurable space that satisfies these conditions (including the existence of a faithful measure) is called *localizable*.
This theorem tells us that if we want to prove anything nontrivial about measurable spaces, we better restrict ourselves to localizable enhanced measurable spaces.
We also have a nice illustration of the claim I made in the first paragraph:
none of these statements would be true without identifying objects that differ on a set of measure 0.
For example, take a nonmeasurable set $G$ and a family of singleton subsets of $G$ indexed by themselves.
This family of measurable sets does not have a supremum in the Boolean algebra of measurable sets, thus disproving a naive version of (1).
But restricting to localizable enhanced measurable spaces does not eliminate all the pathologies:
one must further restrict to the so-called *compact* and *strictly localizable* enhanced measurable spaces,
and use a coarser equivalence relation on measurable maps: $f$ and $g$ are *weakly equal almost everywhere*
if for any measurable subset $B$ of the codomain the symmetric difference $f^\*B⊕g^\*B$ of preimages of $B$ under $f$ and $g$ is a negligible subset of the domain.
(For codomains like real numbers this equivalence relation coincides with equality almost everywhere.)
An enhanced measurable space is *strictly localizable* if it splits as a coproduct (disjoint union) of σ-finite (meaning there is a faithful finite measure)
enhanced measurable spaces.
An enhanced measurable space $(X,M,N)$ is (Marczewski) compact if there is a compact class $K⊂M$
such that for any $m∈M∖N$ there is $k∈K∖N$ such that $k⊂m$.
Here a *compact class* is a collection $K⊂2^X$ of subsets of $X$ such that for any $K'⊂K$ the following *finite intersection property* holds:
if for any finite $K''⊂K'$ we have $⋂K''≠∅$, then also $⋂K'≠∅$.
The best argument for such restrictions is the following [Gelfand-type duality theorem for commutative von Neumann algebras](https://arxiv.org/abs/2005.05284).
**Theorem.**
The following 5 categories are equivalent.
1. The category of compact strictly localizable enhanced measurable spaces with measurable maps modulo weak equality almost everywhere.
2. The category of hyperstonean topological spaces and open continuous maps.
3. The category of hyperstonean locales and open maps.
4. The category of measurable locales (and arbitrary maps of locales).
5. The opposite category of commutative von Neumann algebras and normal (alias ultraweakly continuous) unital \*-homomorphisms.
I actually prefer to work with the opposite category of the category of commutative von Neumann algebras,
or with the category of measurable locales.
The reason for this is that the point-set definition of a measurable space
exhibits immediate connections only (perhaps) to descriptive set theory, and with additional effort to Boolean algebras,
whereas the description in terms of operator algebras or locales immediately connects measure theory to other areas of the central core of mathematics
(noncommutative geometry, algebraic geometry, complex geometry, differential geometry, topos theory, etc.).
Additionally, note how the fourth category (measurable locales) is a *full* subcategory of the category of locales.
Roughly, the latter can be seen as a slight enlargement of the usual category of topological spaces,
for which all the usual theorems of general topology continue to hold (e.g., Tychonoff, Urysohn, Tietze, various results about paracompact and uniform spaces, etc.).
In particular, there is a fully faithful functor from sober topological spaces (which includes all Hausdorff spaces) to locales.
This functor is not surjective, i.e., there are *nonspatial locales* that do not come from topological spaces.
As it turns out, all measurable locales (excluding discrete ones) are nonspatial.
Thus, measure theory is part of (pointfree) general topology, in the strictest sense possible.
The non-point-set languages (2–5) are also easier to use in practice.
Let me illustrate this statement with just one example: when we try to define measurable bundles of Hilbert spaces
on a compact strictly localizable enhanced measurable space in a point-set way, we run into all sorts of problems
if the fibers can be nonseparable, and I do not know how to fix this problem in the point-set framework.
On the other hand, in the algebraic framework we can simply say that a bundle of Hilbert spaces is a Hilbert W\*-module over the corresponding von Neumann algebra.
Categorical properties of von Neumann algebras (hence of compact strictly localizable enhanced measurable spaces)
were investigated by Guichardet in “[Sur la catégorie des algèbres de von Neumann](https://dmitripavlov.org/scans/guichardet.pdf)”.
Let me mention some of his results, translated in the language of enhanced measurable spaces.
The category of compact strictly localizable enhanced measurable spaces admits equalizers and coequalizers, arbitrary coproducts, hence also arbitrary colimits.
It also admits products (and hence arbitrary limits), although they are quite different from what one might think.
For example, the product of two real lines is *not* $\R^2$ with the two obvious projections.
The product contains $\R^2$, but it also has a lot of other stuff, for example, the diagonal of $\R^2$,
which is needed to satisfy the universal property for the two identity maps on $\R$.
The more intuitive product of measurable spaces ($\R\times\R=\R^2$) corresponds to the spatial
tensor product of von Neumann algebras and forms a part of a symmetric monoidal structure on the category of measurable spaces.
See Guichardet's paper for other categorical properties (monoidal structures on measurable spaces, flatness, existence of filtered limits, etc.).
Another property worthy of mentioning is that the category of commutative von Neumann algebras
is a locally presentable category, which immediately allows one to use the adjoint functor theorem to construct commutative
von Neumann algebras (hence enhanced measurable spaces) via their representable functors.
Finally let me mention pushforward and pullback properties of measures on enhanced measurable spaces.
I will talk about more general case of $\L^p$-spaces instead of just measures (i.e., $\L^1$-spaces).
For the sake of convenience, denote $\L\_p(M)=\L^{1/p}(M)$, where $M$ is an enhanced measurable space.
Here $p$ can be an arbitrary complex number with a nonnegative real part.
We do not need a measure on $M$ to define $\L\_p(M)$.
For instance, $\L\_0$ is the space of all bounded functions (i.e., the commutative von Neumann algebra corresponding to $M$),
$\L\_1$ is the space of finite complex-valued measures (the dual of $\L\_0$ in the ultraweak topology),
and $\L\_{1/2}$ is the Hilbert space of half-densities.
I will also talk about extended positive part $\EpL\_p$ of $\L\_p$ for real $p$.
In particular, $\EpL\_1$ is the space of all (not necessarily finite) positive measures on $M$.
*Pushforward for $\L\_p$-spaces.*
Suppose we have a morphism of enhanced measurable spaces $M\to N$.
If $p=1$, then we have a canonical map $\L\_1(M)\to\L\_1(N)$, which just the dual of $\L\_0(N)→\L\_0(M)$ in the ultraweak topology.
Geometrically, this is the fiberwise integration map.
If $p≠1$, then we only have a pushforward map of the extended positive parts, namely, $\EpL\_p(M)→\EpL\_p(N)$, which is nonadditive unless $p=1$.
Geometrically, this is the fiberwise $\L\_p$-norm.
Thus $\L\_1$ is a functor from the category of enhanced measurable spaces to the category of Banach spaces
and $\EpL\_p$ is a functor to the category of “positive homogeneous $p$-cones”.
The pushforward map preserves the trace on $\L\_1$ and hence sends a probability measure to a probability measure.
To define *pullbacks of $\L\_p$-spaces* (in particular, $\L\_1$-spaces) one needs to pass to a different category of enhanced measurable spaces.
In the algebraic language, if we have two commutative von Neumann algebras $A$ and $B$,
then a morphism from $A$ to $B$ is a usual morphism of commutative von Neumann algebras $f\colon A\to B$
together with an operator valued weight $T\colon\Ep(B)\to\Ep(A)$ associated to $f$.
Here $\Ep(A)$ denotes the extended positive part of $A$.
(Think of positive functions on $\Spec A$ that can take infinite values.)
Geometrically, this is a morphism $\Spec f\colon\Spec B\to\Spec A$
between the corresponding enhanced measurable spaces and a choice of measure on each fiber of $\Spec f$.
Now we have a canonical additive map $\EpL\_p(\Spec A)\to\EpL\_p(\Spec B)$,
which makes $\EpL\_p$ into a contravariant functor from the category of enhanced measurable spaces
and measurable maps equipped with a fiberwise measure to the category of “positive homogeneous additive cones”.
If we want to have a pullback of $\L\_p$-spaces themselves and not just their extended positive parts,
we need to replace operator valued weights in the above definition
by finite complex-valued operator valued weights $T\colon B\to A$ (think of a fiberwise finite complex-valued measure).
Then $\L\_p$ becomes a functor from the category of enhanced measurable spaces to the category of Banach spaces (if the real part of $p$ is at most $1$)
or quasi-Banach spaces (if the real part of $p$ is greater than $1$).
Here $p$ is an arbitrary complex number with a nonnegative real part.
Notice that for $p=0$ we get the original map $f\colon A\to B$ and in this (and only this) case we do not need $T$.
Finally, if we restrict ourselves to an even smaller subcategory defined by the additional condition $T(1)=1$
(i.e., $T$ is a conditional expectation; think of a fiberwise probability measure),
then the pullback map preserves the trace on $\L\_1$ and in this case the pullback of a probability measure is a probability measure.
There is also a smooth analog of the theory described above.
The category of enhanced measurable spaces and their morphisms is replaced by the category of smooth manifolds and submersions,
$\L\_p$-spaces are replaced by bundles of $p$-densities,
operator valued weights are replaced by sections of the bundle of relative 1-densities,
the integration map on 1-densities is defined via Poincaré duality (to avoid any dependence on measure theory) etc.
There is a forgetful functor that sends a smooth manifold to its underlying enhanced measurable space.
Of course, the story does not end here, there are many
other interesting topics to consider: products of measurable spaces,
the difference between Borel and Lebesgue measurability, conditional expectations, etc.
An [index of my writings on this topic](https://mathoverflow.net/questions/49426/is-there-a-category-structure-one-can-place-on-measure-spaces-so-that-category-t/49542#49542) is available.
| 231 | https://mathoverflow.net/users/402 | 20820 | 13,815 |
https://mathoverflow.net/questions/20827 | 25 | A group $G$ is [Hopfian](http://en.wikipedia.org/wiki/Hopfian_group) if every epimorphism $G\to G$ is an isomorphism. A smooth manifold is aspherical if its universal cover is contractible. Are all fundamental groups of aspherical closed smooth manifolds Hopfian?
Perhaps the manifold structure is irrelevant and makes examples harder to construct, so here is another variant that may be more sensible. Let $X$ be a finite CW-complex which is $K(\pi,1)$. If it helps, assume that its top homology is nontrivial. Is $\pi=\pi\_1(X)$ Hopfian?
**Motivation.** Long ago I proved a theorem which is completely useless but sounds very nice: if a manifold $M$ has certain homotopy property, then the Riemannian volume, as a function of a Riemannian metric on $M$, is lower semi-continuous in the Gromov-Hausdorff topology. (And before you laugh at this conclusion, let me mention that it fails for $M=S^3$.)
The required homotopy property is the following: every continuous map $f:M\to M$ which induces an epimorphism of the fundamental group has nonzero (geometric) degree.
This does not sound that nice, and I tried to prove that some known classes of manifolds satisfy it. My best hope was that all *essential* (as in Gromov's "Filling Riemannian manifolds") manifolds do. I could not neither prove nor disprove this and the best approximation was that having a nonzero-degree map $M\to T^n$ or $M\to RP^n$ is sufficient. I never returned to the problem again but it is still interesting to me.
An affirmative answer to the title question would solve the problem for aspherical manifolds. A negative one would not, and in this case the next question may help (although it is probably stupid because I know nothing about the area):
**Question 2.** Let $G$ be a finitely presented group and $f:G\to G$ an epimorphism. It it true that $f$ induces epimorphism in (co)homology (over $\mathbb Z$, $\mathbb Q$ or $\mathbb Z/2$)?
| https://mathoverflow.net/users/4354 | Are fundamental groups of aspherical manifolds Hopfian? | The Baumslag-Solitar group $B(2,3)=\langle a,b\vert ba^2b^{-1}=a^3\rangle$ is not Hopfian. But it has a natural $K(\pi,1)$ given by the double mapping cylinder of $S^1 \rightrightarrows S^1$ where the maps are $z\mapsto z^2$ and $z\mapsto z^3$. This is a finite CW complex.
---
Edit: The double mapping cylinder can be constructed like this. Take a circle $S^1$ and a cylinder $S^1\times I$. Glue one end of the cylinder to the circle by the degree 2 map $z\mapsto z^2$, and glue the other end of the cylinder to the circle by the degree 3 map $z\mapsto z^3$. The $a$ in the presentation above is the loop around the circle, while the $b$ is the loop that goes along the cylinder (whose ends have been brought together, forming a loop).
To see that this is a $K(\pi,1)$, you can check that its universal cover is the product $T\_5\times \mathbb{R}$ of the infinite $5$-regular tree $T\_5$ with the line $\mathbb{R}$. (Think about what this CW complex looks like locally: away from the circle where we glued everything together, it's locally a $2$-manifold. At the circle, we have $2+3=5$ half-planes meeting along their edges.) There is a picture of this universal cover on page 3 of Farb-Mosher, "[A rigidity theorem for the solvable Baumslag-Solitar groups](http://www.math.uchicago.edu/~farb/papers.html)", Inventiones 131 2 (1998), 419-451.
| 19 | https://mathoverflow.net/users/250 | 20829 | 13,820 |
https://mathoverflow.net/questions/20826 | 13 | It is straightforward to show, for example, that the set of zero divisors of a (commutative unital) reduced Noetherian ring is precisely the union of its minimal primes. When else can we say that the set of zero divisors is equal to the union of the minimal primes? Are there other useful cases where this is true? Is there a structure theory for such rings? I'm primarily looking for conditions that do not assume that the ring is Noetherian.
Edit: I messed up this question and received a correct answer for the wrong assumptions, so I have accepted the answer. The reducedness assumption was for the example, not for the general question.
The correct question is in there, but I accidentally changed the title, so I apologize for the confusion.
| https://mathoverflow.net/users/1353 | When is the set of zero divisors equal to the union of the minimal primes in a reduced ring? | The answer is: *always*, and argument is pretty simple:
Let $R$ be a reduced commutative unital ring.
If $a\in R$ is a zero divisor, then $ab=0,$ for some $b\neq 0.$ Hence
$b\not \in 0 = \text{nil}(R)= \bigcap \text{Spec}(R) =\bigcap \text{Specmin}(R),$ where $\text{Specmin}(R)$ states for family of all minimal prime ideals of $R.$
So there exists a minimal prime ideal $\mathfrak{p}\triangleleft R$ s.t. $b\not \in \mathfrak{p}.$ But $\mathfrak{p}$ is prime, and $ab\in \mathfrak{p},$ thus $a\in \mathfrak{p}.$
Hence the set of zero divisors is contained in union of minimal prime ideals.
On the other hand it is well-known fact that minimal prime ideals in commutative unital rings consist of zero divisors, so the set of zero divisors in reduced commutative unital ring is exactly the union of minimal prime ideals.
| 23 | https://mathoverflow.net/users/5080 | 20833 | 13,823 |
https://mathoverflow.net/questions/20671 | 36 | This question makes sense for any topological group $G$, but I'd particularly like to know the answer for $G$ a compact, connected Lie group.
$G$ acts on itself by conjugation. One has the equivariant singular cohomology
$H^{\ast}\_G(G) = H^{\ast}( (G\times EG)/G )$, with integer coefficients, say.
>
> What is $H^{\ast}\_G(G)$?
>
Concretely, $H^{\ast}\_G(G)$ is the target of the Serre spectral sequence for the fibration $$G \hookrightarrow (G\times EG)/G \to BG.$$
When $G$ is path-connected, so that $BG$ is simply connected, this spectral sequence has $E\_2^{p,q}=H^p(BG; H^q(G))$ (trivial local system). Does the spectral sequence always degenerate at $E\_2$? It does when $G$ is abelian, because then the conjugation action is trivial.
| https://mathoverflow.net/users/2356 | What is the equivariant cohomology of a group acting on itself by conjugation? | I asked Dan Freed, who gave a very clean general solution to this problem (as expected).
Here it is (all mistakes in the transcription are mine of course).
The claim is that the equivariant cohomology of G acting on G is indeed the tensor product of cohomology of BG with cohomology of G - in other words the Leray spectral sequence for the fibration $G=\Omega BG \to G/G=LBG \to BG$ degenerates at E\_2. To see this we will use the Leray-Hirsch theorem -- i.e. if we can show that every class on the fiber (G) extends to a class on the total space (LBG) then we will be done. Now the cohomology of G is generated (as an exterior algebra) by its primitive classes, and these all come from the generators of the cohomology of BG by transgression. So we just need to show that these transgressed classes actually lift to LBG.
But there is a nice direct construction of these classes on LBG. Namely we use the tautological correspondence $$LBG \leftarrow S^1\times LBG \rightarrow BG$$
where the right arrow is the evaluation map. Thus given a class on BG we can lift it to $S^1 \times LBG$ and then integrate along the circle to get a class on $LBG$. When we restrict these classes to a fiber, i.e. $G=\Omega BG$, we recover the usual transgression construction. (This can be seen very explicitly with differential forms.. the transgression involves the path fibration $$\Omega BG \to P(BG) \to BG$$
and is given by the same kind of tautological/evaluation construction for the map from the interval times the path space to BG, integrating over an interval.. when restricted to $\Omega BG$, ie closed paths, this integration becomes the integration over the circle we had above.)
So we've explicitly lifted all the generators of the cohomology of G to equivariant classes, ie to G/G, hence we're done. (Presumably this can also be seen concretely in the Cartan model, that the generators of H^\*G lift to conjugation-equivariant classes..)
| 42 | https://mathoverflow.net/users/582 | 20849 | 13,831 |
https://mathoverflow.net/questions/15074 | 2 | Assume $M$ is an open 3-manifold which can be deformation retracted to a point. Is it necessarily homeomorphic to $\mathbb R^3$?
(I know Whitehead had an example which is contractible and not homeomorphic to $\mathbb R^3$
Does his counterexample strong deformation retract to a point?)
| https://mathoverflow.net/users/3922 | Must a Strong deformation retractible 3-manifold be homeomorphic to $\mathbb{R}^3$? | Allen Hatcher's comment is actually an answer: "Just to clarify: The original question seems to be about the distinction between a space being contractible and the possibly stronger condition of deformation retracting to a point. For nice spaces (manifolds, CW complexes, ...) the two conditions are in fact equivalent. A textbook reference is Corollary 0.20 in Chapter 0 of my algebraic topology book. (See also Example 0.15 and Proposition 0.16.) In the exercises at the end of the chapter there are some examples of weird spaces that are contractible but do not deformation retract to any point."
| 4 | https://mathoverflow.net/users/66 | 20852 | 13,833 |
https://mathoverflow.net/questions/20850 | 2 | Let $f : X \to Y$ be a morphism of schemes. Is it possible to associate to every closed immersion $i : F \to X$ a closed immersion $f^\* i : G \to Y$, such that in the affine case, $(Spec(A) \to Spec(B))^\*$ is given taking ideals of $A$ to ideals of $B$ via preimages? Probably this is well-known to every algebraic geometer :-)
| https://mathoverflow.net/users/2841 | is there a push-forward of closed subschemes? | It sounds as though what you want is the closure of the image of $F$ under $f$. (That is, the minimal closed subscheme of $Y$ factoring $f$.)
If $X =$ Spec $A$, and $Y =$ Spec $B$, and $F =$ Spec $A/I$, and $f$ corresponds to the ring map $f':B\to A$, then we can consider the preimage $J$ of $I$ under $i'$. Consider the set of primes in $B$ containing $J$. Of course, any prime in the image of $F$ under $f$ must contain $J$, since its preimage under the ring map has to contain $I$. Thus, Spec $B/J$ contains the preimage. You can check that it's the biggest such ideal, noting that in order for us to have a map $B/J \to A/I$, $J$ should be contained in the preimage of $I$.
Being the closure of the image of $F$ under $f$ is a universal property of sorts (in particular, it's unique), which more or less allows us to argue that this construction generalizes to non-affine schemes. (Just apply this local construction, and uniqueness tells us that the local constructions glue together.)
| 6 | https://mathoverflow.net/users/5254 | 20854 | 13,835 |
https://mathoverflow.net/questions/20831 | 20 | It is known (given CFSG) that all non-abelian finite simple groups have small outer automorphism groups. However, it's quite tedious to list all the possibilities. Does anyone know a reference for a statement of the following form?
Let G be a non-abelian finite simple group. Then |Out(G)| < f(|G|) (where f is something straightforward that gives a good idea asymptotically of the worst case).
Also, a related question: How good a bound can be obtained in this case without using the classification? Can one do much better than the bounds one might obtain for the outer automorphism group of an arbitrary finite group?
| https://mathoverflow.net/users/4053 | Estimate for the order of the outer automorphism group of a finite simple group | Check article "Probabilistic generation of wreath products of non-abelian finite simple groups" by Martyn Quick. In Section 3.1 he consider this question and get $|Out G|\leq |G|/30$ for every non-abelian finite simple group $G$, which was enough for his needs.
| 15 | https://mathoverflow.net/users/4408 | 20859 | 13,838 |
https://mathoverflow.net/questions/20856 | 10 | Let $\mathfrak{g}=\mathfrak{n}^-\oplus\mathfrak{h}\oplus\mathfrak{n^+}$ be a triangular decomposition of semisimple Lie algebra. Let $\mathcal{Z}$ be the central of universal envoloping Lie algebra of $\mathfrak{g}$.
Let $\mathcal{C}$ be the category of representations of $\mathfrak{g}$, on which $\mathfrak{n ^+}$ and $\mathfrak{h}$ acts locally finite, and $\mathcal{Z}$ acts semisimplely.
Let $\mathcal{D}$ be the another category of representations of $\mathfrak{g}$, on which $\mathfrak{n ^+}$ and $\mathcal{Z}$ acts also locally finitely, and $\mathfrak{h}$ acts semisimply. If I don't make mistake, $\mathcal{D}$ should be called category $\mathcal{O}$.
**Claim**. $\mathcal{C}$ is equivalent to $\mathcal{D}$.
Do I formulate the problem correctly?
About the proof of this theorem, where is it written?
| https://mathoverflow.net/users/5082 | Twin categories in representation of Lie algebra | The equivalence (or something very close, I haven't checked carefully what is written) follows from Beilinson-Bernstein localization. The two categories can be realized roughly speaking as D-modules on B\G/N and N\G/B, and the equivalence comes from the interchange of the two sides. Slightly more precisely, Beilinson-Bernstein tells us that (assuming we ignore singular infinitesimal characters, where things need to be slightly modified) if we want representations on which Z acts semisimply, we look at twisted D-modules on G/B with twisting given by (a lift from h^*/W to h^* of) the eigenvalues of the Z action.
Equivalently these are D-modules on G/N which are weakly H-equivariant - meaning locally constant along the fibers G/N-->G/B, and h acts with strictly prescribed semisimple monodromy-- ie we presecribe monodromy along these fibers. If we want representations with locally finite Z action, we really just look at D-modules on G/N and ask for them to be locally constant along the fibers but don't strictly presecribe monodromies.
Now the two conditions you give for representations correspond to asking for these D-modules to be N-equivariant in the strict case or B-equivariant in the locally finite case.
In any case the whole picture is symmetric under exchanging left and right, hence the equivalence.
(There are two other categories which are symmetric under exchange of left and right -- if we impose the weak/locally finite conditions on both sides we get the category of Harish-Chandra bimodules -- ie (g+g,G)-Harish Chandra modules -- which correspond to representations of G considered as a real Lie group. If we impose strict conditions on both sides we get the Hecke category, which appears as intertwining functors acting on categories of representations and is the subject of Kazhdan-Lusztig theory. Category O, in these two forms, is some kind of intermediate form -- both of the above are monoidal categories and Category O is a bimodule for them, with your involution exchanging the two actions..)
As for a reference, this is standard but I don't know the proper reference. Similar things are discussed in the Beilinson-Ginzburg-Soergel JAMS paper on Koszul duality patterns in representation theory or the Beilinson-Ginzburg paper on wall-crossing functors (available on the arxiv). I presume Ben Webster will let us know..
| 8 | https://mathoverflow.net/users/582 | 20862 | 13,841 |
https://mathoverflow.net/questions/20863 | 6 | First some simple observations in order to motivate the question:
The functor $Set^{op} \to Set, X \to \{\text{subsets of }X\}, f \to (U \to f^{-1}(U))$ is representable. The representing object is $\{0,1\}$ with the universal subset $\{0\}$. Also the functor $Top^{op} \to Set, X \to \{\text{open subsets of }X\}$ is representable: Endow $\{0,1\}$ with the topology such that $\{0\}$ is the unique nontrivial open subset (Sierpinski space), then it is again the representing object.
But what about schemes. Is the functor $Sch^{op} \to Set, X \to \{\text{open subschemes of }X\}$ representable? Of course, we could also talk about open subsets of $X$. My first idea was to endow the Sierpinski space above with a scheme structure, using DVR, but this does not work properly.
| https://mathoverflow.net/users/2841 | Is the functor of open subschemes representable? | No, it is not representable. If it were, the functor of open subset for schemes over a field $k$ would also be representable by the base change to $\mathop{\rm Spec} k$ of the representing object. Suppose that this last functor is represented by an open embedding of $k$-schemes $S\_1 \subseteq S$. If $T$ is a $k$-scheme, there is a unique morphism $\phi\colon T \to S$ such that $\phi^{-1}S\_1 = T$; this means that there a unique morphism $T \to S\_1$, so $S\_1 = \mathop{\rm Spec}k$. On the other hand, it is easy to see that a $k$-rational point on any $k$ is always closed. So $S\_1$ is closed, which implies that every open subschems on a $k$-scheme is also closed; and this is patently false.
This example must be somewhere in my notes on descent theory.
| 14 | https://mathoverflow.net/users/4790 | 20866 | 13,842 |
https://mathoverflow.net/questions/20399 | 17 | What is the [conjugate prior](http://en.wikipedia.org/wiki/Conjugate_prior) distribution of the Dirichlet distribution?
---
Edit: Since I asked this question many years ago, I've written a [Python library](https://github.com/NeilGirdhar/efax) for working with exponential families. [Maximum likelihood estimation of the Dirichlet distribution](https://github.com/NeilGirdhar/efax/blob/main/examples/maximum_likelihood_estimation.py) can be done using its expectation paramtrization (which is intimately related to the conjugate prior).
| https://mathoverflow.net/users/634 | Conjugate prior of the Dirichlet distribution? | [Neil](https://mathoverflow.net/users/634/neil) sent me an email asking:
===
I read your post at <http://www.stat.columbia.edu/~cook/movabletype/archives/2009/04/conjugate_prior.html> and I was wondering if you could expand on how to update the Dirichlet conjugate prior that you provided in your [paper](http://cvsp.cs.ntua.gr/publications/jpubl+bchap/LefkimmiatisMaragosPapandreou_BayesianMultiscalePoissonIntensityEstimation_ieee-j-ip09.pdf):
S. Lefkimmiatis, P. Maragos, and G. Papandreou,
Bayesian Inference on Multiscale Models for Poisson Intensity Estimation: Applications to Photon-Limited Image Denoising,
IEEE Transactions on Image Processing, vol. 18, no. 8, pp. 1724-1741, Aug. 2009
In other words, given in your paper's notation the prior hyper-parameters (vector $\mathbf{v}$, and scalar $\eta$), and $N$ Dirichlet observations (vectors $\mathbf{\theta}\_n, n=1,\dots,N$), how do you update $\mathbf{v}$ and $\eta$?
===
Here is my response:
Conjugate pairs are so convenient because there is a standard and simple way to incorporate new data by just modifying the parameters of the prior density. One just multiplies the likelihood with its conjugate prior; the result has the same parametric form as the prior, and the new parameters can be readily "read-off" by comparing the likelihood-prior product with the prior parametric form. This is described in detail in all standard texts in Bayesian statistics such as Gelman et al. (2003) or Bernardo and Smith (2000).
In the case of the Dirichlet and its conjugate prior described in our paper and using its notation, after observing $N$ Dirichlet vectors $\mathbf{\theta}\_n$, $n=1,\dots,N$, where each vector $\mathbf{\theta}\_n$ is $D$ dimensional with elements $\theta\_n[t]$, $t=1,\dots,D$, the $D+1$ hyper-parameters should be updated as follows:
* $\eta\_N = \eta\_0 + N$
* $v\_N[t] = v\_0[t] - \sum\_{n=1}^N \ln \theta\_n[t], \quad t=1,\dots,D$, where $\eta\_0$, $\mathbf{v}\_0$ and $\eta\_N$, $\mathbf{v}\_N$ are the initial and updated model parameters, respectively.
You can verify this in a few lines of equations by following the previously described general rule.
Hope this helps!
| 22 | https://mathoverflow.net/users/5258 | 20875 | 13,846 |
https://mathoverflow.net/questions/20874 | 9 | Probably a well-know question, but I haven't solved it, so I'll ask.
I can show that every matrix in $M\_2(\mathbb{R})$ is the sum of two squares of matrices in $M\_2(\mathbb{R})$.
If $n>2$, I can also show that every matrix in $M\_n(\mathbb{R})$ is the sum of three squares of matrices in $M\_n(\mathbb{R})$.
So my question is : Is every matrix in $M\_n(\mathbb{R})$ is the sum of two squares of matrices in $M\_n(\mathbb{R})$ (n>2)?
| https://mathoverflow.net/users/3958 | Waring's problem for matrices | The answer is YES if $n$ is even. But if $n$ is odd, then the answer is NO since $-I$ is not a sum of two squares.
See
Griffin and Krusemeyer, Matrices as sums of squares, *Linear and Multilinear Algebra* **5** (1977/78), no. 1, 33-44
for the proofs of these facts and generalizations.
| 14 | https://mathoverflow.net/users/2757 | 20876 | 13,847 |
https://mathoverflow.net/questions/20147 | 4 | It's all in the title basically. There's an interesting topic called [systolic geometry](http://en.wikipedia.org/wiki/Systolic_geometry) that has grown a lot in the past 30 years, with a (first?) [textbook](http://www.ams.org/bookstore-getitem/item=surv-137) on the subject by M.Katz (AMS 2007).
So I was wondering what would a semester-long graduate course typically cover, assuming knowledge of basic Riemannian geometry.
I haven't been able to find much information online: Katz has a [second semester graduate course](http://u.math.biu.ac.il/~katzmik/egreg826.pdf) which doesn't quite cover the same ground as the book. Has there been other courses on the topic elsewhere?
| https://mathoverflow.net/users/469 | What would a graduate course on systolic geometry typically cover? | This is interesting. I imagine that any course would vary quite a bit depending on who taught it.
Any course should probably contain Gromov's proof of the systolic inequality for essential manifolds. Other than that, I am not sure. The course could dive into systoles on surfaces and some of the arithmetic constructions in Teichmuller theory, or it could develop harmonic maps and scalar curvature rigidity theorems, or it could take a dynamical systems approach and discuss the relationship between volume entropy and closed geodesics.
I have no idea. You should come up with a curriculum and post it here.
| 4 | https://mathoverflow.net/users/5261 | 20878 | 13,848 |
https://mathoverflow.net/questions/15176 | 9 | A sequence $a\_n \in \mathbb{T}$ ($n \geq 1$) satisfying $a\_{mn} = a\_m a\_n$ for all $m, n \geq 1$ defines a Dirichlet series $f(s) = \sum\_n a\_n n^{-s}$, absolutely convergent for $\Re s > 1$. If in addition $\limsup \frac{\log \lvert a\_1 + \ldots + a\_n \rvert}{\log n} = 0$, then the series is (conditionally) convergent for $\Re s > 0$. (For example, if $a\_n$ coincides with a non-principal Legendre character $\mod p$ when $p \nmid n$, and $a\_p$ is set arbitrarily, then the partial sums grow logarithmically and we have convergence for $\Re s > 0$.)
I would like to know whether it is possible to get something just a tiny bit stronger. Is there a completely multiplicative $\mathbb{T}$-valued sequence $a\_n$ such that the function $f(s)$ derived from it can be analytically continued to a region that includes the closed half-plane $\Re s \geq 0$? Or to put it another way, must $f(s)$ have a singularity on the imaginary axis?
The question is motivated by the question of whether there exists a completely multiplicative sequence in $\mathbb{T}$ with bounded partial sums, which arises naturally in relation to the Erdos Discrepancy Problem.
If the question is hard, are there any known results in this direction? For example, if we relax the multiplicativity condition?
| https://mathoverflow.net/users/3755 | Analytic continuation of Dirichlet series with completely multiplicative coefficients of modulus 1 | As a matter of fact, it isn't hard to construct a multiplicative sequence $a\_n$ such that $f(z)$ is an entire function without zeroes. Unfortunately, it is completely useless for the questions that you brought up as "motivation".
Here is the construction.
Claim 1: Let $\lambda\_j\in [0,1]$ ($j=0,\dots,M$). Assume that $|a\_j|\le 1$ and $\sum\_ja\_j\lambda\_j^p=0$ for all $0\le p\le P$. Then, if $P>2eT$, we have $\left|\sum\_j a\_je^{\lambda\_j z}\right|\le (M+1)(eT/P)^P$
Proof: Taylor decomposition and a straightforward tail estimate.
Claim 2: Let $P$ be large enough. Let $\Delta>0$, $M>P^3$ and $(M+1)\Delta<1$. Let $I\_j$ ($0\le j\le M$) be $M+1$ adjacent intervals of length about $\Delta$ each arranged in the increasing order such that $I\_0$ contains $0=\lambda\_0$. Suppose that we choose one $\lambda\_j$ in each interval $I\_j$ with $j\ge 1$. Then for every $|a\_0|\le 1$, there exist $a\_j\in\mathbb C$ such that $|a\_j|\le 1$ and $\sum\_{j\ge 0} a\_j\lambda\_j^p=0$ for $0\le p\le P$.
Proof: By duality, we can restate it as the claim that $\sum\_{j\ge 1}|Q(\lambda\_j)|\ge |Q(0)|$ for every polynomial $Q$ of degree $P$. Now, let $I$ be the union of $I\_j$. It is an interval of length about $M\Delta$, so, by Markov's inequality, $|Q'|\le CP^2(M\Delta)^{-1} A\le CP^{-1}\Delta^{-1}A$ where $A=\max\_I |Q|\ge |Q(0)|$. But then on the 5 intervals $I\_j$ closest to the point where the maximum is attained, we have $|Q|\ge A-5\Delta CP^{-1}\Delta^{-1}A\ge A/2$. The rest is obvious.
Claim 3: Suppose that $a\_0$ is fixed and $a\_j$ ($j\ge 1$) satisfy $\sum\_{j\ge 0} a\_j\lambda\_j^p=0$ for $0\le p\le P$. Then we can change $a\_j$ with $j\ge 1$ so that all but $P+1$ of them are exactly $1$ in absolute value and the identities still hold.
Proof: As long as we have more than $P+1$ small $a\_j$, we have a line of solutions of our set of $P+1$ linear equations. Moving along this line we can make one of small $a\_j$ large. Repeating this as long as possible, we get the claim.
Now it is time to recall that the logarithm of the function $f(z)$ is given by
$$
L(z)=\sum\_{n\in\Lambda}a\_ne^{-z\log n}
$$
where $\Lambda$ is the set of primes and prime powers and $a\_n=m^{-1}a\_p^m$ if $n=p^m$. We are free to choose $a\_p$ for prime $p$ in any way we want but the rest $a\_n$ will be uniquely determined then. The key point is that we have much more primes than prime powers for unit length.
So, split big positive numbers into intervals from $u$ to $e^\Delta u$ where $\Delta$ is a slowly decaying function of $u$ (we'll specify it later). Formally we define the sequence $u\_k$ by $u\_0=$something large, $u\_{k+1}=e^{\Delta(u\_k)}u\_k$ but to put all those backward apostrophes around formulae is too big headache, so I'll drop all indices. Choose also some slowly growing functions $M=M(u)$ and $P=P(u)$ to be specified later as well.
We need a few things:
1) Each interval should contain many primes. Since the classical prime number theorem has the error term $ue^{-c\sqrt{\log u}}$, this calls for $\Delta=\exp\{-\log^{\frac 13} u\}$
Then we still have at least $u^{4/5}$ primes in each interval (all we need is to beat $u^{1/2}$ with some margin).
2) We should have $M\Delta\ll 1$, $M>P^3$, and $u\left(\frac{eT}P\right)^P\le (2u)^{-T-3}$ for any fixed $T>0$ and all sufficiently large $u$. This can be easily achieved by choosing $P=\log^2 u$ and $M=\log^6 u$.
3) At last, we'll need $M(P+\sqrt u)\ll u^{4/5}$, which is true for our choice.
Now it is time to run the main inductive construction. Suppose that $a\_n$ are already chosen for all $n$ in the intervals up to $(u,e^{\Delta}u)$ and we still have almost all primes in the intervals following the current interval free (we'll check this condition in the end). We want to assign $a\_p$ for all $p$ in our interval for which the choice hasn't been made yet or was made badly. We start with looking at all $a\_p$ that are not assigned yet or assigned in a lame way, i.e., less than one in absolute value. Claim 3 (actually a small modification of it) allows us to upgrade all of them but $P+1$ to good ones (having absolute value $1$) at the expense of adding an entire function that in the disk of radius $T$ is bounded by $(2u)^{T} u\left(\frac{eT}P\right)^P\le u^{-3}$ to $L(z)$. Now we are left with at most $\sqrt u$ powers of primes and $P+1$ lame primes to take care of. We need the prime powers participate in small sums as they are and we need the small coefficients to be complemented by something participating in small sums too. For each of them, we choose $M$ still free primes in the next $M$ intervals (one in each) and apply Claim 2 to make a (lame) assignment so that the corresponding sum is again bounded by $u^{-3}$ in the disk of radius $T$. We have at most $u$ such sums, so the total addition will be at most $u^{-2}$. This will finish the interval off. Now it remains to notice that we used only about $\sqrt u+P$ free primes in each next interval and went only $M$ intervals ahead. This means that in each interval only $M(\sqrt u+P)$ free primes will ever be used for compensating the previous intervals, so we'll never run out of free primes. Also, the sum of the blocks we constructed will converge to an entire function. At last, when $\Re z>1$, we can change the order of summation and exponentiate finally getting the Dirichlet series representation that we need.
The end.
| 8 | https://mathoverflow.net/users/1131 | 20888 | 13,855 |
https://mathoverflow.net/questions/20838 | 11 | My question is very basic (I don't know too much of differential geometry):
given a fiber bundle, is there a necessary and sufficient condition for its tangent bundle to be trivial?
I have some ideas, but submitted to some conditions on the cohomology ring of the bundle.
(I apologize if it is trivial.)
| https://mathoverflow.net/users/4770 | Is there a necessary and sufficient condition for the tangent bundle of a fiber bundle to be trivial? | This is far from being a complete answer, but there is a case when one construct a parallelizable bundle (meaning its total space has trivial tangent bundle) from a given (geometric) bundle.
The context is that of $G$-structure, which are a formalization of the concept of geometric structures. A $G$-structure is a set of data including a fiber bundle on a smooth manifold, which shall be thought of as the bundle of admissible frames. For example, in the Riemannian case ($G=O(n)$) the bundle is that of orthonormal frames. If the group $G$ has a finite-order rigidity property, then one can construct a sequence of bundles, the total space of each one being the base space of the next one, so that after a finite number of steps one gets a bundle whose total space is parallelizable. This is a tool to prove that the group of automorphisms of the $G$-structure is a Lie group. As an example, if I remember well the total space of the bundle of orthonormal frames on a Riemannian manifold is parallelizable.
All details are available in Kobayashi's *transformation groups in differential geometry*.
| 10 | https://mathoverflow.net/users/4961 | 20904 | 13,863 |
https://mathoverflow.net/questions/20913 | 3 | Given an arithmetic variety $f: X \rightarrow Spec(\mathbb{Z})$.
Is there a notion of boundedness for families of sheaves on $X$?
I only found the notion for families on the fibers of $f$. But i am interested in sheaves defined on $X$.
All definitions / theorems i found only work when $X$ is defined over some field $k$, where one has the Hilbert polynomial, slope etc, which we don't have in this case. Is there some substitute for these terms?
Or are there even results about moduli spaces of sheaves on arithmetic varieties?
Edit: According to <http://arxiv.org/abs/math/0612268> there is a notion of arithmetic (semi)stability. One even has a Harder Narasimhan filtration. Can one define the notion of boundedness in the Arakelov setting? Are there any results on moduli spaces of vector bundles in Arakelov theory?
| https://mathoverflow.net/users/3233 | Families of sheaves on arithmetic varieties | When doing moduli theory over $\mathbb Z$, or another base scheme, one works with sheaves that are flat over the base; this implies that all the discrete invariants, such as the Hilbert polynomial, are constant in the fibers. Stability is defined fiber by fiber; i.e., a sheaf is (semi)stable when it it (semi)stable on all the fibers. The Quot schemes of sheaves with fixed Hilbert polynomial are defined and projective over $\mathbb Z$; then the standard boundedness results all generalize. Thus one obtains stacks of stable, or semistable, bundles, which are defined over $\mathbb Z$. When the existence of (quasi)projective moduli spaces is obtained via GIT, this also works over $\mathbb Z$ (a result of Seshadri, see *Geometric reductivity over arbitrary base*, Adv. Math. 26 (1977), 225–274).
There is an issue of when the fiber of one of these moduli spaces over a prime $p$ is the moduli space of the corresponding sheaves on the fiber of $X$ over $p$; this is not automatic, because the formation of moduli spaces in positive or mixed characteristic does not, in general, commute with non-flat base chage. If this comes up, it has to be analyzed case by case.
I hope this is what you what. If the question is the construction of a space whose points corresponds to global sheaves on $X$ with metric at infinity, defining stability by some kind of Arakelov-theoretic Hilbert polynomial, then I don't have a clue.
| 3 | https://mathoverflow.net/users/4790 | 20917 | 13,872 |
https://mathoverflow.net/questions/20891 | 13 | Let $M$ be a differentiable manifold, $\Delta$ the closed simplex $[p\_0, p\_1,...,p\_k]$. A differential singular $k$-simplex $\sigma$ of $M$ is a smooth mapping $\sigma:\Delta \to M$.
And we construct a chain complex in the same way we construct the chain complex of singular homology, we gain its homology group.
My question is why this homology group equals the singular homology group? I have tried finding in many books but there is no answer of this question.
| https://mathoverflow.net/users/4621 | singular homology of a differential manifold | This depends on what exactly is a smooth mapping from a simplex to the manifold. The standard definition is that the mapping of a non-open subset $X$ of $\mathbf{R}^n$ to a manifold is smooth iff it can be extended to a smooth mapping of an open neighborhood of $X$. With this definition the comparison theorem is true and a very detailed proof can be found e.g. in the book "Introduction to smooth manifolds" by Lee, chapter 16.
| 16 | https://mathoverflow.net/users/2349 | 20926 | 13,876 |
https://mathoverflow.net/questions/20925 | 3 | Does anyone know a reference for the proof of the finite generation of the Mordell-Weil group over finitely generated fields?
| https://mathoverflow.net/users/nan | finite generation of the Mordell-Weil group over finitely generated fields | It is in Lang's book "Fundamentals of diophantine geometry", chapter 6:
[google book preview](http://books.google.co.uk/books?id=ShRel9iqsmUC&lpg=PA138&dq=lang%2520serge%2520mordell-weil%2520finitely&pg=PA138#v=onepage&q&f=false)
| 3 | https://mathoverflow.net/users/5015 | 20928 | 13,877 |
https://mathoverflow.net/questions/20929 | 18 | This question is something of a follow-up to
[Transformation formulae for classical theta functions](https://mathoverflow.net/questions/19400/) .
How does one recognise whether a subgroup of the modular group
$\Gamma=\mathrm{SL}\_2(\mathbb{Z})$ is a congruence subgroup?
Now that's too broad a question for me to expect a simple answer
so here's a more specific question. The subgroup $\Gamma\_1(4)$
of the modular group is free of rank $2$ and freely generated by
$A=\left(
\begin{array}{cc}
1&1\\\
0&1
\end{array}\right)$
and
$B=\left(
\begin{array}{cc}
1&0\\\
4&1
\end{array}\right)$. If $\zeta$ and $\eta$ are roots of unity there is a
homomorphism $\phi$ from $\Gamma\_1(4)$ to the unit circle group
sending $A$ and $B$ to $\zeta$ and $\eta$ resepectively. Then the kernel $K$
of $\phi$ has finite index in $\Gamma\_1(4)$. How do we determine whether $K$
is a congruence subgroup, and if so what its level is?
In this example, the answer is yes when $\zeta^4=\eta^4=1$. There are
also examples involving cube roots of unity, and involving eighth
roots of unity where the answer is yes. I am interested in this example
since one can construct a "modular function" $f$, homolomorphic on
the upper half-plane and meromorphic at cusps such that $f(Az)=\phi(A)f(z)$
for all $A\in\Gamma\_1(4)$. One can take $f=\theta\_2^a\theta\_3^b\theta\_4^c$
for appropriate rationals $a$, $b$ and $c$.
Finally, a vaguer general question. Given a subgroup $H$ of
$\Gamma$ specified as the kernel of a homomorphism from $\Gamma$ or
$\Gamma\_1(4)$ (or something similar) to a reasonably tractable target group,
how does one determine whether $H$ is a congruence subgroup?
| https://mathoverflow.net/users/4213 | Distinguishing congruence subgroups of the modular group | There is one answer in the following paper, along with a nice bibliography of other techniques:
[MR1343700 (96k:20100)](http://www.ams.org/mathscinet-getitem?mr=1343700)
Hsu, Tim(1-PRIN)
Identifying congruence subgroups of the modular group. (English summary)
Proc. Amer. Math. Soc. 124 (1996), no. 5, 1351--1359.
| 19 | https://mathoverflow.net/users/317 | 20931 | 13,879 |
https://mathoverflow.net/questions/20927 | 8 | Let E be an elliptic curve over the rationals with conductor $Mp^2$ with p>5 and M and p coprime, and let $\rho$ be the Galois representation attached to the p-torsion points of E. Is there a way to consider deformations of $\rho$ to representations that still have "additive reduction at p" with Serre weight 2? Actually in that direction, what is the local condition on p that one has to put down? Assuming that's the case, and assuming this is representable by ring R, is there an R=T result in that direction?
| https://mathoverflow.net/users/92 | Is there an R=T type result for modular forms with additive reduction? | If you fix a prime $\ell$, and consider the Galois action of the decomposition group $D\_p$ on the (rational) $\ell$-adic Tate module (here "rational" means tensored with $\mathbb Q\_{\ell}$),
then (in a standard way, due to Deligne) you can convert this action into a representation
of the Weil--Deligne group, and so in particular of the Weil group. Restricting to the
inertia group, you get a representation of the inertia group $I\_p$, known as the inertial type $\tau$. It is independent of $\ell$. (The only reason to detour through the Weil--Deligne group is to deal with possibly infinite image of tame inertia; if the elliptic curve has potentially good reduction, then we can skip this step and just take the representation of $I\_p$ on the $\ell$-adic Tate module, which has finite image and is independent of $\ell$.)
[Added: In the above, one should insist that $\ell \neq p$. If $\ell = p$, then one can also arrive at a Weil-Deligne representation, and hence
inertial type, which is the same as the one obtained as above for $\ell \neq p$, but to do this one must use Fontaine's theory: one forms the $D\_{pst}$ of the rational $p$-adic Tate module,
which then can be converted into a Weil--Deligne representation in a standard way,
and hence gives an inertial type.]
Now one can look at the deformation ring $R\_{\rho}^{[0,1],\tau}$ parameterizing lifts
of $\rho$ of which at $p$ are of inertial type $\tau$ and Hodge--Tate weights $0$ and $1$. (See Kisin's recent
JAMS paper about potentially semi-stable deformation rings.)
[Added: Here $\ell = p$,
i.e. we are looking at $p$-adic deformations of $\rho$ which are potentially semi-stable
at $p$, and whose inertial type, computed via $D\_{pst}$ as in the above added remark,
is equal to $\tau$. But note that, by the preceding discussion, these deformations do precisely capture the idea of lifts of $\rho$ having the same "reduction type" as
the original elliptic curve $E$.]
Let's suppose that $E$ really does have potentially good reduction. Then Kisin's "moduli of finite flat group schemes" paper shows that any lift parameterized by $R^{[0,1],\tau}$ is modular. This shows that $R^{[0,1],\tau} = {\mathbb T},$ for an appropriately chosen
${\mathbb T}$.
One thing to note: unlike in the original Taylor--Wiles setting, from this statement one doesn't get quantitive information about adjoint Selmer groups, and one doesn't get any simple interpretation of what this $R = {\mathbb T}$ theorem means on the integral level.
(In other words, Artinian-valued points of $R^{[0,1],\tau}$ have no simple interpretation in terms of a ramification condition at $p$; this is related to the fact that the theory
of $D\_{pst}$ only applies rationally, i.e. to ${\mathbb Q}\_p$-representations, not integrally,
i.e. not to representations over $\mathbb Z\_p$ or over Artin rings.)
| 7 | https://mathoverflow.net/users/2874 | 20932 | 13,880 |
https://mathoverflow.net/questions/20922 | 7 | Tonight, a friend of mine give me a concise introduction to Shimura variety . I only get some first impression of it. I think the hodge structure is a generalization of the cohomology ring of Kaehler manifold or algebraic manifold , and i think of the Shimura variety an anologue of the analytic familly of complex manifolds , just as introduced in Kodaira's book Complex manifolds.And i suspect that there maybe some anologue theorem's as what Kodaira had done by deformation of complex structures . I'm just doing some imagination unboundedless , do not laugh at me !Heh!
| https://mathoverflow.net/users/4437 | what is the motivation of Shimura variety? | The theory of Shimura varieties was begun by Shimura, and further developed by Langlands (who introduced the name), and is now a central part of arithemtic geometry and of the theories
of automorphic forms, Galois representations, and motives.
Shimura varieties are certain moduli of Hodge structures; but that is perhaps not the best point of view to understand why people study them. Rather, the primary motivation is the following: Shimura varieties are attached to (certain) reductive linear algebraic groups over $\mathbb Q$, and the geometry of the Shimura variety is closely linked to the theory of automorphic forms over the corresponding reductive group.
Thus Shimura varieties make a natural test-case for investigating the conjectural relations between motives and automorphic forms, since they are geometric objects with a direct link to the theory of automorphic forms. (I'm not sure that it's useful to be more specific about this in this particular answer, but for those to whom it is meaningful: on the one hand, one has an analogue, for any Shimura variety, of Eichler--Shimura theory, relating the cohomology of modular curves to modular forms; and on the other hand, by thinking of cohomology as being etale cohomology, one obtains Galois representations which one would like to show (and in many cases can show) are related to automorphic forms in a manner analogous to the relationship between the Galois representations on the etale cohomology of modular curves and Hecke eigenforms that was established by Deligne.)
| 21 | https://mathoverflow.net/users/2874 | 20950 | 13,892 |
https://mathoverflow.net/questions/18460 | 9 | Sorry for my poor English.
Let $X$ be a reducible projective variety.
**My question is:**
1. How can I compute the dualizing sheaf of $X$ and express it in an explicit way?
2. Is there a method to get dualizing sheaf of whole reducible variety $X$ from the information of dualizing sheaves of its irreducible components?
Currently I'm not concern the general case, but I want a few accessible concrete examples such as:
1. reducible hypersurfaces,
2. union of toric varieties glued at isomorphic orbits(Alexeev calls it stable toric variety).
The reason why I concern is to understand the limit in moduli spaces of stable pairs.
| https://mathoverflow.net/users/4643 | Dualizing sheaf of reducible variety? | With regards part 2.
Let's assume that you have two components $X\_1$ and $X\_2$ (or even unions of components) such that $X\_1 \cup X\_2 = X$=. Let $I\_1$ and $I\_2$ denote the ideal sheaves of $X\_1$ and $X\_2$ in $X$.
Set $Z$ to be the *scheme* $X\_1 \cap X\_2$, in other words, the ideal sheaf of $Z$ is $I\_1 + I\_2$.
It is easy to see you have a short exact sequence
$$0 \to I\_1 \cap I\_2 \to I\_1 \oplus I\_2 \to (I\_1 + I\_2) \to 0$$
where the third map sends $(a,b)$ to $a-b$.
The nine-lemma should imply that you have a short exact sequence
$$0 \to O\_X \to O\_{X\_1} \oplus O\_{X\_2} \to O\_Z \to 0$$
If you Hom this sequence into the dualizing complex of $X$, you get a triangle
$$\omega\_Z^. \to \omega\_{X\_1}^. \oplus \omega\_{X\_2}^. \to \omega\_{X}^. \to \omega\_Z^.[1]$$
You can then take cohomology and, depending on how things intersect (and what you understand about the intersection), possibly answer your question.
If $X\_1$ and $X\_2$ are hypersurfaces with no common components (which should imply everything in sight is Cohen-Macualay) then these dualizing complexes are all just sheaves (with various shifts), and you just get a short exact sequence
$$0 \to \omega\_{X\_1} \oplus \omega\_{X\_2} \to \omega\_{X} \to \omega\_{Z} \to 0$$
Technically speaking, I should also probably push all these sheaves forward onto $X$ via inclusion maps.
| 14 | https://mathoverflow.net/users/3521 | 20954 | 13,896 |
https://mathoverflow.net/questions/20955 | 6 | It is known that if GRH holds there does not exist additional Idoneal numbers. (see www.mast.queensu.ca/~kani/papers/idoneal.pdf this paper puts on the question of correctnes for Wikipedia and Wolfram MathWorld since they state there could only be ONE additional Idoneal number)
What I am interested in is whether there are any certain properties that an additional idoneal number X should satisfy. I am trying to prove a theorem which would only work if X has three odd divisors. Is anything like that known or easy to derive?
I would like to find a precise reference to the fact that any idoneal number not in the currently known finite list has to have at least three odd prime factors (this should hold, as long as the answer by Pete is valid. Pete also suggested looking into the book primes of the form x^2 + ny^2 but my knowledge of number theory is too limited to derive the stated fact from there)
**EDIT**: Removed misinterpreted sentence about GRH and idoneal numbers. Added request for reference. I will give 100 bounty points to the first concise reference of this fact.
| https://mathoverflow.net/users/1737 | The missing Euler Idoneal numbers | If X is an idoneal number, then the class group of discriminant -4X has exponent dividing 2, so the class number is equal to the number of genera (Theorem 6 in Kani's paper: for me, this is the most convenient definition), which is given by an explicit recipe in terms of the number of prime factors of X and its congruence class mod 32 (formula (3) of Kani's paper).
In particular, if X is idoneal and is a prime or twice a prime, then its class number is at most 4. But all discriminants of class number 4 have been calculated. Indeed, all discriminants of class number up to 100 have been calculated (work of M. Watkins), so a new idoneal number should have at least $6$ odd prime divisors, or something like that.
Also see Cox's book *Primes of the Form x^2 + ny^2* for treatment of idoneal numbers.
| 6 | https://mathoverflow.net/users/1149 | 20957 | 13,897 |
https://mathoverflow.net/questions/20960 | 155 | I've asked this question in every math class where the teacher has introduced the Gamma function, and never gotten a satisfactory answer. Not only does it seem more natural to extend the factorial directly, but the integral definition $\Gamma(z) = \int\_0^\infty t^{z-1} e^{-t}\,dt$, makes more sense as $\Pi(z) = \int\_0^\infty t^{z} e^{-t}\,dt$. Indeed Wikipedia says that this function was [introduced by Gauss](http://en.wikipedia.org/wiki/Gamma_function#Pi_function), but doesn't explain why it was supplanted by the Gamma function. As that section of the Wikipedia article demonstrates, it also makes its functional equations simpler: we get $$\Pi(z) \; \Pi(-z) = \frac{\pi z}{\sin( \pi z)} = \frac{1}{\operatorname{sinc}(z)}$$ instead of $$\Gamma(1-z) \; \Gamma(z) = \frac{\pi}{\sin{(\pi z)}}\;;$$ the multiplication formula is simpler: we have $$\Pi\left(\frac{z}{m}\right) \, \Pi\left(\frac{z-1}{m}\right) \cdots \Pi\left(\frac{z-m+1}{m}\right) = \left(\frac{(2 \pi)^m}{2 \pi m}\right)^{1/2} \, m^{-z} \, \Pi(z)$$
instead of
$$\Gamma\left(\frac{z}{m}\right) \, \Gamma\left(\frac{z-1}{m}\right) \cdots \Gamma\left(\frac{z-m+1}{m}\right) = (2 \pi)^{(m-1)/2} \; m^{1/2 - z} \; \Gamma(z);$$
the infinite product definitions reduce from
$$\begin{align}
\Gamma(z) &= \lim\_{n \to \infty} \frac{n! \; n^z}{z \; (z+1)\cdots(z+n)}
= \frac{1}{z} \prod\_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}}
\\
\Gamma(z) &= \frac{e^{-\gamma z}}{z} \prod\_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n} \\
\end{align}$$
to
$$\begin{align}
\Pi(z) &= \lim\_{n \to \infty} \frac{n! \; n^z}{(z+1)\cdots(z+n)}
= \prod\_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}}
\\
\Pi(z) &= e^{-\gamma z} \prod\_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n}; \\
\end{align}$$
and the Riemann zeta functional equation reduces from $$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)$$ to $$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Pi(-s)\ \zeta(1-s).$$
I suspect that it's just a historical coincidence, in the same way $\pi$ is defined as circumference/diameter instead of the much more natural circumference/radius. Does anyone have an actual reason why it's better to use $\Gamma(z)$ instead of $\Pi(z)$?
| https://mathoverflow.net/users/5279 | Why is the Gamma function shifted from the factorial by 1? | From Riemann's Zeta Function, by H. M. Edwards, available as a Dover paperback, footnote on page 8: "Unfortunately, Legendre subsequently introduced the notation $\Gamma(s)$ for $\Pi(s-1).$Legendre's reasons for considering $(n-1)!$ instead of $n!$ are obscure (perhaps he felt it was more natural to have the first pole at $s=0$ rather than at $s = -1$) but, whatever the reason, this notation prevailed in France and, by the end of the nineteenth century, in the rest of the world as well. Gauss's original notation appears to me to be much more natural and Riemann's use of it gives me a welcome opportunity to reintroduce it."
| 127 | https://mathoverflow.net/users/3324 | 20962 | 13,899 |
https://mathoverflow.net/questions/20924 | 8 | Let $A, B\in M\_{n}(\mathbb{R})$ be symmetric positive definite matrices. It is easy to see $Tr(A^2+AB^2A)=Tr(A^2+BA^2B)$. Numerical experiments indicate $$Tr[(A^2+AB^2A)^{-1}]\ge Tr[(A^2+BA^2B)^{-1}],~~(1)$$ but it seems difficult to show it.
Remark. When $n=2,3$, by direct computation, (1) is true. Here is an expriment done by matlab:
for s=1:1000
```
as=randn(4,4);
bs=randn(4,4);
ts=as*as';
rs=bs*bs';
ls=trace(inv(ts^2+ts*rs^2*ts)-inv(ts^2+rs*ts^2*rs))
```
end
{\bf Updated.} What about $A, B\in M\_{n}(\mathbb{C})$ be positive definite Hermitian matrices.
| https://mathoverflow.net/users/3818 | A question on a trace inequality | Note first that $A^2+AB^2A=(A+iAB)(A-iBA)$. The reverse product is $(A-iBA)(A+iAB)=A^2+BA^2B-i(BA^2-A^2B)=X-iC$. Thus, the quantity on the left is $\operatorname{Tr} (X-iC)^{-1}$ and that on the right is $\operatorname{Tr} X^{-1}$. Moreover, the self-adjoint complex matrix $X-iC$ is positive definite (as the product of an invertible operator and its adjoint). Similarly, considering the factorization $A^2+AB^2A=(A-iAB)(A+iBA)$, we can write the quantity on the left as $\operatorname{Tr} (X+iC)^{-1}$. Symmetrizing, we see that it will suffice to show that $(X-iC)^{-1}+(X+iC)^{-1}\ge 2X^{-1}$ in the sense of quadratic forms (then the inequality for traces will hold too). We can multiply by $X^{1/2}$ from both sides to reduce it to $(I-iD)^{-1}+(I+iD)^{-1}\ge 2I$ where $D=X^{-1/2}CX^{-1/2}$ and both operators on the left are positive definite. Diagonalizing the self-adjoint operator $iD$, we see that the inequality reduces to $(1+p)^{-1}+(1-p)^{-1}\ge 2$ for $p\in(-1,1)$.
| 19 | https://mathoverflow.net/users/1131 | 20975 | 13,906 |
https://mathoverflow.net/questions/20978 | 3 | Let $X$ be a quasi-compact scheme and let $\mathcal{A}$ be a quasi-coherent sheaf of $\mathcal{O}\_X$-algebras on $X$. $X$ being quasi-compact, we can write $X = U\_1 \cup \dots \cup U\_n$ with each $U\_i$ affine, say $U\_i =\textrm{Spec}R\_i$, and such that $\mathcal{A}\mid\_{U\_i} \simeq \widetilde{A\_i}$ for some finitely generated $R\_i$ algebras $A\_i$. Suppose we know that each $A\_i$ is the union of its subalgebras which are module finite over $R\_i$. Can we say that $\mathcal{A}$ is globally the union of it's subsheaves which are coherent sheaves of $\mathcal{O}\_X$-algebras?
It seems that the obvious thing to do would be to glue together coherent algebras over each of the $R\_i$'s, but it's not clear to me how this can be done. This question arose from Milne's proof of 'Zariski's Main Theorem' from the beginning of his etale cohomology book.
| https://mathoverflow.net/users/493 | Quasi-coherent sheaves of O_X-algebras | Under some slightly stronger hypothesis (Noetherian is certainly enough) we may write
$\mathcal A$ as the union of its coherent subsheaves. If $\mathcal E$ is a coherent subsheaf, then the subalgebra of $\mathcal A$ that it generates will also be coherent,
because this can be tested locally, where it then follows from your assumptions. Thus in this case, $\mathcal A$ is the union of coherent $\mathcal O\_X$-algebras.
I'm not sure how good a notion coherent is outside of the Noetherian context. If no-one
gives an answer in the non-Noetherian context, then you might want to look at the stacks project, which discusses this kind of "coherent approximation to quasi-coherent sheaves" in some generality, if I remember correctly.
| 3 | https://mathoverflow.net/users/2874 | 20983 | 13,909 |
https://mathoverflow.net/questions/20298 | 10 | This is a failed attempt of mine at creating a contest problem; the failure is in the fact that I wasn't able to solve it myself.
Let $x\_1$, $x\_2$, ..., $x\_n$ be $n$ reals. For any integer $k$, define a real $f\_k\left(x\_1,x\_2,...,x\_n\right)$ as the sum
$\sum\limits\_{T\subseteq\left\lbrace 1,2,...,n\right\rbrace ;\\ \ \left|T\right|=k} \left|\sum\limits\_{t\in T}x\_t - \sum\limits\_{t\in\left\lbrace 1,2,...,n\right\rbrace \setminus T} x\_t\right|$.
We mostly care about the case of $n$ even and $k=\frac n 2$; in this case, $f\_k\left(x\_1,x\_2,...,x\_n\right)$ is a kind of measure for the dispersion of the reals $x\_1$, $x\_2$, ..., $x\_n$ (more precisely, of their $\frac n 2$-element sums).
Now my conjecture is that if $n$ is even and $k=\frac n 2$, then
$f\_k\left(x\_1,x\_2,...,x\_n\right)\geq f\_k\left(\left|x\_1\right|,\left|x\_2\right|,...,\left|x\_n\right|\right)$
for any reals $x\_1$, $x\_2$, ..., $x\_n$.
I think I have casebashed this for $n=4$ and maybe $n=6$; I don't remember anymore - it's too long ago. Sorry. I still have no idea what to do in the general case, although my attempts at big-$n$ counterexamples weren't of much success either.
| https://mathoverflow.net/users/2530 | Sum of difference moduli vs. sum of modulus differences | Hi, Darij!
This is actually quite simple (and also much more appropriate for AoPS than for MO). The idea is to show that for every $t$, the expression $\sum\_T|t+D\_T|$, where $D\_T$ is your difference, goes down if you replace all $x\_k$ by their absolute values ($t=0$ is your claim). The base $n=2$ is rather trivial and boils down to the inequality $|t-a|+|t+a|=2\max(|t|,|a|)\ge 2\max(|t|,|b|)=|t-b|+|t+b|$ when $|a|\ge |b|$. Now, assume that we know the statement for $n$ and want to show it for $n+2$. The trick is to choose a random pair of indices $i,j$ and notice that the full sum is just the average over such choices of $\sum\_T(|t+x\_i-x\_j+D\_T|+|t+x\_j-x\_i+D\_T|)$ where $T$ runs over all $n/2$ element subsets of the set of remaining indices. Now, applying the statement with $n=2$ for fixed $T$, we see that we can replace $x\_i$ and $x\_j$ with their absolute values and our sum (with fixed $i,j$) will go down in each term. After that, we replace everything else by the absolute values using the induction assumption and, again, the sum will go down. But now we are completely done: we showed that for each fixed $i,j$, the sum goes down when we replace everything by the absolute value, so it is true after averaging as well.
The whole thing is just a textbook case of the "inventor's paradox". Strange that you haven't figured it out...
| 5 | https://mathoverflow.net/users/1131 | 20985 | 13,911 |
https://mathoverflow.net/questions/20996 | 4 | I would like to know under what condition the morphism $\mathcal{O}\_Y\longrightarrow f\_\ast \mathcal{O}\_X$ induced by a morphism $f:X\longrightarrow Y$ of schemes is injective.
Let me give an example (which I'm not completely sure about though).
I believe, if $X$ and $Y$ are reduced and $f$ is surjective and closed, the morphism $\mathcal{O}\_Y \longrightarrow f\_\ast \mathcal{O}\_X$ is injective.
(Thus, proper flat morphisms of varieties have this property.)
Maybe one could forget about schemes and give a condition for locally ringed spaces?
| https://mathoverflow.net/users/4333 | Given a morphism from X to Y, when is the morphism from O_Y to the pushforward of O_X injective | If $f$ is quasi-compact and quasi-separated then the kernel of the map $O\_Y\to f\_\*(O\_X)$ consists of locally nilpotent elements if and only if $f(X)\subset Y$ is a dense set.
| 6 | https://mathoverflow.net/users/781 | 21000 | 13,919 |
https://mathoverflow.net/questions/20997 | 3 | I was just going to check something in FGA and I didn't have access to my pdf-copy, so I did what I normally do when in such a circumstance: surf to the Grothendieck circle's webpage.
And what did I find there? All mathematical texts written by Grothendieck himself was removed "per his request".
Does anyone know what this is about and/or what's going on here?
| https://mathoverflow.net/users/2147 | What's up with the Grothendieck circle? | Grothendieck seems to have requested that all his materials be taken town. As pointed out in comments, this has been discussed fairly [thoroughly](http://sbseminar.wordpress.com/2010/02/09/grothendiecks-letter/) on [the blogs](http://golem.ph.utexas.edu/category/2010/02/grothendieck_said_stop.html) and, in fact, [on meta](http://mathoverflow.tqft.net/discussion/205/strange-message-on-yves-laszlos-sga-4-page/#Item_27).
| 5 | https://mathoverflow.net/users/66 | 21005 | 13,922 |
https://mathoverflow.net/questions/20968 | 1 | Hi, my apologies for a rather non-specific question. I wonder if there is a general set of conditions under which operators are commutative in functional analysis. Most that I've found is that "operators are, in general, not commutative". Is there any reference someone could point me to for some kind of review or special cases in which commutativity is established (or forbidden)? Thanks!
| https://mathoverflow.net/users/5282 | rules for operator commutativity? | One obvious but important observation is that, for operators on a $n$-dimensional vector space over a field, if $1 < n < \infty$, we have $AB \neq BA$ *generically*. In other words, consider the commutativity locus $\mathcal{C}\_n$ of all pairs of $n \times n$ matrices $A,B$ such that $AB = BA$ as a subset of $\mathbb{A}^{n^2}$. This is clearly a Zariski closed set -- i.e., defined by the vanshing of polynomial equations. It is also proper: take e.g.
$A = \left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right] \oplus 0\_{n-2}$ and $B =
\left[ \begin{array}{cc} 0 & 1 \\ 0 & 1 \end{array} \right] \oplus 0\_{n-2}$. Since $\mathbb{A}^{n^2}$ is an irreducible variety, $\mathcal{C}\_N$ therefore has dimension less than $N^2$. This implies that over a field like $\mathbb{R}$ or $\mathbb{C}$ where such things make sense, $\mathcal{C}\_N$ has measure zero, thus giving a precise meaning to the idea that two matrices, taken at random, will not commute.
One could ask for more information about the subvariety $\mathcal{C}\_N$: what is its dimension? is it irreducible? and so forth. (Surely someone here knows the answers.)
I would guess it is also true that for a Banach space $E$ (over any locally compact, nondiscrete field $k$, say) of dimension $> 1$, the locus $\mathcal{C}\_E$ of all commuting pairs of bounded linear operators is **meager** (in the sense of Baire category) in the space $B(E,E) \times B(E,E)$ of all pairs of bounded linear operators on $E$.
Kevin Buzzard has enunciated a principle that without further constraints, the optimal answer to a question "What is a necessary and sufficient condition for $X$ to hold?" is simply "X". This seems quite applicable here: I don't think you'll find a necessary and sufficient condition for two linear operators to commute which is nearly as simple and transparent as the beautiful identity $AB = BA$.
Still, you could ask for useful sufficient conditions. Diagonalizable operators with the same eigenspaces, as mentioned by Jonas Meyer above, is one. Another is that if $A$ and $B$ are both polynomials in the same operator $C$: this shows up for instance in the Jordan decomposition.
| 8 | https://mathoverflow.net/users/1149 | 21007 | 13,924 |
https://mathoverflow.net/questions/20963 | 6 | It is known that Namba forcing is stationary-preserving and hence can be used in the setting of Martin's Maximum. Does this result in any striking consequences?
| https://mathoverflow.net/users/4706 | What are the Martin's Maximum consequences of Namba forcing? | I think that I may have found a suitable candidate; namely, the result of
Konig and Yoshinobu that $MM$ implies that there are no $\omega\_{1}$-regressive $\omega\_{2}$-Kurepa trees. The proof seems to have the same relatively direct flavor as those in Baumgartner's $PFA$ article.
| 5 | https://mathoverflow.net/users/4706 | 21008 | 13,925 |
https://mathoverflow.net/questions/20600 | 2 | $X$ = bi-elliptic surface (smooth and over $\mathbb{C}$),
Aut($X$) = the group of automorphisms of $X$,
Aut$^0(X)$ = connected component of the identity in Aut($X$).
Is Aut$^0(X)$ always an affine algebraic group?
| https://mathoverflow.net/users/5197 | Automorphism group of bi-elliptic surface | The answer is always "no". By classification, a bielliptic surface over $\mathbb C$ has the form $(E\times F)/G$ where $E,F$ are elliptic curves, $G=\subset Aut(E,0)$ is an abelian group acting by complex multiplications on $E$ and by translations on $F$. ($G$ is not necessarily cyclic as Tuan correctly points out.)
($X$ maps to an elliptic curve $F/G$ and every fiber is isomorphic to an elliptic curve $E$, hence the name *bielliptic*.)
Then $F$ acts on $E\times F$ by $(x,y)\mapsto (x,y+f)$, and this action commutes with the $G$-action. Thus, $F\subset Aut^0(X)$. As $F$ is a projective variety, $Aut^0(X)$ is not affine.
| 4 | https://mathoverflow.net/users/1784 | 21009 | 13,926 |
https://mathoverflow.net/questions/20867 | 6 | Hi all,
given (a1,...,an) formed by distinct letters, it's a well known problem to count the number of permutations with no fixed element.
I've been trying to solve a generalization of this problem, when we allow repetition of the letters.
I was able only to partially solve the problem when we have only repetition of a single letter.
If we have n letters and only one of them is repeated p times, then the number O(n,p) of permutations with no fixed element is given by the following recursive relation:
$O(n,0)=O(n,1)=\mbox{Derangement}(n)$
$O(p+1,p)=\dots=O(2p-1,p)=0, O(2p,p)=p!$
$O(n,p)={n-p\choose p} p!\sum\_{k=0}^p{p \choose k}O(n-p-k,p-k)$
Does anybody know where this problem have been studied before? Does anybody know a general solution for this problem?
Thanks in advance.
| https://mathoverflow.net/users/1172 | Derangements with repetition | The formula based on Inclusion-Exclusion for the usual number $D(n)$ of derangements of $n$ objects can be generalized. The result is the following.
Fix $k\geq 1$. Let $\mathbb{N}=\lbrace 0,1,2,\dots\rbrace$. For $\alpha=(\alpha\_1,\dots,\alpha\_k)\in\mathbb{N}^k$, let $D(\alpha)$ be the number of fixed-point free permutations of the multiset with $\alpha\_1$ 1's, $\alpha\_2$ 2's, etc. Write $x^\alpha = x\_1^{\alpha\_1}\cdots x\_k^{\alpha\_k}$. Then
$$ \sum\_{\alpha\in\mathbb{N}^k} D(\alpha)x^\alpha =
\frac{1}{(1+x\_1)\cdots (1+x\_k)\left(1-\frac{x\_1}{1+x\_1}-\cdots -
\frac{x\_k}{1+x\_k}\right)} $$
$$ = \frac{1}{1-\sum\_S
(|S|-1)\prod\_{i\in S}x\_i}, $$
where $S$ ranges over all nonempty subsets of $\lbrace 1,2,\dots,n\rbrace$.
This result appears as Exercise 4.5.5 in Goulden and Jackson, *Combinatorial Enumeration*. It can be used to obtain a lot of information about $D(\alpha)$.
| 16 | https://mathoverflow.net/users/2807 | 21019 | 13,928 |
https://mathoverflow.net/questions/21018 | 2 | Let $f:X\longrightarrow \mathbf{P}^1$ be a finite morphism of schemes. If $f$ is etale and $X$ is connected, we can show that $f$ is an isomorphism. That is, the projective line is simply connected. I would like to know if something more general can be said about $f$ if it is not assumed to be etale.
In this generality, it seems quite hard. But what if we assume $X$ to be also the projective line for example? Is it true that $f$ is then of the form $[x:y]\mapsto [x^n:y^n]$ in some sense?
| https://mathoverflow.net/users/4333 | What is known about finite morphisms from X to the projective line | No because $f$ can be ramified pretty much anywhere. Just think of a random rational function $f=p(x)/q(x)$ with $p,q$ coprime polynomials, $p$ non-constant and $q$ non-zero. That gives a finite morphism from the projective line to itself that is in general much more complicated than the map you suggest.
| 6 | https://mathoverflow.net/users/1384 | 21020 | 13,929 |
https://mathoverflow.net/questions/15645 | 4 | I wanted recently to discuss with a fairly elementary mathematics class the kinds of self-maps of Euclidean space that carry triangles to triangles. Obviously linear maps do this, and it seemed just as obvious to me that they were the only ones (up to translation).
I asked a colleague, and he pointed out an obvious counterexample: Project to the $x$-axis, then apply whatever continuous but non-linear map one likes. He quickly suggested a remedy: Is a line-segment-preserving local diffeomorphism necessarily (affine) linear?
Well, surely *that's* enough --but I have been told that Walter Poor's "Differential geometric structures" assigns as an exercise to prove that there exist non-affine, projective vector fields on $\mathbb R^n$, and that a solution to this exercise gives a counterexample to the revised conjecture. (I don't know the words, but it sounds plausible to me.)
Under fairly weak continuity hypotheses (certainly local diffeomorphism is sufficient!), it suffices to prove additivity; and, to prove additivity, it suffices to prove that parallelograms are carried to parallelograms. The only way that this can fail for a local diffeomorphism is if there are some lines carried onto proper subsets of lines; and I just can't seem to see how this can happen without breaking some line segments.
(Another colleague to whom I proposed the problem argued as follows: If the Jacobian matrix isn't constant, then there is a point at which the Hessian is non-$0$. Near this point, the map is approximately a non-linear quadratic, and one can show that such quadratics don't preserve line segments. Of course, it is the ‘approximately’ that bites us here; but I'd be interested in seeing how such a straightforward estimate fails.)
| https://mathoverflow.net/users/2383 | Non-affine, projective vector field on $R^n$ | I don't know what was meant in that exercise, but your revised conjecture is certainly true and well-known. Here is an elementary proof.
The assumptions (local injectivity, continuity and segment-to-segment mapping) imply that the map is injective and preserves collinearity of points and endpoints of segments. By continuity it suffices to show that the midpoint of every segment is sent to the midpoint of its image.
Consider any two points $A,B$. Let $A',B'$ be their images and $D'$ the midpoint of $[A'B']$. By continuity there is a point $D\in[AB]$ mapped to $D'$. We are to prove that $D$ is the midpoint of $[AB]$. Pick a point $C$ such that its image $C'$ is not collinear with $A'$ and $B'$ and choose points $E\in[AC]$ and $F\in[BC]$ such that $CD$, $BE$ and $AF$ have a common point $G$ in the triangle $ABC$. Note that all these points belong to the plane containing $A,B,C$.
By elementary geometry (see e.g. [Ceva's theorem](http://en.wikipedia.org/wiki/Ceva%2527s_theorem)), $D$ is the midpoint of $[AB]$ if and only if $EF$ is parallel to $AB$. Suppose it is not, then the line $EF$ intersects the line $AB$ at some point $H$. This configuration is mapped to a configuration $A',B',\dots,H'$ with the same collinearity relations. Since $E'F'$ is not parallel to $A'B'$ (they intersects at $H'$), the same application of Ceva's theorem shows that $D'$ is not the midpoint of $[A'B']$, a contradiction.
| 7 | https://mathoverflow.net/users/4354 | 21022 | 13,930 |
https://mathoverflow.net/questions/10603 | 21 | Let $E\_1$ and $E\_2$ be two Elliptic curve defined over $\mathbb Q$ . Let $p$ be an fixed given odd prime of $\mathbb Q$ at which both the curves have good ordinary reduction. Moreover p-adic $L$-function of $E\_1$ and $E\_2$ are same . Does it mean that the complex $L$-function of $E\_1$ and $E\_2$ are also same ?
Is there some sufficient criteria on p-adic $L$-functions such that such that the $L$ function is determined?
| https://mathoverflow.net/users/2876 | Does p-adic $L$- function determine the $L$ function | Hmmm. I am not so sure that the answer is *"No"*. In fact I would rather bet on *"Yes"*.
Of course, I totally agree that the characteristic ideal, i.e. the ideal generated by the $p$-adic $L$-function in $\Lambda$, is not enough to determine the elliptic curve. In particular there are plenty of curves for which the $p$-adic $L$-function is a unit in $\Lambda^{\times}$.
The $p$-adic $L$-function can be viewed as a measure $\mu$ on the Galois group of $F\_\infty/\mathbb{Q}$. It is build up from modular symbols of the form $\bigl[\frac{a}{p^k}\bigr]$ as $a$ and $k$ varies over all positive integers. Knowing the measure $\mu$ it is easy to extract the unit root $\alpha$ of the Frobenius at $p$ and hence the value of $a\_p$. Then it is not difficult to see from the definition of $\mu$ that one can compute all the modular symbols $\bigl[\frac{a}{p^k}\bigr]$. It is true that these values do not seem to carry the value of $a\_{\ell}$ for primes $\ell\neq p$ with them needed to reconstruct the complex $L$-function; we would need modular symbols with $\ell$ in the denominator and I can not see immediately how to get them from $\mu$.
Nevertheless, there are plenty of values of $a$ and $k$. And it would be a big surprise to me if there very by chance two elliptic curves such that all the values of the modular symbols $\bigl[\frac{a}{p^k}\bigr]$ would be equal. But I have no clue of how to prove this intuition.
So I ran through some examples. Let $F\_{\infty}$ be the cyclotomic $\mathbb{Z}\_p$-extension of $\mathbb{Q}$.
I picked a few elliptic curves of small conductor such that
* $E$ has good ordinary reduction at $p$.
* There are no torsion points in $E(F\_{\infty})$, simply by making sure that the $\ell$-adic Galois representation is surjective for all $\ell$.
* The Tamagawa numbers are all 1 for $E/F\_\infty$, by imposing that the Kodaira type at all bad places is $I\_1$.
* The curve is not anomalous at $p$, i.e. $a\_p \neq 1$. Actually, I fix $a\_p$.
* The Tate-Shafarevich group of $E/\mathbb{Q}$ is trivial.
* The rank of $E(F\_{\infty})$ is $0$. This will follow from the previous conditions if the rank of $E(\mathbb{Q})$ is $0$, since the $p$-adic $L$-function will be a unit.
Then I computed the $p$-adic $L$-functions $L\_p(T)$ for these with $T$ corresponding to $1+p$ under the cyclotomic character Gal$(F\_{\infty}/\mathbb{Q})$. By what I have imposed the leading term will be equal. For each $p^n$-th root of unity $\zeta$, the value of $L\_p(\zeta-1)$ is, up to a power of $\alpha$ which is the same for all my curves because I fixed $a\_p$, equal to the order of the Tate-Shafarevich group at the $n$-th level; at least if one believes the ($p$-adic version of the) Birch and Swinnerton-Dyer conjecture. From what I imposed, it is clear that the $p$-primary part will be trivial, but there may be different primes appearing in Sha for various curves. So there is no reason to believe that it would be easy to find two curves that have the same $p$-adic $L$-function in these family that I have chosen.
Here are some examples with $p=5$ and $a\_5=-1$.
139a1 $4 + 4 \cdot 5 + 4 \cdot 5^2 + O(5^5) + (1 + 4 \cdot 5 + O(5^2)) \cdot T + (3 + 5 + O(5^2)) \cdot T^2 + (3 + 2 \cdot 5 + O(5^2)) \cdot T^3 + (1 + 5 + O(5^2)) \cdot T^4 + O(T^5)$
141e1 $4 + 4 \cdot 5 + 4 \cdot 5^2 + O(5^5) + (4 + 3 \cdot 5 + O(5^2)) \cdot T + (3 \cdot 5 + O(5^2)) \cdot T^2 + (5 + O(5^2)) \cdot T^3 + (2 + 4 \cdot 5 + O(5^2)) \cdot T^4 + O(T^5) $
346a1 $ 4 + 4 \cdot 5 + 4 \cdot 5^2 + O(5^5) + (2 + 5 + O(5^2)) \cdot T + (4 \cdot 5 + O(5^2)) \cdot T^2 + O(5^2) \cdot T^3 + (1 + 2 \cdot 5 + O(5^2)) \cdot T^4 + O(T^5)$
906i1 $4 + 4 \cdot 5 + 4 \cdot 5^2 + O(5^5) + (3 + O(5^2)) \cdot T + (2 + 5 + O(5^2)) \cdot T^2 + (3 + 5 + O(5^2)) \cdot T^3 + (3 + 2 \cdot 5 + O(5^2)) \cdot T^4 + O(T^5)$
Finally a word why I believe the answer should be *Yes*. A big dream in the direction of BSD is to hope that there is a link between the $p$-adic and the complex $L$-function. Of course, we should believe that the order of vanishing at $s=1$ should be equal for instance (and this is only known when it is at most a simple zero). I would hope that such a link will be bijective. But that is a mere intuition and hence possibly wrong. But it also explains that the answer to this question might well be very difficult.
| 13 | https://mathoverflow.net/users/5015 | 21029 | 13,933 |
https://mathoverflow.net/questions/18833 | 10 | Is there some generalization of the Jordan-Hölder decomposition for group objects in a category $\mathcal{C}$?
If $\mathcal{C}$ is the category Sch$(S)$ of schemes over a base scheme $S$ then (I think) this is true, also probably for other categories of "spaces" like Top or Diff it should be true, but I don't have any idea for general categories.
| https://mathoverflow.net/users/4619 | Jordan Hölder decomposition for group objects | You can start by looking at the paper by P.J. Hilton and W. Ledermann, "On the Jordan-Hölder theorem in homological monoids", where three axioms are needed to establish the decomposition, and the third one is essentially guaranteeing the second isomorphism theorem.
In "Mal'cev, protomodular, homological and semi-abelian categories" by F. Borceux, D. Bourn, there is a chapter devoted to homological categories (which are [pointed](http://ncatlab.org/nlab/show/pointed+category), [regular](http://ncatlab.org/nlab/show/regular+category) and protomodular), these are the categories where certain lemmas of homological algebra hold true (five lemma, nine lemma, snake lemma, Noether isomorphism theorems etc.). The fact that Jordan-Holder holds for these categories is proven in "Jordan-Holder, Modularity and Distributivity in Non-Commutative Algebra" by F. Borceux, M. Grandis.
| 6 | https://mathoverflow.net/users/2384 | 21032 | 13,935 |
https://mathoverflow.net/questions/20705 | 6 | As per [a recent question of mine](https://mathoverflow.net/questions/20228/omega-1-times-beta-mathbbn-normal), $\omega\_1 \times \beta \mathbb{N}$ is not normal. I'm wondering whether there's some sort of "natural" condition that describes when a space has a normal product with $\beta \mathbb{N}$, analagous to Dowker's characterisation that $X$ is countably paracompact iff $X \times [0, 1]$ is normal.
The context here is that I'm looking for something I can weaken to something sensible, as I have a property implied by $X \times \beta \mathbb{N}$ being normal which I am "surprised" isn't an equivalence ($\omega\_1$ has this property) and would like to see if I can show its equivalenct to some slightly weaker topological condition.
| https://mathoverflow.net/users/4959 | Is there a "natural" characterization of when X × βN is normal? | The space $X\times\beta\mathbb{N}$ is normal if and only if $X$ is normal and $\mathfrak{c}$-paracompact. This follows from results of Morita ([*Paracompactness and product spaces*](http://matwbn.icm.edu.pl/ksiazki/fm/fm50/fm50119.pdf), MR[132525](http://www.ams.org/mathscinet-getitem?mr=132525)), where he generalizes Dowker's characterization of countable paracompactness.
First note that $X\times\beta\mathbb{N}$ is normal if and only if $X\times K$ is normal for every separable compact Hausdorff space $K$. This is because every separable compact Hausdorff space is a perfect image of $\beta\mathbb{N}$.
Morita's Theorem 2.2 shows that if $X$ is normal and $\mathfrak{c}$-paracompact, then $X \times K$ is normal for every compact Hausdorff space $K$ of weight at most $\mathfrak{c}$. Hence, $X\times K$ is normal for every separable compact Hausdorff space $K$ since these all have weight at most $\mathfrak{c}$.
Morita's Theorem 2.4 shows that a space $X$ is normal and $\mathfrak{c}$-paracompact if (and only if) $X\times[0,1]^{\mathfrak{c}}$ is normal. Since the space $[0,1]^{\mathfrak{c}}$ is a separable compact Hausdorff space, this closes the implication loop.
| 6 | https://mathoverflow.net/users/2000 | 21033 | 13,936 |
https://mathoverflow.net/questions/21031 | 14 | This question was actually asked by John Stillwell in a comment to an answer to this [question](https://mathoverflow.net/questions/20882/most-unintuitive-application-of-the-axiom-of-choice). I thought I would advertise it as a separate question since no one has yet answered and I am also curious about it.
**Question:** Is the existence of a non-principal ultra-filter on $\omega$ a weaker assumption than the existence of a well-ordering of $\mathbb{R}$?
| https://mathoverflow.net/users/2233 | Ultrafilters vs Well-orderings | It is consistent that there exists a non-principal ultrafilter over $\omega$ while $\mathbb{R}$ is not well-ordered. To see this, suppose that the
partition relation $\omega \to (\omega)^{\omega}$ holds in $L(\mathbb{R})$.
Then forcing with $\mathbb{P}= [\omega]^{\omega}$ adjoins a selective ultrafilter $\mathcal{U}$ over $\omega$ and $\mathcal{P}(\omega)$ cannot be well-ordered in
$L(\mathbb{R})[\mathcal{U}]$. (See Eisworth's paper: Selective ultrafilters and
$\omega \to (\omega)^{\omega}$.) Thus $L(\mathbb{R})[\mathcal{U}]$ is a model of $ZF$ which contains the nonprincipal ultrafilter $\mathcal{U}$ and yet $\mathbb{R}$ cannot be well-ordered in
$L(\mathbb{R})[\mathcal{U}]$.
An update: it is perhaps also interesting to note that in $L(\mathbb{R})[\mathcal{U}]$, the ultraproduct $\prod\_{\mathcal{U}} \bar{\mathbb{F}}\_{p}$ of the algebraic closures of the fields of prime order $p$ is not isomorphic to $\mathbb{C}$.
| 15 | https://mathoverflow.net/users/4706 | 21034 | 13,937 |
https://mathoverflow.net/questions/21028 | 3 | Apologies for the vague title - I couldn't come up with a single sentence that summarised this problem well. If you can, please edit or suggest a better one!
This question is also rather specific and contains lots of annoying technical detail. I must admit to not really expecting an answer unless there's an obvious solution I'm missing (which is very possible - I feel like any solution is either going to be obvious or very deep), but some pointers in plausible sounding directions would be greatly appreciated. I suspect the answer will depend on the combinatorics of $\omega\_1$, which I know relatively little about.
Let $V$ be a normed space. For $A \subseteq V$, define $r(A) = \inf \{ r : \exists V, A \subseteq B(v, r) \}$. Define a *bad sequence* in $V$ to be a sequence $\{ v\_\alpha : \alpha < \omega\_1 \}$ with the properties that:
$\forall \beta, r(\{ v\_\alpha : \alpha < \beta \}) \leq 1$
$\inf\_\beta r(\{ v\_\alpha : \alpha \geq \beta \}) > 1$
An example of a space with a bad sequence is $c\_0(\omega\_1)$ (the set of all bounded real-valued sequences of length $\omega\_1$ such that $\{ \alpha : |x\_\alpha| > 0 \}$ is countable). The sequence $2 \* 1\_{\{\alpha\}}$ is bad. The radius of any tail is $2$ because the center must be eventually 0. The radius of the initial segments is $\leq 1$ because the segment up to $\alpha$ is contained in the closed ball of radius 1 around $1\_{[0, \alpha]}$, which is in $c\_0(\omega\_1)$ because $\alpha < \omega\_1$.
I have two (three depending on how you count it) major examples of spaces which have no bad sequences:
* Any separable space: you can choose centers to lie in the countable dense set, so one center must work as a radius for the initial segment for unboundedly many and thus for all $\alpha$.
* Any space which has what I'm imaginatively calling the chain-radius condition: The union of a chain of sets of radius $\leq r$ has radius $\leq r$. This includes:
+ Any reflexive space: If $U\_\alpha$ forms a chain, the sets $F\_\alpha = \bigcap\_{v \in U\_\alpha} \overline{B}(v, r + \epsilon)$ form non-empty closed and bounded convex sets with the finite intersection property, so compactness in the weak topology implies they have non-empty intersection. Any element of the intersection contains the union of the chain in $\overline{B}(c, r + \epsilon)$
+ any space with the property that $\textrm{diam}(A) = 2 r(A)$ (in particular the $l^\infty$ space on any set) because it's clear that unions of chains of diameter $\leq 2r$ have diameter $\leq 2r$.
So... that's all the backstory for this question. Given that, my actual question is very simple: Does $C(\omega\_1)$ contain a bad sequence?
I feel like the answer "must" be no. In particular note that the projection of any sequence onto the first $\alpha$ entries is not bad (because it's a sequence in a separable space) and that if you drop the restriction for continuity the answer is immediately yes. So it sits right between two classes of examples where there are no bad sequences, and I feel that one really should be able to take advantage of that. But on the other hand, functions in $C(\omega\_1)$ are eventually constant, so maybe you can take advantage of that to construct some sets with arbitrary bad tails.
For bonus kudos, I'd love to know for what compact Hausdorff spaces $K$, $C(K)$ contains a bad sequence.
| https://mathoverflow.net/users/4959 | Shape of long sequences in C(ω_1) | I claim that there are no bad sequences in C(ω1).
Suppose to the contrary that xα is bad. For any countable
ordinal β, there is rβ in
C(ω1) such that the distance between rβ and
xα for α < β is at most 1.
For any countable ordinal β and any positive
rational number ε, there is a smaller ordinal
γ < β such that all
rβ(α) are within ε of
rβ(β) for α in
(γ,β]. For fixed ε, this is a regressive
function on the countable ordinals. Thus, by Fodor's
Lemma, there is a stationary and hence unbounded set of
ordinals on which the function has constant value,
which we may call γε. Since there
are only countably many ε, we may find a countable
ordinal γ above all γε. This
ordinal has the property that for all ordinals β
above γ, we have rβ(α) =
rβ(β) for all α in the interval
(γ,β], since the values are within every
ε of each other. That is, every rβ
function is constant from the same fixed γ up to
β.
Let Cβ be the closed interval of values s
such that the constant sequence s of length β lies
within 1 of all xη(α) for all η
≤ β and all γ < α ≤ β. These
are nested and not empty, since rβ(β)
is in Cβ. By compactness, there is a value
s in all Cβ. Thus, the number s is within
xη(α) for all η and all α
above γ.
Thus, we may form the desired sequence r by finding a center that works
for the sequences up to stage γ, using the separability idea in your question, augmented with the constant value s at the stages above
γ up to ω1. That is, we solve the problem separately on the first γ many coordinates, and then append the constant s sequence up to ω1. This sequence is
continuous, and it lies within 1 of every
xη, as desired.
| 3 | https://mathoverflow.net/users/1946 | 21055 | 13,953 |
https://mathoverflow.net/questions/21051 | 14 | Illusie's article about étale cohomology [available here](http://www.math.u-psud.fr/~illusie/Grothendieck_etale.pdf) (in French) mentions that the standard definition of compactly supported cohomology (and higher direct images with compact support) does not give the right answers in the case of étale cohomology.
He says Grothendieck had the idea of defining them as follows instead:
$$Rf\_! = Rg\_\* \circ j\_!$$
where $f = gj$ is a compactification of $f$: $g$ is proper, and $j$ is an open immersion. It takes a bit of work to see that this is well defined (a theorem of Nagata guarantees the existence of this compactification, and the proper base change theorem shows that the result does not depend on which compactification is chosen).
What is the explanation here for the difference between Grothendieck's definition and the usual one, which amounts to $Rf\_! = R(g\_\* j\_!)$ ? What is it that allows us to say that either is the "right" one?
It seems to me that a satisfactory explanation should be more than just computing the groups in a special case and appealing to intuition (indeed, the compactly supported cohomology groups are often strange, at first sight, even in the topological situation).
For example, I could imagine that an explanation might involve derived categories: after all, one of the main uses of $Rf\_!$ is Verdier duality, which implies many useful properties such as Poincaré duality. This maybe gives some hints as to which definition is preferred.
| https://mathoverflow.net/users/362 | Why does the naive definition of compactly supported étale cohomology give the wrong answer? | It is important in etale cohomology, as it is topology, to define cohomology
groups with compact support --- we saw this already in the case of curves in
Section 14. They should be dual to the ordinary cohomology groups.
The traditional definition (Greenberg 1967, p162) is that, for a manifold
$U$,
$
H\_{c}^{r}(U,\mathbb{Z})=dlim\_{Z}H\_{Z}^{r}(U,\mathbb{Z})
$
where $Z$ runs over the compact subsets of $U$. More generally (Iversen 1986,
III.1) when $\mathcal{F}$ is a sheaf on a locally compact topological space
$U$, define
$
\Gamma\_{c}(U,\mathcal{F})=dlim\_{Z}\Gamma\_{Z}(U,\mathcal{F})
$
where $Z$ again runs over the compact subsets of $U$, and let $H\_{c}%
^{r}(U,-)=R^{r}\Gamma\_{c}(U,-)$.
For an algebraic variety $U$ and a sheaf $\mathcal{F}$ on $U\_{\mathrm{et}}$,
this suggests defining
$
\Gamma\_{c}(U,\mathcal{F})=dlim\_{Z}\Gamma\_{Z}(U,\mathcal{F}),
$
where $Z$ runs over the complete subvarieties $Z$ of $U$, and setting
$H\_{c}^{r}(U,-)=R^{r}\Gamma\_{c}(U,-)$. However, this definition leads to
anomolous groups. For example, if $U$ is an affine variety over an
algebraically closed field, then the only complete subvarieties of $U$ are the
finite subvarieties, and for a finite subvariety $Z\subset
U$,
$
H\_{Z}^{r}(U,\mathcal{F})=\oplus\_{z\in Z}H\_{z}^{r}(U,\mathcal{F}).
$
Therefore, if $U$ is smooth of dimension $m$ and $\Lambda$ is the constant
sheaf $\mathbb{Z}/n\mathbb{Z}$, then
$
H\_{c}^{r}(U,\Lambda)=dlim H\_{Z}^{r}(U,\Lambda)=\oplus\_{z\in U}H\_{z}%
^{r}(U,\Lambda)=\oplus\_{z\in U}\Lambda(-m)$ if $r=2m$, and it is 0 otherwise
These groups are not even finite. We need a different definition...
If $j\colon\ U\rightarrow X$ is a homeomorphism of the topological space $U$
onto an open subset of a locally compact space $X$, then
$
H\_{c}^{r}(U,\mathcal{F})=H^{r}(X,j\_{!}\mathcal{F})
$
(Iversen 1986, p184).
We make this our definition.
From Section 18 of my notes: Lectures on etale cohomology.
| 15 | https://mathoverflow.net/users/930 | 21056 | 13,954 |
https://mathoverflow.net/questions/21062 | 6 | I have never understood the trace map,not even after reading [Geometric Interpretation of Trace](https://mathoverflow.net/questions/13526/geometric-interpretation-of-trace). The problem with many answers in the above discussion is the geometric intuition does not apply to other field.
As I don't want this to be closed, let me make the question more precise. Is there a definition of the trace map which
1) is basis independent, (there was a definition given by Sridhar Ramesh in the old post).
2) explains in an intuitive way why if $L$ is a finite separable extension of $K$, the map $ (x,y) \mapsto Tr(xy) $, where $x,y$ are in $L$, is a non degenerated bilinear form on L?
| https://mathoverflow.net/users/2701 | Is there good intution of the trace map? | I don't know if this is what you're looking for, but there's a basis-free definition of the trace in general, outside of the algebraic number theory context -- a linear transformation $V\to V$ corresponds in a natural way to an element of $V\otimes V^\ast$, and the trace map is the map $V\otimes V^\ast\to k$ induced by the bilinear map $V\times V^\*\to k$ which sends $(v,f)$ to $f(v)$.
But I think the easiest way to see why the trace pairing is nondegenerate for a separable extension is to use bases. Intuitively, nonseparability corresponds to a linear dependency relation among the rows of the matrix because it leads to "repetitions" among the conjugates of some element of the field on top. There's a great (though kind of short) exposition of this stuff in Milne's notes on algebraic number theory: <http://jmilne.org/math/CourseNotes/ant.html>
| 10 | https://mathoverflow.net/users/5281 | 21064 | 13,960 |
https://mathoverflow.net/questions/21014 | 2 | I found the following paragraph in the paper " Intro to symplectic field theory "
which I don't understand what does it mean precisely?
Suppose W is a symplectic (or Kahler) manifold.
D, smooth divisor in it.
then $\tilde{W}= W-D $ is a Weinstein manifold which contains an isotropic deformation retract $\Delta$.
Does any body know what kind of object is $\Delta$ here? and what does isotropic deformation mean?
| https://mathoverflow.net/users/5259 | isotropic deformation retract of Weinstein manifolds? | This story begins with the Lefschetz hyperplane theorem - the fact that if $D$ is a smooth, very ample divisor in a closed Kaehler manifold $X$ then $D$ carries all the homology and homotopy of $X$ below the middle dimension of $X$.
One way to understand this theorem is via Morse theory (the Andreotti-Frankel approach). There are explanations in Milnor's "Morse Theory", Griffiths-Harris and elsewhere. The line bundle $\mathcal{O}(D)$ has a canonical section $s$ whose zeroes cut out $D$. One looks at the function $f= - \log \|s\|^2$ on $M=X-D$ (defined via a hermitian metric on the line bundle). This is a bounded-below, proper, strictly plurisubharmonic function; it makes $M$ a Stein manifold. Its critical points (if non-degenerate) all have index $\leq \dim\_{\mathbb{C}}X$.
Assuming $f$ Morse, the isotropic skeleton of $M$ with respect to $f$ is the union of the unstable manifolds of the critical points of $f$. Each of these unstable manifolds is isotropic with respect to the Kaehler form (their union may be singular). The downward gradient flow of $f$ will (after a bit of tweaking) define a deformation retraction from $M$ to this skeleton. E.g. for $\mathbb{CP}^{n-1}\subset \mathbb{CP}^n$, the skeleton will be a point.
Very ample divisors are very special; none of this applies to more general divisors. Nonetheless, the story generalizes to symplectic manifolds whose symplectic class in $H^2$ is integral. In this case, Donaldson ("Symplectic submanifolds and almost complex geometry") showed that one can find distinguished symplectic divisors $D$ representing high multiples of the symplectic class. The discussion of the hyperplane complement $M$ is valid also for the complement of a Donaldson hypersurface. The "Stein" condition has to be replaced by something more suitable for almost complex manifolds ("Weinstein") but by a really startling theorem of Eliashberg, Weinstein structures can always be deformed to Stein structures.
For further reading, try Paul Biran's paper
[Lagrangian Non-Intersections](http://arxiv.org/abs/math/0412110).
| 4 | https://mathoverflow.net/users/2356 | 21068 | 13,961 |
https://mathoverflow.net/questions/20699 | 4 | Let $f\_1,\ldots f\_m \in k[X]$ have degrees bounded by $l$. and $I(\bar{f})$ be the ideal generated by $\bar{f}$.
If $I(\bar{f})$ is not a prime ideal then its non-primeness is witnessed by polynomials of degree below $h(m,l)$ for some $h$ dependent only on $m$ and $l$. Hrushovski gives two references for the truth of this statement, but the reference is at least for me unreadable (one due to unavailability and one due to length and my laziness).
Question: Is it known that the function $h$ is bounded by a recursive function of (m,l)? A reference for a bounding of $h$ would be great.
| https://mathoverflow.net/users/5036 | Prime-ness checking for polynomial ideals over ACFs( algebraically closed fields). | In his paper *Constructions in algebra* (MR[349648](http://www.ams.org/mathscinet-getitem?mr=349648)), Seidenberg fixes some errors in Grete Hermann's classic paper *Die Frage der endlich vielen Schritte in der Theorie der Polynomideale* (MR[1512302](http://www.ams.org/mathscinet-getitem?mr=1512302)). Most of the work concentrates on two problems:
1. Effectively compute the [primary decomposition](http://en.wikipedia.org/wiki/Primary_decomposition#Irreducible_decomposition_in_rings) of an ideal.
2. Effectively compute the [associated prime](http://en.wikipedia.org/wiki/Associated_prime_ideal) of a primary ideal.
Together, these do give a primality test: first check that the ideal is primary by computing its primary decomposition and then check that it equals its associated prime.
Seidenberg's paper is very well written, but he obviously has a very limited audience in mind. He does prove along the way the existence of primitive recursive degree bounds, but the task to put them together into a precise form would require a fair amount of time. Seidenberg's constructivist point of view gets in the way of pragmatism from time to time.
There has been a lot of water under the bridge since 1974 and much of the contents of Seidenberg's paper must have been redone for practical use in modern computer algebra systems. Since working out the bounds from Seidenberg's paper looks impractical, I would consider looking at the computer algebra literature to see which algorithms are currently used to carry out the two steps above and derive effective degree bounds from these.
I am not very knowledgeable in such practical matters, but there are probably computer algebra experts lurking around here. If they don't see this post, then I suggest you ask another question for algorithms or degree bounds for steps 1 and 2 specifically.
| 1 | https://mathoverflow.net/users/2000 | 21069 | 13,962 |
https://mathoverflow.net/questions/21023 | 7 | Let $k$ be an algebraically closed field of characteristic $p > 0$, $X$ a variety over $k$.
We say $X$ lifts to characteristic zero, if there exists a local ring $R$ containing $\mathbb Z$ with residue field $k$ and a flat scheme $\mathcal X$ over $R$, such that
$$ \mathcal X \otimes\_R k \simeq X.$$
In words: There exists a family over a ring of mixed characteristic, which has our $X$ for a special fibre.
For most classes of surfaces in Kodaira dimension zero, liftability is known: For K3 surfaces, liftability was established by Deligne, and for abelian surfaces, one can use more general theories developed for abelian varieties. Bi-elliptic and quasi-bi-elliptic surfaces can be dealt with explicitly.
As far as I know, there is nothing in the literature about Enriques Surfaces. For this class, the case p = 2 is the most interesting.
The question seems natural, so it would struck me as strange if it were still open. Does anyone around here know anything about this?
Thanks a lot.
| https://mathoverflow.net/users/5273 | Liftability of Enriques Surfaces (from char. p to zero) | This may not be exactly the answer you are looking for: I and Nick
Shepherd-Barron have an unpublished (so far) proof of liftability in
characteristic $2$, the only non-trivial case. To atone for the fact that I
refer to unpublished results I give a quick sketch of proof.
The proof starts by showing that in a family $X/S$ of Enriques surfaces
$\mathrm{Pic}^\tau(X/S)$ is flat (of order $2$) over $S$ and
$\mathrm{Pic}(X/S)/\mathrm{Pic}^\tau(X/S)$ is locally constant. This implies
that the tensor square of any line bundle of an Enriques surface extends along
any formal deformation and hence it is enough to find a formal lifting.
For a surface with $h^2(T\_X)=0$ the deformations are unobstructed so we are
OK. There are two types of surfaces with $h^2(T\_X)=1$ (which is the only other
possibility); surfaces with $\mathrm{Pic}^\tau(X)=\alpha\_2$ and surfaces
$\mathrm{Pic}^\tau(X)=\mathbb Z/2$ having non-trivial vector fields (the latter case
exists). The first case is nicer in that we get a map from deformations of
such surfaces to deformations of $\alpha\_2$ and this map is formally smooth. As
we can lift $\alpha\_2$ (a formal deformation has base $\mathbb
Z\_2[[x,y]]/(xy-2)$) we can also lift the surface, in fact over a base with
absolute ramification of order $2$. The second case we know less
about but as $h^2(T\_X)=1$ the base of the deformation is (at most) a
hypersurface singularity and as one can show that it is a very small family one
can show that a versal deformation is flat over $\mathbb Z\_2$. We know nothing
about the ramification necessary in this case.
| 8 | https://mathoverflow.net/users/4008 | 21070 | 13,963 |
https://mathoverflow.net/questions/21076 | 7 | Let $R$ be a Dedekind domain with fraction field $K$.
Say that a Dedekind domain $R$ has the **Riemann-Roch property** if: for every nonzero prime ideal $\mathfrak{p}$ of $R$, there exists an element $f \in (\bigcap\_{\mathfrak{q} \neq \mathfrak{p}} R\_{\mathfrak{q}}) \setminus R$, i.e., an element of $K$ which is integral at every prime ideal $\mathfrak{q} \neq \mathfrak{p}$ and is not integral at $\mathfrak{p}$.
>
> Do all Dedekind domains have the Riemann-Roch property?
>
>
>
Motivation: for any subset $\Sigma \subset \operatorname{MaxSpec}(R)$, put $R\_{\Sigma} := \bigcap\_{\mathfrak{p} \in \Sigma} R\_{\mathfrak{p}}$. Then the maximal ideals of $R\_{\Sigma}$ correspond bijectively to the maximal ideals $\mathfrak{p}$ of $R$ such that $\mathfrak{p} R\_{\Sigma} \subsetneq R\_{\Sigma}$. Thus $\operatorname{MaxSpec}(R\_{\Sigma})$ may be viewed as containing $\Sigma$. $R$ has the Riemann-Roch property iff for all $\Sigma$,
$\operatorname{MaxSpec}(R\_{\Sigma}) = \Sigma$. Equivalently, the mapping $\Sigma \mapsto R\_{\Sigma}$ is an injection.
Remarks: $R$ has the Riemann-Roch property if its class group is torsion: then for every $\mathfrak{p} \in \operatorname{MaxSpec}(R)$ there exists $n \in \mathbb{Z}^+$ and $x \in R$ such that $\mathfrak{p}^n = (x)$, so take $f = \frac{1}{x}$. Also the coordinate ring $k[C]$ of a nonsingular, integral affine curve $C$ over a field $k$ has the Riemann-Roch property...by the Riemann-Roch theorem. Unfortunately this already exhausts the most familiar examples of Dedekind domains!
| https://mathoverflow.net/users/1149 | Do all Dedekind domains have the "Riemann-Roch property"? | Yes. Given a maximal ideal $P$ there exists $x \in K \backslash R\_P$. Let $S$ be the finite set of maximal ideals $Q$ so that $x \notin R\_{Q}$. For each $Q \in S$ such that $Q \neq P$ let $y\_Q \in Q\backslash P$. The element $f$ given by multiplying $x$ by large positive powers of all the $y\_Q$ has the desired property.
| 7 | https://mathoverflow.net/users/519 | 21079 | 13,969 |
https://mathoverflow.net/questions/21071 | 14 | This question arises when looking at a certain constant associated to (a certain Banach algebra built out of) a given compact group, and specializing to the case of finite groups, in order to try and do calculations for toy examples. It feels like the answer should be (more) obvious to those who play around with finite groups more than I do, or are at least know some more of the literature.
To be more precise: let $G$ be a finite group; let $d(G)$ be the maximum degree of an irreducible complex representation of $G$; and (with apologies to Banach-space theorists reading this) let $K\_G$ denote the order of $G$ divided by the number of conjugacy classes.
Some easy but atypical examples:
* if $G$ is abelian, then $K\_G=1=d(G)$;
* if $G=Aff(p)$ is the affine group of the finite field $F\_p$, $p$ a prime, then $$K\_{Aff(p)}=\frac{p(p-1)}{p}=p-1=d(Aff(p))$$
**Question.**
Does there exist a sequence $(G\_n)$ of finite groups such that $d(G\_n)\to\infty$ while
$$\sup\_n K\_{G\_n} <\infty ? $$
To give some additional motivation: when $d(G)$ is small compared to the order of $G$, we might regard this as saying that $G$ is not too far from being abelian. (In fact, we can be more precise, and say that $G$ has an abelian subgroup of small index, although I can't remember the precise dependency at time of writing.) Naively, then, is it the case that having $K\_G$ small compared to the order of $G$ will also imply that $G$ is not too far from being abelian?
**Other thoughts.**
Since the number of conjugacy classes in $G$ is equal to the number of mutually inequivalent complex irreps of $G$, and since $|G|=\sum\_\pi d\_\pi^2$, we see that $K\_G$ is also equal to the mean square of the degress of complex irreps of $G$. Now it is very easy, given any large positive $N$, to find a sequence $a\_1,\dots, a\_m$ of strictly positive integers such that
$$ \frac{1}{m}\sum\_{i=1}^m a\_i^2 \hbox{is small while} \max\_i a\_i > N $$
so the question is whether we can do so in the context of degrees of complex irreps -- and if not, why not? The example of $Aff(p)$ shows that we can find examples with only one large irrep, but as seen above such groups won't give us a counterexample.
| https://mathoverflow.net/users/763 | Can we bound degrees of complex irreps in terms of the average conjugacy class size? | Yes, take 2-extraspecial group $2^{2n+1}$, plus or minus should not matter. It has $2^{2n}$ irreducible representations of degree 1 and one of degree $2^n$. So your $K\_G$ is about 2 while $d(G)=2^n$.
| 12 | https://mathoverflow.net/users/5301 | 21091 | 13,976 |
https://mathoverflow.net/questions/21086 | 4 | Let $k$ be a field, and let $E$ and $F$ be fields extending $k$, both contained in some single extension of $k$. If $E$ and $F$ are finitely generated (as fields) over $k$, must $E\cap F$ also be finitely generated? If not, is there a simple counterexample?
| https://mathoverflow.net/users/5229 | When are intersections of finitely generated field extensions finitely generated? | As Brian Conrad remarked above, subextensions of finitely generated extensions are also finitely generated. Here is a prove. I wish there would be a simpler one!
* If $L/K$ is a finitely generated field extension and $L'$ an intermediate field, then $L'/K$ is also finitely generated.
Proof: Since $tr.deg\_K(L) = tr.deg\_{L'}(L) + tr.deg\_K(L')$ is finite, the same is true for $tr.deg\_K(L')$. Choose a transcendence basis $B'$ of $L'/K$. Replacing $K$ by $K(B')$, we may asume that $L'/K$ is algebraic.
Now let $B$ be a transcendence basis of $L/K$. Then $L/K(B)$ is algebraic and a finitely generated field extension, thus finite. Let $C \subseteq L'$ be linearly independent over $K$. If we knew that $B$ is also algebraically independent over $L'$, we could conclude that $C$ is linearly independant over $K[B]$ and thus over $K(B)$. This implies $|L':K| \leq |L : K(B)| < \infty$. Thus it remains to prove:
* Let $L/L'/K$ be a tower of fields such that $L'/K$ is algebraic. Let $B \subseteq L$ be algebraically independent over $K$. Then $B$ is also algebraically independent over $L'$.
Proof: Since algebraically independence is of finite character, we may assume that $B$ is finite. Since $L'(B) / K(B)$ is algebraic, we have
$tr.deg\_{L'}(L'(B)) = tr.deg\_K(K(B)) + tr.deg\_{K(B)}(L'(B)) = |B|$
Since $B$ generated $L'(B)/L'$, some subset of $B$ is a transcendence basis of $L'(B)/L'$, but this has cardinality $|B|$. Thus $B$ is itsself this basis.
| 13 | https://mathoverflow.net/users/2841 | 21093 | 13,978 |
Subsets and Splits