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https://mathoverflow.net/questions/29117 | 6 | Let G(q) be the generating function for partitions such that if k is a part, then it occurs once and k+1 is not a part.
Let H(q) be the generating function for partitions with the same condition plus that 1 is not a part.
These are the left-hand sides of the Rogers-Ramanujan Identities.
>
> $G(q)=\displaystyle\sum\_{n=0}^\infty\frac{q^{n^2}}{(q;q)\_n}=\frac{1}{(q;q^5)\_\infty(q^4;q^5)\_\infty}$
>
>
> $H(q)=\displaystyle\sum\_{n=0}^\infty\frac{q^{n^2+n}}{(q;q)\_n}=\frac{1}{(q^2;q^5)\_\infty(q^3;q^5)\_\infty}$
>
>
>
I am intrigued by the following unreferenced statement in the wikipedia [page](http://en.wikipedia.org/wiki/Rogers%E2%80%93Ramanujan_identities#Modular_functions%20%22page%22):
>
> If q = e2πiτ, then q−1/60G(q) and q11/60H(q) are modular functions of τ.
>
>
>
1. Do modular forms shed any light on the Rogers-Ramanujan Identities, or is the connection (as far as we know) a curious coincidence?
2. Is there some class of modular forms whose Fourier series count natural collections of partitions such as those counted by the left-hand sides of the Rogers-Ramanujan Identities? In particular I have in mind the seemingly "non-local" condition that if k is part, then it is distinct and also k+1 is not parts.
3. In general, how does one tell if a certain generating function (that counts partitions of a certain type, say) is related (by a multiplicative factor like above) to a modular form of some weight for some group (maybe even with some character)?
| https://mathoverflow.net/users/3988 | What is the relationship between modular forms and the Rogers-Ramanujan identities? | It's hard to compete with Berndt's former student and Berkovich's active collaborator in providing an exhaustive link of references. I can only indicate my own modest [contribution](http://arxiv.org/abs/1001.1571), joint with Ole Warnaar (who is an expert in the business), in which you can find links to further literature as well as discussion of other (not originally expected!) aspects of Rogers-Ramanujan identities.
As for the original question,
>
> What is the relationship between Modular Forms and the Rogers-Ramanujan Identities?
>
>
>
the answer is straightforward: whenever you see Rogers-Ramanujan-type identities, both sides are *modular* forms. It doesn't however work in the opposite direction: there are plenty of modular forms for which an RR-style interpretation isn't known.
| 3 | https://mathoverflow.net/users/4953 | 29185 | 19,055 |
https://mathoverflow.net/questions/29178 | 10 | Let $p:E\to B$ be a continuous map of topological spaces and set $F\_x=p^{-1}(x)$ for an $x\in B$. Take a commutative ring $A$ and assume for simplicity that each $H^\\*(F\_x,A)$ is a free $A$-module. Let $a\_1,a\_2,\ldots \in H^\\*(E,A)$ be classes that give a basis of $H^\\*(F\_x,A)$ when restricted to any $F\_x$. Assume that the direct image $R^0p\_\ast \underline{A}\_E$ of the constant sheaf on $E$ is constant. The Leray-Hirsch principle says that $H^\\*(E,A)$ is a free $H^\\*(B,A)$-module generated by the $a\_i$'s.
I would like to ask if anyone knows a reference for a similar result for étale cohomology. Ideally I would like to have a statement for $E,B$ varieties over an algebraically closed field $k$ and finite coefficients of order prime to $char (k)$.
| https://mathoverflow.net/users/2349 | Leray-Hirsch principle for étale cohomology | [[ I have added a discussion of when $p$ is smooth or has quotient singularities. ]]
[[ I added a discussion on the cohomology of $[X/G]$. ]]
The étale case follows in a way that is altogether analogous to the topological
case. Let me give a proof that gives a teeny bit of extra information. I assume
that $\alpha\_i$ is homogenous (with respect to cohomological degree) of degree
$d\_i$. Then $\alpha\_i$ gives a map in the derived category $A[-d\_i]\to
Rp\_\ast A$ and combining them a map $\bigoplus\_iA[-d\_i]\to Rp\_\ast A$. If we can
show that this map is an isomorphism then we get an isomorphism
$\bigoplus\_iR\Gamma(B,A)[-d\_i]\to R\Gamma(E,A)$ which on taking
cohomology gives the L-H theorem. That this is an isomorphism can be checked
fibrewise and if the natural map $(R^ip\_\ast A)\_x\to H^i(F\_x,A)$ is an isomorphism for
all geometric points $x$ we are through.
This condition is true under one of the following conditions:
* $p$ is proper (by the proper base change theorem).
* $p$ is locally trivial by the Künneth formula.
* The second case covers the case of a $G$-torsor. If $G$ acts on $X$ with
finite stabiliser scheme (a condition slightly stronger than having finite
stabilisers but which guarantees that $X/G$ exists) and the orders of the
stabilisers are invertible in the ring of coefficients it is still true. This
can be seen by looking at the stack quotient $X\to[X/G]$ which is a
$G$-torsor (though with base a stack) and at $[X/G]\to X/G$ which induces an
isomorphism in cohomology fulfilling the condition. It should also be possible
to do directly imitating Deligne's proof (in SGA 4 1/2 I think) that the
cohomology of $G$ is indendent of the characteristic (it use the sequence of
fibrations $G\to G/U\to G/B$ where $B$ is a Borel subgroup and $U$ its unipotent radical).
**Addendum**: Here are, as requested below by algori, some details on the fact that $\pi\colon[X/G]\to X/G$ induces an isomorphism for coefficients $A$ for which the order of the (group of connected components of the) stabilisers are invertible. (This of course is well-known, so well-known in fact that I don't know if there is a proper reference for it.) I will not use that the stack is a global quotient so we may as well consider $\pi\colon\mathcal X\to X$ where $\mathcal X$ is a stack with finite stabiliser scheme and $X$ is its spatial quotient. For simplicity I will assume that the automorphism groups are reduced (i.e., that $\mathcal X$ is a Deligne-Mumford stack). The general case can be proved along the same lines but would be longer and more technical. What we are going to show is that $R\pi\_\*A=A$. As the construction of the spatial quotient commutes with étale localisation on $X$ we may assume that $X$ is local strictly Henselian and then by the local structure theory of DM-stacks (to be found for instance in Laumon-Moret-Bailly) $\mathcal X$ has the form $[Y/G]$, where $G$ is a finite group which can be assumed to be the stabiliser of a point of $Y$ and hence has order invertible in $A$ and $Y$ is also local strictly Henselian. Now using the usual simplicial resolution $T\_n=G^n\times Y$ of $[Y/G]$ we get that $H^\*([Y/G],A)=H^\*(G,A)=A$ as $H^\*(T\_n,A)=A^G$.
* $p$ is smooth. This is proved as follows: For any $A$-complex $K$ on $B$ we
get a map $\bigoplus\_iK[-d\_i]\to Rp\_\ast p^\ast K$ which we want to show is an
isomorphism. This map is functorial so in particular if is an isomorphism for
two complexes in a distinguished triangle it is so for the third. By Noetherian
induction (assuming for simplicity $B$ is Noetherian) we may assume that $B$ is
local Henselian and that the statement is true for $B$ replaced by the
complement $U$ of the closed point. We start by showing that that implies that
if $K$ is an $A$-complex on $U$ and if $j\colon U\to B$ is the inclusion, then the
result is true for $Rj\_\ast K$. Indeed, this follows directly from smooth base
change which implies that $p^\ast Rj\_\ast K=Rj'\_\ast p^\ast K$, where $j'\colon p^{-1}U\to E$ is the
inclusion, and then the result follows from the induction hypothesis. On the
other hand, if $K$ is supported on the closed point $x$ of $B$, then the map is
$\bigoplus\_iK\_x[-d\_i]\to R\Gamma(E\_x,K\_x)$ which is an isomorphism as it is for
$K=A$. Now, the mapping cone of $K\to Rj\_\ast j^\ast K$ has support at $x$ so the statement
follows.
* The only thing that is used about a smooth map is that $p$ is universally
locally acyclic for the torsion primes of $A$ (SGA IV: Exp. XVI,
Thm. 1.1). Unless I am mistaken this is true for $p$ that locally are of the
form $E\times\_GU\to B\times\_GU$, where $E\to B$ is a smooth $G$-map, $U$ a $G$-scheme and $|G|$
is invertible in $A$.
Another way of dealing with the $(G,X)$ case which I think should be more
efficient and general is, following Deligne, to split $X \to X/G$ up into $X\to
X\times\_GG/U\to X\times\_GG/B\to X/G$ (where $U$ is the unipotent radical of a Borel subgroup
$B$). Then $X\times\_GG/B\to X/G$ is proper and in fact the Leray-Hirsch argument
applies provided a large enough integer is invertible in the coefficients (it is
in general not enough to invert the $|H|$ but one also needs to invert some
primes intrinsincally defined by $G$) and $X\to X\times\_GG/U$ has more or less affine
spaces as fibres and induces an isomorphism if the $|H|$ are invertible. Finally
$X\times\_GG/U\to X\times\_GG/B$ is essentially a torus bundle and the cohomology of $X\times\_GG/U$
can be analysed in terms of the cohomology of $X\times\_GG/B$ and the characteristic
classes of the bundle.
| 11 | https://mathoverflow.net/users/4008 | 29191 | 19,060 |
https://mathoverflow.net/questions/29095 | 7 | I am trying to understand something about curved dg algebras as studied by Positselski, E. Segal. These come up in mirror symmetry and when one wants to study Kozsul duality for algebras that are more general then those that fit into the classical framework.
Let C denote the complex numbers. Suppose we take a cdg-algebra B that is C[x] with a curving x^2, where x is now a variable in odd degree and so does not fit into the matrix factorization framework. As I understand the situation, then the Koszul dual of this should be an algebra A which is C[y]/(y^2-1), where y is in even degree.In the framework of Positselski, as I understand it one can compute A as Ext(C,C) and pass back to B by taking the Cobar construction. The curving arises from the fact that the coalgebra dual to A is no longer co-augmented. I think that in this simple example one equivalently can also take a more down to earth approach, as was taken by Dyckerhoff in his paper on matrix factorizations, using explicit Koszul resolutions.
My question is about a confusion I have about the Hochschild cohomology of this dg-category. As I understand it, the Hochschild cohomology of D(B) should be isomorphic to the center of A which is A. Yet if one uses the complex defined in Segal's paper, I believe one gets C[y]/y^2, with y an even variable. The complex he claims should compute HH\*(D(B)) is given by the usual Hochschild complex with differential on the algebra B + Gerstenhaber bracket with x^2. To compute, I first computed ordinary HH\*(C[x]) and the Gerstenhaber bracket on HH\* and then concluded the answer using a spectral sequence.
Segal's justification seems to be that this is what you get when you regard B as a curved A-infinity algebra, but as I understand Positselski, there seems to be some subtelties with this and that one can often end up with no objects in the category. So I was made a little bit nervous by the justification. Most likely everything is ok and I am missing something stupid, but it would be great if someone would be so kind as to point out where my mistake is.
| https://mathoverflow.net/users/6986 | A question on curved algebras, papers by Positselski and E. Segal | The answer is that you shouldn't believe everything you read on the arxiv...
The result I claim in that paper is wrong, at least in that level of generality. The problem is that I try to use the completion of the bar resolution to compute Hochschild homology, but this isn't a free resolution.
Your example is a good one. What I assume is going on here is the following: there are two simple objects in this category, given by the rank one free module with differential either $x$ or $-x$. Presumably they generate the category. Then the category is equivalent to the derived category of your algebra $A$, and its HH is $A$.
If we only took one simple object it would generate a subcategory equivalent to $D^b(\mathbb{C})$, whose HH is just $\mathbb{C}$. I believe my complex is computing this answer, but I don't have much justification for this!
If you're seriously thinking about this I would love to talk to you, and hopefully resolve this problem.
| 7 | https://mathoverflow.net/users/2454 | 29200 | 19,066 |
https://mathoverflow.net/questions/29167 | 3 | Given a subalgebra E of $M\_n$ (nxn complex valued matrices), what can we say about the subspaces F of $M\_n$ such that $EF \subset F$? Googling for an answer gives me the reference:
Israel Gohberg, Peter Lancaster, and Leiba Rodman (2006). Invariant Subspaces of Matrices with Applications.
However, my library doesn't have this book. Is there a nice survey article available anywhere on this?
Thanks
| https://mathoverflow.net/users/5977 | Invariant subspaces of subalgebras of $M_n(C)$ | This was originally tagged fa.functional analysis, I think. So he's an Operator Algebraic answer. I'm going to make the strong assumption that E is self-adjoint (i.e. closed under taking the hermitian transpose). If not, then really this is an algebraic question, and it's probably irrelevant that you are working with the complex numbers...
Anyway, then E is a finite-dimensional von Neumann algebra. The action of E on M\_n is the same as identifying M\_n with $\mathbb C^n \otimes \mathbb C^n = \ell^2\_n \otimes \ell^2\_n$ and letting E act as $E \otimes 1$. Then invariant subspaces for $E$ correspond to orthongonal projections in the commutant of E, which by Tomita is $E' \otimes M\_n$ where $E' = \{ A\in M\_n : AB=BA (B\in E)\}$ the commutant of $E$ in $M\_n$. We identify $E'\otimes M\_n$ with $M\_n(E')$, and then it's just (ahem!) a case of working out the projections (self-adjoint idempotents) here. In concrete cases, this is probably not too hard...
| 2 | https://mathoverflow.net/users/406 | 29207 | 19,070 |
https://mathoverflow.net/questions/29208 | 3 | In this question F is a field and all algebras are finite dimensional F algebras.
Let X be the set of all F algebras A for which there exist an F algebra B and an F division algebra D such that F is the center of D and the tensor product of A and B over F is isomorphic to M\_n(D) for some n. Can we find all the elements of X?
(M\_n(D) is the algebra of all n-by-n matrices with entries from D.)
It is Obvious that every central simple F algebra is in X. Are there some interesting elements of X?
cheers
| https://mathoverflow.net/users/6941 | Special subalgebras of central simple algebras | You want to know which algebras $A$ are such that $A\otimes B$ is central
simple for some algebra $B$.
(All algebras and tensor products being over $F$.) If $Z(A)$
and $Z(B)$ are the centres of $A$ and $B$ then $Z(A)\otimes Z(B)$
is contained in the centre of $A\otimes B$. Hence $A$ and $B$ must both have
centre $F$. If $I$ is a two-sided ideal of $A$ then $I\otimes B$ is a two-sided
ideal of $A\otimes B$. As $A\otimes B$ is simple, then $I$ is zero or $A$,
that is $A$ is simple.
We conclude: the only algebras in $X$ are the central simple algebras over $F$.
| 6 | https://mathoverflow.net/users/4213 | 29209 | 19,071 |
https://mathoverflow.net/questions/28907 | 22 | We have a circle and two families of $n$ red arcs and $n$ blue arcs, positioned on the circle so that every two arcs of different colors intersect. Can one show that there is a point in the perimeter which is part of at least $n$ arcs?
(The statement sounds very simple. It makes me think the answer should be very simple too, but I've been struggling with this one for a bit and got nowhere.)
After reading Suresh's answer below, I can't help but think that there must be some colorful Helly theorem on manifolds, of which the question above is a special case. At the moment I don't even have a meaningful formulation of what this theorem could be, has it been treated before?
| https://mathoverflow.net/users/2384 | Covering a circle with red and blue arcs | This is the second half of a proof started by Peter Shor.
I assume that the set of arcs is already in a position as in Peter's answer: the red arcs $(L\_1,R\_i)$ are cyclically ordered and all blue arcs are of the form $(R\_i,L\_{i-1})$. For convenience, I also assume that no two of red arcs coincide (and hence their $2n$ endpoints are distinct). This is easy to achieve by perturbation. Let $S$ denote the set of the red endpoints and $r:S\to\mathbb N$ denote the covering multiplicity by red arcs: $r(p)$ is the number of red arcs containing $p$.
Fix the red arcs. Then a configuration is determined by a vector of $n$ multiplicities $T=(t\_1,\dots,t\_n)$ where $t\_i$ is how many copies of a blue arc $(R\_i,L\_{i-1})$ we have. For such $T$ and each $p\in S$, let $b\_T(p)$ be the number of blue arcs (in the configuration defined by $T$) containing $p$. Note that $b\_T$ is linear in $T$. By Peter's observation, we have $b\_{T\_0}(p)=n-r(p)$ for all $p\in S$ where $T\_0$ is the standard vector: $T\_0=(1,\dots,1)$.
We have $\sum t\_i=n$ for every admissible vector $T$. I claim that none of the resulting functions $b\_T:S\to\mathbb R\_+$ strictly majorizes any other. The result follows from this claim applied to $T=T\_0$. To prove the claim, it suffices to find a collection $w:S\to\mathbb R\_+$ of nonnegative weights such that, for every (potentially) blue arc, the sum of weights of points covered by this arc equals 1. Indeed, if such weights are found, we have $\sum\_{p\in S} w(p) b\_T(p)=n$ for any $T$, and the claim follows.
To construct $w$, consider the standard covering map $f:\mathbb R\to S^1$. On the line, we have a discrete set $\tilde S=f^{-1}(S)$ and a collection of segments corresponding to pre-images of the potentially blue arcs (both structures are $2\pi$-periodic). The endpoints of the segments are in $\tilde S$, and the segments are ordered from left to right (none of them is contained in another). Let us first solve the weight problem on the line. Begin with one of the segments and mark is right endpoint. Then take the first segment contained in the open half-line after the marked point, and mark its right end. Repeat for the new marked point, and so on. Then every segment to the right of the first one contains exactly one marked point. Let the marked points have weight 1 and all other points have weight 0. We have solved the weigh problem on a half-line.
Observe that the set of marked points is eventually $2\pi k$-periodic for some integer $k$. This follows from the deterministic nature of the construction: once some $x$ and some $x+2\pi k$ are both marked, the pattern will repeat itself. Hence there is a $2\pi k$-periodic weight function on the whole line. We can make it $2\pi$-periodic by averaging over $2\pi m$-translations, $0\le m<k$. Once it is $2\pi$-periodic, it projects back to the circle.
| 12 | https://mathoverflow.net/users/4354 | 29211 | 19,073 |
https://mathoverflow.net/questions/29174 | 4 | I've been studying isometric quotients (by compact Lie groups) of compact simply connected homogeneous spaces $G/H$ and their inherited curvature. One of the issues that continually arises is the following problem:
Let $\mathfrak{g} = \mathfrak{h}\oplus\mathfrak{p}$ be an orthogonal decomposition with respect to a Ad G invariant inner product on $\mathfrak{g}$. When is there a $g\in G$ such that $Ad(g)\mathfrak{p}\cap \mathfrak{p} = \{0\}$?
Of course, if the dimension of $\mathfrak{p}$ is too large (equivalently, if the dimension of $\mathfrak{h}$ is too small), then they must intersect nontrivially for dimension reasons. This leads us to the general form of the question:
>
> Let $G$ be a compact Lie group and let $V\subseteq \mathfrak{g} = Lie(G)$ be a vector subspace (but not neccesarily a subalgebra). Assume dim($V)\leq \frac{1}{2}$ dim($\mathfrak{g})$. Is there a $g\in G$ such that $Ad(g)V\cap V = \{0\}$?
>
>
>
Of course, if there is one $g$, an entire neighborhood around $g$ will have this property. But
>
> when is there an open and dense set of $g$ such that $Ad(g)V\cap V=\{0\}$?
>
>
>
(If it helps, feel free to assume that $G$ is simple. I'm pretty sure (but haven't been able to prove) that one can reduce the question to simple $G$ anyway.)
Note that for $G=SU(2)$ or $G=SO(3)$, there is an open and dense set of such $g$. This follows because in this case, $Ad:G\rightarrow SO(\mathfrak{g})$ is surjective and, of course the usual $SO(n)$ action on $\mathbb{R}^n$ acts transitively on k-subspaces. Further, the set of all elements of $SO(3)$ that fix a given line in $\mathbb{R}^3$ is isomorphic to an $SO(2)$, which is codimension 2 in $SO(3)$.
However, for any other $G$, this trick doesn't work as $Ad$ will fail to be surjective.
Thank you in advance for your time and feel free to add more tags as appropriate!
| https://mathoverflow.net/users/1708 | Adjoint orbits of small subspaces in Lie algebras | I think that answers to your questions can be obtained along the following lines.
Fix the dimension of $V$ to be $k\leq 1/2\dim V$ and let $n=\dim\mathfrak{g}.$ Condiser the real algebraic Grassmanian variety $X=Gr(k,\mathfrak{g}), \dim X=k(n-k).$ The set $(V\_1,V\_2): \dim(V\_1\cap V\_2)\geq 1$ is a Zariski closed subset of $X\times X$ of codimension $n-2k+1$ and for a fixed $V,$ the set $W: \dim(V\cap W)\geq 1$ is a Zariski closed subset $Z$ of $X$ of codimension $n-2k+1$ containing $V.$ Then $G$ is a linear algebraic group over $\mathbb{R}$ that acts on $X$ and the negation of the first question may be restated as follows:
>
> Is the $G$-orbit $O=G\cdot V\subset X$ contained entirely in $Z$?
>
Note that $O$ is Zariski closed and connected in the real topology (assuming that $G$ is connected). The general answer depends on the stabilizer $G\_V$ of $V$ in $G,$ which coincides with the stabilizer of the Lie subalgebra $\langle V\rangle\subset\mathfrak{g}$ generated by $V.$ For example, if $V$ is in the center of $\mathfrak{g}$ then $G\_V=G$ and $gV=V$ for each $g\in G$ (and more generally, if $V$ nontrivially intersects the center of $\mathfrak{g}$ then $gV\cap V$ won't reduce to $0$ for any $g$). Some easy cases, such as $\dim V=1,$ may be decided by the dimension count.
As for your second question, it is really equivalent to the first one.
The set of $g\in G$ such that $V\cap gV=\{0\}$ is the pre-image under the orbit map $G\to G/G\_V\simeq O$ of the Zariski open subset $O'$ of $O$ which is the complement of $O\cap Z.$ If this set is non-empty (yes to the 1st question) then it is open and dense in the real topology (so yes to the 2nd question).
| 2 | https://mathoverflow.net/users/5740 | 29214 | 19,075 |
https://mathoverflow.net/questions/29219 | 12 | I have been looking for books on cellular automata, and I really can't afford more than one book right now, so I really need to make the right choice. What would be the right book for someone with a Computer Science Masters degree and also feels comfortable with Mathematical Logic and basic Abstract Algebra? (Or, to make the question more useful to others, what are the target readers of the most recent books on cellular automata?)
| https://mathoverflow.net/users/6892 | Book recommendations on cellular automata? | First, there is an unannotated list of books on cellular automata [here](http://uncomp.uwe.ac.uk/genaro/Cellular_Automata_Repository/Books.html).
Second, if you are going to get just one book, then I think it has to be Wolfram's *A New Kind of Science*, which, despite its flaws, is the source of so much of the research in cellular automata
that it must be confronted first. [I see I am concuring with Kevin O'Bryant's just-posted recommendation.] You might read [the review](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.113.7950) by Lawrence Gray in the *Notices of the American Mathematical Society* (February 2003) to make
yourself aware of the controversies surrounding this book. If you can overlook its flaws, it is a remarkable book, and quite fun to work through.
| 14 | https://mathoverflow.net/users/6094 | 29224 | 19,081 |
https://mathoverflow.net/questions/29226 | 3 | Let M a riemannian manifold. How can I show that the hodge-laplace-operator of a function $f$ is the negative of the laplace-operator?
| https://mathoverflow.net/users/7015 | Connection between the Hodge laplacian and the Laplace operator | A rather short proof can be found [here](http://en.wikipedia.org/wiki/Proofs_involving_the_Laplace%E2%80%93Beltrami_operator).
I assume you are interested in the case when $f$ is a *scalar* function. Otherwise the
Hodge Laplacian differs from the Laplace–Beltrami operator not only by a sign due to the Ricci curvature. See the [Weitzenböck identity](http://en.wikipedia.org/wiki/Weitzenb%C3%B6ck_identity).
| 6 | https://mathoverflow.net/users/5371 | 29229 | 19,084 |
https://mathoverflow.net/questions/29197 | 15 | I have read about the existence of functions of the kind described in the title in several places, but never seen an instance of them. Sorry if this is too much an elementary question to be posted here.
| https://mathoverflow.net/users/6466 | Non-computable but easily described arithmetical functions | A function $f:\mathbb{N}\to\mathbb{N}$ is computable if and only if the graph of $f$ is $\Sigma\_1$ definable in the [arithmetic hierarchy](http://en.wikipedia.org/wiki/Rice%27s_theorem), which means that $f(x)=y\iff \exists n\ \varphi(x,y,n)$, where $\varphi$ involves only bounded quantifiers. Thus, the essence of computation is that it is the search for an arithmetic witness $n$ of some primitive property.
Many functions, however, are easy to describe but cannot be expressed in this simple form. Here are some examples:
* The characteristic function of the set of theorems of your favorite axiomatization of mathematics, such as PA or ZFC; this is the function that correctly labels assertions as *theorem* or *non-theorem*. We may view assertions directly as syntactic strings of symbols or we may code them as numbers if you wish (and this is surely a cosmetic difference). While we can recognize theorems by their proofs, we provably have in principle no computable way to recognize a non-theorem.
* The truth function, which correctly labels the statements of arithmetic as *true* or *false*, is not computable. This function is not in the arithmetic hierarchy, but it exists at the entry level $\omega$ to the hyperarithmetic hierarchy.
* The halting problem function, which correctly labels program-input pairs as *halting* or *non-halting*, is easy to describe, but not computable.
* The Tiling function, which given any finite set of polygonal tiles, outputs the size $k$ of the largest $k\times k$ sqaure that can be tiled by them, or $0$ if they tile the entire plane.
* The Conjugate function, which given two words in a finite group presentation, correctly states whether they are conjugate or not.
* The Solve function, which given a polynomial over the integers in several variables, outputs the list of smallest-norm integer solutions (giving the empty list if there are none). This is not computable by the MRDP solution to Hilbert's 10th problem. The positive instances are computable, of course, as they are witnessed by their corresponding calculation, but the empty list provably cannot be witnessed in a finitary way.
* The Tot function, which correctly labels Turing machine programs as *total* or *strictly partial*.
* The Empty function, which labels Turing machine programs as *empty* or *non-empty*, depending on whether they accept an input.
* An uncountable supply of examples is provided by [Rice's Theorem](http://en.wikipedia.org/wiki/Rice%27s_theorem), which asserts that no non-trivial property of the c.e. sets is computable from their programs. Thus, if $W\_e$ is the set enumerated by program $e$, then for any family of sets ${\cal A}$ which contains some but not all $W\_e$, the characteristic function of the set of programs $\{ e | W\_e\in{\cal A}\}$ is not computable. For example, the functions that decide whether program $e$ enumerates a connected graph, or whether this set contains any primes, or whether it is eventually periodic, or whether it exhibits any other property that holds of some but not all programs, are all non-computable.
| 24 | https://mathoverflow.net/users/1946 | 29230 | 19,085 |
https://mathoverflow.net/questions/29215 | 1 | There is a linear function of two variables that I am trying to minimize under an equality constraint. But, the constraint is non linear in the variables. Is there any technique to solve this? or can I use approximations and linearize the constraint?
| https://mathoverflow.net/users/7012 | Minimization under non-linear constraints | In addition to the tip about using Lagrange multipliers, take a look at <http://en.wikipedia.org/wiki/Nonlinear_programming> which has a small paragraph about methods for solving nonlinear optimization problems.
If you can know (or can show) that the problem is convex and you want to learn techniques for convex nonlinear optimization, take a look at the following textbook <http://www.stanford.edu/~boyd/cvxbook/> (pdf available)
| 2 | https://mathoverflow.net/users/1530 | 29242 | 19,092 |
https://mathoverflow.net/questions/29241 | 2 | I would appreciate a reference describing the analysis of PDEs on the Klein bottle and the real projective plane. As an example, is there a reference discussing the existence and uniqueness of the solution of the Poisson equation $\nabla^2 u = f$ on either of these? I would prefer to avoid embedding in a higher dimensional space, if possible.
More specifically, I am interested in a `flat rectangular region having the topology of' a Klein bottle or of the real projective plane. I lifted this language from the answer to a question about wave equations on a Mobius strip (see below) because it is clearer than my original question.
| https://mathoverflow.net/users/7023 | PDEs on the Klein bottle and real projective plane | I do not know of a reference, but maybe the problem can be reduced to studying the problem on the sphere $S^2$ and on the torus $T$ and then looking for solutions with certain symmetries. For instance, if $f$ is a function on $RP^2$ then it comes from a function $g$ on $S^2$ such that $g(p) = g(-p)$. Solve $\nabla^{2} u = f$ on $S^2$, then I think that $u' = (u(p) + u(-p))/2$ is a solution of your problem on $RP^2$.
| 4 | https://mathoverflow.net/users/6658 | 29243 | 19,093 |
https://mathoverflow.net/questions/29218 | 2 | It is well known that we can use the Riemann-Hilbert correspondence to describe holonomic D-modules in terms of a category of perverse sheaves on some variety $X$. And $U|\rightarrow Perv(U)$ is a stack. Therefore, for example. If one has flag variety of $sl\_2$,i.e. $P^1$, given its big cells as open affine covers: i.e. two $A^1$, then one can use this formalism to glue holomomic $A\_1$-modules to holonomic $D-mod\_{P^1}$,where $A\_1$ is first Weyl algebra.
We know Description of simple holonomic D-modules on quasi compact schemes is local.
My question
===========
Is there any machinery that can glue **non-holonomic simple** D-modules from open affine covers to **simple non-holonomic D-modules** on total space? Is there a functorial way which can give a fairly complete list of simple **non-holonomic** $D-mod\_{G/B}$. For example, let's consider the flag variety of $sl\_3$; the big cells are given by 2-dimensional affine space $A^2$. Then we know there are a lot of non-holonomic $A\_2$-modules(simple) from Bernstein-Lunts. In this case, how can we globalize them to non-holonomic D-modules on flag variety of $sl\_3$?
Thanks in advance
| https://mathoverflow.net/users/1851 | Is simple non-holonomic D-module a local concept? | For the last question [edit: this part of the question has now been removed..], the relation of constructible sheaves with the Fukaya category is the subject of the paper [Microlocal branes are constructible sheaves](http://arxiv.org/abs/math/0612399) by David Nadler, and its predecessor [Constructible Sheaves and the Fukaya Category](http://arxiv.org/abs/math/0604379) by Nadler and Eric Zaslow. As for D-modules in general, there are proposals in the physics dictionary, starting in work of Anton Kapustin (see e.g. [here](http://arxiv.org/abs/hep-th/0502212)) and perhaps best summarized in the [seminal paper of Kapustin-Witten](http://arxiv.org/abs/hep-th/0604151). One has to be careful though what exactly you mean: to get nonholonomic D-modules, the idea is to look at coisotropic but non-Lagrangian A-branes, ie not the Fukaya category but an enlarged version that has yet to be defined mathematically. Also even in the Lagrangian case there are two really different things one can mean by Fukaya category of the cotangent bundle, depending on how you treat behavior at infinity. Nadler uses a refined version that corresponds precisely to constructible sheaves. The one that arises in the physics, and in most of the math literature (eg work of Abouzaid) is the "wrapped Fukaya category" -- this corresponds more closely to modules over infinite order differential operators - in particular delta functions at distinct points on the base become identified (by exponentiating translation)! so this is quite far from what you might want in say representation theory.
As to your other question, I'm not sure I understand the issue -- D-modules form a perfectly nice stack, i.e. they glue together, independently of holonomicity (one doesn't need Riemann-Hilbert to see this, it's immediate from the definition as modules over a sheaf of algebras - in fact from my, perhaps naive, point of view it's easier to see perverse sheaves form a stack by thinking of them as D-modules, but that's certainly not necessary either). So the gluing formalism you ask for is just sheaf theory if I understand correctly: D-modules on a cover plus gluing data define a D-module on the total space. Same holds on the (dg) derived category level. This is in fact how you even define what a D-module on a stack is (in terms of smooth covers), as is discussed at great length in the last "chapter" of the text by Beilinson-Drinfeld on Quantization of Hitchin's Hamiltonians.
Now for the question of describing simple modules, that's a seriously tricky issue that I know nothing about -- I believe Toby Stafford was the first to show that there even are simple but nonholonomic modules over rings of differential operators, and I would look at his papers for insight.
| 6 | https://mathoverflow.net/users/582 | 29246 | 19,094 |
https://mathoverflow.net/questions/29249 | 0 | I just need a quick clarification:
Given a sequence of sets $\{a\_n\}\_{n \in \mathbb{N}}$ in some field $\mathbb{K}$, is saying that it satisfies the finite intersection property equivalent to saying $(\forall n\in \mathbb{N})(\exists x\in \mathbb{K})(x \in \cap\_{i=1}^n a\_n)$
If the previous statement is true, then it seems almost reasonable to say that, because $\{\cap\_{i=1}^n a\_n\}\_{n\in\mathbb{N}}$ is a nested sequence of nonempty sets (because of the f.i.p), $\cap\_{i=1}^{\infty} a\_n \neq \emptyset$, but I know that is not necessarily true
Thanks
| https://mathoverflow.net/users/7025 | Quick Finite Intersection Property Question | What you quote is the finite intersection property.
The point of confusion is in what follows. There is no reason why a nested sequence of nonempty sets can't have empty interesection. Just because we can find an $x\_n\in\cap\_{i=1}^n a\_n$ for every $n$, there is no reason to expect that this can be done uniformly in $n$.
The intervals $[n,\infty)$ also form a nested sequence of nonempty sets with empty intersection.
| 4 | https://mathoverflow.net/users/2559 | 29251 | 19,096 |
https://mathoverflow.net/questions/29240 | 4 | I would like to know if there exists a formulation/incarnation of the Fourier-Mukai duality in terms of the corresponding Frobenius manifold constructed by Barannikov and Kontsevich: <http://arxiv.org/pdf/alg-geom/9710032>
Damien
| https://mathoverflow.net/users/7031 | Frobenius manifold formulation of Fourier-Mukai duality | Yes, a derived equivalence will give an equivalence of the corresponding Frobenius manifolds. First a derived equivalence induces an isomorphism of deformation spaces of the two categories: the deformation theory of the derived category is controlled by the Hochschild cochains with its differential graded Lie algebra structure, and this can be recovered intrinsically from the derived category (or better its enhanced versions) and so is preserved under a Fourier-Mukai transform. Next if the category is Calabi-Yau then it intrinsically defines (following Costello, Kontsevich-Soibelman and later Lurie) an extended 2d topological field theory (aka TCFT), and so again a Fourier-Mukai transform induces an isomorphism of TFTs. Finally to get a Frobenius manifold structure you want a topological string theory, in other words you want to extend the field theory to the boundary of Deligne-Mumford space. But again (following Konstevich et al) this amounts to the derived category satisfying the degeneration conjecture (roughly the circle action on Hochschild homology should be "trivial"), but this is known in the geometric setting (the Hodge theorem) and in any case again this is an intrinsic property of the derived category, so preserved by derived equivalence. The structure of (germ of) Frobenius manifold is describing the genus zero part of the Deligne-Mumford action on the Hochschild cohomology (see papers of Getzler identifying the formal Frobenius operad with those of "gravity", aka compactified genus zero moduli spaces).
Anyway in short, the derived category (or better an enhanced, dg or $A\_\infty$) version controls all the other aspects of the B-model TFT, so all the other constructions in mirror symmetry on the B-side can be recovered intrinsically from it (and are therefore invariant under derived equivalence).
| 6 | https://mathoverflow.net/users/582 | 29253 | 19,097 |
https://mathoverflow.net/questions/29175 | 4 | Given a mobius strip, what do the solutions of the wave equation look like qualitatively? How do they differ from solutions on the equivalent strip glued together as a cylinder? Any refs, particularly to symmetry?
| https://mathoverflow.net/users/7002 | Solutions to the wave equation on non orientable surfaces like a mobius strip | Any solution to the wave equation on a Möbius strip lifts to a solution on its orientation double cover, which is a cylinder of equal width but twice the circumference. In order for a solution on the cylinder to descend to the Möbius strip, it is necessary and sufficient that it be invariant under a certain order two symmetry. If the cylinder has coordinates given by $[0,\pi a]$ in the free direction and $[0, 2\pi L]$ in the loop direction, then the solutions are linear combinations of products $AB$, where $A$ has the form $\cos (kn\_a t)\cos(\frac{n\_a}a x)$ or $\sin (kn\_a t)\cos(\frac{n\_a}a x)$, and $B$ has the form $\cos(kn\_Lt)\cos(\frac{n\_L}{L}y)$, $\sin(kn\_Lt)\cos(\frac{n\_L}{L}y)$, $\cos(kn\_Lt)\sin(\frac{n\_L}{L}y)$, or $\sin(kn\_Lt)\sin(\frac{n\_L}{L}y)$. Here, $n\_a$ and $n\_L$ are nonnegative integers, and $k$ is a constant. Invariance under the symmetry is equivalent to $n\_a + n\_L$ being an even number. In contrast, solutions on the cylinder of the same dimensions correspond to solutions on the double cover such that only $n\_L$ is even.
| 6 | https://mathoverflow.net/users/121 | 29257 | 19,100 |
https://mathoverflow.net/questions/28917 | 10 | I have a (noncommutative) division algebra D which is finite dimensional over its center F. I know that every subfield of D which contains F properly is a maximal subfield of D. What can we say about D?
Is there any characterization of such division algebras?
Does anybody know any book or paper that discusses this?
By the way, the set of such division algebras is obviously not empty because, for example, the (real) quaternion algebra (or any division algebra of prime degree) is in that set.
(The degree of D is the square root of the dimension of D over F.)
I hope my question is not too trivial! Thanks.
| https://mathoverflow.net/users/6941 | Division algebras in which every proper subfield is maximal | The short answer is that not too much is known about this situation, beyond the easy observations that I will now list. I will call $D$ *irreducible* if it has the property you are interested in, i.e., every (commutative) subfield that properly contains the center is a maximal subfield.
1. If $D$ has prime degree, then $D$ is irreducible. This is obvious because every subfield is contained in a maximal subfield, and the maximal subfields all have the same dimension over $F$.
2. If the degree of $D$ has at least two prime factors, then $D$ is reducible. In this case you can factor $D$ as a tensor product of two division algebras of relatively prime degrees. Then you just take a maximal subfield in one of the two factors. This reduces us to considering algebras $D$ whose degree is a prime power.
3. If $D$ has composite degree and $D$ is a crossed product, then $D$ is reducible. Recall that $D$ is a crossed product if it has a maximal subfield $L$ that is Galois over $F$. So suppose that the Galois group is $G$, necessarily of composite order. Then there is a nonzero proper subgroup of $G$, hence $D$ is reducible. (This deduction sounds foolish, because the theorem that something is a crossed product is much stronger than what you are asking about. But asking if something is a crossed product is a standard question, so in this way you can connect your question to standard results.)
4. As a consequence of #3, every $D$ of degree 4 is reducible, and every $D$ of degree 8 and exponent 2 is reducible. That is because such algebras are cross products under $Z/2 \times Z/2$ (Albert) and $Z/2 \times Z/2 \times Z/2$ (Rowen) respectively.
5. If $D$ has degree $p^r > p$ for some $p$ prime and it happens that every finite extension of $F$ has dimension a power of $p$, then $D$ is reducible. Indeed, by Galois theory every maximal subfield contains proper subfields.
So the first open case is where $D$ has degree 8 and exponent at least 4 in the Brauer group, and the base field has extensions of degree not a power of 2.
Translation in terms of algebraic groups
----------------------------------------
Your question is closely related to the question of whether the group $SL\_1(D)$ has nonzero, proper connected subgroups. Well, $SL\_1(D)$ always contains maximal tori. So the question is: Are there others? (If your field has nonzero characteristic, probably one should only consider reductive subgroups.) Subfields of $D$ correspond to tori in $SL\_1(D)$, so your question is the same as asking: For what $D$ are maximal tori the only nonzero, proper reductive subgroups of $SL\_1(D)$?
These sorts of questions are addressed in my joint paper with Philippe Gille *Algebraic groups with few subgroups*, J. London Math. Soc., vol 80 (2009), 405-430. <http://dx.doi.org/10.1112/jlms/jdp030> See especially section 4.
Also, the paper *Irreducible tori in semisimple groups* by Gopal Prasad and Andrei Rapinchuk (IMRN 2001, #23, 1229-1242) <http://ams.rice.edu/leavingmsn?url=http://dx.doi.org/10.1155/S1073792801000587> discusses a similar question for tori. Your maximal subfield has no proper intermediate fields if and only if the corresponding torus is irreducible in their sense. This is why I called your $D$ irreducible above.
| 11 | https://mathoverflow.net/users/6486 | 29276 | 19,111 |
https://mathoverflow.net/questions/29279 | 1 | Suppose the unit sphere in ℝ3 has coordinates (*ρ*, *η*) with *ρ* as the "co-latitude" angle (measured from positive *z*-axis) and *η* as the "longitude" angle measured from positive *x*-axis in the *xy* plane. I am given to understand that the metric tensor is
$g = \left[\matrix{1 & 0 \\ 0 & \sin^2\rho}\right]$
and I am further told that this induces a distance metric on the unit sphere.
How can I obtain a distance metric *d*(*x*, *y*) from this? I have seen several definitions similar to [this one](https://planetmath.org/riemannianmanifold) (see the note), but I am unsure how to actually reduce this to a usable form.
**Disclaimer:** Posted by an engineer in over his head.
| https://mathoverflow.net/users/5029 | Distance metric on the unit sphere in R^3? | Looking at the other answers and comments posted so far, I feel compelled to add a different answer. Only the advice by Anton Petrunin makes sense to me. You really should find a helpful mathematician and discuss your question in person. For one thing, if it were me, I would start by asking you what you need this for. Your answer would guide me in how to proceed.
Anyway, here are some thoughts (I'm not providing details until you say more about what you want):
First, if all you need is the distance between two points on the unit sphere, that is very easy to compute using the dot product of the two points. It's the same as in the plane (which isn't surprising since any two points on the unit sphere lie in a plane containing the origin).
Second, if you also want to be able to compute the length of a curve that lies in the unit sphere, then that is also easy, because it is the same as the length of the curve viewed as a curve in $R^3$, so you use the same arclength formula as you would for any curve in 3-dimensional space.
But if you need the "distance metric" for something more than this, you should give us some more details and that will help us help you better. The others are trying to explain to you what a Riemannian metric is and how to use it. You might need this, but I doubt it.
| 7 | https://mathoverflow.net/users/613 | 29284 | 19,116 |
https://mathoverflow.net/questions/19234 | 5 | Let $F : \mathcal{D} \to \mathbf{Top}$ be a diagram of topological spaces. A local system of coefficients $M$ on $\mathrm{colim}\_\mathcal{D} F$ pulls back to a local system $M\_d$ on $F(d)$ for each $d \in \mathcal{D}$, and also a local system $M\_h$ on $\mathrm{hocolim}\_\mathcal{D} F$.
Is there a Bousfield-Kan type spectral sequence of the form
$$E^2\_{s,t} = \mathrm{colim}^s\_{\mathcal{D}} H\_t(F(d);M\_d) \Rightarrow H\_{s+t}(\mathrm{hocolim}\_\mathcal{D} F;M\_h)$$
and if so where can one find it in the literature? I would also be content to know if this is not possible.
| https://mathoverflow.net/users/318 | Bousfield-Kan spectral sequence with local coefficients | Let LOC be the category in which an object is a space plus a local system on it, and a morphism is a map of spaces covered by a map of coefficient systems in the obvious sense. There's an obvious functor $C$ from LOC to CH, the category of chain complexes; one can speak of the hocolim of a diagram of chain complexes; and $C$ commutes with hocolim up to natural equivalence. Your setup yields a functor $\mathcal D\to LOC$, and then your problem becomes the algebraic problem of making a spectral sequence going from $E^2\_{s.t}=colim\_{\mathcal D}^s H\_t(F)$ to $H\_{s+t}(hocolim\_{\mathcal D} F)$ when $F$ is a functor $\mathcal D\to CH$. For this you can, as Tyler suggests, use a simplicial model for $hocolim\_{\mathcal D}F$ related to the nerve of $\mathcal D$ and thus get a two-quadrant double chain complex for which one of the two standard filtrations yields a spectral sequence of the desired kind. ($colim^i\_{\mathcal D}$ means the $ith$ derived functor of the functor $colim\_{\mathcal D}$ from $\mathcal D$-diagrams of abelian groups to abelian groups; it is the same as $ith$ homology of hocolim of the diagram of (abelian groups viewed as) chain complexes.)
| 8 | https://mathoverflow.net/users/6666 | 29292 | 19,122 |
https://mathoverflow.net/questions/29096 | 7 | Suppose we have a finite, 100-uniform system of sets such that any point is contained in at most 3 sets. Is it true that we can color the points such that every set contains 50 red and 50 blue points?
The question is by Thomas Rothvoss. A positive answer would solve the [Three permutations problem of Beck](http://www.math.sc.edu/~cooper/combprob.html), so a simple answer would be a counterexample...
| https://mathoverflow.net/users/955 | If every point is contained in at most 3 sets and all sets are big, then is the discrepancy zero? | No.
Given sets
$$
a\_1,a\_2,\dots,a\_{99},b\_1{\rm\ and\ }a\_1,a\_2,\dots,a\_{99},b\_2
$$
we see that $b\_1$ and $b\_2$ must be the same color, say, red. Then from
$$
b\_1,b\_2,c\_1,c\_2,\dots,c\_{98}{\rm\ and\ }d\_1,d\_2,c\_1,c\_2,\dots,c\_{98}
$$
we see $d\_1$ and $d\_2$ must both be red. Then from
$$
d\_1,d\_2,e\_1,e\_2,\dots,e\_{98}{\rm\ and\ }f\_1,f\_2,e\_1,e\_2,\dots,e\_{98}
$$
we see that $f\_1$ and $f\_2$ must both be red. Dot, dot, dot. You wind up with as many elements as you like, all of which must be red, and none of them are in more than two of the sets. Once you have more than 50 of them, you can put them in another set which will then have more than 50 red points.
Obviously, we can take 100 to be a variable in this problem and solution, provided we restrict its range to the positive even integers and understand 50 to be 100/2.
| 6 | https://mathoverflow.net/users/3684 | 29294 | 19,123 |
https://mathoverflow.net/questions/29173 | 11 | This was of course motivated by [this question](https://mathoverflow.net/questions/29115/a-characterization-of-stationarity).
Suppose $\kappa<\theta$ are uncountable regular cardinals. Given a structure ${\mathcal A}=(H\_\theta,\in,<,\dots)$ where < is a well-ordering, let $C\_{\mathcal A}=${$\sup(M\cap\kappa)\mid M\prec{\mathcal A}\mbox{ and }|M|<\kappa$}. Then $C\_{\mathcal A}$ contains a club. Under what (reasonably general) circumstances can we guarantee that $C\_{\mathcal A}$ actually is (respectively, is not) a club?
| https://mathoverflow.net/users/6085 | Club sets and substructures | The idea in my previous answer can, I think, be upgraded to solve the whole problem, as follows. Again, fix Skolem functions for $\mathcal A$ as given by the well-ordering $<$, and again let $D$ be a set of fewer than $\kappa$ ordinals $\delta$, each of which is $\sup(\kappa\cap M\_\delta)$ for some $M\_\delta\prec\mathcal A$ with $|M\_\delta|<\kappa$. I need to show that $\sup(D)$ is also in $C\_{\mathcal A}$. For each $\delta\in D$, define $N\_\delta$ to be the Skolem hull of $\kappa\cap\bigcup\_{\xi\in D, \xi\leq\delta}M\_\xi$. With appropriate gratitude for the hypothesis that $\kappa$ is regular, note that $N\_\delta$ is an elementary submodel of $\mathcal A$ of size $<\kappa$ and that the sequence $\langle N\_\delta\rangle$ is an elementary chain. Let $N$ be its union, and note that it, too, is an elementary submodel of $\mathcal A$ of size $<\kappa$. So it suffices to show that $\sup(D)=\sup(\kappa\cap N)$. The $\leq$ direction here is obvious, as $N$ includes $\kappa\cap M\_\delta$ for each $\delta\in D$. To complete the proof, suppose the $\geq$ direction failed. Then we would have $\sup(D)<\sup(\kappa\cap N)$, so there would be an ordinal $\alpha\in\kappa\cap N$ with $\sup(D)\leq\alpha$. By construction, we would have some $\delta\in D$ with $\alpha\in N\_\delta$, and so $\alpha$ would be of the form $f(\vec\beta)$ for some Skolem function $f$ and some ordinals $\beta\_i$ in $\kappa\cap M\_{\xi\_i}$ for certain $\xi\_i\leq\delta$. For each $i$, we have $\beta\_i<\xi\_i\leq\delta$, and, since there are only finitely many $i$ (as Skolem functions are finitary), we can find $\gamma<\delta$ with all $\beta\_i<\gamma$. Increasing $\gamma$ if necessary, we can arrange that $\gamma\in M\_\delta$. In $\mathcal A$, we can define the function $g$ sending each ordinal $\nu<\kappa$ to the supremum of all $f(\vec\eta)<\kappa$ for $\eta$ bounded by $\nu$; the values of this function are $<\kappa$ by regularity. As an elementary submodel of $\mathcal A$, $M\_\delta$ is closed under $g$ and, in particular, contains $g(\gamma)$. But (again by elementarity) $g(\gamma)$ majorizes $f(\vec\beta)=\alpha>\sup(D)\geq\delta$. That contradicts the fact that $\delta$ is the supremum of $\kappa\cap M\_\delta$, and this contradiction completes the proof.
| 12 | https://mathoverflow.net/users/6794 | 29303 | 19,127 |
https://mathoverflow.net/questions/29285 | -4 | Suppose I have some sampling distribution g(x,y,z) which has been marginalized over some variables (say y and z) giving us the marginal distribution which we'll call gx(x).
Suppose I now wish to use Bayes Theorem but on the marginalized distribution to obtain the posterior marginal distribution. Suppose I also know the a good prior for all the variables, call it k(x,y,z).... To use Bayes theorem on the marginalized distribution, must I also marginalize the prior? Or does it makes sense to use the full prior, which of course makes, the answer depend on
| https://mathoverflow.net/users/6137 | In Bayesian statistics, must I use a marginalized prior in conjunction with a marginalized distribution?// | No, you cannot marginalize the prior and then multiply the marginal with g(x). Here is why:
The correct way to use Bayes theorem is to do the following (also suggested by John):
$g\_{p1}(x,y,z) \propto g(x,y,z) k(x,y,z)$
Thus,
$g\_{p1}(x) \propto \int\_{y,z} \bigl(g(x,y,z) k(x,y,z) \bigr)$
Your want to do the following:
$g\_{p2}(x) \propto \int\_{y,z} \bigl(g(x,y,z) \bigr) \int\_{y,z} \bigl(k(x,y,z) \bigr)$
In general, $g\_{p1}(x)$ and $g\_{p2}(x)$ will not be identical.
| 1 | https://mathoverflow.net/users/4660 | 29304 | 19,128 |
https://mathoverflow.net/questions/29309 | 5 | **Background**
Let $(M,g)$ be a riemannian manifold and let $G$ be a finite group acting effectively and isometrically on $M$. Recall that this means that for all $x \in G$, the diffeomorphism $\gamma\_x$ is such that $\gamma\_x^\* g = g$ and that if $\gamma\_x(p) = p$ for all $p \in M$, then $x$ is the identity element.
If $G$ acts freely, then $M/G$ is a smooth manifold, otherwise let's call it a (global) orbifold.
Years ago I was told that even when the action of $G$ on $M$ is not free, its lift to an action on the orthonormal frame bundle $F(M)$ is free. This means that the quotient $F(M)/G$ is smooth and is the orthonormal frame bundle of $M/G$.
I can see this when $M = \mathbb{R}^n$ with the standard euclidean inner product, so that the action of $G$ is via orthogonal (hence in particular linear) transformations. Indeed, suppose that $x\in G$ fixes a point in the frame bundle. Such a point consists of a pair $(p,f)$ where $p$ is a point in $M$ and $f$ is a frame for the tangent space $T\_pM$ to $M$ at $p$. The action of $x$ on the pair $(p,f)$ is given by
$$(p,f)\mapsto (\gamma\_x(p),(D\gamma\_x)\_p f),$$
where $(D\gamma\_x)\_p$ is the derivative (i.e., the push-forward) of $\gamma\_x$ at $p$.
Now if $x$ fixes $(p,f)$, then $\gamma\_x(p)=p$ and $(D\gamma\_x)\_p$ is the identity endomorphism of $T\_pM$. But since $\gamma\_x$ is linear, it agrees with its derivative, which means that $\gamma\_x$ itself is the identity. Finally, since $G$ acts effectively, we conclude that $x$ is the identity.
**Question**
>
> Is this still true for $M/G$, where $M$ is a riemannian manifold? And if so, can someone point me in the direction of a reference where this is proved?
>
>
>
Many thanks in advance.
| https://mathoverflow.net/users/394 | Smoothness of frame bundle of (global) orbifolds [reference request] | First, one can clearly assume $M$ is connected by simply applying the argument to each componenet of $M$.
The key fact is a generalization of your argument for $M=\mathbb{R}^n$: that if $f:M\rightarrow M$ is an isometry with $M$ connected and if there is a point $p\in M$ with $f(p) = p$ and $d\_pf = Id$, then $f$ itself is the identity map.
Assuming this fact for the moment, then if $\gamma\_x(p,f) = (p,f)$ one concludes $\gamma\_x = Id$, and then since the action is effective, x itself must have been the identity element. Thus, the only element of $G$ which fixes any element of the frame bundle is the identity, so the action on the frame bundle is free.
Now, why is the fact true?
Set $X = \{p\in M| f(p) = p$ and $D\_p f = Id\}$. $X$ is nonempty by assumption and clearly closed (by, say, a continuity argument). If we can show it's open, then we'll have $X=M$ by connectedness of $M$. Thus, in particular, $f(p) = p$ for all $p\in M$.
To see $X$ is open, let $q\in X$. Let $U\_q$ be a normal neighborhood around $q$. Normal means that every pair of points in $U\_q$ has a unique minimal geodesic between them. Normal neighborhoods always exist (any sufficiently small neighborhood is totally normal), though I don't immediately remember how to prove it, only that it uses the Gauss lemma. Do Carmo's Riemannian Geometry book proves it, if you need a reference for this.
I claim that $U\_q\subseteq X$. For, if $r\in U\_q$, there is a unique minimal geodesic $c$ with $c(0) = q$ and $c(1) = r$. Since $d\_q f = Id$, we must have $f(c(t)) = c(t)$, so in particular, $f(r) = f(c(1)) = c(1) = r$. In short, $f$ fixes all points in $U\_q$
So, we must simply establish that $d\_r f = Id$. But, if there is a $v\in T\_r M$ with $d\_r f v\neq v$, then the we have $f(exp(tv))\neq f(exp(t d\_r f v))$ for all sufficiently small $t$. But for all sufficiently small $t$, both curves lie in $U\_q$ and we know $f$ fixes all points in $U\_q$. Thus, $d\_f v = v$ and the result follows.
| 3 | https://mathoverflow.net/users/1708 | 29312 | 19,134 |
https://mathoverflow.net/questions/29300 | 83 | Of all the constructions of the reals, the construction via the surreals seems the most elegant to me.
It seems to immediately capture the total ordering and precision of Dedekind cuts at a fundamental level since the definition of a number is based entirely on how things are ordered. It avoids, or at least simplifies, the convergence question of Cauchy sequences. And it naturally transcends finiteness without sacrificing awareness of it.
The one "rumor" I've consistently heard is that it is hard to naturally define integrals and derivatives in the surreals, although I have yet to see a solid technical justification of that.
Are there known results that suggest we should avoid further study of this construction, or that show limitations of it?
| https://mathoverflow.net/users/2498 | What's wrong with the surreals? | At a recent conference in Paris on [Philosophy and Model Theory](http://www.u-paris10.fr/79587394/0/fiche___pagelibre/&RH=1257591848904) (at which I also spoke), [Philip Ehrlich](http://oak.cats.ohiou.edu/%7Eehrlich/) gave a fascinating talk on the surreal numbers and new developments, showcasing it as unifying many disparate paths in mathematics. The abstract is available [here, on page 8](http://lumiere.ens.fr/%7Edbonnay/files/abstracts.pdf), and here his article on the [Absolute Arithmetic Continuum](https://web.archive.org/web/20141027065123/http://www.ohio.edu/people/ehrlich/Unification.pdf). The principal new technical development is a focus on the underlying tree.
Philip expressed his frustration that Conway often treated his creation of surreal numbers as a kind of game or just-for-fun project---an attitude reinforced by the excellent Knuth book---whereas they are in fact a profound mathematical development unifying disparate threads of mathematical investigation into a single unifying structure. And he made a very strong case for this position at the conference.
Meanwhile, perhaps exhibiting Philip's point, at a conference on logic and games here at CUNY, I once heard Conway describe the surreal numbers as one of the great disappointments of his life, that they did not seem after all to have the profound unifying nature that he (and many others) thought they might. Philip Ehrlich strove to make the case that Conway was his own worst enemy in promoting the surreals, and that they actually do have the unifying nature Conway thought they did, but that Conway scared people away from this perspective by treating them as a toy. I encourage you to read Philip's articles.
So my answer, supporting Philip, is that *nothing* is wrong with the surreals---please have at them! Of course they have their own issues, which will need to be surmounted, but we shall all benefit from a greater investigation of them.
| 69 | https://mathoverflow.net/users/1946 | 29320 | 19,142 |
https://mathoverflow.net/questions/29333 | 25 | Warmup (you've probably seen this before)
-----------------------------------------
Suppose $\sum\_{n\ge 1} a\_n$ is a conditionally convergent series of real numbers, then by rearranging the terms, you can make "the same series" converge to any real number $x$. To do this, let $P=\{n\ge 1\mid a\_n\ge 0\}$ and $N=\{n\ge 1\mid a\_n<0\}$. Since $\sum\_{n\ge 1} a\_n$ converges conditionally, each of $\sum\_{n\in P}a\_n$ and $\sum\_{n\in N}a\_n$ diverge and $\lim a\_n=0$.
Starting with the empty sum (namely zero), build the rearrangement inductively. Suppose $\sum\_{i=1}^m a\_{n\_i}=x\_m$ is the (inductively constructed) $m$-th partial sum of the rearrangement. If $x\_m\le x$, take $n\_{m+1}$ to be the smallest element of $P$ which hasn't already been used. If $x\_m> x$, take $n\_{m+1}$ to be the smallest element of $N$ which hasn't already been used.
Since $\sum\_{n\in P}a\_n$ diverges, there will be infinitely many $m$ for which $x\_m\ge x$, so $n\_{m+1}$ will be in $N$ infinitely often. Similarly, $n\_{m+1}$ will be in $P$ infinitely often, so we've really constructed a rearrangement of the original series. Note that $|x-x\_m|\le \max\{|a\_n|\bigm| n\not\in\{n\_1,\dots, n\_m\}\}$, so $\lim x\_m=x$ because $\lim a\_n=0$.
---
>
> Suppose $\sum\_{n\ge 1}v\_n$ is a conditionally convergent series with $v\_n\in \mathbb R^k$. Can the sum be rearranged to converge to any given $w\in \mathbb R^k$?
>
>
>
Obviously not! If $\lambda$ is a linear functional on $\mathbb R^k$ such that $\sum \lambda(v\_n)$ converges absolutely, then $\lambda$ applied to any rearrangement will be equal to $\sum \lambda(v\_n)$. So let's also suppose that $\sum \lambda(v\_n)$ is conditionally convergent for every non-zero linear functional $\lambda$. Under this additional hypothesis, I'm pretty sure the answer should be "yes".
| https://mathoverflow.net/users/1 | Can a conditionally convergent series of vectors be rearranged to give any limit? | The Levy--Steinitz theorem says the set of all convergent rearrangements of a series of vectors, if nonempty, is an affine subspace of ${\mathbf R}^k$. There is an article on this by Peter Rosenthal in the Amer. Math. Monthly from 1987, called "The Remarkable Theorem of Levy and Steinitz". Also see Remmert's Theory of Complex Functions, pp. 30--31.
As an example, taking $k = 2$, suppose $v\_n = ((-1)^{n-1}/n,(-1)^{n-1}/n)$. Then the convergent rearrangments fill up the line $y = x$. The linear function $\lambda(x,y) = x-y$ of course kills the series, which makes Anton's observation explicit in this instance.
The Rosenthal article, at the end, discusses Anton's question. Indeed if there is no absolute convergence in any direction then the set of all rearranged series is
all of ${\mathbf R}^k$. Note by the above example that this condition is stronger than saying the series in each standard coordinate is conditionally convergent. Rosenthal said this stronger form of the Levy-Steinitz theorem was in the papers by Levy (1905) and Steinitz (1913). He also refers to I. Halperin, Sums of a Series Permitting Rearrangements, C. R. Math Rep. Acad. Sci. Canada VIII (1986), 87--102.
| 42 | https://mathoverflow.net/users/3272 | 29340 | 19,155 |
https://mathoverflow.net/questions/28153 | 4 | I have to simulate independent draws from a very complicated distribution. They only feasible way appears to be using MCMC. I was considering running thousands of chains in parallel, but that would slow things down from me considerably. So I am running one long MCMC chain. Assuming that MCMC chain run did converge nicely ( I realize even determining this is practically impossible in general). . Can I argue that the thinned sample obtained from selecting every 100th observation (or at a sufficiently high lag at which the observed autocorrelation function is close to 0) effectively approximates ,in some sense, a i.i.d draw from the desired distribution. If yes, any references would be greatly appreciated.
| https://mathoverflow.net/users/6627 | near independence of markov chain observations at high lags | From talking to statisticians, it seems like the standard thing to do is assume a yes. For a specified sequence or thinning, one can create Markov chains which exhibit 0 autocorrelation but a 'very large' amount of dependence despite the thinning, but the idea is that this should be pretty pathological, and so you are probably 'pretty safe'. The examples that come to mind for me are chains like '$X\_{t}$ is iid 0-1 for t not divisible by a million, and equal to the sum of the previous 999,999 $X\_{s}$ mod 2 for t divisible by a million'
So, you won't get any sort of theoretical justification without telling us something about your chain, because there really are bad chains out there, where what you want fails badly... but in the real world, you can probably wave your hands. Note that just increasing your thinning isn't enough to get rid of this - for any finite amount of thinning, there are bad chains which horribly violate independence despite having 0 covariance.
There is a big world of convergence diagnostics out there. Gelman-Rubin, as pointed out, is a standard. Some other standards include the Geweke test, Mykland-Yu's Cusum test, Raftery-Lewis, and Heidelberg-Welch. There is a lot of literature on each of these, but with the slight exception of some easy calculations in Mykland-Yu, it is all extremely handwavey.
| 2 | https://mathoverflow.net/users/7047 | 29341 | 19,156 |
https://mathoverflow.net/questions/29334 | 14 | Let $R$ be the ring of integers in an algebraic number field. There are beautiful descriptions of $K\_0(R)$ and $K\_1(R)$. Namely, $\tilde{K}\_0(R)$ is the class group of $R$ and $K\_1(R)$ is the group of units of $R$. Question : Is there a nice description of $K\_2(R)$ (or at least some reasonable conjectures)? I couldn't find much about this in Milnor's or Rosenberg's books on algebraic K-theory, so I expect that the answer is pretty complicated. Is it maybe at least known in some special cases (say, for $R$ the integers in a quadratic extension of $\mathbb{Q}$)?
| https://mathoverflow.net/users/317 | K_2 of rings of algebraic integers | It's a theorem of Garland that $K\_2(R)$ is finite. Perhaps the best way to get a handle on it is to use Quillen's localization sequence
$$0\rightarrow K\_2(R)\rightarrow K\_2(F)\stackrel{T}{\rightarrow} \oplus\_v k(v)^\*\rightarrow 0,$$
where $F$ is the fraction field and the $k(v)$ are the residue fields. The map $T$ is the sum of the tame symbols, which is surjective by a theorem of Matsumoto. The injectivity on the left follows from the vanishing of $K\_2$ for finite fields.
This isn't much of an answer, but considering $K\_2(R)$ as a subgroup of $K\_2(F)$ seems a reasonable way to start some concrete considerations. For a detailed discussion of an algorithm that proceeds essentially along these lines ('Tate's method), see the paper
Belabas, Karim; Gangl, Herbert
Generators and relations for $K\_2( O\_F)$.
$K$-Theory 31 (2004), no. 3, 195--231.
Added, 8 July:
I'm sure most people know this, but I forgot to mention (for newcomers) the fact that
$$K\_2(F) = F^\times\otimes\_{\mathbf Z} F^\times/\langle a\otimes(1-a)\mid a\not=0,1\rangle,$$
which I suppose motivates the original question, and makes it worthwhile to view $K\_2(R)$ as a subgroup.
| 10 | https://mathoverflow.net/users/1826 | 29346 | 19,161 |
https://mathoverflow.net/questions/29348 | 0 | Let $\Sigma\_g$ be a Riemann surface of genus g, and $C$ is a cut curve of $\Sigma\_g$, i.e. an oriented simple close curve.
What is a right-handed Dehn twist of $C$ of $\Sigma\_g$?
| https://mathoverflow.net/users/5093 | What is a right-handed Dehn twist of a cut curve of a Riemann surface? | Cut the curve with a scalpel, going along the curve (it is oriented), rotate the right side 360 degrees, and glue it back in...
| 2 | https://mathoverflow.net/users/5301 | 29355 | 19,165 |
https://mathoverflow.net/questions/29353 | 0 | We have a map $f \in \mathcal{C}^{\infty}(M, N)$ with two manifolds $M$ and $N$ (with dimensions $m:\dim(M)$ and $n:=\dim(N)$). We define the graph $F: M \to M \times N$ by $F(p)=(p, f(p))$. I wish to prove:
1.) $e(F)=\frac{m}{2}+e(f)$, where $e$ is the energydensity.
2.) $f$ is harmonic iff $F$ is harmonic.
Thank you and best regards.
| https://mathoverflow.net/users/7028 | Harmonic graphs | It is intended that $M\times N$ is endowed with the direct sum Riemann structure. In this case, for *any* smooth map $f:L\to M\times N$ it is true that $\frac{1}{2}|Df|^2=\frac{1}{2}|Df\_1|^2+\frac{1}{2}|Df\_2|^2$, whence (1) since the energy density of *id* is *m/2*. Also, $f$ is a local minimizer of the local energy integral $E(f)=E(f\_1)+E(f\_2)$ if and only of both $f\_1$ and $f\_2$ are, that implies (2).
**rmk**. Here of course "local minimizer" is a map *f* such that for any point *p* of the domain there exists a nbd *U* of *p* such that the integral of the density energy of *f* over U is minimum with respect to variations with compact support in U. This property is equivalent to harmonicity (and of course does not require that the total energy be finite).
| 2 | https://mathoverflow.net/users/6101 | 29357 | 19,167 |
https://mathoverflow.net/questions/28143 | 10 | Ribbon categories are braided monoidal categories with a [twist](http://nlab.mathforge.org/nlab/show/twist) or balance, $\theta\_B:B\to B$, which is a natural transformation from the identity functor to itself. In the string diagram calculus for ribbon categories, the ribbon is represented as a 360˚ twist in a ribbon (*op. cit.*). (See for example Street's *Quantum Groups* or Kassel's *Quantum Groups* for details.)
My questions are:
* Is there work describing what happens if we consider a 180˚ twist?
* If not, what goes wrong if we take this half twist one of the operations of interest on ribbons?
* How are ribbons with a 180˚ twist axiomatised? What if loops are possible and we have twisted tangles? (I believe that *Traced Monoidal Categories* by Verity, Street and Joyal doesn't cover this case.)
| https://mathoverflow.net/users/2620 | 180˚ vs 360˚ Twists in String Diagrams for Ribbon Categories | The completeness result, which I conjectured in "Autonomous categories in which A is isomorphic to A\*" (as cited by Dave above), has been proven last month. I talked about this at QPL 2010 in May, but it is not yet written. It is actually relatively easy to prove, although it took me over a month to realize that this is so. Essentially it is a reduction to the known result for ribbon categories. The absence of Moebius strips is one of the things that makes this possible.
What must be shown is: given two terms (in the half-twist language) that have the same diagram, then the terms can be proved equal by the axioms.
In a nutshell: first, it suffices to show this for terms that use the half-twist map only at object generators (half twists on A tensor B, on I, and on A\* can be immediately reduced using the axioms). Now given two terms t and s that have the same diagram, there are two possibilities:
(1) each ribbon in the diagram has an even number of half-twists on it. In this case, they can all be moved next to each other and replaced by full twists, using the axioms. Then one can simply use coherence for ribbon categories to show that s and t are provably equal.
(2) some ribbon in the diagram has an odd number of half-twists on it. Since there are no Moebius strips, this can only happen if the ribbon has two ends, each of which is either connected to a box or to a source/sink of the diagram. W.l.o.g. assume one side of the ribbon is connected to an output of the box f in the diagram. Using the above trick, we can use the axioms to replace all but one of the half-twists by full twists, and to move the remaining half-twist adjacent to the box f. The key point is that this happens in both terms s and t. In both s and t, now replace this particular occurrence of the half-twist by a new variable H:A->A\*. Note that this is no longer a half-twist graphically, but simply a new box. Call the modified terms s' and t'. Since both s' and t' have the H in the same place, s' and t' still have isomorphic diagrams. But they have one less ribbon with an odd number of twists, so the result follows by induction.
This proof is annoyingly simple, but it's correct.
| 7 | https://mathoverflow.net/users/7055 | 29370 | 19,175 |
https://mathoverflow.net/questions/29323 | 126 | You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.
I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.
So: What are your favorite dinner conversation math puzzles?
I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.
One problem per answer.
If you post the answer, please obfuscate it with something like [rot13](http://www.rot13.com/). Don't spoil the fun for everyone else.
| https://mathoverflow.net/users/27 | Math puzzles for dinner | I really like the following puzzle, called the blue-eyed islanders problem, taken from Professor Tao's blog :
"There is an island upon which a tribe resides. The tribe consists of 1000 people, with various eye colours. Yet, their religion forbids them to know their own eye color, or even to discuss the topic; thus, each resident can (and does) see the eye colors of all other residents, but has no way of discovering his or her own (there are no reflective surfaces). If a tribesperson does discover his or her own eye color, then their religion compels them to commit ritual suicide at noon the following day in the village square for all to witness. All the tribespeople are highly logical and devout, and they all know that each other is also highly logical and devout (and they all know that they all know that each other is highly logical and devout, and so forth).
Of the 1000 islanders, it turns out that 100 of them have blue eyes and 900 of them have brown eyes, although the islanders are not initially aware of these statistics (each of them can of course only see 999 of the 1000 tribespeople).
One day, a blue-eyed foreigner visits to the island and wins the complete trust of the tribe.
One evening, he addresses the entire tribe to thank them for their hospitality.
However, not knowing the customs, the foreigner makes the mistake of mentioning eye color in his address, remarking “how unusual it is to see another blue-eyed person like myself in this region of the world”.
What effect, if anything, does this faux pas have on the tribe?"
For those of you interested, there is a huge discussion of the problem at <http://terrytao.wordpress.com/2008/02/05/the-blue-eyed-islanders-puzzle/>
Malik
| 54 | https://mathoverflow.net/users/1162 | 29372 | 19,177 |
https://mathoverflow.net/questions/29269 | 4 | Suppose we have a convex hull computed as the solution to a linear programming problem (via whatever method you want). Given this convex hull (and the inequalities that formed the convex hull) is there a fast way to compute the integer points on the surface of the convex hull? Or is the problem NP?
There exist ways to bound the number of integer points and to find the number of integer points inside convex hull, but I specifically want the points on the hull itself.
EDIT: Suppose the set of inequalities (the linear program) have integer coefficients /EDIT
| https://mathoverflow.net/users/429 | Finding integer points on an N-d convex hull | Because the facets of your convex hull are themselves polytopes (of one lower dimension—$d{=}21$ in your case), it seems your question is equivalent to asking how to count lattice points in a polytope. One paper on this topic is "[The Many Aspects of Counting Lattice Points in Polytopes](http://www.math.ucdavis.edu/~deloera/RECENT_WORK/semesterberichte.pdf)" by J. A. DeLoera. Section 4 is entitled, "Actually Counting: Modern Algorithms and Software." A key reference in that section is to a paper by Barvinok entitled, "Polynomial time algorithm for counting integral points in
polyhedra when the dimension is fixed" (*Math of Operations Research*, Vol. 19, 769–779, 1994), whose title (*polynomial time*) seems to provide an answer of sorts to your question.
| 3 | https://mathoverflow.net/users/6094 | 29401 | 19,198 |
https://mathoverflow.net/questions/29384 | 6 | "Everyone" knows that for a general $f\in L^2[0,1]$, the Fourier series of $f$ converges to $f$ in the $L^2$ norm but not necessarily in most other senses one might be interested in; but if $f$ is reasonably nice, then its Fourier series converges to $f$, say, uniformly.
I'm looking for similar results about orthogonal polynomial expansions for functions on the whole real line. What I specifically want at the moment is sufficient conditions on a bounded function $f:\mathbb{R} \to \mathbb{R}$ so that the partial sums of its Hermite polynomial expansion are uniformly bounded on compact sets, but I'm also interested in learning what's known about pointwise/uniform/etc. convergence results for Hermite and other classical orthogonal polynomials.
Possibly such results follow trivially from well-known basic facts about Hermite polynomials, but I'm not familiar with that literature and I'm having trouble navigating it. So in addition to precise answers, I'd appreciate literature tips (but please don't just tell me to look at Szegő's book unless you have a specific section to recommend).
| https://mathoverflow.net/users/1044 | Convergence of orthogonal polynomial expansions | Define $\psi\_n(x) = c\_n H\_n(x) e^{-x^2/2}$ as in <http://en.wikipedia.org/wiki/Hermite_polynomials> . Also define the differential operator $H u = - u'' + x^2 u$. Then the $\psi\_n$ form an othonormal basis of $L^2$ and $H \psi\_n = (2n + 1) \psi\_n$.
**Warning:**
As coudy points out below: one needs $\|H f\| < \infty$ and not just $\langle f, Hf\rangle < \infty$. So the computations below need to be changed.
**Rest of original post**
Given now $f$ such that
$$
A = \langle f,Hf \rangle =\int \overline{f(x)} (Hf)(x) dx
$$
is finite. Then by writing $f(X) = \sum\_{n \geq 0} f\_n \psi\_n(X)$, we obtain
$$
A = \langle f,Hf \rangle
=\langle f, \sum\_{n \geq 0} f\_n H\psi\_n(X) \rangle
= \langle \sum\_{n \geq 0} f\_n \psi\_n(X) , \sum\_{n \geq 0} f\_n (2 n + 1)\psi\_n(X) \rangle
$$
Now using orthonormality of the $psi\_n$, we conclude that
$$
A = \sum\_{n \geq 0} |f\_n|^2 (2n + 1).
$$
Now using that the $\psi\_n(x)$ are all bounded by $2$ it follows that the sequence converges uniformly!
Now, what does $\langle f,Hf \rangle < \infty$ mean for $f(x) = e^{-x^2/2} g(x)$. This can be computed to mean
$$
\int |g'(x) + \frac{x}{2} g(x)|^2 e^{-x^2} dx.
$$
On a philosophical level, this is not about the $H\_n$ being orthogonal polynomials, but about them being eigenfunctions of a self-adjoint operator. (well the $\psi\_n$ are).
| 5 | https://mathoverflow.net/users/3983 | 29412 | 19,204 |
https://mathoverflow.net/questions/29417 | 19 | A very naive question :
I just learned that there is a non-split extension of $GL\_3(F\_2)$ by $F\_2^3$ (with standard action). It can be realized as the subgroup of the automorphism group $G\_2$ of Cayley-Graves octaves (edit: octonions) that preserve up to sign the basis $e\_i$, $i=1..7$of imaginary octaves. Does this happen for other values of $(n,q)$ (as in the title) ?
| https://mathoverflow.net/users/6451 | Non-split extensions of $GL_n(F_q)$ by $F_q^n$ ? | This never happens for finite fields $F \neq \mathbb{F}\_2$. If a group $G$ acts on an abelian group $M$, then short exact sequences
$1 \rightarrow M \rightarrow \Gamma \rightarrow G \rightarrow 1$
are classified by elements of $H^2(G;M)$. It is thus enough to show that if $F \neq \mathbb{F}\_2$ is a finite field and $V = F^n$, then $H^2(GL\_n(F);V)=0$. In fact, we will show that $H^k(GL\_n(F);V)=0$ for all $k$.
We have a short exact sequence
$1 \rightarrow F^{\times} \rightarrow GL\_n(F) \rightarrow PGL\_n(F) \rightarrow 1.$
Associated to this is the Hochschild-Serre spectral sequence in cohomology with coefficients in $V$. The $E\_2$-term is $H^p(PGL\_n(F);H^q(F^{\times};V))$. The key fact here is that $H^q(F^{\times};V)=0$ for all $q$.
On page 58 of Brown's book on group cohomology, there is a calculation of the cohomology of finite cyclic groups with nontrivial coefficients. In the case we're considering, it goes as follows. Define $N = \sum\_{x \in F^{\times}} x \in \mathbb{Z}[F^{\times}]$ (of course, $N$ acts as $0$ on $V$, but forget that for the moment). We then get a map $N : V \rightarrow V$ whose image lies in the ring of invariants $V^{F^{\times}}$ and which satisfies $N(gv)=N(v)$ for all $g \in F^{\times}$ and all $v \in V$. Let $V\_{F^{\times}}$ be the ring of coinvariants, ie the quotient of $V$ by the subspace spanned by $\langle g v-v\ |\ g \in F^{\times},\ v \in V\rangle$. We get an induced map $\overline{N}:V\_{F^{\times}} \rightarrow V^{F^{\times}}$. The result then is that $H^0(F^{\times};V) = V^{F^{\times}}$, that $H^i(F^{\times};V) = ker\ \overline{N}$ for $i \geq 1$ odd, and that $H^i(F^{\times};V) = coker\ \overline{N}$ for $i \geq 1$ even. But clearly $V^{F^{\times}} = 0$, and since $F$ is not the field with $2$ elements we also have $V\_{F^{\times}} = 0$. The result follows.
| 19 | https://mathoverflow.net/users/317 | 29422 | 19,210 |
https://mathoverflow.net/questions/29413 | 21 | Are there precise definitions for what a variable, a symbol, a name, an indeterminate, a meta-variable, and a parameter are?
In informal mathematics, they are used in a variety of ways, and often in incompatible ways. But one nevertheless gets the feeling (when reading mathematicians who are very precise) that many of these terms have subtly different semantics.
For example, an 'indeterminate' is almost always a 'dummy' in the sense that the meaning of a sentence in which it occurs is not changed in any if that indeterminate is replaced by a fresh 'name' ($\alpha$-equivalence). A parameter is usually meant to represent an arbitrary (but fixed) value of a particular 'domain'; in practice, one frequently does case-analysis over parameters when solving a parametric problem. And while a parameter is meant to represent a value, an 'indeterminate' usually does not represent anything -- unlike a variable, which is usually a placeholder for a value. But variables and parameters are nevertheless qualitatively different.
The above 2 paragraphs are meant to make the intent of my question (the first sentence of this post) more precise. I am looking for answers of the form "an X denotes a Y".
| https://mathoverflow.net/users/3993 | Defining variable, symbol, indeterminate and parameter | Regarding the status of variables, you probably want to look at Chung-Kil Hur's PhD thesis ["Categorical equational systems: algebraic models and equational reasoning"](http://www.pps.jussieu.fr/~gil/publications/thesis.pdf). Roughly speaking, he extends the notion of formal (as in formal polynomials) to signatures with binding structure and equations. He was a student of Fiore's, and I think they've been interested in giving better models (inspired by the nominal sets approach) to things like higher-order abstract syntax. I've been meaning to read his thesis for a while, to see if his treatment of variables can suggest techniques that could be used for writing reflective decision procedures which work over formulas with quantifiers.
For schematic variables or metavariables, there's a formal treatment of them in MJ Gabbay's (excellently-titled) paper ["One and a Halfth-Order Logic"](http://portal.acm.org/citation.cfm?id=1140359)
| 5 | https://mathoverflow.net/users/1610 | 29423 | 19,211 |
https://mathoverflow.net/questions/29409 | 17 | Let $(W, S)$ be a Coxeter system. Soergel defined a category of bimodules $B$ over a polynomial ring whose split Grothendieck group is isomorphic to the Hecke algebra $H$ of $W$. Conjecturally, the image of certain indecomposable (projective?) bimodules in $B$ is the well-known Kazhdan-Lusztig basis of $H$. Assuming the conjecture, Soergel showed that the coefficients of the Kazhdan-Lusztig polynomials of $W$ are given by the dimensions of certain Hom-spaces in $B$. It follows that these coefficients are non-negative, which was already known by work of Kazhdan-Lusztig in the Weyl group case by linking these coefficients to intersection cohomology of the corresponding Schubert varieties.
Soergel proved this conjecture in 1992 for $W$ a Weyl group, and Härterich proved it in 1999 for $W$ an affine Weyl group. Unfortunately, I can't access the first paper, and the second paper is in German, so I don't know anything about either of these proofs.
**Question:** Do these proofs depend on the relationship of the coefficients of the K-L polynomials to intersection cohomology, or are they independent of the corresponding machinery?
(The reason I ask is that I am potentially interested in relating known combinatorial proofs of positivity to Soergel's work, and I want to get an idea of how much machinery I would need to learn to do this.)
**Edit:** Soergel's 1992 paper is [here](http://www.reference-global.com/doi/abs/10.1515/crll.1992.429.49), if only I had the appropriate journal access. If anybody does and would like to send me this paper, that would be excellent - my contact information is at a link on my profile.
| https://mathoverflow.net/users/290 | Is Soergel's proof of Kazhdan-Lusztig positivity for Weyl groups independent of other proofs? | My understanding is that Soergel's approach applies just to finite Weyl groups and not directly to other finite Coxeter groups (or more generally), since what he can actually prove depends on some of the geometric machinery used to prove the Kazhdan-Lusztig Conjecture. The same must be true of the 1999 thesis work of his student Martin Harterich involving affine Weyl groups, which doesn't seem to have been formally published. In those situations the coefficients of KL polynomials were seen to be nonnegative in the early steps taken by Kazhdan and Lusztig toward understanding their conjecture via Schubert varieties: they occur as dimensions of certain cohomology groups.
Later on, Soergel made his program more explicit for proving the nonnegativity for arbitrary Coxeter groups using his more algebraic/categorical setting of bimodules: MR2329762 (2009c:20009) 20C08 (20F55)
Soergel,Wolfgang (D-FRBG),
Kazhdan-Lusztig-Polynome und unzerlegbare Bimoduln ¨uber Polynomringen. (German.
English, German summaries) [Kazhdan-Lusztig polynomials and indecomposable
bimodules over polynomial rings]
J. Inst. Math. Jussieu 6 (2007), no. 3, 501–525. This is in a French journal but written in German; the helpful review by Ulrich Goertz is however in English if you have access to MathSciNet. (In any case, J. Reine Angew. Math. has become super-expensive for libraries, so print or online access gets tricky.)
A helpful follow-up paper (in English) by Soergel's later student Peter Fiebig
(now at Erlangen) should also be consulted, though it is still unclear to me how far one can get with Soergel's conjectural approach in this spirit: MR2395170 (2009g:20087) 20F55 (20C08)
Fiebig, Peter (D-FRBG),
The combinatorics of Coxeter categories.
Trans. Amer. Math. Soc. 360 (2008), no. 8, 4211–4233. (Fiebig's papers are on arXiv, by the way.)
I'll have to take another look at this literature, but in any case the nonnegativity of coefficients of KL polynomials for arbitrary Coxeter groups (predicted in 1979 by Kazhdan and Lusztig) remains an intriguing question. The
general setting is far from the kind of representation theory or geometry one encounters in Lie theory, but a purely combinatorial approach seems at the moment unlikely to succeed.
ADDED: Special cases where Kazhdan-Lusztig polynomials have been computed are discussed in section 7.12 of my 1990/1992 book on reflection groups and Coxeter groups. In particular, noncrystallographic finite Coxeter groups all yield nonnegative coefficients. For dihedral groups, the polynomials are all 1, while for type $H\_3$ the computer tables found by Mark Goresky are still on his Webpage at IAS. The 1987 paper in J. Algebra by Dean Alvis which I cited involved his unpublished computer results on the polynomials for $H\_4$, for which his current Webpage gives details: <http://mypage.iusb.edu/~dalvis/h4data/index.html>
These polynomials were later recovered by Fokko du Cloux using his computer system *Coxeter*: see his last published paper
MR2255133 (2007e:20010) 20C08 (20F55)
du Cloux, Fokko (F-LYON-ICJ),
Positivity results for the Hecke algebras of noncrystallographic finite Coxeter groups. J. Algebra 303 (2006), no. 2, 731–741.
| 12 | https://mathoverflow.net/users/4231 | 29428 | 19,212 |
https://mathoverflow.net/questions/29424 | 19 | Ordinary cohomology on CW complexes is determined by the coefficients. There are (more than) two nice ways to define cohomology for non-CW-complexes: either by singular cohomology or
by defining $\widetilde H^n(X;G) = [X, K(G,n)]$. Are there standard/easy examples where these
two theories differ?
One idea that comes to mind is the paper by Milnor and Barratt (about Anomolous Singular
Homology) which says that the $n$-dimensional Hawaiian earring $H^n$ has nontrivial singular
homology in arbitrarily high dimensions. But I don't see an easy way to compute
$[H^n, K(G, m)]$.
| https://mathoverflow.net/users/3634 | Difference between represented and singular cohomology? | The Cantor set has exotic zeroth cohomology. Its singular cohomology is the linear dual of its zeroth singular homology, which is the free abelian group on its set of points. Thus its singular cohomology is an uncountable infinite product of $\mathbb Z$. Its represented cohomology is the set of continuous maps to the discrete space $\mathbb Z$, which must factor through a finite quotient. It is a free abelian group on countably many generators.
| 23 | https://mathoverflow.net/users/4639 | 29433 | 19,215 |
https://mathoverflow.net/questions/29427 | 14 | I know of 2(.5) proofs of Ramsey's theorem, which states (in its simplest form) that for all $k, l\in \mathbb{N}$ there exists an integer $R(k, l)$ with the following property: for any $n>R(k, l)$, any $2$-coloring of the edges of $K\_n$ contains either a red $K\_k$ or a blue $K\_l$.
Both the finite and the infinite versions (the latter being--a 2-coloring of the edges of $K\_\mathbb{N}$ contains an infinite monochrome $K\_\mathbb{N}$) are proven on [Wikipedia](http://en.wikipedia.org/wiki/Ramsey%2527s_theorem#Proof_of_the_theorem), and one may deduce the finite version from the infinite one by compactness, or equivalently using Konig's lemma. The infinitary proof does not give effective bounds on $R(k, l)$, but can be converted to one that does as follows (this is the .5 proof):
Consider a $2$-coloring of the edges of a complete graph on $N=2^{k+l}$ vertices, $v\_1, ..., v\_N$. Let $V\_0$ be the set of all vertices, and let $V\_i$ be the largest subset of $V\_{i-1}$ connected to $v\_i$ by edges of a single color, $c\_i$. After $k+l$ steps, at least $k$ of the $c\_i$ are read or $l$ of the $c\_i$ are blue by pigeonhole; let the set of indices for which this happens be denoted $S$. Then $(v\_i)\_{i\in S}$ is the desired subgraph.
My question is:
>
> Does anyone have a fundamentally different proof of this theorem? In particular, I am curious to know if there are any of a less combinatorial flavor.
>
>
>
| https://mathoverflow.net/users/6950 | Noncombinatorial proofs of Ramsey's Theorem? | I hope this is close to what you are asking. The following compactness principle turns out to be useful in certain construction in dynamical systems and in probability (in particular, in the theory of exchangeable random variables), and it may be seen as a topological version of the infinitary Ramsey theorem.
>
> **Lemma.** Let $X,d$ a compact metric space, and
> $u:\mathbb{N}^2\to X$ a double
> sequence in $X$. Then, there exists a
> strictly increasing
> $\sigma:\mathbb{N}\to\mathbb{N}$ such
> that, denoting
> $v(i,j):=u(\sigma(i),\sigma(\, j)),$
> the following limits exist, and
> coincide:
>
>
> $$\lim\_{i\to\infty} \lim\_{j\to\infty} v(i,j) \, = \lim\_{ i< j, \,(i,j)\to\infty} v(i,j). $$
>
>
>
The proof is just routine (iterated) application of the usual diagonal argument for sequences. How does it implies the infinitary Ramsey theorem? Take $X$ a discrete space of colors and $u$ an
$X$-coloring of the complete graph with vertices set $\mathbb{N}$. Then the existence of the limit in the RHS means that the set $\{\sigma(i):i> c\}$ for some c is a monocromatic complete subgraph.
One can even state a more general version for multi-sequences $u(i)$ indicized on increasing $n$-ples $i:=(i\_1< i\_2\dots < i\_n)$ of natural numbers; the game is that any parenthesization
$$(i\_1,\dots,i\_{k\_1}),(i\_{k\_1+1},\dots,i\_{k\_2}),\dots,(i\_{k\_r},\dots,i\_n)$$
produces a different iterated way of letting $i$ go to infinity (like when making the beads sliding from left to right in an abacus: in small clusters, one at a time, or all together ).
The corresponding limits for u(i) may or may not exist and/or coincide; but, up to a selection of indices via a strictly increasing $\sigma:\mathbb{N}\to\mathbb{N}$, all these iterated limits do exist and coincide.
Actually, it may be argued whether it really gives a *fundamentally* different proof of Ramsey theorem as you are asking. Nevertheless, if you try submitting it to an analyst or to a geometer, I think you are much more likely to obtain an immediate proof of it than with the original set theoretic version (please confirm my guess). On the other hand, it may sound quite weird to a pure set theorist (do not necessarily confirm this statement).
| 21 | https://mathoverflow.net/users/6101 | 29436 | 19,217 |
https://mathoverflow.net/questions/29429 | 5 | As a sort of dual question to [this](https://mathoverflow.net/questions/29427/noncombinatorial-proofs-of-ramseys-theorem) question, I am wondering what proofs people know of lower bounds on Ramsey numbers $R(k, k)$. I know of two proofs: there is Erdos's beautiful probabilistic argument, given [here](http://en.wikibooks.org/wiki/Combinatorics/Bounds_for_Ramsey_numbers#Lower_bound) for example, as well as the following:
(This is a sketch; it's worth working out the details.) Represent a two-coloring of the edges of a complete graph on $n$ vertices as the upper triangle (strictly above the diagonal) of an $n\times n$ matrix of zeroes and ones (that is, ${n\choose 2}$ bits). We may rewrite this representation by noting which vertices are contained in our monochrome subgraph and what color it is, as well as including all the remaining edge data, using some special characters to block off this data. If Ramsey numbers are small, this sends each string of bits under an appropriate encoding to a smaller string of bits, which is impossible by pigeonhole. (I am being purposely vague about the encoding--pick your own, anything goes essentially--because it's a bit boring). The bound this argument gives is essentially the same as the probabilistic one, and indeed it seems to me to be essentially a "derandomization" of that argument.
My question is:
>
> Does anyone know a proof of a similarly good lower bound using a fundamentally different method?
>
>
>
| https://mathoverflow.net/users/6950 | Proofs of Lower Bounds for Ramsey Numbers? | The best known *explicit* Ramsey graph construction is in the paper:
>
> Boaz Barak, Anup Rao, Ronen Shaltiel, Avi Wigderson: 2-source dispersers for sub-polynomial entropy and Ramsey graphs beating the Frankl-Wilson construction. STOC 2006: 671-680
>
>
>
Call a graph $K$-Ramsey if it doesn't have a $K$-clique or a $K$-independent set. They prove
>
> There is an absolute constant $\alpha > 0$ and an explicit construction of a $2^{2^{\log^{1−\alpha} n}} = 2^{n^{o(1)}}$-Ramsey graph over $2^n$ vertices, for every large enough $n \in {\mathbb N}$.
>
>
>
Here, "explicit construction" means roughly that there is an efficient algorithm which when given the string of $N$ ones, it outputs an $N$-node $K$-Ramsey graph. (I know this is "stronger" than what you would like, but you should still check these things out for fun.)
Before the above paper, the best known explicit construction was by Frankl and Wilson, who showed that there are $2^n$ node graph that are about $2^{\Omega(\sqrt{n})}$-Ramsey. Noga Alon had an alternative construction but I think it only matched Frankl and Wilson. See the above paper for more details.
All these constructions are very neat and use radically different methods from simple counting arguments, so I hope you enjoy them. You may find that the problem of finding a succinct/effective description of a family of lower bound graphs is indeed interesting.
| 8 | https://mathoverflow.net/users/2618 | 29438 | 19,219 |
https://mathoverflow.net/questions/29442 | 30 | This question is, in some sense, a variant of [this](https://mathoverflow.net/questions/23361/construction-of-opposite-category-as-a-structure), but for certain cases.
The opposite category of an abelian category is abelian. In particular, if $R-mod$ is the category of $R$-modules over a ring $R$ (say left modules), its opposite category is abelian. The Freyd-Mitchell embedding theorem states that this opposite category can be embedded in a category of modules over a ring $S$. This embedding is usually very noncanonical though.
>
> **Question:** Is there any way to choose $S$ based on $R$?
>
>
>
My guess is probably not, since [these notes](http://www.math.ku.dk/~alexb/HomAlg/Category.pdf) cite the opposite category of $R-mod$
as an example of an abelian category which is not a category of modules. I can't exactly tell if they mean "it is (for most $R$) (provably) not equivalent to a category of modules over any ring" or "there is no immediate structure as a module category." If it is the former, how would one prove it?
I also have a variant of this question when there is additional structure on the category.
The module category $H-mod$ of a Hopf algebra $H$ is a tensor category. A finite-dimensional Hopf algebra can be reconstructed from the tensor category of finite-dimensional modules with a fiber functor via Tannakian reconstruction. Now $(H-mod)^{opp}$ is a tensor category as well satisfying these conditions (namely, the hom-spaces in this category are finite-dimensional). The dual of the initial fiber functor makes sense and becomes a fiber functor from $(H-mod)^{opp} \to \mathrm{Vect}$ (since duality is a contravariant tensor functor on the category of vector spaces). In this case, $(H-mod)^{opp}$ is the representation category of a canonical Hopf algebra $H'$.
>
> **Question$\prime$** What is $H'$ in terms of $H$?
>
>
>
| https://mathoverflow.net/users/344 | What is the opposite category of the category of modules (or Hopf algebra representations)? | One can prove that for any non-zero ring $R$ the category $R$-Mod$^{op}$ is not a category of modules. Indeed any category of modules is Grothendieck abelian i.e., has exact filtered colimits and a generator. So for $R$-Mod$^{op}$ to be a module category $R$-Mod would also need exact (co)filtered limits and a cogenerator. It turns out that any such category consists of just a single object.
I believe this is stated somewhere in Freyd's book Abelian Categories but I am not sure exactly where off the top of my head. **Edit** it is page 116.
**Further edit:** For categories of finitely generated modules here is something else, although it is more in the direction of the title of your question than what is in the actual body of the question. Suppose we let $R$ be a commutative noetherian regular ring with unit and let $R$-mod be the category of finitely generated $R$-modules. Then we can get a description of $R$-mod$^{op}$ using duality in the derived category.
Since $R$ is regular every object of $D^b(R) \colon= D^b(R-mod)$ is compact in the full derived category. The point is that
$$RHom(-,R)\colon D^b(R)^{op} \to D^b(R) $$
is an equivalence (usually this is only true for perfect complexes, but here by assumption everything is perfect). So one can look at the image of the standard t-structure (which basically just "filters" complexes by cohomology) under this duality. The heart of the standard t-structure is $R$-mod sitting inside $D^b(R)$ so taking duals gives an equivalence of $R$-mod$^{op}$ with the heart of the t-structure obtained by applying $RHom(-,R)$ to the standard t-structure.
In the case of $R =k$ a field then this just pointwise dualizes complexes so we see that it restricts to the equivalence $k$-mod$^{op}\to k$-mod given by the usual duality on finite dimensional vector spaces.
As another example consider $\mathbb{Z} $-mod sitting inside of $D^b(\mathbb{Z})$ as the heart of the standard t-structure given by the pair of subcategories of $D^b(\mathbb{Z})$
$$\tau^{\leq 0} = \{X \; \vert \; H^i(X) = 0 \; \text{for} \; i>0\}$$
$$\tau^{\geq 1} = \{X \; \vert \; H^i(X) = 0 \; \text{for} \; i\leq 0\} $$
It is pretty easy to check that $RHom(\Sigma^i \mathbb{Z}^n, \mathbb{Z}) \cong \Sigma^{-i} \mathbb{Z}^n$ and $RHom(\Sigma^i \mathbb{Z}/p^n\mathbb{Z}, \mathbb{Z}) \cong \Sigma^{-i-1}\mathbb{Z}/p^n\mathbb{Z}$ so that this t-structure gets sent to
$$\sigma^{\leq 0} = \{X \; \vert \; H^i(X) = 0 \; \text{for} \; i> 0, \; H^{0}(X) \; \text{torsion}\}$$
$$\sigma^{\geq 1} = \{X \; \vert \; H^i(X) = 0 \; \text{for} \; i< 0, H^0(X) \; \text{torsion free}\} $$
using the fact that objects of $D^b(\mathbb{Z})$ are isomorphic to the sums of their cohomology groups appropriately shifted.
So taking the heart we see that
$$\mathbb{Z}-mod^{op} \cong \sigma^{\leq 0} \cap \Sigma\sigma^{\geq 1} = \{X \; \vert \; H^{-1}(X) \; \text{torsion free}, H^0(X) \; \text{torsion}, H^i(X) = 0 \; \text{otherwise}\} $$
with the abelian category structure on the right coming from viewing it as a full subcategory of $D^b(\mathbb{Z})$ with short exact sequences coming from triangles. It is the tilt of $\mathbb{Z}$-mod by the standard torsion theory which expresses every finitely generated abelian group as a torsion and torsion free part.
| 25 | https://mathoverflow.net/users/310 | 29450 | 19,228 |
https://mathoverflow.net/questions/29404 | 2 | Given m units of flow from a source node, and several possible destinations, is there a network flow gadget to force the flow to use only one destination? That is, send all m units to one (unspecified) destination and 0 to all the others?
If m = 1, we can just connect the source to the destinations and use the integral flow theorem, but what about m > 1?
| https://mathoverflow.net/users/19029 | Network flow gadget | No, there is no such gadget.
Suppose to the contrary that you want to allow flow from vertex x to either vertex y or vertex z, but that you want it to remain unsplit. If there exist two flows F1 and F2, both with m units into vertex x but with those units all going to vertex y in flow F1 and all going to vertex z in flow F2, then for any 0 ≤ p ≤ m there exists a flow F3 with p units from x to y and m – p units from x to z: simply let F3 = (p/m)F1 + ((m–p)/m)F2. It's easy to verify that, if F1 and F2 obey the flow constraints at each vertex and edge, then so does F3.
In order to force the flow to be unsplit, you can't remain within the formulation of a maximum flow; you'd have to extend the problem to include additional side constraints, and by doing so most likely make it NP-hard.
There's one possible exception: if every node of the graph has the same value m and you want to quantize flow in units of m rather than in integers. Then you can just divide everything by m and use an integer flow.
| 3 | https://mathoverflow.net/users/440 | 29452 | 19,230 |
https://mathoverflow.net/questions/29434 | 7 | Is there a sequence of topological spaces $X\_n$ (manifolds ideally), where the sum of the Betti numbers of $X\_n$ remains bounded but the Lusternik–Schnirelmann category is unbounded, as $n \to \infty$? What about vice versa?
One might think of both of these numbers as very rough measures of the "complexity" of a space, and it is well known in particular that both quantities are lower bounds on the number of critical points of a Morse function. But it would be nice to hear about any facts governing the relations between them.
| https://mathoverflow.net/users/4558 | L-S category versus Betti numbers | The 2d surfaces have unbounded Betti numbers, but bounded category.
A matrix in $SL\_n(\mathbb Z)$ describes a diffeomorphism of the $n$-torus. We may form the mapping torus, a bundle of tori over the circle, with monodromy the matrix. If the matrix is generic, so that none of its exterior powers have eigenvalues that are roots of unity, then the homology of the manifold will be the same as that of $S^1\times S^n$, hence total Betti number $4$. But its universal cover has cup length about $n$, hence large category.
| 10 | https://mathoverflow.net/users/4639 | 29457 | 19,234 |
https://mathoverflow.net/questions/29462 | 4 | Unfortunately the question I am asking isnt very well-defined. But I will try to make it as precise as possible. Supposed I am given a mod-p representation of $G\_Q$ into $Gl\_2(F\_p)$. I want to check for arithmetic invariants so that I can conclude that the representation comes from a modular form but not an elliptic curve. The whole point of this exercise is to understand the difference between the representations coming from elliptic curves and cusp forms in general. I hope I was able to make the question precise. A few things that one can look at is the conductor of an elliptic curve (i.e. the exponent of 2 in the level of modular form is too high then it cant come from an elliptic curve) or one can look at the Hasse bound for $a\_l$ for different primes. But I want to know some non-trivial arithmetic constraints attached to such invariants. Also if such a representation doesnt come from an elliptic curve then it must come from an abelian variety of $GL\_2$ type. Can anything be said about that abelian variety in general.
| https://mathoverflow.net/users/2081 | Galois representation attached to elliptic curves | Since your representation $\overline{\rho}$ is defined over $\mathbb F\_p$, you can't do things like the Hasse bounds, since
the traces $a\_{\ell}$ of Frobenius elements at unramified primes are just integers mod $p$,
and so don't have a well-defined absolute value.
One thing you can do is check the determinant; this should be the mod $p$ cyclotomic character if $\overline{\rho}$ is to come from an elliptic curve. In general (or more precisely, if $p$ is at least 7), that condition is not sufficient (although it is sufficient if $p = 2,3$ or 5);
see the various results discussed in [this paper](http://www.math.northwestern.edu/~fcale/papers/disc.pdf) of Frank Calegari,
for example. In particular, the proof of Theorem 3.3 in that paper should
give you a feel for what can happen in the mod $p$ Galois representation attached to
weight 2 modular forms that are not defined over $\mathbb Q$, while the proof of Theorem 3.4
should give you a sense of the ramification constraints on a mod $p$ representation imposed
by coming from an elliptic curve.
| 4 | https://mathoverflow.net/users/2874 | 29464 | 19,240 |
https://mathoverflow.net/questions/29420 | 1 | Hi,
I am trying to figure out if there are any functions, and then for which, where one can say that the gradient of the projection is the same as the projection of the gradient.
In this case a projection of the function f(x,y,z) is an integral $p(x,y) = \int f(x,y,z) dz$.
It seems to me that it might not be true in the general case. But are there any such functions?
Another way of looking at it is that I would like to know if there are an analogy to the projection-slice theorem, but for gradients.
I hope the question is correctly formulated, and my apologies if I missed out something very obvious. It does seem to me like this question should have popped up before and there should be some know result. I might have missed it in the textbook.
Helps and pointers appreciated.
| https://mathoverflow.net/users/7068 | Projection of a gradient and the gradient of a projection | I think you need to reformulate your problem.
The standard definition of a vector field being `projectible' [eg Warner.
See $\pi$-projectible ] requires, in the case of the projection $(x,y,z) \to (x,y)$
it to have the form $F\_1 (x,y){{\partial} \over {\partial x}} + F\_2 (x,y){{\partial} \over {\partial y}} + F\_3(x,y,z) {{\partial} \over {\partial z }}$.
For your ``projection'' to be finite, you need your $f(x,y,z)$ to depend on $z$
in such a way that the integral is finite. As a consequence, its gradient will
have $F\_1$ and $F\_2$ either zero, or depending on $z$.
Combining your def. of `projection of a function' with the standard def. of
projection of a vector field you get the result that the only projected vector field
you can get is the zero vector fields.
| 3 | https://mathoverflow.net/users/2906 | 29468 | 19,244 |
https://mathoverflow.net/questions/21010 | 1 | Let $Y$ be a normal projective surface, let $X$ be a smooth projective surface and let $\pi:Y\longrightarrow X$ be a finite morphism. Why are all singularities of $Y$ cyclic quotient singularities? And what does this mean? Furthermore, why are these singularities rational? And again, what does that mean? (I edited the question. So the example of Ekedahl doesn't work anymore.)
Take a minimal resolution of singularities $\rho:Y^\prime\longrightarrow Y$. Then the above apparently shows that $R^0 \rho\_\ast \mathcal{O}\_{Y^\prime} = \mathcal{O}\_Y$ and $R^i \rho\_\ast \mathcal{O}\_{Y^\prime} = 0$ for $i>0$. Is this something special for surfaces?
The reason I ask this question is the following.
If $Y$ is a normal variety (say over the field of complex numbers) with the above data, do we still have $R^0 \rho\_\ast \mathcal{O}\_{Y^\prime} = \mathcal{O}\_Y$ and $R^i \rho\_\ast \mathcal{O}\_{Y^\prime} = 0$ for $i>0$.
**Note**. The case of a surface over the complex numbers is dealt with in *Compact complex surfaces* by Barth, Hulek, Peters and van de Ven. I believe they show that cyclic quotient singularities are rational in this case.
| https://mathoverflow.net/users/4481 | On minimal resolution of singularities and the type of singularities | Re. cyclic quotient singularity. See Kollar's book: `Resolution of Singularities'
book. p. 81, item (3) and explanations that follow.
I also found Durfee. L'enseignement Math. 1979. Tome 25. fasc. 1-2. p. 131.
`Fifteen characterizations of Rational Double Points' helpful in getting myself
oriented with examples regarding isolated singularities for surfaces.
| 2 | https://mathoverflow.net/users/2906 | 29470 | 19,245 |
https://mathoverflow.net/questions/29302 | 29 | There are a number of informal heuristic arguments for the consistency of ZFC, enough that I am happy enough to believe that ZFC is consistent. This is true for even some of the more tame large cardinal axioms, like the existence of an infinite number of Grothendieck universes.
Are there any such heuristic arguments for the existence of Vopenka cardinals or huge cardinals? I'd very much like to believe them, mainly because they simplify a great deal of trouble one has to go through when working with accessible categories and localization (every localizer is accessible on a presheaf category, for instance).
For Vopenka's principle, the category-theoretic definition is that every full complete (cocomplete) subcategory of a locally presentable category is reflective (coreflective). This seems rather unintuitive to me (and I don't even understand the model-theoretic definition of Vopenka's principle).
What reason is there to believe that ZFC+VP (or ZFC+HC, which implies the consistency of VP) is consistent? Obviously, I am willing to accept heuristic or informal arguments (since a formal proof is impossible).
| https://mathoverflow.net/users/1353 | Reasons to believe Vopenka's principle/huge cardinals are consistent | Most of the arguments previously presented take a set-theoretic/logical point of view and apply to large cardinal axioms in general. There's a lot of good stuff there, but I think there are additional things to be said about Vopěnka's principle specifically from a category-theoretic point of view.
One formulation of Vopěnka's principle (which is the one that I'm used to calling "the" category-theoretic definition, and the one used as the definition in Adamek&Rosicky's book, although there are many category-theoretic statements equivalent to VP) is that there does not exist a large (= proper-class-sized) full discrete (= having no nonidentity morphims between its objects) subcategory of any locally presentable category. I think there is a good argument to be made for the naturalness of this from a category-theoretic perspective.
To explain why, let me back up a bit. To a category theorist of a certain philosophical bent, one thing that category theory teaches us is to avoid talking about equalities between objects of a category, rather than isomorphism. For instance, in doing group theory, we never talk about when two groups are equal, only when they are isomorphic. Likewise in doing topology, we never talk about when two spaces are equal, only when they are homeomorphic. Once you get used to this, it starts to feel like an accident that it even makes *sense* to ask whether two groups are equal, rather than merely isomorphic. And in fact, it is an accident, or at least dependent on the particular choice of axioms for a set-theoretic foundation; one can give other axiomatizations of set theory, provably equivalent to ZFC, in which it doesn't make sense to ask whether two sets are equal, only whether two elements of a given ambient set are equal. These are sometimes called "categorial" set theories, since the first example was Lawvere's [ETCS](http://ncatlab.org/nlab/show/ETCS) which axiomatizes the category of sets, but I prefer to call them *structural* set theories, since there are other versions, like [SEAR](http://ncatlab.org/nlab/show/SEAR), which don't require any category theory.
Now there do exist categories in which it does make sense to talk about "equality" of objects. For instance, any set X can be regarded as a discrete category $X\_d$, whose objects are the elements of X and in which the only morphisms are identities. Moreover, a category is equivalent to one of the form $X\_d$, for some set X, iff it is both a groupoid and a preorder, i.e. every morphism is invertible and any parallel pair of morphisms are equal. I call such a category a "discrete category," although some people use that only for the stricter notion of a category *isomorphic* to some $X\_d$. So it becomes tempting to think that one might instead consider "category" to be a fundamental notion, and *define* "set" to mean a discrete category.
Unfortunately, however, what I wrote in the previous paragraph is false: a category is equivalent to one of the form $X\_d$, for some set X, iff it is a groupoid and a preorder and *small*. We can just as well construct a category $X\_d$ when X is a proper class, and it will of course still be discrete. In fact, just as a set is the same thing as a small discrete category, a proper class is the same thing as a *large* discrete category. However, this feels kind of bizarre, because the large categories that arise in practice are almost never of the sort that admit a meaningful notion of "equality" between their objects, and in particular they are almost never discrete. Consider the categories of groups, or rings, or topological spaces, or sets for that matter. Outside of set theory, proper classes usually only arise as the class of objects of some large category, which is almost never discrete. The world would make much more sense, from a category-theoretic point of view, if there were no such things as proper classes, a.k.a. discrete large categories --- then we could define "set" to mean "discrete category" and life would be beautiful.
Unfortunately, we can't have large categories without having large discrete categories, at least not without restricting the rest of mathematics fairly severly. This is obviously true if we found mathematics on ZFC or NBG or some other traditional "membership-based" or "material" set theory, since there we need a proper class of objects before we can even define a large category. But it's also true if we use a structural set theory, since there are a few naturally and structurally defined large categories that are discrete, such as the category of well-orderings and all isomorphisms between them (the [core](http://ncatlab.org/nlab/show/core) of the full subcategory of Poset on the well-orderings).
Thus Vopěnka's principle, as I stated it above, is a weakened version of the thesis that large discrete categories don't exist: it says that at least they can't exist as full subcategories of locally presentable categories. Since locally presentable categories are otherwise very well-behaved, this is at least reasonable to hope for. In fact, from this perspective, if Vopěnka's principle turns out to be inconsistent with ZFC, then maybe it is ZFC that is at fault! (-:
| 34 | https://mathoverflow.net/users/49 | 29473 | 19,247 |
https://mathoverflow.net/questions/29475 | 11 | The proof that I have in mind is as follows -
$\text{Gal }(\overline{\mathbb Q}/\mathbb Q)$ is a proper uncountable subgroup of the group of permutations on countably many symbols, hence the latter is uncountable. .
But it needs a lot of jargon from topology and algebra. Is there a neat proof like Cantor's diagonal argument?
| https://mathoverflow.net/users/2720 | An easy proof of the uncountability of bijections on natural numbers? | Indeed, Cantor's argument is readily adapted. Let $\pi\_1,\pi\_2,\pi\_3, \ldots $ be any countable sequence of permutations of $\mathbb N$ ; let us show that this sequence
does not exhaust all permutations, by constructing a permutation $\pi$ different from all
the $\pi\_i$. We first define $\pi$ on the even integers inductively, then define
$\pi$ on the odd integers.
Let $X$ be any subset of $\mathbb N$ such that both $X$ and ${\mathbb N} \setminus X$
are infinite (e.g. the even integers, the prime numbers ...).
Set $\pi(0)$ to be an integer in $X$ different from $\pi\_1(0)$. Set $\pi(2)$ to be an integer in $X$ not in $\lbrace \pi(0),\pi\_2(2)\rbrace$. Set $\pi(4)$ to be an integer in $X$ not in $\lbrace \pi(0),\pi(2)\pi\_3(4)\rbrace$. Continuing like this inductively, we define an injection from the even integers to $X$,
such that $\pi(2k-2) \neq \pi\_k(2k-2)$ for any $k$ (and hence $\pi \neq \pi\_k$).
Finally, the set A=${\mathbb N} \setminus \pi(2\mathbb N)$ is countably infinite ; setting
$\pi(2k-1)=$ the $k$-th element of $A$ finishes the proof.
| 13 | https://mathoverflow.net/users/2389 | 29476 | 19,249 |
https://mathoverflow.net/questions/29469 | 4 | I have a $d$-uniform hypergraph on $n$ vertices with $k$ hyperedges, where $d << k$ and $n = 4k d^2$ or so.
The hyperedges are placed independently uniformly at random. I would like to have a handle on the behavior of the sizes of the connected components. By "size" I refer to the number of edges in the component, but understanding the number of vertices would be fine too.
For instance, if $X$ is the size of the component containing the first hyperedge, it seems like we should have $\Pr[X > t] < 1/2^t$. This is because each hyperedge has a less than $1/4$ chance of intersecting any other hyperedge, so this seems like some sort of exponentially decaying branching process.
Furthermore, it seems like there should be a negative association among component sizes: the larger one component is, the smaller the other ones are. Suppose I give each component in the graph a unique random label in $[k]$, and let $Y\_i$ be the size of the component labeled $i$ (or 0 if no component has label $i$). Then I expect that $E[Y\_i | Y\_j = t]$ for $j \neq i$ is decreasing in $t$. Moreover, I expect that the random variable $(Y\_i | Y\_j = t)$ is decreasing in $t$: the variable with small $t$ dominates the variable with large $t$.
But I'm not sure how to rigorously show either property.
| https://mathoverflow.net/users/6826 | Negative Association of Component Size in Random Hypergraph | The paper "[The phase transition in a random hypergraph](http://portal.acm.org/citation.cfm?id=586795.586806)"
by Michal Karoskia and Tomasz Luczak
(*Journal of Computational and Applied Mathematics*,
Volume 142, Issue 1, May 2002, Pages 125-135) seems relevant.
They "prove local limit theorems for the distribution of the size of the largest component of
[the random *d*-uniform hypergraph] in the subcritical and in the early supercritical phase."
A second source could be
"[Critical Random Hypergraphs: The emergence of a giant set of identifiable vertices](http://arxiv.org/pdf/math/0401208)"
by Christina Goldschmidt
(*The Annals of Probability*, 2005, Vol. 33, No. 4, 1573–1600),
although she follows the Poisson random hypergraph model.
| 1 | https://mathoverflow.net/users/6094 | 29492 | 19,257 |
https://mathoverflow.net/questions/29499 | 22 | Okay, so I know MO has had a recent proliferation of this kind of question, and I know MO is not really *for* this type of question (though I suspect perhaps this is a phenomenon that is likely to repeat toward the end of every academic year...)- nonetheless I find myself cap in hand and hoping for some guidance.
### Background
**I wasted my undergraduate degree:** following a fairly successful first year and an interest in pretty pictures, I found myself digging around in the region of complex dynamics and fell for it hard. As first loves go it was a great one- I swooned over Montel's theorem and cooed over the simple presentations of iterative dynamics gleaned from the uniformization theorem- but like all first loves; the detail of the thing did not surpass the idea, and pretty soon it had to end. I was disillusioned and reluctant to look for more fish in the sea- my work ethic dropped to zero.
**I fell in love again:** but too late- algebraic topology/ differential geometry hit me in my fourth year like a simplicial arrow from cupid's own bow but by this time, my grades blew and all the people I knew in the department were DS theorists. I got a 2:1 (for all you non-UK MOers- it's a degree class that's basically a rubber stamp with the word 'mediocre' on it).
**I tried teaching school kids:** not enough cohomology.
**I've got myself a year, a jolly good library and a lot of determination:** My aim being to produce something so intriguing/charming/advanced that someone will give me funding to do pure maths.
### Question
>
> So what, if anything, should I try to produce?
>
>
>
**Specifically:** Would I have to solve some grand unsolved problem? Would I get by with just a small one? If so, where would I find it? Perhaps even a complete set of excercises from an advanced book? A digest paper on a difficult topic? [If it helps my research interests are differential geometry, differential topology and gauge theory- but I'm flexible]
**I am aware:** That the above situation is my fault- and I would be grateful if you were restrained in your remonstrations. That the question, as stated, is highly subjective- but the *opinions* of research mathematicians is precisely what I am trying to gauge. That the answer may simply be: 'try some less prestigious universities'- in which case, fair enough- but I don't want to rule anything out just yet.
Thanks in advance for any help you can spare.
| https://mathoverflow.net/users/5869 | Yet another 'roadmap' style request- a second bite of the cherry | I sympathize with your case. A 2.1 is really not bad. You shouldn't denigrate yourself and view your peripatetic interests as requiring redemption.
Taking on a big unsolved problem without guidance or the background of a PhD student seems doomed to fail. Locking yourself in a library with all the world's books is unlikely to produce anything of merit. I have never heard of a case of a student producing something of "intriguing/charming/advanced" and using that to gain graduate admission. The romantic, amateur heroic view of math is largely bunk as pointed out by [Terry Tao.](http://terrytao.wordpress.com/career-advice/does-one-have-to-be-a-genius-to-do-maths/)
There is still hope. I know that undergraduate research is less common in the UK, but I would expect that if you email lots of professors in areas of interest to you and basically offer yourself as cheap or free labor (undergrad student level), there is a good chance that you'll be taken on as an unofficial research student by someone. I know many cases of people in math and science using this sort of informal contact to start research projects that eventually develop into PhD positions. Making yourself known to a tenured professor who can write you a strong recommendation is probably enough to get you a PhD position somewhere (in the US, UK or Europe). It is unlikely that claiming to solve a big problem or do research on your own is going to be trusted by graduate committees. You need recommendations from people trusted in the academic community.
There are MO users who have taken a decade or more off from education and successfully started PhD positions at Princeton and other top research institues. Good luck.
| 21 | https://mathoverflow.net/users/1622 | 29503 | 19,263 |
https://mathoverflow.net/questions/25778 | 24 | Is there a simple numerical procedure for obtaining the derivative (with respect to $x$) of the [pseudo-inverse](http://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_pseudoinverse) of a matrix $A(x)$, without approximations (except for the usual floating-point limitations)? The matrix $\frac{\mathrm{d}}{\mathrm{d}x}A(x)$ is supposed to be known.
In other words, are there analytical formulas that could be numerically evaluated so as to obtain the derivative of the pseudo-inverse? or, what formula would generalize
$$
\frac{\mathrm{d}}{\mathrm{d}x}A^{-1}(x) = -A^{-1}(x) \left(\frac{\mathrm{d}}{\mathrm{d}x}A(x)\right) A^{-1}(x)
$$
for the pseudo-inverse?
I would be happy if this were possible, as this would allow my [uncertainty calculation programming package](http://packages.python.org/uncertainties/) to precisely calculate uncertainties on the pseudo-inverse of matrices whose elements have uncertainties (currently, a numerical differentiation is performed, which may yield imprecise results in some cases).
Any idea would be much appreciated!
| https://mathoverflow.net/users/3810 | Analytical formula for numerical derivative of the matrix pseudo-inverse? | The answer is known [since at least 1973](http://www.jstor.org/stable/2156365): a formula for the derivative of the pseudo-inverse of a matrix $A(x)$ of constant rank can be found in
>
> The Differentiation of Pseudo-Inverses
> and Nonlinear Least Squares Problems
> Whose Variables Separate.
> Author(s): G. H. Golub and V. Pereyra.
> Source: SIAM Journal on Numerical Analysis, Vol. 10, No. 2 (Apr., 1973), pp. 413-432
>
>
>
References 29 and 30 in the above paper contain an earlier formula that can also be used to obtain the same result (papers by P.A. Wedin).
The case of non-constant rank is simple: the pseudo-inverse is not continuous, in this case (see Corollary 3.5 in On the Perturbation of Pseudo-Inverses, Projections and Linear Least Squares
Problems. G. W. Stewart. SIAM Review, Vol. 19, No. 4. (Oct., 1977), pp. 634-662).
Here is the formula for a matrix of constant rank (equation (4.12), in the Golub paper):
$$
\frac{\mathrm d}{\mathrm d x} A^+(x) =
-A^+ \left( \frac{\mathrm d}{\mathrm d x} A \right) A^+
+A^+ A{^+}^T \left( \frac{\mathrm d}{\mathrm d x} A^T \right) (1-A A^+)
+ (1-A^+ A) \left( \frac{\mathrm d}{\mathrm d x} A^T \right) A{^+}^T A^+
$$
(for a real matrix).
For complex matrices, the above formula works if Hermitian conjugates are used instead of transposes. I don't have any reference on this (anyone?), but this is verified by all the numerical tests I did (with matrices of various shapes and ranks).
| 30 | https://mathoverflow.net/users/3810 | 29511 | 19,268 |
https://mathoverflow.net/questions/29507 | 6 | Suppose you want to work with complete flags $\mathbb{F}\_3$ on $\mathbb{C}^3$. Given a flag
$$ \{0\}\leq V\_1\leq V\_2 \leq \mathbb{C}^3$$
you can think of $V\_1$ as the span of a vector $\vec{u}$, and then you can choose a vector
$\vec{v}$ that is Hermitian orthogonal to $\vec{u}$ so that $V\_2=<\vec{u},\vec{v}>$. Finally you
can choose $\vec{w}$ so that it is Hermitian orthogonal to $<\vec{u},\vec{v}>$. This gives an embedding
$$\mathbb{F}\_3\rightarrow \mathbb{C}P(2)^3 .$$
Since the Hermitian inner product involves complex conjugates, this embedding cannot possibly be
holomorphic. For instance if the first line in homogeneous coordinates is $[a,b,c]$ and the second
is $[d,e,f]$ then they satisfy $a\overline{d}=b\overline{e}+c\overline{f}=0$ where the overline
indicates complex conjugate. Is there some way of playing around with complex structures to fix
this? Is there a similar map, that is better behaved?
| https://mathoverflow.net/users/4304 | Coordinates on Flag Manifolds | I think you can use wedge products. Choose $v \in V\_1$, then $u \in V\_2$, which is linearly independent. Map the flag to $([v], [v \wedge u]) \in (CP^{2})^2$. This should be well defined and holomorphic.
| 7 | https://mathoverflow.net/users/6658 | 29514 | 19,270 |
https://mathoverflow.net/questions/29512 | 6 | Let $G$ be a Lie group, and let $\underline{G}$ denote the sheaf of smooth $G$-valued maps, i.e. for a smooth manifold $M$ we have $G(M) = C^\infty(M,G)$.
What are conditions on $G$ that imply that $\underline{G}$ is acyclic, i.e. the sheaf cohomology $H^n(M,\underline{G})=0$ for all smooth manifolds $M$ and all $n>0$?
It is clear that soft, flabby or fine sheaves are acyclic. I am interested in concrete conditions on the group $G$, e.g. like smooth contractibility.
EDIT: Daniel's answer below answers my question in the case that $G$ is abelian, using the classification of abelian Lie groups. So let us concentrate on the case that $G$ is non-abelian. The condition I am looking for is supposed to imply the vanishing of the set $H^1(M,\underline{G})$. This set can be defined for example via Cech cohomology. Its geometrical meaning is that it classifies principal $G$-bundles over $M$ up to isomorphism.
| https://mathoverflow.net/users/3473 | When is a sheaf of smooth functions acylic? | For Abelian $G$ (that is, the product of a torus with $\mathbb{R}^n$), an argument identical to macbeth's comment gives that $H^n(M, \underline{G})=0$ for all $M, n>0$ iff $G\simeq \mathbb{R}^n$).
Explicitly, in the case $G\simeq \mathbb{R}^n$ the sheaf in question is fine; otherwise, if $G\simeq \mathbb{R}^n\times (S^1)^k$ then it fits into an exact sequence $0\to \mathbb{Z}^k\to \mathbb{R}^{n+k}(M)\to \underline{G}\to 0$, giving the claim.
---
Added (7/7/2010): Having thought a bit about the non-Abelian case, I thought I'd add another non-vanishing theorem.
**Theorem.** Let $G$ be a Lie group admitting a faithful unitary representation, with $\pi\_1(G)\neq 0, \mathbb{Z}$. Then there exists $M$ with $H^1(M, \underline{G})\neq 0$.
**Proof.** Let $\rho: G\to U(n)$ be the given faithful unitary representation, and let $M=U(n)/G$. Then $U(n)$ is a $G$-bundle over $M$, and it is non-trivial as $\pi\_1(U(n))=\mathbb{Z}$ wheareas $\pi\_1(G)$ cannot be a factor of $\mathbb{Z}$ by assumption. That is, $U(n)\not\simeq G\times M$ as $\pi\_1(U(n))\not\simeq \pi\_1(G)\times \pi\_1(M)$. $\square$
This holds for e.g. compact Lie groups with the appropriate fundamental group; it seems likely that this argument can be strengthened by e.g. considering higher homotopy groups or using other results on the existence of faithful representations.
---
Added (7/9/2010): I don't know why I didn't mention it before, but replacing "unitary" with "complex" in the theorem above gives the same result for e.g. complex connected semisimple Lie groups, by an identical proof. In this case the manifold $M$ constructed in the proof cannot be guaranteed to be compact however.
| 6 | https://mathoverflow.net/users/6950 | 29516 | 19,272 |
https://mathoverflow.net/questions/29520 | 7 | Looking at <http://en.wikipedia.org/wiki/Triangle_group> I begin to wonder why the definition explicitly excludes the tessellation of the Euclidean plane by 30-30-120 triangles? In terms of the Wallpaper groups, I am thinking of the group p6 ( <http://en.wikipedia.org/wiki/Wallpaper_group#Group_p6> ).
Is it just an exceptional case? The definition of the triangle group asks that the "order" at each vertex to be even, which is natural, as for odd orders only isosceles triangles can "close" under reflections. So if one of the angles of a tessellating triangle is $2\pi / k$ for $k$ odd, it is necessary that there exists some integer $l$ such that $$\frac{1}{k} + \frac{2}{l} = \frac12$$
and the only solution is $k = 3$ and $l = 12$.
But for less rigid geometries (say hyperbolic), this seems to introduce a large number of additional tessellations. (Though not really more groups, I think, since geometrically replacing the obtuse triangle by two congruent acute triangles leads to one of the already defined cases.)
*Motivation: I am trying to describe an arts/craft project for demonstrating hyperbolic geometry. As such it is more natural to start from the tessellation picture, rather than the reflection group picture. Therefore it was a bit surprising to me that despite the introductory paragraph on Wikipedia, the two points of view are not exactly the same.*
My apologies for the somewhat muddy wording of the question.
| https://mathoverflow.net/users/3948 | Why does the triangle groups not include a tiling by 30-30-120 triangles? | The answer is already contained in your question.
You do not describe more reflection groups, since replacing the obtuse triangle by two congruent acute triangles leads to one of the already defined symmetry groups.
Not directly related to your question: here are two cool java applet having to do with hyperbolic tesselations.
[Applet 1](http://www.plunk.org/~hatch/HyperbolicApplet/).
[Applet 2](http://aleph0.clarku.edu/~djoyce/poincare/PoincareApplet.html).
| 6 | https://mathoverflow.net/users/5690 | 29524 | 19,276 |
https://mathoverflow.net/questions/29528 | 15 | Can some square of side length greater than $n$ be covered by $n^2+1$ unit squares? (The unit squares may be rotated. The large square and its interior must be covered.)
| https://mathoverflow.net/users/7086 | covering a square with unit squares | This reference is certainly pertinent, being the second Google hit for "covering a square with squares" (after your question). Just reading it now...
<http://www.uccs.edu/~faculty/asoifer/docs/untitled.pdf>
UPDATE: so far as I can tell from looking at this article, the author regards your question as an unsolved problem. There is a further article by him and Karabash, apparently in (his) journal Geombinatorics, vol. 18, which I cannot access online and which has not been reviewed on MathSciNet.
| 8 | https://mathoverflow.net/users/5575 | 29530 | 19,281 |
https://mathoverflow.net/questions/29281 | 11 | This question is motivated by the ongoing discussion under my answer to [this](https://mathoverflow.net/questions/29271/algebraic-geometry-used-externally-in-problems-without-obvious-algebraic-struc) question. I wrote the following there:
>
> A $(p, q, r)$ Steiner system is a collection of $q$-element subsets $A$ (called blocks) of an $r$-element set $S$ such that every $p$-element subset of $S$ is contained in a unique element of $A$. Good examples come from considering as blocks the set of hyperplanes in $\mathbb{A}^n$ or $\mathbb{P}^n$ over a finite field. For example, $\mathbb{A}^2$ over $\mathbb{F}\_3$ gives a $(2, 3, 9)$ Steiner system: it contains $9$ ($\mathbb{F}\_3$-rational) points, and let the blocks be the lines, each of which consists of $3$ points. Then any $2$ points are contained in a unique line. This is the unique $(2, 3, 9)$ Steiner system.
>
>
>
In general, considering lines in $\mathbb{A}^n$ or $\mathbb{P}^n$ gives an analogous Steiner system, and papers such as [this one](http://www.springerlink.com/content/62157221j2718857/) contain similar constructions.
Loosely, my question is: which Steiner systems come from similar constructions? I'll make this more precise in a bit.
An artificial construction allows us to realize any Steiner system as the points of a variety in $\mathbb{A}^n$ over $\mathbb{F}\_2$, the blocks of which are given by the intersection of the variety with some specified hyperplanes, as follows. (This construction is due to Jeremy Booher.) Say we have a $(p,q, r)$ Steiner system with $k$ blocks; consider the subvariety of $\mathbb{A}^k$ containing the point $y\_j=(a\_i)\_{1\leq i\leq k}$ with $a\_i=0$ if the $j$-th element in our Steiner system is in block $i$ and $1$ otherwise. Then the intersections with the hyperplanes $x\_i=0$ give our blocks. And this subset is a variety as any subset of $\mathbb{A}^k$ is a variety, as it is finite.
>
> So let us try for something harder: Which Steiner systems $(p, q, r)$ come from a subvariety $X$ of $\mathbb{A}^n$ or $\mathbb{P}^n$ containing $r$ ($\mathbb{F}\_s$-rational) points, with the blocks given as the intersections with *all* $p+1$-dimensional hyperplanes? A slightly weaker version: when is there a subvariety $X$ of $\mathbb{A}^n$ or $\mathbb{P}^n$ with $r$ $\mathbb{F}\_s$-rational points such that every $p+1$-plane intersects $X$ at $q$ points? In particular, this requires that any $p$ points in $X$ be in general position, so things like rational normal curves are natural candidates.
>
>
>
This is probably too hard, so perhaps the simpler question is tractable:
>
> Can you prove that some Steiner system does not come from this construction?
>
>
>
EDIT: One way of doing this might be to find a Steiner system with $b$ blocks, where $b$ is not a $q$-binomial coefficient with $q$ a prime power; such systems exist for $p=1$ but I am looking for a non-trivial example.
| https://mathoverflow.net/users/6950 | Which Steiner systems come from algebraic geometry? | It seems that no Steiner System of the form $(2, 3, 25)$ can be represented in this fashion---many such systems do exist; see [here](http://mathworld.wolfram.com/SteinerTripleSystem.html). In particular, such a system would contain $100$ blocks; but no Grassmannian of lines in $\mathbb{A}^n$ or $\mathbb{P}^n$ over a finite field contains $100$ points.
However, this leaves the possibility of the following (less appealing) construction. Say a scheme $S$ is $k$-rigid for a scheme $Y$ if for any $k$ points of $Y$, $S$ embeds uniquely in $Y$ (up to automorphisms of $S$) so that it passes through all $k$ points. (This seems like a natural definition; does it already have a name?) Then we may ask for varieties $S; T\subset U$ with $S$ $p$-rigid for $U$ such that any embedding of $S$ in $U$ intersects $T$ at $q$ points, where $T$ has $r$ points.
| 5 | https://mathoverflow.net/users/6950 | 29535 | 19,284 |
https://mathoverflow.net/questions/29419 | 16 | ``Proofs without words'' is a popular column in the Mathematics magazine.
Question: What would be a nice way to characterize which assertions have such visual proofs? What definitions would one need?
I suspect that in order to make this question precise, one will have to define a computational model for the ``visual verifier'' and postulate the possibility of a visual proof if there is a quick verification algorithm, and the visual proof itself is short.
Rev. 1:
To elaborate: the intriguing and essential feature of a visual proof is that the proof can be ``read'' or verified rather quickly using primitive operations that are different than those involved in reading and verifying textual proofs (e.g., area comparison or counting). These are operations that are native and quick for the visual system. (There may be other operations that could be used as well: color, texture, shape, or any other visual cue could be part of the repertoire.) It is not necessary that for an assertion to have a visual proof, that it involve inequalities, though this feature does lend itself relatively easily to a direct visual proof.
Thus, one could frame the notion of a visual proof as follows: An assertion has a visual proof if there is a map from the space of assertions (strings over an alphabet) to the space of pictures (strings over a visual alphabet) such that assuming a particular computational model of a visual verifier, there is an algorithm for reading the visual proof that is faster than the best one for reading the textual proof.
The computational model of the verifier allows one to specify what visual primitives may be used and how ``costly'' they are in the algorithm. (For example, if a proof is mapped to a picture that only uses length of a line to represent elements in the proof, then reading the proof will likely be slow, whereas if it uses length and area, the proof might be immediately verified.)
This leads to a sort of complexity theory of visualizations. In fact, it implies that visualization is a special case of a more general activity: re-encoding a given problem so that one can exploit a fast algorithm.
There may be some assertions that have no maps that can be verified quicker than the original proof. What might these be?
I hope this provides a better context for my question.
Thank you for your comments in advance. I appreciate your attention and intellectual work.
| https://mathoverflow.net/users/7048 | Characterizing visual proofs | Here is a complexity theory perspective. Be warned that it may differ wildly from someone whose primary focus is logic.
I think the appropriate definition of a "visual proof" would boil down to giving an appropriate definition of what a verifier does with such a proof. Proof systems in complexity theory are measured by (a) how much the prover and verifier "interact", (b) the allowed lengths of potential proofs, and (c) the power of the verifier.
In the framework you are proposing, the prover simply gives the verifier a proof and walks away. So (a) is already determined. Also, the visual proof is presumably written on a small sheet of paper, so (b) is essentially determined (let's say the visual proof can be encoded in length that is at most a fixed polynomial in the length of the claim).
That leaves (c), which is where proof complexity gets interesting. It turns out that verifiers can be surprisingly weak and still verify the proof of any statement which has short proofs (where "short" is "fixed polynomial"). For example, if you require that proofs be written on a two-dimensional grid, then for every theorem with short proofs, there are proofs of the theorem which can be verified by a *two-dimensional finite automaton*, see
>
> J. Hartmanis, D. Ranjan R. Chang, and P. Rohatyi. On IP = PSPACE and theorems with narrow proofs. EATCS-Bulletin, 41:166–174, 1990.
>
>
>
A verifier could be a "streaming" algorithm: it could be randomized, go over the proof in just one pass, and use a tiny amount of workspace relative to the length of the proof, see
>
> Richard J. Lipton: Efficient Checking of Computations. STACS 1990: 207-215
>
>
>
The famous PCP theorem (of Arora et al.) tells us that the verifier could even be a "spot checker" which is randomized and only probes the proof at a constant number of points (which depend on the random coin tosses).
All of these are effectively different ways of characterizing the class **NP**: the polynomial time verifier in the definition of **NP** can be replaced with verifiers of the above kind.
So I believe that a good characterization of "visual proof" would turn out to give yet another way in which a simple verifier can check the proof of a theorem. However it is natural to think that maybe not *all* theorems with short proofs should have short visual proofs, so perhaps it is too ambitious to think that all of **NP** should have "visual proofs". Hence your definition problem will be a delicate combination of figuring out what the verifier should be able to do in a visual proof, and what kinds of true statements should admit such proofs. Good luck!
**Addendum** (added 7/1/10). The following neat paper on "Approximate Testing of Visual Properties" (doi: [10.1007/978-3-540-45198-3\_31](https://doi.org/10.1007/978-3-540-45198-3_31)) by Sofya Raskhodnikova looks very relevant: <http://people.csail.mit.edu/sofya/pixels.pdf>
| 5 | https://mathoverflow.net/users/2618 | 29538 | 19,286 |
https://mathoverflow.net/questions/7857 | 7 | The definition I'm going to give isn't quite the concept I really want, but it's a good approximation. I don't want to make the definition too technical and specific because if there's a standard name for a slightly different definition, then I want to know about it.
Let $(X,\mu)$ be a measure space, and let $\rho$ be a probability measure on $X$. I call a subset $A$ of $X$ **special** if for all measurable $B\subseteq X$,
1. $\mu(B)\leq\mu(A)$ implies $\rho(B)\leq\rho(A)$, and
2. $\mu(B)=\mu(A)$ and $\rho(B)=\rho(A)$ implies $B=A$ up to measure zero (with respect to both $\mu$ and $\rho$).
What is the standard name for my "special" sets? Equivalently, one could stipulate $\mu(A)\leq\beta$ and call $A$ "special" if it is essentially the unique maximizer of $\rho(A)$ given that constraint.
Also equivalently, we could stipulate a particular $\rho$-measure and consider sets achieving that $\rho$-measure having the smallest possible $\mu$-measure. That's probably the most intuitive way to think about this: **we're looking for sets that contain a certain (heuristically: large) fraction of of the mass of $\rho$ but are as small as possible (with respect to $\mu$).** That seems like a completely natural and obvious concept, which is why I think it should have a standard name. But I have almost no training in statistics, so I don't know what the name is.
This example might be far-fetched, but just to illustrate: suppose the FBI has knowledge that somebody is going to attempt a terrorist attack in a certain huge city at a particular hour. They might not know where, but they might have (some estimate of) a probability distribution for the location of the attack. They want to distribute agents strategically throughout the city, but they probably don't have enough agents to cover the entire city. Let's say every agent can forestall an attack if it occurs within a certain radius of his/her position (which is unrealistic, since the number of nearby agents surely also matters, but ignore that); then, to maximize the probability that the attack will be stopped, to an approximation, they should distribute their agents uniformly over a *special* subset of the city's area. To approach this from the other perspective, it could be the case that 99% of the mass of their probability distribution is contained in a region with very small area. (The one with the smallest area will be a *special* set.) Then, to save resources, if they're okay with 99:1 odds (*c'est la vie*), they might only distribute a relatively small number of agents to that small special region.
If $\rho$ has a density $f$ with respect to $\mu$ (when it makes sense to talk about such), then special sets are closely related to the superlevel sets of $f$, i. e., sets of the form $\{x:f(x)\geq c\}$ for $c\geq 0$. (I think they're basically the same, but specialness of $A$ is unaffected by changing $A$ by a set of measure zero, so a superlevel set actually corresponds to an equivalence class of special sets.) I mention this here because (1) the connection to superlevel sets is one of my reasons for caring about specialness, and (2) "superlevel sets of the density" is not the answer I'm looking for.
---
**Example 1**
Here's a very simple example in which special sets can be completely characterized. Let $X=\{x\\_1,\ldots,x\\_n\}$ be a finite set, and let $\mu$ be counting measure on $X$. Let $\rho$ be any probability distribution on $X$, which necessarily has a density function $f:X\to\mathbb{R}\\_+$, so by definition, $f(x\\_1) + \ldots + f(x\\_n) = 1$ and $\rho(A) = \sum\\_{x\in A} f(x)$. Suppose that no two points have the same $f$-value; then, without loss of generality, $f(x\\_1) > f(x\\_2) > \ldots > f(x\\_n)$. It's easy to see that the special sets in this setup are exactly the sets $A\\_k = \{x\\_1,x\\_2,\ldots,x\\_k\}$, i. e., which contain the largest $k$ points as measured by $\rho$, for $k=0,\ldots,n$. (Why: if you have some other candidate special set $B$, then $A\\_{\\#B}$ has the same $\mu$-measure as $B$ but higher $\rho$-measure, so $B$ can't be special.) It's easy to generalize this example to the case in which $f$ isn't necessarily one-to-one: you have to treat all points with the same $f$-value as a block: either all of them are in the special set, or none of them are. (Otherwise, there's no way to satisfy the "uniqueness" part (point 2) of the definition.)
---
**Example 2**
Here's a generalization of the first example that hopefully clarifies what I said above. Let $(X,\mu)$ be some nice measure space on which integration of functions makes sense (like a Riemannian manifold, or just $\mathbb{R}^d$). Let $f:X\to\mathbb{R}\\_+$ be a nonnegative integrable function with $\int\_X f(x) d\mu = 1$, and let $\rho$ be the probability measure $\rho(Y) = \int\_Y f(x) d\mu$, so $f$ is the density of $\rho$ with respect to $\mu$. Fix some $c\geq 0$ and let $A=\{x:f(x)\geq c\}$.
>
> Claim: $A$ is a special set.
>
>
>
Proof: It suffices to show that if $\mu(B) = \mu(A)$, then $\rho(B)\leq \rho(A)$, with equality if and only if $B$ and $A$ differ by a set of measure zero. If $\mu(B) = \mu(A)$, then $\mu(B-A) = \mu(A-B)$. Now we write
$\begin{align\*} \rho(A) - \rho(B) &= \int\\_A f(x) d\mu - \int\\_B f(x) d\mu \\\\
&= \int\\_{A-B} f(x) d\mu - \int\\_{B-A} f(x) d\mu \\\\
&= \int\\_{A-B} f(x) d\mu - \int\\_{A-B}c\\,d\mu - \int\\_{B-A} f(x) d\mu + \int\\_{B-A}c\\,d\mu\\\\
&= \int\\_{A-B} (f(x)-c) d\mu - \int\\_{B-A} (f(x)-c) d\mu.
\end{align\*}$
By construction, $f(x) \geq c$ on $A$ and $f(x) < c$ on $B-A$, so the first integral is nonnegative and the second integral is nonpositive, and is in fact negative unless $\mu(B-A)=0$, in which case $\mu(A-B)=0$ as well. Thus, $\rho(A)-\rho(B)\geq 0$, with strict inequality unless $A$ and $B$ differ by measure zero, QED.
| https://mathoverflow.net/users/302 | What's the standard name for sets of a given size with maximal probability (or a given probability and minimal size)? | In statistics, especially Bayesian statistics, what you are looking for is
called HPD-region or "highest posterior density region", which is meant to be a set with
minimum volume for a given (large) (posterior) probability.
| 5 | https://mathoverflow.net/users/6494 | 29551 | 19,293 |
https://mathoverflow.net/questions/29373 | 13 | (Apologies in advance for any imprecision in the following; I am a computer scientist and regret never having taken an actual course on topology.)
One way to define the pure braid group $P\_n$ is as follows: consider a pure braid to be a set of $n$ non-intersecting arcs in $x,y,t$-space which are monotone in the $t$ direction, such that the $i$th arc connects $(i,0,0)$ to $(i,0,1)$. Sets of arcs which can be deformed continuously into each other without any arcs intersecting are considered equivalent pure braids.
Consider a generalization where the endpoints are not integer positions on two lines, but arbitrary fixed positions in the $t = 0$ and $t = 1$ planes. It seems clear to me that this does not change the topology of the space [1], so I will call such a collection of arcs an *embedded braid*. I'm looking at characterizing "optimal" embedded braids that minimize a certain functional $F$. For concreteness, let $F$ be the total length of the arcs (the actual functional I need to use is slightly different, but this should be close enough to carry over the results).
If the requirement that arcs do not intersect is removed, the space of embedded braids can be given an affine structure over which $F$ is convex, and it is immediately apparent that there is a single local minimum which is the global minimum: connect each pair of endpoints with a straight line. What can we say about the local minima of $F$ if we retain the non-intersection property? My intuition says that each topologically distinct braiding (corresponding to a particular element of $P\_n$) forms a connected component with a unique local minimum, but I cannot tell how to begin proving this. It would be true if one could show that the connected components are convex subsets of the space; this does not hold under the "obvious" affine structure where we treat each arc and each $t$ independently, but that leaves the possibility of some other choice of affine structure which works. Are there some nice proof techniques that would help proving uniqueness of local minima in this context? Would any ideas or analogies from the theory of minimal surfaces help?
As Andrew Stacey mentions, the connected components are *open* subspaces; I believe existence of local minima can be guaranteed by considering the *closure* of the component seen as a subset of the space of embeddings that allow intersection. (Following Kevin Walker's comment, I realize this is called compactification.) This would include embeddings "on the boundary", whose arcs can intersect but which are only an infinitesimal displacement away from non-intersecting embeddings with the right topology. As a concrete example, the following braid,
```
\ /
\ /
\
/ \
\ /
\
/ \
/ \
```
on being pulled tight, approaches a configuration where the arcs are infinitesimally close together in the middle; the minimal length is attained by the intersecting embedding where the two arcs share a common point.
[1] Since the endpoints are fixed, we can associate each such generalized braid uniquely with a pure braid by shrinking the $t$ dimension slightly and composing each arc $i$ with a fixed arc from its endpoints to $(i,0,0)$ and $(i,0,1)$ respectively.
| https://mathoverflow.net/users/nan | Minimal-length embeddings of braids into R^3 with fixed endpoints | UPDATE.
I revisited the question and realized that verifying the local CAT(0) property is not that easy. When I wrote the original answer, I was under impression that removing any collection of codimension 2 subspaces (more precisely, their tubular neighborhoods) from $\mathbb R^n$ leaves one with a locally CAT(0) space. This is true in $\mathbb R^3$ but there are counterexamples in $\mathbb R^4$. *This* particular subset might satisfy the condition but this does not seem to follow from any "generic" argument.
Further, there are some discouraging examples. First, approximating by arcs by "ropes" with fixed-size square sections does not actually work: in this case there are non-unique minimal configurations. So one really needs to deal with zero-width ones. Second, if you consider 3 arcs and allow two of them intersect while deforming the braid (but still disallow intersections with the 3rd one), then again, you can have two distinct minimal configurations in the same equivalence class.
Below is the original answer (and I suggest that it is unaccepted).
---
I can prove uniqueness of a local minimum for another length-like functional
(similar but not equal to the sum of lengths). I believe that it should work the same way
for the sum of lengths, but unfortunately the underlying geometric theory does not
seem to exist (yet?).
First let me reformulate the problem. Let $X$ denote the set of possible horisontal cross-sections of braids.
This is the set of $n$-tuples of distinct points in $\mathbb R^2$. Geometrically this is $(\mathbb R^2)^n$
minus a collection of codimension 2 subspaces corresponding to positions where some two points coincide.
Actually I prefer another formalization: the arcs forming the braid are ropes of nonzero width
and with square cross-sections.
More precisely, the horizontal section of every rope is an $\varepsilon\times\varepsilon$ square (with sides parallel
to the coordinate axes), and these sections should not overlap. So $X$ is $\mathbb R^{2n}$ minus a union of polyhedral
heighborhoods of codimension 2 subspaces. This formalization makes the local structure simpler,
and the original one is the limit as $\varepsilon\to 0$.
An embedded braid is a path $f:[0,1]\to X$. We want to minimize the functional
$$
L = \int\_0^1 \left(\sqrt{(df\_1/dt)^2+1}+\dots+\sqrt{(df\_n/dt)^2+1}\right) dt,
$$
over all paths between two given positions $a,b\in X$, in a given connected component
of the space of such paths. Here $f\_i=f\_i(t)$ are 2-dimensional coordinates.
Another functional,
$$
L' = \int\_0^1 \sqrt{(df\_1/dt)^2+\dots+(df\_n/dt)^2+1}\ \ dt,
$$
is easier to deal with, because this is the Euclidean length of the corresponding
path $(f(t),t)$ in $X\times\mathbb R\subset \mathbb R^{2n+1}$. Let us work with $L'$.
The connected component of the set of paths is the homotopy class. Fixing the homotopy class of a path is the same as
fixing endpoints of its lift to the [universal cover](http://en.wikipedia.org/wiki/Covering_space) of the space.
(In the case of zero-width ropes, you first take the universal cover and then the completion in order to add "boundary cases".)
And local length minimizers are called geodesics. So we want to show that, in the universal cover of $X\times\mathbb R$,
every pair of points is connected by a unique geodesic.
Our space $X$ (and hence $X\times\mathbb R$) is a locally [CAT(0) space](http://en.wikipedia.org/wiki/CAT%28k%29_space).
This can be shown using standard curvature tests for polyhedral spaces.
A globalization theorem says that a complete, simply connected, locally CAT(0) space is globally CAT(0),
in other words, it is a Hadamard space. So the universal cover is a Hadamard space.
Hadamard spaces feature uniqueness of geodesics, hence the result.
Now what about the original functional $L$? It is also a length functional, but for a non-Euclidean norm on $\mathbb R^{2n+1}$.
So, to carry over the proof, one needs to develop an analog of CAT(0) spaces modelled after Finsler spaces rather than Riemannian. I think it should be possible - after all, all this CAT(0) business is about convexity of distance, and this convexity is there in all normed vector spaces.
But I have not heard of such generalizations (maybe this is just my ignorance).
| 7 | https://mathoverflow.net/users/4354 | 29557 | 19,297 |
https://mathoverflow.net/questions/29490 | 39 | It is well-known that the number of surjections from a set of size n to a set of size m is quite a bit harder to calculate than the number of functions or the number of injections. (Of course, for surjections I assume that n is at least m and for injections that it is at most m.) It is also well-known that one can get a formula for the number of surjections using inclusion-exclusion, applied to the sets $X\_1,...,X\_m$, where for each $i$ the set $X\_i$ is defined to be the set of functions that never take the value $i$. This gives rise to the following expression: $m^n-\binom m1(m-1)^n+\binom m2(m-2)^n-\binom m3(m-3)^n+\dots$.
Let us call this number $S(n,m)$. I'm wondering if anyone can tell me about the asymptotics of $S(n,m)$. A particular question I have is this: for (approximately) what value of $m$ is $S(n,m)$ maximized? It is a little exercise to check that there are more surjections to a set of size $n-1$ than there are to a set of size $n$. (To do it, one calculates $S(n,n-1)$ by exploiting the fact that every surjection must hit exactly one number twice and all the others once.) So the maximum is not attained at $m=1$ or $m=n$.
I'm assuming this is known, but a search on the web just seems to lead me to the exact formula. A reference would be great. A proof, or proof sketch, would be even better.
**Update.** I should have said that my real reason for being interested in the value of m for which S(n,m) is maximized (to use the notation of this post) or m!S(n,m) is maximized (to use the more conventional notation where S(n,m) stands for a Stirling number of the second kind) is that what I care about is the rough size of the sum. The sum is big enough that I think I'm probably not too concerned about a factor of n, so I was prepared to estimate the sum as lying between the maximum and n times the maximum.
| https://mathoverflow.net/users/1459 | How many surjections are there from a set of size n? | It seems to be the case that the polynomial $P\_n(x) =\sum\_{m=1}^n
m!S(n,m)x^m$ has only real zeros. (I know it is true that $\sum\_{m=1}^n
S(n,m)x^m$ has only real zeros.) If this is true, then the value of $m$
maximizing $m!S(n,m)$ is within 1 of $P'\_n(1)/P\_n(1)$ by a theorem of
J. N. Darroch, *Ann. Math. Stat.* **35** (1964), 1317-1321. See also
J. Pitman, *J. Combinatorial Theory, Ser. A* **77** (1997), 279-303.
By standard combinatorics
$$ \sum\_{n\geq 0} P\_n(x) \frac{t^n}{n!} = \frac{1}{1-x(e^t-1)}. $$
Hence
$$ \sum\_{n\geq 0} P\_n(1)\frac{t^n}{n!} = \frac{1}{2-e^t} $$
$$ \sum\_{n\geq 0} P'\_n(1)\frac{t^n}{n!} = \frac{e^t-1}{(2-e^t)^2}. $$
Since these functions are meromorphic with smallest singularity at $t=\log 2$,
it is routine to work out the asymptotics, though I have not bothered to
do this.
**Update.** It is indeed true that $P\_n(x)$ has real zeros. This is because
$(x-1)^nP\_n(1/(x-1))=A\_n(x)/x$, where $A\_n(x)$ is an Eulerian polynomial. It
is known that $A\_n(x)$ has only real zeros, and the operation $P\_n(x)
\to (x-1)^nP\_n(1/(x-1))$ leaves invariant the property of having real
zeros.
| 55 | https://mathoverflow.net/users/2807 | 29564 | 19,302 |
https://mathoverflow.net/questions/29562 | 7 | I need a recommendation letter on my teaching. I want to ask the instructor in last semester for which I was a TA, but I don't know how his impression for my teaching. So do I need to ask him for his opinion about my teaching before letting him write the recommendation letter?
| https://mathoverflow.net/users/2391 | recommendation letter for teaching | Yes. Don't be shy because you think you are asking him to write you a better letter, because in fact you are asking him whether he feels he is capable of writing you a good letter. The contrary could mean not that he doesn't like you, but that he doesn't have enough of an opinion to write a letter that would not be damning with faint praise, and if he declines and is a tactful person he would give that as his reason no matter what.
I have both personal and anecdotal experience with this situation. Once, I asked Barry Mazur for a letter when applying for an NSF graduate fellowship, and we did talk, but our contact had been pretty minimal at that time (it still is, really) and he said exactly what I wrote above: that he *could* write a letter but it wouldn't say much specific, so if I had someone who knew me better then I would be better served to ask them.
As for the anecdote, I will just use (possibly incorrect) pronouns as identifiers. At a "seminar" for applying for jobs, this person described how he once agreed to write a letter for a student who was highly regarded by others, and only later realized that he couldn't actually recommend her well. Having agreed to write the letter he produced one which was adequate but no more, and the student didn't get as good a job as it was generally felt she deserved. The writer described his guilt at doing this, because it was not his obligation to agree to write a letter for someone he couldn't give the best recommendation they could get.
| 13 | https://mathoverflow.net/users/6545 | 29566 | 19,304 |
https://mathoverflow.net/questions/29494 | 10 | What algorithms are used in modern and good-quality random number generators?
| https://mathoverflow.net/users/7081 | Pseudo-random number generation algorithms | Don't miss [this wonderful post](http://groups.google.com/group/comp.lang.c/msg/e3c4ea1169e463ae) by Marsaglia. He's not a fan of the Mersenne Twister and offers some strong PRNGs with exceptionally small code footprints. One of his examples is:
```
static unsigned long
x=123456789,y=362436069,z=521288629,w=88675123,v=886756453;
/* replace defaults with five random seed values in calling program */
unsigned long xorshift(void)
{unsigned long t;
t=(x^(x>>7)); x=y; y=z; z=w; w=v;
v=(v^(v<<6))^(t^(t<<13)); return (y+y+1)*v;}
```
| 11 | https://mathoverflow.net/users/25 | 29580 | 19,315 |
https://mathoverflow.net/questions/29591 | 8 | In this recent question [Math puzzles for dinner](https://mathoverflow.net/questions/29323/math-puzzles-for-dinner/29343#29343) we had a nice time as we were asked to provide new maths puzzles for dinners. I suggested the following:
>
> Given three equal sticks, and some
> thread, is it possible to make a rigid
> object in such a way that the three
> sticks do not touch each other? (all
> objects are 1 dimensional; sticks are
> straight and rigid, and the thread is
> inestensible).
>
>
>
I'm not particularly fond of maths puzzles, and I found this one in order to satisfy the party in a dinner of non-mathematic people, when I'm asked for a puzzle. Indeed, it's suitable for a dinner, and people get to the right conclusion in reasonable time, amusing themselves and disputing a bit.
Possibly as a consequence of the fact that I always proposed it to non-mathematicians, who are not interested in the proof, now I realized that I do not have, or I forget, the proof for the answer (a bit embarassing indeed). Can somebody find a quick proof for the answer, without too many technical computations?
| https://mathoverflow.net/users/6101 | Sticks and thread | Instead of a proof, I will provide references. It's called a "tensegrity prism". See especially sections 1.4, 3.5 and 3.6 of [*Dynamics and Control of Tensegrity Systems*](http://books.google.com/books?id=LxItccmYhBkC). Also see ["Review of Form-Finding Methods for Tensegrity Structures"](http://www-civ.eng.cam.ac.uk/dsl/publications/tensegrity.pdf) and the MS thesis [*Kinematic Analysis of Tensegrity Structures*](http://scholar.lib.vt.edu/theses/available/etd-12022002-155829/).
| 9 | https://mathoverflow.net/users/1847 | 29599 | 19,327 |
https://mathoverflow.net/questions/29441 | 13 | I'm interested in a criterion that determines whether a linear scalar PDE (arbitrary order) has a unique solution given vanishing boundary conditions at spatial infinity. I'll try to formulate the question more precisely below.
Consider a PDE of the form $L[u]=0$ where $u(t,x,y,z)$ is a scalar function of one time $(t)$ and three spatial variables $(x,y,z)$, though this choice of dimensionality is not central to the question. The function $u$ is required to vanish "sufficiently" fast if the $(x,y,z)$ variables are taken to infinity, keeping $t$ fixed. [If that's not enough, it can also be required to vanish at infinity along any hyperplane that is space-like with respect to the Lorentzian metric $\mathrm{diag}(-1,1,1,1)$.] However, no requirements are put on the behavior of $u$ as $t\to\pm\infty$ for fixed $(x,y,z)$. The linear differential operator $L$ can be assumed to have constant coefficients, but could by of any order. Though, I'd also like to know how the answer generalizes to the case when the coefficients and the background Lorentzian metric are no longer constant.
>
> So, my question is this: for which operators $L$ does the equation $L[u]=0$ have a unique solution?
>
>
>
Let me give some examples.
* Equation $\partial\_z u=0$ has a unique solution. An arbitrary solution comes from integrating the rhs wrt to $z$ and adding any function that's constant wrt to $z$. From the boundary conditions, it is easy to see that both pieces must be zero. Hence, $u=0$ is the unique solution.
* The same argument does not work for $\partial\_t u=0$. For any given solution, I can get another solution by adding a function of $(x,y,z)$ only that vanishes at infinity, and there are plenty of those.
* The equation $(\partial\_x^2+\partial\_y^2+\partial\_z^2)u=0$ is uniquely solvable: ignore $t$ dependence and invert the Laplacian, with uniqueness given by the same argument as in the first example.
* The equation $(-\partial\_t^2+\partial\_x^2+\partial\_y^2+\partial\_z^2)u=0$ is not uniqely solvable: solutions are parametrized by Cauchy data on, say, the $t=0$ hyperplane, and so are definitely not unique.
These examples make me think that the answer is some version of an ellipticity condition. Unfortunately, I'm only aware of how to formulate this condition for second order systems. Any help appreciated!
**Status Update:** Willie Wong provided some good, relevant information below. Let me summarize my understanding of it here and then sharpen my question in light of it.
If the partial differential operator (PDO) $L$ contains no time derivatives, then the Fourier transform $\hat{u}$ of a nontrivial solution $u$ must be supported on the zero set of $P(\xi)$, the symbol of $L$. If this set is of measure zero, then $\hat{u}$ can only be a distribution. On the other hand, local regularity of $\hat{u}$ is controlled by the decay of $u$ at infinity. In particular, if $u$ is in the Schwarz space, then $\hat{u}$ cannot be sufficiently singular to be a distribution. Hence, $L[u]=0$ would admit only the trivial solution, and hence be uniquely solvable.
In principle, I can use the same argument for any $L$ that is expressible only in terms of derivatives parallel to a given space-like hyperplane, by appealing to the fact that I've imposed the same boundary conditions at infinity (say Schwarz) for each such hyperplane.
In principle, the above reasoning gives a nice large space of PDOs that satisfy my criteria. But are there any more? I think there are (see below).
Now, suppose that I cannot ignore time derivatives. Willie's suggestion is to write the equation $L[u]=0$ in evolution form $\partial\_t v + Av=0$ and exclude $L$'s for which the evolution equation is well posed as an initial value problem. But not all such $L$ can be excluded, since not all initial data generates solutions that satisfy the boundary conditions (decay at infinity along any space-like direction). I'm thinking that there should be a geometric condition involving the background null cone and the characteristics of $L$ or the zero set of $P(\xi)$. For instance, if $L=-c^{-2}\partial\_t^2+\partial\_x^2$ with $c>1$, then the corresponding equation has infinitely many solutions parametrized by Cauchy data on $t=0$. However, these solutions would correspond to waves pulses propagating along the $x$-axis at a speed faster than the background speed of light. On the other hand, the solution $u(t,ct)$ must vanish for large $t$ as $(1,c)$ is a space-like vector. I believe this is enough to show that this $L$ also admits only trivial solutions. My reasoning here is based on the exact representation of solutions via the D'Alambert formula, but that doesn't generalize easily. Any idea what kind of geometric condition could be used for more general operators?
Ultimately, I'd like to know something about the geometry of the space of operators that satisfy my criteria. Say I fix the maximal order to make things easier. The space is clearly not linear, but could it be convex or the complement of a convex set? (These last guesses are probably not right. I'm just throwing out ideas.) I'd be happy if I could understand this space for just first and second order operators, preferably with hints of how this understanding could generalize to higher orders.
| https://mathoverflow.net/users/2622 | PDEs, boundary conditions, and unique solvability | Hi, I am adding another answer because this suggests a rather different approach then what I have outlined before, and this is targeted at the fact you are willing to grant smooth with compact support on any space-like hyperplane.
If you are willing to let your solutions vanish in such a large set, then the proper tool for the analysis is the theory of unique continuation for solutions to linear differential equations. For example, solutions to elliptic equations tend to have the property of strong unique continuation: if the solution vanish on any open set it must vanish everywhere. Using the geometry you can weaken ellipticity to have cases where no derivatives are taken in certain directions ($\partial\_x^2 + \partial\_y^2$, for example).
Similarly you can address parabolic type operators. (They have infinite speeds of propagation as well.)
For hyperbolic type, if the coefficients of the operator are real analytic, then you can use Holmgren's Uniqueness theorem if the inner-most characteristic cone of the operator remains outside of the Minkowskian null cone: the compact-support criterion on space-like (to back ground metric) slices will guarantee that there will be a time-like surface (relative to the differential operator, not to the background metric) to one side of which the solution vanishes, then Holmgren kicks in and tells you that the solution must vanish everywhere.
(A version of Holmgren's also applies to ultrahyperbolic operators, with suitable changes in the geometry.)
Outside of these trivial cases, you need to actually consider the geometry of the characteristic cones and compare them against the geometry of the support set of your function. The keywords you will be looking for are Carleman estimates, unique continuation, and pseudo-convexity. There's a nice, comprehensive treatment in Lars Hormander's *Analysis of Linear Partial Differential Operators* (I think volume III, may be IV?); there are also some nice notes by Daniel Tataru on his website.
| 3 | https://mathoverflow.net/users/3948 | 29609 | 19,328 |
https://mathoverflow.net/questions/29600 | 7 | In Penrose's book (The Road to Reality, chapter 21) he gives an example of Oliver Heaviside's observation that you can treat differential operators like numbers:
The differential equation $(1+D^2)y = x^5$ can be solved by dividing by $(1+D^2)$ then taking the power series expansion: $$y = (1-D^2+D^4-D^6+\cdots)x^5$$ which evaluates to $y = x^5 - 20x^3 + 120x$.
Apparently this can be made perfectly rigorous!
How is this done? and do you know where I can read more details about this idea?
| https://mathoverflow.net/users/4361 | Treating Differential Operators as Numbers | This is just a fact from linear algebra: if $T$ is a nilpotent transformation of a vector space $V$, then $(1-T)^{-1} = 1 + T + T^2 + \dots$. More generally, the same is true in any commutative Banach algebra (such as the endomorphism ring of a normed complex vector space) if $T$ is of norm less than 1.
In your case, the differential operator $D$ is a nilpotent operator on the vector space of polynomials, and consequently this formula applies.
As people have pointed out in the comments, you may want to look at the Fourier transform, which realizes differentiation as multiplication, which allows you to treat differentiation kind of like a number. (More precisely, there is an isomorphism $F:L^2 \to L^2$ such that $F D F^{-1}$ is equal to multiplication by $x$ on sufficiently nice (e.g. Schwarz) functions.)
In higher dimensions, for instance, this means that if $\Delta$ is the Laplacian, then multiplication by $(1 + |x|^2)^k$ corresponds to applying $(I-\Delta)^k$, and it is thus possible to define an operator $(I - \Delta)^r$ for any real $r$.
| 12 | https://mathoverflow.net/users/344 | 29611 | 19,330 |
https://mathoverflow.net/questions/29534 | 13 | I sent the following question to another forum more than a week ago but haven't got any responses. The moderator of that forum suggested that I pose the following question here:
Suppose we have an ellipse $x^2/a^2 + y^2/b^2 = 1$ (centered at the
origin). Let $n>4$. There are $n$ rays going out of the origin, at angles
$0, 2\pi/n, 4\pi/n, 6\pi/n,...,2\pi(n-1)/n$. Let $(x\_1,y\_1),...,(x\_n,y\_n)$ be
the points of intersection of the rays and the ellipse. The product
from $k=0$ to $n-1$ of $(x\_k)^2 + (y\_k)^2$ is equal to one. Can $a$ and $b$ be
rational? Note that this is obviously possible if $a=b=1$, since then the
ellipse becomes a circle of radius 1. But what about if $a$ is not equal
to $b$? Can $a$ and $b$ still be rational?
Craig
| https://mathoverflow.net/users/7089 | The product of n radii in an ellipse | So, let's finish this. Starting from Qiaochu's formula,
$$\prod\_{k=0}^{n-1} (r^2 \cos^2 (2 \pi k/n) + s^2 \sin^2 (2 \pi k)/n))=1$$.
Each factor is
$$\left(\vphantom{\frac{r}{2}} r \cos (2 \pi k/n) + i s \sin(2 \pi k/n) \right) \left(\vphantom{\frac{r}{2}} r \cos (2 \pi k/n) - i s \sin(2 \pi k/n) \right) =$$
$$\left( \frac{r+s}{2} e^{2 \pi i k/n} + \frac{r-s}{2} e^{-2 \pi i k/n} \right) \left( \frac{r+s}{2} e^{2 \pi i k/n} - \frac{r-s}{2} e^{-2 \pi i k/n} \right).$$
Putting $u=(r+s)/2$, $v=(r-s)/2$ and $\zeta=e^{2 \pi k/n}$, we have
$$\prod\_{k=0}^{n=1} (u + v \zeta^{2k}) (u - v \zeta^{2k})$$
If $n$ is odd, this is $(u^n+v^n)(u^n - v^n)= (u^{2n} - v^{2n})$. If $n$ is even, I get $(u^{n/2} - v^{n/2})^2 (u^{n/2} + v^{n/2})^2 = (u^n-v^n)^2$. So either $u^{2n} - v^{2n}=1$ or $u^n-v^n=1$ or $u^n-v^n=-1$. These are all forms of FLT.
| 13 | https://mathoverflow.net/users/297 | 29616 | 19,335 |
https://mathoverflow.net/questions/29624 | 85 | Define a *growth function* to be a monotone increasing function $F: {\bf N} \to {\bf N}$, thus for instance $n \mapsto n^2$, $n \mapsto 2^n$, $n \mapsto 2^{2^n}$ are examples of growth functions. Let's say that one growth function $F$ *dominates* another $G$ if one has $F(n) \geq G(n)$ for all $n$. (One could instead ask for *eventual domination*, in which one works with sufficiently large $n$ only, or *asymptotic domination*, in which one allows a multiplicative constant $C$, but it seems the answers to the questions below are basically the same in both cases, so I'll stick with the simpler formulation.)
Let's call a collection ${\mathcal F}$ of growth functions complete *cofinal* if every growth function is dominated by at least one growth function in ${\mathcal F}$.
Cantor's diagonalisation argument tells us that a cofinal set of growth functions cannot be countable. On the other hand, the set of all growth functions has the cardinality of the continuum. So, on the continuum hypothesis, a cofinal set of growth functions must necessarily have the cardinality of the continuum.
My first question is: what happens without the continuum hypothesis? Is it possible to have a cofinal set of growth functions of intermediate cardinality?
My second question is more vague: is there some simpler way to view the poset of growth functions under domination (or asymptotic domination) that makes it easier to answer questions like this? Ideally I would like to "control" this poset in some sense by some other, better understood object (e.g. the first uncountable ordinal, the nonstandard natural numbers, or the Stone-Cech compactification of the natural numbers).
EDIT: notation updated in view of responses.
| https://mathoverflow.net/users/766 | How many orders of infinity are there? | For asymptotic domination, commonly denoted ${\leq^\*}$ and often called *eventual domination*, this has been answered by Stephen Hechler, *On the existence of certain cofinal subsets of ${}^{\omega }\omega$*, [MR360266](http://www.ams.org/mathscinet-getitem?mr=360266). What you call a *complete set* is usually called a *dominating family*.
As a poset under eventual domination, a dominating family $\mathcal{F}$ must have the following three properties:
1. $\mathcal{F}$ has no maximal element.
2. Every countable subset of $\mathcal{F}$ has an upper bound in $\mathcal{F}$.
3. $|\mathcal{F}| \leq 2^{\aleph\_0}$
Hechler showed that for any abstract poset $(P,{\leq})$ with these three properties, there is a forcing extension where all cardinals and cardinal powers are preserved, and there is a dominating family isomorphic to $(P,{\leq})$.
In particular, one can have a wellordered dominating family whose length is any cardinal $\delta$ with uncountable cofinality. In this case, the restriction $\delta \leq 2^{\aleph\_0}$ is inessential since one can always add $\delta$ Cohen reals without affecting conditions (1) and (2). However, for arbitrary posets, condition (2) could be destroyed by adding reals.
The total domination order is more complex. One can always get a totally dominating family $\mathcal{F}'$ from a dominating family $\mathcal{F}$ by adding $\max(f,n) \in \mathcal{F}'$ for every $f \in \mathcal{F}$ and $n < \omega$. Since $\mathcal{F}$ is infinite, the resulting family $\mathcal{F}'$ has the same size as $\mathcal{F}$. Howerver, there does not appear to be a simple combinatorial characterization of the possibilities for the posets that arise in this way.
| 71 | https://mathoverflow.net/users/2000 | 29626 | 19,341 |
https://mathoverflow.net/questions/29555 | 0 | $A\rightarrow B$ a ring homomorphism of Noetherian rings, where $A$ is local. $M$, $N$ finitely generated and nonzero $A$- and $B$- modules, respectively. Then I seem to get $\mbox{dim}\_ {B}(M\otimes\_{A} N) = \mbox{dim}\_ {B}N$. Could that be true? It seems a little strange that the dimension of $(M\otimes\_{A} N)$ (as a $B$-module) is independent of $M$.
| https://mathoverflow.net/users/5292 | Dimension of tensor product of modules | See Bruns and Herzog A.5(b) and A.11(b).
| 3 | https://mathoverflow.net/users/7103 | 29628 | 19,343 |
https://mathoverflow.net/questions/29635 | 19 | It is easy to prove that a model structure is determined by the following classes of maps (determined = two model structures with the mentioned classes in common are equal).
* cofibrations and weak equivalences
* fibrations and weak equivalences
The second statement follows immediately from the first by duality.
What about the following classes of maps/objects (A short argument would be very helpful)?
1. cofibrations and fibrations
2. **cofibrant objects and weak equivalences**
3. cofibrant objects and fibrations
4. cofibrant objects and fibrant objects
I think each of these classes determine the structure respectively. For the last one I suppose that one has to use framings but I cannot see how to do it.
**Edit:** Thank you all for the illuminative answers.
1. true
2. **?**
3. true
4. false
| https://mathoverflow.net/users/2625 | What determines a model structure? | This is just a flash answer without enough thought:
1. Cofibrations determine trivial fibrations (by lifting) and fibrations determine trivial cofibrations. Any weak equivalence is a composite of a trivial cofibration and a trivial fibration. So cofibrations and fibrations determine the model structure.
2. Cofibrant objects and fibrations (or weak equivalences) should NOT determine the model structure, or so my intuition says. If you know the model structure is left proper, maybe. You might be tempted to argue something like: fibrations determine trivial cofibrations, so we know all trivial cofibrations between cofibrant objects, so we ought to know all weak equivalences between cofibrant objects by something like Ken Brown's Lemma. But I bet this does not in fact work. Even if it does, I'm not sure what to do next. I'd be tempted to look for a counterexample.
3. Cofibrant objects and fibrant objects surely must not determine the model structure. It must not be too hard to come up with a counterexample for this, but I need to think about it.
| 8 | https://mathoverflow.net/users/1698 | 29641 | 19,351 |
https://mathoverflow.net/questions/29644 | 31 | The well known ["Sum of Squares Function"](http://mathworld.wolfram.com/SumofSquaresFunction.html) tells you **the number** of ways you can represent an integer as the sum of two squares. See the link for details, but it is based on counting the factors of the number N into powers of 2, powers of primes = 1 mod 4 and powers of primes = 3 mod 4.
Given such a factorization, it's easy to find the **number** of ways to decompose N into two squares. But how do you efficiently **enumerate** the decompositions?
So for example, given N=2\*5\*5\*13\*13=8450 , I'd like to generate the four pairs:
13\*13+91\*91=8450
23\*23+89\*89=8450
35\*35+85\*85=8450
47\*47+79\*79=8450
The obvious algorithm (I used for the above example) is to simply take i=1,2,3,...,$\sqrt{N/2}$ and test if (N-i\*i) is a square. But that can be expensive for large N. Is there a way to generate the pairs more efficiently? I already have the factorization of N, which may be useful.
(You can instead iterate between $i=\sqrt{N/2}$ and $\sqrt{N}$ but that's just a constant savings, it's still $O(\sqrt N)$.
| https://mathoverflow.net/users/7107 | Enumerating ways to decompose an integer into the sum of two squares | The factorization of $N$ is useful, since $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$$ There are good algorithms for expressing a prime as a sum of two squares or, what amounts to the same thing, finding a square root of minus one modulo $p$. See, e.g., <http://www.emis.de/journals/AMEN/2005/030308-1.pdf>
Edit: Perhaps I should add a word about solving $x^2\equiv-1\pmod p$. If $a$ is a quadratic non-residue (mod $p$) then we can take $x\equiv a^{(p-1)/4}\pmod p$. In practice, you can find a quadratic non-residue pretty quickly by just trying small numbers in turn, or trying (pseudo-)random numbers.
| 23 | https://mathoverflow.net/users/3684 | 29648 | 19,355 |
https://mathoverflow.net/questions/29623 | 4 | The specific question: I've got a projective variety Y and a subvariety X cut out of Y as the zero scheme of a regular section of a vector bundle E on Y. In the end, I'd like to prove that X has rational singularities... and I was hoping to try to use a theorem I found in a paper of Kovacs:
Let $\phi: Y \rightarrow X$ a surjective morphism of projective varieties over $\mathbb{C}$, with dim Y = N and dim X = n. Assume Y has rational singularities. Then X has rational singularities iff X is Cohen-Macaulay and $R^{N-n}\phi\_\* \omega\_Y \cong \omega\_X$.
I think I've got Cohen-Macaulay from the fact that X is cut out of Y as a zero scheme of a regular section. The silly part: how do I compute $R^{N-n}\phi\_\* \omega\_Y$ or any other $R^i\phi\_\* \omega\_Y$? I just realized everything I know from Hartshorne assumes nonsingularity.
(A more general question is whether there might be a better way to prove my variety X has rational singularities!)
| https://mathoverflow.net/users/7101 | Higher direct images and singular varieties | There might be many ways to prove a variety has rational singularities.
I certainly agree with Zsolt's comment above that you should be careful with your notation, Kov\'acs theorem refers to a $Y \to X$ and above you mention $X \subset Y$, I'm assuming you are simply abusing notation.
With regards to your initial question, I would try to put $R^i \phi\_\* \omega\_Y$ into a long exact sequence (so it depends on what $Y$ is mapping to $X$), see #3 below.
Anyway, here are some things I would try for a subvariety $X$ in $Y$ (obtained in some way), some of which don't use the subvariety structure.
1. Is your variety some quotient by a group action (then more info is likely needed).
2. Is $O\_X$ (locally) a summand of something with rational singularities? (Apply Boutot's theorem)
3. Sandor's result, but you may as well try with a resolution $\pi : X' \to X$ and try to show that $\pi\_\* \omega\_{X'} = \omega\_X$ (at which point, this is due to Kempf, not Kovacs) (I assume you have already shown that $X$ is CM), this is easier to compute. If you have some other $Y$ with rational singularities mapping to it in some natural way, then you might be in business via Sandor's theorem. Of course, the quickest hope for computing some higher cohomology like this is sticking it in some long exact sequence.
4. If $X$ is a divisor, you could try to show that the pair $(Y, X)$ is purely log terminal (see also Lazarsfelds's book, and adjoint ideals). In this same direction, if $X$ is NOT a divisor, you could try to show that $X$ is a minimal log canonical center of some log canonical pair and then apply Kawamata's subadjunction theorem. Of course, you could also just try to show that $X$ is log terminal directly.
5. If you have specific equations, you could also try some reduction to characteristic p techniques (like things related to F-splitting and F-rationality, some of these are very effective if you have explicit equations). Even without specific equations, some of these techniques still might be useful.
6. I suppose you could also do some Bertini type tricks if somehow this subvariety is sufficiently general (for example, a general section of a base point free linear system of something with rational singularities still has rational singularities).
7. You can also see this question: [Is there an obvious way for showing singularities are quotient?](https://mathoverflow.net/questions/23091/is-there-an-obvious-way-for-showing-singularities-are-quotient/23137#23137)
8. Does your variety have a small resoluation ($Y \to X$)? If it is also Cohen-Macaulay and normal, then it has rational singularities.
9. Does your variety have a Cartier divisor $D$ on it with log canonical (or maybe Du Bois) singularities such that $X \setminus D$ is has log terminal singularities (or maybe smooth). Then $X$ can be show to have rational singularities. Some things like this appeared in a paper of Koll\'ar and Shepherd-Barron (also see the related work of Karu as well as a paper of mine on Du Bois singularities).
That's all I can think of right now.
| 13 | https://mathoverflow.net/users/3521 | 29649 | 19,356 |
https://mathoverflow.net/questions/29633 | 3 | Define a reflexive relation on the set of zero-simplices of a simplicial set $A$ by saying that $x\sim y$ iff there is a one-simplex $h$ with $0$-face $y$ and $1$-face $x$. This is not an equivalence relation in general but it is so if $A$ is Kan. Define $\pi\_0(A)=A\_0/\sim$ in this case.
Let $x:\Delta^0\to A$ be a zero-simplex, $n\in\mathbb{N}$ and consider the pullback
\begin{array}{rcl}
A(n,x) &\to& Map(\Delta^n,A)\\
\downarrow &&\downarrow\\
Map(\partial\Delta^n,\Delta^0)&\to&Map(\partial\Delta^n,A)
\end{array}
induced by the obvious maps and set $\pi\_n(A,x)=\pi\_0(A(n.x))$ as the simplicial homotopy groups. This works since $A(n,x)$ is Kan if $A$ is so.
An important theorem states that $\pi\_n(A,x)=\pi\_n(|A|,|x|)$ the bars denoting the realization functor adjoint to the singular functor $S$. The homotopy category of the usual model structure on simplicial sets is given by inverting the "weak equivalences", i.e. the maps $f:A\to B$ such that $\pi\_n(S(f),x)$ is an isomorphism for all $n$ and all basepoints. One has to apply $S$ here to make things Kan.
Does one get the same homotopy category if one lets $\pi\_0$ be the equivalence classes of the equivalence relation **generated** by $\sim$? One can define all necessary concepts exactly as above without demanding $A$ to be Kan. Does one get the same homotopy category then?
| https://mathoverflow.net/users/2625 | Closure of the homotopy relation for a simplicial set | Of course if $\pi\_0$ is defined by the equivalence relation generated by ~ on $0$-simplices then it is the usual thing: topological $\pi\_0$ of the realization, or simplicial $\pi\_0$ of a fibrant replacement.
You are saying: What if we define a new $\pi\_n(A,x)$ as $\pi\_0(A(n,x))$? Well, obviously it maps to the usual $\pi\_n(A,x)=\pi\_0(S(A),n,x)$), and clearly this map is rarely an isomorphism if $A$ is not fibrant.
But I don't even see a comparison map between the resulting homotopy category and the usual one, in either direction. Clearly a map of simplicial sets will sometimes induce an isomorphism of usual homotopy groups while inducing a non-isomorphism of yours. But (this is my point) it can also go the other way. For example, you can make lots of examples of simplicial sets $A$ such that the inclusion $V\to A$ of the $0$-skeleton of $A$ induces an isomorphism $V(n,x)\to A(n,x)$.
| 6 | https://mathoverflow.net/users/6666 | 29655 | 19,360 |
https://mathoverflow.net/questions/29657 | 8 | If $p < q$ are primes then there is a nonabelian group of order $pq$ iff $q = 1 \pmod p$, in which case the group is unique. If $p = 2$ we obtain the dihedral group of order $2q$, which generalizes first to the dihedral group of order $2n$ and then even further to the "generalized dihedral group" where the cyclic group of order $n$ is replaced with any abelian group.
What if $p > 2$? Is there a natural generalization of the groups of order $pq$ to a family of groups of order $pn$? Maybe more than one possible generalization? Is it maybe even meaningful to talk about the "generalized $p$-hedral group"?
| https://mathoverflow.net/users/4336 | Generalizations of the nonabelian group of order $pq$ | The nonabelian group of order $pq$ is given by generators $a$, $b$, with relations $a^p=1$, $b^q=1$, $a^{-1}ba=b^r$, where $r$ is chosen so $r^p$ is 1 mod $q$. If there is an element $r$ of order $p$ mod $n$, then there is a nonabelian group of order $pn$ with generators $a$, $b$, and relations $a^p=1$, $b^n=1$, $a^{-1}ba=b^r$.
| 7 | https://mathoverflow.net/users/3684 | 29661 | 19,363 |
https://mathoverflow.net/questions/29656 | 4 | As part of a larger analysis I have a need to break a polygon into it's individual line segments and mark which side is "inside" of the polygon. If your curious this is going to be fed into a big parallel map reduce algorithm to translate the representation of the polygon into a more efficient data structure for fast recall.
The naive solution I've come up with is to first cast a ray towards the calculated centroid of the polygon from each vertex of every line segment and count the number of line intersections between the vertex and the centroid. That way I can determine if the side of each line that is facing the centroid is "internal" or not. Based upon whether or not each ray has an odd or even number of intersections.
The best way to think about this is that I will be creating a triangle at a single line segment with the two rays that are cast towards the centroid. I can determine based upon the internal angles which side of the line segment is "facing" the centroid. There is a distinct difference between facing the centroid, and actually being on the internal side of the polygon.
Unfortunately, this is quite inefficient to calculate. Assume I have a very detailed polygon with tens of millions of vertexes. It would be very slow if for each vertex I created a vector between the centroid and the vertex, then ran a line intersection algorithm against every other line segment in the polygon. It's just not going to finish in a timely manner.
I think I can walk the line segments one by one and determine based upon the internal angles at each vertex which side of each line segment is "internal" to the polygon. Are there any known theorems out there that I am unaware of that will help me answer this problem?
| https://mathoverflow.net/users/7112 | How can I efficiently determine which side of a line segment is internal to the polygon? | Pick the vertex with the highest x value. You can label the edges touching it. Now just walk along your polygon and label the edges consistently.
| 4 | https://mathoverflow.net/users/4391 | 29664 | 19,366 |
https://mathoverflow.net/questions/29522 | 4 | The short version of my question is:
>
> 1)For which positive integers $k, n$ is there a solution to the equation $$k(6k+1)=1+q+q^2+\cdots+q^n$$ with $q$ a prime power?
>
>
> 2) For which positive integers $k, n$ is there a solution to the equation $$(3k+1)(2k+1)=1+q+q^2+\cdots+q^n$$ with $q$ a prime power?
>
>
>
Now for some motivation. In [this question](https://mathoverflow.net/questions/29281/which-steiner-systems-come-from-algebraic-geometry) I ask for an algebra-geometric construction of certain Steiner systems (there's background on what Steiner systems are and on the details and motivation for the construction in that question). In particular, if the construction can be carried out for a $(p, q, r)$ Steiner system, the blocks of the Steiner system would be given by the $p-1$-plane sections of some variety $X$ in affine or projective space over a finite field $\mathbb{F}\_q$. If $X$ is in affine $n+1$-space and $p=2$, the number of blocks would be given by $1+q+\cdots+q^n$. Now the number of blocks in a $(2, 3, 6k+1)$ system is $k(6k+1)$ and the number of blocks in a $(2, 3, 6k+3)$ system is $(3k+1)(2k+1)$ (a $(2, 3, n)$ Steiner system is realizable iff $n=1, 3$ mod $6$, $n > 3$). So the question comes from setting these two quantities equal.
In other words, these equations must be solvable for all $k$ if the algebra-geometric construction of Steiner systems is to go through for all $(2, 3, n)$ systems (the Steiner triple systems) in affine space.
The most general form of this question (which covers both the affine and the projective versions) is:
>
> For integers $n, 1 < p < q < r$, when is there a prime power $q$ such that $$\frac{r!(q-p)!}{q!(r-p)!}=\left[n \atop p-1\right]\_q$$ or $$\frac{r!(q-p)!}{q!(r-p)!}=\left[n \atop p\right]\_q?$$
>
>
>
Of course, I only expect answers to concentrate on the numbered questions (1) and (2) at the top of the page.
EDIT: Note that e.g. $k=4$ has no solutions for the first equation.
| https://mathoverflow.net/users/6950 | Solving a Diophantine equation related to Algebraic Geometry, Steiner systems and $q$-binomials? | Let $t = 1+q+q^2+\dots+q^n $ then each of the equations (1) and (2) implies that $24t+1$ is a square (namely, $24t+1=(12k+1)^2$ and $24t+1=(12k+5)^2$, respectively). For $n=2$ that leads to a Pellian equation (with possibly infinitely many solutions), for $n=3,4$ to an elliptic curve (with finitely many solutions, if any), and for $n>4$ to a hyper-elliptic curve (with no solutions for most $n$).
Cases $n=3,4$ are easy to solve.
For $n=3$, integer solutions are $q=-1, 0, 2, 3, 13, 25, 32, 104, 177$ out of which only $2,3,13,25,32$ are powers of primes.
For $n=4$, integral solutions are $q=-1,0,1,25,132$ out of which only $25$ is a power of prime.
These numerical values are computed in SAGE and MAGMA.
Also, for a fixed value of $k$, it is possible to verify solubility of the given equations by iterating all possible $q$ dividing the l.h.s. minus 1. In particular, equation (1) has solutions only for the following $k$ below $10^6$:
1, 2, 3, 15, 52, 75, 1302, 32552, 813802.
Similarly, equation (2) has solutions only for the following $k$ below $10^6$:
1, 10, 260, 6510, 162760.
| 5 | https://mathoverflow.net/users/7076 | 29667 | 19,368 |
https://mathoverflow.net/questions/29676 | 27 | Let ${\bf N}^\omega = \bigcup\_{m=1}^\infty {\bf N}^m$ denote the space of all finite sequences $(N\_1,\ldots,N\_m)$ of natural numbers. For want of a better name, let me call a family ${\mathcal T} \subset {\bf N}^\omega$ a *blocking set* if every infinite sequence $N\_1,N\_2,N\_3,N\_4,\ldots$ of natural numbers must necessarily contain a blocking set $(N\_1,\ldots,N\_m)$ as an initial segment. (For the application I have in mind, one might also require that no element of a blocking set is an initial segment of any other element, but this is not the most essential property of these sets.)
One can think of a blocking set as describing a machine that takes a sequence of natural number inputs, but always halts in finite time; one can also think of a blocking set as defining a subtree of the rooted tree ${\bf N}^\omega$ in which there are no infinite paths. Examples of blocking sets include
1. All sequences $N\_1,\ldots,N\_m$ of length $m=10$.
2. All sequences $N\_1,\ldots,N\_m$ in which $m = N\_1 + 1$.
3. All sequences $N\_1,\ldots,N\_m$ in which $m = N\_{N\_1+1}+1$.
The reason I happened across this concept is that such sets can be used to pseudo-finitise a certain class of infinitary statements. Indeed, given any sequence $P\_m(N\_1,\ldots,N\_m)$ of $m$-ary properties, it is easy to see that the assertion
>
> There exists an infinite sequence $N\_1, N\_2, \ldots$ of natural numbers such that $P\_m(N\_1,\ldots,N\_m)$ is true for all $m$.
>
>
>
is equivalent to
>
> For every blocking set ${\mathcal T}$, there exists a finite sequence $(N\_1,\ldots,N\_m)$ in ${\mathcal T}$ such that $P\_m(N\_1,\ldots,N\_m)$ holds.
>
>
>
(Indeed, the former statement trivially implies the latter, and if the former statement fails, then a counterexample to the latter can be constructed by setting the blocking set ${\mathcal T}$ to be those finite sequences $(N\_1,\ldots,N\_m)$ for which $P\_m(N\_1,\ldots,N\_m)$ fails.)
Anyway, this concept seems like one that must have been studied before, and with a standard name. (I only used "blocking set" because I didn't know the existing name in the literature.) So my question is: what is the correct name for this concept, and are there some references regarding the structure of such families of finite sequences? (For instance, if we replace the natural numbers ${\bf N}$ here by a finite set, then by Konig's lemma, a family is blocking if and only if there are only finitely many finite sequences that don't contain a blocking initial segment; but I was unable to find a similar characterisation in the countable case.)
| https://mathoverflow.net/users/766 | Is there a name for a family of finite sequences that block all infinite sequences? | Intuitionists use the name "bar" for what you called a blocking set. The relevant context is "bar induction," the principle saying that, if (1) a property has been proved for all elements of a bar and (2) it propagates in the sense that, whenever it holds for all the
one-term extensions of a finite sequence s then it holds for s itself,
then this property holds of the empty sequence. (I'm omitting some technicalities here that distinguish different versions of bar induction.)
There's also a closely related notion in infinite combinatorics, called a "barrier"; this is a collection $B$ of finite subsets of $\mathbf N$ such that no member of $B$ is included in another and every infinite subset of $\mathbf N$ has an initial segment in $B$. This is the subject of a partition theorem due to Nash-Williams: If a barrier is partitioned into two pieces, then there is an infinite $H\subseteq\mathbf N$ such that one of the pieces includes a barrier for $H$ (meaning that every infinite subset of $H$ has an initial segment in that piece).
| 33 | https://mathoverflow.net/users/6794 | 29682 | 19,380 |
https://mathoverflow.net/questions/29700 | 9 | Is the congruence
group $\Gamma(2)$ generated by the upper triangular matrix $(1, 2; 0, 1)$
and the lower triangular matrix $(1, 0; 2, 1)$ or does on need to also
throw in the negation of the identity? To be specific, how do I check that
the negation of the identity is not a word in the above matrices?
| https://mathoverflow.net/users/7120 | Generators for congruence group $\Gamma(2)$ | Yes, you need to throw in $-I$. Check that the set of all matrices
of the form
$$\left(\begin{matrix}
a&b\\\
c&d
\end{matrix}\right)$$
with $b$ and $c$ even and $a\equiv d\equiv1$ (mod $4$) is a subgroup
of the modular group.
| 11 | https://mathoverflow.net/users/4213 | 29702 | 19,390 |
https://mathoverflow.net/questions/29692 | 8 | Let $u(t) = \Sigma\_{k=1}^n c\_k e^{\lambda\_k t} (c\_k \in \mathbb C, \lambda\_k \in \mathbb C) $ be an exponential polynomial of **order** $n$.
Define $E\_n$ to be the collection of all exponential polynomial of order $n$, i.e.,
$$ E\_n:= \{ u : u(t) = \sum\_{k=1}^n c\_k e^{\lambda\_k t}, c\_k \in \mathbb C, \lambda\_k \in \mathbb C \}. $$
Notice, that two elememts of $E\_n$ may have different (disjoint) set of exponents. Only requirement for $u$ to be in $E\_n$ is that it has order at most $n$.
Let $\mathbf{P}\_n$ be the collection of all polynomials of **degree** at most $n$.
Consider a function $f = \sum\_{j=1}^{M} p\_{m\_j}(t) e^{\lambda\_j t}, p\_{m\_j} \in \mathbf{P}\_{m\_j}, \sum\_{j=1}^{M} (m\_j+1) \leq n$.
My question is, can one find a sequence $u\_m \in E\_n$ such that, $$\sup\_{x\in[0,1]}| f(x)- u\_m(x) |\rightarrow 0. $$
If possible, then how should one go about constructing such a sequence ?
| https://mathoverflow.net/users/6766 | Approximation by exponential polynomials | This follows from the fact that the set of $n\times n$ matrices with simple spectrum is dense in the space of all $n\times n$ matrices ${\bf M}\_n(\mathbb C)$ (or that the set of polynomials of degree $n$ with simple roots is dense in the set of all complex polynomials
of degree $n$).
The function $f$ solves the Cauchy problem for a [linear ODE](http://en.wikipedia.org/wiki/Linear_differential_equation) with *constant* complex coefficients
$$y^{(N)}+a\_1y^{(N-1)}+\dots+a\_Ny=0,$$
$$y^{(k)}(0)=f^{(k)}(0),\quad k=0,1\dots,N-1,$$
where $N=\sum\_{j=1}^{M} (m\_j+1) \leq n$. The exponents $\lambda\_j$, $j=1,\dots,M$ are the roots of the corresponding characteristic equation
$$P(\lambda)=\lambda^{N}+a\_1\lambda^{N-1}+\dots+a\_N\lambda=0$$
with the multiplicities, respectively, $m\_j+1$, $j=1,\dots,M$. The coefficients $a\_1,\dots a\_N$ can be found from the relation
$$P(\lambda)=\prod\limits\_{j=1}^{M}(\lambda-\lambda\_j)^{m\_j+1}.$$
Now, a small generic perturbation of the ODE's coefficients will produce an ODE
$$y^{(N)}+a\_1^{\varepsilon}y^{(N-1)}+\dots+a\_N^{\varepsilon}y=0,$$
such that the roots of the corresponding characteristic equation are all simple. This implies that a solution to the latter ODE belongs to $E\_n$. Finally, we may use a standard result that solutions to linear ODEs depend continuously on the coefficients (in the topology of uniform convergence on finite time intervals).
| 7 | https://mathoverflow.net/users/5371 | 29704 | 19,392 |
https://mathoverflow.net/questions/12137 | 16 | Let $G$ be a semisimple algebraic group over an algebraically closed field $k$. In the case that $k$ has characteristic 0, there has been intensive study of the BGG category O of representations of the enveloping algebra of the Lie algebra of $G$ (which is also, in this case, the hyperalgebra of $G$). When $k$ is a field of positive characteristic, though, the enveloping algebra and the hyperalgebra are different, and there has been a lot of study of representations of the enveloping algebra. On the other hand, it seems that there has been much less study of representations of the hyperalgebra, perhaps because it is more complicated (for example, it is not finitely generated).
So let's now assume that the characteristic of $k$ is positive. A seminal paper of Haboush, "Central Differential Operators of Split Semisimple Groups Over Fields of Positive Characteristic," set up the foundations of category O for hyperalgebras in this setting. More precisely, let $U$ denote the hyperalgebra of $G$. In that paper Haboush defined Verma modules for $U$ (which are defined analogously to the characteristic 0 case) and proved that many of the properties one would expect, including relationships with simple modules and certain "integrality" properties, hold. However, this is just the beginning; there are many questions one could ask about whether or not category O in positive characteristic behaves like category O in characteristic 0. E.g, what is the structure of the blocks in this category? What about the structure of projective generators? Are there translation functors? If so, how do they behave? I.e., how much of the (huge) characteristic 0 story carries over to positive characteristic? Haboush's paper is from 1980, and I haven't been able to find any papers that carry on the study started in the paper -- does anyone know of any?
(As a sidebar, I would note that the hyperalgebra/enveloping algebra dichotomy in positive characteristic is mirrored in the study of quantum groups at roots of 1: the De Concini-Kac algebra is the analog of the enveloping algebra, and the Lusztig algebra is the analog of the hyperalgebra).
| https://mathoverflow.net/users/1528 | On Category O in positive characteristic | Maybe I can answer the original question more directly, leaving aside the interesting recent geometric work discussed further in later posts like the Feb 10 one by Chuck: analogues of Beilinson-Bernstein localization on flag varieties and consequences for algebraic groups (Bezrukavnikov, Mirkovic, Rumynin).
The 1979 conference paper by Haboush may be hard to access and also hard to read in detail, but it raises some interesting questions especially about centers of certain hyperalgebras. I tried to give an overview in Math Reviews: MR582073 (82a:20049) 20G05 (14L40 17B40)
Haboush,W. J.,
Central differential operators on split semisimple groups over fields of positive
characteristic.
Séminaire d’Algèbre Paul Dubreil et Marie-Paule Malliavin, 32ème
année (Paris, 1979), pp.
35–85, Lecture Notes in Math., 795, Springer, Berlin, 1980.
The hyperalgebra here is the Hopf algebra dual of the algebra of regular functions on a simply connected semisimple algebraic group $G$ over an algebraically closed field of characteristic $p$, later treated in considerable depth by Jantzen in his 1987 Academic Press book *Representations of Algebraic Groups* (revised edition, AMS, 2003). After the paper by Haboush, for example, Donkin finished the determination of all blocks of the hyperalgebra.
While the irreducible (rational) representations are all finite dimensional and have dominant integral highest weights (Chevalley), the module category involves locally finite modules such as the infinite dimensional injective hulls (but no projective covers). The role of the finite Weyl group is now played by an affine Weyl group relative to $p$ (of Langlands dual type) with translations by $p$ times the root lattice. In fact, higher powers of $p$ make life even more complicated.
The older work of Curtis-Steinberg reduces the study of irreducibles to the finitely many "restricted" ones for the Lie algebra $\mathfrak{g}$. For these and other small enough weights, Lusztig's 1979-80 conjectures provide the best
hope for an analogue of Kazhdan-Lusztig conjectures when $p>h$ (the Coxeter number). The recent work applies for $p$ big enough": Andersen-Jantzen-Soergel, BMR, Fiebig.
Anyway, the hyperalgebra involves rational representations of $G$ including restricted representations of $\mathfrak{g}$, while the usual enveloping algebra of the Lie algebra involves all its representations. But the irreducible ones are finite dimensional. I surveyed what was known then in a 1998 AMS Bulletin paper. Lusztig's 1997-1999 conjectures promised more insight into the *non-restricted* irreducibles and are now proved for large enough $p$ in a preprint by Bezrukavnikov-Mirkovic. This and their earlier work with Rumynin use a version of "differential operators" on a flag variety starting with the usual rather than divided-power (hyperalgebra) version of the universal enveloping algebra of $\mathfrak{g}$.
To make a very long story shorter, Haboush was mainly looking for the center of the hyperalgebra (still an elusive beast unlike the classical enveloping algebra center, due to the influence of all powers of $p$). His weaker version of Verma modules may or may not lead further. But there is no likely analogue of the BGG category for the hyperalgebra in any case. That category depended too strongly on finiteness conditions and well-behaved central characters.
ADDED: It is a long story, but my current viewpoint is that the characteristic $p$ theory for both $G$ and $\mathfrak{g}$ (intersecting in the crucial zone of restricted representations of $\mathfrak{g}$) is essentially *finite dimensional* and requires deep geometry to resolve. True, the injective hulls of the simple $G$-modules with a highest weight are naturally defined and infinite dimensional (though locally finite), but the hope is that they will all be direct limits of finite dimensional
injective hulls for (the hyperalgebras of) Frobenius kernels relative to powers of $p$. Shown so far for $p \geq 2h-2$ (Ballard, Jantzen, Donkin). In particular, the universal highest weight property of Verma modules in the BGG category (and others) is mostly replaced in characteristic $p$ by *Weyl modules* (a simple consequence of Kempf vanishing observed by me and codified by Jantzen). Then the problems begin, as Lusztig's conjectures have shown. The
Lie algebra case gets into other interesting territory for non-restricted modules.
| 11 | https://mathoverflow.net/users/4231 | 29719 | 19,403 |
https://mathoverflow.net/questions/29707 | 6 | Take a time interval $[0,T]$, and a filtered probability space $(\Omega,P,\mathcal{F},\mathcal{F}\_t)$. If $X \in L^1(\mathcal{F}\_T)$, then $M\_t = E [X \ | \ \mathcal{F}\_t]$ is a martingale. If I want the martingale $M$ to have continuous or right continuous paths, is there a condition I can impose on the filtration to ensure this?
A standard result says that if the filtration is right-continuous, meaning that $\cap\_{s>t} \mathcal{F}\_s = \mathcal{F}\_t$, then there exists a modification of $M$ with right continuous paths (in fact right continuous with left limits). However, I want to say something about the original process, and not a modification.
| https://mathoverflow.net/users/2310 | Path continuity for (closed) martingales? | Generally speaking, you cannot do this at the level of conditions on filtration since conditional expectation is defined up to modifications on zero measure sets. For example, take $T=1$, and the probability space be $[0,1]$ with Borel sigma-algebra and Lebesgue measure. Let $X(\omega)=\omega$ and all sigma-algebras from the filtration coincide with Borel. We can choose $M$ to be defined by $M\_t(t)=0$ and $M\_t(\omega)=\omega$ if $t\ne\omega$. Clearly, every trajectory is discontinuous, although the filtration is as regular as possible.
| 3 | https://mathoverflow.net/users/2968 | 29726 | 19,409 |
https://mathoverflow.net/questions/29734 | 41 | If an entire function is bounded for all $z \in \mathbb{C}$, than it's a constant by Liouville's theorem. Of course an entire function can be bounded on lines through the origin $z=r \exp(i \phi), \phi= \text{const.}, r \in \mathbb{R}$ without being constant (e.g. $\cos(z^n)$ is bounded on $n$ lines).
What is the maximum cardinality of the set of "directions" $\phi$ for which an entire function can be bounded without being constant?
From intuition I would expect only finitely many directions. Is this correct?
(Picard's second theorem says that in any open set containing $\infty$ every value with possibly a single exception is taken infinitely often by an entire non-constant function. Here I'm asking a somehow "orthogonal" question, looking for lines through $\infty$ where an entire non-constant function is bounded.)
| https://mathoverflow.net/users/6415 | Must the set of lines through the origin on which a nonconstant entire function is bounded be finite? | Newman gave an example in 1976 of a non-constant entire function bounded on each line through the origin in "[An entire function bounded in every direction](http://www.jstor.org/stable/2977024)".
I like the second sentence of the article:
>
> This is exactly what is needed to confuse students who have just struggled to comprehend the meaning of Liouville's theorem.
>
>
>
Armitage gave examples in 2007 of non-constant entire functions that go to zero in every direction in "Entire functions that tend to zero on every line". For this I have only seen the [MR review](http://www.ams.org/mathscinet-getitem?mr=2290290). (If you don't have MathSciNet access, the link should still give you the publication information to find the article.)
---
**Update:** I just decided to take a look at the Armitage paper, and the introduction was enlightening:
>
> Although every bounded entire (holomorphic) function on $\mathbb{C}$
> is constant (Liouville’s theorem), it has been known for more than a hundred years
> that there exist nonconstant entire functions $f$ such that $f(z) → 0$ as $z →∞$ along
> every line through 0 (see, for example, Lindelöf’s book [10, pp. 119–122] of 1905). And it has been known for more than eighty years that such functions can tend to 0
> along any line whatsoever (see Mittag-Leffler [11], Grandjot [8], and Bohr [4]). Further
> references to related work are given in Burckel’s review [5] of Newman’s note [12].
> Entire functions with radial decay are used by Beardon and Minda [3] and Ullrich [14]
> in studies of pointwise convergent sequences of entire functions.
>
>
>
Armitage goes on to mention that Mittag-Leffler and Grandjot also gave explicit constructions, but states, "The examples given in what follows may nevertheless
be of some interest because of their comparative simplicity." The examples are
$$F(z)=\exp\left(-\int\_0^\infty t^{-t}\cosh(tz^2)dt\right) - \exp\left(-\int\_0^\infty t^{-t}\cosh(2tz^2)dt\right)$$ and
$$G(z)=\int\_0^\infty e^{i\pi t}t^{-t}\cosh(t\sqrt{z})dt\int\_0^\infty e^{i\pi t}t^{-t}\cos(t\sqrt{z})dt .$$
| 60 | https://mathoverflow.net/users/1119 | 29735 | 19,416 |
https://mathoverflow.net/questions/29741 | 9 | Inspired by an [old question by Kevin Lin](https://mathoverflow.net/questions/9171/idea-for-book-translation-project-closed) and the [communal translation of an answer by Laurent Fargues](https://mathoverflow.net/questions/10913/lifting-the-p-torsion-of-a-supersingular-elliptic-curve/11097#11097), I am proposing a communal effort to translate from Russian to English Drinfeld's famous 1988 letter to Schechtman: [typeset](http://math.harvard.edu/~tdp/Drinfeld-Letter.to.Schechtman.on.New.directions.of.work-1988.pdf) and [in the handwritten original](http://math.harvard.edu/~tdp/original.pdf).
I am starting a blank, communal answer below in which to whip up the translation. Thanks in advance for any contributions!
| https://mathoverflow.net/users/307 | Drinfeld's 1988 letter to Schechtman: translation request | Keith Conrad has kindly produced a translation available [here](https://www.doi.org/10.4171/EMSS/5). It can be cited as
>
> Vladimir Drinfeld, *A letter from Kharkov to Moscow.* EMS Surv. Math. Sci. **1** (2014), 241-248. doi:[10.4171/EMSS/5](https://www.doi.org/10.4171/EMSS/5)
>
>
>
| 11 | https://mathoverflow.net/users/307 | 29742 | 19,420 |
https://mathoverflow.net/questions/29740 | 4 | To prove there is an elementary topos with natural numbers object, it should be sufficient to assume ZF has a model. Probably ZF by itself, or IZF, is already sufficient. And probably even this is not necessary. Do we know what is?
| https://mathoverflow.net/users/6787 | Under what assumptions does an elementary topos (+infinity) exist? | This question was studied in the early days of elementary topos theory, and the connection was worked out by Bill Mitchell and J.C. Cole (independently, as far as I know). The MathSciNet references are:
MR0319757 (47 #8299)
Mitchell, William,
Boolean topoi and the theory of sets.
J. Pure Appl. Algebra 2 (1972), 261--274.
MR0357116 (50 #9584)
Cole, J. C.,
Categories of sets and models of set theory. The Proceedings of the Bertrand Russell Memorial Conference (Uldum, 1971), pp. 351--399. Bertrand Russell Memorial Logic Conf., Leeds, 1973.
Essentially, they show that Boolean topos theory with a natural numbers object is equivalent to a version of Zermelo set theory (not ZF, i.e., without the replacement scheme) and with the separation scheme restricted to formulas with all quantifiers bounded (but the bounds can involve power sets, so quite a lot of separation is available). (I'm omitting some technicalities here, not because they're negligible but because I don't remember them.)
If you don't assume Boolean logic for your topos, I don't think it changes the consistency strength of the theory. In any nontrivial topos with a natural numbers object, the subtopos of double-negation sheaves will also be nontrivial and will have a natural numbers object (the associated sheaf of the original one). In other words, the double-negation translation of classical logic into intuitionistic logic works here.
| 7 | https://mathoverflow.net/users/6794 | 29744 | 19,421 |
https://mathoverflow.net/questions/29745 | 8 | Let S be the class of all rings R which have 1 and satisfy this condition:
for every "non-zero" right ideal I of R there exists a "proper" right ideal J of R such that I + J = R. (The + here is not necessarily direct.)
All semisimple rings are in S and (commutative) local rings which are not fields are not in S. The ring of integers Z is also in S and so S properly contains the class of semisimple rings.
My questions:
Will this condition by itself force an element of S to have any (known, interesting) structure?
A more important question:
What about simple rings which are in S? For example, do they have to be semisimple? (Unlikely!)
| https://mathoverflow.net/users/6941 | Semisimple-ish rings! | By Zorn's lemma, each right ideal is contained in a maximal right ideal,
therefore if $I+J = R$ then $I+M = R$ where $M$ is a maximal right ideal.
If $I+M\ne R$ for all maximal right ideals $M$ then $I\subseteq M$ for
all maximal ideals $M$. Thus $I\subseteq J(R)$, the Jacobson radical of $R$
which is the intersection of all maximal right ideals of $R$. Hence
condition $S$ is equivalent to $J(R)=0$.
A ring with vanishing Jacobson ideal is called *semiprimitive*.
As $J(R)$ is also the intersection of the maximal left ideals of $R$
then the property of semiprimitivity is left-right symmetric.
There are plenty of examples of semiprimitive rings which are not
semisimple. For instance every simple ring is semiprimitive
and every subdirect product of semiprimitive rings is semiprimitive
($\mathbb{Z}$ is a subdirect product of finite fields).
As a reference see Section 10.4 of P. M. Cohn
*Algebra* (2nd ed. vol 3) Wiley 1991.
| 16 | https://mathoverflow.net/users/4213 | 29748 | 19,424 |
https://mathoverflow.net/questions/29749 | 10 | Hi,
I've been looking for a clear reference which shows that the matrix exponential is surjective from $M\_{n}(C)$ to $Gl\_{n}(C)$. Wikipedia claims this is true, but I haven't seen it proven... Also, can someone suggest how to create a power series for a function log(x) defined for a given $A\in Gl\_{n}(C)$ thats outside our standard set B(I,1)??? Specifically, what if $A=e^{B}$ with $\det(B)=0$?
Thanks in advance?
Tom Petrillo
| https://mathoverflow.net/users/7133 | How to show the matrix exponential is onto? And, how to create a powerseries for log that works outside B(I,1) | My recollection is that Rossmann's book on Lie groups has a detailed discussion of the exponential map and surjectivity issue. Matrix exponential map is equivariant under conjugation,
$$\exp(gXg^{-1})=g\exp(X)g^{-1},$$
and, as Robin has already remarked, one can easily check that a matrix in Jordan normal form is in the image of $\exp: M\_n(\mathbb{C})\to GL\_n(\mathbb{C}),$ establishing surjectivity for $G=GL\_n(\mathbb{C}).$
As for your second question, you can always (a) rescale (b) shift by scalar matrices and re-center the $\log$ series:
$$(a)\ \exp(nB)=\exp(B)^n\quad (b)\ \exp(B+\lambda I\_n)=e^{\lambda}\exp(B).$$
For topological reasons, there isn't a canonical formula for $\log$ that works locally everywhere.
---
**Warning:**. Exponential map is not always surjective. The following family of matrices is not in the image of the exponential map from the Lie algebra $\mathfrak{g}=\mathfrak{sl}\_2(\mathbb{C})$ (traceless $2\times 2$ matrices) to the Lie group $G=SL\_2(\mathbb{C})$ (unit determinant $2\times 2$ matrices):
$$h\_a=\begin{bmatrix}
-1 & a\\
0 & -1
\end{bmatrix},\ a\ne 0.$$
No preimage in $M\_2(\mathbb{C})$ of $h\_a$ under $\exp$ can have trace $0$. Indeed, if $\exp(X\_a)=h\_a$ then the eigenvalues of $X\_a$ must be $\pi i+2\pi n, -\pi i - 2\pi m (n, m\in \mathbb{Z})$ and the trace condition implies that $n=m,$ so $X\_a$ has distinct eigenvalues, hence it is diagonalizable, but $h\_a$ is not — contradiction.
| 18 | https://mathoverflow.net/users/5740 | 29760 | 19,433 |
https://mathoverflow.net/questions/29762 | 6 | Given two groups $A$ and $B$ and an injective homomorphism $f : A \to B$. When does a homomorphism $g : B \to A$ exist with $g\circ f = \mathrm{id}\_A$ (but not necessarily $f\circ g = \mathrm{id}\_B$)?
| https://mathoverflow.net/users/2672 | When does an injective group homomorphism have an inverse? | If and only if $B$ is a semidirect product of $A$ and another group (the latter is normal). One direction is obvious, and another direction is easy: $B$ is a semidirect product of the image of $f$ and the kernel of $g$.
| 14 | https://mathoverflow.net/users/5301 | 29763 | 19,435 |
https://mathoverflow.net/questions/29759 | 11 | I think the title says it all. If I have a finite map $p:X\to Y$ between schemes, and $F$ is a coherent sheaf on $X$ such that $p\_\*F$ is locally free, can I conclude that $F$ is locally free?
Assumptions I would be happy to make:
1. The map $p$ is flat.
2. $X$ and $Y$ are both $\mathbb{A}^n$.
>
> I would be much more pleased with a reference than a proof, since I would like to use this result in a paper.
>
>
>
| https://mathoverflow.net/users/66 | Is it true that if the pushforward of a coherent sheaf is locally free, then the original sheaf is locally free? | The reference is "Auslander-Buchsbaum formula". What matters is that $X$ and $Y$ are smooth of some common pure dimension $n$ (which forces flatness due to the quasi-finiteness, by the way), not that one has global affine spaces. Then every coherent sheaf on $X$ has stalks with *finite* projective dimension, and so all of projective dimension zero (= locally free) precisely when the depth of each stalk is equal to the dimension of its support on the local ring. A length-$n$ regular sequence at $p(x) \in Y$ exists by the local freeness on the base, and that's such a sequence at $x$ provided that the successive quotients arising in the definition of "regular sequence" really remain *nonzero* when localizing upstairs.
The completion of the $p(x)$-stalk of the pushforward is the product of the completions at the points of $p^{-1}(p(x))$, so each factor module is free and hence we're OK as long as such factor modules are all nonzero since we can use a regular sequence in the completed local ring at $p(x)$. So we have to rule out the possibility of a vanishing completed stalk, or equivalently a vanishing stalk, upstairs. In such cases, upon passing to connected=irreducible components, the coherent sheaf $F$ would vanish on a Zariski-dense open upstairs, and hence on some $p^{-1}(U)$ for a dense open $U$ in the base. Then $p\_ {\ast}(F)$ would vanish on $U$ and hence vanish by local freeness, so $F = 0$. The case $F = 0$ is easy to handle directly.
| 13 | https://mathoverflow.net/users/6773 | 29777 | 19,442 |
https://mathoverflow.net/questions/29687 | 12 | Is there are nice way to prove the primitive element theorem without using field extensions?
The primitive element theorem says that if $x$ and $y$ are algebraic over $F$ and $y$ is separable over $F$, then there exists a $z \in F(x,y)$ such that $F(x,y) = F(z)$. In the case where $F$ is infinite, $z$ can be expressed in the form $x + {\lambda}y$ with $\lambda \in F$. In fact, almost any $\lambda$ will do. There are only a finite number of exceptions. These exceptions are $\frac{\alpha\_i - x}{y - \beta\_j}$ where $\alpha\_i$ and $\beta\_j$ range over the (other) roots of the minimal polynomials of $x$ and $y$ respectively.
But in order to talk about $\alpha\_i$ and $\beta\_j$ we need to build a field extension where the minimal polynomials of $x$ and $y$ split. This is the step I'm hoping to avoid.
Perhaps we can build a polynomial in $F[x]$ whose roots are $\frac{\alpha\_{i\_1} - \alpha\_{i\_2}}{\beta\_{j\_1} - \beta\_{j\_2}}$ and simply avoid picking $\lambda$s which are roots of this polynomial, and then use whatever properties this polynomials has to prove that this works.
Or maybe there is another completely different way of proving the primitive element theorem while avoiding building field extensions.
I found [a nice proof](http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/separable2.pdf) that $x$ is separable over $F$ if and only if every $y \in F(x)$ is separable over $F$ that avoids building field extensions. It uses derivations instead. Now I'm hoping to do the same with the primitive element theorem.
Edit: I'll try to give some motivation. Adding roots of polynomials to fields in constructive mathematics is more difficult than in classical mathematics (because irreducibility is undecidable). It only works for countable fields, and then you have no guarantee that the original field will be a decidable subset of the new field. Yes, it can be done, but it seems like a pain. There is another way too using double negation translations, but it also seems like a pain. Instead I'd rather avoid the whole issue of *building* splitting fields, if I can, and it seems tantalizingly close to possible. After all, the only reason splitting fields are used here is to build a finite set of elements *in the base field* to avoid.
| https://mathoverflow.net/users/4085 | Primitive element theorem without building field extensions | OK, I have a proof which meets your conditions. I relied on [this write up](http://www.math.washington.edu/~greenber/PrimElem.pdf) of the standard proof as a reference.
**Lemma 1:** Let $K/F$ be an extension of fields, and let $f(x)$ and $g(x)$ be polynomials in $F[x]$. Let $d\_F(x)$ be the GCD of $f$ and $g$ in $F[x]$ and let $d\_K(x)$ be the GCD of $f$ and $g$ in $K[x]$. Then $d\_F(x)$ and $d\_K(x)$ coincide up to a scalar factor.
**Proof:** Since $d\_F(x)|f(x)$ and $d\_F(x)|g(x)$ in $K[x]$, we have $d\_F(x)|d\_K(x)$. Now, there are polynomials $p(x)$ and $q(x)$ in $F[x]$ such that $f(x) p(x) + g(x) q(x) = d\_F(x)$. So $d\_K(x) | d\_F(X)$. Since $d\_F(x)$ and $d\_K(x)$ divide each other, they only differ by a scalar factor.
**Lemma 2:** Let $f(x)$ and $g(x)$ be polynomials with $g(0) \neq 0$. Then, for all but finitely many $t$, the polynomials $f(tx)$ and $g(x)$ have no common factor.
**Proof:** Let $f(x) = f\_m x^m + \cdots + f\_1 x + f\_0$ and $g(x) = g\_n x^n + \cdots + g\_1 x + g\_0$. If $f(tx)$ and $g(x)$ HAVE a nontrivial common factor, then there are polynomials $p(x)$ and $q(x)$, of degrees $\leq n-1$ and $\leq m-1$, such that
$$f(tx) p(x)=g(tx) q(x).$$
This is $m+n$ linear equations on the $m+n$ coefficients of $p$ and $q$.
Writing this out in coefficients, the matrix
$$\begin{pmatrix}
f\_m t^m & \cdots & f\_1 t & f\_0 & 0 & 0 & \cdots & 0 \\
0 & f\_m t^m & \cdots & f\_1 t & f\_0 & 0 & \cdots & 0 \\
\ddots \\
0 & 0 & \cdots & 0 & f\_m t^m & \cdots & f\_1 t & f\_0 \\
g\_n & \cdots & g\_1 & g\_0 & 0 & 0 & \cdots & 0 \\
0 & g\_n & \cdots & g\_1 & g\_0 & 0 & \cdots & 0 \\
\ddots \\
0 & 0 & \cdots & 0 & g\_n & \cdots & g\_1 & g\_0
\end{pmatrix}$$
has nontrivial kernel. The determinant of this matrix is a polynomial in $t$, with leading term $(f\_m)^n (g\_0)^n t^{mn} + \cdots$. (Recall that $g\_0 \neq 0$.) So, for all but finitely many $t$, this matrix has nonzero determinant and $f(tx)$ and $g(x)$ are relatively prime. QED.
Now, we prove the primitive element theorem (for infinite fields). Let $\alpha$ and $\beta$, in $K$, be algebraic and separable over $F$, with minimal polynomials $f$ and $g$. We will show that, for all but finitely many $t$ in $F$, we have $F(\alpha - t \beta) = F(\alpha, \beta)$.
Let $f(x) = (x -\alpha) f'(x -\alpha)$ and $g(x) = (x - \beta) g'(x - \beta)$. Since $\beta$ is separable, we know that $g'(0) \neq 0$. Note that $f'$ and $g'$ have coefficients in $K$. By Lemma 2, for all but infinitely many $t$ in $F$, the polynomials $f'(ty)$ and $g'(y)$ have no common factor. Choose a $t$ for which no common factor exists. Set $F' = F(\alpha - t \beta)$; our goal is to show that $F'=F(\alpha, \beta)$.
Set $h\_t(x) = f(tx + \alpha-t \beta)$. Note that $h\_t(x)$ has coefficients in $F'$. We consider the GCD of $h\_t(x)$ and $g(x)$.
Working in $K$, we can write $h\_t(x) = t (x - \beta) f'(t (x- \beta))$ and $g(x) = (x-\beta) g'(x-\beta)$. By the choice of $t$, the polynomials $f'(t (x- \beta))$ and $g'(x-\beta)$ have no common factor, so the GCD of $h\_t(x)$ and $g(x)$, in the ring $K[x]$, is $x-\beta$.
By Lemma 1, this shows that $x - \beta$ is in the ring $F'[x]$. In particular, $\beta$ is in $F'$. Clearly, $\alpha$ is then also in $F'$, as $\alpha= (\alpha - t \beta) + t \beta$. We have never written down an element of any field larger than $K$. QED.
| 7 | https://mathoverflow.net/users/297 | 29783 | 19,446 |
https://mathoverflow.net/questions/29781 | 2 | Given a complex-analytic manifold of dimension $d$, why does the cohomology of coherent sheaves vanish in dimension $> 2d$ (without using GAGA)?
| https://mathoverflow.net/users/6960 | coherent analytic cohomology vanishes for q > 2dim | Note that it is not necessary to say to avoid GAGA, as GAGA has no relevance in the absence of compactness assumptions.
Anyway, something much more general (and satisfying) is true: all topological sheaf cohomology on a (paracompact Hausdorff) analytic space vanishes beyond twice the analytic dimension. Here is a sketch of a proof. By metrization theorems, such spaces are metrizable. Also, connected components of analytic spaces are open, so cohomology is direct product of cohomologies on the connected components. We can therefore restrict attention to the connected case, so the underlying topological space is separable (i.e., countable base of opens); I think this latter fact is stated with reference in the Introduction of the book Theory of Stein Spaces. (If not, assume the given analytic space is separable, a very "practical" assumption!)
By the local analytic "Noether normalization" (really Weierstrass Preparation), twice the analytic dimension equals the "topological dimension" in the sense of dimension theory as in Engelking's marvelous book "General topology" for separable metric spaces. (For separable metric spaces, various notions of topological dimension are proved to agree; all done in that book. For opens in a real Euclidean space it recovers the expected "dimension"!) That book shows open covers of separable metric spaces have refinements whose $(n+1)$-fold overlaps are empty for $n$ beyond the topological dimension (in one of the various *equivalent* senses of dimension: the "covering" dimension!). Now use equality of Cech and derived functor cohomology for paracompact Hausdorff spaces to conclude.
| 13 | https://mathoverflow.net/users/6773 | 29784 | 19,447 |
https://mathoverflow.net/questions/29766 | 8 | I'm looking for a news site for Mathematics which particularly covers recently solved mathematical problems together with the unsolved ones. Is there a good site MO users can suggest me or is my only bet just to google for them?
| https://mathoverflow.net/users/7081 | List of recently solved mathematical problems | As a counter-point to my somewhat flippant previous answer (which only really applies if one is a specialist in the field), if you are looking at a field in which you are not as much a specialist in, I suggest reading the articles from the [Bulletin of the AMS](http://www.ams.org/publications/journals/journalsframework/aboutbull). The articles are designed to be fairly up-to-date and expository in nature, and often gives the state of the art in their reviews.
Of course, a similar caveat as that to Helge's answer applies: the "news" maybe several months out of date. But considering the glacial paces at which a lot of mathematical refereeing takes place, I think it is quite okay.
In the spirit of this answer, you may also find [Which journals publish expository work?](https://mathoverflow.net/questions/15366/which-journals-publish-expository-work) to be useful.
| 8 | https://mathoverflow.net/users/3948 | 29785 | 19,448 |
https://mathoverflow.net/questions/29772 | 5 | Let $M$ be a smooth manifold. Its structure sheaf $\mathcal{O}\_M$ is the sheaf of smooth real-valued functions. Together they form a ringed space $(M,\mathcal{O}\_M)$. The tangent sheaf $\mathcal{T}\_M$ is a sheaf of modules over the structure sheaf. It can be defined as the sheaf of derivations of the structure sheaf.
A smooth map of manifolds $f: M \rightarrow N$ induces a morphism $df: \mathcal{T}\_M \rightarrow f^\*(\mathcal{T}\_N)$ of $\mathcal{O}\_M$-modules, where $f^\*(\mathcal{T}\_N)$ is the inverse image of $\mathcal{T}\_N$. It is called the differential or pushforward of the map $f$.
Does anyone have a reference for the definition of the differential? In which kind of textbook would this be explained? It seems to be somewhere between differential geometry and algebraic geometry but I could not find it in any textbook in neither of these areas.
(I am not looking for the differential between tangent *bundles* which is explained in detail in every basic book on differential geometry.)
Thanks in advance!
| https://mathoverflow.net/users/5631 | Differential between tangent sheaves | You might want to try Ramanan's [Global Calculus](http://www.google.com/url?sa=t&source=web&cd=1&ved=0CBIQFjAA&url=http%253A%252F%252Fwww.amazon.com%252FGlobal-Calculus-Graduate-Studies-Mathematics%252Fdp%252F0821837028&ei=-q0oTO-ZBePesAavhok9&usg=AFQjCNED1_dGrlHkytEUuyDiutnunakcCg&sig2=ZC8MnQVdRimCkJuGrGlVVg) which does a bit of introductory manifold theory, and uses the mechanism of sheaves throughout.
| 4 | https://mathoverflow.net/users/622 | 29795 | 19,456 |
https://mathoverflow.net/questions/29750 | 50 | Riemann-sums can e.g. be very intuitively visualized by rectangles that approximate the area under the curve.
See e.g. [Wikipedia:Riemann sum](https://en.wikipedia.org/wiki/Riemann_sum).
The [Itô integral](https://en.wikipedia.org/wiki/Ito_integral) has due to the unbounded total variation but bounded quadratic variation an extra term (sometimes called Itô correction term). The standard intuition for this is a [Taylor expansion](https://en.wikipedia.org/wiki/Taylor_series), sometimes [Jensen's inequality](https://en.wikipedia.org/wiki/Jensen_inequality#Proofs).
But normally there is more than one intuition for a mathematical phenomenon, e.g. in Thurston's paper, ["On Proof and Progress in Mathematics"](https://arxiv.org/abs/math/9404236), he gives seven different elementary ways of thinking about the derivative.
**My question**
Could you give me some other intuitions for the Itô integral (and/or Itô's lemma as the so called "chain rule of stochastic calculus"). The more the better and from different fields of mathematics to see the big picture and connections. I am esp. interested in new intuitions and intuitions that are not so well known.
| https://mathoverflow.net/users/1047 | Intuition and/or visualisation of Itô integral/Itô's lemma | I find the intuitive explanation in [Paul Wilmott on Quantitative Finance](https://www.amazon.co.uk/Paul-Wilmott-Quantitative-Finance-2nd/dp/0470018704) particularly appealing.
Fix a small $h>0$. The stochastic integral
$$\int\_0^{h} f(W(t))\ dW(t)=\lim\limits\_{N\to\infty}\sum\limits\_{j=1}^{N}
f\left(W(t\_{j-1})\right)\left(W(t\_{j})-W({t\_{j-1}})\right),\quad t\_j= h\frac{j}{N},$$
involves adding up an infinite number of random variables. Let's substitute every term $f\left(W(t\_{j-1})\right)$ with its formal Taylor expansion. Then there are several contributions to the sum: those that are *a sum of random variables* and those that are *a sum of the squares of random variables*, and then there are *higher-order terms*.
Add up a large number of independent random variables and the Central Limit Theorem kicks in, the end result being a normally distributed random variable. Let's calculate its mean and standard deviation.
When we add up $N$ terms that are normal, each with a mean of $0$ and a standard deviation of $\sqrt{h/N}$, we end up with another normal, with a mean of $0$ and a standard deviation of $\sqrt{h}$. This is our $dW$. Notice how the $N$ disappears in the limit.
Now, if we add up the $N$ *squares* of the same normal terms then we get something which is normally distributed with a mean of
$$N\left(\sqrt{\frac{h}{N}}\right)^2=h$$
and a standard deviation which is
$h\sqrt{2/N}.$
This tends to zero as $N$ gets larger. In this limit we end up with, in a sense, our $dW^2(t)=dt$, because *the randomness as measured by the standard deviation disappears leaving us just with the mean $dt$*.
The higher-order terms have means and standard deviations that are too small, disappearing
rapidly in the limit as $N\to\infty$.
| 30 | https://mathoverflow.net/users/5371 | 29800 | 19,458 |
https://mathoverflow.net/questions/29765 | 2 | **Perturbative behaviour of solutions of the solutions of the Dirichlet problem for the Laplacian:**
Lets consider $ B = B(0, 1) \in \mathbb{R}^2$ be the unit circle with center at $0\in\mathbb{R}^2$. Let $u\_0$ be an harmonic function on $B$ also harmonic at the boundary, that is, $u\_0$ is harmonic in the ball $B(0, 1+\varepsilon)$ for $\varepsilon > 0$ small. Then, if we denote by $f = {u\_0}\_{|\partial B}$ we have that $u\_0$ satisies (trivially) the Dirichlet problem
$$
\begin{array} {rcl}
\Delta u\_0(x) & = & 0 \newline
{u\_0}\_{|\partial B}(x) &= &f(x)
\end{array}
$$
Now, let $K\subset B$ be a compact set and $g:K\rightarrow \mathbb{R}$ be a smooth function (real analytic, for instance), and consider the one parameter family of Dirichlet problems
$$
\begin{array} {rcl}
\Delta u\_s(x) & = & 0 \newline
{u\_s}\_{|\partial B}(x) &= &f(x)\newline
{u\_s}\_{|K}(x) &= & {u\_0}\_{|K}(x)+sg(x)\newline
\end{array}
$$
It is clear that for $s=0$ the solution of this problem is the same as the original problem stated above, so we consider this as a perturbative problem.
MY QUESTION IS:
How does $u\_s$ behaves near the compact set $K$? It is known that $u\_s$ is continuous in all the unit ball (also in $K$) but it is hoped that is not differentiable near $K$. It is possible to show that, generically, there exists an $\alpha\in\mathbb{R}$ such that it is satisfied
$$
\lim\_{x\longrightarrow z}\frac{|u\_s(x)-u\_s(z)|}{||x-z||^{\alpha}} = C(s, z) \neq 0,
$$
where $C(s, z)$ is a constant, depending on $s$ and $z\in K$?
Note that for $s=0$, the above limit exists when $\alpha = 1$ and $C(0)$ is the Lipschitz constant of of $u\_0$.
| https://mathoverflow.net/users/5231 | Asymptotic behaviour near the boundary in the Dirichlet problem for the Laplacian. | No, it will not be differentiable in the whole ball. To see this, let $u$ be the zero function and $g$ be nearly anything nonnegative and not identically zero in $K$. For example $g=1$. Then recall Hopf's lemma.
This will also work to show that differentiability fails at any point on the boundary of $K$, at which $g$ achieves its maximum (on the whole of $K$).
However, it will be $C^\alpha$ in the ball, which is the last question you stated. This follows from the smoothness of $g$ and Holder estimates for $u$. For this you also need something about $K$ itself being smooth of course-- all hope is lost if the boundary of $K$ is irregular.
| 1 | https://mathoverflow.net/users/5678 | 29819 | 19,469 |
https://mathoverflow.net/questions/29813 | 22 | For simplicity, I will be talking only about connected groups over an algebraically closed field of characteristic zero.
The basic theorem of affine algebraic groups is that they all admit faithful, finite-dimensional representations. The *fundamental* theorem for semisimple groups is that these representations are all completely reducible, but unfortunately there is no reason that any irreducible summand of a faithful representation should be faithful, only that the kernels of all these representations intersect trivially.
>
> My question is whether such a representation does, in fact, exist.
>
>
> (Answered: iff the center is cyclic.)
>
>
>
This does not hold of general reductive groups for the following reason: if $T$ is any torus of rank $r > 1$, then its irreducible representations are all characters $\chi \colon T \cong \mathbb{G}\_m^r \to \mathbb{G}\_m$, which therefore have nontrivial kernels. More generally, any reductive group $G$ has connected center a torus of some rank $r$, so by Schur's lemma this center acts by a character $\chi$ in any irreducible representation of $G$ and if $r > 1$, therefore does not act faithfully.
The exceptional case $r = 1$ *does* have an example, namely $\operatorname{GL}\_n$, whose standard representation is faithful and irreducible and whose center has rank 1. A more general version of this question might be, then:
>
> Does any reductive group whose center has rank at most 1 have a faithful irreducible representation?
>
>
> (Answered: when not semisimple, iff the center is connected.)
>
>
>
Another special case is that if $G$ is simple and of adjoint type, then its adjoint representation is irreducible and faithful by definition (or, depending on your definition, because the center is trivial). A constructive version of this question for any $G$ (semisimple or reductive of central rank 1) is then:
>
> Can we give a construction of a faithful, irreducible representation of $G$ from its adjoint representation?
>
>
> (Not yet answered!)
>
>
>
This is deliberately a little vague since I don't want to restrict the possible form of such a construction, only that it not start out with "Throw away the adjoint representation and take another one such that..."
Finally, suppose the answer is "no".
>
> What is the obstruction to such a representation existing?
>
>
> (Answered: for $Z$ the center, it is the existence of a generator for $X^\*(Z)$.)
>
>
>
| https://mathoverflow.net/users/6545 | Do semisimple algebraic groups always have faithful irreducible representations? | *Edit*: I now give the argument for general reductive $G$.
Let $G$ be a reductive algebraic group over an alg. closed field $k$ of char. 0. Fix a max
torus $T$ and write $X = X^\*(T)$ for its group of characters. Write $R$ for the
subgroup of $X$ generated by the roots of $G$. Then the center $Z$
of $G$ is the diagonalizable subgroup of $T$ whose character group is $X/R$.
>
> **Claim:** $G$ has a faithful irreducible representation if and only if the character
> group $X/R$ of $Z$ is cyclic.
>
>
>
Note for semisimple $G$, the center $Z$ is finite. Since the characteristic of $k$ is 0, in this case the group of $k$-points of $Z$ is (non-canonically) isomorphic
to $X/R$. Thus $Z$ is cyclic if and only $X/R$ is cyclic.
In general, the condition that $X/R$ is cyclic means either that the group of points $Z(k)$ is finite cyclic, or that $Z$ is a 1 dimensional torus.
As to the proof, for $(\implies)$ see Boyarsky's comment following reb's answer.
For $(\Leftarrow)$ let me first treat the case where $G$ is *almost simple*; i.e. where the root system $\Phi$ of $G$ is irreducible. Supopse that the class of $\lambda \in X$ generates the cyclic group $X/R$. Since
the Weyl group acts on $X$ leaving $R$ invariant, the class of any $W$-conjugate of $\lambda$ is also a generator of $X/R$. Thus we may as well suppose $\lambda$ to be dominant
and non-0 [if $X=R$, take e.g. $\lambda$ to be a dominant root...]
Now the simple $G$-module $L=L(\lambda) = H^0(\lambda)$ with "highest weight $\lambda$"
will be faithful. To see this, note that since $\lambda \ne 0$, $L$ is not the trivial representation. Since $G$ is almost simple, the only proper normal subgroups
are contained in $Z$. Thus it suffices to observe that the action of $Z$ on the
$\lambda$ weight space of $L$ is faithful.
The general case is more-or-less the same, but with a bit more book-keeping.
Write the root system $\Phi$ of $G$ as a disjoint union $\Phi = \cup \Phi\_i$
of its irreducible components. There is an isogeny
$$\pi:\prod\_i G\_{i,sc} \times T \to G$$
where $T$ is a torus and $G\_{i,sc}$ is the simply connected almost simple group
with root system $\Phi\_i$. Write $G\_i$ for the image $\pi(G\_{i,sc}) \subset G$.
The *key fact* is this:
a representation $\rho:G \to \operatorname{GL}(V)$ has
$\ker \rho \subset Z$ if and only if the restriction $\rho\_{\mid G\_i}$ is non-trivial
for each $i$.
Now, as before pick $\lambda \in X$ for which the coset of $\lambda$ generates
the assumed-to-be cyclic group $X/R$. After replacing $\lambda$ by a Weyl group
conjugate, we may suppose $\lambda$ to be dominant. After possibly repeatedly replacing
$\lambda$ by $\lambda + \alpha$ for dominant roots $\alpha$, we may suppose
that $\lambda$ has the following property:
$$(\*) \quad \text{for each $i$, there is $\beta\_i \in \Phi\_i$ with $\langle \lambda,\beta\_i^\vee \rangle \ne 0$}$$
Now let $L = L(\lambda)$ be the simple module with highest weight $\lambda$. Condition
$(\*)$ implies that $G\_i$ acts nontrivially on $L$ for each $i$, so by the "key fact",
the kernel of the representation of $G$ on $L$ lies in $Z$. but since $\lambda$
generates the group of characters of $Z$, the center $Z$ acts faithfully on the
$\lambda$ weight space of $L$.
| 12 | https://mathoverflow.net/users/4653 | 29820 | 19,470 |
https://mathoverflow.net/questions/29822 | 3 | Let G be a Lie group and $H\subseteq G$ be a closed subgroup. It can be shown that $H$ has a unique differentiable structure such that the inclusion map $H\to G$ is an embedding of manifolds.
The functor of points $h\_G=\text{Hom}(-,G)$, $h\_H=\text{Hom}(-,H)$ of the Lie groups G and H define group objects in the category of manifolds. The inclusion map $H\to G$ gives a natural transformation $h\_H\to h\_G$ which clearly makes $h\_H$ into a subfunctor of $h\_G$. Therefore we may take the quotient functor Q which takes each manifold M to the quotient $Q(M):=\text{Hom}(M,G)/\text{Hom}(M,H)$.
Is this functor $Q$ a sheaf on the Grothendieck pretopology where the coverings of an arbitary manifold M are given by the sets of arrows {$\iota\_i:U\_i \to M $ } such that each $\iota\_i$ is a diffeomorphism of $U\_i$ onto an open subset of M and $\bigcup \iota\_i(U\_i)=M$?
(An analogous question is often posed in the context of algebraic groups, however, I am interested in this differentiable setting.)
| https://mathoverflow.net/users/7146 | Is the quotient functor of points of a Lie group with the subfunctor of a closed subgroup a sheaf? | No; in fact any sheaf $F$ in this topology is a sheaf on the open subsets of $M$ in the standard (continuous) topology as open subsets of $M$ map diffeomorphically into $M$, and conversely any manifold mapping diffeomorphically into $M$ is isomorphic to an open subset of $M$ (its image). So if this were true the functor of global sections for sheaves on manifolds would be exact; but we know that this is not the case.
---
Added, (2/20/2011): I figured I would, as per BCnrd's comment, add an example where the quotient fails to be a sheaf, using connected Lie Groups. The construction is surprisingly elementary, and if I didn't make some silly mistake, I think quite nice!
Let $\rho: SO(3)\hookrightarrow U(n)$ be a faithful unitary representation of $SO(3)$; let $M=U(n)/SO(3)$. I first claim that the quotient map $U(n)\to M$ is a non-trivial principal $SO(3)$-bundle. Indeed $\pi\_1(U(n))=\mathbb{Z}$, whereas $\pi\_1(SO(3))=\mathbb{Z}/2\mathbb{Z}$, and thus $U(n)$ does not split topologically as a product of $M\times SO(3)$, as $\mathbb{Z}/2\mathbb{Z}$ is not a factor of $\mathbb{Z}$. So $U(n)\to M$ represents a non-trivial class in $H^1(M, SO(3))$, (which is a pointed set, corresponding to isomorphism classes of principal $SO(3)$-bundles over $M$, via the Cech construction).
But note that this class is the image of the canonical element $i$ of $$H^0(M, U(n)/SO(3))=\operatorname{Hom}(M, U(n)/SO(3))=\operatorname{Hom}(U(n)/SO(3), U(n)/SO(3))$$ (namely, the identity map), through the boundary map $H^0(M, U(n)/SO(3))\to H^1(M, SO(3))$. Thus $i$ maps to a non-trivial class, and so must not be in the image of the map $H^0(M, U(n))\to H^0(M, U(n)/SO(3))$.
Thus this last map is not surjective, and so the sheaf of $U(n)/SO(3)$ valued functions on $M$ is *not* the presheaf $\operatorname{Hom}(-, U(n))/\operatorname{Hom}(-, SO(3))$.
| 1 | https://mathoverflow.net/users/6950 | 29834 | 19,477 |
https://mathoverflow.net/questions/29829 | 5 | Incidentally I've obtained a hypergeometric identity that I've not seen before:
$${}\_3F\_2(-m,-n,m+n; 1, 1; 1) = \frac{m^2+n^2+mn}{(m+n)^2} {\binom{m+n}{m}}^2$$
So, I wonder if it is well-known and possibly represents a particular case of something more general?
P.S. I've tried to simplify() the l.h.s. in Maple but it did not succeed, giving a hope that the identity is not completely trivial. ;)
EDIT: There seems to be a bug in formula rendering, so I'm repeating it below in plain LaTeX:
{}\_3F\_2(-m,-n,m+n; 1, 1; 1) = \frac{m^2+n^2+mn}{(m+n)^2} {\binom{m+n}{m}}^2
| https://mathoverflow.net/users/7076 | A (known?) hypergeometric identity | Your relation is a particular case of the Karlsson--Minton relations (see Section 1.9 in the $q$-Bible by Gasper and Rahman). It's also a contiguous identity to Pfaff--Saalschütz.
**EDIT.**
First of all I apologise for giving insufficient comments on the problem.
I learned from Max a very nice graph-theoretical interpretation of the identity
which makes good reasons for not burring it in the list of "ordinary" problems.
The hypergeometric series (function)
$$
{}\_ {p+1}F\_ p\biggl(\begin{matrix} a\_ 0,\ a\_ 1,\ \dots,\ a\_ p \cr
b\_ 1,\ \dots ,\ b\_ p\end{matrix};x\biggr)
= \sum\_ {k=0}^\infty \frac{(a\_ 0)\_ k(a\_ 1)\_ k\dots (a\_ p)\_ k}{(b\_ 1)\_ k\dots (b\_ p)\_ k}\frac{x^k}{k!},
$$
where
$$
(a)\_ 0=1 \quad\text{and}\quad
(a)\_ k =\frac{\Gamma(a+k)}{\Gamma(a)}= a(a+1)\dots (a+k-1) \quad\text{for } k\in \mathbb Z\_ {>0}
$$
(I consider the ones with finite domain of convergence $|z|<1$), have very nice
history and links to practically everything in mathematics. There are many
*transformation* and *summation* theorems for them, both classical and contemporary.
There are very efficient algorithms and packages for proving them, like
the algorithm of creative telescoping (due to W. Gosper and D. Zeilberger)
and the package HYP
which allows one to manipulate and identify binomial and hypergeometric series
(due to C. Krattenthaler). An example of classical summation theorem is
the Pfaff--Saalschütz sum
$$
{}\_ 3F\_ 2\biggl(\begin{matrix} -m,\ a,\ b \cr
c,\ 1+a+b-c-m\end{matrix};1\biggr)
=\frac{(c-a)\_ m(c-b)\_ m}{(c)\_ m(c-a-b)\_ m}
$$
where $m$ is a negative integer, with a generalisation
$$
{}\_ {p+1}F\_ p\biggl(\begin{matrix} a,\ b\_ 1+m\_ 1,\ \dots,\ b\_ p+m\_ p \cr
b\_ 1,\ \dots ,\ b\_ p\end{matrix};1\biggr)=0
\quad\text{if } \operatorname{Re}(-a)>m\_ 1+\dots+m\_ p
$$
and
$$
{}\_ {p+1}F\_ p\biggl(\begin{matrix} -(m\_ 1+\dots+m\_ p),\ b\_ 1+m\_ 1,\ \dots,\ b\_ p+m\_ p \cr
b\_ 1,\ \dots ,\ b\_ p\end{matrix};1\biggr)=(-1)^{m\_ 1+\dots+m\_ p}
\frac{(m\_ 1+\dots+m\_ p)!}{(b\_ 1)\_ {m\_ 1}\dots (b\_ p)\_ {m\_ p}}
$$
due to B. Minton and Per W. Karlsson (here $m\_ 1,\dots,m\_ p$ are nonnegative integers).
Max's original identity is not a straightforward particular case but a linear combination
of three *contiguous* Pfaff--Saalschütz-summable hypergeometric series.
(Two hypergeometric functions are said to be contiguous if they are alike except
for one pair of parameters, and these differ by unity.) Because of having *three*
hypergeometric functions, I do not see any fun in writing the corresponding details
but indicate a simpler hypergeometric derivation.
Applying Thomae's transformation
$$
{}\_ 3F\_ 2\biggl(\begin{matrix} -m,\ a,\ b \cr
c,\ d\end{matrix};1\biggr)
=\frac{(d-b)\_ m}{(d)\_ m}\cdot{}\_ 3F\_ 2\biggl(\begin{matrix} -m,\ c-a,\ b \cr
c,\ 1+b-d-m\end{matrix};1\biggr)
$$
the problem reduces to evaluation of the series
$$
{}\_ 3F\_ 2\biggl(\begin{matrix} -m,\ n+1,\ m+n \cr
1,\ n\end{matrix};1\biggr).
$$
Writing
$$
\frac{(n+1)\_ k}{(n)\_ k}=\frac{n+k}{n}=1+\frac kn
$$
the latter series becomes
$$
{}\_ 3F\_ 2\biggl(\begin{matrix} -m,\ n+1,\ m+n \cr
1,\ n\end{matrix};1\biggr)
={}\_ 2F\_ 1\biggl(\begin{matrix} -m,\ m+n \cr 1 \end{matrix};1\biggr)
+\frac{(-m)(m+n)}{n} {}\_ 2F\_ 1\biggl(\begin{matrix} -m+1,\ m+n+1 \cr 2 \end{matrix};1\biggr)
$$
and the latter two series are summed with the help of the Chu--Vandermonde summation
(a particular case of the Gauss summation theorem).
As for general forms of Max's identity, I can mention that there is no use of the integrality of $n$
in the last paragraph, and I could even expect something a la Minton--Karlsson in general.
| 14 | https://mathoverflow.net/users/4953 | 29843 | 19,483 |
https://mathoverflow.net/questions/29847 | 7 | Say I have a map of $G$-spaces $f : X \to Y$ and I know it is a homotopy-equivalence in the *plain* sense that there exists a map (maybe not equivariant) $g : Y \to X$ such that the two composites are homotopic to the identity $f\circ g \simeq Id\_Y$, $g\circ f \simeq Id\_X$.
Is there something of an obstruction theory that tells you when you can promote $f$ to a homotopy-equivalence in the category of $G$-spaces?
Ideally the level of generality I care for is $X$ and $Y$ manifolds and $G$ a compact Lie group. But anything in that ballpark interests me.
| https://mathoverflow.net/users/1465 | An obstruction theory for promoting homotopy equivalences that are equivariant maps to equivariant homotopy equivalences? | When you say $f$ is a map of $G$-spaces I am guessing you mean a morphism in the category of $G$-spaces, i.e. a continuous map satisfying $f(ax)=af(x)$ for $a\in G$. If so, then there is a good answer. (But "promoting" the map to a strong $G$-equivalence is a funny way to say it, because it suggests a piece of extra structure rather than a property.)
The fixed point spaces of closed subgroups are the key to much of equivariant homotopy theory. In the most important model structure for $G$-spaces, a morphism $X\to Y$ is called a weak equivalence (resp. fibration) iff for every subgroup $H$ the induced map $X^H\to Y^H$ of fixed point spaces is a weak homotopy equivalence (resp. Serre fibration). Cofibrations are generated by attaching orbits of cells. For cofibrant objects (and this includes manifolds with smooth action) weak equivalences are then the same as maps that actually have an inverse up to $G$-homotopy.
| 3 | https://mathoverflow.net/users/6666 | 29855 | 19,490 |
https://mathoverflow.net/questions/29850 | 2 | Given a discriminant d>0 (make it fundamental if that is easier), when can a prime p be the the $x^2$ coefficient of a reduced indefinite quadratic form?
That is, for what p is there a reduced form $px^2 + bxy + cy^2,$ with $b^2-4pc=d$?
| https://mathoverflow.net/users/695 | Primes as the first coefficient of a reduced indefinite quadratic form | I recommend a book by Duncan A. Buell called "Binary Quadratic Forms."
First, we discard the case where $d$ is a square. In such a case the forms represent entire arithmetic progressions. For example, with $x^2 - y^2$ and $d = 4$ we get
$ (n+1)^2 - n^2 = 2 n + 1.$ Or, with $x y$ and $d=1,$ we have $n \cdot 1 = n.$
Note that we always require $$ d \equiv 0,1 \pmod 4.$$
For what primes $p > 0$ is there any form, reduced or not, having $p$ as a "diagonal" coefficient? With odd $p$ that does not divide $d,$ this answer is essentially quadratic reciprocity. We are demanding
$$ \beta^2 \equiv d \pmod p $$ If we can solve this, that is $(d | p) = 1,$ we can choose either $ b = \beta$ or
$ b = \beta + p$ to arrange
$$ b^2 \equiv d \pmod {4p}. $$ But this is the condition $$ b^2 = d + 4 p c, $$
or $ b^2 - 4 p c = d .$
Note that we also have the form $$ (-p)x^2 + b x y + (-c) y^2 $$ with the same discriminant. So as long as you do not ask whether the two forms are equivalent we are in good shape.
Finally, you asked about "reduced" forms, which is to say coefficients
$$ \langle a,b,c \rangle $$ and discriminant $d$ with
$$ 0 < b < \sqrt{d}, \; \; \mbox{and} \; \; \sqrt{d} - b < 2 | a | < \sqrt{d} + b. $$
Here we have Lagrange's theorem that any represented number $n$ occurs as a coefficient of $x^2$ in a reduced form if $$ | n | < \; \frac{1}{2} \; \sqrt{d}, $$ so you have a simple answer for small primes. It is a bit of a toss-up if you have
$$ \frac{1}{2} \; \sqrt{d} < p < \; \sqrt{d} .$$ Here I suggest creating a form and then checking the entire cycle of reduced forms in its equivalence class. The recipe for doing exactly that is on pages 21-23 of Buell.
EDIT: I'm afraid I was not sufficiently cautious as relates to primes that divide the discriminant. It is true that 2 is represented when $ d \equiv 1 \pmod 8$ and not when
$ d \equiv 5 \pmod 8.$ But as soon as we have even discriminant there is need for care. The trouble is the existence of imprimitive forms, $$ \langle a,b,c \rangle $$ with $$ \gcd(a,b,c) \neq 1 .$$ What follows is from page 75 in Buell. If $ d \equiv 0 \pmod {16}$ then 2 is not represented by a primitive form, but if $ d \equiv 8 \pmod {16}$ it is, by the class of the primitive (but not reduced) form
$$ \langle 2,0,\frac{-d}{8} \rangle .$$ If $ d \equiv 4 \pmod {16}$ then 2 is not represented by a primitive form, but if $ d \equiv 12 \pmod {16}$ it is, by the class of the primitive (but probably not reduced) form
$$ \langle 2,2,\frac{4-d}{8} \rangle. $$ Alright, now that I see Buell's Theorem 4.24, the case of odd primes $p$ dividing the discriminant is comparatively clean. If $$ p^2 | d$$ then only imprimitive forms represent $p.$ If $$ p \parallel d$$ then $p$ is represented by a primitive form, either $$ \langle p,0,\frac{-d}{4 p} \rangle $$ if $d$ is even, or $$ \langle p,p,\frac{p^2-d}{4 p} \rangle $$ if $d$ is odd. My comments about the size of $p$ and reduced forms still apply.
| 6 | https://mathoverflow.net/users/3324 | 29860 | 19,495 |
https://mathoverflow.net/questions/29835 | 5 | Let $\mathcal{C}$ and $\mathcal{D}$ be categories, and suppose $F\colon\mathcal{C}\to\mathcal{D}$ is a functor. It induces two adjoint pairs between $Set~^{\mathcal{C}}$ and $Set~^{\mathcal{D}}$; one is denoted $(F^\star,F\_\star)$ and one is denoted $(F\_!,F^\star)$. One proves easily that the counit to $(F^\star,F\_\star)$ is a natural isomorphism of functors $\mathcal{C}\to Set$ if and only if $F$ is fully faithful.
I am interested in the counit of the other adjunction $F\_!:Set~^{\mathcal{C}}\Longleftrightarrow Set^{\mathcal{D}}:F^\*$. Lets denote it by $$\epsilon\_F\colon F\_!F^\*\to \operatorname {id}\_{Set^{\mathcal{D}}}.$$
Question: Under what conditions on $F$ is $\epsilon\_F^~$ a natural isomorphism?
| https://mathoverflow.net/users/2811 | When is the $(F_!,F^*)$ counit a natural isomorphism? | It appears to me that the condition on $F:\cal C\to\cal D$ would be:
For any morphism $s: a\to b$ of $\cal D$, the following category is connected:
An object consists of a $\cal C$-object $c$ and a factorization $a\to F(c)\to b$ of $s$.
A morphism $c\_1\to c\_2$ is a $\cal C$-morphism such that the induced map $F(c\_1)\to F(c\_2)$ is compatible with the maps from $a$ and to $b$.
I don't recall ever having run into this sort of 'two-sided comma category' before, but it seems to be the answer.
I got this by choosing $G$ from $\cal D$ to Set to be represented by the object $a$ and thinking about the fiber of the map $\epsilon: (F\_!F^\*G)(b)\to G(b)$ over the element $s$.
| 8 | https://mathoverflow.net/users/6666 | 29865 | 19,499 |
https://mathoverflow.net/questions/29866 | 38 | I was wondering what would be the best way to present your paper at a conference, if your paper is selected for "short communication", lasting for about 15 minutes?
Should you concentrate on the main results or the proofs?
And what should a first-time presenter be wary of?
Thanks in advance.
| https://mathoverflow.net/users/7144 | Presenting a paper: Do's and Don'ts? | The first priority is to state your main results and explain why they are interesting (e.g. how they fit in with related work). With only 15 minutes you do not have much time to discuss proofs, but it is nice to give a brief outline of the proof of your main result and what is involved.
As a first-time presenter, I would watch out for the following:
First, as someone commented, it is crucial to not go over time. Don't try to cram too much in.
Second, watch out for the mechanical aspects of the presentation, e.g. is your print large enough to read, and is your voice loud enough to hear.
Third, strive to emphasize the most important points and not get lost in less important details. This is especially important when you only have a short time to talk.
Fourth, know your audience, so that you have some idea what you can assume is known and what you need to review.
| 55 | https://mathoverflow.net/users/6670 | 29867 | 19,500 |
https://mathoverflow.net/questions/29828 | 14 | Does anyone know of a **closed form** for the function on $\mathbb{N}$ which returns the greatest power of two which divides a given integer?
To be more precise, any positive integer $n\in\mathbb{N}$ can be uniquely expressed as $n=2^pq$ where $p,q\in\mathbb{N}$ and furthermore $q\equiv1\mod2$. I am looking for a closed form of the resulting function $f:\mathbb{N}\to\mathbb{N}$ which is such that $f:n\mapsto p$, as defined e.g. on [Wikipedia](http://en.wikipedia.org/wiki/Closed-form_expression).
As a starting point, I constructed a summation which does the job:
$$f(n)=\sum\_{j=1}^{\rho(n)}\left(\prod\_{i=1}^{j}\cos\left[\frac{\pi n}{2^i}\right]\right)^2$$
where $\rho(n)=\lfloor\log\_2n\rfloor$. Sadly, this expression is not very useful, and I would prefer a closed form expression. Using Morrie's Law, the product can be converted to a limit as follows:
$$f(n)=\lim\_{\epsilon\to0}t[\pi(n+\epsilon),\rho(n)]$$
where
$$t[x,m]=\sum\_{j=1}^{m}\left(\frac{2^{-j}\sin[x]\cos[x]}{\sin[2^{-j}x]}\right)^2$$
However, I cannot find a closed form for this summation...
So in summary, I'd be grateful if anyone could give me an expression for $t(x,m)$ which would make my version of $f$ usable, or if anyone could tell me another such $f$.
Thanks!
EDIT: I followed Gerry's answer and derived the following Fourier series for $f$:
$$f(n)=(1+\cos[\pi n])\left(1-2^{-\rho(n)}+\sum\_{j=1}^{\rho(n)}\sum\_{k=1}^\infty\frac{\sin[2\pi k n 2^{-j}]-\sin[2\pi k (n-1) 2^{-j}]}{k}\right)$$
I will try to further simplify this...
| https://mathoverflow.net/users/7154 | Greatest power of two dividing an integer | I suspect this answer will not be found satisfactory, but here goes. Write $[x]$ for the integer part of $x$. Then $[n/2]-[(n-1)/2]$ is 1 if $n$ is a multiple of 2, 0 otherwise. $[n/4]-[(n-1)/4]$ is 1 if $n$ is a multiple of 4, 0 otherwise. Etc. So the function you want is $$[n/2]-[(n-1)/2]+[n/4]-[(n-1)/4]+[n/8]-[(n-1)/8]+\dots$$ where the sum isn't really infinite, it has $r$ terms, where $r$ is something like $\log\_2n$.
| 12 | https://mathoverflow.net/users/3684 | 29871 | 19,503 |
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