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https://mathoverflow.net/questions/35393 | 1 | Can we express the following property of natural numbers as FOL. The property given below is only indicative, I am more interested in knowing how the concepts such as "infinitely many X exists for so and so" can be expressed in FOL. Also these need to be expressed as uninterpreted functions.
>
> "For every natural number n, there are infinitely many other natural numbers such that the greatest common divisor of n and each of these other numbers is 1"
>
>
>
As I said if you can't express this sentence in FOL at least suggest links/material where I can get leads.
| https://mathoverflow.net/users/8246 | Natural number properties as uninterpreted functions in first order logic | If $f(n)$ is a predicate in the first order language of arithmetic,
then "there are infinitely many $n$ such that $f(n)$ holds" can be
expressed as
$$\forall m\in\mathbb{N}\ \exists k\in\mathbb{N}:f(m+k+1).$$
| 2 | https://mathoverflow.net/users/4213 | 35394 | 22,840 |
https://mathoverflow.net/questions/35388 | 6 | Algebraic topologists like to cook up algebraic invariants on topological spaces in order to answer questions, so they are often concerned with how strong those invariants are. Currently, I am concerned with just how much information is lost when moving from a space to `the' chain complex associated to that space.
Now, I should be a bit more specific here. There are many homology theories in which one takes a space, cooks up a chain complex, and takes its homology. I am mainly interested in just singular homology for this question, but if you can only think of an answer using sheaf cohomology or some other homology theory then that's alright. Now, the actual question is:
>
> Do there exist two spaces that have chain homotopic associated chain complexes, but are not, themselves, homotopic?
>
>
>
I imagine that, unless the answer to the question is "no," it would be a bit difficult to show that the two spaces are not homotopic, since we have taken away the tool we usually use to prove such facts. My first thought was that one might be able to cook up a counterexample by looking at two spaces whose compactifications give different homologies... but I wasn't able to come up with anything immediately. (Another method may be looking at homotopy groups... but they're so hard to compute, I didn't even try this approach). I probably didn't give this as much thought as I should have, so if the answer to this question is somewhat trivial, then go ahead and scold me and I'll go put some more effort into thinking about it.
| https://mathoverflow.net/users/6936 | A chain homotopy that does not arise from a homotopy of spaces? | A bounded below complex of free $R$-modules is acyclic if and only if it is contractible (in the sense that the identity map is chain homotopic to zero). Since the singular chain complex of a space is constructed out of free $\mathbb{Z}$-modules, any space with no homology would have to contract to a point, which is not the case (check out the [wikipedia page](http://en.wikipedia.org/wiki/Acyclic_space) for some examples/references).
Thus the chain homotopy exhibiting contractibility of the identity does not come from a topological homotopy.
| 11 | https://mathoverflow.net/users/182 | 35395 | 22,841 |
https://mathoverflow.net/questions/35401 | 2 | I have read a couple of proofs for the undecidability of the post correspondence problem, but neither reference gave a concrete example of two lists of words over a fixed alphabet such that the problem was undecidable for that set of two lists. In other words, the proofs showed the existence of such an example without actually giving the example. Does anybody know a reference where I can find such an example? Thanks.
| https://mathoverflow.net/users/8434 | post correspondence problem | As Tsuyoshi said, it doesn’t make sense to search for an undecidable *instance* of a problem. It’s only the *problem* itself that can be undecidable.
In particular, for *every* instance of PCP (or any other problem for that matter) there trivially exists an algorithm that gives the correct answer for that particular instance. If we’re dealing with the decision version of the problem, it’s either the algorithm that always answers “yes”, or the algorithm that always answers “no” (granted, this is not a constructive proof).
On the other hand, you might find specific instances of PCP without a *known* answer, for example by exploiting any open problem of mathematics and the fact that the halting problem reduces to PCP, say via a many-one reduction *R*.
Consider the Turing machine *M* that searches for a proof of the Riemann hypothesis by enumerating all proofs, and halts when it finds it. If RH is provable, this machine will halt in a finite amount of time, otherwise it will run forever. You can use the reduction from the halting problem to construct a PCP instance *R*(*M*) = *x*. Now, by deciding whether *x* is a positive or negative instance of PCP, you also decide RH. But that’s an open problem, and so the status of *x* also is.
| 9 | https://mathoverflow.net/users/5304 | 35403 | 22,843 |
https://mathoverflow.net/questions/35408 | 9 | In group theory, the single most important piece of information about a group is its cardinality, which is of course either finite, countably infinite, or uncountably infinite. Usually, however, uncountably infinite simply means a cardinality of $\aleph\_{1}$, the same as $\mathbb{R}$. My question is: is there anywhere that groups with cardinality strictly greater than $\aleph\_{1}$ arise naturally? Of course, it is easy enough to construct groups with arbitrarily large cardinality, but I cannot recall ever seeing them used.
| https://mathoverflow.net/users/6856 | Naturally occuring groups with cardinality greater than the reals. | In line with Joel's answer, my favorite "outrageously large group" is the group $G = \operatorname{Aut}(\mathbb{C})$ of field automorphisms of the complex numbers. It has cardinality $2^{2^{\aleph\_0}}$, which is pretty scary. But that's just the beginning of how large it is. For instance, from the study of real-closed fields, one can deduce that the number of conjugacy classes of order $2$ elements of $G$ is also $2^{2^{\aleph\_0}}$. It is also an extension of the absolute Galois group of $\mathbb{Q}$ (a profinite group which is conjectured to have among its quotients every finite group, up to isomorphism) by the huge simple group $\operatorname{Aut}(\mathbb{C}/\overline{\mathbb{Q}})$.
| 30 | https://mathoverflow.net/users/1149 | 35410 | 22,848 |
https://mathoverflow.net/questions/35399 | 2 | In Freedman's series of 3 books on Markov processes, I find that I keep on running into terms like:
P[$\max\_{0 \leq s \leq 1, s \leq t \leq rs}$ | B(t) - B(s) | > $\epsilon$]
in the background of proofs I'm reading. As Freedman mentions in B+D (19), its easy to see that for all fixed $\epsilon > 0$ this goes to 0 as r goes to 1. Its not so easy for me to see how to get any even somewhat-reasonable bounds on this probability for fixed $\epsilon$ and $r$. Does anybody have any suggestions (or out-of-the-box theorems) for some bounds?
PS: I've tried only a few silly things - e.g. trying to approximate the probability for finite sums and use Donsker's principle, and asking some people around here about martingale tricks, but no luck so far.
EDIT: A clarification in response to the first answer. I know that this probability (call it p(r, $\epsilon$) ) goes to 0 as r goes to 1 for ever fixed $\epsilon$, but I'm interested in statements like p(0.04, 0.000000001) < 0.02, or (more optimistically) p(r, $\epsilon) < 500 \frac{\log(\log(r-1))}{\epsilon}$. I don't have any reason to believe those two bounds are true (though both seem plausible to me), they are merely illustrative.
| https://mathoverflow.net/users/8432 | Maximal inequality over two indices | If $B$ is a.s.-continuous (and, therefore, uniformly continuous) then the maximum in question converges to $0$ a.s.
Your statement means convergence in probability, and follows automatically.
| 1 | https://mathoverflow.net/users/2968 | 35411 | 22,849 |
https://mathoverflow.net/questions/35367 | 5 | In universal algebra there is the notion of congruence relation: Consider a (1-sorted) algebraic structure, i.e. a set $A$ with a bunch of finitary operations $f\_i$ satisfying equations.
A *congruence relation* is an equivalence relation $\sim$ on $A$ such that the operations on $A$ produce well-defined operations on the set $A/\sim$ of equivalence classes by applying them to representatives. I.e. for any operation $f$ on $A$ the operation $\bar{f}$ on $A/\sim$ given by $\bar{f}([x\_1],...,[x\_n]):=[f(x\_1,...x\_n)]$ is well-defined, i.e. if $x\_1 \sim y\_1, ... , x\_n \sim y\_n$ then $f(x\_1,...,x\_n) \sim f(y\_1,...,y\_n)$ for all operations $f$ of the given structure.
Thus these relations are the right ones to form quotients inside the given category of algebraic structures.
An *essentially algebraic structure* is a (if 1-sorted) or several (if many-sorted) sets with partially defined operations satisfying equational laws, where the domain of any given operation is a subset defined by equations between previously defined operations (equivalenty: it is a $Set$-model of a finite limit sketch). The standard example are categories, where one has three global operations, identity, source and target, and a partial operation, composition, defined only for certain pairs of morphisms.
**My question is:** Is there a notion of congruence relation for these more general algebraic structures? E.g. one equivalence relation on each set satisfying the analogous properties to the above? If so have these "congruence relations" been studied, do they e.g. form lattices?
**Motivation:** Just curiosity really. I asked myself this question, after reading [this MO-question](https://mathoverflow.net/questions/35335/collapsing-objects-in-a-category) of Colin Tan, which might be a special case. He asks whether there is a way to collapse two objects in a category. If there was a lattice of congruence relations on a category, there might be the congruence relation generated by the relation which identifies just the two objects (this would of course mean to treat categories in an ["evil"](http://ncatlab.org/nlab/show/evil), non-two-categorical way, but that was what the question sounded like to me).
Googling did reveal nothing, so I ask you people...
| https://mathoverflow.net/users/733 | Is there a notion of congruence relation for essentially algebraic structures? | As Finn says, lfp categories have all coequalizers. However, there is a fly in the ointment. In set-models of algebraic theories, the underlying functor preserves the congruences and the quotients of the congruences. This means that a congruence is an equivalence relation on the underlying set and the quotient alegbraic structure has underlying set the set-quotient of the congruence. This is true even in many sorted algebraic theories.
But in models of finite limit sketches the quotient can blow up. The quotient need not be a structure whose underlying set is the quotient of the equivalence relation on the underlying set. An example of this happens in the category of small categories when you merge nonisomorphic objects. All of a sudden arrows may compose that didn't meet each other before, creating new arrows. (So the underlying set of arrows in the quotient is not the quotient of the congruence on the underlying set.)
This is spelled out and proved in Toposes, Triples and Theories, by Michael Barr and Charles Wells, in Theorem 4.1 of Chapter 8. In that theorem, "LE" means "Finite Limits" and "EE" means "effective equivalence relations whose quotients are preserved by the underlying set functor". This book is available for free on the internet -- just google it. Exercises EEPO and ORTHODOX give specific examples of that behavior. I think there is a specific example for small categories somewhere but at the moment I can't find it.
ADDED: The specific example is in section 1.8, exercise CBB. Section 1.8 talks about effective equivalence relations in general and calls them congruences.
| 3 | https://mathoverflow.net/users/342 | 35421 | 22,852 |
https://mathoverflow.net/questions/35420 | 3 | Let $k$ be a field and let $A$ be the polynomial ring over $k$ in $3n$ variables: $A = k[X\_{ij} \vert i=1,2,3 \quad j=1,2,\cdots,n]$.
${\rm SO}\_3(k)$ acts on $A$ in the following way: Given $g \in {\rm SO}\_3(k)$, we define:
$$g(X\_{ij})=g\_{ik}X\_{kj}$$
with respect to the summation convention.
Can the ring of invariants of this action be expressed in terms of generators and relations? I get the feeling that this is a standard exercise in invariant theory, but am not sure where to look.
| https://mathoverflow.net/users/5394 | How to compute the ring of invariants of SO_3(k) acting on a polynomial ring | This is addressed by the classical invariant theory, but the answer is more complicated than for general linear or orthogonal groups (in particular, not all minimal generators are quadratic). Let $k$ be a field of characteristic 0. The group $G=SO\_m$ acts on $m\times n$ matrices by the left multiplication and this induces a $G$-action on $A=k[X\_{ij}].$ Let us view the variables as the entries of the $m\times n$ generic matrix over $k.$ Then the algebra of invariants $A^G$ is generated by:
1 Scalar products of the columns of the matrix $X.$
2 Order $m$ minors of the matrix $X.$
This is the First Fundamental Theorem (FFT) of classical invariant theory for $SO\_m.$ In fact, the elements of the first type generate $O\_m$-invariants and the elements of the second type generate $SL\_m$-invariants ($SO\_m=O\_m\cap SL\_m$).
Moreover, all relations between these generators are also known (the Second Fundamental Theorem, SFT) and there is a good description of a standard monomial basis of $A^G.$ If I am not mistaken, the last part is due to Laskshmibai and coauthors. A comprehensive modern reference is
>
> Laskshmibai and Raghavan, *Standard monomial theory. Invariant theoretic approach.* Encyclopaedia of Mathematical Sciences, vol 137 (Invariant Theory and Algebraic Transformation Groups VIII), Springer.
>
| 5 | https://mathoverflow.net/users/5740 | 35426 | 22,856 |
https://mathoverflow.net/questions/35444 | 1 | This question is related to one [I asked previously](https://mathoverflow.net/questions/35351/minimum-differences-in-vectors-of-naturals). This is probably a little harder. I had a crack at it today, but have become stuck. I suspect the result is buried in the [order statistics](http://mathworld.wolfram.com/OrderStatistic.html) literature somewhere, and perhaps somebody is familiar with it. That, or Peter might insta-solve again :).
Given a vector $s$ of integers let $d(s)$ be the maximum difference between any two integers in $s$ when sorted in ascending order. That is, if we sort $s$ in ascending order to obtain $v$, then
$$d(s) = \max\_{i} (v\_{i+1} - v\_i).$$
For $s$ a vector of length $m$ from $\lbrace 1,2,\dots,n\rbrace^m$ we must have $0 \leq d(s) < n$.
>
> Given $0 \leq k < n$, how may such vectors have $d(s) = k$ ?
>
>
>
Again, I'm more interested in the case where $n$ is much larger than $m$ and if reasonable bounds can be found for $d(s)$, then this would be useful too.
Note: If $N\_k$ is the answer for $k$. Then you should have $n^m = \sum\_{k=0}^{n-1}N\_k$
| https://mathoverflow.net/users/5378 | Maximum differences in sorted vectors of naturals | You are essentially looking at a graph (loops allowed) whose vertex set is $V=\{1,2,\dots,n\}$ with edges $E=\{(i,j) \ | \ |i-j|\le k\}$. These graphs are called *path-schemes*, see my answer [here](https://mathoverflow.net/questions/28649/how-many-hamiltonians-paths-there-are-in-almost-regular-graph/28654#28654).
Let $A$ be the adjacency matrix of this graph, then $\sum\_{r\le k} N\_k$ is equal the sum of all entries in $A^{m-1}$, because $(A^{m-1})\_{ij}$ is the number of walks in the graph of length $m-1$ from $i$ to $j$. I don't expect a simple expression in the end, however since $A$ is a Toeplitz matrix the calculations should be friendly.
| 1 | https://mathoverflow.net/users/2384 | 35448 | 22,865 |
https://mathoverflow.net/questions/35268 | 2 | Does P≠NP over ℝ imply P≠NP ?
where ℝ is for Real number algorithms as described by Smale with a suitable formulation of P≠NP over ℝ.
Complexity Theory and Numerical Analysis, Steve Smale, 2000
<http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.33.4678&rep=rep1&type=pdf>
| https://mathoverflow.net/users/8232 | Does P≠NP over ℝ imply P≠NP ? | Some years ago I read the paper "Computing over the Reals, Where Turing Meets Newton" by Lenore Blum, in which this question and related questions ("transfer results") are addressed: <http://www.ams.org/notices/200409/fea-blum.pdf>
| 7 | https://mathoverflow.net/users/83 | 35450 | 22,866 |
https://mathoverflow.net/questions/34581 | 6 | Denote $P[n]$ as the prime sequence $\{p\_1,p\_2,\cdots,p\_n\}$.
**Conjecture:**
* When $n=2k+1$ is odd, prime list $P[n]$ can be partitioned into two non-overlapping sublists, in which each sublist has equal sum $\operatorname{Total}[P[n]]/2$.
* When $n=2k$ is even, prime list $P[n]$ can be partitioned into two non-overlapping sublists, one sublist's sum is $(\operatorname{Total}[P[n]]-1)/2$, the other's is $(\operatorname{Total}[P[n]]+1)/2$.
For example:
$$
\begin{align\*}
3-2 &= 1 \\
5-3-2 &= 0 \\
7 - 5 - 3 + 2 &= 1 \\
11 - 7 - 5 + 3 - 2 &= 0 \\
13 - 11 - 7 + 5 + 3 - 2 &= 1
\end{align\*}
$$
and so on.
How could I write an efficient program to check it? Any clues to prove or disprove this conjecture?
BTW: I asked this question at mathgroup before, but I didn't describe it clearly.
| https://mathoverflow.net/users/8140 | A prime sequence can be partitioned into two sets of equal or consecutive sum | Scott Carnahan had an interesting idea; let's formalize it into an actual solution. We will show that, given $n \ge 2$ a positive integer, $p\_1, \cdots, p\_n$ the first *n* primes, we have some $e\_1, \cdots, e\_n$ with $e\_i = \pm 1$ such that $|e\_1p\_1 + e\_2p\_2 + \cdots + e\_np\_n| \le 1$. (Note that we may further stipulate that $e\_n = 1$.) A simple parity argument from here suffices to prove the conjecture.
We will prove this by induction on $n$. The cases $2 \le n \le 6$ are trivial to verify, and were provided already by a-boy. We now fix $n \ge 7$.
We first need some asymptotics in the form of the Bertrand-Chebyshev theorem; we use the formulation that for $m > 1$ there is a prime between $m$ and $2m$.
Write $S\_k = e\_np\_n + e\_{n-1}p\_{n-1} + \cdots + e\_{n-k+1}p\_{n-k+1}$, and let $M(k)$ be the minimum of $|S\_k|$ over all tuples $(e\_n, e\_{n-1}, \cdots, e\_{n-k+1})$. We stipulated earlier that $e\_n = 1$, so we have $M(1) = p\_n$. Two facts that will be useful to us in the future are that (1) $M(k+1) \le |M(k)-p\_{n-k}|$ and (2) if $|a| \le |b|$, then $\min{\{|a+b|,|a-b|\}} \le |b|$.
We claim that $M(k) \le p\_{n-k+1}$ for $k = 1, 2, \cdots, n-2$. We prove this by induction on $k$. The claim for $k = 1$ is trivial. Now if $M(k) \le p\_{n-k}$, then we are done, as $M(k+1) \le \min{\{|M(k)+p\_{n-k}|,|M(k)-p\_{n-k}|\}} \le p\_{n-k}$ by fact (2).
Now suppose $p\_{n-k} < M(k) \le p\_{n-k+1}$. Write $2m+1 = p\_{n-k+1} \ge p\_3 = 5$, so that $m > 1$. In this case we know that $m < p\_{n-k} < M(k) \le 2m$. But then $M(k+1) \le M(k) - p\_{n-k} \le 2m-(m+1) = (m-1) < p\_{n-k}$ as desired.
The fact that $M(k) \le p\_{n-k+1}$ is eminently useful.
Indeed, we may use it to dispatch of the even case immediately. Set $k = n-6$. Then we have $M(n-6) \le 17$. As all sums considered in $M(n-6)$ are sums of an even number of odd terms, we in fact have $M(n-6) \le 16$ and even. Now we simply note that all odd numbers between -15 and 15 are realizable as sums and differences of the first 6 primes, which is left as an easy computational exercise.
In the odd case, we consider $k = n-5$. Then $M(n-5) \le 13$. For the same parity reasons as above, we have in fact $M(n-5) \le 12$. And again, we note that all even numbers between -12 and 12 are realizable as sums and differences of the first 5 primes - another easy computational exercise.
The limits of $n-6$ and $n-5$ are the best possible for our small-case analysis.
If we were to establish an algorithm for this, we could just do the greedy algorithm on choosing $e\_n$, then $e\_{n-1}$, and so on, each time choosing $e\_k$ so as to minimize $S\_{k+1}$ (or randomly if $S\_k = 0$). Our claim that $M(k) \le p\_{n-k}$ will continue to be satisfied by the greedy algorithm, as the proof of the claim does not involve changing prior $e\_i$. Thus our greedy-algorithm mimium modulus must satisfy the same inequality, and we continue until we are at $n-6$ or $n-5$, then finish as in our nonconstructive proof.
| 5 | https://mathoverflow.net/users/8345 | 35458 | 22,871 |
https://mathoverflow.net/questions/34695 | 10 | Given a torus $T$ is there way to classify all the toric varieties it gives rise to? That is, classify all toric varieties $X$ whose torus is isomorphic to $T$. Is there a way to construct these toric varieties (i.e. give equations for them)?
Remark: as has been explained in the comments and answers, this question is I'll posed. But it seems to have generated good discussion so I'm leaving the phrasing as is.
| https://mathoverflow.net/users/7 | Counting/constructing Toric Varieties | As far as my understanding goes the answer is no, and I will try to explain why and clarify the list of comments (I have little reputation so I cannot comment there) and give you a partial answer. I hope I do not patronise you, since you may now already part of it.
First of all, as Torsten said, it depends what you understand for classification. In this context a **torus** $T$ of dimension $r$ is always an algebraic variety isomorphic to $(\mathbb{C}^\*)^r$ as a group. A complex algebraic variety $X$ of finite type is *toric* if there exists an embedding $\iota: (\mathbb{C}^\ast)^r \hookrightarrow X$, such that the image of $\iota$ is an open set whose Zariski closure is $X$ itself and the usual multiplication in $T=\iota((\mathbb{C}^\ast)^r)$ extends to $X$ (i.e. $T$ acts on $X$).
Think about all toric varieties. It is hard to find a complete classification, i.e. being able to give the coordinates ring for each affine patch and the morphisms among them for *all* toric varieties.
However, when the toric varieties we consider are normal there is a structure called the fan $\Sigma$ made out of cones. All cones live in $N\_\mathbb{R}\cong N\otimes \mathbb{R}$ where $N\cong \mathbb{Z}$ is a lattice. A cone is generated by several vectors of the lattices (like a high school cone, really) and a fan is a union of cones which mainly have to satisfy that they do not overlap unless the overlap is a face of the cone (another cone of smaller dimension). There is a concept of morphism of fans and hence we can speak of fans 'up to isomorphism' (elements of $\mathbf{SL}(n,\mathbb{Z})$). Given a lattice N, there is an associated torus $T\_N=N\otimes (\mathbb{C}^\*)$, isomorphic to the standard torus.
Then we have a 1:1 correspondence between *separated* normal toric varieties $X$ (which contain the torus $T\_N$ as a subset) up to isomorphism and fans in $N\_\mathbb{R}$ up to isomorphism. There are algorithms to compute the fan from the variety and the variety from the fan and they are not difficult at all. You can easily learn them in chapter seven of the Mirror Symmetry book, [available](http://www.claymath.org/library/monographs/cmim01.pdf) for free. Given any toric variety (even non-normal ones) we can compute its fan, but computing back the variety of this fan many not give us the original variety unless the original is normal. You can check this easily by computing the fan of a $\mathbf{V}(x^2-y^3)$ (torus embedding $(t^3,t^2)$) which is the same as $\mathbb{C}^1$ but obviously they are not isomorphic (the former has a singularity at (0,0)). In fact, since there are only two non-isomorphic fans of dimension 1 (the one generated by $1\in \mathbb{Z}$ and the one generated by 1 and -1) we see that there are only three normal toric varieties of dimension 1, the projective line and the affine line, and the standard torus.
The proof of this statement is not easy and to be honest I have never seen it written down complete (and I would appreciate any reference if someone saw it) but I know more or less the reason, as it is explained in the book about to be published by Cox, Little and Schenck (partly [available](http://www.cs.amherst.edu/~dac/toric.html)) This theorem is part of my first year report which is due by the end of September, so if you want me to send you a copy when it is finished send me an e-mail.
So, yes, in the case of normal varieties there is some 'classification' via combinatorics, but in the case of non-normal I doubt there is (I never worked with them anyways).
[Become a toric fan!](http://www.facebook.com/pages/Become-a-Toric-Fan/109010825806946).
| 12 | https://mathoverflow.net/users/1887 | 35466 | 22,874 |
https://mathoverflow.net/questions/35469 | 18 | It is a well known result that a random walk on a 2D lattice will return to the origin [see Polya's random walk constant](http://mathworld.wolfram.com/PolyasRandomWalkConstants.html). Based on this, it is not a big stretch to conclude that the random walk will visit every point of the plane with probability 1. A bit more surprising is the fact that this is not true in higher dimensions (see the link above).
My question is the following: What is the probability that two random walks with distinct origins will arrive at the same point after the same number of steps?
I think it's pretty clear that the answer will depend on the distance between the origins of the walks. So far, I've been trying reduce this to a problem of one random walk in a higher dimensional lattice, but I'm not sure if this is a good approach.
In case the answer is obvious, this problem is easy generalize (higher dimensional lattices, more random walks, etc.). I'd appreciate a reference or two where I could learn more.
Thanks
| https://mathoverflow.net/users/5593 | What is the probability that two random walkers will meet? | The difference between the positions is another random walk in the same dimension. You can either view the steps as different, or sample a random walk at even times. So, the probability is $0$ if meeting is ruled out by parity, and $1$ in the plane if meeting is possible.
| 30 | https://mathoverflow.net/users/2954 | 35471 | 22,876 |
https://mathoverflow.net/questions/35465 | 3 | In "Brauer groups and quotient stacks", Edidin et. al prove the following theorem:
Theorem 2.7. Let $\mathcal{X}$ be an algebraic stack over a Noetherian base (of finite type). Then the diagonal $\mathcal{X}\to \mathcal{X}\times \mathcal{X}$ is quasi-finite if and only if there is a finite surjective morphism $X\to \mathcal{X}$ for a scheme $X$.
On the other hand, Kresch in "Cycle groups for Artin stacks" proves the following:
Proposition 3.5.7. Let $\mathcal{X}$ be a stack of finite type over a field. The the following are equivalent:
1) The diagonal is quasi-finite;
2) The stabilizer $\mathcal{X}\times\_{\mathcal{X}\times\mathcal{X}}\mathcal{X}\to \mathcal{X}$ is quasi-finite.
Further, if $\mathcal{X}$ has quasi-finite diagonal $\mathcal{X}$ admits a stratification by quotient stacks.
Now, suppose that $\mathcal{X}$ is already a quotient stack $[Y/G]$, say with $Y$ an affine scheme and $G$ some group scheme (both of finite type over a field). Then $\mathrm{id}: Y\to\mathcal{X}$ is a finite surjective morphism, so by 2.7 above have quasi-finite diagonal. Then by 3.5.7 the stabilizer is quasi-finite, but this seems false in general. For instance, take $G=GL(n)$ and then you are almost guaranteed to have non-finite stabilizers.
What am I missing here? It's obvious that there's something here that I've gotten wrong.
| https://mathoverflow.net/users/2147 | Question about global quotient stacks | The morphism $Y \to [Y/G]$ is a $G$-torsor, so it is finite only if $G$ is finite.
| 4 | https://mathoverflow.net/users/4790 | 35477 | 22,880 |
https://mathoverflow.net/questions/35462 | -1 | Suppose there are $ K $ buckets each can be filled upto $ N-1 $ balls. The gain on putting $ i $ balls in the $ k^{th} $ bucket is given by $ \Delta l\_{k,i}, \, i \in [1,N-1] $. The problem is to put $ \lambda $ balls in those buckets to maximize the overall gain.
How do we solve it?
| https://mathoverflow.net/users/8447 | dynamic programming and combinatorics | You can use dynamic programming as you suggest in the title.
Let $w\_{ij}$ be the max gain you can get putting $j$ balls into first $i$ buckets.
Then $w\_{ij}$ has the following recursive relation:
$$
w\_{i,j} = \max\_{0 \le t \le \min(N-1, j)}(w\_{i-1,j-t} + \Delta l\_{i, t})
$$
There $t$ is the number of balls you put into $i$-th bucket.
Hence you can calculate $w\_{K, \lambda}$ (your answer) in time $O(K \lambda N)$.
| 0 | https://mathoverflow.net/users/7079 | 35478 | 22,881 |
https://mathoverflow.net/questions/35479 | 6 | Background
----------
I am interested in the projective classification of reduced curves of degree four in $\mathbb{P}^3(\mathbb{R})$ (and more generally of degree $n+1$ in $\mathbb{P}^n(\mathbb{R})$). More precisely, I am looking at the case where the curve is a union of four distinct lines. I need this classification because I want to make sure that I consider all possible cases in a problem in interpolation theory.
---
For instance, there are two types of configurations of three lines in $\mathbb{P}^2$. Either three lines meet in a single point, or three lines meet in three distinct points. More generally, according to [this integer sequence](http://www.research.att.com/~njas/sequences/A132346), there are 3 configurations of four lines in $\mathbb{P}^2$, 5 configurations of five lines in $\mathbb{P}^2$, and 18 configurations of 6 lines in $\mathbb{P}^2$. These configurations are shown in [this figure](http://www.research.att.com/~njas/sequences/a048872.pdf) (except for the configurations in which all lines are concurrent).
I believe there are six configurations of three lines in $\mathbb{P}^3$: Two configurations for which the three lines lie in a plane, three configurations for which precisely two of the three lines lie in a plane, and one configuration where none of the lines intersect.
My (related) questions are now as follows:
>
> 1. How many configurations are there of four lines in $\mathbb{P}^3$ (and more generally of $n+1$ lines in $\mathbb{P}^n$)?
> 2. Is there a convenient way to enumerate these?
>
>
>
| https://mathoverflow.net/users/1792 | Up to projectivities, which configurations of four lines in $\mathbb{P}^3$ can one distinguish? | Up to projectivities, there are uncountably many configurations. Let's do the naive dimension count: The Grassmannian of lines in $\mathbb{P}^3$ is four dimensional, so the parameter space for four lines is 16 dimensional. The automorphism group of $\mathbb{P}^3$, the projections, is made up of four by four matrices modulo the diagonal matrices, so is dimension 16-1=15. So we should get a whole curve worth of possible configurations.
This is analagous to how you can use cross ratio to distinguish different configurations of four points in $\mathbb{P}^1$.
EDIT: To make this more general and distinguish it from jvp's comment, if you look at $n+1$ lines in $\mathbb{P}^n$, then you have $(n+1)^2-1=n^2+2n$ automorphisms, and the space of lines is $2(n-1)$ dimensional, so in general you have $2(n-1)(n+1)-(n+1)^2+1$ dimensions worth of configurations, and this simplifies to $n^2-2n-2$.
| 5 | https://mathoverflow.net/users/622 | 35490 | 22,888 |
https://mathoverflow.net/questions/35461 | 7 | I'm working on a proof-checker that can verify termination proofs. The fundamental method it provides for constructing such proofs is to translate the program into primitive recursion. Basically, I provide a combinator $\rho$ typed as:
$\rho: \forall A,B:(A\rightarrow Nat \rightarrow A)\rightarrow (A \rightarrow B)\rightarrow A\rightarrow Nat \rightarrow B$
which, in the notation defined [here](http://en.wikipedia.org/wiki/Primitive_recursion), constructs $h$ given $f$ and $g$.
Although the term language contains a fixed-point combinator and is therefore Turing-complete, terms that use it have a "tentative" flag in their type that indicate this. The $\rho$ combinator and the fixed-point combinator are the only two language primitives that allow for recursion or looping of any sort (i.e., without either of these two combinators, all you've got is a finite-state machine). Therefore, all terms that are well-typed and non-tentatively typed are primitive recursive.
What I'm wondering is if there are any interesting complexity classes that you can build by starting with primitive-recursive constructions, and adding a finite number of other functions $Nat \rightarrow Nat$, each of which is in R but not in PR, and allowing composition with these functions. It's easy to come up with *non*-interesting examples of such classes, e.g. "primitive recursion plus the Ackermann function", but I'm looking for any that have sufficiently interesting properties that it would be worth adding the functions which characterize them as admitted axioms in the proof system.
| https://mathoverflow.net/users/8272 | Interesting complexity classes $PR \subsetneq c \subsetneq R$ | First of all, it’s certainly possible to obtain *some* intermediate class by taking a language that only computes PR functions (say, an imperative programming language [using only `for` loops](http://en.wikipedia.org/wiki/BlooP_and_FlooP)) and adding any total computable but non PR function (e.g., Ackermann’s function). The resulting language *L* is non-universal, because it only computes total functions: you can still construct a computable but non-*L*-computable function by diagonalisation. However, *L* is clearly more powerful than the original language.
As for “interesting”, I guess it really depends on what you mean by that.
If “interesting” means “of practical use”, then one could answer that all computable functions of practical use are PR, since a non-PR function requires an amount of time to compute that is not, in turn, PR. Considering that time bounds such as 2*n*, 2*2**n*, 2*2**2*n**, …, are all PR, you see that there isn’t much hope to compute non-PR functions for large values of *n*.
If “interesting” means “logically interesting”, then I think the answer is “yes”. I’m somewhat familiar with Girard’s System F (also called “second order λ-calculus” or “polymorphic λ-calculus”), described for instance in Girard’s *Proofs and Types* (freely available [here](http://www.paultaylor.eu/stable/Proofs+Types.html)). The functions that can be computed in F are “*exactly* those which are provably total in [second order Peano arithmetic]” (page 123), and among these we have Ackermann’s function. There is an explicit λ-term for it on [these slides](http://www.pps.jussieu.fr/~miquel/enseignement/mpri/SystF.pdf) (page 20).
If I recall correctly, the standard calculus of constructions includes System F and only computes total functions, so it also provides an example.
| 7 | https://mathoverflow.net/users/5304 | 35495 | 22,892 |
https://mathoverflow.net/questions/35472 | 1 | Let M be a Riemannian Manifold, $X$ is a smooth vector field on M with isolated zeros.
Is there a one-form $\omega$ with isolated zeros such that $\omega(X)$ has nontrivial zeros? (nontivial zero means that the piont is neither in $X$'s zeros nor in $\omega$'s zeros.)
If this $\omega$ exist, how to construct it?
| https://mathoverflow.net/users/3896 | A question about a one-form on Riemannian manifold | Assuming the dimension of $M$ is at least 2 (otherwise it's false), you can do the following. Let $p\_1,p\_2,\dots$ be isolated points where $X$ does not vanish but where you want $\omega$ to vanish. In a neighborhood $U\_i$ of each $p\_i$, there are coordinates $(x^1,\dots,x^n)$ centered at $p\_i$ on which $X$ has the coordinate representation $X = \partial/\partial x^1$. In each $U\_i$, let $\omega\_i = dx^2 + |x|^2 dx^1$. Then let $U\_0$ be the complement of {$p\_1,p\_2,\dots$}, and let $\omega\_0=X^\flat$ (the 1-form dual to $X$ via the metric). Let {$\phi\_0,\phi\_i$} be a partition of unity subordinate to the cover {$U\_0,U\_i$}, and let $\omega = \sum\_{i\ge 0}\phi\_i\omega\_i$. The fact that $\omega\_i(X)>0$ at points other than $p\_i$ and zeros of $X$ ensures that $\omega(X)$ vanishes only at such points.
| 5 | https://mathoverflow.net/users/6751 | 35497 | 22,894 |
https://mathoverflow.net/questions/35428 | 4 | Let $E$ be a rank two vector bundle on $\mathbb{P}^n$. Assume that $\text{Ext}^1(E, E)=0$. Will $\text{Ext}^2(E, E)$ be zero? Why? Any geometric explanation (in terms of deformation theory?)?
Edit: As pointing out by Angelo, in the case $n=2$, the answer is no. However, I really want to know when $n\geq 4$.
| https://mathoverflow.net/users/2348 | Vanishing of Self-Ext groups of vector bundles | To complement Angelo's answer (which you should accept, as it answers your original question):
If $\mathrm H^1(E^\vee \otimes E) = 0$ then $E$ must be homogeneous, see for instance Theorem 3 [this paper](http://www.math.wustl.edu/~kumar/papers/hydkempf.pdf) by Mohan Kumar.
It is well-known that homogeneous vector bundles on $\mathbb P^n$ of rank less than $n$ splits as a direct sum of line bundles (Theorem 3.2.3 of Okonek-Schneider-Spindler). So if $n>2$ the answer to your question is yes.
| 6 | https://mathoverflow.net/users/2083 | 35498 | 22,895 |
https://mathoverflow.net/questions/35443 | 2 | Let be $\mathcal M(\partial\mathbb D)$ denote the set of all Borel complex probability measures on $\partial\mathbb D$ (unit circle in the complex plane). Define a mapping $\Phi:\mathcal M(\partial\mathbb D)\to \ell^{\infty}(\mathbb N)$ by
$$
\Phi(\mu)=(c\_0,c\_1,\ldots)
$$
where $c\_k$, are the coefficients of the Taylor expansion (around $z=0$) of the function
$$
f(z)=\int\_{\partial\mathbb D}\frac{1}{1-wz} d\mu(w),
$$
**Question:** Who is the set $\Phi(\mathcal M(\partial\mathbb D))$ ? Given a sequence is there a simple criterion to verify if it belongs to this space?
**Motivation:** Let be $A=(a\_0,a\_1,\ldots)\in\Phi(\mathcal M(\mathbb D))$, $g:U\subset\mathbb C\to \mathbb C$ analytic with
$$g(z)=\sum\_{k=0}^{\infty}b\_kz^k$$
Define $A\*g:U\subset\mathbb C\to \mathbb C$ by
$$
A\*h(z)=\sum\_{k=0}^{\infty} a\_kb\_kz^k.
$$
The above integral representation, can be used to give a short proof of
$$
\|A\*h\|\_{U}\leq \|h\|\_{U},
$$
where $\|g\|\_{U}=\sup\_{z\in U}|g(z)|$.
PS:This question it is a generalization of a question that arose in a discussion at Area 51.
**Edition:** I am correcting the question, because in the previous version the integrals could have no meaning. In fact, I was looking for the criterion by a line integral representation.
| https://mathoverflow.net/users/2386 | Coefficients of holomorphic functions defined by Borel probability measures on the unit disc | There's probably something I do not understand about your question, in case just forget my babbling. Anyway let me try: if you expand
$$\frac{1}{1-wz}=\sum\_{k\ge0}z^kw^k$$
and write
$$f(z)=\sum\_{k\ge0}z^k\int\_{\partial D}w^k d\mu(w)=
\sum\_{k\ge0}z^k\int\_0^{2\pi}e^{ik\theta}d\mu(\theta)$$
you see that the $c\_k$ are the Fourier coefficients of positive order of the measure $\mu$. Now I would not go too far and say that any bounded sequence may apply, but this is well documented in several places (see e.g. the [trigonometric moment problem](http://en.wikipedia.org/wiki/Trigonometric_moment_problem)).
EDIT: and maybe [this](https://facultystaff.richmond.edu/~wross/PDF/cauchy.pdf) is a good starting point.
| 3 | https://mathoverflow.net/users/7294 | 35499 | 22,896 |
https://mathoverflow.net/questions/35439 | 10 | Can the Dikgraaf-Witten model for a finite gauge group $G$ [Robbert Dijkgraaf and Edward Witten, Topological Gauge Theories and Group Cohomology, Commun. Math. Phys. 129 (1990), 393] be described in terms of the geometry of moduli spaces $\overline{\mathcal{M}}\_{g,n,\beta}([\*//G])$ of stable maps to the stack $[\*//G]$? I strongly suspect the answer is yes, in view of the classical relation between 3-dimensional topological quantum field theories and complex analytic 2-dimensional modular functors, but I'm unaware of rigorous results in this direction.
| https://mathoverflow.net/users/8320 | The algebro-geometric counterpart of the Dijkgraaf-Witten model | This has been done, in a variety of related ways. A lot of the difficulty is in defining an appropriate notion of a "stable" map to [pt/G].
The earliest mathematical work I know of is Chen & Ruan's "orbifold cohomology", which is done in the symplectic category. (Caveats: Abramovich's lecture notes on orbifold GW theory quote a 1996 letter from Kontsevich, who outlines a lot of the basic ideas in 2 pages. Also, string theorists were looking at **non**-topological sigma models to orbifolds at least as far back as Dixon, Harvey, Vafa, & Witten's 1985 papers.)
In algebraic geometry, this stuff has been studied by Jarvis, Kaufmann, & Kimura, who focused on G-bundles, and by Abramovich, Graber, & Vistoli, who figured out how to deal with D-M stacks.
(You can also carry out these constructions in K-theory for finite-dimensional Lie groups. See, for example, Frenkel, Teleman, & [cough].)
| 6 | https://mathoverflow.net/users/35508 | 35504 | 22,899 |
https://mathoverflow.net/questions/35507 | 13 | Using the duality between locally compact Hausdorff spaces and commutative $C^\*$-algebras one can write down a vocabulary list translating topological notions regarding a locally compact Hausdorff space $X$ into algebraic notions regarding its ring of functions $C\_0(X)$ (see Wegge-Olsen's book, for instance). For example, we have the following correspondences:
$$
\;\;\;\text{open subset of $X$}\quad \longleftrightarrow\quad\text{ideal in $C\_0(X)$}
$$
$$
\;\;\;\;\;\quad\text{dense open subset of $X$}\quad \longleftrightarrow\quad\text{essential ideal in $C\_0(X)$}
$$
$$
\;\;\;\quad\text{closed subset of $X$}\quad \longleftrightarrow\quad\text{quotient of $C\_0(X)$}
$$
$$
\text{locally closed subset of $X$}\quad \longleftrightarrow\quad\text{subquotient of $C\_0(X)$}
$$
$$
\;\;\;\quad\qquad\qquad\qquad\qquad\text{???}\qquad\qquad \longleftrightarrow\quad\text{$C^\*$-subalgebra in $C\_0(X)$}
$$
By ideal I always mean a two-sided closed (and hence self-adjoint) ideal.
Well, I can't quite see **how to reconvert a $C^\*$-subalgebra in $C\_0(X)$ into something topological involving only the space $X$ (and some data describing the subalgebra in topological terms).** Can you come up with something handy?
---
*Example:* A simple example of a subalgebra of a commutative $C^\*$-algebra not being an ideal is
$$
\mathbb C\cdot(1,1)\subset \mathbb C\oplus\mathbb C.
$$
---
*First attempts:* Instead of talking about a subalgebra, we should probably talk about the injective $^\* $-homomorphism given by the inclusion of this subalgebra. But is this inclusion proper (i.e., does it preserve approximate units) in general? Well, at least when we restrict to compact spaces. Then an injective $^\* $-homomorphism $C(Y)\to C(X)$ will induce a surjective continuous map $X\to Y$. How to proceed?
---
*Remark:* Alternatively, we could think about this question within the duality of affine algebraic varieties and finitely generated commutative reduced algebras or even within the duality between affine schemes and commutative rings.
---
**Disclaimer:** I posted this question yesterday on [MSE](https://math.stackexchange.com/questions/2229/what-is-the-commutative-analogue-of-a-c-subalgebra). I also got an interesting answer. However, I'm not yet fully satisfied. If I violate any policy by reposting the question here, please tell me about it.
| https://mathoverflow.net/users/1291 | What is the commutative analogue of a C*-subalgebra? | Let $A$ be a commutative $C^\star$-algebra and $B$ a $C^\star$-subalgebra (in general, this will not preserve approximate units as the example ${\mathbb C} \oplus 0 \subset {\mathbb C} \oplus{\mathbb C}$ shows). This gives also rise to an inclusion of the unitalization $B^+$ into $A^+$. Now you are in the realm of unital $C^\star$-algebras and your own remark applies. If $A$ was isomorphic to the algebra of continuous functions on the locally compact space $X$, then $A^+$ is the algebra of continuous functions on the one-point compactification $X^+$ of $X$. Now, $B^+ \cong C(Y)$, where $Y$ is some quotient of the one-point compactification of $X$. And again, $Y$ is the one-point compactification of some space locally compact space $Z$ such that $B = C\_0(Z)$. What is the relation between $Z$ and $X$? Well, the continuous quotient map $f \colon X^+ \to Z^+=Y$ gives rise to a canonical open subset $U \subset X$, which arises as the pre-image of $Z$. Moreover, one gets a quotient map $f|\_U \colon U \to Z$ and $f|\_U$ is easily seen to be proper. This is everything that can be said.
(Conversely, let $U \subset X$ be open and $g \colon U \to Z$ be a proper surjection. Then
$C\_0(Z) \to C\_0(U) \subset C\_0(X)$ gives rise to a sub-algebra.)
This is explained in more detail in the book by Higson and Roe ([see here](http://books.google.de/books?id=r-Icer50QoIC)).
| 14 | https://mathoverflow.net/users/8176 | 35509 | 22,902 |
https://mathoverflow.net/questions/35491 | 8 | A class of "minimally 2-vertex-connected graphs" - that is, 2-vertex-connected graphs which have the property that removing any one vertex (and all incident edges) renders the graph no longer 2-connected - have come up in my research.
Dirac wrote a paper on "minimally 2-connected graphs" (G. A. Dirac, Minimally 2-connected graphs, J. Reine Angew. Math. 228 (1967),. 204-216), which gives quite a detailed description of the structure of such graphs. However, in his sense, minimal 2-connectivity means that deleting any EDGE leaves a graph which is not 2-connected, which is not an equivalent property to the vertex-deletion one. Does anyone know anything about graphs with the latter property?
In the hope of stimulating some discussion, here is a wildly speculative and vague conjecture: The only graphs satisfying this property are simple cycles, and certain cycles with chords.
| https://mathoverflow.net/users/4078 | Minimally 2-vertex-connected graphs? | Here is a more general family:
Draw your favorite tree in the plane, with circles for the nodes and "thick" lines for the edges. Now turn every circle into a cycle, and every thick line into a pair of parallel paths $p\_1, \ldots, p\_m$ and $q\_1, \ldots, q\_n$ with various crossbraces. The crossbraces just have to follow the rule that if $p\_i$ is connected to $q\_\ell$ and $p\_j$ is connected to $q\_k$, for $i<j$ and $k<\ell$, then $j =i+1$ or $\ell=k+1$.
This is probably still not close to a complete characterization, but at least shows that the class is a lot broader than the small class you posited to promote discussion.
| 2 | https://mathoverflow.net/users/7936 | 35521 | 22,906 |
https://mathoverflow.net/questions/35453 | 2 | Hello,
I'm looking for a formula or algorithm to find the number of cycles of a certain length $k$ in a graph.
I know that $(A^k)\_{ii}$ gives me the number of cycles from vertex $i$ to itself ($A$ is the adjacency matrix), but these are cycles that might contain the same vertex twice.
I have to tried to devise some sort of a recurrence formula but to no avail.
Thanks!
| https://mathoverflow.net/users/8446 | Finding all cycles of a certain length in a graph | Is your graph topologically planar or non-planar, weighted or unweighted, directed or undirected?
Do you want an algorithm and/or a formula/bound?
For bounds on planar graphs, see [Alt et al. On the number of simple cycles in planar graphs](http://www.ams.org/mathscinet-getitem?mr=1731975)
For an algorithm, see the following paper. It incrementally builds k-cycles from (k-1)-cycles and (k-1)-paths without going through the rigourous task of computing the cycle space for the entire graph. It also handles duplicate avoidance.
* Hongbo Liu; Jiaxin Wang; , *"A new
way to enumerate cycles in graph*,"
Telecommunications, 2006. AICT-ICIW
'06. International Conference on
Internet and Web Applications and
Services/Advanced International
Conference on , vol., no., pp. 57-
57, 19-25 Feb. 2006 doi:
10.1109/AICT-ICIW.2006.22 URL: <http://www.ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=1602189&isnumber=33674>
| 4 | https://mathoverflow.net/users/5372 | 35528 | 22,908 |
https://mathoverflow.net/questions/35525 | 1 | Let $L$ be a language in $NP$. Then are there any results on whether there exists a polynomial-time algorithm (polynomial in the length of the description of $L$) to decide whether $L \in P$? Are there any results on the hardness of the search version of the problem vs. the decision version?
The only result I can think of is that $P=NP$ implies the existence of such an algorithm since then an algorithm which outputs $True$ everytime works.
| https://mathoverflow.net/users/1612 | A polynomial-time algorithm for deciding whether a language has a polynomial time algorithm | If there is such an algorithm, then P = NP.
.
L(m) := {s : machine(m) halts within length(s) steps and sat\_instance(s) is true}
If P != NP, then "Is L(m) in P?" is equivalent to "Does machine(m) run forever?".
| 6 | https://mathoverflow.net/users/nan | 35530 | 22,909 |
https://mathoverflow.net/questions/35526 | 4 | Eric Broug in his book *Islamic Geometric Patterns* gives
straightedge and compass construction of some simpler patterns.
It is clear his techniques will provide constructions for many
Islamic patterns.
Looking at formal constructibility, the Wikipedia pages gives Gauss' result that
7, 9, 11, 13, 14, 18... etc sided polygons are not constructible. Hence the pattern
<http://tilingsearch.org/HTML/data160/J43C.html>
is not constructible since it contains a regular
9-pointed star polygon.
I have over 800 Islamic patterns on my web site but
I use a computer and trigonometry to produce my images. It seems that
about 40 Islamic patterns on my site are not constructible.
Given an Islamic pattern that is not excluded from construction by Gauss'
result, it is almost certainly constructible if the following is true:
>
> Given two points on the plane, a polygon ($n$ sides) can be constructed with the two points
> as an edge, provided $n$ is not equal to 7, 9, 11, 13, 14, 18... etc.
>
>
>
This result would allow patterns to be built up piece-by-piece.
EDIT, Will Jagy: from his profile page, the OP's website, in this address preset to display the tilings in a slideshow format on a web browser, is at
<http://www.tilingsearch.org/>
| https://mathoverflow.net/users/8461 | What Islamic tiling patterns are constructible? | Yes, that is right. It seems all you are missing is this: given a number of sides $n$ such that the regular polygon of $n$ sides is constructible (by the results of Gauss and Wantzel), how to force the edge length to be a fixed length, call it $L$?
$$ $$
All you need to do is construct the regular $n$-gon. Draw the perpendicular bisector of any edge, which is then a line that passes through the center of the regular polygon, and stop the line at that center so that it is actually a half-line, a "ray." Call this ray $r.$ The ray $r$ starts in the interior of one central triangle of the regular polygon, call the radii acting as the two othe edges ot that triangle $e\_1$ and $e\_2$
$$ $$
Now, parallel to $r,$ construct the parallel line at distance $$ \frac{L}{2}$$
from $r.$ At some point $P,$ the ray $r$ intersects either $e\_1$ or $e\_2.$ The point $P$ is a vertex of the polygon you want. To find the rest of the polygon just draw a circle with center at the center of the polygon and passing through $P,$ then extend the radii passing through the vertices of your original polygon until they reach the circle.
| 4 | https://mathoverflow.net/users/3324 | 35532 | 22,910 |
https://mathoverflow.net/questions/35512 | 1 | The simplest case of the problem I'm thinking about involves an elliptic differential operator, $Lu = -u'' + qu$, on the interval $(0,1)$, with homogeneous Dirichlet boundary conditions. I want to show that the bilinear form on $H\_0^1 \subset H\_1$ defined by
$a(u,v) = \int\_0^1 u'v' + quv~dx$
is bounded for the $H^1$-norm, i.e., $|a(u,v)| \leq M\|u\|\_1\|v\|\_1$ for some constant $M>0$.
My question: can I assume that the linear coefficient $q$ is $L^1$ or even $L^2$ and still guarantee boundedness?
I was thinking that this is possible, but the only books that I have lying around discussing this consider only the case when $q$ is smooth or $L^\infty$. I've played around with the Cauchy-Schwarz inequality for the term $\int quv$ but am not getting anywhere.
| https://mathoverflow.net/users/8353 | variational formulation: boundedness of the bilinear form | Of course, as Helge says, the $H^1(0,1)$ norm controls the $L^\infty(0,1)$ norm, so you just write
$$ \left| \int quv \right| \le \|q\|\_{L^1} \|u\| \_{L^\infty} \|v\| \_{L^\infty} \le C
\|q\| \_{L^1} \|u\| \_{H^1} \|v\| \_{H^1}.$$
| 1 | https://mathoverflow.net/users/7294 | 35533 | 22,911 |
https://mathoverflow.net/questions/35524 | 14 | In the usual presentation of Goodstein's theorem, the base is bumped up by the "add 1" function. Does the theorem still hold when we replace this function by a fast-growing one (e.g. Ackermann or busy beaver)? How far can we push this? For example, let's define $g\_0(n)$ to be the number of Goodstein iterations needed to reach 0 when we start with base 2 and seed $n$ (so that $g\_0(0)$ = 0). Then we can build a hierarchy of functions by defining $g\_{k+1}(n)$ as the number of Goodstein iterations needed to reach 0 with seed $n$ and base-bumping function $g\_k$ ($k$ = 0, 1, ...), continuing through the ordinals by diagonalization at each limit ordinal. Surely it's got to break down when we go past $\varepsilon\_0$, if not long before that!
| https://mathoverflow.net/users/7458 | How fast can the base-bumping function in Goodstein's theorem grow? | As long as your fast-growing "base-bumping" function still takes every natural number to a natural number (instead of, say, an infinite ordinal)--and the busy beavers do--the Goodstein iterations are still upper-bounded by the strictly-decreasing sequence of ordinals in "base" $\omega$, which must be of finite length as a decreasing sequence in a well-ordered set.
| 10 | https://mathoverflow.net/users/7936 | 35535 | 22,912 |
https://mathoverflow.net/questions/35438 | 1 | I have a simple little analysis question that I'm hoping is well known.
Suppose $D=\lbrace(x,y): x^2+y^2<1\rbrace$ is the unit disk and that $u$ is a harmonic function on $D$. Suppose in addition that $u(0)=0$ and $\nabla u(0)=0$. Lets also assume $u$ has finite $L^2$ norm -- i.e. $||u||\_2<\infty$
If $u\_{xy}=\partial\_{xy}u=0$ on $D$ then it is clear that $u=a(x^2-y^2)$ for some $a$.
Suppose instead that we only know that $|| u\_{xy}||\_2<<1$. I'm pretty sure a compactness argument implies that there is an $a$ so that $||u-a(x^2-y^2)||\_2<<1$ (or at least this is true on any smaller disk).
What I'm most interested in is whether there is a constant $C>0$ so that
$ \inf\_{a\in \mathbb{R}} ||u-a(x^2-y^2)||\_2\leq C ||u\_{xy}||\_2$.
(I'm also happy if the control was only on a smaller disk).
On a related note, does anyone know if $u\_{xy}=0$ in a distributional sense then $u=F(x)+G(y)$. I can't remember now if this is true or not...
| https://mathoverflow.net/users/26801 | Partial $L^2$ control on (part of) the Hessian of a harmonic function. | Okay so I thought about this some more and I believe it is just a (really) straightforward application of the Poincare/Wirtinger inequality and the obvious way you would solve $u\_{xy}=0$ classically.
Lets work on the square $S=(0,2\pi)\_x\times(0,2\pi)\_y$ as it is simpler to work there and doesn't change much.
We assume $u$ is $C^\infty\_0(S)$ (i.e. smooth and of compact support) for the time being but do not need it to be harmonic.
Set $f(s)=\frac{1}{2\pi}\int\_0^{2\pi} u\_x(s, t)dt$ so $\int\_0^{2\pi} u\_{x}(s, t)-f(s) dt=0$ for all $s$.
By Wirtinger's inequality we have
$\int\_0^{2\pi} (u\_x(s,t)-f(s))^2 dt \leq \int\_0^{2\pi} u\_{xy}^2(s,t) dt$.
Now pick a $F$ so that $F'(x)=f(x)$.
Set $G(t)=\frac{1}{2\pi}\int\_0^{2\pi} u(s,t)-F(s) ds$ so $\int\_0^{2\pi} u(s,t)-F(s)-G(t) ds=0$ for all $t$. Since $\frac{d}{ds}\left( u(s,t)-F(s)-G(t)\right)=u\_x(s,t)-f(s)$ we have by Wirtinger's inequality that:
$\int\_0^{2\pi} (u(s,t)-F(s)-G(t))^2 ds \leq \int\_0^{2\pi} (u\_x(s,t)-f(s))^2 $ for all $t$.
Then by Fubini we have that
$\int\_{S} (u(x,y)-F(x)-G(y))^2 \leq \int\_S u^2\_{xy}$
Everything is valid if $u$ is in $H^2(S)$. Notice the $F,G$ are given explicitly in terms of $u$.
If $u$ is harmonic with the normalizing properties at $0$ then the $F=ax^2, G=-ay^2$.
At least I think this works...
| 1 | https://mathoverflow.net/users/26801 | 35543 | 22,918 |
https://mathoverflow.net/questions/35561 | 2 | I am not number theorist, forgive me if this is a stupid question.
Recently I was curious about the ideas behind the transcendence of $\log 2$.
For the number $e$, It seems that the transcendence can be obtained by a argument of fast convergence of the Taylor expansion but the same ideas do not apply to $\log 2$. Talking with number theorists I got an explanation based on the results of Baker. Unfortunately this person can not point me towards a survey or a paper discussing the ideas behind this transcendence so I come here to ask:
*What are the ideas behind the transcendence of this number?*
Any text (paper,book, blog) suitable for non number theorists is also welcome as an answer.
| https://mathoverflow.net/users/2386 | Transcendence of $\log 2$ | This follows from the [Lindemann-Weierstrass theorem](http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem). There is a sketch of a proof and several references at the Wikipedia article.
| 5 | https://mathoverflow.net/users/290 | 35562 | 22,930 |
https://mathoverflow.net/questions/35487 | 0 | Given the following:
* an $(n \times z)$ matrix $A = {(a\_1,a\_2, ... ,a\_n)}^{T}$ where $z \geq n$ and every $a\_i$ is a $z$-dimensional row vector.
* $a\_i = [a\_{i1} a\_{i2} ... a\_{iz}]$ where $a\_{ij} \geq 0 \forall j$
* $\sum\_{i=1}^{z}a\_{ri} = 1, \forall r \in ${$1,2,...,n$}.
* $\sum\_{i=1}^{z}|a\_{pi} - a\_{qi}| \leq \epsilon, \forall p,q$ where $\epsilon << 1$.
Find a provable upper bound on:
* $\sum\_{i,j=1}^{z}|(1/n)\*\sum\_{k=1}^{n}[a\_{ki}.(a\_{f(k)j} - a\_{g(k)j})]|$
where f and g are permutations over the set {$1,2,...,n$} such that $f(i) \neq g(i) \forall i$.
I am expecting the bound to be $\epsilon^2$ but I have no idea how to prove it.
| https://mathoverflow.net/users/8449 | a provable upper bound on the summation | Here is a copy-and-paste of the [answer](https://math.stackexchange.com/questions/2250/a-provable-upper-bound-on-the-summation/2380#2380) I posted to the same question on math.stackexchange.com so that the questioner can close this question.
The sum in question is at most ε2. (We do not need the condition that the row sum equals 1 or the condition f(i)≠g(i) to obtain this.)
**Proof**. Since
$$\sum\_{k=1}^n(a\_{f(k)j}-a\_{g(k)j})=\sum\_{k=1}^na\_{f(k)j}-\sum\_{k=1}^na\_{g(k)j}=0,$$
we have
$$\left|\sum\_{k=1}^na\_{ki}(a\_{f(k)j}-a\_{g(k)j})\right|
=\left|\sum\_{k=1}^n(a\_{ki}-a\_{1i})(a\_{f(k)j}-a\_{g(k)j})\right|$$
$$\le\sum\_{k=1}^n|a\_{ki}-a\_{1i}||a\_{f(k)j}-a\_{g(k)j}|.$$
Therefore, the sum in question is at most
$$\frac1n\sum\_{i,j=1}^z\sum\_{k=1}^n|a\_{ki}-a\_{1i}||a\_{f(k)j}-a\_{g(k)j}|
=\frac1n\sum\_{k=1}^n\left(\sum\_{i=1}^z|a\_{ki}-a\_{1i}|\right)\left(\sum\_{j=1}^z|a\_{f(k)j}-a\_{g(k)j}|\right)$$
$$\le\frac1n\sum\_{k=1}^n\epsilon^2=\epsilon^2.$$
| 1 | https://mathoverflow.net/users/7982 | 35573 | 22,936 |
https://mathoverflow.net/questions/35577 | 15 | The Normal Basis Theorem: If $E/F$ is a finite Galois extension, then there exists $a \in E$ such that the orbit of $a$ under the action of $\mathrm{Gal}(E/F)$ is a basis for $E$ as a vector space over $F.$
Who discovered this?
I've looked through the collected works of Frobenius and Dedekind, which are the earliest works I've seen referring to it, but it looks like the theorem led Dedekind to what is called the group determinant, and he doesn't give a reference. (p. 433 of Dedekind's Gesammelte Werke, v. 2, via Curtis's Pioneers of Representation Theory. See KConrad's answer below.) Among others, I've also looked at some of the correspondence of Hasse and Noether. The works are in German, which is second language to me, so it's possible I missed something. Needless to say, I've searched using Google to no avail. If anyone knows something, I'd be very grateful.
| https://mathoverflow.net/users/8479 | History of the Normal Basis Theorem | The cached page
<http://webcache.googleusercontent.com/search?q=cache:q5q43iNq1SQJ:siba2.unile.it/ese/issues/1/690/Notematv27n1p5.ps+normal+basis+theorem&cd=3&hl=en&ct=clnk&gl=us&client=safari>
gives some information: Eisenstein conjectured it in 1850 for extensions of finite fields and Hensel gave a proof for finite fields in 1888. Dedekind used such bases in number fields in his work on the discriminant in 1880, but he had no general proof. (See the quote by Dedekind on the bottom of page 51 of Curtis's "Pioneers of Representation Theory: Frobenius, Burnside, Schur, and Brauer".) In 1932 Noether gave a proof for some infinite fields while Deuring gave a uniform proof for all fields (also in 1932).
In Narkiewicz's "Elementary and Analytic Theory of Algebraic Numbers" (3rd ed.) he writes on the bottom of p. 186 that the normal basis theorem is due to Noether, but as usual the history is slightly more complicated.
| 14 | https://mathoverflow.net/users/3272 | 35578 | 22,938 |
https://mathoverflow.net/questions/35442 | 2 | If
$$p(x,y) = x^N + a\_{N-1}(y)x^{N-1} + \ldots + a\_0(y), \quad x,y \in \mathbf{C}$$
is a monic polynomial in $x$, and the coefficients $a\_j$ are analytic functions of $y$, then the roots of $p$ have expansions in [Puiseux series](http://en.wikipedia.org/wiki/Puiseux_series) (in powers of $y^{1/m}$ for some $m$) which are convergent for $y$ sufficiently close to 0.
Is this true in an asymptotic sense when the $a\_j$ are only assumed to be smooth? i.e. Do the roots have asymptotic expansions which are formal Puiseux series (not necessarily convergent)?
For $A$ to have an asymptotic expansion $A \sim B\_1 + B\_2 + \ldots$ with respect to some grading $\mathcal{O}(n)$ means that $B\_n \in \mathcal{O}(n)$, and for each $N$, $A - \sum\_{n=1}^N B\_n \in O(N+1)$. Here $\mathcal{O}(n)$ means $\mathcal{O}(|y|^{n/m})$ in the usual big-O notation, as $y\to 0$.
| https://mathoverflow.net/users/4402 | Puiseux series for roots of polynomials with smooth coefficients | The constructive proof of the Newton-Puiseux theorem works *formally*, and a posteriori one can show that if the original coefficients are convergent, then the Puiseux series are convergent (see the details in Casas-Alvero's book, "Singularities of Plane Curves"). So Torsten Ekedahl's comment is right, of course.
However, I wonder how useful this would be when the power series don't converge to the original smooth functions.
| 3 | https://mathoverflow.net/users/1939 | 35579 | 22,939 |
https://mathoverflow.net/questions/35572 | 2 | Let S be any nontrivial blocking set in a projective plane of order q,
such that S not containing any line.
Let A be a set of points in the same projective plane of order q,
raging over all these S.
Is it true that if A$\cap$S != $\phi$ then |A| >= q+1
and equality exists only if A is a line?
| https://mathoverflow.net/users/8478 | Blocking set in a projective plane. | Let $\ell\_1,\ell\_2,\ell\_3$ be three lines in the plane that do not all contain the same point. The *triangle* formed by $\ell\_1,\ell\_2,\ell\_3$ is the set obtained from $\ell\_1 \cup \ell\_2 \cup \ell\_3$ by removing the three pairwise intersections of the lines. Clearly, any *triangle* is a blocking set: every line in the plane meets each line $\ell\_1,\ell\_2,\ell\_3$, and cannot contain all the three points that we are removing from the *triangle*. *Triangles* are the only blocking sets we consider.
We say that $A$ has property (`*`) if the intersection of $A$ with every *triangle* is not empty. Note that if a subset of the plane intersects non-trivially every blocking set, then it has property (`*`). We prove the (*a priori*) stronger statement that a set with property (`*`) and at most $q+1$ elements consists of all the points contained in a line.
**Lemma 1.**
*Let $A$ be a set of at least three points in the plane such that any line intersects $A$ in at most two points; then $A$ cannot have property (`*`).*
*Proof.*
Let $A' \subset A$ be a subset with three elements, and note that the points in $A'$ are not collinear. Let $\ell\_1,\ell\_2,\ell\_3$ be the three lines each containing two of the points of $A'$. By assumption $\ell\_1 \cap A , \ell\_2 \cap A , \ell\_3 \cap A \subset A'$; thus, the *triangle* formed by $\ell\_1 , \ell\_2 , \ell\_3$ is a blocking set disjoint from $A$. $\square$
**Lemma 2.**
*Let $A$ be a set of at most $q+1$ points in the plane that are not collinear and let $\ell$ be a line in the plane such that $|A \cap \ell| \geq 3$; then $A$ cannot have property (`*`).*
*Proof.*
Let $a$ be a point of $\ell \setminus A$: such a point exists, since the points of $A$ are not collinear and $\ell$ contains $q+1$ points. There are $q$ lines different from $\ell$ through $a$ and at most $q-2$ points in $A \setminus \ell$; thus there are two distinct lines $\ell\_1 , \ell\_2$ through $a$ disjoint from $A$. Let $b'$ be any point of $A$ not contained in $\ell$ and let $b$ be any point not contained in $A$ on a line $\ell'$ joining $b'$ to a point of $A$ contained in $\ell$; there are $q$ lines through $b$ different from $\ell'$ and only $q-1$ points in $A \setminus \ell'$; we deduce that there is a line $\ell\_3$ through $b$ disjoint from $A$. Thus the *triangle* formed by $\ell\_1,\ell\_2,\ell\_3$ is a blocking set disjoint from $A$. $\square$
**Corollary.**
*Let $A$ be a set of at most $q+1$ points in the plane having property (`*`); then $A$ consists of all the points contained in a line.*
*Proof.*
By Lemma 1 we know that there must be a line containing three points of $A$. If the points of $A$ were not collinear, then Lemma 2 would imply that $A$ does not have property (`*`). Thus the points of $A$ are collinear. To conclude it suffices to show that if $A$ misses a point of the line containing $A$, then $A$ does not have property (`*`). This is easy: let $a$ be a point of $A$ and let $\ell\_1,\ell\_2$ be distinct lines through $a$; let also $b$ be a point on the line containing $A$ not in $A$, and let $\ell\_3$ be any line through $b$ not containing $A$. The *triangle* formed by $\ell\_1,\ell\_2,\ell\_3$ is a blocking set disjoint from $A$. $\square$
| 2 | https://mathoverflow.net/users/4344 | 35581 | 22,941 |
https://mathoverflow.net/questions/35584 | 26 | More precisely, is Ricci flow with surgery on a 3-dimensional Riemannian manifold M given by the "constant-time" sections of some canonical smooth 4-dimensional Riemannian manifold?
There would be a discrete set of times corresponding to the surgeries, but the 4-dimensional manifold might still be smooth at these points even though its sections would have singularities. The existence of such a 4-manifold is well known if there are no singularities: the problem is whether one can still construct it in the presence of singularities.
Background: Ricci flow on M in general has finite time singularities. These are usually dealt with by a rather complicated procedure, where one stops the flow just before the singularities, then carefully cuts up M into smaller pieces and caps off the holes, and constructs a Riemannian metric on each of these pieces by modifying the metric on M, and then restarts the Ricci flow. This seems rather a mess: my impression is that it involves making several choices so is not really canonical, and has a discontinuity in the metric and topology on M. If the flow were given by sections of a canonical smooth 4-manifold as in the question this would give a cleaner way to look at the surgeries of Ricci flow.
(Presumably if the answer to the question is "yes" this is not easy to show, otherwise people would not spend so much time on the complicated surgery procedure. But maybe Ricci flow experts know some reason why this does not work.)
| https://mathoverflow.net/users/51 | Does Ricci flow with surgery come from sections of a smooth Riemannian manifold? | This is an open question, as far as I know. Perelman makes a comment in one of his papers
to the effect that he would like to achieve some sort of canonical Ricci flow-with-surgery
in space-time (see section 13.2 of [his first, 2002 paper](https://arxiv.org/abs/math/0211159)). There are several unresolved issues having to do with the formation of the singularities in Ricci flow that make this question difficult. On the other hand, recently Angenent, Knopf, and Caputo have shown that one may do a canonical surgery in the rotationally symmetric case: *[Minimally invasive surgery for Ricci flow singularities](https://arxiv.org/abs/0907.0232)*. There are several simplifications in the rotationally symmetric case
which makes their approach possible: the Ricci flow is reduced to a (coupled) ODE, so one may apply the maximum principle to show that there are finitely many singularities, and analyze the asymptotics. Since the Ricci flow becomes rotationally symmetric near a singularity in dimension 3 to first order, there is some hope that their approach should work without the rotationally symmetric hypothesis.
| 17 | https://mathoverflow.net/users/1345 | 35586 | 22,944 |
https://mathoverflow.net/questions/35590 | 18 | k-XORSAT is the problem of deciding whether a Boolean formula $$\bigwedge\_{i \in I} \oplus\_{j=1}^k l\_{s\_{ij}}$$ is satisfiable. Here $\oplus$ denotes the binary [XOR](http://en.wikipedia.org/wiki/Xor) operation, $I$ is some index set, and each clause has $k$ distinct literals $l\_{s\_{ij}}$ each of which is either a variable $x\_{s\_{ij}}$ or its negation.
Equivalently, $k$-XORSAT requires deciding whether a set of linear equations, each of the form $\sum\_{j=1}^k x\_{s\_{ij}}\equiv 1\; (\mod 2)$, has a solution over $\mathbb{Z}\_2 = \mathbb{Z}/2\mathbb{Z}$.
Every decision problem Q has an associated counting problem #Q, which (informally speaking) requires establishing the number of distinct solutions. Such counting problems form the complexity class [#P](http://qwiki.stanford.edu/wiki/Complexity_Zoo%3ASymbols#sharpp). The "hardest" problems in #P are #P-complete, as any problem in #P can be reduced to a #P-complete problem.
The counting problem associated with any NP-complete decision problem is #P-complete. However, many "easy" decision problems (some even solvable in linear time) also lead to #P-complete counting problems. For instance, Leslie Valiant's original 1979 paper [*The Complexity of Computing the Permanent*](http://dx.doi.org/10.1016/0304-3975%2879%2990044-6) shows that computing the permanent of a 0-1 matrix is #P-complete. As a second example, the list of #P-complete problems in the companion paper [*The Complexity of Enumeration and Reliability Problems*](http://dx.doi.org/10.1137/0208032) includes #MONOTONE 2-SAT; this problem requires counting the number of solutions to Boolean formulas in conjunctive normal form, where each clause has two variables and no negated variables are allowed. (MONOTONE 2-SAT is of course rather trivial as a decision problem.)
Andrea Montanari has written about the partition function of $k$-XORSAT in some [lecture notes](http://www.stanford.edu/~montanar/TEACHING/Stat316/handouts/lecture-4.pdf), and his book with Marc Mézard apparently discusses this (unfortunately I do not have a copy available to hand, and the relevant Chapter 17 is not included in Montanari's online draft).
These considerations lead to the following question:
>
> Is #$k$-XORSAT #P-complete?
>
>
>
Note that the formula in Montanari's notes does not obviously appear to answer this question. Just because there is a nice closed form solution, doesn't mean we can evaluate it efficiently: consider the [Tutte polynomial](http://en.wikipedia.org/wiki/Tutte_polynomial).
Some difficult counting problems in #P can still be approximated in a certain sense, by means of a scheme called an FPRAS. Jerrum, Sinclair, and collaborators have linked the existence of an FPRAS for #P problems to the question of whether $NP = RP$. I would therefore also be interested in the subsidiary question
>
> Does #$k$-XORSAT have an FPRAS?
>
>
>
*Edit: clarified second question as per comment by Tsuyoshi Ito. Note that Peter Shor's answer renders this part of the question moot.*
| https://mathoverflow.net/users/7252 | Is #k-XORSAT #P-complete? | The solutions for XOR-SAT form an affine subspace of the vector space GF(2)$^n$. You can see this by realizing that if you add three solutions together, you get another solution. The counting problem for XOR-SAT is then that of deciding how many points are in this affine space over GF(2). This is trivial if you know the rank of a generating matrix for this space (the number is $2^r$ for rank $r$). This rank can be figured out by a standard linear algebra algorithm, so the counting problem is in polynomial time.
| 28 | https://mathoverflow.net/users/2294 | 35595 | 22,949 |
https://mathoverflow.net/questions/35264 | 10 | Let $E/F$ be a quadratic extension of number fields. Let $W$ be a hermitian space over $E$ of dimension $2,$ and let $V$ be a skew-hermitian space of dimension $3$ over $E.$ Consider the associated unitary groups $H:=U(W)$ and $G:=U(V).$ Let $\sigma$ be an irreducible, cuspidal, automorphic representation of $H(\mathbb{A}\_F).$ Let $\pi=\Theta(\sigma,\psi,\gamma)$ be a theta lift of $\sigma$ to $G(\mathbb{A}\_F)$. ($\psi:\mathbb{A}\_F/F\to \mathbb{C}^\times$ and $\gamma:\mathbb{A}\_E^\times/E^\times\to\mathbb{C}^\times$ are the splitting data necessary to define the theta-lift for unitary groups.)
My question is, how do automorphic $L$-functions (standard, adjoint, etc.) for $\pi$ relate to those for $\sigma$?
| https://mathoverflow.net/users/3186 | $L$-functions for $\Theta$-lifts | This question is answered in a paper of Gan, Gross, and D. Prasad. Here's a link:
<http://www.math.ucsd.edu/~wgan/ggp-evidence4-1.pdf>
The relation between L-parameters of representations and their theta-lifts (at least locally) is discussed in section 7 of the paper.
| 3 | https://mathoverflow.net/users/3186 | 35596 | 22,950 |
https://mathoverflow.net/questions/35569 | 8 | Given two integers $m,n$ such that $n < m$, it is easy to construct a ring with global dimension $m$ or weak dimension $n$. But I wonder whether there exists a ring satisfying both the conditions?
| https://mathoverflow.net/users/5775 | How to construct a ring with global dimension m and weak dimension n? | If $R$ is Noetherian then they are equal.
For $n=0$ one can use the fact that any Boolean ring has weak dimension $0$ (any module is flat), but a free Boolean ring on $\aleph\_n$ generators have global dimension $n+1$, see the last paragraph of this [paper](https://projecteuclid.org/journals/nagoya-mathematical-journal/volume-32/issue-none/Upper-bounds-on-homological-dimensions/nmj/1118797385.full).
For any given pair of $(m,n)$ one can perhaps use polynomial rings over the examples for $n=0$ case (The global dimensions do go up properly, but the behavior of weak dimensions seem to be trickier, may be someone who is a real expert can confirm this?)
| 5 | https://mathoverflow.net/users/2083 | 35599 | 22,951 |
https://mathoverflow.net/questions/35618 | 1 | Is it possible to give a nice characterization of matrices $A \in R^{n \times n}$ which are self-adjoint with respect to *some* inner product?
These matrices include all symmetric matrices (of course) and some nonsymmetric ones: for example, the transition matrix of any (irreducible) reversible Markov chain will have this property.
Naturally, all such matrices must have real eigenvalues, though I do not expect that this is a sufficient condition (is it?).
About the only observation I have is that since any inner product
can be represented as $\langle x,y \rangle = x^T M y$ for some positive definite matrix $M$, we are looking for matrices $A$ which satisfy $A^T M = M A$ or $M^{-1} A^T M = A$. In other words, we are looking for real matrices similar to their transpose with a positive definite similarity matrix.
| https://mathoverflow.net/users/4267 | matrices self-adjoint with respect to some inner product | In addition to having real eigenvalues, the matrix will have to be diagonalizable, i.e., there have to be enough eigenvectors to span $R^n$. Once these conditions are satisfied, you can take a basis consisting of $n$ eigenvectors and define an inner product by declaring these basis vectors to be orthonormal. That inner product will make your matrix self-adjoint, because there's an orthonormal basis with respect to which the matrix is diagonal with real entries on the diagonal.
| 6 | https://mathoverflow.net/users/6794 | 35621 | 22,965 |
https://mathoverflow.net/questions/35612 | 6 | Let Prod(C, D) be the set of finite-product preserving functors from C to D. Is it true that for any Lawvere theory L, Prod(L, Set) has small colimits? This seems to be the case, as it is invoked here:
<http://ncatlab.org/nlab/show/database+of+categories>
Where do this colimits come from? In the case of sifted colimits, pointwise calculation works, as sifted colimits commute with finite products in Set. However, it doesn't seem at all obvious that an arbitrary colimit of product-preserving functors would also be product-preserving.
Does anyone know of a reference or proof that such colimits exist, or even better, how to go about computing them?
| https://mathoverflow.net/users/800 | Computing colimits in a Lawvere theory | Let $T$ be a Lawvere theory. Then view the category of $T$-algebras, as Professor Blass wrote, as a class of universal algebras for a signature that corresponds to $T$ and is equationally definable, i.e. a variety. Then do what one does with these kinds of algebras:
To get coequalizers use quotients of algebras by what are called congruences (at least, that is what they are called in Universal Algebra). A congruence is an equivalence relation on the set that corresponds under your functor $A$ to the object 1 in $T$ (that is, the underlying set) that is preserved under all operations of the theory (arrows in $T$).
To get coproducts construct the left adjoint by defining the free $T$-algebra on some set $S$ to be the obvious structure on the set of all formal expressions $f(s\_1,...,s\_n)$ with $f$ an arrow of $T$ and $s\_1,...,s\_n\in S$. It is then clear what the coproduct of a set of free $T$-algebras is, and you can use congruences to get the coproduct on an arbitrary set of $T$-algebras.
| 9 | https://mathoverflow.net/users/7747 | 35624 | 22,966 |
https://mathoverflow.net/questions/35610 | 5 | In [this](https://projecteuclid.org/ebooks/institute-of-mathematical-statistics-lecture-notes-monograph-series/Group%20representations%20in%20probability%20and%20statistics/chapter/Chapter%207:%20Representation%20Theory%20of%20the%20Symmetric%20Group/10.1214/lnms/1215467416) book chapter, the irreducible representations of the symmetric group $S\_n$ is given in terms of polytabloids of a Ferrer's diagram $\lambda$, defined as
$e\_t = \sum\_{\pi \in C\_t} \text{sgn}(\pi) e\_{\pi \lbrace t \rbrace}$.
Here $t$ is a tableau of $\lambda$, $C\_t$ is the column stablizing subgroup for $t$ in $S\_n$. $\text{sgn}$ is the signature of the permutation $\pi$. Finally {t} is the equivalence class of tableau (called tabloid) represented by $t$, where two tableaux are considered equivalent if they have the same row entries.
My question is, how is the definition of polytabloids above independent of the choice of $t$ in its equivalence class? For instance, if $t$ is the tableau {1,2},{3,4}, then it's equivalent to s={2,1},{3,4}, but $e\_t \neq e\_s$. So maybe it's not independent of representative. But then there seems to be too many polytabloids. I would also appreciate if someone could help me establish the connection with Fulton and Harris's book on representation theory problem 4.47. I am not sure what is meant by a standard tableau there. Also in the second construction of the irreps of $S\_n$ in the same problem, I am not sure how the action of $S\_n$ on the polynomials is defined.
| https://mathoverflow.net/users/4923 | some confusion about the explicit construction of irreducible representations of $S_n$ | 1. Yes, equivalent tableaux $t$ may yield different $e\_t$'s. However, equivalent tableaux $t$ yield the equivalent $\pi t$'s for any permutation $\pi$, so that the notation $\pi\left\lbrace t\right\rbrace$ on page 132 is justified. Nobody is claiming that $e\_t$ depends on the tabloid $\left\lbrace t\right\rbrace$ only.
2. The Specht module $S^{\lambda}$ is defined as the vector space generated by $e\_t$ for all Young tableaux $t$ corresponding to the partition $\lambda$. Now it turns out that there is a lot of redundancy in these $e\_t$; that is, they are linearly dependent. One very nice basis of $S^{\lambda}$ is $\left\lbrace e\_t \mid t\text{ is a standard tableau}\right\rbrace$, where a tableau is called standard if the rows are strictly monotonically increasing and the columns are strictly monotonically increasing (the word "strictly" is not of much importance here, because the numbers in our tableaus are pairwise distinct, but sometimes one also considers tableaux where the entries may be equal, and then it matters).
3. Concerning Fulton-Harris' problem 4.47, the first part (about the $E\_T$) is exactly the definition of the Specht module that Diaconis gives. As for the second part (about the $F\_T$), you have to show that there is an $S\_d$-equivariant isomorphism $V\_{\lambda}\to W\_{\lambda}$, where $V\_{\lambda}$ is the Specht module defined by means of the $E\_T$'s, and $W\_{\lambda}$ is the $k\left[S\_d\right]$-submodule ($k\left[S\_d\right]$ is what Fulton-Harris denotes by $\mathbb C\left[\mathfrak{S}\_d\right]$) of $k\left[x\_1,x\_2,...,x\_d\right]$ spanned by the polynomials $F\_T=\prod\limits\_{i < j;\ i\text{ and }j\text{ lie in the same column of }T}\left(x\_i-x\_j\right)$. To construct this isomorphism, let $\Psi$ be the vector space homomorphism $U\_{\lambda}\to k\left[x\_1,x\_2,...,x\_d\right]$ (where $U\_{\lambda}$ is the representation of $S\_d$ with basis the tabloids for the Young diagram $\lambda$) defined by $\Psi\left(\left\lbrace T\right\rbrace\right) = \prod\limits\_{i=1}^{d}x\_i^{\left(\text{number of the row in which }i\text{ lies in the tableau }T\right)-1}$ for every tableau $T$ (this is well-defined since the product on the right hand side depends only the equivalence class $\left\lbrace T\right\rbrace$ of $T$). Besides, $\Phi$ is easily seen to be $S\_d$-equivariant and injective. Now, $\Psi\left(V\_{\lambda}\right)=W\_{\lambda}$, because every tableau $T$ satisfies $\Psi\left(E\_T\right)=F\_T$ or $\Psi\left(E\_T\right)=-F\_T$ (by Vandermonde's determinant, applied to the entries in every column of $T$), and thus the restriction of this homomorphism $\Psi$ to the subspace $V\_{\lambda}$ of $U\_{\lambda}$ is a $G$-equivariant bijective homomorphism $V\_{\lambda}\to W\_{\lambda}$. Thus, $V\_{\lambda}\cong W\_{\lambda}$ as representations of $S\_d$.
4. My first actual source for the representation theory of $S\_n$ were [Etingof's lecture notes](http://www-math.mit.edu/~etingof/replect.pdf), but beware: they are very compressed and don't have much on $S\_n$ (that's not the point of them either). Then, there is Fulton-Harris with a whole chapter on $S\_n$ (but the proofs are mostly exiled into the exercises, which means that you often get hints rather than proofs). "The Representation Theory of the Symmetric Group" by James and Kerber looks very good as a comprehensive reference. There are also typewriter-style lecture notes by James (LNM 682: "The Representation Theory of the Symmetric Groups") which have the advantage of being just 136 pages long. I don't have any experience with them, however.
| 6 | https://mathoverflow.net/users/2530 | 35641 | 22,978 |
https://mathoverflow.net/questions/35643 | 6 | Suppose I have a symmetric $N \times N$ matrix A which has a one-dimensional Nullspace $N$. A is positive definite on $N^\bot$. In my case $N$ is the space of constant vectors (i.e. generated by the all-one vector).
I have to solve the problem $Ax = b$, with $b \in R(A)$ which has infinitely many solutions. I am looking for the minimum norm solution. The matrix $A$ is very large and sparse, direct methods aren't really an option. The rank-deficient least squares algorithms I have seen also appear to be prohibitive.
I was solving a non-singular version of this problem with Conjugate Gradient. Is there anyway I can modify the algorithm to solve this particular problem?
**EDIT**: Boiling the problem down to the bone the question is if $A$ is positive semi-definite with exactly one 0 eigenvalue, does CG work?
Thanks.
| https://mathoverflow.net/users/4047 | Conjugate Gradient for a "slightly" singular system. | I'd suggest you to shift away the singularity: solve $(A+ee^T)y=b$ instead and then orthogonalize $y$ with respect to $e$ to get $x$. $A+ee^T$ is not sparse but you can compute matrix-vector products cheaply, and that's all you need for CG.
EDIT: forgot to define it, $e$ is the vector of all ones
| 7 | https://mathoverflow.net/users/1898 | 35653 | 22,985 |
https://mathoverflow.net/questions/35649 | 10 | I'm looking for an example of a finite abelian group *A* and a finite group *G* acting trivially on *A* such that there are two extensions $E\_1$ and $E\_2$ with base *A* and quotient *G* (i.e., they are both central extensions, and hence both give corresponding elements of $H^2(G,A)$) and:
1. $E\_1$ and $E\_2$ are isomorphic as abstract groups.
2. Under the natural action of $\operatorname{Aut}(G) \times \operatorname{Aut}(A)$ on $H^2(G,A)$ (by pre- and post-composition with 2-cocycles that then descends to action on cohomology classes), the cohomology classes corresponding to $E\_1$ and $E\_2$ are *not* in the same orbit.
Basically condition (2) states that $E\_1$ and $E\_2$ are not only not congruent extensions, they are not even congruent up to a relabeling of the subgroup *A* and the quotient *G*. Another way of putting this is that there is no isomorphism between $E\_1$ and $E\_2$ that sends the *A* inside $E\_1$ to the *A* inside $E\_2$.
The analogous statement with a nontrivial action of *G* on *A* is also of interest to me. In this latter case, though, the entire group $\operatorname{Aut}(G) \times \operatorname{Aut}(A)$ does not act.
I think that examples exist (because of my experience with finding examples for similar specifications) but there may well be a proof to the contrary.
| https://mathoverflow.net/users/3040 | Extensions isomorphic as groups but not congruent or pseudo-congruent | E = SmallGroup(32,28) is the first example. It has two central subgroups A1 and A2 isomorphic to A ≅ 2 with quotient isomorphic to SmallGroup(16,11), but A1 and A2 are not conjugate in Aut(E).
Examples such as this are reasonably common in p-groups.
**Edit:** You can even have such an example with G abelian: G = 4×2, A = 4×2, E = SmallGroup(64,3) = 8⋉8, Z(E) = 4×4. E has two central copies of A=4×2 that are not conjugate in Aut(G), but the quotients are both abelian and isomorphic to G.
**Edit:** [Vipul notes](http://groupprops.subwiki.org/wiki/Series-equivalent_not_implies_automorphic_in_finite_abelian_group) you can even have E abelian of order p7.
| 11 | https://mathoverflow.net/users/3710 | 35662 | 22,991 |
https://mathoverflow.net/questions/35664 | 21 | If this problem is really stupid, please close it. But I really wanna get some answer for it. And I learnt computational complexity by reading books only.
When I learnt to the topic of relativization and oracle machines, I read the following theorem:
>
> There exist oracles A, B such that $P^A = NP^A$ and $P^B \neq NP^B$.
>
>
>
And then the book said because of this, we can't solve the problem of NP = P by using relativization. But I think what it implies is that $NP \neq P$. The reasoning is like this:
First of all, it is quite easy to see that:
$$A = B \Leftrightarrow \forall \text{oracle O, }A^O = B^O$$
Though I think it is obvious, I still give a proof to it:
[A simple proof of NP != P ?](http://www.voofie.com/content/141/a-simple-proof-of-np-p/#Prove)
And the negation of it is:
$$A \neq B \Leftrightarrow \exists \text{oracle O such that } A^O \neq B^O$$
Therefore since there is an oracle B such that:
$$ NP^B \neq P^B$$
we can conclude that $ NP \neq P $
What's the problem with the above reasoning?
| https://mathoverflow.net/users/5217 | Why relativization can't solve NP !=P? | The map $A \to A^O$ does not depend only on the elements contained in the language $O$, so it is not an operation on languages. It depends on the semantic way in which the language $A$ is defined. For instance, $NP^O$ is allowed *both* nondeterminism and access to $O$. $P^O$ is allowed deterministic polynomial time and access to $O$. I believe it is true that $BPP$ can be separated from $P$, even though it is thought that $BPP = P$.
| 23 | https://mathoverflow.net/users/344 | 35666 | 22,994 |
https://mathoverflow.net/questions/35669 | 20 | Let $f:R\_+\to R\_+$ be smooth on $(0,\infty)$, increasing, $f(0)=0$ and
$\lim\_{x\to\infty}=\infty$. Assume also that $f$ is subadditive:
$f(x+y)\le f(x)+f(y)$ for all $x,y\ge 0$. Must $f$ be concave? The converse is obvious.
| https://mathoverflow.net/users/8504 | subadditive implies concave | For not smooth surely not, take $f(x)=2x+|\sin x|$. I am nearly sure that for smooth answer is the same. For example, it looks like function $|\sin x|$ may be changed near points $\pi k$ so that it becomes smooth but still semiadditive.
well, more concrete construction is like follows (some details are however omited)
construct a function $f$ such that $f(x)=|\sin x|$ unless $|x-k\pi|<1/100$ for some positive integer $k$, $\sin 1/100\geq f(x) > \sin 1/100-1/1000000$ for $|x-k\pi| < 1/100$, $f$ is convex on $[k\pi-1.100,k\pi+1/100]$. When may $f(x+y) \le f(x)+f(y)$ fail? If $f(x+y)=|\sin(x+y)|$ then $f(x+y)\le |\sin x|+|\sin y|\le f(x)+f(y)$. If $f(x+y)\ne |\sin (x+y)|$, then $f(x+y)\le \sin(1/100)$, so if $f(x)+f(y) < f(x+y)$, then also $f(x) < \sin(1/100)$ and $f(y) < \sin(1/100)$. If both $x$ and $y$ are greater then $1/100$, then $f(x)+f(y) > 2(\sin(1/100)-1/1000000) > 1/100 > f(x+y)$.
Now without loss of generality $k\pi < y< x+y< k\pi+1/100$. Then $f(x+y)-f(y)=xf'(\theta)$ for some $\theta\in [y,x+y]$, by convexity $f'(\theta)\le \cos 1/100$. So, it suffices to prove that $x\cos(1/100)\le \sin x$, i.e. $\sin x/x\geq \cos 1/100$. Since $\sin t/t$ decreases on $[0,1/100]$, we have $\sin x/x\geq 100\sin 1/100\geq \cos 1/100$ as $\tan t > t$ for $t=1/100$
| 13 | https://mathoverflow.net/users/4312 | 35673 | 22,998 |
https://mathoverflow.net/questions/35680 | 28 | I've seen a couple papers (that I now can't find) that say that in his paper "On irreducible 3-manifolds which are sufficiently large" Waldhausen proved that the data $\pi\_1(\partial (S^3\setminus K)) \to \pi\_1(S^3\setminus K)$ is a complete knot invariant. However, the word "knot" doesn't appear in this paper (although the phrase "to avoid an orgy of notation" does :-). Is the claimed result a straightforward corollary of his main results? Or am I looking at the wrong paper?
| https://mathoverflow.net/users/2669 | Complete knot invariant? | As Ryan says, this follows from Waldhausen's paper, when appropriately interpreted. Sufficiently large 3-manifolds are usually called "Haken" in the literature, and as Ryan says, they are irreducible and contain an incompressible surface (which means that the surface is incompressible and boundary incompressible). An irreducible manifold with non-empty boundary and not a ball (ie no 2-sphere boundary components) is always sufficiently large, by a homology and surgery argument. By [Alexander's Lemma](http://en.wikipedia.org/wiki/Heegaard_splitting), knot complements are irreducible, and therefore sufficiently large (the [sphere theorem](http://en.wikipedia.org/wiki/Sphere_theorem_%283-manifolds%29) implies that they
are aspherical).
Waldhausen's theorem implies that if one has two sufficiently large 3-manifolds $M\_1, M\_2$ with connected boundary components, and an isomorphism $\pi\_1(M\_1) \to \pi\_1(M\_2)$ inducing an isomorphism $\pi\_1(\partial M\_1) \to \pi\_1(\partial M\_2)$, then $M\_1$ is homeomorphic to $M\_2$. This is proven by first showing that there is a homotopy equivalence $M\_1\simeq M\_2$ which restricts to a homotopy equivalence $\partial M\_1\simeq \partial M\_2$. Then Waldhausen shows that this relative homotopy equivalence is homotopic to a homeomorphism by induction on a hierarchy. The peripheral data is necessary if the manifold has essential annuli, for example the square and granny knots have homotopy equivalent complements.
If $K\_1, K\_2\subset S^3$ are (tame) knots, and $M\_1=S^3-\mathcal{N}(K\_1), M\_2=S^3-\mathcal{N}(K\_2)$ are two knot complements, then Waldhausen's theorem applies. However, one must also cite the knot complement problem solved by [Gordon and Luecke](http://www.ams.org/mathscinet-getitem?mr=965210), in order to conclude that $K\_1$ and $K\_2$ are isotopic knots. Otherwise, one must also hypothesize that the isomorphism $\partial M\_1 \to \partial M\_2$ takes the meridian to the meridian (the longitudes are determined homologically). This extra data is necessary to solve the isotopy problem for knots in a general 3-manifold $M$, to guarantee that the homeomorphism $(M\_1,\partial M\_1)\to (M\_2,\partial M\_2)$ extends to a homeomorphism $(M,K\_1)\to (M,K\_2)$, since for example there are knots in lens spaces which have homeomorphic complements by a result of [Bleiler-Hodgson-Weeks](http://www.ams.org/mathscinet-getitem?mr=1734400).
| 27 | https://mathoverflow.net/users/1345 | 35687 | 23,008 |
https://mathoverflow.net/questions/35220 | 4 | It is a basic result of group cohomology that the extensions with a given abelian normal subgroup *A* and a given quotient *G* acting on it via an action $\varphi$ are given by the second cohomology group $H^2\_\varphi(G,A)$. In particular, when the action is trivial (so the extension is a central extension), this is the second cohomology group $H^2(G,A)$ for the trivial action. In the special case where *G* is also abelian, we classify all the class two groups with *A* inside the center and *G* as the quotient group.
I am interested in the following: given a sequence of abelian groups $A\_1, A\_2, \dots, A\_n$, what would classify (up to the usual notion of equivalence via commutative diagrams) the following: a group *E* with an ascending chain of subgroups:
$$1 = K\_0 \le K\_1 \le K\_2 \le \dots \le K\_n = E$$
such that the $K\_i$s form a central series (i.e., $[E,K\_i] \subseteq K\_{i-1}$ for all *i*) and $K\_i/K\_{i-1} \cong A\_i$?
The case $n = 2$ reduces to the second cohomology group as detailed in the first paragraph, so I am hoping that some suitable generalization involving cohomology would help describe these extensions.
Note: As is the case with the second cohomology group, I expect the object to classify, not isomorphism classes of possibilities of the big group, but a notion of equivalence class under a congruence notion that generalizes the notion of congruence of extensions. Then, using the actions of various automorphism groups, we can use orbits under the action to classify extensions under more generous notion of equivalence.
Note 2: The crude approach that I am aware of involves building the extension step by step, giving something like a group of groups of groups of groups of ... For intsance, in the case $n = 3$:
$$1 = K\_0 \le K\_1 \le K\_2 \le K\_3 = G$$
with quotients $A\_i \cong K\_i/K\_{i-1}$, I can first consider $H^2(A\_3,A\_2)$ as the set of possibilities for $K\_3/K\_1$ (up to congruence). For each of these possibilities *P*, there is a group $H^2(P,A\_1)$ and the total set of possibilities seems to be:
$$\bigsqcup\_{P \in H^2(A\_3,A\_2)} H^2(P,A\_1)$$
Here the $\in$ notation is being abused somewhat by identifying an element of a cohomology group with the corresponding extension's middle group.
What I really want is some algebraic way of thinking of this unwieldy disjoint union as a single object, or some information or ideas about its properties or behavior.
| https://mathoverflow.net/users/3040 | Cohomology analogue for central series of length more than two | This looks like a (slightly) non-additive version of Grothendieck's theory of
"extensions panachées" (SGA 7/I, IX.9.3). There he considers objects (in some
abelian category) $X$ together with a filtation $0\subseteq X\_1\subseteq
X\_2\subseteq X\_3=X$. In the first version he also fixes (just as one does for
extensions) isomorphisms $P\rightarrow X\_1$, $Q\rightarrow X\_2/X\_1$ and
$R\rightarrow X\_3/X\_2$. However, in the next version he fixes the isomorphism
class of the two extensions $0\rightarrow P\rightarrow X\_2\rightarrow
Q\rightarrow0$ and $0\rightarrow Q\rightarrow X\_3/X\_1\rightarrow R\rightarrow0$
so that if $E$ is an extension of $P$ by $Q$ and $F$ is an extension of $Q$ by
$R$, then the category $\mathrm{EXTP}(F,E)$ has as objects filtered objects $X$
as above together with fixed isomorphisms of extensions $E\rightarrow X\_2$ and
$F\rightarrow X\_3/X\_1$ and whose morphisms are are morphisms of $X$'s preserving
the given structures. The morphisms of $\mathrm{EXTP}(F,E)$ are necessarily
isomorphisms so we are dealing with a groupoid. Similarly for objects $A$ and
$B$ $\mathrm{EXT}(B,A)$ is the groupoid of extensions of $B$ by $A$.
Grothendieck then shows that $\mathrm{EXTP}(F,E)$ is a torsor over
$\mathrm{EXT}(R,P)$ (in the category of torsors, Grothendieck had previously
defined this notion). The action on objects of an extension $0\rightarrow
P\rightarrow G\rightarrow R\rightarrow0$ is given by first taking the pullback
of it under the map $X/X\_1\rightarrow R$ and then using the obtained action by
addition on extensions of $P$ by $F$. To more or less complete the picture,
there is an obstruction to the existence of an object of $\mathrm{EXTP}(F,E)$:
We have that $E$ gives an element of $\mathrm{Ext}^1(Q,P)$ and $F$ one of
$\mathrm{Ext}^1(R,Q)$ and their Yoneda product gives an obstruction in
$\mathrm{Ext}^2(P,Q)$.
The case at hand is similar (staying at the case of $n=3$ and with the caveat
that I haven't properly checked everything): We choose fixed isomorphisms with
$K\_2$ and a given central extension and with $K\_3/K\_1$ and another given central
extension (assuming that we have three groups $P$, $Q$ and $R$ as before)
getting a category $\mathrm{CEXTP}(F,E)$ of central extensions. We shall shortly
modify it but to motivate that modification it seems a good idea to start with
this. We get as before an action of $\mathrm{CEXT}(R,P)$ on
$\mathrm{CEXTP}(F,E)$ as we can pull back central extensions just as before. It
turns however that the action is not transitive. In fact we can analyse both the
difference between two elements of $\mathrm{CEXTP}(F,E)$ and the obstructions
for the non-emptyness of it by using the Hochschild-Serre spectral sequence. To
make it easier to understand I use a more generic notation. Hence we have a
central extension $1\rightarrow K\rightarrow G\rightarrow G/K\rightarrow1$ and
an abelian group $M$ with trivial $G$-action. There is then a succession of two
obstructions for the condition that a given central extension of $M$ by $G/K$
extend to a central extension of $M$ by $G$. The first is $d\_2\colon
H^2(G/K,M)\rightarrow H^2(G/K,H^1(K,M))$, the $d\_2$-differential of the H-S
s.s. Now, we always have a map $H^2(G/K,M)\rightarrow H^2(G/K,H^1(K,M))$ given
by pushout of $1\rightarrow G\rightarrow G/K\rightarrow1$ along the map
$K\rightarrow \mathrm{Hom}(K,M)=H^1(K,M)$ given by the action by conjugation of
$K$ on the given central extension of $M$ by $K$ (equivalently this map is given
by the commutator map in that extension). It is easy to compute and identify
$d\_2$ but I just claim that it is equal to that map by an appeal to the What Else
Can It Be-principle (which works quite well for the beginnings of spectral
sequences with the usual provisio that the WECIB-principle only works up to a
sign).
This means that we can cut down on the number of obstructions by redefining
$\mathrm{CEXTP}(F,E)$. We add as data a group homomorphism $\varphi\colon
K\_3/K\_1\rightarrow\mathrm{Hom}(Q,P)$ that extends $Q\rightarrow
\mathrm{Hom}(Q,P)$ which describes the conjugation action on $K\_2$ and only look
the elements of $\mathrm{CEXTP}(F,E)$ for which the action is the given
$\varphi$ to form $\mathrm{CEXTP}(F,E;\varphi)$. Now the action of
$\mathrm{CEXT}(R,P)$ on $\mathrm{CEXTP}(F,E;\varphi)$ should make
$\mathrm{CEXTP}(F,E;\varphi)$ a
$\mathrm{CEXT}(R,P)$-(pseudo)torsor. Furthermore, there is now only a single
obstruction for non-emptyness which is given by $d\_3\colon H^2(R,M)\rightarrow
H^3(P,M)$.
Going to higher lengths there are two ways of proceeding in the original
Grothendieck situation: Either one can look at the the two extensions of one
length lower, one ending with the next to last layer (i.e., $X\_{n-1}$) and the
other being $X/X\_1$. This reduces the problem directly to the original case
(i.e., we look at filtrations of length $n-2$ on $Q$). One could instead look at
the successive two-step extensions and then look at how adjacent ones build up
three-step extensions and so on. This is essentially an obstruction theory point
of view and quickly becomes quite messy. An interesting thing is however the
following: We saw that in the original situation the obstruction for getting a
three step extension was that $ab=0$ for the Yoneda product of the two twostep
filtrations. If we have a sequence of three twostep extensions whose three step
extensions exist then we have $ab=bc=0$. The obstruction for the existence of
the full fourstep extension is then essentially a Massey product $\langle
a,b,c\rangle$ (defined up to the usual ambiguity). The messiness of such an
iterated approach is well-known, it becomes more and more difficult to keep
track of the ambiguities of higher Massey products. The modern way of handling
that problem is to use an $A\_\infty$-structure and it is quite possible (maybe
even likely) that such a structure is involved.
If we turn to the current situation and arbitrary $n$ then the first approach
has problems in that the midlayer won't be abelian anymore and I haven't looked
into what one could do. As for the second approach I haven't even looked into
what the higher obstructions would look like (the definition of the first
obstruction on terms of $d\_3$ is very asymmetric).
| 3 | https://mathoverflow.net/users/4008 | 35692 | 23,012 |
https://mathoverflow.net/questions/33913 | 5 | Let $E$ be a closed and convex set of distributions on a finite set $A$. Let $P',Q'\notin E$ and let $P^{\star},Q^{\star}$ be their respective estimates in $E$ with respect to the KL-divergence, i.e., $D(P'\|P^{\star})=\min\_{P\in E}D(P'\|P)$ and similarly for $Q^{\star}$. I am wondering whether $D(P'\|Q')\ge D(P^{\star}\|Q^{\star})$.
| https://mathoverflow.net/users/7699 | Question regarding divergence | The inequality $D(P'|Q') \ge D(P^\star| Q^\star)$ does not need to hold.
Here is an example.
Let $A$ be the set $\{1,2,3,...,n\}$. Let $E$ be the set of measures $P$ on $A$ such that $P(\{1\}) = 0$. Projecting a measure $P$ on $E$ using $D$ is equivalent to conditioning $P$ on $ A- \{1\}$. Choose $P'$ and $Q'$ such that they both put equal and nonzero mass on $\{1\}$. By direct computation one sees: $D(P^\star| Q^\star) = \frac{1}{1-P'(\{1\})} D(P'|Q') > D(P' | Q')$.
The details of the above computation are as follows.
For ease of notation set $n=3$. Let $E$ be the set of measures $P$ with $P(\{1\}) =\epsilon$; to obtain the example above, one sets $\epsilon = 0$. Let us parametrize the measures on $\{1,2,3\}$ as follows: $P(\{1\}) = p\_1$, $P(\{2\}) =p\_2$ and $P(\{3\}) = 1-p\_1 -p\_2$. Our problem is:
$$
\inf\_{ Q \in E}\left[ p\_1 \log \frac{p\_1}{q\_1} + p\_2 \log \frac{p\_2}{q\_2} + (1-p\_1 -p\_2) \log\frac{ 1- p\_1 - p\_2}{ 1- q\_1 - q\_2 } \right].
$$
Let $F$ denote the expression after the $\inf$.
$F$ is strictly convex in $Q$ and therefore will have a unique optimizer. In the above coordinates, the normal to $E$ is the vector $(1,0)$. Then
$$
\frac{\partial F} {\partial q\_1} = -\frac{p\_1}{q\_1} + \frac{1-p\_1-p\_2}{1-q\_1-q\_2} = \lambda
$$
and
$$
\frac{\partial F} {\partial q\_2} = -\frac{p\_2}{q\_2} + \frac{1-p\_1-p\_2}{1-q\_1-q\_2} = 0.
$$
We have the constraint that $Q\in E$, i.e., $q\_1 =\epsilon$.
From the last two equalities one infers:
$$
q\_2 = \frac{(1-\epsilon) p\_2}{ 1-p\_1}.
$$
Going back to the coordinates $(p\_1,p\_2,p\_3)$ to denote a measure on $\{1,2,3\}$,
projecting a measure on $E$ using $D$ corresponds to the following map:
$$
(p\_1,p\_2,p\_3) \rightarrow \left(\epsilon, (1-\epsilon)\frac{p\_2}{p\_2+p\_3}, (1-\epsilon)\frac{p\_3}{p\_2 + p\_3}\right).
$$
For $\epsilon =0$, this is the same as conditioning $P$ on $\{2,3\}$.
One obtains the expression for the relative entropy given above by directly computing it using this formula for the projections.
| 3 | https://mathoverflow.net/users/3370 | 35706 | 23,017 |
https://mathoverflow.net/questions/35710 | 6 | Can any triangle be inscribed in any convex figure? i.e. given a convex figure and a triangle can we transpose and scale and rotate that triangle so that its vertices are on the boundary of the convex figure?
| https://mathoverflow.net/users/2003 | Can any triangle be inscribed in any convex figure? | A more general result is known: if $C$ is any Jordan curve and $T$ is a triangle then there exists a triangle similar to $T$ inscribed in $C.$ Moreover, the vertices of such triangles are dense in $C.$ See the references in the Wikipedia article on the [Inscribed Square Problem](http://en.wikipedia.org/wiki/Inscribed_square_problem#Variants_and_generalizations).
| 12 | https://mathoverflow.net/users/5740 | 35715 | 23,023 |
https://mathoverflow.net/questions/35704 | 11 | What are the applications of theory of fusion systems to finite group theory
or representation theory of finite groups? More concretely, is there any important
result in finite group theory or representation theory of finite groups whose prove
uses fusion systems in the essential way?
| https://mathoverflow.net/users/8257 | Applications of fusion systems | Lots of results in group cohomology only have topological proofs using the techniques of Bob Oliver and his (generalized) collaborators. For instance, many results along the lines of "controls fusion iff controls cohomology" only have topological proofs using the same techniques that Bob Oliver called fusion systems (though I think some of the papers stick to the topological language). "Controls fusion" is a very old term predating fusion systems by 50 years, so I think of this as a pretty pure finite group cohomology result.
I think it is common to call these topological results by the name "fusion" results, even if fusion systems are not used directly, but rather p-completed classifying spaces are used (also known as p-local groups and such), even if the fusion system used to define the topological space is not prominent. Allowing such abuse of language, one of my favorite examples is the topological proof that the famous Z\*-theorem holds for odd primes too: [MR1125010](http://www.ams.org/mathscinet-getitem?mr=1125010). It was proven later using more pure finite group theory, but I suspect lots of people prefer the topological proof.
In modular representation theory, the work of Puig on [nilpotent blocks](http://www.ams.org/mathscinet-getitem?mr=558864) is (as far as I know) literally defined in terms of fusion systems and gets results on the representation theory using only conditions on the fusion systems. One also has some nice descriptions of the representation type of blocks of dihedral defect in terms of the fusion system induced on the block (I believe [Linckelmann's](http://www.maths.abdn.ac.uk/~bensondj/html/archive/linckelmann.html) [introduction](http://web.mat.bham.ac.uk/C.W.Parker/Fusion/fusion-intro.pdf) gives this as an example).
More historically, the classification of finite groups with [dihedral](http://www.ams.org/mathscinet-getitem?mr=177033) (and then for [semi-dihedral/wreathed](http://www.ams.org/mathscinet-getitem?mr=284499)) begins by dividing clearly into cases based on which fusion system occurs. This just organizes the (couple hundred page) papers, so I guess whether it is essential depends on how practical you are.
Perhaps the most "exciting" answer is not quite ready yet, but Michael Aschbacher and collaborators have begun to recast the classification of finite simple groups in terms of fusion systems. It is quite surprising and comforting that several very important steps in the classification are radically easier in the fusion system case. The fusion system theorems would also have as side-effects corresponding theorems in modular representation theory and the theory of p-local groups.
Let me know if you want references for any of the wishy washy terms, but basically read Linckelmann's introduction, read a few papers of BLO (Broto–Levi–Oliver), and check out the slides from any [recent](http://www.maths.abdn.ac.uk/skye2009/) finite group theory conference, and you should have some pretty convincing evidence that fusion systems are opening up a very bright future for finite group theory.
| 12 | https://mathoverflow.net/users/3710 | 35716 | 23,024 |
https://mathoverflow.net/questions/35713 | 18 | I believe there is a straightforward formula for the abelianization of a semi-direct product: if $G$ acts on $H$, and we form the semi-direct product of $G$ and $H$ in the usual way, and the abelianization of this semi-direct product is the product $G^{ab}\times (H^{ab})\_{G}$.
(Here the subscript $G$ denotes taking the coinvariants with respect to $G$. That is, $(H^{ab})\_{G}$ is a the quotient of $H^{ab}$ by the subgroup generated by elements of the form $h^g-h$ for $h$ in $H$ and $g$ in $G$, and where the superscript $g$ denotes the action of $G$ on $H^{ab}$ induced by the action of $G$ on $H$.)
Does anyone happen to know a good reference for this?
| https://mathoverflow.net/users/3513 | Abelianization of a semidirect product | I agree with Ryan and Victor, except that you don't need presentations. The subgroup $[G \ltimes H,G \ltimes H]$ is generated by $[H,H] \cup [G,H] \cup [G,G]$, so you can write
$$(G \ltimes H)^{ab} = (G \ltimes H) / \langle [H,H] \cup [G,H] \cup [G,G] \rangle.$$
If you apply the relators $[H,H]$, you get $G \ltimes H^{ab}$; then if you apply the relators $[G,H]$, you get $G \times (H^{ab})\_G$; then finally if you apply $[G,G]$, you get $G^{ab} \times (H^{ab})\_G$. You can add this as an extra half-paragraph or footnote rather than giving a citation.
I don't think that the referee has the right to demand a longer explanation than this, unless maybe you are writing a textbook.
| 20 | https://mathoverflow.net/users/1450 | 35721 | 23,028 |
https://mathoverflow.net/questions/23747 | 15 | There are several papers which compute zeroes of modular forms for genus 0 congruence subgroups, such as "Zeros of some level 2 Eisenstein series" by Garthwaite et al published in Proc AMS and work of Shigezumi and others in levels 3,5 and 7. However, there don't seem to be generalizations of this to higher genus subgroups.
I know a few examples of modular forms for higher genus subgroups where one can compute all the zeroes; for instance, the unique normalized cusp form of weight 1 and level $\Gamma\_0(31)$ with character the Legendre character modulo 31 has simple zeroes at the two cusps and the two elliptic points because the valence formula forces them to be there. Similar ideas work for levels 17, 19, 21 and 39.
My question is this: is there a more general way to find the zeroes of modular forms in an explicit way for congruence subgroups?
| https://mathoverflow.net/users/4555 | Finding zeroes of classical modular forms | If you are looking for examples of modular forms whose zeros can be described explicitly, then you probably want the zeros to be cusps or imaginary quadratic irrationals. In this case the Gross-Kohnen-Zagier theorem implicitly gives lots of examples, by describing the relations between Heegner points on modular elliptic curves. (Heegner points are closely related to imaginary quadratic numbers in the upper half plane.) Many examples of modular forms with zeros at imaginary quadratic irrationals can also be constructed explicitly as automorphic products.
| 7 | https://mathoverflow.net/users/51 | 35722 | 23,029 |
https://mathoverflow.net/questions/35726 | 8 | Can any rectangle be inscribed in any convex figure?
| https://mathoverflow.net/users/2003 | Can any rectangle be inscribed in any convex figure? | Yes, this follows from a more general result in
Nielsen and Wright, *Rectangles inscribed in symmetric continua*. Geom. Dedicata 56 (1995), no. 3, 285–297 [MR](http://www.ams.org/mathscinet-getitem?mr=1340790)
(This is reference 4 in the [Wikipedia article](http://en.wikipedia.org/wiki/Inscribed_square_problem) I quoted in my [answer](https://mathoverflow.net/questions/35710/can-any-triangle-be-inscribed-in-any-convex-figure/35715#35715) to your previous question.)
In their terminology, a simple closed curve $C$ is *symmetric* if there exists a point $P\notin C$ such that each straight line through $P$ intersects $C$ in exactly 2 points. This condition is trivially satisfied when $C$ is a boundary of a convex region.
| 6 | https://mathoverflow.net/users/5740 | 35732 | 23,033 |
https://mathoverflow.net/questions/35699 | 8 | One very handy (counter)example I often think about is the scheme $Spec(k[a,b,c]/(ab-c^2))$ (where $k$ is a field), which you may also know as $Spec(k[x^2,xy,y^2])$, as $\mathbb A^2/\mu\_2$, or as the $A\_1$ singularity. As with other (counter)examples, I'd like to be able to say as much as possible about it.
There is a finite surjection $f:\mathbb A^2\to Spec(k[x^2,xy,y^2])$ corresponding to the inclusion $k[x^2,xy,y^2]\subseteq k[x,y]$. The question is whether this surjection is in some sense universal.
>
> Suppose $g:Y\to Spec(k[x^2,xy,y^2])$ is finite, surjective, and $Y$ is a smooth $k$-scheme. Must $g$ factor through $f:\mathbb A^2\to Spec(k[x^2,xy,y^2])$?
>
>
>
A couple of remarks:
* The finiteness hypothesis on $g$ is definitely necessary. Otherwise we could take $Y$ to be a resolution of the singularity (by a blow-up). If such a resolution factored through $\mathbb A^2$, you'd get a section of $f$ defined away from the singularity, which would imply that $f$ is a birational equivalence, which it isn't.
* The assumption that $Y$ is a scheme is important. The couple of people I've talked to have pointed out that the smooth stack $[\mathbb A^2/\mu\_2]$ is a finite cover of $X$. If $[\mathbb A^2/\mu\_2]$ factored through $\mathbb A^2$, you'd again get a rational section of $f$.
| https://mathoverflow.net/users/1 | Is $\mathbb{A}²$ the universal smooth scheme which is a finite cover of $\mathbb{A}²/μ₂$? | It seems to me that in the global case the answer should be $no$ because of the following argument.
Set $S:=Spec$ $k[x,y,z]/(z^2-xy)$. Then $S$ is isomorphic to a quadric cone in $\mathbb{A}^3$. The point is that there are plenty of smooth double covers of $S$, which are pairwise non-isomorphic.
To see this, notice first that the morphism $f \colon \mathbb{A}^2 \to S$ corresponds to the restriction of a double cover $\mathbb{P}^2 \to$ (Cone $\subset \mathbb{P}^3$) branched on the vertex of the cone and on a smooth conic contained in the hyperplane at infinity.
Now one can generalize this construction by taking a double cover $f\_k \colon Y\_k \to S$ which is the restriction to $S$ of the projective cover branched on the vertex and on a smooth curve of $even$ degree $2k$ not passing through the vertex. The fact that $f\_k$ is branched at the vertex ensures that $Y\_k$ is smooth.
When $k=1$ we have $Y\_1=\mathbb{A}^2$.
When $k=2$, $Y\_2$ is an affine, open subset of a smooth surface of general type with $p\_g=4, q=0, K^2=5$. These surfaces were studied by Horikawa in his famous paper "On deformations of quintic surfaces"; it turns out that the projective double cover is actually the canonical map.
Of course $f\_2$ does not factor through $f$, since they are covering of the same degree but $Y\_2$, being of general type, is not isomorphic to $\mathbb{A}^2$.
In fact, $f\_k$ does not factor through $f$ except for $k=1$.
| 6 | https://mathoverflow.net/users/7460 | 35740 | 23,036 |
https://mathoverflow.net/questions/35739 | 2 | Suppose I have N real random variables with identical PDF. At every instance of these r.vs, I pick $K$ largest out of $N$. Lets call their sum as $S\_K$. Alternatively, based on some criteria, I pick in an average sense $K$ largest numbers (i.e some times we pick K, K-1,K+1 etc, randomly). Lets call their sum $S\_i$. Obviously $S\_K$ and $S\_i$ are r.vs. Can we say anything about the inequality associated with their means. $E(S\_i) > E(S\_K)$ ?.
What does it depend on? Will the result change for non-identical r.vs.?
| https://mathoverflow.net/users/8447 | sum of order statistics | The inequality $E(S\_K) \geq E(S\_i)$ holds.
To avoid any doubt, let me be more specific. Let $Y\_1, Y\_2, ..., Y\_N$ be a collection of random variables, and write $X\_1 \geq X\_2 \geq ... \geq X\_N$ for their reordering in non-increasing order.
Suppose $K < N$ is fixed and let $S\_K$ be the sum of the $K$ largest of the random variables, that is $S\_K=X\_1+...+X\_K$.
Let $R$ be a random variable taking values in $0,1,...,N$ which is independent of the random variables $Y\_i$. The independence from the $Y\_i$ is important of course (this is how I interpret your "based on some criteria". If the $R$ is allowed to depend on the realisation of the $Y\_i$ then all sorts of different behaviours are possible).
Now let $S\_R$ be the sum of the $R$ largest of the random variables, that is $S\_R=X\_1+...+X\_R$. (In your notation this is $S\_i$).
Suppose that $ER=K$. Then I claim that $E S\_R \leq E S\_K$, with equality iff $R=K$ with probability 1. (Unless the $Y\_i$ are somehow degenerate, in which case equality can occur in other cases as well).
Proof: Write $p\_k=P(R\geq k)$ for $k=1,2,...,N$. We have $\sum p\_k=ER=K$.
Also
$S\_R=\sum\_{k=1}^N X\_k I(R\geq k)$
so
$ES\_R=\sum\_{k=1}^N P(R\geq k) E X\_k = \sum\_{k=1}^N p\_k E X\_k$.
(Here we used the independence of $R$ from the $X\_i$).
Consider maximising this sum subject to the constraints that $\sum p\_k=K$ and that
$1\geq p\_1 \geq p\_2 \geq p\_3\geq ...$.
Since the terms $E X\_k$ are decreasing in $k$,
the maximum is achieved when $p\_k=1$ for $k\leq K$ and $p\_k=0$ for $k>K$.
(Provided the $Y\_i$ are not degenerate, the terms $E X\_k$ are strictly decreasing,
and this is the only way to achieve the maximum. If not, the maximum may be achieved in some other cases too).
That is, the maximum value of $ES\_R$ occurs precisely if $R$ is equal to $K$ with probability 1.
It doesn't matter whether the $Y\_i$ are identically distributed, and also they don't need to be independent. However, it is important that $R$ is independent of the $Y\_i$.
| 1 | https://mathoverflow.net/users/5784 | 35745 | 23,038 |
https://mathoverflow.net/questions/35746 | 76 | In a recent issue of New Scientist (16 Aug 2010), I was surprised to read that a part of Wiles' proof of Taniyama-Shimura conjecture relies on inaccessible cardinals.
Here's the article
* Richard Elwes, *[To infinity and beyond: The struggle to save arithmetic](http://www.newscientist.com/article/mg20727731.300-to-infinity-and-beyond-the-struggle-to-save-arithmetic.html)*, New Scientist, August 2010.
Here's the relevant bit from the article:
>
> "Large cardinals have been studied by logicians for a century, but their intangibility means they have seldom featured in mainstream mathematics. A notable exception is the most celebrated result of recent years, the proof of Fermat's last theorem by the British mathematician Andrew Wiles in 1994 [...]
> To complete his proof, Wiles assumed the existence of a type of large cardinal known as an inaccessible cardinal, technically overstepping the bounds of conventional arithmetic"
>
>
>
Is this true ?
If so, could someone please outline how they are used ?
| https://mathoverflow.net/users/8528 | Inaccessible cardinals and Andrew Wiles's proof | The basic contention here is that Wiles' work uses cohomology of sheaves on certain Grothendieck topologies, the general theory of which was first developed in Grothendieck's SGAIV and which requires the existence of an uncountable [Grothendieck universe](http://en.wikipedia.org/wiki/Grothendieck_universe). It has since been clarified that the existence of such a thing is equivalent to the existence of an [inaccessible cardinal](http://en.wikipedia.org/wiki/Inaccessible_cardinal), and that this existence -- or even the consistency of the existence of an inaccessible cardinal! -- cannot be proved from ZFC alone, assuming that ZFC is consistent.
Many working arithmetic and algebraic geometers however take it as an article of faith that in any use of Grothendieck cohomology theories to solve a "reasonable problem", the appeal to the universe axiom can be bypassed. Doubtless this faith is ~~predicated on~~ abetted by the fact that most arithmetic and algebraic geometers (present company included) are not really conversant or willing to wade into the intricacies of set theory. I have not really thought about such things myself so have no independent opinion, but I have heard from one or two mathematicians that I respect that removing this set-theoretic edifice is not as straightforward as one might think. (Added: here I meant removing it from *general constructions*, not just from applications to some particular number-theoretic result. And I wasn't thinking solely about the small etale site -- see e.g. the comments on crystalline stuff below.)
Here is an article which gives more details on the matter:
* Colin McLarty, *What does it take to prove Fermat’s last theorem? Grothendieck and the logic of number theory*, Bull. Symb. Log. **16** No. 3 (2010) pp. 359–377, doi:[10.2178/bsl/1286284558](https://doi.org/10.2178/bsl/1286284558), [archived author pdf](https://web.archive.org/web/20100528052913/http://www.cwru.edu/artsci/phil/Proving_FLT.pdf).
Note that I do not necessarily endorse the claims in this article, although I think there is something to the idea that written work by number theorists and algebraic geometers usually does not discuss what set-theoretic assumptions are necessary for the results to hold, so that when a generic mathematician tries to trace this back through standard references, there may seem to be at least an apparent dependency on Grothendieck universes.
P.S.: If a mathematician of the caliber of Y.I. Manin made a point of asking in public whether the proof of the Weil conjectures depends in some essential way on inaccessible cardinals, is this not a sign that "Of course not; don't be stupid" may not be the most helpful reply?
| 85 | https://mathoverflow.net/users/1149 | 35749 | 23,039 |
https://mathoverflow.net/questions/35748 | 12 | Pretty straitforward:
If a field has a metric in which it is complete can it have a metric in which it is not complete?
By metric I mean field norm of course
| https://mathoverflow.net/users/4477 | Is completeness of a field an algebraic property? | How about the algebraic closure of the $p$-adics $\mathbb{Q}\_p^{\mathrm{alg}}$.
This is not complete under the $p$-adic metric, but it is isomorphic as
a field to the complex numbers $\mathbb{C}$ which is complete under the
standard metric (as both fields are algebraically closed of characteristic
zero with the same transcendence degree over $\mathbb{Q}$
(assuming AC of course)).
But if you want the second field to have nonarchimedean norm, take
the $p$-adic complex field $\mathbb{C}\_p$
(the completion of $\mathbb{Q}\_p^{\mathrm{alg}}$).
| 20 | https://mathoverflow.net/users/4213 | 35750 | 23,040 |
https://mathoverflow.net/questions/35763 | 5 | Let $\mathfrak{g}$ be a differential graded Lie algebra on a charcteristic zero field, whose underlying chain complex is bounded below and degreewise finite dimensional. Then $\mathfrak{g}$ defines a Set-valued functor on differential graded commutative algebras (also, with some finiteness and boundedness assumption) mapping each cgda $\Omega$ to the set $MC(\Omega\otimes \mathfrak{g})$ of Maurer-Cartan elements of the dgla $\Omega\otimes \mathfrak{g}$ (with the natural dgla structure on the tensor product of a dgla with a cdga). It is well known that this functor is representable: $MC(\Omega\otimes \mathfrak{g})\cong Hom\_{dgca}(CE(\mathfrak{g}),\Omega)$, where the Chevalley-Eilenberg dgca $CE(\mathfrak{g})$ is the free graded commutative algebra on the shifted linear dual of $\mathfrak{g}$ endowed with the differential induced by the dgla structure on $\mathfrak{g}$.
If one looks at the category of commutative graded algebras instead (i.e., one forgets the differential), then $CE(\mathfrak{g})$ represents the functor $\Omega\mapsto (\Omega\otimes\mathfrak{g})^1$.
My question is: is this latter functor representable also in the category of differential graded commutative algebras? if yes, by which dgca?
| https://mathoverflow.net/users/8320 | Maurer-Cartan and representable functors on differential graded commutative algebras | By the [Weil algebra](http://ncatlab.org/nlab/show/Weil+algebra) of $\mathfrak{g}$.
| 7 | https://mathoverflow.net/users/381 | 35767 | 23,053 |
https://mathoverflow.net/questions/35755 | 2 | Say there is a 2D plane (square) with some points inside it.
How to move all the points in such a way that they fill the plane as evenly as possible but every point maintains its neighbors?
In other words, I want the points to be as far from each other as possible but their locality (topology) should be preserved and they should lay in the square.
In other words, I want to kind of zoom-in in the rich-point-populated area and zoom-out in the empty areas.
PS: is there a general solution for higher-dimension spaces? Is there a direct solution or only iterative one?
**Edit**:
Now what, three great answers but I can accept only one. Thank you all!
| https://mathoverflow.net/users/8531 | Topological scaling (?) | The naive approach would be to write down the variance for the pairwise distances as a function of the coordinates of the points and do a gradient descent to spread them out evenly. You probably want to limit which pairwise distances you consider: it might suffice to take, for each point, its three closest neighbors. To spread things out evenly within a square, include the corners and perhaps evenly spaced points on the boundary in your calculation, but don't move them. The strategy would be the same in higher dimensions, but with more neighbors included.
| 2 | https://mathoverflow.net/users/4391 | 35773 | 23,058 |
https://mathoverflow.net/questions/35778 | 7 | Let $G$ be a locally compact group. The group C\*-algebra $C^\* (G)$ is designed to come with a natural bijection between its (nondegenerate) representations and the (strongly continuous, unitary) representations of $G$.
*Question*: Is there a similar statement for the reduced group C\*-algebra $C^\*\_r (G)$?
If the answer is no, I'll probably end up asking for the actual purpose of defining $C^\*\_r (G)$. So far, I know that its isomorphic to $C^\* (G)$ in important cases, and that its construction is in some sense simpler than the one of $C^\* (G)$.
(The definitions and the claims used above can be found in Blackadar's *Operator Algebras*.)
| https://mathoverflow.net/users/1291 | What does the representation theory of the reduced C*-algebra correspond to? | So there is a similar property.
Now $C^\*\_r(G)$ is the $C^\star$-algebra generated by the left-regular rep. It a general theorem that if you have a unitary rep $\pi:G\rightarrow \mathcal{U} (H)$, and if $\rho: G\rightarrow \mathcal{U}(K)$ is another unitary rep that is weakly contained ($\rho\prec\pi$) in $\pi$, then there is a surjective map from the reduced $C^\star$-algebra to the algebra generated by $\rho(G)\subset B(K)$
So $C^\star\_r(G)$ surjects onto all reps that weakly contain the left-regular.
Note: $C^\star\_r(G)\simeq C^\star(G)$ iff G is amenable.
A good source for most of this
<http://perso.univ-rennes1.fr/bachir.bekka/KazhdanTotal.pdf>
This is the pdf of a book about Property (T). Appendix F.4 is about the above questions but the whole book is of interest for people in operator algebras, representation theory, geometric group theory, and many other fields.
EDIT: Another good source, which is directed to Yemon's comment is <http://arxiv.org/PS_cache/math/pdf/0509/0509450v1.pdf>
This is a survey, by Pierre de la Harpe, of groups whose reduced $C^\star$-algebra is simple.
| 7 | https://mathoverflow.net/users/5732 | 35779 | 23,060 |
https://mathoverflow.net/questions/35628 | -1 | Planar graphs with a Hamiltonian loop connecting all faces do not necessarily have a Hamiltonian on their edges, which would make a 3 edge coloring and thus a 4 face coloring easy. However they have a lot of great structure. If you split the graph along the face Hamiltonian using it like an equator,the edges in the northern or southern hemisphere form trees. Thus the problem of coloring Hamiltonian planar graphs with n faces reduces to finding a mutual edge 3 coloring of any two n trees.
Regarding the Hamiltonian graphs as compositions of trees makes them easily counted, and gives a relation between graphs by relating their trees.
Between trees of the same size: Any tree of size n may converted into another of size n by iterated diamond switches, DS, of internal branches. I have proven any n-tree can be converted into any other n-tree in less than 2n DS. Nicely, if 2 trees are within n DS they have a mutual coloring.
Between trees of different sizes: Larger and smaller Hamiltonian graphs can be made by adding or subtracting branches that cross the equator, preserving the Hamiltonian loop. So inductive proofs are encouraged.
Also any tree of n roots has exactly 2^(n-3) colorings , which can be arranged on a hypercube, so you could prove all Hamiltonian graphs with n faces are colorable by proving any two color-hypercubes of n-trees intersect.
Hamiltonian graphs are a good restricted simpler group of graphs to try to prove colorable, interesting in their own right. They are even more important if their colorability implies the colorability of all planar graphs. Is it so?
| https://mathoverflow.net/users/8483 | Are all Hamiltonian planar graphs are 4 colorable? Does this imply all planar graphs are colorable? | If $G$ is a graph all of whose faces are triangles, and $G'$ is its dual, then recall that $G$ is $4$-vertex-colorable if and only if $G'$ is $3$-edge-colorable. It is unclear whether you are asking about the situation of $G$ or of $G'$ having a Hamiltonian cycle. As far as I know (but I am not a graph theorist), there is no relation between these two conditions.
Whitney's theorem says that, modulo some minor hypotheses, $G$ has a Hamiltonian cycle. Specifically:
>
> [**Whitney's Theorem:**](http://www.ams.org/mathscinet-getitem?mr=1503003) Let $G$ be a planar graph all of whose faces are triangles. Suppose that $G$ contains no loops, no multiple edges, and no $3$-cycles which are not faces. Then $G$ has a Hamiltonian cycle.
>
>
>
$4$-colorability for graphs that meet the hypotheses of Whitney's theorem easily implies $4$-colorability of all planar graphs.
If $G'$ has a Hamiltonian cycle, then $G'$ is trivially $3$-edge-colorable. Graphs of the form $G'$ tend to be Hamiltonian. This is true both rigorously -- Hamiltonian graphs have density approaching $1$ among cubic graphs with $n$ labeled vertices (see [Robinson and Wormwald](http://www.ams.org/mathscinet-getitem?mr=1151355)) and as a statement about common experience -- the smallest counter-examples have about 36 vertices and are not obvious.
MathSciNet recommends [this survey](http://ajc.maths.uq.edu.au/pdf/20/ocr-ajc-v20-p111.pdf) for more on Hamiltonianicity of cubic graphs.
| 6 | https://mathoverflow.net/users/297 | 35782 | 23,063 |
https://mathoverflow.net/questions/34784 | 0 | $(A,\mathfrak{m})$ an Artin local ring, $E(A/\mathfrak{m})$ the injective hull of $A/\mathfrak{m}$. How do I see that $E(A/\mathfrak{m})$ is a finite $A$-module?
| https://mathoverflow.net/users/5292 | Finiteness of injective hull of residue field for Artin local ring | This is a proof of $\ell(M)=\ell(\mbox{Hom}(M,\mbox{E}(A/\mathfrak{m}))$ suggested by Mariano:
Induction on $\ell(M)\ $:
If $\ell(M)=0$, $M=0$ so obviously true. Suppose $\ell(M)=n\geq 1$. From a composition series of $M$ choose the submodule N right beneath M so that $\ell(N)=n-1$ and $M/N\simeq A/\mathfrak{m}$. $0\rightarrow N\rightarrow M \rightarrow A/\mathfrak{m}\rightarrow 0$ induces $0\leftarrow \mbox{Hom}(N,E(A/\mathfrak{m}))\leftarrow \mbox{Hom}(M,E(A/\mathfrak{m}))\leftarrow \mbox{Hom}(A/\mathfrak{m},E(A/\mathfrak{m}))\leftarrow 0$.
Now $\mbox{Hom}(A/\mathfrak{m},E(A/\mathfrak{m}))\simeq A/\mathfrak{m}$ since $E(A/\mathfrak{m})$ is an essential extension of $A/\mathfrak{m}$, and $\ell(\mbox{Hom}(N,E(A/\mathfrak{m})))=\ell(N)=n-1$ by the induction hypothesis. $\ell(A/\mathfrak{m})=1$ so $\ell(\mbox{Hom}(M,E(A/\mathfrak{m}))=(n-1)+1=n$
| 0 | https://mathoverflow.net/users/5292 | 35783 | 23,064 |
https://mathoverflow.net/questions/35774 | 5 | Hi,
I'm wondering if there is a some classification of representations of CCR algebras (<http://en.wikipedia.org/wiki/CCR_algebra>), where say the underlying vector space is a separable Hilbert space.
My naive understanding is that for a QFT, one wants a representation of a CCR algebra satisfying certain properties.
| https://mathoverflow.net/users/8533 | Classification of representations of CCR algebras? | The question depends very much on the regularity that you demand. You have to decide before asking the question which operators are supposed to be self-adjoint or merely symmetric as unbounded operators etc. Weyl has solved the problem by exponentiating everything and looking at the resulting relations. This however gives rise to some *unphysical* representations.
Buchholz and Grundling give a new $C^\star$-algebraic approach to the problem in [0705.1988](http://arxiv.org/abs/0705.1988) using the notion of resolvent algebra. This settles the problem very nicely from a mathematical and physical perspective.
| 1 | https://mathoverflow.net/users/8176 | 35785 | 23,065 |
https://mathoverflow.net/questions/35788 | 21 | In this [question](https://mathoverflow.net/questions/22111/extending-vector-bundles-on-a-given-open-subscheme), Ariyan asks about the question of uniqueness of extensions of vector bundles when they exist.
Sasha's answer suggests that extensions of vector bundles don't always exist.
More precisely, if $F$ is a vector bundle on an open subscheme $U$, there does not always exist a vector bundle $F'$ on the ambient space $X$ such that $F'|\_U \cong F$.
Can anyone give me a simple example of such an $F$?
I am mainly interested in the case when $X$ is a variety (over $\mathbb{C}$), and $U$ is an open subvariety. Probably I want $X$ to be smooth.
| https://mathoverflow.net/users/83 | Extending vector bundles on a given open subscheme, reprise | The simplest example is the following. Take $X = A^3$ with coordinates $(x,y,z)$, and let $E = Ker(O\_X \oplus O\_X \oplus O\_X \stackrel{(x,y,z)}\to O\_X)$. Let $U$ be the complement of the point $(0,0,0) \in X$. Then $E\_{|U}$ is a vector bundle. On the other hand, $E$ is not a vector bundle, but $E^{\*\*} \cong E$, hence $E$ is the reflexive envelope of $i\_\*i^\*E$, and thus there is no vector bundle on $X$ extending $E\_{|U}$.
---
[Edit by Anton: I just spent some time digesting some pieces of the above answer, so figured I'd include the results for future readers similar to me.]
**("$E$ is not a vector bundle")** The sequence $O\_X\xrightarrow{\pmatrix{z\\ y \\ x}}O\_X^3\xrightarrow{\pmatrix{y & -z & 0\\ -x & 0 & z\\ 0 &x&-y}}O\_X^3\xrightarrow{\pmatrix{x& y& z}}O\_X$ is exact, so $E$ is the cokernel of the first map. Since taking fibers commutes with taking cokernels, we compute that $E$ has 2-dimensional fibers away from the origin, and 3-dimensional fiber at the origin.
**("$E^{\*\*}\cong E$")** Note that $E$ is $S\_2$ (i.e. sections defined away from codimension 2 extend uniquely) since it is the kernel of a map from an $S\_2$ sheaf to a torsion-free sheaf (the section of $O\_X^3$ extends uniquely, and its image is zero away from codimension 2, so must be zero, so the extended section is in $E$). Note also that the dual of any sheaf is $S\_2$ (if $\phi\colon F\to O\_X$ is defined on an open set $V$ with codimension 2 complement and $s$ is a section, $\phi(s)$ must be the unique extension of $\phi(s|\_V)$ as a section of $O\_X$), so $E^{\*\*}$ is $S\_2$. The canonical map $E\to E^{\*\*}$ is then a map of $S\_2$ sheaves which is an isomorphism away from codimension 2, so it must be an isomorphism.
**("and thus there is no vector bundle on $X$ extending $E|\_U$")** If $F$ is an $S\_2$ extension of $E|\_U$ (i.e. $i^\*F=i^\*E$), then there is a map $F\to i\_\*i^\*E\to (i\_\*i^\*E)^{\*\*}=E$ which is an isomorphism over $U$, so is an isomorphism by the argument in the previous paragraph. A vector bundle extension would be a different $S\_2$ extension.
| 29 | https://mathoverflow.net/users/4428 | 35790 | 23,068 |
https://mathoverflow.net/questions/35765 | 8 | Please imagine a discrete random walk on an N-dimensional rectangular lattice with dimensional lengths $(l\_1, ..., l\_N) \in L$ and total lattice points $P = \prod{l\_i}$, for $i = 1, ..., N$. At each time step, the walker will move to one of it's adjacent lattice points with equal probability. The N-dimensional random walk is non-self-avoiding, the walker must move with each time step, and the boundaries of the lattice are reflecting. However, jump probabilities must be adjusted at edges and corners due to a reduction in the number of adjacent nodes - i.e. jump probabilities will vary from $\frac{1}{2N}$ internal to the lattice to $\frac{1}{N}$ at the edges of the lattice.
Provided the random walk specifications above, what might be the expected step-time distribution for the walker visiting every position in the N-dimensional rectangular lattice with dimensional lengths $L$?
| https://mathoverflow.net/users/3248 | Expected number of steps for a discrete random walk to visit every point on an N-dimensional rectangular lattice | If you look for "cover time of graph" you will find a lot of references, cf. e.g. "Jonasson, Schramm, ON THE COVER TIME OF PLANAR GRAPHS, Elect. Comm. in Probab. 5 (2000) 85-90, <http://www.emis.de/journals/EJP-ECP/_ejpecp/ECP/include/getdocbfb7.pdf>. In this paper you find the following result by Zuckermann (for details pls. cf the paper):
If $G$ is a finite portion of a $d$-dimensional lattice $\mathbb{Z}$ with $n$ vertices, the expected cover time (to visit all vertices) is $\Theta(n^2)$ for $d=1$, $\Theta(n (\log n)^2)$ for $d=2$, and $\Theta(n \log n)$ for $d \ge 3$.
| 10 | https://mathoverflow.net/users/6415 | 35791 | 23,069 |
https://mathoverflow.net/questions/35825 | 4 | I'm trying to learn about forcing, and have heard that there are several equivalent ways to define genericity. For instance, let M be a transitive model of ZFC containing a poset (P, ≤). Suppose G ⊆ P is such that q ∈ G whenever both p ∈ G and q ≥ p. Suppose also that whenever p,q ∈ G then there is r ∈ G such that r ≤ p and r ≤ q.
Then the following are equivalent ways to say that G is generic:
(1) G meets every element of M dense in P. That is, for all D ∈ M, if for all p ∈ P there is q ∈ D such that q ≤ p, then G ∩ D is nonempty.
(2) G is nonempty and meets every element of M dense below some p ∈ G. That is, for all p ∈ G and all B ∈ M, if for each q ≤ p there is r ∈ B such that r ≤ q, then G ∩ B is nonempty.
Proving this equivalence seemed like it would be an easy exercise, but I think I'm missing something. Can someone point me toward a source where I can find a proof? I hope this is an acceptable question; this is my first time posting.
EDIT: Typo and omission fixed.
| https://mathoverflow.net/users/8546 | Equivalent definitions of M-genericity. | If $G$ satisfies (1), then it satisfies (2) because if $p$ is in $G$ and $D$ is dense below $p$, then let $D'$ be the set of conditions $q$ which are either in $D$ or incompatible with $p$. This is dense in $P$ since any condition that is compatible with $p$ will have elements of $D$ below it, and any condition incompatible with $p$ is already in $D'$. But $G$ cannot meet $D'$ in something incompatible with $p$, by your assumption on $G$, and so it must meet it in $D$, as desired.
Conversely, if $G$ satisfies (2), then it will satisfy (1) because if $D$ is dense, then it is dense below any $p$, and so $G$ will meet it.
| 5 | https://mathoverflow.net/users/1946 | 35828 | 23,086 |
https://mathoverflow.net/questions/35793 | 25 | It is clear that each maximal ideal in ring of continuous functions over $[0,1]\subset \mathbb R$ corresponds to a point and vice-versa.
So, for each ideal $I$ define $Z(I) =\{x\in [0,1]\,|\,f(x)=0, \forall f \in I\}$. But map $I\mapsto Z(I)$ from ideals to closed sets is not an injection! (Consider the ideal $J(x\_0)=\{f\,|\,f(x)=0, \forall x\in\hbox{ some closed interval which contains }x\_0\}$)
How can we describe ideals in $C([0,1])$ ? Is it true that prime ideals are maximal for this ring?
| https://mathoverflow.net/users/4298 | prime ideals in C([0,1]) | Here is a way to construct a non-maximal prime ideal: consider the multiplicative set $S$
of all non-zero polynomials in $C[0,1]$. Use Zorn lemma to get an ideal $P$ that is disjoint from $S$ and is maximal with this property. $P$ is clearly prime (for this you only need $S$ to be multiplicative.) On the other hand $P$ cannot be any one of the maximal ideals, since
it does not contain $x-c$ for every $c \in [0,1]$.
| 27 | https://mathoverflow.net/users/3635 | 35832 | 23,090 |
https://mathoverflow.net/questions/24031 | 5 | Does dimension of a module (say, dimension of its support) have anything to do with the supremum length of chains of prime submodules like rings?
Let's restrict to finitely generated modules over Noetherian ring.
Prime submodules are defined analogously to primary submodules: a submodule P in M is prime if P$\neq$M and $M/P$ has no zero divisors, i.e. $am\in P$ implies $m\in P$ or $a \in \mbox{Ann}(M/P)$.
| https://mathoverflow.net/users/5292 | Dimension of module | Let $R$ be an integral domain, then for the module $R^n$ its maximal length of chains of prime submodules is much larger than its dimension (for $n>>0$).
| 2 | https://mathoverflow.net/users/8257 | 35839 | 23,094 |
https://mathoverflow.net/questions/31772 | 7 | Let $(W, S)$ be a Coxeter system, and let $T = \bigcup\_{w \in W, s \in S} wsw^{-1}$. Associated to every element $t \in T$ is a unique positive root $\alpha\_t \in \Phi^{+}$ considered as a vector in the standard geometric representation $V$ of $W$. A total order on $T$ is a **reflection order** if, whenever $\alpha\_{t\_1} < \alpha\_{t\_2}$, it follows that $\alpha\_{t\_1} < x \alpha\_{t\_1} + y \alpha\_{t\_2} < \alpha\_{t\_2}$ whenever the middle term is a positive root with $x > 0, y > 0$. (See, for example, [Bjorner and Brenti's](http://books.google.com/books?id=OsaeruFjCRsC&printsec=frontcover&dq=bjorner+and+brenti&source=bl&ots=mvBZNqSImQ&sig=wNGg-W9sBQa4547x8-g1CGmbJFM&hl=en&ei=De08TOjcEoeglAeek5DqCA&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBcQ6AEwAA#v=onepage&q=bjorner%20and%20brenti&f=false) book.)
Fix a reflection order and let $[u, v]$ be a Bruhat interval. A maximal chain $u = w\_0 \to w\_1 \to ... \to w\_m = v$ in the Bruhat order is what I'll call monotonic if $w\_i w\_{i-1}^{-1} > w\_{i+1} w\_i^{-1}$ in the reflection order.
There is a nonrecursive formula for the Kazhdan-Lusztig polynomials $P\_{u,v}(q)$ which implies that $P\_{u,v}(0)$ is equal to the number of monotonic maximal chains in $[u, v]$. This number is known by other means to be equal to $1$, so I know that there is a unique monotonic maximal chain; however, I can't prove this directly. So far all I've been able to do is use the greedy algorithm to prove that at least one monotonic maximal chain exists.
Does anyone have a direct proof of this fact?
**Edit:** No progress, but now I have a more general conjecture which I no longer know by other means is true. Fix a sequence $a\_1, ... a\_m$ of odd positive integers such that $\sum\_i a\_i = \ell(v) - \ell(u)$. Then there exists at most one monotonic chain (not necessarily maximal) such that $w\_i w\_{i-1}^{-1} \in T$ and such that $\ell(w\_{i-1}) - \ell(w\_i) = a\_i$.
| https://mathoverflow.net/users/290 | Monotonic maximal chains in a Coxeter group | Hi, you may want to try to peruse these two papers:
Dyer, M. J. Hecke algebras and shellings of Bruhat intervals. Compositio Math. 89 (1993), no. 1, 91--115.
Dyer, M. J. Hecke algebras and shellings of Bruhat intervals. II. Twisted Bruhat orders. Kazhdan-Lusztig theory and related topics (Chicago, IL, 1989), 141--165, Contemp. Math., 139, Amer. Math. Soc., Providence, RI, 1992.
Dyer proves that Bruhat intervals are EL-shellable, and this gives the answer to your first question. As to your more general question, there may be answers to those as well (indirectly of course).
| 6 | https://mathoverflow.net/users/8549 | 35841 | 23,095 |
https://mathoverflow.net/questions/35838 | 12 | Hi
So there is an edge-colouring of a complete graph on R (the reals), with countably many colours that as no monochromatic triangle. To find it map R to (0,1) write the numbers in binary and if 2 numbers differ 1st in the kth digit use colour k.
Now this colouring has cycles of length 4. (1/4, 3/4, 1/3, 2/3 for example). You can get rid of cycles of length 4 by considering the 1st 2 binarary digits in which 2 numbers differ (and of course seperate colours if they only differ in 1 digit). Anyway my question is can we avoid cycles completely? i.e. does there exist a colouring of the complete graph on R such that there is no monochromatic cycle.
| https://mathoverflow.net/users/1000 | monochromatic cycle-free colouring of the complete graph on R? | Turns out that the existence of such a coloring is equivalent to the continuum hypothesis. This was proved by Erdos and Kakutani in 1943 in the paper "On non-denumerable graphs". They prove:
>
> A complete graph of cardinal number $m$ (that is, the cardinal number of the vertices is $m$) can be split up into a countable number of trees if and only if $m\le \aleph\_1$.
>
>
>
| 14 | https://mathoverflow.net/users/2384 | 35850 | 23,101 |
https://mathoverflow.net/questions/35834 | 8 | Consider some biconnected graph $G$. Removing any single edge will not disconnect $G$. However, unless $G$ is triconnected, there is some pair of edges whose removal will disconnect $G$. For a cycle of length $l$, the removal of any pair will disconnect the graph (if the edges are adjacent, there will be an isolated vertex).
>
> Conjecture: any biconnected graph on $l$ vertices has at most $\binom{l}{2}$ pairs of edges that disconnect it. Furthermore, the bound is tight only for cycles.
>
>
>
Is the conjecture true? I have a candidate proof but I suspect that there's a simpler one.
| https://mathoverflow.net/users/7732 | How many pairs of edges can disconnect a biconnected graph? | The statement is true. In fact, much more general statements are true. If $G$ is a graph with $n$ vertices and $c$ is the cardinality of a minimum edge cut of $G$, then the number of edge cuts of cardinality $c$ is at most $\binom{n}{2}$, and for every half-integer $k \geq 1$, the number of edge cuts containing at most $kc$ edges is bounded above by $2^{2k-1} \binom{n}{2k}.$
The upper bound of $\binom{n}{2}$ on the number of minimum cuts is attributed to Bixby and Dinitz-Karzanov-Lomonosov. The more general bound on the number of approximate minimum cuts is due to Karger ([Global min-cuts in RNC, and other ramifications of a simple min-cut algorithm](http://portal.acm.org/citation.cfm?id=313559.313605)), who also re-proved the $\binom{n}{2}$ bound on minimum cuts. His appealingly simple proof rests on the analysis of a simple "randomized contraction" algorithm. Here we present the proof that the number of minimum cuts is at most $\binom{n}{2}$.
Suppose that $G$ is a multigraph with $n$ vertices, $c>0$ is the number of edges in a minimum cut of $G$, and $A$ is a specific set of $c$ edges whose removal disconnects $G$. Repeatedly perform the following process to obtain a sequence of multigraphs $G = G\_0, G\_1, \ldots, G\_{n-2}$: choose a uniformly random edge of $G\_t$ and contract it to obtain $G\_{t+1}$. In other words, if $(u,v)$ is the edge chosen in step $t$, then we replace $u$ and $v$ with a single vertex $z$ in $G\_{t+1}$, and we replace every edge of $G\_t$ having exactly one endpoint in $\{u,v\}$ with a corresponding edge of $G\_{t+1}$ with endpoint $z$. (Edges from $u$ to $v$ in $G\_t$ are deleted during this step.) Note that $G\_{n-2}$ has exactly two vertices $a,b$, these vertices correspond to a partition of $V(G)$ into two nonempty sets $A,B$ (those vertices that were merged together to form $a \in V(G\_2)$, and those that were merged together to form $b$), and that the edges of $G\_{n-2}$ are in one-to-one correspondence with the edges of the cut separating $A$ from $B$ in $G$. Denote this random cut by $R$.
Now consider a specific cut $C$ of cardinality $c$. We claim that the probability of the event $R=C$ is at least $1 \left/ \binom{n}{2} \right.$, from which it follows immediately that the number of distinct cuts of cardinality $c$ is at most $\binom{n}{2}$. To prove the upper bound on the probability that $R=C$, observe that for all $t = 0,\ldots,n-2$, every vertex of $G\_t$ has degree at least $c$. (Otherwise, that vertex of $G\_t$ corresponds to a *set* of vertices in $G$ having fewer than $c$ edges leaving it, contradicting our assumption about the edge connectivity of $G$.) Consequently, the number of edges of $G\_t$ is at least $(n-t)c/2$, and the probability that an edge of $C$ is contracted in step $t$, given that no edge of $C$ was previously contracted, is at most $c/|E(G\_t)| \leq 2/(n-t)$. Combining these bounds, we find that the probability that no edge of $C$ is *ever* contracted is bounded below by $\prod\_{t=0}^{n-3} \left(1 - \frac{2}{n-t}\right) = \frac{n-2}{n} \cdot \frac{n-3}{n-1} \cdots \frac{1}{3} = \frac{2}{n(n-1)}.$
| 15 | https://mathoverflow.net/users/8049 | 35865 | 23,113 |
https://mathoverflow.net/questions/35455 | 33 | **Question**
Suppose there is a bijection between the underlying sets of two finite groups $G, H$, such that every subgroup of $G$ corresponds to a subgroup of $H$, and that every subgroup of $H$ corresponds to a subgroup of $G$. Does this imply that $G, H$ are isomorphic? Note that we do not require the bijection to actually be the isomorphism.
**Motivation**
The question is interesting to me because I am considering maps of groups which aren't homomorphisms but preserve the subgroup structure in some sense - given a group, we can forget the multiplication operation and look only at the closure operator that maps a subset of $G$ to the subgroup generated by it. If the question is resolved in the affirmative, then the forgetful functor from the usual category $Grp$ to this category won't create any new isomorphisms. (Note that I didn't precisely specify the morphisms this new category -- you could just use the usual definition of a homomorphism, and say that if the mapping commutes with the closure operator, then its a morphism. The definition I actually care about is, a morphism of this category is a mapping such that every closed set in the source object is the preimage of a closed set of the target object. It doesn't make much difference as far as this question is concerned, the isomorphisms of both categories are the same.)
I asked a friend at Mathcamp about this a few weeks ago, he said a bunch of people started thinking about it but got stumped after a while. The consensus seems to have been that it is probably false, but the only counter examples may be very large. I don't really have any good ideas / tools for how to prove it might be true, I mostly wanted to just ask if anyone knew offhand / had good intuition for how to find a finite counterexample.
---
Edit (YCor): (a) the question has reappeared in the following formulation: does the hypergraph structure of the set of subgroups of a (finite) group determine its isomorphism type? A hypergraph is a set endowed with a set of subsets. The hypergraph of subgroups is the data of the set of subgroups, and therefore to say that groups $G,H$ have isomorphic hypergraphs of subgroups means that there's a bijection $f:G\to H$ such that for every subset $A\subset G$, $f(A)$ is a subgroup of $H$ if and only $A$ is a subgroup of $G$. Several answers, complementing the one given here, have been provided [in this question](https://mathoverflow.net/questions/312177/).
(b) There a weaker well-studied notion for groups, namely to have isomorphic subgroup lattices. Having isomorphic hypergraphs of subgroups requires such an isomorphism to be implemented by a bijection (this is not always the case: take two groups of distinct prime order).
| https://mathoverflow.net/users/8445 | Does the hypergraph structure of the set of subgroups of a finite group characterize isomorphism type? | The answer is *no* in general. I.e, there are finite non-isomorphic groups G and H such that there exists a bijection between their elements which also induces a bijection between their subgroups.
For this, I used two non-isomorphic groups which not only have the same subgroup lattice (which certainly is necessary), but also have the same conjugacy classes. There are two such groups of size 605, both a semidirect product $(C\_{11}\times C\_{11}) \rtimes C\_5$ (see [this site](http://myyn.org/m/article/lattice-of-subgroups/) for details on the construction). In the small group library of [GAP](http://www.gap-system.org/), these are the groups with id [ 605, 5 ] and. [ 605, 6 ]. These are provably non-isomorphic (you can construct the groups as described in the reference I gave, and then use GAPs IdSmallGroup command to verify that the groups described there are the same as the ones I am working with here). With a short computer program, one can now construct a suitable bijection.
First, let us take the two groups:
```
gap> G:=SmallGroup(605, 5);
<pc group of size 605 with 3 generators>
gap> H:=SmallGroup(605, 6);
<pc group of size 605 with 3 generators>
```
The elements of these groups are of order 1, 5 or 11, and there are 1, 484 and 120 of each. We will sort them in a "nice" way (that is, we try to match each subgroup of order 5 to another one, element by element) and obtain a bijection from this. First, a helper function to give us all elements in "nice" order:
```
ElementsInNiceOrder := function (K)
local elts, cc;
elts := [ One(K) ];
cc := ConjugacyClassSubgroups(K, Group(K.1));
Append(elts, Concatenation(List(cc, g -> Filtered(g,h->Order(h)=5))));
Append(elts, Filtered(Group(K.2, K.3), g -> Order(g)=11));
return elts;
end;;
```
Now we can take the elements in the nice order and define the bijection $f$:
```
gap> Gelts := ElementsInNiceOrder(G);;
gap> Helts := ElementsInNiceOrder(H);;
gap> f := g -> Helts[Position(Gelts, g)];;
```
Finally, we compute the sets of all subgroups of $G$ resp. $H$, and verify that $f$ induces a bijection between them:
```
gap> Gsubs := Union(ConjugacyClassesSubgroups(G));;
gap> Hsubs := Union(ConjugacyClassesSubgroups(H));;
gap> Set(Gsubs, g -> Group(List(g, f))) = Hsubs;
true
```
Thus we have established the claim with help of a computer algebra system. From this, one could now obtain a pen & paper proof for the claim, if one desires so. I have not done this in full detail, but here are some hints.
Say $G$ is generated by three generators $g\_1,g\_2,g\_3$, where $g\_1$ generates the $C\_5$ factor and $g\_2,g\_3$ generate the characteristic subgroup $C\_{11}\times C\_{11}$. We choose a similar generating set $h\_1,h\_2,h\_3$ for $H$. We now define $f$ in two steps: First, for $0\leq n,m <11$ it shall map $g\_2^n g\_3^m$ to $h\_2^n h\_3^m$.
This covers all elements of order 1 or 11, so in step two we specify how to map the remaining elements, which all have order 5. These are split into four conjugacy classes: $g\_1^G$, $(g\_1^2)^G$, $(g\_1^3)^G$ and $(g\_1^4)^G$. We fix any bijection between $g\_1^G$ and $h\_1^H$ and extend that to a bijection on all elements of order 5 by the rule $f((g\_1^g)^n)=f(g\_1^g)^n$. With some effort, one can now verify that this is a well-defined bijection between $G$ and $H$ with the desired properties. You will need to determine the subgroup lattice in each case; linear algebra helps a bit, as well as the fact that all subgroups have order 1, 5, 11, 55, 121 (unique) or 605. I'll leave the details to the reader, as I myself am happy enough with the computer result.
**UPDATE**: as pointed out in another answer below by @dvitek (explained by @Ian Agol in comments), there is actually a much simpler example, which I somehow overlooked when I did my computer search. Credit to them, but just in case people want to reproduce their example with GAP, here is an input session doing just that:
```
gap> G:=SmallGroup(16,5);; StructureDescription(G);
"C8 x C2"
gap> H:=SmallGroup(16,6);; StructureDescription(H);
"C8 : C2"
gap> Gelts := ListX([1..8],[1,2],{i,j}->G.1^i*G.2^j);;
gap> Helts := ListX([1..8],[1,2],{i,j}->H.1^i*H.2^j);;
gap> f := g -> Helts[Position(Gelts, g)];;
gap> Gsubs := Union(ConjugacyClassesSubgroups(G));;
gap> Hsubs := Union(ConjugacyClassesSubgroups(H));;
gap> Set(Gsubs, g -> Group(List(g, f))) = Hsubs;
true
```
| 48 | https://mathoverflow.net/users/8338 | 35866 | 23,114 |
https://mathoverflow.net/questions/35860 | 4 | The Lovász Local Lemma (or **LLL**) concerns itself with the probability of avoiding a collection of "bad" events **A**, given that the set of events is "nearly independent" (each bad event *A* ∈ **A** has probability which is bounded above in terms of the number of other events *A*', *A*'', etc. from which it is not independent), there is a non-zero probability of avoiding all of the bad events simultaneously. The original presentation seems to be the Lemma on page 8 of [this pdf](http://www.cs.elte.hu/~lovasz/scans/LocalLem.pdf) (the link to which can be found on [Wikipedia's page on the LLL](http://en.wikipedia.org/wiki/Lov%C3%A1sz_local_lemma#References)); several other papers present it in a similar fashion.
In the article [[arXiv:0903.0544](http://www.arxiv.org/abs/0903.0544)], restricting to the setting where the "bad events" of the LLL are defined in terms of a probability space of independently distributed bits, Moser and Tardos present a probabilistic algorithm for sampling from the event space until an event is found which avoids all bad events, which requires at most polynomially many samples with high probability. However, their characterization of the LLL is significantly different from any presentation of it that I have seen elsewhere. Their version of the LLL is as follows:
>
> **Theorem.** Let **A** be a finite set of events in a probability space. For *A* ∈ **A** let Γ(*A*) be a subset of **A** satisfying that *A* is independent from the collection of events **A** \ ({*A*} ∪ Γ(*A*)). If there exists an assignment of reals x : **A** → (0,1) such that $$ \forall A \in \mathbf A : \Pr[A] \;\leqslant\; \mathrm x(A) \prod\_{B \in \Gamma(A)} (1-\mathrm x(B))$$
>
>
> then the probability of avoiding all events in A is at least $\prod\limits\_{A \in \mathbf A} \;(1 \;−\; \mathrm x(A))$, in particular it is positive.
>
>
>
The proof that their sampling algorithm works seems to depend substantially upon this presentation of the LLL, but I cannot decipher the exact relationship between this and more common presentations of it. It looks as though the product over Γ(*A*) in the bounding condition "wants to be" a bound on the conditional probabilities for events *A*, but I haven't been able to make the link. Could someone help me with the connection between this statement and more familar versions of the LLL?
| https://mathoverflow.net/users/3723 | Can you explain the description of the Lovasz Local Lemma by Moser+Tardos? | The version of the LLL that you wrote out above is stronger than the one on page 8 of the paper you linked to. The one you link to is sometimes known as the "symmetric form" of the LLL and the one you wrote out above as the "general form".
To see that the general form implies the symmetric form: restrict to the case where
$|\Gamma(A)|\leq d$ for all $A$ (as assumed in the symmetric form), and let $x(A)$ be a constant $x$ such that $x(1-x)^d \geq 1/(4d)$. (Check that you can always find such an $x$).
Then the condition $P(A) \leq 1/(4d)$ in the symmetric form implies the condition
$P(A) \leq x(A) \prod\_{B\in\Gamma(A)} (1-x(B))$
which is the hypothesis of the general form.
A standard reference is Chapter 5 of the book of Alon and Spencer, "The Probabilistic Method". Of course it doesn't cover Moser's new approach!
| 8 | https://mathoverflow.net/users/5784 | 35867 | 23,115 |
https://mathoverflow.net/questions/35868 | 13 | Let $T$ be a torus $V/\Gamma$, $\gamma$ a loop on $T$ based at the origin. Then it is easy to see that $$2 \gamma = \gamma \ast \gamma \in \pi\_1(T).$$
Here $2 \gamma$ is obtained by rescaling $\gamma$ using the group law, while $\ast$ denotes the operation in the fundamental group. The way I can check this is rather direct: one lifts the loop (up to based homotopy) to a segment in $V$ and uses the identification of $\pi\_1(T)$ with the lattice $\Gamma$.
Is there a more conceptual way to prove this identity that will extend to more general (real or complex) Lie groups, or maybe to linear algebraic groups? Or is this fact false in more generality?
| https://mathoverflow.net/users/828 | Fundamental group of Lie groups | Yay! It's the [Eckmann-Hilton](http://ncatlab.org/nlab/show/Eckmann-Hilton+argument) argument!
There are *two* group structures on $\pi\_1(G)$ and they commute *with each other*. It turns out that that is sufficient to show that they are the same structure *and* that that structure is commutative.
For a proof of this, using interpretative dance, take a look at the movie in [this seminar](http://www.math.ntnu.no/~stacey/Seminars/pearl.html) that I gave last semester. There's also something on YouTube by The Catsters (see the nLab page linked above).
(Forgot to actually answer your question!) This only depends on the fact that $\pi\_1$ is a representable group functor and that $G$ is a group object in $hTop$. So it will extend to other group objects in $hTop$, such as those that you mention. This also explains why $\pi\_k$ is abelian for $k \ge 2$ since $\pi\_2(X) = \pi\_1(\Omega X)$ and $\Omega X$ is a group object in $hTop$.
| 31 | https://mathoverflow.net/users/45 | 35869 | 23,116 |
https://mathoverflow.net/questions/35855 | 10 | According to the Elephant, and [these notes](http://www.staff.science.uu.nl/~ooste110/syllabi/toposmoeder.pdf), an object X in a category C is *indecomposable* if given an isomorphism $X \cong \coprod\_i U\_i$ there is a unique $i$ such that $X \cong U\_i$ and $U\_j \cong 0$ for $j\neq i$ where 0 is the initial object. If C is [extensive](http://ncatlab.org/nlab/show/extensive+category), then X is indecomposable iff it is [connected](http://ncatlab.org/nlab/show/connected+object) (proof and details [here](http://ncatlab.org/nlab/show/indecomposable+object)).
Lambek and Scott give a different definition: they say that X is indecomposable if given an epi $[k,l] \colon U + V \twoheadrightarrow X$, one of k or l is epi. I suppose this can be generalised to say that X is indecomposable if any jointly epic family into X contains at least one epi.
Perhaps I'm missing something obvious, but I can't see that either definition implies the other. So my question is
>
> Are these definitions equivalent, or does one imply the other, in general or in some specific class of categories? Do you know of a reference that compares or discusses the two?
>
>
>
| https://mathoverflow.net/users/4262 | Indecomposable objects in a category |
>
> **Briefly:** there's a simple difference in how they treat 0. That fixed, still neither implies the other in general. In a regular extensive category, a slight modification of the LS definition implies the Elephant one. I suspect they're not fully equivalent in anything short of a topos. As Mike Shulman points out, even in a topos they are not equivalent.
>
>
>
The simple difference: 0 is always indecomposable by Lambek and Scott's definition (since any map into 0 is epi), but never by the Elephant's (since the uniqueness condition won't hold; or by considering when the coproduct decomposition is empty). So, let's temporarily change one of the definitions to fix this. I'd suggest we add “…and the map $0 \to X$ is not epi.” to Lambek and Scott's definition. (As you noted, their binary condition generalises to a $k$-ary one; this is just the case $k=0$.)
In eg **Top**, however, we can see that the Elephant def still doesn't imply the LS def. $[0,1]$ satisfies the former (it's not decomposable by an iso), but not the latter (it is decomposable by an epi). Even more, it’s decomposable by a *regular* epi (more on this distinction below).
Conversely, the LS definition doesn't imply the Elephant one either; it fails in eg $\mathbf{Set}^\mathrm{op}$, since in $\mathbf{Set}$, $0$ is co-decomposable by iso ($0 \cong A \times 0$) but not co-decomposable by monos (for any map $(f,g) \colon 0 \to A \times B$, not just one but *both* of $f$ and $g$ are mono).
When *do* they imply each other? If we upgrade the LS definition to involve *regular* epis, then in a [regular](http://ncatlab.org/nlab/show/regular+category) [lextensive](http://ncatlab.org/nlab/show/extensive+category) category, it implies the Elephant definition, if I'm not mistaken. For this, suppose $X$ is “indecomposable by reg epis”, and suppose $X \cong A + B$ — WLOG $X = A + B$. The coproduct inclusions are then jointly reg epi, so one of them is reg epi. But it's also mono (in a lextensive category, every coproduct inclusion is a pullback of $1 \to 1 + 1$, so is mono); so it's iso. There's a little more fiddly stuff to check involving messing around with $0$, but it's all the same sort of thing.
**Edit from Mike Shulman's comments:** if moreover we're in a pretopos, all epis are regular, so there the original LS definition will imply the Elephant definition. On the other hand, the Elephant definition doesn't imply the LS even in a topos: the terminal object of $\mathbf{Sh}([0,1])$ is a counterexample, essentially for the same reasons that $[0,1]$ was a counterexample in $\mathbf{Top}$.
However, the two definitions are equivalent for *projective* objects… and I guess that's how this situation has arisen, since a common use of indecomposable objects in topos theory is the theorem that the indecomposable projectives in a presheaf category are exactly the retracts of representables. (This is useful because it lets us recover the idempotent-completion of $\mathbf{C}$, which is very close to $\mathbf{C}$ itself, from $[\mathbf{C}^\mathrm{op},\mathbf{Set}]$.)
| 10 | https://mathoverflow.net/users/2273 | 35884 | 23,126 |
https://mathoverflow.net/questions/35538 | 5 | Given the coefficients $a\_0,\ldots,a\_N$, $b\_1,\ldots,b\_N$ of a real trigonometric polynomial:
$ f(x) = a\_0 + \sum\_{n=1}^N a\_n \cos(nx) + \sum\_{n=1}^N b\_n \sin(nx) $
is there any efficient way to approximately determine $\max\_{x \in R} f(x)$? If so, what is the accuracy versus efficiency tradeoff?
| https://mathoverflow.net/users/8460 | The maximum of a real trigonometric polynomial | It turns out that it is possible to achieve an arbitrarily small additive error using semidefinite programming. This is from the paper:
J.W. McLean, H.J. Woerdeman. Spectral factorizations and sums of squares representations via semidefinite programming. SIAM J. Matrix Anal. Appl., 23(3):646--655, 2001. ([link](http://dx.doi.org/10.1137/S0895479800371177))
The result can be rephrased as follows. Let $f(x)=F(e^{ix})$ where $F(z)= \sum\_{n=-N}^N c\_n z^n$, with $c\_n=\frac{1}{2}(a\_n-i\ b\_n)$ and $c\_{-n}=\bar{c}\_n$. Then $\min\_x f(x)$ is equal to $c\_0$ minus the value of the following semidefinite program:
$ min\_F\ tr(F) $
such that $F \succeq 0$, and $\sum\_{p=k}^N F\_{p,p-k} = c\_k$ for $k=1,\ldots,N$.
Since semidefinite programming can achieve an arbitrarily small additive error, we can approximate the minimum (and thus, the maximum) of $f$ within the same bound.
| 7 | https://mathoverflow.net/users/8460 | 35886 | 23,128 |
https://mathoverflow.net/questions/35900 | 35 | Let $X$ be a differentiable manifold. Its cotangent bundle $T^\*X$ carries a canonical 1-form $
\alpha$ whose exterior differential $\omega = d\alpha$ endows $T^\*X$ with the structure of a symplectic manifold.
But what about the converse question? Which symplectic manifolds are cotangent bundles?
Clearly a necessary condition is that $\omega$ must be exact, so cohomological obstructions are relevant. Is that all? Compact symplectic manifolds have non-trivial de Rham cohomology in grade two, so the cohomological test passes muster for that important class of examples.
I'm also interested in examples of manifolds where different symplectic forms (modulo exact 2-forms) give qualitatively different dynamics with the same Hamiltonian. As symplectic structures on the same manifold are all locally equivalent by Darboux's theorem, one expects such phenomena would occur only on a global scale.
| https://mathoverflow.net/users/2036 | When is a symplectic manifold equivalent to a cotangent bundle? | Using the h-principle, Gromov showed that there is a symplectic form on $\mathbb{R}^6$ which admits $S^3$ as a Lagrangian submanifold. Using holomorphic curves, he showed that the standard symplectic form on $T^\* \mathbb{R}^3$ does not admit any such Lagrangian. There is now a whole industry of building exotic symplectic forms on non-compact manifolds (see papers of Seidel-Smith, Mark McLean, ...).
Probably the only reasonable answer to characterising cotangent bundles uses the existence of a Lagrangian foliation by planes. If you have a foliation parametrised by a manifold which admits a Lagrangian section, then you have yourself an open subset of a cotangent bundle (this is just Weinstein's theorem). You can't drop the condition of the existence of a section precisely because you can add the pull back of a $2$-form on the base. If your symplectic form is "complete" then the existence of a Lagrangian section is a cohomological condition. Pick any section: If the pullback of $\omega$ doesn't vanish, then you don't have a cotangent bundle. If it vanishes in cohomology, you can use a primitive $1$-form to flow your section to a Lagrangian.
Added Remark:
I want to point out that the methods we have for producing different symplectic forms do not proceed by writing down different $2$-forms on the same space. Rather, you find some construction of symplectic manifolds (using some general notion of symplectic surgery) which produces a large class of symplectic manifolds, then you prove that some of these result in the same smooth manifold. The existence of a diffeomorphism is obtained abstractly, so I do not know of examples where we can write down a Hamiltonian whose dynamics for two different symplectic forms can be compared.
| 37 | https://mathoverflow.net/users/6948 | 35901 | 23,137 |
https://mathoverflow.net/questions/35907 | 12 | The page "mapping class groups" on wikipedia says the topological MCG of T^n is GL(n,Z), but does anyone know a reference? Also, is the smooth MCG of T^n known?
| https://mathoverflow.net/users/8565 | What is the topological/smooth mapping class group of an n-dimensional torus? | * Indeed, $MCG(\mathbb T^n)=GL(n,\mathbb Z)$ in dimension $n<4$, but it is not simple. In dimension 2 it was first proved by Earle and Eells using complex analysis.[Edit: As Allen Hatcher points out this was known for a long time, Earle and Eells prove much stronger statement: $\mathbb T^2$ is deformation retraction of $Diff\_0(\mathbb T^2)$]
* I am not sure what happens in dimension 4.
* This is not correct in dimension >4, the MCG is semidirect product of $GL(n,\mathbb Z)$ with another (non-finitely generated group). Let me just quote Hatcher:
>
> If $n\ge 5$ then there are split exact sequences
> $$
> 0\to \mathbb Z\_2^\infty\to\pi\_0(Top(\mathbb T^n))\to GL(n,\mathbb Z)\to 0
> $$
> $$
> 0\to \mathbb Z\_2^\infty\oplus\binom n2\mathbb Z\_2\to\pi\_0(PL(\mathbb T^n))\to GL(n,\mathbb Z)\to 0
> $$
> $$
> 0\to \mathbb Z\_2^\infty\oplus\binom n2\mathbb Z\_2\oplus\sum\_{i=0}^n\binom n i\Gamma\_{i+1}\to\pi\_0(Diff(\mathbb T^n))\to GL(n,\mathbb Z)\to 0
> $$
> where $\Gamma\_i$ are Kervaire-Milnor finite abelian groups of homotopy spheres, $\mathbb Z\_2$ is just the group of order 2 and $\mathbb Z\_2^\infty$ is the group of finite strings.
>
>
>
The above quote is from Hatcher "Concordance spaces, higher simple homotopy theory and application."
| 13 | https://mathoverflow.net/users/2029 | 35909 | 23,142 |
https://mathoverflow.net/questions/32385 | 7 | I often hear that modern computer programs "may prove any theorem in elementary Euclidean geometry". Of course, as stated it is false - say, they can not prove theorems about $n$-gons for arbitrary or large enough $n$, and so on. But I wonder, how really powerful are they in problems with not so many points, lines and circles. [Morley's theorem](https://en.wikipedia.org/wiki/Morley%27s_trisector_theorem) is the often example of hard theorem, which may be easily proved by computer, but this does not impress me much, since Morley theorem has two free parameters, and some equality has to be proved. The question is how number of parameters, conditions on them and inequalities make the process of automatic proof slower. To make it more precise, let consider specific example.
Can any computer program prove [Erdos-Mordell inequality](https://en.wikipedia.org/wiki/Erdos-Mordell_inequality)?
| https://mathoverflow.net/users/4312 | Computer power in plane geometry | First off, I would be skeptical of the claim that computer programs "may prove any theorem in elementary Euclidean geometry", simply because it is so wide and general that is prone to be false. Secondly, I am not directly an expert in this field myself, but I hope my references are not too much off.
However, modern automatic geometric theorem proving is definitely capable of dealing with a large number of geometric problems, including those which involve geometric inequalities. The older methods (going back to Wu), translate a geometric statement is translated into an implication of the form
$$ \bigwedge\_{i=1}^n f\_i(x\_1,\ldots,x\_m) = 0 \implies f\_0(x\_1,\ldots,x\_m) = 0$$
where the $f\_i$ are polynomials. From this, with various methods one then obtains a proof of the statement, or a counter example. I am suppressing here that often you need to specify further side conditions for a proof to be possible, e.g. that a triangle is non-degenerate; in fact, Wu's approach and the Gröbner basis even allow deducing sufficient conditions to make a theorem true in retrospect. The [Wikipedia page for Wu's method](https://en.wikipedia.org/wiki/Wu%27s_method) gives some more details and a few references to relevant papers; you can easily google more.
Anyway, this allows encoding things like multiple points being collinear, points being contained on a circle, intersection of lines, perpendicularity of lines, and so on. However, this does indeed not allow encoding inequalities effectively; e.g. just specifying that a point is 'inside' a triangle, or that one value is less than another, in general is not possible.
But since Wu's original work, there have been many advances. If one looks a bit closer, then one notices that the above techniques actually prove theorems about *complex* geometry, as we are arguing about zeros of polynomials, and this all happens over an algebraically closed field. But we are usually interested in *real* geometry only. There are surprisingly many classical theorems from "real" geometry which remain true in the complex case, and this somewhat surprising (and as far as I know rather mysterious) fact ensures that nevertheless Wu's method and its relatives are quite successful. Still, people have worked on overcoming this limitation, as well as that of inequalities.
One approach is described "A New Approach for Automatic Theorem Proving in Real Geometry", by Dolzmann, Sturm & Weispfenning, who use quantifier elimination (from logic) to prove theorems in real geometry (as the title suggests), using their [Reduce](http://www.reduce-algebra.com/) package [Redlog](https://web.archive.org/web/20171211073409/http://redlog.dolzmann.de:80/). They use that, for example, to prove Pedoe's inequality. I am, however, not sure if this can be used to prove the Erdős–Mordell inequality; one could ask them. I think Sturm wrote his PhD thesis on the subject.
Another paper to look at is ["A Practical Program of Automated Proving for a Class of Geometric Inequalities"](https://link.springer.com/chapter/10.1007/3-540-45410-1_4) by Lu Yang and Ju Zhang. There they describe a Maple package "Bottema" (unfortunately, this does not seem to be available on the net, at least I couldn't find it). They use it to prove a load of inequalities, and give many examples involving inequalities. To give you a flavor, here is an example (which they proved using their package):
>
> **Example 4.** By $m\_a$, $m\_b$, $m\_c$ and $2 s$
> denote the three medians and perimeter
> of a triangle, show that
> $$\frac{1}{m\_a}+\frac{1}{m\_b}+\frac{1}{m\_c}\geq \frac{5}{s}.$$
>
>
>
And here is another excerpt (I included it as it also goes back to Erdős).
>
> A. Oppenheim studied the following inequality in order
> to answer a problem proposed by P. Erdös.
>
>
> **Example 9.** Let $a$, $b$, $c$ and $m\_a$, $m\_b$, $m\_c$ denote the side
> lengths and medians of a triangle,
> respectively. If $c = \min\{a, b, c\}$, then
> $$2m\_a+2m\_b+2m\_c \leq 2a+2b+(3\sqrt{3}-4)c.$$
>
>
>
This all does not answer your question about the Erdős–Mordell inequality. And I am afraid I don't know the answer! But I hope my above explanations at least made it plausible that the answer could be yes, however vague that statement is.
| 13 | https://mathoverflow.net/users/8338 | 35917 | 23,147 |
https://mathoverflow.net/questions/35912 | 8 | This is a sort of Chaitin/Omega constant type question, and so I do not expect this probability to be computable to arbitrary precision. However, it is also a very practical thing to know from the perspective of inductive learning. The motivation to ask this question comes from reading some of Solomonoff's papers on algorithmic probability and resource bounded probability.
In Solomonoff's algorithmic probability, one studies induction / machine learning under the assumption that all observed data is the output of a universal Turing machine with random input. With this prior, expectation maximization via Bayesian methods leads to the natural notion that the best model for a particular observed sequence of data is the one generated by the shortest program (aka minimum description length / Kolmogorov complexity).
One cute thing about this prior is that it beats the so-called "no free lunch theorems" as best as possible. Because Kolmogorov complexity is unique up to an additive constant across Turing machines, as long as the underlying sequence is non-uniformly random it can at least theoretically be learned given a finite number of samples.
The bad thing of course is that Kolmogorov complexity is uncomputable, so this is really more of a theoretical curiosity than an actual tool for machine learning. To get around this, Solomonoff proposed the notion of resource bounded probability, where the resources for the Turing machine are ultimately limited (ie bounded compute time/space). In the case of a space bound, this limitation ultimately transforms the Turing machines into DFAs, which leads to my question:
Given the algorithmic prior (ie observed data is generated by some universal Turing machine fed uniform random inputs), what is the probability that the data observed is the output of a DFA?
Of course if it is a DFA, life is really great since there are a bunch of tricks out there for estimating such a DFA. Also it might be interesting to maybe look at other classes of automata as means to get other variants of resource limited types of results.
| https://mathoverflow.net/users/4642 | What is the probability a random Turing machine is isomorphic to a DFA? | The set of possible answers to this question is a countable dense subset of (0,1), because it depends on your choice of universal Turing machine.
| 6 | https://mathoverflow.net/users/4600 | 35935 | 23,153 |
https://mathoverflow.net/questions/35870 | 16 | Given an elliptic curve $E/\mathbb{Q}$ of conductor $N$, parameterization $\psi : X\_0(N) \rightarrow E$, and a point $P \in E$, take the fiber $\psi^{-1}(P)$. Its points, being on $X\_0(N)$, correspond to equivalence classes of pairs $(E\_x, C\_x)$.
Is there a geometric meaning for these pairs in relation to the point $P$? Something about these elliptic curves that has something to do with $P$? (Other than the j-invariant solving some polynomial equation with coefficients depending only on $P$ (and $E$ of course))
Anything special about the $C\_x$'s in relation to $P$?
Modular parameterization is fascinating, but I just don't understand where it comes from. $X\_0$ is for a whole bunch of elliptic curves. $E$ is a single specific one.What's the connection between points on the modular curve, to a single specific point of $E$?
| https://mathoverflow.net/users/2024 | Geometric meaning of fiber of modular parameterization over a point of an elliptic curve? | As Stankewicz explains, although elliptic curves appear in two guises in the modular parameterization $X\_0(N) \to E,$ first because $E$ is an elliptic curve, and secondly because $X\_0(N)$ parameterizes elliptic curves, it is something of a red herring to think of these two appearances of elliptic curves as having anything to do with one another.
The reason that $X\_0(N)$ appears in the problem of describing elliptic curves is because
elliptic curves have two dimensional $H^1$, and $X\_0(N)$ is the Shimura variety over $\mathbb Q$ associated to the group $GL\_2$. Thus, as Stankewicz notes, Shimura curves (which are the Shimura varieties attached to twisted forms of $GL\_2$) can equally well give parameterizations of elliptic curves.
Now the way we prove things about $X\_0(N)$ (e.g. properties of its Heegner points, as in Pete Clark's answer) is using its moduli interpretation. But there are two things to bear in mind:
First, most (maybe all?) properties of the special points on $X\_0(N)$, such as the Heegner
points, are special cases of general aspects of the theory of Shimura varieties (so although the proofs use the moduli interpretation, the statements can be formulated in a way that
doesn't refer to the moduli-theoretic interpretation, but instead refers to the interpretation
of $X\_0(N)$ as a Shimura variety).
Second, the transfer of information is always from $X\_0(N)$ to $E$. So while Heegner points give certain interesting points on $X\_0(N)$ defined over class fields of quadratic imaginary fields, which can be mapped down to $E$ to give interesting points on $E$ defined over such fields, if one takes a random point on $E$ defined over a class field of an imaginary quadratic field and pulls it back to $X\_0(N)$, it is not so easy to say what is going on with the preimages in general.
Finally, I think remark (3) in Pete Clark's answer is an interesting one. In the Mazur and Swinnerton-Dyer paper that David Hansen refer's to in his first comment, if I am remembering correctly, they also suggest that the images in $E$ of the critical points of the map from $X\_0(N)$ to $E$ that lie on the geodesic arc joining $0$ to $\infty$ in the upper half-plane may be worth studying. As with Birch's suggestion of Weierstrass points, I'm not sure how much has been done on this.
| 19 | https://mathoverflow.net/users/2874 | 35941 | 23,157 |
https://mathoverflow.net/questions/35902 | 13 | Let $f(n) = \theta n^d + a\_{d-1} n^{d-1} + \cdots a\_1 n + a\_0$ be a polynomial with real coefficients, and $\theta$ irrational. Let $S\_N = \sum\_{n=1}^N e^{2 \pi i f(n)}$. Weyl's Equidistribution theorem for polynomials is equivalent to the claim that $S\_N/N \to 0$ as $N \to \infty$. You can read a [nice proof](http://terrytao.wordpress.com/2010/03/28/254b-notes-1-equidistribution-of-polynomial-sequences-in-torii/#vdc-eq) of this theorem on Terry Tao's blog (see Corollary's 5 and 6). I had thought that Weyl's Inequality was supposed to be a more precise version of this bound. However, I can't actually figure out how to get Weyl's Inequality to imply the required claim!
Specifically, let $p/q$ be a rational number in lowest terms with $|\theta - p/q| \leq 1/q^2$. Weyl's Inequality is the bound:
$$S\_N/N \leq 100 \left( \log N \right)^{d/2^d} \left( \frac{1}{q} + \frac{1}{N} + \frac{q}{N^d} \right)^{1/(2^d-1)}$$
Here are I am quoting from Timothy Gowers' [notes](http://www.dpmms.cam.ac.uk/~wtg10/addnoth.notes.ps). (**UPDATE:** George Lowther, below, suggests that Gowers may have a typo.) [Wikipedia](http://en.wikipedia.org/wiki/Weyl%2527s_inequality) has a softer version, with more freedom in choosing parameters; I think my question applies to both versions.
Now, suppose that the convergents $p\_i/q\_i$ of $\theta$ grow so fast that $q\_{i+1} > e^{(d+1) q\_i}$.
And take $N \approx e^{q\_i}$. I get that, for any choice of $q$ with $|\theta - p/q| < 1/q^2$, either $1/q > 1/\log N$ or $q/N^d > 1$. This gives infinitely many $N$'s for which the right hand bound is useless (greater than $1$). So it seems that Weyl's inequality does not prove $S\_N/N \to 0$.
Am I missing something?
The motivation for this question was my attempt to answer [this question](https://math.stackexchange.com/questions/2270/convergence-of-sum-n1-infty-sinnk-n) over at math.SE. So any useful comments you have on that question would be appreciated as well.
| https://mathoverflow.net/users/297 | Does Weyl's Inequality prove equidistribution? | I may be wrong, but it seems to me that in fact the liminf is sufficient since it is obtain via a bound that does not depend on $f$. I do not have time to check this in detail, so I apologize if this is all wrong. The idea is to use the points at which we have a good control on $S\_N$ (simultaneously for $f$ and all its translates, and it is the point that seems dubious to me) and then cut the summation interval into pieces of appropriate size.
Let us denote by $S\_N^t$ the exponential sum corresponding to the function $f(\cdot+t)$. Then if I am not mistaken, for all $\varepsilon$ there is a $N\_\varepsilon$ such that for all $t$, $S\_{N\_\varepsilon}^t\leqslant \varepsilon N\_\varepsilon$. Then given any $K$, one has
$$S\_{KN\_\varepsilon}\leqslant \sum\_{k=1}^K S\_{N\_\varepsilon}^{kN\_\varepsilon}\leqslant K\varepsilon N\_\varepsilon.$$
Then, for all $N$, letting $K=\lfloor N/N\_\varepsilon \rfloor$ we get that $S\_N\leqslant K\varepsilon N\_\varepsilon + N\_\varepsilon\leqslant \varepsilon N+N\_\varepsilon$. It follows that $\limsup S\_N/N = 0$.
| 4 | https://mathoverflow.net/users/4961 | 35944 | 23,158 |
https://mathoverflow.net/questions/35518 | 8 | There have been several recent advances on packing regular tetrahedra in $\mathbb{R}^3$. All the results I've seen have been lower bounds -- first John Conway and Sal Torquato showed that there exists an arrangement of tetraheda filling about 72% of space. This has been improved in a series of papers, and the latest result of which I am aware is Elizabeth Chen's record of 85.63%. (A NYTimes article summarizing the history of the problem can be found [here](http://www.nytimes.com/2010/01/05/science/05tetr.html?_r=1).)
My question is does anyone know of any upper bounds, either published or unpublished? I saw a colloquium by Jeff Lagarias, and he said someone was claiming that they had proved something like $1 - 10^{-26}$, but that it was still unpublished.
(A compactness argument gives that since regular tetrahedra don't tile space the maximum volume is strictly less than one, but this argument does not give a quantitative bound.)
| https://mathoverflow.net/users/4558 | Upper bound for tetrahedron packing? | A paper just appeared on the arXiv that supports Jeff Lagarias's claim: "[Upper bound on the packing density of regular tetrahedra and octahedra](http://arxiv.org/abs/1008.2830)," by Simon Gravel, Veit Elser, and Yoav Kallus (medicine and physics researchers at Stanford and Cornell):
>
> In this article, we obtain an explicit bound to the packing density of regular tetrahedra, namely
> $\phi < 1-\delta$ with $\delta = 2.6...\times 10^{-25}$.
> [...]
> In order to obtain a bound to the packing density, we show the existence, in any tetrahedron packing, of a
> set of disjoint balls whose intersection with the packing is particularly simple, and whose density can be
> bounded below. The construction is such that the density of the packing within each of the balls can be
> bounded away from one. The combination of these two bounds gives the main result.
>
>
>
They modify their argument to obtain a $10^{-12}$ bound on regular octahedra.
| 6 | https://mathoverflow.net/users/6094 | 35955 | 23,163 |
https://mathoverflow.net/questions/35956 | 3 | Assume (M,∊M) is a model of ZF. Assume also that (n,∊n) ∊ M is a model in the sense of M and (N,∊N) is a model in the real world with the property that for all sentences σ
N ⊨ σ if and only if M ⊨ (n ⊨ σ).
By using conjunction ∧ it follows that, if T is a finite set of sentences then
N ⊨ T if and only if M ⊨ (n ⊨ T). (1)
For example, if T consists of only two sentences σ1 and σ2 then
N ⊨ T iff N ⊨ σ1∧σ2 iff M ⊨ (n ⊨ σ1∧σ2) iff M ⊨ (n ⊨ T)
However, for an infinite set of sentences T the same argument to prove (1) does not work. Furthermore I think that (1) is not true for infinite T but the reasons for this are unclear to me.
Let's take T=ZF for example. To prove (1), could we not proceed as follows? First we construct ZF (the set of sentences) inside M. We can surely do this inside a model of ZF. Let's call this set ZFM. Then if we knew that there is some kind of correspondence between ZFM and the real world ZF, then using this correspondence, could we somehow deduce (1) or even one implication of the equivalence (1)?
My question is that what kind of relationship (if any) there is between the set of sentences constructed in the model M, say TM, and the real T. And if they happen to be fundamentally different, then what are the reasons behind this? Furthermore, is there some property of the model M which guarantees that TM and the real T are essentially the same?
| https://mathoverflow.net/users/8584 | Models within a model of set theory | We will always have ${\rm ZF}^M\supseteq {\rm ZF}$. The problem comes
if $M$ contains nonstandard sentences--which is equivalent to the question of whether $M$ contains nonstandard natural numbers since we can identify formulas with their Godel codes.
If there are no nonstandard integers, then they type of externl induction you are doing works fine, but if there aren't it might break down at a nonstandard stage.
| 6 | https://mathoverflow.net/users/5849 | 35961 | 23,166 |
https://mathoverflow.net/questions/35970 | 15 | It is straightforward to construct proper uncountable subgroups of $\mathbb{R}$. One can construst a basis for $\mathbb{R}$ over $\mathbb{Q}$, and then there are many possibilities (just consider the group generated by the basis or the vector subspace generated by some proper uncountable set of the basis).
However, the first step (constructing the basis) requires the axiom of choice.
So does anyone know of any proper uncountable subgroup of $\mathbb{R}$ that does not require choice to construct?
or is this not possible.
Meaning are there models not involving choice where every uncountable subgroup of $\mathbb{R}$ is equal to $\mathbb{R}$.
| https://mathoverflow.net/users/5732 | Construction of a proper uncountable subgroup of $\mathbb{R}$ without Choice. | This [earlier answer of mine](https://mathoverflow.net/questions/23202/explicit-big-linearly-independent-sets/23206#23206) shows how to get an uncountable $\mathbb{Q}$-independent subset of $\mathbb{R}$ in ZF. This set is not a Hamel basis so the $\mathbb{Q}$-span of this set is as required.
| 17 | https://mathoverflow.net/users/2000 | 35974 | 23,175 |
https://mathoverflow.net/questions/35971 | 10 | What is the smallest ordinal alpha which is elementarily equivalent to some smaller ordinal beta with the signature (<)?
What is the corresponding ordinal beta?
What if we instead require that beta be an elementary substructure of alpha?
| https://mathoverflow.net/users/nan | Elementary equivalence of ordinals | The first-order theory of well-orderings was studied in great detail in a paper of Doner, Mostowski, and Tarski, "The elementary theory of well-ordering -- a metamathematical study" [Logic Colloquium '77, edited by A. Macintyre, L. Pacholski, and J. Paris, North-Holland (1978) pp. 1-54]. In particular, their Corollary 44 characterizes (unless their notation is very non-standard --- I haven't checked carefully) when two ordinals are elementarily equivalent. Modulo an apparent typo in the definition just before the corollary (one of the strict inequalities should be non-strict), it seems that the first pair of distinct but elementarily equivalent ordinals is $\omega^\omega$ and $\omega^\omega\cdot2$. A thorough reading of the paper (which I don't have time for right now) should also reveal the answer to your second question, about elementary submodels (probably the same pair of ordinals).
| 11 | https://mathoverflow.net/users/6794 | 35982 | 23,180 |
https://mathoverflow.net/questions/35989 | 10 | I was just curious, since the B-T paradox is a measure theoretic result, if there are any consequences of this paradox in probability theory? Also, is there is a way of stating the B-T paradox in the language of probability theory?
I am ultimately interested in finding an application of the B-T paradox in physics which leads to an experimental prediction.
| https://mathoverflow.net/users/8509 | Applications of Banach-Tarski Paradox to Probability Theory? | I thought the whole point of having a $\sigma$-algebra for your probability space was to avoid non-measurable sets like the ones used in the proof of BT. Hence, it would seem that the BT paradox would be impossible to state in probability theory on account of the sets you need not being present in your algebra... but I might be mistaken, can someone else comment more?
| 7 | https://mathoverflow.net/users/8239 | 35994 | 23,184 |
https://mathoverflow.net/questions/35977 | 7 | A crossed module is a pair of groups $C$ and $G$, an action of $G$ on $C$, and a homomorphism $\partial: C \to G$ that satisfy
* $\partial(g\cdot c)=g(\partial c)g^{-1}$, and
* $cc'c^{-1}=(\partial c)\cdot c'$
Let $(X,A)$ be a pointed pair of spaces. Whitehead proved that, in the homotopy long exact sequence of the pair,
$$
\pi\_2(X,A) \stackrel{\partial}{\rightarrow} \pi\_1(A)
$$
is a crossed module.
Simply put, my question is: what does this give us, other than an extra bit of structure? Does knowing that this is true aid on calculation? Does it aid in distinguishing spaces? Does it give us something really cool that I haven't thought of? (Probably the answer to this one is "yes".)
| https://mathoverflow.net/users/343 | Crossed module structure on homotopy groups | There are numerous calculations that are easily done with crossed module techniques that are much more difficult to obtain using `traditional' homotopy theory. Some of these use the next stage up, that is crossed squares, and the resulting non-Abelian tensor product. A neat sample calculation is of the homotopy type of the suspension of a K(G,1), if I remember rightly. This is given as the kernel of the commutator map from $G\otimes G$ to $G$.
You ask is it good at distinguishing spaces. The answer is most decidedly yes. (But I would say that wouldn't I.) MacLane and Whitehead proved that the crossed module models the homotopy 2-type, extending the classification of homotopy 1-types by groups. (Ok there is a price to pay. The correspondence gives 2-types correspond to equivalence classes of crossed modules but the equivalence relation is algebraic not topological in nature so that is reasonable.) Loday proved that homotopy n-types had algebraic models which were crossed n-cubes, n-fold generalisations of crossed modules.
The cool thing is that conceptually they are one of a linked set of models for low dimensional homotopy information that have geometric significance, and yet are relatively easy to manipulate. I like to say that a crossed module is a normal subgroup that is not a subgroup. Crossed modules 'are' also 2-groups, cat$^1$-groups and various other equivalent formulations.
For some higher dimensional vKT applications, look at higher Hopf formulae in work by Brown and Ellis.
For applications of crossed modules in non-Abelian cohomology etc. look at Larry Breen's work, or for a gentle introduction, my Menagerie notes which you can find on the n-Lab.
I could go on listing things but will stop here. If you (or anyone else needs more detail) ask me or ask here.
| 18 | https://mathoverflow.net/users/3502 | 35995 | 23,185 |
https://mathoverflow.net/questions/35996 | 6 | What is the Ehrhart polynomial of the regular cross-polytope of dimension d? Are there published upper and lower estimates?
| https://mathoverflow.net/users/nan | Ehrhart polynomial | If you mean the polytope with vertices $(0,\ldots,0,\pm1,0,\ldots,0)$
then it is easily seen to be
$$\sum\_{k=0}^d 2^k{d\choose k}{x\choose k}.$$
| 11 | https://mathoverflow.net/users/4213 | 36000 | 23,188 |
https://mathoverflow.net/questions/35986 | 8 | I use $\dim\_H(E)$ to denote the Hausdorff dimension of a set $E \subseteq \mathbb{R}$ and $|E|$ to denote its Lebesgue measure. It is easy to see from the definition of Hausdorff dimension that if $\dim\_H(E) < 1$, then $|E| = 0$. The converse is not true, and there are many cases where $\dim\_H(E) = 1$ yet $|E| = 0$. So the question:
**What was the first (or most elementary) example of this phenomenon?**
After some looking around, I was able to prove that a central Cantor set $C$ with ratio of dissection $r\_k = 1/(2+\frac{1}{k})$ satisfies the condition I want. It is easy to see that $|C| = 0$ since at step $n$ of the process to construct this Cantor set, it has measure $2^n(r\_1 \cdots r\_n)$ which in this case limits to 0, but for the Hausdorff dimension I required a non-trivial result from the paper *Sums of Cantor sets* (Cabrelli, Hare, Molter) that gave the formula
$\dim\_H(C) = \liminf\_n \frac{n \ln 2}{\ln r\_1 \cdots r\_n}$.
This result is fairly recent and sophisticated, and I feel that there should be older and simpler examples.
| https://mathoverflow.net/users/7165 | Measure 0 sets on the line with Hausdorff dimension 1 | Try a countable union of sets (such as Cantor sets) whose Hausdorff dimension tends to 1.
| 15 | https://mathoverflow.net/users/51 | 36005 | 23,193 |
https://mathoverflow.net/questions/36015 | 1 | I know this isn't a research question, so it might get voted off, but here goes:
You know that a couple has two children. You go to the couple's house and one of their children, a young boy, opens the door. What is the probability that the couple's other child is a girl?
If you list all possibilities for the sexes of two children, BB, BG, GB, GG, you see that 2 of the 3 pairs that have B (for boy) in them also have a girl, so the answer one could argue is 2/3.
On the other hand, one could argue that the answer is 1/2, since the probability that any one child is a girl is 1/2, and intuitively (?) should be independent of the gender of its siblings.
Some background to possibly justify posting it here: the question was asked at an interview for an actuarial/insurance type position, and the interviewer was the answer was 2/3, whereas my friend who was being interviewed (and has a masters in math) thought the answer was 1/2, even after the interviewer explained his logic. My friend felt that the interviewer wasn't taking into account the fact that it is not equally likely that a boy will open the door in the BB versus the BG combination, and one has to take into account that fact. I have no idea which is the correct answer, both sound somewhat convincing to me (I have a Ph.D. in math, but I won't mention from where in an effort to avoid embarrassing my degree granting institution!). Anyways, any help would be appreciated and I apologize if this is too simple a question for this forum.
| https://mathoverflow.net/users/8599 | very simple conditional probability question | We will assume all the obvious implicit assumptions (eg. random child being boy of girl is 50/50, boys and girls open the door uniformly, etc.).
If you had a slightly different question, i.e. if you asked the couple if they have at least one boy, and the answer is yes, then the chance of the other one being a girl is 2/3. Intuitively, the probability is not 1/2 because in this case the answer depends on *both* the children, i.e. it is a function of both of them considered together.
However, if you asked the couple to pick a child at random, then she/he bears no information about the other child, and consequently his/her gender does not give you any information about the sibling.
Your case is the second case, where the child opening the door is selected at random, and she happens to be female. This does not bear any information regarding the other child.
So answer is 1/2 and **your friend is correct**.
Mathematically,
$P(Other\ is\ B|G\ opens\ door) = P(BG|G\ opens\ door) =$
$\frac{P(BG\ and\ G\ opens)}{P(GG\ and\ G\ opens\ door) + P(BG\ and\ G\ opens\ door)} = \frac{1/2\*1/2}{1/4+1/2\*1/2} = 1/2$
(Note here, that $P(BG\ and\ G\ opens)=P(G\ opens|BG)\*P(BG)=1/2\*1/2$.)
However as a digression, a twist in the question can be brought about - if you take probabilities for a boy and girl to be different for opening the door.
Eg. suppose boy opens with probability $p$, girl with $q=1-p$, in a family with BG.
Then $P(Other\ is\ B|G\ opened\ door) = P(BG)/P(G\ opened\ door) = $
$\frac{P(BG\ and\ G\ opens)}{P(GG\ and\ G\ opens\ door) + P(BG\ and\ G\ opens\ door)} = \frac{1/2\*q
}{1/4+1/2\*q} = \frac{2q}{2q+1}$.
This means $q = 0 \implies P(BG|G\ opens)=0$. That makes sense, since girls don't open the door if there is a boy, so definitely the other one is girl too.
| 1 | https://mathoverflow.net/users/8602 | 36021 | 23,203 |
https://mathoverflow.net/questions/35984 | 7 | In Appendix B to their *Model Theory*, Chang and Keisler list some problems and conjectures that, at the time of publication, were unsolved. A few of them take imperative form, for instance:
"Develop a theory of models which stresses the order type of the model $\mathfrak{A} = \langle A, <, \dots\rangle$ rather than the cardinality of the set $A$."
Has anyone worked on this?
| https://mathoverflow.net/users/8547 | Model theory stressing order type of universe. | I don't think many model theorists have worked on this. Granted, I'm a little unclear what Chang and Keisler were asking here, but here's one possible precisification:
**Question:** Suppose we are given a (complete?) theory T in a language with a binary relation < such that T proves "< is a strict linear ordering." Try to develop a theory of the "order-type spectrum" $I(\alpha, T)$, which is defined as the number of nonisomorphic models $M$ of $T$ such that $(M, <^M)$ has order type $\alpha$.
You could start by trying to think about what it means for $T$ to be "$\alpha$-categorical" for some order type $\alpha$, meaning, what are necessary and sufficient conditions for $I(\alpha, T)$ to equal $1$? (I have no idea whether anybody has investigated this before.) For example, if $T$ proves that the ordering is dense without endpoints and $\eta$ is the order type of the rationals, then $I(\eta, T) = 1$ if and only if $T$ is $\omega$-categorical, since the complete theory of $(\mathbb{Q}, <)$ itself is $\omega$-categorical.
An immediate complication I see to this project is that I don't know if there is any good analogue of the upward Lowenheim-Skolem theorem. It seems like it would be difficult to answer the question: "Given a theory $T$, for which infinite order types $\alpha$ is $I(\alpha, T) \neq 0$?" (The corresponding question for cardinalities of the universe for a $T$ with infinite models is trivial, by Lowenheim-Skolem.) For example: your theory $T$ could force the order type of any model to not be Dedekind complete (e.g. take the complete theory of an densely-ordered ring with a unary predicate for a proper convex subring).
Are there Morley-like categoricity theorems? If $T$ is countable, say, and $I(\alpha, T) = 1$ for some uncountable order type $\alpha$, can we conclude that $I(\beta, T) \leq 1$ for every uncountable $\beta$? Possibly this could be an interesting question; offhand I have no idea what the answer is.
Illustrating the difficulties of this, here's a paper just on the possible order types of a particular theory, PA:
"Order-types of models of Peano arithmetic," by Andrey Bovykin and Richard Kaye, pp. 275-285 of *Logic and Algebra*, edited by Yi Zhang with a preface by Oleg Belegradek, Contemporary Mathematics 302, AMS.
A different interpretation of the original question would be: given a particular order type $\alpha$, investigate structures with order type $\alpha$. Along these lines, many model theorists (such as Pillay, van den Dries, Wilkie, and others) have been studying expansions of the ordered field of the real numbers under the rubric of "o-minimal theories," though generally the interest has been in definable sets rather than models per se. Chris Miller is an example of an o-minimalist who has done a significant amount of research just on structures expanding the field of reals; check out his webpage for some state-of-the-art papers in this area.
| 7 | https://mathoverflow.net/users/93 | 36022 | 23,204 |
https://mathoverflow.net/questions/36016 | 13 | Suppose I have a topological space $X$. Let $\mathcal{O}(X)$ denote the poset of open subsets. There is a canonical functor $\mathcal{O}(X) \to Top/X$ which sends an open $U \in \mathcal{O}(X)$ to $U \hookrightarrow X$. By left-Kan extension, this produces an adjunction between $Set^{\mathcal{O}(X)^{op}}$ and $Top/X$. The left-adjoint, $L$, is precisely the etale space construction, whereas the right adjoint, $\Gamma$, is the "sheaf of sections" functor. $\Gamma \circ L$ is sheafification. By abstract nonsense, this adjunction restricts to an equivalence between, one one hand, the subcategory where the unit is an iso, and the other hand, the subcategory where the counit is an iso. The former is easily seen to be the category of sheaves over $X$. I know the later is the category of etale spaces over $X$, i.e. maps $Y \to X$ which are local homeomorphisms. This is traditionally proven usually an explicit description of the etale space of a sheaf, topologizing the germs of local sections etc. However, is there a way to see this at a higher level of abstraction, only appealing to the abstract definition of this induced adjunction and abstract properties of local homeomorphisms?
| https://mathoverflow.net/users/4528 | Understanding the etale space construction from a formal viewpoint | Here is a sketch of why I think the condition that $Y$ is a local homeomorphism over $X$ should be sufficient for the counit to be a homeomorphism. I haven't worked out the converse yet.
For a presheaf $F \in Set^{\mathcal{O}(X)^{op}}$, the formula for the left Kan extension should be $$L(F) = \mathrm{colim}\_{y(U) \rightarrow F} U$$ where $y : \mathcal{O}(X) \rightarrow Set^{\mathcal{O}(X)^{op}}$ is the Yoneda embedding. By Yoneda's lemma, the indexing category for the colimit is exactly the *category of elements* of $F$ which I will write as $\int F$.
Now, consider the case where $F = \Gamma\_Y$ for some space $p: Y \rightarrow X$ and assume that $p$ is a local homeomorphism. We have $\Gamma\_Y(U) = \{\sigma : U \rightarrow Y | p \circ \sigma = \mathrm{id}\_U \}$. So the objects in the category $\int \Gamma\_Y$ are exactly the sections over the various open sets of $X$, and the morphisms are given by restriction of sections. I'll write $d(\sigma)$ for the domain of a given section.
Our Kan extension formula becomes $$L(\Gamma\_Y) = \mathrm{colim}\_{\sigma \in \int \Gamma\_Y} d(\sigma)$$ From here it's easy to see what the counit is: since our object is given by a colimit, it suffices to construct a map $d(\sigma) \rightarrow Y$ for each $\sigma \in \int \Gamma\_Y$. But clearly $\sigma$ itself qualifies.
Now choose an open covering $\{V\_\alpha\}$ of $Y$ such that $p$ restricts to a homeomorphism on each $V\_\alpha$. We then have a collection $\{\sigma\_\alpha : p(V\_\alpha) \rightarrow V\_\alpha\}$ of sections by choosing the inverse to each restriction. My claim would be that this collection is cofinal (or final? I can never remember which) in the category $\int \Gamma\_Y$ so that we can restrict our colimit to just this subcategory. Notice that in this case, the components of the counit map above are homeomorphisms.
Moreover, this subcategory should also be cofinal in $\mathcal{O}(Y)$ by associating $\sigma\_\alpha$ with the open set $V\_\alpha$. Then the fact that the counit is a homeomorphism should be the statement that a topological space is the colimit of any of its open coverings.
Is this along the lines of what you were thinking?
| 5 | https://mathoverflow.net/users/4466 | 36028 | 23,209 |
https://mathoverflow.net/questions/36034 | 7 | My question is referred to the statement and proof of Prop. 2.4 of Diamond's
article "An extension of Wiles' Results", in Modular Forms and Fermat Last
Theorem, page 479.
More precisely: fix $l$ and $p$ two distinct primes, with $l$ odd. Let $\sigma$ be an
irreducible, continuous, degree 2 representation of the absolute Galois group
$G\_{p}$ of $Q\_{p}$, with coefficients in $k$, an algebraic closure of the finite
field with $l$ elements.
Proposition 2.4 states that if the restriction of $\sigma$ to the inertia
subgroup of $G\_{p}$ is irreducible and $p$ is odd, then $\sigma$ is isomorphic to the
representation induced from a character of the Galois group of a quadratic
ramified extension $M$ of $Q\_{p}$.
The proof given works if the restriction of $\sigma$ to the wild inertia of $G\_{p}$ is
reducible (I think there's a typo in the first line of the proof). What if
$\sigma$ is irreducible on wild inertia (and $p$ is always odd)? It seems to me that this case is not covered in the proof of the Proposition, but maybe I'm not seeing something obvious.. If such a representation exists, it cannot be induced from a quadratic extension $M$ as above, so how does it fit in the description given by the Proposition? Can one say something about such a representation (for example something about its projective image?).
Thanks
| https://mathoverflow.net/users/8606 | Mod l local Galois representations (l different from p) | The image of wild inertia is a finite $p$-group, and if $d$ is the degree of an irreducible representation of a $p$-group over an algebraically closed field of characteristic $\ne p$, then $d$ is a power of $p$. So for $p$ odd the image of wild inertia is always reducible.
| 8 | https://mathoverflow.net/users/5480 | 36036 | 23,215 |
https://mathoverflow.net/questions/35997 | 0 | A very easy question I can't seem to answer: For a universal r-form on a co-quasi-triangular Hopf algebra why is $r(a \otimes 1) = r(1 \otimes a) = \epsilon(a)$?
| https://mathoverflow.net/users/1095 | Action of Co-quasi-triangular Universal r-form on $a \otimes 1$ | Like David said, the proof is almost identical to the earlier one for $R$-matrices:
$r(x\otimes 1) = r\circ (id\otimes\mu)(x\otimes 1\otimes 1) = (r\_{13}\ast r\_{12})(x\otimes 1\otimes1)= \sum r(x'\otimes 1)r(x''\otimes 1) = (r \ast r)(x\otimes 1).$
Since $r$ is invertible, $r(x\otimes 1)=\epsilon(x)$.
| 1 | https://mathoverflow.net/users/8225 | 36039 | 23,218 |
https://mathoverflow.net/questions/36035 | 5 | What is the most general class of metric spaces for which the closest pair of points in any finite subset can be found in time O(n^(1+eps))? I have studied how to do this in O(n log(n)) in the plane, and I believe I can generalize the same method to some other surfaces, but it does not work in 3-space (maybe it is possible but I suspect not). Are there any interesting examples of metric spaces in which this problem can be solved efficiently?
| https://mathoverflow.net/users/2003 | What is the most general class of metric spaces for which the closest pair of points in a finite subset can be found in time O(n^(1+eps))? | I assume you are aware of the classic paper by Jon Bentley,
"[Multidimensional divide-and-conquer](http://portal.acm.org/citation.cfm?id=358850)"
[*Commun. ACM* **23**(4):214-229 (1980)],
in which he showed how to find the closest pair of points in $\mathbb{R}^3$
in the Euclidean metric in $O(n \log n)$ time.
His algorithm works in arbitrary dimensions in $O(n \log^{d-1} n)$.
I realize I am not answering your question about metric spaces, but it might be worth revisiting
his algorithm to see how heavily it leans on the norm.
Rabin's 1976 randomized algorithm achieves $O(n)$ expected time.
An updated detailed analysis is in the paper
"[A Reliable Randomized Algorithm for the Closest-Pair Problem](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6WH3-45NJK2G-2&_user=10&_coverDate=10%252F31%252F1997&_rdoc=1&_fmt=high&_orig=search&_sort=d&_docanchor=&view=c&_searchStrId=1434955821&_rerunOrigin=google&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=fe7a5e8a4209763de823c9438fb642f0)"
by Martin Dietzfelbinger, Torben Hagerup, Jyrki Katajainen, and Martti Penttonen
[*Journal of Algorithms* **25**(1): 19-51 (1997)].
Again I am not addressing your focus on other metric spaces, but these efficient algorithms
for Euclidean distance would be a place to start.
| 5 | https://mathoverflow.net/users/6094 | 36040 | 23,219 |
https://mathoverflow.net/questions/36055 | 2 | I'm pretty sure this has an easy solution, but I can't seem to find it.
Let $X$ be a contractible $2$-dimensional CW-complex, let $\gamma$ be an embedded loop in $X$, and let $f : D^2 \rightarrow X$ be an embedding of a disc in $X$ which maps the boundary of $D$ to $\gamma$.
My question is the following. Let $f' : D^2 \rightarrow X$ be a continuous map of a disc into $X$ which takes the boundary of $D$ to $\gamma$. Must we then have $f(D^2) \subset f'(D^2)$ ? I'm pretty sure that the answer is yes, but I can't seem to prove it.
Of course, this has an obvious generalization to higher dimensional complexes, and I'd be interested in that too.
| https://mathoverflow.net/users/8614 | Spanning discs in contractible 2-d complexes | One way I can think of is to take a point $x\in f(D)\setminus f'(D)$, assuming on the contrary. One may assume $x$ lies in the interior of $f(D)$ and the interior of some 2-cell. Then you can remove a small disk $U$ in $f(D)$ which still lies in the 2-cell, and a M-V sequence argument shows $\gamma$ is homologically nontrivial in $H\_1(X\setminus U)$, as $[\gamma]=[\partial U]$. This gives a contradiction.
| 3 | https://mathoverflow.net/users/8565 | 36057 | 23,231 |
https://mathoverflow.net/questions/29100 | 43 | This question is predicated on my understanding that real algebraic geometry (henceforth RAG) is the version of algebraic geometry (AG) one gets when replacing (esp. algebraically closed) fields with formally real (esp. real closed) fields. This makes for substantial differences in the theory because such fields can be ordered, and with order comes the notion of a semialgebraic set and a stronger topology.
I am aware that there is a notion of "real spectrum" analogous to the traditional spectrum of a commutative ring, though I'm not terribly familiar with either. I assume this allows one to glue things together and define "real schemes" or some such thing. Or if not, I assume the reason this doesn't work is something one would learn in the study of RAG.
My question: Given the differences in the theories, how well does one need to understand "traditional" AG to study RAG? Are there references (preferably books) which introduce RAG at an abstract level without assuming much knowledge of AG? Or is asking for this like when people ask how they can learn about motives without knowing about AG first?
I already have Basu, Pollack, and Roy's [Algorithms in Real Algebraic Geometry](http://perso.univ-rennes1.fr/marie-francoise.roy/bpr-ed2-posted2.html) but I'm looking for something less algorithmic.
| https://mathoverflow.net/users/5963 | Real algebraic geometry vs. algebraic geometry | Real algebraic geometry comes with its own set of methods. While keeping in mind the complex picture is sometimes useful (e.g. for any real algebraic variety *X*, the Smith-Thom inequality asserts that $b(X(\mathbb{R})) \leq b(X(\mathbb{C}))$, where $b(\cdot)$ denotes the sum of the topological Betti numbers with mod 2 coefficients), most of the technique used are either built from scratch or borrow from other areas, such as singularity theory or model theory.
The literature is a lot smaller for RAG than for traditional AG; the basic reference is the book by Bochnak, Coste and Roy (preferably the English-language edition which is more recent by more than 10 years, and has been greatly expanded). The book covers in particular the real spectrum, the transfer principle (which makes non-standard methods really easy), stratifications and Nash manifolds, among other topics. Michel Coste also has *An Introduction to Semialgebraic Geometry* available on [his webpage](http://perso.univ-rennes1.fr/michel.coste/articles.html#books) a very short treatment of some basic results, enough to give you a first impression.
Other interesting books tend to be shorter and more focused than BCR, dealing with a specific aspect; e.g. Prestel's *Positive polynomials.* (dealing mostly with results such as Schmudgen's theorem), and Andradas-Brocker-Ruiz *Constructible sets in real geometry* (dealing mostly with the minimum number of inequalities required to define basic sets). The book by Benedetti and Risler is very interesting and concrete; I found some passages very useful and some results are hard to find in other books (the sections on additive complexity of polynomials are very thorough), but it is a bit scatterbrained for my taste.
As the name indicates, the book by Basu Pollack and Roy is entirely focused on the algorithmic aspects. It's a very good book, and you may still pick up some of the theory in there, but it does not sound like what you are after right now.
As for o-minimality, there again, Michel Coste's webpage contains an introduction that nicely complements van den Dries's book. I would hesitate to bundle o-minimality with real algebraic geometry. In some respects, the two domains are undoubtedly close cousins, and o-minimality can be seen as a wide-ranging generalization of real algebraic structures; on the other hand, each disciplines has also its own aspects and problems that do not translate all that well into the other.
I'm being verbose as usual. Still, I hope it helps.
| 36 | https://mathoverflow.net/users/8212 | 36060 | 23,233 |
https://mathoverflow.net/questions/36070 | 2 | Dear all,
I would like to know if the Gauss transformation *T(x) = fractional part of 1/x, x in (0,1)* (with the Gauss invariant probability measure) is an exact endomorphism (in the sense of Rokhlin). I have failed to find an answer in the literature, any reference would be welcomed.
| https://mathoverflow.net/users/8623 | exactness of the Gauss transformation | Hi Steven,
the answer to your question is yes and there are several ways of deriving the exactness of Gauss map with respect to Gauss probability: for instance, in this [text](http://w3.impa.br/~viana/out/sdds.pdf) of M. Viana, it is derived as a consequence of the proof of the exponential decay of correlations.
| 2 | https://mathoverflow.net/users/1568 | 36073 | 23,239 |
https://mathoverflow.net/questions/36061 | 1 | I have a probability problem, which I need to simulate in a reasonable amount of time. In its simplified form, I have 30 unfair coins each with a different known probability of being heads. I then want to answer such questions as *what is the probability that exactly 12 will be heads?* and *what is the probability that AT LEAST five will be tails?*
I know basic probability theory, so I know that I can enumerate all ${{30} \choose x}$ possibilities, but that's not efficient. The worst case—${{30} \choose {15}}$—exceeds 150 million combinations. Is there a better computational approach to this problem?
| https://mathoverflow.net/users/8616 | Discrete probability algorithms | Suppose the probability for getting head is $p\_i$ for $i$th coin.
You can easily (and economically) compute the probabilities of exactly $k$ heads using the recursive relation -
$H\_{n,k}=p\_nH\_{n-1,k-1}+(1-p\_n)H\_{n-1,k}$
---
Explanation follows.
Let $H\_{n,k}$ be the probability of getting exactly $k$ heads using the first $n$ coins. For answering the type of questions you want to solve, all you need is a list of $H\_{n,k}$'s.
Note that $H\_{n,k}=\sum\_{\left[over\ e\_i's\in\{0,1\},\sum\_i^n e\_i=k\right]}\prod\_{i=1}^n p\_i^{e\_i}(1-p\_i)^{1-e\_i}$
The sum (as you mentioned) contains $n\choose k$ entries.
However note that $H\_{n,k}=p\_nH\_{n-1,k-1}+(1-p\_n)H\_{n-1,k}$
So you can recursively build up the $H\_{n,k}$'s which should be simple since there are only a few of them. (To be precise, for $N$ coins, there are $N(N+3)/2$ many of $H\_{n,k}$'s since $n\in \{1,...,N\}$ and $k\in \{0,...,n\}$).
As a base for the recursive relation, you can use the following (obvious) identities.
* $H\_{n,k}=0$ for $k\gt n$
* $H\_{n,0}=\prod\_{i=1}^n(1-p\_i)$
* $H\_{n,n}=\prod\_{i=1}^np\_i$
| 4 | https://mathoverflow.net/users/8602 | 36074 | 23,240 |
https://mathoverflow.net/questions/36082 | 3 | Consider a connected, complete and compact Riemannian manifold $M$. Is it correct that the following equality holds: $\text{inj}(x)=\text{dist}\left(x,\text{CuL}(x)\right)$? Or in words that the injectivity radius of a point is the distance from the point to its cut locus.
Here is my explanation: As the manifold is compact and complete, then the cut locus $\text{CuL}(x)$ is compact as well[1]. Thus, there exists a point $y\in \text{CuL}(x)$ such that $\text{dist}\left(x,\text{CuL}(x)\right)=\text{dist}(x,y)$. Since $y$ is a cut point of $x$, there exists a tangent vector $\xi\_0\in T\_x M$ such that $y=\exp\_x\left(c(\xi\_0)\xi\_0\right)$[2], where $c(\xi\_0)$ is the distance from $x$ to its cut point in the $\xi\_0$ direction. This in turn means that $\text{dist}(x,y) = c(\xi\_0)$.
Recall that $\text{inj}(x)=\inf\_{\xi\in T\_x M}(c(\xi))$. This means that $\text{inj}(x) \leq c(\xi\_0) = \text{dist}(x,y)=\text{dist}\left(x,\text{CuL}(x)\right)$. If $\text{inj}(x)< c(\xi\_0)$, then since $M$ is compact, it means that there exists some other tangent vector $\xi\in T\_x M$ with $c(\xi) < c(\xi\_0)$. But this means that $\exp\_x(c(\xi)\xi)$ is a cut point of $x$ closer to it then $y$, and this is a contradiction.
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[1] See *Contributions to Riemannian Geometry in the Large* by W. Klingenberg
[2] Here I'm using the notation of I. Chavel in his book *Riemannian Geometry - Modern Introduction*.
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**Update(@dror)**
Today I finally found a copy of the book \*Riemannian Geometry" by Takashi Sakai, and there the above is stated as proposition 4.13 in chapter 3.
Thanks anyway.
| https://mathoverflow.net/users/8047 | Injectivity radius and the cut locus | The injectivity radius for a point $x$ is the largest distance $r$ such that any geodesic starting from x is length-minimizing for at least distance $r$. So there exists at least one geodesic starting from $x$ that is *not* length-minimizing past distance $r$. On the other hand a point $p$ is in the cut locus of $x$ if a geodesic starting from $x$ and passing through $p$ is *not* length-minimizing for any point past $p$. So the injectivity radius at $x$ is the distance from $x$ to its cut locus.
| 6 | https://mathoverflow.net/users/613 | 36083 | 23,245 |
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