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https://mathoverflow.net/questions/32082 | 6 | I would like to ask for a reference to the following statement (hopefully correct):
Let $M$ be a manifold of sectional curvature at most $1$ and let $\gamma$ be a closed geodesic.
Suppose that $\gamma$ is contractible. Then for any contraction of this geodesic at some point its length will be equal to $2\pi$.
It would be even better if there is a reference for the case when $M$ a locally $CAT(1)$ space (not necessarily manifold)
| https://mathoverflow.net/users/943 | Contracting a geodesic on a space of curvature less than 1 | There is an article by B. Bowditch, Notes on locally CAT(1) spaces, in Geometric Group Theory, R.Charney, M.Davis, and M.Shapiro eds., de Gruyter (1995) that will likely be of help. The article is about curve shortening in locally CAT(1) spaces.
| 8 | https://mathoverflow.net/users/7021 | 32086 | 20,841 |
https://mathoverflow.net/questions/32063 | 6 | Following the notation of [Etingof-Nikshych-Ostrik](http://arxiv.org/abs/0909.3140) what is Out(G-mod) for a finite group G?
That is what are all bimodule cateogries over the fusion category G-mod of complex G-modules which have the property that they're just G-mod as a left (resp. right) G-module up to equivalence of bimodules? I think this is the same as the group of tensor autoequivalences of G-mod which do not come from conjugating by 1-dimensional objects, but ENO only prove that in the case where there are no 1-dimensional objects.
If you have an automorphism of the group G then you get an automorphism of G-mod. I'm not totally sure though if outer automorphisms of G necessarily give outer automorphisms (for example could you have an outer automorphism that fixed all conjugacy classes?) and I also don't know whether all outer automorphisms come up this way.
The reason that I'm asking is that I want to understand the relationship between Out(C) and Out(D) where C and D are Morita equivalent fusion categories. However, I realized I don't have enough examples where I understand what Out(C) actually is.
| https://mathoverflow.net/users/22 | What is Out(G-mod) for a finite group G? | The group $\operatorname{Out}(G\operatorname{-Mod})$ is equivalent to the group $\operatorname{Aut}\_{\textbf{Tw}}(k[G])$ of gauge equivalence classes of twisted automorphisms of the Hopf algebra $k[G]$ defined by Davydov in [*Twisted automorphisms of Hopf algebras*](https://arxiv.org/abs/0708.2757). In Davydov's other paper, [*Twisted automorphisms of group algebras*](https://arxiv.org/abs/0708.2758), he describes this group in the case that $\lvert G \rvert$ is relatively prime to 6. (The proposed description I gave in the earlier version of this answer fails in general because $\operatorname{Out}(G)$ is not in general a normal subgroup of $\operatorname{Aut}\_{\textbf{Tw}}(k[G])$.)
First, we note that since $G\operatorname{-Mod}$ is symmetric, any inner tensor autoequivalence is automatically tensor equivalent to the identity, so it suffices to describe the group of isomorphism classes of tensor autoequivalences. Let $\omega: G\operatorname{-Mod} \to \operatorname{Vect}$ be the usual fiber functor. Any tensor autoequivalence $F: G\operatorname{-Mod} \stackrel{\cong}{\to} G\operatorname{-Mod}$ gives a new (tensor) fiber functor $\omega \circ F: G\operatorname{-Mod} \to \operatorname{Vect}$. This gives $G\operatorname{-Mod}$ the structure of the category of representations of some Hopf algebra; it is shown in the paper [*On families of triangular Hopf algebras*](https://arxiv.org/abs/math/0110043) by Etingof and Gelaki that this Hopf algebra will be a twist of the original one, and thus $F$ induces a twisted automorphism of $H$. Conversely, any twisted automorphism of $F$ gives rise to an autoequivalence of $G\operatorname{-Mod}$. Natural transformations of monoidal functors correspond to gauge equivalences of twisted automorphisms. Thus, we can identify isomorphism classes of monoidal autoequivalences of $G\operatorname{-Mod}$ with gauge equivalence classes of twisted automorphisms of $k[G]$.
| 4 | https://mathoverflow.net/users/396 | 32091 | 20,844 |
https://mathoverflow.net/questions/32083 | 5 | For a finite type morphism $f:X\to S$, $X$ is a regular scheme, should there always exist an open (dense) subscheme $U\subset S$ such that the fibre of $f$ at each Zariski point of $U$ is regular? All schemes are excellent.
If the answer is 'yes', then: could one choose such an $U$ such that the preimage of any regular subscheme of $U$ is regular? Are these conditions on $U$ equivalent?
| https://mathoverflow.net/users/2191 | For a morphism f from a regular scheme, should there exist an open subscheme U of the target such that fibre of f at each point of U is regular | Over a field of characteristic zero, your result is true. This is Corollary III.10.7 in Hartshorne.
---
In characteristic $p$ no. The simplest example is to take $k$ an algebraically closed field and map $\mathbb{A}^1\_k$ to itself by $x \mapsto x^p$. For every $t \in k$, the fiber above $t$ is $\mathrm{Spec}\ k[x]/(x^p-t) \cong \mathrm{Spec} \ k[y]/y^p$ where the isomorphism is $x-t^{1/p} = y$. So every fiber is singular.
There is a more interesting counter-example due to Serre: Let $k$ be an algebraically closed field of characteristic $2$. Consider the planar cubic
$$ F\_{a,b,c}(x,y,z) :=a (y^2 z + y z^2) + b (x^2 z + x z^2) + c (x^2 y + x y^2) \quad (\*)$$
We leave it to the reader to check that $F=0$ is singular at $(\sqrt{a}:\sqrt{b}:\sqrt{c})$. Generically, the singularity is a node cusp. Choose $(a\_1, b\_1, c\_1)$ and $(a\_2, b\_2, c\_2)$ in $k^3$ and try to map $\mathbb{P}^2\_k \to \mathbb{P}^1\_k$ by
$$(x:y:z) \mapsto (F\_{(a\_1,\ b\_1,\ c\_1)}(x,y,z) : F\_{(a\_2,\ b\_2,\ c\_2)}(x,y,z))$$
Then the fiber over $(t\_1:t\_2)$ is $F\_{(t\_1 a\_1+t\_2 a\_2,\ t\_1 b\_1+t\_2 b\_2,\ t\_1 c\_1+t\_2 c\_2)}=0$ which, as we just computed, is singular. More precisely, the above map is undefined at the $9$ points where $F\_{(a\_1,\ b\_1,\ c\_1)} = F\_{(a\_2,\ b\_2,\ c\_2)} =0$. But, if we take $X$ to be $\mathbb{P}^2$ blown up at those $9$ points, then we get a map from the regular $X$ to the regular $S$, where every fiber is a nodal cuspidal cubic or worse.
Remark: Both of these counter-examples are still counter-examples if you replace "algebraically closed" by "perfect", but it would make my exposition messier.
| 14 | https://mathoverflow.net/users/297 | 32096 | 20,849 |
https://mathoverflow.net/questions/32051 | 3 | First, let me explain what I mean by "synthetic" in the title, which is a proof that reasons purely axiomatically and does not explicitly invoke local coordinate charts (either via concrete expansions or Penrose-style abstract tensor notation). For example in Euclidean geometry, one can either prove statements using Euclid's postulates or via vector arithmetic/Cartesian geometry. Euclidean proofs tend to be shorter and more elegant, but the arithmetical proofs tend to be easier to discover.
To better illustrate in the context of geodesic flow, this I shall now sketch out a basic non-synthetic (arithmetical) derivation for the geodesic equation. Given a compact, path-connected Riemannian manifold (M, g) and a pair of designated points A, B, find a path $\gamma : [0, 1] \to M$ to
Minimize: $\int \limits\_0^1 g\_{\gamma(t)}( \dot{\gamma}(t), \dot{\gamma}(t) ) dt$
Subject to: $\gamma(0) = A, \gamma(1) = B$
Now we construct a variation, $\gamma' : (-\epsilon, \epsilon) \times [0,1] \to M$ such that for all s, t, $\gamma'(0, t) = \gamma(t)$, $\gamma(s,0) = A$ and $\gamma(s, 1) = B$. So, a path $\gamma$ is an extremal if for all variations $\gamma'$,
$\left. \frac{d}{d s} \right|\_{s = 0} \int \limits\_0^1 g\_{\gamma(s, t)}( \frac{d}{d t}\gamma(s, t), \frac{d}{d t}\gamma(s, t) ) dt = 0$
So far so good. Now at this point, the usual strategy is to fix a subinterval $(a, b) \subset [0,1]$ such that for all t in (a,b), $\gamma (t)$ is contained within a single coordinate chart, then expand the functional in terms of this basis and apply the Euler-Lagrange equations. If you do this, then lo-and-behold differentiating the metric gives you a second order term and the Christoffel symbols, which corresponds to the Levi-Civita connection. To extend this to a global result requires showing that the Euler-Lagrange equations agree across a boundary and that the metric is locally convex.
Now my question is can we do this last part synthetically? In other words, just using the basic properties of connections can we prove that this has to be an extremum of the length functional? I really would hope so, otherwise why go through all the trouble of defining connections axiomatically, just pick coordinates and Christoffel symbols and be done with it!
| https://mathoverflow.net/users/4642 | "Synthetic" proof of geodesic flow equation? | If $(s,t) \mapsto \Gamma(s,t)$ is a family of curves in Riemannian manifold $M$, where $s \in [0,1]$ is the curve parameter and $t \in (-\delta,\delta)$ is the variation parameter, let $S = \partial\_s\Gamma, T = \partial\_t\Gamma \in T\_{\Gamma(s,t)}M$. Note that $[S,T] = 0$. Then the derivative of the energy functional can be computed as follows:
If we assume that the two endpoints are fixed, then $T(0) = T(1) = 0$. Therefore,
$$\frac{d}{dt}\int\_0^1 |S|^2\,ds = \int\_0^1 \frac{d}{dt}(S\cdot S)\,ds$$
$$= 2\int\_0^1 S\cdot\nabla\_TS\,ds$$
$$= 2\int\_0^1 S\cdot\nabla\_ST\,ds$$
$$= 2\int\_0^1 \partial\_s(S,T) - T\cdot\nabla\_S S\,ds$$
$$= -2\int\_0^1 T\cdot\nabla\_S S\,ds.$$
The second variation formula can be derived using a similar calculation.
This calculation, however, makes no sense on the manifold itself, since the two vector "fields" $S$ and $T$ are defined only on the image of $\Gamma$, which is a possibly degenerate surface. What you need to do is to pull back the tangent bundle of $M$ to a rank $n$ vector bundle over $[0,1]\times(-\delta,\delta)$ and pull back the Riemannian metric and Levi-Civita connection back to an inner product and connection on the vector bundle and view the calculation above as being on that vector bundle. The easiest way to check that this works is, well, write everything with respect to local co-ordinates. Still, I do prefer this approach to doing calculations. Christoffel symbols are to be avoided as much as possible.
| 7 | https://mathoverflow.net/users/613 | 32101 | 20,851 |
https://mathoverflow.net/questions/19946 | 46 | Short version of the question: Presumably, it's poly$(t)$. But what polynomial, and could you provide a reference?
Long version of the question:
I'm sort of surprised to be asking this, because it's such an extremely basic sounding question. Here are some variants on it:
1. How much time does it take to compute $\pi$ to $t$ bits of precision?
2. How much time does it take to compute $\sin(x)$ to $t$ bits of precision?
3. How much time does it take to compute $e^x$ to $t$ bits of precision?
4. How much time does it take to compute $\mathrm{erf}(x)$ to $t$ bits of precision?
5. How much time and how many random bits does it take to generate a (discrete) random variable $X$ such that there is a coupling of $X$ with a standard Gaussian $Z \sim N(0,1)$ for which $|X - Z| < \delta$ except with probability at most $\epsilon$?
In my area of theory of computer science, no one seems to pay much attention to such questions; an algorithm description might typically read
"Generate two Gaussian random variables $Z$ and $Z'$ and check if $\sin(Z \cdot Z') > 1/\pi$"
or some such thing. But technically, one should worry about the time complexity here.
One colleague of mine who's more of an expert on these things assured me that all such "calculator functions" take time at most poly$(t)$. I well believe that this is true. But again, what polynomial (out of curiosity, at least), and what is the reference?
I kind of assumed that the answers would be in every single numerical analysis textbook, but I couldn't find them there. It seems (perhaps reasonably) that numerical analysis cares mainly about getting the answers to within a fixed precision like 32 or 64 bits or whatever. But presumably *somebody* has thought about getting the results to arbitrary precision, since you can type
```
Digits := 5000; erf(1.0);
```
into Maple and it'll give you an answer right away. But it seemed hard to find. After much searching, I hit upon the key phrase "unrestricted algorithm" which led me to the paper "An unrestricted algorithm for the exponential function", Clenshaw-Olver-1980. It's pretty hard to read, analyzing the time complexity for $e^x$ in terms of eight (??!) parameters, but its equation (4.55) seems to give some answers: perhaps $\tilde{O}(t^2)$ assuming $|x|$ is constant?
And really, all that work for little old $e^x$? As for erf$(x)$, I found the paper "The functions erf and erfc computed with arbitrary precision" by Chevillard in 2009. It was easier to read, but it would still take me some time to extract the answer; my first impression was $\tilde{O}(t^{3/2})$. But again, surely this question was not first investigated in 2009, was it?!
(By the way, question #5 is the one for which I really want to know the answer, but I can probably work it out from the answer to question #4.)
| https://mathoverflow.net/users/658 | What is the time complexity of computing sin(x) to t bits of precision? | For state-of-the-art arithmetic algorithms, I'd recommend this book (a work in progress), available online, from Brent and Zimmermann:
<http://www.loria.fr/~zimmerma/mca/pub226.html>
See chapter 4.
As Steve points out, log, exp, and trig functions are $O(M(n) \log n)$ (in fact they're all calculated from log), where $M(n)$ is the cost of multiplication (theoretically $O(n\log n 2^{\log^\*n})$ by Furer's algorithm) and $n$ is the number of bits accuracy. Pi also falls into this complexity. This is just theoretically, however; for less than a billion digits other algorithms with worse asymptotics are faster.
Erf is apparently harder, the book gives some algorithms based on power series and continued fractions, but it avoids giving an explicit computational cost as there are different convergence speeds in different regions.
| 19 | https://mathoverflow.net/users/7106 | 32114 | 20,856 |
https://mathoverflow.net/questions/32109 | 3 | Given an $n$-dimensional convex polytope $P$, one may set into motion a point-mass, starting on one of the facets of $P$, which travels along a straight trajectory inside $P$ except on collision with the walls, when it is subjected to an elastic response (i.e., its direction vector undergoes a reflection in the facet the point-mass has collided with).
The total trajectory covered by the point-mass is its orbit, and such an orbit is periodic if the point-mass eventually returns to its starting spot and its starting velocity. This billiards flow defines a fairly well-studied dynamical system.
I'm wondering if there are conditions on the symmetry group of $P$ which have consequences for the existence of periodic orbits inside $P$. More precisely, are there any conditions on a finite group $G$ which *forbid* the existence of a periodic orbit in any convex polytope having $G$ as its symmetry group?
| https://mathoverflow.net/users/1104 | Connections between a polytope's symmetry group and the existence of periodic orbits | Dear Zach Conn,
I think your problem is very interesting but quite difficult to approach even in dimension 2: for instance, it is a well-known open problem to decide whether every irrational triangular billiard has a periodic orbit. As far as I know, the conjectural answer is yes, but, besides the case of acute triangles (where Fagnano's orbit can be constructed) and sufficiently small perturbations of isosceles triangles (after the work of R. Schwartz and W. P. Hooper), there is few progress in the general case.
Anyway, let me point out that this conjecture (if true) says that we can't hope (in general) to find conditions on the symmetry group of the billiard to avoid periodic orbits: in fact, the existence of periodic orbits in triangular billiards is expected in both irrational ("small symmetry group") and rational ("large symmetry group") cases.
Best,
Matheus
| 2 | https://mathoverflow.net/users/1568 | 32115 | 20,857 |
https://mathoverflow.net/questions/32117 | 13 | Does there exists a subset E of R with positive measure and without containing any midpoints (i.e. x,y distinct in E, (x+y)/2 not in E)?
| https://mathoverflow.net/users/7663 | Set of real numbers with positive measure containing no midpoints | According to James Foran, Non-averaging sets, dimension, and porosity, Canad Math Bull 29 (1986) 60-63, "It follows from the Lebesgue Density Theorem that a measurable, non-averaging subset (of
$(0,1]$) cannot have positive measure."
| 11 | https://mathoverflow.net/users/3684 | 32124 | 20,861 |
https://mathoverflow.net/questions/32126 | 24 | There is an example of a function that is unbounded on every open set. Just take $f(n/m) = m$ for coprime $n$ and $m$ and $f(irrational) = 0$.
I want to generalize this in a way to get a function which is not just unbounded on every open set, but whose range is equal to $\mathbb{R}$ on every open set. The latter construction clearly doesn't work.
I'm interested whether such function exists and if it exists is there any constructive way to define it?
| https://mathoverflow.net/users/7079 | Function with range equal to whole reals on every open set | See [Conway's base 13 function](http://en.wikipedia.org/wiki/Conway_base_13_function).
| 43 | https://mathoverflow.net/users/4213 | 32127 | 20,862 |
https://mathoverflow.net/questions/31350 | 10 | At the time of writing, the most recent blog post over at *What's new* by Terrence Tao is [Cayley graphs and the geometry of groups](http://terrytao.wordpress.com/2010/07/10/cayley-graphs-and-the-geometry-of-groups/), and that (excellent, as with most of Tao's writing) post most immediately inspired this question. Also, a disclaimer: Cayley graphs, and, indeed, abstract group theory generally, are not my area of expertise, so feel free to explain how my question is overly naive and/or needs revision.
It is a classical result that a connected topological group is generated by any open neighborhood of its identity element. (Proof: the subgroup generated by the open neighborhood is a union of opens, and hence open; but then so are all its cosets; so this subgroup is both open and closed; hence it is everything if the group is connected as a topological space.) So I am tempted to take some topological group $G$, and some *very small* neighborhood $S$ of the identity $e$, and construct the corresponding Cayley graph. My intuition says that as $S$ gets smaller and smaller, the Cayley graph better and better approximates the topological space $G$. Notice also that arbitrary small open neighborhoods $S\ni e$ know everything about the topology of $G$, because $G$ is homogeneous.
I don't know how to define a "formal neighborhood" of a topological space, unless it is actually a manifold or algebraic space or .... So maybe I should restrict my attention to Lie or Algebraic groups for this question. But anyway:
>
> Is there a way to define the "infinitesimal generators" or "formal neighborhood" for a topological group in such a way that the corresponding "Cayley graph" *is* the group as a topological space (Edit: which of course doesn't make any sense; see the comments. I mean something like "so that the geometry/topology of the group comes immediately from the graph)? If not in this generality, does it at least work when the group is Lie? Algebraic? Or other regularity conditions on the topology?
>
>
>
| https://mathoverflow.net/users/78 | Is there a way to see a topological group as the "Cayley graph" of its "infinitesimal generators"? | Several people have been working on what might be described as geometric group theory for compactly generated, totally disconnected topological groups.
The main definitions as well as several nice results are in this paper:
Krön, Bernhard; Möller, Rögnvaldur G.
Analogues of Cayley graphs for topological groups.
Math. Z. 258 (2008), no. 3, 637--675.
and if you search for papers which reference this one you'll find some more recent work.
| 10 | https://mathoverflow.net/users/3955 | 32139 | 20,867 |
https://mathoverflow.net/questions/32145 | 1 | I have encountered a few problems regarding the minimal subgroups of a finite group $G$. Any references and/or answers regarding the following questions will be very welcome.
1)If $G$ is a finite group, what can be said about its minimal normal subgroups (under inclusion)?
2) Is there a criterion to decide when an element $g \in G$ belongs to such a minimal subgroup?
3) Is the product of all minimal normal subgroups of a given finite group $G$ the entire group $G$?
| https://mathoverflow.net/users/7670 | Minimal normal subgroups of a finite group | Try reading the [planetmath.org article](http://planetmath.org/groupsocle) on the socle of a group.
| 6 | https://mathoverflow.net/users/6153 | 32148 | 20,872 |
https://mathoverflow.net/questions/32116 | 3 | It is well known that every group of exponent $n=2$ is abelian. I remember having seen that this is also the case for $n=3$. (can someone give a proof). How does this generalize to any $n$ or to any prime $p$.
| https://mathoverflow.net/users/7489 | Exponent of a group | The group defined by $\langle x,y,z; x^3 = y^3 = z^3 = 1, yz = zyx, xy = yx, xz = zx\rangle$ has order 27, exponent 3 and is non-abelian.
(Checking exponent 3 basically comes down to ensuring that $(yz)^3 = (y^2z)^3 = (yx^2)^3 = 1$. Or by using Gap.)
| 16 | https://mathoverflow.net/users/6503 | 32155 | 20,877 |
https://mathoverflow.net/questions/32149 | 1 | I have been studying measure theory of late, and i was stuck up in these two things.
BOUNDED CONVERGENCE THEOREM: It states that $f\_n$ is a sequence of measurable functions, defined on a measurable set $E$ and if $f\_{n} \to f$ pointwise on $E$,and is $f\_{n}$ is uniformly bounded, that is if $|f\_{n}(x)| \leq M$ for each $x$ and $n \in \mathbb{N}$ then $$ \lim \int f\_{n} = \int f$$
Could anyone tell me as to why this will fail if we don't assume the uniformly bounded criterion. Please elaborate.
| https://mathoverflow.net/users/1483 | Fatou's Lemma and the bounded convergence theorem. | I am not sure if the question fits MO standards as it is an elementary measure theory question (and if it's not, expect to be tazered by the MO police). Here goes an answer, anyway.
To expand on Robin Chapman's comment, first, the theorem as stated is false withouth the assumption that $E$ has finite measure. The correct generalization is the Lebesgue dominated convergence where the sequence $f\_{n}$ is such that there is an integrable $g$ such that $\|f\_n{x}\|\leq g$.
To see why, it fails without the boundedness condition consider the sequence of intervals $E\_n= [0, 1/n]$ and take the sequence $(n\chi(E\_n))$ where $\chi(E\_n)$ is the characteristic function (or indicator functions) of $E\_n$. This sequence converges pointwise to $0$ but
```
$\int n\chi(E_n) = 1$
```
so that the sequence of integrals does not converge to $0$. What is happening is that you are shrinking the support of the functions but the same time increasing their "amplitude" so that the two cancel each other out and the integral stays constant while the functions themselves converge to zero. The uniform bound on the sequence, prevents their "amplitudes" of running off to infinity and screwing up the integrals.
Examples can be concocted where the convergence is uniform instead of just pointwise. The idea is to do the reverse of the previous example: shrink the amplitude of the functions (to guarantee their uniform convergence) while enlarging their support. This will need a measure space of infinite measure. I will leave that as an exercise.
Regards,
G. Rodrigues
| 6 | https://mathoverflow.net/users/2562 | 32156 | 20,878 |
https://mathoverflow.net/questions/32150 | 3 | Let $i$, $k$ be integers such that $2 \leq i \leq k$. I would like to show that the sum
$$
\sum\_{j=1}^{i-1} \frac{(-1)^{j-1}(i-j)^k}{(i-j)! (j-1)!}
$$
is positive. I have carried out extensive numerical experiments to check this for small values of $k$. In fact, much more should be true. Define polynomials
$$
U(x)=(x+i-1)^k
$$
and
$$
V(x)=x(x+1)\cdots(x+i-1).
$$
Let $Q$ and $R$ be the quotient and remainder on dividing $U$ by $V$. The above sum is the leading coefficient of $R$. It seems that all the coefficients of $Q$ and $R$ are always positive, and it would be nice to prove this, but I only need the positivity of the above sum. This question has applications for proving the irrationality of certain series.
| https://mathoverflow.net/users/4140 | Positivity of a finite sum | These are [Stirling numbers of the second kind](http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind). More precisely
your sum is S(k,i-1) where $S$ denotes Stirling number of the second kind.
| 11 | https://mathoverflow.net/users/4213 | 32157 | 20,879 |
https://mathoverflow.net/questions/31936 | 8 | I recently heard of a game between two players "Line" and "Point" and wanted to look for more information on it. However, without knowing the name of it (if it has one) finding more information is hard, has anyone heard of it? Is there a winning strategy for one of the players?
The game is as follows, it is played on the unit disk $D^2$ in $\mathbb{R}^2$ with the point $p\_0 = (0,0)$ marked to begin with. Play alternates between L and P (starting with L) and on turn $n$ they do the following:
L chooses a new line $l\_n$ through point $p\_{n-1}$ and then P chooses a new point $p\_n$ on line $l\_n$ inside $D^2$.
This forms a sequence of points $(p\_n)\_{n = 1}^\infty$ in $D^2$. L wins if this sequence converges to a point in $D^2$, P wins if it does not.
As far as I can tell P has a winning strategy, but I my formal proof for this is a sketch at best.
| https://mathoverflow.net/users/3121 | Choosing lines and points in D^2 | Line actually has a winning strategy: it can force a convergent sequence. The problem was posed and solved in the following paper:
J. Maly and M. Zeleny (2006), A note on Buczolich's solution of the Weil gradient problem: a construction based on an infinite game, [*Acta Mathematica Hungarica*, Vol. 113, pp. 145-158.](http://dx.doi.org/10.1007/s10474-006-0096-7)
| 5 | https://mathoverflow.net/users/3376 | 32159 | 20,880 |
https://mathoverflow.net/questions/32069 | 23 | Are deformations of Nakajima quiver varieties also Nakajima quiver varieties ?
In case the answer to this is (don't k)no(w), here are some simpler things to ask for.
1. (If you're a differential geometer) Is any hyperkahler rotation / twistor deformation of a Nakajima quiver variety also a Nakajima quiver variety ?
2. (An example, for algebraic geometers) Consider the Hilbert scheme $H$ of $k$ points on the minimal resolution of $\mathbb C^2/(\mathbb Z/n)$. Or restrict to $n=2=k$ and consider $Hilb^2T^\*\mathbb P^1$. Its exceptional divisor over the symmetric product defines a class in $H^1(\Omega\_H)$ (despite the noncompactness). Using the holomorphic symplectic form, we get a deformation class in $H^1(T\_H)$. (The corresponding deformation is not so far from the twistor deformation, and can be realised as a composition of a deformation of the ALE space followed by a twistor deformation.) Is there a Nakajima quiver variety in the direction of this deformation ?
For instance if I take the quiver $\bullet^{\ \rightrightarrows}\_{\ \leftleftarrows}\bullet$, dimension vector (1,1), and an appropriate stability condition (or value of the real moment map) then I get $T^\*\mathbb P^1$ as the moduli space over the value $0$ of the complex moment map, and the smoothing of the surface ordinary double point over nonzero values.
Now if I take dimension vector (2,2) I can presumably get $Hilb^2$ of these surfaces, for an appropriate stability condition. However, as I vary the value of the complex moment map I simply vary the surface that I take $Hilb^2$ of, rather than getting the deformation I'm after. (The hyperkahler rotation is not a $Hilb^2$, since the exceptional divisor disappears in this deformation.)
But is there another quivery way of producing this deformation ?
| https://mathoverflow.net/users/7653 | Deformations of Nakajima quiver varieties | I do not know a general statement. I just want to give a comment:
>
> Now if I take dimension vector $(2,2)$ I can presumably get $Hilb^2$ of these surfaces, for an appropriate stability condition.
>
>
>
No. You only get the symmetric product of $T^\*P^1$ if you work on quiver varieties with the dimension vector $(2,2)$.
To get a $Hilb^2$ of the surface, one need to put the one-dimensional vector space $W $at the vertex 0, and take a suitable stability condition. (I hope you are familiar with convention for quiver varieties.)
Then we have two dimensional family of quiver varieties from the complex moment map deformation. Thus we get one more dimension from the deformation of the underlying surface.
| 17 | https://mathoverflow.net/users/3837 | 32165 | 20,883 |
https://mathoverflow.net/questions/32135 | 4 | If $f:X\to S$ is a universal homeomorphism, is $f':X\times\_S X\to X$ always a nil-immersion? This seems to be easy, yet possibly I miss something. Should I give references to this fact in a paper?
| https://mathoverflow.net/users/2191 | If $f:X\to S$ is a universal homeomorphism, is $f':X\times_S X\to X$ a nil-immersion? | In general, the answer is no. As Brian Conrad points out in the comments above, purely inseparable field extensions do not have this property. However (and maybe this is what you were getting at?), it is true that if $X\to S$ is a universal homeomorphism, then the diagonal $X\to X\times\_SX$ *is* a nilimmersion.
| 5 | https://mathoverflow.net/users/3049 | 32166 | 20,884 |
https://mathoverflow.net/questions/32163 | 7 | What are the curves of contact on a convex body $B$ rolling down an inclined plane?
Assume $B$ is smooth, and there is sufficient friction to prevent slippage.
Certainly, one can develop a geodesic to a straight line on a plane by rolling $B$ so that the geodesic
is the point of contact, but it doesn't appear that this in general would be the point of contact
in the physical situation of free rolling under gravity. It is for a sphere, which will roll along
great circles. And an ellipsoid should roll along its three simple closed geodesics (although there
are significant instabilities with all but the shortest closed geodesic—I'd prefer to ignore
stability issues). But for other shapes,
I imagine that an off-center center of gravity (and perhaps rotational momentum?) will cause the rolling to deviate from a
geodesic. (But I am uncertain of this. Please correct me if I'm wrong!)
Assuming this is correct (that rolling curves are not always geodesics),
what are the conditions that determine if a curve $\gamma$ is a *rolling curve*:
$\gamma$ is the trace of the point of contact between $B$ and an inclined plane as it rolls,
from some initial position?
Perhaps: If $p \in \gamma$ is a point of contact, then (a) the normal vector $N$ of $\gamma$ at $p$ must be
perpendicular to the plane, and (b) the center of gravity of $B$
must lie in the osculating plane of $\gamma$ at $p$ (the plane containing $N$ and
the tangent $T$ vector at $p$)?
Have what I christened *rolling curves* been studied in the literature? If so, under what name?
My searches have been unsuccessful. Can you think of shapes outside of {sphere, ellipsoid, cylinder} where the rolling curves can be determined? Thanks for any thoughts or pointers!
| https://mathoverflow.net/users/6094 | Rolling a convex body: Geodesics vs. rolling curves | The rolling motion of a convex symmetric body on a *horizontal* plane is a classical problem. In the symmetric case, Chaplygin was the first who showed that the full equations of motion can be reduced to a linear integrable system of two ODEs. A modern exposition of Chaplygin's results can be found in the very recent [book](http://books.google.co.uk/books?id=-zKuPwAACAAJ&dq=Geometry+of+Nonholonomically+Constrained+Systems&hl=en&ei=AmNATJmJHoe24gaAhOnRDg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CDIQ6AEwAA) by Cushman, Śniatycki and Duistermaat.
The problem of rolling motion on an *inclined* plane is, in general, *nonintegrable* (this problem was studied, in particular, by V.V. Kozlov in 1990s).
As for the tracing trajectories of the point of contact, you might be interested in this [article](http://www.turpion.org/php/full/getFT.phtml/rd_8_201.pdf?journal_id=rd&paper_id=237&agree=on&tpdfn=rd_8_201&x=63&y=8) (also [available](http://arxiv.org/abs/nlin/0502039v1) on arXiv) and the references therein. The authors discuss the case of a disc (i.e. a convex body of revolution) rolling on a horizontal plane.
From the Introduction:
>
> It appears that the point of contact performs the composite bounded motion: it periodically traces some closed curve which rotates as a rigid body with some constant angular velocity about the fixed point...
>
>
>
| 5 | https://mathoverflow.net/users/5371 | 32170 | 20,886 |
https://mathoverflow.net/questions/32169 | 22 | On a Riemannian manifold, a coordinate system is called "isothermal" if the Riemannian metric in those coordinates is conformal to the Euclidean metric:
$$g\_{ij} = e^{f} \delta\_{ij}$$
My question is: Why are such coordinate systems called "isothermal"? It must have something to do with classical thermal physics. I tried looking for a reason online, with no success.
It is well known that when the dimension $n=2$, there always exist isothermal coordinates, and this is probably where they were first introduced. So maybe the nomenclature has something to do with heat diffusion in the plane?
(The reason I ask is because I am planning to give a seminar talk next week giving a proof that such coordinates exist when $n=2$, and thought it would be nice to explain to the students where the name comes from...)
| https://mathoverflow.net/users/6871 | Why are they called isothermal coordinates? | Isothermal coordinates are harmonic. In other words, it solves $\triangle\_g u = 0$. So locally it is a stationary solution of the heat equation. In physics, for a steady state distribution of temperatures, each level set is called an isotherm.
| 24 | https://mathoverflow.net/users/3948 | 32172 | 20,888 |
https://mathoverflow.net/questions/32158 | 3 | Hello all,
I need some theoretical pointers (formulas, articles, online links) on how to merge Singular Value Decompositions (SVD) of two matrices (two different sets of observations over the same set of features).
That is, I have two SVDs: $A=U\_A\*S\_A\*V^T\_A$ and $B=U\_B\*S\_B\*V^T\_B$ and want to know SVD $A|B=U\_{A|B}\*S\_{A|B}\*V\_{A|B}$. The original matrices $A$ and $B$ are unavailable, the solution must make use of the $U\_A, S\_A, V\_A, U\_B, S\_B, V\_B$ matrices only.
I need this because I want to implement a distributed version of incremental SVD: have several computation nodes work on different sets of observations independently, and then merge their results into one.
Cheers!
| https://mathoverflow.net/users/7672 | distributed incremental SVD | The people aspiring for the [Netflix prize](http://en.wikipedia.org/wiki/Netflix_Prize) like incremental SVDs. See
1. <https://issues.apache.org/jira/browse/MAHOUT-371>
2. B.M. Sarwar, G.Karypis, J.A. Konstan, and J. Reidl. Incremental singular value deocmposition algorithms for highly scalable recommender systems. In Proceedings of the Fifth International Conference on Computer and Information Technology (ICCIT), 2002. ([ONLINE](http://www.grouplens.org/papers/pdf/sarwar_SVD.pdf))
3. M. Brand. Fast online svd revisions
for lightweight recommender systems.
In Proceedings of the 3rd SIAM
International Conference on Data
Mining, 2003. and [his tech report](http://www.merl.com/publications/TR2002-024/)
Have you tried searching the [ACM digital library](http://portal.acm.org/portal.cfm) for parallel SVD or singular value decomposition?
**EDIT1 based on new input** : See the following two papers by Hall, Marshall and Martin
1. [Merging and Splitting Eigenspace
Models](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.51.804) (Section 3)
2. [On Adding and Subtracting
Eigenspaces with EVD and SVD](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.9.1496)
| 4 | https://mathoverflow.net/users/5372 | 32177 | 20,891 |
https://mathoverflow.net/questions/32179 | 5 | In a paper I read recently, the authors use the fact that if two groups G and H have no nontrivial common quotient, then neither do GxG and HxH. It's unclear from the context whether this is true for all groups, or just groups of the type that are important for this paper, and they don't prove the claim.
I've been trying to prove the statement for general groups without any success. Is it true?
I've also tried restricting to the case I really care about, as follows. Let K, K', M, and M' be normal subgroups of U, with $U/K \simeq U/K'$ and $U/M \simeq U/M'$. Suppose that KK' = KM = KM' = K'M = K'M' = MM' = U. Is it also the case that $(K \cap K')(M \cap M') = U$? The reason this is connected to the above is that $U/(K \cap K')(M \cap M')$ is a common quotient of $U/(K \cap K') \simeq U/K \times U/K' \simeq U/K \times U/K$ and $U/(M \cap M') \simeq U/M \times U/M' \simeq U/M \times U/M$. So if U/K and U/M have no nontrivial common quotients, is the same true of $U/(K \cap K')$ and $U/(M \cap M')$?
Finally, if the result is not true in general for either of these cases, what about if we restrict to finite groups G and H?
| https://mathoverflow.net/users/913 | Common quotients of direct products | If $Q$ is a quotient of $G\times G$, then it has a normal subgroup $A$ (the image of $G\times 1$) such that both $A$ and $Q/A$ are quotients of $G$. If it is also a quotient of $H\times H$, then it has a normal subgroup $B$ such that both $B$ and $Q/B$ are quotients of $H$. Now assume that every common quotient of $G$ and $H$ is trivial. Then $Q/AB$ is trivial, so $AB=Q$. So $Q/A=AB/A=B/(A\cap B)$ is trivial, so $A=Q$. Likewise $B=Q$. So $Q$ is trivial.
| 13 | https://mathoverflow.net/users/6666 | 32185 | 20,895 |
https://mathoverflow.net/questions/32160 | 8 | Let $g$ be a Lie algebra over $\mathbb{C}$. Then the equivalence between the derived category of modules over $U(g)$ and the coderived category of co-modules over it's Chevalley complex $C\_\*(g)$ in which $M\rightarrow C\_\*(g,M)$ is a classical example of Koszul duality. In the case when $g$ is a finite dimensional Lie algebra, it is easier for most humans to dualize everything and discuss modules over the Chevalley cochain complex $C^\*(g)$. Most of the time, people are most interested in various subcategories of modules over a given Lie-algebra. For example the category of semi-simple modules or finite dimensional modules over $g$. I would like to know if for any Lie-algebra $g$ if one can understand what happens to the subcategories of the derived category generated by these objects under the Koszul duality. Are the corresponding subcategories equally natural to describe? My intuition is that the semisimple modules are sent to something like (co) perfect modules based upon the observation that $\mathbb{C}$ is sent to $C\_\*(g) $. But I don't think this is exactly right and I wonder if someone could help me out.
| https://mathoverflow.net/users/6986 | Koszul duality and modules over the Chevalley complex | The derived category of finite-dimensional $g$-modules is not a full subcategory of the derived category of arbitrary $g$-modules for a finite-dimensional Lie algebra $g$, in general (e.g., for a semi-simple Lie algebra $g$). However, one can consider the full subcategory of the derived category of $g$-modules consisting of complexes of $g$-modules with bounded and finite-dimensional cohomology. This triangulated category is equivalent to the homotopy category of DG-modules over $C^\ast(g)$ that are free/projective and finitely generated as graded modules.
The proof (based on the general Koszul duality theorem mentioned in the question) proceeds as follows. The category of DG-modules over $C^\ast(g)$ that are injective/projective as graded modules is a full subcategory of the co/contraderived category of DG-modules. The category of complexes of $g$-modules with bounded and finite-dimensional cohomology is generated by finite-dimensional $g$-modules, and these are transformed by the Koszul duality functor into DG-modules that are free and finitely generated as graded modules. This provides a fully faithful functor in one direction.
To prove that it is essentially surjective, consider the functor assigning to a complex of $g$-modules the underlying complex of vector spaces. Let us presume that our Koszul duality functor is $M\longmapsto C^\ast(g,M)$. Then this Koszul duality functor transforms our functor of forgetting the action of $g$ into the functor assigning to a DG-module over $C^\ast(g)$, projective as a graded module, its quotient complex by the action of the augmentation ideal of $C^\ast(g)$. Thus the complex of $g$-modules corresponding to a DG-module that is projective and finitely generated as a graded module has bounded and finite-dimensional cohomology.
| 11 | https://mathoverflow.net/users/2106 | 32194 | 20,902 |
https://mathoverflow.net/questions/30000 | 18 | The aim of this question is to understand SYZ conjecture ("Mirror symmetry is T-Duality").
I don't expect a full and quick answer but to find a better picture from answers and comments.
The whole idea is to construct the Mirror C.Y $Y$ from $X$ intrinsically as follows.
One considers the moduli of special Lagrangian tori with a flat $U(1)$ bundle on it in $X$.
Then we put a metric on this moduli (plus corrections coming from J-holomorphic disks) and expect that this moduli and the metric given on it is the mirror C.Y we were looking for.
Here are the things I can not understand:
---What is the metric given in the paper "Mirror symmetry is T-Duality"?
Where does it come from? (I can not understand the formulation of metric there).
and more importantly
--How do we deform the metric using J-holomorphic disks(instantons)?
| https://mathoverflow.net/users/5259 | Do you understand SYZ conjecture | Hi-
Just saw this thread. Maybe I should comment. The conjecture
can be viewed from the perspective of various categories:
geometric, symplectic, topological. Since the argument is
physical, it was written in the most structured (geometric)
context -- but it has realizations in the other categories
too.
Geometric: this is the most difficult and vague, mathematically,
since the geometric counterpart of even a conformal field theory
is approximate in nature. For example, a SUSY sigma model with
target a compact complex manifold X is believed to lie in the
universality class of a conformal field theory when X is CY,
but the CY metric does not give a conformal field theory on
the nose -- only to one loop. Likewise, the arguments about
creating a boundary conformal field theory using minimal (CFT) +
Lagrangian (SUSY) are only valid to one loop, as well.
To understand how the corrections are organized, we should
compare to (closed) GW theory, where "corrections" to the classical
cohomology ring come from worldsheet instantons -- holomorphic
maps contributing to the computation by a weighting equal
to the exponentiated action (symplectic area). The "count"
of such maps is equivalent by supersymmetry to an algebraic
problem. No known quantity (either spacetime metric or
Kahler potential or aspect of the complex structure) is
so protected in the open case, with boundary. That's why
the precise form of the instanton corrections is unknown,
and why traction in the geometric lines has been made
in cases "without corrections" (see the work of Leung, e.g.).
Nevertheless, the corrections should take the form of
*some* instanton sum, with known weights. The sums seem
to correspond to flow trees of Kontsevich-Soibelman/
Moore-Nietzke-Gaiotto/Gross-Siebert, but I'm already running
out of time.
Topological: Mark Gross has proven that the dual torus
fibration compactifies to produce the mirror manifold.
Symplectic: Wei Dong Ruan has several preprints which
address dual Lagrangian torus fibrations, which come
to the same conclusion as Gross (above). I don't know
much more than that.
Also-
Auroux's treatment discusses the dual Lagrangian
torus fibration (even dual slag, properly understood)
for toric Fano manifolds, and produces the mirror
Landau-Ginzburg theory (with superpotential) from this.
With Fang-Liu-Treumann, we have used T-dual fibrations
for the same fibration to map holomorphic sheaves
to Lagrangian submanifolds, proving an equivariant version of
homological mirror symmetry for toric varieties.
(There are many other papers with similar results
by Seidel, Abouzaid, Ueda, Yamazaki, Bondal, Auroux,
Katzarkov, Orlov -- sorry for the biased view!)
Reversing the roles of A- and B-models, Chan-Leung
relate quantum cohomology of a toric Fano to the
Jacobian ring of the mirror superpotential via T-duality.
Help or hindrance?
| 18 | https://mathoverflow.net/users/1186 | 32197 | 20,904 |
https://mathoverflow.net/questions/32121 | 0 | I'm trying to write a simple recommendation system. I have a set of products that exist in a set of categories and I know whether a given customer liked a subset of the items. From this I can deduce an "affinity" from each customer to each category (a number 0..1 defining the fraction of the products from that category that they liked).
Given these affinities I can place the user in a vector space in N dimensions (where N==the number of categories), cluster together similar customers and recommend to them the top products in categories that they might like but not know that we carry. There are a very high number of categories (although a much smaller number that are active in any given period), users, and products.
I'm trying to cluster them using the kmeans algorithm ([specifically this Haskell library](http://hackage.haskell.org/packages/archive/kmeans/0.1.1/doc/html/src/Data-KMeans.html#kmeans)). The problem I'm having is deciding how to compute the distance between the vectors of two customers, given their sparseness. Not only am I worried about the computational complexity and storage of having to expand the vector of every customer to a dimensionality that will fit every category (even though a given customer has probably only rated items from ten categories in the common case), I also can't just assume that the value of a missing dimension is 0, because that's a meaningful value (they hated every item from that category that they purchased).
Is there a well-defined way to compare vectors with undefined values?
| https://mathoverflow.net/users/7664 | Clustering sets of sparse vectors with high dimensionality | Let affinity range between -1 and 1, with zero representing indifference/no opinion/haven't tried it. Customers have a right to hate.
Storage is much easier: You need only store the non-zero affinities. e.g. A.J. = [MathOverFlow : 1, Coffee : 1, Decaf: -1]
My consulting fee is $5. I accept cash or check.
| 1 | https://mathoverflow.net/users/35508 | 32205 | 20,910 |
https://mathoverflow.net/questions/32108 | 1 | Let p=(v1,…,vn) be a self-avoiding walk in a graph G. Let d(p) be the number of unique i, 1≤i<n such that there's a self-avoiding walk q that starts at vn and ends at vi without visiting any other vertices in p. Let d(G) be maxp d(p) with maximum taken over all self-avoiding walks in G. Is d(G) related to treewidth of G?
Motivation: [arXiv:math/0701494](http://arxiv.org/abs/math/0701494) gives a way to represent marginal probability of a node on an arbitrary graph (ie, the probability of that node taking a particular color when we randomly pick a proper coloring of G) as the marginal probability of the root of a self-avoiding walk tree built from this graph. Finding marginal probability in this representation seems to scale exponentially in d(G), whereas finding marginal in original representation (junction tree algorithm), scales exponentially in treewidth of G.
This relates to the complexity of counting the number of colorings in a graph, because there's a computationally efficient way to get the number of colorings from marginals (sequential cavity method), so if d(G) grew slower than treewidth(G) for some family of graphs, this would give an algorithm for counting colorings that's faster than the tree-decomposition based one
| https://mathoverflow.net/users/7655 | Is this measure related to treewidth? | If a graph has bounded treewidth, it has $O(n)$ edges. And if it has $O(n)$ edges, it has $2^{O(n)}$ self-avoiding walks, since a self-avoiding walk can be specified as a set of edges. This is in contrast with arbitrary graphs where the number of self-avoiding walks can be exponential in $n\log n$. But it's not really a relation to treewidth since the same thing holds for any sparse graph.
| 1 | https://mathoverflow.net/users/440 | 32214 | 20,917 |
https://mathoverflow.net/questions/32193 | 19 | These days, lots of people are excited by Frobenius algebras because commutative Frobenius algebras are the same thing as 2D topological quantum field theories.
...but this seems like teaching an old dog new tricks. Can anyone sum up (using only diet representation theory :-P), why Frobenius algebras were invented and what was so good about them?
Also, any nice texts and/or papers along this line would be much appreciated. I'm working through the old Nakayama papers now, but perhaps this material exists in a friendlier and more modernised form somewhere?
| https://mathoverflow.net/users/800 | Why did people originally like Frobenius algebras? | Frobenius's original turn-of-the-century perspective was the nonvanishing of a determinant. Brauer–Nesbitt–Nakayama studied some equivalent definitions in the late 30s and early 40s. For instance, an equivalence between the left and right regular representations is a rare and beautiful thing; this gives an equivalence between projectivity and injectivity that is explained in modern language in Lam's Lectures on Modules and Rings. This also gives a "perfect duality" studied by Dieudonné in the late 50s. I added the missing early sources to the [wikipedia article,](http://en.wikipedia.org/wiki/Frobenius_algebra) including the Brauer–Nesbitt announcement in PNAS which is pretty easy to read.
| 10 | https://mathoverflow.net/users/3710 | 32216 | 20,918 |
https://mathoverflow.net/questions/32217 | 2 | Suppose $R$ is a commutative ring, and $S \subset R^{n\times n}$ is an $R$-module. We are given $H\_0,\dots,H\_n \in R^{n\times n}$, and we know that for all $r \in R$,
$$H\_0 + r H\_1 + \dots r^n H\_n \in S$$
The question is: when can we conclude that $H\_i \in S$ for all $i=0,1,\dots,n$ ?
Clearly this is true when $R = \mathbb{R}$, because we can choose real numbers $r\_0,r\_1,\dots,r\_n$ and write:
$$ H\_0 + r\_i H\_1 + \dots r\_i^n H\_n \in S \qquad \text{for $i=0,1,\dots,n$}$$
The corresponding Vandermonde matrix $V$ is invertible provided the $r\_i$ are distinct. Now take a linear combination of the left-hand-sides of the above relation, using coefficients provided by the ith row of $V^{-1}$. Since $S$ is closed under linear combinations, we conclude that $H\_i \in S$, as required.
The same argument won't work for a general commutative ring, but is the result still true? If not, what constraints must be imposed on $R$ to make it so?
| https://mathoverflow.net/users/7667 | For which rings does there exist an invertible Vandermonde matrix? | The fact that the $H\_i$ are matrices is irrelevant; the question is whether a module generated by elements $H\_0\dots,H\_n$ is necessarily generated by all the elements of the form $\Sigma\_{0\le i\le n}r^iH\_i$. If this is true in modules of matrices then it's true in all free modules, and if so then it's true when the $H\_i$ form a basis for a free module, and if it's true in that case then it's true in all modules, in particular modules of matrices.
When the $H\_i$ are a basis for $R^{n+1}$ then the question becomes "Is the ideal generated by all $(n+1)\times (n+1)$ Vandermonde determinants the unit ideal?" That fails if and only if there is a maximal ideal $m$ containing all such determinants, so if and only if there exists $m$ such that over the field $R/m$ there is no invertible Vandermonde matrix, so it fails if and only if some residue field of $R$ has at most $n$ elements.
| 15 | https://mathoverflow.net/users/6666 | 32227 | 20,923 |
https://mathoverflow.net/questions/32221 | 3 | I have a way of generating random parameterized maps from $S^1 \to \mathbb{R}^3$. This method can create very simple knots, such as ellipses, but can also create knots with more crossings than I can count. The images can be knotted as I have seen the method generate a figure eight knot. I know that with a parameterization instead of a diagram one can compute things such as various knot energies. My question is the following: Is having the ability to generate random parameterized knots of use to anyone?
| https://mathoverflow.net/users/7681 | What can one do with randomly generated parameterized knots in 3 space? | Not very likely. Any physical questions about random knotting will come with all sorts of requirements about the random process that generates the knots, and so it's unlikely that whatever method you have in mind would be helpful in any given situation.
Also, it's very easy to produce random knots, and without specifying some nice property your method has it's hard to say much. The simplest method I can think of is to pick n random directions in R^3, and concatenate a series of n unit intervals in those directions, then join the first and last point. Why should your method being any more or less interesting than this one?
You might be interested in reading Dorothy Buck's papers on the arXiv, which include some simple models for transformations of knots that are relevant for knotted DNA, to get an idea of the very particular requirements of specific applications of "random knotting".
| 1 | https://mathoverflow.net/users/3 | 32230 | 20,926 |
https://mathoverflow.net/questions/32231 | 8 | Let $A\_0, \dots, A\_{n-1}$ be upper triangular matrices with ones on the diagonal. Let $B\_{n-1}, \dots, B\_0$ be of the same form.
I am interested in bounding
$$|| A\_0 \dots A\_{n-1} B\_{n-1}^{-1} \dots B\_0^{-1}||$$
and in particular showing that this product is close to the identity when $A\_i$ and $B\_i$ are close (specifically, $||A\_i - B\_i|| \leq \delta \lambda^{- \min(i, n-i)}$ where $\lambda>1$).
In general, this will probably not be possible, but there may be conditions when there are interesting bounds.
The problem is, I don't really know anything about techniques for bounding products of matrices. There is the obvious multiplicativity of the matrix (i.e., operator) norm, but it is far from best possible in this case. For instance, it is easy to check that the norm can grow at most polynomially in $n$ if the norms of the $A\_i, B\_i$ are bounded above by some constant.
Hence:
**Question:** What techniques/tricks are there for bounding these kinds of products of matrices?
Unfortunately I don't know exactly which techniques would be useful, so I'd appreciate any pointers to relevant papers or books.
I can at least explain how the problem is motivated. Namely, we have a Holder-continuous matrix-valued function $A: X \to GL\_m(\mathbb{R})$ for $X$ a compact metric space.
Then the $A\_i$ will be the values $A(T^ix)$ and the $B\_i$ will be $A(T^ip)$ for $p$ close to $x$. The multiplicative ergodic theorem states that the norm products $||A\_0 \dots A\_{n-1}||$ grow like $e^{\lambda n}$ almost everywhere (in $x$) with respect to any invariant measure (where $\lambda$ is the maximal Lyapunov exponent), but in the case of upper triangular matrices, this is actually immediate---we even have polynomial growth if ones are on the diagonal (a corollary of the nilpotence of strictly upper triangular matrices). The multiplicative ergodic theorem and its variants were used in a paper that motivated the problem I'm working on, but it doesn't quite seem to help here (because we make slightly different assumptions), which is why I was curious about other techniques.
| https://mathoverflow.net/users/344 | Techniques to bound products of upper triangular matrices and their inverses | The standard trick to quantify the joint continuity of a product operation such as $(A\_1,\ldots,A\_k) \to A\_1 \ldots A\_k$ (which is essentially what you are trying to do here) is to split a difference such as
$$ A\_1 \ldots A\_k - B\_1 \ldots B\_k $$
as the telescoping sum of $k$ expressions of the form
$$ A\_1 \ldots A\_{i-1} (A\_i - B\_i) B\_{i+1} \ldots B\_k$$
and then estimate each term of the latter separately by various standard inequalities, e.g.
$$ \|AB\|\_{op} \leq \|A\|\_{op} \|B\|\_{op}.$$
This will give some bound on the norm of the difference of the product in terms of the norms of the individual differences $A\_i - B\_i$.
A similar device will also control $A\_1 \ldots A\_{n-1} B\_{n-1}^{-1} \ldots B\_1^{-1} - I$.
| 10 | https://mathoverflow.net/users/766 | 32233 | 20,928 |
https://mathoverflow.net/questions/32196 | 5 | Let all schemes below be excellent.
Let $X\_0$ be a regular (not necessarily smooth, projective) non-empty scheme of finite type over the generic point $\eta$ of a regular connected scheme $S$. As the answers to my question
[For a morphism f from a regular scheme, should there exist an open subscheme U of the target such that fibre of f at each point of U is regular](https://mathoverflow.net/questions/32083/for-a-morphism-f-from-a-regular-scheme-should-there-exist-an-open-subscheme-u-of)
show, there does not have to exist a dense open $U\subset S$ such that $X\_0$ possesses a fibrewise regular model over $U$. Yet, should there always exist a pseudo-finite dominant morphism $j:U\to S$ and some model $X$ of $X\_0$ over $S$ such that $U$ is regular and the reduced scheme associated to $X\_U$ is smooth over $U$? Can we assume that the morphism $U\to S$ is radiciel?
Upd.: it is probably sufficient for my purposes to find such a $U$ for a variety $X\_0$ that is smooth over the perfect closure of $\eta$. It seems that (the first) BCnrd comment solves the problem; thanks!!
| https://mathoverflow.net/users/2191 | Model of a scheme regular over the generic point | To synthersize a bit: let $S$ be an excellent (quasi-excellent is enough) integral scheme, let $X\_0$ be a regular scheme of finite type over $K=K(S)$. I will suppose $X\_0$ separated to avoid possible pathologies.
1. There exists a separated integral noetherian scheme $X$ over $S$ with generic fiber isomorphic to $X\_0$.
2. The singular locus of $X$ is closed (excellence), its projection in $S$ is constructible (Chevalley) and does not contain the generic point of $S$, therefore is contained in a proper closed subset $F$ of $S$. The scheme $X\_U$, where $U=S \setminus F$, is a regular model of $X\_0$ over $U$ (or $S$).
3. If $(\overline{X\_0})\_{\rm red}$ is smooth, then, as explained by BCnrd, there exits a radicial finite flat (hence dominant) morphism $U\to S$ with $U$ integral such that $(X\_U)\_{\rm red}\to U$ is smooth.
[**Edit**] The assumption of smoothness is also **necessary** if $(X\_U)\_{\rm red}\to U$ is smooth for some dominant morphism $U\to S$ (just consider the generic fiber).
4. The above assumption of smoothness is essentially sharp [**Edit**] if we want to have a quasi-finite (i.e. finite type with finite fibers) and dominant base change $U\to S$ such that $(X\_U)\_{\rm red}\to U$ is fiberwise regular. Consider the example $S={\rm Spec} (k[t])$ with $k$ perfect of characteristic $p>2$, and let $X\_0$ be the regular affine curve over $K=k(t)$ defined by the equation $y^2=x^p-t$. Then $X\_0$ is geometrically integral but not smooth over $K$. Let $X\to S$ be a model of $X\_0$ (i.e. finite type separated morphism with generic fiber isomorphic to $X\_0$). Shrinking $S$ if necessary, we can suppose that $X\to S$ is flat with geometrically integral fibers ([EGA], IV.9 or directly compare with the obvious projective model associated to the equation $y^2=x^p-t$ over $S$) [**Edit**] and that the Zariski closure in $X$ of the non-smooth point $(y, x^p-t)\in X\_0$ meets every fiber. Let $U\to S$ be any quasi-finite dominant morphism $U\to S$ (with $U$ integral). Then $X\_U$ is integral. But none of the closed fibers of $X\_U\to U$ is regular because it would be smooth (any finite extension of $k$ is perfect) and then $X\_0$ would be smooth at $(y, x^p-t)$, contradiction.
| 6 | https://mathoverflow.net/users/3485 | 32234 | 20,929 |
https://mathoverflow.net/questions/32147 | 10 | Let $G$ be a locally compact group and let $A(G)$ be the http://eom.springer.de/f/f120080.htm>Fourier algebra of $G$, which we view as the predual of the group von Neumann algebra $\mathcal M(G)$. Let $MA(G)$ be the space of multipliers of $A(G)$, i.e., $\varphi \in MA(G)$ if and only if $\varphi \psi \in A(G)$ for all $\psi \in A(G)$. Then $\varphi \in MA(G)$ induces a bounded operator $m\_\varphi: A(G) \rightarrow A(G)$, and hence also a bounded operator $M\_\varphi = m\_\varphi^\*$ on $\mathcal M(G)$.
$M\_\varphi$ is completely bounded if $\| M\_\varphi \|\_{CB} = \sup\_n \| M\_\varphi \otimes {\rm id}\_n \| < \infty$, where ${\rm id}\_n$ is the identity operator on the $n \times n$ matrices $\mathbb M\_n(\mathbb C)$. $M\_\varphi$ is $n$-positive if $M\_\varphi \otimes {\rm id}\_n$ takes the positive cone $\mathcal M(G)\_+$ into itself, or equivalently $\| M\_\varphi \otimes {\rm id}\_n \| = \varphi(e)$. $M\_\varphi$ is completely positive if it is $n$-positive for every $n \in \mathcal N$.
A well known result is that $G$ is amenable if and only if $A(G)$ has an approximate unit $\{ \varphi\_k \}\_k$ such that $M\_{\varphi\_k}$ is completely positive for all $k$. Haagerup showed that $SL\_2(\mathbb R)$, and all of its lattices have the completely bounded approximation property: For these groups, $A(G)$ has an approximate unit $\{ \varphi\_k \}\_k$ such that $\sup\_k \| M\_{\varphi\_k} \|\_{CB} < \infty$, (in fact he showed that this supremum can be 1 for $SL\_2(\mathbb R)$, and all of its lattices). To contrast, he also showed that $SL\_m(\mathbb R)$, and all of its lattices do not have the completely bounded approximation property whenever $m \geq 3$. De Canniere and Haagerup have also shown that free groups have the $n$-positive approximation property for every $n \in \mathbb N$: For these groups, given any $n \in \mathbb N$, $A(G)$ has an approximate unit $\{ \varphi\_k \}$ of compactly supported functions such that $\varphi\_k$ is $n$-positive.
Recently, I was at a conference and Mikael de la Salle asked me if I knew of any examples of groups which do not have the $n$-positive approximation property. I do not and so I thought I would ask here.
1) What is an example of a group which for some $n$ does not have the $n$-positive approximation property?
2) What is an example of a group which for any $n$ does not have the $n$-positive approximation property?
3) Are there groups which have the $n$-positive approximation property for some $n$, but not for $n + 1$?
| https://mathoverflow.net/users/6460 | Examples of groups without the n-positive approximation property | Haagerup actually proved (lattices of) higher rank Lie groups do not have $1$-positive approximation property, nor bounded approximation property, i.e., there is no uniformly bounded sequence of compactly supported multipliers that converges pointwise to $1$. (It's still open whether the reduced group C$^\*$-algebra of such a group also fails bounded approximation property.) Also, lamplighter groups on non-amenable groups do not have $1$-positive approximation property (bounded approximation property with constant $1$). The reason is that the proofs of no $1$-positive approximation property boil down to non-existence of certain types of multipliers on an amenable group; and for multipliers $M\_{\varphi}$ on an amenable group, cb-norm coincides with norm. (See Cowling et al., Duke Math. J. 127 (2005), 429--486 for a survey.) Problem 3 seems inaccessible.
| 13 | https://mathoverflow.net/users/7591 | 32247 | 20,939 |
https://mathoverflow.net/questions/32137 | 10 | If R is a commutative noetherian ring, M and N are modules with M finite. It is well known in commutative algebra that $AssHom\_R(M,N)=Supp(M)\cap Ass(N)$. But I want to know whether there is a formular for $AssExt\_R^i(M,N) ?$ Thanks!
| https://mathoverflow.net/users/5775 | The associated prime ideals of $Ext^i_R(M,N)$ | The short answer is no, as hinted at in the comments by Karl and Graham. I would argue that even the question of understanding the minimal primes of $\text{Ext}^i(M,N)$ (which is the minimal set of the associated primes, hence an easier question) is intractable. Let's assume that $R$ is Noetherian and $M,N$ are finitely generated.
Since $\text{Ext}$ localizes, one essentially needs to understand when $\text{Ext}^i(M,N)$ vanishes over a local ring $R$. There is no good answer in general. Even in the very nice case when $N$ is the canonical module (assuming it exists), these $\text{Ext}$ are Matlis dual to the local cohomology modules of $M$, and while there are good bounds on the indices when they vanish (below the depth and beyond the dimension), there are no general result which can tell you exactly when.
Here is one more way to see the complexity of the problem even when $i=1$. Take a free cover of $M$ to get an exact sequence $0 \to M\_1 \to F \to M \to 0$. Applying $\text{Hom}(-,N)$ to get:
$$0 \to \text{Hom}(M,N) \to \text{Hom}(F,N) \to \text{Hom}(M\_1,N) \to \text{Ext}^1(M,N) \to 0$$
Assuming some mild condition on $M,N$ (reflexive for example), this shows that the set of associated prime of $\text{Ext}^1(M,N)$ is a subset of the set of primes $p$ such that $\text{depth}(\text{Hom}(M,N)\_p) = 2$. Again, this set is not very well understood in general.
| 5 | https://mathoverflow.net/users/2083 | 32249 | 20,940 |
https://mathoverflow.net/questions/32245 | 9 | One of my favorite theorems is that of Fáry-Milnor, stating that the total curvature of a knot in $\mathbb R^3$ which is not an unknot (an ununknot) is at least $4\pi$.
Can one quantify the way in which knottedness forces an increase in total curvature?
| https://mathoverflow.net/users/1409 | The total curvature of very knotty knots | Fields medalist Michael Freedman got involved with knots using a very simple technique, assign a sort of energy integral that becomes infinite if there is a genuine self-crossing, that is if you try to force the curve to change isotopy class. The results relate, at least, to Ryan's comment "the figure 8 knot is twice as knotted as the trefoil".
<http://www.jstor.org/pss/2946626>
Excerpts from a column by Ian Stewart...the quoting process does not seem to have rendered the mathematical symbols very well, and I do not know what the letter p means, but here is the link:
<http://www.fortunecity.com/emachines/e11/86/knotprob.html>
>
> But it now looks as if the most
> interesting "energy" concept for knots
> is not elastic, but electrostatic, as
> suggested in 1987 by S. Fukuhara of
> Tsuda College, Tokyo. Imagine the knot
> to be a flexible wire of fixed length,
> which can pass through itself if
> necessary and which has a uniform
> electrostatic charge along its length.
> Because like charges repel each other,
> a knot that is free to move will
> arrange itself so as to keep
> neighbouring strands as far apart as
> possible in order to minimise its
> electrostatic energy. This minimum
> energy value is the invariant.
>
>
> But is it a useful one? Does it have
> simple, natural properties that
> mathematicians can exploit? In 1991,
> Jun O'Hara of Tokyo Metropolitan
> University proved that the minimum
> energy of a knot really does increase
> as the knot becomes more complicated.
> Only a finite number of topologically
> different knots exist with energy less
> than or equal to any chosen value.
> This means that topological types of
> knots can be "ordered" in terms of
> their energy. There is a natural
> numerical scale of complexity for
> knots, ranging from simple knots at
> the low energy end to more complicated
> ones higher up.
>
>
> What are the simplest knots? In the
> most recent issue of the Bulletin of
> the American Mathematical Society, a
> team of four topologists - Steve
> Bryson of NASA's Ames Research Center
> in California, Michael Freedman and
> Zhenghan Wang of the University of
> California at San Diego, and Zheng-Xu
> He of Princeton University- prove that
> the simplest knots are exactly what
> you would expect. They are "round
> circles"- that is, circles in the
> everyday sense. Topologists, whose
> "circles" are usually bent and
> twisted, have to append an adjective
> to remind themselves when, as in this
> case, they are not.
>
>
> The energy of a round circle is 4, and
> all other closed loops have higher
> energy. Any loop with energy less than
> $6 \pi + 4$ is topologically unknotted - it
> is a bent circle. More generally, a
> knot with $c$ crossings in some
> two-dimensional picture has energy at
> least $2 \pi c + 4,$ though this bound is
> probably not the best possible, as the
> lowest known energy for an overhand
> knot is about 74. The number of
> topologically distinct knots of energy
> $E$ is at most $(0.264) x (1.658)^E .$
>
>
>
| 10 | https://mathoverflow.net/users/3324 | 32250 | 20,941 |
https://mathoverflow.net/questions/32255 | 7 | In a joint paper that I am working on, we are interested in taking the intersection product $[X] \cap [Y]$ of the fundamental classes of two compact, oriented pseudomanifolds $X$ and $Y$ in a compact, oriented manifold $M$. Now, the usual intersection product takes values in $H\_\*(M)$, but I need an intersection product that takes values in $H\_\*(Y)$. I think that if $X$ and $Y$ are favorably stratified, whatever that means, then such an intersection product should exist. My question is, what is a good approach for such a construction? What is a good proof that it's well-defined? What is a good reference?
The intuitive idea is to wiggle $X$ generically to make it transverse to $Y$, and then take transverse intersections along strata. But there are various different rigorous frameworks that you could use to prove that this is well-defined.
For example, if $X$, $Y$, and $M$ are all PL, then one way to define the right thing is to restrict the Poincaré dual $[X]^\*$ to $H^\*(M,M\setminus N) \cong H^\*(N,\partial N)$, where $N$ is a regular neighborhood of $Y$, then take the Poincaré dual of that to obtain an element of $H\_\*(N) \cong H\_\*(Y)$. This way to define the intersection is based on very standard ideas. You can prove that it works is to use the fact that $N$ is unique up to a relevant isotopy.
In the case that I/we need, $M$ is a smooth complex projective variety and $X$ and $Y$ are singular complex subvarieties. Now, it's easy to suppose that the definition and proof from the PL case still work using the fact that $X$ and $Y$ are Whitney-stratified. However, I no longer know how standard the argument is. For all I know, a regular neighborhood argument is not as standard as a direct argument with transverse position. Or, for all I know, $X$ and $Y$ don't have to be Whitney-stratified; maybe they can be much more general. Certainly it seems like window dressing to demand that they be complex projective.
| https://mathoverflow.net/users/1450 | Intersection product in a manifold, taking values in one factor | I don't think you have to get involved with strata or transversality at all. Poincare duality (Edit: and cap products) will do all the work. Let's assume:
$M$ is a compact oriented $n$-manifold.
$\xi$ is a $p$-dimensional homology class of $M$. (We don't care if it comes as the fundamental class $[X]$ of some pseudomanifold in $M$.)
$Y$ is a subset of $M$ and $[Y]$ is an element of $H\_q(Y)$. (We don't care if $Y$ is a pseudomanifold, or $q$-dimensional, or anything.)
Then duality in $M$ yields an element $\xi^\*\in H^{n-p}(M)$. Restrict it to get an element of $H^{n-p}(Y)$. Cap with $[Y]$ to get an element of $H\_{p+q-n}(Y)$.
(Final paragraph now simplified after comment from Greg.)
| 14 | https://mathoverflow.net/users/6666 | 32258 | 20,946 |
https://mathoverflow.net/questions/32269 | 11 | Consider a game where one player picks an integer number between 1 and 1000 and
other has to guess it asking yes/no questions.
If the second player always gives correct answers than it's clear that in worst
case it's enought to ask 10 questions. And 10 is the smallest such number.
What if the second player is allowed to give wrong answers? I'm interested in a
case when the second player is allowed to give at most one wrong answer.
I know the strategy with 15 guesses in worst case.
Consider a number in range [1..1000] as 10 bits. At first you ask the values of
all 10 bits ("Is it true that $i$-th bit is zero?"). After that you get some number.
Ask if this number is the number first player guessed.
And if not you have to find where he gave wrong answer. There are 11 positions. Using
the similar argument you can do it in 4 questions.
Is it possible to ask less then 15 questions in worst case?
| https://mathoverflow.net/users/7694 | Guess a number with at most one wrong answer | Yes, there is a way to guess a number asking **14** questions in worst case. To do it you
need a linear code with length 14, dimension 10 and distance at least 3. One such code can be built
based on Hamming code (see <http://en.wikipedia.org/wiki/Hamming_code>).
Here is the strategy.
Let us denote bits of first player's number as $a\_i$, $i \in [1..10]$.
We start with asking values of all those bits. That is we ask the following questions: "is it true that i-th bit of your number is zero?"
Let us denote answers on those questions as $b\_i$, $i \in [1..10]$.
Now we ask **4** additional questions:
Is it true that $a\_{1} \otimes a\_{2} \otimes a\_{4} \otimes a\_{5} \otimes a\_{7} \otimes a\_{9}$ is equal to zero? ($\otimes$ is sumation modulo $2$).
Is it true that $a\_{1} \otimes a\_{3} \otimes a\_{4} \otimes a\_{6} \otimes a\_{7} \otimes a\_{10}$ is equal to zero?
Is it true that $a\_{2} \otimes a\_{3} \otimes a\_{4} \otimes a\_{8} \otimes a\_{9} \otimes a\_{10}$ is equal to zero?
Is it true that $a\_{5} \otimes a\_{6} \otimes a\_{7} \otimes a\_{8} \otimes a\_{9} \otimes a\_{10}$ is equal to zero?
Let $q\_1$, $q\_2$, $q\_3$ and $q\_4$ be answers on those additional questions. Now second player calculates $t\_{i}$ ($i \in [1..4]$) --- answers on those questions based on bits $b\_j$ which he previously got from first player.
Now there are 16 ways how bits $q\_i$ can differ from $t\_i$. Let $d\_i = q\_i \otimes t\_i$ (hence $d\_i = 1$ iff $q\_i \ne t\_i$).
Let us make table of all possible errors and corresponding values of $d\_i$:
position of error -> $(d\_1, d\_2, d\_3, d\_4)$
no error -> (0, 0, 0, 0)
error in $b\_1$ -> (1, 1, 0, 0)
error in $b\_2$ -> (1, 0, 1, 0)
error in $b\_3$ -> (0, 1, 1, 0)
error in $b\_4$ -> (1, 1, 1, 0)
error in $b\_5$ -> (1, 0, 0, 1)
error in $b\_6$ -> (0, 1, 0, 1)
error in $b\_7$ -> (1, 1, 0, 1)
error in $b\_8$ -> (0, 0, 1, 1)
error in $b\_9$ -> (1, 0, 1, 1)
error in $b\_{10}$ -> (0, 1, 1, 1)
error in $q\_1$ -> (1, 0, 0, 0)
error in $q\_2$ -> (0, 1, 0, 0)
error in $q\_3$ -> (0, 0, 1, 0)
error in $q\_4$ -> (0, 0, 0, 1)
All the values of $(d\_1, d\_2, d\_3, d\_4)$ are different. Hence we can find where were an error and hence find all $a\_i$.
| 24 | https://mathoverflow.net/users/7079 | 32270 | 20,954 |
https://mathoverflow.net/questions/32268 | 9 | Does anyone know of a monograph/survey on the modern history of (basic or elliptic) hypergeometric functions and their applications?
I haven't had much time to search the literature, and because it is summer it is hard to reach professors or specialists, which is why I am asking the question here. It is also likely that there are obvious choices out there that I am unaware of because of my ignorance in the field. I would appreciate it a lot if along with the suggestions you could give a quick description of what the book/article treats.
| https://mathoverflow.net/users/2384 | Recent work on hypergeometric functions | Gjergji, there are remarkable articles by Richard Askey:
(1) "Ramanujan and hypergeometric and basic hypergeometric series"
in
[*Russian
Math. Surveys* **45**:1 (1990) 37--86](http://dx.doi.org/10.1070/RM1990v045n01ABEH002325);
reprinted in *Ramanujan: essays and surveys*, Hist. Math. **22**
Amer. Math. Soc., Providence, RI, 2001, pp. 277--324;
(2) "A look at the Bateman project"
in *The mathematical legacy of Wilhelm Magnus: groups, geometry
and special functions* (Brooklyn, NY, 1992), 29--43,
Contemp. Math. **169**, Amer. Math. Soc., Providence, RI, 1994.
(I asked Dick exactly this question, maybe without accenting on "modern theory",
some years ago.) The modern theory is mostly multiple hypergeometric functions
related to root systems; for a nice survey on the roots of these functions,
the Selberg integral, see
(3) P. Forrester and S.O. Warnaar,
"The importance of the Selberg integral",
[*Bull. Amer. Math. Soc. (N.S.)* **45**:4 (2008) 489--534.](http://dx.doi.org/10.1090/S0273-0979-08-01221-4)
Elliptic functions are hypergeometric functions of the 21st century:
(4) V.P. Spiridonov,
"Essays on the theory of elliptic hypergeometric functions",
[*Russian Math. Surveys* **63**:3 (2008) 405--472.](http://dx.doi.org/10.1070/RM2008v063n03ABEH004533)
| 8 | https://mathoverflow.net/users/4953 | 32275 | 20,959 |
https://mathoverflow.net/questions/32276 | 4 | Let $Q$ be a smooth conic (the zero set of a homogeneous degree 2 polynomial)
in $\mathbb{P}^2(\mathbb{C})$ and let $j:Q\rightarrow\mathbb{P}^2(\mathbb{C})$ be the
immersion in the projective plane. How can I compute $j\_\*H\_2(Q,\mathbb{Z})\subseteq H\_2(\mathbb{P}^2(\mathbb{C}),\mathbb{Z})$? (in $H\_i(X,\mathbb{Z})$, i means the i-th singular homology group).
I know that $Q$ is homeomorphic to $\mathbb{P}^1(\mathbb{C})$ so their homology groups are isomorphic, I can restrict to the Veronese embedding $\phi:\mathbb{P}^1(\mathbb{C})\rightarrow\mathbb{P}^2(\mathbb{C})$,
$\phi([x\_0:x\_1])=[{x\_0}^2: x\_0x\_1 :{x\_1}^2]$.
I also note that the map $\phi$ is homotopic to $\psi: \mathbb{P}^1(\mathbb{C})\rightarrow\mathbb{P}^2(\mathbb{C})$, $\psi([x\_0:x\_1])=[{x\_0}^2: 0 :{x\_1}^2]$ that is a line "counted two times" so my guess is that $j\_\*H\_2(Q,\mathbb{Z})$ can be the whole $H\_2(\mathbb{P}^2(\mathbb{C}),\mathbb{Z})$ or it can be the subgroup of $H\_2(\mathbb{P}^2(\mathbb{C}),\mathbb{Z})$ generated by two times its generator. How can I proceed? Is there a simpler way to do it?
Thank you in advance.
| https://mathoverflow.net/users/4971 | Homology of a complex projective conic | The subgroup $j\_\* H\_2(Q)$ must be generated by twice the generator of
$H\_2(P^2(\mathbb{C}))$
(I'm dropping the coefficient group from my notation).
To see this, your map $\psi$ decomposes as the embedding from $P^1$ into
$P^2$ (which induces isomorphism on $H\_2$) composed with the map $[x,y]\to[x^2,y^2]$
on $P^1$. This map has degree $2$ and so acts on $H\_2$ by multiplication by $2$.
To see this, think of it as the squaring map on the Riemann sphere. This
is homotopic to the suspension of the double covering of $S^1$ onto itself.
(That is multiplication by $2$ on $H\_1(S^1)$ so suspends to multiplication by $2$
on $H^2(S^2)$).
All this generalizes to degree $d$ rational curves in $P^2$.
| 4 | https://mathoverflow.net/users/4213 | 32278 | 20,961 |
https://mathoverflow.net/questions/32282 | 4 | Let $\mathfrak{F}(0)$ be the set of all bijections $\mathbb{N}\mapsto\mathbb{N}$, and let $\mathfrak{F}(n+1)$ be the set of all bijections $\mathfrak{F}(n)\mapsto\mathfrak{F}(n)$. Given $\alpha\in\mathfrak{F}(n)$, is there a lower bound on the cardinality of $C\_{\mathfrak{F}(n)}(\alpha)$, the centralizer of $\alpha$? Of course, since $C\_{\mathfrak{F}(n)}(I)=\mathfrak{F}(n)$, where $I$ is the identity, the cardinality of the centralizer can be the same as that of $\mathfrak{F}(n)$. So the question is, is it possible to, using the axiom of choice as necessary, exhibit $\alpha\in\mathfrak{F}(n)$ such that $|C\_{\mathfrak{F}(n)}(\alpha)|<|\mathfrak{F}(n)|$? Or, making things a little more concrete, is it possible to create $f:\mathbb{R}\mapsto\mathbb{R}$, a bijection, such that $|C\_{\mathfrak{F}(1)}(f)|<|2^{\mathbb{R}}|$?
| https://mathoverflow.net/users/6856 | What is the cardinality of the centralizer of a bijection? | As far as I understand, the answer is negative (assuming $n>0$). It is easy to prove the following:
**Claim.** Let $X$ be an uncountable set. Let $G$ be the group of all bijections $X\to X$. Then for any
$g\in G$, its centralizer $Z(g)\subset G$ satisfies $|Z(g)|=|G|$. (This implies the claim, since all of $\mathfrak{F}(n)$ are uncountable.)
**Proof.** It suffices to show that $|Z(g)|\ge|G|$. Consider the partition of $X$ into orbits under the action of $g$. Each orbit is identified with a cyclic group ${\mathbb Z}/n{\mathbb Z}$ ($n\ge 0$) with $g$ acting as the shift by one. Since there are countably many isomorphism types of orbits, and each has countably many elements, there exists a particular $n\ge 0$ such that the corresponding set of orbits $X\_n$ of this type has $|X\_n|=|X|$. Now it is easy to see that the map from $Z(g)$ to the bijections of $X\_n$ is surjective. (And hence
the cardinality of $Z(g)$ is at least the cardinality of the group of bijections of $X\_n$.)
| 4 | https://mathoverflow.net/users/2653 | 32284 | 20,963 |
https://mathoverflow.net/questions/32287 | 13 | Suppose that we are given a nice space $X$ and a sheaf of abelian groups $F$ on $X$. Fix an integer $n$. Then We have a contravariant functor from nice spaces over $X$ to abelian groups; Namely, to a space $f: Y \to X$ we associate the abelian group $H^n (Y, f^{\*}F)$ (Sheaf cohomology).
If $X$ is a point, Then this functor is represented, in the homotopy category, by the Eilenberg-Maclane space $K(F,n)$.
My questions are:
1) Can one formalize what will it mean for our functor to be representable, "homotopically"? I am not very sure, but I suspect that the most naive definition of homotopy category of spaces over $X$, and the requirement that the functor is representable in this category, are not right (and I did not check that the functor actually factors to this "homotopy" category).
2) Is it representable?
| https://mathoverflow.net/users/2095 | Representing cohomology of a sheaf à la Eilenberg-Maclane | Sheaf cohomology over $X$ is representable in the homotopy category of oo-stacks over X / spaces over X, yes.
Details, links and references are at <http://ncatlab.org/nlab/show/cohomology>
| 8 | https://mathoverflow.net/users/381 | 32291 | 20,966 |
https://mathoverflow.net/questions/32290 | 2 | I ran into the following algorithmic problem while experimenting with classification algorithms. Elements are classified into a polyhierarchy, what I understand to be a poset with a single root ("largest" element), please correct me if I am mistaken. I have to solve the following problem, which looks a lot like the [set cover problem](http://en.wikipedia.org/wiki/Set_cover_problem).
Let $L$ be a directed, acyclic, unweighted graph of $n \in \mathbb{N}$ vertices $V = \{v\_1, v\_2, \dots, v\_n\}$ with edges $E = \{(v\_i, v\_j) | v\_i, v\_j \in V\}$. $L$ has a poset structure. There is one designated "root" vertex and a direction "down" the graph, following the direction of the edges. Basically a tree where all non-root vertices may have multiple parents.
Let $G = \{g\_1, g\_2, \dots, g\_m\} \subseteq V$ be a subset of $V$ containing vertices to be covered. For each vertex $v \in V$, let $\sigma(v)$ be the verticies of the sub-graph rooted at $v$.
For a given $k \in \mathbb{N}$, $k \leq m \leq n$, select $S = \{s\_1, s\_2, \dots, s\_l\} \subseteq V$, $l \geq k$ such that:
1. $G \subseteq \bigcup\_{i=1}^{l} \sigma(s\_i)$ ($S$ covers $G$).
2. $\forall s\_i, s\_j \in S, i \neq j: \sigma(s\_i) \nsubseteq \sigma(s\_j)$ (no redundant vertices in $S$).
3. $\nexists S' \subseteq V$ that satisfies 1 with $k \leq |S'| < l$ (minimal).
Note that 2 is not implied by 3 because of the lower bound $k$. If we have to to drop 3 in order to get an efficient algorithm, we still want to keep 2.
This problem will in general not have a solution for $l=k$, which is why we need the additional variable $l$: Say you choose $k=2$ and $V$ with $|V|=4$ has the form of a root node with 3 children. Let $G$ be these 3 children. This is solvable for $l=3$ but not for $l=k=2$.
Finding an optimal solution (satisfying 3) may be hard. This looks a lot like the set cover problem (which is NP-hard), but really is a special case of it, given the special structure of the subsets.
Devising an approximation algorithm that satisfies 1 & 2 is quite easy, just start at the vertices of $G$ and "walk up" or start at the root and "walk down". Say you start at the root, iteratively expand vertexes and then remove unnecessary vertices until you have at least $k$ sub-graphs. The approximation bound depends on the number of children of a vertex, which is OK for my application.
However I would be very interested in proofing that this problem is NP-hard. So far my tries with set cover and knapsack failed. Does anyone have a hint which NP-hard problem would lend itself to a reduction? Or maybe the problem isn't NP-hard after all, given the special structure of the subsets?
This is my first post on MathOverflow, so I hope I chose the right tags, please feel free to improve my tagging.
| https://mathoverflow.net/users/7702 | Selecting k sub-posets | You seem to rely on a notion of a vertex preceding another (you use the terms "lattice" and "polyhierarchy", and refer to the direction "down"). So the edge relation $E$ appears to be transitive, forming a strict partial order.
To show why the $k$ parameter is important, you suggest an example where the target set $G$ presumably contains the three children of the root $\mu$. In this case, $k=1$ would clearly allow $S = \lbrace \mu \rbrace$. On the other hand, setting $k=2$ would force each of the children to be in $S$, to avoid condition 2.
This gives the key to the reduction of SET COVER to the decision version of your problem. Given is a set of subsets of a finite universe, and an integer $l$. The problem is to determine whether one can find at most $l$ subsets which cover the entire universe. For each subset create a vertex, and let $E$ be the subset relation between subsets, expressed in terms of the vertices. If the universe itself is one of the subsets, then let $k=1$ (note that in this case the problem is quite trivial, the solution $S$ will simply contain this one element). Otherwise let $k=2$ and add a new root vertex denoting the universe, with edges to every other vertex. Finally, let $G$ be the set of all minimal vertices (corresponding to all minimal subsets).
This instance of your problem has a solution (a set of at most $l \ge k$ vertices, covering all minimal vertices) if, and only if, the set cover instance has a solution.
I hope I have managed to capture your problem correctly.
| 2 | https://mathoverflow.net/users/7252 | 32300 | 20,969 |
https://mathoverflow.net/questions/32133 | 28 | Suppose $A\in R^{n\times n}$, where $R$ is a commutative ring. Let $p\_i \in R$ be the coefficients of the characteristic polynomial of $A$: $\operatorname{det}(A-xI) = p\_0 + p\_1x + \dots + p\_n x^n$.
I am looking for a proof that
$-\operatorname{adj}(A) = p\_1 I + p\_2 A + \dots + p\_n A^{n-1}$.
In the case where $\operatorname{det}(A)$ is a unit, $A$ is invertible, and the proof follows from the Cayley–Hamilton theorem. But what about the case where $A$ is not invertible?
| https://mathoverflow.net/users/7667 | Expressing $-\operatorname{adj}(A)$ as a polynomial in $A$? | Here is a direct proof along the lines of the standard proof of the Cayley–Hamilton theorem. [*This works universally, i.e. over the commutative ring $R=\mathbb{Z}[a\_{ij}]$ generated by the entries of a generic matrix $A$.*]
The following lemma combining Abel's summation and Bezout's polynomial remainder theorem is immediate.
**Lemma** Let $A(\lambda)$ and $B(\lambda)$ be matrix polynomials over a (noncommutative) ring $S.$ Then $A(\lambda)B(\lambda)-A(0)B(0)=\lambda q(\lambda)$ for a polynomial $q(\lambda)\in S[\lambda]$ that can be expressed as
$$q(\lambda)=A(\lambda)\frac{B(\lambda)-B(0)}{\lambda}+\frac{A(\lambda)-A(0)}{\lambda}B(0)=A(\lambda)b(\lambda)+a(\lambda)B(0) \qquad (\*)$$
with $a(\lambda),b(\lambda)\in S[\lambda].$
Let $A(\lambda)=A-\lambda I\_n$ and $B(\lambda)=\operatorname{adj} A(\lambda)$ [*viewed as elements of $S[\lambda]$ with $S=M\_n(R)$*], then
$$A(\lambda)B(\lambda)=\det A(\lambda)=p\_A(\lambda)=p\_0+p\_1\lambda+\ldots+p\_n\lambda^n$$ is the characteristic polynomial of $A$ and $$A(0)B(0)=p\_0 \text{ and } q(\lambda)=p\_1+\ldots+p\_n\lambda^{n-1}$$
Applying $(\*),$ we get
$$q(\lambda)=(A-\lambda I)b(\lambda)-\operatorname{adj} A \qquad (\*\*) $$
for some matrix polynomial $b(\lambda)$ commuting with $A.$ Specializing $\lambda$ to $A$ in $(\*\*),$ we conclude that
$$q(A)=-\operatorname{adj} A\qquad \square$$
| 25 | https://mathoverflow.net/users/5740 | 32303 | 20,971 |
https://mathoverflow.net/questions/32262 | 4 | The endomorphisms of an abelian group form a ring under pointwise group operation and composition. Every ring is isomorphic to a subring of the endomorphism ring of some abelian group (left module over itself).
Is every ring isomorphic to the endomorphism ring of some abelian group? (not just a subring)
| https://mathoverflow.net/users/nan | Representation of rings | Generally it's difficult to characterize rings isomorphic to an endomorphism ring of an abelian group. Interest in such problems was sparked by a problem given by Fuchs in his widely-read monograph *Abelian Groups*, cf. the excerpt below from the introduction to the [paper [1]](http://dx.doi.org/10.1016/j.jalgebra.2006.01.036)
>
> The notion of an E-ring goes back to a
> seminal paper of Schultz [20] written
> in response to Problem 45 in the
> well-known book `Abelian Groups' by
> Laszlo Fuchs [11]. In this paper
> Schultz distinguished between two
> possibly different approaches, the
> first we will continue to call an
> E-ring, while the second we shall
> refer to as a generalized E-ring. Thus
> a ring R is said to be an E-ring if R
> is isomorphic to the endomorphism ring
> of its underlying additive group, R+,
> via the mapping sending an element r
> $\in$ R to the endomorphism given by
> left multiplication by r, whilst R is
> a generalized E-ring if some
> isomorphism, not necessarily left
> multiplication, exists between R and
> its endomorphism ring End(R+). Since
> right multiplication is always an
> endomorphism, it is not difficult to
> see that E-rings are necessarily
> commutative. The existence of a
> non-commutative generalized E-ring has
> recently been established [15], and so
> it follows that the class of
> generalized E-rings is strictly larger
> than the class of E-rings.
>
>
> Since Schultz's original paper there
> has been a great deal of interest in
> E-rings and some natural
> generalizations, see e.g.
> [1,2,4,6,8-10,17,19,21]. A notable
> feature of much of this recent work
> has been the use of so-called
> realization theorems, whereby a
> cotorsion-free ring is realized, using
> combinatorial ideas derived from
> Shelah's Black Box - see e.g. [7] for
> details of this technique - as the
> endomorphism ring of an Abelian group.
> This present work arose from an
> observation of the second author in
> response to a question from the first
> about the existence of generalized
> E-algebras over the ring $J\_p$ of
> p-adic integers; see [16] for further
> details. A natural question which
> arises, is to what extent is it
> necessary for a ring to be
> cotorsion-free in order to be a
> generalized E-ring and the principal
> objective of this work is to
> characterize generalized E-rings
> `modulo cotorsion-free groups.' The
> characterization is quite elementary
> but seems to have been overlooked
> heretofore. It should be noted that
> Bowshell and Schultz showed in [2]
> that a reduced cotorsion E-ring has
> the form $\prod\_{p \in U} {\mathbb
> > Z}(p^{k\_p}) \oplus \prod\_{p\in V} J\_p$
> where $U,V$ are disjoint sets of
> primes.
>
>
>
[1](http://dx.doi.org/10.1016/j.jalgebra.2006.01.036) R. Gobel, B. Goldsmith.
Classifying E-algebras over Dedekind domains
Jnl. Algebra, Vol. 306, 2006, 566-575
| 2 | https://mathoverflow.net/users/6716 | 32305 | 20,972 |
https://mathoverflow.net/questions/32296 | 18 | It is well-known that topological K-theory is blessed with the Bott periodicity theorem, which specifies an isomorphism between $K^2(X)$ and $K^0(X)$ (where $K^n$ is defined from $K^0$ by taking suspensions). I am wondering if other generalized cohomology theories have their own periodicity theorems, and if there is a general framework for conceptualizing them. I am interested in any substantive answer to this question, but there are two specific avenues for generalization that I am particularly curious about.
The first avenue begins with the Clifford algebra approach to Bott periodicity. This approach relates periodicity in K-theory to a certain natural periodicity present in the theory of complex Clifford algebras, and it generalizes the 8-fold periodicity of real K-theory (corresponding to an 8-fold periodicity in real Clifford algebras). Can one fruitfully generalize the notion of a Clifford algebra, associate to it a generalized cohomology theory, and analogously produce a periodicity theorem?
The second avenue involves Cuntz's proof of Bott periodicity for C\*-algebras (which in particular implies topological Bott periodicity by specializing to commutative C\*-algebras). Cuntz proves Bott periodicity for any functor from the category of C\*-algebras to the category of Abelian groups which is stable (i.e. insensitive to tensoring with the C\*-algebra of compact operators on Hilbert space), half exact, and homotopy invariant. The proof uses topological properties of Toeplitz algebras in an essential way. Because of the generality of his approach, I am left wondering if the essential features of his argument can be translated into more general contexts.
Any ideas are welcome!
| https://mathoverflow.net/users/4362 | Periodicity theorems in (generalized) cohomology theories | In the spirit of first approach, there is a conjecture for a Clifford-algebra type proof of the 576-fold periodicity of TMF. This is a generalized cohomology theory constructed by piecing together all the elliptic cohomology theories together in a suitable way. I heard about this conjecture from Andre Henriques, who is working on a geometric approach to TMF using conformal nets.
The idea is that the free fermion conformal net (a introduction can be found in [this article](http://arxiv.org/abs/0912.5307) by Bartels, Douglas and Henriques) is to TMF as the Clifford algebras are to K-theory. For a suitable sense of equivalence, i.e. some generalization of Morita equivalence, $Free(n)$ and $Free(n+576)$ should be equivalent. I believe people are still far from a proof, but the motivation comes from looking at orientations: a manifold is orientable for K-theory if the frame bundle (a principal $SO(n)$-bundle) lifts to a principal $Spin(n)$-bundle. The $Spin(n)$ groups can be defined as a group in the Clifford algebra. For TMF, a manifold is orientable if the frame bundle extends to a principal $String(n)$-bundle, which can be obtained from $Spin(n)$ by killing $\pi\_3$, just like $Spin(n)$ is obtained from $SO(n)$ by killing $\pi\_1$. There is a way to define $String(n)$ using the free fermion conformal nets.
The only reference I know for these ideas is the following [summary](http://www.staff.science.uu.nl/~henri105/Grants/CNTMF-B2.pdf).
| 14 | https://mathoverflow.net/users/798 | 32314 | 20,977 |
https://mathoverflow.net/questions/32311 | 16 | The representation theory of finite groups over the complex numbers is classical und it is usually quite easy to compute the set of isomorphism classes of irreducible representations, at least for small examples. Now, sometimes one is not content with the representation theory over the complex numbers or even over any field, but one wants to consider representations over $\mathbb{Z}$ or $\mathbb{Z}\_{(p)}$. At least for the non-expert it is not so easy to obtain a complete list of isomorphism classes of indecomposable representations in these cases even for small examples. So, what I want to ask is the following:
1) Can one formulate a "guide" how to obtain such a list?
2) Is there a place in the literature where a list of indecomposables/irreducibles is given for some small examples as $S\_3$ (the symmetric group on three elements)? In the last example I am especially interested.
| https://mathoverflow.net/users/2039 | Explicit integral representation theory | You might be asking about four separate types of modules:
* irreducible Z[G] modules,
* Z-forms of irreducible Q[G] modules,
* indecomposable Z[G] modules, or
* indecomposable Z[G] modules that are finitely generated and free as Z-modules.
I'll assume the last is the main concern.
The irreducible modules of ZS3 are all finite and have an elementary abelian p-group as their additive group. For p=2,3 there are 2 each, and for p>3, there are 3 each.
The irreducible CS3 modules are all realizable over Q. Every such module may be realized over Z, but the two-dimensional representation has two distinct Z-forms, giving four total "irreducible" Z-free ZS3 modules, that is, four total Z-forms of irreducible QS3 modules.
Indecomposable ZS3 modules up to isomorphism are more complicated than the human mind can possibly comprehend. Indeed, even those in which S\_3 acts as the identity are much too complex. Luckily they divide up into several types: annihilated by a prime p (then classified by modular representation theory), torsion (more complicated, but basically now p-adic integral reps), Gorenstein projective (Z-free, so covered in the next bullet point), or madness (that is, the rest).
The indecomposable ZS3 modules that are free as Z-modules are classified in:
Lee, Myrna Pike. "Integral representations of dihedral groups of order 2p."
Trans. Amer. Math. Soc. 110 (1964) 213–231.
[MR 156896](http://www.ams.org/mathscinet-getitem?mr=156896)
[doi:10.2307/1993702](http://dx.doi.org/10.2307/1993702)
There are 10 of them, and the Krull-Schmidt theorem fails for them. Not only are indecomposables not completely reducible, the decomposition of a finitely generated Z-free module into indecomposable summands is not unique. In other words, integral representations of even very small groups are quite complicated.
| 25 | https://mathoverflow.net/users/3710 | 32321 | 20,980 |
https://mathoverflow.net/questions/32316 | 2 | This is again a question about forcing. Start in $L$, the constructible universe. CH holds. Let $\lambda$ be an inaccessible cardinal, also let $\lambda$ > $\aleph\_0$. For each $\alpha < \lambda$, let $P\_\alpha$ be the set of all functions such that $dom(p\_\alpha) \subset \aleph\_0$, $|dom(p\_\alpha)|<\aleph\_0$ and $ran(p\_\alpha) \subset \alpha$. $p\_\alpha$ is stronger than $q\_\alpha$ iff $p\_\alpha$ extends $q\_\alpha$.
Now consider $(P,<)$ be the $\kappa$-product of the $P\_\alpha$, $\alpha<\lambda$. Now the conditions of $P$ are functions taking their argument on the set of all subsets of $\lambda \cdot \aleph\_0$ such that the cardinality of the domain of $p$ is strictly smaller than $\aleph\_0$ and such that $p(\alpha,\xi)<\alpha$ for each pair $(\alpha,\xi) \in dom(p)$.
If $G$ is a generic set of conditions then let for each $\alpha$, $G\_\alpha$ be the projection of $G$ on each $P\_\alpha$, each $G\_\alpha$ is a generic filter, let $f\_\alpha= \bigcup G\_\alpha$ is a function fro, $\aleph\_0$ onto $\alpha$ for every $\alpha < \lambda$ we have $|\alpha| \leq \aleph\_0$.
Since the forcing is $<\aleph\_0$-closed so cardinals and cofinalities are preserved and it satisfies the $\lambda$-chain condition so $\lambda$ is a cardinal in the generic extension $L[G]$.
Now we have $\lambda=\aleph\_0^+$. But in light of the basic fact (that I had overlooked in my previous post), the continuum can't be strong limit. So the above is clearly false. Can you help me point my mistakes?
| https://mathoverflow.net/users/3859 | Collapsing cardinals before the first inaccessible | There are several small mistakes in your question, and one big mistake, leading to your confusion.
First, the small mistakes:
* You say that $\lambda$ is inaccessible and also $\lambda>\aleph\_0$. This is redundant, since the usual definition has that $\lambda$ is an *inaccessible cardinal* if and only if it is an uncountable regular strong limit cardinal.
* You say that $P$ is the $\kappa$-product of the $P\_\alpha$, but what you really mean is that $P$ is the *finite support* $\lambda$-product of the $P\_alpha$ (and this is indeed how you specify it just afterwards, by insisting that conditions have finite support). If you used full support, then you won't get the $\lambda$-chain condition, and $\lambda$ itself would be collapsed.
Second, the big mistake:
* You say "Since the forcing is $\lt\aleph\_0$-closed so cardinals and cofinalities are preserved...," but this is completely wrong. As François mentions, every partial order is $\lt\omega$ closed, since this is just saying that finite descending sequences have lower bounds, which is trivial. And in fact, the forcing $P\_\alpha$ clearly collapses $\alpha$ to $\omega$, and so $P$ collapses all $\alpha\lt\lambda$ to $\omega$. The cardinal $\lambda$ itself is not collapsed, because the forcing $P$ is $\lambda$-c.c., a fact which can be proved by a combinatorial $\Delta$-system argument. This fact is surely the key to understanding the Levy collapse, which is how your forcing is known.
The Levy collapse collapses all cardinals below $\lambda$ to $\omega$ and preserves $\lambda$ itself, thereby making $\lambda$ into the $\omega\_1$ of the forcing extension. There is no need to start in $L$ with this forcing, or with CH. It works quite generally. Furthermore, one needn't collapse to $\omega$; by collapsing all ordinals $\alpha\lt\lambda$ to some other fixed cardinal $\delta$, one forces $\lambda$ to be $\delta^+$ in the extension. For example, one can make $\lambda$ equal to $\aleph\_2$ or $\aleph\_3$ in the extension.
Finally, let me point out that $\lambda$ is indeed the continuum in the extension. A chain condition argument shows that every real of $V[G]$ is added by some stage $V[G\_\alpha]$, where we cut off the forcing at $\alpha\lt\lambda$, and so the reals of $V[G]$ are the union of the reals of $V[G\_\alpha]$, each of which are countable in $V[G]$. So the continuum of $V[G]$ has size $\lambda$, which is now $\omega\_1$. In particular, $V[G]$ satisfies CH, even if the ground model $V$ does not. But of course, $\lambda$ is no longer inaccessible in $V[G]$, since it is a successor cardinal there, and so the objection that the continuum cannot be $\lambda$ in $V[G]$ evaporates.
| 6 | https://mathoverflow.net/users/1946 | 32326 | 20,983 |
https://mathoverflow.net/questions/32308 | 4 | Many books describe how one can construct "by hand" a table of ordinals $1,\ 2,\ \ldots,\ \omega,\ \omega +1,\ \omega +2,\ \ldots,\ \omega\cdot 2,\ \omega\cdot 2 +1,\ \ldots,\ \omega^{2},\ \ldots,\ \omega^{3},\ \ldots\ \omega^{\omega},\ \ldots,\ \omega^{\omega^{\omega}},\ \ldots, \epsilon\_{0},\ \ldots$.
But does this span the entire ordinal class? For some reason I can't seem to prove it. Is there an easy way to see that?
Thanks!
| https://mathoverflow.net/users/5292 | Can one really construct an "ordinal table"? | Since ordinal numbers have a unique division, logarithm and subtraction properties, when given an ordinal $\alpha$ you can write any other ordinal as a finite polynomial in $\alpha$, when $\alpha = \omega$ you get what's known as "Cantor normal form of $\gamma$ for the base $\omega$".
I.e., any ordinal $\gamma$ can be written as a finite sum: $$\gamma = \sum\_{i=0}^n \omega^{\alpha\_i}\cdot\beta\_i$$
Where $\alpha\_i$ is a decreasing chain of ordinals, and $\beta\_i$ is finite.
(More generally, you can take some base $\zeta$ and then $\beta\_i < \zeta$)
Thing is that we only have a finite number of symbols, so at most we can represent (uniquely) a countable number of numbers, since we have a proper class of ordinals, which is a mind boggling concept of infinitude, you obviously can't write them all. But still, any given ordinal can be presented as a finite polynomial in $\omega$, thus spanning the table discussed.
| 6 | https://mathoverflow.net/users/7206 | 32330 | 20,986 |
https://mathoverflow.net/questions/32313 | 0 | Non-uniform circuits, according to my understanding, are those which have different circuit depending on the input size. Constant depth circuit are those whose depth is constant in the input size. So if for example we considered an instance *k* of the complexity class *TC0* which is a non-uniform constant-depth circuit, and let *kn* be the circuit instance at input size *n*, does that mean that the following sentence is allowed or not: *depth(ki) ≠ depth(kj)* when *i≠j*.
In other words, does the non-uniformity of a complexity class allow a constant-depth circuit to have different depths for different input size or not ?
| https://mathoverflow.net/users/7685 | Non-uniform constant-depth circuits | Recall "non-uniform circuits" are represented as an infinite sequence {$C\_n$}, where $C\_n$ is the circuit that will handle $n$-bit inputs. The usual definition of "constant depth" means that the depth may vary from circuit $C\_n$ to circuit $C\_{n'}$, but the depth of $C\_n$ is never larger than a fixed number $d$. That is, the maximum depth of any circuit in the family is a constant, independent of the input size to a circuit.
| 7 | https://mathoverflow.net/users/2618 | 32333 | 20,989 |
https://mathoverflow.net/questions/32320 | 8 | Consider the problem of deciding a language $L$; for concreteness, say that this is the [graph isomorphism problem](http://en.wikipedia.org/wiki/Graph_isomorphism_problem). That is, $L$ consists of pairs of graphs $(G, H)$ such that $G\simeq H$. Now the time complexity of deciding this problem as stated depends on how the graphs are encoded. For example, if one were to have a "canonical" encoding of graphs (such that encoding strings are in bijective correspondence with isomorphism classes of graphs) the problem would be $O(n)$, as we could decide whether $G\simeq H$ simply by comparing the string representing $G$ to the string representing $H$.
On the other hand, if we represent a graph via its adjacency matrix, the best known algorithm (according to Wikipedia) gives only a [subfactorial](http://en.wikipedia.org/wiki/Subfactorial) bound. Now consider the time complexity of converting from one language to another. If we let $T\_1, T\_2$ be the time complexity of deciding languages $L\_1$ and $L\_2$ respectively, and $T\_{ij}$ be the time it takes a Turing machine to take a string $S$ and output another string $S'$ which is in language $j$ if and only if $S$ is in language $i$. We have
$$T\_1\leq T\_2+T\_{12}$$
$$T\_2\leq T\_1+ T\_{21}$$
as given a string that we want to test for its belonging to $L\_i$, we may run it through the translation $L\_i \to L\_j$ and then decide language $j$. Indeed, this is a special case of a trivial "triangle inequality" for translation; the time it takes to translate from $L\_1$ to $L\_2$ plus the time it takes to translate from $L\_2$ to $L\_3$ is greater than or equal to the time it takes to translate from $L\_1$ to $L\_3$. (I say it is a special case because a decision problem is the same as converting a language $L$ to the language $\{ 1 \}$.)
What I want to know is:
>
> Can we better quantify the relationship between the time complexity of a decision problem and the nature of the encoding?
>
>
>
So that this question is not prohibitively vague, let us say that I am looking for (1) related references, and (2) a measure of the complexity of an encoding which more tightly relates to time complexity of the "underlying" decision problem.
---
Added (7/19/2010): The answers below, particularly Ryan Williams' excellent survey of the dependence of the time complexity of various problems on their encoding, get at the motivation to my question but not at my question itself. In particular, it's clear that every problem may be re-encoded to allow (say) $O(\log n)$ time complexity, by padding. My question is whether there's a reasonable way to measure this dependence.
For example, say the decision problem for $L\_1$ is reducible to the decision problem for $L\_2$, and vice versa, so that $L\_1$ and $L\_2$ in some sense represent the same problem. Is there a way to formalize this last statement (about "representing the same problem")? I am imagining, for example, a measure $C\_i$ of the complexity of a language so that if $T\_i$ is the time complexity of the language, and $L\_1$ and $L\_2$ are, say, easily reducible to one another, then $T\_1/C\_1\sim T\_2/C\_2$. (Of course $C\_i=T\_i$ works, but ideally $C\_i$ would be somehow a property of the language, rather than the decision problem.) This is unfortunately becoming quite speculative, so again, related references would be a great answer.
| https://mathoverflow.net/users/6950 | What is the relationship between "translation" and time complexity? | When it comes to the time complexity of problems, the encoding of the problem can be totally crucial. In general, the encoding of the problem cannot be separated from the complexity of the problem itself.
The first canonical example of this (as mentioned before in [answering another question](https://mathoverflow.net/questions/31796/exptime-algorithms/31805#31805)) can be seen with the following two problems:
(1) *Given a deterministic Turing machine $M$, string $x$, and integer $k$ **written in binary**, does $M$ accept $x$ within $k$ steps?*
Problem (1) is $EXPTIME$-complete. However the following problem is $P$-complete:
(2) *Given a deterministic Turing machine $M$, string $x$, and integer $k$ **written as a string of $k$ ones**, does $M$ accept $x$ within $k$ steps?*
So already, the way in which $k$ is represented in an instance completely determines the complexity of the problem. (Note if I wrote $k$ in ternary, $4$-ary, etc., problem (1) remains $EXPTIME$-complete.)
Another interesting example comes from circuit complexity. Consider the following two problems:
(3) *Given a truth table of $2^n$ bits for a function* $f:$ {$0,1$}$^n \rightarrow ${$0,1$}*, return a circuit with AND/OR/NOT gates that computes $f$ and contains a minimum number of gates.*
(4) *Given a function* $f:$ {$0,1$}$^n \rightarrow ${$0,1$} *represented as a circuit with AND/OR/NOT gates, return a circuit that also computes $f$ and contains a minimum number of gates.*
Problem (3) can be easily seen to be in $NP$, since the minimum circuit for $f$ needs at most $O(2^n/n)$ gates, and checking that a given circuit works for $f$ takes $2^{O(n)}$ steps. However (3) is not known to be in $P$, nor is it clear that it's $NP$-complete. The curious status of (3) is discussed in
>
> Valentine Kabanets, Jin-yi Cai: Circuit minimization problem. STOC 2000: 73-79
>
>
>
What about problem (4)? It is *not* known to be in $NP$! It is known to be in $\Sigma\_2 P$ of the polynomial time hierarchy, but not known to be complete for that class. However the version where you use the representation of *formulas* instead of circuits is known to be $\Sigma\_2 P$-complete under Turing reductions:
>
> David Buchfuhrer, Christopher Umans: The Complexity of Boolean Formula Minimization. ICALP (1) 2008: 24-35
>
>
>
Examples of this sort are everywhere in complexity theory, simply because the encoding can really matter if the relative sizes of encodings (or the complexities of encodings) are different enough. Luckily, most "natural" encodings (for which there are polynomial time mappings from one encoding to another) do not seem to affect the overall complexity of a problem (e.g. whether or not a problem is in $NP$). This is another reason why the notion of polynomial time is one of the main focuses in complexity. It is a "robust" notion that isn't affected by whether you use e.g. adjacency lists versus adjacency matrices to represent a graph in your graph problem. Related to this, there is a recent and thought-provoking reference that outlines a complexity theory for succinctly represented graphs (graphs whose adjacency matrices are the truth tables of small size circuits):
>
> Sanjeev Arora, David Steurer, Avi Wigderson: Towards a Study of Low-Complexity Graphs. ICALP (1) 2009: 119-131
>
>
>
Finally, concerning your proposed "isomorphism-respecting" encoding of graphs: while it would be very neat to have, it would not be considered natural, since we don't know how to efficiently obtain such an encoding from any of the other encodings that have already been deemed natural.
**UPDATE TO ADDRESS YOUR REVISED QUESTION:** I think it is a neat idea to try to study "problems" as classes of languages that "represent the same thing" in some strong sense. I'm not aware of significant prior work on this (other than the cheap reply that "all NP-complete problems represent the same thing", which I don't think is what you are driving at). The closest reference I can think of is a related attempt to define "algorithm" in a similar way. See Blass, Dershowitz, and Gurevich's cool paper: <http://research.microsoft.com/en-us/um/people/gurevich/Opera/192.pdf>
| 9 | https://mathoverflow.net/users/2618 | 32336 | 20,991 |
https://mathoverflow.net/questions/32071 | 17 | It seems that there is a common theme in mathematics where, if we want to find out about a category C, then we look at $\hat{C}$ (the category of contravariant functors from $C$ to $Set$). There are all sorts of good reasons for this (Yoneda's lemma being a big one, and the fact that this is a topos). There are other versions, sometimes you look at contravariant functors into some other category, like groups (e.g. for algebraic groups). But I have never seen the dual notion: covariant functors from some other category INTO $C$.
Is this notion as useful as the notion of a presheaf? I guess it's like looking at a category "over" $C$ as opposed to a category "under" it. I guess, hidden in this question is the question: Can one learn anything about a category $C$ by looking at presheaves of $C$? For example, does the difference between presheaves of sets and presheaves of ableian groups tell us anything about the differences between the category of Groups and the category of Sets?
EDIT: There seems to be a bit of confusion of what I mean by "dual of a presheaf." I don't mean "copresheaves" (which I didn't know existed before I asked this), I mean what you get by reversing the arrow of the functor not taking the opposite category. So I'm looking at functors into the category of interest, as opposed to out of them. I can see how this is confusing because usually "dual" doesn't turn around functors, just arrows inside categories. So I guess I mean presheaves in $C^{op}$... (but covariant instead of contravariant?)... who knows.
| https://mathoverflow.net/users/6936 | Is the dual notion of a presheaf useful? | Let me try and take a stab at this question. I will give not so much as an answer to your question, but more of a rambling collection of remarks. The post turned out to be *much longer* than I wanted, because as B. Pascal once complained, I did not have the time to write a shorter one. Expect some inconsistencies, a lot of hand-waving, etc.
To put some order in my thoughts, I will try to argue the following four points. Fix a category $\mathcal{A}$, which is the category "of interest". Then:
(1) Functors into $\mathcal{A}$ are interesting.
(2) Functors into $\mathcal{A}$ tell you everything about $\mathcal{A}$.
(3) Part of the perceived asymmetry follows from the fact that in many cases the interesting category $\mathcal{A}$ is equivalent to a presheaf category (or some localization thereof).
(4) There is not really an asymmetry between functors out of and functors into $\mathcal{A}$, but more of a duality.
The first point is easy to settle, as Qiaochu Yuan already gave a class of interesting examples: for a category of interest $\mathcal{A}$ and any category $\mathcal{I}$, a functor $\mathcal{I}\to \mathcal{A}$ is a diagram of shape $\mathcal{I}$ in $\mathcal{A}$, so hell yeah, functors into $\mathcal{A}$ are important. You may think this a rather pedestrian example, but below I give more examples.
To explain (2), instead of working with categories, that is, in the $2$-category of categories, let us go one level down to the category of sets. A set $X$ is determined completely by its elements, that is, maps $\ast\to X$ where $\ast$ is a singleton set (any one will do: for the sake of determinacy, take the singleton set comprised of the element $\emptyset$). Now in a general category, we cannot speak of elements, but we can (and do) speak of *generalized elements*.
**Definition:** Let $\mathcal{A}$ be a category and $a$ an object of $\mathcal{A}$. A *generalized element* of $a$ is a map $x\colon d\to a$. This is also written symbolically as $x\in\_{d} a$.
We do not need generalized elements to develop category theory, but personally, I found them useful to build upon the intuition gained from working with more "concrete" categories. A very nice discussion of generalized elements is in Awodey's book on category theory. Now the kicker: Yoneda's lemma tells us that if we know all the generalized elements of an object $a$ then we know everything about $a$, including of course, the arrows *out of* $a$ (note: and by duality, if we know all the arrows out of $a$ we will know everything about $a$). Two possible objections may be raised:
1. The original example involves a $2$-category: does not make much of a difference, as my reply to Kevin's comment (see above) still applies. And besides, there are $2$-categorial versions of Yoneda lemma.
2. Yoneda's lemma works ok, but you need the knowledge of all generalized elements and these range over a potentially proper class of objects: that is true, but in virtually every interesting category one can trim down this proper class to a (small) set, even a singleton set. I will not spell out the proper definitions; they are intimately tied with "smallness" conditions and the adjoint functor theorems (and yes, the category of (small) categories satisfies them).
For (3), let $\mathcal{C}^{\mathcal{B}}$ be a functor category. For size reasons, $\mathcal{B}$ has to be small (note: if you have no scientifick problems with creation ex nihilo, you can always spawn a larger universe by invoking the axiom of universes and sidestep this particular size issue). Since functor categories inherit most of the good properties of the codomain category, you will want $\mathcal{C}$ to be as good behaved as possible (e.g. complete, cocomplete, abelian, symmetric monoidal closed, etc.). It is not true that the structure of $\mathcal{B}$ is irrelevant for the structure of $\mathcal{C}^{\mathcal{B}}$, but it is true that in general, $\mathcal{B}$ only needs a bare minimum of structure. This asymmetry between the domain and the codomain categories is reinforced by the fact that many of the interesting categories $\mathcal{A}$ are equivalent to functor categories, even presheaf categories (or some localization of them). Here are two examples.
1. Let us consider the category of groups $\operatorname{Grp}$, undoubtebly a category "of interest". There is a category $\operatorname{Th}(\operatorname{Grp})$ that has all finite products such that $\operatorname{Grp}$ is equivalent to the category of product-preserving functors $\operatorname{Th}(\operatorname{Grp})\to \operatorname{Set}$ where $\operatorname{Set}$ is the category of sets. This equivalence can be generalized to a *very large class* of categories of "algebraic flavor" and even some that at first sight do not bear the least resemblance to "algebraic categories". A few extra remarks about this example. First, the category $\operatorname{Th}(\operatorname{Grp})$ is a category constructed to make the equivalence work (it is the free category with products on a group object), in other words, it does not exactly fall within the class of categories "of interest". It's a similar to the example of diagram categories in (1), where the domain, a free category on a graph, is just a categorial construction to make the identification of diagrams with functors, not an interesting category by itself. Second, by replacing $\operatorname{Set}$ by another category $\mathcal{B}$ (with at least finite products), you can now speak of groups in $\mathcal{B}$. This gives another class of examples where functors into a category are interesting.
2. If $(\mathcal{C}, \mathcal{J})$ is a site (a category with a Grothendieck topology), by first taking the category of presheaves and then a suitable localization, one obtains the category of sheaves. This produces a host of geometric categories, like manifolds and schemes.
Much like in example 1, the interest is not so much on $(\mathcal{C}, \mathcal{J})$ and even less in the codomain, which is usually the category of sets (other categories for codomain also work, but the categorial requirements for everything to work smoothly are fairly strong), but in the (pre)sheaf category.
For my last point (4), a lot could be said, but I will just point you to two articles by F. W. Lawvere in the TAC Reprints, "Metric Spaces, Generalized Logic and Closed Categories" and "Taking categories seriously" (google for them, they are available online). In them, Lawvere makes several remarks about the duality between spaces and algebras of functions which are directly relevant to your question. To show that there is not so much an asymmetry but a duality between functors into and out of, let me give you two examples.
1. In your (that is, the OP) last post, you speak about stacks. A stack on a category $\mathcal{A}$ can be defined as a functor with values in $\mathcal{A}$ satisfying some conditions -- this is the fibered category approach to stacks. But a stack can also be defined as weak $2$-functor with values in the $2$-category of categories (satisfying some extra conditions). For reasons that I will not explain, the first approach is better, but nevertheless the point should be clear. There are actually many examples of this "duality", that identifies some category of functors into $\mathcal{A}$ with some category of functors out of $\mathcal{A}$.
2. Let me end up with an example from physics that further illustrates this duality. Quantum field theories are notoriously hard objects to define (let alone study). Several years ago, V. Turaev defined the notion of a Homotopy Quantum Field Theory, HQFT for short (check his papers in the arxiv if you are interested) which is a very simple, "toy" example of a QFT. If $X$ is a topological space, we can define a category that has for objects manifolds $M$ equipped with a homotopy class of maps $g\colon M\to X$ and a morphism $(M, g)\to (N, h)$ is a cobordism $W\colon M\to N$ with a homotppy class of maps into $X$ extending $g$ and $h$. An HQFT is a monoidal functor from this category into another monoidal category (usually, the category of finite-dimensional complex linear spaces). I am omitting lots of details, but the gist is that an HQFT gives us invariants of manifolds $M$ by mapping $M$ into some fixed background space $X$. But we can turn things around, for the category of $X$-HQFT's is a (functorial) invariant of the homotopy type of $X$, an invariant cooked up by mapping manifolds into $X$.
Hope it helps, regards,
G. Rodrigues
| 14 | https://mathoverflow.net/users/2562 | 32358 | 21,003 |
https://mathoverflow.net/questions/32353 | 4 | In Qing Liu's book *Algebraic geometry and arithmetic curves* I came across several confusing definitions. Several times he defines a notion only for a subclass of schemes/morphisms but later he is never *explicitly* mentioning these extra conditions again. Here are some examples:
* Let $X$ be a locally Noetherian scheme, and let $x \in X$ be a point. We say that $X$ is *regular* at $x$ if [...]. We say that $X$ is *regular* if it is regular at all of its points. **Question:** If he later says "Let $X$ be a regular scheme", then is it implicit that $X$ is locally Noetherian? If so, then why doesn't he say "A scheme is called regular if it is locally Noetherian and [...]"?
* Let $X$ be a reduced Noetherian scheme. Let $\xi\_1,\ldots,\xi\_n$ be the generic points of $X$. We say that a morphism of finite type $f:Z \rightarrow X$ is a *birational* morphism if [...]. **Question**: If he later says that a morphism $f:Z \rightarrow X$ of (arbitrary) schemes is birational, then is it implicit that $X$ is reduced Noetherian and that $f$ is of finite type? If so, then why doesn't he say "A morphism $f$ is called birational if it is of finite type, if $X$ is reduced Noetherian and if [...]"?
* Now it gets really confusing: Let $X$ be a reduced locally Noetherian scheme. A proper birational morphism $\pi:Z \rightarrow X$ with $Z$ regular is called a *desingularization* of $X$. **Question:** He defined *birational* only for reduced Noetherian schemes. What is *birational* for reduced locally Noetherian schemes? Is his *desingularization* now automatically of finite type?
Edit:
1. In Liu's book I found the following definition now: We say that a morphism $f:X \rightarrow Y$ is *proper* if it is of finite type, separated and universally closed. So, first of all, I think that this definition is now given in the non-confusing style, and second, this implies that the desingularizations above are of finite type (although it doesn't answer the locally Noetherian/Noetherian question).
2. I was asking "...then why doesn't he say that..." because I wasn't sure (and I'm still not sure) if there is some "higher truth" in this style of definition. Of course nobody except for Liu himself can answer this but perhaps someone else has more experience than I have and can give an explanation for this...
| https://mathoverflow.net/users/717 | Confusing definitions in Liu's Algebraic geometry and arithmetic curves? | Dear Arminius, I'm certainly not going to answer your questions "why doesn't he say...?":
Qing is a frequent and friendly contributor to MO and he will answer himself if he wants to.
Here is what I think is the consensus about your questions.
1) For a scheme *regular* definitely implies locally noetherian: De Jong 19.8.2
2) Birational necessitates neither noetherian nor reducedness conditions on schemes nor finite type assumptions on morphisms: De Jong 20.7.1
3) Qing's definition now makes perfectly good sense in view of 1) and 2). Desingularization is automatically of finite type because a proper morphism is of finite type by definition : De Jong 20.36.1
*Bibliographical note* I didn't want to give a long list of references for the definitions you ask about. I have only quoted De Jong and collaborators' monumental [Stacks Project](http://math.columbia.edu/algebraic_geometry/stacks-git/) which is the most up-to-date reference and which is incredibly well thought-out. Also De Jong is arguably the mathematician who has made the greatest progress on the resolution of singularities for schemes since Hironaka in 1964 .
| 6 | https://mathoverflow.net/users/450 | 32363 | 21,006 |
https://mathoverflow.net/questions/32370 | 4 | How do I see that there is a group of an arbitrary cardinality? Is this also true for abelian groups? Also, given a commutative ring $R\neq 0$ how do I see that there is an $R$-module of arbitrary cardinality?
I'm sure I saw this result somewhere but I can't seem to find it anywhere (books, google,...) Thanks!
| https://mathoverflow.net/users/5292 | Group & modules of arbitrary cardinality | For any algebraic theory that is expressible in first order logic in a countable language, and this includes groups, rings, fields, partial orders, lattices, etc. etc., then the basic fact is expressed by the [Lowenheim-Skolem theorem](http://en.wikipedia.org/wiki/L%C3%B6wenheim%E2%80%93Skolem_theorem), which asserts that if the theory has an infinite model, then it has models of every infinite cardinality. In general, one gets models of the theory of every cardinality above the size of the language (and this covers your $R$-module case).
One needs the Axiom of Choice to prove this, however, and this use is necessary, since the Axiom of Choice is equivalent to the assertion that every set carries a group structure, as explained in [the answer to this MO question](https://mathoverflow.net/questions/12973/does-every-non-empty-set-admit-a-group-structure-in-zf).
| 9 | https://mathoverflow.net/users/1946 | 32371 | 21,009 |
https://mathoverflow.net/questions/32369 | 2 | I would like to ask if there are any set of functions $u\_n(x)$ which is orthogonal to $x^n$?
i.e.:
$\int\_0^1 x^n u\_m(x) dx = \delta\_{n,m}$
Edit: For clarification, this question asked for all non-negative integer m and n.
| https://mathoverflow.net/users/5217 | Functions orthogonal to x^n | If $f \in L^2([0, 1])$ and $\int\_0^1 x^n f(x)\, dx=0$ for all $n\ge N$ where $N$ is a nonnegative integer then $f$ is zero almost everywhere. To see this note that $x\mapsto x^N f(x)$ is an $L^2$ function orthogonal to all polynomials, and the polynomials are dense in $L^2([0,1])$. So the answer to your question is "no" for $L^2$ functions.
| 9 | https://mathoverflow.net/users/4213 | 32372 | 21,010 |
https://mathoverflow.net/questions/32351 | 9 | I want to know what all doubly-transitive groups look like. Do you know some good reference where I can read about it?
| https://mathoverflow.net/users/4246 | Doubly-transitive groups | In Section 7.7 "The Finite 2-transitive Groups" of the book *Permutation groups* by John D. Dixon and Brian Mortimer, the authors describe the complete list of finite 2-transitive groups without proofs but with references.
They list eight infinite families: the alternating, symmetric, affine and projective groups in their natural actions, as well as the less known groups of Lie type: the symplectic groups, the Suzuki groups, the unitary groups and the Ree groups. The symplectic groups have two distinct 2-transitive actions, the last three classes are 2-transitive on the sets of points in their action on appropriate Steiner systems. Additional there are 10 sporadic examples of 2-transitive groups.
| 10 | https://mathoverflow.net/users/970 | 32384 | 21,018 |
https://mathoverflow.net/questions/32396 | 7 | Let $X$ be a projective variety over $\mathbb{Z}$, and suppose that $X$ has everywhere good reduction. Let $Y$ be the blow-up of $X$ at an integral point.
Then is it the case that $Y$ also has everywhere good reduction?
The example situation that I have in mind is the following (my main motivation is del Pezzo surfaces).
Take $X= \mathbb{P}^2$ over $\mathbb{Z}$. This clearly has good reduction everywhere. Next let $Y$ be the blow-up of $\mathbb{P}^2$ at the integral point $(0:0:1)$. This can be realised as the subvariety of $\mathbb{P}^2 \times \mathbb{P}^1$ (with variables $x\_0,x\_1,x\_2$ and $y\_1,y\_2$) given by the equation $x\_1 y\_2 = x\_2 y\_1$. Then $Y$ has everywhere good reduction (at least if my caluclations are correct).
I am curious to know if this happens after successively blowing up more integral points to obtain other del Pezzo surfaces. Note however that I am not claiming that all del Pezzo surfaces have everywhere good reduction!
Thanks in advance!
| https://mathoverflow.net/users/5101 | Good reduction and blow-ups | It is true if the integral point $T$ is actually a section (as in your example), because you then blow-up a smooth scheme $X\to {\rm Spec}(\mathbb Z)$ along a smooth center $T\simeq {\rm Spec}(\mathbb Z)$. In general, as $T$ is flat over $\mathbb Z$, the fiber $Y\_p$ of $Y$ at a prime $p$ is the blow-up of $X\_p$ along $T\_p$. At a $p$ ramified for $T\to {\rm Spec}(\mathbb Z)$, $T\_p$ is non-reduced and in general $Y\_p$ is not smooth.
As an example, take $X=\mathbb P^2={\rm Proj}\ \mathbb Z[x,y,z]$ and $T=V\_+(x, y^2-2z^2)$. Then $Y$ has singular fiber at $2$.
[**Edit**] Sorry, I was a little to optimistic on the compatibility of the blowing-up of $X$ with the base change $X\_p\to X$. However the conclusion is the same. Suppose for simplicity that the generic fiber of $X$ is geometrically connected. Let $p$ be any prime number and let $(X\_p)'$ be the blow-up of $X\_p$ along $T\_p$. Then we have a canonical closed immersion $(X\_p)'\to Y\_p$ which commutes with $(X\_p)'\to X\_p$ and $Y\_p\to X\_p$. Suppose now that $Y$ is smooth, then as $X\_p$ and $Y\_p$ are connected and smooth of the same dimension and $(X\_p)'\to X\_p$ is birational, $(X\_p)'\to Y\_p$ is an isomorphism. Hence $(X\_p)'$ must be smooth too. But in general this is not the case as $T\_p$ is not necessarily reduced (in the above example
$(X\_2)'$ is a normal singular surface).
| 8 | https://mathoverflow.net/users/3485 | 32417 | 21,034 |
https://mathoverflow.net/questions/32413 | 3 | Let $E$ be a Banach space and $f:E\to E$ be a continuous map. By $f^n$ we denote the $n$-th iterate of $f$, i.e. $f^n:=\underbrace{f\circ f\circ\cdots \circ f}\_{\text{n times}}$. Let $x\_0$ denote a fixed point of $f^n$.
1) Is it true that the question as to when $x\_0$ is a fixed point of $f$ has been resolved in the case of $E$ being reflexive? (I would be also thankful for some reference.)
2) What are considered to be the main obstacles in the case of $E$ being not reflexive when dealing with the above question (besides the obvious)? (As far as I understand, the problem is still open for general Banach spaces. Again, I would be thankful for a reference.)
A side remark: I realize that my questions may appear a bit too general. I am also aware of the fact that the field (fixed-point theory in banach spaces) is overflowed with publications of questionable value/quality/contribution like no other, which makes things even harder particularly for someone who is coming from a different field (e.g. number theory). Thus MO seems to be the only reasonable place to get a clear and compact answer from someone who has the overlook over the field.
Thanks!
| https://mathoverflow.net/users/3014 | When is a fixed point of f^n a fixed point of f? | Actually I can't see the role of reflexivity for an answer to the question as it is, unless further properties on $f$ are assumed.
In general, to start with the obvious case: if $f$ is a contraction, it has a fixed point, which is also the unique fixed point of the contraction $f^n$, therefore it is $x\_0$, so $x\_0$ is a fixed point of $f$. Of course, if $f$ is not a contraction, like e.g. the map $x\mapsto -x$, a fixed point of $f^n$ need not be a fixed point of $f$. Also, in a certain sense, the case of contractions essentially covers all cases where the answer is affirmative: by a result of Bessaga of around '60, if a map $f$ on a set $S$ is such that for all $n\in \mathbb{N}$ the iterated $f^n$ has a unique fixed point, then $f$ is a contraction with respect to a suitable complete metric on $S$.
| 7 | https://mathoverflow.net/users/6101 | 32419 | 21,035 |
https://mathoverflow.net/questions/32386 | 4 | Let $k$ be a field which I don't suppose to be algebraically closed. Then, the endomorphism ring of any irreducible representation of a finite group over $k$ is a division ring.
What is known about these division rings? When are they fields? When they are, are these fields Galois over $k$? At least, normal? When not, are the centers of the division rings Galois over $k$? What do we know about these centers?
If $k$ is a perfect field, do the irreducible representations themselves split into *distinct* irreducible representations over the algebraic closure of $k$, or can some be equal?
Sorry for this many questions that are probably well-known, but I can't find an introductory text on representation theory which takes non-algebraically closed fields seriously.
| https://mathoverflow.net/users/2530 | Galois theory of endomorphism rings of irreducible representations | Concerning
>
> are the centers of the division rings Galois over $k$
>
>
>
A finite dim'l $k$-algebra $A$ is
*split* provided $\operatorname{End}\_A(S) = k$
for every simple $A$-module $S$.
[This terminology is consistent with that used
in other contexts -- $A$ is split
just in case the reductive quotient of the "unit group" $A^\times$ is a split reductive algebraic group over $k$.]
Suppose that $k$ is perfect and that $A\_\ell = A \otimes\_k \ell$ is a split $\ell$-algebra for a finite, separable extension $\ell \supset k$. Then we have the following:
>
> $(\*)$ If $S$ is a simple $A$-module, the center $Z$ of the division $k$-algebra $\operatorname{End}\_A(S)$ is a subfield of $\ell$.
>
>
>
To see this, I claim first that $\operatorname{End}\_A(S) \otimes\_k \ell$ is a split semisimple
$\ell$-algebra. Indeed, since $k \subset \ell$ is separable, the $A\_\ell$-module $S \otimes\_k \ell = S\_\ell$ is semisimple, say $S\_\ell = \bigoplus\_i S\_i$ as $A\_\ell$-module, where $S\_i$ is the $T\_i$-isotypic
component of $S\_\ell$ and where $T\_i$ are distinct simple $A\_\ell$-modules. Since
by assumption $\operatorname{End}\_{A\_\ell}(T\_i) = \ell$, we see
that $$\operatorname{End}\_A(S) \otimes\_k \ell \simeq \operatorname{End}\_{A\_\ell}(S\_\ell) = \prod\_i \operatorname{End}\_{A\_\ell}(S\_i)$$
is a product of full matrix algebras over $\ell$. Now observe that the center of a split semisimple $\ell$-algebra
is a a split commutative etale $\ell$-algbra $\ell \times \cdots \times \ell$. Assertion $(\*)$
now follows.
Apply this to $A = kG$ for a finite group $G$. By Torsten's comment (following Emerton's answer) we may suppose $k$ to be perfect.
Then $A \otimes\_k \ell$ is split
for an *Abelian* extension $\ell$ of $k$ -- a suitable $\ell$ can be obtained by adjoining to $k$ enough roots of unity. It follows that $Z$ is always Galois over $k$ (for $A = kG$).
| 3 | https://mathoverflow.net/users/4653 | 32420 | 21,036 |
https://mathoverflow.net/questions/32422 | 2 | $(A\_{\alpha})\_{\alpha\in B}$ a family of sets indexed by a set $B$. Is $\Pi \_ {\alpha\in B} \ A\_{\alpha}$ a set? I can see that it is a set if $A\_{\alpha}=A \ \ \forall\alpha$ because in that case the product is a subset of the power set ${\cal P}\ (A\times B)$. As far as I understand, the axiom of choice only says the product is not the empty set, and it doesn't say if it is a set at all. Or am I wrong?
| https://mathoverflow.net/users/5292 | Is an arbitrary product of sets a set? | The standard definition of $(A\_\alpha)\_{\alpha \in B}$ a family of sets indexed by a set $B$ is that one is given a function $A$ with domain $B$ so that for $\alpha \in B$ we understand $A\_\alpha$ as $A(\alpha)$. Then, yes, the product is a set.
The function $A$ itself is an element of the powerset of $B \times \text{ran}(A)$ (where $\text{ran}(A)$ is the range of $A$) and the product $\prod\_{\alpha \in B} A\_\alpha$ is by definition the set of functions $f$ with domain $B$ having the property that $(\forall \alpha \in B) f(\alpha) \in A(\alpha)$. As such, one proves that the product exists by applying the comprehension axiom to the above defining formula observing that any such $f$ will be an element of the powerset of $B \times \bigcup \text{ran} A$.
[And yes, the Axiom of Choice merely says that the product is nonempty if all of the $A\_\alpha$ are nonempty.]
| 13 | https://mathoverflow.net/users/5147 | 32424 | 21,038 |
https://mathoverflow.net/questions/7493 | 16 | I was asked this years ago, but I don't remember by whom, and have never managed to solve it.
Consider the $2^n \times n$ matrix of all vectors in {-1,1}$^n$.
Someone comes and maliciously replaces some of the entries by zeros.
Show that there still remains a non-empty subset of rows that add up to the all zero vector.
| https://mathoverflow.net/users/2229 | A riddle about zeros, ones and minus-ones | Another answer (I guess they must be equivalent):
* Write each *original* line as a difference of two 0/1 vectors.
* Adapt this representation to the modified lines by changing *only the subtrahends*.
* You now have a function from {0,1}^n to {0,1}^n. Find a cycle.
| 13 | https://mathoverflow.net/users/7732 | 32425 | 21,039 |
https://mathoverflow.net/questions/32236 | 4 | Where can I find a complete classification of the $l$-adic Weil-Deligne representations for a local field $F$ of residual characteristic $p$ (with $p$ different from $l$)? Thanks.
| https://mathoverflow.net/users/7686 | Weil-Deligne representations | Here are the references:
D. Rohrlich. Elliptic curves and the Weil-Deligne group. Elliptic curves and related topics, 125{157, CRM Proc. Lecture Notes, 4, Amer. Math. Soc., Providence, RI, 1994.
or
J. Tate, Number theoretic background.
| 4 | https://mathoverflow.net/users/1816 | 32453 | 21,055 |
https://mathoverflow.net/questions/32450 | 15 | The [wikipedia article](http://en.wikipedia.org/wiki/Chevalley%E2%80%93Shephard%E2%80%93Todd_theorem#Statement_of_the_theorem/%22wikipedia%20article%22) claims that the theorem "was first proved by G. C. Shephard and J. A. Todd (1954) who gave a case-by-case proof. Claude Chevalley (1955) soon afterwards gave a uniform proof". I read the paper by Chevalley and it seems that he only proves the implication: "If the group is generated by pseudo-reflections, then the ring of invariants is polynomial". I wonder whether there is a uniform proof of the inverse implication? Where is it written?
| https://mathoverflow.net/users/6772 | Chevalley–Shephard–Todd theorem | There are indeed many presentations (if I remember correctly Bourbaki has it)
but the proof is very elegant and short so that I find it hard to refrain from giving
it. Let $H$ be the normal subgroup of the finite $G\subset \mathrm{GL}\_n$
generated by the pseudo-reflections. By the other direction $X:=\mathbb{A}^n/H$
is again affine space and in particular is smooth. We have an action of $G/H$ on
$X$ and a moment's thought reveals that it acts freely in codimension $1$
(as a point fixed by a non-identity element would lie below a
reflection hyperplane of $\mathbb{A}^n$ and the fixing element below a
pseudo-reflection). Hence $X \to X/(G/H)=\mathbb{A}^n/G$ is étale in codimension
$1$. If $\mathbb{A}^n/G$ were smooth, purity of the branch locus would imply that
the map were étale. However, that forces $G/H$ to act freely on $X$ but the
image of the origin is fixed by all of $G/H$ and therefore $G=H$.
| 21 | https://mathoverflow.net/users/4008 | 32456 | 21,058 |
https://mathoverflow.net/questions/32318 | 11 | Modern statements of Gödel's incompleteness theorems are usually in terms of first-order predicate logic. However, I've often read the claim that they extend to arbitrary formal systems that can prove basic propositions about numbers. Indeed, according to Wikipedia, the original theorems referred to the type theory of Principia Mathematica, which is apparently not based on predicate logic.
My two questions are:
1. What is the most general concept/definition of a formal system for which Gödel's theorems have been stated?
2. How does their proof differ from the predicate logic variant?
Regarding 1., I could imagine several equally general definitions, for example based on either strings of symbols or abstract syntax trees. Being a little biased, I actually think of formal systems as data structures of more or less arbitrary computer programs, so maybe there is a definition based on Turing machines... In any case, it would need to specify what a "proof" of a "theorem" is, but I would like to do without the concept of "axioms." (See also [Derivation rules and Godel theorem](https://mathoverflow.net/questions/29774/derivation-rules-and-godel-theorem/29989#29989).)
Regarding 2., I'm specifically thinking about the diagonal lemma (or arithmetic fixed-point theorem, or whatever it is really called). The version I know refers to a "formula with one free variable," but that presupposes such concepts as "formula" and "free variable" in the formal system, and I'm wondering how to generalize that to arbitrary formal systems. I know there are proofs of the first incompleteness theorem which take an entirely different route, but AFAIK they don't carry over to the second incompleteness theorem.
I would like to add that I don't doubt the generality of Gödel's incompleteness theorems in any way. I just feel there is a gap between their claimed general nature and the way they are usually presented. A year ago, I devised a nonstandard formal system for a proof assistant. Although it could easily formalize its own concepts and express its own consistency, a translation of Gödel's incompleteness theorems from predicate logic turned out to be surprisingly nontrivial.
| https://mathoverflow.net/users/6904 | Most general formulation of Gödel's incompleteness theorems | Raymond Smullyan gave a very general formulation in terms of representation systems. They appear in his "Theory of Formal Systems", and in the first and last chapters of "Godel's Incompleteness Theorems". They generalise first- and higher-order systems of logic, type theories, and Post production systems.
A representation system consists of:
1. A countably infinite set $E$ of expressions.
2. A subset $S \subseteq E$, the set of sentences.
3. A subset $T \subseteq S$, the set of provable sentences.
4. A subset $R \subseteq S$, the set of refutable sentences.
5. A subset $P \subseteq E$, the set of (unary) predicates.
6. A function $\Phi : E \times \mathbb{N} \rightarrow E$ such that, whenever $H$ is a predicate, then $\Phi(H,n)$ is a sentence.
The system is complete iff every sentence is either provable or refutable. It is inconsistent iff some sentence is both provable and refutable.
We say a predicate $H$ represents the set $A \subseteq \mathbb{N}$ iff $A = \{ n : \Phi(H,n) \in T \}$.
Let $g$ be a bijection from $E$ to $\mathbb{N}$. We call $g(X)$ the Godel number of $X$.
We write $E\_n$ for the expression with Godel number $n$.
Let $\overline{A} = \mathbb{N} \setminus A$ and $Q^\* = \{ n : \Phi(E\_n,n) \in Q \}$.
We have:
1. (Generalised Tarski Theorem) The set $\overline{T^\*}$ is not representable.
2. (Generalised Godel Theorem) If $R^\*$ is representable, then the system is either inconsistent or incomplete.
3. (Generalised Rosser Theorem) If some superset of $R^\* $ disjoint from $T^\*$ is representable, then the system is incomplete.
In case it's not clear: in a first-order system, we can take $P$ to be the set of formulas whose only free variable is $x\_1$, and $\Phi(H,n) = [\overline{n}/x\_1]H$.
| 14 | https://mathoverflow.net/users/7247 | 32463 | 21,065 |
https://mathoverflow.net/questions/32469 | 0 | I need the following sum ( in the sense of principal value):
$$\sum\_{s=-\infty}^{\infty}\frac{(-1)^{s}e^{-2\pi isy}}{x+s}$$
It is possible to show that
$$\sum\_{s=-\infty}^{\infty}\frac{e^{-2\pi isy}}{x+s}=\pi\frac{e^{\pi ix(2FractionalPart[y]-1)}}{\sin(\pi x)}$$
Hence the sum I need is
$$f(y)=\sum\_{s=-\infty}^{\infty}\frac{e^{-2\pi is(y+0.5)}}{x+s}=\pi\frac{e^{\pi ix(2FractionalPart[y+0.5]-1)}}{\sin(\pi x)}$$
But now it is difficult to see that the function f is continuous on $(0,1)$.
How to obtain expression without 0.5?
The second question is what are this series in both cases?
Maybe I wrong, but it seems that it is not Fourier series of $\pi\frac{e^{\pi ix(y-1)}}{\sin(\pi x)}$
| https://mathoverflow.net/users/3589 | series Sum[(-1)^n/(x+n)] | This is in a typical complex analysis text
$$
\frac{1}{x} + \sum\_{s = 1}^{\infty}\left( \frac{1}{x + s} + \frac{1}{x - s}\right) = \pi \cot (\pi x)
$$
When grouped this way, it converges...
Maple says
$$
\frac{1}{x} + \sum\_{s = 1}^{\infty} \left(\frac{\operatorname{e} ^{(-2 i\pi s y)}}{x + s} + \frac{\operatorname{e} ^{2 i \pi s y}}{x - s}\right) =
-\frac{1 - LerchPhi \biggl(\frac{1}{\operatorname{e} ^{2 i \pi y}},1,x\biggr) x + LerchPhi \bigl(\operatorname{e} ^{2 i \pi y},1,-x\bigr) x}{x}
$$
and numerically this seems to agree with what you said.
So your formula looks correct. Why do you say it is not continuous? The complex exponential has period $2\pi i$.
| 3 | https://mathoverflow.net/users/454 | 32475 | 21,073 |
https://mathoverflow.net/questions/32477 | 16 | On a euclidean plane, what is the minimal area shape S, such that for every unit length curve, a translation and a rotation of S can cover the curve.
What are the bounds of the shape's area if this is a open problem?
When I asked this problem few years ago, someone told me it's open. I don't know if this is still open and I can't find any reference on it.
I don't even know what branch of mathematics it falls under. so I can't even tag this question.
| https://mathoverflow.net/users/6886 | Smallest area shape that covers all unit length curve | Whereas I don't know of any recent progress in this problem, let me mention one result for
*closed* curves.
>
> **Theorem.** A closed plane curve of length $L$ and curvature bounded by $K$ can be contained inside a circle of radius $L/4 - (\pi - 2)/2K$.
>
>
>
This was proved in 1974 by H.H. Johnson ([link 1](http://www.ams.org/mathscinet-getitem?mr=0348631)) who used calculus of variations methods. A geometric proof was given a bit later by Chakerian, Johnson and Vogt ([link 2](http://www.ams.org/journals/proc/1976-057-01/S0002-9939-1976-0402611-2/home.html%20)).
---
**Edit.** Apparently the problem is still open. Here's an article ([arXiv link](http://arxiv.org/abs/math.MG/0701391)), which contains a survey of some known results as of 2009. From the Introduction:
>
> In 1966, Leo Moser asked for the region of smallest area which can accommodate
> every planar arc of length one. The problem is known as “Moser’s worm problem” and is a variation of universal cover problems. In Moser’s problem, a cover is a set which contains a copy of any rectifiable planar arc of unit length, and is usually assumed to be convex. Such a minimal cover is known to have area between 0.2194 and 0.2738. However, the
> original problem remains unsolved.
>
>
>
| 16 | https://mathoverflow.net/users/5371 | 32483 | 21,077 |
https://mathoverflow.net/questions/32495 | 1 |
>
> **Possible Duplicate:**
>
> [Cardinality of the permutations of an infinite set](https://mathoverflow.net/questions/27785/cardinality-of-the-permutations-of-an-infinite-set)
>
>
>
Why does the symmetric group on an infinite set X have the cardinality of the power set ${\cal P}(X)$?
| https://mathoverflow.net/users/5292 | Cardinality of symmetric group | I assume you mean an infinite set $X$. You need to use the Axiom of Choice to prove this fact, but I'm not sure to what extent it is necessary.
Since $|X \times X| = |X|$ (uses AC) and $Sym(X) \subseteq \mathcal{P}(X \times X)$, it is clear that $|Sym(X)| \leq 2^{|X \times X|} = 2^{|X|}$.
Since $|X \times 2| = |X|$ (uses AC) we can split $X$ into two disjoint sets $X\_0$ and $X\_1$, each of size $|X|$. Let $a:X\_0 \to X\_1$ be a bijection. For each set $A \subseteq X\_0$ define $\sigma\_A \in Sym(X)$ to be the bijection that exchanges $x$ and $a(x)$ for every $x \in A$ and leaves all other elements unchanged. It is clear that $A \in \mathcal{P}(X\_0) \mapsto \sigma\_A \in Sym(X)$ is an injection. Therefore $|Sym(X)| \geq 2^{|X\_0|} = 2^{|X|}$.
So the equality $|Sym(X)| = 2^{|X|}$ holds unconditionally for all infinite sets such that $|X \times X| = |X|$. The fact that $|X \times X| = |X|$ for all infinite sets $X$ is equivalent to AC by an old result of Tarski.
| 4 | https://mathoverflow.net/users/2000 | 32499 | 21,089 |
https://mathoverflow.net/questions/32427 | 5 | I am sorry for spamming MO with questions I have not thought about for more than 3 hours, but currently I am quite busy with preparing a talk on representations of $S\_n$, and I don't want these to get lost. I hope this one is not quite as vague as the last one.
This here is an attempt to generalize Exercise I.21 in [Kraft-Procesi, *Classical Invariant Theory*](http://www.math.unibas.ch/~kraft/Papers/KP-Primer.pdf).
Let $K$ be a field - say, infinite, since we are going to do classical invariant theory. Let $K\left[\mathrm{SL}\_n K\right]$ denote the $K$-algebra of polynomial functions on $\mathrm{SL}\_n K$, which I define either as
$\left\lbrace f\mid\_{\mathrm{SL}\_n K} \ \mid \ f\in K\left[\mathrm{M}\_n K\right]\right\rbrace$
or as $K\left[\mathrm{M}\_n K\right]\diagup \left(\det-1\right)$ (proving the equivalence of these two definitions is not the matter, it's rather easy - even easier than Kraft and Procesi try to make one believe).
Now, the group $\mathrm{U}\_n K$ of unipotent upper triangular matrices acts on $\mathrm{SL}\_n K$ from the left. What is the invariant ring? It is easily seen that
$\det\left(\text{the submatrix formed by the intersection of the rows }i,i+1,...,n\text{ with the columns }j,i+1,i+2,...,n\right)$
is an invariant for any $i\geq j$. These generate the fraction field of the invariants, but do they also generate the ring of the invariants itself?
(The above-mentioned exercise is the above for $n=2$.)
Arguments using Victorian age methods (as opposed to Zariski-topological or other algebro-geometrical) would be particularly preferred.
**EDIT:** As Allen Knutson has pointed out, my question has a negative answer. However, the (larger) collection of determinants of the form
$\det\left(\text{the submatrix formed by the intersection of the rows }i,i+1,...,n\text{ with the columns }j\_1, j\_2, ..., j\_{n-i+1}\right)$
for $1 < i \leq n$ and $1 \leq j\_1 < j\_2 < ... < j\_{n-i+1} \leq n$ does generate the ring of invariants. When $K$ has characteristic $0$, this can be proven using the standard theory of highest-weight modules and multiplicity-free algebras explained in Kraft-Procesi (see [my errata, "Page 9, Exercise 21"](http://web.mit.edu/~darij/www/algebra/KP-errata-web.pdf) for a proof). I am still wondering whether it holds for arbitrary $K$ and has a more elementary or combinatorial proof.
| https://mathoverflow.net/users/2530 | Left U_n-invariants of SL_n - an exercise in Kraft-Procesi | No, they don't. $U\_n(K)$ is performing upward row operations, so any $m\times m$ minor that uses the last $m$ rows will be $U\_n(K)$-invariant, e.g. any single bottom entry. You won't be able to generate those linear functions using your higher-degree functions. (Victor's disproof is nice too!)
What *is* true is that the invariant ring is generated by those $2^n-1$ many minors (corresponding to nonempty subsets of columns). One nice place to read about them is [Miller-Sturmfels], chapter 14, where they show e.g. that you can degenerate this invariant ring by replacing each minor by the product of its diagonal entries, obtaining the semigroup algebra of the cone of Gel'fand-Cetlin patterns.
| 5 | https://mathoverflow.net/users/391 | 32500 | 21,090 |
https://mathoverflow.net/questions/32401 | 6 | For a partition $\lambda$ let $S^{\lambda}$ be the corresponding Schur functor. Is it true that for every $\lambda$ there exists an irreducible representation $V$ of a finite nonabelian group $G$ such that $S^{\lambda}(V)$ is still irreducible?
This is not obvious to me even for the symmetric and exterior powers (although maybe I'm not thinking hard enough), so any partial results would be appreciated.
| https://mathoverflow.net/users/290 | Can the image of a Schur functor always be made an irreducible representation? | Since the Guralnick and Tiep paper is very long, I thought I would summarize my understanding of it. Note that I only learned about this result from moonface last night, so I am hardly an expert. This answer is community wiki, in case anyone can improve on my summary.
We want to establish the following result: Let $G$ be a finite noncommutative subgroup of $GL(V)$, with $V$ a $\mathbb{C}$ vector space. Let $k \geq 6$. Then $\mathrm{Sym}^k(V)$ is reducible. (If $G$ is commutative and $V$ is one dimensional then, of course, $\mathrm{Sym}^k(V)$ is always one dimensional and hence irreducible.)
Our proof is by induction on $|G|$. Our base case will be when $G$ is a central extension of a simple group.
Choose a nontrivial normal subgroup $H$ of $G$ such that $G/H$ is simple. If we can't do this then $G$ is simple and we are in the base case. Let $V \cong \bigoplus U\_i$ be the decomposition of $V$ into $H$-isotypic components. If there is more than one summand, then $\bigoplus \mathrm{Sym}^k(U\_i)$ is a nontrivial $G$-subrep of $\mathrm{Sym}^k(V)$. So we may assume that $V \cong W \otimes X$, where $W$ is an $H$-irrep and $H$ acts trivially on $X$. Then one can show (proof of lemma 2.5) then $G$ is contained in $GL(W) \times GL(X)$. (This is nontrivial but not deep; you could give it as a problem in a graduate-level representation theory course.)
Then $\mathrm{Sym}^k(W) \otimes \mathrm{Sym}^k(X)$ is a subrepresentation of $\mathrm{Sym}^k(V)$. This subrep is proper unless $\dim X=1$ or $\dim W=1$. If $\dim X=1$, then $V$ is an irrep of $H$ and $\mathrm{Sym}^k(V)$ is irreducible as an $H$-rep, so we are done by induction.
If $W$ is one dimensional, then $H$ acts on $V$ by scalars, so $H$ is central. So we are in our base case: a central extension of a simple group. This case is done by group cohomology and the classification of simple groups. See section 4 for groups of Lie type, section 6 for alternating groups and section 7 for sporadic groups.
| 3 | https://mathoverflow.net/users/297 | 32503 | 21,091 |
https://mathoverflow.net/questions/32502 | 2 | Suppose a bounded sequence $(x\_n)$ converges to $x$ in the Cesaro sense (i.e., $\frac{1}{n}(x\_1 + x\_2 + \dots + x\_n)\rightarrow x$) in a separable Hilbert space $H$. How to prove that some subsequence $(x\_{n\_k})$ converges weakly to $x$?
| https://mathoverflow.net/users/5498 | Cesaro convergence implies weak convergence of a subsequence | If we take $x\_n = (-1)^n x$ then $x\_n$ converges to $0$ in Cesaro sence. But no subsequence of $x\_n$ converges weakly to $0$. $x\_n$ is also a bounded sequence.
Hence your statements seems wrong.
| 8 | https://mathoverflow.net/users/7079 | 32508 | 21,095 |
https://mathoverflow.net/questions/32515 | 2 | Matlab has a set of dot operators, such as .\*, ./, .^. Each of these operators consists of a dot and a normal algebraic operator. They perform element-wise algebraic operations on a matrix. For example, consider the following codes
```
A = [1 2 3; 3 2 1];
x = [1 2 4];
B = A.^2
y = 1./x
```
The result is
```
B =
1 4 9
9 4 1
y =
1.0000 0.5000 0.2500
```
I find these dot operators very convenient. My question is, how to write these dot operators in mathematical expressions? (By mathematical expressions, I mean the expressions used in proofs.)
EDIT -
One obvious way is to define the result matrix element-wise. But is there a way to write this result in a more compact manner?
| https://mathoverflow.net/users/7595 | How to write Matlab's dot operators in mathematical expressions? | Your matrix $B$ is the Hadamard product of $A$ and $A$ which uses the notation $B = A \circ A$. However I don't know of any others, particularly for expressing $y$.
See: <http://en.wikipedia.org/wiki/Matrix_multiplication#Hadamard_product>
| 4 | https://mathoverflow.net/users/3121 | 32522 | 21,105 |
https://mathoverflow.net/questions/32527 | 4 | I am seeking a measure of the "complexity" of a surface $S$,
a quantity that reflects how widely the metric varies from spot to
spot. I am primarily interested in surfaces topologically
equivalent to a sphere in $\mathbb{R}^3$, so measures that rely
on the genus are not useful.
Ideally the measure would achieve its minimum for a (round) sphere,
would be larger but still small for closed convex surfaces,
and large for surfaces with steep mountains and plummeting valleys.
Ultimately I need to discretize the measure, but I would like to
understand what are the alternatives for smooth metrics.
I can concoct reasonable ad hoc measures, but I'd prefer
to start from a more principled foundation.
From its name, the *entropy of a Riemannian manifold* sounds like it might be
appropriate, but I have only a tenuous grasp of this concept,
so it is unclear to me if this aligns with my goals.
I've also looked at the *systolic ratio* and several other geodesic-based
concepts, but none seem to capture what I want.
I'd appreciate pointers to concepts in this general intellectual neighborhood.
Thanks!
**Addendum**. Thanks for the useful suggestions: normalized surface area, Bregman divergence, Gromov-Hausdorff metric, Willmore energy. My question was too vague to permit a definitive answer,
but I'll accept Will Jagy's suggestions on the Willmore energy, which taught me much.
| https://mathoverflow.net/users/6094 | Measures of the complexity of a metric | I think you would be pretty happy with the Willmore functional for, well, compact orientable
$C^\infty$ surfaces in $\mathbb R^3.$ It is just the integral of the square of the *mean* curvature or
$$ \frac{1}{2 \pi} \int\_{M^2} \; \; H^2 \; dS $$
This quantity is at least 2, and is only equal to 2 for a round sphere. The Willmore Conjecture is that the minimum for an imbedded torus is achieved on the torus (sometimes called the Clifford torus, by the Bryant correspondence) created by revolving a circle of radius 1 with its center at distance $\sqrt 2$ from the axis of revolution. Here the functional has value $ \pi.$ Leon Simon proved that the minimum (a priori the infimum) is achieved. Rob Kusner found some rather earlier references (before Willmore) to this problem. $$ $$ See, for example, "Total Curvature in Riemannian Geometry" by Thomas J. Willmore. $$ $$ I do not expect there would be much trouble making a discrete version of this.
$$ $$
NOTE: sometimes Willmore writes with the $2 \pi$ divisor, sometimes not.
$$ $$
I found a nice wiki page and some pdf's with references and other information, one a schedule for an October 2010 seminar at Oberwolfach. Anyway,
<http://en.wikipedia.org/wiki/Willmore_energy> and
<http://www.mfo.de/programme/schedule/2010/43b/programme1043b.pdf> and
<http://www.warwick.ac.uk/~maseq/wmsri.pdf> and
<http://www.math.ethz.ch/~riviere/papers/riviere-tartar.pdf>
$$ $$
I was not aware of this, it seems the discrete version of this has been worked out, a fair amount published, including treatment in a book,
"Discrete differential geometry"
by Alexander I. Bobenko, which can be viewed with google books. I ran google with "discrete willmore functional."
| 8 | https://mathoverflow.net/users/3324 | 32529 | 21,108 |
https://mathoverflow.net/questions/32349 | 1 | Suppose given a d-dimensional Brownian motion $B\_t$ starting from the origin and a centered ball with radius 1. Define T as the first hitting time of the sphere (boundary of the ball). How can one prove that T and $B\_T$ are independent?
| https://mathoverflow.net/users/7713 | Independence of conditional hitting distribution and hitting time | The short, and somewhat heuristic, answer is that rotational invariance implies that given $T$ the distribution of $B\_T$ is uniform on the sphere $S^{d-1}$. Since the conditional distribution does not depend on $T$, this is independence of the two variables.
In more detail, and with more rigor, let $\Omega$ denote the sample space of Brownian paths, chosen so that all paths are continuous. For $B \in \Omega$ and a rotation $R \in SO(d)$ let $R\omega$ denote the path
$$ [R B](t)= R(B\_t).$$
More generally, given an event $E \subset \Omega$ and $R \in SO(d)$, let $RE=\lbrace B \ : \ R^{-1} B \in E \rbrace.$ Wiener measure is rotation invariant, so $\Pr (R E)=\Pr(E)$.
Now let $M\subset S^{d-1}$ and $J\subset [0,\infty)$ be Borel sets and let $E\_M$ and $F\_J$ be the events $\lbrace B\_T \in M\rbrace $ and $\lbrace T \in J \rbrace$ respectively. To prove independence of $B\_T$ and $T$ we must show
$$ \Pr (E\_M \cap F\_J ) = \Pr (E\_M) \Pr (F\_J) \quad \quad (\star)$$
for all such $M$ and $J$.
Let $R\in SO(d)$. Then $R E\_M = E\_{RM}$. Since the exit time
$$ T(B)= \inf \lbrace t \ : \ |B(t)|\ge 1 \rbrace ,$$
we see that $T(R B)=T(B)$ and thus $R F\_J = F\_J$ for $R \in SO(d)$. Thus
$$ \Pr (E\_M \cap F\_J ) = \Pr(E\_{RM} \cap F\_J),$$
so the measure $M \mapsto \Pr (E\_M \cap F\_J)$ is rotation invariant. Since it has total mass $\Pr(F\_J)$, we conclude (e.g., from the uniqueness of Haar measure on $SO(d)$) that
$$\Pr (E\_M \cap F\_J) = |M| \Pr (F\_J),$$
where $|\cdot|$ is normalized Lebesgue measure on $S^{d-1}.$ Taking $J=[0,\infty)$ (for which $\Pr(F\_J)=1$), we see that $\Pr(E\_M)=|M|$ and $(\star)$ follows.
| 5 | https://mathoverflow.net/users/6781 | 32532 | 21,110 |
https://mathoverflow.net/questions/32486 | 9 | First let me give a precise formulation of the question; I'll give some background/motivation at the end.
If X is a projective variety which is Q-factorial (meaning X is normal, and some sufficiently high multiple of every Weil divisor is Cartier), then a *small Q-factorial modification* or *SQM* of X means a birational map φ: X --> Y (where Y is another Q-factorial projective variety) which is an isomorphism in codimension 1. (Examples: flips and flops.) The question is then this:
>
> Is there an example of a Q-factorial projective variety X and φ: X --> Y an SQM of X such that the nef cone Nef(X) is rational polyhedral but Nef(Y) is not rational polyhedral? Or (highly unlikely I think) can one prove that this situation is impossible?
>
>
>
And to push my luck:
>
> Give sufficient conditions to ensure that in the above situation, Nef(X) rational polyhedral implies Nef(Y) rational polyhedral.
>
>
>
Really I think only the first question has a hope of being answered positively, but it would be nice to know if I was wrong.
**Background:** The question was prompted by a theorem of Hu and Keel ("Mori dream spaces and GIT", Michigan Math. J., 2000) which gives a characterisation of so-called *Mori dream spaces* --- certain varieties which behave very well with respect to the operations of the minimal model program. In particular, if X is a Mori dream space, then Nef(X) is rational polyhedral and so too is Nef(Y) for any SQM Y of X. It then seems natural to ask for an example where the first condition here holds but the second fails.
Hu--Keel's theorem gives one answer to my second question above, since Birkar--Cascini--Hacon--McKernan proved that any Fano variety (or more generally, any variety of Fano type) is a Mori dream space. But it would be great to know of any sufficient condition that applies outside the Fano domain.
**Edit (July 23):** Balazs' answer settles the first question in the affirmative: flops can change the nef cone from finite to infinite. But it seems to remain open (and interesting to me) whether there are examples $X$ of this phenomenon with Kodaira dimension $kd(X)=-\infty$.
| https://mathoverflow.net/users/nan | How much can small modifications change the nef cone? | In the world of Calabi-Yau (as opposed to Fano) varieties, one does not expect miracles of Mori dream space type. A specific example which gives a positive answer to your first question is described in the paper <http://xxx.lanl.gov/abs/math/0102055> of Michael Fryers: a Calabi-Yau threefold (a degenerate quintic) which has some small resolutions having finite polyhedral nef cone, and some having an infinitely generated cone, which is locally rational polyhedral away from a single point on the boundary. The geometry, related to the Horrocks-Mumford bundle and abelian surfaces in projective four-space, is very beautiful.
| 8 | https://mathoverflow.net/users/6107 | 32536 | 21,113 |
https://mathoverflow.net/questions/32540 | 4 | If {c(n)} is an arbitrary sequence of irrational numbers converging to 0 then Q + c(n), the set obtained by adding c(n) to the set of rational numbers Q, is clearly disjoint from Q for each n.
Is there an uncountable dense set of the real numbers, say D, for which a sequence {c(n)} converging to 0 exists such that D + c(n) does not intersect D for all n?
If such a set exists, and is borel, then it must have measure 0 since D - D would contain an interval if D had a positive measure.
| https://mathoverflow.net/users/6627 | disjoint translates of a dense uncountable set | Choose your favorite c(n), and define $D$ as any complementary $\mathbb{Q}$-vector space of $\mathrm{span}(\{c(n): n\in\mathbb{N}\})$ in $\mathbb{R}$ as $\mathbb{Q}$ vector space. This $D$ is not measurable, of course (countably many translates of it cover $\mathbb{R}$, so it can't be of measure zero, and on the other hand as you said it can't be of positive measure). It's a version of the construction of the Vitali set (actually the standard Vitali set already works with a sequence of rationals c(n) ).
| 3 | https://mathoverflow.net/users/6101 | 32541 | 21,115 |
https://mathoverflow.net/questions/32528 | 4 | See my previous question [What is the product in the 2-category of spans?](https://mathoverflow.net/questions/32526/what-is-the-product-in-the-2-category-of-spans) for notation. In brief, $\mathcal S$ is a category with finite limits, $\operatorname{Span}(\mathcal S)$ is the 2-category whose 1-morphisms are diagrams in $\mathcal S$ of the form $\bullet \leftarrow\bullet \rightarrow \bullet$, and for want of a better name (is there one?) I call the functor $\{X\to Y\} \mapsto \{X = X \to Y\}$ "spanishization".
>
> Suppose I have an equivalence in $\operatorname{Span}(\mathcal S)$ (a pair of 1-morphisms whose compositions are isomorphic to identities). Is it necessarily isomorphic to the spanishization of an isomorphism in $\mathcal S$?
>
>
>
| https://mathoverflow.net/users/78 | Do all equivalences in the 2-category of spans come from isomorphisms? | Yes. Suppose we have spans $X\_1 \stackrel{f\_1}{\leftarrow} Y \stackrel{f\_2}{\to} X\_2$ and $X\_2 \stackrel{g\_2}{\leftarrow} Z \stackrel{g\_1}{\to} X\_1$ with an isomorphism $Y \times\_{X\_2} Z \stackrel{\omega}{\to} X\_1$ such that $\omega = f\_1 \circ \pi\_1 = g\_1 \circ \pi\_2$, where $\pi\_1$ and $\pi\_2$ are the projections from $Y \times\_{X\_2} Z$ onto its factors. The morphism $f\_1: Y \to X$ yields a 2-morphism from the $Y$ span to the spanishization of $f\_2 \circ \pi\_1 \circ \omega^{-1}: X\_1 \to X\_2$. The map $\pi\_1 \circ \omega^{-1}: X\_1 \to Y$ yields an inverse 2-morphism. Thus, the $Y$ span is isomorphic to the spanishization of $f\_2 \circ \pi\_1 \circ \omega^{-1}$.
The same argument using the other composite gives an isomorphism between the $Z$ span to the spanishization of $g\_1 \circ p\_1 \circ \beta^{-1}: X\_2 \to X\_1$, where $p\_1: Z \times\_{X\_1} Y \to Z$ is the first projection, and $\beta: Z \times\_{X\_1} Y \to X\_2$ is our isomorphism with the identity span. That the two morphisms we have constructed are inverse to each other in $\mathcal{S}$ follows from the fact that their spanishizations compose to something isomorphic to the identity span in both directions.
| 1 | https://mathoverflow.net/users/396 | 32544 | 21,118 |
https://mathoverflow.net/questions/32533 | 49 | Convex Optimization is a mathematically rigorous and well-studied field. In linear programming a whole host of tractable methods give your global optimums in lightning fast times. Quadratic programming is almost as easy, and there's a good deal of semi-definite, second-order cone and even integer programming methods that can do quite well on a lot of problems.
Non-convex optimization (and particularly weird formulations of certain integer programming and combinatorial optimization problems), however, are generally heuristics like "ant colony optimization". Essentially all generalizable non-convex optimization algorithms I've come across are some (often clever, but still) combination of gradient descent and genetic algorithms.
I can understand why this is - in non-convex surfaces local information is a lot less useful - but I would figure that there would at least be an algorithm that provably learns for a broad class of functions whether local features indicate a nearby global optimum or not. Also, perhaps, general theories of whether and how you can project a non-convex surface into higher dimensions to make it convex or almost convex.
Edit: An example. A polynomial of known degree k only needs k + 1 samples to reconstruct - does this also give you the minimum within a given range for free, or do you still need to search for it manually? For any more general class of functions, does "ability to reconstruct" carry over at all to "ability to find global optima"?
| https://mathoverflow.net/users/942 | Is all non-convex optimization heuristic? | If the question is "Are there non-convex global search algorithms with provably nice properties?" then the answer is "Yes, lots." The algorithms I'm familiar with use interval analysis. Here's a seminal paper from 1979: [Global Optimization Using Interval Analysis](http://old.ict.nsc.ru/interval/Library/Thematic/GlobOptim/GlOptUsingIA-1.pdf). And here's a [book on the topic](http://www.crcnetbase.com/isbn/9780203026922). The requirements on the function are that it be Lipschitz or smooth to some order, and that it not have a combinatorial explosion of local optima. These techniques aren't as fast as linear or convex programming, but they're solving a harder problem, so you can't hold it against them. And from a practical point of view, for reasonable functions, they converge plenty fast.
| 37 | https://mathoverflow.net/users/7759 | 32550 | 21,124 |
https://mathoverflow.net/questions/32498 | 4 | Can anyone provide a reference to proofs of statements of the following type: The higher algebric group cohomology of a reductive group $G$ over $\mathbb{C}$ vanishes.
I am interested not just in finite dimensional modules but also "rational representations" for instance the functions on a vector space $\mathbb{C}^{n}$ on which $G$ acts.
| https://mathoverflow.net/users/3396 | Group Cohomology for Reductive Groups | Rational representations are directed unions of finite-dimensional ones, on which all linear representations of $G$ are completely reducible (either by an ad hoc definition of "reductive group" or a theorem applied to a good definition). So the functor of $G$-invariants on the category of rational representations is exact, hence one gets the desired higher vanishing (by whatever reasonable method one chooses to define the higher cohomologies).
| 9 | https://mathoverflow.net/users/3927 | 32551 | 21,125 |
https://mathoverflow.net/questions/32512 | 5 | Assume that M=G/K is a non-compact Hermitian symmetric space, for G the real points of a semisimple (or even simple) algerbaic group and K a maximal compact subgroup. M admits the structure of a complex manifold. Now, if X is a complex subvariety of M then its pre-image in G is a real analytic subvariety of G. My question is: how does one recognize, among all real analytic subvarieties of G, those which project onto complex subvarieties in M?
More concretely, if H is a real Lie subgroup of G (even real algebraic), when is it the case that its image under the projection map (into M=G/K) gives a complex subvariety of M?
| https://mathoverflow.net/users/7753 | Complex subvarieties of hermitian symmetric spaces | I will explain how to check it in any particular case, but I am not sure whether there is a good classification available (perhaps, it is even easy and I did not think enough about it). Helgason's book and papers of Joe Wolf and Alan Huckleberry may contain some relevant information.
Since the complex structure on $M=G/K$ is homogeneous, the image of $H$ is a complex subvariety if and only if its tangent subspace at $e$ is a complex subspace of $T\_e M,$ i.e. invariant under the complex structure operator $J: T\_e M\to T\_e M.$ Let $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$ be the Cartan decomposition, then $T\_e (G/K)$ may be identified with $\mathfrak{p}$ via the projection onto the second summand and $J=\text{ad}(X\_0)$ for a certain element $X\_0$ in the center of $\mathfrak{k}$ (if $G$ is simple then this center is one-dimensional). Thus the answer is "yes" if and only if
$$T\_e(H/K\cap H)=\mathfrak{h}/\mathfrak{k}\cap\mathfrak{h} \subset
\mathfrak{p} \text{ is } \text{ad}(X\_0)\text{-invariant.}$$
Subalgebras $\mathfrak{h}$ with this property would need to be classified modulo conjugation by $K$.
One can proceed a bit further by complexifying $\mathfrak{g}$, so that $\mathfrak{p}\_\mathbb{C}=\mathfrak{p}\_{+}\oplus\mathfrak{p}\_{-}$ is the eigenspace decomposition for $J\_{\mathbb{C}}$ (the eigenvalues are $\pm i$ and $\mathfrak{p}\_{\pm}$ are abelian subalgebras of $\mathfrak{g}\_{\mathbb{C}}$), keeping in mind that a complex Lie subalgebra of $\mathfrak{g}\_\mathbb{C}=\mathfrak{g}\otimes\_{\mathbb{R}}\mathbb{C}$ is a complexification of a Lie subalgebra of $\mathfrak{g}$ if and only if it is invariant under the complex conjugation.
If $H$ is a connected semisimple group of Hermitian type and $f:H\to G$ is an embedding, then after a suitable conjugation, the image of a maximal compact subgroup $L<H$ is contained in $K,$ $f(H)=(f(H)\cap K)(f(H)\cap P)$ and $f(H)$ mod $K$ is isomorphic to a Hermitian symmetric space $H/L,$ which is thus realized in $G/K.$ This allows one to construct many examples by starting with (1) a group $H$ as above and (2) a faithful complex representation of $H$ that preserves an indefinite unitary form, so that $G=SU(p,q)$ if the signature is $(p,q).$
A final elementary observation is that there always exist nontrivial subgroups $H$ that do not arise from the previous construction, for example, $NA$ from the Iwasawa decomposition $G=NAK$ projects *onto* $G/K.$
| 2 | https://mathoverflow.net/users/5740 | 32561 | 21,131 |
https://mathoverflow.net/questions/32526 | 9 | Let $\mathcal S$ be a category with finite limits. The 2-category $\operatorname{Span}(\mathcal S)$ has the same objects as are in $\mathcal S$. For objects $X,Y$, the hom category in $\operatorname{Span}(\mathcal S)$ between $X$ and $Y$ is the category of diagrams in $\mathcal S$ of the form $X \leftarrow \bullet \rightarrow Y$, and morphisms are natural transformations of such diagrams that restrict to $\operatorname{id}\_X,\operatorname{id}\_Y\,\,$ at the endpoints. The 1-composition of 1-morphisms is given by the obvious pull-back diagrams: $$\{X\leftarrow A \rightarrow Y\} \circ \{Y\leftarrow B \rightarrow Z\} = \{X\leftarrow A\underset Y \times B \rightarrow Z\}$$
Thinking of $\mathcal S$ as a 2-category with only identity morphisms, the "spanishization" functor (does this functor have another name?) $\mathcal S \to \operatorname{Span}(S)$ is the identity on objects and takes $\{X \overset f \to Y\}$ to $\{X = X \overset f \to Y\}$. There is also an obvious isomorphism $\mathcal S \cong \operatorname{Hom}\_{\operatorname{Span}}(1,1)$, where $1 \in \mathcal S$ is the terminal object.
I believe that the correct weakened notion of "cartesian product" in a 2-category is that $X\times Y$ is determined up to equivalence (not isomorphism) as the representing object for the 2-functor $Z \mapsto \operatorname{Hom}(Z,X) \times \operatorname{Hom}(Z,Y)$, where on the right-hand side is the usual product of categories. (Incidentally, what's a good reference for $n$-Yoneda's Lemma?) Even if $\mathcal S$ has finite limits, or even all small limits, then I'm not sure whether $\operatorname{Span}(\mathcal S)$ has finite products. But for good enough categories $\mathcal S$, I feel like $\operatorname{Span}(\mathcal S)$ should also be good.
However, I believe that the product in $\operatorname{Span}(\mathcal S)$ is not the product in $\mathcal S$, i.e. spanishization does not respect limits. Provided all my beliefs are correct:
>
> Is there an easy description of the product in $\operatorname{Span}(\mathcal S)$ in terms of $\mathcal S$? Do any products at all exist in $\operatorname{Span}(\mathcal S)$?
>
>
>
| https://mathoverflow.net/users/78 | What is the product in the 2-category of spans? | If $\mathcal S = Set$, it looks to me like the product in $Span(\mathcal S)$ is given by disjoint union.
Define functors $F\_Z: Span(Z,X\sqcup Y) \to Span(Z,X)\times Span(Z,Y)$
$$
F\_Z(Z \leftarrow A \rightarrow X \sqcup Y) = ((Z \leftarrow A \times \_{X\sqcup Y}X \rightarrow X), (Z \leftarrow A \times \_{X\sqcup Y}Y \rightarrow Y))
$$
and $ G\_Z: Span(Z,X)\times Span(Z,Y) \to Span(Z,X\sqcup Y)$
$$
G\_Z((Z \leftarrow B \rightarrow X), (Z \leftarrow C \rightarrow Y)) = (Z \leftarrow A\sqcup B \rightarrow X\sqcup Y).
$$
Basically, if you have a span between $Z$ and $X\sqcup Y$, take the preimage of each component to get a pair of spans. I think it should be clear what happens on morphisms. Now, in Set (or maybe we just require that products distribute over coproducts), I think these form an equivalence. This should also behave well under maps $Z \to Z^\prime$.
This seems pretty weird to me, and it may not be correct (my category theory is pretty rudimentary). I am interested to find out more though, as it seems related to a question I was once asked about limits and colimits in cobordism categories (thinking of cobordisms as certain cospans of manifolds).
| 7 | https://mathoverflow.net/users/7762 | 32567 | 21,136 |
https://mathoverflow.net/questions/32570 | 9 | Suppose we have a recursively enumerable set of polynomials $\mathcal{P}=\{ p\_1({\bf x}), p\_2({\bf x}), \ldots\}, p\_i \in \mathbb{Z}[{\bf x}], {\bf x} = (x\_1, \ldots, x\_n)$. Let $V(\mathcal{P})$ denote the affine variety in $\mathbb{C}^n$ defined by $\mathcal{P}$. Is there an algorithm to compute $V(\mathcal{P})$? By the Nullstellensatz, we know that we need only use finitely many of the polynomials $p\_i$ to cut out $V(\mathcal{P})$. We can recursively compute varieties cut out by $\{p\_1, \ldots, p\_k\}$, for example by computing a Grobner basis for the radical ideal of $(p\_1,\ldots,p\_k)$. But is there a way to compute $k$ such that $V(\mathcal{P})=V(p\_1,\ldots,p\_k)$?
Please let me know if this question needs clarification or if I'm not using the correct notation.
Addendum: This problem was motivated by [this MO question](https://mathoverflow.net/questions/32000/conjugacy-problem-in-a-conjugacy-separable-group). It would follow from:
**If one has a finitely generated group $G$ with solvable word problem, for any $n$ can one compute the representation variety $G\to SL\_n(\mathbb{C})$?**
I view $G$ as being given as the homomorphic image of a free group $\\langle g\_1,\ldots,g\_k\\rangle$. Moreover, there is a Turing machine which takes as input any element $h\in \\langle g\_1,\ldots,g\_k\\rangle$ and tells if $h$ is trivial in $G$. The space of representations $\rho:G\to SL\_n(\mathbb{C})$ is an affine variety, with $kn^2$ variables given by the entries of the matrices of $\rho(g\_i)$. One can recursively generate polynomials
which are the entries of the matrices $\rho(h)-I$ which cut out the representation variety (together with $det(\rho(g\_i))-1$). So the algorithm should depend on how these polynomials are generated, if one wants to be able to compute the representation variety for each $n$.
I suspect that the answer is no, although I'm not sure how to generalize Borcherds or Groves' answers to this context.
If one could compute the representation variety, then one could determine if $G$ has a homomorphism to a finite group.
| https://mathoverflow.net/users/1345 | Variety defined by a recursively enumerable set of polynomials | There is no such algorithm, even if n is zero. Take p\_i to be 0 unless you can find a counterexample of length i to your favorite unsolved math problem (such as "is 2i+4 the sum of 2 primes?"), in which case p\_i is 1.
| 6 | https://mathoverflow.net/users/51 | 32571 | 21,138 |
https://mathoverflow.net/questions/32568 | 25 | Let $(T,X)$ be a discrete dynamical system. By this I mean that $X$ is a compact Hausdorff space and $T: X \to X$ a homeomorphism.
For example, take $X$ to be the sequence space $2^{\mathbb{Z}}$ and $T$ the Bernoulli shift. Then there is a dense set of periodic points, and there is another (disjoint) dense set of points whose orbits are dense (in view of topological transitivity). However, there are also points of $X$ that belong to neither of these sets---for instance, the sequence $(\dots, 1, 1, 1, 0, 0, 0, \dots )$.
This led to me the following:
>
> **Question:** Is there a discrete dynamical system $(T,X)$ such that
> every point has either a finite or a
> dense orbit? (Cf. the below caveats.)
>
>
>
There are a few caveats to add. We want both periodicity and topological transitivity to occur; this rules out examples such as rotations of the circle (where every point is of the same type, either periodic or with a dense orbit). So assume:
1. there is at least a point with dense orbit;
2. there is at least a periodic point;
3. also, assume there is no isolated point.
I've been thinking on and off about this question for a couple of days, and the basic examples of dynamical systems that I learned (shift spaces, toral endomorphisms, etc.) don't seem to satisfy this condition, and intuitively it feels like the compactness condition should imply that there are points which are "almost periodic," but not, kind of like the $(\dots, 1, 1, 1, 0, 0,0, \dots)$ example mentioned earlier. Nevertheless, I don't see how to prove this.
| https://mathoverflow.net/users/344 | Is there a dynamical system such that every orbit is either periodic or dense? | I believe you will find such examples for $X=\mathbb{C}$ and $T$ a rational map in
[Mary Rees](http://www.liv.ac.uk/~maryrees/maryrees.homepage.html), **Ergodic rational maps with dense critical point forward orbit**, *Ergodic Theory and Dynamical Systems* 4 (1984), 311-322. [official version](http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=2123740).
In my Ph.D. thesis, I showed that some of these even support a metric with respect to which these dynamical systems are ``hyperbolic''. This metric gives a notion of length of curves comparable to the usual metric on the Riemann sphere, but is defined by a function which is singular on a dense set of points on the sphere (the forward orbit of the critical point).
| 13 | https://mathoverflow.net/users/3993 | 32574 | 21,141 |
https://mathoverflow.net/questions/32554 | 59 | I'm teaching a short summer course on algebraic groups and it's time to talk about the Killing form on the Lie algebra. The students are all undergrads of varying levels of inexperience, and I try to make everything seem like it has a point (going back to the basic goals of "what is an algebraic group" and "what does this have to do with representation theory"). I am having a hard time justifying the Killing form from anything like first principles: it is useful, and I can prove theorems explaining why it is useful, but I can't think of an explanation of why it is reasonable to invent. The ideal answer to this question will be a "naive" explanation. Other interesting answers (which I would appreciate for myself) can be more sophisticated.
| https://mathoverflow.net/users/6545 | Why the Killing form? | Hi Ryan,
I presume given your description of the students that they know finite groups pretty well, and have seen the averaging idempotent $e=\frac{1}{|G|}\sum\_{g\in G} g$, and how this can be used to construct an invariant inner product on any representation of a finite group. Perhaps you can convince them that compact groups admit the same sort of averaging idempotents via integral, and so perhaps you can construct the invariant inner product on finite dimensional representations of a compact group in more or less direct analogy with finite groups. Then you can derive the properties the Killing form should satisfy on the Lie algebra by setting g=e^tX, and taking derivatives of the axioms of the group's inner product?
This is the closest connection I can think of to finite group theory, which is hopefully well-understood by, or at least familiar to, your students.
What do you think?
-david
| 38 | https://mathoverflow.net/users/1040 | 32576 | 21,143 |
https://mathoverflow.net/questions/32552 | 4 | Suppose a simple graph has $n$ vertices and $m$ edges. If the vertices are labelled, then each edge then corresponds to a transposition in a natural way. A theorem in Godsil and Royle's Algebraic Graph Theory, section 3.10, asserts the following:
If the graph is not a tree, then some product of the $m$ distinct transpositions is not an $n$-cycle.
For example, consider $K\_4-e$, i.e. the graph with vertex set $\[4\]$ and edges $\lbrace 1, 2\rbrace, \lbrace 1, 3\rbrace, \lbrace1, 4\rbrace, \lbrace 2, 3\rbrace$, and $\lbrace2, 4\rbrace$. Then $(1 2)(1 4)(1 3)(2 4)(2 3) = (1 4 2 3)$, a 4-cycle. However, $(1 2)(1 4)(1 3)(2 3)(2 4) = (1 3).$
Unfortunately for me, the proof is left as an exercise, which I cannot solve. Can anyone help?
| https://mathoverflow.net/users/7760 | characterization of trees in terms of products of transpositions | Notation: αβ means apply α then β.
If the graph is not a tree, then either it contains a cycle or it contains less than n−1 edges. In the latter case, we get a contradiction since less than n−1 transpositions cannot multiply to a big cycle.
So suppose that the graph contains a cycle (12...k), and assume that for each order of the edges, you get a big cycle. In particular, this holds for orderings where you take the cycle last. Fix some ordering of the other edges, and denote the interim product (without the edges of the cycle) by π. So for each product σ of the edges of the cycle, πσ is a long cycle.
For each i<j, there is some ordering of the cycle which produces a permutation containing the transposition (ij). Indeed, take (i i+1)(i+1 i+2)...(j-2 j-1) (j j+1) ... (k 1) (1 2) ... (i-1 i) (j-1 j).
Thus for each i<j, π(ij)τ is a big cycle, where τ does not involve i or j. In particular, π(i)≠j. Since this is true for all i≠j, it follows that π must be the identity. We get a contradiction since σ is not a big cycle.
| 4 | https://mathoverflow.net/users/7732 | 32578 | 21,145 |
https://mathoverflow.net/questions/32597 | 31 | John Hubbard recently told me that he has been asking people if there are compact surfaces of negative curvature in $\mathbb{R}^4$ without getting any definite answers. I had assumed it was possible, but couldn't come up with an easy example off the top of my head.
In $\mathbb{R}^3$ it is easy to show that surfaces of negative curvature can't be compact: throw planes at your surface from very far away. At the point of first contact, your plane and the surface are tangent. But the surface is everywhere saddle-shaped, so it cannot be tangent to your plane without actually piercing it, contradicting first contact.
This easy argument fails in $\mathbb{R}^4$. Can the failure of the easy argument be used to construct an example? Is there a simple source of compact negative curvature surfaces in $\mathbb{R}^4$?
| https://mathoverflow.net/users/2510 | Compact surfaces of negative curvature | You will find examples (topologically, spheres with seven handles) in section 5.5 of *Surfaces of Negative Curvature* by E. R. Rozendorn, in *Geometry III: Theory of surfaces*, Yu. D. Burago VI A. Zalgaller (Eds.) EMS 48.
Rozendorn tells us that «from the visual point of view, their construction seems fairly simple.» Well...
| 27 | https://mathoverflow.net/users/1409 | 32598 | 21,161 |
https://mathoverflow.net/questions/32607 | 6 | Shimura (Crelle 221, 1966) considers the elliptic curve $E:y^2+y=x^3-x^2$ (although he doesn't use this equation) of conductor $11$ whose associated modular form is
$$
q\prod\_{k=1}^{+\infty}(1-q^k)^2(1-q^{11k})^2=\sum\_{n=1}^{+\infty}c\_nq^n
$$
where $q=e^{2i\pi\tau}$ and $\tau$ is in the upper half of $\bf C$. For a prime $l$, he denotes by $K\_l$ the extension of $\bf Q$ obtained by adjoining the $l$-torsion points of $E$ and shows that if $l\in[7,97]$, then ${\rm Gal}(K\_l|{\bf Q})$ is isomorphic to ${\rm GL}\_2({\bf F}\_l)$.
**Question.** Is ${\rm Gal}(K\_l|{\bf Q})$ now known to be isomorphic to ${\rm GL}\_2({\bf F}\_l)$ even for $l>97$ ?
Even if the faithful representation ${\rm Gal}(K\_l|{\bf Q})\rightarrow{\rm GL}\_2({\bf F}\_l)$ fails to be surjective for a few $l>97$, does the recent proof of Serre's modularity conjecture not imply the
**Statement**. For every prime $l>5$ and every prime $p\neq11,l$, the characteristic polynomial of ${\rm Frob}\_p$ (thought of as an element of ${\rm GL}\_2({\bf F}\_l)$) is
$\equiv X^2-c\_pX+p \pmod l$ ?
Shimura shows this only for $l\in[7,97]$.
**Addendum.** (2010/07/24) Looking at Shimura's paper beyond the first page shows that he actually proves (Section 3) that the characteristic polynomial for the action of ${\rm Frob}\_p$ on the $l$-adic Tate module $T\_l(E)$ is
$X^2-a\_pX+p\in{\bf Z}\_l[X]$
for all primes $l$ and $p\neq11, l$ and (Section 6) that $a\_p=c\_p$ for all $p\neq11$. And yes, he does use the Eichler-Shimura relation.
| https://mathoverflow.net/users/2821 | Reciprocity law for number fields defined by torsion points of modular elliptic curves | The first question is answered in Serre's
[[Propriétés galoisiennes des points d'ordre fini des courbes elliptiques,
*Invent. Math.* **15**:4 (1972) 259--331]](http://dx.doi.org10.1007/BF01405086), on page 304, section 5.2, exactly for this curve. In general this paper give a good way to determine for which $\ell$ the mod-$\ell$ representation is not surjective. *Sage* can do that efficiently for a given curve.
For every prime $p$ different from $\ell$ and $11$, the characteristic polynomial of $\rm{Frob}\_p$ is indeed $T^2 - c\_p T +p$ in $ \mathbb F\_{\ell}[T]$. The isomorphism
$\rm{Gal}(K \_{\ell}/\mathbb Q) \to \rm{Aut}(E[\ell])=\rm{GL} \_2(\mathbb F \_{\ell})$ sends $\rm{Frob}\_p$ to the Frobenius endomorphism $\phi:E[\ell] \to E[\ell]$ on $E/\mathbb{F}\_p$.
Your $c\_p$ is the trace of $ \phi$ and $p$ is the determinant of it since the Eichler--Shimura relation shows that $c\_p$ is the Fourier coefficent of the associated modular form. See [this answer](https://mathoverflow.net/questions/19390/intuition-behind-the-eichler-shimura-relation/19399#19399) for why it is so.
| 5 | https://mathoverflow.net/users/5015 | 32616 | 21,173 |
https://mathoverflow.net/questions/32437 | 5 | ### Universal codings of integers
A (binary) *coding of the integers* is a prefix-free code of the natural numbers, whose codewords are non-decreasing in size. A coding is *universal* if it is short enough (log n + o(log n)), but that's not important.
Some examples:
* The unary coding 0, 10, 110, ...; code length is n
* Code first the length of the number in unary, then the number itself in binary; code length is about 2log n
* Code first the length of the number using the previous coding, then the number itself in binary; code length is about log n + 2log log n
* ...
* Diagonalize the construction to get code length of log n + log log n + ... + 2log\* n
* Continue this way through the constructible ordinals
The diagonalized code, known as the $\omega$-code, is due to Peter Elias.
### A partial ordering of codes
The sequence of codes above are progressively better, in the following sense:
* A coding a is better than a coding b if |b(n)| - |a(n)| tends to infinity.
There are some natural questions to ask:
* Is there a best code?
* If not, is there an optimal sequence of codes?
As it turns out, not only is there no *best* code, but given any sequence of codes, we can always find a code which is better than *all* of them; the proof from one of Hausdorff's papers (Untersuchungen über Ordnungtypen V from 1907) can be adapted to our setting.
### Scales
The best thing that can be hoped for is a *chain* of codes which is cofinal for the poset of codes, i.e. a set of mutually comparable codings, such that for each arbitrary coding, our scale contains a superior one (such a beast Hausdorff called a Pantachie).
The problem of scales is well-known, and it is easy to show the existence of a scale given CH (following Hausdorff's steps). In other settings (and possibly this one), existence already follows from MA. However, most of the literature deals with somewhat different posets, and it is not clear that their results apply in this case.
Here are some pointers:
* *Hausdorff Gaps and Limits* by Frankiewicz and Zbierski, which deals with the ordering f > g if f(n) > g(n) infinitely often.
* *Gaps in $\omega^\omega$* by Marion Scheepers, which deals with the ordering f > g if f(n) - g(n) tends to infinity.
In their settings, Hechler forcing can be used to produce worlds in which there is no scale.
>
> Is the existence of scale (in the context of monotone codings of integers) independent of set theory?
>
>
>
### Codings and series
Some easy reductions connect our problem with problems involving convergent series and divergent series satisfying some extra conditions, which stem from our monotonicity requirements; the key is Kraft's inequality, stating that a code with codeword lengths wi exists iff the sum $\sum 2^{-w\_i}$ converges.
The reductions are most easily stated if we extend our posets with some equivalence relation. We then say that two posets are *interlacing* if there are two order-preserving mappings (between the two posets in both directions) which are pseudo-inverses, i.e. their composition sends a point to an equivalent one. Given two interlacing posets, one has a scale iff the other one has a scale.
The following posets are interlacing:
* Arbitrary (non-monotone) codes, with a < b if b is better than a, and a ~ b if |a(n)-b(n)| is bounded.
* Convergent positive series, with a < b if b(n) = o(a(n)), and a ~ b if a(n) = O(b(n)) and b(n) = O(a(n)).
* Divergent positive series (reverse definition of <).
Monotonicity complicates the picture (the corresponding series are no longer arbitrary) but seems necessary, since one can give a non-monotone code with the property that no monotone code is better than it.
### Effective and efficient codings
The motivation behind the question is the actual usage of universal codings by computer engineers. New, impractical methods of codings are suggested all the time, but no one seems to have tackled the fundamental question.
This prompts us to ask similar questions for effective (computable) codings.
>
> Can classical recursion theory hierarchies be adapted to the setting of codes?
>
>
>
It would be nice to get an analog of the fast-growing hierarchy, for example.
We could further wonder what happens if we ask our coding procedure to be efficient, for example linear-time computable.
| https://mathoverflow.net/users/7732 | When do cofinal chains of universal codings of the integers exist? | I believe I can answer your first question. But the answer involves forcing, which I cannot explain here (see Kunen's Set Theory. An introduction to independence proofs).
Assuming CH, there is a scale of codes. Why? Enumerates all codes as
$(c\_\alpha)\_{\alpha<\omega\_1}$. Construct a sequence $(a\_\alpha)\_{\alpha<\omega\_1}$
such that for each $\alpha$, the code $a\_\alpha$ is better than $c\_\alpha$ and all
$a\_{\beta}$, $\beta<\alpha$. This is possible since for every countable set of codes
there is one code that is better than all of them, if I understand you correctly.
Now $(a\_\alpha)\_{\alpha<\omega\_1}$ is a scale of codes.
For the other consistency result, namely ZFC is consistent with the non-existence of scales, consider the partial order of finite initial segments of monotone prefix-free codes, where a finite initial segment $c$ is stronger than $d$ ($c\leq d$) if $c$ extends $d$.
This partial order is countable and every element has two extensions that don't have a common
extension. This implies that the partial order is forcing equivalent to the so called
Cohen forcing.
Whenever $c$ is a finite initial segment of a code, $n$ is a natural number and $a$ is a code, then $c$ can be extended to a finite initial segment $d$ of a code such that
for some $m>n$, $d$ already contains the code word of $m$ and this code word for $m$ is longer than the code word for $m$ that $a$ has.
Similarly, $c$ can be extended to a finite $d$ that for some $m>n$ has a shorter code word for $m$ than $a$ has.
This shows that forcing with this partial order adds a code that is incomparable with all
codes in the ground model.
Now we start from a model of set theory that satisfies CH and force over it with a finite support product of $\aleph\_2$ copies of the countable partial order defined above.
This forcing adds a family $(c\_\alpha)\_{\alpha<\omega\_2}$ generic codes.
(CH fails in this generic extension).
I claim that no subfamily of size $\aleph\_1$ of this family of codes has an upper bound.
Why? Let $A\subseteq\omega\_2$ be of size $\aleph\_1$. By the properties of Cohen forcing
(c.c.c. in particular) we may assume, after enlarging $A$ if necessary, that
$A$ is already in the ground model.
Now, whenever $a$ is a code in the generic extension, then $a$ has a name that only depends
on countably many of the $c\_\alpha$'s.
Take $\beta\in A$ outside this countable set of indices.
Then, by the argument above, $c\_\beta$ and $a$ are incomparable as codes
(this is because $c\_\beta$ is generic over a model containing $a$) and hence
$a$ is not an upper bound of the $c\_\alpha$, $\alpha\in A$.
A similar argument shows that in the model of set theory that we have constructed (which is in fact Cohen's original model that refutes CH) no set of codes of size $<\aleph\_2$
is cofinal. (Why? If $C$ is a set of $\aleph\_1$ codes, then there still is some
$\alpha<\omega\_2$ such that $c\_\alpha$ is generic over a model that contains $C$
and now no element of $C$ is an upper bound for $c\_\alpha$.)
We now have a model of set theory in which there is an unbounded set of codes of size $\aleph\_1$ but no cofinal set of size $<\aleph\_2$. This implies that there is no scale.
I hope it is possible to get something out of this answer.
I am fully aware that I am using forcing jargon here, but to really give a complete proof
would take a lot of space and time. I would guess that the code problem can actually be reduced to some partial order which has been studied in the literature.
| 4 | https://mathoverflow.net/users/7743 | 32622 | 21,177 |
https://mathoverflow.net/questions/32626 | 15 | This question is mainly a curiosity, but comes from a practical experience (all players of *Race for the galaxy*, for example, must have ask themselves the question).
Assume I have a deck of cards that I would like to shuffle. Unfortunately, the deck is so big that I cannot hold it entirely in my hands. Let's say that the deck contains $kn$ cards, and that the operation I can perform are: 1. cut a deck into any number of sub-decks, without looking at the cards but remembering for all $i$ where the $i$-th card from top of the original deck has been put; 2. gather several decks into one deck in any order (but assume that we do not intertwin the various decks, nor change the order inside any of them); 3. shuffle any deck of at most $n$ cards. Assume moreover that such a shuffle consist in applying an unknown random permutation drawn uniformly.
Here is the question: is it possible to design a finite number of such operations so that the resulting deck has uniform law among all possible permutations of the original deck? If yes, how many shuffles are necessary, or sufficient, to achieve that ?
The case $k=2$ seems already interesting.
| https://mathoverflow.net/users/4961 | How to shuffle a deck by parts? | A truly uniform distribution, no. (Well, your question is not completely well posed, but I will argue this for most ways of making it so.) There are $(kn)!$ factorial ways to shuffle a deck of $kn$ cards. So you want each permutation to occur with probability $1/(kn)!$. In particular, for every prime $p \leq kn$, you want $p$ to occur in the denominator of the probability that each permutation occurs. Let's look at your operations: Reordering $i$ decks can only introduce primes $\leq i$. Perfectly shuffling an $n$ card deck can only introduce primes $\leq n$. Cutting depends on what mathematical model you use for cutting; if all cut points are equally likely, you only get primes dividing $n(n-1)\ldots (n-i+1)$. I imagine other models of cutting will cause similar problems.
The more commonly studied question is how to get a probability distribution that is extremely close to random. There are lots of good results on this; see [Trailing the Dovetail Shuffle to its Lair](https://projecteuclid.org/journals/annals-of-applied-probability/volume-2/issue-2/Trailing-the-Dovetail-Shuffle-to-its-Lair/10.1214/aoap/1177005705.full).
| 14 | https://mathoverflow.net/users/297 | 32631 | 21,180 |
https://mathoverflow.net/questions/32624 | 21 | Dirichlet's theorem states that for any coprime $k$ and $m$ there exists infinitely many primes $p$ such that $p \equiv k \pmod m$.
Some special cases of this theorem are easy to prove without any analytic methods. Those cases include, for example, $m=4, k=1$ and $m=4, k=3$.
Both cases could be proved by considering first $t$ prime numbers $p\_i \equiv k \pmod m$ and constructing a new number which is proved to have prime divisor $p \equiv k \pmod m$ that is not equal to any $p\_i$.
For case $m=4, k=1$ we can consider number $(p\_1 p\_2 \cdots p\_t)^2 + 1$. And for case $m=4, k=3$ number $4p\_1 p\_2 \cdots p\_t + 3$.
Those constructions could also be applied to some other special cases as well.
Are there any other special cases for which there exists a simple non-analytic proof which don't use any of those two constructions?
| https://mathoverflow.net/users/7079 | Special cases of Dirichlet's theorem | There is a simple non-analytic proof for $p\equiv 1 \bmod n$; see e.g. Proposition $3$ in [this note](https://static1.squarespace.com/static/57bf2a6de3df281593b7f57d/t/57bf68b26a49636398ee2dce/1472161971095/primes1mod4.pdf). The proof gives a (Euclidean) argument that infinitely many primes divide the values of an integer-coefficient polynomial on the integers, and then notes that the prime divisors of the values of the $n$-th cyclotomic polynomial either divide $n$ or have remainder $1$ upon division by $n$. (The proof is well-known; I don't know the originator.) By the way, the note also contains a cute analytic argument for $p\equiv 1 \bmod 4$ giving bounds on the partial sums of the reciprocals of such primes; the argument uses representations via sums of two squares.
Edit: [This paper](https://projecteuclid.org/journals/functiones-et-approximatio-commentarii-mathematici/volume-35/issue-none/Primes-in-Certain-Arithmetic-Progressions/10.7169/facm/1229442627.full) by Murty and Thain discusses obstructions to Euclid-style proofs for various congruence classes. I believe that a proof has been carried out for $p\equiv a\bmod b$ for $(a, b)=1$ for $b= 24$ in the style of Euclid, however.
Here is an open-access [paper](http://www.math.uconn.edu/%7Ekconrad/blurbs/gradnumthy/dirichleteuclid.pdf) by Keith Conrad expositing this impossibility theorem and giving some background.
Edit 2: Here is the [paper](https://www.jstor.org/stable/2310975) I recalled with the Euclidean proof for $b= 24$; unfortunately it is not open-access. It is JSTOR however so many of you likely have institutional access.
| 21 | https://mathoverflow.net/users/6950 | 32635 | 21,184 |
https://mathoverflow.net/questions/32623 | 2 | if $P\_{1}$ and $P\_{2}$ are distinct places of equal degree of the function field F/K, and $\sigma$ is a K-field automorphism, such that $\sigma(P\_{1})=P\_{2}$. then, does $\deg (P\_{1}\cap K(x))=\deg (P\_{2}\cap K(x))$, where K(x) is the rational function field?
in particular, is this true over the hermitian function field?
| https://mathoverflow.net/users/7773 | behavior of places of a function field under automorphism | No, not in general, that is not without particular requirements for $x$:
take $F=\mathbb{R}(y)$, the rational function field in one variable over the reals.
Then the equation $\sigma (y)=y+1$ determines an automorphism of $F/\mathbb{R}$.
Let $P\_1$ be the place associated to the polynomial $y^2+1$; then $\deg (P\_1)=2$.
Let $P\_2 := \sigma (P\_1)$; then $P\_2$ is associated to the polynomial $y^2+2y+2$ and (automatically) $\deg (P\_2)=2$.
Let $x := y^2+1$; then $[F:\mathbb{R}(x)]=2$ and $P\_1|\_{\mathbb{R}(x)}$ has degree $1$.
On the other hand $yP\_2 $ either equals $i-1$ or $-i-1$. In both cases $xP\_2$ is non-real and thus $\deg (P\_2)=2$.
H
| 3 | https://mathoverflow.net/users/3556 | 32636 | 21,185 |
https://mathoverflow.net/questions/32620 | 13 | Lehmer's conjecture for Ramanujan's tau function,
$$
\Delta(q)=q\prod\_{n=1}^\infty(1-q^n)^{24}=\sum\_{m=1}^\infty\tau(m)q^m,
$$
asserts that $\tau(m)$ never vanishes for $m=1,2,\dots$.
In the [recent question](https://mathoverflow.net/questions/31058/)
it was asked why it is important to have the nonvanishing.
I am wondering whether there are upper bounds,
unconditional or conditional (modulo some other
known conjectures), in terms of $x\in\mathbb R\_+$ for the number
of integers $m\le x$ satisfying $\tau(m)=0$ (maybe better,
for the number of primes $p\le x$ satisfying $\tau(p)=0$)?
It looks like the series $\Delta(q)$ is very far from being "lacunary".
But besides Deligne's upper bound $|\tau(m)|\le d(m)m^{11/2}$
(where $d(\ )$ counts the number of divisors) and the lower bound
$$
\operatorname{card}\lbrace\tau(n):n\le x\rbrace\ge \operatorname{const}\cdot x^{1/2}e^{-4\log x/\log\log x}
$$
from
[[M.Z. Garaev, V.C. Garcia, and S.V. Konyagin,
A note on the Ramanujan $\tau$-function, *Arch. Math. (Basel)* **89**:5 (2007) 411--418]](http://dx.doi.org/10.1007/s00013-007-2246-8)
for the distribution of tau values, I cannot find any quantitative progress
towards Lehmer's original question.
| https://mathoverflow.net/users/4953 | Lehmer's conjecture for Ramanujan's tau function | One of the canonical references for questions like this is Serre's "Quelques applications du theoreme de densite de Chebotarev", Publ. Math. IHES 54. He proves, for example, that the number of primes $0\leq p \leq X$ with $\tau(p)=0$ is $\ll X (\log{X})^{-3/2}$ unconditionally, and is $\ll X^{\frac{3}{4}}$ under GRH.
| 5 | https://mathoverflow.net/users/1464 | 32643 | 21,190 |
https://mathoverflow.net/questions/32641 | 17 | Robin Chapman [introduced me](https://mathoverflow.net/questions/32126/function-with-range-equal-to-whole-reals-on-every-open-set/32127#32127) to [Conway's Base 13 Function](http://en.wikipedia.org/wiki/Conway_base_13_function). Now, my real analysis is a tiny bit rusty, so maybe my question has a really simple and quick answer, but here it goes:
Consider the support set of the base-13 function, is the set Lebesgue measurable? And if so, does it have non-zero measure?
| https://mathoverflow.net/users/3948 | Is Conway's base-13 function measurable? | Call the support set $S$. The answer is yes it is Lebesgue measurable and no, it has zero measure. It is even Borel measurable, which would take a tiny bit more effort to prove.
Note that $S$ is included in the set $T$ of numbers in which the "digits" '+','-', and '.' appear finitely many times in the "base 13 expansion". But almost all numbers are normal, hence have all digits appearing infinitely often. So the Borel set $T$ has measure zero, hence $S$ is Lebesgue measurable with measure zero.
| 16 | https://mathoverflow.net/users/5963 | 32647 | 21,193 |
https://mathoverflow.net/questions/32652 | 3 | I am a student almost without background on algebraic geometry (but I do know basic graduate algebra and topology). Now I am trying to understand something about algebraic stacks.
I want to start with this short AMS article first: <http://www.ams.org/notices/200304/what-is.pdf>
This article tries to define algebraic stacks by introducing the notion of the moduli
space of elliptic curves. However, I do not know anything about elliptic curves and
there are just too many possible books.
Can someone give some suggestions about how to proceed to learn enough elliptic curves so as to understand something about algebraic stacks? I also welcome other suggestions. Thank you.
| https://mathoverflow.net/users/7780 | Elliptic curves and algebraic stacks | You should really learn algebraic geometry first. After you're done with that, try reading Mumford's paper "Picard Groups of Moduli Problems", which is a fascinating window into the mind of one of the people who later invented algebraic stacks as he was himself figuring out what you are also trying to figure out. From that, you should have a sketch of how the properties of elliptic curves contribute to the properties of their moduli stack; make sure you work out or look up the various things he says about them. Then you can read a regular treatise on stacks.
| 8 | https://mathoverflow.net/users/6545 | 32653 | 21,196 |
https://mathoverflow.net/questions/24545 | 10 | Intuitively, Gromov-Witten theory makes perfect sense. Via Poincare duality, we look at the cohomology classes $\gamma\_1, \ldots, \gamma\_n$ corresponding to geometric cycles $Z\_i$ on a target space $X$, pull them back and then the integral
$$\langle \gamma\_1 \cdots \gamma\_n\rangle=\int\_{\overline{\mathcal{M}}\_{g,n}(X,\beta)^{vir}}ev\_1^\*\gamma\_1 \smile \cdots \smile ev\_n^\*\gamma\_n$$
should count the number of curves whose intersection with the given cycles is non-empty.
However, we also have the ψ-classes (or "gravitational descendants") arising from the moduli space $\overline{\mathcal{M}}\_{g,n}$ which are the chern classes of the $i$-th cotangent line bundle to a given $(C, x\_1, \ldots, x\_n) \in \overline{\mathcal{M}}\_{g,n}$.
So what, geometrically, do these represent? The fact that they arise from $\overline{\mathcal{M}}\_{g,n}$ means that the inclusion of a ψ-class places restriction on the geometry of the curves which we count; that much is clear. What is this restriction?
The reason that I am curious is that I am trying to evaluate the GW-invariants corresponding to maps which have components collapsing to an A1 singularity (i.e. a $B\mathbb{Z}/2$), but such that not all of the curve collapses. It has been mentioned in passing that including a ψ-class could help with this, and while the little I understand makes this sound plausible, I don't exactly see why.
So what are ψ-classes? Can I use them to split my curve up into parts so that a fixed component lands on my stacky point, while the rest of it does whatever else curves do?
| https://mathoverflow.net/users/1703 | What is the geometry behind psi classes in Gromov-Witten theory? | The following answer is unfortunately not quite correct, but it may be useful anyway. I will of course be ignoring any virtual fundamental class issues.
Imagine that you are computing a Gromov-Witten invariant where you require the i-th marked point to land at a specific point (i.e. your i-th insertion γi is the class of a point), and now lets add aditionally the i-th psi-class as an insertion. You can restrict to the subspace of maps with $f(x\_i) = x$ for some generic choice of $x \in X$. Fixing an arbitrary non-trivial map $\Phi \colon T\_x \to k$ gives you by composition a map from the relative tangent bundle of the universal curve over $M\_{g, n}(X)$ at the section xi to the trivial line bundle, in other words a section $\phi$ of the relative cotangent bundle of the universal curve. It will vanish on curves which are tangent to a hypersurface through x with tangent direction matching the zero-locus of the map $\Phi$.
So you can think of Gromov-Witten invariants with psi-classes as counting maps which additionally satisfy tangency conditions at the marked points.
Why is this not correct? The zero locus of $\phi$ computes the Chern class of the relative cotangent bundle at $x\_i$ over Mg, n(X), which is not the same as the pull-back of the $\psi$-class from Mg, n. Insertions of the former are sometimes called "gravitational ancestors", and the difference to gravitational descendants is described explicitly in alg-geom/9708024.
| 13 | https://mathoverflow.net/users/7437 | 32670 | 21,208 |
https://mathoverflow.net/questions/32637 | 28 | I've been trying for a while to get a real concrete handle on the relationship between representations and modules. To frame the question, I'll put here the standard situation I have in mind:
A ring $R$ lives in the category Ab of Abelian groups as an internal monoid $(\mu\_R, \eta\_R)$. A module is then just an Abelian group $A$ and a map $m : R \otimes A \rightarrow A$ that commutes with the monoid structure in the way you'd expect.
Alternatively, take an Abelian group $A$ and look at its group of endomorphisms $[A,A]$. This has an internal monoid $(\mu\_A, \eta\_A)$ just taking composition and identity. Then a representation is just a monoid homomorphism $(R, \mu\_R, \eta\_R) \rightarrow (A, \mu\_A, \eta\_A)$ in Ab. I.e. a ring homomorphism.
But then, Ab is monoidal closed, so these are the same concept under the iso
$$\hom(R\otimes A, A) \cong \hom(R, [A,A])$$
This idea seems to work for any closed category where one wants to relate a multiplication to composition. So, my question is, since these things are isomorphic in such a general context, why are they taught as two separate concepts? Is it merely pedagogical, or are there useful examples where modules and representations are distinct?
| https://mathoverflow.net/users/800 | When are modules and representations not the same thing? | Here is my representation theorist's perspective: the key difference between representations and modules is that representations are "non-linear", whereas modules are "linear". I'll concentrate on the case of groups as the most familiar, but this applies more generally.
As Greg has already mentioned, in the most general sense, a representation is a homomorphism $f:G\to H,$ and usually there is no linear (or additive) structure on $H$, i.e. the set $f(g)$ need not be closed under sums; in fact, if $H$ is a non-abelian group, e.g. the symmetric group, the notion of sum doesn't even make sense (if $H=GL(V)$ then we may view its elements as endomorphisms of $V$ and add them, but this is unnatural since, by definition, $f$ is compatible with *multiplicative* structure). By contrast, a module involves a *linear* action $G\times V\to V,$ which is then "completed" by allowing arbitrary linear combinations, leading to certain technical advantages.
Here is an example of a construction that is very useful and makes perfect sense module-theoretically, but not representation-theoretically: change of scalars. Given a module $M$ over a group ring $R[G]$ and a commutative ring homomorphism $R\to S,$ one gets a module $S\otimes\_R M$ over the group ring $S[G]$. Common examples involve extensions of scalars (e.g. from $\mathbb{R}$ to $\mathbb{C}$, from a field $K$ of definition to the splitting field, from $\mathbb{Z}$ to $\mathbb{Z}\_p$) and, more to the point, reductions (e.g. from $\mathbb{Z}$ or $\mathbb{Z}\_p$ to $\mathbb{Z}/p\mathbb{Z}$). The module language is, predictably, also very useful in providing categorical descriptions of various operations on representations, such as functors of induction and restriction,
$$Ind\_H^G: H\text{-mod}\to G\text{-mod}\ \text{ and }\ Res\_H^G: G\text{-mod}\to H\text{-mod},$$
where $H$ is a subgroup of $G,$ or the monoidal structure on $G$-mod.
Finally, here are two illustrations of the complementary nature of the two approaches besides the group case, in linear algebra. A single linear transformation $T:V\to V$ on a finite-dimensional vector space $V$ over $K$ is most naturally viewed as a representation (no additive structure); in this case, it's a representation of the quiver with a single vertex and a single loop. From this point of view, classification up to isomorphism is a problem about conjugacy classes of linear transformations,
$$T\to gTg^{-1},\ g\in GL(V).$$
By contrast, in the module style description we associate with $T$ a module over the ring $K[x]$ of polynomials in one variable over $K$ and classification problem reduces to the structure of modules over $K[x]$, which is a PID, with all the usual consequences. (*Here the module picture is more illuminating.*) If we consider a linear operator $S:V\to W$ between two different vector spaces,
$$S\to hSg^{-1},\ g\in GL(V),\ h\in GL(W),$$
and a classification up to isomorphism is accomplished by row and column reduction. The corresponding quiver $\circ\to\circ$ is a single arrow connecting two distinct vertices, but its path algebra is less familiar. (*Here the representation theory picture is more illuminating.*)
| 15 | https://mathoverflow.net/users/5740 | 32677 | 21,215 |
https://mathoverflow.net/questions/32666 | 29 | Hello,
recently, I've been reading some algebra and sometimes I stumble up on the concept of something "being too big" to be a set. An example, is given in (<http://www.dpmms.cam.ac.uk/~wtg10/tensors3.html>) , where he writes, "Let B be the set of all bilinear maps defined on VxW. (That's the naughtiness - B is too big to be a set, but actually we will see in a moment that it is enough to look just at bilinear maps into R.)" (where V and W are vector spaces over R). This is, too big to be a set, but why?
My general question is this, when is something too big to be a set? What is it instead? Why have we put these requirements on the definition? Do we run into any problems if we let, say, B as defined up there be a set? What kind of problems do we run into?
| https://mathoverflow.net/users/7607 | When is something too big to be a set? | I just want to give a refinement of other answers so far, as well as a different point of view (namely, that of a person who knows little about set theory but who also encounters these kinds of issues).
As others have mentioned, the root cause of the problem is that there are big logical problems with considering "the set of all sets". Unless you would like to learn more about set theory, you needn't concern yourself with what these problems are (though Russel's paradox is fairly elementary and kind of fun). It is just one of those facts of life that non-set theorists learn to live with and that set theorists learn to love. The non-existence of the set of all sets forces us to abandon other putative sets, such as "the set of all groups", "the set of all vector spaces", "the set of all manifolds", etc. For example, it is possible to equip any set with the structure of a group and so if we were able to build the set of all groups then we would necessarily have also build the set of all sets. This is almost always what people mean when they claim that a certain construction is "too big to be a set" - the construction invokes a sloppy use of set theory language that taken literally accidentally constructs the set of all sets as a byproduct. In your case, the existence of the set of all bilinear maps on $V \times W$ constructs as a byproduct the set of all vector spaces over $R$ (every bilinear map has to have a target), and if there were a set of all vector spaces over $R$ then there would be a set of all sets.
This is probably not the last time you will encounter this sort of issue. In basically every case, however, there is a trick that swoops in and saves the day. Generally the idea is to observe that you don't actually need all of the flexibility that you tried to give yourself by constructing a non-set, and that it is enough to consider a simpler object (in your case the set of all bilinear maps from $V \times W$ to $\mathbb{R}$) which is small enough to be a set but big enough to have the property that you want (in your case you want it to function as a sort of universal bilinear pairing between $V$ and $W$).
Ultimately I regard these sorts of concerns as analogous to the "end user agreements" that you have to certify you've read whenever you install a Microsoft product or sign up for a gmail account. I'm sure all that fine print is important, but I feel like I would have to become a lawyer to understand it all. And just as in that case, you don't have to be a set theorist to understand how to resolve these sorts of issues most of the time - usually it just requires you to capture the flexibility present in what you are already working on.
Just recently I was reading about an object which was given as the quotient by a certain equivalence relation of the set of all pairs $(T, H)$ where $H$ is a Hilbert space and $T$ is a certain kind of operator on $H$. The book pointed out that one cannot consider the set of all Hilbert spaces, but that the set theoretic difficulties can be resolved by proving that every pair $(T', H')$ is equivalent to a pair $(T, H)$ on a fixed Hilbert space $H$. So the problem was avoided by exploiting some inherent flexibility in the equivalence relation under consideration. This sort of behavior is quite typical.
| 30 | https://mathoverflow.net/users/4362 | 32681 | 21,219 |
https://mathoverflow.net/questions/32690 | 17 | This question is influenced by the following riddle:
>
> You are given a rectangular set in the plane with a rectangular hole cut out (in any orientation). How do you cut the region into two sets of equal area?
>
>
>
*SPOILER ALERT!!* - The answer is that you can cut through the center of both rectangles, and because any line through the center of a rectangle divides it into two pieces of equal area, this cut works.
---
I have been wondering about the following question - What sort of conditions on a set guarantee that it has this property, that any line through the center of mass divides it into two regions of equal area?
The only thing that I have been able to think of is $\pi$ rotational symmetry around the center of mass. For example the rectangle has this symmetry. This symmetry means that in fact the two regions cut by any line through the center of mass are congruent, and not just equal area.
Thus, my question is:
>
> Suppose that we have a planar (measurable) set $A \subset \mathbb{R}^2$ (with positive measure). If there is a point $a\in \mathbb{R}^2$ such that: for any line $\ell \subset \mathbb{R}^2$ through $a$, denoting the regions of $A$ on either side of the line $B$ and $C$ then we have $|B| = |C|$ (Lebesgue measure), then is it necessarily true that
> (1) $a$ is the center of mass of $A$ and (2) that $A$ has $\pi$ rotational symmetry around $A$ in the a.e. sense, i.e. if $\tilde A$ is $A$ rotated by $\pi$ around $a$ then the symmetric difference between $A$ and $\tilde A$ has measure zero, i.e. $ |A \ \Delta \ \tilde A| = 0$.
>
>
>
I feel like (1) should be true, but I'm not so sure about (2). If the answer to (2) is no, then what sort of sufficient conditions are there? I'm mostly just curious about the answer, so by all means feel free to strengthen the assumptions on $A$, like requiring it to be a region bounded by a smooth boundary, etc. Thanks!
| https://mathoverflow.net/users/1540 | Planar sets where any line through the center of mass divides the set into two regions of equal area. | Assume that $A$ is compact and convex. If there is a point $P$ such that any line through it is a bisector of $A$ then $A$ has to be centrally symmetric. In fact a stronger result is known (see the [paper](https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-16/issue-1/A-theorem-on-partitions-of-mass-distribution/pjm/1102995091.full) by V. Menon):
>
> **Theorem.** Let $K$ be a compact convex figure. The following
> four statement are equivalent:
>
>
>
>
> * the point $P$ through which three bisectors of $K$ pass is unique,
> * all bisectors of $K$ are concurrent in $P$;
> * there exists a point $P$ such that any line through it is a bisector of $K$;
> * $K$ is a centrally symmetric figure with $P$ as its centre.
>
>
>
| 19 | https://mathoverflow.net/users/5371 | 32696 | 21,229 |
https://mathoverflow.net/questions/32705 | 35 | If $f: [a,b] \to V$ is a (nice) function taking values in a vector space, one can define the definite integral $\int\_a^b f(t)\ dt \in V$ as the limit of Riemann sums $\sum\_{i=1}^n f(t\_i^\*) dt\_i$, or as the final value $F(b)$ of the solution $F: [a,b] \to V$ to the ODE problem $F'(t) = f(t); F(a) = 0$.
In a similar spirit, given a (nice) function $f: [a,b] \to {\mathfrak g}$ taking values in a Lie algebra $\mathfrak g$ of a Lie group $G$, one can define the multiplicative definite integral, which for sake of discussion I will denote $\Pi\_a^b \exp(f(t)\ dt) \in G$, either as the limit of Riemann products $\prod\_{i=1}^n \exp(f(t\_i^\*) dt\_i)$ (with the product read from left to right), or as the final value $F(b)$ of the solution $F: [a,b] \to G$ of the ODE $F'(t) = F(t) f(t); F(a) = 1$.
Thus, for instance, when the Lie algebra is abelian, the multiplicative integral is just the exponential of the ordinary integral,
$$\Pi\_a^b \exp(f(t)\ dt) = \exp( \int\_a^b f(t)\ dt)$$
but in general the two are a little bit different, though still quite analogous.
This notion arises implicitly in many places (solving ODE, integrating connections along curves, dynamics and random walks on Lie groups (e.g. in the work of Terry Lyons), the "noncommutative Fourier transform" from scattering theory, etc.), but I am sure that it must be studied explicitly in some body of literature (and even vaguely recall seeing such at some point in the past). But I am having difficulty locating this literature because I am not sure I have the correct terminology for this concept. So my questions are:
1. What is the accepted name and notation for this concept in the literature? (Perhaps there is more than one such notation, coming from separate bodies of literature.)
2. What are the references for the theory of this concept?
| https://mathoverflow.net/users/766 | What is the standard notation for a multiplicative integral? | This type of construction also arises in topology and algebraic geometry as "iterated integrals" or "Chen's iterated integrals". There are many sources of which a famous one by Chen himself is: [Link](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-83/issue-5/Iterated-path-integrals/bams/1183539443.full) .
Path-ordered (or time-ordered) exponential, as suggested in the other answer, is the most common term, or at least would get the most hits in a search, but this is due to the usage in physics.
ADDED: this paper by Hain (Chen's student) calls the construction "iterated line integrals". <https://arxiv.org/abs/math.AG/0109204> . Another paper calls it "iterated integrals" in a more specific context matching that of the question: p.21 of <http://www.math.toronto.edu/drorbn/papers/OnVassiliev/OnVassiliev.pdf> .
| 13 | https://mathoverflow.net/users/6579 | 32713 | 21,241 |
https://mathoverflow.net/questions/32478 | 14 | I would like to understand this concept. It seems to be important (for the theory of perverse sheaves), yet I don't know any nice exposition of the properties of smooth sheaves.
| https://mathoverflow.net/users/2191 | A nice explanation of what is a smooth (l-adic) sheaf? | I'll give an answer, only because I'm interested in chasing down these references
myself. But all I'm doing is assembling references. I assume that
BCnrd will keep me honest. [July 21: I've added some remarks about constructibility, to
makes this more useful (at least to me).]
Since I'm a complex geometer rather an arithmetic one, let me start with the first case for intuition. If $X\_{an}$ is a (connected) complex variety endowed with the classical topology then one knows that representations of the usual $\pi\_1(X\_{an},x)$ correspond to locally constant sheaves on $X\_{an}$. This is classical. A good source of examples are as follows:
if $f:Y\to X$ is a surjective smooth proper map, then it is topologically
a fibre bundle (Ereshmann). Therefore $R^if\_\*\mathbb{Z}$ is locally constant. The
corresponding $\pi\_1(X)$-module is the monodromy representation. The most
general statement one can make, without making any assumptions
on $f$, is that the proper direct image $R^if\_!\mathbb{Z}$ is constructible.
Note that constructibility can mean different things in the topological world.
The best notion (from my point of view) is what is sometimes called algebraic constructibility: there exists a partition of the base into Zariski locally closed strata
such that the restrictions of the sheaf are locally constant. The only reference that I know which takes this viewpoint is Verdier, *Classe d'homologie associée à un cycle*. If people are aware of other sources, please let me know.
Remarkably, the analogous results hold in the $\ell$-adic case, although for different reasons. Let $X$ be variety over some field.
A lisse (resp. constructible) $\ell$-adic sheaf is now a prosheaf
$$\ldots \mathcal{F}\_n\to \mathcal{F}\_{n-1}\ldots $$
on the etale site $X\_{et}$ such that each item above is a locally constant (resp. constructible) $\mathbb{Z}/\ell^n$-module etc. (see Freitag-Kiehl, pp 118-131, for the precise conditions). For lisse sheaves, each $\mathcal{F}\_n$ gives a representation of the etale fundamental group
$$\pi\_1^{et}(X,x)\to GL\_N(\mathbb{Z}/\ell^n)$$
($x$ a geom. pt.). So passing to the limit, we get a continuous representation
$$\pi\_1^{et}(X,x)\to GL\_N(\mathbb{Z}\_\ell)$$
This constuction is an equivalence [FK,p 286].
The corresponding result that $R^if\_\*\mathbb{Z}\_\ell$
is lisse, when $f$ is smooth, proper and surjective, should follow from Theorem 20.2 of Milne "Lectures on etale cohomology" from his website. The contrucibility of $R^if\_!\mathbb{Z}\_\ell$ would follow from SGA4 exp XIV 1.1 (It ought to be in [FK,M], but I probably didn't look hard enough.)
When $X$ is defined over $\mathbb{C}$, one can compare cohomology for the classical
and etale topologies with general coefficients by applying SGA4 exp XVI 4.1 and taking inverse limits. A more
general comparison result for the "6 operations" is given in [Beilinson-Bernstein-Deligne
p 150], but the proof seems a bit sketchy.
Remark added July 22: Unfortunately, this part of the story appears to be inadequately
addressed in the literature. See BCnrd's comment below.
| 27 | https://mathoverflow.net/users/4144 | 32717 | 21,243 |
https://mathoverflow.net/questions/32725 | 9 | What are the websites for general position in applied or industrial mathematics(or financial mathematics) related jobs (that is if we have to find a non academic job temporarily) ?Thanks!
| https://mathoverflow.net/users/2391 | website for jobs in applied or industrial mathematics (or financial math) | For finance, I recommend wilmott.com and gloriamundi.org. There's also www.mathjobs.org.
| 6 | https://mathoverflow.net/users/613 | 32727 | 21,250 |
https://mathoverflow.net/questions/32699 | 5 | Problem
-------
Consider two *d* x *d* complex matrices, *R* and *S*, whose entries lie in the unit disk:
$\quad |R\_{i,j}|<1 \quad$ and $\quad |S\_{i,j}|<1 $.
Say that *R* is constructed by randomly choosing complex numbers from the unit disk, but *S* is constructed as
$\quad S\_{i,j} = f(i/d,j/d)$
where $f(x,y)$ is a smooth function for $x,y \in [0,1]$, with $|f(x,y)|<1$. In other words, the entries of *S* are smooth functions of the indices (in the limit of large *d*), but those of *R* are not.
Question
--------
How do the trace norms
$\quad ||R||=Tr[\sqrt{R^\dagger R}] \quad$ and $\quad ||S||=Tr[\sqrt{S^\dagger S}]$
of these matrices behave as $d \to \infty$ ?
Numerical Evidence
------------------
A few lines of Mathematica strongly suggest that
$\quad ||R|| \propto d^{3/2}$
but
$\quad ||S|| \propto d$
for large *d*. (The proportionality constants depend on the probability distribution used to pick numbers from the unit disk for *R* and the function $f(x,y)$ used to pick entries for *S*, respectively.)
What explains this behavior?
Addendum
--------
After Willie's excellent answer below, I thought I'd mention that it's really fast to see the scaling behavior once you discretize the function. Let $F$ be some matrix of discrete values for the function, and let $J\_n$ be the $n \times n$ matrix with all elements equal to unity.
$||F \otimes J\_n|| = \mathrm{Tr} \sqrt {(F^\dagger \otimes J\_n)( F \otimes J\_n)} = \mathrm{Tr} \sqrt {(F^\dagger F) \otimes (J\_n J\_n)} = ||F|| \cdot ||J\_n|| = n ||F||$
Basically, the idea is that once the dimension of $F$ is large enough to capture the important detail in the function, increasing the dimension is really just increasing the dimension of $J$.
| https://mathoverflow.net/users/5789 | Dependence of trace norm on matrix size for smooth vs. random matrices. | To flesh out Helge's answer a bit before I go to bed:
Assume that $f(x,y)$ is a smooth function on the unit square.
Define the functions $f\_n(x,y) = f(\frac{\lfloor nx \rfloor}{n}, \frac{\lfloor ny\rfloor}{n})$. This is a piecewise step function. Observe that the operator $S$ of dimension $d$ is the same if you define it relative to $f$ or $f\_d$. It is elementary to show that $f\_n\to f$ uniformly (as functions).
Define an action of $f\_n$ on $L^2[0,1]$ by
$$ g(x) \mapsto \int\_0^1 f\_n(x,y)g(y) dy $$
and note that for a $n$-vector $v = (v\_1,\ldots, v\_n)$ we can associate
$$ g\_v(x) = \sum v\_i \chi\_{[\frac{i-1}{n},\frac{i}{n})} $$
we observe that
$$ f\_n(g\_v(x)) = \frac{1}{n} g\_{Sv}(x)$$
the $1/n$ factor coming from the fact that the length of the segment $[(i-1) / n, i/n)$ is $1/n$.
Now consider $V^n$ as the subspace of $L^2[0,1]$ spanned by $\chi\_{[\frac{i-1}{n},\frac{i}{n})}$. By definition $f\_n$ annihilates its orthogonal complement, and $f\_n$ restricted to $V^n$ is equivalent to a rescaled version of $S$, so in particular you have that the trace norms of $f\_n$ (acting on $L^2$) is the same as $1/n$ times the trace norm of $S$ (acting on $\mathbb{R}^n$).
To finish you just need to note that via some functional analysis voodoo the corresponding operators $f\_n\to f$, and so the trace norms of $f\_n$ converges. Therefore you have that $1/n$ times the trace norm of $S$ converges to a constant (which may be zero). Note that the regularity for $f$ is only needed in this last step, and you probably just need uniform continuity to assure that the operators converge in a strong enough sense.
[Addendum: all the functional analysis voodoo you need (which is not very much for this problem) can be found in Reed & Simon, volume 1, chapter 6.]
| 5 | https://mathoverflow.net/users/3948 | 32728 | 21,251 |
https://mathoverflow.net/questions/32734 | 7 | Does there exist a notion of Jordan curve homotopy?
In particular, suppose we have two Jordan curves $C\_0 : S^1 \rightarrow \mathbb{R}^2$ and $C\_1 : S^1 \rightarrow \mathbb{R}^2$. When does there exist a continuous function $f: S^1 \times [0,1] \rightarrow \mathbb{R}^2$ such that:
$f(x,0) = C\_0(x)$, $f(x,1) = C\_1(x)$, and for all $t \in [0,1]$, the function $C\_t: S^1 \rightarrow \mathbb{R}^2$ defined by $C\_t(x) = f(x,t)$ is a Jordan curve.
My intuition tells me that such a function always exists, but I'm unsure about how to go about proving this. Also, if this is a known result, are there similar results for manifolds other than $\mathbb{R}^2$?
| https://mathoverflow.net/users/7726 | Jordan Curve Homotopy | Homotopies through embeddings are usually called isotopies.
There is a subtlety called local flatness that comes up in higher dimensions. Let $E$ be any embedding of $\mathbb R$ in $\mathbb R^3$ such that $E(s)=(s,0,0)$ when $s<-1$ or $s>1$. Define a homotopy $H\_t$ with $E\_0(s)=(s,0,0)$ for all $s\in \mathbb R$ and $E\_1=E$, as follows: $E\_t(s)=tE(s/t)$ if $-t\le s\le t$ and otherwise $E\_t(s)=(s,0,0)$. This is a homotopy through embeddings, but it (un)ties the knot. This is easily adapted to apply to examples of embeddings of $S^1$ in $\mathbb R^3$, for example.
The way to fix this problem is to only consider embeddings that are locally flat and isotopies that are locally flat. An embedding $E:M\to N$ is (topologically) locally flat if for every point $p\in M$ there exist charts around $p$ and $E(p)$ such that $E$ looks like $(x\_1,\dots,x\_m)\mapsto (x\_1,\dots,x\_m,0,\dots,0)$. An isotopy $E\_t$ is locally flat if for each point $p\in M$ and time $\tau\in I$ there are charts around $(p,\tau)\in M\times I$ and around $(E\_\tau(p),\tau)\in N\times I$, both of them using projection to $I$ as last coordinate, such that locally $(x,t)\mapsto (E\_t(x),t)$ looks like $(x\_1,\dots,x\_m,t)\mapsto (x\_1,\dots,x\_m,0,\dots,0,t)$. Local flatness is automatic when $m=1$ and $n=2$. The example I gave (with $m=1$ and $n=3$) was such that the isotopy was not locally flat although if the original embedding $E\_1$ was locally flat then for every $t$ the embedding $E\_t$ was, too.
The Alexander horned sphere ($m=2$, $n=3$, not locally flat) can be smoothed out by such a procedure, too.
Another way of limiting oneself to the right kind of isotopies is to use ambient isotopies: to require $E\_t$ to be $H\_t\circ E\_0$ where $H\_t$ is a homeomorphism $N\to N$ depending on $t$. (Local flatness in the case $m=n$ follows from invariance of domain.)
Another way is to limit oneself to smooth embeddings (meaning, as usual, smooth maps that are topological embeddings and that are one to one on the tangent-space level, or equivalently locally flat in the smooth category) and smooth isotopies.
| 10 | https://mathoverflow.net/users/6666 | 32742 | 21,259 |
https://mathoverflow.net/questions/32730 | 4 | Given some number $n$ and a seed number $s$<$n$, I want a random number generator (RNG) that will go through all integers `<$n$ before coming back to $s$. The resulting random number must be roughly uniformly distributed (which of course it will be if you go through the entire sequence, so I mean for any "large" subsequence) and roughly un-autocorrelated. Furthermore I want the RNG to be "efficient" in that it takes up little memory and little computation. Perhaps I can say that it is $O(1)$ w.r.t. $n$ in terms of memory and computation.
For instance, I can think of a RNG right now that will fulfill the former requirements, but not the latter: Create a list of all numbers $0$ through $n-1$. "Mark off" the seed number $s$. Then take a random number $r$ from a Mersenne Twister RNG, move $r$ numbers to the right and if that number hasn't been marked off report it back and then mark it off. Continue the process until you've marked off all numbers in the list. - This method will report back non-repeating, un-autocorrelated integers, but will be super memory and time intensive.
I imagine the ideal answer to be some sort of small equation to provide the next number in the sequence based upon this number (or perhaps the last few).
Can such a RNG be proven to exist? Are such specific RNGs know to exist? Can their existence be disproven?
Editorial note: If anyone reading has super edit power... please feel free to clean up my post to make me sound more mathy (I am but a lowly engineer). Retag me too please.
| https://mathoverflow.net/users/6898 | Does an "efficient" random number generator exist? | Your requirements aren't rigorously stated, so it's hard to say what you can prove exists or doesn't exist. In the strictest sense, a pseudo-random number generator cannot possibly be "roughly uniformly distributed". Every PRNG is an expansion of entropy from its settings to the set of possible sequences, and it is easy to see for reasons similar to your comments about memory requirements that the set of possible sequences has vastly more entropy than the settings. What a PRNG really does is passes certain computationally feasible tests of randomness and not necessarily other tests.
There are conjectures in computer science that "one-way functions exist". Some of these conjecture would imply that there are PRNGs that look random for any polynomial-time test, and for some conjectures permutations that look random for any polynomial-time test. However, these conjectures are harder than the P vs NP problem, so no one is about to prove them. In any case, if you just want a PRNG for your own practical use, it's overkill to look for one that has been analyzed cryptographically.
It is known that modular exponentiation is a pretty good PRNG. If you want something that looks like a permutation, let $p$ be a prime number, and let $a$ be a carefully chosen residue mod $p$. (Carefully chosen means that $a$ should be a primitive residue far away from $0$.) Then the function
$$f\_a(k) = a^k \bmod p$$
is already statistically okay. This is a permutation of the numbers $1 \le k \le p-1$.
Now, the most common way to compute $f\_a$ is to store $f\_a(k)$ and then multiply by $a$ to get the next power. (As Richard Borcherds mentions, the iteration $x \mapsto ax+b \bmod n$ is a similar idea and a major standard, including in Knuth's book and in the Unix RNG "drand48".) However, these days that level of efficiency isn't so important, and it is interesting that you can compute $f\_a$ directly by repeated squaring. So you can improve the strength of $f\_a$ by making a composition such as $f\_b(f\_a(k)+c)$, where the addition is taken mod $p-1$. Or you could insert a more creative transformation. For instance, if $p$ is a Mersenne prime, then permuting the bits of $k$ is a simple transformation that can be inserted between applications of $f\_a$.
If you want a permutation of some $n$ that is not of the form $p-1$, then you can find some prime $p > n$ that is not much larger and use the above same tricks. You can just skip values that are out of range.
Decades ago, I wanted a pseudo-random permutation for a scrambled screen fade in a computer game. I just used consecutive values of $f\_a$ for some convenient modulus (which doesn't have to be prime; there are other variations) and it looked fine.
| 7 | https://mathoverflow.net/users/1450 | 32743 | 21,260 |
https://mathoverflow.net/questions/32720 | 72 | This is a simple doubt of mine about the basics of measure theory, which should be easy for the logicians to answer. The example I know of non Borel sets would be a Hamel basis, which needs axiom of choice, and examples of non Lebesgue measurable set would be [Vitali sets](http://en.wikipedia.org/wiki/Vitali_set), which seems to be unprovable without axiom of choice. Then I saw [the answer of François G. Dorais](https://mathoverflow.net/questions/23202/explicit-big-linearly-independent-sets/23206#23206). His construction of an uncountable $\mathbb{Q}$-independent set in $\mathbb R$ does not require axiom of choice. Which leaves a faint hope for the following:
>
> Is it possible to construct without using the axiom of choice examples of non Borel sets?
>
>
>
There is a classic example of an analytic but non-Borel set due to Lusin, described by Gerald Edgar [here](https://groups.google.com/forum/#!search/not-Borel$20continued-fraction/sci.math/OZE06oaV5RQ/HqTYd6cGFSgJ), but it is not clear to me whether it needs axiom of choice since it seems to be putting restrictions on higher and higher terms in the continued fraction expansion. Since I do not know logic and set theory, I hoped of asking the experts.
| https://mathoverflow.net/users/2938 | Non-Borel sets without axiom of choice | No, it is not possible. It is consistent with ZF without choice that
>
> the reals are the countable union of countable sets. (\*)
>
>
>
From this it follows that all sets of reals are Borel. Of course, the "axiom" (\*) makes it impossible to do any analysis. As soon as one allows the bit of choice that it is typically used to set up classical analysis as one is used to (mostly countable choice, but DC seems needed for Radon-Nikodym), one can implement the arguments needed to show
>
> (\*\*) The usual hierarchy of Borel sets (obtained by first taking open sets, then complements, then countable unions of these, then complements, etc) does not terminate before stage $\omega\_1$ (this is a kind of diagonal argument).
>
>
>
Logicians call the sets obtained this way $\Delta^1\_1$. They are in general a subcollection of the Borel sets. To show that they are all the Borel sets requires a bit of choice (One needs that $\omega\_1$ is regular).
There is actually a nice result of Suslin relevant here. He proved that the Borel sets are precisely the $\Delta^1\_1$ sets: These are the sets that are simultaneously the continuous image of a Borel set ($\Sigma^1\_1$ sets), and the complement of such a set ($\Pi^1\_1$ sets).
That there are $\Pi^1\_1$ sets that are not $\Delta^1\_1$ (and therefore, via a bit of choice, not Borel) is again a result of Suslin. He also showed that any $\Sigma^1\_1$ set is either countable, or contains a copy of Cantor's set and therefore has the same size as the reals. His example of a $\Sigma^1\_1$ not $\Delta^1\_1$ set uses logic (a bit of effective descriptive set theory), and nowadays is more common to use the example of the $\Pi^1\_1$ set WO mentioned by Joel, which is not $\Delta^1\_1$ by what logicians call a boundedness argument.
A nice reference for some of these issues is the book Mansfield-Weitkamp, **Recursive Aspects of Descriptive Set Theory**, Oxford University Press, Oxford (1985).
| 73 | https://mathoverflow.net/users/6085 | 32746 | 21,261 |
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