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https://mathoverflow.net/questions/9737 | 14 | According to [this page](http://ncatlab.org/nlab/show/Yoneda+lemma+for+(infinity,1)-categories) on the nLab, it is currently unclear whether or not the entire Yoneda lemma generalizes to $(\infty ,1)$-categories rather than just the Yoneda embedding. Have there been counterexamples to the stronger statement? If not, what complications are there in generalizing the entire Yoneda lemma? The whole lemma is a very powerful tool in ordinary category theory, and if there is no definitive answer, are there any reasons to believe that the generalization should or shouldn't be true?
By the "entire" Yoneda lemma, I mean the isomorphism $F(X)\cong Nat(h\_X,F)$
| https://mathoverflow.net/users/1353 | The Yoneda Lemma for $(\infty,1)$-categories? | Here is a (tautological) proof in the setting of quasi-categories. Let $A$ be a quasi-category. In ordinary category theory, one can describe the category of presheaves of sets over small category $C$ as the full subcategory of $Cat/C$ spanned by Grothendieck fibrations wih discrete fibers. A quasi-category version would consist to say that a presheaf over a quasi-category $A$ is a Grothendieck fibration (cartesian fibration in Lurie's terminology), whose fibers are $\infty$-groupoids (this the way of being discrete in the higher setting). Such fibrations are simply the right fibrations.
More precisely, a model for the theory of presheaves (or $\infty$-stacks) over $A$ is the model category of simplicial sets over $A$, in which the fibrant objects are right fibrations $X\to A$, while the cofibrations are the monomorphisms. The weak equivalences between two right fibrations over $A$ are simply the fiberwise categorical equivalence (for the different models for the theory of stacks over a quasi-category, see §5.1.1 in Lurie's book). From this point of view, the representable stacks over $A$ are the right fibrations $X\to A$ such that $X$ has a terminal object. If $a$ is an object ($0$-cell) of $A$, there is a canonical right fibration $A/a \to A$ (from the general theory of joins): this is the representable stack associated to $a$. You can also construct a model of $A/a$ by taking a fibrant replacement of the map $a:\Delta[0]\to A$ (seen as an object of $SSet/A$). This model category has the good taste of being a simplicial model category. In particular, you have a simplicially enriched Hom, which I will denote here by $Map\_A$, and which can be described as follows. If $X$ and $Y$ are two simplicial sets over $A$, there is a simplicial set $Map\_A(X,Y)$ of maps from $X$ to $Y$ over $A$: if $\underline{Hom}$ is the internal Hom for simplicial sets, then $Map\_A(X,Y)$ is simply the fiber of the obvious map $\underline{Hom}(X,Y)\to\underline{Hom}(X,A)$ over the $0$-cell corresponding to the structural map $X\to A$. If $Y$ is fibrant (i.e. $Y\to A$ is a right fibration), then $Map\_A(X,Y)$ is a Kan complex (because it is the fiber of a right fibration, hence of a conservative inner Kan fibration), which is the mapping space of maps from $X$ to $Y$ for this model structure on $Sset/A$. $Map\_A$ is a Quillen functor in two variables with value in the usual model category of simplicial sets.
If $a$ is an object of $A$, seen as a map $a: \Delta[0]\to A$, i.e. as an object of $SSet/A$, then for any right fibration $F\to A$, we see that $Map\_A(a,F)$ is isomorphic to $F\_a$, that is the fiber of the map $F\to A$ at $a$ (which is also an homotopy fiber for the Joyal model structure). Considering the weak equivalence from $a: \Delta[0]\to A$ to $A/a\to A$, we also have a weak equivalence of Kan complexes
$$Map\_A(A/a,F)\overset{\sim}{\to}Map\_A(a,F)=F\_a\, .$$
This gives the full Yoneda lemma.
| 20 | https://mathoverflow.net/users/1017 | 9808 | 6,701 |
https://mathoverflow.net/questions/9807 | 7 | *Hi, I have a minor in math and this is not a homework problem - my prof mentioned it 5 years ago and I could not even begin to tackle it until I took a good intro to linear algebra (after work). Please try to adjust the answer to my level.*
A map is a smaller version of the land: rotated and scaled down. The prerequisite is that the map (say of USA) lands entirely inside the land (and not partly in Canada or in the ocean). Another prerequisite might be that the map has no holes (***is that really necessary?***) and perhaps that it must be a convex region (***however I doubt that this is needed***.) Please help me eliminate these doubts and prove that $p \in M, p \in L$. Correct my notation if needed by editing this post.
Here is the notation that I came up with (let's see if LaTex likes me):
* $L$ = land (region in $R^2$)
* $M$ = map (region in $R^2$)
* $T$ = transformation in $R^2$ (scalar times a rotation matrix)
* $\vec{s}$ = shift vector (since the overall transformation is generally non-linear)
* $\vec{p}$ = "pivot" - the point that does not change.
Now, I can solve for p uniquely by picking a non-trivial triangle on the land denoted by vectors $\vec{l\_1}, \vec{l\_2}, \vec{l\_3}$, locating the corresponding vectors $\vec{m\_1}, \vec{m\_2}, \vec{m\_3}$ on the map and writing out:
$\vec{m\_1} = T \vec{l\_1} + \vec{s}$, $\vec{m\_2} = T \vec{l\_2} + \vec{s}$, $\vec{m\_3} = T \vec{l\_3} + \vec{s}$
I solve for T:
$T = [ m\_1 - m\_2 \; \; \; m\_1 - m\_3 ] [ l\_1 - l\_2 \;\;\; l\_1 - l\_3 ]^{-1}$
Because we picked a non-trivial triangle, it's area will be non-zero and the matrix on the right will be invertible. So, we can always solve for T.
We can also solve for s: $\vec{s} = \vec{m\_1} - T \vec{l\_1}$, and finally for the pivot p: $\vec{p} = (I - T)^{-1} \vec{s}$. Since T = c \* rotation matrix, where $c \leq 1$, the only time when (I - T) does not have an inverse is when I = T (details omitted).
***So, it seems that I can solve for a unique p.***
* Now, how do I prove that pivot $\vec{p}$ will be inside both shapes/regions L and M?
* Finally, what assumptions must I make about convexity of M, absence of holes, etc ?
P.S. Which undergraduate classes might have this as a homework problem?
| https://mathoverflow.net/users/2814 | Help me with this proof: Drop a printed map of the land on the land and there must be some common point. | This is usually quoted as an "application" of Brouwer's fixed point theorem:
<http://en.wikipedia.org/wiki/Brouwer%27s_fixed_point_theorem>
which says that a continuous map from a closed ball (in any number of dimensions) to itself must have a fixed point.
You definitely need it to not have holes, because otherwise you could just rotate around the center of the hole and no fixed points will exist. However, convexity is not relevant. The main point is that the United States (minus Alaska and Hawaii and whatever holes it might have) is homeomorphic to the closed disk, so we can apply the homeomorphism, find the fixed point for the closed disk, and then reverse the homeomorphism (apply its inverse) to get the fixed point of the original function (which is your laying of the land on itself).
I'm not really sure how to finish your proof, but I would think this would appear in an algebraic topology course.
| 4 | https://mathoverflow.net/users/321 | 9811 | 6,704 |
https://mathoverflow.net/questions/9835 | 6 | Along the lines of [Polynomial representing all nonnegative integers](https://mathoverflow.net/questions/9731/polynomial-representing-all-nonnegative-integers), but likely well-known question:
>
> is there a polynomial $f \in \mathbb Q[x\_1, \dots, x\_n]$ such that $f(\mathbb Z\times\mathbb Z\times\dots\times\mathbb Z) = P$, the set of primes?
>
>
>
| https://mathoverflow.net/users/65 | Polynomial representing prime numbers | No. Any such polynomial would have the property that any of its restrictions $f(x)$ to one variable consist only of primes, but this is easily seen to be impossible, since if $p(a)$ is prime then $p(k p(a) + a)$ is divisible by $p(a)$. (Even accounting for the coefficients in $\mathbb{Q}$ is straightforward by multiplying by the common denominator and using CRT; in fact, we can show that given an integer polynomial $q(x)$ and a positive integer $n$ there exists $x\_n$ such that $q(x\_n)$ is divisible by $n$ distinct primes.)
However, there do exist multivariate polynomials with the property that their **positive** integer outputs consist of the set of primes. See [the Wikipedia article](http://en.wikipedia.org/wiki/Hilbert%27s_tenth_problem#Applications).
| 19 | https://mathoverflow.net/users/290 | 9839 | 6,718 |
https://mathoverflow.net/questions/9834 | 40 | What is the heuristic idea behind the Fourier-Mukai transform? What is the connection to the classical Fourier transform?
Moreover, could someone recommend a concise introduction to the subject?
| https://mathoverflow.net/users/1547 | Heuristic behind the Fourier-Mukai transform | First, recall the classical Fourier transform. It's something like this: Take a function $f(x)$, and then the Fourier transform is the function $g(y) := \int f(x)e^{2\pi i xy} dx$. I really know almost nothing about the classical Fourier transform, but one of the main points is that the Fourier transform is supposed to be an invertible operation.
The Fourier-Mukai transform in algebraic geometry gets its name because it at least superficially resembles the classical Fourier transform. (And of course because it was studied by Mukai.) Let me give a rough picture of the Fourier-Mukai transform and how it resembles the classical situation.
1. Take two varieties $X$ and $Y$, and a sheaf $\mathcal{P}$ on $X \times Y$. The sheaf $\mathcal{P}$ is sometimes called the "integral kernel". Take a sheaf $\mathcal{F}$ on $X$. Think of $\mathcal{F}$ as being analogous to the function $f(x)$ in the classical situation. Think of $\mathcal{P}$ as being analogous to, in the classical situation, some function of $x$ and $y$.
2. Now pull the sheaf back along the projection $p\_1 : X \times Y \to X$. Think of the pullback $p\_1^\ast \mathcal{F}$ as being analogous to the function $F(x,y) := f(x)$. Think of $\mathcal{P}$ as being analogous to the function $e^{2\pi i xy}$ (but maybe not exactly, see below).
3. Next, take the tensor product $p\_1^\ast \mathcal{F} \otimes \mathcal{P}$. This is analogous to the function $F(x,y) e^{2\pi i xy}$ $=$ $f(x)e^{2\pi i xy}$.
4. Finally, push $p\_1^\ast\mathcal{F} \otimes \mathcal{P}$ down along the projection $p\_2: X \times Y \to Y$. The result is the Fourier-Mukai transform of $\mathcal{F}$ --- it is $p\_{2,\ast} (p\_1^\ast \mathcal{F} \otimes \mathcal{P})$. This last pushforward step can be thought of as "integration along the fiber" --- here the fiber direction is the $X$ direction. So the analogous thing in the classical situation is $g(y) = \int f(x)e^{2\pi i xy}dx$ --- the Fourier transform of $f(x)$!
But to make all of this actually work out, we have to actually use the derived pushforward, not just the pushforward. And so we have to work with the derived categories.
When $X$ is an abelian variety, $Y$ is the dual abelian variety, and $\mathcal{P}$ is the so-called Poincare line bundle on $X \times Y$, then the Fourier-Mukai transform gives an equivalence of the derived category of coherent sheaves on $X$ with the derived category of coherent sheaves on $Y$. I think this was proven by Mukai. I think this is supposed to be analogous to the statement I made about the classical Fourier transform being invertible. In other words I think the Poincare line bundle is really supposed to be analogous to the function $e^{2\pi i xy}$. A more general choice of $\mathcal{P}$ corresponds to, in the classical situation, so-called integral transforms, which have been previously discussed [here](https://mathoverflow.net/questions/2809/intuition-for-integral-transforms). This is probably why $\mathcal{P}$ is called the integral kernel. You may also be interested in reading about [Pontryagin duality](http://en.wikipedia.org/wiki/Pontryagin_duality), which is a version of the Fourier transform for locally compact abelian topological groups --- this is obviously quite similar, at least superficially, to Mukai's result about abelian varieties. However I don't know enough to say anything more than that.
There are some cool theorems of Orlov, I forget the precise statements (but you can probably easily find them in any of the books suggested so far), which say that in certain cases any derived equivalence is induced by a Fourier-Mukai transform. Note that the converse is not true: some random Fourier-Mukai transform (i.e. some random choice of the sheaf $\mathcal{P}$) is probably not a derived equivalence.
I think Huybrechts' book "Fourier-Mukai transforms in algebraic geometry" is a good book to look at.
Edit: I hope this gives you a better idea of what is going on, though I have to admit that I don't know of any good heuristic idea behind, e.g., Mukai's result --- it is *analogous* to the Fourier transform and to Pontryagin duality, and thus I suppose we can apply whatever heuristic ideas we have about the Fourier transform to the Fourier-Mukai transform --- but I don't know of any heuristic ideas that explain the Fourier-Mukai transform in a direct way, without appealing to any analogies to things that are outside of algebraic geometry proper. Hopefully somebody else can say something about that.
But --- there is certainly something deep going on. Just as [CommRing behaves a lot like Setop](https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence/3904#3904), I think there is probably some kind of general phenomenon that sheaves (or vector bundles) behave a lot like functions, which is what's happening here. Pullback of sheaves behave a lot like pullback of functions... Pushforward of sheaves behave a lot like integration of functions... Tensor product of sheaves behave a lot like multiplication of functions...
| 51 | https://mathoverflow.net/users/83 | 9840 | 6,719 |
https://mathoverflow.net/questions/9817 | 3 | I have only recently started exploring this region of homogeneous spaces and its geometry and the question is born from that and given the beginner state of my exploration the questions might sound ill-framed.
All this is motivated by trying to understand how the spectrum of the Dirac Operator is obtained on homogeneous spaces which in physics is a very natural situation.
1. Can anyone explain to me how connections on SU(2) can be labeled by the irreducible representations of SU(2) (what in physics is called "spin") ?
2. Accordingly how are laplacians on SU(2) labelled by irreducible representations of SU(2)?
3. Literature seem to state that symmetric, traceless and divergence less tensors on $S^3$ are also labeled by irreducible representations of SU(2). How is that?
4. What is the precise definition of a "tensor harmonic" on a homogeneous space?
And how do "tensor harmonics" on SU(2) give a basis for expanding sections of the spin-bundle on SU(2)?
5. SU(2) can be thought of as the homogeneous space $SU(2)\times SU(2)/SU(2)$ under the diagonal action and then how does sections of this principle bundle give a basis for tensor valued functions on SU(2)?
6. I am adding another point here. In the above bundle with the diagonal action the projection map to the base space is $(g\_1,g\_2) \mapsto g\_1g\_2 ^{-1}$ which will have as inverse images orbits of the above diagonal action. And hence one sees the fibers of this principle bundle structure of this homogeneous space.
Now some calculations in the literature seem to tell me that choosing a section in this bundle with respect to the above projection map somehow canonically defines me vielbiens in the base $SU(2)$.
The point being that given the usual diagonal metric on $S^3$ one could have guessed the vielbiens upto a sign. But somehow it seems that the choice of a section fixes the signs.
Can anyone kindly explain how the choice of a section over $S^3$ (the fields obviously lying in the bundle-space $SU(2)\times SU(2)$) gives a "natural" vielbien on the base $S^3$?
One of the popular sections here are called "thermal sections" in physics.
| https://mathoverflow.net/users/2678 | Representations of SU(2) and tensors on SU(2) | You may wish to take a look at this [survey article](http://geometrie.math.uni-potsdam.de/documents/baer/dirac_spectrum_survey.pdf) by Christian Bär about the spectrum of the Dirac operator. It contains a section (1.1.2.2) on homogeneous spaces. For the particular case of three dimensions, there is also his [earlier paper](https://doi.org/10.1007/BF01199016).
| 2 | https://mathoverflow.net/users/394 | 9841 | 6,720 |
https://mathoverflow.net/questions/9825 | 2 | k is an algebraically closed field, X is a smooth, connected, projective curve over k. f: X-->P^1 is a finite morphism. Let t be a parameter of P^1, suppose f is etale outside t=0 and t=\infty, and tamely ramified over these two points. Prove that f is a cyclic cover, i.e., K(X)=k(t)[h]/(h^n-ut), u is a unit in field k.
| https://mathoverflow.net/users/2008 | tamely branched cover over P^1 | Here's an alternative way to think about it:
You can easily deduce from Riemann-Hurwitz that the genus of X is 0, i.e. it is just the projective line. Look at the affine patch: t is not infinity. Above t=infinity there's only one point. So take that point out, and call the parameter of the resulting affine line h. Then we have the inclusion of k[t] in k[h]. So t=g(h) where g is a polynomial. Since over t=0 there is one point with ramification n, g has multiplicity n: g=u(h-c)n. So after change of variables g=u\*hn, as required.
| 9 | https://mathoverflow.net/users/2665 | 9845 | 6,723 |
https://mathoverflow.net/questions/9849 | 16 | Let $F\_0 : C \to D$ be a functor. If it exists, let $G\_0 : D \to C$ be its left adjoint. If it exists, let $F\_1 : C \to D$ be **its** left adjoint. And so forth. In situations where the infinite sequence $(F\_0, G\_0, F\_1, G\_1, ...)$ exists, when is it periodic? Aperiodic? (Feel free to replace all "lefts" by "rights," of course.)
| https://mathoverflow.net/users/290 | Iterated adjoint functors | <http://www.springerlink.com/content/pmj5074147116273/> considers sequences of adjoint functors just like you describe.
| 7 | https://mathoverflow.net/users/2384 | 9851 | 6,725 |
https://mathoverflow.net/questions/9703 | 4 | Let $X$ be an algebraic variety and $A$ is a ample divisor on $X$. Let $m$ be a sufficiently large natural number such that $X \overset{\varphi\_{mA}}{\to} \mathbf{P}H^0(X, \mathcal{O}\_X(mA))$ defined by the linear system $|mA|$ is an embedding. Denote by $Y$ the image. What's a effective upper bound of $Y$ in terms of $\text{dim}X$, $A^{\text{dim}X}$ and $m$, or other invariants of $X$ and $A$.
This may not be a concrete question. But I am wondering how can we compute the degree of the image of a variety under the (bi)rational morphism given by a linear system.
| https://mathoverflow.net/users/2348 | Degree of an embedded algebraic variety | I'll expand on Felipe's answer. The degree is defined to be $(\dim X)!$ times the lead term of the Hilbert polynomial of the variety. Now, given an ample line bundle $A$, the Hilbert polynomial of the embedding given by $A$ is $n\mapsto\chi(X,A^n)$. First off, this function is in fact a polynomial (check any standard book, Hartshorne for instance). This term will then be the self intersection $A^{\dim X}$, which can be computed in any number of ways (see any book on intersection theory), and so, for the line bundle $mA$, we'll get $(mA)^{\dim X}=m^{\dim X}A^{\dim X}$.
But as Felipe said, it's pretty much by definition. In a real sense, the definition of $A^{\dim X}$ can be taken to be $(\dim X)!$ times the lead term of the Hilbert polynomial.
| 3 | https://mathoverflow.net/users/622 | 9862 | 6,733 |
https://mathoverflow.net/questions/9844 | 4 | Hello.
Let's say I have a set of input vectors $I = \{\mathbf{x\_1}, \dots, \mathbf{x\_k}\} \subset \mathcal{R}^m$ and a set of output vectors $O = \{\mathbf{y\_1}, \dots, \mathbf{y\_k}\} \subset \mathcal{R}^n$, and I want to obtain a mapping $f : \mathcal{R}^m \to \mathcal{R}^n$ such that
$$ f(\mathbf{x\_i}) = \mathbf{y\_i} + \epsilon\_i, \forall i \in \{1, \dots, k\}$$
where $\epsilon\_i$ is *small*, and this mapping should be continuous at least around the input/output pairs.
There are many ways of doing so.
If we suppose that the input/output pairs won't change, what are the advantages of using an Artificial Neural Network over other methods to approximate functions?
EDIT: when I say **advantages**, I mean the practical advantages on the use of neural networks over other function approximation methods on any domain that use the neural networks.
For example, if we think of (the canonical example of) **handwriting recognition** that mailing services might use to read zip codes, the neural network is nothing more than a function that maps, let's say, [0, 1]^35 (if we think of a 5x7 grid, in which the values are the "intensity" or "amount" of the ink on each cell) to [0, 1]^10 (corresponding to each digit between 0 and 9, the value being the probability of the digit). So, in this case, if we think that we will write a software to do this recognition, and that the patterns will never change, we could simply have used another technique to map the input to the output. If we use any method that produces a continuous mapping, small "variations" on the handwritting wouldn't affect too much the output of the function.
Thanks.
| https://mathoverflow.net/users/1933 | Advantages of a back-propagation neural network over other function approximation methods | Artificial neural networks are mainly useful when:
1. There is no information on the form of the function f(x) in advance and the task of specifying the functional form of f(x) from the data is computationally complex.
2. And, on the other hand there is a representative sample of inputs and outputs to be used as a training set.
The main advantage of the neural network method is that it can fairly approximate a large class of functions f(x). On the other hand if additional information on the function f(x) is known, then other estimation techniques are likely to work better. For example if it is known that f(x) is linear, then linear regression would surely give better estimation errors.
The problem presented in the question seems to be well suited to neural networks.
| 4 | https://mathoverflow.net/users/1059 | 9869 | 6,738 |
https://mathoverflow.net/questions/9879 | 18 | I've started using TikZ for a paper I'm writing and am worrying that a journal might not accept the paper with inline TikZ. I had to update my PGF installation for all the examples to work and I'm not even sure that my collaborator will have an updated version...
So here are the questions.
-Do people know if journals (math and physics) accept inline TikZ?
-Is there a nice way to create a pdf figure from a tikzpicture ? Perhaps I could have a latex file for each figure and only one \begin{tikpicture} \end{tikzpicture} pair and somehow compile the result to get a pdf or eps containing just the figure (and not a whole page with a figure in it...)
Cheers,
Yossi.
| https://mathoverflow.net/users/3578 | Using TikZ in papers | There is a method for doing this in the back of the TIkZ manual; you can put special commmands around a TikZ picture do have it make a separate PDF which you then include as usual graphics. I'll admit, though, that I've had poor luck using it. I'd rather just give an earful to any journal who doesn't like TikZ.
| 15 | https://mathoverflow.net/users/66 | 9883 | 6,748 |
https://mathoverflow.net/questions/9863 | 17 | An interesting mathoverflow question was one due to Philipp Lampe that asked whether [a non-surjective polynomial function on an infinite field can miss only finitely many values](https://mathoverflow.net/questions/6820/). In my interpretation of the question, if $k$ is a starting field and $f$ is a polynomial, you could ask what happens if you repeatedly adjoin a root of $f(x)-a$, except for a finite set of values $a \in S \subset k$ for which you hope a root never appears. You have to adjoin a root for all $a \in \tilde{k} \setminus S$, where $\tilde{k}$ is the growing field. Either a root of $f(x)-a$ for some $a \in S$ will eventually appear by accident, or $f$ as a polynomial over the limiting field $\tilde{k}$ is an example.
(**Edit:** I call this an interpretation rather than a construction, because in generality it is equivalent to Philipp's original question. I also don't mean to claim credit for the idea; it was already under discussion when I posted my answer then. Maybe an answer to the question below was already implied in the previous discussion, but if so, I didn't follow it.)
For some choices of $f$ and a non-value $a$, you can know that you are sunk at the first stage. For instance, suppose that $f(x) = x^n$. When you adjoin a root of $x^n-a$, you also adjoin a root of $x^n-b^na$ for every $b \in k$. You cannot miss $a$ without also missing every $b^na$, which is then infinitely many values when $k$ is infinite.
So let $k$ be an infinite field, and let $f \in k[x]$ be a polynomial. Define an equivalence relation on those elements $a \in k$ such that $f(x)-a$ is irreducible. The relation is that $a \sim b$ if adjoining one root of $f(x)-a$ and $f(x)-b$ yield isomorphic field extensions of $k$. Is any such equivalence class finite? What if $k$ is $\mathbb{Q}$ or a number field?
In my partial answer to the original MO question, I calculated that if $f$ is cubic and the characteristic of $k$ is not 2 or 3, then the equivalence classes are all infinite.
| https://mathoverflow.net/users/1450 | When f(x)-a and f(x)-b yield the same field extension | The answer is that over a number field $k$, an equivalence class can be finite, and in fact it is usually so for $f$ of moderately large degree. Consider $f(x):=x^7+x$ over $\mathbf{Q}$, for example. If the equations $f(x)=1$ and $f(x)=t$ for some other $t \in \mathbf{Q}$ yield the same degree $7$ extension, then in particular the discriminant $D(t)$ of the polynomial $f(x)-t$ in $x$ must equal $D(1)$ times a square. In fact, $D(t) = -823543 t^6 - 46656$, so the necessary condition is $-823543 t^6 - 46656 = -870199 u^2$.
This defines a genus $2$ curve, so Faltings' theorem implies that this equation has only finitely many rational solutions.
Moreover, for a typical $f$ of slightly higher degree, it is reasonable to expect that all the equivalence classes are singletons, although *proving* such a statement would seem to require understanding the rational points on surfaces of general type, which is probably beyond the current state of knowledge.
Reasoning along these lines suggests to me that the iterative procedure you and others have proposed should succeed in constructing a field with a nonsurjective polynomial as in Philipp Lampe's question, even if we can't prove this.
| 24 | https://mathoverflow.net/users/2757 | 9889 | 6,752 |
https://mathoverflow.net/questions/9878 | 2 | Let φ(n) denote Euler's totient function and k $\perp$ n denote that k, n are integers and relatively prime. Let N = φ(n) + 1. If n is not a prime power
$$ \prod\_{\substack{0 < k < n, \\ k \perp n }} \Gamma \left(\frac{k}{n}\right)
= \sqrt{N}\prod\_{ 0 < k < N}\Gamma \left(\frac{k}{N}\right) \quad (n \neq p^a) . $$
This identity was proved [here](http://arxiv.org/abs/0909.1838) as a corollary to some other identities, but the authors asked: ``is there some natural direct proof of this formula?´´
| https://mathoverflow.net/users/2797 | A product of gamma values over the numbers coprime to n. | Denote $$f(n)=\prod\_{k=1}^{n-1}\Gamma \left(\frac{k}{n}\right)$$ and $$ F(n)=\prod\_{1\le k\le n-1, k\perp n}\Gamma \left(\frac{k}{n}\right)$$
We have $f(n)=\prod\_{d|n}F(n)$ and therefore by Mobius inversion $F(n)=\prod\_{d|n}f(d)^{\mu(n/d)}$
By the multiplication theorem we have $f(n)=\frac{1}{\sqrt{n}}(2\pi)^{\frac{n-1}{2}}$, so if $n$ is not a prime power $$F(n)=\prod\_{d|n}\left(\frac{1}{\sqrt{d}}(2\pi)^{\frac{d-1}{2}}\right)^{\mu(n/d)}=(2\pi)^{\frac{1}{2}\varphi (n)}$$
The formula $F(n)=\sqrt{\varphi(n)+1}f(\varphi(n)+1)$ follows.
| 8 | https://mathoverflow.net/users/2384 | 9892 | 6,754 |
https://mathoverflow.net/questions/9898 | 28 | What's the most common way of writing the all-ones vector, that is, the vector, when projected onto each standard basis vector of a given vector space, having length one? The zero vector is frequently written $\vec{0}$, so I'm partial to writing the all-ones vector as $\vec{1}$, but I don't know how popular this is, and I don't know if a reader might confuse it with the identity matrix.
I'm writing for a graph theory audience, if that helps pick a notation.
| https://mathoverflow.net/users/2836 | Notation for the all-ones vector | I have used the notation $\vec{1}$ in a paper. I think that it's a good choice if you help the reader by defining it. I did a Google Scholar such of "vector of all ones", and I found a lot of so-so notation such as $e$, $u$, $\mathbf{e}$, $\mathbf{1}$, and even just plain $1$. I don't think that the literature is loyal to any particular choice. Confusing $\vec{1}$ with a matrix would be a little strange, because a matrix is suggested by a two-headed arrow, or $\stackrel{\leftrightarrow}{1}$.
| 26 | https://mathoverflow.net/users/1450 | 9902 | 6,761 |
https://mathoverflow.net/questions/9921 | 1 | I've been studying universal objects of universal algebra in a quite general setting and try to exhibit the structure of their elements just using the universal property. a very nice example for this is given in Serres *Trees* (normal form for elements in amalgamated sums of subgroups). up to know, it works in all examples I've came across. even tensor products, see: Pierre Mazet, Caracterisation des Epimorphismes par relations et generateurs. but I'm stuck with localizations of rings (or monoids, or modules). rings and monoids are here assumed to be commutative.
so I *define* $S^{-1} A$ to be a ring which represents the subfunctor of $\hom(A,-)$, which maps elements of $S$ to units. here $S$ is a submonoid of a ring $A$. it can be shown with rather general facts that $S^{-1} A$ exists, in several ways. but that's not the point: I want to avoid explicit constructions (I might elaborate the reasons later).
the definition implies that there is a natural homomorphism $A \to S^{-1} A$, which is denoted simply by $a \mapsto a$, and that every element of $S^{-1} A$ has the form $a/s$ ($a \in A, s \in S$). clearly $a/s=b/t$ holds, when $uta=usb$ for some $u \in S$. but how can we prove the converse, *only using the universal property?* I hope my aim is clear. in particular, it would be cheeting applying the universal property to another explicit constructed model of $S^{-1} A$.
here is an example how elements might be described without using any construction: we want to show that in the category of abelian groups, elements of the coproduct $A+B$ (provided it exists) have a unique representation $a+b$, where $a \in A$ and $b \in B$. again we have an abuse of notation here, $a$ also means the image of $a$ in $A+B$. to prove this, observe that $\{a+b : a \in A, b \in B\}$ is a subgroup of $A+B$ which also satisfies the universal property. then it follows that every element has the form $a+b$. now define $A+B \to A$ by extending $id : A \to A$ and $0 : B \to A$. this maps $a+b \mapsto a$. hence $a$ is unique, and similar also $b$.
as already said, this also works in other situations, but it get's more complicated. conclusion: we don't have to invent other objects to study universal objects. for we may apply the universal property to themselves! I hope that this also works for localizations, in order to see that $a/1=0 \in S^{-1} A$ if and only if a is annihilated by some $s \in S$. I've already found out many basic results about localizations just using the universal property (e.g. "coherence isomorphisms", behavior under colimits), and using that I can reduce all to the fact that $S^{-1} A$ is a flat $A$-module, but this also seems to be hard without elements. a major step would be the case of an integral domain.
EDIT: A new improved version of this question can be found [here](https://mathoverflow.net/questions/86923/elements-in-a-localization-category-theoretic-approach).
| https://mathoverflow.net/users/2841 | equality of elements in localization via universal property [unsolved!] | I'm not sure I understand the question. More exactly, I think I disagree with what seems to be an assumption built into it: that there are sharp lines to be drawn between 'only using the universal property' and not, or between working 'without elements' and not.
Generally, a universal property describes how something interacts with the world around it. For example, if you say that an object $I$ of a category $\mathcal{C}$ is initial, that describes how $I$ interacts with other objects of $\mathcal{C}$. If you don't know much about the objects of $\mathcal{C}$, it doesn't tell you much about $I$. Similarly, you're not going to be able to deduce anything about the ring $S^{-1}A$ without using facts about rings. I don't know which of those facts are ones you'd be happy to use, and which aren't.
There's nothing uncategorical about elements. For example, you're dealing with rings, and an element of a ring $A$ is simply a homomorphism $\mathbb{Z}[x] \to A$. (And $\mathbb{Z}[x]$ can be characterized as the free ring on one generator, that is, the result of taking the left adjoint to the forgetful functor $\mathbf{Ring} \to \mathbf{Set}$ and applying it to the terminal set 1.)
So I'm unsure what exactly your task is. But I'd like to suggest a different universal property of localization, which might perhaps make your task easier. Here it is.
Let $\mathbf{Set}/\mathbf{Ring}$ be the category of rings equipped with a set-indexed family of elements. Formally, it's a 'comma category'. An object is a triple $(S, i, A)$ where $S$ is a set, $A$ is a ring, and $i$ is a function from $S$ to the underlying set of $A$. (You might think of $i$ as an including $S$ as a subset of $A$, but $i$ doesn't have to be injective.) A map $(S, i, A) \to (S', i', A')$ is a pair $(p, \phi)$ consisting of a function $p: S \to S'$ and a homomorphism $\phi: A \to A'$ making the evident square commute.
There is a functor $R: \mathbf{Ring} \to \mathbf{Set}/\mathbf{Ring}$ given by
$$
R(A) = (A^\times \to A)
$$
where $A^\times$ is the set of units of $A$ and the arrow is the inclusion. Then $R$ has a left adjoint $L$, given by $L(S, i, A) = (iS)^{-1}A$. In other words, the left adjoint to $R$ is localization.
| 5 | https://mathoverflow.net/users/586 | 9923 | 6,774 |
https://mathoverflow.net/questions/9917 | 12 | I've made the following observation: let V be a vector space over $\mathbb{R}$ with a inner product $\langle , \rangle$. then there is a "natural contravariant" injective map $V \to \hom(V,\mathbb{R})$. if we apply this twice, we get a "natural covariant" injective map $V \to \hom(\hom(V,\mathbb{R}),\mathbb{R}), v \mapsto (\phi \mapsto \phi(v))$. but the same things happen in category theory: let $C$ be a category (which I assume to be locally-small), then $C \to \hom(C,Set), x \mapsto \hom(x,-)$ is a natural contravariant fully-faithful functor and applying this twice yields (up to natural isomorphism) the natural covariant functor $C \to \hom(\hom(C,Set),Set), x \mapsto (F \mapsto F(x))$ (yoneda-lemma).
so how can we unify these two phenomena? perhaps we can make $V$ to a enriched category over $\mathbb{R}$ and hope that the yoneda-lemma for symmetric closed monoidal categories is the common generalization? what is the right structure on $\mathbb{R}$?
| https://mathoverflow.net/users/2841 | yoneda-embedding vs. dual vector space | I don't see the need to try to make vector spaces into categories. I would just say that in each case we have a closed symmetric monoidal category (respectively Vect or Cat), a map f : X ⊗ Y → Z for some objects X, Y, Z (respectively $\langle-,-\rangle$ : V ⊗ V → R and Hom : Cop × C → Set) and we are forming the associated map X → hom(Y,Z) (where hom denotes the internal hom functor). The double dual construction is obtained by setting Y = hom(X,Z) and letting f be the evaluation map; it doesn't depend on anything but X and Z.
That said, there is a great analogy between Vect and Cat, where R and Set play parallel roles: but what corresponds to the construction sending C to Hom(Cop, Set) is the free vector space functor from Set to Vect. The analogy goes something like this. (I am omitting some technical conditions for convenience.)
```
sets categories
vector spaces cocomplete categories (and colimit-preserving functors)
additive structure colimits
free v.s. on S category of presheaves on C
the ground field Set
(comm.) algebras cocomplete closed (symmetric) monoidal categories
A-modules cocomplete V-enriched categories
etc.
```
I am not claiming there is a way to take an object in one column and get a corresponding object in the other column (although under some circumstances that may be possible): rather that it is fruitful to use the left-hand column as a way of thinking about the right-hand column.
See [this nLab page](http://ncatlab.org/nlab/show/free+cocompletion) for an introduction to these ideas.
| 12 | https://mathoverflow.net/users/126667 | 9928 | 6,777 |
https://mathoverflow.net/questions/9901 | 23 | Which are the rigid suborders of the real line?
If A is any set of reals, then it can be viewed as an order structure itself under the induced order (A,<). The question is, when is this structure rigid? That is, for which sets A does the structure (A,<) have no nontrivial order automorphisms?
For example, the positive integers are rigid under the usual order. More generally, any well-ordered subset of R is rigid. Similarly, any anti-well-ordered set, such as the negative integers, is also rigid.
It is also true that the order sum of any well-order plus an anti-well-order is rigid. For example, a sequence converging upward to 0 plus a sequence converging downward to 0 will have order type ω+ω\*, which is rigid. (Whereas it is easy to see that the sum of an infinite anti-well-order and an infinite well-order will not be rigid, since it has a copy of Z in the center.) A more elaborate example will be a well-ordered sum of anti-well-orders, such as the set consisting of k+1/n for any positive integers k and n.
All these examples are countable; are there uncountable examples?
Perhaps there will be some ZFC independence for certain types of examples? I am primarily interested in the situation under ZFC. In ZF without the Axiom of Choice, there can be weird anomalies of uncoutable sets that are Dedekind finite. All such sets are rigid, as I explained in [this question](https://mathoverflow.net/questions/6262/#6725). But if someone can provide a ZF characterization, that would also be interesting.
| https://mathoverflow.net/users/1946 | Which are the rigid suborders of the real line? | Here is a simple example of size continuum. Do the ordinary middle-third construction of the Cantor set, except that whenever you delete the $n$-th (numbered by level and then left to right, say) middle-third interval leave in exactly $n$ points from that interval. Let's call the resulting set $X$. Any automorphism of $X$ must map (resp. left-, right-) isolated points to (resp. left-, right-) isolated points. Since we left a different number of points in each middle-third interval, the automorphism must fix every inner point of each middle-third interval as well as its two endpoints. By density, it follows that the automorphism fixes every point of $X$.
All of your examples are scattered, I checked the Rosenstein's *Linear Orderings* to see if he had anything good to say about which scattered linear orders are rigid. To my dismay, this is what I found: "These considerations seem to make impossible an inductive argument (on $F$-rank or $VD$-rank) to determine which scattered types are rigid." (p. 133) However, he does cite a result of Anne Morel (*Ordering relations admitting automorphisms*, Fund. Math. 54 (1964), 279-284.) which says that a linear order $A$ is not rigid if and only if $A \cong A\_1 + A\_2\times\mathbb{Z} + A\_3$ for some linear orderings $A\_1,A\_2,A\_3$, with $A\_2$ nonempty.
Dense examples of rigid subsets of $\mathbb{R}$ would be interesting to see. This is probably not too difficult to construct under CH. But a ZFC example might have to deal with important barriers such as Baumgartner's result that *All ${\aleph\_1}$-dense subsets of $\mathbb{R}$ can be isomorphic*, Fund. Math. 79 (1973), 101-106. Maybe there are some examples in this classic paper of Sierpinski *Sur les types d'ordre des ensembles linéaires*, Fund. Math. 37 (1950), 253-264.
All three papers can be found [here](http://matwbn.icm.edu.pl/spis.php?wyd=1&jez=en).
---
Addendum (after sdcvvc's comment): For the sake of completeness, I'm including a simplification of the Dushnik-Miller argument that produces a dense subset $X$ of $\mathbb{R}$ which is rigid (though not the stronger result that $X$ has no self-embeddings).
To ensure density, the set $X$ will contain all rational numbers. Note that an automorphism $f$ of $X$ is then completely determined by its restriction to $\mathbb{Q}$. Indeed, since $f[\mathbb{Q}]$ must be dense (in $X$ and) in $\mathbb{R}$, we always have
$f(x) = \sup\{f(q):q \in (-\infty,x)\cap\mathbb{Q}\} = \inf\{f(q):q \in (x,\infty)\cap\mathbb{Q}\}.$
There are only $c = 2^{\aleph\_0}$ increasing maps $f:\mathbb{Q}\to\mathbb{R}$ with dense range. Let $\langle f\_\alpha:\alpha<c \rangle$ enumerate all such maps, except for the identity on $\mathbb{Q}$.
We will define by induction a sequence $\langle (x\_\alpha,y\_\alpha) : \alpha<c \rangle$ of pairs of irrational numbers. The $x\_\alpha$ will be points of $X$ while the $y\_\alpha$ will be in the complement of $X$. For each $\alpha$, we will have $f\_\alpha(x\_\alpha) = y\_\alpha$ (in the sense of the inf/sup definition above).
Suppose we have defined $(x\_\beta,y\_\beta)$ for $\beta<\alpha$. Since $f\_\alpha$ is not the identity, there is a rational $q$ such that $f\_\alpha(q) \neq q$. Let's suppose that $f\_\alpha(q) > q$ (the case $f\_\alpha(q) < q$ is symmetric). Since the real interval $(q,f\_\alpha(q))$ has size $c$ and the extension of $f\_\alpha$ to all of $\mathbb{R}$ is injective, we can always pick
$x\_\alpha \in (q,f\_\alpha(q)) \setminus(\mathbb{Q}\cup\{y\_\beta:\beta<\alpha\})$
such that $y\_\alpha = f\_\alpha(x\_\alpha) \notin \mathbb{Q}\cup\{x\_\beta:\beta<\alpha\}$. Note that $x\_\alpha < f\_\alpha(q) < y\_\alpha$ so $x\_\alpha \neq y\_\alpha$.
In the end, we will have
$\{x\_\alpha: \alpha < c\} \cap \{y\_\alpha : \alpha<c\} = \varnothing$
and any set $X$ such that
$\mathbb{Q}\cup\{x\_\alpha:\alpha<c\} \subseteq X \subseteq \mathbb{R}\setminus\{y\_\alpha:\alpha<c\}$
is necessarily rigid since $f\_\alpha(x\_\alpha) = y\_\alpha \notin X$ for each $\alpha<c$.
| 25 | https://mathoverflow.net/users/2000 | 9932 | 6,780 |
https://mathoverflow.net/questions/9922 | 5 | let's call an object $x$ of a cocomplete category (categorical) finitely generated if $\hom(x,-)$ commutes with filtered colimits of monomorphisms, and finitely presented if $\hom(x,-)$ even commutes with arbitrary filtered colimits. I know that in the literature there are some other definitions for these notions. at least in categories of algebraic structures, it would be nice that these notations coincide with the usual ones. let's concentrate on the category of groups, for these make problems ;).
it's easy to see that categorical finitely generated means the usual finitely generated. but I've figured out that a group $G$ is categorical finitely presented if and only if there is a finitely presented group $H$, over which $id\_G : G \to G$ factors through. is this the same as beeing finitely presented?
the question is equivalent to: let $H$ be a finitely presented group and $p : H \to H$ a projection, i.e. a homomorphism satisfying $p^2 = p$. is then $im(p)$ also finitely presented?
| https://mathoverflow.net/users/2841 | projections of finitely presented groups | Here is a direct proof that "categorically finitely presented" (which I think it is more usual to call "finitely presentable," since no actual presentation is necessarily given) is the same as "finitely presented" in the usual sense. This is a special case of Theorem 3.12 in Adamek and Rosicky's book "Locally presentable and accessible categories," which applies to all finitary varieties of algebras.
First suppose $G$ is finitely presented in the usual sense, with $X$ a finite set of generators and $R \subset F[X]$ a finite set of relations, where $F[-]$ is the free group functor. Then there is a coequalizer diagram $F[R] \rightrightarrows F[X] \twoheadrightarrow G$. Since finite colimits of c.f.p. objects are c.f.p. (since finite limits commute with filtered colimits in Set), it suffices to show that finitely generated *free* groups are c.f.p. But $Hom\_{Grp}(F[X],H) \cong Hom\_{Set}(X,U(H))$ where $U$ is the underlying-set functor, which preserves filtered colimits, and finite sets are c.f.p. in Set (this is easy to see).
Conversely, suppose $G$ is c.f.p. Then in particular it is c.f.g. and hence (as you say) finitely generated in the usual sense. One way to see that is to write $G$ as the directed union (= filtered colimit of monics) of its finitely generated subgroups; then since $G$ is c.f.g. its identity must factor through one of these subgroups, and hence that subgroup must be all of $G$. Now let $X$ be a finite set of generators and consider the diagram of groups whose vertices are all finitely presented quotients $F[X]\twoheadrightarrow H\_i$ through which $F[X]\twoheadrightarrow G$ factors, and whose transition maps $k\_{i,j}\colon H\_i \to H\_j$ are maps under $F[X]$ (and hence over $G$). Then the canonical maps $h\_i\colon H\_i \to G$ make $G$ into the colimit of this diagram, since any relation holding in $G$ must hold in some finitely presented quotient of $F[X]$. Moreover, this diagram is filtered, so $id\_G\colon G\to G$ factors through some $H\_i$ via a map $u\colon G \to H\_i$ with $h\_i u = id\_G$.
I'm guessing this is the point to which you got. The clever argument from A&R then goes as follows. By the first part of the argument, $H\_i$ is c.f.p. But we have two maps $id\_{H\_i}\colon H\_i\to H\_i$ and $u h\_i\colon H\_i\to H\_i$ which become identified in the colimit, i.e. $h\_i id\_{H\_i} = h\_i u h\_i$; thus there must be some $k\_{i,j}\colon H\_i \to H\_j$ such that $k\_{i,j} = k\_{i,j} u h\_i$. It then follows that $h\_j\colon H\_j \to G$ is an isomorphism with inverse $k\_{i,j} u$, for on the one hand, $h\_j k\_{i,j} u = h\_i u = id\_G$, but on the other hand, $k\_{i,j} u h\_j k\_{i,j} = k\_{i,j} u h\_i = k\_{i,j}$ and $k\_{i,j}$ is epic. Thus, $G$ is finitely presented since $H\_j$ is so.
| 6 | https://mathoverflow.net/users/49 | 9939 | 6,786 |
https://mathoverflow.net/questions/9944 | 35 | Let $G$ be a group such that $\operatorname{Aut}(G)$ is abelian. is then $G$ abelian?
This is a sort of generalization of the well-known exercise, that $G$ is abelian when $\operatorname{Aut}(G)$ is cyclic, but I have no idea how to answer it in general. At least, the finitely generated abelian groups $G$ such that $\operatorname{Aut}(G)$ is abelian can be classified.
| https://mathoverflow.net/users/2841 | When is Aut(G) abelian? | From MathReviews:
---
[MR0367059](https://mathscinet.ams.org/mathscinet-getitem?mr=0367059) (51 #3301)
Jonah, D.; Konvisser, M.
Some non-abelian $p$-groups with abelian automorphism groups.
Arch. Math. (Basel) 26 (1975), 131--133.
This paper exhibits, for each prime $p$, $p+1$ nonisomorphic groups of order $p^8$ with elementary abelian automorphism group of order $p^{16}$. All of these groups have elementary abelian and isomorphic commutator subgroups and commutator quotient groups, and they are nilpotent of class two. All their automorphisms are central. With the methods of the reviewer and Liebeck one could also construct other such groups, but the orders would be much larger.
---
FYI, I found this via a google search.
The first to construct such a group (of order $64 = 2^6$) was [G.A. Miller](http://en.wikipedia.org/wiki/George_Abram_Miller)\* in 1913. If you know something about this early American group theorist (he studied groups of order 2, then groups of order 3, then...and he was good at it, and wrote hundreds of papers!), this is not so surprising. I found a nice treatment of "Miller groups" in Section 8 of
<http://arxiv.org/PS_cache/math/pdf/0602/0602282v3.pdf>
(\*): The wikipedia page seems a little harsh. As the present example shows, he was a very clever guy.
| 48 | https://mathoverflow.net/users/1149 | 9946 | 6,791 |
https://mathoverflow.net/questions/9953 | 9 | In another question someone linked to [this video](http://www.math.toronto.edu/~drorbn/Gallery/KnottedObjects/WaistbandTrick/index.html) where the lady ties a knot in a seemingly impossible manner.
What I don't understand is how the end sticking out beyond her right hand gets longer as she pulls through (seems that if it's to change it should get shorter). This suggests to me that there is some footage cut out.
| https://mathoverflow.net/users/2855 | Is this a sleight of hand or a video edit? | Sleight of hand! I learned to do this by watching the video - there is a moment when you let go of the end of the rope and grab another part. Your observation is telling you which end to let go of, when to let go, and where to grab.
| 14 | https://mathoverflow.net/users/1650 | 9955 | 6,796 |
https://mathoverflow.net/questions/9773 | 9 | Let $K$ be the closed unit ball of $C[0,1]$, and let $f$ in $C(K,\mathbb{\, R})$.
Is it true that there exists an infinite dimensional reflexive subspace
$E$ of $C[0,1]$ s.t. $f(K\cap E)$ is bounded ?
If the answer is affirmative, this would be a very weak kind of Weierstrass-type theorem
[and also a very general one, due to the "universality" of $C[0,1]$ (i.e., the Banach-Mazur Embedding Theorem)].
---
One may also replace $C[0,1]$ by $B[0,1]$, the space of *all* bounded functions on $[0,1]$,
endowed with the sup-norm.
| https://mathoverflow.net/users/2508 | Boundedness of nonlinear continuous functionals | Ady, I think there is a counterexample to your question.
To describe it, let $(V\_n)$ be a basis of $[0,1]$ consisting
of non-empy open sets; $K$ stands for the closed unit ball of $C[0,1]$.
For every $n$ let $C\_n$ be the closure of $V\_n$ and define
$U\_n={g \in K: \min{|g(t)|:t \in C\_n} > \|g\| - 1/4}$
where $\|g\|=\sup{|g(t)|:t \in [0,1]}$.
The family $(U\_n)$ is an open cover of $K$. Let $(F\_m)$
be a partition of unity subordinate to $(U\_n)$.
For every m let $n\_m$ be the least integer $n$ such that
$\sup(F\_m)={g \in K: F\_m(g)>0}$ is contained in $U\_n$.
Now define $F:K\to \mathbb{R}$ by
$F(g)= \sum\_{m=1}^{\infty} n\_m\cdot F\_m(g)$
Notice that F is well-defined and continuous.
Finally notice that $F(K\cap E)$ is unbounded for
every infinite-dimensional subspace E of $C[0,1]$.
This follows from the following fact: for every
integer i and every infinte-dimensional subspace
$E$ of $C[0,1]$ there is a norm-one vector $e \in E$ such
$e$ is NOT in $U\_n$ for every $n < i$ (and therefore, if
$m$ is such that $F\_m(e)>0$, then necessarily $n\_m$ is
greater or equal to $i$ which gives that $F(e)$ is
also greater or equal to $i$).
| 4 | https://mathoverflow.net/users/2858 | 9959 | 6,799 |
https://mathoverflow.net/questions/9950 | 6 | The Cohen-Lenstra measure on the set of abelian p-groups assigns $\mathbb{P}(G) = \prod\_{i \geq 1} \left( 1 - \frac{1}{p^i}\right) \cdot |\mathrm{Aut}(G)|^{-1} $. Apparently, this is equivalent to taking cokernels of random maps $f: (\mathbb{Z}\_p)^N \to (\mathbb{Z}\_p)^N$ and letting $N \to \infty$. These are the p-adics and there is a Haar measure on this linear space of maps. Alternatively, choose random maps between the finite groups: $f: (\mathbb{Z}/p^k\mathbb{Z})^N \to (\mathbb{Z}/p^k \mathbb{Z})^N$ and let $k, N \to \infty$.
**Q**: If G is a random abelian p-group according to the Cohen-Lenstra measure and A is a deterministic, why is the expected number of surjections $\phi: G \to A$ equal to 1? In fact, if G were deterministic I don't think this number could ever be 1 unless |G| = 1.
For references see Section 8 of [Homological stability for Hurwitz spaces and the Cohen-Lenstra conjecture over function fields](https://arxiv.org/abs/0912.0325) by Ellenberg, Venkatesh and Westerland or Terry Tao's blog entry [At the AustMS conference](http://terrytao.wordpress.com/2009/10/02/at-the-austms-conference/).
| https://mathoverflow.net/users/1358 | An Expectation of Cohen-Lenstra Measure | Let me spell out the cokernel description of the Cohen-Lenstra distribution in more detail, as my answer will depend on it.
A map $(\mathbb{Z}\_p)^N \to (\mathbb{Z}\_p)^N$ is given by an $N \times N$ matrix of $p$-adic integers. Choose such a map by picking each of the digits of each integer uniformly at random from $\{ 0, 1, ..., p-1 \}$; this is the same as using the additive Haar measure on $\mathrm{Hom}((\mathbb{Z}\_p)^N, (\mathbb{Z}\_p)^N)$. With probability $1$, this map does not have determinant $0$, so its cokernel is a finite abelian $p$-group. Let $\mu\_N$ be the probability measure on isomorphism classes of abelian $p$-groups which assigns each $p$-group the probability that it arises as this cokernel.
The Cohen-Lenstra distribution is the limit as $N \to \infty$ of $\mu\_N$. As shown in several of the references you link to, it is given by the formula
$$\lim\_{N \to \infty} \mu\_N(G) = \frac{1}{ |\mathrm{Aut}(G)|} \prod\_{i=1}^{\infty} (1-1/p^i) .$$
For notational convenience, it will help to distinguish between the domain and range of a map in $\mathrm{Hom}((\mathbb{Z}\_p)^N, (\mathbb{Z}\_p)^N)$. I will call the former $U^N$ and the latter $V^N$.
---
Now, to answer your question. Let $A$ be a fixed finite abelian $p$-group. Let $e\_N(A)$ be the expected number of surjections from an abelian $p$-group $G$, picked according to measure $\mu\_N$, to $A$. Ignoring issues about interchanging limits, we want to show that $\lim\_{N \to \infty} e\_N(A)=1$.
Lets start by considering $H\_N(A) := \mathrm{Hom}(V^N, A)$. The set $H\_N(A)$ has cardinality $|A|^N$, as any map is specified by giving the image of a basis for $V^N$. Inside this set, let $S\_N(A)$ be the surjective maps and $C\_N(A)$ the nonsurjective maps.
For any map $f \in S\_N(A)$, let's consider the possibility that it extends to the cokernel of a random map $U^N \to V^N$. This will occur if and only if the $N$ generators of $U^N$ land in the kernel of $f$. Since $f$ is in $S\_N(A)$, its kernel has index $|A|$. So the probability that $U^N$ is mapped into the kernel of $f$ is $1/|A|^N$.
We want to compute
$$e\_N(A) = |S\_N(A)| \cdot (1/|A|^N) = 1 - |C\_N(A)|/A^N.$$
If $A$ can be generated by $r$ elements, then $|C\_N(A)|/A^N \leq (1-1/|A|^r)^{\lfloor N/r \rfloor}$, so the second term drops out as $N \to \infty$. (To see this bound, group the basis elements of $V^N$ into $N/r$ groups of size $r$; the probability that these $r$ basis elements are not sent to the $r$ generators of $A$ is $(1-1/|A|^r)$. This bound is probably much weaker than the true rate of convergence.)
| 8 | https://mathoverflow.net/users/297 | 9960 | 6,800 |
https://mathoverflow.net/questions/6262 | 21 | Let us say that a set B admits a rigid binary relation, if there is a binary relation R such that the structure (B,R) has no nontrivial automorphisms.
Under the Axiom of Choice, every set is well-orderable, and since well-orders are rigid, it follows under AC that every set does have a rigid binary relation.
My questions are: does the converse hold? Does one need AC to produce such rigid structures? Is this a weak choice principle? Or can one simply prove it in ZF?
(This question spins off of Question [A rigid type of structure that can be put on every set?](https://mathoverflow.net/questions/5920).)
| https://mathoverflow.net/users/1946 | Does every set admit a rigid binary relation? (and how is this related to the Axiom of Choice?) | Update: Joel and I have written an article based on the concepts introduced in this question, which can be seen at <http://arxiv.org/abs/1106.4635>
It looks to me that it is consistent with ZF that there is a set without a rigid binary relation. Use
the standard technique for constructing such wierd sets. First construct a permutation model of ZFA where
the set of atoms A has the desired property and then use the Jech-Sochor theorem to transfer the result to a ZF model. Any sentence that
can be stated just using quantifiers over some fixed iteration of the powerset operation over A can be transferred.
In our case the sentence "A has no binary relation without a nontrivial automorphism" only needs to quantify
over say the fifth iteration of the powerset of A (probably less). The standard reference for these techniques
is Jech's text The Axiom of Choice from 1973. (There the Jech-Sochor theorem is called the First Embedding Theorem).
In our case what is called the basic Fraenkel model is the desired ZFA model. (This and other similar
models are constructed in Chapter 4). Suppose R is a binary relation on A. Then there is a finite set E (called
the support of R in Jech's terminology) such that any permutation of A fixing E pointwise maps R to itself. In
other words such bijections are automorphisms of R. Since E is finite we can without AC find nontrivial such bijections and
it follows that R is not rigid.
In fact examining the proof of the Jech-Sochor theorem shows that Joel's fact about sets of reals is optimal
in the following sense: models of ZFA are simulated by ZF models by transferring the set of atoms to a
set of sets of reals and thus one cannot in ZF go any further up the hierarchy of types than the
powerset of omega and hope to construct rigid binary relations.
| 16 | https://mathoverflow.net/users/2436 | 9966 | 6,804 |
https://mathoverflow.net/questions/7477 | 33 | I can show that if $X$ is a scheme such that all local rings $\mathcal{O}\_{X,x}$ are integral and such that the underlying topological space is connected and Noetherian, then $X$ is itself integral.
This doesn't seem to work without the "Noetherian" condition. But can anyone think about a nice counterexample to illustrate this? So I am looking for a non-integral scheme - with connected underlying topological space - having integral local rings.
| https://mathoverflow.net/users/1107 | Non-integral scheme having integral local rings | Let me try to give a counterexample. (I don't know whether it is 'nice'). First, let us rewrite your properties for an affine scheme $X=Spec(A)$.
Connectedness for $A$ means $A$ has no nontrivial idempotents;
Integrality for $A$ is the usual one ($A$ is a domain);
Local integrality means that whenever $fg=0$ in $A$, every point of $X$ has a neighborhood
where either $f$ or $g$ vanishes.
Let us construct a connected locally integral ring that is not integral.
Roughly speaking, the construction is as follows: let $X\_0$ be the cross (the union of coordinate axes) on the affine plane. Then let $X\_1$ be the (reduced) full preimage of $X\_0$ on the blow-up of the plane ($X\_1$ has three rational components forming a chain). Then blow up the resulting surface at the two singularities of $X\_1$, and let $X\_2$ be the reduced preimage of $X\_1$
(which has five rational components), etc. Take $X$ to be the inverse limit.
The only problem with this construction is that blow-ups glue in a projective line, so $X\_1$ is not affine. Let us correct this by gluing in an affine line instead (so our scheme will be an open subset in what was described above).
Here's an algebraic description:
For every $k\ge 0$, let $A\_k$ be the following ring: its elements are collections of
polynomials $p\_i\in{\mathbb C}[x]$ where $i=0,\dots,2^k$ such that $p\_i(1)=p\_{i+1}(0)$.
Set $X\_k=Spec(A\_k)$. $X$ is a union of $2^k+1$ affine lines that meet transversally in a chain. (It may be better to index polynomials by $i/2^k$, but the notation gets confusing.)
Define a morphism $A\_k\to A\_{k+1}$ by
$$(p\_0,\dots,p\_{2^k})\mapsto(p\_0,p\_0(1),p\_1,p\_1(1),\dots,p\_{2^k})$$
(every other polynomial is constant). This identifies $A\_k$ with a subring of $A\_{k+1}$.
Let $A$ be the direct limit of $A\_k$ (basically, their union). Set $X=Spec(A)$. For every
$k$, we have a natural embedding $A\_k\to A$, that is, a map $X\to X\_k$.
Each $A\_k$ is connected but not integral; this implies that $A$ is connected but not integral. It remains to show that $A$ is locally integral.
Take $f,g\in A$ with $fg=0$ and $x\in X$. Let us construct a neighborhood of $x$ on which one of $f$ and $g$ vanishes. Choose $k$ such that $f,g\in A\_{k-1}$ (note the $k-1$ index).
Let $y$ be the image of $x$ on $X\_k$. It suffices to prove that $y$ has a neighborhood on
which either $f$ or $g$ (viewed as functions on $X\_k$) vanishes.
If $y$ is a smooth point of $X\_k$ (that is, it lies on only one of the $2^k+1$ lines), this is obvious. We can therefore assume that $y$ is one of the $2^k$ singular points, so two components of $X\_k$ pass through $y$. However, on one of these two components (the one with odd index), both $f$ and $g$ are constant, since they are pullbacks of functions on $X\_{k-1}$. Since $fg=0$ everywhere, either $f$ or $g$ (say, $f$) vanishes on the other component.
This implies that $f$ vanishes on both components, as required.
| 35 | https://mathoverflow.net/users/2653 | 9967 | 6,805 |
https://mathoverflow.net/questions/9968 | -4 | Given a 2-dimensional array of MxN heights, how to transform it to a sphere? Every element of this array is just a 3D point (x,y,z) where z represents some height. One has to transform this array into a sphere, twisting it around the origin so, that only minimal distortions will happen.
Representing it by spherical coordinates is not very good, because of the severe distortions. It's probably better if there is no direct one-to-one mapping from 2D plane to a surface of 3D sphere - many plane's points will not be involved. But what is the best possible mapping and how to transform involved points (elements of array)?
This is for a 3D-planet terrain simulation. First, fractal landscape is produced, then, it is to be transformed to 3D sphere.
Thanks in advance!
**SOLUTION**: [Map projection](http://en.wikipedia.org/wiki/Map_projection)
| https://mathoverflow.net/users/2266 | How to transform a plane into a sphere? [SOLVED] | I may be misunderstanding your question but if you are asking how to take a data set of points in R^3 and convert them to a surface of a sphere then you can use the transformations given on wikipedia by associating latitude and longitude to your plane. I wrote code to do this myself for the math modeling competition a couple years ago.
Sorrry for the limited scope of this answer and the lack of hyperlinks I am updating from the iPhone and that makes it a little bit annoying.
| 3 | https://mathoverflow.net/users/348 | 9970 | 6,806 |
https://mathoverflow.net/questions/9957 | 12 | A metric space *(V,d)* will be called distance regular if for every distances *a>0, b, c* a nonnegative integer *p(a,b,c)* is defined, so that whenever *d(B,C)=a*, there are precisely *p(a,b,c)* points *A* such that *d(A,B)=c, d(A,C)=b*.
The Euclidean plane is an example: *p(a,b,c)=0,1,* or *2* when the triangle inequality for *a,b,c*, correspondingly, fails, turns into equality, or is strict.
If we also require that *p(a,b,c)>0* whenever the triangle inequality does not fail, then I conjecture that this is the only possibility for the parameters *p(a,b,c)*. That is, there may be many non-isomorphic examples, but the parameters will be the same for all of them. (Thanks to Heather for this clarification.)
Has anybody formulated/proved/refuted this conjecture before? It looks very natural.
UPD. I should have mentioned this: "for every distances *a>0, b, c*" means all nonnegative reals, and the same for "whenever the triangle inequality does not fail". In particular this means that all positive real distances actually occur.
UPD2. After a week trial, the question seems to be new, open, and interesting. Anton suugested a line of attack, and I believe I can write down the proof of the first step: that *p(a,b,c)=1* when the triangle inequality turns into equality. Fedja produced examples showing that this first step is indeed essential.
I'm adding the "open problem" tag to the question.
| https://mathoverflow.net/users/2795 | distance regular metric spaces | Note that any metric with unique infinite geodesics on $\mathbb R^2$ has this property.
In particular, hyperbolic plane as noted by Heather Macbeth.
It also includes Minkowski plane with smooth ball and all complete negatively curved Riemannian metrics on $\mathbb R^2$.
So it is better to ask:
* Is it true that the function $p$ is the only possible? ($p(a,b,c)=0,1,$ or $2$ when the triangle inequality for a,b,c, correspondingly, fails, turns into equality, or is strict.)
* Is it true that space has to be homeomorphic to $\mathbb R^2$?
**Sketch for locally compact case**
You condition implies existence of line (i.e. infinite minimizing geodesic) through any pair of points.
Such line must be unique; otherwise for some values $p(a,b,a+b)=\infty$ for some values $a,b$.
If space is locally compact then line depends continuousely on the points.
For some points $a\_1, a\_2, a\_3, a\_4$ take geodesic $a\_1a\_2$ connect each point to $a\_3$ and connect each point of this trianle to $a\_4$.
This way we get a cone-map of 3-simplex in your space.
If it is *nondegenerate* (i.e. the image does not coinsides with image of its boundary) then it would be easy to get a triple of small numbers $a,b,c$ such that $p(a,b,c)=\infty$ thus we arrive to a contradiction.
Thus the space must be 2-dimensional and with unique infinite geodesics --- I bet it should be homeomorphic to $\mathbb R^2$. Then it follows that $p(a,b,c)=0,1$ or $2$ as defined above.
| 8 | https://mathoverflow.net/users/1441 | 9974 | 6,809 |
https://mathoverflow.net/questions/9969 | 4 | Consider some number of charged particles on a closed interval, each with possibly different amounts of charge. Assuming some kind of 'friction' to dampen their motion, they will eventually find a stationary equilibrium, where the repelling force from all the other charges is balanced (except in the case of the particles on the end points, which are also acted on by the 'walls'). It seems intuitively clear that this equilibrium depends only on the final ordering of the particles, but how does one show this? What is it about the interaction potential $\frac{q\_aq\_b}{d}$ that makes this this solution unique?
I actually care about a slightly more complicated situation, which the previous example was meant to be a simplification of. Consider some number of charged particles on the unit circle in $\mathbb{R}^2$, each with possibly different amounts of charge. However, the force between them behaves a little weird:
* The repelling force between two particles is not constrained to the unit circle, it cuts across the circle along a chord and tries to push the particle along the straight-line connecting the two points. However, because the particles are constrained to the circle, the effective force is the projection of this vector to the tangent space to the circle.
* The repelling force is proportional to the inverse of the distance between two particles (along the chord), not the inverse squared. Therefore, the potential is of the form $q\_aq\_b\ln(d)$.
It is then not hard to show the force exerted on particle $b$ by particle $a$ is proportional to $\cot\left(\frac{\theta\_b-\theta\_a}{2}\right)$, and the potential is proportional to $\ln\left(\sin\left(\frac{\theta\_b-\theta\_a}{2}\right)\right)$.
Again, it seems intuitively clear that, for a fixed cyclic ordering of the particles, there should be a unique equilibrium solution with all forces balanced (modulo rotation of the entire picture). However, how would one show this?
It seems like there should be well known tricks for how to do this, but I don't know much about this kind of stuff.
| https://mathoverflow.net/users/750 | Uniqueness of Force Balancing Solutions | Monotonicity is too weak a thing for uniqueness, but strict convexity is enough. If $U(X)$ is your potential on the configuration $X$ and you know that $U(tX+(1-t)Y)<tU(X)+(1-t)U(Y)$ for $0<t<1$ and $X\ne Y$, then there exists a unique stationary point, which is the global minimum. Note that $1/x$ is strictly convex on $x>0$ and so is -$\log \sin x$ on $(0,\pi)$, so this trick takes care of both your questions.
| 6 | https://mathoverflow.net/users/1131 | 9977 | 6,811 |
https://mathoverflow.net/questions/9716 | 27 | Qiaochu asked this in the comments to [this question](https://mathoverflow.net/questions/9661/is-semisimple-a-dense-condition-among-lie-algebras). Since this is really his question, not mine, I will make this one Community Wiki. In [MR0522147](http://www.ams.org/mathscinet-getitem?mr=522147), Dyson mentions the generating function $\tau(n)$ given by
$$ \sum\_{n=1}^\infty \tau(n)\,x^n = x\prod\_{m=1}^\infty (1 - x^m)^{24} = \eta(x)^{24}, $$
which is apparently of interest to the number theorists ($\eta$ is Dedekind's function). He mentions the following formula for $\tau$:
$$\tau(n) = \frac{1}{1!\,2!\,3!\,4!} \sum \prod\_{1 \leq i < j \leq 5} (x\_i - x\_j)$$
where the sum ranges over $5$-tuples $(x\_1,\dots,x\_5)$ of integers satisfying $x\_i \equiv i \mod 5$, $\sum x\_i = 0$, and $\sum x\_i^2 = 10n$. Apparently, the $5$ and $10$ are because this formula comes from some identity of $\eta(x)^{10}$. Dyson mentions that there are similar formulas coming from identities with $\eta(x)^d$ when $d$ is on the list $d = 3, 8, 10, 14,15, 21, 24, 26, 28, 35, 36, \dots$. The list is exactly the dimensions of the simple Lie algebras, except for the number $26$, which doesn't have a good explanation. The explanation of the others is in [I. G. Macdonald, *Affine root systems and Dedekind's $\eta$-function*, Invent. Math. 15 (1972), 91--143, MR0357528](http://www.ams.org/mathscinet-getitem?mr=357528), and the reviewer at MathSciNet also mentions that the explanation for $d=26$ is lacking.
So: in the last almost-40 years, has the $d=26$ case explained?
| https://mathoverflow.net/users/78 | What's the status of the following relationship between Ramanujan's $\tau$ function and the simple Lie algebras? | The case of $d=26$ is related to the exceptional Lie algebra $F\_4$. Let me quote from the 1980 [paper](http://dx.doi.org/10.1070/RM1980v035n01ABEH001566) by Monastyrsky which was originally published as a supplement to the Russian translation of the Dyson's paper:
A more careful study of Macdonald's article reveals that the identity for the
26th power of $\eta(x)$ is not really such a mystery. It is related to the exceptional
group $F\_4$ of dimension 52, where the space of dual roots $F\_4^V$ and the space of roots $F\_4$ are not the same. Thus, there are two distinct identities associated
with $F\_4$, one for $\eta^{52} (x)$ and the other for $\eta^{26} (x)$. A similar situation prevails in
the case of the algebra $G\_2$ of dimension 14, which yields identities for $\eta^{14} (x)$
and $\eta^{7} (x)$. The identities for $\eta^{26} (x)$ and $\eta^{7} (x)$ are considerably more complicated.
| 26 | https://mathoverflow.net/users/2149 | 9979 | 6,813 |
https://mathoverflow.net/questions/9924 | 11 | I thought that the order of the [Tate-Shafarevich group](http://en.wikipedia.org/wiki/Tate-Shafarevich_group) should always be a square (it's also supposed to be finite, but for the purposes of this question let's assume we know this) but I don't seem to find a good explanation; Wikipedia is silent on the matter.
While I know it may be an open problem, is there a good argument *pro* or *contra* this?
| https://mathoverflow.net/users/65 | Order of the Tate-Shafarevich group | The first example of an abelian variety with nonsquare Sha was discovered in a computation by Michael Stoll in 1996. He emailed it to me and Ed Schaefer, because his calculation depended on a paper that Ed and I had written. At first none of us believed that it was what it was: instead we thought it must be due to either an error in Stoll's calculations or an error in the Poonen-Schaefer paper. Stoll and I worked together over the next few weeks to develop a theory that explained the phenomenon, and this led to the paper <http://math.mit.edu/~poonen/papers/sha.ps> - that paper contains a detailed answer to your question.
To summarize a few of the key points: If the abelian variety over a global field $k$ has a principal polarization coming from a $k$-rational divisor (as is the case for every elliptic curve), then the order of Sha is a square (if finite), because it carries an alternating pairing - this is what Tate proved, generalizing Cassels' result for elliptic curves. For principally polarized abelian varieties in general, the pairing satisfies the skew-symmetry condition $\langle x,y \rangle = - \langle y,x \rangle$ but not necessarily the stronger, alternating condition $\langle x,x \rangle=0$, so all one can say is that the order of Sha is either a square or twice a square (if finite). Stoll and I gave an explicit example of a genus 2 curve over $\mathbf{Q}$ whose Jacobian had Sha isomorphic to $\mathbf{Z}/2\mathbf{Z}$ unconditionally (in particular, finiteness could be proved in this example).
If the polarization on the abelian variety is not a *principal* polarization, then the corresponding pairing need not be even skew-symmetric, so there is no reason to expect Sha to be even within a factor of $2$ of a square. And indeed, William Stein eventually found explicit examples and published them in the 2004 paper cited by Simon.
A final remark: Ironically, my result with Stoll quantifying the *failure* of Sha to be a square is used by Liu-Lorenzini-Raynaud to prove that the Brauer group $\operatorname{Br}(X)$ of a surface over a finite field *is* a square (if finite)!
| 35 | https://mathoverflow.net/users/2757 | 9982 | 6,814 |
https://mathoverflow.net/questions/9980 | 3 | ? A graph is four colorable if and only if it is planar.
Is this true, I know that if a graph is planar it is four colorable, but is it true that if a graph is four colorable it must be a planar graph.
(EDIT) The following would have been a better way for me to have ask the question.
What are the requirements for a graph to be planar?
What are the requirements for a graph to be 4 colorable?
Is there a simplification of the intersection of not planar and four colorable?
| https://mathoverflow.net/users/2869 | ? A graph is four colorable if and only if it is planar. | A graph is planar if and only if it does not have $K\_5$ or $K\_{3,3}$ as a minor. As Hunter's comment points out, $K\_{3,3}$ is bipartite, ie two-colourable.
| 12 | https://mathoverflow.net/users/1463 | 9983 | 6,815 |
https://mathoverflow.net/questions/9984 | 18 | Minhyong Kim's [reply](http://minhyongkim.wordpress.com/2007/12/14/specz-and-three-manifolds/) to a question John Baez once asked about the analogy between $\text{Spec } \mathbb{Z}$ and 3-manifolds contains the following snippet:
> Finally, regarding the field with one element. I'm all for general theory building, but I think this is one area where having some definite problems in mind might help to focus ideas better. From this perspective, there are two things to look for in the theory of $\mathbb{F}\_1$.
>
> 1) A theory of differentiation with respect to the ground field. A well-known consequence of such a theory could include an array of effective theorems in Diophantine geometry, like an effective Mordell conjecture or the ABC conjecture. Over function fields, the ability to differentiate with respect to the field of constants is responsible for the **considerably stronger theorems of Mordell conjecture type** [emphasis mine], and makes the ABC conjecture trivial.
>
>
>
>
What is the strongest such theorem? Does anyone have a reference? (A preliminary search led me to results that are too general for me to understand them. I'd prefer to just see effective bounds on the number of rational points on a curve of genus greater than 1 in the function field case.)
| https://mathoverflow.net/users/290 | What does Faltings' theorem look like over function fields? | The "function field analogues" of Faltings' theorem were proved by Manin, Grauert and Samuel: see
<http://archive.numdam.org/ARCHIVE/PMIHES/PMIHES_1966__29_/PMIHES_1966__29__55_0/PMIHES_1966__29__55_0.pdf>
especially Theorem 4. (The quotation marks above are because all of this function field work came first: the above link is to Samuel's 1966 paper, whereas Faltings' theorem was proved circa 1982.)
The statement is the same as the Mordell Conjecture, except that there is an extra hypothesis on "nonisotriviality", i.e., one does not want the curve have constant moduli. For some discussion on why this hypothesis is necessary, see e.g. p. 7 of
[http://alpha.math.uga.edu/~pete/hassebjornv2.pdf](http://alpha.math.uga.edu/%7Epete/hassebjornv2.pdf)
An effective height bound in the function field case is given in Corollaire 2, Section 8 of
---
Szpiro, L.(F-PARIS6-G)
Discriminant et conducteur des courbes elliptiques. (French) [Discriminant and conductor of elliptic curves]
Séminaire sur les Pinceaux de Courbes Elliptiques (Paris, 1988).
Astérisque No. 183 (1990), 7--18.
---
Note that effectivity on the height is much better than effectivity on the *number* of rational points (Faltings' proof does give the latter). This is not to be confused with uniform bounds on the number of rational points, for which I believe there are only conditional results known in any case.
| 14 | https://mathoverflow.net/users/1149 | 9985 | 6,816 |
https://mathoverflow.net/questions/9987 | 18 | I imagine this is pretty much standard, but surely someone here will be able to provide useful references...
Suppose $X$ is a topological space. Let me say that two triangulations $T$ and $T'$ of $X$, are *homotopic* if there is a triangulation on $X\times [0,1]$ which induces $T$ and $T'$ on $X\times\{0\}$ and on $X\times\{1\}$, respectively.
* This rather natural equivalence relation has probably been studied. Does it have a standard name? Are there standard references?
* In particular, can this equivalence be redefined by using local moves, *ie*, as the transitive symmetric closure of a (small) set of local changes in triangulations?
| https://mathoverflow.net/users/1409 | Homotopies of triangulations | The standard name for this type of relation between two structures on $X$ is concordance rather than homotopy. If two structures on $X$ are isotopic (with the respect to the appropriate homeomorphism group), then they are concordant, but not necessarily vice versa. In some cases you can also assign a separate meaning to homotopy, but I don't think that it means the same thing as concordance.
There are then two levels to your question. A triangulation $T$ of $X$ induces a piecewise linear structure $\mathcal{P}$. You could ask whether the PL structures $\mathcal{P}$ and $\mathcal{P}'$ are isotopic or concordant, without worrying about the original triangulations. For simplicity suppose that $X$ is a closed manifold. Then at least in dimension $n \ge 5$, Kirby-Siebenmann theory says that the set of PL structures on $X$ up to isotopy are an affine space (or torsor) of $H^3(X,\mathbb{Z}/2)$. I think that the concordance answer is the same, because the Kirby-Siebenmann invariant comes from the stable germ-theoretic tangent bundle of $X$. In other words, two PL structures give you two different sections of a bundle on $X$ whose fiber is $\text{TOP}(n)/\text{PL}(n)$, the group of germs of homeomorphisms of $\mathbb{R}^n$ divided by the subgroup of germs of PL homeomorphisms. Stabilization in this case means replacing $n$ by $\infty$ by adding extra factors of $\mathbb{R}$. If $n = 4$, then up to isotopy there are lots of PL structures on many 4-manifolds, as established by gauge theory. But I think that the concordance answer is once again Kirby-Siebenmann. (I learned about this stuff in a seminar given by Rob Kirby — I hope that I remembered it correctly! You can also try [the reference by Kirby and Siebenmann](http://press.princeton.edu/titles/648.html), although it is not very easy to read.)
There is a coarser answer than the one that I just gave. I tacitly assumed that the triangulations not only give you a PL structure (which always happens), but that they specifically give you a PL manifold structure, with the restriction that the link of every vertex is a PL sphere. These are called "combinatorial triangulations". It is a theorem of Edwards and Cannon that $S^5$ and other manifolds also have non-combinatorial triangulations. If your question is about these, then it is known that they are described by some quotient of Kirby-Siebenmann theory, but it is not known how much you should quotient. It is possible that every manifold of dimension $n \ge 5$ has a non-combinatorial triangulation, and that PL structures are always concordant in this weaker sense. It is known that you should quotient more than trivially, that there are manifolds that have a non-combinatorial triangulation but no PL structure. (I think.)
The other half of the question is to give $X$ a distinguished PL structure $\mathcal{P}$, and to look at triangulations $T$ and $T'$ that are both PL with respect to $\mathcal{P}$. In this case there are two good sets of moves to convert $T$ to $T'$. First, you can use stellar moves and their inverses. A stellar move consists of adding a vertex $v$ to the interior of a simplex $\Delta$ (of some dimension) and supporting structure to turn the star of $\Delta$ into the star of $v$. The theorem that these moves suffice is called the stellar subdivision theorem. (The theorem is due to Alexander and Newman and it is explained pretty well in [the book by Rourke and Sanderson](http://www.worldcat.org/oclc/264145137).) The other set of moves are specific to manifolds and they are the bistellar moves or Pachner moves. One definition is that a bistellar move is a stellar move that adds a vertex $v$ plus a different inverse stellar move that removes $v$ (hence the name). But a clearer definition is that in dimension $n$, a bistellar move replaces $k$ simplices by $n-k+1$ simplices in the minimal way, given by a local $n+1$ concordance that consists of attaching a single $n+1$-dimensional simplex. The theorem that these moves work is [due to Pachner](http://portal.acm.org/citation.cfm?id=107898). Pachner's moves in particular give you a shellable triangulation of $X \times [0,1]$.
Even though the bistellar moves are motivated by concordance and the stellar moves are not, it is not hard to make concordance triangulations for stellar moves as well.
| 24 | https://mathoverflow.net/users/1450 | 9995 | 6,823 |
https://mathoverflow.net/questions/9990 | 2 | This is the simplest case of a question that's been bugging me for a while: say we have a Riemannian metric in polar coordinates on a (2-d) surface:
g=dr2+f2(r, θ)dθ2, such that the θ parameter runs from 0 to 2π. Assume that f is a smooth function on (0,∞)X S1 such that f(0, θ)=0.
Define the cone angle at the pole to be
$ C=\lim\_{r\rightarrow 0^{+}} \frac{L(\partial B(r))}{r} $, where B(r) is the geodesic disc of radius r centered at the origin. Then it's fairly easy to see(by switching into Cartesian coordinates) that a necessary condition for the metric to be smooth is that C=2π. If C<2π, there is a cone point at the origin. One can write out a cone metric, and show that the triangle inequality holds, so there is a singular metric, but which still induces a metric space structure.
Now, if C>2π, it seems pretty clear that we'll end up with a space which violates the triangle inequality; it will be shorter to take a broken segment through the origin than to follow the shortest geodesic(in the sense of a curve γ(t) such that Dγ'γ'=0.) One can show this directly for some simple cases, eg a flat metric with a cone angle greater than 2π.
But there must be an elementary proof of the general case! I can't seem to find one though, and I spent the afternoon playing around with the Topogonov and Rauch comparison estimates to no avail. The basic problem I'm having is that the cone angle condition is essentially a condition on metric balls, but we expect a violation of the triangle inequality, which is a condition on distances.
This is not really related to anything I'm working on, but it's driving me crazy, so I'd appreciate any insight.
| https://mathoverflow.net/users/2497 | Cone angles for Riemannian metrics in polar coordinates | Although this isn't quite the answer you want, the distinction between the 'shortest path' an the 'shortest geodesic' (in the "no acceleration" sense) has been observed in discrete settings.
Your limiting conditions are akin to the case of the pole being a flat, convex or "saddle" point. It's been known for a while that on convex polytopes (specifically where there are no points with C > 2 pi), the shortest path between two points can be found by identifying the relevant facets the shortest path goes through, unfolding these facets, and then drawing a straight line. This is precisely the geodesic criterion you're referring to. And it's also known in general that this might not be true for non-convex polyhedra
The paper that does is by Sharir and Schorr: "On shortest paths in polyhedral spaces, SIAM J. Comp. 15, 193-215, 1986". Although this is all in discrete land, the underlying phenomenon is likely the same.
| 0 | https://mathoverflow.net/users/972 | 9999 | 6,825 |
https://mathoverflow.net/questions/9997 | 4 | I'm trying to understand the definition of a Beta process, as given in the paper:
www.ece.duke.edu/~lcarin/Paisley\_BP-FA\_ICML.pdf
The problem is that from the definition it follows that every Dirichlet process is also a beta process, which seems, ahm, wrong. Can you help me figure out what I don't understand?
This is the definition from the paper:
"Let $H\_0$ be a continuous probability measure and α a positive scalar. Then for all disjoint, infinitesimal partitions, $B\_1 ,\ldots, B\_K$, the beta
process is generated as follows,
$H(B\_k) \sim Beta(\alpha H\_0 (B\_k ), \alpha(1 − H\_0(B\_k )))$
with K → ∞ and $H\_0(B\_k)$ → 0 for $k = 1,\ldots,K$. This process is denoted $H \sim BP(\alpha H\_0 )$."
This is the definition for a Dirichlet Process (DP):
If $X \sim DP(\alpha H\_0)$ where $\alpha$ is a scalar and $H\_0$ is a probability distribution, then for every finite partition $A\_1,\ldots,A\_K$ it follows that
$(X(A\_1),\ldots,X(A\_K)) \sim DIR(\alpha H\_0(A\_1), \ldots,\alpha H\_0(A\_K))$.
So let's assume that I have $X\sim DP(\alpha H\_0)$. Given any partition $B\_1 ,\ldots, B\_K$, and any $k = 1 \ldots K$, I can define a partition $A\_1 = B\_k, A\_2 = \Omega -B\_k$ and from the DP definition it follows that
$(X(A\_1),X(A\_2)) \sim DIR(\alpha H\_0(A\_1), \alpha H\_0(A\_2))$
which is equivalent to saying that
$X(B\_k) \sim Beta(\alpha H\_0(B\_k), \alpha(1-H\_0(B\_k)))$
hence $X\sim BP(\alpha H\_0)$. Where is my mistake?
| https://mathoverflow.net/users/2873 | Difference between Beta Process and Dirichlet process | One cannot conclude $X\sim BP(\alpha H\_0)$ just by knowing the marginal distribution of each $X(B\_k)$, separately.
Your calculation is not wrong as the univariate marginal distribution and conditional distribution of a Dirichlet distribution are [Beta distributed](http://en.wikipedia.org/wiki/Dirichlet_process#Stick-breaking_construction).
In particular, in a Dirichlet process, samples correspond to the density function $$f(\theta)=\sum \beta\_i \delta\_{\theta\_i}$$
(Here the $\beta\_i$ are constructed as $\beta\_i=\beta\_i'\prod\_{j < i}(1-\beta\_j')$, and $\beta\_j'\sim \text{Beta}(1,\alpha)=Y$)
While in a Beta process, given an **infinitesimal** partition $(B\_1,...,B\_K)$ with $K\to \infty$ and $H(B\_k)\to 0$ the samples correspond to the density function
$$H(B)=\sum \pi\_i\delta\_{B\_i}$$ where $\pi\_i\sim \text{Beta}(\alpha H\_0(B\_i),\alpha (1-H\_0(B\_i)))$
I hope you can see the difference. One more thing, the reason why the Dirichlet process is defined in terms of finitely dimensional distributions is because [Kolmogorov extension theorem](http://en.wikipedia.org/wiki/Kolmogorov_extension_theorem) guarantees that it defines a stochastic process. Unfortunately the Beta process, does not verify the conditions of this theorem, and as a continuous time Levy process must be defined directly in the infinitesimal limit.
| 2 | https://mathoverflow.net/users/2384 | 10003 | 6,828 |
https://mathoverflow.net/questions/10002 | 7 | In [Lemma 2.1.3.4](http://books.google.com/books?id=CTe68E8wK4QC&lpg=PP1&ots=o8qXwh__pq&dq=higher%20topos%20theory&pg=PA67#v=onepage&q=&f=false) of Higher Topos Theory, the statement of the lemma requires that the fibers are not only nonempty but contractible. However, in the proof, I don't see where contractibility is directly used, only the fact that the fibers are nonempty. There is one other place where contractibility is mentioned, "Since the boundary of this simplex maps entirely into the contractible kan complex $S\_t$, it is possible to extend $f'$ to $X(n+1)$." However, I don't see how contractibility directly factors in, since that would only attest to the uniqueness of the extension. The existence of the extension comes from the fact that the inclusion $\partial \Delta^n \times \Delta^1 \subseteq X(n+1)$ is left anodyne and $S\_t$ is a nonempty Kan complex and the fact that the map f' factors through the inclusion of $S\_t$.
Please correct me if I'm wrong. Also, there is a [relevant post](http://mathoverflow.tqft.net/discussion/118/appropriate-place-to-ask-these-types-of-questions/#Item_2) on meta where I first asked if this question is appropriate, and I was greenlighted by Anton.
| https://mathoverflow.net/users/1353 | Is every left fibration of simplicial sets with nonempty fibers a trivial kan fibration? | The inclusion $\partial \Delta^n \times \Delta^1 \subseteq X(n+1)$ isn't any kind of anodyne extension, though. It's formed by attaching an n-simplex to $\partial \Delta^n \times \Delta^1$ with boundary $\partial \Delta^n \times 0$. So extending a map to $S\_t$ from $\partial \Delta^n \times \Delta^1$ to $X(n+1)$ is exactly the same as extending a map from $\partial \Delta^n$ to $\Delta^n$, and being able to do this for all $n$ is exactly the same as $S\_t$ being a contractible Kan complex (since the maps $\partial \Delta^n \to \Delta^n$ as $n$ varies form generating cofibrations for the Kan model structure).
| 7 | https://mathoverflow.net/users/126667 | 10004 | 6,829 |
https://mathoverflow.net/questions/9991 | 11 | Given an equation of a parametric surface, is there a general way to sample of points uniformly distributed on that surface?
I'm interested in this problem for purposes of visualisation - rather than attempting to attempt to triangulate the surface and display with polygons, display a dense sample of points. This makes it easier to generalise to >3d.
Here's an example of a surface I'd like to display: the Klein bottle.
```
u = [-pi, pi]
v = [-pi, pi]
x1 = (r * cos(v) + a) * cos(u),
x2 = (r * cos(v) + a) * sin(u),
x3 = r * sin(v) * cos(u/2),
x4 = r * sin(v) * sin(u/2)
```
(where r and a are parameters that control the shape of the overall surface)
| https://mathoverflow.net/users/2871 | How can I sample uniformly from a surface? | This paper may be of interest to you:
J. Arvo, [Stratified Sampling of 2-Manifolds](http://www.ics.uci.edu/~arvo/papers/notes2001.pdf)
It directly answers your question, though you may need to do some of the computations numerically depending on how complicated your surfaces are. Moreover, stratifying your samples will help the sampling *look* uniform -- sampling uniformly over the entire domain tends to look blotchy.
| 7 | https://mathoverflow.net/users/1557 | 10005 | 6,830 |
https://mathoverflow.net/questions/10023 | 10 | I want a ring $R$ of "numbers" such that:
For any sequence of congruences $x\equiv a\_1 \pmod{n\_1}, x\equiv a\_2 \pmod{n\_2},\dots$ with $a\_i\in \mathbb{Z}$ and $n\_i\in \mathbb{N}$ such than any finite set of these congruences has a solution $x\in\mathbb{Z}$, there is a $r\in R$ such that $r\equiv a\_1 \pmod{n\_1}, r\equiv a\_2 \pmod{n\_2},\dots$
and
For any $r\in R$ and $n\in\mathbb{N}$ there is a $a, 0\leq a< n $ such that $r\equiv a \pmod{n}$.
I think that $R$ has to be the product set of the p-adic integers over all primes p, but what do you call this ring?
(Perhaps there should be a "terminology" tag? **Edit:** It already exists but it is called "names")
| https://mathoverflow.net/users/2097 | What do you call this ring? | The standard notation is $\widehat{\mathbb{Z}}$. The names I know are "the profinite completion of $\mathbb{Z}$" and "$\mathbb{Z}$-hat".
| 22 | https://mathoverflow.net/users/297 | 10024 | 6,843 |
https://mathoverflow.net/questions/9864 | 25 | Presburger arithmetic apparently proves its own consistency. Does anyone have a reference to an exposition of this? It's not clear to me how to encode the statement "Presburger arithmetic is consistent" in Presburger arithmetic.
In Peano arithmetic this is possible since recursive functions are representable, so a recursive method of assigning Godel numbers to formulas and proofs means that Peano arithmetic can represent its own provability relation (of course, showing all that requires a lot of work). In particular, we can write a Peano arithmetic sentence which says "there is no natural number which encodes a proof of $\bot$".
On the other hand, Presburger arithmetic can't represent all recursive functions. It can't even represent all the primitive recursive ones, so this same trick doesn't work. If it did, the first incompleteness theorem would apply.
| https://mathoverflow.net/users/2693 | Presburger Arithmetic | Presburger arithmetic does NOT prove its own consistency. Its only function symbols are addition and successor, which are not sufficient to represent Godel encodings of propositions.
However, consistent self-verifying axiom systems do exist -- see the work of Dan Willard (["Self-Verifying Axiom Systems, the Incompleteness Theorem and Related Reflection Principles"](http://www.cs.albany.edu/~dew/m/jsl1.pdf)). The basic idea is to include enough arithmetic to make Godel codings work, but not enough to make the incompleteness theorem go through. In particular, you remove the addition and multiplication function symbols, and replace them with subtraction and division. This is enough to permit representing the theory arithmetically, but the totality of multiplication (which is essential for the proof of the incompleteness theorem) is not provable, which lets you consistently add a reflection principle to the logic.
| 37 | https://mathoverflow.net/users/1610 | 10027 | 6,845 |
https://mathoverflow.net/questions/10010 | 9 | It might be a stupid question.
Suppose There is a category of categories,denoted by CAT,where objects are categories, morpshims are functors between categories
Take multiplicative system S={category equivalences}. Then we take localization at S. Then we get localized category S^(-1)CAT
Another Category, denoted by |CAT|,where objects are categories, morphism are isomorphism classes of functors between categories.
I want to prove this two categories are equivalent. I want to prove this problem in two ways:
1. Naive prove
Because either |CAT| or S^(-1)CAT has same objects as CAT. So one can prove the morphism class of these two category has same equivalent relationship. For the S^(-1)Cat, suppose F and G are two functors in the same equivalent class, we have sF=sG,for s belongs to S. Then I can prove F and G belongs to the same equivalent class in |CAT|. But on the other hand, if F and G are isomorphic functor. I can not prove sF=sG,for some s belongs to S. What I can only prove is sF is isomorphic to sG.
2. Using adjoint functors
Because I want to prove the projection functor CAT--->|CAT| is a localization functor at S. So I want to construct an adjoint functor |CAT|--->CAT which is fully faithful. But what is this functor. It seems I can not make it well defined.
This is an exercise in Toen's lecture: Lectures on DG-categories
I attach the lecture notes on DG-categories here [Toen notes](http://www.math.univ-toulouse.fr/~toen/swisk.pdf)
page 5,exercise 2
| https://mathoverflow.net/users/1851 | is localization of category of categories equivalent to |Cat| | There are set theoretic issues that will hinder any proof of this statement. In particular I don't believe you can construct the adjunction in part (2) without applying the axiom of choice to CAT, and it's not clear the localization S^(-1)CAT makes sense either.
That said, let's ignore these issues and pretend that CAT is a *small* category of (small) categories.
I was originally confused by this. I believe that there are some problems in the formulation of this question. In particular if S consists of those functors which are equivalences of categories, then it **does not form a [multiplicative system](http://ncatlab.org/nlab/show/calculus+of+fractions)**. Specifically, the right Ore condition fails. I'll give an example later. What this means though, is that the property your are trying to check in part (1) actually fails. To see this pick a group G and view it as a category with one object. Then the functors from H to H are the homomorphisms, and the equivalence classes of this functors are the orbits of Hom(H,H) under the conjugation action. Choose H such that there are distinct automorphisms of H which are equivalent under conjugation.
Claim: If F,G: H --> H are two such automorphisms, then there is no equivalence s:H --> C such that sF = sG. I think this is easy to check, so I'll leave it at that.
So the second part of (1) is impossible to prove.
I first thought that we can answer this question be going to skeletal categories. This is not correct, but let's look at why it is not correct. Let SKEL be the full subcategory of CAT consisting of the [Skeletal categories](http://ncatlab.org/nlab/show/skeleton), i.e. those categories in which $a \cong b$ implies $a = b$, i.e. those categories which have a single object in each isomorphism class.
Every object in CAT is equivalent to one in SKEL. This is an easy exercise you should work out for yourself. By applying the axiom of choice to all of CAT we can construct an equivalence between CAT and SKEL, and in particular we have L: CAT ---> SKEL with an equivalence $C \cong L(C)$ for every category C.
Notice that a functor between skeletal categories is an equivalence if and only if it is an isomorphism on the nose. So we are part way there. This made me think that SKEL was the localizing subcategory we were after.
However two functors into a skeletal category can be naturally isomorphic without being equal. The group example above is actually an example of this. The functors F and G are naturally isomorphic, but not equal.
Let J denote the "Joyal interval", the category with two objects and an isomorphism between them. J can be used to say when two functors are naturally isomorphic. F,G: C --> D are nat. isomorphic if they extend to a functor $C \times J \to D$. There is a quotient of J where we identify the two objects. This is in fact the category Z (the group of integers viewed as a category). Let W be the set of morphisms consisting of the single morphism J --> Z. Notice that SKEL consists of exactly the "W-local" objects. But this is not quite what we want.
---
Okay, now for the rest. First, I promised an example where the right Ore condition fails. This is given by considering the two inclusions of pt into J. Both of these are equivalences, but this cannot be completed to an appropriate square. So the class of equivalences in not a multiplicative system.
Next, Toen doesn't assume that S is a multiplicative system. He (correctly) defines the localization in terms of a (weak) universal property. Namely, there exists a locization functor:
$ L: CAT \to S^{-1}CAT$
such that fro any other category D, $L^\* : Fun( S^{-1}CAT, D) \to Fun(CAT, D)$ is fully-faithfull and the essential image consists of those functors which send elements of S to isomorphisms in D. In particular the quotient functor from CAT to |CAT| does this, so this gives us functor from S^(-1)CAT to |CAT| defined up to unique natural isomorphism.
On the other hand, |CAT| is also defined by a universal property. This gives us maps between |CAT| and S^(-1)CAT and by the usual sort of general nonsense this is an equivalence. The details aren't too hard, so I'll leave them to you.
Now the second part of your question asks about whether we can realize |CAT| as a full subcategory of CAT such that the quotient $ q:CAT \to |CAT|$ is left adjoint to the inclusion. The answer is no. Let's first look at whether q can have a right adjoint at all and what this would mean.
Suppose that R: |CAT| --> CAT is right adjoint to the quotient q. Be abuse, we will identify the objects of CAT and |CAT|. This means that for all categories X and Y we have:
$CAT(X, R(Y)) = |CAT|(X, Y) = |CAT|(X \times J, Y) = CAT( X \times J, R(Y))$
So $R(Y)$ must be cartesian local with respect to the map J --> pt. In particular we see, taking X=pt, that R(Y) has no non=identity automorphisms, i.e. R(Y) is rigid.
Now let's look at an example. Consider the group Z viewed as a category with one object. Then the endomorphisms of Z in |CAT| is the monoid Z (under multiplication). Now if R is fully-faithful then we need:
$ \mathbb{Z} = |CAT| (\mathbb{Z}, \mathbb{Z}) = CAT (R(\mathbb{Z}), R(\mathbb{Z}))$
but this last is equal to the automorphism of some set R(Z), hence we have a contradiction. So there is no such fully-faithful right adjoint.
| 9 | https://mathoverflow.net/users/184 | 10028 | 6,846 |
https://mathoverflow.net/questions/10038 | 6 | The [wikipedia article on absolute continuity](https://en.wikipedia.org/wiki/Absolute_continuity) gives a delta-epsilon definition for a measure $\mu$ defined on the Borel $\sigma$-algebra on the real line, with respect to the Lebesgue measure $\lambda$:
$\mu\ll\lambda$ if and only if for every $\epsilon>0$ and for every bounded real interval $I$ there is a $\delta>0$ such that for every (finite or infinite) sequence of pairwise disjoint sub-intervals [$x\_i,y\_i$] of $I$ with
$\sum\_{i} |y\_i - x\_i| < \delta$
it follows that
$\sum\_{i} |\mu((-\infty, y\_i])-\mu((-\infty, x\_i])| < \epsilon$.
My questions are: Does this connection generalise? What would be a topological reformulation of $\mu\ll\nu$? If the notion does not generalise for arbitrary measures, does it generalise for the Lebesgue measure on $R^n$? How would a sketch of the proof look like?
| https://mathoverflow.net/users/1313 | What is the "continuity" in "absolute continuity", in general? | For every $\varepsilon>0$, there exists $\delta>0$ such that every measurable set of $\nu$-measure less than $\delta$ has $\mu$-measure less than $\varepsilon$. There are some technical assumptions to be made to have this equivalent to $\mu\ll\nu$ (say, that both measures are finite) but otherwise it is as simple as that. In many decent measure spaces, it suffices to check the inequality just for some nice sets $E$ (the definition you quoted is just this imequality for finite unions of half-open intervals).
| 8 | https://mathoverflow.net/users/1131 | 10040 | 6,852 |
https://mathoverflow.net/questions/10036 | 30 | By this I mean the specialisation of the quantum group Uq(g) with q a root of unity, and the 'correct' meaning of 'correct' (enclosed in quotations since there isn't necessarily a correct answer) is likely to mean that its category of representations is the 'correct' one.
My suspicion is that I want to take an integral form of Uq(g) defined over ℤ[q±1/2] and base change to the appropriate ring of integers in a cyclotomic field. Having heard of the 'small quantum group' and Lusztigs algebra U dot (notation in his quantum groups book), I suspect the existence of multiple approaches, which diverge at least when an integral form is desired, and hence turn to mathoverflow for clarification and enlightenment.
Afficionados of this type of question can consider it as a continuation in a sequence initiated by [this question](https://mathoverflow.net/questions/7112/which-is-the-correct-universal-enveloping-algebra-in-positive-characteristic) on the universal enveloping algebra in positive characteristic.
| https://mathoverflow.net/users/425 | Which is the correct version of a quantum group at a root of unity? | There are (at least) five interesting versions of the quantum group at a root of unity.
**The Kac-De Concini form:**
This is what you get if you just take the obvious integral form and specialize q to a root of unity (you may want to clear the denominators first, but that only affects a few small roots of unity). This is best thought of as a quantized version of jets of functions on the Poisson dual group. It's most important characteristic is that it has a very large central Hopf subalgebre (generated by the lth powers of the standard generators). In particular, its representation theory is sits over Spec of the large center, which is necessarily a group and turns out to be the Poisson dual group. It also has a small quotient Hopf algebra when you kill the large center.
The main sources for the structure of the finite dimensional representations are papers by subsets of Kac-DeConcini-Procesi (the structure of the representations depends on the symplectic leaf in G\*, in particular there are "generic" ones coming from the big cell) as well as some more recent work by Kremnitzer (proving some stronger results about the dimensions of the non-generic representations) and by DeConcini-Procesi-Reshetikhin-Rosso (giving the tensor product rules for generic reps). The main application that I know of this integral form is to invariants of knots together with a hyperbolic structure on the compliment and to invariants of hyperbolic 3-manifolds due to Kashaev, Baseilhac-Bennedetti, and Kashaev-Reshetikhin. The hope is that these invariants will shed some light on the volume conjecture.
**The Lusztig form:**
Here you start with the integral form that has divided powers. Structurally this has a small subalgebra generated by the usual generators (E\_i, F\_i, K\_i) since E^l = 0. The quotient by this subalgebra gives the usual universal enveloping algebra via something called the quantum Frobenius map. The main representation that people look at are the "tilting modules." Tilting modules have a technical description, but the important point is that the indecomposable tilting modules are exactly the summands of the tensor products of the fundamental representations. Indecomposable tilting modules are indexed by weights in the Weyl chamber. The "linkage principle" tells you that inside the decomposition series of a given indecomposable tilting module you only need to look at the Weyl modules with highest weights given by smaller elements in a certain affine Weyl group orbit.
It is the Lusztig integral form (not specialized) that is important for categorification. The Lusztig form at a root of unity is important for relationships between quantum groups and representations of algebraic groups and for relationships to affine lie algebras. The main sources are Lusztig and HH Andersen (and his colaborators). I'm also fond of a [paper of Sawin](http://arxiv.org/abs/math/0308281)'s that does a very nice job cleaning up the literature.
The Lusztig integral form is also the natural one from a quantum topology point of view. For example, if you start with the Temperley-Lieb algebra (or equivalently, tangles modulo the Kauffman bracket relations) and specialize q to a root of unity what you end up with is the planar algebra for the tilting modules for the Lusztig form at that root of unity.
**The small quantum group:**
This is a finite dimensional Hopf algebra, it appears as a quotient of the K-DC form (quotienting by the large central subalgebra) and as a subalgebra of the Lusztig form (generated by the standard generators). I gather that the representation theory is not very well understood. But there has been some work recently by Roman Bezrukavnikov and others. I also wrote a [blog post](http://sbseminar.wordpress.com/2008/06/02/representations-of-the-small-quantum-group/) on what the representation theory looks like here for one of the smallest examples.
**The semisimplified category:**
Unlike the other examples, this is not the category of representations of a Hopf algebra! (Although like all fusion categories it is the category of representations of a weak Hopf algebra.) You start with either the category of tilting modules for the Lusztig form or the category of finite dimensional representations of the small quantum group and then you "semisimplify" by killing all "negligible morphisms." A morphism is negligible if it gives you 0 no matter how you "close it off." Alternately the negligible morphisms are the kernel of a certain inner product on the Hom spaces. The resulting category is semisimple, its representation theory is a "truncated" version of the usual representation theory. In particular the only surviving representations are those in the "Weyl alcove" which is like the Weyl chamber except its been cut off by a line perpendicular to l times a certain fundamental weight (see Sawin's paper for the correct line which depends subtly on the kind of root of unity).
This example is the main source of modular categories and of interesting fusion categories. Its main application is the 3-manifold invariants of Reshetikhin-Turaev (where this quotient first appears, I think) and Turaev-Viro. For those invariants its very important that your braided tensor category only have finitely many different simple objects.
**The half-divided powers integral form:**
This appears in the work of Habiro on universal versions of the Reshetikhin-Turaev invariants and on integrality results concerning these invariants. This integral form looks like the Lusztig form on the upper Borel and like the K-DC form on the lower Borel. The key advantage is that in the construction of the R-matrix via the Drinfeld double you should be looking at something like U\_q(B+) \otimes U\_q(B+)\* and it turns out that the dual of the Borel without divided powers is the Borel with divided powers and vice-versa. There's been very little work done on this case beyond the work of Habiro.
| 59 | https://mathoverflow.net/users/22 | 10045 | 6,856 |
https://mathoverflow.net/questions/9829 | 7 | This is a question a friend of mine asked me some time ago. I suspect the answer is "no" but can't prove it.
Every free complex of abelian groups is isomorphic to the reduced cellular complex of some finite CW-complex; in fact, one can take a wedge of balls and Moore spaces. The question is whether there is a similar result for complexes equipped with a basis.
Namely, let $C\_{\*}$ be a finite chain complex (i.e., the differential decreases the degree) of free abelian groups. Suppose a basis $B\_i$ of each $C\_i$ is chosen.
1. Is it true that $(C\_{\*},(B\_i))$ is isomorphic as a complex with a basis to the (possibly shifted) reduced cellular complex of some CW-complex with the basis given by the cells?
2. What happens if one works with complexes of finite-dimensional vector spaces over $\mathbb{Z}/2$ instead?
| https://mathoverflow.net/users/2349 | Realizing complexes with bases as cellular complexes | Here is a sketch of an argument to show that all based chain complexes are realizable. (This might end up being pretty similar to Tyler's argument.)
First one gives an algebraic argument that by a change of basis the chain complex can be put in a standard "diagonal" form. Moreover, the change of basis can be achieved by a sequence of elementary operations, as in linear algebra, but now over the integers rather than a field, using the fact that the group $GL(n,Z)$ is generated by elementary matrices, including signed permutations. The most important of the elementary operations is to add plus or minus one basis element to another. Doing such an operation in $C\_i$ changes the boundary maps to and from $C\_i$ by multiplication by an elementary matrix and its inverse.
The "diagonalized" chain complex can easily be realized geometrically, so it remains to see that the elementary basis change operations can be realized geometrically. In the special case of top-dimensional cells, one can slide a part of one such cell over another to achieve the elementary operation of adding plus or minus one column of the outgoing boundary matrix to another. For lower-dimensional cells one wants to do the same thing and then extend the deformation over the higher cells. It should be possible to do this directly without great difficulty. The slide gives a way of attaching a product $\{cell\}\times I$, and this product deformation retracts onto either end, so one can use the deformation retraction to change how the higher-dimensional cells attach.
The argument should work for 1-cells as well as for higher-dimensional cells, so it shouldn't be necessary to assume that $C\_1$ is trivial.
An alternative approach would be to temporarily thicken the cell complex into a handle structure on a smooth compact manifold-with-boundary of sufficiently large dimension, with one i-handle for each i-cell. Sliding an i-cell then corresponds to sliding an i-handle, and there is well-established machinery on how to do this sort of thing, as one sees in the proof of the h-cobordism theorem for example. Or one can use the language of morse functions and gradient-like vector fields as in Milnor's book on the h-cobordism theorem. Either way, after all the elementary basis changes have been realized by handle slides, one can collapse the handles back down to their core cells to get the desired based cellular chain complex.
There are plenty of details to fill in here in either the cell or handle approach. I don't recall seeing this result in the classical literature, but I wouldn't be too surprised if it were there somewhere, maybe in some paper or book on J.H.C.Whitehead's simple homotopy theory where elementary row and column operations play a big role.
| 4 | https://mathoverflow.net/users/23571 | 10050 | 6,860 |
https://mathoverflow.net/questions/10053 | 5 | In the wikipedia entry for 'frames and locales', pains are taken to distinguish between the category of locales - defined to be the opposite of the category of frames - and the category whose objects are the complete Heyting algebras but whose arrows are the adjoints of the frame arrows. The two are clearly isomorphic as categories. How far are they from being identical, though? I became acquainted with locales through Borceux's excellent handbook and he defines arrow in locales to be the adjoints. So this is a little worrisome. Ok, I know this isn't precise... Let me put it this way: What are concrete examples of how these two differently defined categories actually differ?
Thank you in advance
| https://mathoverflow.net/users/2884 | Definition of Category of Locales | Well, as you say, these two categories are isomorphic, so it's going to be hard to say how they differ! They only differ in the names the maps are given.
Maybe it would help to recap the definitions. I'll take them from p.39-40 of Peter Johnstone's book *Stone Spaces*.
A **frame** is a complete lattice $A$ satisfying the infinite distributive law
$$
a \wedge \bigvee S = \bigvee \{ a \wedge s | s \in S \}
$$
($a \in A, S \subseteq A$). A **homomorphism of frames** is a function preserving finite meets and arbitrary joins. This defines the category **Frm** of frames.
Note (as you did) that every homomorphism of frames has a right adjoint.
The category **Loc** of **locales** is the opposite of the category of frames. Morphisms in **Loc** are called **continuous maps**.
Then Johnstone says: "We adopt the convention that if $f: A \to B$ is a continuous map of locales, we shall write $f^\*: B \to A$ for the corresponding frame homomorphism, and $f\_\*: A \to B$ for the right adjoint of $f^\*$."
So in Johnstone's convention (which is the one I know), the elements of **Loc**$(A, B)$ are identified with frame homomorphisms $B \to A$. In Borceux's convention, the elements of **Loc**$(A, B)$ are identified with order-preserving maps $A \to B$ that are right adjoint to frame homomorphisms.
I guess the other thing to say is that when you're dealing with ordered sets, adjoints are *genuinely* unique (not just unique up to isomorphism). So taking the right adjoint of a frame homomorphism is a bijective process.
I don't know what else to say. It's really just a matter of naming.
| 9 | https://mathoverflow.net/users/586 | 10060 | 6,868 |
https://mathoverflow.net/questions/10041 | 8 | Let $a,b,c$ be integers which are the sides of a triangle with integral area, a so called Heronian triangle. [This](http://www.mathpages.com/home/kmath474.htm) website attributes to Gauss the result that there must then exist integers $m,n,p,q$ such that
$a = mn(p^2+q^2)$
$b = (mp)^2+(nq)^2$
$c = (m+n)(mp^2-nq^2)$
(where I left out a $4pq$ factor designed to make the radius of the circumscribed circle integral as well). It's not hard to see that the triangle defined by these formulas is indeed Heronian, however I could neither prove nor find a reference for the fact that this parametrization is exhaustive.
Can someone do one of these two things?
Thanks!
(Note: I'm communicating this question on behalf of my dad, who is really the person who looked into that but is not easily capable of asking it himself over here. I may be slow to respond on his behalf if questions come up).
| https://mathoverflow.net/users/25 | A parametrization of Heronian triangles | Let your triangle $\triangle{ABC}$ have side lengths $a,b,c \in \mathbb{Q}$ and rational area. Assume WLOG that $c$ is the longest side and drop the altitude from $C$ with length $h\in Q$. The triangle is divided into two right triangles one with hypotenuse $a$ and legs $d,h$, and one with hypotenuse $b$ and legs $e,h$. We have $d+e=c\in \mathbb{Q}$. Also notice that $$d-e=\frac{d^2-e^2}{d+e}=\frac{a^2-b^2}{c}\in \mathbb{Q}$$ so we conclude that $d,e$ are rational. From the pythagorean triples we have relations $$a-d=r,\quad a+d=\frac{h^2}{r},\quad b-e=s,\quad b+e=\frac{h^2}{s}$$ and therefore $$a=\frac{1}{2r}(h^2+r^2),\quad b=\frac{1}{2s}(h^2+s^2),\quad c=\frac{r+s}{2rs}(rs-h^2)$$
This is exactly your parametrization up to scaling.
| 5 | https://mathoverflow.net/users/2384 | 10061 | 6,869 |
https://mathoverflow.net/questions/3820 | 49 | I asked a related question [on this mathoverflow thread](https://mathoverflow.net/questions/3274/how-hard-is-it-to-compute-the-euler-totient-function). That question was promptly answered. This is a natural followup question to that one, which I decided to repost since that question is answered.
So quoting myself from that thread:
>
> How hard is it to compute the number of prime factors of a given integer? This can't be as hard as factoring, since you already know this value for semi-primes, and this information doesn't seem to help at all. Also, determining whether the number of prime factors is 1 or greater than 1 can be done efficiently using Primality Testing.
>
>
>
| https://mathoverflow.net/users/1042 | How hard is it to compute the number of prime factors of a given integer? | There is a folklore observation that if one was able to quickly count the number of prime factors of an integer n, then one would likely be able to quickly factor n completely. So the counting-prime-factors problem is believed to have comparable difficulty to factoring itself.
The reason for this is that we expect any factoring-type algorithm that works over the integers, to also work over other number fields (ignoring for now the issue of unique factorisation, which in principle can be understood using class field theory). Thus, for instance, we should also be able to count the number of prime factors over the Gaussian integers, which would eventually reveal how many of the (rational) prime factors of the original number n were equal to 1 mod 4 or to 3 mod 4.
Using more and more number fields (but one should only need polylog(n) such fields) and using various reciprocity relations, one would get more and more congruence relations on the various prime factors, and pretty soon one should be able to use the Chinese remainder theorem to pin down the prime factors completely.
More generally, the moment one has a way of extracting even one non-trivial useful bit of information about the factors of a number, it is likely that one can vary this procedure over various number fields (or by changing other parameters, e.g. twisting everything by a Dirichlet character) and soon extract out enough bits of information to pin down the factors completely. The hard part is to first get that one useful bit...
[EDIT: The above principle seems to have one exception, namely the parity bit of various number-theoretic functions. For instance, in the (now stalled) Polymath4 project to find primes, we found a quick way to compute the parity of the prime counting function $\pi(x)$, but it has proven stubbornly difficult to perturb this parity bit computation to find other useful pieces of information about this prime counting function.]
| 87 | https://mathoverflow.net/users/766 | 10062 | 6,870 |
https://mathoverflow.net/questions/10066 | 20 | In dimension 2 we know by the Riemann mapping theorem that any simply connected domain ( $\neq \mathbb{R}^{2}$) can be mapped bijectively to the unit disk with a function that preserves angles between curves, ie is conformal.
I have read the claim that conformal maps in higher dimensions are pretty boring but does anyone know a proof or even a intuitive argument that conformal maps in higher dimensions are trivial?
| https://mathoverflow.net/users/2888 | Conformal maps in higher dimensions | I think you're looking for [Liouville's theorem](http://en.wikipedia.org/wiki/Liouville%27s_theorem_(conformal_mappings)). This theorem states that for $n >2$, if $V\_1,V\_2 \subset \mathbb{R}^n$ are open subsets and $f : V\_1 \rightarrow V\_2$ is a smooth conformal map, then $f$ is the restriction of a higher-dimensional analogue of a Mobius transformation.
By the way, observe that there are no assumptions on the topology of the $V\_i$ -- they don't have to be simply-connected, etc.
---
EDIT : I'm updating this ancient answer to
[link](http://lamington.wordpress.com/2013/10/28/liouville-illiouminated/) to a blog post by Danny Calegari which contains a sketch of a beautifully geometric argument for Liouville's theorem.
| 32 | https://mathoverflow.net/users/317 | 10068 | 6,873 |
https://mathoverflow.net/questions/9645 | 4 | For which manifolds or varieties is quantum cohomology known to converge? Are there any manifolds for which quantum cohomology is known to not converge? I seem to have the impression that quantum cohomology is known to converge for Fano manifolds or maybe toric Fano manifolds, but I don't know if this is actually true.
By "convergence" I mean convergence of, e.g., the genus 0 Gromov-Witten potential.
| https://mathoverflow.net/users/83 | Convergence of quantum cohomology | See Dmitri's comment.
| 0 | https://mathoverflow.net/users/83 | 10071 | 6,876 |
https://mathoverflow.net/questions/10084 | 1 | Update based on Michael's answer (thanks again!) - Can the LLL or PSLQ algorithms provide a (knowably - i.e. not just incidental) unique solution for the set of integer multiplicative factors? Are there other algorithms (perhaps with worse run-time complexities) that can?
Imagine that one has a set of 'n' finite-precision real numbers - (r\_1, r\_2, ..., r\_n), each with an associated positive integer multiplicative factor, (i\_1, i\_2, ..., i\_n). Here, one only has access to the values in the set of real numbers, as well as the sum of the real numbers multiplied by their corresponding multiplicative factors, 'S', i.e. - Sum(i\_1\*n\_1, i\_2\*n\_2, ..., i\_n\*r\_n).
What's the most efficient way to test that the sum of a particular set of real numbers will (or, obviously, will not) always allow us to extract a unique solution for the set of integer multiplicative factors? Or to find/do this with fewest restrictions on the values of the integer multiplicative factors?
In the case that this has a very simple answer, apologies.
(Edit - Changed "each with an associated "set of" positive integer multiplicative factors" --> "each with an associated positive integer multiplicative factor". One finite-precision real number 'r\_k' has one integer multiplicative factor 'i\_k'.)
| https://mathoverflow.net/users/2891 | Extracting integer multiplicative factors from the sum of certain sets of (finite-precision) real numbers? | You want [integer relation algorithms](http://en.wikipedia.org/wiki/Integer_relation_algorithm), such as the [LLL algorithm](http://en.wikipedia.org/wiki/Lenstra%E2%80%93Lenstra%E2%80%93Lov%C3%A1sz_lattice_basis_reduction_algorithm).
| 2 | https://mathoverflow.net/users/143 | 10085 | 6,887 |
https://mathoverflow.net/questions/10033 | 19 | Is there a standard example of two abelian varieties $A$, $B$ over some number field $k$ which are $k\_v$-isomorphic for every place $v$ of $k$ but not $k$-isomorphic ?
| https://mathoverflow.net/users/2821 | Everywhere locally isomorphic abelian varieties | (If you upvote this answer, please consider upvoting the answers by Felipe Voloch and David Speyer too, since this answer builds on their ideas.)
The smallest examples are in dimension $2$. Let $E$ be any elliptic curve over $\mathbf{Q}$ without complex multiplication, e.g., $X\_0(11)$. We will construct two twists of $E^2$ that are isomorphic over $\mathbf{Q}\_p$ for all $p \le \infty$ but not isomorphic over $\mathbf{Q}$.
Let $K:=\mathbf{Q}(\sqrt{-1},\sqrt{17})$. Let $G:=\operatorname{Gal}(K/\mathbf{Q}) = (\mathbf{Z}/2\mathbf{Z})^2$. Let $\alpha \colon G \to \operatorname{GL}\_2(\mathbf{Z}) = \operatorname{Aut}(E^2)$ be a homomorphism sending the two generators to the reflections in the coordinate axes of $\mathbf{Z}^2$, and let $A$ be the $K/\mathbf{Q}$-twist of $E^2$ given by $\alpha$. Define $\beta$ and $B$ similarly, but with the lines $y=x$ and $y=-x$ in place of the coordinate axes. The representations $\alpha$ and $\beta$ of $G$ on $\mathbf{Z}^2$ are not conjugate: only the former is such that the lattice vectors fixed by nontrivial elements of $G$ generate all of $\mathbf{Z}^2$. Thus $A$ and $B$ are not isomorphic over $\mathbf{Q}$.
On the other hand, every decomposition group $D\_p$ in $G$ is smaller than $G$ since $-1$ is a square in $\mathbf{Q}\_{17}$ and $17$ is a square in $\mathbf{Q}\_2$. Also, the restrictions of $\alpha$ and $\beta$ to any proper subgroup of $G$ are conjugate: any *single* line spanned by a primitive vector in $\mathbf{Z}^2$ can be mapped to any other by an element of $\operatorname{GL}\_2(\mathbf{Z})$. Thus $A$ and $B$ become isomorphic after base extension to $\mathbf{Q}\_p$ for any $p \le \infty$. $\square$
**Remark:** The abelian surfaces $A$ and $B$ constructed above are *isogenous* even over $\mathbf{Q}$, because the $\mathbf{Z}^2$ with one Galois action can be embedded into the $\mathbf{Z}^2$ with the other Galois action: rotate $45^\circ$ and dilate.
**Remark:** The nonexistence of examples in dimension $1$ follows from these two well-known facts:
1) Twists of an elliptic curve over a field $k$ of characteristic $0$
are classified by
$H^1(k,\mu\_n)=k^\times/k^{\times n}$ where $n$ is 2, 4, or 6.
2) If $n<8$, the map $k^\times/k^{\times n} \to \prod\_v k\_v^\times/k\_v^{\times n}$ is injective.
[**Edit:** This answer was edited to simplify the construction and to add those remarks at the end.]
| 26 | https://mathoverflow.net/users/2757 | 10088 | 6,889 |
https://mathoverflow.net/questions/10086 | 7 | Do people study the category of representations of a compact ~~finite~~ group (not just irreducible ones)? I'm more interested in small cases like S\_3 and SU(2) but I'd be curious about general cases like $S\_n, SU(n)$. These must be tensor categories since - well... they admit tensor products and direct sums. Can these representations be considered a ring?
\*\* What does this category look like in the case of S\_3?
| https://mathoverflow.net/users/1358 | The Category of Representations of a Group | The category Rep(G) is a symmetric tensor category, and it is a theorem that this structure determines G (Tannaka-Krein duality, but I'm not familiar with it). Each object is dualizable because there is a dual representation, from which appropriate evaluation and coevaluation maps can be constructed. The unital object is the trivial representation.
(Incidentally, a symmetric tensor category is a kind of categorification of a commutative ring.)
In the finite case, It is a fusion category because there are finitely many simple objects, there is rigidity as stated above, and because $\mathbb{C}[G]$ (or more generally $k[G]$ for $k$ a field of characteristic prime to the order of $G$) is a semisimple algebra (Maschke's theorem). Semisimplicity is also true for continuous finite-dimensional representations of compact groups by the same "averaging" argument used in Maschke's theorem, though the group algebra is not necessarily semisimple. In the finite group case, the number of simple objects is equal to the number of conjugacy classes in G. In the infinite group case, for instance, the rotation group $SO(n)$ has infinitely many irreducible finite-dimensional representations obtained by the action on the spherical harmonics of various degrees (i.e. harmonic polynomials restricted to the sphere).
In the case of S\_n, generators of this ring can be indexed by the Young diagrams of size n. The relations are given by the tensor product rules, and while the Pieri rule gives a special case of this, as far as I know, there is no a simple general way to express the tensor product of two representations associated to Young diagrams as a sum of Young diagrams.
However, there are [apparently](http://www.cs.ucsb.edu/~omer/DOWNLOADABLE/symmetric_group85.pdf) algorithms to do this.
| 11 | https://mathoverflow.net/users/344 | 10094 | 6,894 |
https://mathoverflow.net/questions/10083 | 2 | The A\_2 affine Weyl group is the symmetry group of the triangulation of the plane by equilateral triangles. As Sean points out, it may be generated by reflections $r\_1, r\_2, r\_3$ about the edges of a single equilateral triangle. Since A\_n is a Coxeter group, every element $\alpha \in A\_2$ may be assigned a length $l(\alpha)$ with respect to the generators $r\_i$. How might one compute this length with respect to this presentation?
Any generating set has to contain a reflection about a line in each of the three directions ($0, 2\pi/3, 4\pi/3$) of the lattice. Is that condition sufficient?
| https://mathoverflow.net/users/1358 | computing lengths in the A_2 affine weyl group | Ok, here's a reply in "answer" format.
I asked first whether you knew about the Coxeter group presentation because I didn't want to write a bunch of stuff that you already knew (and therefore would not answer your question). I also didn't (and still don't) understand how strong of a concept you mean by "compute". It would have been helpful if you said something to clear up these two issues in your most recent reply.
Anyway here's some basic stuff I can hammer out quickly...
The presentation of a Coxeter group (by definition) is that has a finite list of order 2 generators (called "simple reflections" in the Weyl group case) and otherwise the only relations are to specify the orders of products of pairs of generators, and some pairs can be omitted (=infinite order). Affine Weyl groups are Coxeter groups so all this applies.
An affine Weyl group $W$ is an **infinite** group and there are arbitrarily long reduced expressions (=shortest words). Even though there are only a finite number of *generating* reflections, there are an infinite number of reflections and the hyperplanes they fix (in your $A\_2$ case, they are lines) divide the space up into connected pieces usually called "alcoves" (in your $A\_2$ case, they are the triangles). A fundamental fact is that $W$ acts simply-transitively on the set of these alcoves, which means if you pick a "basepoint" alcove $\mathcal{A}$, you can associate (bijectively) to each element $ w \in W$ the alcove $ w(\mathcal{A})$.
Again, I'm not sure what you mean by "compute", but here's an attempt at an answer anyway:
A "geometric" way to compute the length $\ell(w)$ of $w \in W$ is to count the number of hyperplanes separating $\mathcal{A}$ from $w(\mathcal{A})$. In a $2$-dimensional (or perhaps even $3$-dimensional) case, e.g. $A\_2$, I guess you could quickly compute shortest words for a particular "geometrically defined" element $w$ as follows:
* print out a drawing of the triangulated space and label $\mathcal{A}$ (this is probably the longest step)
* figure out which alcove $w(\mathcal{A})$ is (you should already know this by definition of "geometrically defined")
* count hyperplanes to determine $\ell(w)$ (this is quick and easy)
* label all the alcoves with their lengths, up to and including $\ell(w)-1$ (quick and easy)
* check all 3 simple reflections until you find one sending $w(\mathcal{A})$ to an alcove whose labeled length is $< \ell(w)$ (quick and easy)
* write down that simple reflection on a piece of paper
* now repeat the previous 2 steps for the new alcove
* when you get to the base alcove, you're finished and the list of simple reflections you wrote down is a reduced word
It's a long list of steps because I like to be explicit but it really doesn't take that long.
| 8 | https://mathoverflow.net/users/2047 | 10097 | 6,897 |
https://mathoverflow.net/questions/10091 | 7 | Here's a definition for homotopy limits that isn't quite right, but seems salvageable. Does anyone know how to fix it?
Suppose the category $C$ is some reasonable setting for homotopy theory, say it's enriched over some kind of category of spaces (e.g. chain complexes, simplicial sets, ...).
Def: Let F: Dop $\to$ C be a diagram (functor). An object X together with a map η from X to the diagram is a LIMIT for the diagram iff the induced natural transformation of functors HomC(-, X) $\to \lim$ HomC(-,F) is an isomorphism. A pair (X,η) is a *homotopy* limit for the diagram F iff the induced transformation of functors HomC(-,X) $\to \lim$ HomC(-, F) is a weak equivalence.
This definition doesn't quite cut it since, in most of the motivating examples I know, though the homotopy limit object X does come equipped with a morphism to each object in the diagram, these do not commute with the morphisms in the diagram---they only commute up to homotopy. So a homotopy limit won't even come with a map to the diagram, so it doesn't come with an induced natural transformation. How then can I characterize the object X by a similar universal property as the (strict) limit?
| https://mathoverflow.net/users/2536 | Definition of homotopy limits | Reid's answer is quite right, but long before "quasicategories" became fashionable, algebraic topologists were doing exactly the same thing using the "simplicial bar construction" and plain old topological or simplicial enriched categories.
For a fixed x, the limit $\lim \hom\_C(x,F)$ is equivalent to the set of natural transformations from the constant functor $\Delta\_1\colon D\to Set$ at $1$ to the functor $\hom\_C(x,F(-))$. So to replace it by something "coherent" we need a notion of "homotopy coherent transformation." Now the set of natural transformations from a functor $G\colon X\to Top$ to a functor $H\colon Y\to Top$ can be defined as an "end," and computed as an equalizer of the two maps $\prod\_x \hom(G x, H x) \rightrightarrows \prod\_{x \to y} \hom(G x, H y)$. But these two maps are the first two coface maps of a cosimplicial object that continues with $\prod\_{x\to y\to z} \hom(G x, H z)$ and so on, so we can define the space of "homotopy coherent transformations" to be its *totalization* (the dual of geometric realization).
Now a homotopy limit can be defined as a representing object for the space of homotopy coherent transformations from $\Delta\_1$ to $\hom\_C(x,F(-))$ (where now $C$ is topologically enriched, so that these functors take values in spaces). Moreover, if $C$ admits "totalizations" as a topologically enriched category (a sort of "weighted limit"---it suffices to have ordinary limits and "cotensors"), then the homotopy limit can be constructed by "internalizing" the above construction.
| 15 | https://mathoverflow.net/users/49 | 10100 | 6,899 |
https://mathoverflow.net/questions/10103 | 36 | When I was a teenager, I was given the book [Men of Mathematics](http://en.wikipedia.org/wiki/Men_of_Mathematics) by E. T. Bell, and I rather enjoyed it. I know that this book has been criticized for various reasons and I might even agree with some of the criticism, but let's not digress onto that. E. T. Bell made a reasonable list of 34 of the greatest mathematicians from the ancient Greek period to the end of the 19th century. His list isn't perfect — maybe he should have included Klein or skipped Poncelet — but no such list can be perfect anyway. I think that his selection was good. He also made the careers of these mathematicians exciting and he addressed why mathematicians care about them today. It helped me learn what achievements and topics in mathematics are important.
(Just to head off discussion, the standard complaints include that the title is sexist and so is the book, that Bell was loose with biographical facts, and that he "chewed the scenery".)
After the period covered by Bell, there is a 50-70 year gap, followed by the the Fields Medals. The list of Fields Medalists has its own limitations, but it is another interesting, comparably long list of great mathematicians. Somewhat accidentally, this list also orients and motivates advanced mathematics students today.
But who are the great mathematicians in the gap itself, that is, those born between 1850 and 1900, or say between 1860 and 1910? (Or 1920 at the latest; that is what I had before.) There is a good expanded list of mathematicians with biographies [at the St. Andrews site](http://www-history.mcs.st-and.ac.uk/BiogIndex.html). However, it is too long to work as a sequel to Bell's book. (Not that I plan to write one; I'm just asking.) If you were to make a list of 20 to 50 great mathematicians in this period, how would you do it or who would they be? Presumably it would include Hilbert, but who else? (Poincaré was born in 1854 and is the latest born in Bell's list.)
For example, I know that there was the IBM poster "Men of Modern Mathematics" that also earned some criticism. But I don't remember who was listed, and my recollection is that the 1850-1900 period was somewhat cramped.
I decided based on the earliest responses to convert this list to community wiki. But I am not just asking people to throw out names one by one and then vote them up. If a good reference for this question already exists, or if there is some kind of science to make a list, then that would be ideal. If you would like to post a full list, great. If you would like to list one person who surely should be included and hasn't yet been mentioned, then that's also reasonable. It may be better to add years of birth and death in parentheses, for instance "Hilbert (1862-1943)".
**Note:** To make lists, you should either add two spaces to the end of each line, or " - " (space dash space) to the beginning of each line.
| https://mathoverflow.net/users/1450 | Great mathematicians born 1850-1920 (ET Bell's book ≲ x ≲ Fields Medalists) | The St. Andrews site is an invaluable resource. From that list, I picked (usually) at most one great mathematician born in each year from 1860 to 1910:
$\textbf{EDIT: By popular demand, the list now extends from 1849 to 1920.}$
1849: Felix Klein, Ferdinand Georg Frobenius
1850: Sofia Vasilyevna Kovalevskaya
1851: honorable mention: Schottky
1852: William Burnside
1853: honorable mentions: Maschke, Ricci-Curbastro, Schoenflies
1854: Henri Poincare
1856: Emile Picard (honorable mention: Stieltjes)
1857: honorable mention: Bolza
1858: Giuseppe Peano (honorable mention: Goursat)
1859: Adolf Hurwitz (honorable mention: Holder)
1860: Vito Volterra
1861: honorable mention: Hensel
1862: David Hilbert
1864: Hermann Minkowski
1865: Jacques Hadamard (honorable mention: Castelnuovo)
1868: Felix Hausdorff
1869: Elie Cartan
1871: Emile Borel (honorable mentions: Enriques, Steinitz, Zermelo)
1873: honorable mentions: Caratheodory, Levi-Civita, Young
1874: Leonard Dickson
1875: Henri Lebesgue (honorable mentions: Schur, Takagi)
1877: Godfrey Harold Hardy
1878: Max Dehn
1879: honorable mentions: Hahn, Severi
1880: Frigyes Riesz
1881: Luitzen Egbertus Jan Brouwer
1882: Emmy Amalie Noether (honorable mentions: Sierpinski, Wedderburn)
1884: George Birkhoff, Solomon Lefschetz
1885: Hermann Weyl (honorable mention: Littlewood)
1887: Erich Hecke (honorable mentions: Polya, Ramanujan, Skolem)
1888: Louis Joel Mordell (honorable mention: Alexander)
1891: Ivan Matveevich Vinogradov
1892: Stefan Banach
1894: Norbert Wiener
1895: honorable mention: Bergman
1896: Carl Ludwig Siegel (honorable mention: Kuratowski)
1897: honorable mention: Jesse Douglas
1898: Emil Artin, Helmut Hasse (honorable mentions: Kneser, Urysohn)
1899: Oscar Zariski (honorable mentions: Bochner, Krull, Ore)
1900: Antoni Zygmund
1901: Richard Brauer
1902: Alfred Tarski (honorable mention: Hopf)
1903: John von Neumann (hm's: Hodge, Kolmogorov, de Rham, Segre, Stone, van der Waerden)
1904: Henri Cartan (honorable mentions: Hurewicz, Whitehead)
1905: Abraham Adrian Albert
1906: Kurt Godel, Andre Weil (honorable mentions: Dieudonne, Feller, Leray, Zorn)
1907: Lars Ahlfors, Hassler Whitney (honorable mentions: Coxeter, Deuring)
1908: Lev Pontrjagin
1909: Claude Chevalley, Saunders Mac Lane (honorable mentions: Stiefel, Ulam)
1910: Nathan Jacobson (honorable mention: Steenrod)
1911: Shiing-shen Chern (honorable mentions: Birkhoff, Chow, Kakutani, Witt)
1912: Alan Mathison Turing (honorable mentions: Eichler, Zassenhaus)
1913: Samuel Eilenberg, Paul Erdos, Israil Moiseevich Gelfand (dis/honorable mention: Teichmuller)
1914: honorable mentions: Dantzig, Dilworth, Kac
1915: Kunihiko Kodaira (honorable mentions: Hamming, Linnik, Tukey)
1916: Claude Elwood Shannon (honorable mention: Mackey)
1917: Atle Selberg (honorable mentions: Iwasawa, Kaplansky)
1918: Abraham Robinson
1919: honorable mention: Julia Robinson
1920: Alberto Calderon
| 33 | https://mathoverflow.net/users/1149 | 10114 | 6,909 |
https://mathoverflow.net/questions/10107 | 2 | I am interested in programmatically working with constructible numbers (the closure of the rational numbers under square roots). In order to perform comparisons between numbers I believe I would need a unique (symbolic) representation for them. Does such a thing exist, or what are relevant references for this kind of thing?
| https://mathoverflow.net/users/1074 | Unique representation of constructible numbers | A representation of construcible numbers together with algorithms suitable for mechanized computations is given in chapter 4 of [these lecture notes on computer science](http://books.google.com/books?id=nX9En3pwqWsC&pg=PA279&lpg=PA279&dq=representation+of+constructible+numbers&source=bl&ots=2BWAueNw41&sig=mGPSlh0ltREnq6DitnbAhxSSxoY&hl=en&ei=yRs7S53mEommnQObz4TuCA&sa=X&oi=book_result&ct=result&resnum=2&ved=0CAoQ6AEwAQ#v=onepage&q=representation%2520of%2520constructible%2520numbers&f=false)
| 2 | https://mathoverflow.net/users/1059 | 10119 | 6,912 |
https://mathoverflow.net/questions/6419 | 9 | Two rings $A$ and $B$ are said to be Morita equivalent if the category of modules over $A$ and $B$ are equivalent as additive categories. (Here I'm considering left modules).
Ex: $M\_n(R)$ (the algebra of matrices over a ring $R$) is morita equivalent to $R$.
In fact more generally whenever $A$ is a ring and $e$ is an idempotent in $A$ and $AeA = A$ then the follwing functor is a morita equivalence:
$A$ modules $\rightarrow$ $eAe$ modules
$M$ $\mapsto$ $eM$
Now under nice conditions the categories of $A$ modules (resp $B$ modules) might have a moduli description. Then can you say anything about the induced map on the moduli spaces? I'm asking for situations in which this is known.
Also is there a nice book or paper which talks about Morita equivalence and has lots of examples?
| https://mathoverflow.net/users/1710 | Morita equivalence and moduli problems | In the setting you are interested in, that is, finitely generated k-algebras and GIT-quotients of closed or semistable orbits, Morita equivalence induces isomorphism on the moduli spaces provided one scales dimension(vectors) and stability structures accordingly. That is, if B=M\_n(A) one should compare Mod(A,k) to Mod(B,nk) moduli. If A and B have a complete set of orthogonal idempotents e\_i resp. f\_i(that is, a quiver-like situation) and if the Morita equivalence induces rank(f\_i) = n\_i rank(e\_i), then one should compare A-reps of dimension vector alpha=(alpha\_i) to B-reps with dimension vector beta=(n\_i alpha\_i).
Geometrically, the module varieties of Morita-equivalent algebras are related via associated fibre bundle constructions. In the example above, Mod(B,kn) = GL(nk) x^GL(k) Mod(A,k). In general,one had such a description locally in the Zariski topology (coming roughly from the fact that a vectorbundle (or projective) is locally free). Anyway, this gives a natural and geometric one-to-one correspondence between GL(nk)-orbits (isoclasses) of B-reps and GL(k)-orbits of A-reps inducing the desired isos on the quotient variety level (isos of semi-simples). In quiver-like situations one should adjust the dim vectors as above.
Now as to semi-stability. Observe that semi-stable reps are just ordinary reps of a universal localization of your algebra(s), so one can reduce to the closed case by covering the variety of semi-stables by Zariski open (affine) pieces. In quiver-like situations when the Morita-data is as above and your stability structure mu for A is given by the vector (mu\_i), then the corresponding stability structure for B is mu'=(1/n\_i x mu\_i) (or multiply it with a common factor if you want it to have integral components.
| 6 | https://mathoverflow.net/users/2275 | 10130 | 6,917 |
https://mathoverflow.net/questions/10126 | 17 | Let $G$ be a finite group and $\chi$ be an irreducible character of
$G$ (characteristic zero algebraically closed base field). If $H$ is
the kernel of $\chi$ then the irreducible representations of $G/H$
are exactly all the irreducible constituents of all tensor powers
$\chi^n$.
1. Do you know any reference for this theorem?
2. Is it also working in positive characteristic?
3. Is it also working for some infinite groups? (maybe some special
classes: reductive, Lie type, etc)
Thank you very much!
| https://mathoverflow.net/users/2805 | Reference for this theorem in representation theory? | I am not quite sure about the reference :( I always thought of this fact as follows.
Matrix elements of tensor powers of a representation U are all possible monomials in matrix elements of U, so the space of all their linear combinations are values of all possible polynomials in the matrix elements of U. Now, by definition of H, values of matrix elements of U separate elements of G/H, so every function on G/H (including all irreducible characters) can be written as a polynomial in the matrix elements of U in the case of finite groups, or can be approximated by polynomials with arbitrary precision in the case of compact infinite groups and the ground field being R or C (Stone-Weierstrass).
Now, to complete the proof, we may use orthogonality of matrix elements: if E\_{ij} are matrix elements of an irrep V, and F\_{ij} --- matrix elements of an irrep W (all thought of as functions on the group), then for the standard bilinear form on the ring of functions C(G) we have (E\_{ij},F\_{kl})=0 unless V is isomorphic to W and, in the latter case, i=l, j=k (in which case the value is 1) - here I probably want the order of the group to not be divisible by char(k) in the finite case, or the group to be compact, and the field be real/complex in the infinite case. Since irreducible characters can be approximated by polynomials in matrix elements, such a character cannot be orthogonal to all matrix elements of tensor powers and is, therefor, contained in one of them.
| 15 | https://mathoverflow.net/users/1306 | 10139 | 6,923 |
https://mathoverflow.net/questions/10131 | 2 | let $X : Ring \to Set$ be a functor (a Z-functor in the language of demazure, gabriel) and $V \subseteq X$ a locally closed subfunctor. assume that $U \subseteq V$ is an open subfunctor. does then exist an open subfunctor $W \subseteq X$ such that $U = V \cap W$? if $X$ and $V$ are schemes, this should be true.
| https://mathoverflow.net/users/2841 | subspace topology for functors | I don't think this is true even when $X$ and $V$ are schemes if you only require that the
map $V\to X$ is an embedding of functors (rather than a locally closed embedding). Example:
Take $X=Spec(k[x])$, $V=Spec(k[x,x^{-1}]\times k)$, $U=Spec(k)$.
That is, $X$ is an affine line, $V$ is the disjoint union of a punctured line and origin,
and $U$ is a point, which we view as the connected component of $V$. The natural map
$V\to X$ (corresponding to stratification of $X$) is a categorical monomorphism: it induces an embedding of functors.
EDIT:
Now we add the assumption that $V\hookrightarrow X$ is locally closed. I think the statement still fails,
here's a counterexample:
$X$ will be an ind-scheme: so there is a sequence of schemes $X\_n$ related by closed embeddings
$X\_n\hookrightarrow X\_{n+1}$ and
$$X(A)=\lim\_{\to} X\_n(A).$$
In other words, every point of $X(A)$ factors through one of $X\_n$'s.
For ind-schemes, a locally closed subfunctor $V\subset X$ is given by a compatible
family of locally closed $V\_n\subset X\_n$ (so that it is a locally closed sub-ind-scheme). Problem
is, the statement fails for ind-schemes.
Let's take
$X\_n={\mathbb A}^n$, with the embedding $X\_n\hookrightarrow X\_{n+1}$ being the coordinate embedding.
Now let $V\_n$ be the union of the origin $0$ and $n-1$ punctured lines
$$l\_k:=\lbrace(k,0,\dots,0,x,0,\dots,0)|x\ne 0\rbrace,$$
where $x$ is in the $k$-th position, and $k$ varies from $2$ to $n$. (Let's say I work over
a field of characteristic $0$, so all integers are distinct.)
Finally, $U\_n\subset V\_n$ is the origin, which is a component of each $V\_n$, so it is both open
and closed.
It is easy to see that it is impossible to find a compatible family of open subsets $W\_n\subset X\_n$
such that $U\_n=V\_n\cap W\_n$. Indeed, $W\_1$ contains $0$, so it is non-empty, and thus contain
$n\in{\mathbb A}^1=X\_1$ for some $n$. But then $W\_n$ must contain $(n,0,\dots,0)$ without meeting
$l\_n$.
| 3 | https://mathoverflow.net/users/2653 | 10143 | 6,927 |
https://mathoverflow.net/questions/8772 | 19 | There is a "folk theorem" (alternatively, a fun and easy exercise) which asserts that a 2D TQFT is the same as a commutative Frobenius algebra. Now, to every compact oriented manifold $X$ we can associate a natural Frobenius algebra, namely the cohomology ring $H^\ast(X)$ with the Poincare duality pairing. Thus to every compact oriented manifold $X$ we can associate a 2D TQFT.
Is this a coincidence? Is there any reason we might have expected this TQFT to pop up?
When $X$ is a compact *symplectic* manifold, perhaps the appearance of the Frobenius algebra can be explained by the fact that the quantum cohomology of $X$, which comes from the A-twisted sigma-model with target $X$, becomes the ordinary cohomology of $X$ upon passing to the "large volume limit".
But for a general compact oriented $X$? I don't see how we might interpret the appearance of the Frobenius algebra in some quantum-field-theoretic way. Maybe there is an explanation via Morse homology?
| https://mathoverflow.net/users/83 | Cohomology rings and 2D TQFTs | These 2D TQFTs do not come from extended theories (unless X is discrete). I interpret this as saying that these theories are non-local (in the 2D bordism) and so you will have trouble interpreting them in a traditional QFT framework. You will have to do something funny and non-local, like squashing your circles to points and surfaces to graphs, as in the Cohen work mentioned by Tim.
| 14 | https://mathoverflow.net/users/184 | 10144 | 6,928 |
https://mathoverflow.net/questions/10146 | 28 | I would want all book tips you could think of regarding problem solving and books in general, in elementary mathematics, with a certain flavour for "advanced problem solving". An example would be the books from [The Art of Problem Solving](https://www.newyorker.com/culture/persons-of-interest/richard-rusczyks-worldwide-math-camp), Arthur Engel's book and Paul Zeit's book. Books on certain topics, say geometry, are also appreciated!
| https://mathoverflow.net/users/2903 | Good books on problem solving / math olympiad | Polya's "How to Solve It" is a good one. When prepping for the Putnam, I used "Problem Solving Through Problems"
| 13 | https://mathoverflow.net/users/622 | 10148 | 6,930 |
https://mathoverflow.net/questions/10118 | 10 | 1. I'm looking for a definition of Chern class (at least the first one) for a torsion-free sheaf $F$ (not necessarily locally free) on a singular curve (for simplicity can assume all the singularities are planar).
The Chern class can be, of course, extracted from an exact sequence relating $F$ to some locally free sheaves. But I would like some more direct definition, like the one given by Hartshorne (Generalized divisors on Gorenstein curves and a theorem of Noether. J. Math. Kyoto Univ. 26 (1986), no. 3, 375--386).
At least for the first Chern class.
1.5 Even if one wants to define $c\_1(F)$ from some resolution: torsion free sheaves on singular curves sometimes have no finite locally free resolutions. What would you do in this case?
1. I'm looking for the Riemann-Roch for torsion-free sheaves on a singular curve (can assume the singularities to be planar). For example Hartshorne in the paper above does it for rank one.
Of course, if the only definition of the first Chern class is from the exact sequence, then Riemann-Roch is tautological (an alternative way to define $c\_1(F)$). So this question is meaningful modulo the first question.
Somehow I do not find all this in classical textbooks.
Thanks to everybody!!!!
| https://mathoverflow.net/users/2900 | on chern classes and Riemann Roch theorem for torsion-free sheaves on singular (possibly multiple) curve | In the affine case, there is a sweet way to define the first Chern class as follows:
Let $R$ by the coordinate ring and $M$ the $R$-module correspond to our sheaf. As $M$ is torsion-free, one can embed $M$ in to a free module:
$ 0\to M \to F \to N \to 0$
( you need $M$ to be of constant rank, and that rank would be the rank of $F$). In this $N$ would be torsion, so the support is finite. Take the cycle $c(N) = \sum length(N\_p)[p]$ where $p$ runs over the support of $N$. Then define $c(M)= -c(N)$.
In general, one could get codimension 1 cycles by picking them from any prime filtration of $M$(you needs to show that what you get from 2 different filtrations are rationally equivalent). This [paper](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6WH2-45JK6K9-1T&_coverDate=09%2F01%2F1999&_alid=197271557&_rdoc=1&_fmt=&_orig=search&_qd=1&_cdi=6838&_sort=d&view=c&_acct=C000055513&_version=1&_urlVersion=0&_userid=1942757&md5=bb66ad332e7312fd296eaad64313e9ce) contains a treatment of that result.
| 7 | https://mathoverflow.net/users/2083 | 10151 | 6,933 |
https://mathoverflow.net/questions/9083 | 9 | Suppose $f$ and $g$ are two newforms of certain levels, weights etc. If we know that
L(f,n)=L(g,n) for all sufficiently large $n$, can we conclude that $f=g$?
Same question when the forms have the same weight and $n$ runs over critical points.
| https://mathoverflow.net/users/2344 | How many L-values determine a modular form? | The answer to the first question is "yes". The standard proof of the uniqueness of a Dirichlet series expansion actually generalizes to show the following.
Theorem. Suppose that $A(s) = \sum\_n a\_n n^{-s}$ and $B(s) = \sum\_n b\_n n^{-s}$ are Dirichlet series with coefficients $a\_n, b\_n$ bounded by a polynomial. If there exists a sequence of complex numbers $s\_k$ with real part approaching infinity such that $A(s\_k) = B(s\_k)$ for all $k$, then $a\_n = b\_n$ for all $n$.
Proof (sketch). Proceed by induction. For $k$ big we have $A(s\_k) = a\_1 + O(2^{-\sigma\_k})$ where $\sigma\_k$ is the real part of $s\_k$. Similarly, $B(s\_k) = b\_1 + O(2^{-\sigma\_k})$. Since $A(s\_k) = B(s\_k)$, we conclude that $a\_1 = b\_1$. A similar argument shows $a\_2 = b\_2$, $a\_3 = b\_3$, etc.
| 5 | https://mathoverflow.net/users/2627 | 10153 | 6,935 |
https://mathoverflow.net/questions/10172 | 9 | This question might turn out not to make any sense, but here it is: Witten's (and Reshetikhin and Turaev's) 3-manifold invariant can be "defined" as an integral over the space of connections on the trivial SU(2) bundle over the 3-manifold M, modulo gauge transformation. In the stationary phase approximation as the level k-->infinity, we write this integral as the sum of functions f\_i(k), where the sum is over the set {c\_i} of critical points of the chern-simons functional (which is basically the log of the integrand). The critical points are the flat connections. In his analysis of the function f\_tr(k) corresponding to the trivial flat connection (product connection), Rozansky (as well as previous authors, I believe) found the Casson invariant (as the second coefficient if you write f\_i as a power series in 1/k, or something like that).
Now, at first glance it strikes me as odd that the casson invariant shows up here, because the casson invariant can, by definition, be thought of as a (signed) sum over ALL flat connections, EXCEPT the trivial connection. So my question is: what gives? Why does the contribution from the trivial connection give the casson invariant? Why isn't it instead, say, the sum of the 2nd coefficients of the contributions from all other flat connections? is there some explanation for this? Have contributions from other connections, say reducible connections for a homology 3-sphere, been analyzed? Does the casson invariant show up there as well?
If you've gotten this far, here's another question: right now, i can write |H\_1(M)|\*(Casson-Walker invariant of M), say for lens spaces M=L(p,q) as the sum over certain INTEGERS, one per conjugacy class of reducible flat connection on SU(2)XM, PLUS the signature of the 2-bridge knot b(p,q). I'm trying to identify this decomposition of the casson invariant with others in the literature, of which there are many, but I'm having the problem that none of these are (always) integral decompositions, even though they ADD UP to an integer. Well this is a long shot, but any ideas would be appreciated!
| https://mathoverflow.net/users/492 | Casson's invariant and the trivial connection contribution to witten's 3-manifold invariant | The Casson invariant is not the same sum or integral over connections that you would derive from the perturbative expansion Cherns-Simons quantum field theory at all flat connections. There is more than one way to rigorously interpret that expansion; one method uses Kontsevich's configuration space integrals. Dylan Thurston and I proved (in *[Perturbative 3-manifold invariants by cut-and-paste topology](https://arxiv.org/abs/math/9912167)*) that the configuration space invariant using only the flat connection is proportional to the Casson invariant, for any simple Lie group as the gauge group. (The configuration space integral is known as the theta invariant, because the Feynman-Jacobi diagram is a theta.) By contrast, Casson's invariant has been interpreted by Witten as a gauge theory with a certain Lie supergroup, whose underlying Lie group is SU(2).
So what gives? One answer is that the Casson invariant is the unique [finite-type invariant](http://en.wikipedia.org/wiki/Finite_type_invariant) of homology 3-spheres of degree 1, up to a scalar factor. (The Wikipedia link discusses finite-type invariants of knots, but as the page mentions briefly, there is an extension to 3-manifolds due to Garoufalidis, Habiro, Levine, and Ohtsuki.) As such, you should expect it to show up many times. The corresponding phenomenon at the level of link invariants is that all of the standard quantum link invariants that are polynomials in $q$ have the same second derivatives at $q=1$, again up to a scalar factor. In fact, Dylan and I didn't do anything directly with the Casson invariant; instead we showed that the theta invariant has finite-type degree 1.
| 6 | https://mathoverflow.net/users/1450 | 10178 | 6,955 |
https://mathoverflow.net/questions/9756 | 7 | I've gotten stuck in a project I have been working on, essentially on the following combinatorial question about the symmetric group.
One can obtain a 1-dimensional representation $M^n\_c$ of the algebra $T\_n := S\_n \rtimes \mathbb{C}[y\_1, \dots, y\_n] $ by letting each $y\_i$ act by $c$ and $S\_n$ act trivially.
Given a partition $\pi$ of $n =n\_1 + \dots + n\_k$ and $c\_1, \dots, c\_k \in \mathbb{C}$, we can consider the standard induced module
$$ M^{\pi}\_c = \mathrm{Ind} ( M^{n\_1}\_{c\_1} \otimes \dots \otimes M^{n\_k}\_{c\_k}) $$
where the induction is from the subalgebra $T\_{n\_1} \otimes \dots \otimes T\_{n\_k} \subset T\_n$ to $T\_n$.
As far as I know, the simple quotients of these standard modules form a complete class for the simple representations of $T\_n$. (I think this is a general fact about semidirect products of a finite group with a commutative algebra, together with the fact that representations of the symmetric group $S\_n$ can be obtained by taking quotients induction of the trivial representation of a Young subgroup $S\_{n\_1} \times \dots \times S\_{n\_k}$.)
My question is how to represent this explicitly in terms of the simple objects in the semisimple category $Rep(S\_n)$. First of all, the $S\_n$-module $M^{\pi}\_c$ can be described in a combinatorial manner in terms of the simple objects in $Rep(S\_n)$ using the Kostka numbers. The action of the $y\_i$'s on $M^{\pi}\_c$ can be given in terms of a suitable morphism $y: \mathfrak{h} \otimes M^{\pi}\_c \to M^{\pi}\_c$, where $\mathfrak{h} \in Rep(S\_n)$ is the regular representation.
The Pieri rule allows one to compute the decomposition in irreducibles (as parametrized by Young diagrams, of course) of $\mathfrak{h} \otimes M^{\pi}\_c $, so we can view $y$ as a bunch of matrices based upon this decomposition (matrices w.r.t. the simple objects in $Rep(S\_n)$, not as vector spaces).
Is there an approach to compute these matrices?
I am interested in this because it may allow a way to directly interpolate the construction of these modules to complex rank, via Pavel Etingof's [program](http://www.newton.ac.uk/programmes/ALT/seminars/032716301.html). (I believe one can interpolate the construction in another way, by reasoning more directly on the definition of the category $Rep(S\_t)$ given by Deligne, but this seems to be useless as far as explicit computations--which might be helpful to study the "degeneracy phenomena" that Etingof has suggested might exist--are concerned.) In this case we have tensor categories $Rep(S\_t)$ for $t$ not necessarily an integer, and while the interpretation in terms of vector spaces fails, the one in terms of Young diagrams does not.
**Edit** (12/27) I added a bounty today and here is some additional information that may be useful:
It should come out that the matrices representing the $y$-morphism $\mathfrak{h} \otimes M^{\pi}\_c \to M^{\pi}\_c$ are polynomials in the dimension $n$. When increasing $n$, we change the partition $\pi$ by adding to the first (largest) element and leaving the rows below fixed. Since a simple object in $Rep(S\_t)$ for $t$ not an integer can be represented as a normal Young diagram (of size, say, $N$) with a "very long line" of "size" $t-N$ at the top, this kind of a polynomial interpolation will allow for an interpolation of the $M^{\pi}\_c$ to complex rank. My claim that it should come out as a polynomial was based upon studying induction directly on these categories and finding it was interpolable. However, I don't know how to compute the $y$'s directly as matrices via the simple object decomposition.
My hope was that there is a clean not-too-computationally-intensive way to do this, but unfortunately I'm not yet sufficiently comfortable with the theory of the symmetric group to have any ideas as to how to proceed.
I am also interested in the degenerate affine Hecke algebra of type A, where these kinds of standard induced modules can be defined similarly. Their simple quotients form a complete collection of irreducible modules for the Hecke algebra according to a theorem of Zelevinsky, and I know that these, too, can be interpolated by reasoning on the definitions in Deligne's paper (so one gets objects in the interpolated category $Rep(H\_t)$, which is defined in Etingof's talk). But I am curious here too how it is possible to compute the $y$-morphisms as matrices using the decomposition into irreducibles in $Rep(S\_t)$.
| https://mathoverflow.net/users/344 | Explicit computation of induced modules of semidirect products with the symmetric group | I suspect I know the answer, but I don't yet have a proof (not because I think it would be hard to prove, but because I didn't try really; when you see my guess, you'll likely want to believe it). The answer is stated not in the basis of simples, because I didn't compute the decomposition of $\mathbb{C}[S\_n/S\_{pi}]$. However, it is stated in the tensor category S\_n-mod, so that given that decomposition, you can easily adjust what I write here.
>
> **Fact:** Let H be a finite dimensional semi-simple Hopf algebra (e.g. H=\mathbb{C}[S\_n]), and let $V\in H$-mod be an irrep. Let us regard H as an H-module via the left action. Then $V\otimes H\cong H^{\oplus dim(V)}$.
>
>
>
The proof of this fact is given as follows (see <http://www-math.mit.edu/~etingof/tenscat.pdf>, or Akhil's comments below):
$Hom\_H(V\otimes H,W)=Hom\_H(H,^\ast V\otimes W) = \widetilde{^\ast V\otimes W}$
On the other hand, $Hom\_H(\tilde{V}\otimes H,W) = \tilde{V}\otimes Hom\_H(H,W) = \widetilde{V\otimes W}$,
where $\tilde{M}$ means we forget the module $M$ down to a vector space, which we use as a multiplicity space (just because the direct sum decomposition I asserted originally isn't canonically given, you just know that there's this multiplicity space)
(above we took right duals since I didn't assume $H$ is commutative or co-commutative; for $C[G]$ there is no need to distinguish.) One could (and should) be uncomfortable that we got duals on the one hand and not on the other. However, the standard representation for $S\_n$ is special in that it is isomorphic to its own dual, by sending $e\_i$ to $e^i$ (the point is that the standard rep for $S\_n$ has a basis build into its definition).
The general fact above about Hopf algebras is used to relate Frobenius-Perron dimension for representations of Hopf algebras to ordinary dimension of the underlying vector space; indeed the regular representation is the unique eigenvector which realizes the Frobenius Perron dimension as an eigenvalue.
Okay so now we are considering $\mathfrak{h}\otimes \mathbb{C}[S\_n/S\_{\pi}]\to \mathbb{C}[S\_n/S\_{\pi}]$. This is then isomorphic to $(\mathfrak{h}\otimes \mathbb{C}[S\_n])\otimes\_{S\_\pi}\mathbf{1}$, where we tensor the trivial $S\_\pi$-module on the right. This is because $C[S\_n]$ is a $S\_n-S\_\pi$ bi-module, so that the map
$\mathfrak{h}\otimes \mathbb{C}[S\_n] \to \mathfrak{h}\otimes \mathbb{C}[S\_n/S\_\pi]$ given by right multiplying with the symmetrizer $a\_\pi=\sum\_{g\in S\_\pi} g$ is an $S\_n$-morphism, and allows us to identify $\mathfrak{h}\otimes \mathbb{C}[S\_n]\otimes\_{S\_\pi}\mathbf{1}$ with $\mathfrak{h}\otimes \mathbb{C}[S\_\pi]$.
Morally, this is just because $S\_\pi$ acts on the right, while the other action is on the left.
[edited an error from preceding paragraph]
Together with the Fact, this implies that $\mathfrak{h}\otimes \mathbb{C}[S\_n/S\_\pi]$ is in fact just isomorphic to $\mathbb{C}[S\_n/S\_\pi]^{\oplus dim(\mathfrak{h})},$ which I should really write as
$\mathbb{C}[S\_n/S\_\pi]\otimes \tilde{\mathfrak{h}^\ast}$.
Well, now we have this function $c: \mathfrak{h}\to \mathbb{C}$. We will project $\mathbb{C}[S\_n/S\_\pi]^{\oplus dim(\mathfrak{h})}$ (or rather $\mathbb{C}[S\_n/S\_\pi]\otimes \tilde{\mathfrak{h}}$) to $\mathbb{C}[S\_n/S\_\pi]$ by just applying $c$ to the multiplicity space.
I haven't really proved that this last paragraph is what happens, but once one has applied "Fact" above, this seems like the only natural guess. I imagine verifying it would be pretty straightforward.
Note that it doesn't seem to matter how $\mathbb{C}[S\_n/S\_{\pi}]$ decomposes into simples, since they all get lumped together.
| 2 | https://mathoverflow.net/users/1040 | 10183 | 6,959 |
https://mathoverflow.net/questions/10182 | 6 | Can someone please clarify if there always exist regular neighborhoods for a properly embedded surface in a 3-manifold? More precisely, if $F$ is a properly embedded surface in a 3-manifold $M$ and I give a simplicial complex structure to $M$, then will $F$ automatically recieve a subcomplex structure (after probably finitely many barycentric subdivisions of the triangulation of $M$) ? If this is so, then we can obviously construct a regular neighborhood of $F$. But I am feeling unsure about it now due to the following example where $F$ might be too big to allow such a compatible subdivision of the triangulation of $M$:
**Example:** Consider a mobius band without a contractible neighborhood in a genus 1 handlebody. One can construct one as follows: First take a solid cylinder $D^2\times{[-1,1]}$ and look at the central strip. Now glue the ends of the cylinder with $180$ degrees twist to make it a genus 1 handlebody. This will glue the central strip with a twist to make it a mobius band and it has no small neighborhood in the ambient genus 1 handlebody. It seems to me that this mobius band will not have a subcomplex structure for any given simplicial complex structure on the genus 1 handlebody.
In case the answer to my question is yes, I would also like to ask if there are no issues regarding the orientability of $F$ and $M$ while considering regular neighborhoods. That is whether it matters for constructing such a neighborhood if $F$ is non-orientable but $M$ is orientable or vice-versa or other combinations.
I am unable to clarify this by looking at Hempel's book on 3-manifolds. It would be great if someone could elucidate this.
| https://mathoverflow.net/users/2533 | Does a regular neighborhood always exist for a properly embedded surface in a 3-manifold? | The answer is yes, with the correct technical hypothesis of "local flatness". (Local flatness rules out, for example, the sort of behavior shown by the [Alexander horned sphere](http://en.wikipedia.org/wiki/Alexander_horned_sphere).) You are correct to think that this is a foundational issue in three-manifolds. The reference you mention (Hempel) is also the correct one. You could perhaps look at Bing's [book](http://books.google.co.uk/books?id=NBCQ0a_cdXcC&printsec=frontcover&dq=inauthor%3A%2522R.+H.+Bing%2522&client=firefox-a&cd=2#v=onepage&q=&f=false).
In your example the Mobius band does have a regular neighborhood. Note that the open regular neighborhood is homeomorphic to the normal bundle of the surface F in the three-manifold M. The regular neighborhood need not (and in your example, is not) be homeomorphic to the product FxI.
| 4 | https://mathoverflow.net/users/1650 | 10185 | 6,960 |
https://mathoverflow.net/questions/10187 | 5 | let $T$ be a triangulated category and $A \to B \to C \to A[1]$ a triangle in $T$ such that for every $A\_0 \in T$ the induced long sequence
$... \to \hom(A\_0,A) \to \hom(A\_0,B) \to \hom(A\_0,C) \to \hom(A\_0,A[1]) \to \hom(A\_0,B[1]) \to ...$
is exact. is then $A \to B \to C \to A[1]$ exact? I'm a beginner, so this could be rather trivial. I've checked it in special cases, but I can't translate the proof into $T$. I would like the result because it would imply that the homological algebra you know for abelian groups takes over to triangulated categories. comparable to the fact that you can work with group objects in arbitrary categories as with ordinary groups. you can do it directly with the axioms, but this is a big mess.
in many answers I've seen here so far, even standard facts are enriched with wonderful, ample insights. so you are also invited to dish durt about these basics of triangulated categories because I'm just beginning to learn them.
| https://mathoverflow.net/users/2841 | exactness in triangulated categories is reflected by hom-functor | no: consider changing the sign of one of your maps.
about your philosophical question concerning homological algebra for abelian groups carrying over to triangulated categories: actually the fundamental triangulated category is not of abelian groups, but of spectra (in algebraic topology). to make correct and precise statements illustrating this claim one should work with enhanced versions of triangulated categories, my favorite being stable (\infty,1)-categories in the sense of Lurie's DAG I. then for instance one can (probably?) say that any stable (\infty,1)-category is canonically enriched and tensored in spectra, formulate an analog of the mitchell embedding theorem, etc...
| 5 | https://mathoverflow.net/users/2908 | 10189 | 6,962 |
https://mathoverflow.net/questions/10128 | 87 | This is totally elementary, but I have no idea how to solve it: let $A$ be an abelian group such that $A$ is isomorphic to $A^3$. is then $A$ isomorphic to $A^2$? probably no, but how construct a counterexample? you can also ask this in other categories as well, for example rings. if you restrict to Boolean rings, the question becomes a topological one which makes you think about fractals: let $X$ be Stone space such that $X \cong X + X + X$, does it follow that $X \cong X + X$ (here $+$ means disjoint union)?
**Edit**: In the answers there are already counterexamples. but you may add others in other categories (with products/coproducts), especially if they are easy to understand :).
| https://mathoverflow.net/users/2841 | When is $A$ isomorphic to $A^3$? | The answer to the first question is no. That is, there exists an abelian group $A$ isomorphic to $A^3$ but not $A^2$. This result is due to A.L.S. (Tony) Corner, and is the case $r = 2$ of the theorem described in the following Mathematical Review.
[MR0169905](https://mathscinet.ams.org/mathscinet-getitem?mr=169905)
Corner, A.L.S., On a conjecture of Pierce concerning direct decomposition of Abelian groups. 1964 *Proc. Colloq. Abelian Groups (Tihany, 1963) pp.43--48 Akademiai Kiado, Budapest.*
It is shown that for any positive integer $r$ there exists a countable torsion-free abelian group $G$ such that the direct sum of $m$ copies of $G$ is isomorphic to the direct sum of $n$ copies of $G$ if and only if $m \equiv n \pmod r$. This remarkable result is obtained from the author's theorem on the existence of torsion-free groups having a prescribed countable, reduced, torsion-free endomorphism ring by constructing a ring with suitable properties. It should be mentioned that the question of the existence of algebraic systems with the property stated above has been considered by several mathematicians. The author has been too generous in crediting this "conjecture" to the reviewer.
Reviewed by R.S. Pierce
| 91 | https://mathoverflow.net/users/586 | 10194 | 6,966 |
https://mathoverflow.net/questions/10184 | 12 | Let $G$ be a group, say finitely presented as $\langle x\_1,\ldots,x\_k|r\_1,\ldots,r\_\ell\rangle$. Fix $n\geq 1$ a natural number. Then there exists a scheme $V\_G(n)$ contained in $GL(n)^k$ given by the relations. This scheme parameterizes $n$ dimensional representations of $G$.
Now, I've known this scheme since I first started learning algebraic geometry (one of the first examples shown to me of an algebraic set was $V\_{S\_3}(2)$) but I've never found a good reference for this. So my first question is:
```
Is there a good reference for the geometry of schemes of representations?
```
Now, I have some much more specific questions. The main one being a point I'd been wondering about idly and tangentially since reading about some open problems related to the Calogero-Moser Integrable System:
Are there natural conditions on $G$ that will guarantee that $V\_G(n)$ be smooth? Reduced? Now, this is on the affine variety, I know that the projective closure will generally be singular, but in the case of $V\_{S\_3}(2)$, I know that the affine variety defined above is actually smooth, of four irreducible components.
Finally, for any $G$ and $n$, we have $V\_G(n)\subset V\_G(n+1)$ (By taking the subscheme where the extra row and column are zeros, except on the diagonal, where it is 1). We can take the limit and get an ind-scheme, $V\_G$. What is the relationship between $V\_G$ and the category $Rep(G)$? Can the latter be realized as a category of sheaves on the former? I know nothing here, and as I said, most of these questions are the result of idle speculation while reading about something else.
Edit: It occurs to me that as defined, $V\_G(n)$ and $V\_G$ may not be invariants of $G$, but really of the presentation. So two things to add: one, $V\_G(n)$ is intended to really be the scheme $Hom(G,GL(n))$ (there's some issues I want to sweep under the rug with finitely generated infinite groups here, which is part of why I was thinking in presentations), and second, the situations that I'm thinking of are often the data of group with a presentation, so for that situation, $V\_G(n)$ as defined should be good enough.
| https://mathoverflow.net/users/622 | Schemes of Representations of Groups | Charlie, as Dmitri pointed out there is a big difference between compact Kaehler and non-Kaehler manifolds as far as the structure of the representation varieties of their fundamental groups are concerned.
By the way, by a theorem of Taubes, every finitely presentable group is the fundamental of a compact complex three dimensional complex manifold so you don't really get any restriction by saying that you want your group to be the fundamental group of a complex manifold. Taubes' manifolds however are constructed as twistor spaces of anti-self dual real 4-manifolds and are never Kaehler. The condition that your group can be realized as the fundamental group of a compact Kaehler manifold is a serious condition and puts many constraints on the representation variety.
Another comment is that the representation scheme does not depend on the presentation of your group. You only need to know that the group is finitely generated.
There is a huge literature on this subject. I will list just a few of the landmarks:
1. A very nice classical source is the Lubotzky-Magid book ["Varieties of representations of finitely generated groups"](http://books.google.com/books?id=3UC5EKcYyjIC&dq=Lubotzky,+Magid&printsec=frontcover&source=bl&ots=wama5qwKgo&sig=gaPTJVGlEoY_IWqw3WigW3YylXI&hl=en&ei=7e87S-fUL9XNlAe5yrWaBw&sa=X&oi=book_result&ct=result&resnum=2&ved=0CAoQ6AEwAQ#v=onepage&q=&f=false)
2. The paper of Goldman-Milson that Dmitri mentioned is a must-read.
3. There are two fundamental papers of Simpson that I mentioned in this [MO post](https://mathoverflow.net/questions/7212/moduli-space-of-flat-bundles/7221#7221).
4. On the topic of Kaehler fundamental groups you can start with this [book](http://www.ams.org/cgi-bin/bookstore/booksearch?pg1=&s1=&op1=ADJ&co1=AND&pg2=&s2=&op2=ADJ&co2=AND&pg3=CN&op3=ADJ&s3=amoros&co3=AND&pg4=CN&op4=ADJ&s4=burger&co4=AND&pg5=AS&op5=ADJ&s5=&co5=AND&pg6=CMSC&s6=&op6=or&co6=AND&pg7=%40PYR&arg1=eq&s7=&fn=100&d=BOOK&p=1&u=&r=0&l=100&f=S) by Amoros et al. and with this [article](http://www.google.com/url?sa=t&source=web&ct=res&cd=1&ved=0CAcQFjAA&url=http%3A%2F%2Fwww.msri.org%2Fpublications%2Fbooks%2FBook28%2Ffiles%2Farapura.pdf&ei=BPQ7S-KQH8fVlAf0s62SBw&usg=AFQjCNGLzeHE9DBEPc3rssgguUacavDbBw) by Arapura.
| 9 | https://mathoverflow.net/users/439 | 10203 | 6,970 |
https://mathoverflow.net/questions/10216 | 1 | Consider a real symmetric matrix $A\in\mathbb{R}^{n \times n}$. The associated quadratic form $x^T A x$ is a convex function on all of $\mathbb{R}^n$ iff $A$ is positive semidefinite, i.e., if $x^T A x \geq 0$ for all $x \in \mathbb{R}^n$.
Now suppose we have a convex subset $\Phi$ of $\mathbb{R}^n$ such that $x \in \Phi$ implies $x^T A x \geq 0$. Is $x^T A x$ a convex function on $\Phi$ (even if $A$ is not positive definite)? Of course, the answer in general is "no," but we can still ask about the most inclusive conditions under which convexity holds for a given $A$ and $\Phi$. In particular I'm interested in the question:
**Suppose we have a quadratic form $Q:\mathbb{R}^{n \times n} \rightarrow \mathbb{R}$. What is the *weakest* condition on $Q$ that guarantees it will be convex when restricted to the set of positive semidefinite matrices?**
| https://mathoverflow.net/users/1557 | If a quadratic form is positive definite on a convex set, is it convex on that set? | $x^2-y^2$ is positive on $[2,3]\times [-1,1]$ but not convex there. This creates problems for any convex sets not containing the origin. You are, probably, after something else not so obviously false. Why don't you just tell us what it is?
Edit: Even then it is false: just take $B\_{11}B\_{22}$. By the way, for a pure quadratic form, convexity on an open set and convexity on the entire space are the same thing.
| 8 | https://mathoverflow.net/users/1131 | 10218 | 6,980 |
https://mathoverflow.net/questions/10212 | 5 | Reading 2007 paper [A tour of theta dualities on moduli spaces of sheaves](http://arxiv.org/abs/0710.2908) by [Alina Marian](http://arxiv.org/find/math/1/au%3a+Marian_A/0/1/0/all/0/1) and [Dragos Oprea](http://arxiv.org/find/math/1/au%3a+Oprea_D/0/1/0/all/0/1).
>
> Why is any moduli space of coherent sheaves on a K3 surface deformation equivalent to a moduli space of sheaves on an elliptic K3?
>
>
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(The authors consider a space of "Gieseker H-semistable sheaves", if that is important)
| https://mathoverflow.net/users/65 | Moduli spaces of coherent sheaves on K3s | This follows from a result of Yoshioka. In Theorem 8.1 of this [paper](http://arxiv.org/abs/math/0009001) Yoshioka showed that every moduli space of coherent sheaves on a K3 surface $X$ is deformation equivalent to an appropriate Hilbert scheme of points of $X$. Since every K3 is deformation equivalent to an elliptic K3 it follows that their Hilbert schemes are deformation equivalent and so you get the statement that you wanted.
| 9 | https://mathoverflow.net/users/439 | 10220 | 6,981 |
https://mathoverflow.net/questions/10227 | 28 | So, say we are working with non-CH mathematics. This means, AFAIK, that there is at least one set $S$ in our non-CH mathematics, whose cardinality is intermediate between $|\mathbb{N}|$ (card. of naturals) and $|\mathbb{R}|=2^\mathbb{N}$, the continuum.
Question: what kind of objects would we find in this set $S$?
Also: is this mathematics radically different from the one where CH holds?
Specifically, are there results that are used in everyday math , at a relatively introductory level, which do not hold on our non-CH math.?. What results that we find in everyday math would not hold in our new math? Would there, e.g., still exist non-measurable sets? Maybe more specifically: what results depend on the CH?
| https://mathoverflow.net/users/2915 | In set theories where Continuum Hypothesis is false, what are the new sets? | The question of what happens when CH fails is, of course, intensely studied in set theory. There are entire research areas, such as the area of [cardinal characteristics of the continuum](https://mathoverflow.net/questions/8972#9027), which are devoted to studying what happens with sets of reals when the Continuum Hypothesis fails.
The lesson of much of this analysis is that many of the most natural open questions turn out to be themselvesd independent of ZFC, even when one wants ¬CH. For example, the question of whether all sets that are intermediate in size between the natural numbers and the continuum should be Lebesgue measure 0, is independent of ZFC+¬CH. The question of whether only the countable sets have continuum many subsets is independent of ZFC+¬CH. There are a number of cardinal characteristics that I mention [here](https://mathoverflow.net/questions/8972#9027), whose true nature becomes apparant only when CH fails. For example, must every unbounded family of functions from ω to ω have size continuum? It is independent of ZFC+¬CH. Must every dominating family of such functions have size continuum? It is independent of ZFC+¬CH. Those question are relatively simple to state and could easily be considered part of "ordinary" mathematics.
However, much of the rest of what you might think of as ordinary mathematics is simply not affected by CH or not CH. In particular, the existence of non-measurable sets that you mentioned is provable in ZFC, whether or not CH holds. (This proof requires the use of the Axiom of Choice, however, unless large cardinals are inconsistent, a result proved by Solovay and Shelah.)
Nevertheless, there is a growing body of research on some sophisticated axioms in set theory called forcing axioms, which have powerful consequences, and many of these new axioms imply the failure of CH. This topic began with Martin's Axiom MAω1, and has continued with the Proper Forcing Axiom, Martin's Maximum and now many other variations.
Lastly, in your title you asked what are the new sets like. The consistency of the failure of the Continuum Hypothesis was proved by Paul Cohen with the method of [forcing](http://en.wikipedia.org/wiki/Forcing_(mathematics)). This highly sophisticated and versatile method is now used pervasively in set theory, and is best thought of as a fundamental method of constructing models of set theory, sharing many affinities with construction methods in algebra, such as the construction of algebraic or transcendental field extensions. Cohen built a model of ZFC+ ¬CH by starting with a model V of ZFC+CH, and then using the method of forcing to add ω2 many new real numbers to construct the forcing extension V[G]. Since V and V[G] have the same cardinals (by a detailed combinatorial argument), it follows that the set of reals in V[G] has size at least (in fact, exactly) ω2. In particular, the old set of reals from V, which had size ω1, is now one of the sets of reals of intermediate size. Thus, these intermediate sets are not so mysterious after all!
| 37 | https://mathoverflow.net/users/1946 | 10229 | 6,987 |
https://mathoverflow.net/questions/10233 | 8 | Hi all,
I want to compute in $\mathbb{F}\_q (x)((y))$ i.e. a Laurent series ring over the rational functions over $\mathbb{F}\_q$. The computations are fairly basic, but they involve raising to the qth power a lot. I thought that this would be easy (I thought that it will merely shirt powers around), so I tried it in SAGE. I have to say that I am highly impressed with the ease of programming in SAGE, but I think it is too big (and slow) for the calculation I need (I know that SAGE has lots of components (PARI, GAP, etc.) some of them may be what I need).
So I wanted to ask the people who have more experience then me for a recommendation. Which algebra system is good at Laurent series over rational function fields in char p if you need to do a lot of raising to the qth power.
~AP
| https://mathoverflow.net/users/2917 | Choosing a fast computer algebra system that works in characteristic p? | My personal experience is a few years old, but I don't think things have changed much. Sage is (or actually, was) more about ease of use then about performance. The only three CAS's you want to consider are
* Singular (Macaulay 2 uses Singular's engine)
* Cocoa.
* Magma.
Back then the fastest of the bunch was Magma, but not by much. Regarding ease of use, it was a tie between Macaulay 2 and Magma.
And now to some criticism: I never looked at Magma's code (proprietary), but I did look at both Singular and Cocoa. None of them uses SSE/GPGPU, which could probably give you an acceleration factor of 10-100.
| 3 | https://mathoverflow.net/users/404 | 10247 | 6,998 |
https://mathoverflow.net/questions/10223 | 10 | Is there an integer $n$ with an infinite number of representations of the form
$n=2q-p$, where $p$ and $q$ are both primes?
Given a positive integer $k>1$, I would like to know for which (if any) integers $n$ the linear equation $q-kp=n$ admits an infinite number of solutions, where $p$ and $q$ are primes.
(I'm not including $k=1$ because it reduces to well know open problems, $k=1$ and $n=2$ would be the twin primes conjecture)
The density of the prime numbers implies that at least there are integers $n$ with an arbitrarily large number of representations.
| https://mathoverflow.net/users/1019 | Linear equation with primes | Assuming the Hardy-Littlewood prime tuples conjecture, any n which is coprime to k will have infinitely many representations of the form q-kp.
Assuming the Elliot-Halberstam conjecture, the work of Goldston-Pintz-Yildirim on prime gaps (which, among other things, shows infinitely many solutions to 0 < q-p <= 16) should also imply the existence of some n with infinitely many representations of the form q-kp for each k (and with a reasonable upper bound on n). [UPDATE, much later: Now that I understand the Goldston-Pintz-Yildirim argument much better, I retract this claim; the GPY argument (combined with the more recent methods of Zhang) would be able to produce infinitely many $m$ such that at least two of $m + h\_i$ and $km + h'\_i$ are prime for some suitably admissible $h\_i$ and $h'\_i$, but this does not quite show that $q-kp$ is bounded for infinitely many $p,q$, because the two primes produced by GPY could both be of the form $m+h\_i$ or both of the form $km+h'\_i$. So it is actually quite an interesting open question as to whether some modification of the GPY+Zhang methods could give a result of this form.]
Unconditionally, I doubt one can say very much with current technology. For any N, one can use the circle method to show that almost all numbers of size o(N) coprime to k have roughly the expected number of representations of the form q-kp with q,p = O(N). However we cannot yet rule out the (very unlikely) possibility that as N increases, the small set of exceptional integers with no representations covers all the small numbers, and eventually grows to encompass all numbers as N goes to infinity.
| 15 | https://mathoverflow.net/users/766 | 10248 | 6,999 |
https://mathoverflow.net/questions/10239 | 52 |
>
> Is it true that, in the category of $\mathbb{Z}$-modules, $\operatorname{Hom}\_{\mathbb{Z}}(\mathbb{Z}[x],\mathbb{Z})\cong\mathbb{Z}[[x]]$ and $\operatorname{Hom}\_{\mathbb{Z}}(\mathbb{Z}[[x]],\mathbb{Z})\cong\mathbb{Z}[x]$?
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The first isomorphism is easy since any such homomorphism assigns an integer to $x^i, \forall\ {i>0}$ which defines a power series. For the second one might think similarly that if $S$ is the set of all power series with non-zero constant term then $\mathbb{Z}[[x]]=0\oplus{S}\oplus{x}S\oplus{x^2}S\dots$, but it doesn't quite work since it is not clear how to map $S$.
| https://mathoverflow.net/users/2533 | Is it true that, as $\Bbb Z$-modules, the polynomial ring and the power series ring over integers are dual to each other? | Yes: this is an old chestnut. Let me write $\oplus\_n\mathbf{Z}$ for what you call $\mathbf{Z}[x]$ and $\prod\_n\mathbf{Z}$ for what you call $\mathbf{Z}[[x]]$ (all products and sums being over the set {$0,1,2,\ldots$}). Clearly the homs from the product to $\mathbf{Z}$ contain the sum; the issue is checking that equality holds. So say I have $f:\prod\_n\mathbf{Z}\to\mathbf{Z}$ and let me prove $f$ is in the sum.
Let $e\_n$ ($n\geq0$) be the $n$th basis element in the product (so, what you called $x^n$).
First I claim that $f(e\_n)=0$ for all $n$ sufficiently large. Let's prove this by contradiction. If it were false then I would have infinitely many $n$ with $f(e\_n)\not=0$, so by throwing away the $e\_n$ such that $f(e\_n)=0$ (this is just for simplicity of notation; otherwise I would have to let this infinite set of $n$ be called $n\_0$, $n\_1\ldots$ and introduce another subscript) we may as well assume that $f(e\_n)=c\_n\not=0$ for *all* $n=0,1,2,\ldots$. Now choose any old integers $d\_i$ such that $\tau:=\sum\_{i\geq0}2^id\_ic\_i$, a 2-adic integer, is not in $\mathbf{Z}$ (this can easily be done: infinitely many $d\_i$ are "the last to change a binary digit of $\tau$" and hence one can recursively rule out all elements of $\mathbf{Z}$), and consider the integer $t:=f(\sum\_{i\geq0}2^id\_ie\_i)\in\mathbf{Z}$. The point is that $\sum\_{i\geq N}2^id\_ie\_i$ is a multiple of $2^N$ in the product, and hence its image under $f$ must be a multiple of $2^N$ in $\mathbf{Z}$. So one checks easily that $t-\tau$ is congruent to zero mod $2^N$ for all $N\geq1$ and hence $t=\tau$, a contradiction.
[Remark: in my first "answer" to this question, I stopped here. Thanks to Qiaochu for pointing out that my answer wasn't yet complete.]
We deduce that $f$ agrees with an element $P$ of the sum on the subgroup $\oplus\_n\mathbf{Z}$ of $\prod\_n\mathbf{Z}$. So now let's consider $f-P$; this is a map from the product to $\mathbf{Z}$ which is zero on the sum, and our job is to show that it is zero. So far I have used the fact that $\mathbf{Z}$ has one prime but now I need to use the fact that it has two. Firstly, any map $(\prod\_n\mathbf{Z})/(\oplus\_n\mathbf{Z})\to\mathbf{Z}$ is clearly going to vanish on the infinitely $p$-divisible elements of the left hand size for any prime $p$ (because there are no infinitely $p$-divisible elements of $\mathbf{Z}$ other than $0$). In particular it will vanish on elements of $\prod\_n\mathbf{Z}$ of the form $(c\_0,c\_1,c\_2,\ldots,c\_n,\ldots)$ with the property that $c\_n$ tends to zero $p$-adically. Call such a sequence a "$p$-adically convergent sequence". But using the Chinese Remainder Theorem it is trivial to check that *every* element of $\prod\_n\mathbf{Z}$ is the difference of a 2-adically convergent sequence and a 3-adically convergent sequence, and so now we are done.
Remark: I might be making a meal of this. My memory of what Kaplansky writes is that he uses the second half of my argument but does something a bit simpler for the first half.
| 52 | https://mathoverflow.net/users/1384 | 10249 | 7,000 |
https://mathoverflow.net/questions/10251 | 10 | If two elliptic curves over $\mathbb{Q}$ are $\mathbb{Q}\_p$-isogneous for almost all primes $p$, then they are $\mathbb{Q}$-isogenous.
This follows from the fact that they have the same number of $\mathbb{F}\_p$-points for almost all $p$, hence their $L$-functions have the same local factors at all these $p$, therefore a combination of "multiplicity one" and Faltings' isogeny theorem implies that they are $\mathbb{Q}$-isogenous. Correct me if I'm wrong.
Here $\mathbb{Q}$ can be replaced by any number field $k$.
Question : Does the same argument work for any two abelian varieties $A$, $B$ over $k$ ? It should, since $H^i$ is $\wedge^i H^1$ for $i>0$.
If so, this explains why Poonen's abelian surfaces $A,B$ [(everywhere locally isomorphic but not isomorphic)](https://mathoverflow.net/questions/10033/everywhere-locally-isomorphic-abelian-varieties) are $\mathbb{Q}$-isogenous.
| https://mathoverflow.net/users/2821 | A local-to-global principle for isogeny | Yes, the local-global principle for isogenies is valid for all abelian varieties over all number fields, as a consequence of Faltings' isogeny theorem [and, as Kevin Buzzard points out, of the semisimplicity of the Galois action on the Tate module, also proved by Faltings.]
The proof for abelian varieties is almost the same as that for elliptic curves. You just need to observe that if A, A' are two g-dimensional abelian varieties over F\_q, then the following are equivalent:
(i) They are $F\_q$-rationally isogenous.
(ii) They have the same characteristic polynomial of Frobenius.
(iii) For all $1 \leq i \leq g, \ |A(F\_{q^i})| = |A'(F\_{q^i})|$.
(iv) The Hasse-Weil zeta functions of A and A' coincide.
Although I have not checked in order to answer this question, I think it is likely that proofs -- or references to proofs -- of this fact can be found in at least one of the papers
---
Waterhouse, William C. Abelian varieties over finite fields. Ann. Sci. École Norm. Sup. (4) 2 1969 521--560.
Waterhouse, W. C.; Milne, J. S.
Abelian varieties over finite fields. 1969 Number Theory Institute (Proc. Sympos. Pure Math., Vol. XX, State Univ. New York, Stony Brook, N.Y., 1969), pp. 53--64. Amer. Math. Soc., Providence, R.I., 1971.
---
| 4 | https://mathoverflow.net/users/1149 | 10253 | 7,003 |
https://mathoverflow.net/questions/10193 | 18 | The classic handshake puzzle goes something like this:
* "Given that everyone has a different skin disease, how can you safely shake hands with 3 people when you have only 2 gloves?"
Its common variations are:
* "How can a man engage in safe sex with 3 women using 2 condoms?"
* "How can a doctor operate on 3 patients with only 2 gloves while avoiding skin-blood contact between any two people"
Let's say N is the number of other people (patients/women...etc) and K is the number of gloves (or condoms). The above case of N=3 and K=2 is not hard (and its solution readily available on the net).
**QUESTION 1:** In general, what can we say about the feasible N's and K's? It seems like (2K >= N+1) is a necessary condition (K gloves has 2K sides and there are a total of N+1 people involved). Is this also sufficient?
While researching on Google, I came across a posting that claimed the generalization of this similar puzzle is an open problem:
* "Two couples get together for an evening of hetero swinging. What is the minimum number of condoms necessary for safe sex in all of the male-female pairings?"
<http://mathematicsontribe.tribe.net/thread/d0f5c284-762d-4045-be74-21e6ede7e31e>
**QUESTION 2:** I assume the general form of the question would study the feasibility of N couples and K condoms. What is known about the general problem? Is it still open?
---
(Qiaochu Yuan:) Based on the downvotes, which I would guess are directed at the way in which the problem is stated rather than its content, here is a "cleaned up" version appropriate for mathematicians:
You have a collection of $K$ tokens which have two sides, each of which can be marked. There are two families of marks, $N$ of which are of the first family and $M$ of which are of the second family. For each pair $(i, j)$ of a mark of the first family and the second family, attempt the following:
* Stack a collection of tokens from left to right. (Tokens may be rotated.)
* If two tokens are adjacent, the adjacent sides share marks.
* Mark the left side of the leftmost token with mark $i$ and the right side of the rightmost token with mark $j$. This move is only possible if each of the sides to be marked is either initially unmarked or is marked **only** with the mark you are trying to mark it with and with no other marks.
For which values of $K, N, M$ is this possible?
| https://mathoverflow.net/users/2910 | Generalization of the shakehands/condom puzzle? | This problem is well-known as "glove problem" or, indeed, "condom problem". It was almost solved by Hajnal and Lovasz in 1978, with final touches put by Vardi in 1991.
<http://mathworld.wolfram.com/GloveProblem.html>
<http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/condoms-n-m>
| 14 | https://mathoverflow.net/users/2795 | 10256 | 7,005 |
https://mathoverflow.net/questions/10267 | 6 | let $M$ be the set of natural numbers such that there is a group of this order, which is not solvable. what is the minimal distance $D$ of two numbers in $M$?
the examples $660$ and $672$ show $D \leq 12$. the famous theorem of feit-thompson implies $D>1$.
| https://mathoverflow.net/users/2841 | distribution of non-solvable group orders | By the Euclidean algorithm, the answer is the gcd of all orders of all non-abelian finite simple groups. I believe that this is 4 (looking at the groups listed in Wikipedia, one can see that it is at most 4 since once can get down to 12 on the tables of low order groups, and the Suzuki groups have order not divisible by 3). My recollection is that a finite simple group actually cannot have cyclic 2-Sylow, and thus must have order divisible by 4.
| 13 | https://mathoverflow.net/users/66 | 10269 | 7,013 |
https://mathoverflow.net/questions/10268 | 5 | Let $p$ be a prime number. Let $n,m \geq 1$ be such that the topological spaces $\mathbb{Q}\_p^n$ and $\mathbb{Q}\_p^m$ are homeomorphic. Can we conclude $n=m$?
For $\mathbb{Z}\_p$ it's false: In fact, Brouwer's theorem implies that $\mathbb{Z}\_p$ is homeomorphic to the Cantor set $C$, which of course satisfies $C^n \cong C^m$ for all $n,m$.
| https://mathoverflow.net/users/2841 | p-adic noninvariance of dimension | $\mathbb{Q}\_p$ is homeomorphic to a countable direct sum of copies of the Cantor set $C$. Indeed, because the valuation is discrete, for each $n \geq 1$ the "annulus"
$A\_n =$ {$x \in \mathbb{Q}\_p \ | \ p^{n-1} < ||x|| \leq p^{n}$}
is closed and homeomorphic to the Cantor set $C$. (Take of course $A\_0 = \mathbb{Z}\_p$.)
Since as you observed above, $C \times C \cong C$, it follows that $\mathbb{Q}\_p^n$ and
$\mathbb{Q}\_p^m$ are homeomorphic for all $m, \ n \in \mathbb{Z}^+$ (and the homeomorphism type is independent of $p$).
| 10 | https://mathoverflow.net/users/1149 | 10273 | 7,015 |
https://mathoverflow.net/questions/10272 | 2 | Can anyone give me a reference which explain the derivation of the partial differential operator expression for the laplacian on the euclidean n-dimensional space and on $S^n$ ?
One generally writes the laplacian on the n-dim euclidean space as a sum of a operator on the radial coordinate and $\frac{1}{r^2}$ times the laplacian on $S^n$.
And very often the laplacian on $S^n$ is written through a recursion relation.
I am looking for a reference which shows me the derivations of these.
| https://mathoverflow.net/users/2678 | Looking for a reference for the laplacian operator | The Laplacian can be defined on any Riemannian manifold as div grad. Here grad f for f a smooth function is the vector field dual to the 1-form df via the bilinear form of the metric. Div of a vector field X corresponds to taking the covariant derivative $\nabla X$, which is a (1,1) tensor, and taking the trace of that. In local coordinates one can give a formula using the symbols for the metric, which should yield what you are looking for.
Another way to define div is to take the Lie derivative of the volume form: that is, $L\_X V = (div X) V$. The volume form depends on an orientation, which can be locally chosen. This way is actually probably easier for computing in local coordinates since you don't need to worry about a covariant derivative or Christoffel symbols.
For a reference, see e.g. Taylor's *Partial Differential Equations,* Vol. 1. In Folland's *Introduction to Partial Differential Equations,* there isn't much about Riemannian manifolds, but Folland does talk about how the Laplacian changes with respect to new coordinates.
| 3 | https://mathoverflow.net/users/344 | 10274 | 7,016 |
https://mathoverflow.net/questions/10236 | 5 | So I have this manifold $M$, along with a metric $g\_{\mu\nu}(x)$ and metric-compatible covariant derivative $\nabla\_\mu$ (which is not necessarily the one corresponding to the Levi--Civita connection).
When dealing with the action principle, we can ignore boundary terms. My question is, which of the following is a boundary term? Let $g(x) = \mathrm{det}(g\_{\mu\nu}(x))$.
$$\int\_M \partial\_\mu A^\mu \sqrt{|g(x)|} \, \mathrm{d} x$$
or
$$\int\_M \nabla\_\mu A^\mu \sqrt{|g(x)|} \, \mathrm{d} x \ ?$$
(Or both, or neither? Maybe the latter, but only when $\nabla$ is Levi--Civita?)
I guess my question reduces to, how does Stokes's theorem work for general covariant derivatives, and with the natural volume element $\sqrt{|g(x)|} \, \mathrm{d} x$?
| https://mathoverflow.net/users/2918 | Help me understand boundary terms in actions over nontrivial manifolds | The second expression should be correct. The Stokes theorem per se does not "know" about covariant derivatives. However, the differential forms have certain transformation properties under the changes of local coordinates. To get the boundary term, you need an exact $n$-form under the integral sign, and $\left(\partial\_\mu A^\mu\right) \sqrt{|g(x)|}$ just does not transform the right way (assuming that $A\_\mu$ are components of a vector field), so in the first case the expression under the integral sign can't be an exact $n$-form while in the second case it is if $\nabla$ is the Levi-Civita connection for $g$, and that's that.
Namely, in the *second* case the integral can (up to an inessential constant factor) be rewritten as $$\int\_M \partial\_\mu \left(\sqrt{|g(x)|} A^\mu\right) \mathrm{d} x,$$ and you can use the Stokes theorem.
**Important warning:** If $\nabla$ is a completely generic connection rather than the Levi-Civita connection, our $n$-form is not exact, and the argument fails because the Stokes theorem does not apply anymore.
Now, if $\nabla$ is compatible with $g$ and ${}^g\nabla$ is the Levi-Civita connection for $g$, then $\nabla g$=0 and it can be shown that there exists a (1,2)-tensor field $T$ such that $$\nabla={}^g\nabla+T.$$.
In particular, we have $$\nabla\_\mu A^\mu={}^g\nabla\_\mu A^\mu+T\_{\rho\nu}^\nu A^\rho.$$
While ${}^g\nabla\_\mu A^\mu$ is proportional to $\partial\_\mu \left(\sqrt{|g(x)|} A^\mu\right)$, and we can apply the Stokes theorem to see that the contribution of this term to the integral vanishes, we get
$$\int\_M \nabla\_\mu A^\mu \sqrt{|g(x)|} \mathrm{d} x=\int\_M T\_{\rho\nu}^\nu A^\rho \sqrt{|g(x)|} \mathrm{d} x.$$
However, the algebraic conditions on $T$ that follow from compatibility of $\nabla$ with the metric $g$ appear to yield $T\_{\rho\nu}^\nu=0$ (have no time to write this out in detail, sorry), so the above integral vanishes and the above arguments work even if $\nabla$ is just compatible with $g$. Apologies for not pointing this out in the earlier version of my answer.
| 8 | https://mathoverflow.net/users/2149 | 10275 | 7,017 |
https://mathoverflow.net/questions/10271 | 8 | According to a result of Higman and Sims (which I learned about in [this paper](http://arxiv.org/abs/math/0608491) of Poonen's) the typical p-group is 3-step nilpotent of a particular form. In particular the typical group is a 3-step nilpotent 2-group of a particular form. By typical here I mean that eventually the number of these groups dominate.
Is anything known about what the typical non-solvable group looks like? Probably some sort of modification of PSL\_2(F\_p)?
| https://mathoverflow.net/users/22 | What does the typical non-solvable group look like? | It seems like, if solvable groups dominate everything else, then groups with a single $A\_5$ factor and all other factors cyclic should likewise dominate every other type of nonsolvable group.
| 8 | https://mathoverflow.net/users/297 | 10283 | 7,023 |
https://mathoverflow.net/questions/10290 | 20 | Can a topos ever be a nontrivial abelian category? If not, where does the contradiction lie? If a topos can be an abelian category, can you give a (notrivial!) example?
| https://mathoverflow.net/users/1353 | Can a topos ever be an abelian category? | No. In fact no nontrivial cartesian closed category can have a zero object 0 (one which is both initial and final), as then for any X, 0 = 0 × X = X. (The first equality uses the fact that – × X commutes with colimits and in particular the empty colimit, and the second holds because 0 is also the final object.)
| 46 | https://mathoverflow.net/users/126667 | 10291 | 7,029 |
https://mathoverflow.net/questions/10293 | 3 | This is a follow-up to [this question](https://mathoverflow.net/questions/10290/can-a-topos-ever-be-an-abelian-category). Since an abelian category cannot be cartesian closed, clearly the hom functor is not right adjoint to the product (by an object). However, does the product (by an object) admit a right adjoint for some objects? If so, what is it? Does it exist in general? Does it only hold for finite products?
| https://mathoverflow.net/users/1353 | Does the product (by an object) in an abelian category ever have a right adjoint? | In an additive category the functor F(–) = – × A = – ⊕ A cannot have a right adjoint unless A = 0. If F had a right adjoint then it would preserve coproducts and in particular A = F(0) = F(0 ⊕ 0) = F(0) ⊕ F(0) = A ⊕ A via the fold map. This means Hom(A, K) = Hom(A, K) × Hom(A, K) for every K, but Hom(A, K) is nonempty (we have zero maps) so Hom(A, K) = • and thus A = 0 by Yoneda.
| 8 | https://mathoverflow.net/users/126667 | 10294 | 7,031 |
https://mathoverflow.net/questions/10279 | 7 | A very interesting [Robertson-Seymour (graphs minors) theorem](http://en.wikipedia.org/wiki/Robertson-Seymour_theorem) says:
>
> Any infinite collection of graphs $C$ with the property that if $G\in C $ then its minors also are has the form $\{$graphs $G$ that don't contain any $E\_i\}$ for some *finite* collection $E = \{E\_i\}$.
>
>
>
So, the theorem says that you could create a list of forbidden minors to find out if the graph is torically embeddable, but this doesn't help much, since the list is both not fully known and large.
I wonder whether the above difficulty is because
1. it is indeed hard to test this property of a graph
2. the theorem does turn easily testable properties into long lists
3. it is not known how to effectively turn easily testable properties into lists
Here's the formal question:
>
> Consider a polynomial algorithm $P$ that returns a yes/no question given a graph as an input and which always returns yes for minors of any graph for which it returns yes. There exists $E$, the exceptional list of a collection defined by $P$. What is known about the computability of the map $P\mapsto E$?
>
>
>
| https://mathoverflow.net/users/65 | How unhelpful is graph minors theorem? | For a reference concerning this problem, see [Cattell et al, "On computing graph minor obstruction sets", Theor. Comput. Sci. 2000](http://dx.doi.org/10.1016/S0304-3975%2897%2900300-9).
I think the answer to your specific question is that it's recursively enumerable (one can test all graphs using P to see whether they belong to E) but not recursive (without further information there is no way of knowing that one has found all obstructions).
Here's a specific construction that shows this: given an instance h of the halting problem, construct an algorithm A that either recognizes all graphs if h is a non-halting instance, or that recognizes the graphs with no K\_s minor if h halts in s steps. It's not hard to do this in such a way that A is always a polynomial time algorithm, but one can't tell whether E is empty or non-empty without solving the halting problem.
| 8 | https://mathoverflow.net/users/440 | 10302 | 7,036 |
https://mathoverflow.net/questions/10301 | 3 | What is the number $N^d\_k$ of real-valued parameters that are needed to specify a k-dimensional subspace of $\mathbb{R}^d$? And how can these parameters be interpreted?
---
I know: $N^d\_1 = N^d\_{n-1} = d - 1 = \binom{d}{1} - 1$.
The parameters can be interpreted as the d components of a vector spanning the 1-dimensional subspace minus its (arbitrary) length.
I know: $N\_1^3 = N^3\_2 = 2 = \binom{3}{2} - 1$.
The parameters can be interpreted as two angles or as the three components of a normal vector of the 2-dimensional subspace minus its (arbitrary) length.
I know: $N^d\_2 = N^d\_{d-2} = \binom{d}{2} - 1$
I believe this, because a d-dimensional rotation has $\binom{d}{2}$ degrees of freedom, one for the rotation angle, the remaining $\binom{d}{2} -1$ ones for the (d-2)-dimensional (hyper)plane of rotation which also defines a 2-dimensional hyperplane as its orthogonal complement.
Question: How do I know that $\binom{d}{2}$ is the number of degrees of freedom of a d-dimensional rotation?
How can these $\binom{d}{2}$ parameters of a rotation or the $\binom{d}{2} - 1$ parameters of a 2-dimensional hyperplane be interpreted (maybe even intuitively)?
---
I guess that $N^d\_k$, the number of parameters that are needed to specify a k-dimensional subspace of $\mathbb{R}^d$, is given by $\binom{d}{k} -1$. How can this be shown? Only formally by mathematical induction or more directly, using e.g. the observation, that there are $\binom{d}{k}$ k-dimensional subspaces of $\mathbb{R}^d$ spanned by k of d elements of an orthonormal basis of $\mathbb{R}^d$?
| https://mathoverflow.net/users/2672 | How many parameters are needed to specify a k-dimensional subspace of R^d? | This answer is similar to what Ben said about $d \times k$-matrices, but maybe a little more visual. Instead of computing the dimension "as a whole", I'll just compute the dimension of subspaces neighboring a given one.
Take some fixed $k$-dimensional subspace $P$ with complementary space $P^\perp$ of dimension $n-k$. Any sufficiently nearby subspace $P'$ to $P$ looks like a graph of a linear function $A : P \to P^\perp$. And on the other hand, any such linear function defines a unique subspace. So you only need to count linear maps from $P$ to $P^\perp$, which are $k(n-k)$-dimensional.
---
Thinking about the issue locally also helps avoid a mistaken dimension count like $n \choose k$. The problem with the $n \choose k$ count is that you are enumerating some random points on the Grassmannian, which doesn't tell you anything about the dimension. For example, take the case of 2-dimensional subspaces of $\mathbb{R}^3$. A subspace here is determined by its normal vector, so there is a bijection between 2-dimensional (oriented) subspaces and the unit sphere.
The $3 \choose 2$ planes in your count correspond to the north pole and two equatorial points $90^\circ$ apart. But you wouldn't conclude that the sphere is 3-dimensional just because it has three points!
| 7 | https://mathoverflow.net/users/2510 | 10305 | 7,039 |
https://mathoverflow.net/questions/10246 | 15 | There is a model category structure on Set in which the cofibrations are the monomorphisms, the fibrations are maps which are either epimorphisms or have empty domain, and the weak equivalences are the maps $f : X \rightarrow Y$ such that $X$ and $Y$ are both empty or both nonempty.
In order for the lifting axioms to hold we need the axiom of choice. Suppose we want to avoid the axiom of choice. One option seems to be to replace "epimorphism" with "map which has a section" everywhere. Can we instead leave the definition of fibration unchanged and change the definition of cofibration?
Note that if $A$ is a cofibrant set in this hypothetical model structure then any surjection $X \rightarrow A$ has a section. So it would be necessary that every set admits a surjection from a set $A$ of this type, which seems rather implausible to me. Perhaps the notion of model category needs to be modified in a setting without the axiom of choice.
(Apologies if this question turns out to be meaningless or trivial; I have not thought about it much nor do I often try to avoid using the axiom of choice.)
---
Let me explain the motivation behind this question. I am trying to get a better picture of what category theory without the axiom of choice looks like.
My rough understanding is that in the absence of the axiom of choice we should use [anafunctors](http://ncatlab.org/nlab/show/anafunctor) in place of functors in some if not all contexts. An anafunctor from $C$ to $D$ is a span of functors $C \leftarrow E \rightarrow D$ in which the left leg $E \leftarrow C$ is a surjection on objects and fully faithful. The point is that even under these conditions $E \rightarrow C$ may not have a section. For instance, suppose $C$ is a category with all binary products. There may not be a product functor $- \times -: C \times C \rightarrow C$, because we cannot simply choose a distinguished product of each pair of objects. However, if we define $Prod(C)$ to be the category whose objects are diagrams of the shape $\bullet \leftarrow \bullet \rightarrow \bullet$ in $C$ which express the center object as the product of the outer two, there is a forgetful functor $Prod(C) \rightarrow C \times C$ remembering only the outer objects, and it is surjective on objects (because $C$ has binary products) and fully faithful. Furthermore there is another forgetful functor $Prod(C) \rightarrow C$ which remembers only the center object. Together these define a product anafunctor $C \times C \leftarrow Prod(C) \rightarrow C$.
"Classically" (:= under $AC$) the following paragraph holds: There is a model category structure on $Cat$ in which the cofibrations are functors which are injective on objects, fibrations are the functors with the right lifting property w.r.t. the inclusion of an object into the contractible groupoid on two objects, and weak equivalences are equivalences of categories. In particular the acyclic fibrations are the functors which are surjective on objects and fully faithful, exactly the functors we allow as left legs of anafunctors. Since every category is fibrant in this model structure, we can view an anafunctor from $C$ to $D$ as a representative of an element of $RHom(C, D)$, i.e., a functor from a cofibrant replacement for $C$ to $D$. Of course, every category is cofibrant too so for our cofibrant "replacement" we can just take $C$, and we learn that anafunctors from $C$ to $D$ are the same as functors when we consider both up to natural isomorphism (homotopy).
I would like to understand anafunctors as a kind of $RHom$ also without the axiom of choice. But I cannot use the same definition of the model category structure, because the lifting axioms require $AC$. I would like to keep the same acyclic fibrations, since they appear in the definition of anafunctor, and I would like every object to be fibrant. I cannot really imagine what a cofibrant replacement could look like in this model structure, but then I am not accustomed to working without $AC$.
My original question is related to this one via the functor which assigns to a set $S$ the codiscrete category on $S$, and it seems to contain the same kinds of difficulties.
| https://mathoverflow.net/users/126667 | Model category structure on Set without axiom of choice | Indeed, COSHEP (more traditionally called the "presentation axiom" by constructivists) does seem to be what you need in order to get a model structure on Set, or Cat. That's true in a lot of similar cases: cofibrant objects are always "projective" in some sense, and you can't expect to get many projective objects in a category constructed from sets unless you started with enough projective objects in Set itself. So I don't expect that ordinary model category theory will be much good for anything at all if you don't have at least COSHEP. But, thankfully, model categories are not the be-all and end-all of homotopy theory, so we can still formally invert the weak equivalences (functors that are fully faithful and essentially surjective) to obtain anafunctors as a derived hom.
| 7 | https://mathoverflow.net/users/49 | 10307 | 7,040 |
https://mathoverflow.net/questions/10314 | 16 | I have sometimes hard time reading papers that are written in the language of schemes being replaced by the functors they represent (I have especially homotopy scheme theory in mind).
I think the topic is connected to topoi and Grothendieck topologies, but for now I'm looking for something simple, just the working overview of the language of representable functors, what is a scheme, etc.
| https://mathoverflow.net/users/65 | "Every scheme as a sheaf" references? | You can start with [these](http://homepage.sns.it/vistoli/descent.pdf) notes by Vistoli, which talk about that stuff in the direction of doing stacks and descent theory. The other articles in FGA explained might be useful, as they do a lot of moduli space construction (ie, prove that a given functor is representable). Another place to look would be in Eisenbud and Harris's "Geometry of Schemes" where they characterize schemes among sheaves on CommRing, towards the end of the book.
| 7 | https://mathoverflow.net/users/622 | 10316 | 7,042 |
https://mathoverflow.net/questions/10255 | 46 | Eric Mazur has a wonderful [video](https://www.youtube.com/watch?v=WwslBPj8GgI) describing how physics is taught at many universities and his description applies word for word to the way I learned mathematics and the way it is still being taught, i.e. professors lecture to students and sketch some proofs. Suffice it to say I'm not a fan of the current methods and I don't think it would be too far from the truth to say that I do all the actual learning outside the classroom. Has anyone tried anything different and seen any difference in student understanding and comprehension in graduate or undergraduate courses?
Some background motivation: I'm a TA and my current method of doing things is to just write some problems on the board and then go through their solutions. This is fine and it's what the students expect but sometimes I feel guilty because I'm just teaching them problem/solution patterns and reinforcing all the bad stereotypes about what mathematics is instead of showing them the underlying conceptual tapestry and helping them rethink their attitudes toward mathematics. It's kinda like the old saying “Give a man a fish; you have fed him for today. Teach a man to fish; and you have fed him for a lifetime”. So basically I throw a bunch of fish at the students hoping it will feed them for the semester.
| https://mathoverflow.net/users/nan | effective teaching | The topic you touch upon is vast, but I wanted to comment on this phrase: "problem/solution patterns which is very different from showing them the underlying conceptual tapestry".
If for some reason you have to use this format (department restrictions or whatnot) choosing your problems well will simultaneously introduce some of the conceptual tapestry. Rather than introducing a mathematical tool and then the problem that goes with it, you introduce the problem first (just out of range of the student ability) and bring it to the point where things get stuck, where something new is needed to go further. Then the motivation is clear for the new tool.
| 24 | https://mathoverflow.net/users/441 | 10318 | 7,044 |
https://mathoverflow.net/questions/10315 | 8 | The Wikipedia article on (real) [Grassmannians](http://en.wikipedia.org/wiki/Grassmannian#Schubert_cells) gives a simple argument that the Euler characteristic satisfies a recurrence relation $$\chi G\_{n,r} = \chi G\_{n-1,r-1} + (-1)^r \chi G\_{n-1,r}$$. This implies that the euler characteristic is zero if and only if $n$ even and $r$ odd. So $\chi G\_{6,3} = 0$, for example. Basic obstruction theory on manifolds tells us that if $\chi M=0$ then there is a non-vanishing vector field on $M$.
Does anyone have any simple, explicit examples of such vector fields? Let's say $G\_{n,1}$ for $n$ even does not count. Not to say those aren't interesting examples -- I'd love solutions that simple. But it's not immediately clear how they can be generalized.
| https://mathoverflow.net/users/1465 | Formulas for vector fields on Grassmannians? | Identify $\mathbb R^2$ with $\mathbb C$ and consider the $S^1$ action on $\mathbb R^{2n} \simeq \mathbb C^n$ induced by cordinatewise complex multiplication. These of course lead to the trivial examples on $G\_{2n,1}$. For $n$ even and $r$ odd the very same examples do the trick. One has just to observe that these $S^1$ actions have no invariant odd dimensional subspaces, and therefore induce $S^1$-actions without fixed points on $G\_{n,r}$.
| 11 | https://mathoverflow.net/users/605 | 10321 | 7,046 |
https://mathoverflow.net/questions/10335 | 5 | It seems to me that Hilbert modular varieties (forms) are generalization from Q to totally real fields. While Siegel modular varieties (forms) are generalization from 1 dimensional to higher dimensional abelian varieties. But they should both be some kind of Shimura variety (automorphic forms), right?
According to Milne's note of Shimura varieties, Siegel modular varieties are Shimura varieties coming from the Shimura datum (G, X) where G is the symplectic similitude group of a symplectic space (V, \phi). So what is the corresponding Shimura datum for the Hilbert modular variety? Or am I asking a wrong question?
Also, in the definition of Shimura varieties G(Q)\G(A\_f)X/K, why we only consider Q and its adele group? why not general number fields and their adeles? (again, maybe a wrong question)
Thank you.
| https://mathoverflow.net/users/1238 | about Hilbert and Siegel modular varieties (forms) | Both of your questions have the same answer: if $K$ is an arbitrary number field, it is not necessary to consider separately reductive groups over $K$, because if $G\_{/K}$ is a reductive group, then $Res\_{K/\mathbb{Q}} G$ is a reductive group over Q. Here
$Res\_{K/\mathbb{Q}}$ denotes [Weil restriction](http://en.wikipedia.org/wiki/Weil_restriction).
In particular, if K is a totally real field, then the reductive group whose Shimura variety is the corresponding Hilbert modular variety is $Res\_{K/\mathbb{Q}} \operatorname{GL}\_2$.
| 5 | https://mathoverflow.net/users/1149 | 10336 | 7,058 |
https://mathoverflow.net/questions/10334 | 77 | I am a non-mathematician. I'm reading up on set theory. It's fascinating, but I wonder if it's found any 'real-world' applications yet. For instance, in high school when we were learning the properties of *i*, a lot of the kids wondered what it was used for. The teacher responded that it was used to describe the properties of electricity in circuits. So is there a similar practical app of set theory? Something we wouldn't be able to do or build without set theory?
---
Edit: Actually, I'm asking about the practicality of the knowledge of the properties of infinite sets, and their cardinality. I'm reading Peter Suber's [A Crash Course in the Mathematics Of Infinite Sets][1] ([Wayback Machine](https://web.archive.org/web/20110703003113/https://earlham.edu/~peters/writing/infapp.htm)). The properties of infinite sets seem unintuitive, but of course, the proofs show that they are true.
My guess is that whoever came up with the square root of -1 did so many years before it 'escaped' from mathematics and found a practical use. Before then perhaps people thought it was clever, but not necessarily useful or even 'true'. So then, if you need to understand electricity, and you can do it best by using *i*, then even someone who thinks it's silly to have a square root of negative -1 would have to grudgingly admit that there's some 'reality' to it, despite its unintuitiveness, because electricity behaves as if it 'exists'.
Seeing as how there was so much resistance to infinite sets at the beginning, even among mathematicians, I wonder: has the math of infinite sets been 'proven worthwhile' by having a practical application outside of mathematics, so that no one can say it's just some imaginative games?
| https://mathoverflow.net/users/2929 | What practical applications does set theory have? | The purpose of set theory is not practical application in the same way that, for example, Fourier analysis has practical applications. To most mathematicians (i.e. those who are not themselves set theorists), the value of set theory is not in any particular theorem but in the **language** it gives us. Nowadays even computer scientists describe their basic concept - Turing machines - in the language of set theory. This is useful because when you specify an object set-theoretically there is no question what you are talking about and you can unambiguously answer any questions you might have about it. Without precise definitions it is very difficult to do any serious mathematics.
I guess another important point here is that it is hard to appreciate the role of set theory in mathematics without knowing some of the history behind the [crisis of foundations](https://en.wikipedia.org/wiki/Foundations_of_mathematics#Foundational_crisis) in mathematics, but I don't know any particularly good references.
---
Your second question is more specific, so I'll give a more specific answer: to thoroughly understand the mathematics behind, say, modern physics does in fact require (among many other things) that you understand the properties of infinite sets because **topology** has become an important part of this mathematics, and understanding general topology depends heavily on understanding properties of infinite sets. Whether this means that set theory has any bearing on "reality" depends on how much faith you have in topological spaces as a good model for the real world.
As a specific example, the mathematics behind general relativity is called [differential geometry](https://en.wikipedia.org/wiki/Differential_geometry). I think it's fair to say the development of general relativity would have been impossible without the mathematical language to express it. Differential geometry takes place on special kinds of manifolds, which are special kinds of topological spaces. So to understand differential geometry you need to understand at least some topology. And I don't think I need to justify the usefulness of general relativity!
| 60 | https://mathoverflow.net/users/290 | 10337 | 7,059 |
https://mathoverflow.net/questions/10344 | 7 | I am a Computer Science undergraduate who does a lot of other tinkering in his free time. Right now, I'm tinkering with n-spheres. Specifically, I'm looking at the distances between a collection of points on n-sphere surfaces. Euclidean distances are trivial (but in this particular application still interesting). I would like to look at "great-circle" distances between points on an n-sphere, but unfortunately I am not familiar with Riemannian Geometry or anything of the sort.
How can one go about calculating the distance between two points on an n-sphere? Can you make this digestible for an undergraduate student who is unfamiliar with the literature?
| https://mathoverflow.net/users/2930 | The orthodrome of n-spheres. | The two-dimensional formula applies (why?): the great-circle distance is $\cos^{-1}(\vec u\cdot \vec v)$ where $\vec u$ and $\vec v$ are position vectors of the points.
| 9 | https://mathoverflow.net/users/2912 | 10345 | 7,066 |
https://mathoverflow.net/questions/10349 | 11 | Here's something that's been bothering me, and that's come up again for me recently while reading some stuff about Hilbert schemes of points (Nakajima's lectures, specifically):
Let $C$ be an algebraic curve. Define $S^nC$ to be $C\times\ldots\times C/S\_n$, the symmetric power.
Now, over $\mathbb{C}$, I can show that $C$ a complex manifold implies that $S^nC$ is, and that if $X$ is a variety with $S^nX$ smooth, then $X$ is one dimensional, but the argument I have involves looking in analytic open sets and reducing to the case of $\mathbb{C}^n$, and additionally is fairly unhelpful for identifying the total space (ie, that $S^n\mathbb{P}^1\cong\mathbb{P}^n$)
So here's my question: how can we, in a fairly quick and natural way, show that
1. If $C$ is a smooth, 1-dimensional variety over an algebraically closed field $k$,, then $S^nC$ is smooth.
2. If $X$ is a smooth variety over an algebraically closed field $k$, and $S^nX$ is smooth, then $X$ is one dimensional.
Now, I don't want any projectivity hypotheses here, and I'm curious, with the more arithmetically inclined, is this still true over an arbitrary field?
| https://mathoverflow.net/users/622 | Smoothness of Symmetric Powers | Just convince yourself that if $(C,P\_1,\dots,P\_n)$ and $(D,Q\_1,\dots,D\_n)$ are analytically isomorphic -- $C$ at the point $P\_i$, $D$ at the point $Q\_i$ -- then $Sym^n C$ at the point $(P\_1,\dots,P\_n)$ is analytically isomorphic to $Sym^n D$ at the point $(Q\_1,\dots,Q\_n)$. For this, you need to see that completion commutes with taking $S\_n$-invariants. Let me hedge and say that this is clear if $P\_2,\dots,P\_n$ are smooth points.
(When I say "analytically isomorphic", I don't mean working over $\mathbb C$. Instead, as customary, I mean that the completion of the corresponding rings are isomorphic, for example $\widehat{\mathcal O}\_{C,P\_i}\simeq \widehat{\mathcal O}\_{D,Q\_i}$.)
Thus for 1 you can reduce to $D=\mathbb A^1$, and you are done.
1 is true over any field, without restrictions on characteristic. Indeed, the ring corresponding to $(\mathbb A^1)^n/S\_n$ is the ring of invariants $k[x\_1,\dots, x\_n]^{S\_n}$, and it is a polynomial ring on $n$ variables, the elementary symmetric polynomials in $x\_i$. That is true over any ring $k$ (commutative, with identity).
Part 2 is reduced to the case of $\mathbb A^2$ by the same trick, and then it is an explicit computation. I will illustrate the $(\mathbb A^2)^n/S\_n$ case. We need to find the ring of invariants $k[x\_1,y\_1,x\_2,y\_2,\dots,x\_n,y\_n]^{S\_n}$. Now there are $2n$ elementary polynomials in $x\_i$, resp. in $y\_i$, of course. But there is more. For example $x\_1y\_1 + \dots + x\_ny\_n$ is an invariant.
To prove that $X=(\mathbb A^2)^n/S\_n$ is singular at the point $P=(0,\dots,0)$ is equivalent to showing that $\dim T\_{P,X} = \dim m/m^2 > 2n = \dim X$, where $m$ is the maximal ideal in $\mathcal O\_{P,X}$. It is an exercise that the above $2n+1$ polynomials are linearly equivalent in $m/m^2$, which would complete the proof.
| 12 | https://mathoverflow.net/users/1784 | 10351 | 7,069 |
https://mathoverflow.net/questions/10347 | 8 | Let $u$ be a nonconstant real-valued harmonic function defined in the open unit disk $D$. Suppose that $\Gamma\subset D$ is a smooth connected curve such that $u=0$ on $\Gamma$. Is there a universal upper bound for the length of $\Gamma$?
Remark: by the Hayman-Wu theorem, the answer is yes if $u$ is the real part of an injective holomorphic function; in fact, in this case there is a universal upper bound for the length of the entire level set in $D$. For general harmonic functions, level sets can have arbitrarily large length, e.g. $\Re z^n$.
| https://mathoverflow.net/users/2912 | Level set of a harmonic function | It can get arbitrarily ugly. Indeed, approximate $1/z$ by a polynomial $p$ in the domain $K\subset\mathbb D$ whose complement is connected but goes from $0$ to the boundary along a long winding narrow path. Then each connected component of the set $\mbox{Re}p=A$ with large $A$ will have to escape the circle along essentially the same path and there are only finitely many $A$ for which we have branching points in these sets ($p'$ has finitely many roots).
| 12 | https://mathoverflow.net/users/1131 | 10357 | 7,073 |
https://mathoverflow.net/questions/10374 | 2 | I'm studying the proof of the **Riesz Representation Theorem** as it appears in Ch. 6 of Royden's *Real Analysis*. When I looked on the web I noted there are a few different theorems that go by the name "Riesz Representation Theorem" so I'll state the one I'm looking at:
>
> Let F be a bounded linear functional
> on $L^p$, $1 \leq p < \infty$. Then
> there is a function $g \in L^q \ni$,
> $F(f) = \int fg$.
>
>
>
The proof starts by showing that g exists for the characteristic functions $\chi\_s = \chi\_{[0,s]}$. Then we can write any step function as the sum of $\chi\_{s\_{i}}$. So based on an earlier theorem, if f is a bounded measurable function in [0,1] we can write a sequence of step functions, $<\psi\_n>$ that converge to f almost everywhere. This is the sentence that confuses me:
>
> "Since the sequence $<|f- \psi\_n|^p>$
> is uniformly bounded and tends to zero
> almost everywhere, the bounded
> convergence theorem implies that
> $||f-\psi\_n||\_p \rightarrow 0$."
>
>
>
But, when I look at the **bounded convergence theorem**, it would require $\mathop{\lim}\limits\_{n \to \infty} |f-\psi\_n|^p = 0$. Period. Not just almost everywhere, to get $\mathop{\lim}\limits\_{n \to \infty} \int\_{[0,1]}|f-\psi\_n|^p = \int\_{[0,1]} 0 = 0$.
So, that's where I'm stuck. I just don't see how the bounded convergence theorem can work here. (Side question: I also, don't feel I really know what Royden means by "uniformly bounded" is that just saying there is one bound that works for the whole set? How is that different from regular bounded?)
| https://mathoverflow.net/users/2907 | How can we use the bounded convergence theorem in this proof of the Riesz Representation Theorem? | Such questions should be really asked on AoPS rather than here, but, once you've already posted it on MO, I'll answer.
1) The set of zero measure can always be ignored when performing Lebesgue integration, so to say $g\_n\to 0$ everywhere or almost everywhere is practically the same: just drop the measure zero set where the convergence fails and apply the bounded convergence theorem as you know it to the integral over the rest.
2) Yes, "uniformly bounded" means here that there is one bound for all functions simultaneously. In this context there is any difference between saying "uniformly bounded sequence" and "bounded sequence" but there is a clear difference between saying "a sequence of bounded functions" and "a sequence of uniformly bounded functions".
| 2 | https://mathoverflow.net/users/1131 | 10376 | 7,086 |
https://mathoverflow.net/questions/10358 | 26 | Just because a problem is NP-complete doesn't mean it can't be usually solved quickly.
The best example of this is probably the traveling salesman problem, for which extraordinarily large instances have been optimally solved using advanced heuristics, for instance sophisticated variations of [branch-and-bound](http://en.wikipedia.org/wiki/Branch_and_bound). The size of problems that can be solved exactly by these heuristics is mind-blowing, in comparison to the size one would naively predict from the fact that the problem is NP. For instance, a tour of all 25000 cities in Sweden has been solved, as has a VLSI of 85900 points (see [here](http://www.tsp.gatech.edu/pla85900/index.html) for info on both).
Now I have a few questions:
1) There special cases of reasonably small size where these heuristics either cannot find the optimal tour at all, or where they are extremely slow to do so?
2) In the average case (of uniformly distributed points, let's say), is it known whether the time to find the optimal tour using these heuristics is asymptotically exponential n, despite success in solving surprisingly large cases? Or is it asymptotically polynomial, or is such an analysis too difficult to perform?
3) Is it correct to say that the existence of an average-case polynomial, worst-case exponential time algorithm to solve NP problems has no importance for P=NP?
4) What can be said about the structure of problems that allow suprisingly large cases to be solved exactly through heuristic methods versus ones that don't?
| https://mathoverflow.net/users/942 | Solving NP problems in (usually) Polynomial time? | This phenomenon extends beyond the traveling salesman problem, and even beyond NP, for there are even some *undecidable* problems with the feature that most instances can be solved very quickly.
There is an emerging subfield of complexity theory called [generic-case complexity](http://en.wikipedia.org/wiki/Generic-case_complexity), which is concerned with decision problems in the generic case, the problem of solving most or nearly all instances of a given problem. This contrasts with the situtation in the classical complexity theory, where it is in effect the worst-case complexity that drives many complexity classifications. (And even for approximate solutions in NP-hard problems, the worst-case phenomenon is still present.)
Particularly interesting is the **black-hole** phenomenon, the phenomenon by which the difficulty of an infeasible or even undecidable problem is concentrated in a very tiny region, outside of which it is easy. (Here, tiny means tiny with respect to some natural measure, such as asymptotic density.) For example, many of the classical decision problems from combinatorial group theory, such as the word problem and the conjugacy problem are linear time solvable in the generic case. This phenomenon provides a negative answer to analogue of your question 1 for these problems. The fact that the problems are easily solved outside the black hole provides a negative answer to the analogue of question 2. And I think that the fact that these problems are actually undecidable as total problems suggests that this manner of solving almost all cases of a problem will not help us with P vs. NP, in your question 3.
For question 4, let me mention that an extreme version of the black-hole phenomenon is provided even by the classical halting problem. Of course, this is the most famous of undecidable problems. Nevertheless, Alexei Miasnikov and I proved that for one of the standard Turing machine models with a one-way infinite tape, there is an algorithm that solves the halting problem on a set of asymptotic measure one. That is, there is a set A of Turing machine programs, such that (1) almost every program is in A, in the sense of asymptotic density, (2) A is linear time decidable, and (3) the halting problem is linear time decidable for programs in A. This result appears in (J. D. Hamkins and A. Miasnikov, The halting problem is decidable on a set of asymptotic probability one, Notre Dame J. Formal Logic 47, 2006. <http://arxiv.org/abs/math/0504351>). Inside the black hole, the complement of A, of course, the problem is intractible. The proof, unfortunately, does not fully generalize to all the other implementations of Turing machines, since for other models one finds a black hole of some measure intermediate between 0 and 1, rather than measure 0.
| 47 | https://mathoverflow.net/users/1946 | 10379 | 7,088 |
https://mathoverflow.net/questions/10383 | 1 | I did not understand number theory or characteristic p-algebraic geometry at all. I just know a little about frobenius homomorphism between two schemes. On the other hand, when I learned something on triangulated category. I found there was also a definition of "frobenius morphism" the definition is as follows:
There are two categories C and D. f\_*:D--->C, f^* :C--->D is left adjoint to f\_*, we call f\_* is a Frobenious morphism if there exists an auto-equivalence G of C such that composition f^\* G is right adjoint to f\_\*.
First question is:is there any relationship between this two frobenius morphism?
Second question is:does frodenius category play roles in algebraic geometry?
All the comments related to this are welcomed.
| https://mathoverflow.net/users/1851 | Any relationship of frobenius homomorphism and frobenius category? | That notion of Frobenius morphism between categories is a generalization of
[Frobenius algebras](http://en.wikipedia.org/wiki/Frobenius_algebra) (those which have a non-degenerate mulplicative bilinear form) to triangulated categories.
This is quite unrelated to the Frobenius morphism on a scheme. There are lot of things named ater Frobenius!
On the other hand, Frobenius categories show up all the time in geometrical contexts. They provide a nice way to construct triangulated caegories (and the triangulated categories so constrcted are particularly nice: they are ´algebraic´) For example, they are used to construct one of the categories equivalent to the derived category of coherent sheaves on projective space in the canonical example of why derived categories are relevant to geometry! This is explained in the book by Gelfan'd and Manin on homological algebra, if I recall correcty.
| 5 | https://mathoverflow.net/users/1409 | 10384 | 7,091 |
https://mathoverflow.net/questions/10388 | 6 | I'm learning about representable functors from [Vistoli notes](http://homepage.sns.it/vistoli/descent.pdf) thanks to [Charles Siegel's answer](https://mathoverflow.net/questions/10314/every-scheme-as-a-sheaf-references/10316#10316).
I see that any category $\mathcal C$ can be embedded into $\text{Hom}\\,(\mathcal C^{op}, \mathcal Set)$ by means of Yoneda embedding. I wonder if there are examples where the latter category would be interesting in itself, other then for these purposes?
| https://mathoverflow.net/users/65 | Yoneda embedding target | Lots! Categories of that form (when C is small) are often called "presheaf categories". Many interesting categories are presheaf categories, such as simplicial sets, cubical sets, symmetric sets, etc. In particular, any presheaf category is a topos, and many interesting toposes are presheaf categories. The category of G-sets for any discrete group G is another nice example, since G can be regarded as a groupoid, hence as a category. Presheaves on a topological space are also interesting, if only as a means to the construction of sheaves. And *simplicial* presheaves on a category C (which are the same as presheaves on $C\times \Delta$) are sometimes easier to work with (once you put a nice model structure on them) than simplicial sheaves.
Many other interesting categories are full subcategories of some presheaf category; in fact a category is a full subcategory of a presheaf category as soon as it has a small dense subcategory. Thus, in particular, any accessible category is a subcategory of a presheaf category. This includes almost any "algebraic" category, such as groups, rings, fields, Lie algebras, etc.
| 13 | https://mathoverflow.net/users/49 | 10390 | 7,095 |
https://mathoverflow.net/questions/10377 | 0 | Hi everyone
This is my first post...
I do mathematics from home... ie., not attached with any institution...
I have deduced some results...
$\lim \inf\_{n\to\infty} \frac{d\_n}{\log p\_n} = 0$
and, for constants $A,B$
$\lim\_{n\to\infty} \log p\_n - \sum\_{i=1}^{n-1} \frac{d\_i}{p\_{i+1}} = A$
$\lim\_{n\to\infty} \log p\_n - \sum\_{i=1}^{n-1} \frac{d\_i}{p\_{i}} = B$
Where, $p\_n$ is the nth prime... $d\_n = p\_{n+1} - p\_n$
My question is:
Do you think these results are good ?
| https://mathoverflow.net/users/2865 | A result on prime numbers | Disclaimer: I am no specialist in Analytic Number Theory, nor did I read the whole paper under the link. I just looked into the end of the argument, and there is a limit computation (10) there.
From what I know from Analysis, this computation is clearly wrong, not in the sense that the answer is necessarily wrong, but in the sense that the premises do not justify the conclusion. The author attempts to compute the lower limit of the product $$\liminf\_{n\to\infty}\left(\frac{p\_n}{\log p\_n}\log\frac{p\_{n+1}}{p\_n}\right)$$
as the product of the limits. He replaces the second factor with $log(1)=0$ and proceeds to claim that the lower limit of the product is $0$. However, even though the (lower) limit of the second factor may well be $0$, the limit of the first factor is clearly $\infty$, so one cannot compute the lower limit of the product in this way.
| 7 | https://mathoverflow.net/users/2106 | 10391 | 7,096 |
https://mathoverflow.net/questions/10364 | 6 | let $hTop\_\*$ denote the homotopy category of pointed spaces. I believe that it has no pushouts, in general. the reason is that you can't expect the involved homotopies to be compatible. can anyone give an explicit example, with proof? I know that homotopy colimits are related to this, but they don't seem to be categorical colimits, so I don't think that they fit here.
especially I'm interested in the following special case: let $G= \langle X | R \rangle$ a presentation of a group and consider the resulting map $\omega : \vee\_{r \in R} S^1 \to \vee\_{x \in X} S^1$. does the cokernel of $\omega$ exist in $hTop\_\*$? ~~in $Top\_\*$, the cokernel is just~~ consider the 2-dimensional CW-complex $Q$, which is optained from $\vee\_{x \in X} S^1$ via the attaching map $\omega$. now if $f : \vee\_{x \in X} S^1 \to T$ is a pointed map such that $f \omega$ is nullhomotopic, it is easy to see that it extends to a map $\overline{f} : Q \to T$. but I think that we cannot expect that $\overline{f}, \overline{g}$ are homotopic, when $f,g$ are homotopic: the homotopies between $f$ and $g$ don't have to be compatible. can you give an example for that? probably it already works for $\omega : S^1 \to S^1, z \mapsto z^2$, thus $Q = \mathbb{R} P^2$.
anyway, this only would show that $Q$ is not the cokernel in the category $hTop\_\*$. the proof, that the cokernel does not exist at all, will be even more difficult and I don't know how to approach it.
you may also replace the category by $hCW\_\*$ (CW-complexes), $hCG\_\*$ (compacty generated spaces) etc., if it's useful.
| https://mathoverflow.net/users/2841 | categorical homotopy colimits | Your example (the "cokernel" of the multiplication by 2 map) also works.
Consider the diagram $S^1 \leftarrow S^1 \rightarrow D^2$ in the based homotopy category of CW-complexes, where the left-hand map is multiplication by 2. Suppose it had a pushout $X$ in the homotopy category. Then for any $Y$, $[X,Y]$ is isomorphic to the set of 2-torsion elements in $\pi\_1(Y)$.
Taking $Y = S^0$, we find $X$ is connected.
Taking $Y = K(\pi,1)$, we find that $\pi\_1(X)$ must be isomorphic to $\mathbb{Z}/2$. This means that there is a map from ${\mathbb{RP}^2}$ to $X$ inducing an isomorphism on $\pi\_1$, and that there is a map $X \to K(\mathbb{Z}/2,1)$ that also induces an isomorphism on $\pi\_1$.
Net result, we get a composite sequence of maps $\mathbb{RP}^2 \to X \to \mathbb{RP}^\infty \to \mathbb{CP}^\infty$. The final space is simply connected, so the map from $X$ would be nullhomotopic and hence so would the map from $\mathbb{RP}^2$.
However, the composite of the first two maps is an isomorphism on $\pi\_1$, hence on $H\_1$. Looking at induced maps on the second cohomology group $H^2$, we get the sequence of maps:
$$\mathbb{Z}/2 \leftarrow H^2(X) \leftarrow \mathbb{Z}/2 \leftarrow \mathbb{Z}$$
The rightmost map is surjective, the composite of the two leftmost maps is an isomorphism by the universal coefficient theorem, and the composite of the two rightmost maps is supposed to be nullhomotopic and hence zero. Contradiction.
| 13 | https://mathoverflow.net/users/360 | 10399 | 7,102 |
https://mathoverflow.net/questions/10405 | 5 | [Wikipedia](http://en.wikipedia.org/wiki/Special_values_of_L-functions) says:
>
> this circle of ideas is distinct from the Bloch–Kato conjecture of K-theory, extending the Milnor conjecture, a proof of which was announced in 2009
>
>
>
What exactly is the K-theory conjecture of Bloch-Kato and has it been proven?
| https://mathoverflow.net/users/65 | Proof of Bloch-Kato conjecture of K-theory? | [Here](http://www.math.rutgers.edu/~weibel/papers-dir/Bloch-Kato.pdf) are some lectures by Charles Weibel. Early on, they discuss Milnor Conjecture and Bloch-Kato, and they should go through the proof. My understanding is that there were a bunch of people involved in the proof, though a few were a bit reticent to actually write up their parts of it, and so Weibel drew the short straw and is the one writing it up.
EDIT: Adding a bit, [here's](http://www.math.rutgers.edu/~weibel/motivic2006.html) Weibel's 2006 page where he notes the status as of then, and to make sure that this is roughly self-contained, here's the statement:
>
> For an odd prime $\ell$, and a field $k$ containing $1/\ell$, the Milnor K-theory $K^M\_n(k)/\ell$ is isomorphic to the étale cohomology $H^n\_{ét}(k,μ\_\ell^n)$
> of the field $k$ with coefficients in the twists of $μ\_\ell$.
>
>
>
| 11 | https://mathoverflow.net/users/622 | 10407 | 7,104 |
https://mathoverflow.net/questions/10406 | 2 | I was reading about the internal hom functor for simplicial sets, and the construction is very "localized" (nothing to do with localization, just the english word). It seems like there should be a general construction for any presheaf category that would be similar to this. That is, an actual construction, not just the existence of the functor provided by the theorem that every Grothendieck topos is a Lawvere topos and therefore cartesian closed. Does such a construction exist, and if so, can you give a reference?
| https://mathoverflow.net/users/1353 | General construction for internal hom in a presheaf category | The formula for the internal hom between presheaves $F\colon C^{op}\to Set$ and $G\colon C^{op}\to Set$ can be derived from the Yoneda lemma. Given $c\in C$, we know that we must have $G^F(c) \cong Hom(y(c), G^F) \cong Hom(y(c) \times F, G)$ so we can simply define $G^F(c) = Hom(y(c) \times F, G)$, which is evidently a presheaf on $C$. The isomorphism $Hom(H,G^F)\cong Hom(H\times F, G)$ for non-representable $H$ then follows from the fact that every presheaf $H$ is canonically a colimit of representables, and $Hom(-,G^F)$ and $Hom(-\times F,G)$ both preserve colimits (the former by definition of colimits, and the latter by that and since limits and colimits in presheaf categories are computed pointwise and products in $Set$ preserve colimits).
This is Proposition I.6.1 in "Sheaves in geometry and logic."
| 11 | https://mathoverflow.net/users/49 | 10410 | 7,105 |
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