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https://mathoverflow.net/questions/36737 | 15 | By the EGA definition, a morphism of schemes of finite presentation is required to be quasi-separated. As far as I can see, removing this requirement does not prevent from proving the basic properties such as stability of the notion under composition, products, etc. So my question is :
>
> where exactly in proving important theorems involving morphisms of finite presentation is the quasi-separated assumption crucial ?
>
>
>
Note that a morphism of finite type is *not* required to be quasi-separated.
All kinds of examples and counter-examples will be appreciated.
| https://mathoverflow.net/users/17988 | Why does finitely presented imply quasi-separated ? | One of the main interests in finitely presented morphisms comes from the various theorems in EGA IV,8. They show that for many questions about morphisms of schemes and sheaves on them, the condition of finite presentation allows one to reduce to a noetherian situation. For these theorems the assumption of quasi-separatedness is crucial.
Let me quickly try to explain why. The heart of the reduction to the noetherian case are theorems like the following: Let X over Spec A be a finitely presented scheme. Then there is a subring $A\_0$ of $A$ which is a finitely generated $\mathbb{Z}$-algebra (and in particular noetherian) and an $A\_0$-scheme $X\_0$ of finite presentation such that $X$ arises from $X\_0$ via the base-change $A\_0\to A$. If $X$ is affine, this is pretty clear, as $X$ is definied by finitely many equations in an affine space over $A$. In order to pass from the affine case to the general case, it does NOT suffice to know that we can cover $X$ by finitely many affine pieces (which would be the assumption of quasi-compactness), but we also need that the glueing data for the affine pieces are somehow described by a finite number of equations. This is ensured by the assumption that the intersection of two affine pieces is quasi-compact which corresponds precisely to the assumption that $X$ is quasi-separated over A.
I guess that these theorems were the reason for Grothendieck to include this condition in the definition of finitely presented.
| 29 | https://mathoverflow.net/users/2308 | 36741 | 23,636 |
https://mathoverflow.net/questions/36604 | 19 | In the (whimsically written) article [Division by three](http://arxiv.org/abs/math/0605779), Doyle and Conway describe a proof, (apparently) not using Choice, that an isomorphism $A \times 3 \simeq B\times 3$ of sets (where $3$ is a given three-element set) gives an isomorphism $A \simeq B$. The result is easy for well-ordered $A$ and $B$, but clearly assuming this isn't constructive. However, the authors seem (to me) to use excluded middle. Also, I don't know if I'm entirely convinced they avoid all forms of choice - it may be they are relying on some weak form of choice (aside from excluded middle).
Has anyone given this any thought? The article is from the mid-90s despite its ArXiV date, and purports to have discovered a lost proof of Lindenbaum from the 1920s which predates a different proof given by Tarski in 1949.
| https://mathoverflow.net/users/4177 | How constructive is Doyle-Conway's 'Division by three'? | The construction in the paper seems to rely on two non-constructive assumptions:
>
> 1. We can decide whether two elements in a set (involved in the division by 3) are equal.
> 2. A countable subset of $\mathbb{N}$ is infinite or not infinite.
>
>
>
(By "infinite" I mean "contains an infinite sequence of pairwise distinct elements".)
In computability theory the second assumption corresponds to Turing degree $0''$.
To see where the second assumption is used, consider the proof of Lemma 2 on page 26. There one is given an infinite sequence of 0's, 1's and 2's and one must decide whether there are infinitely many 0's in the sequence (and if not, whether there are infinitely many 1's). Let us show that such a decision procedure is equivalent to the second assumption. In one direction, a ternary sequence contains infinitely many 0's iff the set of indices at which the 0's appear is infinite. Conversely, given a countable subset $A \subseteq \mathbb{N}$ enumerated by $e$, consider the sequence $a\_0, a\_1, \ldots$ where
>
> $a\_k = 0$ if $e$ enumerates a new element at step $k$, and
> $a\_k = 1$ otherwise.
>
>
>
The sequence has infinitely many 0's iff $A$ is infinite.
Another kind of a typical construction in the paper requires one to determine whether the orbit of an element $x$ under a bijection is finite or infinite. We can do this using our assumptions as follows. Given $x \in A$ and a bijection $f : A \to A$, construct a sequence $a\_0, a\_1, \ldots$ as follows:
>
> $a\_k = 0$ if there is $m \leq k$ such that $x = f^m(x)$, and
> $a\_k = 1$ otherwise.
>
>
>
Here we assumed that $A$ has decidable equality. The sequence $a\_k$ contains a zero iff it contains infinitely many zeroes (thus we only require a $0'$ oracle to perform the following steps). If it contains a zero, say $ak = 0$ then the orbit $\lbrace x, f(x), f^2(x), \ldots \rbrace$ is finite with at most $k$ elements. If it does not contain a zero then the orbit is infinite because $f^k(x) \neq f^j(x)$ for all $k \neq j$, as otherwise we would have $f^{|k-j|+1}(x) = x$.
Let me just remark that one might be tempted to consider a non-constructive assumption such as:
>
> *A group generated by a single element is either infinite or finite.*
>
>
>
(This would certainly allow us to determine whether orbits of elements under bijections are finite or infinite.) In fact division by 3 follows, but for a rather strange reason, namely that such a principle implies the law of excluded middle. Suppose $p$ is an arbitrary thruth value, and consider the Abelian group $G$ freely generated by the generators $p$ and $\top$ (true). Consider the subgroup of $G$ generated by the element $\top - p$. If it is infinite then there is $n \in \mathbb{N}$ such that $n \top \neq n p$, hence $\top \neq p$ and so $\lnot p$ holds. If the subgroup is finite then there is $n > 0$ such that $n \top = n p$, hence $\top = p$ and $p$ holds. (I hope I got this right, we have to be careful because finite sets need not have definite sizes.)
This hopefully gives us some idea about how non-constructive are the techniques employed in the proof of division by three (essentially it looks like a $0''$ oracle). It seems hard to quantify how non-constructive the theorem itself is. Knowing that I can divide by 3 does not obviously allow me to derive non-constructive consequences. For all I know, some suitably constructivized version of division by 3 might actually be constructively valid.
Can I have a green flag now, please?
| 12 | https://mathoverflow.net/users/1176 | 36746 | 23,639 |
https://mathoverflow.net/questions/36730 | 4 | I would like to write down explicitly the generating cocycles of this second cohomology group, $H^2(Z\_n \times Z\_n, k^\*)$. Here $k$ is an algebraically closed field of characteristic zero and $Z\_n$ is the cyclic group with $n$ elements.
I need to know what resolution to use and how to get the formulas!
| https://mathoverflow.net/users/7670 | Describe the second cohomology group $H^2(Z_n \times Z_n. k^*)$. | I would take the standard cyclic resolution of $G = Z/nZ$:
$$
\dots \stackrel{1-t}\to Z[G] \stackrel{\sum t^i}\to Z[G] \stackrel{1-t}\to Z[G] \to Z \to 0,
$$
where $t$ is the generator of $G$, and then take the tensor square of two such --- this would give a resolution
$$
\dots \to Z[G\_1\times G\_2]^3 \stackrel{d\_2}\to Z[G\_1\times G\_2]^2 \stackrel{d\_1}\to Z[G\_1\times G\_2] \to Z \to 0,
$$
where $G\_1 = G\_2 = Z/nZ$ and the maps are given by
$$
d\_1 = (1-t\_1,1-t\_2),
\qquad
d\_2 = \left(\begin{array}{ccc}
\sum t\_1^i & 1-t\_2 & 0 \cr
0 & 1-t\_1 & \sum t\_2^i
\end{array}\right)
$$
($t\_1$ and $t\_2$ are the generators of $G\_1$ and $G\_2$ respectively).
I think you can use this for the calculations.
| 9 | https://mathoverflow.net/users/4428 | 36750 | 23,643 |
https://mathoverflow.net/questions/36743 | 5 | I would like to solve a series of linear systems Ax=b as quickly as quickly as possible. However, the systems are related. Specifically, each matrix A is given by:
cI + E
where E is a fixed sparse, symmetric positive definite real matrix (unchanged in all the linear systems), I is the identity matrix, and c is a varying complex number.
In other words, I am wondering how to quickly solve a series of complex linear systems which are all identical except for complex perturbations along the diagonal. I should say that the resulting matrices are not necessarily Hermitian, so currently I compute the LU decomposition. This works, but given the large number of rather closely related systems to be solved, I wonder if there is a better way to solve the problem, perhaps by using a more expensive (e.g. QR) decomposition up front.
(Edit for Jiahao: Yes, the bs are all the same.)
(Edit for J. Mangaldan: The matrices are of order n=10^5 ~ 10^6, with about 10 times that many nonzeros.)
Update:
I'd like to thank everyone here for their suggestions. My implementation is ugly, but in the end interpolation was the key to a reasonable (10x) speedup. Since the c are quite close (imagine a small region of the complex plane, small in the sense that the spectrum of the matrix E is much larger) I could get away with computing solutions for a subset of the values of c and interpolating a solution for a given value of c using the precomputed values. It isn't elegant at all but it's something.
| https://mathoverflow.net/users/8105 | solving series of linear systems with diagonal perturbations | You want the resolvent of $E$ (at $z:=-c$). Recall it's an analytic function of $z$ defined on the resolvent set, $\mathbb{C}\setminus\operatorname{spec}(E)$. According to the complexity of the matrix $E$, and with the number and the location of the $c$ you need to consider, it may be worth computing a power series expansion at various centers so as to cover the set $\{c\}$ of the data. For $|z|$ larger than the spectral radius you have of course the Laurent expansion $(z-E)^{-1}=1/z+ E/z^2+ E^2/z^3+\dots$
| 2 | https://mathoverflow.net/users/6101 | 36753 | 23,646 |
https://mathoverflow.net/questions/36701 | 11 | Let $X$ be a finite (CW or simplicial - doesn't matter) complex and consider the space of all compact subspaces of $R^\infty$ which are homotopy equivalent to $X$, topologized say as a subspace of the space of all compact subspaces of $R^\infty$ with the Hausdorff metric. What is known about this space? I vaguely recall that it is related to Waldhausen's $A$-theory - are there any references for that?
Motivation: if one instead looks at the space of all subspaces homeomorphic to or respectively diffeomorphic (if $X$ is a manifold) to $X$ one has a model for $BHomeo(X)$ or $BDiff(X)$ (since the corresponding space of embeddings of $X$ in $R^\infty$ is weakly contractible and this space is the quotient of the space of embeddings by the homeo/diffeo group). I have been having fun lately thinking about these models when $X$ is a finite set of points or a surface. This space of subspaces homotopy equivalent to $X$ is different - in particular I do not see it (directly anyways) as a model for $BHomotopySelfEquiv(X)$ - so I'd like to know more what is known about it.
| https://mathoverflow.net/users/4991 | The space of compact subspaces of $R^\infty$ homotopy equivalent to a given finite complex. | To me, Hausdorff metric is an unaccustomed way of making such a space of spaces. I think I don't trust it because fixing a homotopy type gives you a set that is neither closed nor open in general.
But yes I believe the picture is that some kind of "space of spaces of homotopy type $X$" is closely related to $A(X)$.
Let's start with smooth manifolds, but of codimension zero. For a fixed $n$ and a finite complex $K\subset \mathbb R^n$, let $M\_n(K)$ be the space of smooth compact $n$-manifolds $N\subset \mathbb R^n$ containing $K$ in the interior as a deformation retract. (Let's say, the simplicial set where a $p$-simplex is a suitable thing in $\Delta^p\times \mathbb R^n$ such that the projection to $\Delta^p$ is a smooth fiber bundle.) You can map $M\_n(K)\to M\_{n+1}(K)$ by crossing with $[-1,1]$ (and doing something about corners), and you can consider the (homotopy) colimit over $n$. Using the classification of $h$-cobordisms you can work out that the set of components is the Whitehead group of $K$. The loopspace of one component is the smooth stable pseudoisotopy space of $K$. To get the idea, think of the case when $K$ is a point: the space $M\_n(K)$ is then, after you discard extraneous components corresponding to cases where the boundary is not simply connected -- which were going to go away anyway upon stabilizing over $n$ -- your quotient of {embeddings $D^n\to\mathbb R^n$}~$O(n)$ by {diffeomorphisms $D^n\to D^n$}. It's also a kind of "space of all $h$-cobordisms on $S^{n-1}$, and thus a delooping of the (unstable) pseudoisotopy space of $S^{n-1}$.
When $K$ is more complicated than a point, it's important to distinguish between the space of all blah blah blah containing $K$ as a deformation retract and the space of all blah blah homotopy equivalent to $K$; they differ by the space of homotopy equivalences $K\to K$.
There is a similar story for the piecewise linear or topological case.
The piecewise linear manifold version of this construction can, I believe, be shown to be equivalent to a non-manifold construction more like what you asked about: some kind of "space of compact PL spaces in $\mathbb R^n$ containing a fixed $K$ as deformation retract".
Waldhausen tells us that the stable smooth construction above and the stable PL construction above are respectively (the underlying spaces of spectra which are) the fiber of a map from the suspension spectrum of $K\cup{point}$ to $A(K)$, and the fiber of a map from $A(\*)\wedge (K\cup{point})$ to $A(K)$.
| 4 | https://mathoverflow.net/users/6666 | 36756 | 23,648 |
https://mathoverflow.net/questions/36734 | 10 | Suppose X is an inner product space, with Hilbert space completion H (actually, I'm interested in the *real* scalar case, but I doubt there's any difference). If H is separable, then so is X, and I can find a (countable or finite) orthonormal basis of H inside X. Indeed, start with some countable subset Y of X which is dense in H. Then, by induction, we can move to a linearly independent subset of Y, and then apply Gram–Schmidt, again by induction. The point (to me, anyway) is that at any stage, we never take limits, and so we never leave X.
Now, what happens if H is not assumed separable? I've tried to use a Zorn's Lemma argument, but I keep end up wanting to take limits (or, rather, infinite sums) which gives me an orthonormal basis (in the generalised, non-countable, sense) in H, but I cannot ensure that it's in X. Am I just missing something obvious, or is there a slight technicality here...?
| https://mathoverflow.net/users/406 | Orthonormal basis for non-separable inner-product space | This is Problem 54 in Halmos' "A Hilbert Space Problem Book".
However, I think [this](http://www.angelfire.com/journal/mathematics/innerproduct.pdf) is a concrete counterexample. [Please let me know if not viewable.]
| 5 | https://mathoverflow.net/users/2508 | 36759 | 23,650 |
https://mathoverflow.net/questions/36760 | 1 | Suppose $G\_1,G\_2$ and *A* are abelian groups. Consider the cohomology groups for trivial action:
$$H^2(G\_1 \times G\_2,A), H^2(G\_1,A), H^2(G\_2,A)$$
We have projection maps $G\_1 \times G\_2 \to G\_1$ and $G\_1 \times G\_2 \to G\_2$, and these induce maps in the opposite direction on the cohomology groups. Combining these, we get a map:
$$p:H^2(G\_1,A) \oplus H^2(G\_2,A) \to H^2(G\_1 \times G\_2,A)$$
We can further see that this homomorphism is injective.
We also have natural inclusions $G\_1 \to G\_1 \times G\_2$ and $G\_2 \to G\_1 \times G\_2$ and so we get a map:
$$i:H^2(G\_1 \times G\_2,A) \to H^2(G\_1,A) \oplus H^2(G\_2,A)$$
Moreover, the maps *p* and *i* are one-sided inverses of each other. In particular, the kernel of *i* can be identified with the quotient $H^2(G\_1 \times G\_2,A)/(H^2(G\_1,A) \oplus H^2(G\_2,A))$. This latter quotient can also be computed using the Kunneth formula, according to which (if I'm reading it correctly) the quotient should be $\operatorname{Hom}(G\_1,A) \otimes \operatorname{Hom}(G\_2,A)$.
My question: is there a natural bijection between the kernel of the map *i* described above and what the quotient "should" be by the Kunneth formula. And can this bijection be proved directly?
Note: Concretely, the kernel of i is those cohomology classes that, if you restrict to either the embedded $G\_1$ or the embedded $G\_2$, you get the trivial cohomology class on that piece.
| https://mathoverflow.net/users/3040 | The splitting for the cohomology version of Kunneth formula | The classifying space $B(G\_1\times G\_2)$ is homotopy equivalent to $BG\_1\times BG\_2$. Since the Leray spectral sequence of $B(G\_1\times G\_2)\to BG\_1$ degenerates, there exists a filtration on $H^2(B(G\_1\times G\_2),A)=H^2(G\_1\times G\_2,A)$ where the consecutive quotients are $H^2(G\_1,A), H^1(G\_1,H^1(G\_2,A)), H^0(G\_1,H^2(G\_2,A))$. The first group is just the image of $p^2\_1$. The third group is isomorphic to $H^2(G\_2,A)$ and it lifts to $H^2(G\_1\times G\_2)$ as the image of $p\_2^2$. Here $p\_k^l$ is the map of $H^l$'s induced by the $k$-th projection.
So the kernel of the map $i$ of the posting is isomorphic to $H^1(G\_1,H^1(G\_2,A))$ (and for similar reasons to $H^1(G\_2,H^1(G\_1,A))$). This is isomorphic to $Hom(G\_1,Hom(G\_2,A))$. If $G\_1$ and $G\_2$ are abelian, we get $Hom(G\_1\otimes G\_2,A)$ as per Tom's comment. However this is not necessarily the same as $Hom(G\_1,A)\otimes Hom(G\_2,A)$: take e.g. $G\_1=G\_2=\mathbf{Z},A=\mathbf{Q}/\mathbf{Z}$, then one group is $\mathbf{Q}/\mathbf{Z}$ and the other is 0. If $A$ is a ring, we can take the tensor product of the Hom's over $A$ and we have a natural map $F:Hom(G\_1,A)\otimes\_A Hom(G\_2,A)\to Hom(G\_1\otimes G\_2,A)$ given by $F(f\_1\otimes f\_2)(g\_1\otimes g\_2)=f\_1(g\_1)f\_2(g\_2)$. However, this again is not necessarily an isomorphism: morally, the tensor product of the duals can be smaller that the dual of the tensor product.
upd: the first version contained an incorrect example.
| 3 | https://mathoverflow.net/users/2349 | 36768 | 23,654 |
https://mathoverflow.net/questions/36762 | 3 | how many injective homomorphism between two lie algebra $sl\_2 $and $sp\_6$ up to conjugate by$Sp\_6$ ?
| https://mathoverflow.net/users/3945 | how many injective homomorphism between two lie algebra sl2 and sp6 up to conjugate by Sp6? | As a follow-up to Jim's answer (which came in as I was typing an inferior answer), let me add that the 7 possible embeddings are given in the $C\_3$ entry of Table VI in the paper: *[Classification of semisimple subalgebras of simple Lie algebras](http://www.ams.org/mathscinet-getitem?mr=310139)* by Lorente and Gruber. It's of course based on Dynkin, but they work out the details up to rank 6.
---
**Added**
The defining vectors for the 7 embeddings are given by: (1,0,0), (1,1,0), (1,1,1), (2,2,0), (3,1,0), (3,1,1) and (5,3,1). Recall that the embedding with defining vector $(a,b,c)$ is one for which the Cartan generator $H$ of the $\mathfrak{sl}(2)$ subalgebra is given by $H = a H\_1 + b H\_2 + c H\_3$, where $(H\_i)$ is an orthonormal basis of a Cartan subalgebra of $C\_3$ containing $H$.
| 5 | https://mathoverflow.net/users/394 | 36774 | 23,658 |
https://mathoverflow.net/questions/36766 | 6 | Given an adjunction, we get a monad on one side and a comonad on the other side. What is the connection between their algebra and coalgebra categories? Are they allways equivalent?
The example i have in mind is the starting point of algebraic geometry (or more general: The fundamental theorem of formal concept analysis) ~ the relation between polynomials and points in affine space given by fRx iff f(x)=0 induces an order preserving isomorphism between radical ideals and "algebraic sets".
| https://mathoverflow.net/users/1261 | Adjunctions: Algebras of the induced monad VS. Coalgebras of the induced comonad. | No, they are not generally equivalent. Suppose for example $U: C \to D$ is monadic; this means $U$ has a left adjoint $F: D \to C$ such that the canonical comparison functor $C \to Alg(UF)$ is an equivalence, so that $C$ "is" in effect the category of algebras and $U$ is the forgetful functor. Then you'd be asking whether $Coalg(FU)$ is equivalent to $Alg(UF)$, i.e., whether $C$ is equivalent to the category of $FU$-coalgebras over $C$.
For a simple example where this fails, take $C = Set^G$ (category of sets equipped with an action of a group $G$), and $U: Set^G \to Set$ the forgetful functor. The left adjoint is $G \times -: Set \to Set^G$, and one may check that the category of coalgebras is equivalent to $Set$. But $Set$ and $Set^G$ are not equivalent (e.g., $Set^G$ has lots of indecomposable objects given by transitive actions).
There is a fairly large class where the canonical comparison functor $Coalg(FU) \to Alg(UF)$ is an equivalence, namely when either $U$ is fully faithful or $F$ is fully faithful. This occurs for example when $C$ and $D$ are posets [edit: this obviously includes the algebraic geometry situation you were considering], and for various "completion processes" (such as sheafification, or taking Cauchy completions on the category of metric spaces and contractive maps, among many others).
One might inquire whether or under what circumstances adjunctions $F \dashv U$ lift to adjunctions $Coalg(FU)$ and $Alg(UF)$. This ought to be easy to investigate, but just now my children are rattling around and I'm finding it difficult to concentrate. :-)
Edit, now that my children have quieted down: there seems to be no reason for there to be an adjunction between $Coalg(FU)$ and $Alg(UF)$ in general: one can write down various functors in either direction, but I don't see any adjoint pairs cropping up this way.
One interesting situation where there is an equivalence between a category of coalgebras and a category of algebras is where a monad has a right adjoint. One may then exhibit a canonical comonad structure on this right adjoint, such that its category of coalgebras is canonically equivalent to the category of algebras of the original monad. This comes up in topos theory, for example, where for a category of presheaves $Set^{C^{op}}$, the evident forgetful functor $Set^{C^{op}} \to Set^{Ob(C)}$ is both monadic and comonadic.
| 12 | https://mathoverflow.net/users/2926 | 36775 | 23,659 |
https://mathoverflow.net/questions/36735 | 20 | In Peter J. Cameron's book "Permutation Groups" I found the following quote
>
> It is a slogan of modern enumeration theory that the ability to count a set is closely related to the ability to pick a random element from that set (with all elements equally likely).
>
>
>
Indeed, one can count and sample uniformly from labeled trees, spanning trees, spanning forests, dimer models, young tableaux, plane partitions etc. However one can't do either of these very efficiently with groups, for example. My question is if one can make this into a rigorous statement, perhaps through complexity theory. That is, if I have an algorithm to produce a uniform sample from a set of objects, can I somehow come up with an efficient way to count them or vice-versa?
Does this slogan have a standard name? Are there any references?
| https://mathoverflow.net/users/2384 | Enumeration and random selection | Yes, there is formal way of saying this using complexity theory. I think the statement is something like: For all self-reducible relations, the problems of approximate sampling and approximate counting are equivalent (with polynomial time reductions). More specifically, for such problems, the existence of an FPRAS (fully polynomial-time randomized approximation scheme) implies the existence of an FPAUS (fully polynomial-time almost uniform sampler) and vice versa.
References:
* Jerrum, Valiant, and Vazirani, "Random generation of combinatorial structures from a uniform distribution"
* Sinclair and Jerrum, "Approximate counting, uniform generation and rapidly mixing Markov chains"
Alternately, just search for "approximate counting", "approximate sampling" and "self-reducible" on google, and you'll find lots of lecture notes and presentations explaining the ideas.
| 9 | https://mathoverflow.net/users/8075 | 36776 | 23,660 |
https://mathoverflow.net/questions/36777 | 12 | It occurred to me that a subgroup of the modular group $\Gamma$ is a congruence subgroup iff it contains a subgroup of the form $\Gamma(N)$, while a subgroup of a general topological group is open iff it contains an open subgroup. This suggests making a topology on the modular group $\Gamma$ with the subgroups $\Gamma(N)$ as a basis of open neighborhoods of the origin so that $\Gamma$ becomes a topological group. It would then follow that a subgroup of $\Gamma$ is a congruence subgroup iff it is open.
Furthermore, for any $\gamma \in \Gamma$ not equal to the identity, there exists $N$ such that $\gamma \notin \Gamma(N)$, so this topology is Hausdorff, even totally disconnected.
I was inspired in part by [this thread](https://mathoverflow.net/questions/20929/distinguishing-congruence-subgroups-of-the-modular-group) and looked at [this paper](http://www.ams.org/journals/proc/1996-124-05/S0002-9939-96-03496-X/S0002-9939-96-03496-X.pdf) but could not find anything about this idea.
Has anyone considered this topology? Does it provide insight into the problem of determining whether a group is a congruence subgroup?
| https://mathoverflow.net/users/1355 | Congruence Subgroups as Open Subgroups of the Modular Group Under the Right Topology | To expand Henry Wilton's concise answer, the Congruence Subgroup Problem has a distinguished history including important work by Serre and a number of others (exploiting effectively the congruence topology). See for example:
[MR0272790](https://mathscinet.ams.org/mathscinet-getitem?mr=0272790) (42 #7671) 14.50,
Serre, Jean-Pierre,
Le problème des groupes de congruence pour SL2. (French)
Ann. of Math. (2) 92 1970 489–527.
This sort of topology on a group originates earlier, but the application here is highly original.
ADDED: Like many other journal articles, the one mentioned here by Serre is available in PDF format but only through JSTOR (or other library resource). There is a lot of literature, including my 1980 Springer Lecture Notes 789 *Arithmetic Groups* which cover some of the background as well as an expository account of Matsumoto's thesis.
| 7 | https://mathoverflow.net/users/4231 | 36782 | 23,663 |
https://mathoverflow.net/questions/36653 | 7 | Elon Lindenstrauss explains in his [talk](http://www.msri.org/communications/vmath/VMathVideos/VideoInfo/4043/show_video) at the MSRI in Fall 2008 (the relevant comment is at minute 41 of the video) that the set of large Fourier coefficients of a probability measure $\mu$ on the torus ${\mathbb T}^n$ respects the additive structure. More precisely, he defines
$$A\_{\delta} := \lbrace b \in {\mathbb Z}^n \mid |\hat \mu(b)| \geq \delta \rbrace$$
and says that it is "morally" true that $A\_{\delta} - A\_{\delta} \subset A\_{\delta^2}$. (Here, the difference of two subsets is defined to be the set of all possible differences of elements in the respective subsets.) The precise statement (according to Lindenstrauss) is a consequence of the Balog-Szemeredi-Gowers Lemma.
Can someone provide the precise statement or give some hint how the lemma can be used to obtain bounds on Fourier coefficients?
| https://mathoverflow.net/users/8176 | Additive combinatorics and large Fourier coefficients | I think I figured it out myself. What was meant is that for every finite subset $S$ of $A\_{\delta}$ one has
$$| \lbrace (n,m) \in S \times S \mid n-m \in A\_{\delta^2/2} \rbrace | \geq \frac{\delta^2 |S|^2}{2}.$$
This follows from the proof of the second part of Lemma 4.37 in *Tao and Vu, Additive Combinatorics*. (There, the inequality is stated incorrectly as $\leq$.)
| 2 | https://mathoverflow.net/users/8176 | 36790 | 23,668 |
https://mathoverflow.net/questions/36758 | 23 | What is really the conceptual difference between a calculus and an algebra.
Eg. Is SKI combinator calculus really a calculus?
A friend claims that free variables are fundamental for a calculus, and as such that SKI is not a calculus, but an algebra.
| https://mathoverflow.net/users/8797 | Difference between a 'calculus' and an 'algebra' | In logic, the terminology seems to have been influenced by two factors. The very early development of various deductive systems was done by people who were more philosophers than mathematicians and who seem to have used "calculus" to refer to anything that looked mathematical. Also, that development took place before "algebra" had acquired all of its current meanings.
My impression is that the use of "calculus" in logic is restricted to the meaning of "formal deductive system" --- and usually rather old systems. As for the SKI system of combinators, I would call it a calculus if you're talking about rules of inference. But if you mean the system of all combinators, with the operation of application, generated by S and K (I is redundant), then this is an algebra.
| 15 | https://mathoverflow.net/users/6794 | 36805 | 23,676 |
https://mathoverflow.net/questions/36778 | 3 | I'm curious if anyone knows a reference for the following. It seems like someone must have done this somewhere, but I couldn't find a reference.
Recall that an excellent reduced noetherian ring $R$ is called *weakly normal* if any finite birational map $R \subset S$ with $S$ also reduced such that
1. The induced map on spectra is a bijection.
2. Every induced residue field extension $R\_q/q \subset S\_p/p{ }$ is purely inseparable.
is an isomorphism.
(In other words, there can be no such proper, finite birational extension of $R$ satisfying conditions (i) and (ii) ).
Suppose now that $R$ is weakly normal and local with maximal ideal $\mathfrak{m}$.
**Question** Is the completion of $R$ at the maximal ideal also weakly normal?
For the case of seminormality (replace inseparable in (ii) with isomorphism), this is known, see Greco-Traverso. It may be that one can mimic that argument (although there are a couple possibly small issues that I see).
| https://mathoverflow.net/users/3521 | Is weak normality stable under completion? | Here is a partial solution: modulo a problem of constructing "sufficiently generic" elements in the maximal ideal of a reduced noetherian local ring of dimension > 1 (in a sense made precise at the end in terms of associated primes and vanishing loci, and which might require some care when the residue field is finite), I claim it suffices to handle passage to residually trivial local-\'etale extensions (or what comes to the same thing, as we'll see below, henselization instead of completion). I'm not sure if this is any easier than completion, but it certainly can't be harder! I will use Artin-Popescu approximation (which wasn't known in 1980, the time of the Greco-Traverso paper). Let $R$ be a weakly normal reduced excellent local ring. Its henselization $R^{\rm{h}}$ not only inherits reducedness, but also inherits excellence (EGA IV$\_4$, 18.7.6).
Let's reformulate the definition of weak normality in more convenient terms: the condition is that any finite radiciel surjection $X \rightarrow Y := {\rm{Spec}}(R)$ that is an isomorphism over a dense open is an isomorphism when $X$ is reduced. Now consider a finite radiciel surjection $X\_1 \rightarrow Y\_1 := {\rm{Spec}}(R^{\rm{h}})$ that is an isomorphism over a dense open, with $X\_1$ reduced. Since $R^{\rm{h}}$ is a direct limit of residually trivial local-\'etale neighborhoods of $R$, by "spreading out" techniques from EGA IV$\_3$, sections 8 & 9 there is a residually trivial local-\'etale cover $Y' \rightarrow Y$ and finite radiciel surjection $X' \rightarrow Y'$ that is an isomorphism over a dense open and induces $X\_1 \rightarrow Y\_1$ after base change, with $X'$ reduced. Thus, if all such $Y'$ were weakly normal then $X' \rightarrow Y'$ would be an isomorphism and so $X\_1 \rightarrow Y\_1$ would be too. In other words, if we knew weak normality were inherited by residually trivial local-\'etale extensions then the henselization would be weakly normal. I will now assume we knew such inheritance does hold, so we can replace $R$ with its henselization.
Suppose the completion $\widehat{R}$ (which we know is reduced, and of course excellent) were *not* weakly normal. Then there would be a finite radiciel surjection $T \rightarrow \widehat{Y} := {\rm{Spec}}(\widehat{R})$ that is an isomorphism over a dense open but *not* an isomorphism, with $T$ reduced. Thus, there is some infinitesimal fiber whose structure map to the base is not an isomorphism, say $n$th-order with $n \ge 1$. By A-P approximation for the excellent henselian $Y$, there is a local-smooth cover $Y' \rightarrow Y$, a $Y$-map $s:\widehat{Y} \rightarrow Y'$, and a finite radiciel surjection $T' \rightarrow Y'$ which is an isomorphism over a dense open $U' \subseteq Y'$ such that $T'$ is reduced and the pullback $T'\_{\widehat{Y}}$ coincides with $T$ on $n$th-order fibers. Thus, if we pull back $T'$ along a $Y$-point $s'$ of $Y'$ which approximates $s$ to $n$th order (as we may certainly do, from the local structure theorem for smooth morphisms) then we arrive at a finite radiciel surjection $T'' \rightarrow Y$ which is an isomorphism over the $s'$-pullback of $U'$ and *not* an isomorphism on some infinitesimal special fiber. Since $Y$ is reduced and in fact weakly normal, to get a contradiction it suffices to arrange that ${s'}^{-1}(U')$ is not only dense in $Y$ (i.e., contains all generic points) but also that the corresponding open in $T''$ is schematically dense (so $T''$ inherits the reducedness from that open subscheme).
In other words, we are now faced with a problem having nothing to do with weak normality and everything to do with making a good choice for $s'$ to not lose contact with "dense open" downstairs and "reducedness" upstairs. Since $T' \rightarrow Y'$ is a finite radiciel surjection onto a reduced local ring with $T'$ also reduced, we can find a global function $f$ on $Y'$ that is non-vanishing at all generic points of $Y'$ and $T'$, hence nowhere a zero-divisor in both places. The issue is to find the "slice" $s'$ of the local-smooth (residually trivial) $Y$-scheme $Y'$ so that $s'$ is specified to $n$th order (from the $\widehat{Y}$-point) but $s'$-pullback preserve that property of $f$ being nowhere a zero-divisor both downstairs and upstairs.
If the base is a field there is nothing to do in the original problem, so we can assume the base $Y$ has positive dimension, so the local-smooth $Y'$ has dimension at least 2 (otherwise $Y' = Y$ and again there's nothing to do). In a reduced noetherian local ring with dimension > 1,
If we build up the slice $s'$ one step at a time (going through the "smooth coordinate" specializations into the base $Y$ approximating the given $\widehat{Y}$-point one at a time), standard arguments with associated primes and symmetry of ${\rm{Tor}}^1$ turn this into a task of prime avoidance: since we're allow to tweak things (for approx. purposes) beyond $n$th order in the closed point of the base, we are left having to solve the following general problem: if $A \hookrightarrow B$ is a module-finite radiciel extension of reduced noetherian local rings of dimension at least 2 and with the same total rings of fractions, we need to find $a \in A$ which is a non-zero-divisor in both $A$ and $B$ such that (i) its associated primes in both rings lie outside some specified nowhere-dense closed subset $Z$ of their "common" Spec and (ii) for any other given $a'$ in the maximal ideal of $A$ and chosen generating set $\{a\_1,\dots,a\_n\}$ of an ideal for $Z$ a "generic" linear combination of the $a\_i$ using a fixed set of representatives for the residue field can be added to $a'$ to get an element non-vanishing at the generic points and the associated points for $a$ viewed in each of the two rings. (So you see, the case of finite residue field needs more care; but when you unravel where this is coming from, you may find a better way to express it to avoid that problem.)
| 4 | https://mathoverflow.net/users/3927 | 36820 | 23,684 |
https://mathoverflow.net/questions/36832 | 5 | This may be subjective, but does anyone have any insight into why this is the case? This struck me while considering that it's also the eigth Mersenne prime (2^31-1=2147483647).
>
> I'm now wondering why this might be the case.
>
>
>
**UPDATE:**
It's been pointed out that the relationship doesn't necessarily hold for larger storage classes, e.g., 2^63 - 1 is not prime.
| https://mathoverflow.net/users/8812 | Why is the largest signed 32 bit integer prime? | Why is $3$ prime? I don't really know that there are meaningful answers to these kinds of questions. The best I can think of is some reasons it is not obviously composite, e.g. since $5$ is prime $2^5 - 1 = 31$ is not obviously composite (and it turns out to be prime) hence $2^{31} - 1$ is not obviously composite. This is two applications of the "lemma" that if $p$ is prime then $2^p - 1$ is not obviously composite.
Note that any prime factor of $2^p - 1$ has to be congruent to $1 \bmod p$ by Fermat's little theorem, so it is "easier" for such numbers to be prime.
| 7 | https://mathoverflow.net/users/290 | 36835 | 23,693 |
https://mathoverflow.net/questions/23475 | 8 | This property seems to be used both in the context of differential equations and several kinds of discrete equation systems or automata.
It seems to be related in certain case to the Painlevé Property first discovered for Painlevé equations and their solutions.
I have seen several definitions, notations, criteria, which do not always seem to match each other or to be easily applicable to all context. I would appreciate some clarification about it.
I am mostly interested in the discrete applications but all explanations about the continuous cases are welcome especially if there are links to the discrete setting.
| https://mathoverflow.net/users/5387 | What is exactly the (singularity) confinement property ? | The singularity confinement property refers to a property of discrete integrable systems. I am unaware of this property in the context for continuous systems. I can understand why you might have difficulty in getting a definition, since it is rather oddly defined all too often. Since the paper of Goriely and La Fortune have been referenced I will assume that you are interested in this property for maps rather than partial difference equations.
Let me start with an example that hopefully illustrates what is going on, before generalizing: Let us take a simple second order difference equation, which we right as a sequence, satisfying the recurrence relation:
$x\_{n+1}x\_{n-1} = 1+ x\_n$
If we let $x\_1 = -1$, then $x\_2 = 0$, regardless of what $x\_0$ is. Then $x\_3 = -1$ and $x\_4 = \infty$. Worst of all, you have also lost your initial conditions, namely $x\_0$. Suppose, instead of $x\_1 = -1$, we take $x\_1 = -1+e$, where e is assumed very small. Then
$x(2) = \frac{e}{x\_0}$
$x(3) = \frac{e+x\_0}{(e-1)x\_0}$
$x(4) = \frac{1+x\_0}{e-1}$
Notice that in the limit as $e \to 0$, $x\_2 \to 0$, $x\_3 \to -1$, $x\_4 \to -1 -x\_0$. Somehow, in the limit around a "bad" point, or "singularity", we recover initial conditions, namely $x\_0$. We say that the singularity is confined because, despite a loss of information at $x\_2$ and $x\_3$ in the limit, we can regain the information again at $x\_4$ in the same limit. We say the equation has the singularity confinement property if we can always regain the initial conditions in the limit somewhere along the line. This is the usual, however you may want something more general.
The generalization of the above is a map
$f: (x\_n, x\_{n-1}) \to (x\_{n+1},x\_n) = \left(\frac{x\_n+1}{x\_{n-1}},x\_n\right)$,
however, we generally take a map of $R^n$ and then take the corresponding projective space, $P^n$, then we consider a map
$f: P^n \to P^n$.
This function now describes a discrete dynamical system, via $y\_{n+1} = f(y\_n)$. A singularity in this context is where the function, $f$, fails to be invertible. For example, in the above, we had all initial conditions of the for $(x\_0,-1)$ being sent to $(-1,0)$. However, we say the singularity is confined if some power of $f$ is invertible. Note that in the above example, the third power of the map can be continuously extended to the map $(x\_0,-1) \to (-1,-1-x\_0)$, hence the singularity, $(-1,0)$ is confined. A system has the singularity confinement property if all singularities are confined.
| 9 | https://mathoverflow.net/users/8817 | 36849 | 23,701 |
https://mathoverflow.net/questions/36791 | 18 | The 2x3 and 3x4 chessboard complexes (form a square grid of vertices and make a simplex for any set of vertices no two of which are in the same row or column) are a 6-cycle and a triangulated torus with 24 triangles, respectively. The 4x5 chessboard complex is only a pseudomanifold — each vertex has the 3x4 torus as its link, rather than a spherical link that a proper manifold would have — but if you delete its 20 vertices you get a bona-fide cusped hyperbolic 3-manifold, triangulated by 120 regular ideal tetrahedra. It sort of looks like the kind of manifold you might get as the complement of a 20-component link in Euclidean space. Is it a link complement? And if so, which link is it the complement of?
Edit: here are a couple of general references on chessboard complexes.
Ziegler, G. M. (1994). Shellability of chessboard complexes. [Israel J. Math. 87: 97–110](http://dx.doi.org/10.1007/BF02772986).
Björner, A.; Lovász, L.; Vrecica, S. T.; Zivaljevic, R. T. (1994). Chessboard complexes and matching complexes. [J. London Math. Soc. 49: 25–39](http://web.cs.elte.hu/~lovasz/morepapers/chess.pdf).
They also have important applications to the proof of colored Tverberg theorems in discrete geometry: see, e.g.
Pavle V. M. Blagojević, Benjamin Matschke, Günter M. Ziegler (2009). Optimal bounds for the colored Tverberg problem. [arXiv:0910.4987](http://arxiv.org/abs/0910.4987).
| https://mathoverflow.net/users/440 | Is the 4x5 chessboard complex a link complement? | I met this manifold before. It is a normal cover of the orbifold $\mathbb{H}^3/\mathrm{PSL}(2,\mathbb{Z}[\zeta])$ where $\zeta=e^{\pi i/3}$.
I suspect that it actually is $\mathbb{H}^3/\ker\left(\mathrm{PSL}(2,\mathbb{Z}[\zeta])\rightarrow \mathrm{PSL}(2,\mathbb{Z}[\zeta]/I)\right)$ where $I$ is the ideal $\langle 2+2\zeta\rangle$.
For $I=\langle 2+\zeta\rangle$, the manifold is a link complement, see Ian Agol's paper on the Thurston Congruence Link for a drawing.
For $I=\langle 3\rangle$, I could show that it is a link complement, but failed at constructing the link.
| 19 | https://mathoverflow.net/users/47710 | 36858 | 23,708 |
https://mathoverflow.net/questions/36771 | 18 | Is there an explicit formula expressing the [power sum symmetric polynomials](http://en.wikipedia.org/wiki/Power_sum_symmetric_polynomial)
$$p\_k(x\_1,\ldots,x\_N)=\sum\nolimits\_{i=1}^N x\_i^k = x\_1^k+\cdots+x\_N^k$$
of degree $k$ in $N < k$ variables entirely through the power
sum symmetric polynomials $p\_j(x\_1,\ldots,x\_N)$ of degrees $ j \le N $?
Examples:
$$N=1,\ k=2: \quad p\_2=x^2=x\times x=p\_1^2$$
$$N=2,\ k=3: \quad
p\_3 = x^3 + y^3 = [3(x^2+y^2)(x+y)-(x+y)^3]/2 = (3 p\_2 p\_1-p\_1^3)/2$$
>
> What is the general formula?
>
>
>
I am looking for a formula similar to that for the expansion of the [Schur functions](http://en.wikipedia.org/wiki/Schur_polynomial#Relation_to_representation_theory)
$s\_\lambda$ in terms of the symmetric power sums:
$$ s\_\lambda=\sum\_{\rho=(1^{r\_1},2^{r\_2},3^{r\_3},\dots)}\chi^\lambda\_\rho \prod\_j \frac{p^{r\_j}\_j}{r\_j!},$$
where the coefficients $\chi^\lambda\_\rho$ are the characters of the representation of the symmetric group indexed by the partition $\lambda$ evaluated at elements of cycle type indexed by the partition $\rho=(1^{r\_1},2^{r\_2},3^{r\_3},\dots)$, which contains
$ r\_j $
parts of length $j$.
Clearly, the power sums of degree higher than $N$ can be expanded in a similar manner:
$$
p\_k=\sum\_{\rho}a\_{k;\rho}\prod\_{j=1}^N p\_j^{r\_j},
$$
where $\rho=(1^{r\_1},2^{r\_2},\dots,N^{r\_N})$ is the partition of $k$ such that
$k=r\_1+2r\_2+3r\_3+...+Nr\_N$.
In the above example for $N=2,\ k=3$ one has $a\_{3;\ (1^{1},2^{1}) }=3/2$ and
$a\_{3;\ (1^{3},2^{0})}=-1/2$.
>
> My question can be thus reformulated as follows:
> given $r\_1,...,r\_N$ what is the
> explicit formula for $a\_{k;\rho}$?
>
>
>
---
**Note Added**
Actually, [Wikipedia](http://en.wikipedia.org/wiki/Newton%27s_identities#Expressions_as_determinants) tells us how to construct a certain explicit formula for $p\_k$.
It gives the following expressions for $p\_n$ with $n=N$ in terms of $ e\_j, $
$$
p\_n =
\begin{vmatrix}
e\_1 & 1 & 0 & \cdots & \\\
2e\_2 & e\_1 & 1 & 0 & \cdots & \\\
3e\_3 & e\_2 & e\_1 & 1 & \cdots & \\\
\vdots &&& \ddots & \ddots & \\\
ne\_n & e\_{n-1} & \cdots & & e\_1 &
\end{vmatrix},
$$
and for $e\_n$ with $n=N$ in terms of $ p\_j, $
$$
e\_n=\frac1{n!}
\begin{vmatrix}p\_1 & 1 & 0 & \cdots\\\ p\_2 & p\_1 & 2 & 0 & \cdots \\\ \vdots&& \ddots & \ddots \\\ p\_{n-1} & p\_{n-2} & \cdots & p\_1 & n-1 \\\ p\_n & p\_{n-1} & \cdots & p\_2 & p\_1
\end{vmatrix}.
$$
As far as I can see from the derivation described in Wikipedia, these determinant expressions are also valid for $p\_n$ with $ n > N $ and for $e\_n$ with $ n < N $.
For $p\_n$ with $n>N$ one should take into account that all $ e\_k=0 $ for $ k > N $, so that the resulting matrix has zeros in both right-upper and left-lower corners.
Substituting the determinants for $e\_j$ into the determinant for $p\_k$, one gets the
explicit formula which seems to solve the problem.
However, I still don't know how to obtain the coefficients $a\_{k;\rho}$ in the expansion of $ p\_k $ in terms of the first $N$ power sums which would be the desired (really explicit) formula.
| https://mathoverflow.net/users/8802 | Expressing power sum symmetric polynomials in terms of lower degree power sums | Assuming you have $n$ variables then for $k\geq n$, Robin Chapman's identity above can be written as
$$(p\_n,p\_{n-1},\dots, p\_1)\begin{pmatrix}
e\_1 & 1 & \cdots & 0 \\\
-e\_2 & 0 & \ddots & \vdots \\\
\vdots & \vdots & \ddots & 1 \\\
(-1)^{n-1}e\_n & 0 & \cdots & 0
\end{pmatrix}^{k-n}=(p\_k,p\_{k-1},\dots, p\_{k-n+1})$$
Now to finish the job you need to express the $e\_i$'s in terms of the power sum symmetric functions too. This is given by $$e\_n=\sum\_{|\lambda|=n}(-1)^{|\lambda|-l(\lambda)} z\_{\lambda}^{-1}p\_{\lambda}$$ where $|\lambda|$ is the size of the partition $\lambda$ and $l(\lambda)$ is its length, $p\_{\lambda}=p\_{\lambda\_1}p\_{\lambda\_2}\cdots$ and
$$z\_{\lambda}=\prod\_{i\geq 1}\left(i^{m\_i}\cdot m\_i!\right)$$
where $m\_i$ is the number of parts of $\lambda$ equal to $i$.
---
I thought I'd remark that the formulas you are quoting are all valid in $\Lambda\_{\mathbb{Q}}$, the ring of symmetric functions in *infinitely* many variables while the one you are searching for is not, because the $p\_\lambda$'s are an orthogonal basis in this ring with $\langle p\_{\lambda},p\_{\mu}\rangle =\delta\_{\lambda \mu}z\_{\lambda}$. This is also the same reason why the formula for Schur polynomials may contain arbitrary $p\_{\lambda}$'s in it. In fact the reason why that formula is important is because it gives you the transition from the basis of Schur polynomials to that of power sum polynomials in $\Lambda\_{\mathbb{Q}}$.
| 6 | https://mathoverflow.net/users/2384 | 36873 | 23,718 |
https://mathoverflow.net/questions/36859 | 3 | Hello, do you know more about, or some exposition of [Morava's talk](https://www.ams.org/amsmtgs/2110_abstracts/1046-55-2092.pdf "pdf")?
| https://mathoverflow.net/users/451 | Morava's "Motives and cell bundles"? | Google knows about this preprint ("A theory of base motives") which seems related:
<http://folk.uio.no/rognes/yff/morava.pdf>
| 5 | https://mathoverflow.net/users/8824 | 36874 | 23,719 |
https://mathoverflow.net/questions/36851 | 1 | [ERNIE](http://en.wikipedia.org/wiki/Premium_Bond#ERNIEBlockquote) is a hardware random number generator used to generate winning Premium Bond numbers in the UK. Wikipedia says: "ERNIE's output is independently tested each month by an independent actuary appointed by the government, and the draw is only valid if the output passes tests that indicate it is statistically random." But doesn't it mean that, if ERNIE output was truly and fully random (i.e. generated from a unitary distribution over a set of integers from 1 to N), rejecting some of the draws would actually distort the randomness?
| https://mathoverflow.net/users/1580 | Is ERNIE output skewed by statistical tests? | No.
The final 'it' in the quoted section refers to the word 'output' *not* to the word 'draw'. Thus the "independent actuary" tests the output of the machine to check that it is working properly. If so, the draw is made. They do *not* test the actual draw - that would be pointless!
Note: I know nothing about ERNIE or the process itself. I am inferring this from the wikipedia page together with a basic assumption - which may well be unfounded - that the British government couldn't be *that* stupid. Combining this with the fact that *if* the sentence is grammatically correct - perhaps another unfounded assumption - then the 'it' refers, as I said, to the 'output' not to the 'draw', I conclude that the 'output' and the 'draw' are two separate things.
| 2 | https://mathoverflow.net/users/45 | 36880 | 23,725 |
https://mathoverflow.net/questions/36887 | 5 | I'm interested in conditions on a metric space $X$ which imply that boundedness is equivalent to total boundedness (or, assuming that $X$ is complete, that compactness is equivalent to precompactness). If $X$ is a normed space, then we know that this is true if and only if $X$ is finite-dimensional. Is there some suitable concept of dimension for general metric spaces such that boundedness implies total boundedness if and only if the space has finite dimension in this sense? You may assume any reasonable hypotheses on $X$ (e.g. that $X$ is a Polish metric space).
| https://mathoverflow.net/users/7392 | Conditions on a metric space so that boundedness implies total boundedness | If $X$ is locally compact, then it has this Heine-Borel-property. For topological vector spaces locally compactness is equivalent to finite dimension if I remember correctly.
But there are other examples even vector spaces that have the Heine-Borel-property without being locally compact. The space $H(U)$ of holomorphic functions on an open set $U\subseteq\mathbb{C}$ with the topology of locally uniform convergence of all derivatives. This is Montel's theorem and therefore such spaces are called Montel spaces (well a certain additional condition is needed, but that's not the point) Another example is the Schwartz-Space $\mathcal{S}(\mathbb{R}^n)$ of rapidly decreasing functions. Because being Montel is stable under taking strong duals, the space of tempered distributions $\mathcal{S}'(\mathbb{R}^n)$ has the Heine-Borel-property too (but is not metrizable).
| 5 | https://mathoverflow.net/users/3041 | 36892 | 23,732 |
https://mathoverflow.net/questions/36897 | 10 | Suppose there is a function $f(x)$ which is the "probability" that the integer $x$ is prime. The integer $x$ is prime with probability $f(x)$, and then divides the larger integers with probability $1/x$; so as $x$ changes from $x$ to $x+1$, $f(x)$ changes to (roughly)
$$f(x)\left(1-f(x)/{x} \right).$$
How do I show that? I can go on to show
$$\frac{df}{dx} + \frac{f^2}{x}=0$$
and thus $\frac{1}{\log{x}} + c$ is solution but I can't show that step on how $f(x)$ changes. please advise
| https://mathoverflow.net/users/8826 | Probabilistic interpretation of prime number theorem | First of all, I assume you understand that this is meant to be a nonrigorous argument, so there will be a limit to how rigorous I can make my answer.
The intuition here is that $n$ is prime if and only if it is not divisible by any prime $<n$. So we "should" have
$$f(n) \approx \prod\_{p < n} \left( 1-1/p \right).$$
Similarly
$$f(n+1) \approx \prod\_{p<n+1} \left( 1-1/p \right) = \prod\_{p < n} \left( 1-1/p \right) \cdot \left\{ \begin{matrix} \left( 1-1/n \right) \ \mbox{if}\ n\ \mbox{is prime} \\ 1 \ \mbox{if}\ n\ \mbox{is not prime} \end{matrix} \right.$$
$$\approx f(n) \cdot \left\{ \begin{matrix} \left( 1-1/n \right) \ \mbox{if}\ n\ \mbox{is prime} \\ 1 \ \mbox{if}\ n\ \mbox{is not prime} \end{matrix} \right. .$$
Since $n$ is prime with "probability $f(n)$", we interpolate between the two cases next to the brace by writing:
$$f(n+1) \approx f(n) \left( 1-f(n)/n \right).$$
One might argue that it would be better to interpolate with a factor of $(1-1/n)^{f(n)}$, but this will make no difference in the asymptopics as $(1-1/n)^{f(n)} = 1-f(n)/n+O(1/n^2)$.
This argument is famously fishy, because it gives the right answer, but the intermediate point is wrong! The actual asymptopics of $\prod\_{p<n} \left( 1-1/p \right)$ do not look like $1/\log n$, but like $e^{-\gamma} /\log n$. I've never seen a good intuitive explanation for why we get the wrong estimate for $\prod\_{p<n} \left( 1-1/p \right)$, but the right estimate for the density of the primes.
| 22 | https://mathoverflow.net/users/297 | 36902 | 23,739 |
https://mathoverflow.net/questions/36893 | 7 | Hello everybody
There is a nice classical result in linear algebra: if $A, B$ are two matrices in $M\_n(k),$ where $k$ is a field, and $B$ commutes with every element of $M\_n(k)$ which commutes with $A$, then $B = f(A)$ for some polynomial $f(x)$ in $k[x].$
I was wondering if anybody knows any (important) theorem which is proved using this result. Thank you.
| https://mathoverflow.net/users/8828 | Looking for applications of a nice result in linear algebra | Tate's famous "Endomorphisms of Abelian Varieties over Finite Fields," which proves the Tate conjecture in the finite field case, uses the full force of the theorem of bicommutation in a reduction lemma. As KConrad mentions in the comments, the result you've cited is the special case of this theorem where one works with the subalgebra generated by one element.
| 4 | https://mathoverflow.net/users/1018 | 36906 | 23,742 |
https://mathoverflow.net/questions/36903 | 1 | I have read in a few places that $\mathbf{PH}$ can be interpreted in terms of the complexity of determining the winner in two-player games. I would like to know a) the original reference for this result and/or b) a concise explanation of it that requires little to no background in complexity theory (e.g., less than [Goldreich's book](http://books.google.com/books?id=EuguvA-w5OEC)).
| https://mathoverflow.net/users/1847 | Where does the game-theoretic characterization of PH come from? | The answer to part (a) of your question is this reference:
>
> A. Meyer and L. Stockmeyer. The equivalence problem for regular expressions with squaring requires exponential space. In *Proceedings of the 13th IEEE Symposium on Switching and Automata Theory*, pages 125-129, 1972. [[pdf]](http://people.csail.mit.edu/meyer/rsq.pdf)
>
>
>
What an amazing paper this was! Two later papers that discuss refinements of the result include these:
>
> C. Wrathall. Complete sets and the polynomial-time hierarchy. *Theoretical Computer Science* 3:23-33, 1977.
>
>
> A. Chandra, D. Kozen, and L. Stockmeyer. Alternation. *Journal of the ACM* 28(1):114-133, 1981.
>
>
>
| 7 | https://mathoverflow.net/users/7641 | 36911 | 23,747 |
https://mathoverflow.net/questions/36219 | 1 | For a co-quasi-triangular Hopf algebra $H$, with universal $r$-form $r$, there exists an important map $Q$ defined by
$$
Q:H \otimes H \to k, ~~~~~~h \otimes g \mapsto r(g\_{(1)}\otimes h\_{(1)})r(h\_{(2)}\otimes g\_{(2)}).
$$
The map is usually called the *quantum Killing form*.
In some papers I have read, it seems that the authors have tacitly assumed that the kernel of $Q$ is a right ideal. Is this true? If so, why?
| https://mathoverflow.net/users/1095 | The Killing Form for Co-Quasi-Triangular Hopf Algebras | I doubt this can be true. I claim that:
**Lemma.** Let $k$ be a commutative ring, $A$ be a $k$-algebra, and $Q:A\to k$ be a $k$-linear map such that $Q\left(1\right)=1$. Then, the following four assertions (1), (2), (3), (4) are pairwise equivalent:
(1) The kernel $\mathrm{Ker} Q$ is a two-sided ideal of $A$.
(2) The kernel $\mathrm{Ker} Q$ is a right ideal of $A$.
(3) The kernel $\mathrm{Ker} Q$ is a left ideal of $A$.
(4) The map $Q$ is a $k$-algebra homomorphism.
*Proof of Lemma.* Clearly, (4) $\Longrightarrow$ (1) $\Longrightarrow$ (2). Now let us prove that (2) $\Longrightarrow$ (4): Assume that (2) holds. That is, we assume that $\mathrm{Ker} Q$ is a right ideal of $A$. Clearly, every $a\in A$ satisfies $Q\left(a\right)\cdot 1-a\in\mathrm{Ker} Q$ (since the $k$-linearity of $Q$ yields $Q\left(Q\left(a\right)\cdot 1-a\right)=Q\left(a\right)\cdot \underbrace{Q\left(1\right)}\_{=1}-Q\left(a\right)=0$). Thus, every $a\in A$ and $b\in A$ satisfy $\underbrace{\left(Q\left(a\right)\cdot 1-a\right)} \_ {\in\mathrm{Ker} Q} b \in\mathrm{Ker} Q$ (since $\mathrm{Ker} Q$ is a right ideal), so that
$0=Q\left(\left(Q\left(a\right)\cdot 1-a\right)b\right)=Q\left(Q\left(a\right)b-ab\right)=Q\left(a\right)Q\left(b\right)-Q\left(ab\right)$
(by the $k$-linearity of $Q$), so that $Q\left(a\right)Q\left(b\right)=Q\left(ab\right)$. Together with the $k$-linearity of $Q$ and $Q\left(1\right)=1$, this yields that $Q$ is a $k$-algebra homomorphism, so that assertion (4) holds. Thus we have shown that (2) $\Longrightarrow$ (4), which completes the (4) $\Longrightarrow$ (1) $\Longrightarrow$ (2) $\Longrightarrow$ (4) circle. Thus, (4) $\Longleftrightarrow$ (1) $\Longleftrightarrow$ (2). Similarly (4) $\Longleftrightarrow$ (1) $\Longleftrightarrow$ (3). This proves that all four assertions (1), (2), (3), (4) are pairwise equivalent, and the lemma is proven.
---
The Lemma shows that as long as you want the kernel of $r$ to be an ideal (one- or two-sided), $r$ will be forced to be a $k$-algebra homomorphism. Considering the main example of co-quasi-triangular Hopf algebras, namely the group algebra with a bicharacter, the counterexample we gave in the comments above will hold.
Maybe the "right ideal" that your references claimed refered to a different algebra structure? One of my main sources of confusion in the advanced Hopf algebra theory has always been the presence of many conflicting multiplications, comultiplications, actions etc. on one and the same set.
| 2 | https://mathoverflow.net/users/2530 | 36917 | 23,751 |
https://mathoverflow.net/questions/34454 | 3 | I've been looking at proof techniques in formal systems like Coq and Agda recently, and encountered the `newring` tactic described [here](http://www.lix.polytechnique.fr/~assia/Publi/ring.ps) for proving equalities over arbitrary (semi)rings. It does this by reducing the generated polynomials to a Horner normal form and comparing equality over that.
That got me wondering whether a similar approach is possible for inequalities over general "ordered semirings" (where the order fits with the semiring operations). Someone pointed me to the `omega` tactic in Coq, for proving arbitrary statements (including inequalities) about Presburger arithmetic (with some restricted simulated multiplication). Anyway, it's clearly possible for Presburger arithmetic because all statements are decidable, but I don't really care about all statements, and was wondering whether it was possible for a particular algebraic structure without relying on properties of particular structures like the integers or reals.
Does anyone know whether it is possible to ? I'd be happy to even get some pointers to look more deeply into the problem, as I'm not even sure what to search for at this point.
Thanks!
Edit: there's more on the question in [these slides](http://www.andrew.cmu.edu/user/avigad/Teaching/practical/slides_cmu.pdf). So I can do it on integral domains and fields through a connection with the reals. I'd really like to do it for arbitrary semirings, though.
| https://mathoverflow.net/users/202 | Proving inequalities over algebraic structures | The desired degree of generality seems to vary greatly among the various parts of the question and the subsequent comment. Here's a simple answer to one aspect of the question. There is no algorithm for computably enumerating all true inequalities between polynomials with natural number coefficients. (Here "true" means identically true, for all natural number values of the variables.) In particular, there is no deductive system that can prove exactly those true inequalities.
The reason is that any statement of the form "$P(x\_1,\dots,x\_n)$ has no integer solutions," with $P$ a polynomial over the integers, can be rewritten as an inequality between polynomials with natural number coefficients, as follows: Replace each of the integer variables $x\_i$ with the difference $y\_i-z\_i$ of two natural number variables (so that we can talk about solutions in $\mathbb N$ instead of $\mathbb Z$); then express the absence of zeros of the polynomial as $P^2 \geq 1$; and finally, transpose any terms with negative coefficients to the right side.
Now suppose, toward a contradiction, that we could enumerate the true inequalities over the semiring $\mathbb N$. So, we could enumerate all true statements of the form "$P(x\_1,\dots,x\_n)$ has no integer solutions," considered above. Then we could decide whether any given Diophantine equation $P=0$ has a solution. Just run both the enumeration (to see whether it turns up a proof that there's no solution) and a systematic search for a solution. Eventually, you'll get the answer. But such a decision algorithm is impossible, by the solution of Hilbert's 10th problem.
| 2 | https://mathoverflow.net/users/6794 | 36918 | 23,752 |
https://mathoverflow.net/questions/36916 | 0 | Suppose *G*, *A*, and *B* are abelian groups with $i:A \to B$ an injective homomomorphism. Consider the groups $H^2(G,A)$ and $H^2(G,B)$ for the trivial action of *G* on *A* and *B*. *i* induces an injective homomorphism:
$$i\_\*: H^2(G,A) \to H^2(G,B)$$
The map $i\_{\ast}$ is not always injective. For instance, setting *A* as the integers embedded in *B* the rationals, we see that the target group of $i\_\*$ is always trivial but the source group is often nontrivial (for instance, when *G* is finite cyclic).
Question: Under what conditions is $i\_\*$ injective and under what conditions is it an isomorphism? More specifically I am interested in the case where:
*G* is a finite abelian *p*-group and *A* and *B* are both finite cyclic *p*-groups with the order of *A* dividing the order of *B*. Here, *p* is a prime number.
Generic note: Note that the induced map at the level of cocycles (before going down to cohomology) is injective, but the trouble arises because it seems possible that some cocycles that were not coboundaries become coboundaries on enlarging the group of coefficients.
| https://mathoverflow.net/users/3040 | Under what conditions does the second cohomology preserve injectivity? | Sasha's comment is correct. In your case you have the trivial action on $p$-power order cyclic modules. So let me write the map $B \to B/A$ as $B = \mathbb Z/p^n \to
\mathbb Z/p^m = B/A,$ where $m \leq n$, and the map is the natural one (reduce a mod $p^n$ class to a mod $p^m$ class).
Now $H^1$ of $G$ against a trivial module is just homs of $G^{ab}$ into this module, so
we have to look at
$$Hom(G^{ab},\mathbb Z/p^n ) \to Hom(G^{ab},\mathbb Z/p^m).$$
Now $G^{ab}$ is (in your setting) itself an abelian $p$-power order group, so is a product
of cyclic $p$-power order groups, so (since $Hom$ from a finite product is the product
of the individual $Hom$s) we see that it is enough to consider whether
$$Hom(\mathbb Z/p^r,\mathbb Z/p^n) \to Hom(\mathbb Z/p^r,\mathbb Z/p^m)$$ is surjective.
Assuming that $m < n$ (i.e., in the original terms of the problem, that $A$ is non-zero,
so that the question is non-trivial), then this map is surjective if and only
if $n \leq r.$ (This is not hard to check; see below for a careful explanation.)
Putting this together for all $r$, we get the following: assuming that $A$ is non-zero, that $B$
is cyclic of $p$-power order, and that $G$ is a $p$-group, then the map of $H^2$ is injective precisely when each cyclic direct summand of $G^{ab}$ has order at least
that of $B$.
Proof of surjectivity fact:
The $Hom$ space $Hom(\mathbb Z/p^r,M)$ is equal to $M[p^r],$ the $p^r$-torsion subgroup of $M$,
for any abelian group $M$ (just look at the image of $1$ mod $p^r$). So we have to consider
the surjectivity (or non-surjectivity) of
$(\mathbb Z/p^n)[p^r] \to (\mathbb Z/p^m)[p^r],$ which
is the map
$$p^{\max(0,n-r)}\mathbb Z/p^n\mathbb Z \to p^{\max(0,m-r)}\mathbb Z/p^n \mathbb Z.$$
This is surjective if $n - r \leq 0,$ or if $m = n$, but otherwise is not.
| 7 | https://mathoverflow.net/users/2874 | 36928 | 23,759 |
https://mathoverflow.net/questions/36891 | 8 | Hello everybody, I would like to know about the work of Élie Cartan of PDE's that relate to the theory of foliations and differential forms.
I am interested in the subject and will be happy to receive basic references on the subject (articles) as well as explanations on the importance of the subject in mathematics today.
| https://mathoverflow.net/users/8671 | Differential forms, PDE's and Élie Cartan | Robert Bryant is the reigning expert on this. An excellent book on the subject (later than the one mentioned) is:
Exterior Differential Systems and Euler-Lagrange Partial Differential Equations, Chicago Lectures in Mathematics (2003), University of Chicago Press (vii+213 pages, ISBN: 0-226-07794-2.) by
R. Bryant, Phillip Griffiths and Dan Grossmann.
I just recalled, Bryant has a very nice set of nine introductory lectures on the subject. It may be just what you are looking for! They are available online here:
<https://services.math.duke.edu/~bryant/Introduction_to_EDS.pdf>
| 19 | https://mathoverflow.net/users/7311 | 36931 | 23,761 |
https://mathoverflow.net/questions/36936 | 4 | Let $\Lambda(n)$ be the von Mangoldt function. The prime number theorem is equivalent to the statement that $\sum\_{n \leq N} \Lambda(n) \approx N$. Defining $\lambda\_{\*}(n)= \Lambda(n)-1$ we may rewrite this as $S(N) = \sum\_{n \leq N} \Lambda\_{\*}(n) =o(N)$. Now it is known that $|S(N)| \gg |N|^{1/2}$ infinitely often. Moreover, on the RH we have that $|S(N)| \ll N^{1/2}ln^2(N)$. Not that these estimates differ by a factor of $ln^2(N)$.
My question is the following: What do probabilistic considerations suggest the correct error term to be?
Let me suggest a model: Let $X\_n$ be a sequence of independent random variables such that $X\_n = \ln(n)-1$ with probability 1/ln(n) and $-1$ with probability $1-1/ln(n)$, and form the sum $T(N)= \sum\_{n=1}^{N} X\_n$. Is there an elementary function $E(N)$ such that $ \lim sup\_N |T(N)|/E(n) = 1 $ holds almost surely?
Notice that if the primes had positive density in the integers and we adjusted our model accordingly the law of the iterated logarithm would allow us to take $E(N)$ to be a multiple of $|N|^{1/2}\ln\ln(N)$.
(More generally, I'm interested in understanding sums of the above form (that is independent random variables with slowly increases variance) if you know of an appropriate reference.)
| https://mathoverflow.net/users/630 | What does the probabilistic model suggest the error term in the PNT should be? | Let $P\_n$ be independent variables which are 1 with probability $1/\log n$ and $0$ with probability $1-1/\log n$ and let
$$ \Pi(x) = \sum\_{n\leq x} P\_n.$$
Then Cram\'{e}r showed that, almost surely,
$$ \limsup\_{x\rightarrow \infty} \frac{|\Pi(x)-\ell i(x)|}{\sqrt{2x}\sqrt{\frac{\log\log x}{\log x}}} = 1 $$
where
$$\ell i (x) = \int\_2^x \frac{dt}{\log t}.$$
See page 20 here: <http://www.dms.umontreal.ca/~andrew/PDF/cramer.pdf>
Edit: H. L. Montgomery has given an unpublished probabilistic argument that suggests
$$ \limsup\_{x\rightarrow \infty} \frac{|\psi(x)-x|}{\sqrt{x} (\log\log\log x)^2} = \frac{1}{2\pi}.$$
This is announced in: H.L. Montgomery, "The zeta function and prime numbers," Proceedings of the Queen's Number Theory Conference, 1979, Queen's Univ., Kingston, Ont., 1980, 1-31.
| 7 | https://mathoverflow.net/users/3659 | 36937 | 23,766 |
https://mathoverflow.net/questions/36821 | 8 | How can one compute the Schubert variety (by compute I mean having actual polynomials that define it) for SL(n)? If this is well known forgive my ignorance and just point me to the right book/paper.
**EDIT:** Sorry I did not return here for quite some time. It is kind of amusing that the way I learned about Schubert varieties is not even mentioned. Here is how I learned it:
>
> 1. $G$ an algebraic group with Lie algebra $\mathbf{g}$.
> 2. $L(\Lambda)$ is an *integrable* highest weight module for $\mathbf{g}$.
> 3. For $w$, an element of the Weyl group, consider the 1-dimensional root space $L(\Lambda )\_{w \cdot \Lambda}$.
> 4. Denote the vector space $U(\mathbf{b}) \bullet L(\Lambda)\_{w \cdot \Lambda}$ by $E\_w(\Lambda)$ (you take the 1-dimensional root space and act on it by all the raising operators plus the Cartan). Then $E\_w(\Lambda) \subset L(\Lambda)$.
>
>
> Now we are ready: since $L(\Lambda )\_{w \cdot \Lambda}$ is 1-dimensional it becomes a single point in $\mathbf{P} \left (E\_w(\Lambda) \right)$. We look at the orbit $B \bullet L(\Lambda)\_{w \cdot \Lambda} \subset \mathbf{P} \left (E\_w(\Lambda) \right)$. We call its closure the ***Schubert variety associated to $w$ and $\Lambda$*** and denote it by $S\_{w, \Lambda}$.
>
>
>
I don't know if this is a good way of computing things but in principle it should give you any Schubert variety you need.
| https://mathoverflow.net/users/8811 | Explicit equations for Schubert varieties | If one is learning about this, computing directly with matrices seems like the easiest way (though not as powerful as standard monomials and the toric degenerations that result). Alex Woo's references are a good source for this point of view; I'd also add the first couple chapters of the book *Schubert varieties and degeneracy loci* by Fulton and Pragacz. And a quick example, in that spirit: to find equations for $X\_{2143}$ inside $SL\_4/B$, do the following:
(1) Form the rank matrix $(r\_{ij})$ for the permutation:
$$\left[\begin{array}{cccc} 0 & 1 & 0 & 0 \\\
1 & 0 & 0 & 0 \\\
0 & 0 & 0 & 1 \\\
0 & 0 & 1 & 0 \end{array}\right] \to
\left[\begin{array}{cccc} 0 & 1 & 1 & 1 \\\
1 & 2 & 2 & 2 \\\
1 & 2 & 2 & 3 \\\
1 & 2 & 3 & 4 \end{array}\right].$$
(2) Write down the equations on the $4 \times 4$ generic matrix that say "upper-left $i\times j$ submatrix has rank at most $r\_{ij}$". These are your polynomials cutting out the matrix Schubert variety, e.g.,
$$x\_{11}=0, \\ \det( x\_{ij} )\_{1\leq i,j\leq 3} = 0.$$
(3) If you want equations in an open affine patch, set appropriate $x\_{ij}$'s to $0$ or $1$, e.g., the opposite cell would have free variables in the $\*$ positions:
$$\left[\begin{array}{cccc} \* & \* & \* & 1 \\\
\* & \* & 1 & 0 \\\
\* & 1 & 0 & 0 \\\
1 & 0 & 0 & 0 \end{array}\right].$$
(4) If you want equations in the Plucker embedding, the determinants are built-in to the definition of Plucker coordinates, and you just intersect with appropriate linear subspaces.
In general, these equations are highly redundant, and a big part of the work of Lakshmibai et al, Fulton, and Woo-Yong (as I understand) is to find *minimal* sets of equations. For the matrix Schubert varietes for $SL\_n/B$, the Fulton paper cited by Alex gives a simple answer.
| 11 | https://mathoverflow.net/users/5081 | 36940 | 23,768 |
https://mathoverflow.net/questions/36926 | 10 | There is an enormous amount of information about the common applied math problem of minimizing a function.. software packages, hundreds of books, research, etc.
But I still have not found a good reference for the case where the function to be sampled is extremely expensive.
My specific problem is an applied one of computer science, where I have a simulation which has databases with a dozen parameters that affect voxel sizes, cache distribution, tree branching, etc. These numeric parameters don't affect the algorithm correctness, just runtime. I want to minimize runtime.
Thus, I can treat the problem like a black box minimization. My cost function is runtime, which I want to minimize. I don't have derivatives, and I can treat it like a black box.
I have a decent starting point and even rough scales of each parameter. There can be interations and correlations between parameters and even noise in time measurements (luckily small.)
So why not just throw this into a standard least-squares minimization tool, using any package out there? Because my timing samples each take 8 hours to run.. so each data point is precious, and the algorithms I find tend to ignore this cost. A classic Levenberg-Marquand procedure freely "spends" samples and doesn't even remember the full history of each sample taken (instead updating some average statistics).
So my question is to ask for a pointer to iterative function minimization methods which use the minimum number of samples of the function. Ideally it would work where I could pass in a set of already-sampled locations and the value at each location, and the algorithm would spit out a single new location to take the next sample (which may be an exploratory sample, not a guess at a best minimum location, if the algorithm thinks it's worthwhile to test.)
I can likely take hundreds of samples, but only hundreds, and most multidimensional minimization methods expect to take millions.
Currently I am doing doing the minimization manually daily, using my own ad-hoc invention. I have say 40 existing timing samples to my 15-parameter model. I fit all my existing samples to a sum of independent quadratics (making the big initial assumption that each parameter is independent) then look at each of the N\*(N-1)/2 ~=100 possible correlation coefficients of the full quadratic matrix. I find the few single matrix entries that when allowed to change from 0.0, give the best quadratic fit to my data, and allow those few entries to be their best least-squares fit. I also give locations with small (faster) values higher weight in the fit (a bit ad hoc, but useful to throw out behavior distant from the minimum) Once I have this matrix, I manually look at graphs in each of the major eigenvalue directions and eyeball locations which seem to need better sampling. I recombine all these guesses back into a new sample location. Each day, I tend to generate 4 new points, set up a run to test them over the next day, and repeat the whole thing again after the computation is done. Weekends get 10 point batches!
Thanks for any ideas! This question likely doesn't have a perfect "best" answer but I'm stuck at what strategy would work best when the evaluation cost is so huge.
| https://mathoverflow.net/users/7107 | Robust black box function minimization with extremely expensive cost function | I've read the paper, but never used the approach.
["Efficient Global Optimization of Expensive Black-Box Functions"](https://doi.org/10.1023/A:1008306431147 "Journal of Global Optimization 13, 455–492 (1998). zbMATH review at https://zbmath.org/?q=an:0917.90270")
by: Donald R. Jones, Matthias Schonlau, William J. Welch
PDF available from one of the [page of an author](http://www.schonlau.net/publication.html).
| 4 | https://mathoverflow.net/users/8838 | 36941 | 23,769 |
https://mathoverflow.net/questions/36929 | 7 | If the sequence $x\_1,x\_2,\dots$ is periodic, the unweighted averages $(\sum\_{i=1}^n x\_i)/n$ converge to the asymptotic average of the $x\_n$'s with error $O(1/n)$, but the weighted averages $(\sum\_{i=1}^n i(n+1-i)x\_i)/(n(n+1)(n+2)/6))$ converge even more quickly, with error $O(1/n^2)$.
This fact is easy to prove (e.g. first prove it for $(x\_n) = (\zeta^n)$ with $\zeta$ an arbitrary root of unity and then appeal to linearity), but it's something I stumbled upon on my own, and I don't really understand what's going on. Can anyone provide a context for this fact? My guess is that it must be well-known to people who study series-convergence (and acceleration thereof), and also well-known to Fourier analysts, though possibly in disguised form. (Speaking of disguises: This question is related to my earlier question [A specific Dedekind-esque sum](https://mathoverflow.net/questions/26608/a-specific-dedekind-esque-sum) ; in my earlier post, the relevant sequence is almost-periodic rather than periodic, and the discrepancy goes down like $O((\log n)/n^2)$ rather than $O(1/n^2)$.)
I suspect that $O(1/n^2)$ is the end of the line, in the sense that no weighted average of $x\_1,\dots,x\_n$ with fixed coefficients will differ from the asymptotic average of the $x\_n$'s by $O(1/n^c)$ for any $c>2$, and I might even try to give a proof using the geometry of numbers, but I suspect this is old stuff and would appreciate some pointers.
Thanks!
Jim Propp
| https://mathoverflow.net/users/3621 | accelerated convergence to the mean using quadratic weights | Just think a bit of what the Poisson summation formula gives you for the function $\varphi\_n(x)=\varphi(x/n)$ where $\varphi$ is some not too bad compactly supported function (you can view the weighted $n$-th sum for the periodic sequence as the finite weighted sum of several infinite sums of values of $\varphi\_n$ over arithmetic progressions). The real end of line is almost exponential in $n$.
Edit: Suppose that $\varphi$ is reasonably smooth and has integral $1$ (the characteristic function of an interval is not falling under this argument formally but the function $[x(1-x)]\_+$ already is). Now, let $P$ be the period and let the sequence be $a\_0,a\_1,\dots,a\_{P-1},a\_0,a\_1,\dots$. Then the $\varphi$ weighted sum
$$
S\_n=\frac 1n \sum\_k\varphi\_n(k)a\_k=\frac 1P\sum\_{k=0}^{P-1}a\_k\sigma\_k
$$
where
$$
\sigma\_k=\frac Pn\sum\_m\varphi\_n(k+mP)= \sum\_m\widehat\varphi(mn/P)e^{2\pi i mk/P}
$$
by the Poisson summation formula.
Now, $|\sigma\_k-1|\le\sum\_{m\ne 0}|\widehat\varphi(mn/P)|$ and, if $\widehat\varphi$ decays fast (which you can always achieve by making $\varphi$ smooth enough), this bound decays fast with $n$.
You may object that one has to divide not by $n$ but by $\sum\_k \varphi\_n(k)$ but it is close to $n$ with the same relative precision (just run the same argument for the sequence consisting of all ones).
| 3 | https://mathoverflow.net/users/1131 | 36944 | 23,770 |
https://mathoverflow.net/questions/36822 | 8 | In his really nice thesis, Tobias Dyckerhoff proved the following theorems about matrix factorizations(of possibly infinite rank) over a regular local k-algebra R with a function w and residue field k such that the Tyurina algebra, T= $R/(w,dw)$ is finite dimensional. This last condition says that w has an isolated singularity. For further reference, let S denote the ring R/(w).
1) The homotopy category of matrix factorizations has a compact generator as a triangulated category, which he denotes as $k^{stab}$.
2) As a consequence of 1), he derives that there is a natural complex which represents the identity functor thought of as an element of MF($R\otimes R,1\otimes w-w\otimes 1)$ which he denotes as the stabilization of the diagonal $R^{stab}$.
3) MF(R,w) is a Calabi Yau dg-category.
Now my question is how much of the above remains true for when the singularity is non-isolated? In some writings, Kontsevich, while not explicitly saying so, writes as if the homotopy category always has a compact generator and that the category is there by "dg-affine", e.g. equivalent to D(A), the derived category of modules over a dg-algebra. Is this indeed known to be true or false? If not, is there a way to prove 2) without making reference to 1)? I'm asking because I haven't found anything about this stuff in the literature, but a lot of things in this field are not written or written in physics literature that I'm not familiar with.
I've checked a few examples with non-isolated singularities and it appears that for example in the category of factorizations $(k[[x,y]], xy^2)$, that while $k^{stab}$ doesn't generate as Dyckerhoff proves, one has $(k\oplus k[[x]])^{stab}$, which I think does generate. The way I want to argue this is Dyckerhoff's theorem 3.6 that it is enough to show that $Tor\_S(k\oplus k[[x]], M)$ implies that $Tor\_S(N,M)=0$, where N is a finitely generated T module and M is any S module.Then one does an analysis of finitely generated modules over T(I didn't think about the characteristic 2 case) and does some devissage with short exact sequences. Please let me know if this sounds off. I also think that with a bit more calculation one can prove similarly that in (k[[x,y,z]], xyz) the module $(k[[x]]\oplus k[[y]]\oplus k[[z]]\oplus k)^{stab}$ is a compact generator.
**Added:** I think the right generalization of the above two examples is the following, in <http://websupport1.citytech.cuny.edu/faculty/hschoutens/pdf/finiteprojdim.pdf>, the author introduces the notion of a "net". The above method should give a compact generator, whenever the net of finitely generated modules over T is generated as a net by finitely many modules. This happens for example when T has finitely prime ideals. The modules A/p, where p is a prime, generate the net of finitely generated projective modules over T, which is enough to prove the vanishing above. In particular, this should take care of the case when T has dimension 1. A question is what are some conditions on (R,w) which lead to the net of finitely generated modules over T being generated by finitely many objects?
Assuming that this right, I think that to derive 2 and 3 for these examples becomes a formality in view of Dyckerhoff's section 5. One just replaces his compact generator with the new one.
| https://mathoverflow.net/users/6986 | Matrix factorization categories beyond the isolated singularity case | The answer to (1) is yes for any local abstract hypersurface $S$ whose singular locus is closed (which is barely a hypothesis, and free in the case of interest). Let us write $\mathrm{Sing} \;S$ for the singular locus, which we can write as a union finitely many irreducible components corresponding to primes $\mathfrak{p}\_i$ for $i=1,\ldots,n$. Then the image of the object
$$ \bigoplus\_{i=1}^n S/\mathfrak{p}\_i $$
in the category of matrix factorizations is a compact generator. The point is that there is a notion of support for objects of the category of all matrix factorizations. For a hypersurface this support gives a classification of the localizing subcategories in terms of subsets of the singular locus. The image of the given object is supported everywhere so must generate.
The statement about the classification of localizing subcategories is not actually published anywhere (Iyengar has announced this as part of a more general result on complete intersections and I have an (independent and different) proof, also for complete intersections, which also works for certain singular schemes - but Iyengar has not yet released a preprint and the results of mine are in my thesis which isn't publicly available yet). Although there is [this paper](http://arxiv.org/abs/0908.0107) by Takahashi, the main result of which is sufficient to give the statement about the generator above.
The generator above actually works for any noetherian ring which is locally a hypersurface if one works in one's favourite infinite completion of $D\_{\mathrm{Sg}}(S)$.
**Edit:** As promised in the comments (well sort of, it didn't turn out to be soon) [here](http://arxiv.org/abs/1105.4698) is the link to the preprint concerning subcategories of singularity categories from which one can deduce the existence of compact generators in certain cases as above.
| 9 | https://mathoverflow.net/users/310 | 36947 | 23,772 |
https://mathoverflow.net/questions/36942 | 4 | I recall that the Manin constant for a strong elliptic curve is a rational integer $c\_E$ such that, for a modular parametrization $\phi: X\_1(N) \to E$, one has $\phi^\*(\omega\_E)= 2\pi i c\_E f(z)\mathrm{d}z$ ($f$ is the modular form associated to $E$). The Manin constant is supposed to equal +/-1, but it is not proved yet.
I know that the prime divisors of $c\_E$ are well known, but is there any bound for $c\_E$, as the conducteur varies? If any, is there any reference for a proof? I heard that Edixhoven proved that $c\_E$ is bounded (independently of $N$), but the article is forthcoming...
| https://mathoverflow.net/users/8786 | A bound for the Manin constant | A useful reference on this topic (which maybe you know?) is "The Manin Constant",
by Agashe, Ribet, and Stein, available [here](http://modular.math.washington.edu/papers/ars-manin/agashe-ribet-stein-the_manin_constant.pdf). On p. 3 they write
>
> B. Edixhoven also has unpublished results (see [Edi89]) which assert that the
> only primes that can divide $c\_E$ are 2, 3, 5, and 7; he also gives bounds that are
> independent of $E$ on the valuations of $c\_E$ at 2, 3, 5, and 7. His arguments rely
> on the construction of certain stable integral models for $X\_0(p^2 )$.
>
>
>
The reference [Edi89] is to Edixhoven's thesis, which is unpublished, but
which can be found [here](http://www.math.leidenuniv.nl/~edix/public_html_rennes/publications/prschr.html)
at Edixhoven's website. At the end of the introduction to his thesis, Edixhoven writes
>
> Finally,
> in the last section, we derive some results concerning the constant
> "c" attached to a strong Weil curve $E$. Manin conjectured that $c=1$. It
> is known that $c$ is a positive integer and Mazur proved that only 2 and
> primes where $E$ has additive reduction can divide $c$. Our methods show
> that primes $p>7$ where $E$ has additive reduction divide $c$ at most once,
> and in fact, for most of the possible reduction types (=Kodaira symbols), not (Theorem 4.6.3 and the remarks following this theorem). It might well
> be that a computation involving the period lattices of normalized newforms
> can solve the problem in the case of potentially good, ordinary reduction of type $II, III$ or $IV$. It should also be tried to get bounds on the exponents
> of 2, 3, 5 and 7 in $c$.
>
>
>
This suggests that the statement of Agashe, Ribet, and Stein is perhaps a little strong,
in that Edixhoven does not give uniform bounds at all primes, but only at primes $p > 7$,
with the suggestion that one could also hope to obtain such bounds at 2, 3, 5, and 7.
You might try writing to Edixhoven for clarification, or perhaps also to one of Agashe, Ribet, or Stein to find out more precisely what they had in mind.
| 16 | https://mathoverflow.net/users/2874 | 36951 | 23,774 |
https://mathoverflow.net/questions/36938 | 7 | Around 1998, I encountered a (forgotten) reference to a particularly strange space-filling curve.
Consider a foliation as a collection of continuous nonintersecting curves that start at $(0,0)$ and end at $(1,1)$ and collectively fill the unit square, such as the graphs of functions $f\_t(x) = x^t$ where $t \ge 0$. Supposedly there exists a continuous curve G that starts at $(1,0)$, ends at $(0,1)$, fills the unit square, and crosses each $f\_t$ curve only once.
This initially sounds even more impossible than the Cantor curve. But intuitively a space-filling curve could trace back and forth over the $f\_t$ curves and only cross at the corners $(0,0)$ and $(1,1)$. Can someone please explain a construction of such a space-filling curve?
| https://mathoverflow.net/users/6769 | Unusual space-filling curve | The space filling curve you are looking for does not exist.
Assume by contradiction that such a space filling curve $\gamma:I\rightarrow [0,1]^2$ exists.
Since $\gamma$ intersects each curve $f\_t\subset [0,1]^2$ only once, the preimage $\gamma^{-1}(f\_t)$ is either a point or an interval. The curve $\gamma$ being space-filling, that preimage can't be a point. It is therefore an interval and, in particular, of positive measure.
Letting $t$ vary,
we have constructed an uncountable family of disjoint subsets of $[0,1]$, all of whom have positive measure: contradiction!
| 18 | https://mathoverflow.net/users/5690 | 36952 | 23,775 |
https://mathoverflow.net/questions/8358 | 13 | The drawing on the last page of [Yoshida's notes](http://www.dpmms.cam.ac.uk/~ty245/Yoshida_2007_colloquim.pdf) make me puzzle, perhaps you can help? It shows a "landscape" featuring the monodromy-weight conj., the general Ramanujan-conj., the Langlands correspondence and some connections between them. However, I guess the Ramanujan-conj. "is" the monodromy-weight-conj. expressed via Langlands correpondence and the Rapoport-Zink spectral sequence is a tool for showing the MWC in some cases and reducing the computation of L-functions to semi-simple L-functions. That seems not to fit to the drawing. Can you can clearify that?
Edit: A fascinating [new survey](http://de.arxiv.org/abs/1003.0927) on the "philosophy of weights" with many conjectures by Jannses. (Posted here because I don't know an entry where it would fit better)
| https://mathoverflow.net/users/451 | The monodromy-weight-, Ramanujan-, Langlands-landscape | Dear Thomas,
This "landscape" is, I think, a sketch of the proof of the following theorem of Taylor and Yoshida: if $\Pi$ is a self-dual cuspdidal automorphic form on $GL\_n$ over $E$ (a CM field)
(satisfying some further technical conditions) and $\rho$ is the associated $n$-dimensional
Galois representation (constructed by Harris and Taylor in their book "The geometry and cohomology of some simple Shimura varieties"), then the local factors of $\Pi$ at any prime
$p$ of $E$ matches, via local Langlands, with the restriction of $\rho$ to a decomposition group at $p$.
What was proved in Harris--Taylor was that this matching is correct, up to the question
of matching the $N$ on each side. What Taylor and Yoshida verified is that the $N$ on each side
also matches.
The top part of the landscape represents the reduction to the unipotent situation (i.e. the context in which $\Pi$, locally at $p$, has an Iwahori fixed vector --- the analogue
of classical level $\Gamma\_0(p)$), by
base-changing to an extension of $E$. On the automorphic side, this is the process
labelled by the names Arthur--Clozel--Selberg; on the Galois side, this is just restricting
from $Gal(\overline{E}/E)$ to $Gal(\overline{E}/E')$ for some appropriately chosen
extension $E'$ of $E$. If one thinks of $\rho$ as appearing in the cohomology of a Shimura
variety (which is, after all, how it is constructed), then this just corresponds to
base-changing the variety from $E$ to $E'$, which is why this process is labelled as
geometric base change in the "landscape".
On the lower right of the diagram, one has the generalized Ramanujan conjecture, stating that the local factor of $\Pi$ at $p$ should be tempered,
which in turn implies that the local (at $p$) Galois representation (more precisely,
Weil--Deligne representation) attached to this local factor satisfies the monodromy
weight conjecture. This temperedness was proved by Harris--Taylor.
Now we already know that $\rho$ locally at $p$ matches with the local factor of $\Pi$ at
$p$ up to $N$, and it is not hard to show that there is a unique way to add $N$ so as
to obtain a local Weil--Deligne representation satisfying the monodromy weight conjecture.
So to prove their theorem, Taylor and Yoshida are reduced to proving that $\rho$, locally
at $p$, satisfies the monodromy weight conjecture.
They do this via an application of the Rapoport--Zink spectral sequence and a careful analysis of the bad reduction of the Shimura variety in whose cohomology $\rho$ lives. This is represented in the lower left part of the diagram.
The very bottom part of the diagram on the left represents the fact that a priori one knows
that $\rho$ is mixed, i.e. its semi-simplification has Frobenius eigenvalues that are
Weil numbers of various weights. But the main part of the landscape here is the monodromy weight conjecture, which describes the precise relationship between the Frobenius eigenvalues and the $N$ operator.
To prove the MWC in their context, Taylor and Yoshida
use the interplay between the geometry of the
special fibre of the Shimura variety and the representation theory of $\Pi$ that is the main subject of Harris and Taylor's book, as well as Section 2 of Taylor and Yoshida's article. In particular, the fact that $\Pi$, locally at $p$, is tempered unitary, and is assumed to have an Iwahori fixed vector, puts strong restrictions on its structure (using the known classification of unitary reps. of $GL\_n$ over local fields; this is why Tadic's name appears on the right hand side of the landscape), which when fed over to
the geometric side implies that the RZ spectral sequence degenerates at $E\_1$, giving the desired monodromy weight conjecture.
My suggestion is that if you want to understand this in more detail, you should read [Taylor and Yoshida's article](http://www.math.harvard.edu/~rtaylor). Start at the very end of the paper, where the main theorem is proved (i.e. the paragraph beginning "We can now conclude ... "). The lemmas referred to in section 1 are more or less elementary. Theorem 3.2 is the heart of the argument; it is where degeneration of the RZ spectral sequence is proved. One can try to read it more or less formally, taking various assertions as a black box,
at least to see what role the temperedness of $\Pi$ locally at $p$ plays. The earlier parts of Section 3 are just establishing notation, and making contact with the book of Harris--Taylor. Section 2 is devoted to establishing the basic properties of the semi-stable models of the Shimura varieties in whose cohomology $\rho$ lives; I recommend treating it as a black box on first reading.
Added: First, I should note that the references I make are to the version of Taylor--Yoshida that is currently posted on Taylor's web-page, which I am told may differ quite substantially in its organization from the published version of the paper.
Also, let me add something about the role of the classification of unitary representations
of $GL\_n$, which will help illuminate the structure of the lower part of the landscape:
First, one uses the general result that geometrically obtained Galois representations are mixed (i.e. have Frobenius eigenvalues that are Weil numbers), coupled with the local-global compatibility without the $N$, to deduce that the Frobenius eigenvalues
of the Weil--Deligne representation attached to the local factor of $\Pi$ at $p$ are Weil numbers. Second, results of Tadic on the classification of unitary representations of $GL\_n$ of $p$-adic fields greatly restrict the structure of this
local factor (since it is both unitary and generic, being the local factor of a cuspidal
automorphic representation for $GL\_n$). Finally, when this is combined with the fact that
the Weil--Deligne rep'n associated to it via local Langlands is mixed, ones sees that this
local factor is forced to be tempered. This is how Harris--Taylor deduce temperedness;
it is an analogue at $p$ of an argument at the archimedean prime made by Clozel in his Ann Arbor paper. (See Lemma 4.9 on p.144 of volume I of the Ann Arbor conference. A scan
is available on Jim Milne's web-page [here](http://www.jmilne.org/math/Books/AA1988a.pdf),
but note that it is quite a big file.)
Finally, let me now explain how to correctly read the landscape: one starts in the lower left, following the arrow. When you come to a bridge, you cross the river from the motivic/Galois side to the automorphic side, or back again, continuing to follow the arrows. The areas marked with
x's are impenetrable (or, at least, you don't try to cross them directly --- e.g. you don't prove GRC directly, you don't prove WMC directly, you don't study the general semi-stable reduction problem directly); the only way to proceed is by crossing the river. This then gives
the structure of the Taylor--Yoshida argument.
| 24 | https://mathoverflow.net/users/2874 | 36953 | 23,776 |
https://mathoverflow.net/questions/36965 | 11 | Let $M$ be a smooth paracompact manifold. I think that the ring $C^{\infty}(M)$ contains many (possibly almost all?) geometric or topological information about $M$.
(e.g. Let $E$ be a vector bundle over $M$,$\Gamma(E)$ be a set of smooth section of $E$. Then, $\Gamma(E)$ is a $C^\infty(M)$-module. (Actually, I think $\Gamma(E)$ is projective $C^\infty(M)$-module because every a short exact sequence of vector bundle splits.))
But I have a feeling that $C^\infty(M)$ is too large to change the problem of Manifold theory into an algebraic problem or Ring theoretic problem.
Are there any well-known concrete description about the ring $C^\infty(M)$ for some manifold $M$ with simple topology?
| https://mathoverflow.net/users/7776 | The ring $C^{\infty}(M)$? | You are correct: $C^\infty(M)$ does contain all the geometry and topology
of $M$ (at least when it is considered as an $\mathbb{R}$-algebra).
For example when $M$ is compact the points of $M$ correspond to the maximal
ideals of $C^\infty(M)$ (this is quite easy to prove). If $M$ is not compact
there are maximal ideals $I$ not corresponding to points, but these can be distinguished
since $C^\infty(M)/I$ is a proper extension of $\mathbb{R}$ for such $I$.
See the book [Smooth Manifolds and Observables](http://books.google.co.uk/books?id=N5mHmFiovgkC&printsec=frontcover&dq=Jet+nestruev&source=gbs_similarbooks_s&cad=1#v=onepage&q&f=false) by "Jet Nestruev" to see these
ideas fully worked out.
| 9 | https://mathoverflow.net/users/4213 | 36966 | 23,785 |
https://mathoverflow.net/questions/36914 | 1 | I hope this is an appropriate forum for this question, and I asked on math.stackexchange as well. If it doesn't belong, I don't mind closing this. If my questions is not clear, please just let me know and I'll try to add information/explanation where I can.
Say I have a set of data points that I run through a [FLAME clustering algorithm](http://en.wikipedia.org/wiki/FLAME_clustering), and I get my set of cluster supporting objects, and the fuzzy memberships among them.
Next, let's say I get an additional set of data, for argument's sake, let's say it doubles the amount of data. If I want to add that to my clustering, without having to reprocess the old data (i.e. just using the cluster supporting objects, and adding these new observations to the cluster), will I be introducing some kind of bias? To get the most "accurate" picture, would it be better to reprocess ALL the data, re-establish CSO's, re-compute distances, etc.?
To add to the complexity, let's say I'm going to be periodically adding similar quantities of data points. Does the answer to the above change?
And I guess a final question - is there a clustering algorithm (or family of algorithms) that is not subject to extreme bias in the beginning, such that added data points can be added relatively pain-free? I'm sort of thinking that FLAME clustering, with its nice fuzziness properties is a reasonable way of going about this, but any more suggestions would be helpful.
Thanks!
| https://mathoverflow.net/users/6854 | Is any bias introduced from initial clustering | It is intrinsic to the finite sampling of a dataset to have a "variance" that affect clustering results, and this is not a matter of the algorithm.
Algos have parameters, like the K number of neighbors in FLAME, that adjust for this variance but, on the other side, introduce a "bias".
Thus what happens usually is that you choose a K that is large enough to be robust to your sample, but not too large for estimating an "informative" model. (Obviously you cannot do better than noise) (see for example:<http://www.math.princeton.edu/~amits/publications/laplacian_ACHA.pdf>)
An interesting criterion to use is that of stability: if samples with the same cardinality of your data give similar clustering results then you can, heuristically, avoid recomputations. On the other hand, if your data are too few you should recompute your model each time new data arrives.(or wait you have enough data) (link:<http://arxiv.org/abs/1007.1075>)
| 1 | https://mathoverflow.net/users/885 | 36969 | 23,786 |
https://mathoverflow.net/questions/36972 | 2 | Hi, I met several times the expressions "predicative definition" and "unpredicative definiton" in texts about logic. What these expressions do mean ? I precise I'm a french student, thanks for your help.
| https://mathoverflow.net/users/nan | Predicative definition | A definition of an object X is called impredicative if it quantifies over a collection Y to which X itself belongs (or at least could belong). The classic example is the set occurring in Russell's paradox, defined by "the members of X are all sets s that are not members of themselves". This quantifies over all sets, including X itself.
But impredicative definitions occur (without paradox) in ordinary mathematics also. For example, one might define a real number r as the supremum of a set A that might have r itself among its members. Unraveling the definition of "supremum" we would find quantification over A (and indeed quantification over the set of all real numbers).
Russell proposed to eliminate the set-theoretic and logical paradoxes by eliminating impredicative definitions, and "Principia Mathematica" (by Russell and Whitehead) develops an elaborate mechanism for this. Unfortunately, too much of ordinary mathematics was unprovable in that system, so Russell and Whitehead found it necessary to add the so-called axiom of reducibility, whose principal effect is to counteract the predicativity-enforcing mechanism and make impredicative mathematics available again.
| 14 | https://mathoverflow.net/users/6794 | 36977 | 23,788 |
https://mathoverflow.net/questions/36970 | 4 | Hi I'm trying to understand the most general conditions under which I can conclude finite time blow up of an ODE of the form $\dot{x} = f(x)$ with initial condition $x\_0 > 0$ and $f(x) \geq 0$ for all $x \geq 0$.
If I re-write this in a separable way so that $dt = \frac{dx}{f(x)}$ then I want to determine if there is some finite time $T$ for which $x(T) = \infty$. Ok so it is clear to me that if $\int\_{x\_0}^{\infty} \frac{dx}{f(x)} = \infty$ then there is **no finite time blowup** for the initial condition $x\_0$ (since we would need some finite $T$ for which $x(T)=\infty$ and this precludes that). I'm confused however if this is a necessary condition. In other words, if I know that $\int\_{x\_0}^{\infty} \frac{dx}{f(x)} < \infty$ does this necessarily tell me that I **DO** get finite time blowup? This is my question. It's not entirely clear to me why this should *prove* finite time blowup.
Two examples to keep in mind in all of this are $f(x) = x^2$ (finite time blowup) and $f(x) = x^{1/2}$ (no finite time blowup).
| https://mathoverflow.net/users/8755 | Criterion for finite time blowup of an ODE | (This used to be a comment, but I think it deserves to be an answer, after mulling over it a bit.)
I don't think your criteria are quite correct. Some counterexamples:
Let $f(x) = - x^2$, and $x(0) = -1$. This ODE blows up in finite time toward $-\infty$. But $\int\_{-1}^\infty dx / f(x) $ diverges due to the singularity at $x = 0$.
Similarly, for any $f(x)\geq 0$ such that $f(0) = 0$, for any initial value $x(0) < 0$ we must have $x(0) \leq x(t) \leq 0$ for any $t > 0$. Hence we have global existence (no blow up). But if we take $f(x) = \sqrt{|x|}$ if $|x| \leq 1$ and $f(x) = x^2$ if $|x| > 1$, then $1/f(x)$ is integrable, and in particular $\int\_{x\_0}^\infty dx/f(x) < \infty$ for any $x\_0$.
One key point used in your examples $f(x) = x^2$ and $f(x) = \sqrt{x}$ with initial data positive, is that there are no stationary points. And so for any positive initial datum the evolution eventually goes toward $x\to \infty$, and so the question of blowup reduces to a question of how fast that happens. And for that the integral test is a good one.
Another way of saying this is that you wanted to use the equality
$$ \int\_0^s dt = \int\_{x(0)}^{x(s)} \frac{dx}{f(x)} $$
but you falsely presupposed that the end state is necessarily $x(s) = \infty$.
| 4 | https://mathoverflow.net/users/3948 | 36981 | 23,791 |
https://mathoverflow.net/questions/36976 | 9 | As it is more or less well-know, and as it has come up on MO a couple of times, the $\mathbb R$-algebra $C^\infty(M)$ of smooth functions on a (say) compact manifold contains essentially everything there is to know about $M$ itself.
>
> Does one *really* need to know the $\mathbb R$-algebra structure, though? Can we manage with $C^\infty(M)$ given as an abstract ring?
>
>
>
For comparison, if I recall correctly one can reconstruct (as an analytic manifold) an open subset $U\subseteq\mathbb C$ from the field of meromorphic functions on it, given as an abstract field. (It follows, of course, that the ring of holomorphic functions also can be used to reconstruct $U$.)
| https://mathoverflow.net/users/1409 | On the $\mathbb R$-algebra structure on $C^\infty(M)$. | If $M$ is connected then one can determine $\mathbb{R}$
(the constant functions) within
$C^\infty(M)$ (the ring of smooth real-valued functions on $M$).
One can certainly determine $\mathbb{Q}$ within $C^\infty(M)$.
If $f$ is not in $\mathbb{Q}$ then it's a constant function iff
$f-a$ is a unit in $C^\infty(M)$ for all $a\in\mathbb{Q}$. The key
is that all nonconstant functions must take a rational value.
This doesn't work for non-connected $M$, since a function $f$
may take distinct constant values on a clopen partition on $M$.
Maybe one can use idempotents...
| 11 | https://mathoverflow.net/users/4213 | 36985 | 23,795 |
https://mathoverflow.net/questions/36967 | 1 | I'm concerned with a generic uniformly elliptic operator $L$ on $\mathbb{R}^n$. If $L$ is uniformly elliptic and I am studying the equation $Lu=f$ then the way I can deduce regularity on $\mathbb{R}^n$ is via the Fourier transform: $\hat{Lu} = \hat{f}$ which leads to $P(\xi)\hat{u} = \hat{f}$. From this finally I use the assumption that $P(\xi) \geq c |\xi|^2$ to deduce along with Parseval that $\|u\|\_{H^2} \lesssim \|f\|\_{L^2} + \|u\|\_{L^2}$. My question is, why does this become so complicated on bounded domains?
**Question:** Why can't we simply write $u = \sum\_k \phi\_k \hat{u}(k)$ as a Fourier series and deduce from the equation that $\|k\|^2|\hat{u}(k)|^2 = |f(k)|^2$ (the Fourier coefficients) and use this to deduce that $$\sum\_k (1+|k|^2)^2 |\hat{u}(k)|^2 \lesssim \|u\|\_{L^2} + \|f\|\_{L^2}$$ where again I've used Parseval's identity. In other words, doesn't everything from the $\mathbb{R}^n$ case just get converted into statements about the Fourier *series*? (as opposed to tranform).
Hope this is clear!
Thanks!
| https://mathoverflow.net/users/8755 | Elliptic regularity on bounded domains | Dorian, aren't you messing up things a little? Surely you can expand any $L^2$ function in a series of eigenfunctions for the elliptic operator, but please notice that this simple fact already requires quite a detailed theory of elliptic operators on bounded domains. In order to prove the existence of eigenfunctions you must be able to solve the equation $Lu=\lambda u$, and if you want to use your expansion for regularity results, you need to study the properties of the eigenfunctions, their growth etc.
But actually you are right, in a sense. It is indeed possible to prove existence and regularity of solution at the interior of the domain by using essentially the same methods as on the whole space. However, if you want to control the properties of the solution at the boundary, then this requires new tools. As a minimum, in the lucky situation of a smooth boundary, you can reduce to the case of a half space, but no less than that. If you are not convinced, think of the fact that some results cease to be true if you drop the assumption that the boundary is Lipschitz or satisfies some suitable cone condition.
As you suspect, it is also possible to do frequency space analysis much in the same way as on the whole space, but I would not call this easy. There is a beautiful set of notes "Lectures on semiclassical analysis" available on the web, by Evans and Zworsky, see Theorem 3.17 there (they prove interior Schauder estimates using Paley-Littlewood, apparently following a suggestion of H.Smith). I repeat: this is interior regularity, the behaviour at the boundary is substantially more difficult.
| 6 | https://mathoverflow.net/users/7294 | 37001 | 23,802 |
https://mathoverflow.net/questions/36968 | 5 | Let $G$ be a simple unweighted graph. The *distance* between two vertices $u,v$ in $G$ is the length of a shortest path in $G$ between $u$ and $v$. The *diameter* of $G$, denoted $diam(G)$, is the largest distance between two vertices in $G$. For a natural number $k$, The $k^{\mathrm{th}}$ power of $G$, denoted $G^{k}$, is the graph obtained from $G$ by adding edges between every two vertices $u,v$ where the distance between $u,v$ in $G$ is at most $k$.
>
> Let $\mathcal{F}\_{k}$ be the family of all graphs that are the $k^{\mathrm{th}}$ powers of graphs of diameter $k+1$. That is,
>
>
>
> >
> > $\mathcal{F}\_{k} = \{G^{k}\mid diam(G)=k+1\}$
> >
> >
> >
>
>
>
$\mathcal{F}\_{1}$ is the set of all graphs of diameter $2$, and for all $k\ge2$, $\mathcal{F}\_{k}\subseteq\mathcal{F}\_{2}$.
Is something more known about the structure of the graphs in $\mathcal{F}\_{k}$ for $k\ge2$? In particular, are these containments strict? Do we get more structure the higher the value of $k$?
Thanks for any ideas, pointers, and/or references on this.
| https://mathoverflow.net/users/1667 | What is known about the $k^{\mathrm{th}}$ powers of graphs of diameter $k+1$? | The containments must be strict for the trivial reason that each $\mathcal F\_k$ contains a graph with $k+2$ nodes (the case $G$ = path) but no graph with $k+1$ nodes (if $G$ has diameter $k+1$, it must have at least $k+2$ nodes).
For further insight into the structure of $\mathcal F\_k$, I'd study the structure of the *complements* of the graphs in $\mathcal F\_k$. Each edge in the complement of $G^k$ corresponds to a diameter of $G$; that is, if you have an edge $\{u,v\}$ in the complement of $G^k$, the distance between $u$ and $v$ is exactly $k+1$ in $G$. Thus your question is really about the structure of the diameters of graphs.
| 4 | https://mathoverflow.net/users/7170 | 37009 | 23,807 |
https://mathoverflow.net/questions/33704 | 3 | I am new to categories and I found in a book that it is possible to construct a category in which the following are true: there exist morphisms $f:A \to B$ and $g:B \to C$, and monomorphisms $\alpha:A' \to A$, $I:B' \to B$ and $J:C' \to C$ such that
(1) $I$ is an image of $\alpha$ under $f$.
(2) $J$ is an image of $I$ under $g$.
(3) $J$ is not an image of $\alpha$ under $g \circ f$.
This is rather counterintuitive, but I kind of see that what is counterintuitive is the definition of "image".
So, here comes my question: What are conditions to impose on the category to ensure that (3) does not occur?
| https://mathoverflow.net/users/7991 | Image of composite morphisms | This cannot happen in a regular category. Below I give a proof using the sequent calculus of subobjects in a regular category. It can be deciphered using the book 'Sketches of an Elephant Volume 2' by Peter T. Johnstone, in particular chapter D1.
I write $\beta:=I$ and $\gamma:=J$. I hope the definition of image given in your book is the same as mine, namely the image of a subobject (~mono) $S$ under a morphism $\phi$ is the least subobject of the codomain of $\phi$ through which $\phi\circ\overline{S}$ factors, where $\overline{S}\in S$.
Assume we know
$\exists x(\alpha(x)\wedge f(x)=y) \dashv\vdash\_{y:Y}\quad \beta(y)$ and $\exists y(\beta(y)\wedge g(y)=z) \dashv\vdash\_{z:Z}\quad \gamma(z)$. We then want to prove two things. The first is that $\exists x(\alpha(x)\wedge g(f(x))=z)\vdash\_{z:Z}\quad \gamma(z)$, the second that $\gamma(z)\vdash\_{z:Z} \quad \exists x(\alpha(x)\wedge g(f(x))=z)$.
For the first we have the following.
$\alpha(x)\wedge g(f(x))=z$
$\vdash\_{x:X,z:Z} \quad\alpha(x)\wedge g(f(x))=z \wedge f(x)=f(x)$
$\vdash\_{x:X,z:Z}\quad \alpha(x)\wedge g(f(x))=z \wedge \beta(f(x))$
$\vdash\_{x:X,z:Z}\quad \gamma(g(f(x)))$. Therefore $\alpha(x)\wedge g(f(x))=z\vdash\_{x:X,z:Z}\quad \gamma(z)$ and hence $\exists x(\alpha(x)\wedge g(f(x))=z)\vdash\_{z:Z}\quad \gamma(z)$.
The second also holds. First note that $\beta \wedge g(y)=z$
$\vdash\_{y:Y,z:Z}\quad \exists x(\alpha(x)\wedge f(x)=y)\wedge g(y)=z$
$\vdash\_{y:Y,z:Z}\quad \exists x(\alpha(x)\wedge f(x)=y\wedge g(y)=z)$
$\vdash\_{y:Y,z:Z}\quad \exists x(\alpha(x)\wedge g(f(x))=z)$ from which we may conclude that $\gamma(z)\vdash\_{z:Z} \quad \exists y(\beta(y)\wedge g(y)=z)\vdash\_{z:Z} \quad \exists x(\alpha(x)\wedge g(f(x))=z)$.
| 1 | https://mathoverflow.net/users/7747 | 37019 | 23,814 |
https://mathoverflow.net/questions/37006 | 2 | Let $p: C\to D$ be a functor, and let $f:y\to x$ be a morphism of $C$. We say that $f$ is *cartesian* if the canonical map $Q:(C\downarrow f) \to P:=(C\downarrow x)\times\_{(D\downarrow p(x)} (D\downarrow p(f))$ is a surjective (on objects) equivalence of categories. However, if we write out what the (strict 2-) pullback means, the objects are precisely the pairs of morphisms $g: z\to x$ and $h:p(z)\to p(y)$ such that $p(g)=p(f) \circ h$. If we look at the fibres of $Q$ over objects of $A$, we see that that they are contractible groupoids.
Using the more common definition of a cartesian morphism, we must show that any pair of morphisms $(g, h)$ as above *uniquely* determines an arrow $\ell:z\to y$ such that $f\circ \ell= g$ and $p(\ell)=h$.
I see how the first definition implies the existence of such a map, but how does it determine the map's uniqueness (up to more than a contractible space of choices)?
| https://mathoverflow.net/users/1353 | Equivalence of definitions of cartesian morphisms | The only morphisms in the fibers of $Q$ are identity maps, so it is actually an *isomorphism* of categories. To see this, suppose $\ell,\ell'\colon z\to y$ both induce $g\colon z\to y$. What would a morphism from $\ell$ to $\ell'$ *in the fiber* of $Q$ be? It would be a morphism $\varphi\colon z\to z$ over $y$ (the first $z$ is over $y$ via $\ell$ and the second via $\ell'$) which induces *the identity morphism* on $g$ in $(C\downarrow x)$. But $\varphi$ induces the morphism
```
φ
z ----> z
\ /
g\ /g
v v
x
```
The only way this is the identity morphism of $g$ is if $\varphi=id\_z$.
| 3 | https://mathoverflow.net/users/1 | 37020 | 23,815 |
https://mathoverflow.net/questions/37015 | 3 | Why is Beta(1,1) the maximum entropy distribution over the bias of a coin expressed as a probability given that:
* If we express the bias as odds (which is over the support $[0, \infty)$), then Beta-prime(1,1) is the corresponding distribution to Beta(1,1). Isn't the maximum entropy distribution over the positive reals the exponential distribution (which is not Beta-prime(1,1))?
* If we express the bias in log odds (which is over the support of the reals), then the logistic distribution (with mean 0 and scale 1) is the corresponding distribution to Beta(1,1).
Beta(1,1) makes sense as maximum entropy because it's flat over its support. The other distributions are not flat. If we had chosen a different parametrization, we should clearly arrive at the corresponding distribution (not something else). How are the other two distributions the maximum entropy distributions over their support? There must be some other requirement that I'm missing. What is it?
| https://mathoverflow.net/users/634 | Why is Beta the maximum entropy distribution over Bernoulli's parameter? | I think there are two separate things going on here. One is the issue of a maximum entropy distribution. The other is of whether or not distributions are invariant under different parameterizations. Regarding the second matter, I think your statement "if we had chosen a different parameterization, we should clearly arrive at the corresponding distribution" is probably not quite right (I say probably because I may be interpreting you wrong). Only particular distributions have this property and sometimes are not probability distributions. See <http://en.wikipedia.org/wiki/Jeffreys_prior> if this is what you're interested in.
ps I'd have preferred to leave this as a comment, but can't yet I guess.
| 2 | https://mathoverflow.net/users/8719 | 37027 | 23,818 |
https://mathoverflow.net/questions/36956 | 3 | If a certain property of graphs cant not be expressed by a first order logic sentence $\phi$ over $\Sigma$ then can we say with confidence that such as property can not be expressed even by a an infinite family of FOL sentences $\eta$ over $\Sigma$ ?
$\Sigma$ is the vocabulary {E,=} used to represent graph where E is a binary predicate and = has the usual meaning.
EDIT:
What if the graph is directed?
| https://mathoverflow.net/users/8246 | Graph properties and FOL | I think none of the two answers so far really addresses the edited version of the question
which deals with infinite families of first order sentences.
The following graph properties (among others) can be expressed by infinitely many first order sentences, but not by finitely many (note that finitely many is equivalent to a single one):
(1) The graph is infinite: Take the collection of all sentences $\phi\_n$ where $\phi\_n$ states that the graph has at least $n$ vertices. A compactness argument shows that this is not finitely axiomatizable. (This is Kaveh's comment.)
(2) The graph is bipartite. Take the collection of sentences $\psi\_n$ where $\psi\_n$ says that the graph has no cycles of length $2n+1$. (This is Dave Marker's comment.)
It is possible to construct infinitely many different properties of graphs or directed graphs of this nature. Just take the property saying that for a fixed finite graph $H$,
the graph $G$ contains infinitely many disjoint copies of $H$.
It is worth pointing out that "the graph is finite" i.e., the opposite of (1), cannot be expressed by even infinitely many first order sentences.
(Given a first order theory $\Phi$ that is satisfied by all finite graphs, add infinitely many constants to your vocabulary. Then, by compactness, the theory consisting of $\Phi$ together with infinitely many sentences saying that the constants are pairwise distinct has a model, an infinite graph satisfying $\Phi$.)
Similarly, not being bipartite is not axiomatizable by a first order theory
(take an ultraproduct of the graphs $C\_{2n+1}$, $n\in\mathbb N$, where $C\_{2n+1}$ is the cycle of length $2n+1$).
Being connected is also not expressible even by an infinite first order theory.
I currently don't know whether non-connectedness is first order axiomatizable.
I asked this [here](https://mathoverflow.net/questions/37034/is-non-connectedness-of-graphs-first-order-axiomatizable).
| 6 | https://mathoverflow.net/users/7743 | 37029 | 23,820 |
https://mathoverflow.net/questions/36670 | 16 | Let $X$ be a commutative H-space. A group completion is an H-map $X\to Y$, where $Y$ is another H-space, such that
* $\pi\_0(Y)$ is a group
* The Pontrjagin ring $H(Y; R)$ is the localization of the Pontrjagin ring $H\_\*(X; R)$ at the multiplicative submonoid $\pi\_0(X)$ for every coefficient ring $R$.
Perhaps most interesting is the case where $X$ is a commutative monoid or more generally an $E\_\infty$-space. In this case, May gives a functorial group completion $B\_0$ via the two-sided bar construction which is an infinite-loop space. If I understand it right, it is defined as $colim\_j \Omega^j|\Sigma^j(C\_j\times C')^\bullet X|$, where $C\_j\times C'$ denotes the monad associated to the product of the little $j$-cube operad and an $E\_\infty$-operad and $\bullet$ the simplicial variable. I have some question concerning group completions:
1) Are all group completions equivalent? That is, does there always exist a homotopy equivalence of H-spaces between them?
2) Does the group completion preserve homotopy limits? For example, does the May functor preserve homotopy limits of $E\_\infty$-spaces?
3) Suppose one knows that all loop spaces of $X$ are infinite loop spaces. Is there a simple relationship between the infinite-loop space $B\_0 X$ and the loop spaces of $X$? Especially, I am interested in the homotopy groups of $B\_0 X$.
An answer to any of these question would be helpful to me.
| https://mathoverflow.net/users/2039 | Group Completions and Infinite-Loop Spaces | A well-written discussion of the group completion can be found on pp. 89--95 of
J.F. Adam: Infinite loop spaces, Ann. of Math. studies 90 (even though he only
discusses a particular group completion of a monoid). In particular you
assumption of commutativity comes in under the assumption that $\pi\_0(M)$ is
commutative which makes localisation with respect to it well-behaved
(commutativity is not the most general condition what is needed is some kind of
Øre condition).
In any case if you really want conclusions on the homotopy equivalence level I
think you need to put yourself in some nice situation for instance requiring
that all spaces be homotopy equivalent to CW-spaces. If you don't want that you
should replace homotopy equivalences by weak equivalences, if not you will
probably find yourself in a lot of trouble. In any case I will assume that we
are dealing with spaces homotopy equivalent to CW-complexes.
Starting with 1) a first note is that your conditions does not have to involve
an arbitrary ring $R$. It is enough to have $R=\mathbb Z$ and one should
interpret the localisation in the way (for instance) Adams does:
$H\_\ast(X,\mathbb Z)=\bigoplus\_\alpha H\_\ast(X\_\alpha,\mathbb Z)$, where
$\alpha$ runs over $\pi\_0(X)$, and a $\beta$ maps $H\_\ast(X\_\alpha,\mathbb Z)$
to $H\_\ast(X\_{\alpha\beta},\mathbb Z)$. Then your group completion condition is
that the natural map $\mathbb Z[\pi\_0(Y)]\bigotimes\_{\mathbb Z[\pi\_0(X)]}
H\_\ast(X,\mathbb Z)\rightarrow H\_\ast(Y,\mathbb Z)$ should be an isomorphism.
This then implies the same for any coefficient group (and when the coefficient
group is a ring $R$ you get your condition). (Note that for this formula to even make sense we need at least associativity for the action of $\pi\_0(X)$ on the homology. This is implied by the associativity of the Pontryagin product of $H\_\*(X,\mathbb Z)$ which in turn is implied by the homotopy associativity of the H-space structure.)
Turning now to 1) it follows from standard obstruction theory. In fact maps into
simple (hope I got this terminology right!) homotopy types, i.e., spaces for
which the action of the fundamental groups on the homotopy groups is trivial (in
particular the fundamental group itself is commutative). The reason is that the
Postnikov tower of such a space consists of principal fibrations and the lifting
problem for maps into principal fibrations is controlled by cohomology groups
with ordinary coefficients. Hence no local systems are needed (they would be if
non-simple spaces were involved). The point now is that H-spaces are simple so
we get a homotopy equivalence between any two group completions and as
everything behaves well with respect to products these equivalences are H-maps.
**Addendum**:
As for 2) it seems to me that this question for homotopy limits can only be solved under supplementary conditions. The reason is that under some conditions we have the Bousfield-Kan spectral sequence (see Bousfield, Kan: Homotopy limits, completions and localizations, SLN 304) which shows that $\varprojlim^s(\pi\_s X\_i)$ for *all* $s$ will in general contribute to $\pi\_0$ of the homotopy limit. As the higher homotopy groups can change rather drastically on group completion it seems difficult to say anything in general (the restriction to cosimplicial spaces which the OP makes in comments doesn't help as all homotopy limits can be given as homotopy limits over $\Delta$. Incidentally, for homotopy colimits you should be in better
shape. There is however an initial problem (which also exists in the homotopy limit case): If you do not assume that the
particular group completions you choose have any functorial properties it is not
clear that a diagram over a category will give you a diagram when you group
complete. This can be solved by either assuming that in your particular
situation you have enough functoriality to get that (which seems to be the case
for for instance May's setup) or accepting "homotopy everything" commutative
diagrams which you should get by the obstruction theory above. If this problem
is somehow solved you should be able to conclude by the Bousfield-Kan spectral
sequence $\injlim^\ast H\_\*(X\_i,\mathbb Z)\implies
H\_\*(\mathrm{hocolim}X\_i,\mathbb Z)$. We have that localisation is exact and
commutes with the higher derived colimits so that we get upon localisation a
spectral sequence that maps to the Bousfield-Kan spectral sequence for $\{Y\_i\}$
and is an isomorphism on the $E\_2$-term and hence is so also at the convergent.
As for 3) I don't altogether understand it. Possibly the following gives some
kind of answer. For the H-space $\coprod\_n\mathrm{B}\Sigma\_n$ which is the
disjoing union of classifying spaces of the symmetric groups its group
completion has homotopy groups equal to the stable homotopy groups of spheres
which shows that quite dramatic things can happen to the homotopy groups upon
group completion (all homotopy groups from degree $2$ on of the original space
are trivial).
| 11 | https://mathoverflow.net/users/4008 | 37030 | 23,821 |
https://mathoverflow.net/questions/37034 | 14 | A recent
[question](https://mathoverflow.net/questions/36956/graph-properties-and-fol)
asked for graph properties that are first order axiomatizable but not finitely axiomatizable.
Connectedness was mentioned in the context. Connectedness can be axiomatized in infinitary logic, but not in ordinary first order logic. Just take an ultraproduct of the paths
$P\_n$ of length $n$, $n\in\mathbb N$. The paths are connected, but the ultraproduct has exactly two vertices of degree 1 and these two are not joined by a path of any finite length.
If connectedness was axiomatizable by a first order theory $\Phi$, then all the $P\_n$ would satisfy $\Phi$ and hence the ultraproduct satisfies $\Phi$. But the ultraproduct is not connected, a contradiction.
I was wondering whether non-connectedness is first order axiomatizable. I guess it is not,
but I don't have an argument for this right now.
An attempted proof goes as follows: Let $G\_n$ be the disjoint union of two cycles of length $n$. Take an ultraproduct $H$ of the $G\_n$. Now all vertices of $H$ have degree $2$ and
there are no finite cycles. In other words, $H$ is the disjoint union of a family of
infinite (in both directions) paths.
We were done if we could show that $H$ is elementary equivalent to the bi-infinite path
(is there a notation for this graph?). I assume that this is the case, but I don't see why.
A different proof that non-connectedness is not first order axiomatizable would also be welcome (or an axiomatization).
| https://mathoverflow.net/users/7743 | Is non-connectedness of graphs first order axiomatizable? | Stefan's original idea is realized in the following observation, which shows that one $\mathbb{Z}$-chain is elementary equivalent to two such chains.
Theorem. The theory of nontrivial cycle-free graphs where every vertex
has degree $2$ is complete.
Proof. All models of uncountable size $\kappa$ consist of
$\kappa$ many $\mathbb{Z}$ chains, and hence are isomorphic. Thus,
the theory is $\kappa$-categorical, and hence complete. QED
Thus, all cycle-free graphs with every vertex of degree
$2$ have the same first order theory. In particular, the graph consisting of one $\mathbb{Z}$-chain is
elementary equivalent to the graph consisting of any number
of such $\mathbb{Z}$ chains. Since the first graph is connected and
the latter are not, it follows that neither connectivity
nor disconnectivity are first-order expressible as theories
in the language of graph theory.
| 11 | https://mathoverflow.net/users/1946 | 37049 | 23,834 |
https://mathoverflow.net/questions/36995 | 25 | Some time ago, while putting my nose in the Sloane's Online Encyclopedia of Integer Sequences, I came over the sequence [A019568](http://oeis.org/A019568) defined as follows:
>
> $a(n):=$ the smallest positive integer $k$ such
> that the set $\{1^n, 2^n, 3^n,\dots k^n\}$ can be partitioned into two
> sets with equal sum.
>
>
>
In other words, $a(n)$ is the smallest $k$ such that there is a choice of signs **+** or **-** in the expression $$1^n\pm2^n\pm\dots\pm k^n \qquad\qquad(1) $$ that makes it vanish. In order to show that this $a(n)$ is a well-defined integer (that is: that at least one such $k$ does exist), a simple observation gives in fact a bound $$a(n)<2^{n+1}.$$ Reason: $(1-x)^{n+1}$ divides the polynomial
$$Q(x):=(1-x)(1-x^2)(1-x^4)\dots(1-x^{2^n})=+1-x-x^2+x^3-\dots +(-1)^n x^{2^{n+1}-1},$$
therefore, if $S$ is the shift operator on sequences, the operator $Q(S)$ has the $(n+1)$-th discrete difference $(I-S)^{n+1}$ as factor, hence annihilates any sequence that is polynomial of degree not greater than $n$. In particular, the algebraic sum (1) with the signs of the coefficients of $Q(x)$ vanishes (incidentally, the sequence of signs is the so called Thue-Morse sequence [A106400](http://oeis.org/A106400), $+--+-++--++-+--+\dots$.
However, looking at the reported values of $a(n)$ for $n$ from $0$ to $12:$
$$2,\ 3,\ 7,\ 12,\ 16,\ 24, \ 31,\ 39,\ 47,\ 44,\ 60,\ 71,\ 79,$$
it looks like the growth of $a(n)$ is much below $2^{n+1}$ (I have a weakness for sequences that grow slowly, here's possibly the main motivation of this question).
>
> **Question**: Does anybody have a reference for the
> above sequence? Can you see how to
> prove an asymptotics, or a more
> realistic bound than $a(n)<2^{n+1}$?
>
>
>
| https://mathoverflow.net/users/6101 | Asymptotic growth of a certain integer sequence | Since there are about 2k possible sums, with typical order of magnitude about kn, it seems reasonable to guess that the first case when one of these sums is 0 will occur when these 2 numbers are about equal, which is when k is about n log(n)/log(2). This incredibly crude estimate is somewhat smaller than the numerical data, but maybe suggests about the right growth rate,
| 13 | https://mathoverflow.net/users/51 | 37053 | 23,836 |
https://mathoverflow.net/questions/37052 | 10 | This is my first question with mathOverflow so I hope my etiquette is up to par here.
My question is regarding a $3\times3$ magic square constructed using the la Loubère method (see [la Loubère method](http://en.wikipedia.org/wiki/Magic_square#Method_for_constructing_a_magic_square_of_odd_order))
Using the method, I have constructed a magic square and several semimagic squares (where one or both of the diagonals do not add up to a magic sum) with a program on written on my graphing calculator. After playing around with the program, I was shocked that the determinants of these $3\times3$ magic squares are all the same (specifically -360). Why is this so? (I am still an undergraduate so please go easy on the math :] )
| https://mathoverflow.net/users/8866 | Determinant of a $3\times3$ magic square | I don't have an explanation, but here is an outline of a proof (I've checked all the details myself, but it's laborious to write up correctly) that what you claim to happen actually does happen.
**Result: Let $M$ be a 3x3 integer matrix whose columns, rows, diagonal, and anti-diagonal each total 15. The following are equivalent:**
1. **The entries of $M$ are distinct and positive;**
2. **The determinant of $M$ is $\pm 360$;**
3. **Calling $h,i$ the (3,2) and (3,3) entries, $$(h,i)\in \{(1,6), (1,8), (3,4), (3,8), (7,2), (7,6), (9,2), (9,4)\}.$$**
The proof is just writing down the equations for the sums, rows, etc., to be 15 and solving, and then writing down the formula for the determinant and simplifying (it turns out $det = 45(h-5)(h+2i-15)$) and solving the resulting diophantine equation.
The explanation may well be just that there aren't very many 3x3 magic squares (basically, it looks like there's just one that we rotate and flip to get eight). For 4x4, the solution space will be 7d instead of 2d, and that is a *lot* of extra freedom. If you find a simply-described structure in the determinants of those, I'd be surprised.
| 12 | https://mathoverflow.net/users/935 | 37056 | 23,838 |
https://mathoverflow.net/questions/36999 | 22 | I'd like to learn a bit about uniform spaces, why are they useful, how do they arise, what do they generalize, etc., without getting away from the context of general topology. I have to prepare an 1h30min talk on the subject, for an audience formed in standard general topology (i.e. Munkres), not so much in abstract algebra (so I'd like not to use topological groups).
The references I have are Kelley or Willard texts on Topology, Isbell's "Uniform Spaces" and James's "Topological and Uniform Spaces".
I discarded the last one because of its heavy use of filters from the beginning. I don't know about any other good references. What I'd like to see is the subject treated as Munkres does in his book: he gives good motivations, pictures, and is gentle to the reader.
| https://mathoverflow.net/users/6249 | A good place to read about uniform spaces | I would motivate them as follows: if topological spaces were invented to give a general meaning to "continuous function", then uniform spaces were invented to give a general meaning to "uniformly continuous function". It is clear what "uniformly continuous" should mean for metric spaces and topological groups, but how should the general notion be formalized?
Once this is formalized, one can define the notion of Cauchy net in a uniform space (which is something you cannot do for general topological spaces). This leads to the notion of completeness of course (every Cauchy net converges to *at least* one point), although the theory is much cleaner for complete *Hausdorff* uniform spaces, where you have convergence to *at most* one point as well.
To illustrate this: the Cauchy completion of a uniform space $X$ can be defined in the usual way via equivalence classes of Cauchy nets. It is a complete Hausdorff uniform space $\bar{X}$ together with a map $i: X \to \bar{X}$ which satisfies a universal property: given a complete Hausdorff uniform space $Y$ and a *uniformly* continuous function $f: X \to Y$, there is a unique uniformly continuous map $\bar{f}: \bar{X} \to Y$ such that $\bar{f} \circ i = f$. (If you omit "Hausdorff" or "uniformly", you lose the universal property, which is arguably the point of the completion.)
The nLab has an article on uniform spaces with some material not included in the Wikipedia article.
| 15 | https://mathoverflow.net/users/2926 | 37057 | 23,839 |
https://mathoverflow.net/questions/37044 | 5 | Is the following statement, refining classical Cauchy-Davenport Theorem (that states that for sets $A$, $B$ of residues modulo prime $p$, $|A+B|\geq |A|+|B|-1$ provided that RHS does not exceed $p$) true/known?
Let $A$, $B$ be two subsets of $\mathbb{F}\_p$, $p$ being prime, and $|A|+|B|\leq p+1$. Then a complete bipartite graphs with parts $A$, $B$ (rigorously speaking, disjoint copies of $A$ and $B$, say $A\times \{0\}$ and $B\times\{1\}$) have a spanning tree such that all $|A|+|B|-1$ edgesums are different. (for edge $e=(a,b)$ its edgesum is defined as $\,a+b$).
| https://mathoverflow.net/users/4312 | Cauchy-Davenport strengthening? | I believe that your statement follows from Cauchy-Davenport via matroid intersection theorem. (Matroid intersection theorem is stated in Chapter 41 of Alexander Schrijver's "Combinatorial optimization" book and can be also found [here](http://www-math.mit.edu/~goemans/18997-CO/co-lec13.ps).)
You want to find a "rainbow" spanning tree in a complete bipartite graph you define, where colors correspond to edgesums. "Rainbow" spanning trees, in fact, seem to be commonly used as an example of matroid intersection.
By matroid intersection it suffices to show that for any set of edges $U$ in your graph
$r\_1(U)+r\_2(E \setminus U) \geq |A|+|B|-1,$
where:
$E$ is the set of all $|A||B|$ edges,
$r\_1(U)$ is the rank of $U$ in the cycle matroid, and is equal to $|A|+|B|-c(U)$ where $c(U)$ is the number of connected components in the graph induced by $U$, and
$r\_2(E \setminus U)$ is the number of edgesums obtained by the edges not in $U$.
If $c(U)=1$ then we are done. Otherwise, let $A' \subseteq A$, $B' \subseteq B$ be obtained from $A$ and $B$ by choosing one element from each component of the graph induced by $U$, so that both are non-empty. Then the edges between $A'$ and $B'$ are not in $U$ and thus by Cauchy-Davenport
$r\_2( E\setminus U) \geq c(U)-1$,
as desired.
| 4 | https://mathoverflow.net/users/8733 | 37059 | 23,841 |
https://mathoverflow.net/questions/37055 | 1 | Given vectors $V$ of length $d$, construct a graph $G = (V, E)$ where $\{u, v\} \in E$ iff the Pearson correlation between $u$ and $v$ is larger than some threshold $t > 0$. Is $G$ chordal? It seems like it should be, because a long chordless cycle like $a$ correlates with $b$, $b$ with $c$, $c$ with $d$ but nothing else seems difficult to construct. However, I cannot find a simple proof or a reference.
| https://mathoverflow.net/users/7016 | Is the graph of a thresholded correlation matrix chordal? | It was described in [this](https://mathoverflow.net/questions/19406/constructing-bernoulli-random-variables-with-prescribed-correlation) previous question how to obtain a correlation matrix whose entries come from the scalar product of certain vectors $u\_1, u\_2, \dots,u\_n$. If we let the vectors be $$u\_i=(1, \cos(\frac{2\pi i}{n}), \sin(\frac{2\pi i}{n}),0,\dots, 0)$$ we can set a high enough threshold so that the corresponding graph is a cycle of length $n$ and thus not chordal.
| 2 | https://mathoverflow.net/users/2384 | 37060 | 23,842 |
https://mathoverflow.net/questions/37084 | 2 | This question is related to this [Question](https://mathoverflow.net/questions/36956/graph-properties-and-fol).
Above questions revealed that even though FOL is not expressive enough to describe properties such as Connectivity, Bipartite etc. It is possible to express these properties as infinite FOL sentences.
Now the question I got in my mind is: is it possible to express ALL properties of graph using such infinite families of FOL sentences?
An example showing that it is not is what I want.
| https://mathoverflow.net/users/8246 | Graph properties and infinite FOL sentences | Let me suppose for simplicity at first that we are speaking here just of countable graphs. There are continuum many isomorphism types of countable graphs, and any collection of such isomorphism types would seem to constitute a *property* of countable graphs. Thus, there are $2^{2^\omega}$ many properties of countable graphs.
If your infinitary logic does not have this many sentences, therefore, then clearly not every property can be described by an infinitary sentence. For example, if as you seem to, you only allow countable conjunctions, disjunctions and quantifications, so that every sentence is countable, then there are only continuum many sentences, but more than continuum many properties.
If you are more generous with the infinitary sentences, by allowing larger iterated disjunctions (but still just countable quantifications in any one block of quantifiers), then the answer is yes. The reason is that every countable graph $G$ is uniquely specified up to isomorphism by a sentence $\sigma\_G$ in infinitary logic saying: there are such-and-such vertices connected in exactly such-and-such way and no other vertices (according to what is true in the particular graph $G$). This assertion, which is expressible provided that we have countable conjunctions and quantifications, is true in a graph exactly when it is isomorphic to $G$. Now for any property $P$, let $\Gamma\_P$ be the disjunction over all $\sigma\_G$ among the $G$ with property $P$, which will be a size at most continuum disjunction, if we use at most one $G$ of each isomorphism type. The sentence $\Gamma\_P$ is true of a countable graph exactly when that graph has property $P$. (The same idea works for uncountable $G$ by allowing larger quantifier blocks and disjunctions.)
I take this answer to show that one needs to be more specific about which infinitary logic is to be used. If the logic is too generous, then the infinitary language doesn't seem really to give any insight to the nature of the graphs, since it amounts to describing the property just by listing the graphs that have that property, and we can already do this in our ordinary mathematical treatment of graphs, without any formal language. But when the logic is restricted, for example, if you consider a specific infinitary logic $L\_{\kappa,\lambda}$ and focus in detail on what is describable in that logic, then interesting answers may emerge. The answer above shows that larger logics will be more expressive.
| 5 | https://mathoverflow.net/users/1946 | 37088 | 23,857 |
https://mathoverflow.net/questions/37087 | 1 | Is it true that if an isogeny between two principally polarized abelian varieties respects the polarization, then it is in fact an isomorphism?
| https://mathoverflow.net/users/2008 | Morphism between polarized abelian varieties | That should be true, yes.
A polarization of $A$ is given by a bilinear form on $H\_1(A, Z)$; this is equivalent to a map $H\_1(A,Z) \to H\_1(A,Z)^\vee$, which is an isomorphism if the polarization is principal.
A map between two abelian varieties is given by a corresponding linear map $H\_1(A\_1, Z) \to H\_1(A\_2, Z)$. The map between the varieties is an isomorphism if the map on $H\_1$ is.
The map induced on the bilinear form then is the composition
$$
H\_1(A\_1,Z)\to H\_1(A\_2,Z) \to H\_1(A\_2,Z)^\vee \to H\_1(A\_1,Z)^\vee
$$
If the form is respected by this map, then this is an isomorphism. Consequently, the left-hand map must be as well, as claimed.
| 3 | https://mathoverflow.net/users/1703 | 37089 | 23,858 |
https://mathoverflow.net/questions/37083 | 14 | The (uniform) word problem for groups can be stated in several equivalent ways:
**Word Problem for Groups (WP)**
*Instance*: A finite presentation of a group G and an element w of G as a product of generators and their inverses.
*Question*: Does every linear representation of G in a (not necessarily finite-dimensional) Hilbert space map w to the identity operator?
*Equivalently*: Does every unitary representation of G in a Hilbert space map w to the identity operator?
*Equivalently*: Does every normal subgroup of G contain w?
*Equivalently*: Is w=1 in G?
As is well-known, WP is undecidable; more specifically, it belongs to RE∖coRE, where [RE](http://qwiki.stanford.edu/wiki/Complexity_Zoo:R#re) is the class of recursively enumerable languages and [coRE](http://qwiki.stanford.edu/wiki/Complexity_Zoo:C#core) is the class of their complements. (In fact, it is known that WP remains undecidable even if we fix the group G and its finite presentation suitably, but that is not the topic of this question.)
We can consider the finite-dimensional version of WP.
**Finite-Dimensional Word Problem for Groups (FWP)**
*Instance*: Same as WP.
*Question*: Does every *matrix* representation of G map w to the identity matrix?
*Equivalently*: Does every unitary *matrix* representation of G map w to the identity matrix?
*Equivalently*: Does every normal subgroup of G *of finite index* contain w?
(The equivalence is based on a result by Malcev; see [this post](https://mathoverflow.net/questions/9628/finitely-presented-sub-groups-of-gln-c/9635#9635) by Greg Kuperberg and the comments attached to it.)
The two problems have different answers for some instances; that is, sometimes w≠1 in G but every matrix representation maps w to the identity matrix. There even exists a finitely presented infinite group which does not have a nontrivial matrix representation. See the answers to the question “[Finitely presented sub-groups of GL(n,C)](https://mathoverflow.net/questions/9628/finitely-presented-sub-groups-of-gln-c)” by Dmitri.
In fact, FWP is in coRE unlike WP: if we also give the dimension as part of the input, the problem becomes decidable (because it is a system of algebraic equations over ℂ), and therefore FWP can be put in coRE by trying all dimensions.
>
> **Question**: Is FWP decidable?
>
>
>
I do not even know whether FWP is NP-hard or not. As far as I know, FWP can be in P or undecidable, or anywhere in between!
(Note: FWP can be viewed as a (very) special case of the problem discussed in the question “[Decidability of matrix algebra](https://mathoverflow.net/questions/33879/decidability-of-matrix-algebra)” by Ricky Demer.)
| https://mathoverflow.net/users/7982 | Finite-dimensional version of the word problem for groups | FWP is undecidable by a [result of Slobodskoi.](https://doi.org/10.1007/BF01735740) Slobodskoi shows that the "Universal theory" of finite groups is undecidable. What you are asking for is whether the "Q-theory" of the pseudovariety of finite groups is decidable. The universal theory and Q-theory are equivalent for the pseudovariety of finite groups (see [Kharlampovich-Sapir](https://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.12.766), section 2.4 for a discussion), so the Q-theory is undecidable too.
| 19 | https://mathoverflow.net/users/1345 | 37095 | 23,862 |
https://mathoverflow.net/questions/37082 | 7 | I have recently made the following observation:
Let $v\_i := (v\_{i1}, v\_{i2})$, $1 \leq i \leq k$, be non-zero *positive* elements of $\mathbb{Q}^2$ such that no two of them are proportional. Let $M$ be the $k \times k$ matrix whose entries are $m\_{ij} := \max${$v\_{ik}/v\_{jk}: 1 \leq k \leq 2$}. Then $\det M \neq 0$.
The above statement is equivalent to the basic case of a result I recently discovered about pull back of divisors under a birational mapping of algebraic surfaces. I was going to include it as a part of another paper, then noticed the equivalent statement stated above and found it a bit amusing. My question is: is it worthwhile to try to publish it in a journal (as an example of an application of algebraic geometry to derive an arithmetic inequality), and if it is, then which journal(s)?
It is of course also very much possible that it is already known, or has a trivial proof (or counterexample!) - anything along those directions would also be appreciated.
**Edit:** Let me elaborate a bit about the geometric statement. In the 'other' paper, I define, for two algebraic varieties $X \subseteq Y$, something called "linking number at infinity" (with respect to $X$) of two divisors with support in $Y \setminus X$. I can show that when $Y$ is a surface, (under some additional conditions) the matrix of linking numbers at infinity of the divisors with support in $Y \subseteq X$ is non-singular. In a special (toric) case, the matrix of linking numbers takes the form of $M$ defined above. So the question is if the result about non-singularity of the matrix and its corresponding implication(s) are publishable anywhere.
| https://mathoverflow.net/users/1508 | Appropriate journal to publish a determinantal inequality | It seems to be true, but relatively simple (unless I made a mistake). Let's see:
First of all, scaling any pair $(v\_{i,1},v\_{i,2})$ by a constant $c$ does not change the determinant (one row of the matrix is multiplied by $c$, and one column is divided by $c$). We can therefore assume without losing generality that $(v\_{i,1},v\_{i,2})=(v\_i,1)$.
Also, permutation of $v\_i$'s does not change the determinant; therefore, we may assume that
$v\_i$'s are strictly increasing. In this case, the matrix has the form
$$m\_{ij}=\begin{cases} 1,& i\le j\cr v\_i/v\_j,& i>j\end{cases}.$$
Now subtract the top row from all others, and shift it to the bottom. You end up with a triangular matrix whose determinant is non-zero.
| 11 | https://mathoverflow.net/users/2653 | 37113 | 23,873 |
https://mathoverflow.net/questions/37115 | 76 | It seems to me like every book on representation theory leaps into groups right away, even though the underlying ideas, such as representations, convolution algebras, etc. don't really make explicit use of inverses. This leads to the fairly natural question of how much of representation theory still works for monoids (or even semigroups)?
I would imagine that irreducible representations probably exist in some form, since all you really need is to have some way to reduce monoid-modules over the complex numbers. Of course it could also be that monoid representation theory is in some form fatally flawed and reduces down to something boring for reasons that are not obvious to me at this point.
| https://mathoverflow.net/users/4642 | Why aren't representations of monoids studied so much? | Certainly irreducible representations exist; one can still construct the monoid algebra of a monoid and consider modules over the algebra. But Maschke's theorem is false in general for finite monoids. Indeed, consider the monoid $M = \langle x | x^3 = x^2 \rangle$. Complex (for the sake of argument) representations of $M$ are the same as representations of the monoid algebra $\mathbb{C}[x]/(x^3 - x^2)$ and this algebra is not semisimple, so its finite-dimensional representations are not completely reducible. (Note that the usual proof of Maschke's theorem fails miserably; you can't average inner products over a monoid without inverses, and any unitary representation of a monoid has to factor through the Grothendieck group).
So part of the answer may just be that the representation theory of monoids is inherently more complicated. Although this isn't doesn't seem to be stressed much in textbooks, having inverses is a pretty important structural property of groups; it endows group algebras with an antipode and endows the category of representations of a group with duals.
| 73 | https://mathoverflow.net/users/290 | 37116 | 23,874 |
https://mathoverflow.net/questions/37051 | 2 | Let $f=(\varphi,\theta):X\longrightarrow S$ a morphism of preschemes whith $\varphi$ surjective. Let $\theta(S):\Gamma(S,O\_S)\longrightarrow \Gamma(S,f\_\* O\_X)=\Gamma(\varphi^{-1}(S),O\_X)=\Gamma(X,O\_X)$.
What conditions can we put on $f$ in order to get that the morphism $\theta(S)$ is
(1) injective
(2) surjective if it is already injective (and so bijective)?
| https://mathoverflow.net/users/8736 | When does the global sections of a prescheme X over an other S equals those of S?n | Obligatory tautological answer: since $f$ includes the data of $\theta$, it's necessary and sufficient to require that $\theta(S)$ is injective (resp. surjective).
Since you're trying to extract information about $\theta(S)$, my guess is that you actually want conditions on $\varphi$, $S$, and/or $X$, which don't specifically refer to $\theta$.
Injectivity
-----------
Suppose $I$ is the kernel of $\theta(S)$. Then the morphism $X\to Spec(O\_S(S))$ factors through the closed subscheme $Spec(O\_S(S)/I)$. Since $X\to S$ is surjective, this closed subscheme must contain the image of $S$ (set theoretically). The closure of the image of $S$ in $Spec(O\_S(S))$ is all of $Spec(O\_S(S))$,† so $Spec(O\_S(S)/I)$ must be set-theoretically equal to $Spec(O\_S(S))$, so $I$ is contained in every prime of $O\_S(S)$, so it is in the nilradical.
>
> In particular, if $S$ is reduced, $\theta(S)$ is injective.
>
>
>
Note that we didn't actually need $\varphi$ surjective, just that the closure of the image is all of $S$. Since composing $X\to S$ with a nilpotent thickening of $S$ doesn't change $X$ or $\phi$ at all, I imagine you can't get a better condition than this without saying something directly about $\theta$.
† If $S$ is quasi-compact and
its image in $Spec(O\_S(S))$ is contained in the complement of some basic open neighborhood $D(f)$, then $f\in O\_S(S)$ vanishes at every point of $S$, so $f$ is nilpotent on any affine open subscheme of $S$. Since $S$ is quasi-compact, there is a single $n$ such that $f^n$ is identically zero on $S$. Since $f$ is nilpotent, $D(f)=D(f^n)\subseteq Spec(O\_S(S))$ is empty. So the image of $S$ is not in the complement of any non-empty open subset of $Spec(O\_S(S))$.
There must be a way to show this without assuming quasi-compactness of $S$, but I don't see it right now. Suppose $S\to Spec(O\_S(S))$ *set theoretically* factors through a closed subset $Z$. Why must there be a *closed subscheme* structure on $Z$ so that $S$ factors scheme theoretically through $Z$? A scheme theoretic factorization of $S$ through $Spec(O\_S(S)/I)$ amounts to a factorization of the identity map $O\_S(S)\to O\_S(S)$ through the quotient $O\_S(S)\to O\_S(S)/I$, which cannot happen unless $I=0$.
Surjectivity
------------
I can't think of a good condition to ensure surjectivity that isn't essentially tautological.
There's no condition you can put on $\varphi$ or the underlying topological space of $X$. If $S=Spec(k)$ with $k$ a field, you could take $X$ to be a projective or affine curve over $S$. These two curves are homeomorphic, yet you have a surjection in one case but not the other.
A necessary but insufficient condition for isomorphism is that $S$ have as many connected components as $X$, since connected components correspond to irreducible idempotents in the ring of regular functions.
Note that if $X$ and $S$ are arbitrary projective schemes over a field, then any morphism between them induces an isomorphism on global regular functions. Any Stein morphism $f\colon X\to S$ obviously makes $\theta(S)$ an isomorphism. Is there an example of a surjective non-Stein $f$ which makes $\theta(S)$ an isomorphism?
| 0 | https://mathoverflow.net/users/1 | 37117 | 23,875 |
https://mathoverflow.net/questions/37114 | 5 | For definitions of graph minors and topological minors, see wikipedia's article on [graph minors](http://en.wikipedia.org/wiki/Graph_minor).
Theorem: For every graph H, there is a finite set of graphs, say S(H), such that G contains H as a minor if and only if G contains some graph from S(H) as a topological minor.
Can anyone point me to a paper/book where this is proved? (I know how to prove it, I just want a reference to cite.)
| https://mathoverflow.net/users/8075 | Ref request: A graph G contains H as a minor iff it contains one of finitely many graphs as a topological minor | Hello !!! You will find this theorem as results 2.2 and 2.3 in Graph minors VIII : A Kuratowski theorem for general surfaces.
Nathann
| 5 | https://mathoverflow.net/users/1715 | 37120 | 23,876 |
https://mathoverflow.net/questions/36795 | 16 | I would like the simplest example of the failure of an ODE to be locally diffeomorphic to its linearization, despite being locally homeomorphic to it. More precisely, consider x' = f(x) with f(0) = 0 in R^n. Let A = f'(0) so that the local linearization is x' = Ax. Suppose the eigenvalues of A all have nonzero real part (i.e., 0 is a hyperbolic critical point).
The Hartman-Grobman theorem tells us that there is a homeomorphism of a neighborhood of 0 which conjugates the system x'=f(x) to its linearization x'=Ax. If one reads the elementary `differential equations from the dynamical systems point of view' literature, however, you will gain the false impression that there is a *diffeomorphism* h : U --> U of a nhood of 0 which does this, and that further, one can even do this with h'(0) = I, the identity matrix. The point of this is to ensure that the trajectories of the nonlinear system are tangent to the trajectories of the linearization: if $h'(0)\neq I$ then this may be false.
Smale's stable manifold theorem gives partial information in this direction, saying that the stable manifolds of the system and its linearization are tangent, and similarly for the unstable manifolds. In 2D, at a saddle, this is sufficient to imply that the separatrices of the original system are tangent to those of the linearization. For a node in 2D, or in higher dimensions, I am under the impression this need not hold. I even think I had worked out an example many years ago, which I no longer recall.
Any enlightenment on this issue would be much appreciated. I am not at all expert in these matters, so welcome any corrections, if I have distorted the facts. I am hoping for a 2 dimensional example.
**Added later:**
Yuri: resonances and normal forms are definitely relevant. When I get time I will look into the references you suggest.
Here's an example of what I am trying to avoid. Consider a flow on the unit disk with trajectories the radial lines y = mx. Conjugate by $(r,\theta) \mapsto (r,f(r,\theta))$ where $f(0,\theta)$ is constant on, say, $[-\pi/2,\pi/2]$, e.g. $f(r,\theta) = r\theta$ on $[-\pi/2,\pi/2]$ and $(2\theta-\pi) + r(\pi-\theta)$ in the left half plane. Now all the trajectories leaving the unit circle in the right half plane approach 0 along the positive x-axis. So, conjugating with such a homeomorphism has replaced a single trajectory with horizontal tangent by an entire interval of such.
I strongly suspect that this sort of pathology doesn't happen with polynomial flows. Perhaps the normal forms will show this. What I would hope is that for each slope, there is a 1-1 correspondence between the trajectories in the original flow and its linearization approaching or leaving the singular point at that slope. In particular, that you can't have a single trajectory in the linearization but a whole interval of them in the nonlinear flow. A counterexample to this hope would be disappointing, but would settle the matter.
| https://mathoverflow.net/users/6872 | Local linearization of ODE at singular point | In three dimensions, Hartman gave the example $dx/dt=ax$, $dy/dt=(a-b)y+cxz$, $dz/dt=-bz$ where $a>b>0$ and $c \neq 0$. On the other hand, any $C^2$ *planar* flow is $C^1$ linearizable (another result by Hartman), so you will not find any polynomial examples in the plane. See [Linearization via the Lie Derivative](http://www.emis.de/journals/EJDE/Monographs/02/abstr.html) by Chicone & Swanson for references and more details.
(Strengthening the smoothness assumption will not give any more than $C^1$ linearizability even in the planar case, as shown by the $C^{\infty}$ example $dx/dt=-x$, $dy/dt=-2y+x^2$, which is easily solved explicitly: $x(t)=x\_0 e^{-t}$, $y(t)=(y\_0+x\_0^2 t) e^{-2t}$. The solution curves are of the form $y=Ax^2-B x^2 \ln|x|$, and are therefore only $C^1$ at the origin, whereas the linearized system of course has smooth solution curves.)
| 5 | https://mathoverflow.net/users/4678 | 37122 | 23,878 |
https://mathoverflow.net/questions/37119 | 3 | The question arose while comparing the notions of compactness, countable compactness, local compactness, and "Lindelofness" in Hausdorff spaces. It is straightforward to show that compactness implies any of the other properties. I found ready counterexamples (I will be glad to provide them if asked) for all but one of the other possible implications, namely the question of whether countable compactness implies local compactness. The paper "On Countably Compact Nonlocally Compact Spaces" by T. B. Rushing shows that, in general, countable compactness does not imply local compactness. The examples he provides, however, are not Hausdorff.
| https://mathoverflow.net/users/8871 | Does countable compactness imply local compactness in Hausdorff spaces? | Examples abound: take for instance a $\Sigma$-product of two-point spaces. To be specific let $X$ be the set of points in $\lbrace0,1\rbrace^{\omega\_1}$ that have only countably many coordinates that are $1$. This set is dense but not open in the product, hence not locally compact but it is countably compact as each countably infinite subset sits in a compact subset of $X$.
| 8 | https://mathoverflow.net/users/5903 | 37124 | 23,879 |
https://mathoverflow.net/questions/37132 | 5 | Given a [CAT(0) space](http://en.wikipedia.org/wiki/CAT%2528k%2529_space) $X$ and a compact, convex subset $A$ of $X$. One can define its midpoint $m(A)$ as the point, at which the following function attains its minimum.
$f:A\rightarrow \mathbb{R}\qquad x\mapsto \sup\{d(x,y)|y\in A\}$
One can show, that there is a unique such point. So my question is:
Given two compact,convex subset $A\_1,A\_2\subset X$. Is $d(m(A\_1),m(A\_2))$ less or equal to the [Hausdorff distance](http://en.wikipedia.org/wiki/Hausdorff_distance) between $A\_1$ and $A\_2$?
| https://mathoverflow.net/users/3969 | Stability of midpoints in CAT(0) spaces | No, even if $X=\mathbb R^2$.
Let $A\_1$ be (the convex hull of) 4 points with coordinates $(\pm 1,\pm 1)$. Then $m(A\_1)=(0,0)$, as the 4 points are on the circle $S\_1$ of radius $\sqrt 2$ centered at $(0,0)$. Shift $S\_1$ a small distance $\varepsilon$ in the horizontal direction, denote the resulting circle by $S\_2$. For each vertex of $A\_1$, mark its nearest point on $S\_2$. The marked points are vertices of a convex quadrangle $A\_2$ inscribed in $S\_2$ and containing its center $(\varepsilon,0)$. Hence $m(A\_2)=(\varepsilon,0)$ but the Hausdorff distance between $A\_1$ and $A\_2$ is $\approx\varepsilon/\sqrt 2$.
| 7 | https://mathoverflow.net/users/4354 | 37134 | 23,884 |
https://mathoverflow.net/questions/37107 | 7 | Can integrals of the form
$$
\int\_{-\infty}^{\infty}{\exp\left(-\left[x - c\right]^{2}\right) \over 1 + x^{2}}\, {\rm d}x
$$
be computed in closed form using contour integration (or any other technique)? If
$c = 0$, the integral is $\pi{\rm e\ erfc}\left(1\right)$, but I'm interested in
$c$ real and non-zero.
( In probability terms, the integrand is a product of normal and Cauchy densities. )
| https://mathoverflow.net/users/136 | Contour integration problem from probability | $$ J(c)=\int\_{-\infty}^{\infty}\frac{\exp[-(x-c)^2]}{1+x^2}dx=e^{-c^2}\int\_{-\infty}^{\infty}\frac{\exp[-x^2]}{1+x^2} e^{2cx}dx $$
The integral on the right can be treated as the Fourier transform $\mathcal{F}(\exp[-x^2]/(1+x^2))$, with the transform parameter equal to $\mbox{i}2c$. The function is actually symmetric wrt $x$, thus, it is the cosine Fourier transform we are talking about. The necessary transform is available in Vol. 1 of Bateman & Erdelyi's "Tables of Integral Transforms" (1954). I used a shortcut and computed the transform using Maple. The resulting expression is:
$$ J(c)=\frac{\pi\mbox{e}}{2}\left( \mbox{erfc}(1+\mbox{i}c)e^{\mbox{i}2c}+\mbox{erfc}(1-\mbox{i}c)e^{-\mbox{i}2c} \right) $$
It is easy to check that this answer satisfies the ODE obtained by fedja. Written as the sum of conjugate terms, the function $J(c)$ is clearly real-valued for real $c$. It remains an open question whether this is a "nicer" form compared to what you had originally!
| 6 | https://mathoverflow.net/users/8670 | 37138 | 23,886 |
https://mathoverflow.net/questions/37130 | 2 | In [this](http://gdz.sub.uni-goettingen.de/no_cache/en/dms/load/img/?IDDOC=213742) and [this](http://www.jstor.org/pss/1971047) papers Thurston proves that every distribution is homotopic to an integrable one (in the first one for codimension greater than one and in the other for codimension one).
Recently, I've came up with a [nice paper](http://www.pdmi.ras.ru/~svivanov/papers/coherence.pdf) by Burago and Ivanov which with some other hypothesis manage to start with a distribution and construct a foliation tangent to arbitrarily small cone field around it.
I am far from foliations, but I was wondering if there are examples of distributions which are not integrable and can not be perturbed in the $C^0$ topology in order to become integrable (Edit: That is, a distribution is $\epsilon$ close to other if it is contained pointwise in a cone of angle $\epsilon$ of the original one.)
References appreciated!
| https://mathoverflow.net/users/5753 | Integrability of distributions close to a given one. | No smooth non-integrable distribution can be $C^0$ approximated by integrable ones.
For example, consider the following 2-dimensional distribution in $\mathbb R^3$: the plane at $(x,y,z)\in\mathbb R^3$ is spanned by vectors $(1,0,0)$ and $(0,1,x)$. Perturb this distribution within a small $C^0$ distance $\varepsilon\ll 1$. Consider the square in $\mathbb R^2$ with vertices $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$ and let $\gamma$ be its boundary (counter-clockwise). This $\gamma$ has a "lift", that is a curve $\tilde\gamma$ in $\mathbb R^3$ which is tangent to the distribution and whose projection to the horizontal plane is $\gamma$. The lift is found by solving an o.d.e., so it is unique if the distribution is smooth but may be non-unique if it is only $C^0$. In the non-perturbed case, the unique lift ends at $(0,0,1)$, hence in the perturbed case all lifts end near $(0,0,1)$. This implies that the distribution is not integrable - if it was integrable, there would be at least one lift (the one contained in a leaf of a foliation) that ends near the origin.
The proof in the general case is similar.
| 4 | https://mathoverflow.net/users/4354 | 37139 | 23,887 |
https://mathoverflow.net/questions/37097 | 3 | It can be seen [here](http://www-groups.dcs.st-and.ac.uk/~john/Zagier/Solution1.1.html) that the only numbers for which $n^{m+1}\equiv n \bmod m$ is true are 1, 2, 6, 42, and 1806. Through experimentation, it has been found that $\displaystyle\sum\_{n=1}^{m}{n^m}\equiv 1 \bmod m$ is true for those numbers, and (as yet unproven) no others. Why is this true?
If there is a simple relation between $n^{m+1} \bmod{m}$ and $n^m \bmod{m}$, that would probably make this problem make more sense. It is obvious that $n^m \equiv 1 \bmod{\frac{m}{d}}$ (dividing out $n$ from both sides gives this result) for all $n$ on the interval $[1,m]$ where $d$ is a divisor of $m$. As a result of this, $n^m \bmod{m}$ takes on only values of the form $1+k \frac{m}{d} \bmod m$ where $k = -1, 0, 1$. How can it be shown that the sum of those values is equivalent to $1 \bmod{m}$?
I have a proof somewhat in the works [here](http://mathbin.net/51297), but it's not anywhere near complete (it doesn't even relate the 5 integers to the actual problem). Am I missing something in going from the first relation to the second?
| https://mathoverflow.net/users/8875 | Why are the only numbers $m$ for which $n^{m+1}\equiv n \bmod m$ also the only numbers such that $\displaystyle\sum_{n=1}^{m}{n^m}\equiv 1 \bmod m$? | Let $m > 2$ be an integer such that $S\_m(m) = \sum\_{n=1}^{m-1} n^m\equiv 1 \bmod{m}$. (Taking away $m^m$ does not harm the question, of course). Then $S\_m(m)$ has the following expression in terms of Bernoulli numbers:
\begin{equation\*}
S\_m(m) = \sum\_{k=0}^{m}\binom{m}{k}B\_{m-k}\frac{m^{k+1}}{k+1}
= B\_m \cdot m + B\_{m-1} \frac{m^2}{2} + \binom{m}{2} B\_{m-2} \frac{m^3}{3} + \cdots
\end{equation\*}
By the theorem of Clausen and von Staudt, the denominator of $B\_k$ is the product of all primes $p$ such that $p-1$ divides $k$. In particular, they are square-free and hence we have the last displayed term and all that follow are conruent to $0$ modulo $m$. If the term
$B\_{m-1} \frac{m^2}{2}$ were not congruent to $0$ by the same argument, then $m$ would be even. But then $B\_{m-1}=0$ unless $m=2$.
So $S\_m(m) \equiv B\_m \cdot m\bmod{m}$. It is now clear that $m$ must be square -free as otherwise $S\_m(m) \not\equiv 1\bmod{m}$. Let $p$ be a prime dividing $m$. If now $B\_m\cdot m$ is $p$-integral, then $p-1$ must divide $m$. Now look at the answer to the question about the "wrong little Fermat". One derives that $m$ has to be in the list $\{1,2,6,42,1806\}$ just as before. Then one checks by hand that it is true for these integers.
| 12 | https://mathoverflow.net/users/5015 | 37141 | 23,888 |
https://mathoverflow.net/questions/37128 | 2 | To explain my problem, I must give a lemma:
>
> Let $X$, $Y$, $Z$ be curves over $k$ (of characteristic 0) such that the genus of $Z$ is greater than 2, and $\pi : X \to Y$, $\phi : X \to Z$ two non-constant morphisms.
> If $\phi^\star(H^0(Z,\Omega))\subseteq\pi^\star(H^0(Y,\Omega))$, where $\Omega$ denotes the sheaf of regular 1-forms in each case, then there exists a non-constant morphism $u: Y \to Z$ such that $\phi = u \circ \pi$.
>
>
>
Now, in a proof, I saw the use of this lemma, except that the hypothesis was the inclusion $\mathrm{Image}(\mathrm{Jac}(Z) \to \mathrm{Jac}(X)) \subseteq \mathrm{Image}(\mathrm{Jac}(Y) \to \mathrm{Jac}(X))$, instead of $\phi^\star(H^0(Z,\Omega))\subseteq\pi^\star(H^0(Y,\Omega))$. I can guess it is equivalent, but why? Is it related to Grothendieck's duality? Did I miss something obvious?
| https://mathoverflow.net/users/8786 | About a non-obvious (?) link between the jacobians of curves and differentials | Since you are in char zero, you can assume the ground field is the complex numbers. The inclusion of jacobians follows from the inclusion of spaces of differentials via the description in terms of periods and calculus.
| 3 | https://mathoverflow.net/users/2290 | 37146 | 23,893 |
https://mathoverflow.net/questions/37147 | 2 | As well known, Perelman proved Poincare conjecture by proving Thurston's Geometrization conjecture.
Somebody says that we can understand part of the universe from Poincare conjecture.
As a purely topological viewpoint, why do you think the poincare conjecutre is important and how about Smooth poincare conjecture in dimension 4?
One may simply answer that because it is extremely difficult to prove or because it made 100 years of development of history of Topology or geometry....
But I want to know your professional, new, own viewpoint.
| https://mathoverflow.net/users/7776 | The importance of Poincare Conjecture or SPC4? | Mine is no profesional, and certainly don't believe its new nor own, but I'll give it a try.
In my opinion, algebraic topology tries to caracterize nice topological spaces (say CW complexes) modulo homotopy equivalence (which is the reasonable equivalence given the fact that the invariants used are usually invariants under homotopy equivalence). This caracterization has a good important theorem (for me) which is [Whitehead's theorem](http://en.wikipedia.org/wiki/Whitehead_theorem).
When studying manifold topology, one would like to get clasification modulo *homeomorphisms* so, the above study it is not enough. This gives great importance to theorems such as the clasification of spheres by homotopy type.
I think it is interesting that the result from this point of view had been solved for tori (which are much simpler from the point of view of higher homotopy groups) by [Hsiang and Wall](http://blms.oxfordjournals.org/cgi/content/citation/1/3/341) (and maybe others).
In brief, I believe that Poincare Conjecture it is one of the central and most natural questions one can pose in manifold topology (or geometric topology?).
| 5 | https://mathoverflow.net/users/5753 | 37149 | 23,895 |
https://mathoverflow.net/questions/37150 | 0 | Let $X$ be a finite CW-complex with only even cells $x\_1,\ldots, x\_k$ and let $Y$ be the complex obtained by attaching one more even cell to $X$, call it $y$. Assume both $X$ and $Y$ are connected. The quotient complex $Y^n/X^n$ has the cell structure with one cell for each product of cells $e\_1\times\cdots\times e\_n$ where the $e\_i$ are either equal to $x\_i$ or $y$ and at least one $e\_i$ is equal to $y$. I want to say that this amounts to a wedge of products of spheres each product of spheres depending on how many copies of $y$ there is in that cell. Is this correct?
| https://mathoverflow.net/users/8658 | What Is This Quotient Space? | Suppose Y is obtained by attaching a zero-cell, so $Y = X \cup \{\ast\}$. Then $Y^2$ is $$(X \times X) \cup (X \times \{\ast\}) \cup (\{\ast\} \times X) \cup (\{\ast\} \times \{\ast\})$$
and so $Y^2/X^2$ is homeomorphic to
$$
\{\ast\} \cup X \cup X \cup \{\ast\}.
$$
This can be arbitrarily complicated depending on X.
ADDENDUM: By request, a connected example is the inclusion $\mathbb{CP}^1 \subset \mathbb{CP}^2$. The quotient $Y^2/X^2$, in this case, has a nonzero cohomology operation $Sq^2$ from H6 to H8 with ℤ/2-coefficients, and there is no wedge of products of spheres that can have this cohomology. (You should work out the details.)
| 5 | https://mathoverflow.net/users/360 | 37159 | 23,902 |
https://mathoverflow.net/questions/37157 | 0 | I recently came across a system of PDEs
$\frac{\partial S}{\partial z}= f\_1(x,y,z,w,t)$,
$\frac{\partial S}{\partial w}= f\_2(x,y,z,w,t)$,
$\frac{\partial S}{\partial t}= f\_3(x,y,z,w,t)$,
$S(x,y,1,1,1)=f\_4(x,y)$,
where $S$ is an unknown function of five variables $x,y,z,w,t$ and $f\_i$ are known.
The question is how to obtain a general solution for $S$?
| https://mathoverflow.net/users/6594 | Solution for a system of PDEs | What you have is a family of PDEs labelled by $x,y$ and for fixed $x,y$ you have an equation
$$dS = f\_1 dz + f\_2 dw + f\_3 dt$$
for a function $S:\mathbb{R}^3 \to \mathbb{R}$ of the three variables $z,w,t$. The first thing to check is that the equation is integrable: namely, that the 2-form $d(dS) = 0$. If that is the case, then $S$ exists up to a locally constant function, which you fix from the fourth condition.
To find the explicit solution, choose any path $\gamma$ from $(1,1,1)$ to $(z,w,t)$ and integrate the right-hand side of the equation:
$$S(x,y,z,w,t) = S(x,y,1,1,1) + \int\_\gamma (f\_1 dz + f\_2 dw + f\_3 dt).$$
| 7 | https://mathoverflow.net/users/394 | 37164 | 23,906 |
https://mathoverflow.net/questions/37172 | 58 | What are the open big problems in algebraic geometry and vector bundles?
More specifically, I would like to know what are interesting problems related to moduli spaces of vector bundles over projective varieties/curves.
| https://mathoverflow.net/users/8234 | What are some open problems in algebraic geometry? | A few of the more obvious ones:
\* [Resolution of singularities in characteristic p](http://en.wikipedia.org/wiki/Resolution_of_singularities)
\*[Hodge conjecture](http://en.wikipedia.org/wiki/Hodge_conjecture)
\* [Standard conjectures on algebraic cycles](http://en.wikipedia.org/wiki/Standard_conjectures) (though these are not so urgent since Deligne proved the Weil conjectures).
\*Proving finite generation of the canonical ring for general type used to be open though I think it was recently solved; I'm not sure about the details.
For vector bundles, a longstanding open problem is the classification of vector bundles over projective spaces.
(Added later) A very old major problem is that of finding which moduli spaces of curves are unirational. It is classical that the moduli space is unirational for genus at most 10, and I think this has more recently been pushed to genus about 13. Mumford and Harris showed that it is of general type for genus at least 24. As far as I know most of the remaining cases are still open.
| 39 | https://mathoverflow.net/users/51 | 37173 | 23,912 |
https://mathoverflow.net/questions/20071 | 83 | I am very interested in reading some and skimming through the list of invited talks at the International Congress of Mathematicians. Since the proceedings contain talks supposedly by top experts in each area, even the list of invited talks would hopefully provide some picture of how mathematics changed throughout the last century or so.
I looked it up but [wikipedia](http://en.wikipedia.org/wiki/International_Congress_of_Mathematicians) only provides links to the proceedings of ICM since 1998. So it excludes many talks I really want to read, like those by Serre, Grothendieck, Auslander, Quillen, etc.
**Does anyone know how to find the rest of the ICM proceedings, hopefully online?** Thanks.
UPDATE: The whole collection of *all* the proceedings of the ICMs is available [here](https://www.mathunion.org/icm/proceedings)!
| https://mathoverflow.net/users/2083 | How to find ICM talks? | **Update:** (Oct. 2018) For the first time, all ICM 2018 lectures (plenary, invited and special) as well as panels and special events are [presented (by good-quality videos) on the ICM 2018 You tube channel](https://www.youtube.com/channel/UCnMLdlOoLICBNcEzjMLOc7w/videos?disable_polymer=1).
**Update:**(Dec 2017) The ICM launched a new website. All previous ICM proceedings are available [here](https://www.mathunion.org/icm/proceedings). However, I cannot find the old page with access to individual papers and search options.
Just recently the International Mathematical Union (IMU) put online all the previous proceedings on the ICM's! [Here](http://www.mathunion.org/ICM/): (update:broken). This webpage is based on joint work by R. Keith Dennis (Ithaca) and Ulf Rehmann (Bielefeld). So you can read for free all the articles (including Hilbert's famous problem paper; Martin Grötschel demonstrated it at the opening ceremony of ICM 2010). I don't know when the proceedings of ICM2010 will be added. **UPDATE:** the 2010 articles have now been added! **Update** The ICM2014 papers are available [here](http://www.icm2014.org/en/vod/proceedings.html) (you can download each of 4 volumes; the IMU site does not contain these papers yet.)
As for the talks themselves, there is a page with all the ICM 2010 plenary talks [here](http://www.icm2010.in/from-the-venue/online-streaming-archives); links for videos from earlier ICM's (plenary talks and other events) can be found [here](http://www.mathunion.org/activities/icm).
**Update (August, 6 2014)** Many (46 for now) of the ICM 2014 proceedings contributions are already available on arXiv, via [this search](http://arxiv.org/find/all/1/co:+icm/0/1/0/all/0/1). (I got it from Peter Woit's blog.) Videos of lectures [can be found here](http://www.icm2014.org/en/vod/vod).
**Update (May 2014):** Starting 1992 there is also every four years the European Congress of Mathematics (ECM) that the European Mathematical Society (EMS) is running.
The proceedings of the first three ECMs are now freely open. These volumes are available [here](http://ada00.math.uni-bielefeld.de/ECM/).
(Digitising the proceedings of the first three ECMs, published by Birkhäuser was a task carried out by the EMS Electronic Publishing Committee.)
Starting with the 4ECM (2004), the Proceedings are published by the EMS Publishing House. The EMS decided to make them freely available online too. I expect that this will happen soon and I will keep you posted. **Further update (June 2014)** The ECM Proceedings are now available [here](http://www.euro-math-soc.eu/ECM/)! I was told that in a few months, the EMS will put also the 6ECM volume.
| 60 | https://mathoverflow.net/users/1532 | 37175 | 23,914 |
https://mathoverflow.net/questions/36866 | 1 | Hey There,
i have a simple question:
What are $\kappa$-categories?
Do you have something related for further reading?
thx and greetings,
frosch03
edit:
There is actually a paper called "Closed Freyd- and kappa-categories" by A. John Power and Hayo Thielecke (portal.acm.org/citation.cfm?id=646229.681558)
| https://mathoverflow.net/users/8820 | What are κappa-categories? | The intuitive explanation is that $\kappa$-categories are to first-order functions what cartesian closed categories are to higher-order functions.
This all started with Lambek's work on polynomial categories; the best reference for that is
*[J. Lambek. Functional completeness of cartesian categories. Annals of Mathematical Logic, 6:251–292, 1973](http://dx.doi.org/10.1016/0003-4843%2874%2990003-5)*. That paper introduced the choice of the letter $\kappa$. A polynomial category is what you get when you take a category with a terminal object, pick an object $X$, and then freely adjoin a new morphism $f:1\to X$ and close under composition. This is very much like the ring of polynomials $R[X]$ over a ring $R$ arrived at by adjoining an indeterminate element and closing under the ring operations. Lambek shows that this property can be stated in universal terms -- as the unique category admitting a particular kind of functor from the original category. This is formally more satisfying but not the best route for beginners.
Later, Hasegawa developed this idea much further in [*M. Hasegawa. Decomposing typed lambda calculus into a couple of categorical programming languages. Lecture Notes in Computer Science, 953, 1995.*](http://dx.doi.org/10.1007/3-540-60164-3_28) He showed that just as the $\lambda$-calculus can be used as a "syntax" for specifying morphisms in a cartesian closed category, so too can the $\kappa$-calculus -- roughly the $\lambda$-calculus without first-class functions -- be used as a "syntax" for specifying morphisms. I recommend studying the figure immediately after the first paragraph of section 3 in his paper (very carefully). It conveys both the essence of these categories and their relevance to the study of programming languages.
To wrap up, cartesian closed categories have been an immensely useful tool in understanding programming languages with first-class functions. Unfortunately they can only be used to study languages with the property that for every pair of types $B$ and $C$ there is also a type of functions from $B\Rightarrow C$ and the ways of getting a $B\Rightarrow C$ from an $A$ are in one-to-one-correspondence with the ways of getting a $C$ from a (cartesian) pair of an $A$ and a $B$. Lambek-Hasegawa $\kappa$-categories are an elegant way to extend these techniques to languages in which this assumption does not hold.
Lastly, as a postscript, both Lambek and Hasegawa assume that the underlying monoidal structure of their categories is Cartesian. Some of the most fascinating results arise when you repeat their constructions in categories which are merely binoidal or premonoidal -- you'd be surprised how few modifications are required.
| 12 | https://mathoverflow.net/users/2361 | 37180 | 23,918 |
https://mathoverflow.net/questions/37182 | 9 | Let $\phi$ define a $\*$-automorphism from the matrix algebras $M\_n(\mathbb{C})$ to $M\_n(\mathbb{C})$ such that $\phi(I) = I$. Is it true that any such map $\phi$ can be represented as $\phi(x) = U x U^{\dagger}$ (where $U$ is a suitable unitary matrix)? If not, what is the most general expression?
| https://mathoverflow.net/users/8890 | Representation of $*$-automorphism on finite dimensional matrix algebras | Here is one generalization:
>
> Every $\*$-automorphism of the algebra of compact operators on a Hilbert space is conjugation by a unitary operator on that space.
>
>
>
Using the fact that the algebra of compact operators is irreducible, this can be seen as a special case of:
>
> Every irreducible $\*$-representation of the algebra of compact operators on a Hilbert space is unitarily equivalent to the identity representation.
>
>
>
A proof can be found for instance in Section 1.4 of Arveson's *[An invitation to C`*` algebras](http://books.google.com/books?id=zRep59fhoGkC&lpg=PP1&dq=arveson%2520invitation&pg=PA20#v=onepage&q&f=false)*. Another proof of the first assertion that gives more information can be found in Proposition 1.6 of Raeburn and Williams's *[Morita equivalence and continuous trace C`*`-algebras](http://books.google.com/books?id=KV_cC5uYA7EC&lpg=PP1&pg=PA5#v=onepage&q&f=false)*.
The first part is still true if you take all bounded operators instead of only the compact ones. (And these are the same thing in the finite dimensional case.)
| 6 | https://mathoverflow.net/users/1119 | 37187 | 23,924 |
https://mathoverflow.net/questions/28329 | 6 | Given a weighted directed graph $G=(V,E, w)$, suppose we generate a new graph $G'=(V,E,w')$ with the same vertices and edges, but now letting the weight of edge $(i,j)$ be
an exponential random variable with mean $w\_{ij}$. My question is: what is the
expected diameter of $G'$?
Why I'm interested in this: I was intrigued by the observation that the expected diameter of $G'$ can be
quite different from the diameter of $G$. Indeed, consider the following example: define $G$ by taking the complete graph $K\_{n+1}$, picking an arbitrary vertex $a$, and assigning weight $n$ to any edge incident on $a$, and weight $1$ to
every other edge. Then, the diameter of $G$ is $n$. On the other hand, the expected diameter
of $G'$ is O(1) since we can expect one of the edges incident on $a$ to have small weight.
| https://mathoverflow.net/users/1407 | diameter of a graph with random edge weights | For the special case of the complete graph $K\_n$ which you mention in your post, Svante Janson answered your question in [this paper](http://portal.acm.org/citation.cfm?id=971602); the answer is that the weighted diameter grows like $3 \log n$ in probability.
There is also some very nice [work by Bhamidi et. al](http://www.win.tue.nl/~rhofstad/ERRG_fin.pdf) on this question when the underlying graph is the giant component of Erdos--Renyi random graph $G\_{n,c/n}$ with $c>1$ fixed, although they only prove lower bounds. [Amini et. al.](http://www.di.ens.fr/~amini/Publication/Flooding.pdf) (link is to a PDF) have found the asymptotics of the weighted diameter for random graphs with a given degree sequence, under some conditions, for degree sequences which in particular result in graphs which are with high probability connected. [Ding et.al.](http://arxiv.org/abs/0906.1840) (Theorems 3.7 and 3.8 of the linked paper) prove quite refined estimates, and tail bounds, for the weighted diameter of random $d$-regular graphs, for $d \geq 3$. (Since random regular graphs are a special case of random graphs with a given degree sequence, the results of Amini et. al. and Ding et.al. have some overlap).
There is also related work, on the *hopcount* of randomly weighted graphs. The hopcount is what you get if you count the number of edges on the smallest-weight path. The primary interest of Bhamidi et. al. in fact seems to be hopcounts rather than weighted path lengths.
| 4 | https://mathoverflow.net/users/3401 | 37190 | 23,926 |
https://mathoverflow.net/questions/37111 | 8 | Is $\Psi^0(\mathbb{R})$ (pseudodifferential operators with symbols obeying
$
|\partial^\alpha\_x \partial^\beta\_\xi a(x,\xi)| \leq C\_{\alpha,\beta} (1+|\xi|)^{-|\beta|}
$
) a $C^\*$-algebra?
In other words, is $\Psi^0(\mathbb{R})$ is closed in $\mathcal{L}(L^2(\mathbb{R}))$ in the operator norm topology?
---
If not, then is there any nice characterization by the $C^\*$-algebra generated by $\Psi^0$? Alternatively, what is the strongest (or just a reasonable) topology on $\mathcal{L}(L^2(\mathbb{R}))$ such that $\Psi^0$ is a closed subspace?
---
**Edit:** Per Yemon Choi's comments below, the above question seems somewhat hopeless. As described [here](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=MR&pg5=TI&pg6=PC&pg7=ALLF&pg8=ET&r=1&review_format=html&s4=MR90d%3A47016&s5=&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq), $\Psi^0(\mathbb{R})$ is a Fréchet $\*$-algebra with a topology stronger than the operator topology. I assume that this is the topology given by the seminorms on symbols:
$$
\Vert a \Vert\_{\alpha,\beta} = \sup\_{x,\xi \in \mathbb{R}} (1+|\xi|)^{|\beta|} |\partial^\alpha\_x \partial^\beta\_\xi a(x,\xi)|.
$$
So, in addition to the above question, I am adding the following question, to make it so that there might be an answer:
Is there a reasonable description of the smallest $C^\*$-algebra containing $\Psi^0$?
| https://mathoverflow.net/users/1540 | What is the smallest $C^*$-algebra containing the "standard" pseudodifferential operators? | I have to confess to being more confused by the theory of pseudodifferential operators than I should be, but I think an answer to a question at least related to yours is briefly discussed in chapter 2 of Higson and Roe's Analytic K-homology.
Begin with an open subset $U$ of $\mathbb{R}^n$ and consider a smooth complex valued function $\sigma$ on $T^\*(U)$ with the following properties:
* $\sigma(x, t \xi) = \sigma(x, \xi)$ for $t \geq 1$, $|\xi| \geq 1$
* $\sigma(x, \xi)$ vanishes for $x$ outside of some compact subset of $U$
Define the pseudodifferential operator associated to $\sigma$ to be the operator
$D\_\sigma f(x) = \frac{1}{(2 \pi)^n} \int \sigma(x, \xi) \hat{f}(\xi) e^{i(x,\xi)} d\xi$
The first condition on $\sigma$ above implies that $\sigma$ gives rise to a function on the cosphere bundle $S^\*M$ (the symbol of $D\_\sigma$). $D\_\sigma$ extends to a bounded operator on $L^2(U)$, so consider the C\*-algebra $\mathcal{B}(U)$ generated by the $D\_\sigma$. Higson and Roe assert that the map sending $D\_\sigma$ to its symbol extends to a surjective $\*$-homomorphism from $\mathcal{B}(U)$ to $C\_0(S^\* U)$ whose kernel is precisely the C\* algebra of compact operators on $L^2(U)$.
Thus a certain class of pseudodifferential operators generates a C\* algebra extension of $C\_0(U)$ by the compact operators. I make no claim about the relationship between $\mathcal{B}(U)$ and your $\Psi^0$, but maybe this statement is still useful to you.
| 6 | https://mathoverflow.net/users/4362 | 37200 | 23,930 |
https://mathoverflow.net/questions/36987 | 8 | Let $M^3$ be a rational homology 3-sphere. (i,e, $M^3$ is closed 3-manifold with
$H\_{\*}(M;Q)=H\_{\*}(S^3;Q)$
As beautifully explained in Ranicki's Algebraic and Geometry surgery book and Davis-Kirk's Lecture notes in Algebraic toplogy book, we have a $Q/Z$ valued linking form, $\lambda\colon H\_{1}(M;Z)\times H\_{1}(M;Z)\to Q/Z$ defined by the adjoint to following isomorphism.
(Actually we can define linking form in more general setting, e.g.) odd dimensional manifold without the restriction such as rational homology sphere condition. Because I want to just intuitive idea about linking form, I restricted the case)
$H\_1(M;Z)\cong H^2(M;Z)\cong H^1(M;Q/Z)=Hom(H\_1(M;Z),Q/Z)$
First isomorphism : Poincare duality,
Second isomorphism : Inverse of Bockstein homomorphism $\delta$ induced form $0\rightarrow Z \rightarrow Q\rightarrow Q/Z\rightarrow 0$. More precisely, induced long exact sequence is that $\ldots\rightarrow H^1(M;Q)\rightarrow H^1(M;Q/Z)\rightarrow H^2(M;Z)\rightarrow H^2(M;Q)$ and here $H^1(M;Q)$ and $H^2(M;Q)$ vanishes. Therefore, the long exact sequence shows that $H^1(M;Q/Z)$ and $H^2(M;Z)$ are naturally isomorphic (if $M$ is rational homology 3-sphere) and we call that homomorphism as Bockstein homomorphism.
Third isomorphism : Universal Coefficient theorem
In short, $\lambda\colon H\_{1}(M;Z)\times H\_{1}(M;Z)\to Q/Z$ is defined by $\lambda(x,y)=<\tilde{x}\cup\delta^{-1}(\tilde{y}),[M]>$, where $\tilde{x},\tilde{y}$ are Poincare dual to $x,y$ and $\delta\colon H^1(M;Q/Z)\to H^2(M;Z)$ is a Bockstein homomorphism as defined above. Cup products are defined in $H^1(M;Z)\times H^2(M;Q/Z) \to H^3(M;Q/Z)$ induced from the Bilinear map $Z \times Q/Z \to Q/Z$.
I understand this linking form only algebraic viewpoint. Therefore, I can play with this form only by using dilluminating algebraic topology.
I'm struggle to find geometric interpretation but I have no idea to express the Q/Z valued in terems of geometric language which seems to be highly algebraic. (Feeling like Injective, divisible, Ext or something linke that )
Are there any intuitve and geometric (clean) interpretation about this linking form? (e.g.)such as Alexander duality, like the argument that .........we can find a dual basis which represents meridian......)
| https://mathoverflow.net/users/7776 | Intutive interpretation about Linking forms | This may be a good place to explain the well-known principle
$$\text{intersection in the interior = linking in the boundary}$$
in an oriented $m$-dimensional manifold with boundary $(M,\partial M)$. Let
$$f~:~(K,\partial K)\subset (M,\partial M)~,~g~:~(L,\partial L) \subset (M,\partial M)$$
be embeddings of oriented manifolds with boundary, such that
$${\rm dim}(K)~=~k~,~{\rm dim}(L)~=~\ell~,~k+\ell~=~m~,~
f(\partial K) \cap g(\partial L)~=~\emptyset \subset \partial M~.$$
Assume there exists an isotopy (= homotopy through embeddings) rel $\partial K$
$$f\_t~:~K \to M~~(0 \leqslant t \leqslant 1)$$
such that $f\_0=f$ and $f\_1(K)\subset \partial M$, with each $f\_t(K), g(L)$ intersecting transversely in $M$, so that $f\_t(K) \cap g(L)\subset M$ is a finite set with an
intersection index $I(x)\in \{\pm 1\}$ at each point $x \in f\_t(K) \cap g(L)$
according to the orientations. A continuity argument shows that the function
$$\lambda~:~[0,1] \to {\mathbb Z}~:~t\mapsto \lambda(t)=
\sum\limits\_{x \in f\_t(K) \cap g(L)}I(x)$$
is constant, so that
$${\rm intersection}(f(K),g(L)\subset M)~=~\lambda(0)~=~\lambda(1)$$
$$=~{\rm linking}(f(\partial K),g(\partial L) \subset \partial M) \in {\mathbb Z}~.$$
This is best seen by drawing pictures for $(M,\partial M)=(D^2,S^1)$.
The localization exact sequence in algebraic $L$-theory is based on an abstract homological version of this principle.
| 11 | https://mathoverflow.net/users/732 | 37217 | 23,940 |
https://mathoverflow.net/questions/37195 | 10 | Given a topological space X one can define several notion of compactness:
X is **compact** if every open cover has a finite subcover.
X is **sequentially compact** if every sequence has a convergent subsequence.
X is **limit point compact** (or Bolzano-Weierstrass) if every infinite set has an accumulation point.
X is **countably compact** if every countable open cover has a finite subcover.
X is **σ-compact** if it is the union of countably many compact subspaces.
X is **pseudocompact** if its if its image under any continuous function to $\mathbb{R}$ is
bounded.
X is **paracompact** if every open cover admits an open locally finite refinement (i.e. every point of X has a neighborhood small enough to intersect only finitely many members of the cover).
X is **metacompact** if every open cover admits a point finite open refinement (i.e. if every point of X is in only finitely many members of the refinement).
X is **orthocompact** if every open cover has an interior preserving open refinement (i.e. given an open cover there is a open subcover such that at any point, the intersection of all open sets in the subcover containing that point is also open).
X is **mesocompact** if every open cover has a compact-finite open refinement (i.e. given any open cover, we can find an open refinement such that every compact set is contained in finitely many members of the refinement).
So, there are quite a few notions of compactness (there are surely more than those I quoted up here). The question is: where are these definitions systematically studied? What I'm interested in particular is knowing when does one imply the other, when does it not (examples), &c.
I can fully answer the question for the first three notions:
Compact and first-countable --> Sequentially compact.
Sequentially compact and second-countable --> Compact.
Sequentially compact --> Limit-point compact.
Limit point compact, first-countable and $T\_1$ --> Sequentially compact.
but I'm absolutely ignorant about the other cases. Has this been systematically studied somewhere? If so, where?
| https://mathoverflow.net/users/6249 | Different forms of compactness and their relation | I don't have Munkres' book. So I don't know what is done there. You should probably consult the "Counterexamples in Topology" as mentioned above.
My favourite book for questions of this type is "General Topology" by Ryszard Engelking.
It has a diagram in the back with interrelations between different properties of topological spaces. It has all the notions you are interested in, except the last two. Local compactness and various additional versions of paracompactness are also discussed in the book.
I assume, but haven't checked in every case, that the relations in the diagram are proved in the book.
The book also contains plenty of examples, showing that certain implications do not hold, sometimes in the form of exercises with hints.
| 7 | https://mathoverflow.net/users/7743 | 37221 | 23,944 |
https://mathoverflow.net/questions/37188 | 13 | (For information on cardinal characteristics of the continuum aka cardinal invariants see Joel David Hamkins' MO answer [here](https://mathoverflow.net/questions/8972#9027); Andreas Blass's [handbook article](http://www.math.lsa.umich.edu/~ablass/hbk.pdf) is an excellent reference.)
Problem 2.3 of Shelah's ["On What I Do Not Understand (and Have Something to Say), Part I"](http://arxiv.org/abs/math/9906113) (published in 2000 in Fundamenta Mathematicae) states, "Investigate cardinal invariants of the continuum showing $\geq 3$ may have prescribed order". One major barrier to such an investigation is the fact that countable support iteration of proper forcings yields models where the continuum is $\aleph\_2$. In such models given any three cardinal characteristics at least two will have to be equal.
My question is the following. To what extent has such an investigation been pursued? In either the literature or folklore are there any results proving the consistency of inequalities $\mathfrak{c}\_0<\mathfrak{c}\_1<\mathfrak{c}\_2$ where the $\mathfrak{c}\_i$ are cardinal characteristics?
| https://mathoverflow.net/users/2436 | Consistency results separating three cardinal characteristics simultaneously | There is a paper of Shelah and Goldstern devoted to the separation of many simple cardinal invariants (this is a technical term): *Many simple cardinal invariants* ([Sh:448](https://shelah.logic.at/papers/448/)). There are more recent papers on this subject by Kellner and Shelah, if I remember correctly.
An easy case that I am very familiar with are the so called localization numbers.
A closed set $S\subseteq\omega^\omega$ is $n$-ary if in the tree $T(S)$ of finite initial segments of elements of $S$ every node has at most $n$ immediate successors.
For $n\geq 2$ let $\ell\_n$ be the least size of a family of $(n-1)$-ary sets that covers all of $n^\omega$.
Any finitely many $\ell\_n$ can be separated from each other simultaneously.
This is shown in [Geschke, Kojman, [Convexity numbers of closed subsets in R^n](https://doi.org/10.1090/S0002-9939-02-06437-7),
Proc. Am. Math. Soc. 130, No. 10, 2871-2881 (2002)], which is [here](https://www.math.uni-hamburg.de/home/geschke/publikationen.html.en) ([Wayback Machine](http://web.archive.org/web/20200104133729/https://www.math.uni-hamburg.de/home/geschke/publikationen.html.en)).
Proofs of such statements usually involve forcing with large countable support products over a model of GCH rather than iterated forcing. However, there are also some examples that use iterated forcing. See for example the paper *The covering numbers of Mycielski ideals are all equal*, by Shelah and Steprans ([Sh:665](https://shelah.logic.at/papers/665/)).
| 10 | https://mathoverflow.net/users/7743 | 37233 | 23,949 |
https://mathoverflow.net/questions/22869 | 7 | Let $X$ be a compact oriented manifold, and $A$ and $B$ closed oriented submanifolds intersecting cleanly. Then I've always been under the impression that pushing forward a cohomology class from $A$ to $X$ and then pulling back from $B$ should have a base change formula where instead one pulls back to $A\cap B$ and pushes forward to $B$.
Of course, this couldn't possibly be right if $A$ and $B$ aren't transverse. I think in the non-transverse case, one should correct by the Euler class of the excess bundle $T\_{A\cap B}X/(T\_{A\cap B}A+T\_{A\cap B}B)$.
All of my intuition for algebraic topology tells that this true and easy to prove, but of course, one can't write that in a paper.
>
> Does anyone know a convenient reference for this fact? I've tried to find it via Google, but apparently can't find the right keywords, and a quick scan of Hatcher came up negative.
>
>
>
| https://mathoverflow.net/users/66 | Reference for base change of cohomology pull-push for clean intersections. | Apologies if this is too late, but the canonical reference for this is Quillen's seminal paper
"Elementary proofs of some results of cobordism theory using Steenrod operations" Advances in Math. **7** 1971 29--56 (1971).
The proof given there is for complex cobordism and is entirely geometric. Presumably Quillen learned this from Bott, who gives a clean intersection formula in his paper
"On the iteration of closed geodesics and the Sturm intersection theory"
Comm. Pure Appl. Math. **9** (1956), 171--206.
If you are interested in the generalisation to immersions, see
F. Ronga, "On multiple points of smooth immersions" Comment. Math. Helv. 55 (1980), no. 4, 521--527.
| 5 | https://mathoverflow.net/users/8103 | 37240 | 23,955 |
https://mathoverflow.net/questions/37231 | 7 | Suppose that we have a parametrization via polynomials as follows:
$$t\longrightarrow (f\_1(t),\ldots,f\_n(t)),$$
where $t$ is a vector in $\mathbb{C}^r$ and $f\_i$ are polynomials of arbitrary degree.
Can we always find equations such that the image is an affine algebraic variety?
The question is motivated by Exercise 1.11 in Hartshorne:
>
> Let $Y\subseteq A^3$ be the curve given parametrically by $x = t^3, y= t^4, z = t^5$. Show
> that $I(Y)$ is a prime ideal of height 2 in $k[x,y,z]$ which cannot be generated by
> 2 elements.
>
>
>
I am not interested in the exercise in particular. Finding the variety is easy sometimes, for instance $t\rightarrow (t^2,t^3)$ is given by $I(x^3-y^2)$.
I am looking for a result which says that the image is always an affine algebraic variety AND a procedure to find the ideal.
| https://mathoverflow.net/users/1887 | Are all parametrizations via polynomials algebraic varieties? | I can't comment (b/c I'm not a registered user) but let me add: in case the dimension of the domain is 1 (as in your motivating example) the image is in fact an affine variety. To see this, note that the map can always be extended to a map from the projective line to projective space by homogenizing things (compare with Dan's example---if you tried homogenizing his map, you'd get $[x:y:z] \mapsto [xy:yz:z^2]$ which is not defined if $x=z=0$ or $y=z=0$), and use the fact that the image of a projective variety by a regular map is closed. Finally, observe that the image of the affine line you started with is precisely the intersection of the image of the homogenized map with the affine coordinate patch determined by your homogenization. Therefore the image of the map you started with *is* an affine variety. So Hartshorne's example is not an accident.
| 7 | https://mathoverflow.net/users/8552 | 37243 | 23,958 |
https://mathoverflow.net/questions/37239 | 2 | Does anyone know a continuous group (not necessarily locally compact) with dense cyclic subgroup other than a torus?
| https://mathoverflow.net/users/8906 | Dense cyclic subgroup | You already have some examples in the other answers. Groups which have a dense cyclic subgroup are called *Monothetic* groups. In the article "On monothetic groups" by P.R. Halmos and H. Samelson, you can find many of their properties, such as
>
> Every compact connected separable (abelian) group is monothetic.
>
>
>
| 5 | https://mathoverflow.net/users/2384 | 37245 | 23,960 |
https://mathoverflow.net/questions/37223 | 20 | Let G be the (non-principal) ultraproduct of all finite cyclic groups of orders n!, n=1,2,3,... . Is there a homomorphism from G onto the infinite cyclic group?
| https://mathoverflow.net/users/nan | Ultraproducts of finite cyclic groups | I think the answer is no. The ultraproduct $U$ is naturally a quotient of ${\mathbb Z}^{\infty}$, the direct product of countably many copies of ${\mathbb Z}$. In the obvious quotient map, the image of the direct sum is zero. Now, it is enough to show that:
Claim: Any homomorphism $ \phi: {\mathbb Z}^{\infty} \to {\mathbb Z}$ that vanishes on the direct sum is identically zero.
Proof: (I learned this from a book by T.Y.Lam):
For a prime number $p$, let $A\_p$ be the set of elements in ${\mathbb Z}^{\infty}$ of the form $(a\_0, pa\_1, p^2a\_2,...)$, i.e. the elements whose $i$-th coordinate is divisible by $p^i$. Any element $x \in A\_p$ can be decomposed as
$x= y+z= (a\_0, pa\_1, \dots, p^{n-1}a\_{n-1}, 0,0, \dots ) + p^n (0,0,..,0, a\_n, pa\_{n+1},..)$
Now, $y$ is in the direct sum, hence $\phi(y)=0$. Also $\phi(z) \in p^n {\mathbb Z}$, which
implies that $\phi (x) \in \cap\_{n=1}^{\infty} p^n {\mathbb Z} = \{ 0 \}$
Now, choose two distinct primes $p$ and $q$. Since $\gcd(p^n,q^n)=1$, it is easy to see that
$A\_p+A\_q= {\mathbb Z}^{\infty}$. This implies that $\phi \equiv 0$.
| 27 | https://mathoverflow.net/users/3635 | 37249 | 23,962 |
https://mathoverflow.net/questions/37248 | 0 | Hi, is it possible to give an explicit formula for the function G(s) defined for positive s as
$G(s) := \lim\_{N\to\infty} \sum\_{k=1}^N \frac{1}{k}{N\choose k} \left(\frac{s}{N}\right)^k \left(1-\frac{s}{N}\right)^{N-k}$.
Wolfram Mathematica says the sum for finite $N$ is some hypergeometric function, the limit of it as $N\to\infty$ is not able to express explicitely.
Thanks, Ian
| https://mathoverflow.net/users/8908 | Looking for an explicit formula for a limit of a binomial-like expression | Unless I've made some horrible miscalculation your limit is the same as
$$\lim\_{N\to \infty}\left(1-\frac{s}{N}\right)^N \int\_{0}^{\frac{s}{N-s}}\frac{(1+x)^N-1}{x} \ dx$$
which is equal to
$$e^{-s} \lim\_{N\to \infty} \int\_{0}^{\frac{Ns}{N-s}}\frac{\left(1+\frac{y}{N}\right)^N-1}{y} \ dy=e^{-s}\int\_{0}^s \frac{e^x-1}{x} \ dx.$$
And so you get the final answer
$$e^{-s}\left(Ei(s)-\gamma -\log s\right)$$
where $Ei(x)$ is the [Exponential integral](http://en.wikipedia.org/wiki/Exponential_integral) and can not be "simplified" further.
| 6 | https://mathoverflow.net/users/2384 | 37251 | 23,964 |
https://mathoverflow.net/questions/37253 | 1 | Whilst trying to solve a combinatorics problem I am faced with summing this series:
1+ 2C\_1 2/(3^2) + 4C\_2 (2^2)/(3^4) + 6C\_3 (2^3)/(3^6)+ ... + 2nC\_n (2^n)/(3^(2n))+...
Where 4C\_2 is 4 choose 2.
Any idea how to approach this problem?
| https://mathoverflow.net/users/8826 | Trying to sum a series (related to catalan numbers perhaps) | The generating function of the [central binomial coefficients](http://en.wikipedia.org/wiki/Central_binomial_coefficient) is
$$\sum\_{n=0}^{\infty}\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}$$ and so the value of your series is 3.
| 4 | https://mathoverflow.net/users/2384 | 37254 | 23,965 |
https://mathoverflow.net/questions/37246 | 2 | Is there a classification of embeddings of SL\_2 into SP\_6 as algebraic groups over Q and R respectively?
see also the link:mathoverflow.net/questions/36762,
| https://mathoverflow.net/users/3945 | Is there a classification of embeddings of SL_2 into SP_6 as algebraic groups over Q and R respectively? | You probably want the Jacobson–Morozov theorem, which says that homomorphisms of the Lie algebra sl2 over a field of characteristic 0 to a semisimple Lie algebra g can be classified in terms of the nilpotent elements of g. More precisely, if e, f, h, is the usual basis of sl2 then you can choose the image of e to be any nilpotent element of g, and the images of f and h are then determined up to conjugation by the centralizer of e.
For details see [Jacobson's book on Lie algebras](http://books.google.com/books?isbn=9780486638324).
| 5 | https://mathoverflow.net/users/51 | 37256 | 23,967 |
https://mathoverflow.net/questions/37214 | 30 | Much of modern algebraic number theory can be phrased in the framework of group cohomology. (Okay, this is a bit of a stretch -- much of the part of algebraic number theory that I'm interested in...). As examples, Cornell and Rosen develop basically all of genus theory from cohomological point of view, a significant chunk of class field theory is encoded as a very elegant statement about a cup product in the Tate cohomology of the formation module, and Neukirch-Schmidt-Wingberg's fantastic tome "Cohomology of Number Fields" convincingly shows that cohomology is the principal beacon we have to shine light on prescribed-ramification Galois groups.
Of course, we also know that group cohomology can be studied via topological methods via the (topological) group's classifying space. My question is:
>
> Question: Why doesn't this actually happen?
>
More elaborately: I'm fairly well-acquainted with the "Galois cohomology for number theory" literature, and not once have I come across an argument that passes to the classifying space to use a slick topological trick for a cohomological argument or computation (though I'd love to be enlightened). On the other hand, for example, are things like Tyler's answer to my question
[Coboundary Representations for Trivial Cup Products](https://mathoverflow.net/questions/15375/coboundary-representations-for-trivial-cup-products)
which strikes me as saying that there may be plenty of opportunities to carry over interesting constructions and/or lines of reasoning from the topological side to the number-theoretic one.
Maybe the classifying spaces for gigantic profinite groups are too hideous to think about? (Though there's plenty of interesting Galois cohomology going on for finite Galois groups...). Or maybe I'm just ignorant to the history, and that indeed the topological viewpoint guided the development of group cohomology and was so fantastically successful at setting up a good theory (definition of differentials, cup/Massey products, spectral sequences, etc.) that the setup and proofs could be recast entirely without reference to the original topological arguments?
(**Edit**: This apparently is indeed the case. In a comment, Richard Borcherds gives the link [Link](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-82/issue-1/Topology-and-logic-as-a-source-of-algebra/bams/1183537593.full) and JS Milne suggests MacLane 1978 (Origins of the cohomology of groups. Enseign. Math. (2) 24 (1978), no. 1-2, 1--29. MR0497280)., both of which look like good reads.)
| https://mathoverflow.net/users/35575 | Why aren't there more classifying spaces in number theory? | Classifying spaces are widely used in algebraic number theory, but in slightly disguised form. A classifying space is really just an approximation to the classifying topos of a group. However the classifying topos is just the category of G-sets, which is exactly what one uses in defining group cohomology and so on. Or to put it another way, all the useful information in the classifying space is already contained in the category of G-sets.
The comment at the end of the question is correct: group cohomology was discovered as the cohomology of the classifying space, and the topological constructions were then turned into algebraic constructions and removed from the theory. So in some sense all the group cohomology calculations are implicitly using the classifying space.
| 23 | https://mathoverflow.net/users/51 | 37259 | 23,969 |
https://mathoverflow.net/questions/37230 | 1 | My third question about Shishikura's result :
Shishikura (1991) proved that the Hausdorff Dimension of the boundary of the Mandelbrot set equals 2, in [this paper](http://arxiv.org/abs/math/9201282)1. The Mandelbrot set is defined by iterating to infinity the z^2+c map.
Does his result also apply for higher powers, such as z^8 + c ?
Thanks again.
| https://mathoverflow.net/users/8779 | Hausdorff dimension of higher powers of the Mandebrot set ? | Yes, it does. See the full statement of Theorem 2 on page 6. The assumptions of the theorem are:
>
> Suppose that a rational map $f\_0$ of degree $d\ (> 1)$ has a parabolic fixed
> point ζ with multiplier exp(2πip/q) ($p, q \in\mathbb{Z}, \mathit{gcd}(p, q) = 1$) and that the immediate parabolic basin of ζ contains only one critical point of $f\_0$.
>
>
>
This is the case for $z^d+c$.
| 3 | https://mathoverflow.net/users/3993 | 37261 | 23,970 |
https://mathoverflow.net/questions/37260 | 9 | Throughout, by finite triangulation I mean a triangulation consisting of a finite number of triangles.
Suppose $T$ and $T'$ are finite triangulations of a 3-manifold $M$. We will say that $T'$ is simpler than $T$ iff $T'$ consists of the same number or fewer triangles than $T$ and that $T'$ is a simplest triangulation of $M$ iff $\forall$ triangulation $T$ of $M$, $T'$ is simpler than $T$.
Note: If a 3-manifold $M$ has a finite triangulation, then clearly it has a simplest triangulation.
By a theorem of Pachner (Theorem A.1.1. in 'The geometry of dynamical triangulations') any two triangulations of a manifold can be transformed from one to another by a finite number of stellar subdivisions. As we are only dealing with 3-manifolds, there are only 4 stellar subdivisions; known as the $1 \to 4$, $2 \to 3$, $3 \to 2$ and $4 \to 1$ moves as described in <http://at.yorku.ca/t/a/i/c/45.pdf> and hereafter called the Pachner moves. So clearly, there exists a finite sequence of Pachner moves from any finite triangulaiton $T$ of $M$ to $T'$, a simplest triangulation of $M$.
>
> If $T$ is a finite triangulation of $M$, does the greedy algorithm of just applying as many $4 \to 1$ and $3 \to 2$ Pachner moves to $T$ as possible always result in a simplest triangulation of $M$?
>
>
>
Or alternatively,
>
> Is there a finite triangulation $T$ of a 3-manifold $M$ such that repeatedly applying only the $4 \to 1$ and $3 \to 2$ Pachner moves does not eventually result in a simplest triangulation of $M$?
>
>
>
| https://mathoverflow.net/users/3121 | Simplifying triangulations of 3-manifolds | There are many such examples, depending upon what you mean by "triangulation". If a triangulation is just a glueing of tetrahedra along faces, then the simplest one is probably the following: the 3-sphere has a triangulation with 1 tetrahedron, and a triangulation with 2 tetrahedra (it is a nice excercise to find them). However, you cannot apply any $4 \to 1$ or $3 \to 2$ move to a triangulation having only 2 tetrahedra.
If by "triangulation" you mean a "simplicial triangulation", then it suffices to construct a triangulation of the 3-sphere such that every edge meets at least 4 distinct tetrahedra. In this case, no $4 \to 1$ and no $3 \to 2$ moves can apply.
For instance, you can take a triangulation of the 2-sphere such that every vertex is incident to at least 4 triangles (as an example, the octahedron). By doing a suspension of this triangulation you get a triangulation of the 3-sphere such that every edge is incident to at least 4 tetrahedra.
| 10 | https://mathoverflow.net/users/6205 | 37268 | 23,975 |
https://mathoverflow.net/questions/37272 | 33 | The question is the title.
Working in ZF, is it true that: for every nonempty set X, there exists a total order on X ?
If it is false, do we have an example of a nonempty set that has no total order?
Thanks
| https://mathoverflow.net/users/8913 | Are all sets totally ordered ? | In the paper [Dense orderings,
partitions and weak forms of choice, by Carlos G. Gonzalez FUNDAMENTA MATHEMATICAE 147 (1995)](http://matwbn.icm.edu.pl/ksiazki/fm/fm147/fm14712.pdf), the author states the following theorem, where AC is the Axiom of Choice, DO is the assertion that every infinite set has a dense linear order, O is the assertion that every set has a linear order, and DPO is the assertion that every infinite set has a (nontrivial) dense partial order.
**Theorem 1.** AC implies DO implies O implies DPO. Moreover, none of the implications
is reversible in ZF and DPO is independent of ZF.
Thus, in particular, the assertion that every set has a total order is strictly weaker than AC.
(Also, it would seem that Gonzalez means to assume Con(ZF) for the latter claims of his theorem.)
| 38 | https://mathoverflow.net/users/1946 | 37281 | 23,980 |
https://mathoverflow.net/questions/37118 | 1 | Let $I$ be an ideal of $k[x\_1, \ldots, x\_m, y\_1, \ldots, y\_n]$, $k$ being a field. Does any of the computer algebra systems implement any algorithm to calculate the generators of the 'bi-homogenization' $\tilde I$ of $I$ with respect to $x$ and $y$ variables?
(Recall that the 'bi-homogenization' of a polynomial $f = \sum a\_{\alpha, \beta} x^\alpha y^\beta$ is by definition $\tilde f := \sum a\_{\alpha, \beta} x^\alpha y^\beta x\_0^{d - |\alpha|} y\_0^{e- |\beta|}$, where $x\_0$ and $y\_0$ are two new variables, $d := \deg\_x(f)$ and $e := \deg\_y(f)$. Then $\tilde I := ${$\tilde f: f \in I$}.)
My motivation is geometric: to find the closure $\overset{-}{V}$ of a subvariety $V$ of $k^{m+n}$ in $\mathbb{P}^m \times \mathbb{P}^n$. Of course I could as well calculate the Segre embedding of $\overset{-}{V}$ in $\mathbb{P}^{mn + m +n}$, but I would like to have something computationally less expensive.
I can think of an algorithm which involves introducing $n$ (or $m$, whichever is the smaller) new variables $t\_1, \ldots, t\_n$ and computing the monomial basis of an ideal $J$ in $k[x,y,t]$, where $J$ is to be constructed from $I$. But I was wondering if someone had already implemented some (possibly better) algorithm which would do this job.
| https://mathoverflow.net/users/1508 | Any implemented algorithm to compute the closure of an affine variety in a product of projective spaces? | This can be done in a few steps in probably any computer algebra package. You take the generators of your original ideal $I$, and bi-homogenize them, as described in the question. Then saturate with respect to the two hyperplanes at infinity, which are defined by the equation $x\_0 y\_0$.
For example, the diagonal in $\mathbb A^3 \times \mathbb A^3$ is defined by $x\_1 - y\_1$, $x\_2 - y\_2$, and $x\_3 - y\_3$. If I wanted to use this to compute the ideal of the diagonal in $\mathbb P^3 \times \mathbb P^3$, I would use the following commands in Macaulay2:
```
r = QQ[x0,x1,x2,x3,y0,y1,y2,y3]
i = ideal(x1*y0-y1*x0, x2*y0-y2*x0, x3*y0-y3*x0)
saturate(i, x0*y0)
```
The code in Singular would be:
```
ring r = 0, (x0,x1,x2,x3,y0,y1,y2,y3), dp;
ideal i = x1*y0-y1*x0, x2*y0-y2*x0, x3*y0-y3*x0;
LIB "elim.lib";
sat(i, x0*y0);
```
| 6 | https://mathoverflow.net/users/8914 | 37284 | 23,983 |
https://mathoverflow.net/questions/37270 | 2 | Suppose we have a set of quadratic forms $Q\_i (x\_1, \dots, x\_n)$ for $1 \leq i \leq k$ in $n$ variables, defined over $\mathbb{R}$. We suppose these are 'collectively nondegenerate' in the sense that there does not exist a change of variables which takes us into a set of quadratic forms with less than $n$ variables.
I am looking at linear combinations of these forms: $$ Q\_{\boldsymbol{\lambda}}(\textbf{x})=\sum\_i \lambda\_i Q\_i(x\_1, \dots, x\_n)$$ for $\boldsymbol{\lambda} = (\lambda\_1, \dots , \lambda\_k) \in \mathbb{R}^k$. My question is whether we are guaranteed a set of $\lambda$s which gives us a quadratic form of full rank i.e. $n$? **Edit:**: this has been shown to be untrue, so...
Is there anything we can do to guarantee a 'high' rank, say bigger than 5? For example by taking $n \gg k$?
| https://mathoverflow.net/users/4426 | Rank of a linear combination of quadratic forms | The answer to the first part (about finding a linear combination which has full rank) is no. A counterexample with $n=3$ and $k=2$ is given by the quadratic forms $xy$ and $xz$. A general linear combination of these two is of the form $\lambda\_1 xy + \lambda\_2 xz = x(\lambda\_1 y + \lambda\_2 z)$, which obviously has rank 2.
The equivalent formulation in terms of symmetric matrices is that any linear combination of
\begin{equation\*}
\begin{pmatrix}
0 & 1 & 0 \\\
1 & 0 & 0 \\\
0 & 0 & 0
\end{pmatrix}
\quad\mbox{and}\quad
\begin{pmatrix}
0 & 0 & 1 \\\
0 & 0 & 0 \\\
1 & 0 & 0
\end{pmatrix}\end{equation\*}
is singular, but if we put the matrices side by side, then
\begin{equation\*}
\begin{pmatrix}
0 & 1 & 0 & 0 & 0 & 1 \\\
1 & 0 & 0 & 0 & 0 & 0 \\\
0 & 0 & 0 & 1 & 0 & 0
\end{pmatrix}
\end{equation\*}
has full rank.
| 3 | https://mathoverflow.net/users/8914 | 37294 | 23,990 |
https://mathoverflow.net/questions/37258 | 6 | Let $\Delta$ be 2-disk. Let $C(\Delta;n)$ be a configuration space.
i.e.) $C(\Delta;n)= \lbrace (z\_1,\ldots,z\_n)\in \Delta\times\ldots\Delta | z\_i\neq z\_j ~\textrm{if}~ i\neq j \rbrace $
Then, it is well known or direct to see that $\pi\_1(C(\Delta;n))= PB\_n$, where $PB\_n$ is original pure braid group of n-strands.
I heard that by using this configuration space we can easily generalize Braid group in arbitrary topological space $X$.
i.e.) We can define $PB\_n(X)$ (Pure Braid group of n-strands in $X$) by $PB\_n(X)=\pi\_1(C(X;n))$.
Also, we can study some exact sequences analogous to original braid group situation if topology of $X$ is good.
Here, but I heard that if $X$ is manifold with dimension greater than 2. Then, the structure of $PB\_n(X)$ is somewhat trivial because there are enough rooms to move
I think that this phenomena occurs because we are observing too small objects in large space. Roughly speaking, this seems to be the same situation that Knot theory is trivial in codimension greater than 2.
Instead of this formulation, I hope that we could modify or generalize this configuration space set up by observing some space containg information like distribution or foliation or something like that?
Can I do this business? In short, Can we think nontrivial generalization of Braid group in higher dimensional manifold?
| https://mathoverflow.net/users/7776 | Higher-dimensional braid group? | While I don't know about the **braid group**, there are certainly generalizations of **braids** to higher dimensions. In fact there is a huge literature on the subject. Perhaps the place to start is in Lee Rudolph's
*Braided surfaces and Seifert ribbons for closed braids* Comment. Math. Helv. **59** (1983), 1-37.
He defines a *braided surface* as follows. Let $D\_1\times D\_2$ be a product of 2-disks, and $\mathrm{pr}\_i:D\_1\times D\_2\rightarrow D\_i$ be the $i$th factor projection. Then a properly embedded, locally flat surface S in $D\_1\times D\_2$ is called a braided surface if:
* $\mathrm{pr}\_2|\_S:S\rightarrow D\_2$ is a branched covering.
* S is oriented so that $\mathrm{pr}\_2|\_S$ is orientation-preserving away from the (finite number of) branch points.
Rudolph proves that every ribbon surface in a 4-ball may be deformed into a braided surface. Braided surfaces are related to the (classical) braid group and to knot theory.
Oleg Viro modified the above definition, and defined a *2-dimensional braid*, in order to obtain a concept which relates to higher-dimensional knot theory. Viro requires that $\partial S=P\_n\times \partial D\_2$, where $P\_n$ is a set of $n$ distinct interior points in $D\_1$. For this more restrictive definition, he proves that every surface link type is represented by a closure of a 2-dimensional braid. I don't know if Viro ever published his work, but [here is a reference](http://books.google.com/books?id=vjHK3VrcPvIC&pg=PA197&lpg=PA197&dq=surface+braids+topology&source=bl&ots=3Jn7E7ZLQ8&sig=MqEP1MzJJAVo_3NcbE3qoYNopas&hl=en&ei=j1F9TJmaFMmrnAe2oIidCw&sa=X&oi=book_result&ct=result&resnum=10&ved=0CEgQ6AEwCQ#v=onepage&q=surface%20braids%20topology&f=false).
| 4 | https://mathoverflow.net/users/2051 | 37296 | 23,991 |
https://mathoverflow.net/questions/37277 | 5 | I think this is basically the inverse question of [Matrices whose exponential is stochastic](https://mathoverflow.net/questions/33230/matrices-whose-exponential-is-stochastic).
i.e. what are sufficient conditions on the matrix representation of an evolution operator of a (finite) discrete Markov chain for it to be embeddable in a continuous Markov chain?
I've found some old paper that may answer this (something about embeddability criteria) but I can't access it as it published in a closed-access journal.
I hope this is a sane question.
| https://mathoverflow.net/users/8916 | (Stochastic) matrix for which a stochastic matrix logarithm exists? | Steve Hunstman's link above is good:
See the part leading up to Theorem 9 for something relevant to applications:
>
> The main application of the following
> theorem may be to establish that
> certain Markov matrices arising in
> applications are not embeddable, and
> hence either that the entries are not
> numerically accurate or that the
> underlying process is not autonomous.
> The theorem is a quantitative
> strengthening of Lemma 8. It is of
> limited value except when n is fairly
> small, but this is often the case in
> applications.
>
>
>
Also the part on regularization for best compromises when matrices are not embeddable.
| 5 | https://mathoverflow.net/users/8916 | 37297 | 23,992 |
https://mathoverflow.net/questions/37303 | 3 | The game of Nimble is played as follows. You have a game board consisting of a line of squares labelled by the nonnegative integers. A finite number of coins are placed on the squares, with possibly more than one coin on a square. A move consists of picking up one of the coins and placing it on a square somewhere to the left of its original position. Players alternate moves. The game ends when no moves are possible, i. e. when all the coins are on the square labelled 0. The last player to move wins.
As described, this is just [Nim](http://en.wikipedia.org/wiki/Nim) in disguise, where a coin in position $n$ corresponds to a pile containing $n$ coins. The winning strategy is well-known. Define the nim-sum of some positive integers as the result when they're written in binary and addition is performed without carrying. Then the only winning positions are those where the nim-sum of all the pile sizes (or coin positions) are zero.
But what is the winning strategy if we don't allow coins to be stacked? This corresponds to Nim where we don't allow two piles of the same size, but this restriction feels a lot more natural in Nimble than in Nim.
There are two reasonable terminal positions. One is to allow stacking on the square marked zero only, and say the game terminates when all the coins are at zero; the other is to not allow stacking and say the game terminates when coins are on $0, 1, \ldots, k-1$ if there are $k$ coins. Either one would be of interest.
If someone gives me an answer by Friday, I might talk about it in my class.
| https://mathoverflow.net/users/143 | The game of "nimble" with no stacking | This is Welter's game, analysed in detail in Conway's
[On Numbers and Games](http://books.google.co.uk/books?id=tXiVo8qA5PQC&lpg=PA24&dq=conway%2520on%2520number&pg=PA153#v=onepage&q&f=false) chapter 13. It gets quite complicated....
| 8 | https://mathoverflow.net/users/4213 | 37308 | 23,995 |
https://mathoverflow.net/questions/37298 | 10 | Limits and colimits have very nice definitions in terms of Kan extensions, and therefore enjoy very nice adjointness properties. Mac Lane's *Categories for the Working Mathematician* gives a construction called the subdivision category of a category $C$, which allows one to reduce the theory of ends and coends to the theory of limits and colimits (and therefore the theory of Kan extensions). This construction feels a bit artificial and messy, although it is very useful for quick and dirty proofs of many of the details about ends and coends.
Can we give a definition of the end and coend as some sort of Kan extension without invoking the subdivision category construction?
| https://mathoverflow.net/users/1353 | Ends and coends as Kan extensions (without using the subdivision category of Mac Lane)? | Ends and coends should be thought of as very canonical constructions: as Finn said, they can be described as weighted limits and colimits, where the weights are hom-functors.
Recall that if $J$ is a (small) category, a *weight* on $J$ is a functor $W: J \to Set$. The limit of a functor $F: J \to C$ with respect to a weight $W$ is an object $lim\_J F$ of $C$ that represents the functor
$$C^{op} \to Set: c \mapsto Nat(W, \hom\_C(c, F-)).$$
Dually, given a weight $W: J^{op} \to Set$, the weighted colimit of $F: J \to C$ with respect to $W$ is an object $colim\_J F$ that represents the functor
$$C \to Set: c \mapsto Nat(W, \hom\_C(F-, c)).$$
Then, as Finn notes above, the end of a functor $F: J^{op} \times J \to C$ is the weighted limit of $F$ with respect to the weight $\hom\_J: J^{op} \times J \to Set$, and the coend is the weighted colimit of $F$ with respect to $\hom\_{J^{op}}: J \times J^{op} \to Set$.
The ordinary limit of $F$ is the weighted limit of $F$ with respect to the terminal functor $t: J \to Set$. Ordinary limits suffice for ordinary ($Set$-based) categories, but they are inadequate for enriched category theory. The concept of weight was introduced to give an adequate theory of enriched limits and colimits (replacing $Set$ by suitable $V$, and functors as above by enriched functors, etc.)
Weighted colimits and weighted limits (in particular coends and ends) can be expressed in terms of Kan extensions. For any weight $W$ in $Set^{J^{op}}$, the weighted colimit of $F: J \to C$ (if it exists) is the value of the left Kan extension of $F: J \to C$ along the Yoneda embedding $y: J \to Set^{J^{op}}$ when evaluated at $W$, in other words
$$(Lan\_y F)(W)$$
A similar statement can be made for weighted limits, as values of a right Kan extension.
| 12 | https://mathoverflow.net/users/2926 | 37310 | 23,997 |
https://mathoverflow.net/questions/37302 | 1 | I have a Markov chain $\mathbf{A} = (A\_0, A\_1, \ldots)$ with state space $\{0, \ldots, n\}$ which converges towards a stationary distribution $\pi$. There are a lot of well-known results on upper-bounding the convergence rate. However, I'm interested in getting a lower bound.
---
In detail, the problem looks like this: The transition probability is given as
$p\_{ij} = {n \choose i}\left(1-({1\over 2})^j\right)^i\left(({1\over 2})^j\right)^{n-i}$ for $j\neq 0$,
$p\_{ij} = {n \choose i}\left(1-({1\over 2})^n\right)^i\left(({1\over 2})^n\right)^{n-i}$ for $j = 0$,
and the initial distribution $A\_0$ is $(1, 0, \ldots, 0)^T$, i.e., the chain starts in state $0$ with probability $1$.
Given this information, is it possible to derive a lower bound on the convergence rate? Since I'm particularly interested in state 0, I would like to come up with something like this:
$\lvert \mathbb{P}(A\_k = 0) - \pi\_0 \rvert \geq \ldots$ some function of $k$ and $n$.
Any hints on how to approach this would be appreciated. Please also speak up if you think that it is unlikely that such a closed-form expression exists and I should stop wasting my time on this problem. I'm a computer scientist, not a mathematician, so it's quite possible that I've overlooked something obvious.
| https://mathoverflow.net/users/8921 | Lower bound on the convergence rate of a specific Markov chain | See Chapter 7 of [*Markov Chains and Mixing Times*](http://www.uoregon.edu/~dlevin/MARKOV/) by Levin, Peres, and Wilmer.
| 2 | https://mathoverflow.net/users/1847 | 37312 | 23,999 |
https://mathoverflow.net/questions/37301 | 8 | What are the intuitive and historical reasons for choosing the word "exterior" for the concept of an exterior derivative of a form?
The reasoning I've heard about it is the following: let p(t) be a continuous parametric curve, then if you fix t\_0, the tangent line to the curve p(t) at t\_0 lies "exterior" of the curve p(t), since it is an approximation of p(t) itself.
| https://mathoverflow.net/users/5841 | Why did the word "exterior" get chosen for the idea of "exterior derivative"? | I) The term exterior multiplication ("äussere Multiplication") is due to Grassmann, who introduced the term in his book (written in 1844)
*Die Wissenschaft der extensiven Grösse oder die Ausdehnungslehre, eine neue Mathematische Disciplin"*
As you can check in the table of contents of the book (on page 276), paragraphs §§34,35 are called *Grundgesetze der äusseren Multiplication* (Basic laws of exterior multiplication).
Here is the [scan](http://ia341214.us.archive.org//load_djvu_applet.php?file=3/items/dielinealeausde00grasgoog/dielinealeausde00grasgoog.djvu) of this book by Google .
II) The terminology exterior differential ("différentielle extérieure") was introduced in the 1930's by articles of Elie Cartan, inspired by Grassmann.
[Here](http://books.google.fr/books?id=GtaDSFURUNAC&pg=PA449&lpg=PA449&dq=cartan+elie+produit+exterieur&source=bl&ots=5LCsbUx44Q&sig=QaslOOOe4OvcSZZwjfWcfFRjNoo&hl=fr&ei=P299TPioFYre4Abk4u2nDA&sa=X&oi=book_result&ct=result&resnum=2&ved=0CBkQ6AEwAQ#v=onepage&q&f=false) is a secondary reference from an Analysis course by Chatterji and [another](http://www.ams.org/journals/bull/1952-58-02/S0002-9904-1952-09588-4/S0002-9904-1952-09588-4.pdf) by Chern and Chevalley, in their analysis of Elie Cartan's mathematical contributions (cf. in particular pages 229 and 230 )
| 6 | https://mathoverflow.net/users/450 | 37317 | 24,003 |
https://mathoverflow.net/questions/37299 | 5 | The question can be generalized, but we might as well restrict to this case.
Let $X \rightarrow Y$ be a morphism between nonsingular surfaces (say over $\mathbb{C}$). Let $R\_1$ be an irreducible component of the ramification divisor (in $X$). Let $n$ be how much $R\_1$ ramifies generically, and let $S$ be the finite set of points of $R\_1$ that ramify to a degree which is not $n$. Is it true that the set $S$ is contained in the set of the points where $R\_1$ intersects the other irreducible components of the ramification divisor (and maybe where $R\_1$ is singular)?
The intuitive answer is yes, but I'm still somewhat skeptical.
| https://mathoverflow.net/users/5309 | Ramification in morphisms of surfaces | Let me expand jvp's answer, giving a picture of the situation in the case of a $general$ flat triple cover $f \colon X \to Y$.
Let $R \subset Y$ be the ramification divisor and $B \subset Y$ the branch divisor, that is $B = f(R)$. Then $R$, $B$ are both reduced and irreducible, and $B$ has only a finite number of ordinary cusps $q\_1, \ldots, q\_t$ as singularities. These cusps are exactly the points over which $f$ is $totally$ $ramified$. Moreover $R$ is isomorphic to the normalization of $B$, in particular it is $smooth$.
One has the equality of divisors
$f^\*(B)=2R + R'$,
where $R'$ is another irreducible curve, isomorphic to $R$, which meets $R$ in a finite number of points $p\_1, \ldots, p\_t$. Notice that $R'$ is $not$ a component of the ramification locus, since the latter consists of $R$ alone.
Moreover
* $R$ and $R'$ are *tangent* at $p\_1, \ldots, p\_t$;
* $p\_1, \ldots ,p\_t$ are the preimages of the cusps $q\_1, \ldots, q\_t$.
Summing up, in this case your $S$ is the set whose elements are the points $p\_1, \ldots ,p\_t$. They correspond to the points where the ramification divisor $R$ meets the curve $R'=f^\*(B) \setminus R$. In other words, they come from the singular points of the *branch* divisor $B$ (whereas the ramification divisor $R$ is smooth).
This is easy to see; a good reference is Miranda's paper "Triple covers in algebraic geometry".
Anyway, the crucial fact here is that a general triple cover is not a Galois cover, so over the branch locus $B$ there are both points where $f$ is ramified (the curve $R$) and points where it is not (the curve $R'$).
If you consider instead any Galois cover, say with group $G$, then every preimage of a branch point is a ramification point (and the stabilizers of points lying on the same fibre are conjugated in $G$). In this case there are formulae relating the ramification number of a point on $X$ with the ramification numbers of the components of the ramification locus passing through it.
See Pardini's paper "Abelian covers of algebraic varieties" for more details.
| 5 | https://mathoverflow.net/users/7460 | 37326 | 24,009 |
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