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https://mathoverflow.net/questions/19282 | 20 | That's the question. Recall that a morphism $f\colon A\to B$ of commutative rings is integral if every element in $B$ is the root of a monic polynomial with coefficients in the image of $A$ and that $f$ is an epimorphism if and only if the multiplication map $$B\otimes\_A B\to B$$ is an isomorphism.
If we make the additional assumption that $B$ is finitely generated as an $A$-algebra, then it is true. This can be proven by Nakayama's lemma, for example.
This came up not so long ago when I was trying to show that the Witt vector functor (of finite length) preserves separatedness of algebraic spaces. In this application I was able to reduce things to the finitely generated case and could therefore use the weaker statement above, but I still wonder about the general case.
| https://mathoverflow.net/users/1114 | Is every integral epimorphism of commutative rings surjective? | If I'm not mistaken, there is a counter-example. Have a look at Lazard's second counter-example in:
["Deux mechants contre-exemples" in Séminaire Samuel, Algèbre commutative, 2, 1967-1968](http://archive.numdam.org/ARCHIVE/SAC/SAC_1967-1968__2_/SAC_1967-1968__2__A8_0/SAC_1967-1968__2__A8_0.pdf).
For any field $k$, Lazard provides a **non-surjective epimorphism** of local $k$-algebras $C\to D$, both of Krull dimension zero, and both of residue field equal to $k$. It is then easy to show that $D$ is also integral over $C$, which is what we need here. Indeed, every $d\in D$ can be written as $d=a+b$ with $a\in k$ and $b$ in the maximal ideal (and unique prime) of $D$, which is therefore nilpotent $b^n=0$, hence trivially integral. Since $a\in k$ is also in $C$, our $d$ is the sum of two integral elements. (Or simply, $D$ is a $k$-algebra, hence a $C$-algebra, generated by nilpotent, hence integral, elements.)
In cash, for those who don't want to click, the rings are constructed as follows:
Consider the local ring in countably many pairs of variables $S=(k[X\_i,Y\_i]\_{i\geq 0})\_M$ localized at $M=\langle X\_i,Y\_i\rangle\_{i\geq0}$.
For every $i\geq 0$ choose an integer $p(i) > 2^{i-1}$. Define $J=\langle Y\_i-X\_{i+1} Y\_{i+1}^2 \ ,\ X\_i^{p(i)}\rangle\_{i\geq0}\subset S$ and define $D=S/J$. Note immediately that $D$ is a local $k$-algebra, say with maximal ideal $m$ and with residue field $D/m\cong S/M\cong k$. Finally, he defines $C$ to be the localization (at $C\_0\cap m$) of the subalgebra $C\_0:=k[x\_i,x\_iy\_i]\_{i\geq 0}\subset D$ where the $x\_i$ are the classes of the $X\_i$ in $D$ and I let you guess what the $y\_i$ are. By construction, the residue field of $C$ is an extension of $k$ which is also a subfield of $D/m=k$, so the residue field of $C$ must be $k$ and we are in the announced situation.
| 11 | https://mathoverflow.net/users/5417 | 21718 | 14,363 |
https://mathoverflow.net/questions/21717 | 7 | Background and motivation:
I am teaching the "covering space" section in an introductory algebraic topology course. I thought that, in the last five minutes of my last lecture, I might briefly sketch how to compute the "fundamental group of a field," primarily as a way of illustrating the analogy between Galois theory and covering space theory, but also because this is the sort of thing that makes me eager to learn more about a subject when I am on the receiving end of a lecture.
Unfortunately, as yet, I myself know little more about the etale fundamental group than the definition and a bit of motivation (although I am certainly planning to learn more). In particular, I realized that there is an obvious question I don't know how to answer. When defining the fundamental group in an algebraic setting, we define it as an inverse limit because there is, in general, no natural analogue of a universal cover. However, if we are looking at Spec k, there is an obvious candidate, namely the Spec of the algebraic closure of k.
Question:
Let $k$ be a field, with algebraic closure $K$. How does the etale fundamental group of Spec $k$ compare to the automorphism group of $K$ over $k$? Assuming they are different, what (in very general terms) are the reasons for working with one rather than the other?
| https://mathoverflow.net/users/5094 | The etale fundamental group of a field | Essentially by definition, the etale fundamental group of a field $k$ is the group of $k$-automorphisms of a separable algebraic closure $k^{\operatorname{sep}}$ over $k$. If $k$ is perfect, then this is (well-defined up to an inner automorphism) $\operatorname{Aut}(\overline{k}/k)$. On the other hand, any $k$-automorphism of $k^{\operatorname{sep}}$ extends uniquely to an automophism of the algebraic closure $\overline{k}$. Therefore in every case the etale fundamental group of $k$ is isomoprhic to $\operatorname{Aut}(\overline{k}/k)$.
| 9 | https://mathoverflow.net/users/1149 | 21719 | 14,364 |
https://mathoverflow.net/questions/21744 | 7 | Hello everyone, I am a newbie to the category theory. While reading some paper about infinite Galois theory, the 'anti-equivalence' of two categories showed up. Could anyone give me a 'good' explanation what this means? Thanks a lot.
| https://mathoverflow.net/users/3849 | What is an antiequivalence of two categories? | All it means that one of the categories is equivalent to the opposite of the other.
Wikipedia has informative pages on opposites of categories and equivalences
of categories:
<http://en.wikipedia.org/wiki/Opposite_%28category_theory%29> ,
<http://en.wikipedia.org/wiki/Equivalence_of_categories> .
| 16 | https://mathoverflow.net/users/4213 | 21746 | 14,378 |
https://mathoverflow.net/questions/15351 | 6 | Let $\Lambda$ be an $n$-dimensional lattice with basis $b\_1,\ldots,b\_n$. The problem of finding a "good" basis for $\Lambda$, or reducing a "bad" basis into a good one, is a very active area of research. Most basis reduction schemes try to optimize the norms of the basis vectors and their inner products. The goal is to have a basis that is as nearly orthogonal as possible. A more ambitious goal might be for the basis to make the specification of the Voronoi region (i.e. its face vectors) as simple as possible. The expression "Voronoi reduction" is taken from a publication by Conway and Sloane (Proc. Royal Soc. London A, 436 (1992), 55-68) where it is applied to lattices in dimensions $n\le 3$.
>
> What exactly should be the definition of a Voronoi reduced basis?
>
>
>
Central to the construction of the Voronoi region of $\Lambda$ are the $2^n$ cosets of $2\Lambda$. In particular, for any $x\in \Lambda$ one is interested in the set of minimal norm elements in $x+2\Lambda$. Call this set $S(\Lambda,x)$. One definition of a Voronoi reduced basis might therefore be the following:
A basis $b\_1,\ldots,b\_n$ for $\Lambda$ is *Voronoi reduced* if, for any $x\in \Lambda$ and any $y\in S(\Lambda,x)$ the integers $y\_1,\ldots,y\_n$ in the expansion $y=\sum\_{i=1}^n y\_i b\_i$ always satisfy the bound $|y\_i|\le c\_n$.
>
> If this is a good definition, then what is the best constant $c\_n$?
>
>
>
The Conway-Sloane paper shows that $c\_n=1$ for $n\le 3$. In other words, in dimension three and lower there always exists a basis such that the minimal norm vectors of the $2\Lambda$ cosets can be expressed as sums with coefficients limited to $-1, 0, 1$. How fast does $c\_n$ grow with $n$? Does it grow at all?
| https://mathoverflow.net/users/40739 | How to define a Voronoi reduced basis? | Excellent question. I don't know the answer and perhaps what I am suggesting is obvious. Nevertheless, I think it's on the right track.
Let $B$ be a basis for $\Lambda$ and let $P$ be the corresponding (origin centred) fundamental parallelepiped. That is, $P$ is the region given by $Bu$ where $u \in [-0.5,0.5]^n$. Clearly $2P \cap \Lambda$ contains representatives of $\Lambda/2\Lambda$. However, the elements in $2P \cap \Lambda$ are generally not the minimal length representatives for $\Lambda/2\Lambda$, those are contained in $2\mathcal{V} \cap \Lambda$ where $\mathcal{V}$ is the (closed) Voronoi cell of $\Lambda$.
Let $d$ be the smallest real number such that $\mathcal{V} \subseteq dP$. Clearly $2dP \cap \Lambda$ is a superset of $2\mathcal{V} \cap \Lambda$ and a suitable value of $c$ is therefore $\lfloor d \rfloor$. We can refine this a little by asking for the diagonal matrix $D$ such that $\mathcal{V} \subseteq DP$. Then we have $n$ different $c$'s, specifically $c\_i = \lfloor d\_i \rfloor$ where $d\_i$ are the diagonal elements of $D$.
The problem is now to find the value $d$ (or matrix $D$). That is, how much do we need to scale the fundamental parallelepiped so that it completely contains the Voronoi cell? Perhaps if $R$ is suitably reduced, say LLL reduced or Korkine–Zolotareff reduced, then bounds on $d$ can be found?
| 2 | https://mathoverflow.net/users/5378 | 21747 | 14,379 |
https://mathoverflow.net/questions/21742 | 7 | I'm interested in the following collection of questions: Let $S^n\_k = \sqcup\_k S^n$ be a disjoint union of $k$ distinct $n$-dimensional spheres. Write $Emb(S\_k^n, S^{n+2})$ for the space of embeddings of these spheres into $S^{n+2}$. Pick your favorite embedding $e: S\_k^n \to S^{n+2}$, and let $X\_e = S^{n+2} \setminus im(e)$ be the complement of the image of the embedding.
1. What is $\pi\_1(Emb(S\_k^n, S^{n+2}), e)$? Since this is probably unknown, what is known?
2. How is this related to the mapping class group $\pi\_0(Diff(X\_e))$ of $X\_e$?
I ask #2 because in dimension $n=0$, they are the same: the space of embeddings is the configuration space of points in the sphere. Its fundamental group is the (spherical) braid group, which is the same as the mapping class group of the punctured sphere. My guess is that life is not so simple in higher dimensions. Lastly, does any of this simplify out when you get into the range of dimensions where surgery theory starts working well?
| https://mathoverflow.net/users/4649 | Knot complement diffeomorphism groups and embedding spaces | There is a locally-trivial fibre bundle
$$ Diff(S^{n+2}, L) \to Diff(S^{n+2}) \to Emb\_L(\sqcup\_k S^n, S^{n+2})$$
here $Emb\_L(\sqcup\_k S^n, S^{n+2})$ is the component of the link $L$ you're interested in the full embedding space $Emb(\sqcup\_k S^n,S^{n+2})$ and to simplify technicalities, assume $Diff(S^{n+2})$ is the group of orientation-preserving diffeomorphisms of $S^{n+2}$. The bundle is given by restricting a diffeomorphism of $S^{n+2}$ to $L$. $Diff(S^{n+2}, L)$ is the subgroup of $Diff(S^{n+2})$ which preserves the link $L$.
First observation is that the map $Diff(S^{n+2}) \to Emb\_L(\sqcup\_k S^n,S^{n+2})$ is null-homotopic. It's a simple argument -- isotope your link $L$ to sit in a hemi-sphere of $S^{n+2}$. Then apply a linearization process to linearize (simultaneously) all the diffeomorphisms of $S^{n+2}$ on that hemi-sphere. What I'm claiming is that $Diff(S^{n+2})$ has as a deformation-retract the subgroup that is linear on a fixed hemi-sphere -- so it gives a product decomposition $Diff(S^{n+2}) \simeq SO\_{n+3} \times Diff(D^{n+2})$ (first observed by Morlet, or Cerf, I would guess) among other things.
So now you have a fibration:
$$\Omega Emb\_L(\sqcup\_k S^n, S^{n+2}) \to Diff(S^{n+2}, L) \to Diff(S^{n+2})$$
where the induced maps on homotopy groups are a short-exact sequence. In particular the fundamental group of your link space injects into $\pi\_0 Diff(S^{n+2}, L)$, and its cokernel is precisely $\pi\_0 Diff(S^{n+2})$. This group is frequently non-trivial as it is the group of exotic $n+3$-sphere provided $n \geq 3$.
$Diff(S^{n+2}, L)$ is somewhat closely related to $Diff(X\_e)$, especially in high dimensions -- the spherical normal bundle to $L$ is particularly symmetric in low dimensions which causes trouble there. In general the sphereical normal bundle is equivalent to a disjoint union of $S^n \times S^1$, so to make the comparison between $Diff(S^{n+2}, L)$ and $Diff(X\_e)$ you'd need to ask what kind of automorphisms $Diff(X\_e)$ allows on the spherical normal bundle to $L$. There's probably a decent answer to that which doesn't take too much work but the above is a start.
edit: by Cerf's pseudoisotopy theorem, the kernel of the map $\pi\_0 Diff(S^{n+2}, L) \to \pi\_0 Diff(X\_e)$ contains the exotic sphere "part" of $\pi\_0 Diff(S^{n+2}, L)$.
| 8 | https://mathoverflow.net/users/1465 | 21760 | 14,384 |
https://mathoverflow.net/questions/21765 | 2 | Is it true that if a module has a free resolution of length $d$ then any of its submodule has a free resolution of length $\leq d$?
| https://mathoverflow.net/users/5292 | Projective dimension | No. A submodule of a free module need not have finite projective dimension.
As a simple example let $R=\mathbb{Z}/p^2\mathbb{Z}$. The free module $R$
has a submodule $p\mathbb{Z}/p^2\mathbb{Z}\cong\mathbb{Z}/p\mathbb{Z}$
which has no finite projective resolution.
| 8 | https://mathoverflow.net/users/4213 | 21766 | 14,389 |
https://mathoverflow.net/questions/21491 | 0 | I'm looking to efficiently zero-test "sparse integers", i.e. integers of the form $\sum C\_i \cdot A\_i^{X\_i}$ (where $A\_i, C\_i, X\_i$ are integers); equivalently test if a given (integer or rational) point is a zero of a sparse polynomial. For example, a randomised algorithm would be to compute the sum modulo a random prime since Fermat's Little Theorem reduces computations to a "manageable" level and with good probability there won't be a false positive. I'm wondering if this can be derandomised.
So beside the obvious does anyone know anything relevant about the general problem, I have more of a first step question:
Is (are) there (infinitely many) positive integer(s) $M$ such that there are integers $A,B,C,D,E,F,X,Y,Z$
with
$|A|,|B|,|C| < M$, $D,E,F,X,Y,Z>M$ and
$$ 0 < A\cdot D^X+B\cdot E^Y+C \cdot F^Z< \log M?$$
| https://mathoverflow.net/users/5398 | abc-conjecture meets Catalan conjecture? | Regarding Paul's "first step question", I believe the answer may well be that the set of such M is finite. This follows under some coprimality hypothesis from the n-term abc conjecture of Browkin and Brzezinski [Math. Comp. 62 (1994), 931--939]. This states that, if we have
$a\_1, a\_2, \ldots, a\_n$ integers, for $n \geq 3$, with $\gcd(a\_1, \ldots, a\_n)=1$, $a\_1 + \cdots + a\_n=0$, and no vanishing subsums, then
$$
\limsup \frac{\log \max |a\_i|}{\log \mbox{Rad} (a\_1 a\_2 \cdots a\_n)} = 2n-3.
$$
There may be examples with small M (like 3 or 4).....I'm not sure.
| 2 | https://mathoverflow.net/users/3533 | 21773 | 14,395 |
https://mathoverflow.net/questions/14384 | 15 | Warning: This one of those does-anyone-know-how-to-fix-this-vague-problem questions, and not an actual mathematics question at all.
If $X$ is a scheme of finite type over a finite field, then the zeta function $Z(X,t)$ lies in $1+t\mathbf{Z}[[t]]$. We can calculate the zeta function of a disjoint union by the formula $Z(X\amalg Y,t)=Z(X,t)Z(Y,t)$. There is also a formula for $Z(X\times Y,t)$ in terms of $Z(X,t)$ and $Z(Y,t)$, but this is slightly more complicated. In fact, these two formulas are precisely the standard big Witt vector addition and multiplication law on the set $1+t\mathbf{Z}[[t]]$. (Actually, there's more than one standard normalization, so you have to get the right one. I believe this ring structure was first written down by Grothendieck in his appendix to Borel-Serre, but I don't know who first made the connection with the ring of Witt vectors as defined earlier by Witt.) If we let $K\_0$ be the Grothendieck group on the isomorphism classes of such schemes, where addition is disjoint union and multiplication is cartesian product, then we get a ring map $K\_0\to 1+t\mathbf{Z}[[t]]$. We could also do all this with the L-factor $L(X,s)=Z(X,q^{-s})$ (where $q$ is the cardinality of the finite field) instead of the zeta function. This is because they determine each other.
This is all good. The problem I have is when there is bad reduction. So now let $X$ be a scheme of finite type over $\mathbf{Q}$ (say). Then the L-factor $L\_p(X,s)$ is defined by
$$L\_p(X,s)=\mathrm{det}(1-F\_p p^{-s}|H(X,\mathbf{Q}\_{\ell})^{I\_p}),$$
where $I\_p$ is the inertia group at $p$. (Sorry, I'm not going to explain the rest of the notation.) If $I$ acts trivially (in which case one might say $X$ has good reduction), then taking invariants under $I$ does nothing, and so as above, the L-factor of a product and sum of varieties is determined by the individual L-factors. If $I$ does not act trivially, then the L-factor of a sum is again the product of the individual L-factors, but for products there is no such formula! (The following should be an example showing this. Take $X=\mathrm{Spec}\ \mathbf{Q}(i)$, $Y=\mathrm{Spec}\ \mathbf{Q}(\sqrt{2})$. The we have the following Euler factors at 2: $L\_2(X,s)=L\_2(Y,s)=L\_2(X\times Y,s)=1-2^{-s}$ and $L\_2(X\times X,s)=(1-2^{-s})^2$. So the L-factors of two schemes do not determine that of the product.) Therefore the usual Euler factor cannot possibly give a ring map defined on the Grothendieck ring of varieties over $\mathbf{Q}$.
So, is there a way of fixing this problem? I would guess the answer is No, because while some people might allow you to scale Euler factors by numbers, I don't think anyone will let you change them by anything else. But maybe there is some "refined L-factor" that determines the usual one (and maybe incorporates the higher cohomology of the inertia group?) Assuming there is no known way of repairing things, I have a follow-up question: Is there some general formalism that handles this failure? And if so, how does that work?
| https://mathoverflow.net/users/1114 | Can the failure of the multiplicativity of Euler factors at bad primes be corrected? | When I asked Niranjan Ramachandran this question a few days ago, he pointed out that you can indeed fix the problem if you work with integral models instead of varieties over $\mathbf{Q}$: Let $X$ be a scheme of finite type over $\mathbf{Z}$ and define the Euler factor $L\_p(X,s)$ to be $P\_p(X,p^{-s})$, where $P\_p(X,t)$ is the Zeta-function of the fiber of $X$ over $p$. Then the multiplicativity of the product $\prod\_p L\_p(X,s)$ follows immediately from that of Zeta-functions for varieties over finite fields. Lo and behold, in the example I gave above, the products of the minimal integral models are different from the minimal integral models of the products. (This is all a little embarrassing, not just because the solution is really easy, but because I take a certain amount of pleasure in telling people that it's better to work directly with integral models! Also, apologies to the people who spent time thinking about this. I probably gave the impression that working with varieties over $\mathbf{Q}$ was non-negotiable!)
| 4 | https://mathoverflow.net/users/1114 | 21777 | 14,398 |
https://mathoverflow.net/questions/21793 | 18 | An entertaining topological party trick that I have seen performed is to turn your pants inside-out while having your feet tied together by a piece of string. For a demonstration, check out this [video](http://www.youtube.com/watch?v=O70156EDGuY).
I have heard some testimonial evidence that it is also possible to turn your pants *backwards*, again with the constraint of having your feet tied together. This second claim seems pretty dubious to me.
**Question.** Is it indeed possible to turn your pants backwards, while having your feet tied together by a piece of string? A set of instructions or a video demonstration would suffice for a yes answer. A precise mathematical formulation of the problem together with a proof of impossibility would suffice for a no answer.
| https://mathoverflow.net/users/2233 | Turning pants inside-out (or backwards) while tied together | I think that the answer is no, by consideration of [linking numbers.](http://en.wikipedia.org/wiki/Linking_number)
First simplify the human body plus cord joining the ankles to a circle, and
assign it an orientation. Also assign an orientation to each pant cuff.
This can be done, e.g., so that each cuff has linking number +1 with the
"body" (in which case the two cuffs are oppositely oriented).
Now suppose that there is an isotopy of the pants that turns them backwards.
This means the left cuff is now on the right ankle and vice versa. But this
also reverses the linking numbers, which is impossible.
| 25 | https://mathoverflow.net/users/1587 | 21798 | 14,403 |
https://mathoverflow.net/questions/21781 | 14 | I am trying to figure out when a closed, oriented manifold admits an orientation reversing diffeomorphism. My naive argument that the orientation cover should allow you to switch orientations is apparently wrong, since not every manifold admits such a diffeomorphism.
Can anyone give me some criteria for when such a morphism should exist, or why some of the standard counterexamples (such as $\mathbb{P}^{2n}$) fail to admit one?
Thanks
| https://mathoverflow.net/users/3261 | Oriention-Reversing Diffeomorphisms of a Manifold | Such an endomorphism of $M$ gives an automorphism of the cohomology ring that acts by $-1$ on top cohomology. The cohomology ring of your example $M = {\mathbb C \mathbb P}^{2n}$ doesn't have such automorphisms.
| 18 | https://mathoverflow.net/users/391 | 21799 | 14,404 |
https://mathoverflow.net/questions/21800 | 11 | For reasons which are hard to articulate (due to they not being very clear in my mind), but having to do with the eprint [*From Matrix Models and quantum fields to Hurwitz space and the absolute Galois group*](http://arxiv.org/abs/1002.1634) by Robert de Mello Koch and Sanjaye Ramgoolam, I have been wondering whether there is a notion of integration over the algebraic numbers $\overline{\mathbb{Q}}$, as there is over the p-adic completions of the rationals.
| https://mathoverflow.net/users/394 | Is there a notion of integration over the algebraic numbers? | You can definitely talk about integration on $\overline{\mathbb{Q}}$. The question is "with respect to what measure?" The reason integration theory works so well over a completion of $\mathbb{Q}$ is that such a field is, as an additive group, locally compact, and so possesses a Haar measure (a non-zero, translation invariant Radon measure) which is in fact regular because any $p$-adic field is second countable (think of Lebesgue measure on $\mathbb{R}$). In less technical language, the completion has a canonical "nice" measure against which to integrate. The (additive) group of algebraic numbers is definitely not complete in any of its (many) valuations (obtained by embedding it into the algebraic closure of some $\mathbb{Q}\_p$ or into $\mathbb{C}$), and therefore is not locally compact. The main tool for constructing interesting measures on topological spaces (that interact nicely with the topology) is the Riesz representation theorem, but this is specifically for locally compact Hausdorff spaces, so doesn't apply to the algebraic numbers. Really the reason people care about completions of $\mathbb{Q}$ and of number fields in general, is because the completions are amenable to lots of tools for analysis, whereas $\mathbb{Q}$ (and I guess also $\overline{\mathbb{Q}}$) just isn't. Completing fields allows one to bring analysis to the study of, say, number-theoretic problems.
Now, in another direction, integration definitely works on *the Galois group* of $\mathbb{Q}$, $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$. This is a profinite group (meaning it is compact and totally disconnected) in its natural topology, and so, like the completions of number fields, admits a Haar measure.
In yet another direction, one can view the *multiplicative group* of algebraic numbers, $\overline{\mathbb{Q}}^\times$, modulo torsion (roots of unity) as a (multiplicative) vector space over $\mathbb{Q}$. J. Vaaler and D. Allcock sort of initiated the study of this group (as such) and worked out its completion as a certain subspace of $L^1(Y,\mu)$, where $Y$ is a totally disconnected, locally compact Hausdorff space whose underlying set is related to the places of $\mathbb{Q}$ and where $\mu$ is a Radon measure (got using the theorem I mentioned above) that satisfies some kind of relative invariance with respect to the absolute Galois group of $\mathbb{Q}$.
I guess maybe I went way off topic with this. My point is that interesting measures on $\overline{\mathbb{Q}}$ are maybe difficult to come by, and without an interesting measure, there's no reason to do integration.
| 19 | https://mathoverflow.net/users/4351 | 21803 | 14,407 |
https://mathoverflow.net/questions/21785 | 7 | We know that for any topos E, and for any object A in E, the subobjects of A, Sub(A), form a Heyting lattice.
Does anybody know of any sort of modification of the definition of a topos that makes Sub(A) a different type of lattice? Could we get an incomplete lattice, or maybe a quantum lattice?
I'm curious because I know a lot(all?) of logical systems can be realized as a lattice, and I think this may be an interesting way to look at some alternative logics.
| https://mathoverflow.net/users/3664 | `Topos' with alternate subobject lattice? | Different types of categories lead to different types of [internal logics](http://ncatlab.org/nlab/show/internal+logic). Here is a very short list:
```
Regular Logic Regular Category
Coherent Logic Coherent Category
Geometric Logic Infinitary Coherent Category/Geometric Category
First-Order Logic Heyting Category
Dependent Type Theory Locally Cartesian Closed Category
Higher-Order Logic Elementary Topos
```
Some of the names are somewhat standard by now, but be warned that Johnsone ([Sketches of an Elephant](http://ncatlab.org/nlab/show/Elephant)), Freyd & Scedrov (*Categories, Allegories*), the nLab, and many others all use slightly different terminology. I think Johnstone's presentation in the Elephant is very nice, though you can certainly find friendlier and more localized accounts elsewhere.
| 14 | https://mathoverflow.net/users/2000 | 21804 | 14,408 |
https://mathoverflow.net/questions/21783 | 1 | Can one use the real lie algebra so(3) to get knot polynomials? If so, do they have a skein relation (I presume they would, if they come from R-matrices in some standard way. If so, is the R-matrix written down anywhere?) I've only ever seen complex Lie algebras used to get knot polynomials, but I'm not sure if this is for some fundamental reason, or just that no one has bothered to investigate the real ones.
| https://mathoverflow.net/users/492 | SO(3) knot polynomials | I don't see how taking a real form would *change* your knot invariants at all. Whatever it meant would be a number which would then not change when you base extend to the complex numbers.
From the quantum group perspective U\_q(so\_3) and U\_q(su\_2) mean the same thing (as far as I understand it), but for so\_3 the natural representation is 3-dimensional, whereas for su\_2 the natural representation is 2-dimensional.
So the R-matrix you want is just the R-matrix for the *three dimensional* representation of U\_q(su\_2). I'm not sure if that's been written down somewhere that's easy to find, but it's a good exercise to work out yourself. From the diagrammatic perspective there's something called the "Yamada polynomial" which gives the so\_3 knot polynomial (just as the Jones polynomial gives the su\_2 knot polynomial).
The other thing people mean when they say SO(3) instead of SU(2) is that you only look at the category of representations of U\_q(su\_2) which come from deformations of representations of su\_2 which lift to the lie group SO(3), that is only the representations whose highest weights are in the root lattice. The best place I know to learn about this perspective (and in particular what it means in terms of 3-manifold invariants which is what Jose mentioned)
are the papers of Steve Sawin, for example [this one](http://arxiv.org/abs/math/9905010).
| 6 | https://mathoverflow.net/users/22 | 21807 | 14,410 |
https://mathoverflow.net/questions/21816 | 5 | I have a simple question. Let $C$ be a compact Riemann surface of genus, say $g >= 2$, to avoid silly cases.
I think it should be true, but I want to prove the following concretely:
"there exists a divisor $D$ on $C$ of degree $g-1$, that is non-special."
(For those who do not know what special divisors are: a divisor is called special if it has $h^0 (D) >0$ and $h^1 (D) >0$.)
Notice that by the Riemann-Roch, for this degree $g-1$ case we immediately have $h^0 (D) = h^1 (D) = 0$. This is, in fact, equivalent to $D$ being non-special, when $\deg D = g-1$.
Is there an interesting (or any) way to prove this? I believe it should be fairly easy, and maybe I am very dumb so that I can't immediately produce a proof.
More generally, if this is possible, if the degree is a given $d$, when do we see that there exists a non-special or special divisor of given degree $d$ on a given compact Riemann surface?
| https://mathoverflow.net/users/3168 | Proving existence of non-special divisors of a given degree d on compact Riemann surfaces | Take $g+1$ general points $p\_1, \dots, p\_{g+1}$ on your curve. The divisor $p\_1+ \cdots +p\_g - p\_{g+1}$ is non-special. The proof is easy from the following lemma: if $D$ is a divisor such that $\mathrm{h}^0(D) > 0$, then $\mathrm{h}^0(D - p) = \mathrm{h}^0(D) - 1$ for all but finitely many points $p$. First you use Riemann-Roch to deduce that $\mathrm{h}^0(p\_1+ \cdots +p\_g) = 1$, then you apply the lemma once again to $p\_1+ \cdots +p\_g$. This also works for $g = 0$ and $g = 1$.
| 10 | https://mathoverflow.net/users/4790 | 21817 | 14,414 |
https://mathoverflow.net/questions/21796 | 4 | I was wondering, suppose I have a non-compact Kähler manifold $M$ and suppose that outside some compact subset $A\subset M$, there exists a smooth function $f:M\backslash A\longrightarrow\mathbb{R}$ such that $i\partial\bar{\partial}f>0$. Is it always possible for me to find a smooth function $h:M\longrightarrow\mathbb{R}$ such that $h|\_{M\backslash A}=f$ and $i\partial\bar{\partial}h>0$ on the whole of $M$? If not in general, are there sufficient conditions on $M$ that will allow me to do this? Many thanks!
| https://mathoverflow.net/users/4317 | Extension of strictly plurisubharmonic functions on a Kähler manifold | It is instructive to conisder the case of Kahler metrics invariant under torus action. In this case your question becomes a certain (nontivial) question on convex functions.
Recall first, that Kahler metrics on $(\mathbb C^\*)^n$ invariant under the action of $(S^1)^n$ have global potential that is given by a convex function $F$ on $\mathbb R^n$. Here $\mathbb R^n$ is identified with the quotient
$(\mathbb C^\*)^n/(S^1)^n$ and we take coordinates $log|z\_i|$ on $\mathbb R^n$.
So we can translate your original question as follows
QESTION. Suppose you have a smooth convex function $F$, defined on $\mathbb R^n$ outside compact $\Omega$. Is it possible to extend $F$ to a smooth convex function on the whole $\mathbb R^n$?
It easy to construct an example of a non-convex $\Omega$ on $\mathbb R^2$, with convex $F$ defined on $\mathbb R^2\setminus \Omega$, so that $F$ can not be extended. For the moment I don't see how to make such an example when $\Omega$ is the unite disk, but it sounds plausible that such examples exist.
| 5 | https://mathoverflow.net/users/943 | 21824 | 14,418 |
https://mathoverflow.net/questions/21811 | 4 | In JP May's [Concise Course in Algebraic Topology](http://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf "Concise Course in Algebraic Topology"), on page 143 he says that the left- and right-multiplication-by-identity maps $\lambda:X\rightarrow X$ and $\rho:X\rightarrow X$ specify a map $X\vee X\rightarrow X$ that is homotopic to the codiagonal map. This doesn't seem obvious to me, because the homotopies $H^L:\lambda\simeq 1\_X$ and $H^R:\rho\simeq 1\_X$ need not agree on how they move $\lambda(e)=\rho(e)$ to $e$; a priori, the paths $H^L\_t(e)=H^L(e,t)$ and $H^R\_t(e)=H^R(e,t)$ need not be homotopic. My intuition is failing me, because the only H-spaces I can think of are Lie groups and loop spaces. In both of those cases the multiplication already satisfies $e\times e \mapsto e$, and in the former case already $e\times x = x = x\times e$, while in the latter case the obvious homotopies $H^L$ and $H^R$ fix $e$ for all $t$...
| https://mathoverflow.net/users/303 | Can H-space multiplication always be straightened so that mult.-by-id. is the identity on the nose? | There are three possible definitions of an H-space, according to whether the "identity" element is a strict left and right identity, or only an identity up to basepoint-preserving homotopy, or just up to a non-basepoint-preserving homotopy. The three notions turn out to be equivalent, assuming the space is nice enough that one can apply the homotopy extension property when needed. The proof of equivalence was left as an exercise in my book (Exercise 1 in section 3.C), but it's actually a rather tricky argument, not really fair for an exercise, so a couple weeks ago I wrote up the proof and posted it here:
<http://www.math.cornell.edu/~hatcher/AT/ATsolution3C.1.pdf>
It's a one-page argument. If anyone knows a simpler proof, or a reference for this fact in the literature, I'd be happy to hear about it.
| 14 | https://mathoverflow.net/users/23571 | 21828 | 14,421 |
https://mathoverflow.net/questions/21833 | 5 | Background: By Chow's theorem, if a complex manifold can be embedded holomorphically into complex projective space, then this complex manifold must be algebraic.
Question: Suppose X is a compact complex manifold (not necessarily algebraic). Let $f:X--> {\mathbb{CP}}^n$ be a meromorphic map that is injective on its domain. Does this imply that X is algebraic?
| https://mathoverflow.net/users/nan | Can a non-algebraic complex manifold be embedded meromorphically into projective space? | If $X$ is not compact there are loads of problems so I follow you in adding compactness as a condition.
Under the assumption of the question $X$ is bimeromorphic to a projective variety and hence by a result of Artin it is an algebraic space. Hence if you accept algebraic spaces as being algebraic the answer is yes. As there are smooth proper algebraic spaces that are not schemes if you don't the answer is no.
| 11 | https://mathoverflow.net/users/4008 | 21835 | 14,427 |
https://mathoverflow.net/questions/19210 | 21 | The question.
-------------
Let $(X, J)$ be a complex manifold and $u$ a holomorphic vector field, i.e. $L\_uJ = 0$. The holomorphicity of $u$ implies that the Lie derivative $L\_u$ on forms preserves the (p,q) decomposition and also that it commutes with $\bar{\partial}$. From this it follows that $u$ acts infinitesimally on the Dolbeault cohomology groups $H^{p,q}(X)$ of $X$. My question is, does anyone know of an example in which this action is non-trivial?
Some context.
-------------
To give some context, first note that the analgous action for de Rham cohomology is always trivial: If $M$ is any smooth manifold and $v$ any vector field, then the formula $L\_v = d \circ i\_v + i\_v \circ d$ shows that the infinitesimal action of $v$ on de Rham cohmology is trivial. (This is an instance of the more general fact that homotopic maps induce the same homomorphisms on singular cohomology. The field $v$ generates diffeomorphisms which are by construction isotopic to the identity map.)
Returning to Dolbeault cohomology, suppose we know that each Dolbeault class is represented by a $d$-closed form. (For example, this is true if $X$ is a compact Kähler manifold, by Hodge theory.) Then the action is necessarily trivial. The proof is as follows. Let $\alpha$ be a $\bar{\partial}$-closed (p,q)-form which is also $d$-closed. Then we know that $L\_u \alpha = d(i\_u \alpha)$ is also of type (p,q). So,
$$
L\_u\alpha = \bar{\partial}\left((i\_u\alpha)^{p, q-1}\right) + \partial\left((i\_u \alpha)^{p-1, q}\right)
$$
and the other contributions $\bar{\partial}((i\_u\alpha)^{p-1,q}$) and $\partial((i\_u\alpha)^{p,q-1})$ vanish. Now the fact that $\bar\partial((i\_u\alpha)^{p-1,q}) = 0$ and our hypothesis imply that there is a (p-1, q-1)-form $\beta$ such that
$$
(i\_u\alpha)^{p-1,q}+ \bar\partial \beta
$$
is closed. Hence
$$
\partial \left((i\_u\alpha)^{p-1,q}\right) = \bar\partial \partial \beta
$$
and so
$$
L\_u\alpha = \bar \partial \left( (i\_u \alpha)^{p,q-1} + \partial \beta\right)
$$
which proves the action of $u$ on $H^{p,q}(X)$ is trivial.
| https://mathoverflow.net/users/380 | Holomorphic vector fields acting on Dolbeault cohomology | Take a complex nilpotent or solvable group $G$ with the right action by a co-compact lattice $\Gamma$ and conisder the quotient $G/\Gamma$. On this quotient right-invariant $1$-forms give a subspace of $H^{1,0}$. The group $G$ is acting on $G/\Gamma$ on the left and if it would presrve all the $1$-forms, $G$ would be abelian.
Torsten Ekedahl expained that what is following IS NOT CORRECT (the article of Hasegawa tells something different)
In fact, the simplest example of this kind is given by primary Kodaira surfaces (<http://en.wikipedia.org/wiki/Kodaira_surface>), they have two holomorphic $1$-forms.
These surfaces are described as quotinets of sovlable groups, for example, in an article of Keizo Hasegawa <http://arxiv.org/PS_cache/math/pdf/0401/0401413v1.pdf>
| 6 | https://mathoverflow.net/users/943 | 21840 | 14,429 |
https://mathoverflow.net/questions/21839 | 2 | I'm a newcomer to the realm of queueing theory, so please bear with me :)
I'd like to model web server traffic with a modified M/M/1 queue.
In the simple case we have two parameters - $\lambda$ for the arrival rate and $\mu$ for the departure (or service) rate.
If I understand correclty, the general way to get the performance evaluation equations (average number of requests in the queue, for example) is to draw a flow diagram, and solve the equlibrium equation system, namely for the M/M/1 model:
$0 = -\lambda p\_{0}$ + $\mu p\_{1}$
$0 = \lambda p\_{n-1} - (\lambda + \mu) p\_{n} +\mu p\_{n+1}$, n = 1, 2, ...
I don't know how I could extend the model the fit the real-world scenario the most. Each normal request induces a number of image requests, for example, let it be $u$ on average, and let it's service rate be $\sigma$. How can I factor these into the equations?
| https://mathoverflow.net/users/4976 | "Induced" arrivals in an M/M/1 queue? | Just introduce extra states. The total description of a state will include the length of the queue and the stage of service for the current customer. For instance, if each initial service may result in the second stage service with probability $q$ and the departure rate for this second stage service is $\sigma$, then you'll get 2 equations corresponding to 2 possible states with queue length $n$: $\mu p\_n(1)+\lambda p\_n(1)-\mu p\_{n+1}(1)(1-q)-\sigma p\_{n+1}(2)-\lambda p\_{n-1}(1)=0$ and $\sigma p\_n(2)+\lambda p\_n(2)-\mu p\_{n+1}(1)q-\lambda p\_{n-1}(2)=0$ (Just look at how you can depart from the state and put the corresponding terms with plus and then look at how you can arrive to the state and put the corresponding terms with minus. For instance, the terms in the first equation correspond to having been served at stage 1, new arrival to the queue serving a stage 1 customer, completely finishing serving a stage 1 customer in a queue with $n+1$ customers, finishing serving a stage 2 customer in that queue, and arrival of a new customer to the queue of length $n-1$ serving a stage 1 customer).
Sometimes you can simplify resulting big systems to smaller ones but the general idea is always to start with the set of states that fully describes everything that may happen in the system, not just the parameters you want in the end.
| 3 | https://mathoverflow.net/users/1131 | 21845 | 14,434 |
https://mathoverflow.net/questions/21852 | 3 | Hello to all,
I have been looking quite recently at the following theorem:
Let $X$ be a projective variety and $T$ a tilting object for $X$. If $A:=End(T)$ is the associated endomorphism algebra, then the functor
$RHom(T -): D^b(X) \rightarrow D^b(A)$ is in fact an equivalence. Now, this is proven (as in the claasical Bondal paper) by showing that the functor is fully faithful and essentially surjective. But in I have noticed another version where one defines a functor $-\otimes^L\_A T: D^b(A) \longrightarrow D^b(X)$. My question is could someone maybe give a definition of this functor (I of course know what all types of tensor-products are, but I'm not really sure how to "tensor" a module over a noncommutative ring $A$ together with a sheaf to obtain another sheaf.
| https://mathoverflow.net/users/4863 | Tensor product of sheaves and modules | What I've seen is a construction of a quasi-inverse for RHom(T,-), defined as $-\otimes\_A^L T$, as you wrote.
This last symbol should be interpreted as follows.
Given a left A-module M, we define a presheaf which with each U associates
$M\otimes^L\_A T(U)$.
Finally $M\otimes^L\_A T$ is defined as the sheafification of the former.
I should have a reference for this, let me check.
**Reference Added:**
A. A. Beilinson. Coherent sheaves on $\mathbb{P}^n$ and problems in linear
algebra. Funktsional.Anal. i Prilozhen., 12(3):68–69, 1978.
(it's this one, if I remember correctly, but I might not...)
| 3 | https://mathoverflow.net/users/3701 | 21856 | 14,440 |
https://mathoverflow.net/questions/21854 | 20 | I have a question about vector bundles on the algebraic surface $\mathbb{P}^1\times\mathbb{P}^1$. My motivation is the splitting theorem of Grothendieck, which says that every algebraic vector bundle $F$ on the projective line $\mathbb{P}^1$ is a direct sum of $r$ line bundles, where $r$ is the rank of $F$.
My question is: to what extent is this true for $\mathbb{P}^1\times\mathbb{P}^1$, if at all? At first glance I imagine this is classical, but I haven't had luck working it out, nor do I have a good reference for where this might be done.
Irrespective of the content of the answer, would it follow that the answer would be the same if the question were asked for $(\mathbb{P}^1)^k$?
Thanks!
| https://mathoverflow.net/users/5395 | Vector bundles on $\mathbb{P}^1\times\mathbb{P}^1$ | The splitting theorem is most certainly false for vector bundles on $\mathbb{P}^1\times\mathbb{P}^1$. In fact, the theory of vector bundles on quadric surfaces is probably as complicated as the theory of vector bundles on $\mathbb{P}^2$ (that is, very complicated).
Here is a simple example of an indecomposable rank 2 bundle. By the Künneth formula, we see that $\mathrm H^1(\mathcal O(1,-2))$ is not 0; hence there is a non-split extension
$$
0 \to \mathcal O(1,-2) \to E \to \mathcal O \to 0.
$$
I claim that $E$ is not split. Suppose that it splits as the sum $L\_1 \oplus L\_2$ of two line bundles. If $L\_1$ were to map to 0 in $\mathcal O$, the map $E \to \mathcal O$ would factor through $L\_2$, and then the map $L\_2 \to \mathcal O$, being surjective, would have to be an isomorphism, and the sequence would split. Hence both $L\_1$ and $L\_2$ admit a non-zero map to $\mathcal O$; this means that they have to be of the form $\mathcal O(m,n)$ with $m,n \leq 0$. But $L\_1\otimes L\_2$, which is the determinant of $E$, is $\mathcal O(1,-2)$, and this gives a contradiction.
I know that there have been people studying the moduli theory of vector bundles on $\mathbb{P}^1\times\mathbb{P}^1$(see for example [arXiv:0810.4392](https://arxiv.org/abs/0810.4392 "Sukmoon Huh: Moduli of Stable Sheaves on a Smooth Quadric in P_3"); but there must be lots of earlier papers).
| 27 | https://mathoverflow.net/users/4790 | 21863 | 14,443 |
https://mathoverflow.net/questions/21859 | 2 | Hello.
Background
----------
Consider a weighted graph $G=(V,E,w)$. We are given a family of $k$ disjoint subsets of vertices $V\_1, \cdots, V\_k$.
A Steiner Forest is a forest that for each subset of vertices $V\_i$ connects all of the vertices in this subset with a tree.
Example: only one subset of vertices $V\_1 = V$. In this case a Steiner forest is a spanning tree of the whole graph.
Question
--------
>
> Finding such a forest with minimal weight is difficult (NP-complete). Do you know any quicker approximate algorithm to find such a forest with non-optimal weight?
>
>
>
| https://mathoverflow.net/users/5466 | An approximate algorithm for finding Steiner Forest in a graph. | There is a 2-approximation algorithm, see e.g.
>
> A General Approximation Technique For Constrained Forest Problems, Michel Goemans, David P. Williamson, SIAM Journal on Computing 1992.
>
>
>
For special kind of graphs, better bounds can be obtained: for planar graphs there is a PTAS,
>
> Approximation Schemes for Steiner Forest on Planar Graphs and Graphs of Bounded Treewidth, MohammadHossein Bateni, MohammadTaghi Hajiaghayi, Dániel Marx, STOC '10.
>
>
>
| 4 | https://mathoverflow.net/users/4248 | 21865 | 14,445 |
https://mathoverflow.net/questions/21857 | 11 | This semester I am attending a reading seminar on non-archimedean analytic geometry (a subject I know nothing about), roughly following the [notes of Conrad](http://math.arizona.edu/~swc/aws/07/speakers/index.html).
Reading Conrad's notes (and e.g. those of Bosch) it struck me that the prime spectrum of affinoid algebras never seems to appear, only the maximal spectrum. Can somebody explain the reason for this?
| https://mathoverflow.net/users/nan | Why is the prime spectrum not useful in non-archimedean analytic geometry? | I am surprised that Brian got to this one first without making what I thought was another obvious comment: affinoids are Jacobson rings! A function which is zero at all points of an affinoid rigid space corresponds to an element of your affinoid algebra which is in all maximal ideals and hence (by Jacobson-ness) is nilpotent. For a general ring this certainly isn't true: the intersection of all prime ideals is the nilpotent elements, but the intersection of all maximal ideals might be bigger (think of a 1-dimensional local ring, for example).
| 13 | https://mathoverflow.net/users/1384 | 21876 | 14,451 |
https://mathoverflow.net/questions/21879 | 3 | This question stems from Dick Lipton's [recent blog post](http://rjlipton.wordpress.com/2010/04/12/socks-shoes-and-the-axiom-of-choice/) on the Axiom of Choice. I asked there but got no takers. I promise I'm not an inept Googler, but I couldn't find a satisfactory answer. I suspect universal in this context means computable by a universal Turing machine, or something close to that, but I'd like to know for sure.
| https://mathoverflow.net/users/175 | What is a universal function? | In that argument, he just means that g is defined on all the two-element subsets that may arise in the argument, that is, for a given family F of four-element sets, g(A) should be defined on any two-element set A that is a subset of a four-element set in F.
The reason he needs to assume that is that he cannot allow that we need to *choose* the choice function g itself to be used with each separate A. There are many choice functions that work for families of 2-element sets, and different choices of g will give rise to different functions on the families of four-element sets. The way the argument works is that you make *one* choice of the function g that works on all the two element sets that arise, and then you define the choice function on the given family of four-element sets by the clever construction in the article.
In particular, he is not using some technical meaning of *universal*.
| 5 | https://mathoverflow.net/users/1946 | 21884 | 14,454 |
https://mathoverflow.net/questions/3330 | 19 | I've been thinking a lot lately about random permutations. It's well-known that the mean and variance of the number of cycles of a permutation chosen uniformly at random from Sn are both asymptotically log n, and the distribution is asymptotically normal.
I want to know what a typical permutation of [n] with k(n) cycles "looks like" (in terms of cycle structure), where k(n)/(log n) → ∞ as n → ∞. The special case I have in mind is permutations of [n] with n1/2 cycles, since I've come across such permutations in another context, but I'm also curious about the more general problem. In order to do this I would like an algorithm that generates permutations of n with k cycles uniformly at random -- that is, it generates each one with probability 1/S(n,k) where S(n,k) is a Stirling number of the first kind -- so that I can experiment on them. (I'd be willing to settle for a Markov chain that converges to this distribution if it does so reasonably quickly.)
Unfortunately the only way I know to do this is to take a permutation of [n] uniformly at random (this is easy) and then throw it out if it doesn't have k cycles. If k is far from log(n) this is very inefficient, since those permutations are rare.
A few references I've come across that are related: [This paper of Granville](http://www.combinatorics.org/Volume_13/JOC/v13i1r107p.pdf) looks at permutations with o(n1/2-ε) cycles or Ω(n1/2+ε) cycles and shows that their cycle lengths are "Poisson distributed", but right around n1/2 is a transitional zone. And [this paper of Kazimirov](http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=dm&paperid=212&option_lang=eng) studies "the asymptotic behavior of various statistics" under the distribution I've claimed, but I haven't read it yet because I can't read Russian and I'm waiting for the English translation. Finally, the algorithm I'm looking for might be in one of the fascicles of volume 4 of Knuth, but our library doesn't have them.
| https://mathoverflow.net/users/143 | How can I generate random permutations of [n] with k cycles, where k is much larger than log n? | Youll find the answer on page 38 of my lecture notes East Side, West Side, which are a free download from my web site. It's a complete, short Maple program. The problem was originally solved in 1978 in Combinatorial Algorithms, by Albert Nijenhuis and myself.
Herb Wilf
| 21 | https://mathoverflow.net/users/5477 | 21896 | 14,461 |
https://mathoverflow.net/questions/21745 | 11 | I would like to know which is the real difference between the Recursive topos (in the sense of Mulry) and the Effective topos (in the sense of Hyland). Especially what is related to recursive functions. Do they have the same semantic power?
I will be gratefull with some hints about texts related to this.
Thanks in advance.
PS: I don't give definitions because of their big extension, but I could give them if anybody wants.
| https://mathoverflow.net/users/3338 | The difference between the Recursive and the Effective topos. | Your question is a bit unclear, but an obvious difference between these two toposes is that the Recursive Topos is a topos of sheaves, hence cocomplete, wherease the Effective topos only has finite (non-trivial) coproducts. For example, the natural numbers object in the Effective topos is *not* a countable coproduct of 1's.
If you are looking for a deeper explanation, then perhaps it is fair to say that the Recursive Topos models computability a la Banach-Mazur (a map is computable if it takes computable sequences to computable sequences) and the Effective topos models computability a la Kleene (a map is computable if it is realized by a Turing machine). In many respects Kleene's notion of computability is better, but you'll have to ask another question to find out why :-)
| 14 | https://mathoverflow.net/users/1176 | 21898 | 14,463 |
https://mathoverflow.net/questions/21775 | 5 | My collaborators and I are studying certain rigidity properties of hyperbolic toral automorphisms.
These are given by integral matrices A with determinant 1 and without eigenvalues on the unit circle.
We obtain a result under two additional assumptions
1) Characteristic polynomial of the matrix A is irreducible
2) Every circle contains no more than two eigenvalues of A (i.e. no more than two eigenvalues have the same absolute values)
We feel that the second assumption holds for a "generic" matrix. Is it true?
To be more precise, consider the set X of integral hyperbolic matrices which have determinant 1 and irreducible characteristic polynomial.
What are the possible ways to speak of a generic matrix from X? Does assumption 2) hold for generic matrices?
Comments:
* Assumption 1) doesn't bother us as it is a necessary assumption.
* Probably it is easier to answer the question when X is the set
off all integral matrices. In this case we need to know that
hyperbolicity is generic, 2) is generic and how generic is
irreducibility.
| https://mathoverflow.net/users/2029 | Spectrum of a generic integral matrix. | Yes, a generic integer matrix has no more than two eigenvalues of the same norm. More precisely, I will show that matrices with more than two eigenvalues of the same norm lie on a algebraic hypersurface in $\mathrm{Mat}\_{n \times n}(\mathbb{R})$. Hence, the number of such matrices with integer entries of size $\leq N$ is $O(N^{n^2-1})$.
Let $P$ be the vector space of monic, degree $n$ real polynomials. Since the map "characteristic polynomial", from $\mathrm{Mat}\_{n \times n}(\mathbb{R})$ to $P$ is a surjective polynomial map, the preimage of any algebraic hypersurface is algebraic.
Thus, it is enough to show that, in $P$, the polynomials with more than two roots of the same norm lie on a hypersurface. Here are two proofs, one conceptual and one constructive.
**Conceptual:** Map $\mathbb{R}^3 \times \mathbb{R}^{n-4} \to P$ by
$$\phi: (a,b,r) \times (c\_1, c\_2, \ldots, c\_{n-4}) \mapsto (t^2 + at +r)(t^2 + bt +r) (t^{n-4} + c\_1 t^{n-5} + \cdots + c\_{n-4}).$$
The polynomials of interest lie in the image of $\phi$. Since the domain of $\phi$ has dimension $n-1$, the Zariski closure of this image must have dimension $\leq n-1$, and thus must lie in a hyperplane.
**Constructive:** Let $r\_1$, $r\_2$, ..., $r\_n$ be the roots of $f$. Let
$$F := \prod\_{i,j,k,l \ \mbox{distinct}} (r\_i r\_j - r\_k r\_l).$$
Note that $F$ is zero for any polynomial in $\mathbb{R}[t]$ with three roots of the same norm. Since $F$ is symmetric, it can be written as a polynomial in the coefficients of $f$. This gives a nontrivial polynomial condition which is obeyed by those $f$ which have roots of the sort which interest you.
| 6 | https://mathoverflow.net/users/297 | 21906 | 14,467 |
https://mathoverflow.net/questions/21881 | 72 | I am a TA for a multivariable calculus class this semester. I have also TA'd this course a few times in the past. Every time I teach this course, I am never quite sure how I should present curl and divergence. This course follows Stewart's book and does not use differential forms; we only deal with vector fields (in $\mathbb{R}^3$ or $\mathbb{R}^2$). I know that div and curl and gradient are just the de Rham differential (of 2-forms, 1-forms, and 0-forms respectively) in disguise. I know that things like curl(gradient f) = **0** and div(curl F) = 0 are just rephrasings of $d^2 = 0$. However, these things are, understandably, quite mysterious to the students, especially the formula for curl, given by $\nabla \times \textbf{F}$, where $\nabla$ is the "vector field" $\langle \partial\_x , \partial\_y , \partial\_z \rangle$. They always find the appearance of the determinant / cross product to be quite weird. And the determinant that you do is itself a bit weird, since its second row consists of differential operators. The students usually think of cross products as giving normal vectors, so they are lead to questions like: What does it mean for a vector field to be perpendicular to a "vector field" with differential operator components?! Incidentally, is the appearance of the "vector field" $\nabla = \langle \partial\_x , \partial\_y , \partial\_z \rangle$ just some sort of coincidence, or is there some high-brow explanation for what it really is?
>
> Is there a clear (it doesn't have to necessarily be 100% rigorous) way to "explain" the formula for curl to undergrad students, within the context of a multivariable calculus class that doesn't use differential forms?
>
>
>
I actually never quite worked out the curl formula myself in terms of fancier differential geometry language. I imagine it's: take a vector field (in $\mathbb{R}^3$), turn it into a 1-form using the standard Riemannian metric, take de Rham d of that to get a 2-form, take Hodge star of that using the standard orientation to get a 1-form, turn that into a vector field using the standard Riemannian metric. I imagine that the appearance of the determinant / cross product comes from the Hodge star. I imagine that one can work out divergence in the same way, and the reason why the formula for divergence is "simple" is because the Hodge star from 3-forms to 0-forms is simple. Is my thinking correct?
Stewart's book provides some comments about how to give curl and divergence a "physical" or "geometric" or "intuitive" interpretation; the former gives the axis about which the vector field is "rotating" at each point, the latter tells you how much the vector field is "flowing" in or out of each point. Is there some way to use these kinds of "physical" or "geometric" pictures to "prove" or explain curl(gradient f) = **0** and div(curl F) = 0? Is there some way to explain to undergrad students how the formulas for curl and div do in fact agree with the "physical" or "geometric" picture? Though such an explanation is perhaps less "mathematical", I would find an explanation of this sort satisfactory for my class.
Thanks in advance!
| https://mathoverflow.net/users/83 | How should one present curl and divergence in an undergraduate multivariable calculus class? | To me, the explanation for the appearance of div, grad and curl in physical equations is in their invariance properties.
Physics undergrads are taught (aren't they?) Galileo's principle that physical laws should be invariant under inertial coordinate changes. So take a first-order differential operator $D$, mapping 3-vector fields to 3-vector fields. If it's to appear in any general physical equation, it must commute with with translations (and therefore have constant coefficients) and also with rotations. Just by considering rotations about the 3 coordinate axes, you can then check that $D$ is a multiple of curl.
If I want to devise a "physical" operator which has the same invariance property - and therefore equals curl, up to a factor - I'd try something like "the mean angular velocity of particles uniformly distributed on a very small sphere centred at $\mathbf{x}$, as they are carried along by the vector field." (This is manifestly invariant, but not manifestly a differential operator!)
[Here I should admit that, having occasionally tried, I've never convinced more than a fraction of a calculus class that it's possible to understand something in terms of the properties it satisfies rather than in terms of a formula. That's unsurprising, perhaps: it's not an obvious idea, and it's entirely absent from the standard textbooks.]
| 50 | https://mathoverflow.net/users/2356 | 21908 | 14,469 |
https://mathoverflow.net/questions/21890 | 6 | Background and Motivation
-------------------------
Local Class Field Theory says that abelian extensions of a finite extension $K/\mathbb{Q}\_p$ are parametrized by the open subgroups of finite index in $K^\times$. The correspondence takes an abelian extension $L/K$ and sends it to $N\_{L/K}(L^\times)$, and this correspondence is bijective.
If one starts instead with a galois extension $L/K$ that isn't abelian, one can then ask "What abelian extension does $N\_{L/K}(L^\times)$ correspond to?" The answer is the maximal abelian extension of $K$ contained in $L$.
---
The hypothesis of being galois isn't necessary in the statement of the non-abelian theorem: both the question and the answer still make sense. I am thus asking
>
> Assume that $L/K$ as above. Is the abelain extension of $K$ corresponding to $N\_{L/K}(L^\times)$ the maximal abelian extension of $K$ contained in $L$?
>
>
>
A couple examples to illustrate this problem (including the example that I was told would sink this):
If $p > 2$, then consider $L = \mathbb{Q}\_p(\sqrt[p]{p})$. The norm subgroup that I am anticipating is all of $\mathbb{Q}\_p^\times$. Moving into the galois closure and using the theorem, one gets that $N\_{L/\mathbb{Q}\_p}(L^\times)$ contains $p^\mathbb{Z} \times (1 + p\mathbb{Z}\_p)$. Moreover, the norm of a number $a \in \mathbb{Z}\_p$ will just be $a^p$, which is congruent to $a$ mod $p$, so the norm will also hit something congruent to any given root of unity in $\mathbb{Q}\_p$, and that was all that I was missing from the earlier note.
One can also, using the same idea (moving into the galois closure and getting a lot of information from that, and then using the explicit structure of the field that one started with) show that this works for $K(\sqrt[n]p)$ with $(n, p^{f(K/\mathbb{Q}\_p)}) = (n,p) = 1$ as well.
| https://mathoverflow.net/users/5473 | The norm of a non-Galois extension of local fields | Yup, it is. This is Theorem III.3.5 (norm limitation theorem) of Milne's Class field theory notes (available [here](http://jmilne.org/math/CourseNotes/CFT.pdf)). The global analogue is Theorem VIII.4.8.
| 4 | https://mathoverflow.net/users/1021 | 21909 | 14,470 |
https://mathoverflow.net/questions/21901 | 5 | In a review of a book by Ferdinand Gonseth in the Spring-Summer 2006 (Volume XI, Issue 1) of the HOPOS Newsletter it is said that Gonseth was the
"successor on Jerome's (sic) Franel's chair for Mathematics in French language at the ETH"
1) Was this a chair that Franel held or was it a chair Franel endowed? If the former, does the chair have a name?
2) What is the nature of a chair "for Mathematics in the French language"? Is it a history of French mathematics chair or is it a chair for a mathematician that writes in French?
Thanks for any insight.
Cheers, Scott
| https://mathoverflow.net/users/4111 | Jerome Franel's Chair at ETH | Franel was a number theorist who gave introductory calculus lectures in French
at the ETH. Maybe the ETH offered courses in different languages, because of its location in a multilingual country. Does anybody know? Anyway, for a little more information, see [this excerpt](http://books.google.com/books?id=OuHrR_6WEKsC&lpg=PA41&ots=Dc_XS3I3yz&dq=%22jerome%20franel%22%20eth&pg=PA41#v=onepage&q=%22jerome%20franel%22%20eth&f=false) from Jerry Alexanderson's *The Random Walks of George Polya.*
[Added later] At [this site](http://www.ethistory.ethz.ch/materialien/professoren/listen/alle_profs) there is a list of all professors in the history of the ETH, between 1855 and 2005. Franel is not the only professor for "Mathematik in französischer Sprache" -- there were some before Franel, and others in more recent times were Beno Eckmann and Armand Borel. There were also professors of geometry, mechanics and statistics "in französischer Sprache". So, Franel was not at all special in this respect, which strengthens my suspicion that the ETH made a point of offering courses in French.
| 6 | https://mathoverflow.net/users/1587 | 21914 | 14,474 |
https://mathoverflow.net/questions/21877 | 4 | Given $n,k\in\mathbb{N}$ where $k\leq n$, I want to compute the minimum subset of the set of partitions of $N$={$1,\ldots,n$}, satisfying these properties:
1. Each block of every partition has at most $k$ elements.
2. Every pair of elements of $N$ is in the same block in exactly one partition.
Anyone has a clue?
| https://mathoverflow.net/users/961 | Minimum cover of partitions of a set | (Edit: Sorry, my original restatement was incorrect.)
This problem is equivalent to decomposition a complete graph $K\_n$ into a collection of cliques $C:=\{K\_s\}$ where each $s \leq k$, such that $C$ can be resolved (i.e. partitioned) into a set of resolution classes $\mathcal{P}$ (the vertices of the graphs within a resolution class partition $\{1,2,\ldots,n\}$).
If each $s=k$, then $C$ is a [Steiner system](http://en.wikipedia.org/wiki/Steiner_system) S(2,k,n), a special type of [block design](http://en.wikipedia.org/wiki/Block_design), and we say $K\_k$ divides $K\_n$. In this case \[|\mathcal{P}|=\frac{k}{n}|C|=\frac{k}{n}\frac{n \choose 2}{k \choose 2}\] and in fact, this is always a lower bound on $|\mathcal{P}|$. Although, it's not always known when a Steiner system exists (or does not exist). The case $k=3$ (and each $s=k$) gives rise to the well-known Steiner triple system which exist if and only if $n \equiv 1$ or $3 \pmod 6$. For the resolution classes to exist, we must have $n \equiv 3 \pmod 6$, whence we have a [Kirkman triple system](http://designtheory.org/library/encyc/sts/g/).
You could find an upper bound by a greedy algorithm (starting with $K\_n$, pick the largest clique $K\_s$ with $s \leq k$ from the unused vertices, delete those edges and continue until you run out of edges, starting a new part when necessary).
| 7 | https://mathoverflow.net/users/2264 | 21921 | 14,478 |
https://mathoverflow.net/questions/21899 | 28 | I've tried in vain to find a definition of an algebra over a *noncommutative* ring. Does this algebraic structure not exist? In particular, does the following definition from <http://en.wikipedia.org/wiki/Algebra_(ring_theory)> make sense for noncommutative $R$?
>
> Let $R$ be a commutative ring. An algebra is an $R$-module $A$ together with a binary operation
> $$ [\cdot,\cdot]: A\times A\to A $$
> called $A$-multiplication, which satisfies the following axiom:
> $$ [a x + b y, z] = a [x, z] + b [y, z], \quad [z, a x + b y] = a[z, x] + b [z, y] $$
> for all scalars $a$, $b$ in $R$ and all elements $x$, $y$, $z$ in $A$.
>
>
>
So, is there a common notion of an algebra over a noncommutative ring?
| https://mathoverflow.net/users/1291 | Definition of an algebra over a noncommutative ring | The commutative notion of an (associative or not) algebra $A$ over a commutative ring $R$ has two natural generalization to the noncommutative setup, but the one you list with defined **left** $R$-linearity in both arguments is neither of them; in particular your multiplication does not necessarily induce a map from the tensor product, unless the image of $R$ is in the center. Most useful is the notion of an $R$-ring $A$ (or a ring $A$ over $R$), which is just a monoid in the monoidal category of $R$-bimodules: in other words the multiplication is a map $A\otimes A\to A$ which is *left* linear in first and *right* linear in the second factor. If we drop the associativity for the multiplication all works the same way, but I do not know if there is a common name (maybe descriptive like magma internal to the monoidal category of $R$-bimodules; or one may try a rare term nonassociative $R$-ring).
In the commutative case, the mutliplication is both left and right linear in each factor, what is here possible only if $R$ maps into the center of $A$. (Edit: I erased here one additional nonsense sentence clearly written when tired ;) ). Thus the two useful concepts in the noncommutative case are $R$-rings (possibly nonassociative!) and, well, the subclass with that property: $R$ maps into $Z(A)$, deserving the full name of "algebra". There is also a notion of $R$-[coring](http://ncatlab.org/nlab/show/coring), which is a comonoid in the monoidal category of $R$-bimodules, generalizing the notion of an $R$-coalgebra to a noncommutative ground ring.
Edit: I suggest also this [link](http://golem.ph.utexas.edu/category/2008/12/a_quick_algebra_quiz.html).
| 23 | https://mathoverflow.net/users/35833 | 21927 | 14,481 |
https://mathoverflow.net/questions/21688 | 7 | How does one study Krull dimension of some I-adic completion of a ring or, more generally, a module? I know that in case of Noetherian local ring Krull dimension of its completion equals Krull dimension of the ring, but what can we say in general case?
| https://mathoverflow.net/users/5432 | Krull dimension of a completion | For a Noetherian ring R, the Krull dimension of its $I$-adic completion, $\hat{R}$ is given by $\sup h(J)$, where $J$ ranges over all maximal ideals of $R$ containing $I$ and $h(J)$ is the [height](http://en.wikipedia.org/wiki/Height_(ring_theory)) of $J$. Therefore $\dim \hat R\le \dim R$ with equality only when $I\subset \operatorname{rad} R$. A reference is "Topics in $\mathfrak m$-adic topologies" by S.Greco, P.Salmon
| 13 | https://mathoverflow.net/users/2384 | 21930 | 14,483 |
https://mathoverflow.net/questions/21929 | 43 | We're all use to seeing differential operators of the form $\frac{d}{dx}^n$ where $n\in\mathbb{Z}$. But it has come to my attention that this generalises to all complex numbers, forming a field called fractional calculus which apparently even has applications in physics!
These derivatives are defined as fractional iterates. For example, $(\frac{d}{dx}^\frac{1}{2})^2 = \frac{d}{dx}$ or $(\frac{d}{dx}^i)^i = \frac{d}{dx}^{-1}$
But I can't seem to find a more meaningful definition or description. The derivative means something to me; these just have very abstract definitions. Any help?
| https://mathoverflow.net/users/5486 | What is the actual meaning of a fractional derivative? | I understand where Ryan's coming from, though I think the question of how to interpret fractional calculus is still a reasonable one. I found this paper to be pretty neat, though I have no idea if there are any better interpretations out there.
<http://people.tuke.sk/igor.podlubny/pspdf/pifcaa_r.pdf>
| 22 | https://mathoverflow.net/users/1916 | 21933 | 14,485 |
https://mathoverflow.net/questions/21935 | 8 | Requesting: a good reference for formal manipulation of limits of diagrams, with respect to maps of index diagrams.
As an example, consider the following result, for some "nice enough" category C (say, Top), there is a natural isomorphism
$$(A \times\_B C) \times\_C D \cong A \times\_B D.$$
This is a nice, intuitively true result that if we stick two pullback squares next to each other, we get another pullback square. I can easily convince myself of this on paper, chasing around the requisite number of arrows -- and it's a simple enough result that no wizardry is required; all maps could be named and the necessary commutativity relations and existence/uniqueness conditions can be checked.
However, it's a bit unsatisfactory. There's a blow-up of notation required to prove a relatively simple result. What happens when the diagrams get more complicated? It becomes less clear how proving basic results about limits can be written up with an acceptable amount of rigor, while remaining concise.
A general question of this type: let $C$ be a complete category, and let $D,E$ be small categories. A functor $f : D \to E$ induces a functor $f^\* : C^E \to C^D$. For which $f$ is it the case that $\lim\_E = \lim\_D \circ f^\*$ as functors $C^E \to C$?
It is at least intuitively clear that if $f(D)$ sufficiently "sits above" $E$, then this will hold. I am interested in formalizing this intuition.
I'm sure this question must have a standard answer in category theory, but a sufficient reference escapes me at the moment.
| https://mathoverflow.net/users/2532 | How to formally -- and cleanly -- express relationships of limits of diagrams? | As was mentioned in the comments, such general isomorphisms can be reduced by Yoneda to the case of sets, and in your example $((a,c),d) \mapsto (a,d)$ is an isomorphism, simply because $c$ is already determined by $d$. As for me, I always manipulate arrows without caring about single-use names *and* use Yoneda. This worked very good in the past years. For example, for every test object $T \in C$, we have canonical bijections
$\{T \to (A \times\_B C) \times\_C D\} = \{T \to A \times\_B C \to C = T \to D \to C\}$
$= \{T \to A \to B = T \to C \to B, T \to C = T \to D \to C\}$
$=\{T \to A \to B = T \to D \to C \to B\} = \{T \to A \times\_B D\}$
You may argue that it's not clear at all what is given etc., but there is only one plausible interpretation: Everything not given before belongs to the data of the morphism sets. Every equation is a condition on this data. It's even nicer to draw everything in commutative diagrams (I don't know how to draw them here). This method also works when you want to simplify a universal object without knowing the result. You just reduce the diagrams as above. Every step is almost forced. Of course, this is just another way of writing down the proof mentioned first.
Regarding your second question: Your intuition is absolutely correct. A sufficient and useful condition is that $f$ is a final functor. You can read about them in Mac Lane, Categories for the working mathematician, IX.3.
| 3 | https://mathoverflow.net/users/2841 | 21948 | 14,491 |
https://mathoverflow.net/questions/21940 | 5 | Let $I$ be a normal ideal on $P\_{\kappa} (\lambda)$. Let $V$ denote our ground model. Now we force with the $I$-positive sets, and if $G$ is the resulting generic filter, it can be shown that $G$ is a $V$-ultrafilter on $P\_{\kappa} (\lambda)$ extending the dual filter of $I$, which is normal if $I$ is normal.
We can build now the so called generic ultrapower, i.e. we construct inside $V[G]$, $Ult\_{G} (V)$ the class of all functions in $V$ with domain $\kappa$ and the usual binary relation $\in^{\ast}$. We assume that the generic ultrapower is well-founded, so we identify it with its transitive collaps $M \cong Ult\_{G} (V)$
My question now is: Is this $M$ closed under sequences from $V[G]$ of length $\lambda$, i.e. $M^{\lambda} \cap V[G] = M^{\lambda} \cap M$?
$M$ is closed under sequences from $V$ of length $\lambda$, this is clear to me but I don't have a good argument for the sequences of $V[G]$.
| https://mathoverflow.net/users/4753 | The closure of a generic ultrapower | If your ideal is normal, fine, precipitous and has the disjointing property (a consequence of saturation), then the answer is yes. As you likely know, you need more assumptions than you had stated, just in order to know that the ultrapower is well-founded. The difference in closure for the ultrapower that you mentioned appears to be related to the difference between having the disjointing property or not.
For a reference, I recommend Matt Foreman's chapter for the Handbook of Set Theory, which states the following theorem (it is Theorem 2.25 in the preliminary version I have here, but the published number may differ).
**Theorem.** Suppose I is a normal, fine, precipitous ideal on $Z\subset P(X)$, where $|X|=\lambda$. Let $G\subset P(Z)/I$ be generic, and $M$ the generic ultrapower of $V$ by $G$. Then $P(\lambda)\cap V\subset M$. Further, if $I$ has the disjointing property, then $M^\lambda\cap V[G]\subset M$.
Note that this theorem covers your case of $Z=P\_\kappa(\lambda)$.
To prove the first part, you simply observe that $[id]$ represents $j " \lambda$, and then for any $A\subset\lambda$ you can get $j"A$ using the function $g(z)=z\cap A$. Now, from $j"\lambda$ and $j"A$ you can easily build $A$ in $M$.
For the second part, the part you were interested in, you use the disjointing property in order to know that a term for a $\lambda$-sequence of elements of $M$ can be transformed into a $\lambda$-sequence of terms in $M$. That is, if $\langle\dot a\_\alpha :\alpha<\lambda\rangle$ is a $\lambda$-sequence of terms for objects in $M$, then disjointing allows us to find in $V$ a sequence of functions $\vec g = \langle g\_\alpha: \alpha<\lambda\rangle$ such that $[g\_\alpha]^G = \dot a\_\alpha^G$. From this, it follows that the function $g(z) = \langle g\_\alpha(z) | \alpha\in z\rangle$ represents $j(\vec g)(j"\lambda)$, which is $\langle j(g\_\alpha)\_\beta(j"\lambda) | \beta\in j"\lambda\rangle$, from which we can construct $\langle j(g\_\alpha)(j"\lambda) | \alpha <\lambda\rangle$, which is the desired $\lambda$-sequence.
| 5 | https://mathoverflow.net/users/1946 | 21963 | 14,500 |
https://mathoverflow.net/questions/21959 | 10 | Let G be a Lie group and M a smooth manifold. Suppose that P is a principal G-bundle over M. Then by Yoneda, this corresponds to a smooth map $p:M \to [G]$, where $[G]$ is the differentiable stack associated to G. If P is equipped with a connection, how does this fit in this picture?
I was thinking of using the fact that we can obtain $P$ as the weak pullback of the classifying map p along the universal principal G-bundle $\* \to [G]$, and then using that the tangent functor preserves finite weak limits. Then, if we are given a connection in terms of a G-invariant subbundle of $TP$, we could express its inclusion map $H \to TP$ as a map $H \to TM$ together with a certain 2-cell between the composite of the canonical map $H \to [TG]$ (the universal TG-bundle composed with the unique map to the terminal object) and T(p) composed with $H \to TM$. But then of course, we need to put conditions on our map $H \to TM$ to make sure that the induced map $H \to TP$ is an inclusion of vector bundles that moreover defines a G-invariant Ehresmann connection. Maybe this is not the way to go...
Does anyone know of a nice way of encoding a connection in this stacky language?
| https://mathoverflow.net/users/4528 | Connections on principal bundles via stacks? | The only thing which has to be replaced in the representation of a principal bundle as a suitable class of a maps $M\to [\*/G]$ in order to introduce a flat connection is to replace $M$ by its fundamental 1-groupoid $\Pi\_1(M)$, or, if the connection is not flat, by the thin homotopy version $P\_1(M)$ of it, cf. [nlab:path groupoid](http://ncatlab.org/nlab/show/path+groupoid). The same way it works for higher categorical generalizations, see [Schreiber:differential nonabelian cohomology](http://ncatlab.org/schreiber/show/Differential+Nonabelian+Cohomology) and for details also Sec. 7.4 (from page 27 on in version 1) in [arxiv/1004.2472](http://arxiv.org/abs/1004.2472).
| 5 | https://mathoverflow.net/users/35833 | 21967 | 14,504 |
https://mathoverflow.net/questions/21820 | 3 | Let $A \subseteq \mathcal B(\mathcal H)$ be a unital C\*-algebra in its universal representation. The GNS representation $\pi\_\mu\colon A \rightarrow \mathcal B(\mathcal H\_\mu)$ with base state $\mu$ extends uniquely to a normal $\ast$-homomorphism $\pi\_\mu''\colon A'' \to \mathcal B(\mathcal H\_\mu)$. Since $A''$ is a von Neumann algebra, there exists a unique projection $p\in A''$ such that $\mathbf{ker} \\ \pi\_\mu''\ = A''p$. Is $p$ the least upper bound of operators $a \in A$ such that $\mu(a) = 0$ and $0 \leq a \leq 1$?
EDIT: Jonas provided a simple counterexample. For non-commutative C\*-algebras, $\mu(a)=0$ of course does not imply that $\pi\_\mu(a)=0$. The naive question is therefore whether $p$ is the least upper bound of operators $a\in A$ such that $0 \leq a \leq 1$ and $\mu(c^\ast a c)=0$ for all $c \in A$. More generally, I would be grateful for any such "intrinsic" characterization of the projection $p$.
| https://mathoverflow.net/users/2206 | Kernel projections in the universal representation. | So, it seems like the new question is: Is $\ker\pi\_\mu \subseteq A$ dense in $\ker\pi\_\mu'' \subseteq A''$. As we're talking about the universal representation, $A''=A^{\*\*}$, the bidual of A.
So, suppose that $\mu$ is a *faithful* state on A, so $\ker\pi\_\mu=\{0\}$. I don't think it's necessary that $\ker \pi\_\mu''=\{0\}$: this is equivalent to $\pi\_\mu'':A^{\*\*}=A''\rightarrow B(H\_\mu)$ being injective, and hence an isomorphism onto its range.
For example, let $G$ be an infinite discrete group, let $A=C^\*\_r(G)$, and let $\mu$ be the canonical trace on $A$. Then $\pi\_\mu(A)'' = VN(G)$ the group von Neumann algebra (as $H\_\mu$ can be identified with $\ell^2(G)$), and the predual of $VN(G)$ is $A(G)$ the Fourier algebra. The dual of A is $B(G)$ the Fourier-Stieljtes algebra; as $G$ is not compact, $A(G) \subsetneq B(G)$. Then $A^{\*\*} = B(G)^\* = W^\*(G)$ (in common notation) and so the map $\pi\_\mu'':W^\*(G) \rightarrow VN(G)$ is the quotient induced by the adjoint of the inclusion $A(G) \rightarrow B(G)$. In particular, it's not injective.
I could, of course, have misunderstood the question...
| 3 | https://mathoverflow.net/users/406 | 21970 | 14,507 |
https://mathoverflow.net/questions/21916 | 0 | I am fairly new at linear programming/optimization and am currently working on implementing a linear program that is stated like this:
max $\sum\_{i=1}^{k}{p(\vec \alpha \cdot \vec c\_i)}$
$s.t. $
$|\alpha\_j| \le 1$
Where p(x) = 2x if x < 0, x otherwise, and $ \vec c$ is a constant
The p(x) function is what's troubling me, since one can only determine x's sign after an assignment of $\vec \alpha$. How can I remove the function p from the objective and express this objective equivalently as a linear combination of the variables?
Thank you!
| https://mathoverflow.net/users/5483 | Linear programming piecewise linear objective | Sorry for making you wait 14 hours unnecessarily but you are partially guilty yourself: if you posted a correct and full version of the question from the beginning, you would get the answer in 5 minutes. Keep it in mind when you ask a question on a public forum next time.
Your problem is equivalent to maximizing the linear expression $\sum\_i y\_i$ under the linear restrictions $\alpha\_j\ge -1$, $\alpha\_j\le 1$, $y\_i\le c\_i\cdot\alpha$, $y\_i\le 2c\_i\cdot\alpha$. It is as simple as that but it is crucial that your $p$ is concave and that you maximize.
| 3 | https://mathoverflow.net/users/1131 | 21974 | 14,510 |
https://mathoverflow.net/questions/21960 | 1 | Suppose to have a Lie algebra L with a reductive lie subalgebra G. Let l an element of L such that [l,g] is in G for every g in G, is it true that l is an element of G?if not, there are some restriction on G that makes it true?
| https://mathoverflow.net/users/4821 | reductive Lie subalgebra | Sorry, this started as a comment, but got too long.
If $G$ is semisimple, then every derivation of $G$ is inner, so
that the normalizer $N\_L(G)=C\_L(G)+G$ where $C\_L(G)$
is the centralizer. In this situtaion Michele's condition holds
if and only if the centralizer is trivial.
However the case where $G$ is reductive isn't so easy.
A reductive $G$ is the direct sum of an Abelian and a semisimple
Lie algebra. The Abelian part can have (if at least two-dimensional)
non-trivial derivations. These will lift to non-inner derivations
of $G$. If we let $L$ be the semidirect product of $G$ with
a one-dimensional Lie algebra using this derivation, then
$L$ normalizes $G$, $L\ne G$ but $C\_L(G)$ is trivial.
| 4 | https://mathoverflow.net/users/4213 | 21978 | 14,512 |
https://mathoverflow.net/questions/21947 | 17 | In [this post](https://mathoverflow.net/questions/21745/the-difference-between-the-recursive-and-the-effective-topos) about the difference between the recursive and effective topos, Andrej Bauer said:
>
> If you are looking for a deeper explanation, then perhaps it is fair to say that the Recursive Topos models computability a la Banach-Mazur (a map is computable if it takes computable sequences to computable sequences) and the Effective topos models computability a la Kleene (a map is computable if it is realized by a Turing machine). In many respects Kleene's notion of computability is better, but you'll have to ask another question to find out why :-)
>
>
>
So I'm asking:
1) What is "computability a la Banach-Mazur"? I would guess it has something to do with Baire spaces and computable analysis, but I don't really know.
2) Why is Kleene's notion of computability better?
| https://mathoverflow.net/users/1610 | Why is Kleene's notion of computability better than Banach-Mazur's? | This answer requires a bit of background.
**Definition 1:** a *numbered set* $(X,\nu\_X)$ is a set $X$ together with a partial surjection $\nu\_X : \mathbb{N} \to X$, called a *numbering* of $X$. When $\nu\_X(n) = x$ we say that $n$ is a *code* for $x$.
Numbered sets are a generalization of Gödel codes. Some typical examples are:
* $\mathbf{N} = (\mathbb{N}, \mathrm{id}\_\mathbb{N})$ is the standard numbering of natural numbers.
* $\mathbf{P} = (P, \phi)$ where $P$ is the set of partial computable maps and $\phi$ is a standard enumeration of partial computable maps.
* $\mathbf{R} = (R,\nu\_R)$ where $R$ is the set of computable reals and $\nu\_R(n) = x$ when, for all $k \in \mathbb{N}$, $\phi\_n(k)$ outputs (a code of) a rational number $q$ such that $|x - q| < 2^{-k}$.
Numbered sets can be used to give effective structure to many mathematical structures. What should we take as a morphism between numbered sets? Presumably a map $f : X \to Y$ should be considered a morphism from $(X,\nu\_X)$ to $(Y,\nu\_Y)$ when it is "computable" in a suitable sense. We understand fairly well what it means to have a computable map $\mathbb{N} \to \mathbb{N}$, namely computed by a Turing machine, so let us take that for granted. It is easy to extend computability of sequences of numbers to computability of arbitrary sequences:
**Defintion 2:** A map $s : \mathbb{N} \to X$ is a *computable sequence* in $(X,\nu\_X)$ when there exists a computable map $f : \mathbb{N} \to \mathbb{N}$ such that $s(n) = \nu\_X(f(n))$ for all $n \in \mathrm{dom}(\nu\_X)$.
Now suppose we think a bit like analysts. One way to define a continuous map is to say that it maps convergent sequences to convergent sequences. We could mimick this idea to define general computable maps.
**Definition 3:** A function $f : X \to Y$ where $(X,\nu\_X)$ and $(Y,\nu\_Y)$ are numbered sets is *Banach-Mazur computable* when $f \circ s$ is a computable sequence in $(Y,\nu\_Y)$ whenever $s$ is a computable sequence in $(X,\nu\_X)$.
How good is this notion? And how does it compare to the following notion, which is taken as the standard one nowadays?
**Definition 3:** A function $f : X \to Y$ where $(X,\nu\_X)$ and $(Y,\nu\_Y)$ are numbered sets is *Markov computable*, or just *computable*, when there exists a partial computable map $g : \mathbb{N} \to \mathbb{N}$ such that $f(\nu\_X(n)) = \nu\_Y(g(n))$ for all $n \in \mathrm{dom}(\nu\_X)$.
In other words, $f$ is *tracked* by $g$ in the sense that $g$ does to codes what $f$ does to elements. (Note: in [this MO](https://mathoverflow.net/questions/21745) I attributed this notion of computability to Kleene, but I think it's better to attach Markov's name to it, if any.)
Every Markov computable function is Banach-Mazur computable. In some cases the converse holds as well. For example, every Banach-Mazur computable map $\mathbf{N} \to \mathbf{N}$ is Markov computable. However, this is not the case in general:
1. R. Friedberg demonstrated that there is a Banach-Mazur computable map $\mathbf{N}^\mathbf{N} \to \mathbf{N}$ which is not Markov computable. [R. Friedberg. 4-quantifier completeness: A Banach-Mazur functional not uniformly partial recursive. Bull. Acad. Polon.
Sci. Sr. Sci. Math. Astr. Phys., 6:1–5, 1958.]
2. P. Hertling constructed a Banach-Mazur computable map $\mathbf{R} \to \mathbf{R}$ which is not Markov computable. [P. Hertling. A Banach-Mazur computable but not Markov computable function on the computable real numbers. In Proceedings ICALP 2002, pages 962–972. Springer LNCS 2380, 2002.]
3. A. Simpson and I showed that there is a Banach-Mazur computable $\mathbf{X} \to \mathbf{R}$ that is not Markov computable when $\mathbf{X}$ is any inhabited computable complete separable metric space computably without isolated points. [A. Bauer and A. Simpson: [Two Constructive Embedding-Extension Theorems with Applications to Continuity Principles and to Banach-Mazur Computability](http://math.andrej.com/2004/07/27/two-constructive-embedding-extension-theorems-with-applications-to-continuity-principles-and-to-banach-mazur-computability/), Mathematical Logic Quarterly, 50(4,5):351-369, 2004.]
What this says is that Banach-Mazur computability is too general because it admits functions that cannot be computed in the standard sense of the word, i.e., computed by Turing machine (in terms of codes).
| 23 | https://mathoverflow.net/users/1176 | 21979 | 14,513 |
https://mathoverflow.net/questions/21954 | 8 | Can someone please tell me where I can find a citeable reference for the following result:
Call the metric space $(X, d)$ "locally complete" if for every $x \in X$ there a neighbourhood of $x$ which is complete under $d$.
If $(X,d)$ is locally complete and separable then there exists a metric $d'$ on $X$ such that $(X,d) \to (X,d')$ is a homeomorphism and $(X, d')$ is complete.
This result follows immediately from Alexandrov's theorem that a $G\_\delta$ subset of a Polish space is Polish, but I'd rather find a statement in the literature of the above more straightforward result if there is one.
| https://mathoverflow.net/users/3676 | Locally complete space is topologically equivalent to a complete space | For a reference: [this paper](http://www.chimiefs.ulg.ac.be/SRSL/newSRSL/modules/FCKeditor/upload/File/73_1/Bella%20TSIRULNIKOV%20p%209-19.pdf) has a reference [30] that has a proof. The author cites your result and refers to it. I don't have access to these papers, so I cannot verify exactly.
| 3 | https://mathoverflow.net/users/2060 | 21983 | 14,516 |
https://mathoverflow.net/questions/21977 | 1 | Can any one give an example of a 3-fold X which contains an embedded ample divisor $D \cong CP^2$
with normal bundle $O(3)$ in X?
| https://mathoverflow.net/users/5259 | Looking for a 3-fold with some property? | Let me give an attempt of a proof of the fact that such example does not exist.
Proof. Suppose $X$ is such a $3$-fold and let $L$ be the line bundle corresponding to the divisor $\mathbb CP^2$. First we will prove that $Pic(X)=\mathbb Z$ and then will get a contradiction.
Notice that $L$ has a lot of sections. In particular in a neighbourhood of $\mathbb CP^2$ for every two points $x,y$ there is section of $L$ that contains $x$ but does not contain $y$.
And also notice that for every point of $X$ there is a section that does not contain it. Hence we have a morphism $X\to P(H^0(L)^\*)$. This morphism can not contract anything since $L$ is ample. The map is an embedding on the neighbourhood of $D$ and so the image can have singularities at most in codimension $3$.
From this it should follow (I guess), that we can apply Lefshetz that says $Pic (X)=Pic(D)=\mathbb Z$. So $X$ is a Fano with $Pic=\mathbb Z$. Torsten Ekedahl explained how now one can deduce contrudiction, see his comment.
| 4 | https://mathoverflow.net/users/943 | 21984 | 14,517 |
https://mathoverflow.net/questions/21903 | 24 | I'm curious if the term localization in ring theory comes from algebraic geometry or not. The connection between localization and "looking locally about a point" seems like it should be the source for the notion of localization. It seems plausible, but it seems like we would have had to wait until Zariski defined the Zariski topology for the connection to become apparent. That seems hard to believe given the amount of work done in commutative algebra before 20th century, especially given the importance of localization in commutative algebra.
Then this raises the question: Where and when was the term 'localization' first used to describe the adjunction of inverses, and does it originate from algebraic geometry or from somewhere else? Was the notion of localization used regularly with a different name before it was given this name?
| https://mathoverflow.net/users/1353 | Origin of the term "localization" for the localization of a ring | I'm looking at the paper "On the theory of local rings" by Chevalley (Annals of Math. 44 (1943)). In this paper he explains how to localize at a multiplicative set $S$ of non-zero divisors, and calls this the *ring of quotients* of the set $S$.
There is no question that Chevalley was motivated by algebraic geometry.
The paper "Generalized semi-local rings", by Zariski (Summa Brasiliensis Math. 1 (1946))
attributes the theory of local rings to Krull (in a paper called "Dimensionstheorie in Stellenringen", Crelle 179 (1938), which I don't have a copy of at hand) and Chevalley (in the above mentioned paper), so it seems that the Chevalley reference above is a reasonable guide to the situation.
Of course none of these references quite address the origin of the term *localization* at $S$,
but (based on my prior preconceptions, and bolstered by having looked at these two papers) I am fairly confident that it was indeed motivated by algebraic geometry.
| 14 | https://mathoverflow.net/users/2874 | 21987 | 14,519 |
https://mathoverflow.net/questions/21961 | 13 | In the Euclidean plane, is the circle the only simple closed curve that has an axis of symmetry in every
direction?
| https://mathoverflow.net/users/4423 | Can the circle be characterized by the following property? | A slightly different argument is as follows. Choose two symmetries $\sigma,\tau$
with axes
intersecting at a point $P$ and forming an angle of $2\pi \lambda$ with $\lambda$ irrational.
The composition $\rho=\sigma\circ \tau$ is then a rotation of infinite order generating
a dense subgroup of the group of all rotations centered at $P$.
Any closed subset left invariant under $\rho$ is thus a union of concentric circles centered at $P$. A simple closed curve invariant under $\rho$ is thus such a circle.
| 18 | https://mathoverflow.net/users/4556 | 21989 | 14,520 |
https://mathoverflow.net/questions/21939 | 31 | So sometime ago in my math education I discovered that many mathematicians were interested in moduli problems. Not long after I got the sense that when mathematicians ran across a non compact moduli they would really like to compactify it.
My question is, why are people so eager to compactify things? I know compactness is a great property of a space to have because it often makes other results much easier to prove. So I think my question is better stated as: what are some examples of nice/good/cool results related to a moduli spaces that were (only) able to be proved once there was a compactification of the space?
| https://mathoverflow.net/users/7 | What can you do with a compact moduli space? | The answers here are all excellent examples of things that can only be proved once a moduli space is compactified. I would like to add a perhaps more basic reason for compactifying moduli spaces, involving something simpler than theoretical applications such as defining enumerative invariants. The moral is the following:
>
> If you study families of geometric objects then either you are almost certain to encounter the boundary of the moduli space, or you must have some very good reason to rule it out.
>
>
>
For example, to find a non-trivial compact family of smooth complex curves is actually quite awkward and such families are very rare. (The first examples were due to Atiyah and Kodaira.) From this point of view the "ubiquity of the compactification" amounts to the fact that the boundary divisor of singular curves in the compactified moduli space is positive in a certain sense, so it intersects almost all curves in the moduli space. It is this positivity of the boundary which forces us to study it!
Some more examples explain - I hope! - the way compactification enters when considering pseudoholomorphic curves as in Gromov-Witten theory, without ever coming close to trying to define an enumerative invariant. Just by looking at a conic in $\mathbb{CP}^2$, which degenerates into two lines, one sees that when moving a pseudoholomorphic curve around, one is almost certain to encounter bubbling, unless one has a very good reason to know otherwise. Understanding how to compactify the moduli space, we see that this bubbling phenomenon is the main thing which can go wrong. What is interesting here is that often one tries to prove this compactification is *not actually necessary*, by ruling out bubbling somehow. Two examples follow - taken from Gromov's original use of pseudoholomorphic curves in his Inventiones paper - which exploit this idea.
Firstly, Gromov's proof of his non-squeezing theorem. Here the key point in the argument is that one can find a certain pseudoholomorphic disc for a standard almost complex structure on $\mathbb{C}^n$. One would like to know that as one deforms the almost complex structure the disc persists so that we have such a disc for a special non-standard almost complex structure. It is standard in this kind of "continuity method" that you can always deform the disc for a little while because the problem is elliptic. But to push the deformation indefinitely you need to show compactness - why doesn't the disc break up? Thanks to our knowledge of the compactification of the moduli space, we understand that the only thing that can go wrong is bubbling and in this case bubbles cannot form because the symplectic structure is exact.
The second example is of the following type: suppose one knows the existence of *one* pseudoholomorphic curve in a symplectic manifold; then one can try and use it to investigate the ambient space, moving it around and trying to sweep out as much of the space as possible. In this way you can prove, for example, that any symplectic structure on $\mathbb{CP}^2$ which admits a symplectic sphere with self-intersection 1 must be the standard symplectic structure. The reason is you can find an almost complex structure which makes this sphere a pseudoholomorphic curve. Then you move the curve around until is sweeps out the whole space, doing it carefully enough to give a symplectomorphism with the standard $\mathbb{CP}^2$. Here you can push the curve wherever you want because it wont break. Bubbles can't form because the curve has symplectic area 1 and so there is no "spare area" to make bubbles with.
| 19 | https://mathoverflow.net/users/380 | 21997 | 14,525 |
https://mathoverflow.net/questions/21998 | 4 | Hello, I'm interested in the case when all varieties are projective threefolds over the complex numbers.
Start with a flopping contraction $f:Y \rightarrow X$, with corresponding flop $f^+: Y^+ \rightarrow X$. It has been [proved](http://www.tombridgeland.staff.shef.ac.uk/papers/flop.pdf) that there then exists an equivalence $\Phi : D^b(Y^+) \rightarrow D^b(Y)$.
Is there a way to understand this (or any other) equivalence explicitly?
I've heard there is a way to find an equivalence by considering a common resolution of $Y$ and $Y^+$ and then using derived pullback and pushforward, is it true?
I am mostly interested in what happens to sheaves on $Y^+$ supported on the exceptional locus.
Thanks.
| https://mathoverflow.net/users/3701 | Is there a nice way to characterise the derived equivalence induced by a flop? | As always, it depends on what you think "explicitly" means. It's a Fourier-Mukai transform; see, for example, [Van den Bergh and Hille's expository article](http://alpha.uhasselt.be/Research/Algebra/Publications/hille.pdf). It can also be explained in terms of so-called non-commutative crepant resolutions, [see Van den Bergh](http://alpha.uhasselt.be/Research/Algebra/Publications/flops.ps).
| 2 | https://mathoverflow.net/users/460 | 22000 | 14,527 |
https://mathoverflow.net/questions/22001 | 6 | I am teaching a graduate "classical" course on modular forms. I try to achieve the most elementary level for presenting modular polynomials. Serge Lang's "Elliptic functions" cover the topic quite well, except for some inconsistence (from my point of view) in using left/right cosets for actions of the modular group on the set of matrices of determinant $n$. I overcome this trouble by using one more simple arithmetic lemma which relates left and right cosets. I wonder whether there exist another version of the Kronecker--Weber approach for modular polynomials, or maybe even another *elementary* proof.
| https://mathoverflow.net/users/4953 | References for modular polynomials | I'm not quite sure what exactly you're asking for, but you might find the third part of David Cox's *Primes of the form x^2 + n y^2* useful for an elementary approach to modular polynomials.
| 4 | https://mathoverflow.net/users/422 | 22005 | 14,530 |
https://mathoverflow.net/questions/22009 | 8 | I'm looking for an anecdote about a mathematician who studied random walks. I'm attempting to write an article and hope to include the story (but only if I can get the details correct). I'll try to do my best describing it in hopes someone else has heard it and knows a name or the full story.
A mathematician was walking through the park and entertaining mathematical whims. He noticed that he kept running into this same couple as he wandered around aimlessly. He wasn't sure if this behaviour was expected by chance or whether perhaps the female of the couple thought that he was cute. He rushed home to analyse the situation in terms of random walks in two dimensions.
| https://mathoverflow.net/users/3737 | Random Walk anecdote. | The anecdote is about Polya, and it is in his contribution, Two incidents, to the book, Scientists at Work: Festschrift in Honour of Herman Wold, edited by T Dalenius, G Karlsson, and S Malmquist, published in Sweden in 1970. It was recently quoted on page 229 of David A Levin and Yuval Peres, Polya's Theorem on random walks via Polya's Urn, Amer Math Monthly 117 (March, 2010) 220-231.
| 12 | https://mathoverflow.net/users/3684 | 22010 | 14,533 |
https://mathoverflow.net/questions/22012 | 15 | I have some understanding that vector bundles provide a basic, familiar example of what I should call a stack. Namely, consider the functor $Vect$ that assigns to a space $X$ the *set* of isomorphism classes of vector bundles on $X$. This functor isn't local, in the sense that the isoclass of a vector bundle isn't determined by its restriction to an open cover, but rather by *gluing data* on overlapping sets in a cover. Since for any space $Y$ a map $X \to Y$ *is* determined by what it does when restricted to a cover of $X$, this tells us there is no space $Y$ that represents the functor $Vect$ in this fashion. However, I can also consider $Vect$ as a stack, which assigns to $X$ the *groupoid* of vector bundles on $X$. This gadget is fancy enough to understand how vector bundles glue together, and so recovers the locality missing from our earlier functor.
In K-theory, we attach to a space $X$ a ring $K(X)$ whose underlying group is the the free abelian group on the set of isoclasses of vector bundles on $X$, mod short exact sequences. It turns out that one can describe $K(X)$ as the set of homotopy classes of maps from $X$ to $\mathbb{Z} \times BU(\infty)$.
At this point my meager understanding of K-theory seems to be contradict what I said in the first paragraph. The fact that $K(X)$ has a classifying space seems at odds with the observation that vector bundles aren't determined by their restrictions to open covers, whereas maps to another space are. Is something wrong with what I've said so far? If not, perhaps there isn't a contradiction because either 1) $K(X)$ isn't quite the set of isoclasses of vector bundles, but rather a group completion thereof, or 2) we're looking at *homotopy classes* of maps to $\mathbb{Z} \times BU(\infty)$, so what I said in the first paragraph doesn't apply?
| https://mathoverflow.net/users/361 | K-Theory and the Stack of Vector Bundles | [Read this](http://ncatlab.org/nlab/show/moduli+space#because)
| 7 | https://mathoverflow.net/users/83 | 22018 | 14,535 |
https://mathoverflow.net/questions/17565 | 9 | Quite a long ago, I tried to work out explicitly the content of the Newlander-Nirenberg theorem. My aim was trying to understand wether a direct proof could work in the simplest possible case, namely that of surfaces. The result is that the most explicit statement I could get is a PDE I was not able to solve.
Assume a quasi-complex structure $J$ is given on the surface $S$; we want to prove that this is induced by a complex one (in this case there are no compatibility conditions). This can be easily transformed in the problem of local existence for a second order PDE, as follows.
We look for local charts on $S$ which are holomorphic (with respect to the quasi-complex structure on $S$). Two such charts are then automatically compatible. So the problem is local.
Fix a small open set $U \subset S$ and identify it with a neighboorhood of $0 \in \mathbb{R}^2$ via a differentiable chart. Locally we can write $J = \left(\begin{matrix}[a | b] \\ [c | d ]\end{matrix}\right)$ for some functions $a, \cdots, d$ (pretend it is a two by two matrix).
A chart is given by a complex valued function $f = u + iv$. The condition that the differential is $\mathbb{C}$-linear can be verified on a basis of the tangent space; moreover if it is true for a vector v, it remains true for Jv, which is linearly independent. Here we have used that $J^2 = -1$.
So we need only to check it for the vector $\partial\_x$. Since $J \partial\_x = a \partial\_x + c \partial\_y$, the condition says
$-v\_x = a u\_x + c u\_y$
$u\_x = a v\_x + c v\_y$
Hence we need to solve this system, with $f = u + i v$ non singular ($f$ will be then locally invertible). Since $a$ and $c$ do not vanish simultaneously, we can assume $c(0) \neq 0$, hence $c \neq 0$ on $U$ provided $U$ is small.
We can then solve for $u\_y$ and get the equivalent system
$u\_x = a v\_x + c v\_y$
$-u\_y = \frac{1 + a^2}{c}v\_x +a v\_y$
Moreover the Jacobian $J\_f = u\_x v\_y + u\_y v\_x = \frac{1}{c}(v\_x^2 + (a v\_x + c v\_y)^2)$, so $f$ is nonsigular if $v$ is. By Poincaré's lemma, the system admits a local solution if and only if
$\frac{\partial}{\partial\_y} \left( a v\_x + c v\_y \right) - \frac{\partial}{\partial\_x} \left( \frac{1 + a^2}{c}v\_x +a v\_y \right) = 0$.
Hence we are looking for a local solution of the last equation with $(v\_x(0), v\_y(0)) \neq (0, 0)$.
So my question is:
>
> Is there a simple way to prove local existence for a nonsingular solution of the last displayed equation?
>
>
>
I should make clear that I'm not looking for a proof of Newlander-Nirenberg; of this there are plenty. I am more interested in seeing what Newlander-Nirenberg becomes in terms of PDE in the simplest possible case, and then see that the PDE thus obtained is solvable. According to the answer of Andy, the equation which comes out is the Beltrami equation, so I will have a look at it. Still, I'm curious if any standard PDE technique can solve the equation I derived in the most stupid way above.
| https://mathoverflow.net/users/828 | Newlander-Nirenberg for surfaces | I'm not sure if the following is elementary enough, but it does only use standard PDE machinery (plus some basic Riemannian geometry). It's also nice in that it suggests an approach to proving the uniformization theorem (via metrics of constant curvature).
Say you have an almost-complex structure on the unit disk. Your goal is to find a conformal isomorphism of this disk (or at least some neighborhood of the origin) with an open subset of the complex plane. To do it, first choose a metric $g$ on the disk which is compatible with the given complex structure. Let $K$ be the curvature of this metric. If you can find a flat metric $\tilde{g}$ on on the disk which is conformally equivalent to $g$ then you'll be done, since the exponential map with respect to $\tilde{g}$ will be an isometry, hence also a conformal isomorphism.
So, multiply $g$ by an arbitrary positive function $e^f$, and compute the curvature of the new metric. You'll find that it's given by the formula:
$\tilde{K} = e^{-2f}(K - \Delta f) $
where $\Delta$ is the Laplacian with respect to the metric $g$. Setting the left hand side equal to zero, you have reduced to solving the Laplace equation, which can be done locally using standard PDE techniques.
As for a more "direct" approach...
The equation you wrote down should reduce to solving the Laplace equation as well, using the notion of conjugate harmonic functions. However, solving your equation will inevitably be a bit more subtle due to your requirement that the solution have nonvanishing differential at the origin. There is a proof along the lines you're suggesting in Taylor's PDE book, chapter 5, section 11, and I think there's a similar one in Jost's "Postmodern analysis". Basically the idea is to rescale your coordinate system so that the metric is nearly flat, in which case you should have a conformal map that is close to the identity map in a high enough sobolev space, and therefore has a nonvanishing derivative at the origin.
| 8 | https://mathoverflow.net/users/5499 | 22019 | 14,536 |
https://mathoverflow.net/questions/22017 | 11 | Let $S\_{n}$ denote the permutation group on $n$ letters and $G\subset S\_{n}$ a transitive subgroup. The inclusion of $G$ in $S\_{n}$ defines an action of $G$ on $\mathbb{C}^{n}$. By finding a generating set of invariant polynomials and relations among these polynomials we may realize the quotient space $\mathbb{C}^{n}/G$ as an algebraic variety.
I have noticed that for $n=2,3,4$ such quotients are always complete intersections. That is, the difference between the order of a minimal generating set for the invariant polynomials, and the order of a minimal generating set for the relations, is equal to $n=\mathrm{dim}(\mathbb{C}^{n}/G)$.
I would like to know if this fact persists for all $n$. If so, what is the property of the group action which makes the quotient a complete intersection? If not what is a counterexample?
| https://mathoverflow.net/users/5124 | When Are Quotients Complete Intersections? | This paper of Kac and Watanabe may be of interest:
Kac, Victor; Watanabe, Keiichi, *Finite linear groups whose ring of invariants is a complete intersection*
<http://www.ams.org/mathscinet-getitem?mr=640951>
From the review:
"The following theorems are proved: Let $k$ be a field and $G$ be a finite subgroup of $\text{GL}(n,k)$. The group $G$ acts naturally on the polynomial ring $S=k[x\_1,\cdots,x\_n]$. Let $R=S^G$.
**Theorem A**: If $R$ is a complete intersection then $G$ is generated by the set $\lbrace g\in G\colon\text{rank}(g-I)\leq 2\rbrace$.
**Theorem B**: Let $k={\bf C}$. If $R$ has $m$ generators such that their ideal of relations is generated by $m-n+s$ elements, then $G$ is generated by the set $\lbrace g\in G\colon\text{rank}(g-I)\leq s+2\rbrace$.
**Theorem C**: If $R$ is a complete intersection then each stabilizer $G\_x$, $x\in k^n$, is generated by $\lbrace g\in G\_x\colon\text{rank}(g-I)\leq 2\rbrace$. "
In this case, the cyclic group of $S\_4$ generated by the long cycle acting on ${\bf C}^4$ should be a counterexample. I should probably calculate the ring of invariants, but I'll leave it at that for now.
| 14 | https://mathoverflow.net/users/321 | 22020 | 14,537 |
https://mathoverflow.net/questions/22014 | 5 | I was reading a proof that used the following result
Let $C$ be a hyperelliptic of genus $\ge 3$ and $\tau \colon C \to C$ the hyperelliptic involution. If $D$ is an effective divisor of degree $g-1$ such that $h^0(D)>1$ then $D = x + \tau(x) + D'$ where $D'$ is an effective divisor.
My question is, how is this result proved? It seems equivalent to showing that $|D|$ contains the unique $g^1\_2$ and this made me think of Clifford's theorem but this didn't lead to much. For $g = 3$ the result holds because then $|D| = g^1\_2$. But already for $g = 4$ I'm stuck. I tried playing around with the Riemann-Roch theorem but didn't get far.
| https://mathoverflow.net/users/7 | Special divisors on hyperelliptic curves | Suppose that $D=x\_1+x\_2+\cdots+x\_{g-1}$. We may assume that $\tau(x\_i)\neq x\_j$
for all $i\neq j$. Now, assume that $D'=y\_1+\cdots+y\_{g-1}$ is an element of
$|K-D|$. Then $x\_1+x\_2+\cdots+x\_{g-1}+y\_1+\cdots+y\_{g-1}$ is an element of $|K|$
but we know that any such element is of the form
$z\_1+\tau(z\_1)+\cdots+z\_{g-1}+\tau(z\_{g-1})$. After possibly renumbering the $z\_i$ (as
well as possibly replacing $z\_i$ by $\tau(z\_i)$) we may assume $x\_1=z\_1$ and
then $x\_2=z\_2$ and so on. This gives $D'=\tau(D)$ which implies
$h^0(K-D)\leq 1$. However, R-R and gives $h^0(D)=h^0(K-D)\leq1$ which
contradicts assumptions.
| 12 | https://mathoverflow.net/users/4008 | 22022 | 14,539 |
https://mathoverflow.net/questions/22013 | 4 | Given a completely metrizable space, say that it has property X if it can be embedded in some metric space such that its image is not closed. For example, the real line R can be embedded, topologically, in itself as (0,1) which is not closed. A compact space such as S^1, however, clearly cannot be embedded in any metric space in this way.
1. Is it true that a space has property X iff it can be embedded in itself in this way? My intuition says no, but I can't think of a counterexample offhand, partly because I don't know any interesting non-compact metric spaces.
2. Is property X equivalent to not being compact? If every non-compact (completely metrizable) space has a metrizable compactification then this is easy, but I don't know whether that is the case, although I suspect it might be. Which leads onto the following weaker question:
3. Given a space which does not have property X, can it be embedded in a space which does? Here my intuition says yes, this should be the case, but I can't think of a more general approach for constructing a suitable embedding space than compactification which, as I said, I can't prove will give a metrizable space. I don't think I know enough topology (I've only taken a couple of basic undergraduate courses).
Edit: 3. above should read "given a space which has property X, can it be embedded in a space which does not?
| https://mathoverflow.net/users/4336 | Topological embeddings of non-compact, complete metric spaces | The answer to 2 is yes. Let $(X,d)$ be complete but not compact, then for some $r>0$ it has a countable r-separated subset $S=\{s\_i\}:i=1,2,\dots$. Let $d'$ be the maximal metric on $X$ satisfying $d'\le d$ and $d'(s\_i,s\_j)\le r/\min(i,j)$ for all $i,j$. Then $(X,d')$ is homeomorphic to $(X,d)$ - in fact, the identity map is a local isometry - but $d'$ is not complete. So we have embedded the space into the completion of $(X,d')$ as a non-closed subset.
The metric $d'$ can be constructed explicitly: $d'(x,y)$ is the minimum of $d(x,y)$ and the infimum of sums $d(x,s\_i)+d(x,s\_j)+r/\min(i,j)$ over all pairs of $i,j$. Verifying the triangle inequality is straightforward.
As for 3, the answer is no, because you cannot embed any complete space into a compact space. For example, a non-separable Banach space cannot be so embedded, as Qiaochu Yuan explained in comments.
**Update.** It seems that I misunderstood Q3. As stated, it asks whether every compact space can be embedded into a complete non-compact one. The answer is of course yes, as Ady noticed.
| 7 | https://mathoverflow.net/users/4354 | 22031 | 14,546 |
https://mathoverflow.net/questions/21589 | 9 | Even searching for " ['number of trees' leaves](https://oeis.org/search?q=%22number+of+trees%22+leaves) " didn't reveal what I am looking for: an approach for calculating the (approximate) number of trees with exactly n nodes and m leaves. Any hints from MO?
| https://mathoverflow.net/users/2672 | Number of trees with n nodes and m leaves | The answer to this (very natural) question depends on your notion of "tree" (e.g. free, rooted) and the equivalence relation you employ (e.g. labelled, unlabelled). I haven't gone into the nitty-gritty details of all these results, but here's what I've found so far. There's likely published results I haven't found yet, but hopefully this helps to get you started.
We can compute $T\_{m,n}$, the number of non-isomorphic free trees with $m$ leaves and $n$ vertices, for small $m$ and large $m$. For example, (a) $T\_{3,n}$ is the number of partitions of $n-1$ into $3$ positive integer parts ([Sloane's A001399](https://oeis.org/A001399)), (b) $T\_{n-2,n}=\lfloor (n-2)/2 \rfloor$ and (c) $T\_{n-3,n}=\sum\_{j=0}^{n-5} \lfloor (n-3-j)/2 \rfloor$. The first result can be observed by deleting the vertex of degree 3 and the last two can be observed by colouring each non-leaf vertex by the number of adjacent leaves, then deleting the leaves.
Yu (8) seems to have given an algorithm for generating rooted trees with $m$ leaves. Wang (6) and Liu (3,4) considered the number of "structurally different" trees with $m$ leaves (according to MathSciNet). Bergeron, Labelle and Leroux (1) consider the expected number of leaves in trees that admit a certain automorphism. Lam (2) discusses embeddings of trees with $m$ leaves and discusses trees with $(d+1)d^{r+1}$ leaves for integers d and r.
Wilf (7. p. 163) gave a generating function for $\sum\_k T\_{k,n}^{\text{lab}}$ where $T\_{k,n}^{\text{lab}}$ is the number of labelled free trees with $m$ leaves and $n$ vertices. He also gives a formula for the average number of leaves in a labelled tree with $n$ vertices.
There is also this: K. Yamanaka, Y. Otachi, S.-I. Nakano [Efficient Enumeration of Ordered Trees with k Leaves](https://doi.org/10.1007/978-3-642-00202-1_13), which I haven't looked at yet.
(1) F. Bergeron, G. Labelle, and P. Leroux, [Computation of the expected number of leaves in a tree having a given automorphism, and related topics](https://doi.org/10.1016/0166-218X(91)90078-B), Discrete Appl. Math., 34 (1991), pp. 49-66.
(2) P. C. B. Lam, [On number of leaves and bandwidth of trees](https://doi.org/10.1007/BF02677426), Acta Math. Appl. Sinica (English Ser.), 14 (1998), pp. 193-196.
(3) B. L. Liu, The enumeration of directed trees with a given number of leaves and the enumeration of free trees, Kexue Tongbao, 32 (1987), pp. 244-247. In Chinese.
(4) B. L. Liu, Enumeration of oriented trees and free trees with a given number of leaves, Kexue Tongbao (English Ed.), 33 (1988), pp. 1577-1581.
(5) Q. Q. Nong, The degree sequence and number of leaves in a tree, J. Yunnan Univ. Nat. Sci., 24 (2002), pp. 167-171. In Chinese.
(6) Z. Y. Wang, An enumeration problem on ordered trees, J. Math. (Wuhan), 6 (1986), pp. 201-208.
(7) H. C. Wilf, Generatingfunctionology, Academic Press, 1990.
(8) Q. L. Yu, An algorithm for lexicographically generating ordered rooted trees with constraints on the number of leaves, Chinese J. Oper. Res., 6 (1987), pp. 71-72
| 11 | https://mathoverflow.net/users/2264 | 22035 | 14,548 |
https://mathoverflow.net/questions/22028 | 4 | Given two $n-$dimensional convex compact sets $A,B$, we define $d(A,B)$ as $\log({\mathrm{Vol}}(\alpha\_2(A)))-\log(\mathrm{Vol}(\alpha\_1(A)))$ where $\alpha\_1,\alpha\_2$ are two affine bijections such that $\alpha\_1(A)\subset B\subset\alpha\_2(A)$ and such that the ratio $\mathrm{Vol}(\alpha\_2(A))/\mathrm{Vol}(\alpha\_1(A))$ is minimal. The function
$d$ is symmetric, satisfies the triangle inequality, is well-defined for orbits of
convex sets under affine bijections and $d(A,B)=0$ if and only if $A$ and $B$ are in the same orbit under affine bijections.
The function $d$ defines thus a distance on the set $\mathcal C\_n$ of orbits under affine bijections of $n-$dimensional convex compact sets.
What is the diameter of the metric set $\mathcal C\_n$?
(It is easy to see that $\mathcal C\_n$ is of bounded diameter.) A natural guess is that
the diameter is achieved by the distance of (the orbit of) an $n-$dimensional ball to
(the orbit of) the $n-$dimensional simplex.
| https://mathoverflow.net/users/4556 | Diameter of a metric on orbits under affine bijections of $n-$dimensional convex compact sets | I assume you also want your compact sets to have non-empty interior, hence positive volume.
The literature mostly deals with the related [Banach-Mazur](http://en.wikipedia.org/wiki/Banach-Mazur_compactum) metric $d\_{BM}(A,B)$, in which it is assumed that $\alpha\_1(A)$ and $\alpha\_2(A)$ are homothetic, so $d\_{BM}(A,B) \ge d(A,B)$. (Here I'm following your convention and making $d\_{BM}$ a metric, as opposed to the usual definition which makes its logarithm a metric.) Here's a little of what's known about that related to your question:
If $B$ is a Euclidean ball, then $d\_{BM}(A,B) \le \log n$, with equality achieved exactly when $A$ is a simplex. Thus the diameter of $(\mathcal{C}\_n, d\_{BM})$ is at most $2\log n$. I believe the exact diameter is an open question.
Let $\mathcal{C}\_n^0$ be the set of affine equivalence classes of centrally symmetric convex bodies. Then if $B$ is a Euclidean ball, $d\_{BM}(A,B) \le \log \sqrt{n}$, with equality achieved when $A$ is a cube or a crosspolytope (but not only then); therefore the diameter of $\mathcal{C}\_n^0,d\_{BM})$ is at most $2\log\sqrt{n} = \log n$. Gluskin proved that the diameter of $(\mathcal{C}\_n^0,d\_{BM})$ is at least $\log n - c$ for a constant $c$ independent of $n$, by in fact proving the same lower bound for the diameter of $(\mathcal{C}\_n^0,d)$.
| 3 | https://mathoverflow.net/users/1044 | 22042 | 14,551 |
https://mathoverflow.net/questions/22039 | 5 | Let $S$ be a K3 surface over $\mathbb{C}$, that is, $S$ is a simply connected compact smooth complex surface whose canonical bundle is trivial. I recall reading somewhere that any two such surfaces are diffeomorphic, however I can't for the life of me remember where, or how the proof goes.
Does anybody know a good reference to a proof, or can provide a proof?
Thanks,
Dan
| https://mathoverflow.net/users/5101 | Are any two K3 surfaces over C diffeomorphic? | I think this was first proved by Kodaira. See [On the structure of complex analytic surfaces, 1](http://www.jstor.org/pss/2373157). There Kodaira proves that any K3 surface is a deformation of a non-singular quartic surface in $\mathbb{CP}^3$. In particular, they are all diffeomorphic.
| 15 | https://mathoverflow.net/users/380 | 22043 | 14,552 |
https://mathoverflow.net/questions/22036 | 4 | Do pullbacks exist in the category of sets and partial functions?
Are the 'the same' as they are in Sets? That is, given two partial functions $f : A \to C$ and $g : B \to C$, is the pullback given by $\{ (a,b) \in A\times B ~|~ f(a)=g(b) \}$?
If not, what is a simple description of the pullback?
| https://mathoverflow.net/users/2620 | Pullbacks in Category of Sets and Partial Functions | Pullbacks exists but are not what you describe.
The answer is as follows:
The category $\mathcal{C}$ of sets and partial functions is equivalent to the category of based sets and based functions, by sending the set $A$ to $A$ disjoint union a base point $\*$ and sending $f$ to the obvious based function which sends everything on which $f$ was not defined to the base-point.
The pullback in based sets are well-known and for example the product $\times$ in based sets translates back through this equivalence to $\mathcal{C}$ and becomes:
$A \times\_\mathcal{C} B \approx (A \times\_{\textrm{set}} B) \sqcup\_{\textrm{set}} A \sqcup\_{\textrm{set}} B$
From the purely sets and partial function point of view this is also explainable. Indeed, any morphism from $Z$ to this product is given by a choice for each point in $Z$ of either: a point in $A$ and a point in $B$, or a point in $A$, or a point $B$, or nothing.
| 7 | https://mathoverflow.net/users/4500 | 22046 | 14,555 |
https://mathoverflow.net/questions/22040 | 16 | Let $X$ be a connected affine variety over an algebraically closed field $k$, and let $X \subset Y$ be a compactification, by which I mean $Y$ is a proper variety (or projective if you prefer), and $X$ is embedded as an open dense subset.
I am guessing that it is not always the case that $Y\setminus X$ is a divisor, one could imagine it being a single point with a horrible singularity. But if $Y$ is smooth or even normal, is it the case that $Y\setminus X$ is always a divisor? Does anybody know a proof of such a result?
Thanks,
Dan
| https://mathoverflow.net/users/5101 | Is the complement of an affine variety always a divisor? | It it true for any $Y$: see Corollaire 21.12.7 of EGAIV.
| 18 | https://mathoverflow.net/users/4790 | 22052 | 14,560 |
https://mathoverflow.net/questions/21703 | 10 | Greg Muller, in a post called [Rational Homotopy Theory](http://cornellmath.wordpress.com/2008/04/27/rational-homotopy-theory/) on the blog "The Everything Seminar" wrote
"I tend to think of homotopy theory a little bit like ‘The One That Got Away’ from mathematics as a whole. Its full of wistful fantasies about how awesome it would have been if things could only have worked out. Imagine if homotopy groups of spaces and homotopy classes of maps were as easy to compute as homology groups…"
What are the wonderful consequences that he is referring to?
| https://mathoverflow.net/users/4692 | What would be the ramifications of homotopy theory being as easy as homology theory? | Homology groups and homotopy groups are two sides of the same story. Homotopy groups tell us all the ways we can have a map Sn → X, and in particular describe all ways we can attach a new cell to our space. On the other side, the homology groups of a space change in a very understandable way each time we attach a new cell, and so they tell us all the ways that we could build a homotopy-equivalent CW-complex. In cases where we can understand both of them, we can get things like complete theorems about classification of spaces.
Here's an example where we *can* compute: classification of the homotopy types of compact, orientable, simply-connected 4-manifolds. (I originally saw this is Neil Strickland's [bestiary of topological spaces](http://neil-strickland.staff.shef.ac.uk/courses/bestiary/bestiary.pdf).)
Poincare duality tells us that the homology groups are finitely generated free in degree 2, ℤ in degrees 0 and 4, and zero elsewhere. We can cut out a closed ball, and get an expression of the manifold as obtained from a manifold-with-boundary N by attaching a 4-cell. The Hurewicz theorem tells us that we can construct a map from a wedge of copies of S2 to N which induces an isomorphism on H`*`, and by the (homology) Whitehead theorem this is a homotopy equivalence. So our original manifold is obtained, up to homotopy equivalence, by attaching a 4-cell to $\bigvee S^2$.
How many ways are there to do this? It is governed by $\pi\_3 (\bigvee S^2)$, which we can compute because it's low down enough. This homotopy group is naturally identified with the set of symmetric bilinear pairings $H^2(\bigvee S^2) \to \mathbb{Z}$, and this identification is given by seeing how the cup product acts after you attach a cell. So these 4-manifolds are classified up to homotopy equivalence by the nondegenerate symmetric bilinear pairing in their middle-dimensional cohomology.
Some of what we used here is general and well-understood machinery about homology, homotopy, and their relationships. Wouldn't it be nice if the standard tools were always so effective? But the real meat is that we have a complete understanding of homology and homotopy in the relevant ranges. It turns our questions about classification into questions about pure algebra. For questions that require specific knowledge about higher homotopy groups of spheres (or even lower homotopy groups of complicated spaces), it is much harder to get answers. There aren't a lot of spaces where we have complete understanding of both the homology groups and the homotopy. We have tools for reconstructing the former from the latter but their effectiveness wears down the farther out you try to go.
There *are* categories that are somewhat like the homotopy category of spaces where we can get an immediate and specific understanding of both sides of the coin.
One such example is the category of chain complexes over a ring R. There, our fundamental building block is R itself. The homology of any chain complex tells us both how R can be mapped in modulo chain homotopy, and how complicated any construction of the underlying chain complex must be. A more complicated example would be the category of differential graded modules over a DGA, where the divide between how things can be constructed and how things can have new cells attached is, at the very least, governed by the complexity of H`*` A as a ring, and then by the secondary algebraic operations if A is far from being anything like formal.
Another such example is the rational homotopy theory of simply-connected spaces you mentioned. There, homology and homotopy are roughly something like the difference between a ring's underlying abelian group structure and how you build it using generators and relations.
So you might think of the complexity of homotopy groups as telling us how much more complicated spaces are than chain complexes.
| 12 | https://mathoverflow.net/users/360 | 22057 | 14,564 |
https://mathoverflow.net/questions/22050 | 9 | Is there a common name for the complement $\widehat{X} \setminus X$ of a metric space $X$ in its metric completion $\widehat{X}$? Since $X$ is not necessarily open in $\widehat{X}$, the term *boundary* is out of the question (without additional qualifiers). *Metric remainder* seems appropriate but I did not find it in the literature.
| https://mathoverflow.net/users/2000 | Is there a common name for the complement of a metric space in its completion? | Remainder. I agree with that. But I don't find it on-line. Maybe "remainder" is primarily used for $\beta X \setminus X$ ? But it should be OK in your setting if you say the first time you use it: "the remainder of $X$ in its completion" or something.
| 4 | https://mathoverflow.net/users/454 | 22059 | 14,566 |
https://mathoverflow.net/questions/22080 | 17 | It is well-known that if $X$ is an integral scheme, then there is an isomorphism $CaCl(X)\to Pic(X)$ taking $[D]$ to $[\mathcal{O}\_X(D)]$. Does anyone know any simple examples where the above map fails to be surjective, i.e., a line bundle on a scheme $X$, not isomorphic to $\mathcal{O}\_X(D)$ for any Cartier divisor D?
| https://mathoverflow.net/users/3996 | Line bundles vs. Cartier divisors on a non-integral scheme | An example is given in this [note](http://webs.uvigo.es/martapr/Investigacion/Divisors.pdf) (it was credited to Kleiman).
| 12 | https://mathoverflow.net/users/2083 | 22085 | 14,581 |
https://mathoverflow.net/questions/22062 | 12 | A *polyhedron* is the intersection of a finite collection of halfspaces. These halfspaces are not assumed to be *linear*, i.e. their bounding hyperplanes are not assumed to contain the origin. The *support* Supp(M) of a collection M of polyhedra is the union of the polyhedra in M.
I can prove the following theorem:
**Theorem.** Let M be a finite set of n-dimensional polyhedra in Rn. Suppose:
(i) The interior of Supp(M) is path-connected; and
(ii) For every x in the boundary of Supp(M), there exists a closed
halfspace H+ bounded by a hyperplane H such that x is in H, and such that H+ contains every P in M such that x is in P.
Then Supp(M) is convex.
(Acknowledgment: I proved a characterization of coarsenings of a given polyhedral complex and Ezra Miller remarked that part of my argument amounted to some sort of local criterion for convexity. The theorem above is that criterion.)
The point here is that you only need to check, at each point x of the boundary, that Supp(M) looks sufficiently like a convex set near x, and (ii) says exactly what "sufficiently like a convex set" means in this case.
The question is:
>
> Is this a special case of some general theorem that says that convexity is
> somehow a local condition?
>
>
>
I suspect that I'm asking for a reference to something known. One convexity person that I asked about felt that it is "highly likely..., that this result is a special case
of a result in functional analysis, once properly understood." The same person suggested that there might be a connection to the theory of tight manifolds in topology. For that reason I have added the tags fa.functional-analysis and gt.geometric-topology. My apologies if these tags turn out not to be appropriate.
| https://mathoverflow.net/users/5519 | To what extent is convexity a local property? | This is known as Tietze theorem: if $A$ is an open connected set such that for every boundary point there is a locally supporting hyperplane, then $A$ is convex. I don't know what is the standard reference, internet search gave me the following one:
F.A.Valentine. Convex sets. McGraw-Hill, New York, 1964, pp. 51-53.
There are easier proofs than fedja's (at least to geometer's eye). My favorite one is the following. Take $a,b\in A$, there is a polygonal path from $a$ to $b$ in $A$. Suppose this path has $N$ edges. Then there is a shortest polygonal path of at most $N$ edges in the closure of $A$. If it is not a straight segment, the first edge must touch the boundary of $A$ (otherwise one can shorten the path). The first point where it meets the boundary has obvious problems with local supporting hyperplane.
**Update.** As Zsbán Ambrus pointed out in a comment, vertices on the boundary cause problems, so one should restrict to polygonal paths contained in the interior. But then it is not clear why a minimum exists. Rather that considering a minimum, one can do shorthening by hand: begin with any polygonal path in the interior, choose consecutive serments $[a,b]$ and $[b,c]$ and move $b=b(t)$ to $a$ along the segment. If a segment $[a,b(t)]$ touches the boundary at some moment $t$, observe a contradiction. If not, then the final segment $[a,c]$ is also contained in the interior, so we get a polygonal line with fewer edges. Repeat this procedure until it becomes a single segment.
| 19 | https://mathoverflow.net/users/4354 | 22091 | 14,583 |
https://mathoverflow.net/questions/22087 | 9 | In two different books I found these two related statements.
* The book by Jost defines a ``locally symmetric space" as one for which the curvature tensor is constant and which is geodesically complete.
* Andreas' book attributes to Cartan a theorem that a space is locally symmetric if and only if its curvature tensor is constant. Where he defines a space to be locally symmetric if about every point there is a geodesic reflecting isometry.
I could not trace this theorem of Cartan anywhere else nor does Andreas' book give a reference. I would like to know some reference which would prove this stuff and resolve the apparent definition conflict between the above two statements.
Further I came across these 3 statements about classifying constant curvature spaces,
* If the isometry group is transitive on points on the manifold, then the scalar
curvature is constant.
* If the isometry group is transitive on all one-dimensional subspaces
of tangent spaces, then the Ricci curvature tensor is a scalar multiple of the metric.
tensor
* If the isometry group is transitive on all two-dimensional subspaces
of tangent spaces, then the sectional curvature is constant on all
two-dimensional subspaces of tangent spaces.
I have an understanding of why the first statement is true but I would like to know some reference which explains the other two statements.
Further call a space to be homogeneous if it is quotient of some Lie Group mod a closed subgroup. Then in physics literature one finds a statement of the kind that an "isotropic and homogeneous space-time is of constant curvature". I would like to know what is the precise mathematical meaning of this statement.
Further I would also like to know a reference for the fact that a maximally symmetric space always has constant curvature and how being maximally symmetric fits in with being "isotropic and homogeneous".
Many sources refer me to the volumes by Kobayashi and Nomizu or the book by Besse or Spivak for discussion on constant curvature spaces but unfortunately I don't have access to these books. It would be great if some online reference like lecture notes/expository review paper on constant curvature spaces could be linked which clarifies the above questions.
In the books on Riemannian Geometry that I have access to like the ones by Gallot et al,Andreas or Jost, I can't see much of any discussion on these topics.
{In all the above statements I suppose the Riemann Christoffel connection is being assumed. It would be interesting to know how much of the things hold for a general affine connection.}
| https://mathoverflow.net/users/2678 | Constant curvature manifolds | It is difficult to reconcile your first two statements, because they are actually wrong as written!
A riemannian manifold is *locally symmetric* if and only if the Riemann curvature tensor is parallel with respect to the Levi-Civita connection. This condition was studied by Élie Cartan, who classified them using his classification of real semisimple Lie algebras. My favourite reference for this is Besse's *Einstein manifolds*, Chapter 7F, from where I quote the following theorem:
>
> **10.72 Theorem** (Elie Cartan). For a Riemannian manifold (not necessarily complete) the following conditions are equivalent:
>
>
> (i) $DR = 0$;
>
>
> (ii) the geodesic symmetry $s\_p$ around any point $p$ (which is defined only locally) is an isometry.
>
>
> For a complete Riemannian manifold $(M,g)$ the two following conditions are equivalent:
>
>
> (iii) for every point $p$ in $M$ the geodesic symmetry around $p$ is well defined and is an isometry;
>
>
> (iv) the manifold $M$ is a homogeneous space $G/H$ where $G$ is a connected Lie group, $H$ a compact subgroup of $G$, and where there exists an involutive automorphism $\sigma$ of the group $G$ for which, if $S$ denotes the fixed point set of $\sigma$ and $S\_e$ its connected component of the identity one has $S\_e \subset H \subset S$. Moreover the Riemannian metric under consideration on $G/H$ is invariant under $G$.
>
>
> Furthermore: if $(M, g)$ satisfies (iii) or (iv), then it satisfies (i) and (ii). If $(M, g)$ satisfies (i) or (ii) and if it is simply connected and complete then it satisfies (iii) and (iv).
>
>
> **10.76 Definition**. A Riemannian manifold is said to be *locally symmetric* if it satisfies (i) or (ii) above; it is said to be *symmetric* if it satisfies (iii) or (iv).
>
>
>
I suspect the source of the confusion might be with the meaning of "constant curvature", which usually means constant sectional curvature (or perhaps parallel Riemann curvature) and not constant scalar curvature, which is a much weaker condition.
Concerning your final parenthetic question, there is a theorem of Ambrose and Singer, reformulated by Kostant, which says that a riemannian manifold $(M,g)$ is *locally homogeneous* if and only if it admits a metric connection with parallel torsion and parallel curvature. It is locally symmetric if (and only if) the connection is torsion-free.
| 11 | https://mathoverflow.net/users/394 | 22092 | 14,584 |
https://mathoverflow.net/questions/22032 | 62 | I read the article in wikipedia, but I didn't find it totally illuminating. As far as I've understood, essentially you have a morphism (in some probably geometrical category) $Y \rightarrow X$, where you interpret $Y$ as being the "disjoint union" of some "covering" (possibly in the Grothendieck topology sense) of $X$, and you want some object $\mathcal{F'}$ defined on $Y$ to descend to an object $\mathcal{F}$ defined on $X$ that will be isomorphic to $\mathcal{F}'$ when pullbacked to $Y$ (i.e. "restricted" to the patches of the covering). To do this you have problems with $Y\times\_{X}Y$, which is interpreted as the "disjoint union" of all the double intersections of elements of the cover.
I'm aware of the existence of books and notes on -say- Grothendieck topologies and related topics (that I will consult if I'll need a detailed exposition), but I would like to get some ideas in a nutshell, with some simple and maybe illuminating examples from different fields of mathematics.
I also know that there are other MO questions related to descent theory, but I think it's good that there's a (community wiki) place in which to gather instances, examples and general picture.
So,
* What is descent theory in general? And what are it's unifying abstract patterns?
* In which fields of mathematics does it appear or is relevant, and how does it look like in each of those fields? (I'm mostly interested in instances within algebraic geometry, but having some picture in other field would be nice).
* Could you make some examples of theorems which are "typical" of descent theory? And also mention the most important and well known theorems?
| https://mathoverflow.net/users/4721 | What is descent theory? | Suppose we are given some category (or higher category) of "spaces" in which
each space $X$ is equipped with a *fiber*, i.e. a category $C\_X$ of objects
of some type over it.
For example, a space can be a smooth manifold and the fiber
is the category of vector bundles over it;
or a space is an object of the category dual to the category of rings
and the fiber is its category of left modules.
Given a map $f: Y\to X$, one often has an induced
functor $f^\* : C\_X\to C\_Y$ (pullback, inverse image functor,
extension of scalars).
The basic questions of classical descent theory
are:
* When an object $G$ in $C\_Y$ is in the image via $f^\*$ of some object in $C\_X$ ?
* Classify all *forms* of object
$G\in C\_Y$, that is find all $E\in C\_X$ for which $f^\*(E)\cong G$.
Grothendieck introduced pseudofunctors and
[fibred categories](http://ncatlab.org/nlab/show/fibered+category)
to formalize an ingenious method to deal with descent questions.
He introduces additional data on an object $G$ in $C\_Y$
to have a chance of determining an isomorphism class of an object in
$C\_X$. Such an enriched object over $X$ is called a ``descent datum''.
$f$ is an *effective descent morphism* if the morphism $f$
induces a canonical equivalence of
the category of the descent data (for $f$ over $X$) with $C\_X$.
It is a nontrivial result that in the case of rings and modules,
the effective descent morphisms are preciselly
*pure morphisms* of rings. Grothendieck's
flat descent theory tells a weaker result
that faithfully flat morphisms are of effective descent.
In algebraic situations one often introduces a (co)[monad](http://ncatlab.org/nlab/show/monad) $T\_f : C\_X\to C\_X$
(say with the multiplication $\mu: T\_f \circ T\_f \to T\_f$)
induced by the morphism $f$.
The category of descent data is then nothing else than the
Eilenberg-Moore category $T\_f-\mathrm{Mod}$ of (co)modules (also called (co)algebras) over $T\_f$.
Then, by the definition, $f$ is of an effective descent if and only if
the *comparison map* (defined in the (co)monad theory)
between $C\_X$ and $T\_f-\mathrm{Mod}$ is an equivalence.
Several variants of [Barr-Beck theorem](http://ncatlab.org/nlab/show/monadicity+theorem) give conditions
which are equivalent or (in some variants) sufficient to
the comparison map for a monad induced by a pair of adjoint functors
being an equivalence. Generically
such theorems are called monadicity (or tripleability) theorems.
One can describe most of (but not all)
situations of 1-categorical descent theory via the monadic approach.
There are numerous generalizations of monadicity theorems, higher cocycles and descent, both in monadic and in fibered category setup
in higher categorical context (Giraud, Breen, Street, K. Brown, Hermida, Marmolejo, Mauri-Tierney, Jardine, Joyal, Simpson, Rosenberg-Kontsevich, Lurie...); the theory of stacks, gerbes and of general [cohomology](http://ncatlab.org/nlab/show/cohomology) is almost the same as the general descent theory, in a point of view.
For examples, it is better to consult the literature. It takes a while to treat them.
| 34 | https://mathoverflow.net/users/35833 | 22098 | 14,590 |
https://mathoverflow.net/questions/22015 | 19 |
>
> **Definition.** A locally finitely presented morphism of schemes $f\colon X\to Y$ is *smooth* (resp. *unramified*, resp. *étale*) if for any **affine** scheme $T$, any closed subscheme $T\_0$ defined by a square zero ideal $I$, and any morphisms $T\_0\to X$ and $T\to Y$ making the following diagram commute
>
>
>
> ```
>
> g
> T0 --> X
> | |
> | |f
> v v
> T ---> Y
>
> ```
>
> there exists (resp. exists at most one, resp. exists exactly one) morphism $T\to X$ which fills the diagram in so that it still commutes.
>
>
>
For checking that $f$ is unramified or étale, it doesn't matter that I required $T$ to be affine. The reason is that for an arbitrary $T$, I can cover $T$ by affines, check if there exists (a unique) morphism on each affine, and then "glue the result". If there's at most one morphism locally, then there's at most one globally. If there's a unique morphism locally, then there's a unique morphism globally (uniqueness allows you to glue on overlaps).
But for checking that $f$ is smooth, it's really important to require $T$ to be affine in the definition, because it could be that there exist morphisms $T\to X$ locally on $T$, but it's impossible to find these local morphisms in such a way that they glue to give a global morphism.
>
> **Question:** What is an example of a smooth morphism $f\colon X\to Y$, a square zero nilpotent thickening $T\_0\subseteq T$ and a commutative square as above so that there **does not** exist a morphism $T\to X$ filling in the diagram?
>
>
>
I'm sure I worked out such an example with somebody years ago, but I can't seem to reproduce it now (and maybe it was wrong). One thing that may be worth noting is that the set of such filling morphisms $T\to X$, if it is non-empty, is a torsor under $Hom\_{\mathcal O\_{T\_0}}(g^\*\Omega\_{X/Y},I)=\Gamma(T\_0,g^\*\mathcal T\_{X/Y}\otimes I)$, where $\mathcal T\_{X/Y}$ is the relative tangent bundle. So the obstruction to finding such a lift will represent an element of $H^1(T\_0,g^\*\mathcal T\_{X/Y}\otimes I)$ (you can see this with Čech cocycles if you want). So in any example, this group will have to be non-zero.
| https://mathoverflow.net/users/1 | Example of a smooth morphism where you can't lift a map from a nilpotent thickening? | Using some of BCnrd's ideas together with a different construction, I'll give a positive answer to Kevin Buzzard's stronger question; i.e., there is a counterexample for *any* non-etale smooth morphism.
Call a morphism $X \to Y$ *wicked smooth* if it is locally of finite presentation and for every (square-zero) nilpotent thickening $T\_0 \subseteq T$ of $Y$-schemes, every $Y$-morphism $T\_0 \to X$ lifts to a $Y$-morphism $T \to X$.
>
> **Theorem:** A morphism is wicked smooth if and only if it is etale.
>
>
>
**Proof:**
Anton already explained why etale implies wicked smooth.
Now suppose that $X \to Y$ is wicked smooth. In particular, $X \to Y$ is smooth, so it remains to show that the geometric fibers are $0$-dimensional. Wicked smooth morphisms are preserved by base change, so by base extending by each $y \colon \operatorname{Spec} k \to Y$ with $k$ an algebraically closed field, we reduce to the case $Y=\operatorname{Spec} k$. Moreover, we may replace $X$ by an open subscheme to assume that $X$ is etale over $\mathbb{A}^n\_k$ for some $n \ge 0$.
Fix a projective variety $P$ and a surjection $\mathcal{F} \to \mathcal{G}$ of coherent sheaves on $P$ such that some $g \in \Gamma(P,\mathcal{G})$ is not in the image of $\Gamma(P,\mathcal{F})$. (For instance, take $P = \mathbb{P}^1$, let $\mathcal{F} = \mathcal{O}\_P$, and let $\mathcal{G}$ be the quotient corresponding to a subscheme consisting of two $k$-points.) Make $\mathcal{O}\_P \oplus \mathcal{F}$ an $\mathcal{O}\_P$-algebra by declaring that $\mathcal{F} \cdot \mathcal{F} = 0$, and let $T = \operatorname{\bf Spec}(\mathcal{O}\_P \oplus \mathcal{F})$. Similarly, define $T\_0 = \operatorname{\bf Spec}(\mathcal{O}\_P \oplus \mathcal{G})$, which is a closed subscheme of $T$ defined by a nilpotent ideal sheaf. We then may view $g = 0+g \in \Gamma(P,\mathcal{O}\_P \oplus \mathcal{G}) = \Gamma(T\_0,\mathcal{O}\_{T\_0})$.
Choose $x \in X(k)$; without loss of generality its image in $\mathbb{A}^n(k)$ is the origin. Using the infinitesimal lifting property for the etale morphism $X \to \mathbb{A}^n$ and the nilpotent thickening $P \subseteq T\_0$, we lift the point $(g,g,\ldots,g) \in \mathcal{A}^n(T\_0)$ mapping to $(0,0,\ldots,0) \in \mathbb{A}^n(P)$ to some $x\_0 \in X(T\_0)$ mapping to $x \in X(k) \subseteq X(P)$. By wicked smoothness, $x\_0$ lifts to some $x\_T \in X(T)$. The image of $x\_T$ in $\mathbb{A}^n(T)$ lifts $(g,g,\ldots,g)$, so each coordinate of $x\_T$ is a global section of $\mathcal{F}$ mapping to $g$, which is a contradiction unless $n=0$. Thus $X \to Y$ is etale.
| 24 | https://mathoverflow.net/users/2757 | 22101 | 14,593 |
https://mathoverflow.net/questions/22089 | 11 | Fix numbers n,k. Is there a closed formula known for the number of k-regular graphs consisting of n edges? I have a method of enumerating k-regular graphs on n edges, and would like to have a number to compare the algorithm against.
| https://mathoverflow.net/users/5002 | Enumeration of Regular Graphs | I think the answer is no, but I would consult the following link:
<http://www.mathe2.uni-bayreuth.de/markus/reggraphs.html>
which contains tables of the sums of the numbers you are interested in. The author is very generous with sharing data that is not posted online.
| 1 | https://mathoverflow.net/users/4542 | 22109 | 14,597 |
https://mathoverflow.net/questions/22105 | 7 | Reading Perelman's preprint(1991) Alexandrov space II now. Got confused about the last section 6.4, which contains an example which indicate that the statement ".... manifold is diffeomorphic to the normal bundle over soul" in Cheeger-Gromoll's Soul Theory won't hold for Alexandrov spaces. I also read BBI's book(A course in Metric Geometry) (page 400-401) which also contains the description of this example. I am using the notation in BBI's book, the example goes as follows:
Let $\pi: K\_0(\mathbb{CP}^2)\to K\_0(\mathbb{CP}^1)$ be the projection. $\bar{B}\_0(1)$ be the unit ball in $\mathbb{CP}^1$ (Note: here should be $K\_0(\mathbb{CP}^1$), right?). Let $X^5=\pi^{-1}(\bar{B}\_0(1))$. Take double of $X^5$ and it will be the example
The picture in my mind is $K\_0(\mathbb{CP}^1)$ is sub-cone of $K\_0(\mathbb{CP}^2)$, so the projection is the projection on the second factor if we write the coordinate in cone as $(t, x)$ for $t\in \mathbb R$ and $x\in \mathbb{CP}^2$.
My question is what is the topology of $X^5$?
1) Is $\bar{B}\_0(1)$ a close ball? If so then $\bar{B}\_0(1)$ will have boundary $\mathbb{CP}^1$, right? and $X^5$ will be a closed cone over $\mathbb{CP}^2$, right?
2) Is $X^5$ compact?
| https://mathoverflow.net/users/3922 | Details of Perelman's example about soul of Alexandrov space | No $X^5$ is not a cone over $CP^2$ and is not compact. The projection has nothing to do with the cone structure. In fact, it's better to forget about the cone structure altogether (until you ask what is the topology of the thing).
The spaces are just quotients of $\mathbb C^3$ and $\mathbb C^2$ by the standard circle action, and the projection is induced by the coordinate projection $\mathbb C^3\to \mathbb C^2$. For example, there is a whole half-line in the pre-image of the origin.
I suggest you visualize a similar construction with $\mathbb R$ in place of $\mathbb C$ or lower the dimensions by 1 (or both) to see what is going on.
| 9 | https://mathoverflow.net/users/4354 | 22112 | 14,599 |
https://mathoverflow.net/questions/22111 | 24 | Let $U$ be a dense open subscheme of an integral noetherian scheme $X$ and let $E$ be a vector bundle on $U$. Suppose that the complement $Y$ of $U$ has codimension $\textrm{codim}(Y,X) \geq 2$. Let $F$ be a vector bundle on $X$ extending $E$, i.e., $F|\_{U} = E$.
Is any extension of $E$ to $X$ isomorphic to $F$?
| https://mathoverflow.net/users/4333 | Extending vector bundles on a given open subscheme | This is true if $X$ satisfies Serre's condition $S\_2$, i.e. $\mathcal O\_X$ is $S\_2$. Then a vector bundle is $S\_2$ since locally it is isomorphic to $\mathcal O\_X^n$.
More generally, a coherent sheaf $F$ on a Japanese scheme (for example: $X$ is of finite type over a field) which is $S\_2$ has a unique extension from an open subset $U$ with $\operatorname{codim} (X\setminus U)\ge 2$. This follows at once from the cohomological characterization of $S\_2$.
Thus, another name for the $S\_2$-sheaves: they are sheaves which are *saturated in codimension 2*, and another name for the $S\_2$-fication: *saturation in codimension 2*.
P.S. Of course, by Serre's criterion, normal = $S\_2+R\_1$. So the above statement is true for any normal (e.g. smooth) variety.
P.P.S. And of course, Gorenstein implies Cohen-Macaulay implies $S\_2$. So the statement is also true for hypersurfaces and complete intersections, which could be very singular and non-reduced.
---
Edit to define some terms:
1. A Japanese (or Nagata) ring is a ring obtained from a ring finitely generated over a field or $\mathbb Z$ by optionally applying localizations and completions. The property used here is that for a Japanese ring $R$, its integral closure (normalization) $\tilde R$ is a finitely generated $R$-module. This is important because the $S\_2$-fication $S\_2(R)$ lies between $R$ and $\tilde R$.
2. A coherent sheaf $F$ satisfies $S\_n$ if for any point $x\in Supp(F)$, one has
$$ depth\_x (F) \ge \min(\dim\_x Supp(F),n) $$
If $F$ locally corresponds to an $R$-module $M$, and $x$ to a prime ideal $p$, then the depth is the length of a maximal regular sequence $(f\_1,\dots, f\_k)$ of elements of $R\_p$ for $M\_p$ (so, $f\_1$ is a nonzerodivisor in $M\_p$, etc.).
| 37 | https://mathoverflow.net/users/1784 | 22124 | 14,606 |
https://mathoverflow.net/questions/22120 | 11 | More specifically, I was wondering if there are well-known conditions to put on $X$ in order to make $K\_0(X)\simeq K^0(X)$. Wikipedia says they are the same if $X$ is smooth. It seems to me that you get a nice map from the coherent sheaves side to the vector bundle side (the hard direction in my opinion) if you impose some condition like "projective over a Noetherian ring". Is this enough? In other words, is the idea to impose enough conditions to be able to resolve a coherent sheaf, $M$, by two locally free ones $0\to \mathcal{F}\to\mathcal{G}\to M\to 0$?
| https://mathoverflow.net/users/14672 | What is the difference between Grothendieck groups K_0(X) vs K^0(X) on schemes? | Imposing that you can resolve by a length $2$ sequence of vector bundles is too strong. What you want is that there is some $N$ so that you can resolve by a length $N$ sequence of vector bundles. By Hilbert's syzygy theorem, this follows from requiring that the scheme be regular. (Specifically, if the scheme is regular of dimension $d$, then every coherent sheaf has a resolution by projectives of length $d+1$.)
Here is a simple example of what goes wrong on singular schemes. Let $X = \mathrm{Spec} \ A$ where $A$ is the ring $k[x,y,z]/(xz-y^2)$. Let $k$ be the $A$-module on which $x$, $y$ and $z$ act by $0$. I claim that $k$ has no finite free resolution. I will actually only show that $A$ has no *graded* finite free resolution. Proof: The hilbert series of $A$ is $(1-t^2)/(1-t)^3 = (1+t)/(1-t)^2$. So every graded free $A$-module has hilbert series of the form $p(t) (1+t)/(1-t)^2$ for some polynomial $p$; and the hilbert series of anything which has a finite resolution by such modules also has hilbert series of the form $p(t) (1+t)/(1-t)^2$. In particular, it must vanish at $t=-1$. But $k$ has hilbert series $1$, which does not.
There is, of course, a resolution of $k$ which is not finite. If I am not mistaken, it looks like
$$\cdots \to a[-4]^4 \to A[-3]^4 \to A[-2]^4 \to A[-1]^3 \to A \to k$$
| 13 | https://mathoverflow.net/users/297 | 22137 | 14,613 |
https://mathoverflow.net/questions/22122 | 9 | Consider a complete $C^\infty$ Riemannian metric on $\mathbb R^2$ of positive sectional curvature.
1. Is the metric embeddable as the boundary of a convex subset of $\mathbb R^3$?
2. Is the embedding unique?
3. Are there generalizations of 1-2 to complete noncompact surfaces of nonnegative sectional curvature?
4. What are good references for these matters?
UPDATE:
$\bullet\ $ after doing some reading on the subject I found that the assertion 1 is true in the sense that the surface is isometric, **as a metric space**, to the boundary of a convex body in $\mathbb R^3$ (as proved by Alexandrov back in 1942). The matter of uniqueness is well-understood.
$\bullet\ $ However, one should not expect the boundary to be smooth, e.g. there are examples of $C^\infty$ metrics of nonnegative curvature on $S^2$ which cannot be isometrically $C^3$-embedded into $\mathbb R^3$.
$\bullet\ $ If the curvature is **positive**, then smoothness can be achieved as proved by Pogorelov and Nirenberg (independently in the 1950s).
$\bullet\ $ **Local** smooth isometric embedding for nonnegatively curved surfaces was established by Lin in 1985.
$\bullet\ $ A more recent reference for these matters is the book by Burago and Zalgaller, Geometry III, Encyclopedia of Mathematical Sciences.
| https://mathoverflow.net/users/1573 | On Alexandrov embedding theorem | *Is the metric embeddable as the boundary of a convex subset of 3?*
YES, it is a limit case of standard Alexandrov's theorem. Moreover one can choose any embedding of cone at infinity and construct the embedding. This is a [theorem of Olovyanishnikov](http://www.mathnet.ru/php/journal.phtml?wshow=paper&jrnid=sm&paperid=6287&year=1946&volume=60&issue=3&fpage=429&lpage=440&option_lang=rus) --- one of three students of Alexandrov who died in the war.
*Is the embedding unique?*
NO, but I suspect it is unique once you fixed the convex embedding of the cone at infinity. It might follow from the proof of Pogorelov's theorem but I was not able to check his proof.
*Are there generalizations of 1-2 to complete noncompact surfaces of nonnegative sectional curvature?*
I'm not sure what you mean --- if it has strictly positive curvature at one point then it is automatically $\mathbb R^2$. If it is $\mathbb R^2$ then it is all the same.
| 10 | https://mathoverflow.net/users/1441 | 22143 | 14,615 |
https://mathoverflow.net/questions/22138 | 15 | It's a result of low-dimensional topology that in dimensions 3 and lower, two manifolds are homeomorphic if and only if they are diffeomorphic. Milnor's 7-spheres give nice counterexamples to this result in dimension 7, and exotic $\mathbb{R}^4$'s give nice counterexamples in dimension 4. But I don't know about dimensions 5 and 6. Is the result true or false in dimensions 5 and 6? And, if false, what are some classic counterexamples, and do stronger constraints -- say compactness or closedness -- happen to make it true?
| https://mathoverflow.net/users/3092 | exotic differentiable structures on manifolds in dimensions 5 and 6 | It is false in dimension 5 and 6. Spheres happen to be standard, but some other (compact and closed) manifolds happen to admit different smooth (and PL) structures.
Simple example are tori. For example, $\mathbb T^5$ admits 3 different PL structures that give rise to 3 different differentiable structures. See, e.g., Hsiang, Shaneson "Fake tori" or Wall's book on surgery.
| 17 | https://mathoverflow.net/users/2029 | 22147 | 14,618 |
https://mathoverflow.net/questions/22134 | 1 | What is meant by an "ample class" in general? **Motivation:** In the document I am reading, the phrase in question is "fix an ample class $\alpha\in H^1(X,\Omega^1\_X)$." I know what ampleness of a line bundle is. I have checked the only Wikipedia article that could be related (<http://en.wikipedia.org/wiki/Ample_line_bundle>). And I looked in Hartshorne.
Thanks.
| https://mathoverflow.net/users/5395 | Terminology issue: meaning of 'ample class' ? | Charles' and Pete's answer are (almost) the same: First there is a map
$\mathrm{dlog}\colon \mathcal{O}\_X^\ast \rightarrow \Omega^1\_X$ taking $f$ to
$df/f$ (just to show that it also makes algebraic sense) which indeed induces a
group homomorphism $H^1(X,\mathcal{O}\_X^\ast)\rightarrow H^1(X,\Omega^1\_X)$
giving one version of the Chern class. In the other version we have an exact
sequence $0\rightarrow 2\pi i\mathbb Z\rightarrow \mathcal O\_X\rightarrow
\mathcal{O}\_X^\ast\rightarrow0$ which gives a map $H^1(X,\Omega^1\_X) \rightarrow
H^2(X,2\pi i\mathbb Z)$. Combined with the inclusion $2\pi i\mathbb
Z\subseteq\mathbb C$ and the projection on the $(1,1)$-part it gives the
previous Chern class. Of course the sheaf $2\pi i\mathbb Z$ is isomorphic to
$\mathbb Z$ but using the latter forces one to use the map $\mathbb Z
\rightarrow \mathbb C$ taking $1$ to $2\pi i$. It is better to use the sheaf
$2\pi i\mathbb Z$. One other reason for that is to keep track of complex
conjugation. If $X$ comes from a real algebraic variety so that it has an
antiholomorphic involution $\overline{(-)}$. Then we have
$\overline{c\_1(L)}=c\_1(\overline L)$ when we let complex conjugation do what it
should do on $2\pi i\mathbb Z$ (if one uses $\mathbb Z$ one has to throw in a
sign). This is completely analogous to the case of étale cohomology where the
first Chern class takes value in $H^2\_{et}(X,\mathbb Z\_\ell(1))$, where $\mathbb
Z\_\ell(1)$ is the inverse limit of $\{\mu\_{\ell^n}\}$. Similarly the $n$'th
Chern class lies most naturally in cohomology of $(2\pi i)^n\mathbb Z=(2\pi
i\mathbb Z)^{\otimes n}$ resp. $\mathbb Z\_\ell(n):=(\mathbb Z\_\ell(1))^{\otimes n}$.
| 7 | https://mathoverflow.net/users/4008 | 22148 | 14,619 |
https://mathoverflow.net/questions/22142 | 14 | Today I found myself at the Wikipedia page on Vaught's Conjecture,
<http://en.wikipedia.org/wiki/Vaught_conjecture>
and it says that Prof. Knight, of Oxford, "has announced a counterexample" to the conjecture. The phrasing is odd; I interpret it as suggesting that there is some doubt as to whether Prof. Knight attained his goal. Another Wikipedia page uses similar language, "it is thought that there is a counterexample" or something like that. (I forget which page now.)
Prof. Knight's page, which you can easily find through the link above, certainly gives the impression that he himself harbors no doubts about his achievement. Since the counterexample is a 117-page construction and not immediately perspicuous, I thought I'd ask here what the situation is. Is the paper being refereed? Apparently it's from 2002, so something unusual must be going on.
| https://mathoverflow.net/users/4367 | Has Vaught's Conjecture Been Solved? | As far as I understand, no, Vaught's Conjecture has not been resolved.
We held a reading seminar on Robin Knight's proposed counter-example closely following each of his drafts and simplified presentations here at Berkeley some years ago and were ultimately convinced that the draft of January 2003 does not contain a correct disproof of Vaught's Conjecture and requires more than minor emendations to produce a complete proof. We did not discover any essential error, though there were important points in the argument where it seemed to us that even the author had not worked out the technical details.
That said, it is possible that revisions he has posted since then are sufficient, though in view of how much time it would require to enter into the details of the argument, I am not willing to work through the later papers until the basic architecture of the proof is certified by some other expert.
To be fair to Robin Knight, his work in set theoretic topology is well-respected and his construction takes into account the relevant features required for a counter-example to Vaught's Conjecture. If you would like to know whether or not he believes that his proof works, you should ask him directly. If he says that he does believe the proof to be valid, then you can attempt to check the proof yourself. The difficulty in reading his manuscript is not the amount of background material one must know in order to follow it, but exactly the opposite: almost everything is developed from scratch so that one must hold the entire construction in one's mind without having the usual anchors of established theorems.
| 29 | https://mathoverflow.net/users/5147 | 22149 | 14,620 |
https://mathoverflow.net/questions/22065 | 11 |
>
> Does the algebra of continuous
> functions from a compact manifold to
> $\mathbb{C}$ satisfy any specific
> algebraic property?
>
>
>
I'm not sure what kind of algebraic property I expect, but I feel that because of the Gel'fand transform, it may not be unreasonable to expect something. We can drop the compactness condition if we switch to continuous functions to $\mathbb{C}$ that vanish at infinity.
I'm really hoping for some necessary and sufficient condition, but if anybody knows of any sort of condition, that would be appreciated.
| https://mathoverflow.net/users/3664 | Algebraic properties of the algebra of continuous functions on a manifold. | I found a reference for a necessary property that might be called algebraic.
Browder [proved a theorem](http://projecteuclid.org/euclid.bams/1183524331) relating the number of generators of a complex commutative Banach algebra to the Čech cohomology with complex coefficients of the maximal ideal space, and as a corollary concluded that if $M$ is a compact orientable $n$-dimensional manifold, then $C(M)$ cannot be generated as a Banach algebra by fewer than $n+1$ elements. The paper is very short, but for an even shorter summary [here's the MR review](http://www.ams.org/mathscinet-getitem?mr=130580).
---
Just some comments, added later:
One obtains the compact Hausdorff space $X$ (up to homeomorphism) from $C(X)$ by considering the maximal ideal space of $C(X)$ with Gelfand topology, but clearly you want something less tautological than "the maximal ideal space is a manifold." A small step in this direction would be to try to formulate the topological properties of $X$ in terms of the closed ideals of $C(X)$. As alluded to in Qiaochu's comment, there is an analogue of Nullstellensatz: each closed ideal in $C(X)$ consists of all functions vanishing on a (uniquely determined) closed subset of $X$. So for example, the locally Euclidean property could be reformulated for a commutative C\*-algebra $A$ as follows: There is an $n$ such that for every maximal ideal $M$ of $A$ there is a closed ideal $I$ of $A$ such that $I$ is not contained in $M$ and $I$ is $\*$-isomorphic to $C\_0(\mathbb{R}^n)$. Second countability of the maximal ideal space [is equivalent to](http://en.wikipedia.org/wiki/Separable_space#Properties) $A$ being separable in the norm topology; that's not algebraic, but might be considered more intrinsic to the C\*-algebra.
But this only leads to another, more specific question: Is there a useful or interesting (C\*-)algebraic characterization of $C\_0(\mathbb{R}^n)$?`
| 4 | https://mathoverflow.net/users/1119 | 22151 | 14,622 |
https://mathoverflow.net/questions/22154 | 6 | Let $f$ be a function analytic on an open subset $D\subset \mathbb{C}$, and let $\gamma:[0,1] \to D$ be a line segment. $g = f\circ\gamma$ is another curve in the complex plane; is it possible to for $g$ to cross a straight line infinitely often, where the crossing points accumulate towards a point? That is, does there exist a point $\alpha$ and a ray $R$ emanating from the point $\alpha$ such that for all $\epsilon > 0$, $g$ crosses the ray $R$ infinitely many times in the $\epsilon$-ball around $\alpha$?
We're trying to show something about analytic continuation, but we cannot rule out pathological beasts like these.
Thanks!
| https://mathoverflow.net/users/5534 | Can curves induced by analytic maps wiggle infinitely across a line? | The image of $[0,1]$ is compact and so must contain the purported
accumulation point. It makes no loss to assume that $\gamma(t)=t$,
$f(0)=0$ is the accumulation point, and the line in question
is the real axis. Then $f(z)=a\_n z^n+a\_{n+1}z^{n+1}+\cdots$ where
$a\_n$ is nonzero and $n$ is a positive integer.
At this stage I'll assume there is a sequence of reals $t\_1>t\_2>\cdots$
tending to zero with each $f(t\_j)$ real. we want to show that all the $a\_k$
are real. Then considering $f(t\_j)/t\_j^n$ we get $a\_n$ real. Now consider
$f(z)-a\_n z^n$ in place of $f(z)$. We get $a\_{n+1}$ real etc. So $f$
takes reals to reals so there's no "crossing" of the real axis.
In general, there must be a sequence of distinct points $(t\_j)$ in $[0,1]$
where the curve crosses the line and which tends (by Bolzano-Weierstrass)
to a point $t\in[0,1]$. Replace $[0,1]$ by $[0,t]$ or $[t,1]$
(one of these has infinitely many $t\_j$) and rescale the interval
to $[0,1]$.
| 8 | https://mathoverflow.net/users/4213 | 22156 | 14,625 |
https://mathoverflow.net/questions/22161 | 7 | I am looking for the correct technical term in German for the notion of *catenary ring* in commutative algebra.
Does anyone know?
>
> For those who don't know what a catenary ring is but would like to: A Noetherian commutative ring A is called *catenary* if the following codimension formula holds for irreducible closed subsets T ⊆ Y ⊆ Z of Spec A:
>
>
>
> >
> > codim(T, Z) = codim(T, Y) + codim (Y, Z).
> >
> >
> >
>
>
>
| https://mathoverflow.net/users/1841 | What is the German translation of "catenary ring"? | It should be "Kettenring", see for example p. 148 in [Brodmann, Algebraische Geometrie](http://books.google.com/books?id=agw3I3IwUDMC&pg=PA256&lpg=PA256&dq=kettenring+primideal&source=bl&ots=M3d7_XqnGb&sig=mFjpisRgp5tnpwnRj0ub45HzTTQ&hl=en&ei=vgTQS8zKNYOQOI7xnZ0P&sa=X&oi=book_result&ct=result&resnum=3&ved=0CBcQ6AEwAg#v=onepage&q=kettenring&f=false).
| 13 | https://mathoverflow.net/users/5537 | 22168 | 14,634 |
https://mathoverflow.net/questions/22174 | 29 | When teaching Measure Theory last year, I convinced myself that a finite measure defined on the Borel subsets of a (compact; separable complete?) metric space was automatically regular. I used the [Borel Hierarchy](http://en.wikipedia.org/wiki/Borel_hierarchy) and some transfinite induction. But, typically, I've lost the details.
So: is this true? Are related questions true? What are some good sources for this sort of questions? As motivation, a student pointed me to <http://en.wikipedia.org/wiki/Lp_space#Dense_subspaces> where it's claimed (without reference) that (up to a slight change of definition) the result is true for finite Borel measures on any metric space.
(I'm normally only interested in Locally Compact Hausdorff spaces, for which, e.g. Rudin's "Real and Complex Analysis" answers such questions to my satisfaction. But here I'm asking more about metric spaces).
To clarify, some definitions (thanks Bill!):
* I guess by "Borel" I mean: the sigma-algebra generated by the open sets.
* A measure $\mu$ is "outer regular" if $\mu(B) = \inf\{\mu(U) : B\subseteq U \text{ is open}\}$ for any Borel B.
* A measure $\mu$ is "inner regular" if $\mu(B) = \sup\{\mu(K) : B\supseteq K \text{ is compact}\}$ for any Borel B.
* A measure $\mu$ is "Radon" if it's inner regular and locally finite (that is, all points have a neighbourhood of finite measure).
So I don't think I'm quite interested in Radon measures (well, I am, but that doesn't completely answer my question): in particular, the original link to Wikipedia (about L^p spaces) seems to claim that any finite Borel measure on a metric space is automatically outer regular, and inner regular in the weaker sense with K being only closed.
| https://mathoverflow.net/users/406 | Regular borel measures on metric spaces | The book *Probability measures on metric spaces* by K. R. Parthasarathy is my standard reference; it contains a large subset of the material in *Convergence of probability measures* by Billingsley, but is much cheaper! Parthasarathy shows that every finite Borel measure on a metric space is regular (p.27), and every finite Borel measure on a complete separable metric space, or on any Borel subset thereof, is tight (p.29). Tightness tends to fail when separability is removed, although I don't know any examples offhand.
(Definitions used in Parthasarathy's book: $\mu$ is regular if for every measurable set $A$, $\mu(A)$ equals the supremum of the measures of closed subsets of $A$ and the infimum of open supersets of $A$. We call $\mu$ tight if $\mu(A)$ is always equal to the supremum of the measures of *compact* subsets of $A$. Some other texts use "regular" to mean "regular and tight", so there is some room for confusion here.)
| 23 | https://mathoverflow.net/users/1840 | 22177 | 14,639 |
https://mathoverflow.net/questions/22180 | 8 | Is the following statement true?
Given a skew symmetric matrix M, among all of its largest rank sub-matrix, there must be one that is the principal submatrix of M.
| https://mathoverflow.net/users/5543 | Largest rank submatrix of a skew symmetric matrix | Yes. Suppose that you have a general $n \times n$ matrix *M*. The coefficient of $(-x)^k$ in the characteristic polynomial is the sum of all the principal minors of the matrix of size *n-k*. From this it follows that the "rank computed by the principal minors" can differ from the actual rank of the matrix only if the matrix is not diagonalizable. Since skew-symmetric matrices are diagonalizable, the result follows.
| 10 | https://mathoverflow.net/users/4344 | 22183 | 14,643 |
https://mathoverflow.net/questions/22189 | 52 | There are many "strange" functions to choose from and the deeper you get involved with math the more you encounter. I consciously don't mention any for reasons of bias. I am just curious what you consider strange and especially like.
Please also give a reason why you find this function strange and why you like it. Perhaps you could also give some kind of reference where to find further information.
As usually: Please only mention one function per post - and let the votes decide :-)
| https://mathoverflow.net/users/1047 | What is your favorite "strange" function? | A **[Brownian motion sample path](http://en.wikipedia.org/wiki/Wiener_process#Some_properties_of_sample_paths)**.
These are about the most bizarrely behaved continuous functions on $\mathbb{R}^+$ that you can think of. They are nowhere differentiable, have unbounded variation, attain local maxima and minima in every interval... Many, many papers and books have been written about their strange properties.
Edit: As commented, I should clarify the term "sample path". Brownian motion is a stochastic process $B\_t$. We say a sample path of Brownian motion has some property if the function $t \mapsto B\_t$ has that property almost surely. So, run a Brownian motion, and with probability 1 you will get a function with all these weird properties.
| 35 | https://mathoverflow.net/users/4832 | 22206 | 14,656 |
https://mathoverflow.net/questions/22118 | 10 | Classical theorems attributed to Levi, Mal'cev, Harish-Chandra for a finite
dimensional Lie algebra over a field of characteristic 0 state that it has a Levi decomposition (semisimple subalgebra plus solvable radical) and that all such semisimple subalgebras (Levi factors) are conjugate in a strong sense: see Jacobson, *Lie Algebras*, III.9, for example. This carries over to connected linear algebraic
groups, but in prime characteristic there are counterexamples going back perhaps
to Chevalley that involve familiar group schemes like $SL\_2$ over rings of Witt
vectors. Recent posts here have somewhat ignored that difficulty, having just characteristic 0 in mind. Borel and Tits redefined "Levi factor" to be a reductive complement to the unipotent radical, which is makes no real difference in characteristic 0 but allows them to concentrate on positive answers for parabolic subgroups of reductive groups in general. Other familiar subgroups of reductive groups like the identity component of the centralizer of a unipotent element require much more subtle treatment, as in work of George McNinch.
Whether or not the characteristic $p$ question is important, it has remained
open for many decades (say over an algebraically closed field). I gave up after one forgettable paper (Pacific J. Math. 23, 1967). The problem is still
easy to state:
>
> Are there effective necessary or sufficient conditions for existence or uniqueness of Levi factors in a connected linear algebraic group over an algebraically closed field of prime characteristic?
>
>
>
It's clear that a scheme-theoretic viewpoint may be needed. Possibly the known
counterexamples using Witt vectors suggest in some way all possible counterexamples? (Or is the question hopeless to resolve completely?)
EDIT: For online access to my 1967 paper, via Project Euclid, see
[Link](https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-23/issue-3/Existence-of-Levi-factors-in-certain-algebraic-groups/pjm/1102991730.full). Here
Chevalley's counterexample is mentioned only in the abstract, but in remarks
later on it is noted that Borel-Tits (III.15) gave an example involving two
Levi subgroups which fail to be conjugate; see NUMDAM link to PDF version of
Publ. Math. IHES 27 (1965) at <http://www.numdam.org:80/?lang=en>
In April
1967 Tits responded to my inquiry with a letter outlining the behavior of the
group scheme $SL\_2$ over the ring of Witt vectors of length 2, which gives a
6-dimensional algebraic group over the underlying field with unipotent radical of dimension 3 but no Levi factor. He remarked that he got this counterexample from P. Roquette but had also been told about Chevalley's counterexample.
ADDED: The question as formulated probably doesn't have a neat answer, but meanwhile George McNinch has delved much deeper (over more general fields) in his new arXiv preprint
1007.2777. Some technical steps rely on the forthcoming book *Pseudo-reductive groups* (Cambridge, 2010) by Conrad-Gabber-Prasad.
| https://mathoverflow.net/users/4231 | Are there "reasonable" criteria for existence/non-existence of Levi factors or their conjugacy in prime characteristic? | [See **Edit** below.]
This isn't really an answer, but I believe it is relevant.
Work geometrically, so $k$ is alg. closed. Let $G$ reductive over $k$, and let
$V$ be a $G$-module (linear representation of $G$ as alg. gp.).
If $\sigma$ is a non-zero class in $H^2(G,V)$, there is a non-split extension
$E\_\sigma$ of $G$ by the vector group $V$ -- a choice of 2-cocyle representing
$\sigma$ may be used to define a structure of alg. group on the variety
$G \times V$. Here "non-split" means "$E\_\sigma$ has no Levi factor".
And if $H^2(G,V) = 0$, then any $E$ with reductive quotient $G$ and
unipotent radical that is $G$-isomorphic to $V$ has a Levi factor.
You can look at the $H=\operatorname{SL}\_2(W\_2(k))$ example from this viewpoint;
$H$ is an extension of $\operatorname{SL}\_2$ by the first Frobenius twist
$A = (\mathfrak{sl}\_2)^{[1]}$ of its adjoint representation. Of course, this point of view doesn't really help to see that $H$ has no Levi factor; the fact that $H^2(\operatorname{SL}\_2,A)$ is non-zero only tells that it *might* be interesting (or rather: that there *is* an interesting extension).
The extension $H$ determines a class in that cohomology group, and the argument
in the pseudo-reductive book of Conrad Gabber and Prasad -- or a somewhat clunkier representation theoretic argument I gave some time back -- shows this class to be non-zero, i.e. that $H$ has no Levi factor.
So stuff you know about low degree cohomology of linear representations comes up. And this point of view can be used to give examples that don't seem to be related to Witt vectors.
A complicating issue in general is that there are actions of reductive $G$ on a product of copies of $\mathbf{G}\_a$ that are not linearizable, so one's knowledge of the cohomology of linear representations of $G$ doesn't help...
**Edit:** It isn't clear I was correct last April about that "complicating issue". See [this question.](https://mathoverflow.net/questions/32973/can-a-reductive-group-act-non-linearly-on-a-vector-group/32982#32982)
**Also:** the manuscript [arXiv:1007.2777](https://arxiv.org/abs/1007.2777) includes a "cohomological" construction
of an extension $E$ of SL$\_3$ by a vector group of dim $(3/2)(p-1)(p-2)$ having no Levi factor in char. $p$, and an example of a group having Levi factors which aren't geometrically conjugate.
| 3 | https://mathoverflow.net/users/4653 | 22210 | 14,660 |
https://mathoverflow.net/questions/22182 | 3 | I'm a biologist in the process of modeling a fairly simple biological system using a system of ODEs. To verify the simulations, I'm attempting to obtain an analytical steady-state solution that I can check the simulations against. My attempts so far haven't borne fruit, so I thought I'd toss the question out to mathematicians. This is my first post, so apologies if the question isn't right for this site.
The equation is of the form:
$${dS\_3\over dt} = 2Xv\_{max} {S\_1 - {S\_3^2S\_7^4\over K\_{eq,3}}\over K\_m+S\_1+{S\_3^2S\_7^4\over K\_{eq,3}}} + D(S\_{3,in} - S\_3)$$
------------------------------------------------------------------------------------------------------------------------------------------
$${dS\_4\over dt} = Xv\_{max} {S\_1 - {S\_4S\_7^2\over K\_{eq,4}}\over K\_m+S\_1+{S\_4S\_7^2\over
K\_{eq,4}}} + D(S\_{4,in} - S\_4)$$
--------------------------------------------------------------------------------------------------------------------------------------
$${dS\_1\over dt} = -Xv\_{max} \Bigg[{S\_1 - {S\_3^2S\_7^4\over K\_{eq,3}}\over K\_m+S\_1+{S\_3^2S\_7^4\over K\_{eq,3}}} + {S\_1 - {S\_4S\_7^2\over K\_{eq,4}}\over K\_m+S\_1+{S\_4S\_7^2\over
K\_{eq,4}}}\Bigg] + D(S\_{1,in}-S\_1)$$
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
$${dX\over dt} = Xv\_{max}Y \Bigg[4{S\_1 - {S\_3^2S\_7^4\over K\_{eq,3}}\over K\_m+S\_1+{S\_3^2S\_7^4\over K\_{eq,3}}} + 3{S\_1 - {S\_4S\_7^2\over K\_{eq,4}}\over K\_m+S\_1+{S\_4S\_7^2\over
K\_{eq,4}}}\Bigg] + D(X\_{in}-X)$$
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
$${dS\_7\over dt} = Xv\_{max} \Bigg[4{S\_1 - {S\_3^2S\_7^4\over K\_{eq,3}}\over K\_m+S\_1+{S\_3^2S\_7^4\over K\_{eq,3}}} + 2{S\_1 - {S\_4S\_7^2\over K\_{eq,4}}\over K\_m+S\_1+{S\_4S\_7^2\over
K\_{eq,4}}}\Bigg] + D(S\_{7,in}-S\_7)$$
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Where S1, S3, S4 and S7 and X are variables
and
Km, Keq,3, Keq,4, vmax, S1,in, S3,in, S4,in, S7,in, Xin, D and Y are constants.
This system models the change in the substrate Sn or the microbial population X in a perfectly-stirred vessel with microbes acting upon a substrate S1 to produce S3, S4 and S7 when the kinetics of the chemical reactions are thermodynamically reversible.
Sn,in is the input concentration of Sn. Km and vmax are constants that describe the "affinity" of the microbe to S1 and the maximum rate of the reaction respectively and Keq,n is the thermodynamic equilibrium constant for the reaction S1 -> A Sn + B S7. I need to solve this system for Sn where n=1,3,4,7.
Is this even possible, or am I barking up the wrong tree here?
| https://mathoverflow.net/users/5544 | Analytical steady-state solution of a complex ODE | As Andrey mentioned, you shouldn't expect an analytical solution since you're dealing with a system of algebraic equation in several variables. (Just to be clear, what we're envisioning here is the system of equations you get by setting all the left-hand-sides to be zero). In your case, I believe you have 5 variables (X, $S\_i$) and 5 equations, whose denominators can be cleared to make everything polynomial.
Such systems can be solved numerically (once you specify numerical values for your constants). One tool that I've used before is [PHCpack](http://www.math.uic.edu/~jan/download.html), though for your purposes maybe Mathematica or something similar will be just fine.
Perhaps an expert can describe how to calculate resultants of your system of polynomials which will give you information on when the nature of the roots of this system change as you change your parameters...
| 1 | https://mathoverflow.net/users/353 | 22217 | 14,663 |
https://mathoverflow.net/questions/22125 | 1 | I read that the Euclidean Minimum Spanning Tree (EMST) of a set of points is a subgraph of any Delaunay triangulation. Apparently the easiest/fastest way to obtain the EMST is to find the Deluanay triangulation and then run Prim's algorithm on the resulting edges. However, I have a set of 3D points, and the Delaunay triangulation doesn't include many of the "interior" points.
Here are my points:
<http://rpi.edu/~doriad/bunny.jpg>
And here are the edges of the tetrahedra that are produced by the Delaunay triangulation:
<http://rpi.edu/~doriad/bunnyDelaunay.jpg>
You can see that many of the points are not vertices of any of the tetrahedra. How, then, would finding the MST of these edges produce the EMST on the points, since the EMST must go through ALL of the points?
Thanks in advance for any help,
Dave
| https://mathoverflow.net/users/5523 | 3D Delaunay Triangulation -> Euclidean Minimum Spanning Tree | Sorry for the confusion - the "wireframe" view of the tetrahedralization of the points was only showing the outside surface (convex hull). Of course all of your comments are correct - the Delaunay triangulation indeed includes ALL points. Then the EMST is a subgraph of it.
All is well - thanks for the confirmations.
Dave
| 0 | https://mathoverflow.net/users/5523 | 22227 | 14,670 |
https://mathoverflow.net/questions/22188 | 44 | I have studied some basic homological algebra. But I can't send to get started on spectral sequences. I find Weibel and McCleary hard to understand.
Are there books or web resources that serve as good first introductions to spectral sequences? Thank you in advance!
| https://mathoverflow.net/users/nan | introductory book on spectral sequences | Many of the references that people have mentioned are very nice, but the brutal truth
is that you have to work **very hard** through some basic examples before it really makes
sense.
Take a complex $K=K^\bullet$ with a two step filtration $F^1\subset F^0=K$, the spectral
sequence contains no more information than is contained in the long exact sequence associated
to
$$0 \to F^1\to F^0\to (F^1/F^0)\to 0$$
Now consider a three step filtration $F^2\subset F^1\subset F^0=K$, write down all the short
exact sequences you can and see what you get. The game is to somehow relate $H^\*(K)$
to $H^\*(F^i/F^{i+1})$. Suppose you know these are zero, is $H^\*(K)=0$? Once you've mastered
that then ...
| 75 | https://mathoverflow.net/users/4144 | 22234 | 14,674 |
https://mathoverflow.net/questions/22232 | 11 | A simplicial set $X$ is a a combinatorial model for a topological space $|X|$, its realization, and conversely every topological space is weakly equivalent to such a realization of a simplicial set. I am wondering, which properties of finite simplicial sets can be effectively (or even theoretically) computed using a computer program. Maybe there are some implementations and I am not aware of their existence, so I would be very pleased about any information.
The first problem is how to input the simplicial set (maybe that's not really a problem).
Does one know terminating algorithms for the following problems then? Are there other problems for which anything is known in this direction?
1. ~~Is a given finite simplicial set a Kan complex?~~
2. Given two finite simplicial sets $X$ and $Y$, are $|X|$ and $|Y|$ homeomorphic?
3. Given two finite simplicial sets $X$ and $Y$, are $|X|$ and $|Y|$ homotopy equivalent?
4. ~~What are the homotopy/homology groups of a given Kan complex~~?
5. **Given a finite simplicial set $X$, what are the homotopy/homology groups of $|X|$**?
etc.
| https://mathoverflow.net/users/4676 | Which properties of finite simplicial sets can be computed? | For homeomorphism equivalence and homotopy equivalence, the associated problems are recursively unsolvable. This fact dates back to Markov in the 1950s, and relies on the unsolvability of the word problem for finitely presented groups. Apparently it was proved in Markov [1], which is in Russian. That paper has an English review in the Journal of Symbolic Logic [2] that explicitly states the unsolvability of the homeomorphism and homotopy problems. For more modern work, see Nabutovsky and Weinberger [3]. There is also a paper by Soare [4] that discusses some recursion-theoretic aspects of differential geometry and has a sketch of the proof that the homeomorphism problem is recursively unsolvable.
[1] MARKOV, AA: 'On the unsolvability of certain problems in topology', Dokl. Akad Nauk SSSR 123, no. 6 (1958), 978-980
[2] The Journal of Symbolic Logic, Vol. 37, No. 1 (Mar., 1972), p. 197
[3] Alexander Nabutovsky and Shmuel Weinberger, "Algorithmic aspects of homeomorphism problems", <http://arxiv.org/abs/math/9707232>
[4] Robert I. Soare, "Computability theory and differential geometry", <http://people.cs.uchicago.edu/~soare/res/Geometry/geom.pdf>
| 9 | https://mathoverflow.net/users/5442 | 22245 | 14,681 |
https://mathoverflow.net/questions/22247 | 30 | I don't understand wedge product and Grassmann algebra. However, I heard that these concepts are obvious when you understand the geometrical intuition behind them. Can you give this geometrical meaning or name a book where it is explained?
| https://mathoverflow.net/users/5555 | Geometrical meaning of Grassmann algebra | For a brief explanation of the geometric meaning of exterior product,
interior product of a k-form and l-vector, Hodge dual etc. see my answer here:
[When to pick a basis?](https://mathoverflow.net/questions/4648/when-to-pick-a-basis/4900#4900)
The best reference for this stuff is Bourbaki, Algebra, Chapter 3.
| 10 | https://mathoverflow.net/users/402 | 22254 | 14,685 |
https://mathoverflow.net/questions/22266 | 2 | Someone asked me, and I told them I would try to find out... what is the meaning of this symbol:
B'L or BL'
(I'm not sure if the tick comes before or after the L. It was found on a "nerd clock". The value of this symbol, by the way, is 1.
| https://mathoverflow.net/users/5562 | What is the mathematical meaning of this symbol? | Apparently, it is supposed to be [Legendre's constant](http://en.wikipedia.org/wiki/Legendre%27s_constant), also known as $1$.
| 11 | https://mathoverflow.net/users/1409 | 22268 | 14,692 |
https://mathoverflow.net/questions/21911 | 10 | To ask this question in a (hopefully) more direct way:
Please imagine that I take a freely moving ball in 3-space and create a 'cage' around it by defining a set of impassible coordinates, $S\_c$ (i.e. points in 3-space that no part of the diffusing ball is allowed to overlap). These points reside within the volume, $V\_{cage}$, of some larger sphere, where $V\_{cage}$ >> $V\_{ball}$. Provided the set of impassible coordinates, $S\_c$, is there a computationally efficient and/or nice way to determine if the ball can ever escape the cage?
---
Earlier version of question:
In Pachinko one shoots a small metal ball into a forest of pins, then gravity then pulls it downwards so that it will either fall into a pocket (where you win a prize) or the sink at the bottom of the machine. The spacing and distribution of the pins will help to insure that one only wins certain prizes with low probability.
Now imagine that we have a more general game where:
(1) - The ball is simply diffusing in 3-space (like a molecule undergoing Brownian motion). I.e. there is no fixed downward trajectory due to gravity.
(2) - You win a prize if the ball diffuses over a particular coordinate, just like one of the pockets in regular pachinko.
(3) - We generalize he pins as a set of impassible coordinates.
(4) - We define a 'sink' as an always accessible coordinate.
(5) - We define a starting coordinate for the sphere.
Given access the 3-space coordinates for (2), (3), (4), & (5), what's the most efficient way to find whether the game is 'winnable', or if the ball will fall into the 'sink' with a probability of unity? How can we find the minimum set from (3) that prevents the ball from reaching the pocket?
| https://mathoverflow.net/users/3248 | When can a freely moving sphere escape from a 'cage' defined by a set of impassible coordinates? | Replace the pins by balls of radius $R\_{ball}$ and the ball by a point. This is a logically equivalent formulation. The question, then, is: given a finite set of balls, $B\_1$, $B\_2$, ...., $B\_k$ in $\mathbb{R}^n$, and a point $x$, how to determine where $x$ is in the unbounded component of $\mathbb{R}^n \setminus \bigcup B\_i$.
I don't know the answer to this, but here is an easy way to compute the number of connected components of $\mathbb{R}^n \setminus \bigcup B\_i$. In other words, I can determine whether there is some place from which a ball cannot escape.
By [Alexander duality](http://en.wikipedia.org/wiki/Alexander_duality), the number of bounded components of $\mathbb{R}^n \setminus \bigcup B\_i$ this is the dimension of $H\_n(\bigcup B\_i)$.
Cover $\bigcup B\_i$ by the $B\_i$. Every intersection of finitely many $B\_i$ is convex, hence contractible. So $\bigcup B\_i$ is homotopic to the nerve of this cover. That is a simplicial complex, so it is easy to compute its homology.
One final practical idea: I have used painting software where I could click on a point and it would color every point which was connected to that one. Maybe the algorithms used to make that software could solve this problem as well?
| 22 | https://mathoverflow.net/users/5563 | 22271 | 14,695 |
https://mathoverflow.net/questions/22255 | 13 | If you have a function $V: L \rightarrow \mathbb{R}$, where $L$ is an infinite dimensional topological vector space, there are multiple notions of differentiability. For $x,u \in L$, $V$ is Gateaux differentiable at $x$ in the direction $u$ if the limit $\underset{t \rightarrow 0}{\lim} \frac{V(x + tu) - V(x)}{t}$ exists. Supposing that you are in a Banach space, $V$ is Frechet differentiable if the above limit exists for all $u$ in a ball around $x$, and importantly, with the convergence being uniform over this neighborhood.
The question is, what difference does it make for a function to be Frechet differentiable versus Gateaux differentiable, maybe with respect to proving theorems that generalize the finite-dimensional setting, where the two notions of differentiability more or less agree. What kind of pathological behavior can functions exhibit that are merely Gateaux differentiable in every direction? There are also intermediate forms of differentiability between Frechet and Gateaux, defined in terms of uniform convergence of the difference quotients over some preferred family of sets (a bornology). Are there any intermediate kinds of differentiability that are important?
| https://mathoverflow.net/users/2310 | Usefulness of Frechet versus Gateaux differentiability or something in between. | For Lipschitz functions in finite dimensional spaces, Gateaux and Frechet differentiability are the same, but there are huge differences when the domain is infinite dimensional. Lipschitz functions are Gateaux differentiable off a null set when the domain is a separable Banach space and the range has the Radon Nikodym property (e.g., is reflexive), where "null set" can have any of several different meanings. On the other hand, it is rare for a Lipschitz function on an infinite dimensional space to have a Frechet derivative anywhere. A great theorem of David Preiss says that a real valued Lipschitz function on an Asplund space has a point of Frechet differentiability, but it is not known whether a complex valued Lipschitz function on an Asplund space has a point of Frechet differentiability (although some wonderful progress was made recently by Lindenstrauss and Preiss).
If a bi-Lipschitz equivalence from a Banach space X to a Banach space Y has a Frechet derivative at some point, then the derivative is an isomorphism from X onto Y. If it is only Gateaux differentiable, then the derivative is only an into isomorphism. It gets worse when you look at Lipschitz quotients (introduced in my paper with Bates, Lindenstrauss, Preiss, and Schechtman): The Frechet derivative of a Lipschitz quotient is a surjective linear operator, while the Gateaux derivative can be anything!
Look at the book by Benyamini and Lindenstrauss [BL], Geometric nonlinear functional analysis, to learn more.
For a useful concept that is weaker than Frechet differentiability, take a look at the sections in [BL] that treat $\epsilon$-Frechet differentiability. There are much better existence theorems about Lipschitz functions having for all $\epsilon > 0$ points of $\epsilon$-Frechet differentiability than having points of Frechet differentiability, and for many applications it is just as good to have such points as to have points of Frechet differentiability.
| 15 | https://mathoverflow.net/users/2554 | 22273 | 14,696 |
https://mathoverflow.net/questions/22272 | 1 | This may be pretty trivial, but I can't figure it out. Suppose that $S$ is any scheme, and $f\_1:X\_1\to Y\_1$ and $f\_2:X\_2\to Y\_2$ are two morphisms of $S$-schemes, such that the closed image of each one exists. That is, there is a smallest closed subscheme $Z\_i$ of $Y\_i$ over which $f\_i$ factorizes (for $i=1,2$).
Is it true then that the closed image of $f\_1\times\_S f\_2:X\_1\times\_S X\_2\to Y\_1\times\_S Y\_2$ exists? In this case, is it equal to $Z\_1\times\_S Z\_2$ (with its natural closed immersion into $Y\_1\times\_S Y\_2$)?
| https://mathoverflow.net/users/5564 | Closed image of a product of morphisms | This is already false in the affine case.
The closed image of $Spec(B) \to Spec(A)$ is the spectrum of $A/K$, where $K$ is the kernel of $A \to B$. Let $A',B',K$ be analogously defined. The closed image of $Spec(B \otimes\_R B') \to Spec(A \otimes\_R A')$ is the spectrum of $(A \otimes\_R A')/L$, where $L$ is the kernel of $A \otimes\_R A' \to B \otimes\_R B'$. We have to compare this ring with $A/K \otimes\_R A'/K = (A \otimes\_R A')/\langle K,K' \rangle$. If $K=K'=0$, this asks if the tensor product of two injective rings maps is injective, which is false in general.
This discussion shows that everything is fine if, in your notation, $X\_1,Y\_2$ or $X\_2,Y\_1$ are flat over $S$.
| 3 | https://mathoverflow.net/users/2841 | 22283 | 14,702 |
https://mathoverflow.net/questions/22289 | 8 | On [page 168](http://books.google.com/books?id=M6DvzoKlcicC&lpg=PP1&dq=mathematical%2520fallacies%2520and%2520paradoxes&pg=PA168#v=onepage&q&f=false) of *Mathematical Fallacies and Paradoxes*, it states that the fact that the series
$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots $
has a sum depends on the Axiom of Choice. Where does the AC come in to play? I know that if the terms are permuted, we can get any sum we want, and I can see how the AC might be involved there, but just the fact that $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots $ converges?
| https://mathoverflow.net/users/175 | Why does this sum depend on the Axiom of Choice? | It doesn't seem to me that you need any choice principle at all to prove that this series converges. The Alternating Series Test that appears in any elementary calculus books seems to do the job, and doesn't seem to require any amount of AC. If $\Sigma\_n (-1)^n a\_n$ is an alternating series, with $a\_n$ descending to $0$, then the finite partial sums up to positive term are descending and the partial sums up to a negative term are increasing, and that these two sequences converge to the same limit.
So I'm not sure which theorem had been "called upon to show" that the series converges, but unless I am mistaken, it must have been something other than what our students would call the Alternating Series Test.
---
Edit. More generally, I claim that the convergence of a series can never depend on the Axiom of Choice. Suppose that $r=\langle a\_0,a\_1,\ldots\rangle$ is a sequence of real numbers, and we are consider the series $\Sigma a\_n$. The assertion "$\Sigma a\_n$ converges" is a statement about $r$ having complexity $\Sigma^1\_1(r)$, that is, an analytic fact about $r$, and therefore has the same truth value in the set theoretic universe $V$ as it has in the relativized constructible universe $L[r]$, where AC holds. In particular, the series converges in $V$ if and only if it converges in $L[r]$.
Conclusion. If you have a series $\Sigma a\_n$ defined in a sufficiently concrete manner and you can prove in ZFC that it converges, then you can prove in ZF that it converges.
(The technical requirement here about sufficiently concrete is that the description of $r=\langle a\_0,a\_1,\ldots\rangle$ should be absolute from $V$ to $L[r]$. This would be true of any arithmetically definable series as in the question, defined by an arithmetic formula, or even a Borel definition or a $\Sigma^1\_2$ definition.)
| 21 | https://mathoverflow.net/users/1946 | 22290 | 14,705 |
https://mathoverflow.net/questions/22195 | 2 | These questions might be elementary for I just started to learn affine Kac-Moody algebra.
It is well known that if we consider finite dimensional Lie algebra, we have the folloing projection:
$R(\lambda)\otimes R(\nu)\rightarrow R(\lambda+\nu)$
where $\lambda$ and $\nu$ are dominant highest weights, $R(\lambda)$ and $R(\nu)$ are fnite dimensional irreducible representations.
**First question**:
I wonder whether this projection(or some analogue)still survives in affine Kac-Moody case? To be precisely, if we have $R(\lambda)$ and $R(\nu)$ as irreducible integral highest weight representations. Do we have the projection in the same form? (in classical case, we can prove this by using Weyl Character formula or equivalent multiplicity formula, does the affine case follows from the Weyl-Kac formula. Of course I can try to write the argument, but if somebody can point out the reference, I will be happy.
Second question:
It is related to the question I asked several days ago on extreme vector here:[tensor product of extreme vector](https://mathoverflow.net/questions/21875/multiplication-of-extreme-vector), Now I think I know the proof based on the comment of Peter Tingley and Jim.
**Argument based on Peter**
$e\_{w\lambda}\otimes e\_{w\nu}$ has the weight $w(\lambda+\nu)$ which belongs to the same weight space as $e\_{w(\lambda+\nu)}$. So if one need to show $e\_{w\lambda}\otimes e\_{w\nu}=ke\_{w(\lambda+\nu)}$ for some nonzero invertible elements $k$, it is sufficient to show the multiplicity of weight space is 1($dimV[w(\lambda+\nu)]$=1). But this just follows from multiplicity of weight space of highest weight is equal to 1 and the fact that multiplicity of weight space is invariant under action of Weyl group.
**My second question is:**
Can this argument extend to affine Kac-Moody case? Jim made a comment [here](https://mathoverflow.net/questions/21656/what-is-extreme-extremal-vector-according-to-some-weights), he pointed out the extreme vector can be extend to the integral representations of affine Kac-Moody algebra. So if we pick up two extreme vector of intergal highest weight, then take the tesnor product, can we do the similar argument as above(for finite dimensional Lie algebra)to obtain the same result?
Comments: Jim pointed out the reference by S.Kumar on solutions of PRV conjectures. But I think the situation there is slightly different.(maybe much more difficult), because the situation there, one need to consider multiplicity of $V(\lambda+w\nu)$, where just one weight moved by $w$, but in the situation here, it is much simpler.
All comments are welcome.
Thanks!
| https://mathoverflow.net/users/1851 | Several question on Affine Lie algebra | First question: Peter's (and Emerton's) argument works not only for affine algebras, but for arbitrary Kac-Moody algebras. Decompose your integrable representations into weights, and take the tensor product as the sum of tensor products of weight spaces. It is straightforward to check that the result is still an integrable representation. The tensor products live in Category O, where irreducible quotients are unique for each highest weight. See Kac, *Infinite dimensional Lie algebras*, sections 9.1-9.3.
Second question: Yes. See previous paragraph.
| 5 | https://mathoverflow.net/users/121 | 22297 | 14,708 |
https://mathoverflow.net/questions/22295 | 4 | What is the difference between:
($\infty,1$) categories - in which have for two objects you have an ($\infty,0$) category of morphisms (i.e. a space of morphisms)
and
categories weakly enriched over spaces - by that I mean categories such that hom(x,y) is always a space and composition is defined only up to (coherent) homotopy
?
| https://mathoverflow.net/users/1681 | (∞,1) vs Category weakly enriched over spaces | If "space" means the same thing in the two cases, as seems to be implied, then there is no difference, at least not at that level of precision.
There are many different models of $(\infty,1)$-categories. Many of these, like $A\_\infty$-categories, Segal categories, complete Segal spaces, and simplicial categories, try to make explicit the intuition that they are "categories weakly enriched over spaces," though in various different ways. Others, like quasicategories, are less directly related to the enrichment point of view, although they yield a provably equivalent theory (in a certain appropriate sense).
| 10 | https://mathoverflow.net/users/49 | 22298 | 14,709 |
https://mathoverflow.net/questions/22302 | 9 | This is a spur of the moment algebraic number theory question prompted by a side remark I made in a course I'm teaching:
Let $K$ be a number field. The (Hilbert) **class field tower** of $K$ is the sequence defined by $K^0 = K$ and for all $n \geq 0$, $K^{n+1}$ is the Hilbert class field of $K^n$. Put $K^{\infty} = \bigcup\_n K^n$. We say that the class field tower is infinite if $[K^{\infty}:K] = \infty$ (equivalently $K^{n+1} \supsetneq K^n$ for all $n$). Golod and Shafarevich gave examples of number fields with infinite class field tower, and thus which admit everywhere unramified extensions of infinite degree. It is now known that a number field with "sufficiently many ramified primes" has infinite class field tower.
My question is this: is there a known algorithm which, upon being given a number field, decides whether the Hilbert class field tower of $K$ is infinite?
| https://mathoverflow.net/users/1149 | Algorithm for the class field tower problem? | Not in the slightest! The answer is not even known for quadratic imaginary number fields. In fact, the *only* known way to show that the Hilbert class field tower of a number field is infinite is to invoke one of a variety of different forms of Golod-Shafarevich, and I don't think it's even seriously conjectured (more like "wondered") that every infinite Hilbert class field tower arises by applying Golod-Shafarevich to some step in the tower (or to some cleverly chosen subfield).
Incidentally, the "sufficiently many primes ramified" business is a bit of a red herring, in my opinion. The *real* condition is that the $p$-rank of the class group is large for some prime $p$. When $K$ is cyclic of degree $p$, it is only the fact that genus theory relates the $p$-rank of the class group to the number of ramified primes that brings ramified primes into the picture. (For example, the standard Golod-Sharevich examples come from showing the 2-class field tower is infinite by using Gauss' result that many primes ramifying in a quadratic extension imply a large 2-rank). For non-cyclic extensions, the link is more tenuous, and it becomes much more natural to talk strictly in terms of the class group.
| 13 | https://mathoverflow.net/users/35575 | 22306 | 14,714 |
https://mathoverflow.net/questions/22309 | 4 | Some of the fundamental results in analysis (inverse function theorem, existence and uniqueness of solutions to ODEs) have slick proofs using the idea of a contraction. So, it seems plausible to me that one might be motivated to study a "contraction space":
I'll define a contraction space as a metric space $(X,d)$ such that there is at least one fixed point of any function $f:X\rightarrow X$ with the property that for all $x,y$ we have $d(f(x),f(y)) \le \frac12 d(x,y)$.
Here's the question: is every contraction space complete?
I feel like the answer should be yes. The only approach I can see for proving this, though, is to find some way to take a Cauchy sequence $(a\_n)$ and construct a contraction on the whole space taking each $a\_i$ to some $a\_j$ with $j > i$, but even for the case of Cauchy sequences in the rationals I don't see an obvious approach.
A related question: suppose we take an arbitrary metric space, and add a fixed point for every contraction the same way we add limits to Cauchy sequences. Is the resulting space then a contraction space?
| https://mathoverflow.net/users/2363 | Is a "contraction space" always complete? | There is a "contraction space" which is not complete. For example, consider a metric $d$ on $[1,+\infty)$ such that for $x,y\in[n,n+1]$ where $n\in\mathbb N$ one has $d(x,y)=2^{-n}|x-y|^{1/n}$ (other distances are defined by gluing the segments together). The completion is obtained by adding one point at $+\infty$.
But there is no map contracting this space to $+\infty$. Indeed, any Lipschitz map $f$ with $f(x)>n+1$ for some $x\in[n,n+1]$ must be a constant on $[n,n+1]$, so there will be a finite fixed point.
**Added.** Here is a counter-example to the second question.
Let $X=(\mathbb R,d)$ from the above example. Define $Y=(\mathbb R,d')$ similarly by setting $d'(x,y)=2^{-n}|x-y|$ for $x,y\in[n,n+1]$. Then $Y$ is isometric to $[0,1)$. Consider the disjoint union of $X$ and $Y$. For $t\in [0,+\infty)$, denote by $t\_X$ and $t\_Y$ the copies of $t$ in $X$ and $Y$. For every $t$, attach an arc $\gamma\_t$ of length $\ell\_t:=10\cdot 2^{-t}$ connecting $t\_X$ to $t\_Y$. Let $Z$ denote the union of $X$, $Y$ and all these arcs. We have yet defined distances on $X$, on $Y$ and on every arc $\gamma\_t$. The metric on $Z$ is defined as the maximal metric bounded above by these metrics on these subsets. It is easy to see that $X$ and $Y$ are embedded into $Z$ isometrically, as well as sufficiently short intervals of the arcs $\gamma\_t$ (e.g. their halves are sufficiently short).
The counter-example is the space $Z\setminus Y$. Its completion is $Z\cup\{+\infty\}$. The space cannot be contracted to $+\infty$ for the same reasons as above: the arcs $\gamma\_t$ added to $X$ form a tree and don't help to go around weird parts of $X$.
On the other hand, $Z\setminus Y$ can easily be contracted to any point of $Y$ because $\gamma\_t$ nearby $t\_Y$ is isometric to a straight line segment $(0,\epsilon\_t)$. Just map every point of $Z\setminus X$ to a point in this segment in such a way that the distance to $t\_Y$ gets multiplied by a small positive constant (like $\epsilon\_t/100$).
Thus adding "contraction points" to $Z\setminus Y$ yields $Z$. But $Z$ can be contracted to $+\infty$ - just project everything to $Y$ and contract there.
| 9 | https://mathoverflow.net/users/4354 | 22314 | 14,720 |
https://mathoverflow.net/questions/22319 | 5 | If $f:X \to Y$ is a flat and proper surjective morphism between smooth schemes over an algebraically closed field, and $f$ has connected fibers, does it imply that
$$f`\_\*\mathcal O\_X = \mathcal O\_Y?$$
| https://mathoverflow.net/users/5464 | Is the direct image of the structure sheaf on X isomorphic to the structure sheaf on Y when X->Y is flat and proper with connected fibers between smooth schemes over an algebraically closed field? | This follows from Zariski's main theorem if the characteristic is zero and it is false in positive characteristics: consider the the morphism $\mathbb{A}^1 \to \mathbb{A}^1$ given by $x \mapsto x^p$ where $p$ is the characteristic. The statement would also be true in char p if you assume that the general fibre is reduced.
(Note that it suffices to assume that X is integral and Y is normal. $f$ should of course also be surjective.)
| 11 | https://mathoverflow.net/users/519 | 22333 | 14,731 |
https://mathoverflow.net/questions/22327 | 9 | We say that a subset $X$ of $\mathbb{R}$ is midpoint convex if for any two points $a,b\in X$ the midpoint $\frac{a+b}{2}$ also lies in $X$.
My question is: is it possible to partition $\mathbb{R}$ into two midpoint convex sets in a non-trivial way?
(trivial way is $\mathbb{R}=(-\infty,a]\cup(a,+\infty)$ or $\mathbb{R}=(-\infty,a)\cup[a,+\infty)$)
| https://mathoverflow.net/users/5572 | Partition of R into midpoint convex sets | Yes if you assume AC:
With AC let $\{v\_\alpha\}$ be a $\mathbb{Q}$-basis for $\mathbb{R}$ then the following two sets satisfies your property:
$A = \{q\_1v\_{\alpha\_1}+\cdots+q\_nv\_{\alpha\_n} \mid q\_i \in \mathbb{Q} , \sum q\_i \geq 0 \}$
and
$B = \{q\_1v\_{\alpha\_1}+\cdots+q\_nv\_{\alpha\_n} \mid q\_i \in \mathbb{Q} , \sum q\_i < 0 \}$
So in fact these are $\mathbb{Q}$ convex (in the obvious sense).
| 11 | https://mathoverflow.net/users/4500 | 22338 | 14,734 |
https://mathoverflow.net/questions/22320 | 4 | As the title says I would like to know if $K\_1(k)=k^\*$ uniquely determines a field $k$.
For finite fields this is clearly the case, but I suspect it is not ture in general. However I guess cooking up a counterexample is not so easy.
| https://mathoverflow.net/users/2837 | Is a field uniquely determined by its multiplicative group/how much knows K_1 about fields? | Let $K$ and $L$ be two algebraically closed fields of characteristic $0$. Then $K^{\times} \cong L^{\times}$ iff $K$ and $L$ have the same cardinality.
The forward direction is clear. Conversely, if $K$ is algebraically closed, then consider
the short exact sequence
$1 \rightarrow K^{\times}[\operatorname{tors}] \rightarrow K^{\times} \rightarrow Q \rightarrow 1$.
The first term is isomorphic to $\mathbb{Q}/\mathbb{Z}$. In particular it is divisible, hence injective, hence the sequence splits. The group $Q$ is a uniquely divisible abelian group of cardinality equal to that of $K$, hence isomorphic to a $\mathbb{Q}$ vector space of dimension $\# K$. Thus we recover the structure of $K^{\times}$ from $\# K$.
If $K$ is uncountable, then $\# K$ also determines the isomorphism class of $K$, but if $K$ is countable there is another invariant: the transcendence degree. This gives $\aleph\_0$ pairwise nonisomorphic fields with isomorphic multiplicative groups.
I believe that this construction can be modified to construct, for any cardinal $\kappa$, $\kappa$ pairwise nonisomorphic fields with isomorphic multiplicative groups: for instance, instead of algebraically closed, the argument goes through with solvably closed fields containing all the roots of unity.
| 5 | https://mathoverflow.net/users/1149 | 22341 | 14,736 |
https://mathoverflow.net/questions/22281 | 5 | I would like to express the integral of the absolute value of a real-valued function $f$ (over a finite interval) in terms of the Fourier coefficients of $f$. Failing that, I would like to know of any constraints or statistical correlations (in a sense explained in the motivation) relating these quantities.
Motivation: This comes from a biophysics application, but is perhaps best explained as follows. If a rubber band of tension $t$ is stretched along the $x$ axis from $0$ to $L$, then it is easy to calculate the thermal fluctuations of its arc-length by letting $z(x)$ be the (small) deviation from the $x$ axis, and then writing the energy (arc-length times tension) in terms of the Fourier coefficients of $z(x)$. The Boltzmann weight turns out to be a Gaussian since in the limit of small deviations the arc-length becomes a sum of squares of the Fourier coefficients. My problem is more complicated: We have two rubber bands stretched over the same interval, with deviations $z\_{1}(x)$ and $z\_{2}(x)$. The energy includes not only the stretching of the rubber bands, but also a term proportional to the (positive) area enclosed between them, which is
$\int\_{0}^{L}|z\_{1}(x) - z\_{2}(x)|dx$
Hence my question. So it would be nice to know how this area can be related to the Fourier coefficients of $z\_{1}$ and $z\_{2}$ or perhaps just to the arc-lengths of the rubber bands. By "statistical correlations" I am referring to the Boltzmann probability distribution with energy equal to the stretching energy plus the area-energy.
Edit: Specifics on the Boltzmann probability distribution, more motivation.
The state of the system is the pair of functions $z\_{1}(x)$ and $z\_{2}(x)$ describing the deviation of the two rubber bands from the x axis. Let's say it's the set of pairs of functions defined on [0,L] and that these functions are identified with a finite number of Fourier coefficients - I am a physicist and would like to avoid nasty functions or mathematically honest discussions of path integrals.
The probability of occurrence of a state (z\_{1}, z\_{2}) is (before normalization)
$\exp(-\beta E\left[z\_{1},z\_{2}\right])$
where $E\left[z\_{1},z\_{2}\right]$ is the energy of the system, which in this case is the functional
$E\left[z\_{1},z\_{2}\right] = \frac{t}{2}\int\_{0}^{L}\left[(\frac{dz\_{1}}{dx})^{2}+(\frac{dz\_{2}}{dx})^{2}\right]dx + \kappa \int\_{0}^{L}|z\_{1}(x)-z\_{2}(x)|dx$
Where the tension t and the "surface tension" $\kappa$ are just numbers; set them equal to 1 if you wish. The first integral is the energy cost of stretching the rubber bands (in a linearized regime) and the second is the strange term proportional to the area enclosed between them. Without the second term, it is easy to diagonalize this functional in terms of the Fourier series of the two functions. That is why I was interested in writing the second term in terms of Fourier coefficients. That may be too much to ask, but perhaps it is still possible to calculate some quantities such as the statistical average of $z\_1(x)^{2}$ - that is the kind of thing I ultimately want to know.
I realize this is an unnatural-looking problem, so I will just mention that it's not really about rubber bands, but rather about fluctuating interfaces which occur in lipid bilayers with coexisting phases. There are two phase boundaries (one for each monolayer) with their respective "tensions" but there is also a term proportional to the area between them.
| https://mathoverflow.net/users/3291 | Can I relate the L1 norm of a function to its Fourier expansion? | This sounds difficult. For instance, a long-standing open problem of Littlewood used to be this. Let A be a set of integers of size n, and let f be the characteristic function of A. How small (up to a constant) can the sum of the absolute values of the Fourier coefficients of f be? The conjecture was that the smallest was $C\log n$, which is what you get when A is an arithmetic progression. This conjecture is now known to be correct, but plenty of closely related questions are still open. So at least sometimes the relationship between the $L\_1$ norm of a function and the Fourier transform of that function is quite hard to understand.
| 8 | https://mathoverflow.net/users/1459 | 22367 | 14,755 |
https://mathoverflow.net/questions/22364 | 0 | In comparing the norm of two operators, I come across the following problem.
Let $S\in M\_{n}(\mathbb{R})$ be a symmetric matrix. $D\_1=diag(\alpha\_1,\cdots,\alpha\_n)$, $D\_2=diag(\beta\_1,\cdots,\beta\_n)$, with $\alpha\_1\ge\cdots\ge\alpha\_n\ge0, ~\beta\_1\ge\cdots\ge\beta\_n\ge 0$. Is it true that $Tr[(D\_1S^2D\_1D\_2+I)^{-1}]\le Tr[(SD\_1^2SD\_2+I)^{-1}]$?
**Updated** The above statment is not true as shown by Gerald Edgar. Now with the same notation, define $A=D\_2SD\_1^2SD\_2, B=D\_2D\_1S^2D\_1D\_2$ and
$$X\_{k+1}=\frac{1}{2}(X\_k+AX\_k^{-1}),~~ X\_0=I.$$
$$Y\_{k+1}=\frac{1}{2}(Y\_k+BY\_k^{-1}), ~~Y\_0=I.$$
Is it true $Tr X\_{k}\le Tr Y\_{k}$ for all $k\ge 1$?
When $k=1$, it is true.
| https://mathoverflow.net/users/3818 | Is this trace inequality true? | I just wrote down some random 2x2 example, and it had the wrong sense for the inequality.
A = [2,1;1,0], D1=D2=diag(2,1) .
| 8 | https://mathoverflow.net/users/454 | 22376 | 14,760 |
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