most recent 30 from http://mathoverflow.net 2013-05-19T17:16:31Z http://mathoverflow.net/feeds/user/9189 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40791/finite-covers-of-complex-varieties-all-but-two-questions-answered Nikita 2010-10-01T21:11:11Z 2010-10-03T04:26:43Z

<p>EDIT: Thanks to several people I almost have a complete answer. BCnrd pointed out that the generalized Riemann Existence Theorem shows that $\tilde{Y}$ can be uniquely given the structure of a complex variety $\tilde{Y}'$. Also, Georges pointed out that the Kodaira embedding theorem implies that $\tilde{Y}'$ is projective if $Y'$ is projective.</p> <p>This leaves two parts open to my original question. If $Y'$ is quasiprojective, then must $\tilde{Y}'$ be quasiprojective? Also, if $Y'$ is affine, then must $\tilde{Y}'$ be affine?</p> <p>The impression I get from reading the comments is that the answers are "yes" and that everyone but me is able to easily prove them given what has already been said. Can anyone give me a hint or a reference as to how to proceed?</p> <p>Thanks to everyone for all your help so far.</p> <hr> <p>ORIGINAL QUESTION:</p> <p>Let $Y$ be a complex manifold that can be given the structure of a complex variety $Y'$. Let $\pi:\tilde{Y} \rightarrow Y$ be a finite, unramified cover of $Y$. Can $\tilde{Y}$ be given the structure of a complex variety $\tilde{Y}'$ such that there is a finite map $\pi' : \tilde{Y}' \rightarrow Y'$ making the obvious diagram commute? If the answer is yes, then can we take $\tilde{Y}'$ to be projective/quasiprojective/affine if $Y'$ is projective/quasiprojective/affine?</p> <p>This kind of thing is true for Riemann surfaces, but even there I don't know how to prove it except by going through the whole machinery showing that all compact Riemann surfaces are projective varieties. Since such things are not available in higher dimensions, I'm stuck.</p> <p>I should maybe remark that I don't even know how to do the above for affine varieties.</p> <p>Thanks!</p>

http://mathoverflow.net/questions/38844/cohomological-dimension-of-a-group-acting-on-a-cellular-complex/38854#38854 Nikita 2010-09-15T18:22:40Z 2010-09-15T18:22:40Z

<p>If you don't require $X$ to be acyclic, then you can't say anything. Indeed, every group acts freely on its <a href="http://en.wikipedia.org/wiki/Cayley_graph" rel="nofollow">Cayley graph</a>, which is a 1-dimensional cell complex. </p> <p>You're not even saved by assuming the $X$ is highly connected. By attaching cells to the Cayley graph in an equivariant manner, you can obtain a $k$-connected $(k+1)$-dimensional complex on which the group acts freely.</p>

http://mathoverflow.net/questions/38763/is-lp-mathbbr-minus-the-zero-function-contractible Nikita 2010-09-15T02:15:05Z 2010-09-15T05:54:02Z

<p>Is $L^p(\mathbb{R}) \setminus 0$ contractible? My intuition says that the answer is yes, but I'm afraid that this is based on thinking of this as somehow similar to a limit of $\mathbb{R}^n \setminus 0$ as n approaches $\infty$, which is of course nonsense. In any case, every contraction I've tried ends up making some function pass through $0$.</p>

http://mathoverflow.net/questions/38499/monomorphisms-of-sheaves Nikita 2010-09-12T18:33:57Z 2010-09-12T19:44:20Z

<p>The following is surely pretty standard, but I have been unable to prove it or find a proof in the literature (many sources assert it without proof).</p> <p>Let $\phi : \mathcal{F} \rightarrow \mathcal{G}$ be a monomorphism in the category of sheaves of sets on some space $X$ (by this, I mean a monomorphism in the categorical sense : if $\psi_1,\psi_2 : \mathcal{H} \rightarrow \mathcal{F}$ are morphisms of sheaves on $X$ such that $\phi \circ \psi_1 = \phi \circ \psi_2$, then $\psi_1=\psi_2$). Question : must $\phi$ induce an injection on all stalks?</p> <p>The converse is pretty trivial, but this seems harder in the sense that without knowing much about $\mathcal{F}$ and $\mathcal{G}$, I don't know how to construct many interesting morphisms into $\mathcal{F}$ to test injectivity on stalks.</p> <p>Thanks for any help!</p> <p>EDIT : Let me make a few comments up here in reply to Dan's comment. He points out that the result I want is a trivial consequence of the exactness of the inverse image functor. However, I want to emphasize that I am working merely with sheaves of sets! These do not form an abelian category, so it doesn't make sense for a functor to be exact.</p> <p>Of course, I expect that the inverse image functor still takes monomorphisms and epimorphisms to monomorphisms and epimorphisms, which is a weak for of exactness. But the only way I know to prove this (at least for monomorphisms) is to use the result I ask about above!</p>

http://mathoverflow.net/questions/40791/finite-covers-of-complex-varieties-all-but-two-questions-answered Nikita 2010-10-01T22:09:37Z 2010-10-01T22:09:37Z

Thanks Donu! I look forward to seeing your complete answer once you have a chance to write it.

http://mathoverflow.net/questions/40791/finite-covers-of-complex-varieties-all-but-two-questions-answered Nikita 2010-10-01T21:24:47Z 2010-10-01T21:24:47Z

@David : Whoops! I'll add quasiprojective to my hypotheses to make it ok...

http://mathoverflow.net/questions/40576/where-can-i-find-noam-elkies-phd-thesis Nikita 2010-09-30T05:28:02Z 2010-09-30T05:28:02Z

There's this magic thing called email...

http://mathoverflow.net/questions/40576/where-can-i-find-noam-elkies-phd-thesis Nikita 2010-09-30T05:25:32Z 2010-09-30T05:25:32Z

Why don't you ask him? Also, why do you keep posting such poorly thought out questions? Getting your questions downvoted and deleted is a pretty pathetic thrill...

http://mathoverflow.net/questions/40240/elementary-reference-for-algebraic-groups/40244#40244 Nikita 2010-09-28T19:50:39Z 2010-09-28T19:50:39Z

Waterhouse somehow managed to write an entire book on algebraic groups without discussing reductive or semisimple groups. I suppose that he was trying to make some kind of philosophical point, but the result is not helpful for most people who use algebraic groups.

http://mathoverflow.net/questions/40090/prof-murtys-b-sc-thesis Nikita 2010-09-27T15:06:36Z 2010-09-27T15:06:36Z

If you know a library that has it, then you can probably request it via interlibrary loan.

http://mathoverflow.net/questions/39818/the-higher-van-kampen-theorems-and-computation-of-the-unstable-homotopy-groups-of Nikita 2010-09-24T13:53:59Z 2010-09-24T13:53:59Z

I think that it's worth pointing out that algorithms existed long before Brown's work for computing homotopy groups of spheres. The problem is that those algorithms are too slow to be of much practical use. I expect that the same is true for any answer coming from some kind of &quot;higher van Kampen&quot; theorem. There are patterns in the homotopy groups of spheres (eg the chromatic picture of the stable stems), but they are too complicated to arise from any simple description.

http://mathoverflow.net/questions/39538/what-is-the-naming-reason-of-poles-in-complex-analysis Nikita 2010-09-22T19:28:02Z 2010-09-22T19:28:02Z

While the jokes here are pretty tame, the reason that there is whole genre of &quot;Polish jokes&quot; is related to ethnic stereotypes of Poles as stupid or unsophisticated. I think it is pretty offensive to make even non-insulting Polish jokes given the history (for instance, would you tell Jewish or African-American jokes, even ones that weren't insulting?).

http://mathoverflow.net/questions/39572/publishing-with-undergraduates Nikita 2010-09-22T15:09:56Z 2010-09-22T15:09:56Z

R1 and R2 refer to the old Carnegie classification of universities. See here: <a href="http://en.wikipedia.org/wiki/Research_I_university" rel="nofollow">en.wikipedia.org/wiki/Research_I_university</a>

http://mathoverflow.net/questions/39538/what-is-the-naming-reason-of-poles-in-complex-analysis Nikita 2010-09-21T19:58:22Z 2010-09-21T19:58:22Z

Because poles stick up. This seems a bit off-topic for MO. It would be better suited to math.stackexchange.com

http://mathoverflow.net/questions/39308/should-i-check-with-collaborators-before-presenting-unpublished-material/39393#39393 Nikita 2010-09-20T19:05:53Z 2010-09-20T19:05:53Z

I was saying that more because I was wondering about the relevance of that discussion to the question at hand. There's a lot more money involved in medical research. It's a completely different kind of culture.

http://mathoverflow.net/questions/39308/should-i-check-with-collaborators-before-presenting-unpublished-material/39393#39393 Nikita 2010-09-20T15:52:17Z 2010-09-20T15:52:17Z

I don't think there exist any pure math journals that don't allow you to speak about your work before it is published.

http://mathoverflow.net/questions/38639/thinking-and-explaining/38693#38693 Nikita 2010-09-17T19:37:06Z 2010-09-17T19:37:06Z

@Hans : The Cayley-Hamilton theorem is (or, at least, should be) taught in every rigorous undergraduate course in linear algebra, and hence essentially all working mathematicians know about it. It's easier to prove than you might think. Here's one possible set of hints, at least for matrices over C. First, prove it for diagonal matrices (it's easy!). Next, prove it for diagonalizable matrices (almost as easy!). Finally, use the fact that every matrix can be approximated arbitrarly well by a diagonalizable matrix to deduce the general case.

http://mathoverflow.net/questions/38844/cohomological-dimension-of-a-group-acting-on-a-cellular-complex/38854#38854 Nikita 2010-09-16T16:10:33Z 2010-09-16T16:10:33Z

Well, EG is in general infinite-dimensional, but you could take its k-skeleton.

http://mathoverflow.net/questions/38844/cohomological-dimension-of-a-group-acting-on-a-cellular-complex/38859#38859 Nikita 2010-09-15T19:01:22Z 2010-09-15T19:01:22Z

Your link seems to go through your university proxy, so I can't use it. Can you fix that?