Timeline for Analytic tools in algebraic geometry
Current License: CC BY-SA 3.0
29 events
| when toggle format | what | by | license | comment | |
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| S Apr 26, 2017 at 6:53 | history | bounty ended | Gil Kalai | ||
| S Apr 26, 2017 at 6:53 | history | notice removed | Gil Kalai | ||
| Apr 20, 2017 at 23:51 | answer | added | Will Sawin | timeline score: 5 | |
| S Apr 19, 2017 at 20:21 | history | bounty started | Gil Kalai | ||
| S Apr 19, 2017 at 20:21 | history | notice added | Gil Kalai | Draw attention | |
| Oct 3, 2013 at 15:35 | answer | added | Koushik | timeline score: 5 | |
| Apr 20, 2011 at 21:46 | history | edited | Tom Goodwillie | CC BY-SA 3.0 |
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| Sep 22, 2010 at 16:56 | comment | added | Tom Goodwillie | That works. A regular point $x$ in an $n$-dimensional variety $X$ has a Zariski nbhd $U$ admitting an etale map $U\to\mathbb A^n$ with $x\mapsto 0$. If $(Y,y)$ is another $n$-dimensional variety with regular point then do the same with a suitable nbhd $V$ of $y$. Now form the fiber product of $U$ and $V$ over $\mathbb A^n$. (If you want it irreducible, take the component containing the point you care about.) Something like that. | |
| Sep 22, 2010 at 15:50 | comment | added | Michael Bächtold | The definition of locally isomorphic I had in mind, is that there is a third variety which is etale over both and has the two points under consideration in the image of the maps. But maybe that is to strong? | |
| Sep 21, 2010 at 14:05 | comment | added | Tom Goodwillie | Yes, under a reasonable definition of "locally isomorphic". Maybe even under more than one reasonable definition of "locally isomorphic". I have to do some work now, but will be happy to continue this conversation less cryptically later. | |
| Sep 21, 2010 at 12:09 | comment | added | Michael Bächtold | Tom: thanks for your answer (and sorry for coming back only now to this thread). I feel like asking an even more basic question after your answer: does it become true in the etale topology, that any two varieties of the same dimension are locally isomorphic near regular points? | |
| Aug 14, 2010 at 12:51 | history | edited | Tom Goodwillie | CC BY-SA 2.5 |
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| Aug 13, 2010 at 19:35 | history | edited | Tom Goodwillie | CC BY-SA 2.5 |
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| Aug 8, 2010 at 21:48 | answer | added | Daniel Pomerleano | timeline score: 7 | |
| Aug 6, 2010 at 23:13 | comment | added | Steve Huntsman | mathoverflow.net/questions/10667/… | |
| Aug 6, 2010 at 21:27 | comment | added | Simon Pepin Lehalleur | Although this is a very broad question, I hope some people will give it a try... I am more and more convinced that the importation of concepts from analysis (smooth on the one hand and complex analytic on the other) into algebraic geometry is one of its most subtle aspect. | |
| Aug 6, 2010 at 18:01 | comment | added | Tom Goodwillie | Michael: No. Write $dy=ydx$. A transcendental curve $y=e^x$ is not going to become algebraic when you pull it back along an etale surjection to the $x,y$ plane. | |
| Aug 6, 2010 at 17:18 | comment | added | Michael Bächtold | I also like this question. Regarding the ODE example here is one related question: does the Frobenius theorem become true in the etale topology for algebraic varieties: i.e. are any two involutive distributions of the same dimensions "locally" isomorphic? | |
| Aug 6, 2010 at 8:07 | answer | added | algori | timeline score: 36 | |
| Aug 6, 2010 at 5:48 | comment | added | Eugene Eisenstein | I really hesitate to try to answer any of this. All I'll say is, in birational geometry, the analog of Sard's theorem and moving lemmas works well in the vanilla Zariski topology. The notion of "general" and "very general" seems adequate so far. For uncountable fields, the complement of a countable union of hypersurfaces is dense is an algebraic Baire category theorem. Of course, over countable fields like $\overline{\mathbb{Q}}$ the matter is very different. But then some of the things that follow from arguments about very general points are expected to fail in that setup, eg cycles. | |
| Aug 6, 2010 at 5:05 | comment | added | Kevin H. Lin | I mean, I think the algebraic dR complex is not a resolution of the constant sheaf, since for instance $\log z = \int \frac{1}{z} dz$ is transcendental. So, right--the Poincare lemma does not hold. But we can view the Poincare lemma as saying "the dR complex is a resolution of the constant sheaf, so we can use it to compute cohomology of the constant sheaf". So defining algebraic dR cohomology via hypercohomology of the dR complex is kind of like saying "pretend the Poincare lemma is true--pretend that the algebraic dR complex is a resolution of the constant sheaf". Theorem ~> Definition. | |
| Aug 6, 2010 at 4:43 | comment | added | Tom Goodwillie | some: I wonder what I mean, too. | |
| Aug 6, 2010 at 4:42 | comment | added | Tom Goodwillie | Kevin: That kind of algebraic de Rham cohomology doesn't satisfy Poincare Lemma unless you're over $\mathbb Q$. | |
| Aug 6, 2010 at 4:35 | comment | added | some guy on the street | I wonder what you mean for the ODEs, because we know we can't stay in the algebraic category; but then there is all that differential Galois theory, that you get by expanding the language and repeating the thm => def'n ploy. | |
| Aug 6, 2010 at 4:24 | comment | added | Kevin H. Lin | I think the "Poincare Lemma" in algebraic geometry is done via the "French trick": We simply define algebraic de Rham cohomology to be the hypercohomology of the de Rham complex $(\Omega_X^\bullet,d)$. | |
| Aug 6, 2010 at 4:23 | comment | added | Kevin H. Lin | I have always wondered whether and how one can integrate vector fields in algebraic geometry... | |
| Aug 6, 2010 at 3:29 | comment | added | B. Bischof | Beautiful question, touches on a question I have been brewing for weeks. | |
| Aug 6, 2010 at 2:05 | history | edited | Tom Goodwillie | CC BY-SA 2.5 |
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| Aug 6, 2010 at 1:55 | history | asked | Tom Goodwillie | CC BY-SA 2.5 |