Timeline for Algebraic K-theory of complex varieties
Current License: CC BY-SA 3.0
21 events
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| Apr 11, 2015 at 9:28 | history | edited | Andrei Halanay | CC BY-SA 3.0 |
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| Apr 11, 2015 at 8:47 | comment | added | Andrei Halanay | @DavidSpeyer You're right. Once again it's clear that the analytic topology is not good for algebraic geometry. I am interested in fact by complex manifolds, but given that a similar question: mathoverflow.net/questions/199210/… was left answered I was hoping that in a more "algebraic" setting there was some work done. | |
| Apr 10, 2015 at 18:18 | comment | added | David E Speyer | But then isn't this just asking the question in the Zariski topology? | |
| Apr 10, 2015 at 17:55 | comment | added | Andrei Halanay | @DavidSpeyer I assume $U$ to be also a variety. | |
| Apr 10, 2015 at 14:47 | comment | added | David E Speyer | It seems that you have attracted experts much more knowledgeable than I am, but I am still curious: If $U$ is an open subset of $\mathbb{P}^1$, for example, $\{ (x:y) : |x| < |y| \}$, what is the algebraic $K$-theory of $U$? | |
| Apr 10, 2015 at 10:10 | review | Close votes | |||
| Apr 10, 2015 at 14:39 | |||||
| Apr 10, 2015 at 9:04 | history | edited | Andrei Halanay | CC BY-SA 3.0 |
Expanded the question
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| Apr 9, 2015 at 20:34 | comment | added | Denis Nardin | @Andrei You could try to look at the proof of the localization theorem in Thomason-Trobaugh and see if it carries through in the analytic setting. I cannot think of any obvious obstruction, so that's at least a positive sign. The proof is fairly complicated though, so maybe it isn't worth it. | |
| Apr 9, 2015 at 19:34 | comment | added | Andrei Halanay | @DenisNardin Indeed the étale and complex topologies are in some sense equivalent (this is Th. 21. 2 from Milne's Lectures on Etale Cohomology. But K-theory satisfies Mayer-Vietoris for the Nisnevich topology which is finer than the Zariski one and it is not clear to me the relationship between Nisnevich and complex topologies. | |
| Apr 9, 2015 at 19:18 | comment | added | Andrei Halanay | @DavidSpeyer I was thinking about the complex topology as defined in Neeman's book Algebraic and Analytic Geometry Chap. 4 (seems like (1)-the topology induced by the embedding in $\mathbb{P}^n$). | |
| Apr 9, 2015 at 18:03 | comment | added | David E Speyer | As I (and I think most people) use terminology, "analytic subset" means "closed in topology (2)". GAGA (or rather, Chow's theorem en.wikipedia.org/wiki/… ) states that (2) and (3) agree for projective varieties. I think this might be the source of the confusion. | |
| Apr 9, 2015 at 18:01 | comment | added | David E Speyer | See Defn 5.8 in massey.math.neu.edu/Massey/Massey_preprints/sing_notes.pdf for a source that uses these terms in this way. | |
| Apr 9, 2015 at 18:01 | comment | added | David E Speyer | In your comment to Alan, you write "by GAGA, an analytic subset of $X$ will be algebraic." If $X$ is an algebraic complex manifold, there are THREE topologies on $X$: (1) The topology where we consider $X$ as an ordinary real manifold, usually called the analytic topology (2) What I've sometimes heard called the "analytic Zariski topology", where closed sets must locally be zeroes of analytic functions (3) The Zariski topology, sometimes called "algebraic Zariski topology" for emphasis, where closed sets are locally zeroes of algebraic functions. | |
| Apr 9, 2015 at 17:56 | comment | added | Denis Nardin | I would be doubtful of the thesis, since algebraic K-theory does not satisfy étale descent, which should be weaker than descent for the analytic topology. | |
| Apr 9, 2015 at 17:39 | comment | added | Andrei Halanay | @David Speyer Sorry, I was asking about projective varieties, but somehow the word was left out. I edited the question accordingly. | |
| Apr 9, 2015 at 17:39 | comment | added | David E Speyer | I see you edited your question to say that you are thinking about projective varieties, but an open subset of a projective variety is almost never projective. So any interesting example of Mayer-Vietores takes you out of the projective world (or else I am confused.) | |
| Apr 9, 2015 at 17:36 | history | edited | Andrei Halanay | CC BY-SA 3.0 |
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| Apr 9, 2015 at 17:29 | comment | added | David E Speyer | @AndreiHalanay I usually understand "analytic topology" to mean the classical topology where we consider $X$ as a real manifold. So, if $X = \mathbb{C}$ with the standard algebraic structure, then the open disc of radius $1$ is open for the analytic topology, but doesn't come with an algebraic structure. If I am confused, could you spell out your definitions? | |
| Apr 9, 2015 at 16:44 | comment | added | Andrei Halanay | I think that by GAGA an analytic subset will be also algebraic. | |
| Apr 9, 2015 at 15:31 | comment | added | Allen Knutson | I'm not sure the question makes sense -- if I take an analytic but not algebraic open set, what is its algebraic K-theory supposed to be? | |
| Apr 9, 2015 at 14:16 | history | asked | Andrei Halanay | CC BY-SA 3.0 |