Timeline for Question about Hodge number
Current License: CC BY-SA 4.0
13 events
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Jun 10, 2021 at 16:55 | history | edited | Stefan Kohl♦ | CC BY-SA 4.0 |
Removed broken link which meanwhile pointed to a dubious-looking website.
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Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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Oct 21, 2010 at 6:54 | comment | added | Sándor Kovács | @YCho: the upper semicontinuity of Hodge numbers is in for example C.Voisin, Hodge Theory and Complex Algebraic Geometry, I. I don't have the book with me right now, but it is at around p.230. I think there may even be a section named after this. As for being a topological invariant, I am not entirely sure how your proof would go, but it is not true. It is not even a differentiable invariant. See this post: mathoverflow.net/questions/42744/… | |
Oct 21, 2010 at 2:37 | comment | added | Yunhyung Cho | Dear Gunnar and Sándor, I found some statement about the question 1. Salamon-McDuff "J-holomorphic curves and symplectic topology" In the middle of page 335, the beginning of third paragraph, They state that "On any Kahler manifold, the set of Kahler forms is a convex cone, and so any two such forms can be joined by a deformation" Then does it mean that Hodge numbers (classical) are topological invariant? ----------- And could you tell me the reference about the upper-semicontinuity of Hodge numbers? | |
Oct 20, 2010 at 5:35 | history | edited | Sándor Kovács | CC BY-SA 2.5 |
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Oct 19, 2010 at 20:39 | comment | added | Johannes Ebert | If all fibres are Kähler, if the metric varies continuously and if the base is connected, then the Hodge numbers are constant. Proof: $h^{p,q}$ is the dimension of the kernel of the Fredholm operator $\Delta$ from $(p,q)$-forms to itself. In particular, the Hodge number is upper semi-continuous. But the sum of the Hodge numbers is the Betti number, i.e. a topological invariant. So the Hodge numbers are continuous. | |
Oct 19, 2010 at 8:44 | comment | added | Donu Arapura | Sándor, I agree with you. If the base is connected, a locally constant function is constant. So for an everywhere Kähler family, there should be no problem. | |
Oct 19, 2010 at 7:24 | history | edited | Sándor Kovács | CC BY-SA 2.5 |
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Oct 19, 2010 at 6:47 | comment | added | Sándor Kovács | Dear Gunnar, it seems that it still follows that if all fibers of a family are Kähler, then the Hodge numbers are constant and a jump can happen only at non-Kähler points in the moduli space. Right? | |
Oct 19, 2010 at 6:26 | history | edited | Sándor Kovács | CC BY-SA 2.5 |
added 33 characters in body; added 3 characters in body
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Oct 19, 2010 at 6:22 | comment | added | Sándor Kovács | Dear Gunnar: Yes, of course. Thanks. I was sloppy (and I am not Hodge theorist) and I will edit it. It may not have been clear, but what I was trying to say was that even though there are cases when the Hodge numbers are constant I can't believe they would be topological invariants. After all the miracle about the Hodge decomposition is that it connects topological information (the Betti numbers) to analytic information (the Hodge numbers). Anyway, this just adds to the argument that they are not topological invariants. Thanks! | |
Oct 19, 2010 at 5:50 | comment | added | Gunnar Þór Magnússon | Dear Sándor: The Hodge numbers of manifolds in a family are constant in a small neighborhood of the moduli of the given Kahler manifold. In general the Hodge numbers are only upper-semicontinuous as functions of the moduli, so a priori they could jump when deformed too much. I don't have an example at hand (I'll see if I can find one later), but I strongly suspect this falls into Murphy's law of deformation theory: anything that can go wrong, will go wrong in a properly chosen nicely behaved example. | |
Oct 19, 2010 at 4:34 | history | answered | Sándor Kovács | CC BY-SA 2.5 |