Timeline for Hodge decomposition and degeneration of the spectral sequence
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10 events
| when toggle format | what | by | license | comment | |
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| Sep 24, 2022 at 7:38 | comment | added | Doug Liu | Personally I think there is a typo in the example of Hopf surface. Its Hodge number should be $h^{0,1}=1$ and $h^{1,0}=0$. In particular, $H^0(X,\Omega_X^1)=0$. | |
| Oct 1, 2018 at 4:33 | comment | added | Alexander Braverman | I don't mind functional analysis. What I in some sense do mind is that the theory of harmonic forms requires not only a choice of the Kahler but also working with a particular (Dolbeault) resolution of the de Rham complex. My feeling is that in a "good" theory no particular resolution should appear. The statement which implies the resolution is very clear - it just says that the Hodge filtration can be split by means of its complex conjugate. It seems very unfortunate to me that you can't prove such a fact without working with a particular resolution. | |
| Sep 30, 2018 at 19:12 | history | edited | gdb | CC BY-SA 4.0 |
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| Sep 30, 2018 at 6:06 | comment | added | gdb | Also look at this question mathoverflow.net/questions/28265/…. It seems that rather deep (functional) analysis is required in any of these approaches. | |
| Sep 30, 2018 at 5:55 | comment | added | gdb | ...canonical decomposition $H_{et}^n(X\otimes_{K} \bar{K}, \mathbf Q_p)\otimes \mathbf C_p \cong \oplus_{i+j=n}H^i(X,\Omega^j_{X/K})\otimes_{\mathbf Q_p}\mathbf C_p(-j)$ but there is no such canonical decomposition for schemes over $\mathbf C_p$. In this situation the action of Galois group helps to obtain a splitting. So, it is unlikely that one can prove the Hodge decomposition without any further input from analysis (or reduction to the $p$-adic case. But this will not prove hodge symmetry). | |
| Sep 30, 2018 at 5:52 | comment | added | gdb | @AlexanderBraverman Sorry, I partially misunderstood your question. I thought that your main question is "to what extent the 1st statement implies the 2nd". I tried to say that it unlikely that there might be an algebro-geometric proof of the Hodge decomposition. One needs some inpute outside of algebraic geometry to split the Hodge filtration (like harmonic forms in complex geometry or Galois representation in the $p$-adic version). For example, for proper smooth schemes over a finite extension $K/\mathbf Q_p$ one has a canonical decomposition ... | |
| Sep 30, 2018 at 2:01 | comment | added | Alexander Braverman | Thank you. The first part (what is written before the Theorem) is more or less what I wrote originally (in particular the Hopf surface is precisely the example that I meant when I wrote "If $X$ is not necessarily Kahler then it might happen that the spectral sequence degenerates but the Hodge decomposition fails). So, it is clear that the degeneration of the spectral sequence alone doesn't imply Hodge decomposition, but the question is whether it is possible to prove the latter in reasonable generality without using harmonic forms. I mentioned non-projective varieties just as an example. | |
| Sep 30, 2018 at 0:24 | history | edited | gdb | CC BY-SA 4.0 |
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| Sep 30, 2018 at 0:14 | history | edited | gdb | CC BY-SA 4.0 |
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| Sep 29, 2018 at 22:22 | history | answered | gdb | CC BY-SA 4.0 |