Timeline for Are induced morphisms on cohomology strict with respect to the hodge filtration in the non Kähler case?
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| Apr 14, 2017 at 6:13 | comment | added | HYL | You can take $U = X-D$ where $X$ is smooth projective of dimension $n$ and $D$ is a very ample smooth divisor such that $j: H^{n-1}(D) \to H^{n+1}(X)$ is not injective. On the one hand since $U$ is affine, the filtration on $H^n(U)$ induced by the naive filtration on the classical de Rham complex is trivial. On the other hand, the naive filtration on the log-complex defines a mixed Hodge structure (MHS) on $H^n(U)$. Since the residue map $r: H^n(U) \to H^{n-1}(D)$ is a morphism of MHS's and since $Im(r) = \ker (j) \ne 0$, we see that the Hodge filtration on $H^n(U)$ is non-trivial. | |
| Apr 14, 2017 at 4:50 | comment | added | math no more | Do you know an example where the naive filtration on the log $D$ de Rham complex on $X$ doesn't coincide with the naive filtration on the (usual) de Rham complex of $U$ (I think I can see in principle how it could happen)? I think taking the easiest examples $U = \mathbb{A}^1$ or $\mathbb{G}_m$ doesn't seem to work | |
| Sep 19, 2016 at 11:39 | comment | added | jorst | Thank you for this answer! So I guess the "right" way to phrase the question for open manifolds is to consider complex manifolds equipped with an equivalence class of compactifications (and morphisms respecting these) and taking the appropriate definition of the Hodge filtration. For already compact manifolds this would come down to my original question. I will continue to think about this and maybe ask something more specific in another question. | |
| Sep 19, 2016 at 11:34 | vote | accept | jorst | ||
| Sep 17, 2016 at 19:56 | history | answered | HYL | CC BY-SA 3.0 |