# hypercohomology of logarithmic de Rham complex of complement of smooth divisor in smooth variety

Let $V$ be a complex manifold and $D \subset V$ a smooth divisor.

Question 1 Is $H^i(V \setminus D, \mathbb{C}) \simeq \mathbb{H}^i ( V, \Omega^{\bullet}_V(\log D))$ ?

Question 2(Edited) Ok, 1 is true. Is it possible to define naturally a homomorphism $H^2(V \setminus D, \mathbb{C}) \rightarrow H^1(V, \Omega_V^1 (\log D))$?

(In my case $V$ is of the form $U \setminus p$ where $U$ is a $3$-dimensional smooth Stein space and $D$ is of the form $\Delta \setminus p$ where $\Delta \subset U$ is a divosor with an isolated singularity at $p$. Then $H^1(V, \Omega_V^1 (\log D))$ is the set of 1st order deformations of the pair $(U, \Delta)$. Since $\Delta$ has only isolated singularities, this is finite dimensional.)

I think it is true when $V$ is compact. How about non-compact case?

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Q1: yes. See Deligne Theorie de Hodge II for the proof. Q2: I don't understand this. What maps of complexes??? Anyway, in some cases, e.g. if $V$ is compact Kahler, you can get something along the lines of what you're suggesting, but for more complicated reasons. –  Donu Arapura Mar 29 '12 at 12:33
Thank you for the comment. I edited Question 2. I realized that the map I considered is not a complex homomorphism. If you have comment for a new question, please let me know. – –  tarosano Mar 29 '12 at 13:07
Thanks for the comment. Do you mean $\ker d: H^1(\Omega^1_V(\log D)) \rightarrow \ldots$? Is it really $H^0$? And I want to ask you one more. Is Question 2 treatable if $H^2( V, \mathcal{O}_V) =0$? –  tarosano Mar 29 '12 at 14:20
Thanks. If you have comment and time, please let me know later. –  tarosano Mar 29 '12 at 16:04
I deleted my earlier comments which were not quite accurate. Anyway, it looks you have what you need modulo the assumption of vanishing of $H^i(X,\mathcal{O})$. –  Donu Arapura Mar 30 '12 at 13:04
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Question 1: Sure, this is true. Another reference (beyond what Donu pointed out) is chapter 8 of Claire Voisin's Hodge theory of .... But the point is $\Omega_X^{\bullet}(\text{log} D)$ is quasi-isomorphic to the pushforward of a resolution of $\mathbb{C}$ on $X \setminus D$. By the way, this holds not just for smooth $D$, but also for normal crossing $D$.

Question 2: I don't think you have maps of complexes as you describe. For example, why do we have the map $\Omega_V^1(\text{log} D) \to \Omega^{\bullet}_V(\text{log} D)$? If I had a map of complexes, then the image of $d : \Omega_V^0(\text{log} D) \to \Omega_V^1(\text{log} D)$ would be zero (ie, the diagram would commute).

EDIT: Whoops, it looks like Donu beat me to this in the comments.

Revised Question 2: I don't see why this should hold in general. However, if you write down the relevant spectral sequence, and enough terms vanish (maybe the spectral sequence degenerates), you can be ok.

EDIT (Response to the comment below): No, there isn't a map in general. Even for a projective variety and $D = 0$, we only have an $E^1$ degeneration of the spectral sequence. This means that in some sense, $H^2(X, O_X)$, $H^1(X, \Omega_X^1)$ and $H^0(X, \Omega_X^2)$ make up ${H}^2(X, \mathbb{C})$ (there is a filtration of the latter such that these terms make up the filtration). But we have maps:

$$H^2(X, \mathbb{C}) \to H^2(X, O_X), \text{ and } H^0(X, \Omega_X^2) \to H^2(X, \mathbb{C}).$$

There isn't going to be a map to the $H^1(X, \Omega_X^1)$ term in general, unless for some reason $H^2(X, O_X) = 0$ (in the non-compact/non-Kahler setting, things get more complicated as Donu mentioned above). Anyways, if you read a little about spectral sequences and do a couple examples from that perspective, I'm sure you'll see what's going on.

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Thank you for the comment. If you some comments on a new question, please let me know. –  tarosano Mar 29 '12 at 13:07
Thank you for the revise. Actually, what I really want to know is the construction of the homomorphism. I don't mind about the direct summand question. Is that homomorphism not natural? –  tarosano Mar 29 '12 at 13:49
Thanks. In my case, I can assume $H^2(V, \mathcal{O}_V) =0$. So it's OK for me. –  tarosano Mar 29 '12 at 20:47
Great! Just notice that I need some conditions in order to use this (to know that the spectral sequence degenerates). –  Karl Schwede Mar 30 '12 at 0:20
Moreover, I can assume that $H^1(V, \mathcal{O}_V) =0$. Then there is a map $H^2(V \setminus D, \mathbb{C}) \rightarrow {\rm Gr}_F^1 H^2 \subset H^1(V, \Omega^1_V(\log D))$. $\subset$ is implied by $H^1(\mathcal{O})=0$. Do you mean such a condition? –  tarosano Mar 30 '12 at 10:06