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Let $X$ be a smooth complex algebraic variety and let $\overline{X}$ be a compactification by a divisor $D$ with normal crossings. Then there is a non-canonical isomorphism

\begin{equation} (1) \quad \quad \quad \quad H^k(U, \mathbb{C})=\bigoplus_{p+q=k} H^q(\bar{X}, \Omega^p_{\bar{X}}(\log D)) \end{equation}

where $\Omega^\bullet_{\bar{X}}(\log D)$ is the complex of logarithmic differentials. It is filtered by weight and you see that the weight $m$ Hodge numbers of $H^k(U, \mathbb{C})$ are

$$ \dim H^q(\bar{X}, \mathrm{Gr}^W_m \Omega^p_{\bar{X}}(\log D)) $$

for $p+q=k$. I wonder how this extend to cohomology with compact supports.

I guess in that case one should look at

$$ H^q(\bar{X}, \Omega^p_{\bar{X}}(\log D)(-D)) $$

Is there still a decomposition like (1)? If so, how to prove it?

How does one read compactly supported Hodge numbers from the picture?

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Easiest is maybe to just use Poincaré duality, in the form of the statement that $H^k_c(U) \otimes H^{2d-k}(U) \to H^{2d}_c(U) \cong \mathbf Q(-d)$ (where $d = \dim_\mathbf{C} U$) is a perfect pairing, compatible with the mixed Hodge structures. Then $\dim \mathfrak{gr}_F^p\mathfrak{gr}^W_m H^k_c(U)= \dim \mathfrak{gr}_F^{d-p}\mathfrak{gr}^W_{2d-m}H^{2d-k}(U)$, in other words, the Hodge numbers of $H^k_c$ and $H^{2d-k}$ are related by the transformation $(p,q) \leftrightarrow (d-p,d-q)$.

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Thanks Dan! Could you also explain how tensor by $\mathcal{O}_X(D)$ enters in the story? – arrabal Jul 1 '13 at 11:18
Because $H^q(\bar X, \Omega^q(\log D)(-D))$ and $H^{n-q}(\bar X, \Omega^{n-q}(\log D))$ are Serre dual, and this is compatible with Poincar\'e. – Donu Arapura Jul 1 '13 at 16:35
Thanks Donu! Could you explain a bit more (or give some reference) why this are Serre dual? – arrabal Jul 1 '13 at 20:46

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