Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer 1

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 $(p,q) \leftrightarrow (d-p,d-q)$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.