Assume that we know the mixed Hodge structure of a complex manifold $X$. Is it possible to compute the mixed Hodge structure of unramified covering $Y$ of $X$ if we know the deck transformation of the covering? 


This is possible if we pass it to the Euler characteristics. If this is not true in general, what information do we need to compute the mixed Hodge structure? 



Is it possible to compute the mixed Hodge structure of a unramified covering space?
If $p\colon Y\to X$ is the projection, the image $p\_\*\mathcal{O}\_Y$ splits into $\deg p$ line bundles $\mathcal{L}\_i$, roughly corresponding to the eigenvalues of $p$. Then the Hodge structure upstairs (split by the eigenvalues) is formed by $H^p(X;\Omega^q(X)\otimes\mathcal{L}\_i)$; you need these data.


If you want mixed Hodge structure (singular or quasi-projective varieties), you should replace $\Omega^q$ with the appropriate sheaves; the rest remains unchanged.


David's remark: of course, this works if the covering is Galois and the deck translation group is abelian; in the non-abelian case, you have more complicated vector bundles corresponding to the irreducible representations of the deck translation, and for an irregular (= non-Galois) covering this seems to be becoming wilds: all one can say is that the higher direct images vanish and you can compute the cohomology downstairs.


Just one last remark: if the deck translation is cyclic, say, $\mathbb{Z}\_n$, then a covering is pretty much the same as an $n$-torsion line bundle $\mathcal{L}$, and then one has $\mathcal{L}\_i=\mathcal{L}^{\otimes i}$, so it's all pretty straightforward. Essentially this goes back (at least) to Zariski (cyclic multiple planes), although he didn't use this language (as i hadn't been developed at the time).


