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I would like to know if there are explicit formulas for the Hodge-Deligne structure or the Hodge-Deligne polynomials for quotients X/G for finite groups G acting on a (smooth, projective) scheme X.

The only formula that I know is lemma 2.6 on this paper

http://arXiv.org/abs/math/0701642v1

by Munoz, Ortega and Vazquez-Gallo.

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Thank you all for the answers! –  Matteo Feb 2 '12 at 11:15

3 Answers 3

up vote 4 down vote accepted

It's hard to give a very precise answer to such a general question. But perhaps I can try to complement the answers already given with some more specific examples/tricks. As Sándor and algori have pointed out, $X/G$ is essentially smooth for what you seem to be after. In principle there are many known formulas, although they may occur in a disguised form.

  1. Perhaps the simplest example to think about is when $X$ is a smooth projective curve then the only thing one needs to know is the genus of the quotient $X/G$. This can be written down easily using Riemann-Hurwitz. For example, when $G$ acts effectively without fixed points $1-g(X)= |G|(1-g(X/G))$, and in fact the virtual character of $H^0(X,O_X)-H^1(X,O_X)$ is $1-g(X/G)$ times the regular representation.

  2. In general, you want to compute the invariant part of $H^q(X,\Omega_X^p)$ as algori has already suggested. To do this, you can try to use a combination of holomorphic Lefschetz trace and Riemann-Roch to get you hands on the virtual character $\chi(X,\Omega_X^p)\in K_0(\mathbb{C}[G])$. In the fixed point free case, it should again be a multiple of the regular representation. At the risk of self advertising, you can look at my recent post to the arXiv for further details.

  3. For some more concrete calculations, you can take a look at Dolgachev's article on weighted projective spaces (in Springer LN 956). He gives formulas for Hodge numbers of hypersurfaces in them. These can be realized as quotients of ordinary hypersurfaces by finite groups. Although he doesn't base his calculations on this fact, one could try to do it that way as an exercise.
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The rational cohomology of the quotient of a variety $X$ by an action of a finite group $G$ is the $G$-invariant part of $H^*(X,\mathbb{Q})$; this is a Hodge substructure if $G$ acts by automorphisms. So one can deduce the Hodge polynomial of the quotient from the $G$-equivariant Hodge polynomial of $X$.

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Moreover, it seems that $\dots\to X\times G\times G\to X\times G\to X$ is a proper simplicial hyperresolution of $X/G$. So there is a spectral sequence $H^i(G^j\times X)\implies H^{i-j}(X/G)$, that also works on the level of Hodge structures. On the rational level one obtains the $G$-invariant part of the cohomology of $X$. –  Mikhail Bondarko Jan 31 '12 at 6:34
    
But if there is such a spectral sequence, then the Hodge structure of $X/G$ would depend only on $X$ and $G$ and not on the action of $G$ on $X$. This sounds quite strange for me. –  Matteo Feb 1 '12 at 11:44
    
Matteo -- unless I'm mistaken, this spectral sequence computes the equivariant cohomology rather than the ordinary cohomology (since all stabilizers are finite, there is no difference between the two if the coefficients are rational). If so, the second term of the spectral sequence will contain exactly one non-zero column, the 0-th one, which will contain the invariants of the action. –  algori Feb 1 '12 at 12:10
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Also the maps of the simplicial scheme $X\times G\ldots$, and therefore the differentials of the spectral sequence, do depend on the action. So there is no contradiction. –  Donu Arapura Feb 1 '12 at 14:38
    
Ok, that convince me! It was my fault, sorry! –  Matteo Feb 2 '12 at 11:13

It is possible to do Hodge theory on singular varieties. A good reference for that is LNM 1335. An interesting related area is the study of Du Bois singularities, which are (approximately) defined by requiring that the 0th step of the Hodge filtration is given the same way as in the smooth case. It turns out that rational singularities are Du Bois and quotients by finite groups are rational, so your space has Du Bois singularities.

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