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.



Are there any known formulas about the Hodge-Deligne structure of quotients by actions of groups?
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.


