# Chern classes of free quotient manoflds

Let $X$ be a compact complex manifold. Assume that a finite group acts on $X$ freely. Then the quotient $X/G$ is again a compact complex manifold. I wonder if there is a good way to compute Chern classes of $X/G$ in terms of $X$ and $G$.

The top Chern class is simply the topological Euler number and thus we have $e(X/G)=e(X)/|G|$. What about others? I am particularly interested in the case $\dim X=3,4$.

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Since the tangent bundle $X$ is the pullback of that of $X/G$, you can use the functoriality of the Chern classes, no? – Mariano Suárez-Alvarez Feb 4 '13 at 8:04
Yes, but that does not give us much information except for the top class. – M. K. Feb 4 '13 at 8:49
at least rationally it does give you some information right? the problem is only that in general we do not know what $p^*(c_i(X/G))$ is since we don't know what the projection $p$ induces on cohomology. But rationally $p^*: H^*(X/G;\mathbb{Q}) \to H^*(X;\mathbb{Q})$ is just the inclusion of the $G$-invariants, right? Similarly, Chern Weyl theory should also tell you that (using an equivariant connection) you should be able to compute the chern classes quite explicitely? – mland Feb 4 '13 at 10:41
Nitpick: You need to specify that the finite group acts holomorphically and freely, or else $X/G$ need not be a complex manifold. For example, $S^2$ is a complex manifold, but $S^2/(\mathbb{Z}/2) \cong \mathbb{R}P^2$ is not because $\mathbb{R}P^2$ is not orientable. – Jason DeVito Feb 4 '13 at 13:52
>Jason Thank you for pointing the error. Yes, I need to assume holomorphicity of the action. – M. K. Feb 5 '13 at 23:20