If $\pi: C \rightarrow \mathbb{P}^{1}$ is a cyclic cover of $\mathbb{P}^{1}$ with Galois group $\mathbb{Z}/m \mathbb{Z}$ and thus with the (affine) formula $y^{m}= (x_{1}a_{1})^{t_{1}}....(x_{n}a_{n})^{t_{n}}$ there is a Galois group decomposition on the cohomolgy $H^{1}(C, \mathbb{C})$ (or more generally on the sheaf $R^{1} \pi_{*} \mathbb{C}$ ) and the dimension of the eigenspace with respect to the character $j \in \mathbb{Z}/m \mathbb{Z}$ is given by the formula $1+ \sum_{i=1}^n < jt_{i}/m> $ (where $< x>$ denotes the franctional part of the number $x$ ) . Now if the Galois group of the covering $\pi: C \rightarrow \mathbb{P}^{1}$ is not cyclic, is there a formula that computes the dimension of the eigenspaces? Or generally how can one compute the dimension of the eigenspaces? For simplicity you can always assume that the covering $\pi$ is abelian (i.e. the Galois group is abelian) and thereof the fiber product of cyclic covers.
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1$\begingroup$ C. Chevalley, A. Weil. Uber das Verhalten der Integrale 1. Gattung bei Automorphismen des Funktionenkorpers, Abh. Math. Sem. Univ. Hamburg 10 (1934), 358361. $\endgroup$ – Felipe Voloch Aug 8 '12 at 9:00

$\begingroup$ Thank you very much, I have seen that paper(though not read it carefully). But isn't it that they treat only the cyclic case? $\endgroup$ – Jack Aug 8 '12 at 9:03

1$\begingroup$ The character of a representation is determined by its restriction to its cyclic subgroups. $\endgroup$ – Angelo Aug 8 '12 at 9:18

1$\begingroup$ There are modern treatments of the ChevalleyWeil formula which might be clearer. You can find them by searching on Mathscinet or Google. $\endgroup$ – Felipe Voloch Aug 8 '12 at 11:09

5$\begingroup$ You can find explicit formulae for the abelian case in [R. Pardini, Abelian covers of algebraic varieties, J. reine angew. Math. 417 (1991), 191213.]. By the way, it is not true that all abelian covers are fiber products of cyclic ones. Usually one has to normalize the fiber product. $\endgroup$ – rita Aug 8 '12 at 14:02