MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a compact Riemann surface and $G = Aut(X)$ be its group of automorphisms (biholomorphisms between $X$ and $X$). It is known that $G$ acts on the space $Harm(X)$ of all harmonic forms and also on the space $Omega(X)$ of all holomorphic forms.

We know that $Harm(X)$ is a direct sum of $Omega(X)$ and its conjugate.

Now, if we know the representation of $G$ on $Harm(X)$ (by this I mean we have a matrix for each element of $G$), how can we find matrices for the representation of $G$ on $Omega(X)\ ?$

ADDED: I am assuming we have no information about $Omega(X)$ or $Harm(X)$. We just know the genus of $X$, $G$ (with a multiplication table) and the representation of $G$ on $Harm(X)$ (the matrices associated to each element of the group).

share|cite|improve this question
1. Is it obvious that the isomorphism class of Omega(X) as a C[G]-module is determined by the isomorphism class of Harm(X)? 2. On the other hand: when you say you have "matrices associated to each element of the group," it sounds like you have more than a description of the isomorphism class of Harm(X) as C[G]-module, but rather have the action of G on an explicit basis of harmonic forms. Or is what you have the action of G on an explicit basis for the first homology? 3. In any event, I believe this issue was discussed in De Jong & Noot's paper "Jacobians with complex multiplications." – JSE Jul 22 '10 at 19:18
1. I don't know but I saw this claim in Streit's paper "Homology, BelyÄ­ functions and canonical curves" 2. I really just have the matrices. I don't have a basis for any of the spaces. 3. Thanks for the reference. I will take a look at it. – expmat Jul 22 '10 at 19:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.