Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

I have a little problem in representation and/or invariant theory which I need help with.

Let's assume $G \leq \mathbb{C}^{n\times n}$ is a finite complex matrix group which operates linearly via $gp\mapsto p \circ g^{-1}$ on the vectorspace of polynomials over $\mathbb{C}$ in $n$ variables and I am given a natural number $k$. How do I compute the character of the corresponding representation if I restrict the operation to the finitely generated vectorspace of homogeneous polynomials of degree $k$? I already know the character of the given $n$-dimensional representation.

Any help would be greatly appreciated.

share|cite|improve this question
Are the elements of $G$ diagonalizable? – Eric O. Korman Feb 21 '12 at 15:44
Since $G$ is a finite matrix group over an algebraicaly closed field of characteristic $0$ the elements should be diagonalizable by default. I should note that I have an explicit example but just computing the character is not an option since $n=248$ and the order of the group is $2^{15}\cdot 3^{10}\cdot ...$. – Sebastian Schoennenbeck Feb 21 '12 at 16:01
is it not enough to give the character in terms of the eigenvalues of the element? – Eric O. Korman Feb 21 '12 at 16:49
If $g\in G$ has eigenvalues $\lambda_i$, then its action on the polynomials has eigenvalues $\lambda_i^{-1}$, and its eigenvalues on the $k$th homogeneous piece are just $k$-fold combinations of these: $\lambda_{i_1}^{-1}\cdots\lambda_{i_k}^{-1}$, where the $i_j$ are not necessarily distinct. – user641 Feb 21 '12 at 18:34
That is definitely correct but I was hoping to a achieve a (perhaps polynomial) term in the character of the $n$-dimensional representation. – Sebastian Schoennenbeck Feb 22 '12 at 10:36

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.