Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am studying the action of the dihedral group on polynomials and I cannot find an answer to the following question.

Let $V=\mathbb{R}^2$ be the standard representation of the dihedral group where it acts by reflections, rotations etc. What are the irreducible components of $\operatorname{Sym}^k (V^*)$ for $k \ge 1$?

I believe that $V^*$ is just isomorphic to $V$ because it's a reflection group. I also believe that there is an easy way to calculate this, but I don't know it. I tried looking in Fulton & Harris, and a few other representation theory books, but couldn't find an answer. I'd be really grateful for some hints or pointers to relevant literature/textbooks.

share|improve this question

1 Answer 1

up vote 7 down vote accepted

First, you are right that $V \cong V^*$ since this is true for any real reflection group.

My answer will apply to any complex reflection group, but one way to get this is to use a theorem of Chevalley. For notation, let $J_+$ be the ideal generated by all homogeneous positive degree $G$-invariant polynomials.

Theorem (Chevalley). Let $G \subset GL(V)$ be a complex reflection group. Then the ring of invariants $Sym(V^*)^G$ is a polynomial algebra, and the coinvariant algebra $Sym(V^*) / J_+$ is a copy of the regular representation.

We can write $Sym(V^*) = Sym(V^*)^G \otimes (Sym(V^*) / J_+))$, so that we just need 2 pieces of information: what are the degrees of the generators of $Sym(V^*)^G$, and what is the graded decomposition of the coinvariant algebra. The first is well-known: there is a table on Wikipedia and many other places http://en.wikipedia.org/wiki/Complex_reflection_group

For the dihedral group of order $2m$, the degrees are 2 and $m$. For the coinvariant algebra and an irreducible character $\chi$, let $f_\chi(T)$ be the polynomial that incodes the degrees in which $\chi$ appears (this is called fake degree). These have the following expression:

$f_\chi(T) = |G|^{-1} \prod_i (1-T^{d_i}) \sum_{g \in G} \frac{\chi(g)}{\det(1 - Tg)}$

where the determinant is with regards to the action on $V$ and the $d_i$ are the degrees of the generators of the ring of invariants. I couldn't find a more explicit form for the dihedral groups, but I imagine that it simplifies a lot with some thought (which I didn't try to do).

share|improve this answer
    
Thank you! That is exactly what I was looking for. I think I wrote a paper on the coinvariant algebra, so it's a bit embarrassing that I didn't know that. –  Flounderer Jul 17 '11 at 1:23

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.