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.


1 Answer 1


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).

  • $\begingroup$ 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. $\endgroup$
    – Flounderer
    Commented Jul 17, 2011 at 1:23

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