# Invariant polynomials under dihedral group action

I'm trying to solve the following problem:

Find a generating set for the algebra of invariant polynomials $$\mathbb C[x_1, x_2]^\Gamma$$, where $$\Gamma$$ is a dihedral group $$D_n$$, generated by matrices $$\begin{pmatrix}\zeta && 0 \\ 0 && \zeta^{-1}\end{pmatrix}$$ and $$\begin{pmatrix} 0 && 1 \\ 1 && 0 \end{pmatrix}$$. $$\zeta$$ is a primitive $$n$$th root of unity.

Here is my solution:

Polynomial is invariant under the action of $$\Gamma$$ iff it's invariant under the action of generators. Second generator of $$\Gamma$$ just permutes $$x_1$$ and $$x_2$$. So any invariant polynomial is in fact a symmetric polynomial of two variables. Any such polynomial is of the form $$\sum a_i(x_1 + x_2)^{k_i} + \sum b_i (x_1x_2)^{k'_i} + \sum c_i (x_1 + x_2)^{k''_i}(x_1 x_2)^{k'''_i} + d$$.

It is clear that $$x_1 x_2$$ is invariant under the action of the first generator of $$\Gamma$$. So we only want to check for which $$k$$ it is true that $$(\zeta x_1 + \zeta^{-1} x_2)^k = (x_1 + x_2)^k$$. If we expand both expressions, we can see that it means $$\zeta^{k - 2 l} = 1$$ for all $$l \in [0, k]$$. Since $$\zeta$$ is a primitive $$n$$th root of unity, it follows that $$k - 2 l$$ is a multiple of $$n$$. If $$n = 2$$, $$k$$ can be any even number. In this case, generator set for $$\mathbb C[x_1, x_2]^\Gamma$$ is $$((x_1 + x_2)^2, x_1 x_2, 1)$$.

If $$n > 2$$, such $$k$$ does not exist. Indeed, if $$k = r_1 n$$ and $$k - 2 = r_2 n$$ then $$(r_1 - r_2)n = 2$$ which is impossible. Thus generating set for $$\mathbb C[x_1, x_2]^\Gamma$$ in this case is $$(x_1 x_2, 1)$$.

Am I right? Am I missing something?

Thanks!

• Looks OK to me. Apr 4, 2020 at 18:32

I don't think you are fully right because there is no need for $$(x_1+x_2)^k$$ to be an invariant polynomial.
Since $$D_n$$ is a finite Coxeter Group, By Chevalley's theorem, $$\mathbb{C}[x_1,x_2]^\Gamma$$ is a polynomial algbera over $$\mathbb{C}$$ with two homogeneous generators say $$f_1$$ and $$f_2$$. If $$f_1$$ and $$f_2$$ are with degree $$d_1$$ and $$d_2$$, then $$d_1d_2=2n$$ ( the order of the Dihedral Group). Which is met in the first case but not for the others.
Or course $$f_1=x_1x_2$$ is invariant under $$D_n$$.
Let $$f_2=x_1^n+x_2^n$$. Then it's easy to check that $$f_2$$ is also $$D_n$$ invariant.
Now, since the Jacobian of $$f_1$$ and $$f_2$$, $$J(f_1,f_2)=n(x_2^n-x_1^n)\neq 0$$, it follows that $$f_1$$ and $$f_2$$ are algebraically independent over $$\mathbb{C}$$ satisfying $$d_1d_2=2n$$.
Now by Proposition of 3.12 ( Page no-67) of 'Reflection Groups and Coxeter Groups' by James E Humphreys, it follows that $$\mathbb{C}[x_1,x_2]^\Gamma=\mathbb{C}[f_1,f_2]$$ for any $$n$$.