Is the respresentation irreducible?

Question

Consider the action of orthogonal matrices of determinant 1, $SO(n)$, on the vector space of $n$ by $n$ symmetric traceless matrices:

$$ R\bullet \Sigma = R \Sigma R^T $$

I am wondering if, or for what $n$, this respresentation is irreducible. Is this a standard result?

EDIT:

I can show that there are no invariant subspaces of dimension 1. Indeed, assume that there is $\Sigma \neq 0$ such that $\forall R\in SO(n)$, $$ R\Sigma R^T = \alpha \Sigma $$ Since the action of rotation on matrices is an isometry for the Froebenius norm, we have $\alpha=1$. Then we get $R\Sigma=\Sigma R$ for all $R$. This means that $\Sigma$ is proportional to the identity. Since I consider the null trace matrices, $\Sigma=0$, and we have a contradiction.

What about the irreducibility in general?

Thanks for your help

Answer 1

It is, indeed, irreducible for any $n$. Let me set up notation, and then I'll sketch a proof.

I'll write $V$ for the vector space of $n\times n$ symmetric traceless matrices. Let $W\subseteq V$ be non-trivial and $SO(n)$-invariant. Our goal is to show that $W=V$. Let $w\in W$ be non-zero.

Proposition 1: (Spectral theorem): There is an $R\in SO(n)$ for which $D:=R\bullet w$ is diagonal. In particular, $D\in W$.

Proposition 2: Suppose $D'$ is any diagonal matrix obtained by permuting the diagonal entries of $D$. Then there is an $R'\in SO(n)$ with $R'\bullet D = D'$. In particular, $D'\in W$.

Proposition 3: The span of all such $D'$ (which is contained in $W$) consists of all traceless diagonal matrices. (Hint: since $Tr(D) = 0$, $D$ cannot be a multiple of $I$. Now use this MSE question.)

These three propositions give the result as follows: given $v\in V$, the spectral theorem implies there is some $R''\in SO(n)$ with $R''\bullet v$ is in the span in Proposition 3. In particular, $R''\bullet v\in W$, so $v\in W$ by $SO(n)$-invariance.