Is the respresentation irreducible? 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
 A: 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.
