Let $V$ be a finite dimensional $\mathbb{C}$-representation of a finite group $G$. Then $G$ acts on $V\otimes V$ by $g(v\otimes w)=gv\otimes gw$.

Consider subspaces $\mbox{Sym}^2(V)$ and $\wedge^2(V)$.

These are eigenspaces of the operator $T:V\otimes V \rightarrow V\otimes V$, which takes $v\otimes w$ to $w\otimes v$, with eigenvalues $1,-1$.

It is well-known that these subspaces are $G$-invariant; show that $G$ takes each basis element of the subspace into the subspace.

Question: Is there a basis-free proof of the above result that these subspaces are $G$-invariant?


1 Answer 1


Yes: it's a consequence of the fact that $gT(m)=T(gm)$ for all $g\in G$ and $m\in V\otimes W$. Now take $m \in Sym^2(V)$ and $g\in G$. Then $T(gm)=gT(m)=gm$, hence $gm$ is fixed under $T$, hence in $Sym^2(V)$. The argument for the exterior power is the same: in that case, $T(gm)=gT(m)=-gm$.

Really what's happening is that $T$ and $g$ can be seen as elements of $GL(V\otimes W)$ which commute, and then it's the classic fact that commuting operators have the same eigenspaces.


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