# $G$-invariant subspaces of tensor square

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?

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.