I have a problem: Let $V$ be an $n$-dimensional complex vector space and let $B=\{e_1,e_2,...,e_n\}$ denote the elements of a chosen basis. Let $\rho:G \to GL(V)$ be an irreducible representation. Let $T:V \otimes V \to V \otimes V$ be the automorphism defined by $T(e_i \otimes e_j)=e_j \otimes e_i$. Let $Sym^2(V)=\{z \in V \otimes V: T(z) = z\}$ and $Alt^2(V)=\{z \in V \otimes V : T(z) = -z\}$.

Prove that $Sym^2(V)$ is a vector subspace of $V \otimes V$ of dimension $\frac{n(n+1)}{2}$. I have no idea how to prove the latter part (dim is $\frac{n(n+1)}{2}$). I know that in general, the dimension of a vector space is the number of basis vectors. So, in this case, the dimension is the number of basis vectors of $Sym^2(V)$. I have the following hint: $e_1 \otimes e_2 + e_2 \otimes e_1 \in Sym^2(V)$. I'm not asking for a handout, but can anyone point me in a direction/give me a tool that I can use to get this done? Thanks in advance.

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    $\begingroup$ The elements $s_{ij} := e_i \otimes e_j + e_j \otimes e_i$, $t_i := e_i \otimes e_i$, and $u_{ij} := e_i \otimes e_j - e_j \otimes e_i$ together form a basis of $V \otimes V$. We have $s_i, t_i \in {\rm Sym}^2(V)$, and $u_i \in {\rm Alt}^2(V)$. Use this and the fact that ${\rm Sym}^2(V) \cap {\rm Alt}^2(V) = \{0\}$. $\endgroup$ – Derek Holt May 3 '13 at 10:59

The idea is to decompose $V \otimes V$ as a sum of eigenspaces of $T$. Since $T^2$ is the identity map, the only possible eigenvalues of $T$ are $\pm 1$. The $+1$ eigenspace is precisely $Sym^2(V)$ and the $-1$ eigenspace is precisely $Alt^2(V)$. The elements described in Derek's comment above give bases of these two eigenspaces. Once you check that there are exactly $n^2$ of these elements (and they are linearly independent), you know that they give bases (since the dimension of $V \otimes V$ is $n^2$). Then one simply has to count basis elements to get the dimension you're interested in.


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