3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.