$$a(h, a(g, x))=a(h g, x) \quad \text { and } a(e, x)=x .$$

Problem 1.

For any $g \in G$,
$\chi_{\operatorname{Sym}^2 V}(g) & =\left[\chi_V(g)^2+\chi_V\left(g^2\right)\right] / 2$，
$\chi_{\Lambda^2 V}(g) & =\left[\chi_V(g)^2-\chi_V\left(g^2\right)\right] / 2 .$
\end{aligned}


Proof. Having fixed $g$, we choose a basis of $V$ on which $g$ acts diagonally, with eigenvalues $\lambda_1, \ldots, \lambda_d$, repeated as necessary. From our bases of $\operatorname{Sym}^2 V, \Lambda^2 V$, we see that the eigenvalues of $g$ on the two spaces are $\lambda_p \lambda_q$, with $1 \leq p \leq q \leq d$ on Sym ${ }^2$ and with $1 \leq p<q \leq d$ on $\Lambda^2$. Then, the proposition reduces to the obvious identities
$2 \sum_{p \leq q} \lambda_p \lambda_q & =\left(\sum \lambda_p\right)^2+\sum \lambda_p^2$,
$2 \sum_{p<q} \lambda_p \lambda_q & =\left(\sum \lambda_p\right)^2-\sum \lambda_p^2$.

Problem 2.

1) Infinite dimensional Schur lemma. Let $A$ be a countable dimensional algebra over an uncountable algebraically closed field $\mathbb{F}$. Let $V$ be an infinite dimensional irreducible $A$-module. We want to prove that $\operatorname{End}_A(V)=\mathbb{F}$.
1) Show that $V$ is countable dimensional .
2) Show that $\operatorname{End}_A(V)$ is at most countable dimensional.
3) Now let $\varphi \in \operatorname{End}_A(V)$ be a non-constant element. Show that $\varphi-z$ is invertible for any $z \in \mathbb{F}$. Show that the elements $(\varphi-z)^{-1}, z \in \mathbb{F}$, are linearly independent.
4) Prove that $\operatorname{End}_A(V)=\mathbb{F}$ .

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