## A fact about the Alt map on a tensor power

Let $F$ be a field in which $k!$ is a unit and let $V$ be an $F$-vector space. Recall that $S_k$ acts on the tensor power $\mathcal{T}^k(V)$ by permuting the components, and that $\mathsf{Alt}_k$ is defined on $\mathcal{T}^k(V)$ by $\mathsf{Alt}_k(z) = \sum_{\sigma \in S_k} \epsilon(\sigma) \sigma z$. Prove that $\mathsf{im}\ \mathsf{Alt}_k$ is the unique largest subspace of $\mathcal{T}^k(V)$ on which each permutation $\sigma \in S_k$ acts as multiplication by $\epsilon(\sigma)$.

Suppose $z \in \mathcal{T}^k(V)$ such that $\sigma z = \epsilon(\sigma) z$ for all $\sigma \in S_k$. Then $\mathsf{Alt}_k(z) = \frac{1}{k!} \sum_{\sigma \in S_k} \epsilon(\sigma) \sigma z$ $= \frac{1}{k!} \sum_{\sigma \in S_k} \epsilon(\sigma) \epsilon(\sigma) z$ $= \frac{1}{k!} \sum_{\sigma \in S_k} z$ $= z$, so that $z \in \mathsf{im}\ \mathsf{Alt}_k$.

In particular, any subspace of $\mathcal{T}^k(V)$ upon which every permutation $\sigma \in S_k$ acts as scalar multiplication by $\epsilon(\sigma)$ is contained in $\mathsf{im}\ \mathsf{Alt}_k$.

Note also that $\sigma \in S_k$ acts on $\mathsf{im}\ \mathsf{Alt}_k$ as multiplication by $\epsilon(\sigma)$ since $\mathsf{im}\ \mathsf{Alt}_k \cong_F \bigwedge^k(V)$.