Show that $V_{\lambda'} \cong V_\lambda \otimes U$ where $V_\lambda$ is the Specht module and $U$ is the sign representation of $S_n$.

Question

This is Exercise 4.4(c) in Fulton/Harris Representation Theory. Most of the definitions can be found here: https://en.wikipedia.org/wiki/Young_symmetrizer. Let $\lambda \vdash n$. The Specht module $V_\lambda$ is the irreducible representation of $S_n$ defined via the image $\mathbb{C}[S_n]c_\lambda$ of $c_\lambda$ is $\mathbb{C}[S_n]$ where $c_\lambda$ is the Young symmetrizer. I want to show that $$ V_{\lambda'} \cong V_\lambda \otimes U $$ where $U$ is the sign representation and $\lambda'$ is the partition of $n$ conjugate to $\lambda$. As a submodule of the group algebra $\mathbb{C}[S_n]$, we have $U = \mathbb{C} \sum_{\pi \in S_n} sgn(\pi) \pi$.

Can anyone help me show this? I need to define a $\mathbb{C}[S_n]$ module isomorphism but I don't really understand what the elements of the Specht module look like. I have seen a proof using polytabloids, e.g. in James, but I am trying to understand the construction of the Specht modules as submodules of the group algebra instead.

Answer 1

Following Abdelmalek Abdesselam's hint, first define $\Phi:\mathbb{C}[S_n] \to \mathbb{C}[S_n]$ via $\sigma \mapsto \text{sgn}(\sigma)\sigma$ and linear extension. It is an algebra automorphism that maps $V_\lambda \to V_{\lambda'}$. For all $\pi,\sigma \in S_n$, we have \begin{align*}\Phi(\pi\sigma) & = \text{sgn}(\pi) \pi \text{sgn}(\sigma) \sigma\\ & = \text{sgn}(\pi)\Phi(\sigma). \end{align*}

Now define a map $\rho:V_\lambda \otimes U \to V_{\lambda'}$ via $\tau c_\lambda \otimes u \mapsto \Phi(\tau c_\lambda)u$ and linear extension. The codomain is correct since $\Phi(\tau c_\lambda) \in V_{\lambda'}$. This map is $S_n$-equivariant since for any $\pi,\tau \in S_n$ and $u\in U$ we have \begin{align*}\rho(\pi(\tau c_\lambda \otimes u)) & = \rho( \pi\tau c_\lambda \otimes \pi u) & (\text{def. of $S_n$-action on inner tensor product})\\ & = \Phi(\pi \tau c_\lambda) \pi u\\ & = \Phi(\pi \tau c_\lambda)\text{sgn}(\pi) u & (\text{since $U$ is the sign representation})\\ & = \text{sgn}(\pi) \pi \Phi(\tau c_\lambda)\text{sgn}(\pi) u & (\text{def. of $\Phi$})\\ & = \pi \Phi(\tau c_\lambda) u\\ & = \pi \rho(\tau c_\lambda \otimes u). \end{align*}

Answer 2

Let $\Phi:\mathbb{C}[S_n]\rightarrow \mathbb{C}[S_n]$ be the linear map which sends a permutation $\sigma$ to ${\rm sgn}(\sigma)\ \sigma$. It is an algebra automorphism of $\mathbb{C}[S_n]$ which maps the Specht module for $\lambda$ to that for the transpose $\lambda'$. Hint: Is this map $S_n$-equivariant? If not, how can you correct for that?