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I have a question on a proof from the following book :"Symmetry, representations and invariants by Roe Goodman and Nolan R. Wallach. Specifically on section 4.2.4 on Schur Weyl duality and the proof of the following statement

There are irreducible mutually inequivalent $\mathfrak{S}_k$-modules $E^\lambda$ and irreducible mutually inequivalent $G :=GL(n,\mathbb{C})$-modules $F^\lambda$ such that $$ \bigotimes^k \mathbb{C}^n \cong \bigoplus_{\lambda} E^\lambda \otimes F^\lambda$$ as a representation of $\mathfrak{S}_k \times G$

In the book this is presented as a specific case of the the general duality theorem

I know from this theorem that there are irreducible, mutually inequivalent $\mathcal{R}^G$-modules $E^\lambda$ and irreducible, mutually inequivalent $G := GL(n,\mathbb C)$-modules $F^\lambda$ such that $$ L := \bigotimes^k \mathbb C^n \cong \bigoplus_\lambda E^\lambda \otimes F^\lambda$$ as an $\mathcal R^G \otimes \mathbb CG$-module where $$ \mathcal R^G := \{ T \in End(L) \vert \rho_k(g)T = T \rho_k(g) \quad \forall g \in G \}$$ is the commutant of $\rho_k(G)$ in $End(L)$ i.e. $R^G = Comm(\rho_k(G))$. Note that $\rho_k$ denotes the representation of $G$ on $L$

This is quite close to the final theorem. I need only prove that an $R^G$-module is the same as an $\mathbb C\mathfrak{S}_k$ module. Using the following theorem from the book

Theorem 4.2.10 Let $\rho_k$ and $\sigma_k$ denotes the representations of $GL(n,\mathbb C)$ and $\mathfrak S_k$ on $\otimes^k \mathbb C^n$ respectively. Then $$ Comm(Span(\rho_k(G))) = Span(\sigma_k\mathfrak S_k)$$ and $$ Comm(Span(\sigma_k \mathfrak S_k)) = Span \rho_k(G)$$

I can see that that $\mathcal R^G = Span( \sigma_k \mathfrak S_k)$ which is almost what I need. How can I conclude from here ?


Attempts:

A first idea would be to prove that $Span \; \sigma_k \mathfrak S_k \cong \mathbb C \mathfrak S_k$ as algebras. To do this I can consider the map

$$ \sum_{s \in \mathfrak S_k} \mu_s s \mapsto \sum_s \mu_s \sigma_k(s)$$ and show that it is an algebra isomorphism. It is clearly surjective but I don't see how I could easily show injectivity. Suppose that $\sum_s \mu_s \sigma_k(s) = 0$ then to prove injectivity it suffices to show that the linear maps $\sigma_k(s)$ are linearly independent. But I don't see why this would be obvious.


A second idea would be to use the $R^G$ module structure on $E^\lambda$ to define a representation of $\mathfrak S_k$ on $E^\lambda$ by letting $s$ act on $E^\lambda$ by mapping $v$ to $\sigma_k(s)v$. This does turn $E^\lambda$ into an irreducible $\mathbb C\mathfrak S_k$-module but how can I be sure that these new $\mathbb C \mathfrak S_k$ modules are still mutually inequivalent ?

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1 Answer 1

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To complete the proof one must construct an $\mathbb C\mathfrak S_k$-module structure on $E^\lambda$. This can be done by using the $Span\; \sigma_k(\mathfrak S_k)$-module structure on $E^\lambda$.

$$ \mathbb C \mathfrak S_k \rightarrow End(E^\lambda) : \sum_{s \in \mathfrak S_k} \lambda_s s \mapsto \sum_{s \in \mathfrak S_k} \lambda_s \sigma_k(s).$$

In this way the $E^\lambda$ are irreducible and non isomorphic as representation of $\mathfrak S_k$ since

$$ E^\lambda \cong E^\mu \text{ as }\mathbb C \mathfrak S_k\text{-modules} \implies E^\lambda \cong E^\mu \text{ as Span } \sigma_k(\mathfrak{S}_k)\text{-modules.}$$

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