## Product of Representations

Recall that the Frobenius map gives an isomorphism of abelian groups:

Let us compute what the product $\Lambda^{(d)} \times \Lambda^{(e)} \to \Lambda^{(d+e)}$ corresponds to on the RHS.

For that, we take $h_\lambda \in \Lambda^{(d)}$ and $h_\mu \in \Lambda^{(e)}$ where $\lambda\vdash d$ and $\mu\vdash e.$ Multiplication gives $h_\lambda h_\mu = h_{\lambda | \mu}$ where $\lambda|\mu$ is the partition obtained by sorting $(\lambda_1, \lambda_2, \ldots, \mu_1, \mu_2, \ldots).$ Next, we use the following standard result.

Lemma. Let $G$ act transitively on non-empty set $X$. Pick any $x_0 \in X$ and let $H = \{g\in G : gx_0 = x_0\}$ be its stabiliser. Then:

$\mathbb{C}[X] \cong \text{Ind}_H^G \mathbb{C},$

the trivial module on $H$ induced to $G.$

Proof

The RHS is $\mathbb{C}[G] \otimes_{\mathbb{C}[H]} \mathbb{C}$ by definition. Now we map $gx_0\in X$ on the LHS to $g\otimes 1$ on the RHS. Now:

$gx_0 = g'x_0 \iff g^{-1}g' \in H \iff g^{-1}g' \otimes 1 = 1\otimes 1 \iff g\otimes 1 = g'\otimes 1.$

Since the action on X is transitive, this gives an isomorphism of vector spaces, which clearly commutes with the group action. ♦

In particular $U_\lambda = \mathbb{C}[X_\lambda]$ is isomorphic to $\text{Ind}_H^G \mathbb{C}$ where $G = S_d$ and $H$ is the stabiliser group of any partition of [d] into disjoint sets of sizes $\lambda_i.$ Clearly $H$ is isomorphic to the product $\prod_i S_{\lambda_i}$ so we have:

$\displaystyle U_\lambda \cong \text{Ind}_{S_{\lambda_1} \times S_{\lambda_2}\times \ldots}^{S_d}\, \mathbb{C},\quad U_\mu \cong \text{Ind}_{S_{\mu_1} \times S_{\mu_2}\times \ldots}^{S_e}\, \mathbb{C},$

$\displaystyle U_{\lambda |\mu} \cong \text{Ind}_{S_{\lambda_1} \times S_{\lambda_2} \times\ldots \times S_{\mu_1} \times S_{\mu_2}\times\ldots}^{S_{d+e}} \,\mathbb{C}.$

Proposition. The product $\Lambda^{(d)} \times \Lambda^{(e)} \to \Lambda^{(d+e)}$ corresponds to:

$(V, W) \mapsto \text{Ind}_{S_d \times S_e}^{S_{d+e}} V \otimes W$

where $S_d \times S_e$ acts on $V\otimes W$ via $(g,h) : v\otimes w\mapsto (gv)\otimes (hw).$

Proof

First use the following result, whose proof is an easy exercise if you know tensor products (here, tensor products without subscripts are over C).

• Suppose $B, B'$ are $\mathbb{C}$-algebras, $A\subseteq B, A'\subseteq B'$ are subalgebras, and $V, W$ are $A, A'$-modules respectively. Then $(B \otimes_{A} V) \otimes (B'\otimes_{A'} W) \cong (B\otimes B') \otimes_{A \otimes A'} (V\otimes W).$

In particular, take

$B=\mathbb{C}[G],\quad B' = \mathbb{C}[G'],\quad A=\mathbb{C}[H],\quad A'=\mathbb{C}[H']$

for subgroups $H'\le G'$ and $H\le G$. Since $\mathbb{C}[G\times G'] \cong \mathbb{C}[G]\otimes \mathbb{C}[G']$ as $\mathbb{C}$-algebras, we have:

$\left(\text{Ind}_H^G V\right) \otimes \left( \text{Ind}_{H'}^{G'} W\right) \cong \text{Ind}_{H\times H'}^{G\times G'} V\otimes W.$

This gives $U_\lambda \otimes U_\mu \cong \text{Ind}_{S_{\lambda_1} \times \ldots \times S_{\mu_1} \times \ldots }^{S_d \times S_e} \mathbb{C}.$ So we have:

$\text{Ind}_{S_d\times S_e}^{S_{d+e}} (U_\lambda \otimes U_\mu) \cong U_{\lambda | \mu},$

corresponding to $h_\lambda h_\mu = h_{\lambda|\mu}$ as desired. ♦

## Consequences

Ring Isomorphism

Thus we obtain a graded ring isomorphism between $\Lambda$ and the set of virtual representations of all $S_d$ up to isomorphism. And we have:

$\displaystyle s_\mu s_\nu = \sum_\lambda c^{\lambda}_{\mu\nu} s_\lambda \implies \text{Ind}_{S_d \times S_e}^{S_{d+e}} (V_\mu \otimes V_\nu) = \bigoplus_\lambda V_\lambda^{\oplus c^{\lambda}_{\mu\nu}}.$

Pieri’s Formulae

Consider the case where W is trivial and e=1, so the product operation gives

$V \mapsto \text{Ind}_{S_d}^{S_{d+1}}\, V.$

On the other hand the trivial representation for $S_1$ corresponds to $h_1 = e_1 \in \Lambda^{(1)}.$ By Pieri’s formula$h_1 s_\lambda = \sum_\mu s_\mu$ where $\mu$ runs through all partitions obtained from $\lambda$ by adding 1 box. Hence:

$\displaystyle \text{Ind}_{S_d}^{S_{d+1}} V_\lambda = \bigoplus_\mu V_\mu.$

For example,

$\text{Ind}_{S_{14}}^{S_{15}} V_{533111} \cong V_{633111} \oplus V_{543111} \oplus V_{533211} \oplus V_{5331111}.$

Restriction

By Frobenius reciprocity, we have, for any $\lambda \vdash d$ and $\nu \vdash d+ 1$,

$\left< V_\lambda, \text{Res}_{S_{d+1}}^{S_d} V_\nu\right> = \left< \text{Ind}_{S_d}^{S_{d+1}} V_\lambda, V_\nu\right> = \begin{cases} 1, \quad &\text{if } \nu \text{ is obtained from } \lambda \text{ by adding 1 box,}\\0, \quad &\text{else.}\end{cases}.$

Hence, $\text{Res}_{S_{d+1}}^{S_d} V_\nu = \oplus_\lambda V_\lambda$ where the sum is over all partitions $\lambda$ obtained from $\nu$ by removing one box.

Exercise

Express $\text{Res}_{S_{17}}^{S_{15}} V_\lambda$ and $\text{Ind}_{S_{17}}^{S_{19}} V_\lambda$ as a direct sums of irreps where $\lambda = (6,3,3,3,1,1).$

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