Polynomials and Representations XXII

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.


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).


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. ♦



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 formulah_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}.


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.


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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