Branching rule for Specht modules: adding or removing boxes in distinct columns

Let $K\mathrm{/}F$ be a field extension. By [F1], $K{\otimes }_F{S}_F^{\mu }\cong S_K^{\mu }$; by [F4] and [F5], induction commutes with extension of scalars, so $K{\otimes }_F{\mathrm{Ind}}_{S_n\times S_k}^{S_{n+k}}(S_F^{\lambda }⊠F)\cong {\mathrm{Ind}}_{S_n\times S_k}^{S_{n+k}}(S_K^{\lambda }⊠K)$. Moreover, for finite-dimensional $F$-modules $U,W$ carrying $G$-actions, ${\dim }_K{\mathrm{Hom}}_{KG}(K{\otimes }_{F}U,K{\otimes }_{F}W)={\dim }_F{\mathrm{Hom}}_{FG}(U,W)$: choose bases, write the intertwining conditions as a finite linear system over $F$, and note that its rank is unchanged by extending scalars.

Over $\mathbb{C}$, [F3] with $m=n$, $r=k$, $\mu =\lambda$ gives $M_{\mathbb{C}}:={\mathrm{Ind}}_{S_n\times S_k}^{S_{n+k}}(S_{\mathbb{C}}^{\lambda }⊠\mathbb{C})\cong {⨁}_{\mu }{S}_{\mathbb{C}}^{\mu }$, summed over the partitions $\mu ⊢n+k$ obtained from $\lambda$ by adding one box in each of $k$ different columns, each multiplicity one; by [F1, F2] this is the decomposition of $M_{\mathbb{C}}$ into irreducibles.

Let $N$ be a finite-dimensional $F(S_{n-k}\times S_k)$-module and let $\nu ⊢n-k$. Since $\mathrm{char}F=0$, averaging over the finite group $S_k$ identifies the coinvariants $N_{S_k}$ with the invariants $N^{S_k}$ [F6], and because $S_{n-k}$ and $S_k$ commute this identification is $F{S}_{n-k}$-equivariant; hence ${\mathrm{Hom}}_{F{S}_{n-k}}(S_F^{\nu },N_{S_k})\cong {\mathrm{Hom}}_{F(S_{n-k}\times S_k)}(S_F^{\nu }⊠F,N)$, where $S_k$ acts trivially on the first factor. Composing with the adjunction of [F4] gives, for any $F{S}_n$-module $W$, ${\mathrm{Hom}}_{F{S}_{n-k}}(S_F^{\nu },({\mathrm{Res}}_{S_{n-k}\times S_k}^{S_n}W{)}_{S_k})\cong {\mathrm{Hom}}_{F{S}_n}({\mathrm{Ind}}_{S_{n-k}\times S_k}^{S_n}(S_F^{\nu }⊠F),W)$.

Let $M_{\mathbb{Q}}:={\mathrm{Ind}}_{S_n\times S_k}^{S_{n+k}}(S_{\mathbb{Q}}^{\lambda }⊠\mathbb{Q})$. By step 1.1, $\mathbb{C}{\otimes }_{\mathbb{Q}}{M}_{\mathbb{Q}}\cong M_{\mathbb{C}}$, and for each $\mu ⊢n+k$ the multiplicity of $S_{\mathbb{Q}}^{\mu }$ in $M_{\mathbb{Q}}$, which is ${\dim }_{\mathbb{Q}}{\mathrm{Hom}}_{\mathbb{Q}{S}_{n+k}}(S_{\mathbb{Q}}^{\mu },M_{\mathbb{Q}})$ by [F2], equals the multiplicity of $S_{\mathbb{C}}^{\mu }$ in $M_{\mathbb{C}}$. By step 1.2 those multiplicities are $1$ for the horizontal-strip shapes $\mu$ and $0$ otherwise. Since $M_{\mathbb{Q}}$ is finite-dimensional and semisimple over $\mathbb{Q}$ [F2] and its irreducibles are classified, these multiplicities determine it: $M_{\mathbb{Q}}\cong {⨁}_{\mu }{S}_{\mathbb{Q}}^{\mu }$ over the same shapes.

Applying $F{\otimes }_{\mathbb{Q}}-$ to step 2.1 and using step 1.1, [F5] and the base-change isomorphism $F{\otimes }_{\mathbb{Q}}{S}_{\mathbb{Q}}^{\mu }\cong S_F^{\mu }$ of [F1] gives ${\mathrm{Ind}}_{S_n\times S_k}^{S_{n+k}}(S_F^{\lambda }⊠F)\cong F{\otimes }_{\mathbb{Q}}{M}_{\mathbb{Q}}\cong {⨁}_{\mu }{S}_F^{\mu }$ over exactly the horizontal-strip shapes, with all multiplicities one. This is (i).

Take $W=S_F^{\lambda }$ in step 1.3 and apply (i) from step 3.1 to ${\mathrm{Ind}}_{S_{n-k}\times S_k}^{S_n}(S_F^{\nu }⊠F)$: it is ${⨁}_{{\lambda }^{\mathrm{\prime }}}{S}_F^{{\lambda }^{\mathrm{\prime }}}$, summed over the partitions ${\lambda }^{\mathrm{\prime }}⊢n$ obtained from $\nu$ by adding one box in each of $k$ different columns. By [F2] and the classification, ${\dim }_F{\mathrm{Hom}}_{F{S}_n}(S_F^{\lambda },{⨁}_{{\lambda }^{\mathrm{\prime }}}{S}_F^{{\lambda }^{\mathrm{\prime }}})$ is $1$ if $\lambda$ occurs among the shapes ${\lambda }^{\mathrm{\prime }}$ and $0$ otherwise. The condition that $\lambda$ is obtained from $\nu$ by adding one box in each of $k$ different columns is equivalent to its reverse, that $\nu$ is obtained from $\lambda$ by removing one box from each of $k$ different columns; hence the multiplicity $m_{\nu }:={\dim }_F{\mathrm{Hom}}_{F{S}_{n-k}}(S_F^{\nu },({\mathrm{Res}}_{S_{n-k}\times S_k}^{S_n}{S}_F^{\lambda }{)}_{S_k})$ is that $0\mathrm{/}1$ value.

The module $({\mathrm{Res}}_{S_{n-k}\times S_k}^{S_n}{S}_F^{\lambda }{)}_{S_k}$ is finite-dimensional over $F$ and semisimple by Maschke [F2], and its irreducible constituents are among the $S_F^{\nu }$ by the classification; by step 4.1 and [F2] its multiplicity at $S_F^{\nu }$ is $m_{\nu }$, which is $1$ exactly when $\nu$ is obtained from $\lambda$ by removing one box from each of $k$ different columns and $0$ otherwise. Hence $({\mathrm{Res}}_{S_{n-k}\times S_k}^{S_n}{S}_F^{\lambda }{)}_{S_k}\cong {⨁}_{\nu }{S}_F^{\nu }$ over those $\nu$, with all multiplicities $0$ or $1$. This proves (ii) and completes the lemma.