The complex Specht restriction branching rule

[construct] By [F1] there is a chain of $S_{n-1}$-submodules $0=V_0⊊V_1⊊\cdots ⊊V_m=S_{\mathbb{C}}^{\lambda }$ whose successive quotients are $V_i\mathrm{/}{V}_{i-1}\cong S_{\mathbb{C}}^{{\lambda }^{(i)}}$. Since $\mathrm{char}\mathbb{C}=0$ does not divide $\mathrm{\mid }{S}_{n-1}\mathrm{\mid }=(n-1)!$, [F3] applies to each subrepresentation $V_{i-1}\le V_i$: there is an $S_{n-1}$-submodule $U_i\le V_i$ with $V_i=V_{i-1}\oplus U_i$.

By [F5] the projection $V_i\to V_i\mathrm{/}{V}_{i-1}$ restricts to an $S_{n-1}$-isomorphism $U_i\to V_i\mathrm{/}{V}_{i-1}$; composing with the isomorphism of step 1.1 gives an $S_{n-1}$-isomorphism $U_i\cong S_{\mathbb{C}}^{{\lambda }^{(i)}}$.

Since $V_0=0$ we have $V_1=U_1$, and $V_i=V_{i-1}\oplus U_i$ for every $i$ by step 1.1, so $S_{\mathbb{C}}^{\lambda }=V_m=U_1\oplus U_2\oplus \cdots \oplus U_m$; in particular ${\dim }_{\mathbb{C}}{S}_{\mathbb{C}}^{\lambda }={\sum }_{i=1}^m{\dim }_{\mathbb{C}}{U}_i$.

Combining steps 2.1 and 2.2 gives the asserted $\mathbb{C}{S}_{n-1}$-isomorphism ${\mathrm{Res}}_{S_{n-1}}^{S_n}{S}_{\mathbb{C}}^{\lambda }\cong S_{\mathbb{C}}^{{\lambda }^{(1)}}\oplus \cdots \oplus S_{\mathbb{C}}^{{\lambda }^{(m)}}$. By [F2] the irreducible summands of a decomposition into Specht modules are classified up to isomorphism by their shapes, and by [F4] the shapes ${\lambda }^{(i)}$ are pairwise distinct, so each $S_{\mathbb{C}}^{{\lambda }^{(i)}}$ occurs exactly once and, for $\mu ⊢n-1$, the multiplicity of $S_{\mathbb{C}}^{\mu }$ is the number of corners $x$ with $\lambda -x=\mu$. This proves the Statement.

Boundary and choice audit. For $n=1$ one has $\lambda =(1)$, $m=1$, $x_1$ the unique box, ${\lambda }^{(1)}=\mathrm{\varnothing }$ and $S_{\mathbb{C}}^{\mathrm{\varnothing }}=\mathbb{C}$, so the isomorphism reads ${\mathrm{Res}}_{S_0}^{S_1}{S}_{\mathbb{C}}^{(1)}\cong S_{\mathbb{C}}^{\mathrm{\varnothing }}$, both sides being the one-dimensional trivial representation of the trivial group; the filtration has length one and no splitting choice is needed beyond $U_1=V_1$. If $\lambda$ is a row or a column there is exactly one removable corner and the restriction is irreducible; in general $m$ is the finite number of removable corners of $\lambda$. The complement $U_i$ furnished by [F3] is produced by averaging over the finite group $S_{n-1}$ and involves no choice principle, and by step 3.1 the isomorphism type of the resulting direct sum does not depend on those complements. This proves the corollary.