Littlewood's identity - the internal product against an outer product

For fixed $\alpha ⊢r$, $\beta ⊢n-r$ write ${\phi }_{\alpha \beta }:={\chi }^{\alpha }⊠{\chi }^{\beta }$ and $V_{\alpha \beta }:={\mathrm{Ind}}_H^{S_n}(S^{\alpha }⊠S^{\beta })$, so that ${\chi }_{V_{\alpha \beta }}={\mathrm{Ind}}_H^{S_n}{\phi }_{\alpha \beta }$ by [F2]. By [F1] the multiplicity of $S^{\lambda }$ in $V_{\alpha \beta }$ is $c_{\alpha \beta }^{\lambda }$. By [F5] the multiplicity of ${\phi }_{\alpha \beta }$ in ${\mathrm{Res}}_H^{S_n}{\chi }^{\lambda }$ is $⟨{\mathrm{Res}}_H^{S_n}{\chi }^{\lambda },{\phi }_{\alpha \beta }{⟩}_H$, which by Frobenius reciprocity [F3] equals $⟨{\mathrm{Ind}}_H^{S_n}{\phi }_{\alpha \beta },{\chi }^{\lambda }{⟩}_{S_n}$; by [F5] again this is the multiplicity of $S^{\lambda }$ in $V_{\alpha \beta }$, namely $c_{\alpha \beta }^{\lambda }$. Since the ${\phi }_{\alpha \beta }$ form a nonnegative integral basis of the honest characters of $H$ by [F5] and ${\mathrm{Res}}_H^{S_n}{\chi }^{\lambda }$ is an honest character, the expansion ${\mathrm{Res}}_H^{S_n}{\chi }^{\lambda }={\sum }_{\alpha ⊢r,\beta ⊢n-r}{c}_{\alpha \beta }^{\lambda }{\phi }_{\alpha \beta }$ follows.

Put $\xi :={\phi }_{\tau \theta }={\chi }^{\tau }⊠{\chi }^{\theta }$. Multiplying step 1.1 pointwise on $H$ by $\xi$ and using the value formula [F2] for both factors gives $({\chi }^{\alpha }⊠{\chi }^{\beta })(\sigma ,\rho )({\chi }^{\tau }⊠{\chi }^{\theta })(\sigma ,\rho )={\chi }^{\alpha }(\sigma ){\chi }^{\beta }(\rho ){\chi }^{\tau }(\sigma ){\chi }^{\theta }(\rho )=({\chi }^{\alpha }{\chi }^{\tau })(\sigma )({\chi }^{\beta }{\chi }^{\theta })(\rho )$, so the product of the two external products is the external product $({\chi }^{\alpha }{\chi }^{\tau })⊠({\chi }^{\beta }{\chi }^{\theta })$; hence ${\mathrm{Res}}_H^{S_n}({\chi }^{\lambda })\cdot \xi ={\sum }_{\alpha ,\beta }{c}_{\alpha \beta }^{\lambda }({\chi }^{\alpha }{\chi }^{\tau })⊠({\chi }^{\beta }{\chi }^{\theta })$.

Projection formula: for every $g\in S_n$, ${\mathrm{Ind}}_H^{S_n}({\mathrm{Res}}_H^{S_n}({\chi }^{\lambda })\cdot \xi )(g)={\chi }^{\lambda }(g){\mathrm{Ind}}_H^{S_n}\xi (g)$. Indeed, by [F4] the left side is $\frac{1}{\mathrm{\mid }H\mathrm{\mid }}{\sum }_{x\in S_n:x^{-1}gx\in H}{\chi }^{\lambda }(x^{-1}gx)\xi (x^{-1}gx)$; the character ${\chi }^{\lambda }$ is constant on conjugacy classes, so ${\chi }^{\lambda }(x^{-1}gx)={\chi }^{\lambda }(g)$ for every term, and the sum factors as ${\chi }^{\lambda }(g)\cdot \frac{1}{\mathrm{\mid }H\mathrm{\mid }}{\sum }_{x:x^{-1}gx\in H}\xi (x^{-1}gx)={\chi }^{\lambda }(g){\mathrm{Ind}}_H^{S_n}\xi (g)$ by [F4] again. Applying the $\mathbb{C}$-linear induction operator of [F2] to step 2.1 and using this identity for the left factor gives, as class functions on $S_n$, ${\chi }^{\lambda }\cdot ({\chi }^{\tau }\circ {\chi }^{\theta })={\sum }_{\alpha ,\beta }{c}_{\alpha \beta }^{\lambda }({\chi }^{\alpha }{\chi }^{\tau })\circ ({\chi }^{\beta }{\chi }^{\theta })$, since ${\mathrm{Ind}}_H^{S_n}\xi ={\chi }^{\tau }\circ {\chi }^{\theta }$ and ${\mathrm{Ind}}_H^{S_n}(({\chi }^{\alpha }{\chi }^{\tau })⊠({\chi }^{\beta }{\chi }^{\theta }))=({\chi }^{\alpha }{\chi }^{\tau })\circ ({\chi }^{\beta }{\chi }^{\theta })$ by [F2].

Translate step 3.1 through $\mathrm{ch}$. By [F7] and [F6], $\mathrm{ch}({\chi }^{\tau }\circ {\chi }^{\theta })=s_{\tau }{s}_{\theta }$, so ${\mathrm{ch}}^{-1}(s_{\tau }{s}_{\theta })={\chi }^{\tau }\circ {\chi }^{\theta }$; by the definition [F6] of the internal product, ${\mathrm{ch}}^{-1}(s_{\lambda }\ast (s_{\tau }{s}_{\theta }))={\chi }^{\lambda }\cdot {\mathrm{ch}}^{-1}(s_{\tau }{s}_{\theta })={\chi }^{\lambda }\cdot ({\chi }^{\tau }\circ {\chi }^{\theta })$. Likewise ${\mathrm{ch}}^{-1}(s_{\alpha }\ast s_{\tau })={\chi }^{\alpha }{\chi }^{\tau }$ and ${\mathrm{ch}}^{-1}(s_{\beta }\ast s_{\theta })={\chi }^{\beta }{\chi }^{\theta }$ by [F6], and [F7] gives ${\mathrm{ch}}^{-1}((s_{\alpha }\ast s_{\tau })(s_{\beta }\ast s_{\theta }))=({\chi }^{\alpha }{\chi }^{\tau })\circ ({\chi }^{\beta }{\chi }^{\theta })$. Step 3.1 therefore says that ${\mathrm{ch}}^{-1}$ of the two sides of the claimed identity are equal; since $\mathrm{ch}$ is injective [F8], the claimed identity holds in ${\mathrm{\Lambda }}^n$.