Young's rule for complex permutation modules

[construct] The restrictions ${\theta }_T{\mathrm{\mid }}_{S^{\lambda }}:S^{\lambda }\to M^{\mu }$ of the semistandard fillings $T$ of content $\mu$ form a basis of the complex vector space ${\mathrm{Hom}}_{S_n}(S^{\lambda },M^{\mu })$: they span by [F4], they are linearly independent by [F3], and by [F5] there are exactly $K_{\lambda ,\mu }$ of them. Hence ${\dim }_{\mathbb{C}}{\mathrm{Hom}}_{S_n}(S^{\lambda },M^{\mu })=K_{\lambda ,\mu }$, and in particular a nonzero intertwiner $S^{\lambda }\to M^{\mu }$ exists exactly when $K_{\lambda ,\mu }\mathrm{\ne }0$.

By [F6] the finite-dimensional complex representation $M^{\mu }$ is completely reducible, so there are irreducible subrepresentations $U_1,\dots ,U_r$ with $M^{\mu }=U_1\oplus \cdots \oplus U_r$; here $r\ge 1$ because the tabloid basis of [F1] is nonempty. By [F7] each $U_i$ is isomorphic to $S^{\sigma (i)}$ for exactly one partition $\sigma (i)⊢n$, and $S^{\lambda }\cong S^{\sigma }$ holds only for $\sigma =\lambda$.

Let $W$ be a $\mathbb{C}{S}_n$-module and let $M=U_1\oplus \cdots \oplus U_r$ be a direct sum of subrepresentations with projections ${\pi }_i:M\to U_i$ along the other summands. Then ${\pi }_i$ is $S_n$-linear, so the rule $f\mapsto ({\pi }_{1}f,\dots ,{\pi }_{r}f)$ maps ${\mathrm{Hom}}_{S_n}(W,M)$ into ${\mathrm{Hom}}_{S_n}(W,U_1)\oplus \cdots \oplus {\mathrm{Hom}}_{S_n}(W,U_r)$; this map is $\mathbb{C}$-linear, it is injective because $f(w)={\sum }_i{f}_i(w)$ is determined by its components $f_i={\pi }_{i}f$, and it is surjective because a tuple of intertwiners $(f_1,\dots ,f_r)$ defines the intertwiner $w\mapsto {\sum }_i{f}_i(w)$ of which it is the tuple of components. Hence ${\mathrm{Hom}}_{S_n}(W,U_1\oplus \cdots \oplus U_r)\cong {⨁}_{i=1}^r{\mathrm{Hom}}_{S_n}(W,U_i)$.

For irreducible $S^{\lambda }$ and $S^{\sigma }$ one has ${\dim }_{\mathbb{C}}{\mathrm{Hom}}_{S_n}(S^{\lambda },S^{\sigma })=1$ when $\sigma =\lambda$ and $0$ otherwise. If $\sigma \mathrm{\ne }\lambda$, then $S^{\lambda }$ and $S^{\sigma }$ are non-isomorphic by the irredundancy in [F7], so every intertwiner between them is zero by [F8]. If $\sigma =\lambda$, the same fact of [F8] makes every endomorphism of the irreducible $S^{\lambda }$ a scalar multiple of ${\mathrm{i}\mathrm{d}}_{S^{\lambda }}$, so ${\mathrm{End}}_{S_n}(S^{\lambda })=\mathbb{C}{\mathrm{i}\mathrm{d}}_{S^{\lambda }}$ has dimension one.

Applying step 1.3 with $W=S^{\lambda }$ to the decomposition of step 1.2 gives ${\mathrm{Hom}}_{S_n}(S^{\lambda },M^{\mu })\cong {⨁}_{i=1}^r{\mathrm{Hom}}_{S_n}(S^{\lambda },U_i)$, and substituting the isomorphism $U_i\cong S^{\sigma (i)}$ of step 1.2 into step 1.4 gives ${\dim }_{\mathbb{C}}{\mathrm{Hom}}_{S_n}(S^{\lambda },U_i)=1$ when $\sigma (i)=\lambda$ and $0$ otherwise. Hence ${\dim }_{\mathbb{C}}{\mathrm{Hom}}_{S_n}(S^{\lambda },M^{\mu })$ equals the number $\mathrm{\#}\{i:\sigma (i)=\lambda \}$ of summands of this decomposition isomorphic to $S^{\lambda }$.

By step 1.1 the dimension in step 2.1 is $K_{\lambda ,\mu }$, so the decomposition of step 1.2 contains exactly $K_{\lambda ,\mu }$ summands isomorphic to $S^{\lambda }$. Steps 1.2 and 2.1 apply verbatim to every decomposition of $M^{\mu }$ into irreducible subrepresentations, and the quantity they compute, ${\dim }_{\mathbb{C}}{\mathrm{Hom}}_{S_n}(S^{\lambda },M^{\mu })$, depends only on $M^{\mu }$ and $\lambda$; hence every such decomposition contains exactly $K_{\lambda ,\mu }$ summands isomorphic to $S^{\lambda }$ and the multiplicity $[M^{\mu }:S^{\lambda }]$ is well defined and equal to $K_{\lambda ,\mu }$. Grouping the summands of step 1.2 by their isomorphism classes gives the asserted isomorphism $M^{\mu }\cong {⨁}_{\lambda ⊢n}(S^{\lambda }{)}^{\oplus K_{\lambda ,\mu }}$.

Boundary, degenerate, characteristic and choice audit. For $n=0$ the only partition is $\mathrm{\varnothing }$, and $M^{\mathrm{\varnothing }}=\mathbb{C}$, $S^{\mathrm{\varnothing }}=\mathbb{C}$, $K_{\mathrm{\varnothing },\mathrm{\varnothing }}=1$ by [F1] and [F5], so claim 1 reads $M^{\mathrm{\varnothing }}\cong S^{\mathrm{\varnothing }}$ and steps 1.2 and 2.1 give $r=1$ with $\sigma (1)=\mathrm{\varnothing }$. If $K_{\lambda ,\mu }=0$, the summand $(S^{\lambda }{)}^{\oplus 0}=0$ is omitted and claim 2 says that $S^{\lambda }$ does not occur in $M^{\mu }$; this happens for instance when $l(\lambda )>l(\mu )$, since every entry of a semistandard filling of content $\mu$ lies in $\{1,\dots ,l(\mu )\}$ by [F5] while the first column of $[\lambda ]$ has $l(\lambda )$ boxes carrying strictly increasing entries, and also for $\mu =(n)$ with $\lambda \mathrm{\ne }(n)$, where all entries of a filling of content $(n)$ are equal to $1$ and a column of length at least two cannot strictly increase. If $K_{\lambda ,\mu }=1$, the summand is a single copy of $S^{\lambda }$: by [F3] this happens for $\lambda =\mu$, so every $M^{\lambda }$ contains exactly one copy of $S^{\lambda }$; and it happens for the one-row shape $\lambda =(n)$ for every $\mu ⊢n$, since a semistandard filling of the single-row diagram $(n)$ with content $\mu$ is exactly the weakly increasing word $1^{{\mu }_1}{2}^{{\mu }_2}\cdots$ of content $\mu$, which exists and is unique. The argument is particular to $\mathbb{C}$: [F6] uses that $\mathrm{char}\mathbb{C}=0$ does not divide $\mathrm{\mid }{S}_n\mathrm{\mid }=n!$ and [F8] uses that $\mathbb{C}$ is algebraically closed, and no analogue over a field of positive characteristic is asserted. The only selection made is the decomposition of the finite-dimensional module $M^{\mu }$ into finitely many irreducible summands, whose existence is supplied by [F6]; step 3.1 shows the multiplicities do not depend on this selection, and no choice principle is invoked. This proves claims 1 and 2.