Schur-Weyl decomposition and highest weights

[construct] By [F3] the $S_n$-module $E$ is completely reducible, so by [F5] and [F4] its isotypic decomposition $E={⨁}_{\lambda ⊢n}{E}_{\lambda }$ over the classes $S^{\lambda }$ is defined and unique, with $E_{\lambda }$ a (possibly zero) direct sum of copies of $S^{\lambda }$. If $\psi :S^{\lambda }\to E$ is $S_n$-linear, then $\ker \psi$ is a submodule of the irreducible $S^{\lambda }$, so $\psi =0$ or $\psi$ is injective with image isomorphic to $S^{\lambda }$, hence contained in $E_{\lambda }$; therefore ${\mathrm{Hom}}_{S_n}(S^{\lambda },E)={\mathrm{Hom}}_{S_n}(S^{\lambda },E_{\lambda })=M_{\lambda }$. Applying [F6] to the isotypical module $E_{\lambda }$ gives an $S_n$-isomorphism $S^{\lambda }\otimes M_{\lambda }\to E_{\lambda }$, $s\otimes \psi \mapsto \psi (s)$, so ${\dim }_{\mathbb{C}}{E}_{\lambda }={\dim }_{\mathbb{C}}{S}^{\lambda }\cdot {\dim }_{\mathbb{C}}{M}_{\lambda }$ and $E_{\lambda }=0$ if and only if $M_{\lambda }=0$.

Fix a basis $s_1,\dots ,s_k$ of $S^{\lambda }$ and consider the evaluation map $\mathrm{e}\mathrm{v}:M_{\lambda }\to E^k$, $\psi \mapsto (\psi (s_1),\dots ,\psi (s_k))$. It is $\mathbb{C}$-linear and injective, because an $S_n$-linear map is determined by its values on a basis; and it intertwines the postcomposition action on $M_{\lambda }$ with the componentwise diagonal action on $E^k$, since $(g\cdot \psi )(s_l)=g^{\otimes n}\psi (s_l)$ for every $l$.

Fix the basis $e_1,\dots ,e_d$ of $V$ and write $g{e}_j={\sum }_i{g}_{ij}{e}_i$. By the multilinearity of the tensor product in [F1], for all $b_1,\dots ,b_n\in \{1,\dots ,d\}$ one has $g^{\otimes n}(e_{b_1}\otimes \cdots \otimes e_{b_n})=(g{e}_{b_1})\otimes \cdots \otimes (g{e}_{b_n})={\sum }_{a_1,\dots ,a_n}\left({\prod }_{l=1}^n{g}_{a_l{b}_l}\right)e_{a_1}\otimes \cdots \otimes e_{a_n}$; hence each matrix entry of $g^{\otimes n}$ in the elementary tensor basis [F2] is either $0$ or a monomial ${\prod }_{l=1}^n{g}_{a_l{b}_l}$ of degree $n$ in the entries $g_{ij}$.

The evaluation isomorphism of step 1.1 intertwines the postcomposition action of $\mathrm{GL}(V)$ on $M_{\lambda }$ with the diagonal action on $E_{\lambda }$: for $g\in \mathrm{GL}(V)$, $s\in S^{\lambda }$ and $\psi \in M_{\lambda }$ one has $E_{\lambda }(s\otimes g\cdot \psi )=(g\cdot \psi )(s)=g^{\otimes n}\psi (s)=g^{\otimes n}{E}_{\lambda }(s\otimes \psi )$. It also intertwines the action of $S_n$, which is ${\rho }_{\lambda }\otimes \mathrm{i}\mathrm{d}$ because $E_{\lambda }(\sigma s\otimes \psi )=\psi (\sigma s)=\sigma \psi (s)=\sigma E_{\lambda }(s\otimes \psi )$; equivalently the conjugation action $\sigma \cdot \psi =\sigma \psi {\sigma }^{-1}$ is trivial on $M_{\lambda }$, since $\psi$ is $S_n$-linear. Hence $E_{\lambda }\cong S^{\lambda }\otimes M_{\lambda }$ as $(S_n\times \mathrm{GL}(V))$-modules.

Let $\psi \in M_{\lambda }$ and ${\psi }^{\ast }\in M_{\lambda }^{\ast }$. Since $\mathrm{e}\mathrm{v}$ of step 1.2 is injective, it has a linear retraction: choosing a basis of the image $\mathrm{e}\mathrm{v}(M_{\lambda })$ and extending it to a basis of the finite-dimensional space $E^k$, define $r:E^k\to M_{\lambda }$ on the basis by $r(v)={\mathrm{e}\mathrm{v}}^{-1}(v)$ for $v$ in that basis of the image and $r=0$ on the added vectors, so that $r\circ \mathrm{e}\mathrm{v}=\mathrm{i}\mathrm{d}$. Then $g\cdot \psi =r(\mathrm{e}\mathrm{v}(g\cdot \psi ))$ and the matrix coefficient is ${\psi }^{\ast }(g\cdot \psi )=({\psi }^{\ast }\circ r)(g^{\otimes n}\psi (s_1),\dots ,g^{\otimes n}\psi (s_k))$. The vectors $\psi (s_l)\in E$ are fixed, so their coordinates in the elementary tensor basis [F2] are constants, and by step 1.3 the numbers ${\psi }^{\ast }(g\cdot \psi )$ are constant-coefficient linear combinations of monomials of degree $n$ in the entries $g_{ij}$: they are homogeneous polynomials of degree $n$. This holds for the matrix coefficients of the action with respect to any basis of $M_{\lambda }$, so $M_{\lambda }$ is a homogeneous polynomial $\mathrm{GL}(V)$-module of degree $n$. This proves claim 5.

By [F9], $M_{\lambda }\mathrm{\ne }0$ exactly when $\mathrm{\ell }(\lambda )\le d$; combined with step 1.1 and step 2.1 this gives the $(S_n\times \mathrm{GL}(V))$-isomorphism $E={⨁}_{\mathrm{\ell }(\lambda )\le d}{E}_{\lambda }\cong {⨁}_{\mathrm{\ell }(\lambda )\le d}{S}^{\lambda }\otimes M_{\lambda }$, the omitted components being exactly the zero ones, and proves claim 1.

An $S_n$-endomorphism $F\in {\mathrm{End}}_{S_n}(E)$ maps each isotypic component $E_{\lambda }$ into itself: $E_{\lambda }$ is a sum of copies of $S^{\lambda }$ by [F5], and the image under $F$ of such a copy is either $0$ or, by irreducibility of $S^{\lambda }$, a copy of $S^{\lambda }$, hence lies in $E_{\lambda }$. Restriction gives an isomorphism of $\mathbb{C}$-algebras ${\mathrm{End}}_{S_n}(E)\to {\prod }_{\lambda ⊢n}{\mathrm{End}}_{S_n}(E_{\lambda })$ (injective, since $F$ is determined on the direct sum, and blockwise surjective, with componentwise composition). For each $\lambda$ with $M_{\lambda }\mathrm{\ne }0$, [F6] identifies ${\mathrm{End}}_{S_n}(E_{\lambda })$ with $\mathrm{End}(M_{\lambda })$: every $F_{\lambda }\in {\mathrm{End}}_{S_n}(E_{\lambda })$ is uniquely $E_{\lambda }(1\otimes a_{\lambda })E_{\lambda }^{-1}$ with $a_{\lambda }\in \mathrm{End}(M_{\lambda })$ and the identification preserves composition. Therefore, by [F8], $B={\mathrm{End}}_{S_n}(E)\cong {\prod }_{\mathrm{\ell }(\lambda )\le d}\mathrm{End}(M_{\lambda })$ as $\mathbb{C}$-algebras, the product being over the $\lambda$ with $M_{\lambda }\mathrm{\ne }0$ (and $B=0$ when there are none).

Under the identification of step 4.1, an element $b\in B$ acts on $E_{\lambda }$ as $E_{\lambda }(1\otimes a_{\lambda })E_{\lambda }^{-1}$, where $a_{\lambda }$ is its $\lambda$-component in $\prod \mathrm{End}(M_{\lambda })$; hence for $\psi \in M_{\lambda }$ one has $(b\cdot \psi )(s)=b(\psi (s))=E_{\lambda }(s\otimes a_{\lambda }\psi )=(a_{\lambda }\psi )(s)$, that is, $b$ acts on $M_{\lambda }$ by $a_{\lambda }$. Since $M_{\lambda }\mathrm{\ne }0$ for $\mathrm{\ell }(\lambda )\le d$ by [F9], $\mathrm{End}(M_{\lambda })$ is the full matrix algebra $M_{m_{\lambda }}(\mathbb{C})$ with $m_{\lambda }={\dim }_{\mathbb{C}}{M}_{\lambda }\ge 1$, so $B\cong {\prod }_{\mathrm{\ell }(\lambda )\le d}{M}_{m_{\lambda }}(\mathbb{C})$ is a product of full matrix algebras over the field $\mathbb{C}$; by [F11] every simple left $B$-module is supported on exactly one factor and is isomorphic to that factor's column module, and distinct factors have non-isomorphic column modules. The postcomposition module $M_{\lambda }$ is the column module of the $\lambda$-th factor $\mathrm{End}(M_{\lambda })$ (the other factors acting as zero, as step 4.1 shows the action factors through the $\lambda$-component), so $M_{\lambda }$ is a simple $B$-module and $M_{\lambda }\cong M_{\mu }$ as $B$-modules implies $\lambda =\mu$.

Because $B$ is spanned by the operators $g^{\otimes n}$, a $\mathrm{GL}(V)$-stable subspace of $M_{\lambda }$ is stable under every $b\in B$; because $B$ is the unital subalgebra generated by the operators $\mathrm{\Delta }(T)$, $T\in \mathrm{End}(V)$, a subspace stable under all $\mathrm{\Delta }(T)$ (that is, a $\mathfrak{g}\mathfrak{l}(V)$-submodule, [F12]) is also $B$-stable. By step 5.1 the space $M_{\lambda }$ is a simple $B$-module and nonzero, so it has no proper nonzero subspace of either kind: it is irreducible as a $\mathrm{GL}(V)$-module and as a $\mathfrak{g}\mathfrak{l}(V)$-module. This proves claim 2, the case $\mathrm{\ell }(\lambda )>d$ being [F9].

Let $\lambda ,\mu ⊢n$ satisfy $\mathrm{\ell }(\lambda )\le d$ and $\mathrm{\ell }(\mu )\le d$, so that $M_{\lambda }\mathrm{\ne }0$ and $M_{\mu }\mathrm{\ne }0$ by [F9], and let $T:M_{\lambda }\to M_{\mu }$ be an isomorphism of $\mathrm{GL}(V)$-modules. For $b={\sum }_g{c}_g{g}^{\otimes n}\in B$ and $\psi \in M_{\lambda }$ one has $T(b\cdot \psi )={\sum }_g{c}_{g}T(g^{\otimes n}\psi )={\sum }_g{c}_g{g}^{\otimes n}T(\psi )=b\cdot T(\psi )$, so $T$ is a $B$-module isomorphism and, both modules being nonzero, step 5.1 forces $\lambda =\mu$. Likewise an isomorphism of $\mathfrak{g}\mathfrak{l}(V)$-modules intertwines every $\mathrm{\Delta }(T)$, and since finite sums and products of such operators span $B$ by [F8], it is a $B$-module isomorphism and again forces $\lambda =\mu$. If exactly one of $\mathrm{\ell }(\lambda ),\mathrm{\ell }(\mu )$ is at most $d$, then exactly one of $M_{\lambda },M_{\mu }$ is zero by [F9], so the two are not isomorphic even as vector spaces. This proves claim 3.

Assume $\mathrm{\ell }(\lambda )\le d$, so $M_{\lambda }\mathrm{\ne }0$ by [F9] and $M_{\lambda }$ is irreducible over $B$ by step 5.1. The highest weight lemma [F10] then supplies a nonzero $\phi \in M_{\lambda }$ with $\mathrm{\Delta }(E_{ii})\circ \phi ={\lambda }_i\phi$ (with ${\lambda }_i=0$ for $i>\mathrm{\ell }(\lambda )$) and $\mathrm{\Delta }(E_{ij})\circ \phi =0$ for $i<j$, and shows that every nonzero $\psi \in M_{\lambda }$ killed by all raising operators and of weight $\mu$ satisfies $\mu =\lambda$ and $\psi \in \mathbb{C}\phi$. This proves claim 4.

Boundary and choice audit. For $n=0$ one has $E=\mathbb{C}$, the only partition is $\mathrm{\varnothing }$ with $\mathrm{\ell }(\mathrm{\varnothing })=0\le d$, $M_{\mathrm{\varnothing }}=\mathrm{Hom}(\mathbb{C},\mathbb{C})=\mathbb{C}$, and claim 1 reads $E\cong S^{\mathrm{\varnothing }}\otimes M_{\mathrm{\varnothing }}$; $B=\mathbb{C}\mathrm{i}\mathrm{d}$, $M_{\mathrm{\varnothing }}$ is a one-dimensional simple $B$-module, claim 4 holds with $\lambda =(0,\dots ,0)$ and claim 5 with degree $0$ polynomials, the constants. For $d=0$ and $n\ge 1$ one has $V=0$, $E=0$ and no partition of $n$ satisfies $\mathrm{\ell }(\lambda )\le 0$, so the sum in claim 1 is empty and $E=0$, the assertions of claims 2, 3, 4 and 5 are vacuous since all $M_{\lambda }=0$, and $B={\mathrm{End}}_{S_n}(0)=0$ consistently. For $d\ge 1$ and $n\ge 1$ there are finitely many partitions of $n$ and all spaces are finite-dimensional. The argument uses that $\mathbb{C}$ has characteristic $0$ not dividing $n!$ ([F3]), that $\mathbb{C}$ is algebraically closed ([F6], [F7]), and the fixed basis of $V$, the fixed basis of $S^{\lambda }$, the fixed tableau $t$ and the finite-dimensional retraction $r$ of step 2.2; finite sums over the partitions of $n$ and over the coordinate index sets occur throughout, and no choice principle is invoked. This proves claims 1, 2, 3, 4 and 5.