The Schur-Weyl mutual centralizer theorem on tensor powers

[construct] Recall that $A$ is a unital subalgebra of $\mathrm{End}(E)$, because it is the image of a unital algebra homomorphism, and that $B$ is a unital subalgebra, because $g^{\otimes n}{h}^{\otimes n}=(gh{)}^{\otimes n}$ for $g,h\in \mathrm{GL}(V)$; both contain ${\mathbf{1}}_E$ (for $B$ take $g=1_V$, and for $n=0$ note $g^{\otimes 0}={\mathrm{i}\mathrm{d}}_{\mathbb{C}}$ for every $g$ and $\mathrm{GL}(V)$ is nonempty).

By [F3] the $\mathbb{C}[S_n]$-module $E$ is completely reducible, so by [F4] and the uniqueness of the isotypic decomposition in [F6] there are subspaces $E_{\lambda }$ for $\lambda$ in the finite set $\mathrm{\Lambda }:=\{\lambda ⊢n:{\mathrm{Hom}}_{S_n}(S^{\lambda },E)\mathrm{\ne }0\}$ with $E={⨁}_{\lambda \in \mathrm{\Lambda }}{E}_{\lambda }$, each $E_{\lambda }$ is $\lambda$-isotypical, and the evaluation maps of [F6] give $S_n$-isomorphisms $E_{\lambda }\cong S^{\lambda }\otimes M_{\lambda }$ with $M_{\lambda }:={\mathrm{Hom}}_{S_n}(S^{\lambda },E)$ carrying the trivial action; consequently the action of $\sigma \in S_n$ on $E_{\lambda }$ corresponds to ${\rho }_{\lambda }(\sigma )\otimes {\mathbf{1}}_{M_{\lambda }}$, where ${\rho }_{\lambda }$ is the action on the irreducible $S^{\lambda }$.

The simple left $\mathbb{C}[S_n]$-modules are, by [F7], the column modules of the factors of a decomposition $\mathbb{C}[S_n]\cong {\prod }_{i=1}^r{M}_{n_i}(\mathbb{C})$, one class per factor and each supported on one factor; by [F8] the irreducible complex representations of the group $S_n$ are exactly the simple left $\mathbb{C}[S_n]$-modules, so by [F4] the isomorphism classes of simple left $\mathbb{C}[S_n]$-modules are exactly the classes $[S^{\lambda }]$, $\lambda ⊢n$. Hence the factors are indexed by the partitions of $n$, with $n_i={\dim }_{\mathbb{C}}{S}^{\lambda }$ for the factor attached to $\lambda$, and ${\dim }_{\mathbb{C}}\mathbb{C}[S_n]={\sum }_{\lambda ⊢n}({\dim }_{\mathbb{C}}{S}^{\lambda }{)}^2$.

An endomorphism $T\in \mathrm{End}(S\otimes M)$ of a tensor product, $S,M$ finite-dimensional over $\mathbb{C}$ with $M\mathrm{\ne }0$, commuting with ${\mathbf{1}}_S\otimes b$ for all $b\in \mathrm{End}(M)$ is of the form $T=x\otimes {\mathbf{1}}_M$ for a unique $x\in \mathrm{End}(S)$: choose bases $s_1,\dots ,s_p$ of $S$ and $m_1,\dots ,m_q$ of $M$ (both finite, with $q\ge 1$), write $T(s\otimes m_1)={\sum }_{j=1}^q{x}_j(s)\otimes m_j$ with well-defined linear maps $x_j:S\to S$; for each $j$ let $b_j\in \mathrm{End}(M)$ be given on the basis by $b_j(m_1)=m_j$ and $b_j(m_i)=0$ for $i\ge 2$; then $T(s\otimes m_j)=T((\mathbf{1}\otimes b_j)(s\otimes m_1))=(\mathbf{1}\otimes b_j)T(s\otimes m_1)=x_1(s)\otimes m_j$, so $T=x_1\otimes {\mathbf{1}}_M$ on all basis vectors, and conversely every $x\otimes {\mathbf{1}}_M$ commutes with the operators $\mathbf{1}\otimes b$.

The first displayed identity of claim 1 holds: a map $F\in \mathrm{End}(E)$ commutes with every $a\in A$ exactly when it commutes with $\sigma$ for every $\sigma \in S_n$, because $A$ is the linear span of the place operators $\sigma$; hence the centralizer of $A$ is ${\mathrm{End}}_{S_n}(E)$, which equals $B$ by [F2]. The same fact of [F2] identifies $B$ with the image of the diagonal action of $U(\mathfrak{g}\mathfrak{l}(V))$, so claim 2 holds; in particular $B$ is a subalgebra, in agreement with step 1.1.

The canonical algebra homomorphism $\mathrm{\Phi }:\mathbb{C}[S_n]\to {\prod }_{\lambda ⊢n}\mathrm{End}(S^{\lambda })$, $\mathrm{\Phi }(x)=({\rho }_{\lambda }(x){)}_{\lambda }$, is an isomorphism. It is injective: if $x$ acts as $0$ on every $S^{\lambda }$, then under the product decomposition of step 1.3 the element $x$ has a component in each factor which annihilates that factor's own column module, and so is $0$, so $x=0$; here [F8] identifies the column module of the factor attached to $\lambda$ with the simple module $S^{\lambda }$, on which $x$ acts as ${\rho }_{\lambda }(x)$ by definition of $\mathrm{\Phi }$. It is then bijective because, by step 1.3 and ${\dim }_{\mathbb{C}}\mathrm{End}(S^{\lambda })=({\dim }_{\mathbb{C}}{S}^{\lambda }{)}^2$, source and target have the same finite dimension ${\sum }_{\lambda ⊢n}({\dim }_{\mathbb{C}}{S}^{\lambda }{)}^2$.

The algebra ${\mathrm{End}}_{S_n}(E)$ is ${\prod }_{\lambda \in \mathrm{\Lambda }}\mathrm{End}(M_{\lambda })$: an $S_n$-endomorphism of $E$ maps each isotypic component $E_{\lambda }$ into itself, since the image of a copy of $S^{\lambda }$ is $0$ or a copy of $S^{\lambda }$ by [F5] and [F4]; on $E_{\lambda }$ it is, by [F6] and step 1.2, exactly the operator induced by a unique $a_{\lambda }\in \mathrm{End}(M_{\lambda })$ on the second tensor factor; maps between distinct components $E_{\lambda }\to E_{\mu }$ are zero by [F5] and [F4] because $S^{\lambda }$ and $S^{\mu }$ are non-isomorphic for $\lambda \mathrm{\ne }\mu$; and the identifications preserve composition, so this is an algebra isomorphism under which $a_{\lambda }$ corresponds to the operator ${\mathbf{1}}_{S^{\lambda }}\otimes a_{\lambda }$ on $E_{\lambda }$. In particular, by step 2.1, $B$ is exactly the set of operators ${⨁}_{\lambda \in \mathrm{\Lambda }}({\mathbf{1}}_{S^{\lambda }}\otimes b_{\lambda })$ with $b_{\lambda }\in \mathrm{End}(M_{\lambda })$.

Under the isomorphism $\mathrm{\Phi }$ of step 2.2 and the isomorphisms $E_{\lambda }\cong S^{\lambda }\otimes M_{\lambda }$ of step 1.2, the action of $x\in \mathbb{C}[S_n]$ on $E={⨁}_{\lambda \in \mathrm{\Lambda }}{E}_{\lambda }$ is ${⨁}_{\lambda \in \mathrm{\Lambda }}({\rho }_{\lambda }(x)\otimes {\mathbf{1}}_{M_{\lambda }})$, the factors with $\lambda \mathrm{\notin }\mathrm{\Lambda }$ acting on no summand; hence the image $A$ of this action map is exactly $A={⨁}_{\lambda \in \mathrm{\Lambda }}(\mathrm{End}(S^{\lambda })\otimes {\mathbf{1}}_{M_{\lambda }})\subseteq \mathrm{End}(E),$ the direct sum of the indicated block subalgebras indexed by $\mathrm{\Lambda }$: indeed the projection ${\prod }_{\lambda ⊢n}\mathrm{End}(S^{\lambda })\to {\prod }_{\lambda \in \mathrm{\Lambda }}\mathrm{End}(S^{\lambda })$ is surjective and $\mathrm{\Phi }$ is an isomorphism.

Claim 1's second identity holds. Let $F\in \mathrm{End}(E)$ commute with every element of $B$; decompose $F$ as a block matrix $(F_{\mu \lambda }{)}_{\mu ,\lambda \in \mathrm{\Lambda }}$ with $F_{\mu \lambda }\in \mathrm{Hom}(S^{\lambda }\otimes M_{\lambda },S^{\mu }\otimes M_{\mu })$ using the decomposition of step 1.2. Commuting with the operator ${\mathbf{1}}_{S^{\lambda }}\otimes b_{\lambda }$ on the $\lambda$-block and ${\mathbf{1}}_{S^{\mu }}\otimes b_{\mu }$ on the $\mu$-block gives $F_{\mu \lambda }({\mathbf{1}}_{S^{\lambda }}\otimes b_{\lambda })=({\mathbf{1}}_{S^{\mu }}\otimes b_{\mu })F_{\mu \lambda }$. If $\mu \mathrm{\ne }\lambda$, use $b_{\lambda }={\mathbf{1}}_{M_{\lambda }}$, $b_{\mu }=0$ (available in $B$ by step 3.1 and $\lambda ,\mu \in \mathrm{\Lambda }$), obtaining $F_{\mu \lambda }=0$. Thus every off-diagonal block vanishes. For each $\lambda$, the diagonal block $F_{\lambda \lambda }$ commutes with ${\mathbf{1}}_{S^{\lambda }}\otimes b$ for every $b\in \mathrm{End}(M_{\lambda })$, since $B$ contains the operators supported on that single block by step 3.1.

By steps 4.1 and 1.4 the centralizer of $B$ consists exactly of the elements ${⨁}_{\lambda \in \mathrm{\Lambda }}(x_{\lambda }\otimes {\mathbf{1}}_{M_{\lambda }})$ with $x_{\lambda }\in \mathrm{End}(S^{\lambda })$, which is exactly $A$ by step 3.2; this proves the second identity of claim 1, and claim 2 was proved in step 2.1.

Boundary and choice audit. For $n=0$ one has $E=V^{\otimes 0}=\mathbb{C}$, $A=B=\mathbb{C}{\mathrm{i}\mathrm{d}}_{\mathbb{C}}=\mathrm{End}(E)$ and both centralizers equal the whole of $\mathrm{End}(E)$, in agreement with claims 1 and 2; the case is covered by the argument with $\mathrm{\Lambda }=\{(\mathrm{\varnothing })\}$ when ${\mathrm{Hom}}_{S_0}(S^{\mathrm{\varnothing }},\mathbb{C})=\mathbb{C}$. For $V=0$ and $n\ge 1$ one has $E=0$, so $\mathrm{End}(E)=0=A=B$ by [F2] and all assertions hold; here $\mathrm{\Lambda }=\mathrm{\varnothing }$, the products over $\mathrm{\Lambda }$ are the zero algebra, and steps 4.1 and 1.4 are vacuous. For $V\mathrm{\ne }0$ and $n\ge 1$, every $\lambda \in \mathrm{\Lambda }$ has $M_{\lambda }\mathrm{\ne }0$ by definition, so step 1.4 applies with $M=M_{\lambda }$. All bases and decompositions used are attached to finite-dimensional spaces and to the finitely many partitions of $n$; the decomposition of $E$ and the factors of $\mathbb{C}[S_n]$ are canonical, and no choice principle is invoked.