Commuting symmetric-group and linear actions on a tensor power

Let $V$ be a finite-dimensional complex vector space and let $n\ge 0$. Write $\mathrm{End}(V)$ for the $\mathbb{C}$-vector space of all linear maps $V\to V$, and $\mathrm{GL}(V)\subseteq \mathrm{End}(V)$ for the invertible ones (A finite-dimensional representation $\rho :G\to \mathrm{GL}(V)$ over a field, and its degree). All tensor products below are over $\mathbb{C}$.

The tensor power. For $n\ge 1$ fix the parenthesization $E_n:=V^{\otimes n}:=\underset{n\text{ factors}}{\underbrace{V{\otimes }_{\mathbb{C}}\cdots {\otimes }_{\mathbb{C}}V}},V^{\otimes 0}:=\mathbb{C},$ and write $v_1\otimes \cdots \otimes v_n$ for the elementary tensor of $(v_1,\dots ,v_n)\in V^n$; for $n=0$ the unique elementary tensor is $1\in \mathbb{C}$. By Finite iterated tensor products represent multilinear maps independently of parenthesization the assignment $(v_1,\dots ,v_n)\mapsto v_1\otimes \cdots \otimes v_n$ is $\mathbb{C}$-multilinear and every element of $E_n$ is a finite $\mathbb{C}$-linear combination of elementary tensors; for $n=0$ this means $c=c\cdot 1$ for $c\in \mathbb{C}$. If $e_1,\dots ,e_d$ is a $\mathbb{C}$-basis of $V$, then by iterating The elementary tensors of two bases form the product basis of the tensor product the $d^n$ tensors $e_{i_1}\otimes \cdots \otimes e_{i_n}$ form a $\mathbb{C}$-basis of $E_n$, so $E_n$ is finite-dimensional.

Place permutations: the left $S_n$-action. Let $S_n$ be the symmetric group of the set $\{1,\dots ,n\}$ (The symmetric group $\mathrm{Sym}(X)$: the bijections of a set $X$ under composition). For $\sigma \in S_n$ define, on elementary tensors, $\sigma \cdot (v_1\otimes \cdots \otimes v_n):=v_{{\sigma }^{-1}(1)}\otimes \cdots \otimes v_{{\sigma }^{-1}(n)}\mathrm{.}$ The right-hand side depends $\mathbb{C}$-multilinearly on $(v_1,\dots ,v_n)$, so by the universal property in Finite iterated tensor products represent multilinear maps independently of parenthesization there is a unique $\mathbb{C}$-linear map $E_n\to E_n$ with this value on every elementary tensor. Uniqueness is what makes the rule well defined on all of $E_n$, since the elementary tensors span. The identity permutation acts trivially, and for $\sigma ,\tau \in S_n$ and every elementary tensor, $(\sigma \tau )\cdot (v_1\otimes \cdots \otimes v_n)=v_{(\sigma \tau {)}^{-1}(1)}\otimes \cdots \otimes v_{(\sigma \tau {)}^{-1}(n)}=v_{{\tau }^{-1}({\sigma }^{-1}(1))}\otimes \cdots \otimes v_{{\tau }^{-1}({\sigma }^{-1}(n))},$ while $\tau \cdot (v_1\otimes \cdots \otimes v_n)=v_{{\tau }^{-1}(1)}\otimes \cdots \otimes v_{{\tau }^{-1}(n)}$ and hence $\sigma \cdot (\tau \cdot (v_1\otimes \cdots \otimes v_n))=v_{{\tau }^{-1}({\sigma }^{-1}(1))}\otimes \cdots \otimes v_{{\tau }^{-1}({\sigma }^{-1}(n))}\mathrm{.}$ The two agree, so the assignments constitute a left action of the group $S_n$ on $E_n$; equivalently $E_n$ is a left module over $S_n$ (A finite-dimensional representation $\rho :G\to \mathrm{GL}(V)$ over a field, and its degree). For $n=0$ the group $S_0$ is trivial and acts on $E_0=\mathbb{C}$ by the identity.

Diagonal linear action. For $g\in \mathrm{GL}(V)$ define, on elementary tensors, $g^{\otimes n}\cdot (v_1\otimes \cdots \otimes v_n):=g{v}_1\otimes \cdots \otimes g{v}_n,$ again first on elementary tensors by multilinearity and then uniquely on $E_n$. Since $g$ and $h$ are linear, the two assignments compose in the expected way: on elementary tensors, $(gh{)}^{\otimes n}\cdot (v_1\otimes \cdots \otimes v_n)=g(h{v}_1)\otimes \cdots \otimes g(h{v}_n)=g^{\otimes n}\cdot (h^{\otimes n}\cdot (v_1\otimes \cdots \otimes v_n)),$ and $({\mathrm{i}\mathrm{d}}_V{)}^{\otimes n}$ is the identity. In particular $(g^{-1}{)}^{\otimes n}$ is the inverse of $g^{\otimes n}$. For $n=0$ every diagonal operator is the identity on $\mathbb{C}$. Thus $g\mapsto g^{\otimes n}$ is a group homomorphism $\mathrm{GL}(V)\to \mathrm{GL}(E_n)$, that is, a finite-dimensional representation of the group $\mathrm{GL}(V)$ on $E_n$.

The two actions commute. For $\sigma \in S_n$, $g\in \mathrm{GL}(V)$ and an elementary tensor, $\sigma \cdot (g^{\otimes n}\cdot (v_1\otimes \cdots \otimes v_n))=g{v}_{{\sigma }^{-1}(1)}\otimes \cdots \otimes g{v}_{{\sigma }^{-1}(n)}=g^{\otimes n}\cdot (\sigma \cdot (v_1\otimes \cdots \otimes v_n)),$ so the two linear maps $\sigma \cdot (-)$ and $g^{\otimes n}\cdot (-)$ commute; equivalently, every $g^{\otimes n}$ is an $S_n$-equivariant endomorphism of $E_n$ (Intertwiners, the spaces ${\mathrm{Hom}}_G(V,W)$ and ${\mathrm{End}}_G(V)$, equivalent representations, and faithful representations).

The diagonal infinitesimal operator. For $T\in \mathrm{End}(V)$ put $\mathrm{\Delta }(T):={\sum }_{i=1}^n{\mathbf{1}}^{\otimes (i-1)}\otimes T\otimes {\mathbf{1}}^{\otimes (n-i)}\in \mathrm{End}(E_n),$ where $\mathbf{1}={\mathrm{i}\mathrm{d}}_V$ and the sum is $0$ for $n=0$. Each summand lies in $\mathrm{End}(E_n)$, and $\mathrm{\Delta }:\mathrm{End}(V)\to \mathrm{End}(E_n)$ is $\mathbb{C}$-linear. It is the first coefficient of the diagonal action along the line $t\mapsto \mathbf{1}+tT$. In this polynomial calculation, $(\mathbf{1}+tT{)}^{\otimes n}$ is the tensor power of the endomorphism $\mathbf{1}+tT$; it agrees with the $\mathrm{GL}(V)$ action whenever that endomorphism is invertible. For every $t\in \mathbb{C}$ one computes, on elementary tensors, $(\mathbf{1}+tT{)}^{\otimes n}\cdot (v_1\otimes \cdots \otimes v_n)={\sum }_{k=0}^n{t}^k{\sum }_{1\le i_1<\cdots <i_k\le n}{w}_{i_1,\dots ,i_k},(w_{i_1,\dots ,i_k}{)}_j=\{\begin{array}{ll}T{v}_j,& j\in \{i_1,\dots ,i_k\},\\ v_j,& \text{otherwise}\end{array}$ the coefficient of $t$ being $\mathrm{\Delta }(T)\cdot (v_1\otimes \cdots \otimes v_n)$ (zero when $n=0$); both sides are polynomials in $t$ with values in the finite-dimensional space $E_n$ described on a spanning set. Finally let $A_n:=\text{the unital }\mathbb{C}\text{-subalgebra of }\mathrm{End}(E_n)\text{ generated by }\{\mathrm{\Delta }(T):T\in \mathrm{End}(V)\};$ this is the image of the diagonal action of the universal enveloping algebra $U(\mathfrak{g}\mathfrak{l}(V))$ on $E_n$, defined here as that generated algebra, with no Lie-theoretic input.

• Why the inverse is in the place action. Using $v_{\sigma (1)}\otimes \cdots \otimes v_{\sigma (n)}$ would give a right action, $(\sigma \tau )\cdot x=\tau \cdot (\sigma \cdot x)$, because then the permutation acts on the positions by $i\mapsto \sigma (i)$. The convention above is arranged so that $S_n$ acts on the left, which is the direction needed for the permutation-module and Specht-module conventions of this library.

• Degenerate cases. For $n=0$, $E_0=\mathbb{C}$ is the trivial representation of the trivial group $S_0$ and of $\mathrm{GL}(V)$, $\mathrm{\Delta }(T)=0$ is the empty sum, and $A_0=\mathbb{C}{\mathrm{i}\mathrm{d}}_{\mathbb{C}}$ is one-dimensional. For $n=1$, $E_1=V$, the group $S_1$ is trivial, $g^{\otimes 1}=g$, and $\mathrm{\Delta }(T)=T$; thus $A_1=\mathrm{End}(V)$. If $V=0$ then $E_n=0$ for $n\ge 1$ and $E_0=\mathbb{C}$, and all statements below about these spaces remain true with the zero space.

• $\mathrm{\Delta }$ is not multiplicative. In general $\mathrm{\Delta }(T{T}^{\mathrm{\prime }})\mathrm{\ne }\mathrm{\Delta }(T)\mathrm{\Delta }(T^{\mathrm{\prime }})$ for $\dim V>1$: the product expands to include cross terms ${\mathbf{1}}^{\otimes (a-1)}\otimes T\otimes {\mathbf{1}}^{\otimes (b-a-1)}\otimes T^{\mathrm{\prime }}\otimes {\mathbf{1}}^{\otimes (n-b)}$ with $a<b$, and the commutator relation is $[\mathrm{\Delta }(T),\mathrm{\Delta }(T^{\mathrm{\prime }})]=\mathrm{\Delta }([T,T^{\mathrm{\prime }}])$, which is not used below. What is used is that $A_n$ contains $\mathrm{\Delta }(T)$ for every $T$, hence every polynomial in these operators; the centralizer statement that uses this algebra is proved later on this page.

• No choice. All sums are finite sums over $n$ places and over the finite group $S_n$, and the multilinear universal property produces the maps directly; no selection principle is used.