Diagonal tensor operators span the symmetric centralizer

[construct] Fix $T_1,\dots ,T_n\in W$. For fixed $(v_1,\dots ,v_n)\in V^n$ the assignment $T_i\mapsto T_i{v}_i$ is linear in each variable, so $(T_1,\dots ,T_n)\mapsto T_1{v}_1\otimes \cdots \otimes T_n{v}_n$ is multilinear in the $T_i$ and its values in $E$ are linear in $(v_1,\dots ,v_n)$ separately in each $v_i$; hence there is a unique linear map $\mathrm{\Phi }(T_1,\dots ,T_n)\in \mathrm{End}(E)$ with the displayed value on every elementary tensor by [F3], and $\mathrm{\Phi }$ is multilinear in the $T_i$ because both sides of every identity are checked on the spanning elementary tensors by [F3]. By the representing property of the tensor product [F3] there is a unique linear $\mathrm{\Psi }:W^{\otimes n}\to \mathrm{End}(E)$ with $\mathrm{\Psi }(T_1\otimes \cdots \otimes T_n)=\mathrm{\Phi }(T_1,\dots ,T_n)$, which is the map of the Statement; in particular $\mathrm{\Psi }$ is linear and is given on elementary tensors as displayed. For $n=0$ one has $W^{\otimes 0}=\mathbb{C}$ and $\mathrm{\Psi }(1)={\mathrm{i}\mathrm{d}}_{\mathbb{C}}$ by the same universal property.

Let $p:=\frac{1}{n!}{\sum }_{\sigma \in S_n}\sigma \in \mathrm{End}(W^{\otimes n})$, a well-defined operator because $n!\mathrm{\ne }0$ in $\mathbb{C}$; then $p^2=p$, since each group element occurs $n!$ times in the product sum, and $p$ commutes with the action, since left multiplication permutes the summands. Its image is exactly the invariant subspace $(W^{\otimes n}{)}^{S_n}$: $pX$ is invariant for every $X$, and $pX=X$ whenever $X$ is invariant. Thus the invariant tensors are the image of the symmetrization operator. Pure tensors span $W^{\otimes n}$, so their symmetrizations span the image of $p$. For $T_1,\dots ,T_n\in W$, inclusion-exclusion gives ${\sum }_{\sigma \in S_n}{T}_{\sigma (1)}\otimes \cdots \otimes T_{\sigma (n)}={\sum }_{J\subseteq \{1,\dots ,n\}}(-1{)}^{n-\mathrm{\mid }J\mathrm{\mid }}({\sum }_{j\in J}{T}_j{)}^{\otimes n}\mathrm{.}$ Indeed, after expanding the right side, an ordered tensor survives exactly when every index $1,\dots ,n$ occurs, which in a tensor with $n$ factors means each occurs once. Hence each symmetrized pure tensor is a linear combination of powers $T^{\otimes n}$, and $(W^{\otimes n}{)}^{S_n}=\mathrm{span}\{T^{\otimes n}:T\in W\}$. For $n=0$ this equality is immediate in $W^{\otimes 0}=\mathbb{C}$.

The two spans of the Statement are equal. One inclusion is clear, since $\mathrm{GL}(V)\subseteq W$. For the reverse let $T\in W$ and consider the $W^{\otimes n}$-valued polynomial $t\mapsto (T+t{\mathbf{1}}_V{)}^{\otimes n}$; expanding the tensor power of a sum gives $(T+t{\mathbf{1}}_V{)}^{\otimes n}={\sum }_{k=0}^n{t}^k{c}_k$ with $c_0=T^{\otimes n}$, so it is a polynomial of degree at most $n$ in $t$. The function $t\mapsto \det (t{\mathbf{1}}_V+T)$ is a monic polynomial of degree $d={\dim }_{\mathbb{C}}V$ by [F5], hence nonzero with at most $d$ roots by [F5], so by [F6] there exist $n+1$ distinct scalars $t_1,\dots ,t_{n+1}$ with $T+t_i{\mathbf{1}}_V$ invertible for all $i$ by [F5] (if $d=0$ then $W=\{0\}$ and $T=0$ is invertible, so any distinct scalars work). For the Lagrange polynomials $L_i(t):={\prod }_{j\mathrm{\ne }i}(t-t_j)\mathrm{/}(t_i-t_j)$ one has $t^k={\sum }_i{t}_i^k{L}_i(t)$ for every $t$ and every $0\le k\le n$, because it holds at the $n+1$ distinct points and both sides have degree at most $n$; multiplying by $c_k$ and summing gives $T^{\otimes n}=c_0={\sum }_i{L}_i(0)(T+t_i{\mathbf{1}}_V{)}^{\otimes n}\in \mathrm{span}\{g^{\otimes n}:g\in \mathrm{GL}(V)\}$. Hence the two spans are equal.

The matrix units of $W$ with respect to a basis $e_1,\dots ,e_d$ of $V$ form a basis of $W$: define $E_{ab}\in W$ by $E_{ab}{e}_c:={\delta }_{bc}{e}_a$; every $S\in W$ satisfies $S={\sum }_{a,b}{s}_{ab}{E}_{ab}$ with $s_{ab}$ the coefficients in $S(e_b)={\sum }_a{s}_{ab}{e}_a$, because the two sides agree on each basis vector of $V$, and a relation ${\sum }_{a,b}{\lambda }_{ab}{E}_{ab}=0$ evaluated at $e_b$ gives ${\sum }_a{\lambda }_{ab}{e}_a=0$, so all ${\lambda }_{ab}=0$. Likewise the endomorphisms $f_{ab}$ of $E$ defined by $f_{ab}(e_{c_1}\otimes \cdots \otimes e_{c_n}):={\delta }_{b_1{c}_1}\cdots {\delta }_{b_n{c}_n}{e}_{a_1}\otimes \cdots \otimes e_{a_n}$ for words $a,b\in [d{]}^n$ form a basis of $\mathrm{End}(E)$, indexed by the finite set $[d{]}^n\times [d{]}^n$; for $n=0$ this is the single endomorphism ${\mathrm{i}\mathrm{d}}_{\mathbb{C}}$ of the one-dimensional space $\mathbb{C}$, and for $d=0$ and $n\ge 1$ all these sets of words are empty and both spaces are zero. By [F2] and [F3] the $d^{2n}$ tensors $E_{a_1{b}_1}\otimes \cdots \otimes E_{a_n{b}_n}$ form a basis of $W^{\otimes n}$, and $\mathrm{\Psi }$ sends such a tensor to $f_{ab}$ by the formula of step 1.1; a linear map that carries a basis bijectively onto a basis is an isomorphism, so $\mathrm{\Psi }$ is a linear isomorphism.

$\mathrm{\Psi }$ is $S_n$-equivariant for the place action on $W^{\otimes n}$ and conjugation on $\mathrm{End}(E)$: for $\sigma \in S_n$ and $T_1,\dots ,T_n\in W$ the place action gives $\sigma \cdot (T_1\otimes \cdots \otimes T_n)=T_{{\sigma }^{-1}(1)}\otimes \cdots \otimes T_{{\sigma }^{-1}(n)}$ by [F1], so by step 1.1 both $\mathrm{\Psi }(\sigma \cdot (T_1\otimes \cdots \otimes T_n))$ and $\sigma \mathrm{\Psi }(T_1\otimes \cdots \otimes T_n){\sigma }^{-1}$ map $v_1\otimes \cdots \otimes v_n$ to $T_{{\sigma }^{-1}(1)}{v}_1\otimes \cdots \otimes T_{{\sigma }^{-1}(n)}{v}_n$; two linear maps agreeing on the spanning elementary tensors agree, so the identity $\mathrm{\Psi }(\sigma \cdot X)=\sigma \mathrm{\Psi }(X){\sigma }^{-1}$ holds for all $X\in W^{\otimes n}$ by linearity.

For $T\in W$ one has $\mathrm{\Psi }(T^{\otimes n})\in A_n$. For $n=0$ this is ${\mathrm{i}\mathrm{d}}_{\mathbb{C}}\in A_0$, since $A_0$ is unital. Assume $n\ge 1$ and put $T_i:={\mathbf{1}}^{\otimes (i-1)}\otimes T\otimes {\mathbf{1}}^{\otimes (n-i)}\in \mathrm{End}(E)$ for $1\le i\le n$; these operators commute pairwise and $\mathrm{\Psi }(T^{\otimes n})=T_1{T}_2\cdots T_n$, while ${\sum }_{i=1}^n{T}_i^k=\mathrm{\Delta }(T^k)$ for every $k\ge 1$, all by the tensor formula of step 1.1 and [F1]. By [F7] with $R=\mathbb{C}$ (where $n!$ is a unit) there is a polynomial $Q$ in $n$ variables over $\mathbb{C}$ with $e_n(x_1,\dots ,x_n)=Q(p_1,\dots ,p_n)$ in $\mathbb{C}[x_1,\dots ,x_n]$, where $p_k={\sum }_i{x}_i^k$. The operators $T_1,\dots ,T_n$ commute, so the $\mathbb{C}$-subalgebra they generate is commutative, and the universal property of the polynomial ring [F8] gives a $\mathbb{C}$-algebra homomorphism sending $x_i\mapsto T_i$; it sends $e_n(x_1,\dots ,x_n)$ to $T_1\cdots T_n$ by [F7] and $p_k$ to $\mathrm{\Delta }(T^k)$, so $T_1\cdots T_n=Q(\mathrm{\Delta }(T),\dots ,\mathrm{\Delta }(T^n))\in A_n$, since each $\mathrm{\Delta }(T^k)$ lies in the generating algebra $A_n$ and $Q$ has coefficients in $\mathbb{C}$.

For $T\in W$ put $d_T:={\sum }_{i=1}^n{\mathbf{1}}^{\otimes (i-1)}\otimes T\otimes {\mathbf{1}}^{\otimes (n-i)}\in W^{\otimes n}$ (the empty sum for $n=0$, giving $d_T=0\in \mathbb{C}$). Then $\mathrm{\Psi }(d_T)=\mathrm{\Delta }(T)$ by step 1.1, and $d_T$ is invariant under the place action: a permutation sends the $i$-th summand, which has $T$ in position $i$, to the same kind of summand with $T$ in position $\sigma (i)$ by [F1], and $\sigma$ permutes the index set, so $\sigma \cdot d_T=d_T$. By the equivariance of step 2.2, $\mathrm{\Delta }(T)=\mathrm{\Psi }(d_T)$ is fixed by conjugation by every $\sigma \in S_n$, that is $\mathrm{\Delta }(T)\in {\mathrm{End}}_{S_n}(E)$; since ${\mathrm{End}}_{S_n}(E)$ is closed under addition and composition and $A_n$ is generated by the $\mathrm{\Delta }(T)$, this gives $A_n\subseteq {\mathrm{End}}_{S_n}(E)$.

Combining steps: $\mathrm{\Psi }$ induces a bijection from $(W^{\otimes n}{)}^{S_n}$ onto ${\mathrm{End}}_{S_n}(E)$ by step 2.2 and step 2.1, the invariant subspace is $\mathrm{span}\{T^{\otimes n}\}=\mathrm{span}\{g^{\otimes n}\}$ by steps 1.2 and 1.3, and $\mathrm{\Psi }(\mathrm{span}\{T^{\otimes n}\})\subseteq A_n\subseteq {\mathrm{End}}_{S_n}(E)$ by steps 3.1 and 2.3, so all four subspaces of the Statement coincide. For $n=0$: $W^{\otimes 0}=E=\mathbb{C}$, $\mathrm{\Psi }$ is the identity, $S_0$ is trivial, $T^{\otimes 0}=1$ for every $T$, $\mathrm{\Delta }(T)=0$ and $A_0=\mathbb{C}{\mathrm{i}\mathrm{d}}_{\mathbb{C}}$, and the displayed chain reads $\mathbb{C}=\mathbb{C}=\mathbb{C}=\mathbb{C}$. For $V=0$ and $n\ge 1$: $W=0$, $E=0$, both sides of the identity are the zero space, and $A_n=0$. This proves both claims.