Classical tensor centralizer from polarization and Newton identities

Under [F1], identify ${\mathrm{End}}_{\mathbb{Q}}(W)$ with $M_N(\mathbb{Q}{)}^{\otimes n}$. An endomorphism commutes with every place permutation exactly when it is fixed by conjugation by each such permutation. By [F2], this action permutes the tensor-unit basis, so its invariant subspace has as a basis the orbit sums of that basis: invariance says precisely that coefficients are constant on each finite orbit. Thus ${\mathrm{End}}_{B_0}(W)$ is spanned by the symmetrized tensors of $n$ matrix units.

For any $a_1,\dots ,a_n\in M_N(\mathbb{Q})$, multilinear expansion and inclusion–exclusion give ${\sum }_{J\subseteq \{1,\dots ,n\}}(-1{)}^{n-\mathrm{\mid }J\mathrm{\mid }}({\sum }_{j\in J}{a}_j{)}^{\otimes n}={\sum }_{\sigma \in S_n}{a}_{\sigma (1)}\otimes \cdots \otimes a_{\sigma (n)}$. Indeed, a term in the expansion indexed by a function $f:\{1,\dots ,n\}\to \{1,\dots ,n\}$ occurs with total coefficient ${\sum }_{J\supseteq \mathrm{im}f}(-1{)}^{n-\mathrm{\mid }J\mathrm{\mid }}$, which is $1$ when $\mathrm{im}f$ is all of $\{1,\dots ,n\}$ and $0$ otherwise; surjective functions between these equal finite sets are permutations. Taking the $a_j$ to be the matrix units in any multiset orbit makes the right side the orbit sum multiplied by the product of the factorials of its multiplicities. That integer is nonzero and invertible in $\mathbb{Q}$, so every orbit sum is a rational linear combination of pure powers $a^{\otimes n}$.

Fix $a\in M_N(\mathbb{Q})$ and put $x_r=1^{\otimes (r-1)}\otimes a\otimes 1^{\otimes (n-r)}$. The $x_r$ commute by [F3], and $p_m={\sum }_r{x}_r^m={\rho }_n(a^m)\in A_0$. For $E(t)={\prod }_{r=1}^n(1+x_{r}t)={\sum }_{m=0}^n{e}_m{t}^m$, formal differentiation and $E^{\mathrm{\prime }}(t)\mathrm{/}E(t)={\sum }_{j\ge 1}(-1{)}^{j-1}{p}_j{t}^{j-1}$ give, by comparing coefficients, $m{e}_m={\sum }_{j=1}^m(-1{)}^{j-1}{e}_{m-j}{p}_j$. This induction puts each $e_m$ in $A_0$ because $m$ is invertible in $\mathbb{Q}$. In particular, $e_n=x_1\cdots x_n=a^{\otimes n}$ belongs to $A_0$. Step 1.2 now gives ${\mathrm{End}}_{B_0}(W)\subseteq A_0$. Conversely each ${\rho }_n(a)$ commutes with place permutations because conjugation merely permutes its summands, so $A_0\subseteq {\mathrm{End}}_{B_0}(W)$ and equality follows.

The map $a\mapsto {\rho }_n(a)$ is a Lie representation: expanding the commutator, terms in different positions cancel by [F3], leaving $[{\rho }_n(a),{\rho }_n(b)]={\rho }_n([a,b])$. The adjacent matrix units and the diagonal units generate every matrix unit under brackets: for $r<s$, recursively $E_{r,s}=[E_{r,s-1},E_{s-1,s}]$, and for $r>s$ use $E_{r,s}=[E_{r,r-1},E_{r-1,s}]$; each bracket uses a shorter interval and starts with an adjacent unit, while the $E_{jj}$ supply the diagonal basis. Their ${\rho }_n$ images therefore generate all ${\rho }_n(M_N(\mathbb{Q}))$ as an associative algebra, hence generate $A_0$.