Vust's graded double centralizer for the type-A nilpotent centralizer

The first centralizer. An operator $T\in \mathrm{End}(E)$ commutes with the action of $B$ if and only if it commutes with the individual $e$-operators and with the place permutations: the first condition says $T\in A^{\otimes d}$ by [F2], and then the second says $T\in (A^{\otimes d}{)}^{S_d}$, which is spanned by the elements $a^{\otimes d}$, $a\in A$. The image of $U({\mathfrak{g}}^e)$ under the diagonal representation is contained in ${\mathrm{End}}_B(E)$: the diagonal action of $A$ commutes with the individual $e$'s and with places. Conversely, for $a\in A$ the element $a^{\otimes d}$ equals the product of its individual-position operators; For the commuting individual operators $a_i$, put $Q(t)={\prod }_i(1+a_{i}t)={\sum }_r{q}_r{t}^r$ and $p_k={\sum }_i{a}_i^k$. Formal differentiation and the geometric series for each factor give $Q^{\mathrm{\prime }}(t)=Q(t){\sum }_{k\ge 1}(-1{)}^{k-1}{p}_k{t}^{k-1}$, so $q_0=1$ and $q_r=r^{-1}{\sum }_{k=1}^r(-1{)}^{k-1}{p}_k{q}_{r-k}$. Each $p_k={\mathrm{\Delta }}_d(a^k)$ belongs to the enveloping image because $A$ is an associative algebra; here $a^k$ is the matrix power, regarded as an element of the Lie algebra $A$. The recurrence therefore puts $q_d={\prod }_i{a}_i=a^{\otimes d}$ in that image. Hence $a^{\otimes d}\in \phi (U({\mathfrak{g}}^e))$ and $\phi (U({\mathfrak{g}}^e))={\mathrm{End}}_B(E)$.

The second centralizer. An operator $T$ commutes with the diagonal $A$-action if and only if it is $G_e$-fixed, as follows. If $T$ is $G_e$-fixed, commuting with $(1+ta{)}^{\otimes d}$ for all but finitely many $t$ and comparing its linear coefficient gives $[T,{\mathrm{\Delta }}_d(a)]=0$. Conversely, if $T$ commutes with every ${\mathrm{\Delta }}_d(a)$, it commutes with ${\mathrm{\Delta }}_d(a^k)$ for all $k$. Newton identities express $a^{\otimes d}$ in these power sums, so $T$ commutes with $a^{\otimes d}$, in particular for every unit $a\in G_e$. By [F3] every $G_e$-fixed operator is $\mathrm{\Phi }(e)$ for an equivariant polynomial covariant $\mathrm{\Phi }(X)={\sum }_h{c}_h(X){\sigma }_h(X^{h_1}\otimes \cdots \otimes X^{h_d})$; evaluating at $X=e$ replaces each $X^{h_i}$ by $e^{h_i}$, which is a polynomial in the individual $e$-operators of degree less than $l$, so $\mathrm{\Phi }(e)\in \psi (B)$ (the truncated algebra $B$ contains exactly the polynomials $e^0,\dots ,e^{l-1}$ in each factor and the places). Conversely every element of $\psi (B)$ commutes with the diagonal $A$-action: for $a\in A$ put ${\mathrm{\Delta }}_d(a)={\sum }_{k=1}^d{a}^{(k)}$; each individual $e^{(j)}$ satisfies $[{\mathrm{\Delta }}_d(a),e^{(j)}]=[a,e{]}^{(j)}=0$, and each place permutation $P_{\sigma }$ satisfies $P_{\sigma }{\mathrm{\Delta }}_d(a)P_{\sigma }^{-1}={\mathrm{\Delta }}_d(a)$. These operators generate $\psi (B)$, so their products and linear combinations also commute with every ${\mathrm{\Delta }}_d(a)$. Therefore ${\mathrm{End}}_{{\mathfrak{g}}^e}(E{)}^{\mathrm{o}\mathrm{p}}=\psi (B)$.

Stable injectivity. Suppose $e$ has at least $d$ Jordan strings of the maximum length $l$, with top vectors $w_1,\dots ,w_d$ on distinct strings. The vectors $w_1^{r_1}\otimes \cdots \otimes w_d^{r_d}$ with $0\le r_i<l$ (where $w^r=e^{r}w$ is the $r$-th vector of the string) are $l^d$ distinct members of the Jordan tensor basis of $E$; applying the $d!$ place permutations gives $l^{d}d!$ distinct basis vectors, since the string label of each tensor factor is recovered from the vector and the place permutation is recovered from the ordering of the labels. These vectors span a right $B$-submodule isomorphic to the regular module $B$ (the basis vectors $e^{r_1}{w}_1\otimes \cdots \otimes e^{r_d}{w}_d\sigma$, $\sigma \in S_d$, multiply exactly as the standard basis of the truncated polynomial/permutation algebra), so the right action $\psi$ is injective. For $d=0$, $E=\mathbb{C}$, $B=\mathbb{C}$ and both identities and injectivity reduce to scalar identities.