The deformed tensor space for the type-A finite W-algebra

The column-trace function ${\kappa }_c$ kills $[\mathfrak{p},\mathfrak{p}]$: the diagonal column block of the commutator of two block-upper-triangular matrices is the commutator of their diagonal blocks, and its trace is zero. Therefore $X\mapsto X+{\kappa }_c(X)1$ respects Lie brackets and extends to the unital automorphism ${\eta }_c$ of $U(\mathfrak{p})$, with inverse ${\eta }_{-c}$. It preserves the column grading, since only degree-zero generators receive degree-zero scalars. The coproduct gives ${\rho }_d{\eta }_c(X)={\rho }_d(X)+{\kappa }_c(X)\mathrm{id}$, exactly the action on ${\mathbb{C}}_c\otimes E$; no factor $d$ multiplies the character. Its restriction to $W$ is the stated filtered left action.

Use the actual isomorphism ${\mu }_{{\mathbb{C}}_c,V^{\otimes d}}$ of F3, composed with coherent transported associativity, to identify ${\mathbb{C}}_c⊛V^{⊛d}$ with ${\mathbb{C}}_c\otimes E$. The left action is the one in step 1.1, because the balanced map intertwines $\mathrm{\Delta }\eta (w)$ on Whittaker vectors with $\mathrm{\Delta }(w)$ on the target. Indeed $(\eta \otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(w)=\mathrm{\Delta }\eta (w)$ and the evaluation map obeys $\psi (\eta (u)q)=u\psi (q)$. The ambient universal operators ${\mathrm{\Omega }}_{0j}+{\sum }_{k<j}{P}_{kj}$ and $P_{j,j+1}$ preserve the Whittaker subspace and commute with its $W$-action; transporting them through this actual map gives a right Hecke action commuting with $W$. Naturality transports the places to ordinary tensor permutations. This argument makes no claim about commutation with the full target $\mathfrak{p}$-action.

Each $e_{i,j;r}$ has degree $r$. If $r>0$ it contains no diagonal matrix units; if $r=0$ it contains diagonal matrix units only when $i=j$, and then precisely one in each column $1,\dots ,p_i$. Consequently ${\kappa }_c(e_{i,j;r})={\delta }_{ij}{\delta }_{r0}{\sum }_{b=1}^{p_i}{c}_b$. Since ${\mathfrak{g}}^e\subset \mathfrak{p}$, the character of step 1.1 restricts to a Lie character on ${\mathfrak{g}}^e$. Thus ${\eta }_c$ restricts to the graded unital automorphism ${\tau }_c$ of $U({\mathfrak{g}}^e)$, $X\mapsto X+{\kappa }_c(X)1$, with inverse ${\tau }_{-c}$. No assertion that ${\eta }_c$ preserves $W$ is needed.

To compute the transported $x_1$, write $Q_c=Q_{\chi }{\otimes }_W{\mathbb{C}}_c$ and ${\pi }_c(\eta (u)1_{\chi }\otimes 1_c)=u{1}_c$. The evaluation is balanced by F3. For a tensor basis vector $v_{\mathbf{j}}$, write its actual Whittaker preimage as $z={\sum }_{\mathbf{i}}{z}_{\mathbf{i}}\otimes v_{\mathbf{i}}$; then ${\pi }_c(z_{\mathbf{i}})={\delta }_{\mathbf{i},\mathbf{j}}$. For $E_{ab}\in \mathfrak{p}$, ${\pi }_c(E_{ab}{z}_{\mathbf{i}})={\delta }_{ab}(c_{\mathrm{col}(a)}-{\rho }_{\mathrm{col}(a)}){\delta }_{\mathbf{i},\mathbf{j}}$. For $E_{ab}\in \mathfrak{m}$, the actual Whittaker equation instead reads $E_{ab}{z}_{\mathbf{i}}=\chi (E_{ab})z_{\mathbf{i}}-{\sum }_{t:i_t=a}{z}_{\mathbf{i}[a\mapsto b\text{ at }t]}$. These two identities compute every coefficient of $({\pi }_c\otimes 1){\mathrm{\Omega }}_{01}z$.

The fixed loop chart F2 identifies the graded inclusion of $W$ in $U(\mathfrak{p})$ with the ordinary inclusion of $U({\mathfrak{g}}^e)$. The ordinary tensor representation ${\rho }_d$ and ${\eta }_c$ are graded maps. Therefore taking symbols of their restriction gives $\mathrm{gr}{\mathrm{\Phi }}_{d,c}={\phi }_{d,0}{\tau }_c\zeta$. This is equality of the full graded representation, not merely equality of its image. It proves ${\phi }_{d,c}(X)={\sum }_t{X}^{(t)}+{\kappa }_c(X)\mathrm{id}$ and the complete displayed $T_{i,j}^{(r+1)}$ symbol dictionary. The term ${\sigma }_i(c)$ survives in degree zero; for positive degree it is absent. The coefficient index, sign $(-1{)}^r$, finite $K$-range and physical box labels are unchanged.

In that computation the nonnegative-column terms give $(c_{\mathrm{col}(j_1)}+{\sum }_{h\ge \mathrm{col}(j_1)}{q}_h-n)v_{\mathbf{j}}$. The character terms from negative matrices give $v_{L_1(\mathbf{j})}$, since $\chi (E_{ab})=1$ exactly when $b=L(a)$. In the remaining negative sum, replacement in the first tensor position gives $-{\sum }_{h>\mathrm{col}(j_1)}{q}_h{v}_{\mathbf{j}}$, and replacement in position $k>1$ gives $-v_{\mathbf{j}\cdot (1k)}$ exactly when $\mathrm{col}(j_k)<\mathrm{col}(j_1)$. Adding yields $v_{\mathbf{j}}{x}_1=v_{L_1(\mathbf{j})}+a_{\mathrm{col}(j_1)}{v}_{\mathbf{j}}-{\sum }_{k>1:\mathrm{col}(j_k)<\mathrm{col}(j_1)}{v}_{\mathbf{j}\cdot (1k)}$, precisely the matrix-unit display in the Definition. A nonexistent left neighbour contributes zero. Recursion with the place permutations gives all higher $x_j$, and their affine relations already hold by transport in step 2.1. Thus the displayed action is the transported action, rather than an unrelated action satisfying a recursion alone.

For $d\ge 1$ and $l\ge 1$, let $C_k$ be the span of tensors whose first box has column at most $k$. The preceding formula preserves every $C_k$ and gives $C_k(x_1-a_k)\subseteq C_{k-1}$. Successively applying the factors from column $l$ down to column $1$ kills $C_l=E$; they are polynomials in one operator and commute, so ${\prod }_{b=1}^l(x_1-a_b)=0$. This proves the quotient factorization directly for arbitrary complex origins, including repeated roots. In the total grading $\mathrm{deg}{v}_a=l-\mathrm{col}(a)$, the left-shift part has degree one; the diagonal part and each two-position exchange have total degree zero. Hence ${\mathrm{gr}}_1{x}_1=e^{(1)}$. Since the longest Jordan string has length $l$, $(e^{(1)}{)}^k\mathrm{\ne }0$ for $0\le k<l$ on this nonzero tensor space. A monic annihilating polynomial of degree $k<l$ would have that nonzero degree-$k$ symbol. Therefore the displayed degree-$l$ polynomial is the exact minimal polynomial, consistent with F4.

The right action is filtered with $x_j$ of degree one and places of degree zero: this holds for $x_1$ by step 5.1 and for each $x_{j+1}$ by $x_{j+1}=s_j{x}_j{s}_j+s_j$. Taking degree-one symbols gives $e^{(j+1)}=s_j{e}^{(j)}{s}_j$; the extra $s_j$ has degree zero. By F4 the resulting graded right map is the ordinary map $X_j\mapsto e^{(j)}$, $w\mapsto P_w$ on $B^0$. Together with step 3.2 this identifies the entire graded bimodule with the character-twisted Vust module in the Definition.

Since ${\tau }_c$ is an automorphism, ${\phi }_{d,c}(U({\mathfrak{g}}^e))={\phi }_{d,0}(U({\mathfrak{g}}^e))=A^0$ as unital graded subalgebras of $\mathrm{End}(E)$. Equivalently each shifted generator differs from its untwisted operator by the explicitly known scalar multiple of the identity, and either set generates the other together with the common unit. Consequently their commutants are identical as well. By F5, $A^0$ and the unchanged right image $D^0$ are mutual centralizers. This proves the required Vust comparison without identifying the two fixed-chart modules.

Let $C={\mathrm{\Phi }}_{d,c}(W)$ and $D$ be the algebra of right Hecke operators, with filtrations induced from $\mathrm{End}(E)$. They commute by step 2.1. The symbol maps in steps 3.2 and 6.1 have images $A^0,D^0$, respectively, which are contained in $\mathrm{gr}C,\mathrm{gr}D$. For any commuting filtered operator algebras, taking the largest homogeneous terms of a commutator gives the inclusions $\mathrm{gr}C\subseteq \mathrm{gr}Z(D)\subseteq Z(\mathrm{gr}D)$, where $Z$ denotes the commutant in the relevant endomorphism algebra. Therefore $A^0\subseteq \mathrm{gr}C\subseteq \mathrm{gr}Z(D)\subseteq Z(\mathrm{gr}D)\subseteq Z(D^0)=A^0$. The same argument with $C,D$ interchanged gives $D^0\subseteq \mathrm{gr}D\subseteq \mathrm{gr}Z(C)\subseteq Z(\mathrm{gr}C)\subseteq Z(A^0)=D^0$. Every inclusion is equality. This also proves strict equality of the graded image algebras, rather than presuming that an arbitrary filtered representation is strict.

Since $C\subseteq Z(D)$ and their induced associated graded spaces agree by step 8.1, a member of $Z(D)$ can have its leading homogeneous term matched by a member of $C$ of the same filtration degree. Subtraction lowers the integer degree. The finite grading of $E$ bounds the endomorphism filtration below, so iteration terminates and gives $Z(D)=C$. Similarly $Z(C)=D$. Restoring the right-action opposite convention gives ${\mathrm{\Phi }}_{d,c}(W)={\mathrm{End}}_{H_d(\mathrm{\Lambda })}(E_c)$ and ${\mathrm{End}}_W(E_c{)}^{\mathrm{o}\mathrm{p}}={\mathrm{\Psi }}_d(H_d(\mathrm{\Lambda }))$. All arguments hold for arbitrary $c$ and every $d$; no stable injectivity or admissible-origin Morita assertion was used.

For $d=0$, $E_c={\mathbb{C}}_c$, ${\phi }_{0,c}(X)={\kappa }_c(X)$ and ${\mu }_0$ is the identity; the right algebra is $\mathbb{C}$. Both unital operator images and commutants are scalar, even when the fixed-chart character is nonzero. For an empty pyramid the positive tensor powers are zero and both operator image algebras and both commutants in $\mathrm{End}(0)$ are zero; degree zero is scalar. Thus the exceptional cases stated in the Definition are covered without evaluating a nonexistent $x_1$ or invoking a positive-string Vust hypothesis.

In the one-box case $p=(1)$, $q_1=n=1$, $c_1=1$, $d=1$, the fixed-chart loop generator is $T_{11}^{(1)}=E_{11}$, and step 3.2 gives its operator $1+1=2$; the untwisted Vust operator is $1$. Both unital images remain $\mathbb{C}$, exactly as step 7.1 proves. Separately, for ${\mathfrak{g}\mathfrak{l}}_2$, $p=(2)$, $c=(0,0)$ and $d=1$, the displayed $x_1$ is $E_{12}$ and $H=E_{11}-E_{22}\in \mathfrak{p}$ obeys $[H,x_1]=2E_{12}\mathrm{\ne }0$. This witnesses why the correct commutation claim is the transported $W$-commutation established in steps 2.1, 3.1 and 4.1. The full filtered bimodule, coherent ${\mu }_d$, exact graded character chart and all-origin mutual centralizers have now all been proved.