The degenerate cyclotomic Hecke algebra

Bounded spanning in every variable. For each $r$, put $w_r=s_{r-1}\cdots s_1$ (and $w_1=1$). In the affine algebra the cross relations give $w_r{x}_1{w}_r^{-1}=x_r+$ terms of polynomial degree zero. Hence $w_{r}f(x_1)w_r^{-1}=x_r^l+$ terms of degree at most $l-1$. This element is zero in $A$. Every affine ordered monomial with an exponent at least $l$ can therefore have a factor $x_r^l$ replaced by terms of strictly smaller total polynomial degree; straightening those terms never increases that degree. Induction on total degree, applied to every such replacement, expresses every monomial as a linear combination of $x_1^{r_1}\cdots x_d^{r_d}w$ with $0\le r_i<l$. The same induction proves that $F_{m}A$ is spanned by bounded monomials of degree at most $m$.

A realization for arbitrary complex roots. For $d\ge 1$ choose an integer $n\ge d$ and the rectangular partition $\lambda =(l,\dots ,l)$ with $n$ rows, so every column has height $q_j=n$ and $e$ has $n$ strings of length $l$. On $\mathfrak{p}$ define the one-dimensional module ${\mathbb{C}}_{\mathbf{a}}$ by $E_{ab}\mapsto {\delta }_{ab}{a}_{\mathrm{c}\mathrm{o}\mathrm{l}(a)}$. This is a Lie character: it vanishes on the strictly positive nilradical and is $a_j$ times the trace on the $j$th Levi block. In particular no integrality or coset condition on $a_j$ is needed. Restrict this module to $W$ and use F3 with $X=V^{\otimes d}$. Restrict the affine Hecke operators on $(Q_{\chi }{\otimes }_W{\mathbb{C}}_{\mathbf{a}})\otimes V^{\otimes d}$ to Whittaker vectors, then transport through F3. This defines an affine Hecke action on $V_{\mathbf{a}}^{\otimes d}={\mathbb{C}}_{\mathbf{a}}\otimes V^{\otimes d}$; the $s_i$ are place permutations. The following step computes $x_1$ before asserting any quotient factorization.

The transported first operator. Write $Q_c=Q_{\chi }{\otimes }_W{\mathbb{C}}_{\mathbf{a}}$ and define ${\pi }_c:Q_c\to \mathbb{C}$ by ${\pi }_c(\eta (u)1_{\chi }\otimes 1)=u\cdot 1$. It is well defined: the right $W$-relations become multiplication by the restricted character, and PBW gives representatives $\eta (u)1_{\chi }$. For a tensor basis vector $v_{\mathbf{j}}$ write its unique Whittaker preimage under F3 as $z={\sum }_{\mathbf{i}}{z}_{\mathbf{i}}\otimes v_{\mathbf{i}}$, with $z_{\mathbf{i}}\in Q_c$. Then ${\pi }_c(z_{\mathbf{i}})={\delta }_{\mathbf{i},\mathbf{j}}$. If $\mathrm{c}\mathrm{o}\mathrm{l}(a)\le \mathrm{c}\mathrm{o}\mathrm{l}(b)$, then $E_{ab}\in \mathfrak{p}$ and ${\pi }_c(E_{ab}{z}_{\mathbf{i}})={\delta }_{ab}(a_{\mathrm{c}\mathrm{o}\mathrm{l}(a)}+{\sum }_{h\ge \mathrm{c}\mathrm{o}\mathrm{l}(a)}{q}_h-n){\delta }_{\mathbf{i},\mathbf{j}}$, since ${\eta }^{-1}(E_{ab})=E_{ab}-{\delta }_{ab}(n-{\sum }_{h\ge \mathrm{c}\mathrm{o}\mathrm{l}(a)}{q}_h)$. If $\mathrm{c}\mathrm{o}\mathrm{l}(a)>\mathrm{c}\mathrm{o}\mathrm{l}(b)$, the Whittaker equation for $E_{ab}$ instead gives $E_{ab}{z}_{\mathbf{i}}=\chi (E_{ab})z_{\mathbf{i}}-{\sum }_{\mathbf{k}}{z}_{\mathbf{k}}$, where $\mathbf{k}$ runs over tensors obtained from $\mathbf{i}$ by replacing an occurrence of $a$ by $b$. Apply ${\pi }_c\otimes 1$ to ${\mathrm{\Omega }}_{01}z={\sum }_{a,b,\mathbf{i}}{E}_{ab}{z}_{\mathbf{i}}\otimes E_{ba}^{(1)}{v}_{\mathbf{i}}$. The terms with $\mathrm{c}\mathrm{o}\mathrm{l}(a)\le \mathrm{c}\mathrm{o}\mathrm{l}(b)$ give $(a_{\mathrm{c}\mathrm{o}\mathrm{l}(j_1)}+{\sum }_{h\ge \mathrm{c}\mathrm{o}\mathrm{l}(j_1)}{q}_h-n)v_{\mathbf{j}}$. The $\chi$ term gives $v_{L_1(\mathbf{j})}$, because $\chi (E_{ab})=1$ exactly when $b=L(a)$. In the negative sum, replacing the first occurrence gives $-{\sum }_{h>\mathrm{c}\mathrm{o}\mathrm{l}(j_1)}{q}_h{v}_{\mathbf{j}}$; replacing an occurrence in position $k>1$ gives $-v_{\mathbf{j}\cdot (1k)}$ exactly when $\mathrm{c}\mathrm{o}\mathrm{l}(j_k)<\mathrm{c}\mathrm{o}\mathrm{l}(j_1)$. Consequently the transported right action is $v_{\mathbf{j}}{x}_1=v_{L_1(\mathbf{j})}+(a_{\mathrm{c}\mathrm{o}\mathrm{l}(j_1)}+q_{\mathrm{c}\mathrm{o}\mathrm{l}(j_1)}-n)v_{\mathbf{j}}-{\sum }_{k>1:\mathrm{c}\mathrm{o}\mathrm{l}(j_k)<\mathrm{c}\mathrm{o}\mathrm{l}(j_1)}{v}_{\mathbf{j}\cdot (1k)}\mathrm{.}$ Here a nonexistent left-shift term is zero. In our rectangle $q_j=n$, so the diagonal coefficient is precisely $a_{\mathrm{c}\mathrm{o}\mathrm{l}(j_1)}$.

The cyclotomic relation and the leading symbols. Let $C_k$ be the span of tensors whose first entry has column at most $k$, with $C_0=0$. Step 1.3 gives $C_k(x_1-a_k)\subseteq C_{k-1}$: both the left shift and every exchange strictly lower the first column, while $x_1$ preserves $C_{k-1}$. Thus successively applying $(x_1-a_l),\dots ,(x_1-a_1)$ kills $C_l=V^{\otimes d}$. The factors commute, so $f(x_1)=0$ and the action factors through $A$. Grade $V$ by $\mathrm{deg}{v}_b=l-\mathrm{c}\mathrm{o}\mathrm{l}(b)$ and give its endomorphism algebra the induced filtration. In step 1.3 the left shift has degree one and every other term has total tensor degree zero. Hence ${\mathrm{gr}}_1{x}_1=e_1$. The recursion $x_{j+1}=s_j{x}_j{s}_j+s_j$ shows by induction that all $x_j$ are filtered of degree one with leading symbol $e_j$, and that $s_j$ have degree zero with symbols the place permutations.

Independence and the associated graded algebra. The cross relations and the conjugated relations in step 1.1 give a surjection $B=\mathbb{C}[X_1,\dots ,X_d]\mathrm{/}(X_1^l,\dots ,X_d^l)⋊\mathbb{C}[S_d]\twoheadrightarrow \mathrm{gr}A$. Its composite with the associated graded tensor representation is $X_j\mapsto e_j$, $w\mapsto$ place permutation. Choose the rightmost box in each of $d$ distinct rows and let $v$ be the tensor of those boxes. For bounded $\mathbf{r}$ and $w$, the vectors $v{e}_1^{r_1}\cdots e_d^{r_d}w$ are distinct tensor basis vectors: their row labels identify the permutation, and their columns identify all $r_i$. Therefore this composite is injective. It follows that $B\cong \mathrm{gr}A$. Taking leading symbols of a hypothetical linear relation among bounded monomials, first in its largest total degree and then inductively, proves their independence. Along with step 1.1 this gives the claimed PBW basis, its filtration compatibility, and $\dim A=l^{d}d!$. The longest strings have length $l$, the number of columns, and there are $n=q_l\ge d$ of them; no column height is used as their length. For $d=0$ the algebra, basis and associated graded are all $\mathbb{C}$, so the assertions hold directly.

Integral roots and the pyramid dictionary. For integral roots write $m_i=\mathrm{\#}\{j:a_j=i\}$; then $f(x_1)={\prod }_i(x_1-i{)}^{m_i}$, the quotient attached to $\mathrm{\Lambda }={\sum }_i{m}_i{\mathrm{\Lambda }}_i$. For arbitrary complex roots the same algebraic definition applies, with integer-chain bookkeeping separately in each coset of $\mathbb{C}\mathrm{/}\mathbb{Z}$. The rectangle above has $c_j=a_j$ and $q_j=n$, hence $a_j=c_j+q_j-n$; the general pyramid dictionary is the displayed diagonal term of step 1.3. No general complex root is identified with an integer-only fundamental-weight index.

Level one and the anti-involution. For $l=1$, $x_1=a_1$ and $x_{j+1}=s_j{x}_j{s}_j+s_j$ give $x_j=a_1+{\sum }_{k<j}(kj)$ by induction; the basis of step 3.1 identifies the quotient with $\mathbb{C}[S_d]$. The generator-fixing anti-involution fixes $f(x_1)$ and preserves its two-sided ideal, so descends to $A$.

The graded top pairing. Put $D=d(l-1)$ and $T=x_1^{l-1}\cdots x_d^{l-1}$. Let ${\tau }_0:B\to \mathbb{C}$ be the coefficient of $X_1^{l-1}\cdots X_d^{l-1}\cdot 1$. For graded monomials $X^{\mathbf{r}}w,X^{\mathbf{s}}v$, this coefficient is one exactly when $wv=1$ and $\mathbf{r}+w(\mathbf{s})=(l-1,\dots ,l-1)$, and is zero otherwise. The equivalent reversed condition is $vw=1$ and $\mathbf{s}+w^{-1}(\mathbf{r})=(l-1,\dots ,l-1)$; thus ${\tau }_0$ is symmetric. Each basis element pairs with its unique complementary element, so the pairing is perfect, including between degrees $m$ and $D-m$. Define $\tau :A\to \mathbb{C}$ as the coefficient of $T\cdot 1$ in the PBW basis. By filtration compatibility $\tau ={\tau }_0\circ {\mathrm{gr}}_D$ on $A=F_{D}A$, and it vanishes on $F_{D-1}A$. Therefore, if $a\in F_{m}A$ and $b\in F_{D-m}A$, then $\tau (ab)={\tau }_0({\mathrm{gr}}_m(a){\mathrm{gr}}_{D-m}(b))=\tau (ba)$. This establishes cyclicity only under the stated degree bound so far.

Full generator cyclicity. Since permutations have degree zero, step 4.3 gives $\tau (aw)=\tau (wa)$ for every $a\in A$, $w\in S_d$. To check $\tau (x_{i}b)=\tau (b{x}_i)$ for all bounded basis monomials $b$, step 4.3 suffices when $\mathrm{deg}b<D$. The remaining ones are exactly $b=Tw$. The polynomial $T$ is central: it commutes with the $x_j$, and the cross relations directly give $s_j(x_j{x}_{j+1})=(x_j{x}_{j+1})s_j$, while $s_j$ commutes with the other variables. Hence $s_{j}T=T{s}_j$. Applying the already proved permutation cyclicity twice gives $\tau (x_{i}Tw)=\tau (w{x}_{i}T)=\tau (wT{x}_i)=\tau (Tw{x}_i)$. This proves cyclicity with every generator on every element; successively moving generators of a word from the beginning to the end proves $\tau (ab)=\tau (ba)$ for all $a,b\in A$. It also covers $l=1$ (where $T=1$ and $D=0$).

Nondegeneracy and block restriction. For $0\mathrm{\ne }a\in A$, take its largest filtration degree $m$, so ${\mathrm{gr}}_m(a)\mathrm{\ne }0$. By the perfect graded pairing of step 4.3 choose homogeneous $b_0\in B_{D-m}$ with ${\tau }_0({\mathrm{gr}}_m(a)b_0)\mathrm{\ne }0$, and lift it to $b\in F_{D-m}A$. Then $\tau (ab)\mathrm{\ne }0$. Thus the pairing has zero left radical, and symmetry gives zero right radical; this proves nondegeneracy without claiming that products of excess total degree have zero trace. If $e$ is a central idempotent and $0\mathrm{\ne }a\in eA$, choose $b$ as above; then $\tau (aeb)=\tau (ab)\mathrm{\ne }0$, so the restriction to $eA=eAe$ is nondegenerate and symmetric as well.