The universal cyclotomic Hecke canonical trace is perfect and symmetric

The trace and its dual basis. Define $\tau \text{\hspace{0.17em}⁣}:H_n\to R$ by $\tau (\mathcal{B})=1$ at the zero-dot identity word and $0$ at every other PBW basis element, extended $R$-linearly; by [F1] this is well defined over $R$ and after base change. Put $J_1=1$ and $J_i=q_H^{-1}{T}_{i-1}{J}_{i-1}{T}_{i-1}$ in the finite Hecke subalgebra; the $J_i$ are units. Define the second family $\mathcal{C}=\{T_v{J}_1^{\delta (c_1)}\cdots J_n^{\delta (c_n)}{X}_1^{c_1}\cdots X_n^{c_n}:0\le c_i<\mathrm{\ell },v\in S_n\},\delta (0)=0,\delta (c)=-1(c>0)\mathrm{.}$ The standard anti-involution fixing the presentation generators fixes each $X_i$, so ${\mathcal{B}}^{\ast }=\{T_v{X}^c\}$ is a basis; at each fixed profile $c$, multiplying the finite Hecke factor by the unit $J_1^{\delta (c_1)}\cdots J_n^{\delta (c_n)}$ is an invertible $R$-linear change of the finite Hecke basis. Hence $\mathcal{C}$ is an $R$-basis of $H_n$.

Cyclicity for $X_1$ via the tower. At rank one the algebra is commutative, so cyclicity is immediate; for the induction assume $n\ge 2$. Identify $H_{n-1}$ with the subalgebra on $X_1,\dots ,X_{n-1},T_1,\dots ,T_{n-2}$; its bounded words are a subset of $\mathcal{B}$. Put $v_k=s_{n-1}\cdots s_k$ for $1\le k<n$ and $v_n=1$; the decomposition $w=u{v}_k$ with $u\in S_{n-1}$ is length additive, so $\{X_n^a{T}_{v_k}:0\le a<\mathrm{\ell },1\le k\le n\}$ is a left $H_{n-1}$-basis of $H_n$. Define the left-linear projection $E_n\text{\hspace{0.17em}⁣}:H_n\to H_{n-1}$ to be $1$ on $X_n^0{T}_{v_n}$ and $0$ on the other basis elements, and ${\tau }_n={\tau }_{n-1}\circ E_n$. With $V=T_{n-1}\cdots T_1$ and $U=(T_{n-1}-q_H+1)\cdots (T_1-q_H+1)$, the mixed relation gives $V{X}_1=X_{n}U$, and [F2] gives $V{X}_1^m=X_n^{m}U+J_m$, where every term of $J_m$ has total dot degree $m$ and $X_n$-degree between $1$ and $m-1$ (the leading branch transfers the full exponent, each correction transfers a strictly smaller positive exponent, and no variable above $n$ occurs). For $m\le \mathrm{\ell }$ every lower exponent in $J_m$ is below $\mathrm{\ell }$, so $E_n(J_m)=0$; likewise $E_n(X_n^{m}U)=0$ for $1\le m<\mathrm{\ell }$. Multiplying the first-dot relation $f(X_1)=0$ by $V$ gives $0=X_n^{\mathrm{\ell }}U+J_{\mathrm{\ell }}+{\sum }_{m=1}^{\mathrm{\ell }-1}{c}_m(X_n^{m}U+J_m)+c_{0}V$, and applying $E_n$ yields $E_n(X_n^{\mathrm{\ell }}U)=0$, which completes the missing case $m=\mathrm{\ell }$. To verify right $X_1$-linearity, test the displayed left tower basis $X_n^a{T}_{v_k}$. For $k\ge 2$, $T_{v_k}$ and $X_n$ commute with $X_1$; its projection is zero unless $a=0,k=n$, where the equality is immediate. For $k=1$, $V{X}_1=X_{n}U$ gives $E_n(X_n^{a}V{X}_1)=E_n(X_n^{a+1}U)=0$ for $0\le a<\mathrm{\ell }$: bounded positive exponents project to zero, and the endpoint $a+1=\mathrm{\ell }$ was just established. Left $H_{n-1}$-linearity then extends the equality to every $h$. Hence $E_n(h{X}_1)=E_n(h)X_1$ for all $h$, and by left linearity $E_n(X_{1}h)=X_1{E}_n(h)$; induction on $n$ gives ${\tau }_n(X_{1}h)={\tau }_n(h{X}_1)$.

Cyclicity for the finite generators. Let $p=X^a$ be a bounded dot monomial and $w\in S_n$. By [F2], $T_{j}p=(s_{j}p)T_j+(1-q_H)X_{j+1}\frac{p-s_{j}p}{X_j-X_{j+1}};$ for $p=X_j^a{X}_{j+1}^b$ times the other variables, the correction monomials have the same total dot degree and both displayed exponents at most $\max (a,b)<\mathrm{\ell }$ (the exponents run as $(a-1-k,b+1+k)$ or $(a+k,b-k)$), so when the exponent profile is positive every term on either side of $\tau (T_{j}p{T}_w)=\tau (p{T}_w{T}_j)$ has a positive bounded profile and evaluates to $0$. When $p=1$, the finite Hecke multiplication formula [F1] gives coefficient of $1$ in $T_j{T}_w$, respectively $T_w{T}_j$, equal to $q_H$ if $w=s_j$ and $0$ otherwise. This proves $\tau (T_{j}h)=\tau (h{T}_j)$ without using the trace one is proving.

Generation and symmetry. The elements $X_1,T_1,\dots ,T_{n-1}$ generate $H_n$ as an $R$-algebra: the constant term $c_0=(-1{)}^{\mathrm{\ell }}{\prod }_a{Q}_a$ is a unit, so $f(X_1)=0$ makes $X_1^{-1}$ a polynomial in $X_1$; each $T_j^{-1}=q_H^{-1}(T_j-q_H+1)$; and $X_r=q_H^{1-r}{T}_{r-1}\cdots T_1{X}_1{T}_1\cdots T_{r-1}$, with inverses expressed likewise. Moving the generators of a word one at a time around $\tau$, using steps 1.2 and 1.3, gives $\tau (ab)=\tau (ba)$ for every pair of words, hence for all $a,b$ by $R$-bilinearity; the pairing $(a,b)\mapsto \tau (ab)$ is associative and symmetric.

Restricted-profile zero values. For $0\le j\le n$ let $U_j$ be the span of the PBW words whose dot entries above $j$ vanish, all finite permutations included. It is a right finite-Hecke module. Multiplying its dot part by any polynomial in $X_1,\dots ,X_j$ stays in $U_j$: transport the first cyclotomic relation successively only through $T_1,\dots ,T_{j-1}$, so each resulting relation reduces a power of $X_i$, $i\le j$, using variables of index at most $i$ and finite words of indices below $i$; repeated Bernstein straightening and the bounded reduction of [F1] therefore never introduce a variable above $j$. In particular $U_j$ is the dot algebra on its first $j$ variables times the finite Hecke algebra. For $1\le k<\mathrm{\ell }$ and $h\in U_j$, $\tau (X_{j+1}^{k}h)=0$, directly from the bounded basis, since the new exponent is positive and bounded. This statement concerns $h\in U_j$, rather than arbitrary $h\in H_n$.

Transferred restricted-profile powers. Write $q=q_H$. For $1\le i\le j+1\le n$, $1\le k\le \mathrm{\ell }$ and $h\in U_j$ we claim $\tau (X_{j+1}^{k}h)=q^{i-j-1}\tau (T_j\cdots T_i{X}_i^k{T}_i\cdots T_{j}h)$, with empty chains when $i=j+1$. The mixed relations give $X_i^k=q^{-1}{T}_{i-1}{X}_{i-1}^k{T}_{i-1}+q^{-1}(q-1){\sum }_{c=1}^{k-1}{X}_i^c{X}_{i-1}^{k-c}{T}_{i-1}$, by induction on $k$ using that $T_{i-1}$ commutes with $X_{i-1}{X}_i$. To descend from $i$ to $i-1$, it suffices to show every correction has trace zero. Put $A=T_j\cdots T_i$, $B=T_i\cdots T_j$. For $0<c<k$, the mixed relation and the formula $X_{j+1}=q^{i-j-1}A{X}_{i}B$ give the exact identity $A{X}_i^c{X}_{i-1}^{k-c}{T}_{i-1}Bh=q^{j-i+1}A{X}_i^{c-1}{X}_{i-1}^{k-c-1}{T}_{i-1}{A}^{-1}{X}_{j+1}{X}_{i-1}h$: substitute $B=q^{j-i+1}{X}_i^{-1}{A}^{-1}{X}_{j+1}$ and use $X_i{T}_{i-1}{X}_i^{-1}=X_{i-1}^{-1}{T}_{i-1}{X}_{i-1}$. Expand $X_i^{c-1}{X}_{i-1}^{k-c-1}{T}_{i-1}{A}^{-1}$ in the reversed PBW order $\sum a_{w,b}{T}_w{X}^b$ using $T_a^{-1}=q^{-1}(T_a+1-q)$. Bernstein straightening swaps the two exponents or replaces them by intermediate exponents, so every exponent stays nonnegative and at most $\max (c-1,k-c-1)\le k-2$; variables have index at most $j+1$. Cyclicity from step 2.1 makes the trace of each correction a sum of traces of $X_{j+1}^{1+b_{j+1}}(X_1^{b_1}\cdots X_j^{b_j})X_{i-1}hA{T}_w$. The parenthesized remaining factor lies in $U_j$ by step 2.2, and $1\le 1+b_{j+1}\le k-1<\mathrm{\ell }$, so each trace is zero. Downward induction proves the claim, including $k=1$, whose correction sum is empty.

The cyclotomic endpoint. Apply step 3.1 with $i=1$, $k=\mathrm{\ell }$. In $\tau (q^{-j}{T}_j\cdots T_1{X}_1^{\mathrm{\ell }}{T}_1\cdots T_{j}h)$ replace $X_1^{\mathrm{\ell }}$ using its first cyclotomic relation. Each positive power $X_1^k$, $k<\mathrm{\ell }$, transfers back to $X_{j+1}^k$ by step 3.1 and contributes zero by step 2.2. The constant term contributes $\alpha \tau (J_{j+1}h)$, where $\alpha =-f(0)=(-1{)}^{\mathrm{\ell }+1}{Q}_1\cdots Q_{\mathrm{\ell }}$ and $J_{j+1}=q^{-j}{T}_j\cdots T_1{T}_1\cdots T_j$. Thus $\tau (X_{j+1}^{\mathrm{\ell }}h)=\alpha \tau (h{J}_{j+1})$ for every $h\in U_j$. For $j=0$ this follows directly from the first polynomial; for $\mathrm{\ell }=1$ there are no positive-power intermediate terms. The restriction on $h$ is essential.

The full Gram blocks. Let ${\mathcal{B}}_j$ and ${\mathcal{C}}_j$ be the parts of the two bases with entries above $j$ zero, and let $G_j=(\tau (bc))$ be their matrix, of size $N_j={\mathrm{\ell }}^{j}n!$. For new profiles $b,c$ at position $j+1$, cyclicity and commuting dots rewrite an entry as $\tau (X_{j+1}^{b+c}h{J}_{j+1}^{\delta (c)})$, where $h=X_1^{b_1+c_1}\cdots X_j^{b_j+c_j}{T}_w{T}_v{J}_1^{\delta (c_1)}\cdots J_j^{\delta (c_j)}\in U_j$ by step 2.2; earlier combined exponents may exceed $\mathrm{\ell }-1$, and that same reduction argument is why this remains in $U_j$. If $b=c=0$ the block is $G_j$. If $0<b+c<\mathrm{\ell }$ it is zero by step 2.2. If $b+c=\mathrm{\ell }$, then $c>0$ and step 4.1 cancels the last $J_{j+1}^{-1}$, giving the block $\alpha G_j$. Order rows by $b=0,1,\dots ,\mathrm{\ell }-1$ and columns by $c=0,\mathrm{\ell }-1,\dots ,1$. The row and column at zero have only $G_j$ at their intersection. Among positive rows and columns the blocks above the diagonal have $b+c<\mathrm{\ell }$ and vanish, the diagonal has $b+c=\mathrm{\ell }$, and blocks below may be arbitrary. Hence $G_{j+1}$ is block lower triangular with diagonal $G_j,\alpha G_j,\dots ,\alpha G_j$, so $\det G_{j+1}=\pm {\alpha }^{(\mathrm{\ell }-1)N_j}(\det G_j{)}^{\mathrm{\ell }}$. At $j=0$, $G_0=(\tau (T_w{T}_v))$ has entry $q^{\mathrm{length}(w)}{\delta }_{v,w^{-1}}$ by [F1], and its determinant is a unit. Induction gives a unit determinant for $G_n$; since $\mathcal{B}$ and $\mathcal{C}$ are bases, the pairing identifies $H_n$ with its $R$-dual.

Base change and boundary cases. Presentation base change commutes with forming $H_n$, and the universal PBW basis is free, so the trace coefficients and the inverse Gram matrix base-change to any commutative ring in which all parameters remain units; in particular they do so for the LM deformation DVR with $q_{H}t$ and $Q_a{t}^{na}$. For $n=0$ use $H_0=R$ with $\tau (1)=1_R$; for $\mathrm{\ell }=1$ the family $\mathcal{C}$ consists of finite $T_v$ and the same proof applies. The separate level-zero positive-rank quotient is $0$, with the zero perfect pairing, and no claim $\tau (1)=1$ is made for it. No generic faithfulness or imported Bremke–Malle trace statement is used.