Integral cyclotomic straightening and bounded PBW spanning

If $r=1$, $X_1=u_1$ and the affine recursion expresses every $X_i$ in the finite Hecke generators, so [F3] gives the claimed span. For $r\ge 2$ set $R^{\mathrm{\prime }}=R[v,w]\mathrm{/}(v^2-Q,w^r-(-1{)}^{r+1}{e}_r)$. Successive division by these monic polynomials gives the $R$-basis $v^i{w}^j$ with $0\le i<2$, $0\le j<r$, containing $1$; thus $R^{\mathrm{\prime }}$ is free of positive rank. Also $v^{-1}=v{Q}^{-1}$ and $w^{-1}=(-1{)}^{r+1}{e}_r^{-1}{w}^{r-1}$.

Set $a=v-v^{-1}$, $s_j=v^{-1}{T}_{j-1}$ for $2\le j\le n$, and $z=w^{-1}{X}_1$. Scaling gives $s_j^2=a{s}_j+1$, the type-A braid relations, and $z{s}_j=s_{j}z$ only for $j\ge 3$. The affine relation gives $z{s}_{2}z{s}_2=s_{2}z{s}_{2}z$: before scaling the two sides are $X_1{T}_1{X}_1{T}_1=Q{X}_1{X}_2$ and $T_1{X}_1{T}_1{X}_1=Q{X}_2{X}_1$. Expanding ${\prod }_{c=1}^r(wz-u_c)=0$ and using $w^r=(-1{)}^{r+1}{e}_r$ gives $z^r=b_1{z}^{r-1}+\cdots +b_{r-1}z+1$, where $b_k=(-1{)}^{k+1}{e}_k{w}^{-k}$. Hence powers of $z$ reduce to $0\le j<r$, and $z,s_2,\dots ,s_n$ generate the extended algebra.

Put $s=s_2$, $Y=szs$, and ${\mathcal{N}}_2={\mathrm{Span}}_{R^{\mathrm{\prime }}}\{z^c,z^{c}s{z}^d,z^{c}s{z}^{d}s:0\le c,d<r\}$. The rank-four braid gives $Yz=zY$, and expansion of $Y^2=sz{s}^{2}zs$ using $s^2=as+1$ gives $Y^2=a^{2}zY+azsz+s{z}^{2}s$, hence $s{z}^{2}s=Y^2-a^{2}zY-azsz$. Inductively every $Y^m$, $1\le m<r$, is in ${\mathcal{N}}_2$: multiply a term of each of the three displayed forms by $Y$, use $sY=aY+zs$, $z^{d}szs=szs{z}^d$, and $s^2=as+1$, then reduce every excess $z$ power by the monic relation. In particular $z^{c}s{z}^{d}sY=a{z}^{c+d}Y+z^{c+1}s{z}^d$, and $z^{c}s{z}^{d}sY$ with a final $s$ first expands its middle $s^2$ into this expression and $z^{c}s{z}^{d+1}s$. Also $Ys=aY+sz$ and $Yz=zY$, so induction gives $Y^{m}s$ as a sum of $Y^m,z{Y}^{m-1},\dots ,z^{m-1}Y,s{z}^m$, all in ${\mathcal{N}}_2$.

For $U_m=s{z}^{m}s$, $1\le m<r$, induction gives $U_m\in \mathrm{Span}\{z^c,z^{c}s,z^c{Y}^q,z^{c}s{z}^d:c<r,1\le q\le m,1\le d\le m-1\}$. The base terms are $U_1=Y$ and $U_2=Y^2-a^{2}zY-azsz$. For $m\ge 3$, inserting $s^{-1}s$ between $z^{m-1}$ and $z$ gives $U_m=U_{m-1}Y-as{z}^{m-1}Y$, while the rank-four braid and quadratic relation give $s{z}^{m-1}Y=a{z}^{m-1}Y+zs{z}^{m-1}$. Multiplication of the four inductive term types by $Y$ uses $z^{c}Y$, $sY=aY+zs$, $z^c{Y}^{q}Y=z^c{Y}^{q+1}$, and $z^{c}s{z}^{d}Y=a{z}^{c+d}Y+z^{c+1}s{z}^d$; every resulting exponent at least $r$ is reduced by the monic polynomial. This proves the induction. It also proves $U_m{z}^d,U_m{z}^{d}s\in {\mathcal{N}}_2$: commute $z^d$ across $Y^q$, expand $Y^q$ and $Y^{q}s$ by step 1.3, and reduce excess powers. The corrected coefficient identity in the rank-two induction is $a{z}^{c}s{z}^{c^{\mathrm{\prime }}}szs=a^2{z}^{c+c^{\mathrm{\prime }}}Y+a{z}^{c+1}s{z}^{c^{\mathrm{\prime }}}$. Left multiplication by $s$ on $z^c$, $z^{c}s{z}^d$, $z^{c}s{z}^{d}s$ gives respectively $s{z}^c$, $U_c{z}^d$, $U_c{z}^{d}s$; for $c=0$ use $s^2=as+1$. Thus ${\mathcal{N}}_2$ is stable under $z,s$, contains $1$, and equals the rank-two algebra.

For $i\ge 2$ put ${\mathrm{\Lambda }}_1=\{z^c:0\le c<r\}$, $D_i(j)=s_i\cdots s_j$ for $2\le j\le i$, and $D_i(i+1)=1$. Put $L_i(c)=D_i(2)z^c$ for $1\le c<r$, $L_i(0)=D_i(2)$, and $S_i(c,j)=D_i(2)z^c{s}_2\cdots s_j$ for $1\le c<r$, $2\le j\le i$. Let ${\mathrm{\Lambda }}_i$ contain $1$, all $D_i(j)$, all $L_i(c)$ with $c>0$, and all $S_i(c,j)$. Then $\mathrm{\mid }{\mathrm{\Lambda }}_1\mathrm{\mid }=r$, $\mathrm{\mid }{\mathrm{\Lambda }}_i\mathrm{\mid }=ir$, and step 2.1 gives the base $M_2=H_{r,2}{\otimes }_R{R}^{\mathrm{\prime }}$ for $M_n={\mathrm{Span}}_{R^{\mathrm{\prime }}}({\mathrm{\Lambda }}_1\cdots {\mathrm{\Lambda }}_n)$. If $M_{n-1}=H_{r,n-1}{\otimes }_R{R}^{\mathrm{\prime }}$, the older generators preserve $M_n=M_{n-1}{\mathrm{\Lambda }}_n$. Write each prefix as $h^{\mathrm{\prime }}{a}_{n-1}$ with $h^{\mathrm{\prime }}\in H_{r,n-2}{\otimes }_R{R}^{\mathrm{\prime }}$, $a_{n-1}\in {\mathrm{\Lambda }}_{n-1}$; $s_n$ commutes with $h^{\mathrm{\prime }}$, so it remains to reduce $s_n(a_{n-1}{a}_n)$. For $a_n=1$, $s_n{D}_{n-1}(j)=D_n(j)$, $s_n{L}_{n-1}(c)=L_n(c)$, and $s_n{S}_{n-1}(c,j)=S_n(c,j)$. If $a_{n-1}=1$, use $s_n{D}_n(j)=a{D}_n(j)+D_{n-1}(j)$ and, for loops or sandwiches, $s_n{D}_n(2)=a{D}_n(2)+D_{n-1}(2)$ followed by the finite coset factorization proved next. These terms are in $H_{r,n-1}{\mathrm{\Lambda }}_n$.

The finite group has the left-coset decomposition ${\mathfrak{S}}_n={\mathfrak{S}}_{n-1}\{D_n(j):2\le j\le n+1\}$, where $D_n(n+1)=1$: the representatives place the value $n$ in distinct positions, and left multiplication by ${\mathfrak{S}}_{n-1}$ reorders the other values. Their lengths are $n-j+1$, the inversions involving $n$; inversions among the other values are exactly those of the left factor, so lengths add. By [F3], every finite Hecke word therefore expands into $H_{r,n-1}\{D_n(j)\}$. This proves the repeated-square propagation as well: $D_n(3)s_2^2{s}_3\cdots s_i$ is finite for $2\le i\le n$, and expanding each square by $s_m^2=a{s}_m+1$ leaves finite words to which the same factorization applies. The braid sweep used below is $D_n(3)s_2{s}_3\cdots s_j=s_2{s}_3\cdots s_{j-1}{D}_n(2)$ for $2\le j\le n$: the base is $D_n(3)s_2=D_n(2)$, and $D_n(2)s_m=s_{m-1}{D}_n(2)$ for $3\le m\le n$, obtained by commuting the rightmost $s_m$ past $s_{m-2},\dots ,s_2$, applying the adjacent braid, then commuting $s_{m-1}$ past $s_n,\dots ,s_{m+1}$.

A last-stage word in ${\mathrm{\Lambda }}_n$ contains no $X$ or one factor $z^k=w^{-k}{X}_1^k$ with $0\le k<r$, so its total nonnegative $X$-degree is below $r$. Repeatedly apply [F2] to move polynomial factors left of Hecke generators; the divided-difference correction preserves nonnegative exponents and total degree. Thus each last-stage word becomes a sum of $X_1^{a_1}\cdots X_n^{a_n}{T}_y$ with $\sum a_i<r$.

The first three final-factor cases are as follows. For descending/descending, $s_n{D}_{n-1}(i)D_n(j)=D_{n-1}(j-1)D_n(i)$ if $i<j$, and $=a{D}_{n-1}(i)D_n(j)+D_{n-1}(j)D_n(i+1)$ if $i\ge j$: slide the leading $s_n$ through the overlapping descending string by the adjacent braid; at the common $s_i$ use $s_i^2=a{s}_i+1$ to obtain the second branch. For descending/loop, $s_n{D}_{n-1}(i)D_n(2)z^k=a{D}_{n-1}(i)L_n(k)+L_{n-1}(k)D_n(i+1)$; in the second term commute $z^k$ left past $s_n,\dots ,s_3$. For descending/sandwich, $s_n{D}_{n-1}(i)S_n(k,j)=a{D}_{n-1}(i)S_n(k,j)+S_{n-1}(k,j)D_n(i+1)$ if $j<i$, and $=a{D}_{n-1}(i)S_n(k,j)+S_{n-1}(k,j-1)D_n(i)$ if $j\ge i$; after the quadratic term, move the remaining suffix $s_2\cdots s_j$ left past $D_n(i+1)$ when $j<i$, and past $D_n(i)$ stopping at $s_{j-1}$ when $j\ge i$. In the second branch $S_{n-1}(k,1)=L_{n-1}(k)$.

For loop/descending, $s_n{L}_{n-1}(k)D_n(j)=D_{n-1}(2)S_n(k,2)$ if $j=2$, and $=D_{n-1}(j-1)L_n(k)$ if $j>2$; the first is the tail $D_n(2)z^k{s}_2$, and the second moves $z^k$ left past $s_n,\dots ,s_3$. For loop/loop, the $j=2$ sweep leaves $D_{n-1}(2)D_n(3)(s{z}^{k}s{z}^l)$; step 2.1 expands the parenthesized tail into $z^c$, $z^{c}s{z}^{c^{\mathrm{\prime }}}$, and $z^{c}s{z}^{c^{\mathrm{\prime }}}s$ (with $c^{\mathrm{\prime }}>0$ in the last type), giving respectively $L_{n-1}(c)D_n(3)$, $L_{n-1}(c)L_n(c^{\mathrm{\prime }})$, and $L_{n-1}(c)S_n(c^{\mathrm{\prime }},2)$ because $D_n(3)s=D_n(2)$. For loop/sandwich, append $s_2\cdots s_j$ to those three tails. The first is finite after the old prefix. The second is $D_n(2)z^{c^{\mathrm{\prime }}}{s}_2\cdots s_j=S_n(c^{\mathrm{\prime }},j)$ when $c^{\mathrm{\prime }}>0$, and finite when $c^{\mathrm{\prime }}=0$. In the third, expand the final $s^2$: its $a$-term is the preceding sandwich, while the other term has tail $D_n(3)s{z}^{c^{\mathrm{\prime }}}{s}_3\cdots s_j$; commute $z^{c^{\mathrm{\prime }}}$ across $s_3,\dots ,s_j$ and use step 4.1 to obtain an old-algebra prefix followed by $L_n(c^{\mathrm{\prime }})$.

Factor $y=h{D}_n(j)$ length-additively with $h\in {\mathfrak{S}}_{n-1}$. Since $X_n$ commutes with $X_1,\dots ,X_{n-1}$ and $T_1,\dots ,T_{n-2}$, each term lies in $H_{r,n-1}{X}_n^{a_n}{T}_{D_n(j)}$, with $a_n<r$. Induction on $n$ therefore proves bounded PBW spanning over $R^{\mathrm{\prime }}$.

For sandwich/descending, $s_n{S}_{n-1}(k,i)D_n(j)=D_{n-1}(i+1)S_n(k,i+1)$ if $j=i+1$, $=D_{n-1}(j-1)S_n(k,i)$ if $i<j-1$, and $=D_{n-1}(j)S_n(k,i+1)$ if $i\ge j$. These are obtained by moving the descending string through the ascending suffix: at each index use $s_m{s}_{m-1}{s}_m=s_{m-1}{s}_m{s}_{m-1}$ and commute letters at distance at least two; the three stopping points are respectively after $s_{i+1}$, before the suffix, and after crossing $s_i$. For sandwich/loop, the $j=2$ sweep gives $D_{n-1}(2)D_n(3)(s{z}^{k}s{z}^l)s_3\cdots s_{i+1}$. Expand the rank-two tail as $z^c$, $z^{c}s{z}^{c^{\mathrm{\prime }}}$, $z^{c}s{z}^{c^{\mathrm{\prime }}}s$. The first is finite after an old-algebra prefix. In the second, $c^{\mathrm{\prime }}=0$ leaves a finite tail; for $c^{\mathrm{\prime }}>0$, commute $z^{c^{\mathrm{\prime }}}$ through the suffix and use $D_n(3)s_2\cdots s_{i+1}=s_2\cdots s_i{D}_n(2)$ to obtain an old prefix times $L_n(c^{\mathrm{\prime }})$. In the third, $D_n(3)s{z}^{c^{\mathrm{\prime }}}s=D_n(2)z^{c^{\mathrm{\prime }}}s$, giving the old prefix times $S_n(c^{\mathrm{\prime }},i+1)$. For sandwich/sandwich, append $s_2\cdots s_j$ to the same expansion; the three terms are $D_{n-1}(2)z^c{D}_n(4)s_3^2{s}_4\cdots s_{i+1}(s_2\cdots s_j)$, $D_{n-1}(2)z^c{s}_2\cdots s_i{D}_n(2)z^{c^{\mathrm{\prime }}}(s_2\cdots s_j)$, and $D_{n-1}(2)z^c{D}_n(2)z^{c^{\mathrm{\prime }}}{s}_2\cdots s_{i+1}(s_2\cdots s_j)$. The first is finite after an old prefix. The second is an old prefix times $S_n(c^{\mathrm{\prime }},j)$ for $c^{\mathrm{\prime }}>0$, and finite for $c^{\mathrm{\prime }}=0$. In the third, for $j\le i$ the mixed braid sweep is $D_n(2)z^{c^{\mathrm{\prime }}}{s}_2\cdots s_{i+1}{s}_2\cdots s_j=(s_2\cdots s_j)D_n(2)z^{c^{\mathrm{\prime }}}{s}_2\cdots s_{i+1}$; for $j>i$ it is $(s_2\cdots s_{j-1})D_n(2)z^{c^{\mathrm{\prime }}}{s}_2\cdots s_i$. At the $s_2$ boundary use $z^d{s}_{2}z{s}_2=s_{2}z{s}_2{z}^d$: it is the rank-four relation for $d=1$, and if it holds for $d$, then $z^{d+1}{s}_{2}z{s}_2=z^d(z{s}_{2}z{s}_2)=z^d(s_{2}z{s}_{2}z)=(z^d{s}_{2}z{s}_2)z=s_{2}z{s}_2{z}^{d+1}$. All higher crossings use the ordinary adjacent braid. Thus $c^{\mathrm{\prime }}>0$ gives an old prefix times $S_n(c^{\mathrm{\prime }},i+1)$ or $S_n(c^{\mathrm{\prime }},i)$, and $c^{\mathrm{\prime }}=0$ leaves a finite word. All nine cases land in $H_{r,n-1}{\mathrm{\Lambda }}_n$, proving that $M_n$ is stable under every generator and contains $1$.

Let $M$ be the $R$-span of the proposed PBW family and $C=H_{r,n}\mathrm{/}M$. Right exactness identifies $R^{\mathrm{\prime }}{\otimes }_{R}C$ with the quotient of the extended algebra by that family, which is zero by step 5.3. Because $R^{\mathrm{\prime }}$ is free with basis containing $1$, $R^{\mathrm{\prime }}{\otimes }_{R}C$ contains $C$ as a direct summand; hence $C=0$.

Steps 1.1–4.1 prove integral spanning, including $r=1$ and faithful-free descent. The cyclotomic relation is used only on $X_1$; higher generators enter through their actual affine transport and the explicit normal-word reductions.