The commuting left and right length operators and their Hecke relations

For $x\in W$ and $s\in S$, [F1] gives $\mathrm{\ell }(sx)=\mathrm{\ell }(x)\pm 1$, so exactly one clause of the displayed definition of $P_s$ applies to $e_x$, and likewise exactly one clause of the definition of $Q_s$ applies to $e_x$; each basis vector therefore has exactly one prescribed image, and extending $R$-linearly defines $P_s,Q_s\in {\mathrm{End}}_R(E)$ ([F6], [F7]). Both clauses have the form $e_x\mapsto e_{\sigma x}+c{e}_x$ with $\sigma$ the multiplication by $s$ from the appropriate side and $c$ the appropriate unit difference, so no selection is involved.

Write $u_s:=v_s-v_s^{-1}$. If $\mathrm{\ell }(sw)=\mathrm{\ell }(w)+1$, then $P_s^2(e_w)=P_s(e_{sw})=e_w+u_s{e}_{sw}=e_w+u_s{P}_s(e_w)$, because $\mathrm{\ell }(s\cdot sw)=\mathrm{\ell }(w)=\mathrm{\ell }(sw)-1$. If $\mathrm{\ell }(sw)=\mathrm{\ell }(w)-1$, then $P_s(e_w)=e_{sw}+u_s{e}_w$ and $P_s(e_{sw})=e_w$, so $P_s^2(e_w)=e_w+u_s{P}_s(e_w)$; in both cases $P_s^2=u_s{P}_s+{\mathrm{id}}_E$ on each basis vector, hence on $E$ ([F6], [F7]). Therefore $P_s(P_s-u_s{\mathrm{id}}_E)={\mathrm{id}}_E=(P_s-u_s{\mathrm{id}}_E)P_s$, so $P_s$ is invertible with inverse $P_s-u_s{\mathrm{id}}_E$. Multiplying all length data on the right gives the same computation for $Q_s$.

(two-length lemma) Let $x\in W$ and $s,t\in S$ satisfy $\mathrm{\ell }(sxt)=\mathrm{\ell }(x)$ and $\mathrm{\ell }(sx)=\mathrm{\ell }(xt)$; then $sx=xt$. Indeed, let $x=s_1\cdots s_q$ be a reduced expression. If $\mathrm{\ell }(xt)=q+1$, then $(s_1,\dots ,s_q,t)$ is a reduced expression of $xt$ of length $q+1=\mathrm{\ell }(xt)$, and $\mathrm{\ell }(s\cdot xt)=\mathrm{\ell }(x)=q$, so exchange [F1] applied to this expression and the letter $s$ gives an index $i\in \{1,\dots ,q+1\}$ with $s{s}_1\cdots s_{i-1}=s_1\cdots s_i$, where $s_{q+1}:=t$. If $i=q+1$, this reads $sx=xt$ and we are done; if $i\le q$, then $sx=(s{s}_1\cdots s_{i-1})s_i{s}_{i+1}\cdots s_q=(s_1\cdots s_i)s_i{s}_{i+1}\cdots s_q=s_1\cdots s_{i-1}{s}_{i+1}\cdots s_q$ is represented by a word of $q-1$ letters, so $\mathrm{\ell }(sx)\le q-1<q+1=\mathrm{\ell }(xt)=\mathrm{\ell }(sx)$, a contradiction. If $\mathrm{\ell }(xt)=q-1$, put $x^{\mathrm{\prime }}:=xt$; then $x^{\mathrm{\prime }}t=x$, so $\mathrm{\ell }(x^{\mathrm{\prime }}t)=\mathrm{\ell }(x)=q=\mathrm{\ell }(xt)+1=\mathrm{\ell }(x^{\mathrm{\prime }})+1$, $\mathrm{\ell }(s{x}^{\mathrm{\prime }}t)=\mathrm{\ell }(sx)=q-1=\mathrm{\ell }(x^{\mathrm{\prime }})$ and $\mathrm{\ell }(s{x}^{\mathrm{\prime }})=\mathrm{\ell }(sxt)=\mathrm{\ell }(x)=q=\mathrm{\ell }(x^{\mathrm{\prime }}t)$, so $x^{\mathrm{\prime }}$ satisfies the hypotheses of the case just treated with the roles of the lengths interchanged, and that case yields $s{x}^{\mathrm{\prime }}=x^{\mathrm{\prime }}t$; multiplying by $t$ on the right gives $sx=xt$.

Fix $s,t\in S$ and $x\in W$ and expand both $P_s{Q}_t(e_x)$ and $Q_t{P}_s(e_x)$ from the definitions. In the four length configurations (i) $\mathrm{\ell }(sx)=\mathrm{\ell }(xt)=\mathrm{\ell }(x)+1$, $\mathrm{\ell }(sxt)=\mathrm{\ell }(x)+2$, where both sides equal $e_{sxt}$; (ii) $\mathrm{\ell }(sx)=\mathrm{\ell }(xt)=\mathrm{\ell }(x)-1$, $\mathrm{\ell }(sxt)=\mathrm{\ell }(x)-2$, where both sides equal $e_{sxt}+u_t{e}_{sx}+u_s{e}_{xt}+u_s{u}_t{e}_x$; (iii) $\mathrm{\ell }(sx)=\mathrm{\ell }(x)-1$, $\mathrm{\ell }(xt)=\mathrm{\ell }(x)+1$, $\mathrm{\ell }(sxt)=\mathrm{\ell }(x)$, where both sides equal $e_{sxt}+u_s{e}_{xt}$; (iv) $\mathrm{\ell }(sx)=\mathrm{\ell }(x)+1$, $\mathrm{\ell }(xt)=\mathrm{\ell }(x)-1$, $\mathrm{\ell }(sxt)=\mathrm{\ell }(x)$, where both sides equal $e_{sxt}+u_t{e}_{sx}$; the two expansions agree, where $u_s=v_s-v_s^{-1}$ and $u_t=v_t-v_t^{-1}$.

(reduced products at $e_1$) Let $w=s_1\cdots s_k$ be a reduced expression. Every contiguous subword is reduced: replacing a subword by a shorter representative would shorten the entire expression of $w$, contradicting $\mathrm{\ell }(w)=k$ ([F5]). Put $z_i:=s_i\cdots s_k$ for $1\le i\le k$ and $z_{k+1}:=1$; then $\mathrm{\ell }(z_i)=k-i+1$ and $s_i{z}_{i+1}=z_i$, so $P_{s_i}(e_{z_{i+1}})=e_{z_i}$. Applying the operators from right to left gives $P_{s_1}\cdots P_{s_k}(e_1)=e_w$. For right multiplication put $w_i:=s_1\cdots s_i$, $w_0:=1$; the same reduced-subword argument gives $\mathrm{\ell }(w_i)=i$ and $w_{i-1}{s}_i=w_i$, so $Q_{s_i}(e_{w_{i-1}})=e_{w_i}$. Thus $Q_{s_k}\cdots Q_{s_1}(e_1)=e_w$. Both identities also hold for the empty expression.

In the two remaining length configurations, (v) $\mathrm{\ell }(sx)=\mathrm{\ell }(xt)=\mathrm{\ell }(x)-1$, $\mathrm{\ell }(sxt)=\mathrm{\ell }(x)$, the two expansions are $P_s{Q}_t(e_x)=e_{sxt}+u_t{e}_{sx}+u_s{u}_t{e}_x$ and $Q_t{P}_s(e_x)=e_{sxt}+u_s{e}_{xt}+u_s{u}_t{e}_x$; (vi) $\mathrm{\ell }(sx)=\mathrm{\ell }(xt)=\mathrm{\ell }(x)+1$, $\mathrm{\ell }(sxt)=\mathrm{\ell }(x)$, the two expansions are $P_s{Q}_t(e_x)=e_{sxt}+u_s{e}_{xt}$ and $Q_t{P}_s(e_x)=e_{sxt}+u_t{e}_{sx}$. In both, $\mathrm{\ell }(sxt)=\mathrm{\ell }(x)$ and $\mathrm{\ell }(sx)=\mathrm{\ell }(xt)$, so $sx=xt$ by 1.3. Hence $s$ and $t$ are conjugate in $W$ (with $x$ as conjugating element) and therefore $v_s=v_t$ by the parameter rule of the definition ([F2]), so $u_s=u_t$ and the two expansions coincide: also $P_s{Q}_t(e_x)=Q_t{P}_s(e_x)$. Together with the four configurations of 1.4 this proves $P_s{Q}_t(e_x)=Q_t{P}_s(e_x)$ for every basis vector $e_x$, so $P_s{Q}_t=Q_t{P}_s$ because $(e_x)$ spans $E$ ([F6]).

Let $s\mathrm{\ne }t$ with $m:=m(s,t)<\mathrm{\infty }$, and put $a:=P_s{P}_t{P}_s\cdots$ and $b:=P_t{P}_s{P}_t\cdots$, each product having $m$ alternating factors. Let $w_1$ be the alternating product of $m$ factors beginning with $s$ and $w_2$ that beginning with $t$. Each successive factor is length-increasing along the alternating word by [F3] (the partial alternating words of length $\le m$ are reduced), so the computation of 1.5 gives $a(e_1)=e_{w_1}$ and, with the roles of $s,t$ exchanged, $b(e_1)=e_{w_2}$. In $W$ one has $w_1=w_2$: since $(st)(ts)=stts=s^2=1$ in $W$, we have $(ts)=(st{)}^{-1}$ ([F5]), and the relator $(st{)}^m=1$ gives $(st{)}^{-k}=(st{)}^{m-k}$ for $0\le k\le m$; hence if $m=2k+1$ then $w_2=(ts{)}^{k}t=(st{)}^{-k}t=(st{)}^{k+1}t=(st{)}^{k}s=w_1$, while if $m=2k$ then $w_2=(ts{)}^k=(st{)}^{-k}=(st{)}^k=w_1$. Now let $w=u_1\cdots u_k$ be any reduced expression. By 1.5, $Q_{u_k}\cdots Q_{u_1}(e_1)=e_w$, and by 2.1 every $P_s$ commutes with every $Q_t$, so $a(e_w)=a{Q}_{u_k}\cdots Q_{u_1}(e_1)=Q_{u_k}\cdots Q_{u_1}a(e_1)=Q_{u_k}\cdots Q_{u_1}b(e_1)=b(e_w)$; since $(e_w{)}_{w\in W}$ is a basis, $a=b$ ([F6]). The same argument with $P$ replaced by $Q$ throughout gives the braid relations for the $Q$'s.

By [F4] any two reduced expressions of $w$ are braid-equivalent, and a braid move replaces a block $P_s{P}_t{P}_s\cdots$ of $m(s,t)$ factors by $P_t{P}_s{P}_t\cdots$ with the same product by 3.1, so the product $P_{s_1}\cdots P_{s_k}$ depends only on $w$; writing $P_w$ for this common endomorphism, $P_w(e_1)=e_w$ by 1.5. Hence part 4 holds, and with part 1 (2.1), part 2 (1.2) and part 3 (3.1) all four assertions are proved. No finiteness of $W$ and no regularity of $R$ was used, and no choice: the operators are defined by explicit length conditions and every product above runs along a reduced expression that exists by the definition of $\mathrm{\ell }$.