Distant Soergel generators commute

Since $\mathrm{\mid }i-j\mathrm{\mid }>1$ the reflections $s_i$ and $s_j$ commute, each fixes the indeterminates moved by the other, and they move disjoint pairs of indeterminates. Hence $s_i$ preserves the subring $R^{s_j}$ (it maps it onto $R^{s_i{s}_j{s}_i}=R^{s_j}$) and $s_j$ preserves $R^{s_i}$; neither transposition fixes the other invariant subring pointwise, since for instance $s_1(x_1)=x_2$ and $x_1\in R^{s_3}$. The block factorization below uses the disjoint coordinate pairs: writing $R=A{\otimes }_{\mathbb{Q}}B{\otimes }_{\mathbb{Q}}C$ with $A:=\mathbb{Q}[x_i,x_{i+1}]$ and $B:=\mathbb{Q}[x_j,x_{j+1}]$, one has $R^{s_i}=A^{s_i}{\otimes }_{\mathbb{Q}}B{\otimes }_{\mathbb{Q}}C$ and $R^{s_j}=A{\otimes }_{\mathbb{Q}}{B}^{s_j}{\otimes }_{\mathbb{Q}}C$; these independent actions permit the two rank-one factors to be interchanged while retaining both outer $R$-actions.

Collapsing the middle: the tensor product $B_i{\otimes }_R{B}_j$ is $\left(R{\otimes }_{R^{s_i}}R\right){\otimes }_R\left(R{\otimes }_{R^{s_j}}R\right)$ with the outer shift $(2)$, and the map $(a\otimes b)\otimes (c\otimes d)\mapsto a\otimes bc\otimes d$ into the threefold tensor $T:=R{\otimes }_{R^{s_i}}R{\otimes }_{R^{s_j}}R$ is a degree-zero isomorphism of graded $(R,R)$-bimodules: the ${\otimes }_R$ relation identifies $(b,c)$ with the single element $bc$, so the relations of the fourfold tensor (the two balanced relations and the middle $R$-bilinearity) are exactly the relations of $T$ (the $R^{s_i}$-relation on the first two slots and the $R^{s_j}$-relation on the last two), and the identification $R{\otimes }_{R}R\cong R$ is bijective; the left action multiplies the first slot and the right action the last, unchanged by the collapse.

A rank-one block lemma, to be applied twice below. Let $\mathrm{\Lambda }\subseteq R$ be a polynomial subring of the form $\mathrm{\Lambda }=\mathbb{Q}[y_1,y_2]$ on which a simple reflection $s$ acts by swapping $y_1,y_2$, and let $E\subseteq R$ be a polynomial ring with $R=\mathrm{\Lambda }{\otimes }_{\mathbb{Q}}E$ on which $s$ acts trivially, so that $R^s={\mathrm{\Lambda }}^s{\otimes }_{\mathbb{Q}}E$. Then $\mathrm{\Theta }:R{\otimes }_{R^s}R⟶\left(\mathrm{\Lambda }{\otimes }_{{\mathrm{\Lambda }}^s}\mathrm{\Lambda }\right){\otimes }_{\mathbb{Q}}E,\mathrm{\Theta }(r_1\otimes r_2)=(a_1\otimes a_2)\otimes d_1{d}_2(r_k=a_k\otimes d_k,a_k\in \mathrm{\Lambda },d_k\in E),$ is a well-defined degree-zero isomorphism of graded $(R,R)$-bimodules with two-sided inverse ${\mathrm{\Theta }}^{-1}((p_1\otimes p_2)\otimes d)=(p_1\otimes 1)\otimes (p_2\otimes d)$. Well-definedness: for a pure element $f=g\otimes e$ of $R^s={\mathrm{\Lambda }}^s{\otimes }_{\mathbb{Q}}E$ one has $\mathrm{\Theta }((r_{1}f)\otimes r_2)=(a_{1}g\otimes a_2)\otimes d_{1}e{d}_2=(a_1\otimes g{a}_2)\otimes d_{1}e{d}_2=\mathrm{\Theta }(r_1\otimes (f{r}_2))$, since $g\in {\mathrm{\Lambda }}^s$ may cross the balanced tensor $\mathrm{\Lambda }{\otimes }_{{\mathrm{\Lambda }}^s}\mathrm{\Lambda }$, and ${\mathrm{\Theta }}^{-1}$ is well defined for the same reason. The two composites are the identity on the pure tensors, which span: $\mathrm{\Theta }{\mathrm{\Theta }}^{-1}((p_1\otimes p_2)\otimes d)=(p_1\otimes p_2)\otimes d$, while ${\mathrm{\Theta }}^{-1}\mathrm{\Theta }(r_1\otimes r_2)=(a_1\otimes 1)\otimes (a_2\otimes d_1{d}_2)=(a_1\otimes d_1)\otimes (a_2\otimes d_2)$, the last equality moving the element $d_1\in E\subseteq R^s$ from the first to the second slot by the $R^s$-balancing. Both sides carry their standard $(R,R)$-bimodule structures — the left action of $R=\mathrm{\Lambda }{\otimes }_{\mathbb{Q}}E$ multiplies the first slot on the left, which on the right hand side means the first $\mathrm{\Lambda }$-slot by the $\mathrm{\Lambda }$-component and the $E$-tensorand by the $E$-component, and dually on the right — and $\mathrm{\Theta }$ intertwines them; degrees are additive on both sides because $\mathrm{deg}{r}_k=\mathrm{deg}{a}_k+\mathrm{deg}{d}_k$.

Factorization of the threefold tensor. By [F2] the hypothesis of step 2.2 holds for $s=s_i$ with $\mathrm{\Lambda }=A:=\mathbb{Q}[x_i,x_{i+1}]$ and $E=B{\otimes }_{\mathbb{Q}}C$, and for $s=s_j$ with $\mathrm{\Lambda }=B:=\mathbb{Q}[x_j,x_{j+1}]$ and $E=A{\otimes }_{\mathbb{Q}}C$; note that $B{\otimes }_{\mathbb{Q}}C\subseteq R^{s_i}$ and $A{\otimes }_{\mathbb{Q}}C\subseteq R^{s_j}$ and that $s_i$ and $s_j$ act trivially on the complementary factors, since $\mathrm{\mid }i-j\mathrm{\mid }>1$ makes the two blocks disjoint. Put $P:=A{\otimes }_{A^{s_i}}A$ and $Q:=B{\otimes }_{B^{s_j}}B$. Applying step 2.2 twice to the four-slot presentation of step 2.1 gives $T\cong \left(P{\otimes }_{\mathbb{Q}}B{\otimes }_{\mathbb{Q}}C\right){\otimes }_R\left(A{\otimes }_{\mathbb{Q}}C{\otimes }_{\mathbb{Q}}Q\right),$ deg-zero as $(R,R)$-bimodules. Here the first factor is $P{\otimes }_{A}R$ and the second is $R{\otimes }_{B}Q$ as $(R,R)$-bimodules, because $R=A{\otimes }_{\mathbb{Q}}B{\otimes }_{\mathbb{Q}}C$: on $P{\otimes }_{A}R$ the left action multiplies the first $A$-slot of $P$ and the $B$- and $C$-tensorands by their respective components, while the right action multiplies the second $A$-slot of $P$ and those same $B,C$ tensorands, which is exactly the action on $P{\otimes }_{\mathbb{Q}}B{\otimes }_{\mathbb{Q}}C$; dually for $R{\otimes }_{B}Q$ and $A{\otimes }_{\mathbb{Q}}C{\otimes }_{\mathbb{Q}}Q$. Associativity of the tensor product and the unit isomorphisms $A{\otimes }_A-\cong -$, $-{\otimes }_{B}B\cong -$ over the polynomial rings $A,B,C$ (all modules occurring are free, so no flatness question arises) therefore give a degree-zero $(R,R)$-bimodule isomorphism $\rho :T⟶P{\otimes }_{\mathbb{Q}}Q{\otimes }_{\mathbb{Q}}C,$ and chasing a pure tensor through the chain gives $\rho (a_1\otimes a_2{b}_1\otimes b_2\gamma )=(a_1\otimes a_2)\otimes (b_1\otimes b_2)\otimes \gamma$ for $a_1,a_2\in A$, $b_1,b_2\in B$, $\gamma \in C$: the block maps of step 2.2 read $a_1,a_2$ and $b_1,b_2$ off the first and second slots, the $R$-balancing of the tensor over $R$ absorbs the middle $A$-component of $b_2\gamma$ and the middle $B$-component of $a_2{b}_1$ into the outer $A$- and $B$-actions, and the $C$-components multiply. Consequently $\tau :={\rho }^{-1}:P{\otimes }_{\mathbb{Q}}Q{\otimes }_{\mathbb{Q}}C⟶T,(a_1\otimes a_2)\otimes (b_1\otimes b_2)\otimes \gamma ⟼a_1\otimes a_2{b}_1\otimes b_2\gamma ,$ is a well-defined degree-zero $(R,R)$-bimodule isomorphism. Its bimodule structure is read off from $\rho$: the left action of $r=r_A\otimes r_B\otimes r_C\in R$ multiplies the first $A$-slot of $P$ by $r_A$, the first $B$-slot of $Q$ by $r_B$ and the $C$-tensorand by $r_C$, while the right action multiplies the second $A$-slot by $r_A$, the second $B$-slot by $r_B$ and the $C$-tensorand by $r_C$.

Reversal: the same two applications of step 2.2 with the two colours exchanged give a degree-zero $(R,R)$-bimodule isomorphism ${\tau }^{\mathrm{\prime }}:Q{\otimes }_{\mathbb{Q}}P{\otimes }_{\mathbb{Q}}C\to T^{\mathrm{\prime }}$ with $T^{\mathrm{\prime }}:=R{\otimes }_{R^{s_j}}R{\otimes }_{R^{s_i}}R$, of the same shape as $\tau$. The flip $\sigma :P{\otimes }_{\mathbb{Q}}Q{\otimes }_{\mathbb{Q}}C⟶Q{\otimes }_{\mathbb{Q}}P{\otimes }_{\mathbb{Q}}C,\sigma (p\otimes q\otimes \gamma ):=q\otimes p\otimes \gamma ,$ is $k$-balanced and degree zero, and it is an $(R,R)$-bimodule map: by step 3.1 the left action of $R$ is determined by which tensor factor carries which label — $A$ acts on the $P$-factor, $B$ on the $Q$-factor, $C$ on the $C$-factor, on the left through the first $A$- resp. $B$-slot and on the right through the second — and $\sigma$ exchanges the positions of the $P$- and $Q$-factors without changing their labels, so it commutes with both actions. Hence $\phi :={\tau }^{\mathrm{\prime }}\circ \sigma \circ {\tau }^{-1}:T\to T^{\mathrm{\prime }}$ is a degree-zero $(R,R)$-bimodule isomorphism which interchanges the two balanced tensor factors and does not simply reverse the order of the three slots, and $T\cong T^{\mathrm{\prime }}$.

Shifts: $B_i{\otimes }_R{B}_j\cong T(2)$ and $B_j{\otimes }_R{B}_i\cong T^{\mathrm{\prime }}(2)$, each factor contributing its own $(1)$; since $\phi$ is degree-zero, the composite $B_i{\otimes }_R{B}_j\to T(2)\to T^{\mathrm{\prime }}(2)\to B_j{\otimes }_R{B}_i$ is a degree-zero isomorphism of graded bimodules.

Consequently $B_i{\otimes }_R{B}_j\cong B_j{\otimes }_R{B}_i$ as graded $(R,R)$-bimodules; passing to split Grothendieck classes, where the class of a tensor product is the product of the classes, gives $[B_i][B_j]=[B_j][B_i]$. ∎