Rouquier complexes satisfy far commutativity

Put $A=\mathbb{Q}[x_i,x_{i+1}]$, $B=\mathbb{Q}[x_j,x_{j+1}]$ and let $C$ be the polynomial ring in the remaining coordinates. The disjoint transpositions give $R=A{\otimes }_{\mathbb{Q}}B{\otimes }_{\mathbb{Q}}C$ and $R^{s_i}=A^{s_i}\otimes B\otimes C$, $R^{s_j}=A\otimes B^{s_j}\otimes C$. For $E=B\otimes C$, the map $(a\otimes e)\otimes (a^{\mathrm{\prime }}\otimes e^{\mathrm{\prime }})\mapsto (a\otimes a^{\mathrm{\prime }})\otimes e{e}^{\mathrm{\prime }}$ identifies $R{\otimes }_{R^{s_i}}R$ with $(A{\otimes }_{A^{s_i}}A)\otimes E$; its inverse sends $(a\otimes a^{\mathrm{\prime }})\otimes e$ to $(a\otimes 1)\otimes (a^{\mathrm{\prime }}\otimes e)$. Balancing over $A^{s_i}$ and $E$ verifies both maps and their inverse identities. Exchanging the blocks gives the analogous identification for $j$. Under these maps multiplication and root insertion act only in their own block. Thus $F_i^{ϵ}=P_i^{ϵ}\otimes B\otimes C$ and $F_j^{\delta }=A\otimes P_j^{\delta }\otimes C$ as complexes, with the shifts inherited from the generator definitions.

In each bidegree the map $(p\otimes b\otimes c)\otimes (a\otimes q\otimes d)\mapsto pa\otimes bq\otimes cd$ identifies the balanced product with $P_i^{ϵ}{\otimes }_{\mathbb{Q}}{P}_j^{\delta }{\otimes }_{\mathbb{Q}}C$; the inverse sends $p\otimes q\otimes c$ to $(p\otimes 1\otimes 1)\otimes (1\otimes q\otimes c)$. The $A,B,C$ balancing relations verify these inverse identities and preservation of both outer actions. Each differential is a bimodule map acting in its own block, so these identifications intertwine the signed total differentials for every choice of signs, including negative cohomological degrees.

On the external tensor product, define the flip of a term of cohomological bidegree $(p,q)$ by $x\otimes y\otimes c\mapsto (-1{)}^{pq}y\otimes x\otimes c$. It preserves both outer actions because their block labels move with the blocks. For the component $d_{P}x\otimes y$, its image has sign $(-1{)}^{(p+1)q}$, which equals $(-1{)}^{pq+q}$ on the corresponding component of the target differential. For $(-1{)}^{p}x\otimes d_{Q}y$, its image has sign $(-1{)}^{p+p(q+1)}=(-1{)}^{pq}$, again the target sign. It is therefore a chain isomorphism, and its square is the identity.

Transport this flip through the block identifications of step 2.1. It yields $F_i^{ϵ}{\otimes }_R{F}_j^{\delta }\cong F_j^{\delta }{\otimes }_R{F}_i^{ϵ}$ for all signs, with zero internal degree and no shift. Its component on $B_i{\otimes }_R{B}_j$ realizes the distant bimodule isomorphism of F1 with the required differential compatibility. This proves every displayed case in the category of complexes and hence in its homotopy category. The map is the independent-block flip, not an arbitrary flip of balanced bimodule tensors.