A closed MOY resolution's Koszul complex computes Hochschild homology of its Soergel bimodule

Compute $H{H}_{\bullet }(R^{\mathrm{\prime }},B^{\mathrm{\prime }}(D))$ as the homology of $K(D)$. By [L1] the Koszul complex of all $(r+1)m$ elements is quasi-isomorphic to the diagonal Koszul complex of the $R^{\mathrm{\prime }}$-bimodule $B^{\mathrm{\prime }}(D)$, whose homology is $H{H}_{\bullet }(R^{\mathrm{\prime }},B^{\mathrm{\prime }}(D))$ by [L4] applied to $M=B^{\mathrm{\prime }}(D)$; this proves (a), with the naturality of [L4] supplying the compatibility with coefficient maps.

Change the diagonal generators. In the $R^{\mathrm{\prime }}$-bimodule $B^{\mathrm{\prime }}(D)$ the diagonal elements $u_{x_1},u_{x_2},\dots ,u_{x_m}$ are $\mathbb{Q}$-linear combinations of $u_t,u_{y_2},\dots ,u_{y_m}$ and conversely, the change matrix being the invertible matrix of the coordinate change $(x_1,\dots ,x_m)\leftrightarrow (t,y_2,\dots ,y_m)$; by [L3] the two Koszul complexes are isomorphic. The element $t$, the average of all coordinates, is invariant under every simple reflection, hence balances in every tensor factor and satisfies $t\otimes 1=1\otimes t$, so $u_t=0$ on $B^{\mathrm{\prime }}(D)$.

Split off the trivial coordinate. By [L3] the Koszul complex of the sequence $(u_t,u_{y_2},\dots ,u_{y_m})$ over the commutative ring $B^{\mathrm{\prime }}(D)$ is the tensor product of the two-term complex $K(u_t;B^{\mathrm{\prime }}(D))=[B^{\mathrm{\prime }}(D)\{2\}\stackrel{0}{\to }{B}^{\mathrm{\prime }}(D)]$ with $K(u_{y_2},\dots ,u_{y_m};B^{\mathrm{\prime }}(D))$. Since the differential of the first factor is zero by step 1.2 and steps 1.1 and 1.2 identify $H{H}_{\bullet }(R^{\mathrm{\prime }},B^{\mathrm{\prime }}(D))$ with the homology of this Koszul complex, the tensor product is the direct sum of two copies of $K(u_{y_2},\dots ,u_{y_m};B^{\mathrm{\prime }}(D))$, placed in homological degrees $0$ and $1$, so that $H{H}_h(R^{\mathrm{\prime }},B^{\mathrm{\prime }}(D))\cong H_h(K(u_{y_2},\dots ,u_{y_m};B^{\mathrm{\prime }}(D)))\oplus H_{h-1}(K(u_{y_2},\dots ,u_{y_m};B^{\mathrm{\prime }}(D)))\{2\}$, the two summands being the two copies.

Identify the reduced complex. By [L2] $B^{\mathrm{\prime }}(D)=B(D){\otimes }_{\mathbb{Q}}\mathbb{Q}[t]$ and each $u_{y_j}$ acts on $B(D)$ through the bimodule structure and trivially on $\mathbb{Q}[t]$; therefore $K(u_{y_2},\dots ,u_{y_m};B^{\mathrm{\prime }}(D))=K(u_{y_2},\dots ,u_{y_m};B(D)){\otimes }_{\mathbb{Q}}\mathbb{Q}[t]$, and because $\mathbb{Q}[t]$ is flat over $\mathbb{Q}$ its homology is $H_{\bullet }(K(u_{y_2},\dots ,u_{y_m};B(D))){\otimes }_{\mathbb{Q}}\mathbb{Q}[t]$.

Compute the reduced homology. The elements $y_j=x_j-x_1$ $(2\le j\le m)$ are an invertible linear combination of the standard generators $x_1-x_2,\dots ,x_{m-1}-x_m$ of $R$ (explicitly $x_j-x_1=-(x_1-x_2)-\cdots -(x_{j-1}-x_j)$), so by [L3] their Koszul complex is isomorphic to the diagonal Koszul complex of the polynomial ring $R$; by [L4] applied to the polynomial ring $R$ and the $\mathbb{Q}$-central $R$-bimodule $B(D)$ its homology is $H{H}_{\bullet }(R,B(D))$. Substituting into the two-copy formula of step 2.1 and the tensor decomposition of step 3.1 gives the displayed formula of (b).

Recover the original $a$-theory. In the full closed-graph factorization, the sum of all linear entries is zero: the differences telescope between layers and around the closure, and each wide edge preserves the sum of its two coordinates. The row operation summing the linear rows therefore gives a distinguished row $(a,0)$, with all other first entries zero. Its odd cohomology is $\mathbb{Q}\{-1,1\}$ and its even cohomology is zero; polynomial division in $a$ splits off the contractible pairs, leaving the remaining Koszul complex with this universal shift and parity. When $a=0$ instead, that distinguished row becomes $(0,0)$ and supplies two copies. After the regular-layer quotient, its zero relation is the closure of the common invariant coordinate $t$: summing the linear relations gives $m(t^L-t^R)=0$. The nonzero scalar $m$ is absorbed by a basis change, so this is exactly the zero diagonal row split in step 2.1. Removing that row and the polynomial coordinate $t$ leaves the reduced diagonal Koszul complex of $B(D)$, whose homology is $H{H}_{\bullet }(R,B(D))$ by step 4.1. Thus the original reduced resolution contribution agrees after accounting for the universal $\{-1,1\}$ shift with one of the two copies in (b); setting $a=0$ without removing the zero row retains both copies. This proves (c) and explains the grading correction used by the next lemma. The only use of AC is [L4].