Invariance under the braid-like Reidemeister III move

A common curved row and invariant-average split. Put $A=\mathbb{Q}[y_1,y_2]$, with $y_1=x_1-x_2$ and $y_2=x_2-x_3$, and at the $j$th marked three-strand level put $t_j=(x_{j,1}+x_{j,2}+x_{j,3})\mathrm{/}3$. The coordinate change is invertible over $\mathbb{Q}$, with $x_{j,1}=t_j+(2y_{j,1}+y_{j,2})\mathrm{/}3$, $x_{j,2}=t_j+(-y_{j,1}+y_{j,2})\mathrm{/}3$, and $x_{j,3}=t_j-(y_{j,1}+2y_{j,2})\mathrm{/}3$; hence $R^{\mathrm{\prime }}=A[t_j]$. Both simple reflections fix $t_j$ and act on $A$, so $R^{\mathrm{\prime }{s}_i}=A^{s_i}[t_j]$. Thus the full uncurved type-A Soergel bimodule splits as $(A{\otimes }_{A^{s_i}}A)(1){\otimes }_{\mathbb{Q}}\mathbb{Q}[t_j]$, with the common invariant $t_j$ acting equally on both outer sides. In the curved KR diagram the endpoint averages are instead retained as separate variables in the common factor $X$ below; only the reduced difference coordinates enter the coefficient bimodules. This is the invariant-average split used here, and it does not identify the two endpoint averages $t_0,t_3$. The six boundary variables therefore give $U\cong S[a,t_0,t_3]$, where $S=A_L{\otimes }_{\mathbb{Q}}{A}_R$, and $w=3a(t_0-t_3)$. At layer $j$ the total-sum row is $(a,3(t_{j-1}-t_j))$. A homogeneous Koszul row operation on the three such rows extracts the single row $X=(a,3(t_0-t_3))$ and leaves first-entry-zero rows for the internal averages $t_1,t_2$; remove those rows by [F2], substituting their linear relations. The remaining coordinates are the two differences at each layer. Removing redundant straight-arc marks by [F2] leaves every resolution in the form $X\otimes K_{ϵ}$, where $K_{ϵ}$ is the parity folding of the ordinary Koszul complex of its remaining layer relations over $S$. Thus $d_X^2=3a(t_0-t_3)=w$, and the common curved factor is the same on both sides.

Explicit local map homotopies. In a two-strand layer put $z=x_1+x_2-x_3-x_4$, $y_L=x_4-x_3$, $y_R=x_1-x_2$, $u=(y_L-y_R)\mathrm{/}2$ and $v=(y_L+y_R)\mathrm{/}2$. Then $x_2-x_3=u+z\mathrm{/}2$ and $x_4-x_2=v-z\mathrm{/}2$. In the common first row $X_0=(a,z)$, replace the second exterior generator by $e_1-e_0\mathrm{/}2$ for the arc resolution, and by $e_1+(-(v-u)\mathrm{/}2+z\mathrm{/}4)e_0$ for the wide resolution. These are the row operations making the second entries $u$ and $uv$. In the resulting exterior bases, ${\chi }_0$ sends $1\mapsto v-z\mathrm{/}2$, $e_0\mapsto (v-z\mathrm{/}2)e_0$, $e_1\mapsto e_1-u{e}_0\mathrm{/}2$, $e_0\wedge e_1\mapsto e_0\wedge e_1$. Its difference from ${\mathrm{I}\mathrm{d}}_{X_0}\otimes {\psi }^{\mathrm{\prime }}(v)$ is $d{H}_0+H_{0}d$, where $H_0(1)=-e_0\mathrm{/}2$ and $H_0$ is zero on the other three basis elements. Similarly, ${\chi }_1$ sends $1\mapsto 1$, $e_0\mapsto e_0$, $e_1\mapsto (v-z\mathrm{/}2)e_1+u{e}_0\mathrm{/}2$, $e_0\wedge e_1\mapsto (v-z\mathrm{/}2)e_0\wedge e_1$. Its difference from ${\mathrm{I}\mathrm{d}}_{X_0}\otimes \psi (v)$ is $d{H}_1+H_{1}d$, where $H_1(e_1)=-e_0\wedge e_1\mathrm{/}2$ and $H_1$ is zero on the other basis elements. Substitution in $d=a(e_0\wedge -)+z{\iota }_0+u{\iota }_1$ or its target version with $uv{\iota }_1$ verifies all four equations; for example the differences on $e_1$ are $-u{e}_0\mathrm{/}2$ for ${\chi }_0$ and $-z{e}_1\mathrm{/}2+u{e}_0\mathrm{/}2$ for ${\chi }_1$. After the crossing shifts, both homotopies have bidegree $(-1,-1)$.

Ordered regular layer resolutions. At each level write the reduced incoming and outgoing coordinates as $(y_1,y_2)$ and $(y_1^{\mathrm{\prime }},y_2^{\mathrm{\prime }})$, after separating the average. An identity resolution has the two straight-strand difference relations $y_1^{\mathrm{\prime }}-y_1=0$ and $y_2^{\mathrm{\prime }}-y_2=0$; ordered by $y_1^{\mathrm{\prime }},y_2^{\mathrm{\prime }}$, they are monic linears in fresh output variables. For a wide edge at simple root $\rho =y_i$, let $\zeta$ be the adjacent invariant coordinate ($\zeta =y_{i+1}+y_i\mathrm{/}2$ for $i=1$, or $\zeta =y_{i-1}+y_i\mathrm{/}2$ for $i=2$), and let ${\rho }^{\mathrm{\prime }}$ and ${\zeta }^{\mathrm{\prime }}$ denote the outgoing root and invariant coordinate. The straight-strand difference after subtracting the average change is exactly $-\frac{2}{3}({\zeta }^{\mathrm{\prime }}-\zeta )$, so its vanishing is the linear invariant-coordinate relation ${\zeta }^{\mathrm{\prime }}-\zeta =0$. The wide-edge relation is $Q=({\rho }^{\mathrm{\prime }}{)}^2-{\rho }^2=0$. Order these as ${\zeta }^{\mathrm{\prime }}-\zeta ,Q$: the first is monic linear in the fresh invariant coordinate, and the second is monic quadratic in the fresh root coordinate. The invariant ring is $A^{s_i}=\mathbb{Q}[\zeta ,{\rho }^2]$, so this quotient is precisely $A{\otimes }_{A^{s_i}}A$ and has basis $1,{\rho }^{\mathrm{\prime }}$. Each later layer introduces fresh outgoing difference coordinates; concatenate the identity or wide-edge lists in layer order. Modulo earlier rows every new monic relation is a non-zero-divisor: the highest fresh-variable coefficient of its product with a nonzero polynomial cannot vanish. Thus the residual sequence is regular. Its Koszul exactness follows by induction on the rows. Adjoining a row $f$ gives the cone of multiplication by $f$ on the previous Koszul complex; writing a cycle as a pair and subtracting lifts of boundaries gives zero homology above degree one, degree-one homology $\ker (f:M\to M)$, and degree-zero homology $M\mathrm{/}fM$, where $M$ is the preceding quotient. Injectivity of $f$ kills the kernel, proving the induction. Polynomial division gives one basis element at an identity layer and $1,{\rho }^{\mathrm{\prime }}$ at a wide edge; adjoining the shift $(1)$ gives the Soergel bimodule of [F4]. This proves the quotient description with both boundary difference actions retained and the exactness of $K_{ϵ}\to M_{ϵ}$ as an augmented free $S$-resolution: its terms are free over $S$ on the internal-variable monomials times the exterior bases. The endpoint averages stay in $X$ and are not quotiented here. A wedge for a degree-$d$ relation has usual internal degree $d$; regrading Koszul degree $p$ and usual degree $d$ to $(-p,d-p)$ gives the KR row shifts $(-1,1)$ for linears and $(-1,3)$ for quadratics.

The coefficient maps and the full-strand row. The extra straight strand is an identity tensor factor in the local calculation. Combining its linear row with the pair-sum row into the total-average row uses the same basis change on both resolutions and preserves the remaining row operations; consequently each crossing edge in step 1.1 is ${\mathrm{I}\mathrm{d}}_X$ times the comparison map of the regular quotients in step 2.1. Write ${\bar{B}}_i=A{\otimes }_{A^{s_i}}A$ for the unshifted KR wide-edge bimodule and ${\beta }_i=x_i-x_{i+1}$. On the degree-zero quotient modules, step 1.2 sends $1$ to $v=({\beta }_i^L+{\beta }_i^R)\mathrm{/}2$; the KR positive map $r{b}_i$ of [F5] sends $1$ to ${\beta }_i^L+{\beta }_i^R=2v$. Thus the comparison from the original ${\chi }_0$ edge to the KR positive edge contributes the rational unit $2$ (equivalently, scale the wide-resolution vertex by $2$). The KR negative map is multiplication ${\bar{B}}_i\to A$. The type-A Rouquier bimodule is $B_i={\bar{B}}_i(1)$, so after the library shift convention the positive KR complex is $F_i^{-1}\{1\}$ and the negative KR complex is $F_i\{-1\}$: the former has terms $A\{2\}\to {\bar{B}}_i$ in degrees $-1,0$, and the latter has ${\bar{B}}_i\{-2\}\to A\{-2\}$ in degrees $0,1$. The multiplication differential is unchanged. For the positive differential, ${\eta }_i(1)={\alpha }_i^L+{\alpha }_i^R={\sigma }_i({\beta }_i^L+{\beta }_i^R)$, so it differs from the KR map by the unit ${\sigma }_i$; relative to the original local map of step 1.2, the positive Rouquier differential is $2{\sigma }_i$ times that map. On the three-crossing cube, rescale each resolution vertex by the product of $2{\sigma }_{i_k}$ over its positive crossings resolved wide; from an arc vertex to its wide target the scalar ratio is exactly $2{\sigma }_{i_k}$, while negative edges need no rescaling. These scalars commute around every cube square, so this is an isomorphism of the signed totalizations.

A functorial free resolution. For a graded $S$-module $M$, use the normalized bar resolution $P_n(M)=S{\otimes }_{\mathbb{Q}}{\bar{S}}^{\otimes n}{\otimes }_{\mathbb{Q}}M$, where $\bar{S}$ is the augmentation ideal of the positive-degree polynomial ring $S$. Its differential multiplies adjacent factors, with alternating signs, and the last factor acts on $M$; its augmentation is $s\otimes m\mapsto sm$. The usual bar contraction over $\mathbb{Q}$ sends $m$ to $1\otimes m$ and inserts $1\otimes \overline{s_0}$ at the beginning of a higher tensor. Expanding the alternating differential cancels its adjacent terms in pairs, leaving $ds+sd=1$ on the augmented complex. Thus it is a resolution. By the ordered polynomial division in step 2.1, each $M_{ϵ}$ has the explicit $\mathbb{Q}$-basis of left polynomial monomials times one basis element at each wide-edge layer, so every $P_n(M_{ϵ})$ is free over $S$. The construction is additive and functorial on all coefficient maps and homotopies. In a fixed usual internal degree, its normalized tensor factors have positive degree, so only finitely many homological degrees and basis tensors occur.

Comparison maps and their homotopies. Compare any $K_{ϵ}$ with $P(M_{ϵ})$ by augmentation-preserving maps. On a free $S$-basis of degree zero, lift its augmentation to the target resolution; on a basis vector $b$ in degree $n>0$, define the lift recursively by $f_n(b)=s_{\mathrm{t}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{t}}{f}_{n-1}d(b)$ and extend $S$-linearly. The target's augmented $\mathbb{Q}$-contraction exists for the bar complex by step 3.2 and for $K_{ϵ}$ by homogeneous Gaussian elimination on each finite-dimensional usual-degree complex, which is exact by step 2.1; choose the first pivot in the fixed monomial order. The displayed recurrence is a chain map because its argument is a cycle and $ds$ is the identity on cycles. Construct $f$ and $g$ in both directions. For $T=gf-1$ define, on the same free basis, $H_n(b)=s_K(T_n(b)-H_{n-1}d(b))$, starting with $H_{-1}=0$; its argument is a cycle by induction, and this gives $dH+Hd=T$. The identical formula gives $fg\simeq 1$ on the bar resolution. It also shows that two lifts of any coefficient map are homotopic: use their difference for $T$. All maps preserve usual internal degree; $f,g$ preserve Koszul degree, while $H$ raises it by one, so after the regrading in step 2.1 the maps have bidegree $(0,0)$ and homotopies have bidegree $(-1,-1)$. The fixed homogeneous pivots avoid any arbitrary choice of lifts.

Pass to the curved factor. Extend these free resolutions from $S$ to $U$ and tensor with $X$. The signed extension of an inner homotopy is $\tilde{H}(x\otimes k)=(-1{)}^{\mathrm{\mid }x\mathrm{\mid }}x\otimes H(k)$; the two cross terms involving $d_X$ cancel, giving $d\tilde{H}+\tilde{H}d=1_X\otimes (dH+Hd)$ even though $d_X^2=w$. Thus step 4.1 yields actual homotopy equivalences in ${\mathrm{h}\mathrm{m}\mathrm{f}}_w$. Its uniqueness of lifts makes the comparison squares for every crossing commute in that category. Therefore the two diagram complexes in $K({\mathrm{h}\mathrm{m}\mathrm{f}}_w)$ are isomorphic to $X\otimes P(F(\sigma ))$ and $X\otimes P(F({\sigma }^{\mathrm{\prime }}))$, respectively, where $F$ denotes the coefficient complexes of [F5]. No averaging coordinate, Koszul degree or crossing degree has been discarded.

Lift the complete survivor homotopies and check the no-shift normalization. The two three-letter words are $(i,i+1,i)$ and $(i+1,i,i+1)$, and in each braid-like diagram the three crossings have the same sign. The positive no-shift braid equivalence and the two-sided inverse contractions in [F6] give the corresponding equivalence for the inverse triple. The supplier homotopies are bimodule maps, so base-changing their two outer actions along $R^{\mathrm{\prime }}=A[t]\stackrel{t\mapsto 0}{\to }A$ preserves every chain and homotopy equation and gives the relation on the reduced coefficient ring; the endpoint averages $t_0,t_3$ remain in $X$. Its explicit inclusion, projection and contractions give maps $\mathrm{\Phi },\mathrm{\Psi }$ and homotopies $\mathrm{\Psi }\mathrm{\Phi }-1=dh+hd$, $\mathrm{\Phi }\mathrm{\Psi }-1=d{h}^{\mathrm{\prime }}+h^{\mathrm{\prime }}d$ on the coefficient complexes. If all three crossings are negative, the KR complex of either word is the corresponding positive Rouquier triple shifted by $\{-3\}$, so the same equivalence has no additional internal or cohomological shift. If all three crossings are positive, it is the inverse Rouquier triple shifted by $\{3\}$; the root-unit and rational-normalization rescalings of step 3.1 have products $(2{\sigma }_i{)}^2(2{\sigma }_{i+1})=8{\sigma }_{i+1}$ and $(2{\sigma }_{i+1}{)}^2(2{\sigma }_i)=8{\sigma }_i$ on the two all-wide vertices. Their ratio remains ${\sigma }_i{\sigma }_{i+1}=-1$ after the common factor $8$ cancels. This residual scalar is a unit, absorbed by rescaling the comparison map; both words still have the same shift $\{3\}$. Apply the additive bar functor to the four homotopy equations; it preserves compositions and sums. Tensoring with $X$ preserves them with the signed totalization, since the two cross terms of each extended inner homotopy cancel as in step 5.1. Combining these maps with the comparison equivalences and homotopies of steps 4.1–5.1 gives homotopy inverses between $C(D_1)$ and $C(D_2)$ in $K({\mathrm{h}\mathrm{m}\mathrm{f}}_w)$. The Rouquier equivalence, vertex scalars, bar comparison maps and curved factor all have internal degree zero after the equal shifts just computed; the homotopies have cohomological degree $-1$. Hence the isomorphism has no trigrading shift, as claimed.