Rouquier Complexes and Categorical Braid Relations — Examples

The degree-$0$ term is $B_1{\otimes }_R{B}_2$, of cohomological degree $0$; the degree-$1$ term is $F^0\otimes G^1\oplus F^1\otimes G^0=B_1{\otimes }_{R}R(1)\oplus R(1){\otimes }_R{B}_2\cong B_1(1)\oplus B_2(1)$; the degree-$2$ term is $R(1){\otimes }_{R}R(1)\cong R(2)$. The four basis monomials $u_1\otimes u_2$, $u_1\otimes (1\otimes {\alpha }_2)$, $(1\otimes {\alpha }_1)\otimes u_2$, $(1\otimes {\alpha }_1)\otimes (1\otimes {\alpha }_2)$ of $B_1{\otimes }_R{B}_2$ have internal degrees $-2,0,0,2$; the basis $u_i,1\otimes {\alpha }_i$ of $B_i(1)$ has degrees $-2,0$; and the generator of $R(2)$ has degree $-2$.

Under the identifications of step 1.1 the Koszul differentials are $d^0(x\otimes y)=(x\cdot {\epsilon }_2(y),{\epsilon }_1(x)\cdot y)\in B_1(1)\oplus B_2(1)$ and $d^1(a,s)={\epsilon }_1(a)-{\epsilon }_2(s)$ computed in $R(2)$, i.e. $d^1(a\otimes r,s\otimes b)={\epsilon }_1(a)r-s{\epsilon }_2(b)$ in the notation of the display; the minus sign is the Koszul sign on the differential from bidegree $(1,0)$ to $(1,1)$, where the first factor $R(1)$ sits in cochain degree $1$.

For every simple tensor $x\otimes y\in B_1{\otimes }_R{B}_2$ one computes $d^1{d}^0(x\otimes y)={\epsilon }_1(x){\epsilon }_2(y)-{\epsilon }_1(x){\epsilon }_2(y)=0$: the first component of $d^0(x\otimes y)$ contributes ${\epsilon }_1(x){\epsilon }_2(y)$ and the second contributes the same product with the Koszul sign $-1$, so the two cancel. Since the differentials are balanced and $R$-bilinear, this extends to all elements, so $d^1{d}^0=0$.

The degree bookkeeping of step 1.1 shows that $d^0$ and $d^1$ are homogeneous of internal degree zero on the displayed generators, hence on all elements; this records the cohomological and internal degree of every generator of the three terms and completes the verification of the displayed formula.

Put $S=B_s$, $T=B_t$, $L=B_{sts}$. The terms of $K=F_s{F}_t{F}_s$ in degrees $0,1,2,3$ are $STS$, $U\oplus V\oplus W=(TS)(1)\oplus (SS)(1)\oplus (ST)(1)$, $A\oplus B\oplus C=S(2)\oplus T(2)\oplus S(2)$, and $R(3)$. The complex $K^{\mathrm{\prime }}$ has this list with $s,t$ exchanged. With $m_s,m_t$ denoting multiplication, $d_K^0=(m_s\otimes 1\otimes 1,1\otimes m_t\otimes 1,1\otimes 1\otimes m_s),d_K^2=(m_s,-m_t,m_s),$ and $d_K^1=\left(\begin{array}{ccc}-m_t\otimes 1& m_s\otimes 1& 0\\ -1\otimes m_s& 0& m_s\otimes 1\\ 0& -1\otimes m_s& 1\otimes m_t\end{array}\right)\mathrm{.}$ In particular every component of $d^1{d}^0$ cancels in pairs, and the same holds for $d^2{d}^1$.

Write $x=x_1$, $y=x_2$, $z=x_3$, ${\beta }_s=x-y$, ${\beta }_t=y-z$ and $D_s(f)=(f-s(f))\mathrm{/}(2{\beta }_s)$; these are coordinate roots, rather than the balanced roots used for negative generators. Define $J(r\otimes r^{\mathrm{\prime }})=-r\otimes {\beta }_t\otimes 1\otimes r^{\mathrm{\prime }}-r\otimes 1\otimes {\beta }_t\otimes r^{\mathrm{\prime }},p(r\otimes f\otimes g\otimes r^{\mathrm{\prime }})=r{D}_s(fg)\otimes r^{\mathrm{\prime }}\mathrm{.}$ The coefficient operator $D_s$ is $R^s$-linear, so $p$ is balanced across the outer $R^s$ dividers and middle multiplication is $R^t$-balanced. Unit insertion into $SS$ is balanced, and ${\beta }_t\otimes 1+1\otimes {\beta }_t$ is central in $T$ because $R=R^t\oplus {\beta }_t{R}^t$ and ${\beta }_t^2\in R^t$; hence $J$ is a bimodule map. The shifts make both maps degree zero. The identity $s({\beta }_t)={\beta }_t+{\beta }_s$ gives $D_s({\beta }_t)=-1\mathrm{/}2$ and $pJ=1_S$.

The map $j:L\to STS$, $j(r\otimes r^{\mathrm{\prime }})=r\otimes 1\otimes 1\otimes r^{\mathrm{\prime }}$, is balanced since $R^{S_3}$ slides across all dividers, and $pj=0$. To identify its image, expand the second middle slot in the $R^t$-basis $\{1,z\}$ and slide invariant coefficients into the first middle slot; expand that slot in the $R^s$-basis $\{1,x\}$ and slide coefficients left. Since $z\in R^s$ slides right, $STS$ is generated as a bimodule by $g_0=1\otimes 1\otimes 1\otimes 1$ and $g_x=1\otimes x\otimes 1\otimes 1$. For $e=Jp$ and $u=x+y-z$, balancing gives $(1-e)g_0=g_0,(1-e)g_x=\frac{1}{2}(u{g}_0+g_{0}u)\mathrm{.}$ Indeed $p(g_x)=\frac{1}{2}(1\otimes 1)$ and the two middle ${\beta }_t$ tensors sum to $u{g}_0+g_{0}u-2g_x$. Therefore $\ker p=\mathrm{im}(1-e)=\mathrm{im}j$. By F2 and $pJ=1$, $L$ and $\ker p$ have equal finite dimensions in every graded degree, so $j$ is an isomorphism onto $\ker p$. Thus the specific splitting is $K^0=j(L)\oplus J(S)$.

Split $V=(SS)(1)=V_{+}\oplus V_{-}=S(2)\oplus S$ by F1 using the coordinate middle root ${\beta }_s$, a unit multiple of the balanced root. Projection to $V_{-}$ is $r\otimes f\otimes r^{\mathrm{\prime }}\mapsto r{D}_s(f)\otimes r^{\mathrm{\prime }}$, whose composite with the $V$ component of $d_K^0$ is $p$. It is the identity on $J(S)$ and zero on $j(L)$. Cancel the identity pair $S\to V_{-}$ by F3. The surviving $L$ components into $U,V_{+},W$ are the outer-unit inclusions with signs $+,+,+$.

The second pivot is $V_{+}\to A=1_{S(2)}$, while its component into $C$ is $-1_{S(2)}$. Gaussian elimination deletes this pair and replaces the old $C$ row by the sum of the old $C$ and $A$ rows. In degrees $0,1,2,3$ the survivor is $H:L\stackrel{d_H^0}{\to }(TS)(1)\oplus (ST)(1)\stackrel{d_H^1}{\to }S(2)\oplus T(2)\stackrel{d_H^2}{\to }R(3),$ with $d_H^0=(j_{ts},j_{st}),d_H^1=\left(\begin{array}{cc}-m_t\otimes 1& 1\otimes m_t\\ -1\otimes m_s& m_s\otimes 1\end{array}\right),d_H^2=(m_s,-m_t),$ where $j_{ts}(r\otimes r^{\mathrm{\prime }})=r\otimes 1\otimes r^{\mathrm{\prime }}$ and likewise for $j_{st}$. These are the remaining $U,W$ components of the preceding differential and $C,B$ components of the following one.

Apply steps 1.2–4.1 to $K^{\mathrm{\prime }}$ with $s,t$ exchanged. Here $D_t({\beta }_s)=-1\mathrm{/}2$; the complement calculation is transported by $x\leftrightarrow z$, negating both coordinate roots and leaving $Jp$ unchanged. Reorder its survivor $H^{\mathrm{\prime }}$ into the displayed order for $H$. Its differentials are $d_{H^{\mathrm{\prime }}}^0=d_H^0$, $d_{H^{\mathrm{\prime }}}^1=-d_H^1$, $d_{H^{\mathrm{\prime }}}^2=-d_H^2$, by exchanging the rows and columns of the displayed middle matrix. Thus the degreewise map $q=(1,1,-1,1):H\to H^{\mathrm{\prime }}$ is a chain isomorphism. Every term and map has the same written internal shifts.

For a pivot block $\left(\begin{array}{cc}a& b\\ c& 1\end{array}\right)$, the Gaussian chain isomorphism $T$ has components $T^n=\left(\begin{array}{cc}1& 0\\ c& 1\end{array}\right)$, $T^{n+1}=\left(\begin{array}{cc}1& -b\\ 0& 1\end{array}\right)$ and identity elsewhere. Its inclusion, projection and homotopy are $\iota =T^{-1}\mathrm{in}$, $\pi =\mathrm{pr}T$, $h=T^{-1}kT$, where $k$ sends the pivot target back to its source by the identity and vanishes elsewhere. Compose the two eliminations by $i={\iota }_1{\iota }_2$, $p={\pi }_2{\pi }_1$, $h=h_1+{\iota }_1{h}_2{\pi }_1$, and similarly for $K^{\mathrm{\prime }}$. F3 gives $pi=1_H$ and $1_K-ip=dh+hd$, and the primed identities. All entries are the neighboring blocks displayed above. Thus the explicit chain maps $\mathrm{\Phi }=i^{\mathrm{\prime }}qp:K\to K^{\mathrm{\prime }},\mathrm{\Psi }=i{q}^{-1}{p}^{\mathrm{\prime }}:K^{\mathrm{\prime }}\to K$ satisfy $\mathrm{\Psi }\mathrm{\Phi }=ip$ and $\mathrm{\Phi }\mathrm{\Psi }=i^{\mathrm{\prime }}{p}^{\mathrm{\prime }}$, with the written contractions. This verifies the relation of F5 with no grading shift.

The three words represent the same braid by [F1], so the three maps ${\gamma }_{t,u},{\gamma }_{u,w},{\gamma }_{t,w}$ are defined as elements of one-dimensional degree-zero homotopy Hom spaces; their derived images are the comparisons $c_{t,u},c_{u,w},c_{t,w}$, each obtained by composing the multiplication maps from the shifted graph word tensors through $R_{\pi (v)}(-3)$.

Let $M_t,M_u,M_w$ be the three tensor graph models and ${\mu }_a:M_a\to R_{\pi (v)}(-3)$ their iterated multiplication maps, with shifts adding to $(-3)$. The cancelling pair in $u$ contributes $R_{s_2}(-1){\otimes }_R{R}_{s_2}(1)\cong R$, by $b\otimes c\mapsto b{s}_2(c)$, and its inverse sends $1\mapsto 1\otimes 1$. Thus the typed comparisons are $c_{t,u}={\mu }_u^{-1}{\mu }_t:M_t\to M_u$, $c_{u,w}={\mu }_w^{-1}{\mu }_u:M_u\to M_w$ and $c_{t,w}={\mu }_w^{-1}{\mu }_t:M_t\to M_w$. In particular $c_{u,w}{c}_{t,u}={\mu }_w^{-1}{\mu }_u{\mu }_u^{-1}{\mu }_t={\mu }_w^{-1}{\mu }_t=c_{t,w},c_{w,t}{c}_{t,w}={\mu }_t^{-1}{\mu }_w{\mu }_w^{-1}{\mu }_t=1_{M_t}\mathrm{.}$ Transporting each comparison by its source and target multiplication maps gives the identity of the common graph model.

Since $c_{u,w}{c}_{t,u}=c_{t,w}$ by step 2.1 and the derived images of the two sides of the claim are these comparisons, and since the homotopy Hom space is one-dimensional in degree $0$ by [F2], the composite ${\gamma }_{u,w}{\gamma }_{t,u}$ has the same derived image as ${\gamma }_{t,w}$ and therefore coincides with it. The two displayed composites therefore have the asserted unique normalized degree-zero lifts.

The classes agree. By [F1] the generator complex has class $\chi (F_i)=[B_i]-[R(1)]$, and $Z_i$ has the same two terms in the same cohomological degrees with zero differential, so its alternating class is the same alternating sum, $\chi (Z_i)=[B_i]-[R(1)]=\chi (F_i)$; under $\mathrm{\Phi }$ both correspond to $H_i-v$.

The cohomology differs. By [F3] $H^0(Z_i)=B_i$ and $H^1(Z_i)=R(1)$, whereas by [F2] $H^0(F_i)\cong R_{s_i}(-1)$ and $H^1(F_i)=0$; in particular $H^1(Z_i)$ is nonzero while $H^1(F_i)=0$, and the two left $R$-modules $H^0(Z_i)$ and $H^0(F_i)$ have different ranks.

They are not homotopy equivalent. By [F4] the left $R$-module $B_i$ has rank two while $R_{s_i}(-1)$ has rank one, so $H^0(Z_i)\cong ̸H^0(F_i)$; independently $H^1(Z_i)\cong ̸H^1(F_i)$. If $Z_i$ and $F_i$ were homotopy equivalent, [F5] would make their cohomology objects isomorphic, a contradiction; hence $Z_i\simeq ̸F_i$ although their classes agree, which refutes the statement.