Categorical Braid Actions and Decategorification — Examples

The data. Since $F_g={\mathrm{Id}}_Q$ for all $g\in G$, the composite functors satisfy $F_f{F}_g=F_{fg}={\mathrm{Id}}_Q$ on the nose for every pair $(f,g)\in G\times G$, and $F_1={\mathrm{Id}}_Q$; in particular the functor assignment is a weak action of $G$ on $Q$ in the sense of [L2] whenever the pairwise isomorphisms exist.

The scalar comparisons. By [L3] a natural isomorphism $F_f{F}_g\to F_{fg}$ between identity functors is uniquely a nonzero scalar, so the following choices are well-defined natural isomorphisms: ${\mu }_{a,b}:=(-1)\cdot \mathrm{id}$, that is, multiplication by $-1$ on every complex vector space, and ${\mu }_{f,g}:=\mathrm{id}$ (multiplication by $1$) for every other pair $(f,g)\in G\times G$, including the pairs $(a,a)$, $(a,ab)$ and $(1,b)$.

The pentagon at $(a,a,b)$. The two composites of the pentagon for the triple $(a,a,b)$ map $F_a{F}_a{F}_b={\mathrm{Id}}_Q$ to $F_{a^{2}b}=F_b={\mathrm{Id}}_Q$ and are, by [L3], multiplication by the scalars ${\mu }_{1,b}{\mu }_{a,a}=1\cdot 1=1$ and ${\mu }_{a,ab}{\mu }_{a,b}=1\cdot (-1)=-1$ respectively; the first composite uses ${\mu }_{a,a}\text{\hspace{0.17em}⁣}:F_a{F}_a\to F_{a^2}=F_1$ followed by ${\mu }_{1,b}$, and the second uses $F_a{\mu }_{a,b}\text{\hspace{0.17em}⁣}:F_a{F}_a{F}_b\to F_a{F}_{ab}$ followed by ${\mu }_{a,ab}$, and composition of scalar natural transformations is multiplication. Since $1\mathrm{\ne }-1$ in ${\mathbb{C}}^{\times }$, the two composites are distinct natural transformations, so the pentagon diagram does not commute.

Conclusion. The functor assignment $F_g={\mathrm{Id}}_Q$ together with the chosen pairwise isomorphisms of step 2.1 satisfies the letter of the weak-action definition of [L2] but not the pentagon of a coherent action, by step 3.1. Hence pairwise invertible comparisons do not automatically supply coherence data, the refuted statement is false, and the weak Khovanov–Seidel action of this page must not be assumed coherent without further argument.

The four terms and their degrees. By [L4] the terms of $N=R_i{\otimes }_{A_m}{R}_i^{-1}$ are the direct sums over $p+q=n$ of $R_i^p{\otimes }_{A_m}(R_i^{-1}{)}^q$; the nonzero pairs are $(p,q)=(-1,0),(0,0),(-1,1),(0,1)$, giving the terms $U_i$, $A_m$, $U_i{\otimes }_{A_m}{U}_i\{-1\}$ and $U_i\{-1\}$ in homological degrees $-1,0,0,1$ as displayed, with the two degree-$0$ summands ordered as $U_i{\otimes }_{A_m}{U}_i\{-1\}$ and $A_m$; the tensor-unit isomorphisms of [L4] identify the first and last with $U_i$ and $U_i\{-1\}$. This corrects the term count: there are four terms counted with multiplicity, not three $A_m$'s.

The differentials. After negating the natural identification of $U_i{\otimes }_{A_m}{U}_i\{-1\}$ with $P_i{\otimes }_{\mathbb{Z}}Q{\otimes }_{\mathbb{Z}}{}_{i}P\{-1\}$ and the tensor-unit identifications, the Koszul-signed differentials of [L4] take the form ${\mathrm{\partial }}^{-1}(u)=(\tau (u),{\beta }_i(u))$ and ${\mathrm{\partial }}^0(z,a)={\gamma }_i(a)-\delta (z)$ with the maps $\tau ,\delta$ of the display: the component into $A_m$ is ${\beta }_i$, the component into the middle bimodule is $\tau$, and the component out of $A_m$ is ${\gamma }_i$, while the component out of the middle is $-\delta$ after this coordinate change. Before that change the Koszul rule gives $(-\tau ,{\beta }_i)$ and $(\delta ,{\gamma }_i)$, as required.

The two contractible summands. The $u_2$ component of $\tau$ is the identity on $U_i$, so ${\mathrm{\partial }}^{-1}$ is injective and its restriction to its image is an isomorphism with inverse the $u_2$-coefficient projection on the middle summand, so $T_{-1}$ is contractible with contracting homotopy $({\mathrm{\partial }}^{-1}{)}^{-1}$ by [L5]; the restriction $-\delta \text{\hspace{0.17em}⁣}:P_i\otimes \mathbb{Z}{u}_1\otimes {}_{i}P\{-1\}\to U_i\{-1\}$ is an isomorphism with inverse $x\otimes y\mapsto -x\otimes u_1\otimes y$, so $T_1$ is contractible by [L5]. The map $a\mapsto (\xi (a),a)$ identifies the remaining graph in degree $0$ with $A_m$ with zero differential, because ${\mathrm{\partial }}^0(\xi (a),a)={\gamma }_i(a)-\delta \xi (a)=0$.

Conclusion of the example. By the direct sum decomposition of the generator lemma [L2], $N\cong T_{-1}\oplus A_m\oplus T_1$; by step 3.1 the outer summands are contractible, so $N$ is homotopy equivalent to the diagonal bimodule and $R_i{R}_i^{-1}\cong {\mathrm{Id}}_{C_m}$, while $N$ itself is a four-term complex and not equal to $A_m$; the same argument applies to $R_i^{-1}{\otimes }_{A_m}{R}_i$. The two contractible summands are exactly the cancellation cost, and their contracting homotopies are the displayed inverses.

The matrix of $[R_1]$ for $m=2$. By [L2] with $i=1$ and $m=2$: $[R_1][P_1]=-q[P_1]$, $[R_1][P_2]=[P_2]-[P_1]$, and $[R_1][P_0]=[P_0]-q[P_1]$, there being no $P_{-1}$ term; no $j\in \{0,1,2\}$ satisfies $\mathrm{\mid }1-j\mathrm{\mid }>1$. Reading these as columns in the basis $[P_0],[P_1],[P_2]$ gives the displayed matrix $[R_1]$ with columns $(1,-q,0{)}^{\mathsf{T}}$, $(0,-q,0{)}^{\mathsf{T}}$ and $(0,-1,1{)}^{\mathsf{T}}$.

The change of basis. For $m=2$ the matrix $C$ of [L2] is $C_{0,0}=C_{0,1}=(-q{)}^2=q^2$, $C_{1,1}=C_{1,2}=(-q{)}^1=-q$ and $C_{2,2}=1$, which is the displayed matrix; it is upper triangular with diagonal entries $q^2,-q,1$, all units of $\mathbb{Z}[q,q^{-1}]$, so it is invertible over $\mathbb{Z}[q,q^{-1}]$.

The matrix identity. Multiplying out, $C[R_1]=$ the matrix with rows $(q^2(1-q),-q^3,-q^2)$, $(q^2,q^2,0)$, $(0,0,1)$ and $C^{-1}$ has rows $(q^{-2},q^{-1},1)$, $(0,-q^{-1},-1)$, $(0,0,1)$; the product $C[R_1]C^{-1}$ has first row $(1-q,q,0)$, second row $(1,0,0)$ and third row $(0,0,1)$, which is exactly $B_1$ with $t=q$ as in [L3]. Hence $C[R_1]C^{-1}=B_1$.

Conclusion. The three-dimensional instance of the decategorification proposition is the displayed matrix computation: the operator $[R_1]$ in the vertex-projective basis is conjugate by the explicit invertible matrix $C$ to the unreduced Burau generator at $t=q$. No choice principle is used.

The witness braid and its categorical action. Take $m=4$ and let $\psi \in B_5$ be the nontrivial braid of [L1], so $\psi \mathrm{\ne }1$ and ${\rho }_5^{\mathrm{m}\mathrm{a}\mathrm{t}}(\psi )=I_5$. By [L2] applied to the nontrivial braid $\psi$, the endofunctor $R_{\psi }$ is not isomorphic to the identity functor of $C_4$.

Its action on $K_0$ is trivial. By [L3] the operator $[R_{\psi }]$ on $G(A_4)$ corresponds, in the explicit basis of the decategorification proposition, to the matrix ${\rho }_5^{\mathrm{m}\mathrm{a}\mathrm{t}}(\psi )=I_5$; hence $[R_{\psi }]$ is the identity operator on $G(A_4)$. Thus the exact autoequivalence $R_{\psi }$ acts as the identity on the Grothendieck group while, by step 1.1, it is not isomorphic to the identity functor.

Conclusion. The map from derived autoequivalences of $C_4$ to operators on $K_0$ has a nontrivial kernel, containing the class of $R_{\psi }$; the refuted statement is false. AC is inherited from the kernel lemma and the faithfulness theorem; no additional choice is made.