Matrix Factorizations and Khovanov–Rozansky Link Homology — Examples

The products are factorizations. Multiplying out over $R$ gives $P_1{P}_0=\left(\begin{array}{cc}a(x_1+x_2-x_3-x_4)& 0\\ 0& a(x_1+x_2-x_3-x_4)\end{array}\right)=w\cdot I_2$ and $Q_1{Q}_0=\left(\begin{array}{cc}w& 0\\ 0& w\end{array}\right)=w\cdot I_2$, because the off-diagonal entries $(x_1-x_4)(x_3-x_2)+(x_2-x_3)(x_1-x_4)$ and $(x_1+x_2-x_3-x_4)(x_3{x}_4-x_1{x}_2)+(x_1{x}_2-x_3{x}_4)(x_1+x_2-x_3-x_4)$ vanish; hence both presentations are factorizations with potential $w$, in accordance with [F1].

The morphism and its bidegree. The matrix products $Q_0{U}_0^0=U_0^1{P}_0$ and $Q_1{U}_0^1=U_0^0{P}_1$ of the definition of ${\chi }_0$ show that the displayed $U$ matrices commute with the differentials, and the entries $x_4,-x_2$ and the shifts $R\{-1,1\}\to R\{-1,3\}$, $R\{-2,2\}\to R\{-2,4\}$ combine to bidegree $(0,2)$; on the source shifted by $\{0,2\}$ the differential has bidegree $(0,0)$.

The two complexes side by side. The positive complex has terms $C({\mathrm{\Gamma }}_0)\{0,2\}$ in degree $-1$ and $C({\mathrm{\Gamma }}_1)$ in degree $0$ with differential ${\chi }_0$, and the negative complex has terms $C({\mathrm{\Gamma }}_1)\{0,-2\}$ in degree $0$ and $C({\mathrm{\Gamma }}_0)\{0,-2\}$ in degree $1$ with differential ${\chi }_1$; in both cases the differential is a morphism of factorizations of bidegree $(0,0)$ on the shifted terms by step 1.2 and [F2], so each displayed two-term complex is a complex of objects of ${\mathrm{h}\mathrm{m}\mathrm{f}}_w$. The shift asymmetry is exactly the one recorded in The positive and negative Khovanov-Rozansky crossing complexes: $\{0,2\}$ on the positive source versus the overall $\{0,-2\}$ normalization on the negative complex.

The factorization. The circle carries one mark and one arc whose two endpoint labels are both $x$, so the arc factors of [F1] specialize to $(a,x-x)=(a,0)$: the differentials are multiplication by $a$ and by $0$, the middle term carries the shift $\{-1,1\}$, and the square of the differential is $0\cdot a=0$, which is the potential $a(x-x)$ of the closed graph. Hence the complex is exactly $\mathbb{Q}[a,x]\stackrel{a}{\to }\mathbb{Q}[a,x]\{-1,1\}\stackrel{0}{\to }\mathbb{Q}[a,x]$.

Cohomology. In odd inner-factorization parity the cohomology is $\ker (0\text{\hspace{0.17em}⁣}:\mathbb{Q}[a,x]\{-1,1\}\to \mathbb{Q}[a,x])\mathrm{/}\mathrm{im}(a\text{\hspace{0.17em}⁣}:\mathbb{Q}[a,x]\to \mathbb{Q}[a,x]\{-1,1\})=\mathbb{Q}[a,x]\{-1,1\}\mathrm{/}a\mathbb{Q}[a,x]\{-1,1\}\cong \mathbb{Q}[x]\{-1,1\}$, since $\mathbb{Q}[a,x]\mathrm{/}a\mathbb{Q}[a,x]\cong \mathbb{Q}[x]$ and multiplication by $a$ is injective on $\mathbb{Q}[a,x]$. In even inner-factorization parity the cohomology is $\ker (a)\mathrm{/}\mathrm{im}(0)=0$ because multiplication by the nonzerodivisor $a$ is injective. So $H(\mathrm{\Gamma })\cong \mathbb{Q}[x]\{-1,1\}$, all of it in odd inner parity and outer cochain degree $0$, and $a$ acts as zero on it.

Euler characteristic. The graded pieces of $H(\mathrm{\Gamma })$ have $(k,l)=(-1,1+2m)$ for $m\ge 0$, each of dimension $1$ over $\mathbb{Q}$, all in cohomological degree $0$; substituting into the Euler characteristic of [F2] gives $⟨D⟩={\sum }_{m\ge 0}{t}^{-1}{q}^{1+2m}=t^{-1}q\mathrm{/}(1-q^2)$. Since $q^{-1}-q=(1-q^2)\mathrm{/}q$, this is $t^{-1}\mathrm{/}(q^{-1}-q)$; and with $\alpha =-t^{-1}{q}^{-1}$ one has $\alpha \mathrm{/}(1-q^{-2})=-t^{-1}{q}^{-1}\cdot q^2\mathrm{/}(q^2-1)=t^{-1}q\mathrm{/}(1-q^2)$, the same value. This is the unknot normalization used as the base case of the categorification theorem, and it is read off the displayed representative, finite free over $\mathbb{Q}[a,x]$ and infinite free over $\mathbb{Q}[a]$.

The four resolutions and all signed edges. Write $C_{ab}=C({\mathrm{\Gamma }}_{ab})$ for the resolution with choices $a,b\in \{0,1\}$ at the two positive crossings, including all unchanged arc factors. The outer terms are $C^{-2}=C_{00}\{0,4\},C^{-1}=C_{10}\{0,2\}\oplus C_{01}\{0,2\},C^0=C_{11},$ and zero otherwise. In this summand order the outer differentials are ${\mathrm{\partial }}^{-2}=({\chi }_0\otimes 1,-1\otimes {\chi }_0),{\mathrm{\partial }}^{-1}=(1\otimes {\chi }_0,{\chi }_0\otimes 1)\mathrm{.}$ The minus sign comes from the first source factor having cochain degree $-1$. Each edge has internal bidegree $(0,0)$ after the indicated shifts and raises outer degree by $1$. The edge maps commute before the signs, since they act on different tensor factors, so ${\mathrm{\partial }}^{-1}{\mathrm{\partial }}^{-2}=0$. All four terms retain their inner two-periodic factorization differentials, with the same total potential.

The potential. Before closure, the boundary points of the braid diagram carry labels; every internal label occurs in exactly two local factors with opposite signs, so the sum of the local potentials is $a{\sum }_p{ϵ}_p{x}_p$ over the boundary points, and the square of the inner factorization differential is that element; the outer crossing differential still squares to zero by step 1.1. After closing the braid the closure arcs identify the boundary points in pairs with opposite orientations, so each boundary label occurs once with sign $+1$ and once with sign $-1$; the total potential is $0$ and every outer term of $C(D)$ is a genuine two-periodic complex of bigraded $\mathbb{Q}[a]$-modules.

The induced differential and the cohomology. For each term $C^j(D)$ the termwise cohomology $C{H}^j(D)$ is computed after removing the contractible summands, and $\mathrm{\partial }$ commutes with the internal differentials, so it descends to maps $\mathrm{\partial }\text{\hspace{0.17em}⁣}:C{H}^j(D)\to C{H}^{j+1}(D)$; the four resolutions contribute the four summands of $CH(D)$ and the induced maps are the sums of the four edge maps of the cube with Koszul signs. The resulting cohomology $H(D)$ is the trigraded cohomology of the diagram, and the trigradings of the four summands are inherited from the terms; no invariance statement is made here, and the representative is taken up to contractible summands in $K({\mathrm{h}\mathrm{m}\mathrm{f}}_0)$.

The moves used, with their roles. In the proof of [F1] far commutations ${\sigma }_i{\sigma }_j\leftrightarrow {\sigma }_j{\sigma }_i$ are the signed tensor flips of disjoint crossing factors, the braid-like IIa move covers inverse cancellations ${\sigma }_i{\sigma }_i^{-1}$, ${\sigma }_i^{-1}{\sigma }_i$; the coherent-orientation III move covers the braid relation; conjugation covers the change of cyclic order; the kink computations cover the two oriented stabilizations/destabilizations with their shifts $\{1,1\}[1]$ and none; and marking changes are absorbed by the marking-independence theorem. These are the source's Markov moves (a)-(c), with marking independence supplying the auxiliary marking changes. The separate six defining properties of $F$ include a skein relation and an unknot normalization, so they are not a move list.

Why IIb is absent and why this is not a gap. The source restricts the construction to braid diagrams because the oriented IIb move lies outside the braid-diagram calculus and its invariance is not proved there. By [F2], an ambient isotopy between two braid closures can be replaced by a finite Markov sequence, so IIb is never invoked in the invariance argument; conversely the example claims no invariance under IIb and no failure of it, only that the proof's scope is the braid-diagram calculus. The only choice-theoretic input is Markov's theorem inside [F1], as recorded there.

Conclusion. The tabulation of step 1.1 is complete: every move used in the invariance proof is one of far commutation, braid-like IIa cancellation, coherent-orientation III, conjugation, the two oriented stabilizations, or a marking change composed along a Markov sequence, and the IIb move is neither used nor claimed. This is a scope illustration, not a counterexample or a failure statement.