Khovanov homology and Jones categorification — Examples and Counterexamples

The raw zero-crossing complex of $U$ is $\mathcal{A}$ in cohomological degree $0$ with zero differential by [F3]. Its generators $1,X$ have quantum degrees $1,-1$ by [F1], so its chain Euler polynomial is $q+q^{-1}$.

In the positive curl, the $0$-state contributes $(q+q^{-1}{)}^2$ and the $1$-state contributes $-q(q+q^{-1})$, because the state coordinate raises quantum degree by one and homological degree by one. Hence ${\chi }_q^{\mathrm{r}\mathrm{a}\mathrm{w}}(U_{+})=(q+q^{-1}{)}^2-q(q+q^{-1})=1+q^{-2}=q^{-1}(q+q^{-1})$. In the negative curl, the $0$-state contributes $q+q^{-1}$ and the $1$-state contributes $-q(q+q^{-1}{)}^2$. Hence ${\chi }_q^{\mathrm{r}\mathrm{a}\mathrm{w}}(U_{-})=(q+q^{-1})-q(q+q^{-1}{)}^2=-q^3-q=-q^2(q+q^{-1})$. These computations use only the state modules and gradings before the writhe normalization.

For each quantum degree, tensor the bounded finite-rank free complex with $\mathbb{Q}$. Clearing denominators shows that kernels and images localize to the kernels and images of the rationalized maps; applying the same argument to a kernel modulo the preceding image identifies rationalized cohomology with integral cohomology tensored with $\mathbb{Q}$. Its dimension is therefore the integral rank of cohomology. In the bounded rational complex, rank-nullity gives $\dim C^i=\dim H^i+\dim \mathrm{im}{d}^i+\dim \mathrm{im}{d}^{i-1}$; alternating over $i$ cancels the image terms. Thus the raw cohomology Euler polynomials equal the chain Euler polynomials in steps 1.1–1.2. The distinct Laurent polynomials $q+q^{-1}$, $q^{-1}(q+q^{-1})$, and $-q^2(q+q^{-1})$ come from diagrams of the same unknot by [F4]. The raw groups are consequently not invariant, and the raw unknot factor is not the claimed $q+q^{-1}$.

By [F1], the zero-crossing cube has just $C^0(U)=\mathcal{A}$ and differential zero. Therefore its cohomology equals $\mathcal{A}$ in cohomological degree $0$. By [F2], the two basis elements give one free rank-one group in quantum degree $1$ and one in degree $-1$, with no other groups.

In the based quotient, the image of multiplication by $X$ is $\mathbb{Z}\{X\}$, so the quotient of $\mathcal{A}=\mathbb{Z}\{1,X\}$ is freely generated by $[1]$. By [F3] its reduced quantum degree is $1-1=0$; the differential remains zero, so its only cohomology is $\mathbb{Z}$ in bidegree $(0,0)$. Summing the unreduced ranks with their quantum powers gives $q+q^{-1}$, and the reduced rank sum is $1$.