Khovanov homology detects the unknot — Examples and Counterexamples

Define $V$ by $V^{0,0}=\mathbb{Q}$ and all other groups zero. Define $W$ by $W^{0,0}=W^{0,1}=W^{1,1}=\mathbb{Q}$ and all other groups zero; give both complexes zero differentials.

By [F1], the graded Euler characteristics are ${\chi }_q(V)=1$ and ${\chi }_q(W)=1+q-q=1$, but their total dimensions are $1$ and $3$. Therefore equal graded Euler characteristics do not determine total rank.

The crossingless diagram has one based circle, hence its reduced state module is $M=\mathbb{Z}\{1\}$ in degree $0$ by [F2]–[F3]. There are no crossings and therefore no differential, so its cohomology is $\mathbb{Z}$ in degree $0$.

First compute the auxiliary positive two-strand closure, whose $A$-smoothing is the identity tangle and whose $B$-smoothing is the cap-cup tangle. This is only an auxiliary raw cube; it does not change the specified negative diagram. With zero, one, two, or three cap-cup smoothings, the closure has respectively two, one, two, or three circles: composing cap-cups creates one additional internal circle each time after the first. Thus its four raw cochain degrees $0,1,2,3$ have modules $M\otimes \mathcal{A}$, $M^3$, $(M\otimes \mathcal{A}{)}^3$, and $M\otimes \mathcal{A}\otimes \mathcal{A}$, of ranks $2,3,6,4$.

Label the three two-circle states in crossing order and identify the two unbased factors in the all-$B$ state from left to right. The last differential is $(a,b,c)⟼a\otimes X-\mathrm{\Delta }(b)+X\otimes c\mathrm{.}$ Here the first and third edges split the based circle, while the middle edge splits the unbased circle; the signs are the cube signs from [F2]. Its image is exactly the subgroup generated by $1\otimes X$, $X\otimes 1$, and $X\otimes X$, and its kernel consists of triples with equal coefficients of $1$ and with coefficients of $X$ satisfying $a_X-b_X+c_X=0$.

Order crossings from top to bottom. The first differential sends $1\otimes 1$ to $(1,1,1)$ and $1\otimes X$ to zero, so its image is the diagonal copy of $\mathbb{Z}$ and its kernel is $\mathbb{Z}\{1\otimes X\}$. After suppressing the common based factor $M$, the next differential is the signed incidence matrix $\left(\begin{array}{ccc}-1& 1& 0\\ -1& 0& 1\\ 0& -1& 1\end{array}\right)$ with each integer entry multiplying the label $X$, because each edge splits the based circle and the reduced split sends $1$ to $1\otimes X$. Its image has rank two and its kernel is the diagonal copy of $\mathbb{Z}$.

These auxiliary raw differentials have ranks $1,2,3$. Their images are primitive subgroups of their target free groups: the first image has a coordinate $1$, the incidence matrix in step 2.1 has a $2$ by $2$ minor $1$, and the last image is the coordinate span of $1\otimes X,X\otimes 1,X\otimes X$. In each case elementary integral row and column elimination therefore gives only unit nonzero diagonal entries. Hence all auxiliary homology groups are free. The kernel/image calculations in steps 2.1 and 1.3 give raw cohomology $\mathbb{Z}$ in degrees $0,2,3$ and zero in degree $1$.

Now return to the specified negative diagram. At each crossing its A-smoothing is cap-cup and its B-smoothing is through, so the state $v$ has exactly the same circles as the auxiliary positive state $w=1-v$. Put the perfect integral pairing $⟨a,b⟩=ϵ(ab)$ on each unbased circle, with matrix $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, and pair the based module $M$ with itself by $⟨1,1⟩=1$. Multiplication and comultiplication are adjoints for these pairings, directly on $1,X$; the reduced merge and split are adjoints too, since $m_M(1\otimes 1)=1$, $m_M(X\otimes 1)=0$ and ${\mathrm{\Delta }}_M(1)=X\otimes 1$. Use the same actual circle order on the two complementary states. For an edge in coordinate $j$, the old transposed sign is $(-1{)}^{j-1-{\sum }_{k<j}{v}_k}$ whereas the negative cube requires $(-1{)}^{{\sum }_{k<j}{v}_k}$. Multiplying vertex $v$ by $J(v)=(-1{)}^{{\sum }_j(j-1)v_j}$ corrects this ratio: across that edge its ratio is $(-1{)}^{j-1}$. Thus the negative raw cube is integrally isomorphic to the degree-reversed transpose of the auxiliary cube, with this explicit diagonal sign correction. No spatial reflection or change to the given negative crossing is used.

The reversed transpose has ranks $4,6,3,2$ in raw degrees $0,1,2,3$, and differential ranks $3,2,1$. Transposition preserves the unit diagonal entries proved in step 3.1, as do the unimodular pairings and $J$ in step 3.2. Each image is therefore primitive; its quotient in the kernel is torsion-free because it is a subgroup of the free quotient of the whole target. Rank-nullity now gives free homology of ranks $1,1,0,1$ in raw degrees $0,1,2,3$. The negative diagram has three negative crossings, so the normalization $i=\mathrm{\mid }v\mathrm{\mid }-3$ gives $\mathbb{Z}$ exactly in degrees $-3,-2,0$, with zero in degree $-1$. Together with step 1.1 this proves all stated ranks for the actual two diagrams.