Khovanov Homology

Question

I'm reading this and trying to understand how he computed the Khovanov homology of the Hopf link. The construction of the chain complexes and the maps look fine to me but the only problem is, I do not know how to get the table on page 3 (with homological grading and quantum grading). Can anyone give an explanation?

Answer 1

Each generator $x$ comes equipped with two gradings: the (co)homological grading $gr(x)$ and the quantum grading $q(x)$. A generator is a choice of marking $v_+$ or $v_-$ for each component of one of the Kauffman states. See the picture below for the situation with the Hopf link.

The homological grading $gr(x)$ of any generator only depends on the Kauffman state it is associated to. It is the number of $1$-resolutions in the Kauffman state minus the number of negative crossings in the link diagram. Generators associated to the leftmost state above have $gr(x)=-2$, generators associated to the middle two states have $gr(x)=-1$, and generators associated to the rightmost state have $gr(x)=0$.

The quantum grading $q(x)$ can be computed as follows. Let $p(x)$ be the difference in the number of $v_+$'s and $v_-$'s. So if $x=v_+\otimes v_+$, then $p(x)=2$, and if $x=v_+\otimes v_-$, then $p(x)=0$, etc. Let $n_+(D)$ and $n_-(D)$ be the number of positive and negative crossings in the link diagram $D$. Finally let $c_1(x)$ be the number of $1$'s in the Kauffman state associated to $x$. Then $$q(x) = p(x) +c_1(x) +n_+(D) - 2n_-(D).$$

Let $D$ be the diagram pictured above. Consider the state $x=v_-\otimes v_-$ associated to the leftmost Kauffman state. Then $gr(x) = - 2$ and $$q(x) = -2 + 0 + 0 - 2(2) = -6.$$