Invariance of Khovanov homology under first Reidemester move

Question

I am studying Khovanov homology from five lectures on Khovanov homology

and I want to try to show Khovanov homology is invariant under first Reidemester move but I cannot understand how we can write

$C^{î,*}(D)=C^{î,*}(D_0) \oplus C^{î-1,*}(D_{1})$.

where $D,D_0,D_1$ as in the picture below

For example again in this paper it takes $D$ as Hopf link and I'm trying to write it as a sum but I cannot because if we take $i=-2$ then $C^{-2,*}(D_0)=0 $ and $C^{-3,*}(D_{1})=0$ because they only have 1 negative cross, but if I take $C^{î,*}(D)=C^{î+1,*}(D_0) \oplus C^{î,*}(D_{1})$ it is true. Can you say where is my mistake?

Answer 1

I do not understand your example of Hopf link, since one cannot remove a crossing by the first Reidemeister move on this diagram. As a simple exercise to check the result, you can write down the chain complex of $D$, where $D$ is the knot diagram which contains only one (positive) crossing. By comparing the chain complex of $D_0$ (which is the disjoint union of two circles, hence $C^{i, \ast}(D_0)=0$ if $i\neq0$) and $D_1$ (one circle, also have $C^{i, \ast}(D_1)=0$ if $i\neq0$), you can see that $C^{i, \ast}(D)=C^{i, \ast}(D_0)\oplus C^{i-1, \ast}(D_1)$ $(i=0, 1)$. Note that here the shift for $D_1$ comes from $i=r_\alpha-n_-$ and the fact that $D_1$ is obtained from $D$ by 1-smoothing.