Degree shift operation when constructing chain complexes of the Khovanov Homology I'm reading this paper by Dror Bar-Natan, On Khovanov’s categorification of the Jones polynomial (here). On chapter 3 (categorification), he wrote: With every vertex $\alpha\in \{0,1 \}^\chi$ of the cube $\{0,1 \}^\chi$, we associate the graded vector space $V_\alpha(L)=V^{\otimes k}\{r\}$ where $k$ is the number of cycles in the smoothing of $L$ corresponding to $\alpha$ and $r$ is the height $|\alpha|=\sum_{i}\alpha_i$ of $\alpha$. 
My question is, why do we need the degree shift operation $\cdot \{r \}$ on $V^{\otimes k}$? In most of the computation, we kind of ignore this degree shift but only care about $V^{\otimes k}$.
 A: The degree shift operator ensures that the differential preserves polynomial/quantum grading. If we ignore an overall shift depending on the number of positive and negative crossings in the diagram, then the polynomial grading is essentially the number of $v_+$'s minus the number of $v_-$'s plus the height $r$.
Let's look at the maps used to construct the differential. The multiplication $m:V\otimes V \to V$ is defined by
$$m(v_+\otimes v_-)=m(v_-\otimes v_+) = v_-,~m(v_+\otimes v_+)=v_+,~\text{and}~m(v_-\otimes v_-)=0.$$
The comulitplication $\Delta:V\to V\otimes V$ is defined by
$$\Delta(v_+) = v_+\otimes v_- + v_-\otimes v_+~\text{and}~\Delta(v_-)=v_- \otimes v_-.$$
In the multiplication map, there is one fewer $v_+$ in each output and the same number of $v_-$'s. But the height of the output is one more than the height of the input. Thus the polynomial grading is preserved. Similarly, in the comultiplication map, there is one more $v_-$ in the output than the input and the same number of $v_+$'s. Again the height of the output is one greater than the height of the input causing the polynomial grading to be preserved.
