Quantum degree of Khovanov Complex Reading Turner's 5 Lectures on Khovanov Homology (here) , I am trying to understand how to do computations using the long exact sequence defined on Lecture $3$. 
I have realized that I cannot understand what the star parametrizes when we right $C^{i,*}$
I would be glad if , along with an explanation, someone could give an example on Example 3.7, Lecture 1 the Hopf Link: Is it true that $C^{-2,j}=\mathbb{Q}$ for $j=-4,-6$ and $0$ otherwise ?
Also, If I understand correctly, the Khovanov Homology of the Unknot should be $C^{0,1}=C^{0,-1}=\mathbb{Q}$ and $0$ otherwise. Is this true ?
 A: The term $C^{i,*}$ means $\bigoplus_{j\in\mathbb{Z}}C^{i,j}$. In other words, $C^{i,*}$ is the summand of the chain complex $C$ in homological grading $i$ and in any quantum/Jones/polynomial grading. 
In Example 3.7, Turner chooses an orientation on the Hopf link so that both crossings are negative. Thus $n_+=0$ and $n_-=2$. The homological grading $i=-2$ corresponds to the all-$0$ state on the far left of the cube. Since there are two components in that state, there are four possible ways to label that state with $1$'s and $x$'s: $1\otimes 1$, $1\otimes x$, $x\otimes 1$ and $x\otimes x$. The degrees of those four states are $2, 0, 0,$ and $-2$ respectively. Therefore, the $j$-gradings of those states are $-2, -4, -4$ and $-6$ respectively. So 
$$C^{-2,*} = C^{-2,-2}\oplus C^{-2,-4}\oplus C^{-2,-6}$$
where
\begin{align*}
C^{-2,-2} \cong &\; \mathbb{Q},\\
C^{-2,-4} \cong & \; \mathbb{Q}^2,~\text{and}\\
C^{-2,-6} \cong & \; \mathbb{Q}.
\end{align*}
The differential sends $1\otimes 1$ to $(1,1)$ (meaning a $1$ on the top middle state and a $1$ on the bottom middle state), sends both $1\otimes x$ and $x\otimes 1$ to $(x,x)$, and sends $x\otimes x$ to $0$. Thus $1\otimes x - x \otimes 1$ generates homology in the summand $H^{-2,-4}$ and $x\otimes x$ generates homology in the summand $H^{-2,-6}.$ 
You are correct about the Khovanov homology of the unknot.
