Khovanov Bracket So I've been reading about Khovanov Homology in: https://arxiv.org/pdf/math/0201043.pdf, Bar-Natan's paper. I understand the cube complex construction but I'm having a little bit of trouble understanding how Khovanov's bracket construction creates a double complex which he flattens in the third axiom of the bracket. He briefly mentions the bracket but the rest of the paper as far as I can tell just discusses the cube complex.
Can anyone explain to me:


*

*How is Khovanov's bracket creating a double complex?

*How is this bracket definition related to the cubic complex?


Edit: I thought about $(1)$ a little more, perhaps this comes because $V$ is a graded vector space. Then when we tensor $V$ with another graded vector space this will induce a grading on the tensor. Then of course we will have the horizontal grading coming from the third axiom of the Khovanov bracket?
 A: To answer your questions, first let's fix some notation that if you start with a link $L$ and apply the Khovanov bracket at a crossing the links you get are $L_0$ and $L_1$ respectively. I've broken down my answer into two parts responding to your two questions.

*

*The goal of the Khovanov bracket is to define a chain complex $\textrm{CKh}(L)$ associate to a link $L$ (or really a diagram of the link). The third axiom of the bracket that roughly speaking we can define $\textrm{CKh}(L)$ as follows $$\textrm{CKh}(L) = \textrm{CKh}(L_0) \xrightarrow{f} \textrm{CKh}(L_1)\{1\}$$ for some map $f$. Notice that the terms $\textrm{CKh}(L_0)$ and $\textrm{CKh}(L_1)$ are chain complexes with their own homological gradings. There is also a new grading that records for us that $\textrm{CKh}(L_0)$ comes before $\textrm{CKh}(L_1)$ in our definition. The fact that we have these two gradings is what Bar-Natan means when he says "double complex". We want to combine these two gradings into a single homological grading on $ \textrm{CKh}(L)$ and this combination is what Bar-Natan refers to as flattening.


*The relationship between this bracket and the cube complex is an important idea in the definition of Khovanov homology. I've written more details below but the main idea is that we want to iterate using the Khovanov bracket. If we iterate using it as much as possible we will end up with a cube of resolutions.
So far we've said that if we choose a crossing of $L$ we can define  $ \textrm{CKh}(L)$ in terms of $\textrm{CKh}(L_0)$ and $\textrm{CKh}(L_1)$. If we pick a second crossing of $L$, then we can define four more links $L_{0,0}$, $L_{0,1}$,$L_{1,0}$, and $L_{1,1}$. Remembering the same bracket construction, we can define $\textrm{CKh}(L_0)$ in terms of $\textrm{CKh}(L_{0,0})$ and $\textrm{CKh}(L_{0,1})$ like this $$\textrm{CKh}(L_0) = \textrm{CKh}(L_{0,0}) \xrightarrow{f} \textrm{CKh}(L_{0,1})\{1\}$$ Similarly we can define $\textrm{CKh}(L_1)$ in terms of $\textrm{CKh}(L_{1,0})$ and $\textrm{CKh}(L_{1,1})$ like this $$\textrm{CKh}(L_1) = \textrm{CKh}(L_{1,0}) \xrightarrow{f} \textrm{CKh}(L_{1,1})\{1\}$$
Putting these three definitions together gives us a definition of $ \textrm{CKh}(L)$ in terms of $\textrm{CKh}(L_{0,0})$, $\textrm{CKh}(L_{0,1})$, $\textrm{CKh}(L_{1,0})$, and $\textrm{CKh}(L_{1,1})$ which looks like the following square
\begin{array}
\textrm{CKh}(L_{0,0}) & \longrightarrow & \textrm{CKh}(L_{0,1}) \\
\downarrow & & \downarrow \\
\textrm{CKh}(L_{1,0}) & \longrightarrow & \textrm{CKh}(L_{1,1})  
\end{array}
Now there is no reason to stop at just using two crossing of $L$ we could keep doing this same process with more and more crossing of $L$ to define $ \textrm{CKh}(L)$ in terms of the Khovanov chain complexes of simpler and simpler links. If we do this process for every crossing of $L$ then we would end up with a cube of resolutions that we can use to define $ \textrm{CKh}(L)$.
