Intuition for Szabo's geometric spectral sequence In https://arxiv.org/abs/1010.4252, Szabo defines a link invariant $\hat{H}(L)$ which can be computed combinatorially from a link diagram and shows that there is a spectral sequence from Khovanov homology to $\hat{H}(L)$. Conjecturally, this spectral sequence is isomorphic to the spectral sequence (proved by Ozsvath-Szabo) from Khovanov homology to $\hat{HF}(\Sigma(L))$ (the hat version Heegaard-Floer homology for the double branched cover). In particular, one should have $\hat{H}(L)\cong \hat{HF}(\Sigma(L))$.
To define $\hat{H}(L)$, Szabo divides the resolution diagrams into many types and writes down formula for the differential explicitly, case by case. While it is very concrete, I do not see how he obtains these formulas. I understand that he tries to add in differentials corresponding to higher dimensional faces of the resolution cube. But it seems that there are some deeper motivation to help him finding the right formula for these higher differentials. He does not explain much about these motivations in the paper. 
My question is: Any one could explain the intuition underlying Szabo's construction?
 A: *

*The first intuition is that the differentials should depend on a choice of orientation of the surgery arcs (the first differential eventually turns out to be independent, anyways). 


Let's do a split, with a single circle $C$ splitting into two, $C_1$ and $C_2$, by an elementary saddle $S$. Then the monopoles for $C$, $C_1\cup C_2$, and $S$ are $M(C)=S^1$, $M(C_1\cup C_2)=S^1\times S^1$ and $M(S)=S^1$ respectively, and the restriction maps are $M(S)\to M(C)$ is identity and $M(S)\to M(C_1\cup C_2)$ is the diagonal map. The split map differential $d_1\colon H_*(M(C))\to H_*(M(C_1\cup C_2))$ is the pull-push map; at the space level, the pull-push map is not cellular (it is the diagonal embedding), so it you want to get a cellular chain level map, you need to perturb the map off the diagonal. So you need to choose whether you are moving the diagonal towards the $S^1$ factor corresponding to $C_1$ or $C_2$, and that is the choice of an orientation of the surgery arc joining $C_1$ to $C_2$. (Of course the $d_1$ differential doesn't depend on how you chose your perturbation, but you did choose one. If $1$ and $x$ are the top (1-dim) and bottom (0-dim) generators, then the split map becomes $x\mapsto x\otimes x, 1\mapsto 1\otimes x+x\otimes 1$.)


*The second intuition is what should be homotopy $H$ relating these different choices of orientations (perturbations)? 


Once again, for the case of a split, there are two cellular perturbations of the diagonal $S^1$ inside $S^1\times S^1$. These two perturbations are homotopic, and the homotopy is the map $1\mapsto 1\otimes 1$.


*Now you combine these two facts, and try to come up a second differential $d_2$ in bigrading $(2,2)$. It should satisfy the following:


*

*$d_2$ depends on the orientations of the surgery arcs.

*$[d_1:d_2]=0$.

*$d_2^2$ is null-homotopic.

*And if we change the orientation of the arcs at a single crossing $c$, then $d_2$ changes as: $d'_2-d_2=[H_c:d_1]$, where $H_c$ is the above homotopy for the crossing $c$.



And you will come up with Szabo's $d_2$.


*And then you proceed. $d_3$ is forced to satisfy $d_3^2=[d_1:d_2]$, so try to come up with such as a $d_3$, and you will get Szabo's $d_3$. And if you are Szabo, at this point, you figure out the pattern.

