Euler characteristic for pair of checkerboard surfaces of a link

Question

A knot/link diagram can be checkerboard colored (as in, the regions between the strands can be two-colored), and a checkerboard surface is what you get by taking all the regions of a particular color and attaching twisted bands at the crossings; the boundary of a checkerboard surface is the link. Every link diagram has two checkerboard surfaces $\Sigma_1$ and $\Sigma_2$. (Mind that the diagram is in $S^2$, so one of them contains $\infty$ as it were.)

In a paper by Joshua Howie ("A characterization of alternating knot exeriors"), he says that by "a simple Euler characteristic argument" $$\chi(\Sigma_1)+\chi(\Sigma_2)+n=2$$ where $n$ is the number of crossings in the diagram.

I am not sure what argument he had in mind. I came up with one which I will post as an answer, but I am afraid I am missing something about the Euler characteristic, and I would appreciate another argument.

Answer 1

I ran into Josh Howie a few months ago and he told me the argument he had in mind. At this point, I don't exactly remember what he said, but the following shouldn't be too far off:

1. Consider triangulations of $\Sigma_1$ and $\Sigma_2$ which are compatible at the crossings in that they both have an edge spanning the crossing.

2. The union of $\Sigma_1$ and $\Sigma_2$ is homotopy equivalent to a sphere, so $\chi(\Sigma_1\cup\Sigma_2)=2$.

3. $\chi(\Sigma_1)+\chi(\Sigma_2)$ double counts those $n$ spanning edges, and since the number of edges is subtracted in the formula for Euler characteristic, $\chi(\Sigma_1)+\chi(\Sigma_2)+n=2$. (It also double counts the link, but Euler characteristic of a link is $0$.)

Answer 2

At least for a space $X$ with a finite triangulation with two subcomplexes $A$ and $B$ having $A\cup B=X$, it follows from the short exact sequences for the Mayer-Vietoris sequences that $\chi(A)+\chi(B)=\chi(X)+\chi(A\cap B)$. The space $\Sigma_1\cup\Sigma_2$ has a finite triangulation, so $\chi(\Sigma_1)+\chi(\Sigma_2)=\chi(\Sigma_1\cup\Sigma_2)+\chi(\Sigma_1\cap\Sigma_2)$.

The union of the checkerboard surfaces is almost $S^2$ except for some singularities at the crossings. However, the region around a crossing is homotopy equivalent to a disc, so $\chi(\Sigma_1\cup\Sigma_2)=\chi(S^2)=2$.

The intersection of the checkerboard surfaces is the link itself, except that at the crossings there is an additional edge connecting the two strands. These edges can be retracted to a point, so $\Sigma_1\cap \Sigma_2$ is homotopy equivalent to the projection of the link. This is a $4$-regular graph, except perhaps for some disconnected loops --- the loops do not contribute to Euler characteristic, so we may ignore them. Euler characteristic of a graph is $\left|V\right|-\left|E\right|$, and $\left|E\right|=2\left|V\right|$ by $4$-regularity, so $\chi(\Sigma_1\cap \Sigma_2)=-n$, where $n$ is the number of crossings.

Then, $\chi(\Sigma_1)+\chi(\Sigma_2)=2-n$.