# How does the topology of the graphs' Riemann surface relate to its knot representation?

Let me give a worked-out example: The following cubic planar non-simple graph

$\hskip2.3in$

has the adjacency matrix $A=\pmatrix{0&3\\3&0}$. The graph has three faces, so the rank of $G$ is $\chi(G)=2$. The reciprocal of Ihara's $\zeta$ function can be evaluated $$\frac{1}{\zeta_G(u)}={(1-u^2)^{\chi(G)-1}\det(I - Au + 2u^2I)}\\ ={(1-u^2) (4u^4-5u^2+1)}$$

Looking at the orientation of the edges around the vertices, it is obvious that left and right are oriented opposite.

Now blow up every edge like in a ribbon or fat graph. Including the change of orientation the resulting graph looks like:

$\hskip1.7in$

where I stuck to the convention that I flipped every fat graph edge in the same direction. The resulting knot is a trefoil.

Further, the bicubic planar graphs can be related to Riemann surfaces (see here and references therein). Is there a relation between the Riemann surfaces and the knot?

How does the topology of the graphs' Riemann surface relate to its knot representation?

Concerning the construction of Riemann Surfaces, I refer to the one given in "Random Construction of Riemann Surfaces", Robert Brooks and Eran Makover say :

Definition 2.1 A left-hand turn path on $(\Gamma, \mathcal O)$ is a closed path on [the cubic graph] $\Gamma$ such that, at each vertex, the path turns left in the orientation $\mathcal O$.

The genus of $S^O(\Gamma,\mathcal O)$ is given by $$\text{genus}=1+\frac{n-l}2$$ [$l$ is the number of left-hand paths.]

• The computation of the rank and Ihara zeta function for this graph seem to be irrelevant to the actual question. I recommend editing that out and adding a self-contained description of the construction of a Riemann surface that you have in mind. – j.c. May 17 '18 at 14:52