Let me give a worked-out example: The following cubic planar [non-simple](http://mathworld.wolfram.com/SimpleGraph.html) graph $\hskip2.3in$[![enter image description here][1]][1] has the adjacency matrix $A=\pmatrix{0&3\\3&0}$. The graph has three faces, so the rank of $G$ is [$\chi(G)=2$](https://math.stackexchange.com/questions/582616/of-face-and-circuit-rank). 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][5]. Including the change of orientation the resulting graph looks like: $\hskip1.7in$[![enter image description here][2]][3] 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][6] 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"](http://arxiv.org/abs/math/0106251), 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.] [1]: https://i.sstatic.net/eQYHS.png [2]: https://i.sstatic.net/ixMNDm.jpg [3]: http://www.win.tue.nl/~vanwijk/seifertview/tutorial5.htm#trefoilseifert [4]: https://en.wikipedia.org/wiki/Trefoil_knot#Invariants [5]: https://ncatlab.org/nlab/show/ribbon+graph [6]: https://math.stackexchange.com/q/1626800/19341