This sounds like a simple piece of math (which got a long story over time, thanks for reading!) and the consequence seems surprising. At least to me. Here it is:
It boils down to comparing two expression of the reciprocal/inverse of the Ihara zeta function $\frac{1}{\zeta_G(u)}$ for planar graphs and compare the resulting polynomials: $$ \prod_{p} ({1 - u^{L(p)}}) \overset{!}{=}{(1-u^2)^{\chi(G)-1}\det(I - Au + (k-1)u^2I)} $$
The Wiki page on the Ihara $\zeta$ function of a graph $G$ with $n$ vertices gives two definitions:
- $$ \frac{1}{\zeta_G(u)} = \prod_{p} ({1 - u^{L(p)}}) \tag{1} $$ This product is taken over all prime walks $p$ of the graph $G = (V, E)$... $L(p)$ is the length of cycle $p$ (i.e. prime walks on $G$).
- If we restrict ourselves to $k$-regular graphs, we also have: $$ \frac{1}{\zeta_G(u)} = {(1-u^2)^{\chi(G)-1}\det(I - Au + (k-1)u^2I)} \tag{2}$$ where $χ$ is the circuit rank, which is the number of edges deleted from $G$ to form a spanning tree and alternativley,the rank of the fundamental group of the graph.
Let's compare the degree of the polynomials in $u$ in both expressions.
We can use $\chi=F-1$, for planar graphs. $F$ is the number of faces. So $(1-u^2)^{F-2}$ has degree $2(\color{red}{F-2})$.
Now the determinant in the latter expression can be written as product over the eigenvaues $\lambda_m$ of the adjacency matrix $A$: $$ P_A(u)=\prod_{m=1}^n (1-\lambda_mu+(k-1)u^2) \tag{3} $$ This gives a degree of $2\color{blue}n$. A more compact form for cubic graphs is given below...
So the total degree is $2(\color{blue}n+\color{red}{F-2})$. If we apply Euler formula for planar graphs we get $2E$, twice the number of edges. Therefore we have: $$\sum_p L(p)=2E. \tag{4}$$
This is also stated here: The coefficients of the Ihara zeta function, p.219 and seems to have an easy explanation: All prime walks are faces and contribute each edge twice.
But then this means: All Ihara $\zeta$ functions for planar $k$-regular graphs with a given set of faces are equivalent, irrespective of their actual arrangement and although their spectra of $A$ might differ...
Is this correct?
It is true that there are references that state that there are infinitely many prime walks $p$ and then the above said, doesn't make sense. Maybe I got something wrong, so I would be glad if you could point out my error...
EDIT sorry, the numerics looked strange, I gotta think about it...
EDIT2: If we assume $k$-regular graphs, we set $u=(k-1)^{-s}=q^{-s}$ (see The Riemann hypothesis for graphs), this simplifies to $$ q^{-ns}\prod_{m=1}^n ((q^s+q^{1-s})-\lambda_m ) \tag{3$^*$} $$ which further simplifies for $k$-regular Ramanujan graphs, i.e. $\lambda_1\leq 2\sqrt{k-1}$, since the Riemann hypothesis holds (see proof in link) on this case and $s=1/2+ib$, therefore $q^s+q^{1-s}=2\sqrt q \cos(b \log q)$
Like Chris Godsil says in the comments, for regular graphs, the Ihara zeta function contains the same information as the (multi)set of eigenvalues of the adjacency matrix. So two $k$-regular graphs have the same Ihara zeta function iff they're isospectral.