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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:

  1. $$ \frac{1}{\zeta_G(u)} = \prod_{p} ({1 - u^{L(p)}}) \tag{1} $$ 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 $G$).
  2. 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 $q^s+q^{1-s}=2\sqrt q \cos(b \log q)$

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draks ...
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    For a regular graph, the characteristic polynomial of its adjacency matrix and its zeta function give the same information. – Chris Godsil Dec 15 '15 at 00:42

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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.

Qiaochu Yuan
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  • Ok, but then why is $\sum_p L(p)=2E$, which perfectly matches the sum over the edges of the faces independent of the (planar cubic) graph? Just a happy coincidence? – draks ... Dec 22 '15 at 07:18
  • An answer to the question: Are the prime walks $p$ on a planar (cubic) graph identical to the faces of the graph? Since this seems the straight forward interpretation of eq. $(4)$... – draks ... Dec 22 '15 at 07:28
  • @draks: in general a graph will have infinitely many prime walks; the product in the first definition of an Ihara zeta function is an infinite product. So your identity (4) doesn't make sense. – Qiaochu Yuan Dec 22 '15 at 07:36
  • I cite from The coefficients of the Ihara zeta function: In fact, $Z_X(u)$ is always the reciprocal of a polynomial of maximum degree $2|E|$. And I deal with reciprocals of zeta all the time. Example 18 in the linked paper says $Z_{C_n}(u)=1-2u^n+u^{2n}=(1-u^n)^{2}$, so two cycles of length $n$: a finite number of prime walks, isn't it? – draks ... Dec 22 '15 at 07:49
  • @draks: look, I still don't understand what kind of answer you're expecting here. The linked paper also says, immediately after defining the Ihara zeta function, that "[t]ypically, this is an infinite product." So again, your identity (4), in general, doesn't make sense. – Qiaochu Yuan Dec 22 '15 at 07:54
  • I think my confusion comes right from what you quoted: "Typically, [$Z_X(u)$] is an infinite product [of factors $(1-u^{l(c)})^{-1}$]; however, the function is always rational. In fact, $Z_X(u)$ is always the reciprocal of a polynomial of a polynomial of maximum degree $2|E|$." How can this work? An infinite product has a finite maximum degree? – draks ... Dec 22 '15 at 19:14
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    @draks: there are only finitely many prime walks of a certain length, so the product converges in the $u$-adic topology. Here's an example of such an infinite product converging to a rational function: https://qchu.wordpress.com/2009/11/03/the-cyclotomic-identity-and-lyndon-words/ – Qiaochu Yuan Dec 22 '15 at 19:24
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    @draks... I don't have much intelligent to add, but you can try the following experiment to convince yourself that an infinite product can equal a polynomial. Consider the 4-regular graph with one vertex and two loops. This graph has $A=\begin{bmatrix}4\end{bmatrix}$ and has three faces, so $$\frac{1}{\zeta_G(u)}=(1-u^2)(1-4u+3u^2).$$ With some work, one finds that the numbers of equivalence classes of primitive, nonbacktracking circuits of lengths 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... are 4, 4, 8, 18, 48, 116, 312, 810, 2184, 5880, 16104, ... Evaluating the product ... – Will Orrick Dec 23 '15 at 06:13
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    $$\begin{aligned}&(1-u)^4(1-u^2)^4(1-u^3)^8(1-u^4)^{18}(1-u^5)^{48}(1-u^6)^{116}(1-u^7)^{312}\ &\quad\times(1-u^8)^{810}(1-u^9)^{2184}(1-u^{10})^{5880}(1-u^{11})^{16104}\ ,\end{aligned}$$ one finds that it equals $(1-u^2)(1-4u+3u^2)+O(u^{12})$. The exponents are computed as follows: label one of the loops $1$ and the other $2$. The loops may be traversed in either direction, so we arbitrarily choose one of the directions on each loop to be the forward direction, and denote the reverse directions $\bar{1}$ and $\bar{2}$. Two circuits, written as words ... – Will Orrick Dec 23 '15 at 06:25
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    ... on the alphabet ${1,\bar{1},2,\bar{2}}$, are equivalent if they are cyclic shifts of each other. A word is nonprimitive if it equals a nonzero cyclic shift of itself. To be nonbacktracking, the letters $1$ and $\bar{1}$ may not be adjacent, with a similar restriction on the letters $2$ and $\bar{2}$. So the circuits of length $1$ are $1$, $\bar{1}$, $2$, $\bar{2}$, and the circuits of length $2$ are $12$, $1\bar{2}$, $\bar{1}2$, $\bar{1}\bar{2}$. – Will Orrick Dec 23 '15 at 06:25
  • @WillOrrick thanks for your example. Does that mean that, when you extend the product to infinity, the result will converge to $\zeta(u)^{-1}$? Is it this what Qiaochu meant by converging in the $u$-adic topology? – draks ... Dec 23 '15 at 08:45
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    @draks... I have not verified anything, but $(1-u^2)(1-4u+3u^3)$ has zeroes at $-1$, $1$, and $\frac{1}{3}$, so I expect that the product converges in the usual power-series sense to $\left[\zeta(u)\right]^{-1}$ for $\lvert u\rvert<\frac{1}{3}$. The "$u$-adic topology" presumably means that two polynomials are considered to be close when their difference is proportional to a high power of $u$. My expectation is that the product converges to $(1-u^2)(1-4u+3u^3)$ in the sense that their difference is proportional to arbitrarily high powers of $u$ as the number of factors goes to infinity. – Will Orrick Dec 23 '15 at 15:44
  • @WillOrrick are these words, Lyndon words with restrictions? Very illustrative example, thanks for that... – draks ... Dec 23 '15 at 21:00
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    @draks... yes, in order to obtain only primitive circuits and in order to specify a unique equivalence class representative, I have required that words be strictly less than all of their cyclic shifts, which makes them Lyndon words on a four-letter alphabet. The requirement that $1$ not be adjacent to $\bar{1}$ and $2$ not be adjacent to $\bar{2}$ is an additional restriction. – Will Orrick Dec 24 '15 at 02:15
  • @WillOrrick how did you put all this into a counting function? Your example shows that you obviously have one... – draks ... Jan 06 '16 at 23:36
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    @draks... You give me too much credit. I obtained these numbers by stupid brute force. The algorithm is as follows: generate all of the $4^\ell$ length $\ell$ sequences on the alphabet ${1,\bar{1},2,\bar{2}}$. Then delete all sequences that are not lexicographically less than all of their cyclic shifts. This accomplishes two things: (1) it eliminates walks that are not primitive, and (2) it selects a canonical... – Will Orrick Jan 07 '16 at 02:36
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    ... equivalence class representative from the walks that are primitive. Once you have these, delete any sequences with illegal adjacencies, i.e. $1$ next to $\bar{1}$ or $2$ next to $\bar{2}$. – Will Orrick Jan 07 '16 at 02:36
  • A smarter algorithm would do things in the reverse order: generate only sequences with legal adjacencies. Then delete those that are not lexicographically less than their cyclic shifts. The advantage is that instead of $4^\ell$ sequences, you only generate $4\cdot3^\ell$ sequences. Better still is to count walks without generating them, but I haven't given that much thought. – Will Orrick Jan 07 '16 at 03:08
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    @draks... My last comment was wrong. Accounting for the requirement that the adjacency where the end of the word wraps around to the beginning be legal, the correct number of words is $3^\ell+2+(-1)^\ell$. Perhaps the easiest way to derive this is using the adjacency matrix $$B=\begin{bmatrix}1 & 0 & 1 & 1\ 0 & 1 & 1 & 1\ 1 & 1 & 1 & 0\ 1 & 1 & 0 & 1\end{bmatrix},$$ where rows and columns are both indexed by the letters $1$, $\bar{1}$, $2$, $\bar{2}$. The number of legal cyclic words of length $\ell$ is then $\mathrm{Tr},B^\ell=\sum \lambda_j^\ell$, where $\lambda_j$ are the... – Will Orrick Jan 07 '16 at 11:47
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    ...eigenvalues of $B$, namely $3$, $-1$, $1$, $1$. The number of primitive, legal words can then be obtained using Möbius inversion. Let $\phi(\ell)$ be the number of primitive walks satisfying the adjacency rules. Then $\sum_{d\mid \ell}\phi(d)=3^\ell+2+(-1)^\ell$. The Möbius inversion formula then implies that $\phi(\ell)=\sum_{d\mid\ell}\mu(\ell/d)(3^d+2+(-1)^d)$. Since equivalence classes of primitive, legal length-$\ell$ walks are of size $\ell$, the counting function you are looking for is $\phi(\ell)/\ell$. You can verify that this produces... – Will Orrick Jan 07 '16 at 14:14
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    ... the sequence $4$, $4$, $8$, $18$, $48$, $116$, $312$, $\ldots$ in the comment above. – Will Orrick Jan 07 '16 at 14:14
  • @WillOrrick this looks related to necklace polynomials... – draks ... Jan 08 '16 at 09:39
  • @Will if this extension of your B matrix is correct we could extend $\varphi(l)$ to graphs with more loops: n loops -> $\varphi_n(l)=\sum_{d|l} ((2n-1)^d+n+(n-1)(-1)^d)$, correct? – draks ... Jan 08 '16 at 11:56
  • @draks... yes. The matrix $B$ is a specialization of the path matrix defined in Section 4 of What are zeta functions of graphs and what are they good for? by Matthew D. Horton, H. M. Stark, and Audrey A. Terras. – Will Orrick Jan 08 '16 at 16:18
  • @WillOrrick how did you specialize the path matrix to get B? I'm struggling to make a general recipe out of it... – draks ... Jan 18 '16 at 15:26
  • @draks... all I meant by "specialization" is that all the complex variables $z_{ij}$ get set to $1$. – Will Orrick Jan 19 '16 at 12:15
  • @WillOrrick Why is the number of legal cyclic words of length $l$ then $\text{Tr}(B^l)$? – draks ... Jan 20 '16 at 00:15
  • This discussion has been moved to chat, but I wanted to make one correction: the matrix $B$ is the 0,1 edge matrix $W_1$ of Definition 2.2 in the Horton, Stark, and Terras paper, which is a specialization of the edge matrix, not the path matrix. In the case of the one-vertex, two-loop graph discussed in the comments above, the two matrices coincide, but not in general. Incidentally, Eqn. 2.4 of the paper is relevant: $1/\zeta_G(u)=\det(I-uW_1)$. – Will Orrick Jan 21 '16 at 18:42