It all started with Chris' answer saying returning paths on cubic graphs without backtracking can be expressed by the following recursion relation:
$$p_{r+1}(a) = ap_r(a)-2p_{r-1}(a)$$
$a$ is an eigenvalue of the adjacency matrix $A$. Chris mentions Chebyshev polynomials there. It was Will who found the generating function for the given recursion to be:
$$G(x,a)=\frac{1-x^2}{1-ax+2x^2} $$
and just recently Hamed put Chebyshev back on the table:
$$ \frac{1-x^2}{1-ax +2x^2} \xrightarrow{x=t/\sqrt{2}}\frac{1-\frac{t^2}2}{1-2\frac{a}{\sqrt 8} t+t^2}=\left[1-\frac{t^2}{2}\right]\sum_{r=0}^\infty U_r\left(\frac{a}{\sqrt{8}}\right)t^r\\ =\sum_{r=0}^\infty \left(U_r\left(\frac{a}{\sqrt{8}}\right)-\frac12 U_{r-2}\left(\frac{a}{\sqrt{8}}\right)\right)t^r\\ $$ $$ \Rightarrow p_r(t)=\begin{cases}1 & \text{if $r=0$,}\\ 2^{r/2}\left(U_r(t/\sqrt{8})-\frac12U_{r-2}(t/\sqrt{8})\right) & \text{if $r\ge1$.}\end{cases} $$$$ \Rightarrow p_r(a)=\begin{cases}1 & \text{if $r=0$,}\\ 2^{r/2}\left(U_r(a/\sqrt{8})-\frac12U_{r-2}(a/\sqrt{8})\right) & \text{if $r\ge1$.}\end{cases} $$ The final line as taken from Will's community answer...
My question how to relate Ihara's $\zeta$ function and Chebyshev seems therefore mostly settled, but...:
Is it just a funny coincidence that the scaling factor of $\sqrt 8$ coincides with $\lambda_1\leq 2\sqrt 2$, which is the definition of cubic Ramanujan graphs.
And, there is another interesting thing:
As observed by Sunada, a regular graph is a Ramanujan graph if and only if its Ihara zeta function satisfies an analogue of the Riemann hypothesis.
What does this connection between Chebyshev, Ramanujan, Ihara and Riemann mean?
EDIT
I thought maybe something like a corollary could be possible:
- For Ramanujan graphs, the Ihara $\zeta$ function can be related to Chebyshev functions of the second kind, since the scaled eigenvalues of $A$ lie inside the range of convergence.
- A Ramanujan graph $G$ obeys the Riemann Hypothesis.
- Roots of the Ihara $\zeta$ function lie on the critical strip.
- The bunch of people above have contributed to $1\leftarrow 2$.
- $2 \leftrightarrow 3$ is proven here: Eigenvalues are of the form $\lambda=2\sqrt 2\cos(b\log 2)$
- $3\overset{\rightarrow ?}{\leftarrow} 1$ would be nice...