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Let $\Gamma$ be an undirected finite graph. Write $M_r$ for the number of non-backtracking closed walks of length $r$ in $\Gamma$, i.e., walks where a step is never followed by its inverse. Let $A$ be the adjacency matrix of $\Gamma$, and $\lambda_1,...,\lambda_N$ its eigenvalues.

If $\Gamma$ is regular of degree $d$, then

$$M_{2 k} = (d-1)^k \sum_{n=1}^N \left(U_{2 k}\left(\frac{\lambda_n}{2 \sqrt{d-1}}\right) - \frac{1}{d-1} U_{2 k-2}\left(\frac{\lambda_n}{2 \sqrt{d-1}}\right) \right),$$ where $U_m$ is the $m$th Chebyshev polynomial of the second kind. For a proof, see, e.g., section 1.4 of the Davidoff-Sarnak-Valette monograph.

In particular, if many ($\geq \epsilon N$, say) of the eigenvalues $\lambda_i$ are large ($\geq \Lambda$, where $\Lambda > 5 \sqrt{d}$, say) then, for $k$ not too small, $M_{2 k}$ is large (in fact, roughly in the order of $\epsilon N\cdot \Lambda^{2 k}$).

Is an expression such as the above possible when $\Gamma$ is not regular? Can we at least obtain the conclusion that, if many of the eigenvalues are large ($\geq \Lambda$, where $\Lambda \ggg \sqrt{D}$, and $D$ is the maximal degree of the vertices of $\Gamma$) then $M_{2 k}$ must be large?

(Related, vaguer question: Closed geodesics and eigenvalues in a non-regular graph)

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  • $\begingroup$ I do not know what "an expression such as the above possible" means but did you try the graph which is the path of length $n$ or the tree of height 2 with $n+1$ vertices? $\endgroup$
    – markvs
    Commented Jan 12, 2021 at 20:35
  • $\begingroup$ Well, one can't have many large eigenvalues $\lambda_i$ then, since $\textrm{Tr} A^{2 k}$ is not then large: on a tree where every non-root node has $r$ children, the number of closed walks of length $2 k$ is $O(r)^k$. $\endgroup$ Commented Jan 12, 2021 at 21:20
  • $\begingroup$ Yes, and what I am saying is that neither of those (nor any other tree) contradict the conclusion that I mention at the end. I said that the number of closed walks is $O(r)^k$, not that the number of closed, non-backtracking walks is $O(r)^k$ (though of course it is, being $0$). $\endgroup$ Commented Jan 12, 2021 at 23:04

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