Zeta functions abound in mathematics. Audrey Terras describes in Zeta Functions and Chaos three zeta functions--the zeta fct. of a projective non-singular algebraic variety; the Artin-Mazur zeta function; and a special Reulle (aka dynamical systems or Smale) zeta function, the Ihara zeta function for a graph $G$--all can be expressed in the same basic form:

$$\zeta(u)=\exp\left ( \sum_{m\geq 1} \frac{N_mu^m}{m} \right ).$$

For graph zeta functions $\zeta(u,G_n)$ typically $N_m$ is the number of closed walks of $m$ steps (with some qualifications) on the graph $G$ with $n$ vertices and is related to the trace of the power of an edge adjacency matrix. For a vertex adjacency matrix $A_n$, also $N_m = \operatorname{tr}[A_n^m]$ (e.g., A054878 and A092297). (Edited per draks' comment.)

You can use the general heuristic $O=KPK^{-1}\Leftrightarrow P=K^{-1}OK$ to obtain

$$\operatorname{tr}(A)=\ln[\operatorname{det}[\exp(A)]] \Leftrightarrow \operatorname{det}(A)=\exp[\operatorname{tr}[\ln(A)]]$$

and then

$$\operatorname{det}(I-uA_n)=\exp[\operatorname{tr}[\ln(I-uA_n)]]=\exp\left( -\sum_{m\geq 1} \frac{\operatorname{tr}(A_n^m)u^m}{m} \right)$$ $$=\exp\left (-\sum_{m\geq 1} \frac{N_mu^m}{m} \right ),$$


$$\zeta(u;G_n)=\frac{1}{\operatorname{det}(I-uA_n)}=\exp\left(\sum_{m\geq 1} \frac{\operatorname{tr}(A_n^m)u^m}{m} \right)=\exp\left(-:\ln(1-ua): \right).$$ where $a^k=a_k=\operatorname{tr}(A_n^k)$ for $k>0$.

This last expression is the umbral form for the exponential generating function for the cycle index polynomials (OEIS-A036039) for the symmetric group (mod signs).

The Appell sequence in MO-Q111165 incorporating the Riemann zeta function reverses the last relation in some sense:

$$\exp\left (-\beta p_{.}(z)\right )=\exp\left [-(z+\gamma)\beta -\sum_{k=2}^{\infty } \frac{\zeta (k)\beta ^k}{k} \right ]=\exp\left [ :\ln(1-b\beta ) :\right ]$$ where $b^1=b_{1}=(z+\gamma)$ and $b^k=b_k=\zeta(k)$ for $k>1$.

For easy reference: $$p_{0}(x)=1$$ $$p_{1}(x)=x+\gamma$$ $$p_2(x)=(x+\gamma)^2-\zeta(2)$$ $$p_3(x)=(x+\gamma)^3-3\zeta(2)(x+\gamma)+2\zeta(3)$$ $$p_4(x)=(x+\gamma)^4-6\zeta(2)(x+\gamma)^2+8\zeta(3)(x+\gamma)+3[\zeta^2(2)-2\zeta(4)]$$

These polynomials are the first few cycle index polynomials for the symmetric group. I'd like to relate each $p_n(x)$ to the characteristic polynomial of a matrix with a null main diagonal.

For example, for such a 3x3 matrix the char polynomial is

$$ \sigma^3-(a_{12}a_{21}+a_{13}a_{31}+a_{23}a_{32})\sigma+(a_{12}a_{23}a_{31}+a_{13}a_{32}a_{21}).$$

Picture a triangle with the vertices ($v$) labelled 1 to 3. Make an orbit/cycle/closed loop, or path, traversing the triangle from $v_1$ through $v_2$ and $v_3$ and then to $v_1$. Denote this path of three steps and length three by $a_{12}a_{23}a_{31}$ and assign it the "moment/transition amplitude" of $\zeta(3)$. Likewise, assign the amplitude $\zeta(2)$ to paths of two steps and length one $a_{12}a_{21}$, an amplitude of $\sigma=x+\gamma$ to a self- or null-loop, and so on. This generates $p_3(x)$.

Similarly, consider a square with labeled vertices and edges between all pairs of vertices. With cycles/orbits/closed paths of opposing circulation considered distinct cycles, the associated 4x4 determinant generates six paths each with four steps and length four, e.g., $a_{12}a_{24}a_{43}a_{31}$, that can be assigned an amplitude of $\zeta(4)$ each and three sets of two paths of two steps and length one, e.g., $a_{13}a_{31}a_{24}a_{42}$, that can be assigned an amplitude of $\zeta^{2}(2)$. The algorithm can be continued to the other terms to generate $p_4(x)$.

How to prove that the algorithm will work for all $p_n(x)$, i.e., that each $p_n(x)$ can be generated in the above manner from an $n$ by $n$ "adjacency" matrix?

[Nov. 15, 2013 update: Replacing $p_1(x)=x+\gamma$ by $x$ and the $\zeta(n)$ by $1$ gives the characteristic polynomials (mod signs) of the adjacency matrix of the complete n-graph (see A055137).]

  • $\begingroup$ Aren't your "adjacency" matrices just symmetric matrices? In that case, a polynomial can be generated if and only if its roots are all real. $\endgroup$
    – Will Sawin
    Nov 8, 2012 at 2:23
  • $\begingroup$ By "adjacency" matrix I really mean the matrix of indeterminates $a_{ij}$ with $a_{ii}=0$ on the main diagonal. It can be thought of as an "adjacency" matrix from which the char polynoms, in $\sigma$ and the indeterminates, can be formed, which will not have a $\sigma^{n-1}$ term since the trace is zero (neither do the p_n(x)). The question is really whether I can make the transformation as indicated from these indeterminates in the char polynoms to the appropriate $\zeta(j)$ amplitudes to obtain the $p_n(x)$. That might lead to physical/geometric interpretations of these zeta terms. $\endgroup$ Nov 8, 2012 at 11:02
  • $\begingroup$ To assign values to the $a_{ij}$ indeterminates before the transformation is made would be like taking the derivative of a function by first assigning a numerical value to the independent variable at some point. The derivative would always return a zero then. However, we are free to fix parameters to determine the function we are interested in, same as making the main diagonal null. $\endgroup$ Nov 8, 2012 at 11:24
  • $\begingroup$ Call it a pseudo-adjacency matrix since assigning ones to all the off-diagonal elements would give the adjacency matrix for a complete n-graph and a characteristic polynomial with the coefficients oeis.org/A055137, which appear in the $p_n(x)$ as noted in mathoverflow.net/questions/111165/…. $\endgroup$ Nov 8, 2012 at 11:55
  • 1
    $\begingroup$ @Draks, see oeis.org/A263916 and oeis.org/A127672 along with the Damianou and Damianou and Evripidou links. $\endgroup$ Nov 10, 2015 at 0:49

1 Answer 1


I think the validity of the algorithm is corroborated by the relation between the trace and determinant of $m$-dimensional square matrices $A$ inherent in the Cayley-Hamilton theorem applied to the characteristic polynomial of $A$ as explained in Wikipedia.

The relation between the $\det A$ and $(\operatorname{tr} A^k)^j$ for $k,j<m+1$ is precisely that given by the cycle index partition polynomials, and the cycle mapping is clearly shown by Mark Dominus in the link in OEIS/A036039. Substitute $\zeta(k)^j$ for $(\operatorname{tr}(A^k))^j$ in the Wikipedia entry, just as above, but how to formally prove the relation between the indices mapping above and the cycle mapping still is a mystery to me.

Edit Oct. 9, 2020

(The Wikipedia article has changed quite a lot.) I've finally written up a draft compiling some old notes of mine on this and related topics and posted it on my blog as "Appells and Roses: Newton, Leibniz, Euler, Riemann and Symmetric Polynomials." In looking for further reading material, I found Qiaochu Yuan's excellent blog post "GILA VI: The cycle index polynomials of the symmetric groups," which elucidates the combinatorial interpretation of the generating function as enumeration of cycles, an instance of Polya's counting theorem.

  • $\begingroup$ Related: arxiv.org/abs/1502.05771 $\endgroup$ Mar 17, 2016 at 20:54
  • $\begingroup$ Related: "Combinatorial aspects of elliptic curves" by Musiker arxiv.org/abs/0707.3179 $\endgroup$ Dec 25, 2016 at 7:04
  • $\begingroup$ Article in March comment is "Seiberg Duality, Quiver Gauge Theories, and Ihara Zeta Function" by Da Zhou, Yan Xiao, Yang-Hui He $\endgroup$ Dec 25, 2016 at 7:06
  • $\begingroup$ See also p. 38 of "Algebraic and geometric methods in enumerative combinatorics" by Ardila (arxiv.org/abs/1409.2562). $\endgroup$ Jan 2, 2017 at 22:25
  • $\begingroup$ Related: pg. 44 of arxiv.org/abs/1707.01770 "Notes on the Riemann Hypothesis" by Ricardo Pérez-Marco $\endgroup$ Aug 12, 2018 at 20:25

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