Complexity of the circulant foliation over a graph

Abstract

In the present paper, we investigate the complexity of infinite family of graphs $H_n=H_n(G_1,\,G_2,\ldots ,G_m)$ obtained as a circulant foliation over a graph H on m vertices with fibers $G_1,\,G_2,\ldots ,G_m.$ Each fiber $G_i=C_n(s_{i,1},\,s_{i,2},\ldots ,s_{i,k_i})$ of this foliation is the circulant graph on n vertices with jumps $s_{i,1},\,s_{i,2},\ldots ,s_{i,k_i}.$ This family includes the family of generalized Petersen graphs, I-graphs, sandwiches of circulant graphs, discrete torus graphs and others. We obtain a closed formula for the number $\tau (n)$ of spanning trees in $H_n$ in terms of Chebyshev polynomials, investigate some arithmetical properties of this function and find its asymptotics as $n\rightarrow \infty .$

1 Introduction

Let G be a finite connected graph. By the complexity $\tau (G)$ of the graph G, we mean the number of its spanning trees. The complexity is very important algebraic invariant of a graph. Various approaches to its computation are given in the papers [2, 6,7,8, 11, 14, 24]. For an infinite family of graphs $G_n,\, n\in \mathbb {N}$, one can introduce complexity function $\tau (n)=\tau (G_n).$ In statistical physics [12, 16, 22, 25], it is important to know the behavior of the function $\tau (n)$ for sufficiently large values of n.

The aim of the present paper is to investigate analytical, arithmetical and asymptotic properties of complexity function for circulant foliation over a given graph. We note that this family is quite rich. It includes circulant graphs, generalized Petersen graphs, I-, Y-, H-graphs, discrete tori and others.

The structure of the paper is as follows. Some preliminary results and basic definitions are given in Sect. 2. In Sect. 3, we define the notion of circulant foliation over a graph. In Sect. 4, we present explicit formulas for the number of spanning trees of graphs $H_n=H_n(G_1,\,G_2,\ldots ,G_m)$ obtained as a circulant foliation over a graph H on m vertices with fibers $G_1,\,G_2,\ldots ,G_m.$ Each fiber $G_i=C_n(s_{i,1},\,s_{i,2},\ldots ,s_{i,k_i})$ of this foliation is the circulant graph on n vertices with jumps $s_{i,1},\,s_{i,2},\ldots ,s_{i,k_i}$. The formulas will be given in terms of Chebyshev polynomials. In Sect. 5, we provide some arithmetical properties of the complexity function for the family $H_n.$ More precisely, we show that the number of spanning trees in the graph $H_n$ can be represented in the form $\tau (n)=p \,n\,\tau (H) \,a(n)^2,$ where a(n) is an integer sequence and p is a prescribed natural number depending on jumps and the parity of n. In Sect. 6, we use explicit formulas for the complexity in order to produce its asymptotic. In the last section, we illustrate the obtained results by a series of examples.

2 Basic definitions and preliminary facts

Consider a connected finite graph G,  allowed to have multiple edges but without loops. We denote the vertex and edge set of G by V(G) and E(G),  respectively. Given $u, v\in V(G),$ we set $a_{uv}$ to be the number of edges between vertices u and v. The matrix $A=A(G)=\{a_{uv}\}_{u, v\in V(G)}$ is called the adjacency matrix of the graph G. The degree $d_v$ of a vertex $v \in V(G)$ is defined by $d_v=\sum _{u\in V(G)}a_{uv}.$ Let $D=D(G)$ be the diagonal matrix indexed by the elements of V(G) with $d_{vv} = d_v.$ The matrix $L=L(G)=D(G)-A(G)$ is called the Laplacian matrix, or simply Laplacian, of the graph G. Let $X=\{x_v,\,v\in V(G)\}$ be the set of variables and let X(G) be the diagonal matrix indexed by the elements of V(G) with diagonal elements $x_v.$ Then, the generalized Laplacian matrix of G,  denoted by L(G, X),  is given by $L(G,X)=X(G)-A(G).$ In the particular case, $x_v=d_v,$ we have $L(G,X)=L(G).$ In what follows, by $I_n$ we denote the identity matrix of order n.

We call an $n\times n$ matrix circulant and denote it by $circ(a_0, a_1,\ldots ,a_{n-1})$ if it is of the form

$$\begin{aligned} circ(a_0, a_1,\ldots , a_{n-1})= \left( \begin{array}{ccccc} a_0 &{}\quad a_1 &{}\quad a_2 &{}\quad \ldots &{}\quad a_{n-1} \\ a_{n-1} &{}\quad a_0 &{}\quad a_1 &{}\quad \ldots &{}\quad a_{n-2} \\ &{}\quad \vdots &{}\quad &{}\quad \ddots &{}\quad \vdots \\ a_1 &{}\quad a_2 &{}\quad a_3 &{}\quad \ldots &{}\quad a_0\\ \end{array}\right) . \end{aligned}$$

Recall [9] that the eigenvalues of matrix $C=circ(a_0,a_1,\ldots ,a_{n-1})$ are given by the following simple formulas $\lambda _j=p(\varepsilon ^j_n),\,j=0,1,\ldots ,n-1$ where $p(x)=a_0+a_1 x+\cdots +a_{n-1}x^{n-1}$ and $\varepsilon _n$ is an order n primitive root of the unity. Moreover, the circulant matrix $C=p(T_n ),$ where $T_{n}=circ(0,1,0,\ldots ,0)$ is the matrix representation of the shift operator $T_{n}:(x_0,x_1,\ldots ,x_{n-2},x_{n-1})\rightarrow (x_1, x_2,\ldots ,x_{n-1},x_0).$ For any $i=0,\ldots , n-1$, let ${\mathbf{v}_i} =(1,\varepsilon _n^i ,\varepsilon _n^{2i},\ldots ,\varepsilon _n^{(n-1)i})^t$ be a column vector of length n. We note that all $n\times n$ circulant matrices share the same set of linearly independent eigenvectors $\mathbf{v} _0, \mathbf{v} _1, \ldots , \mathbf{v} _{n-1}.$ Hence, any set of $n\times n$ circulant matrices can be simultaneously diagonalizable.

Let $s_1,s_2,\ldots ,s_k$ be integers such that $1\le s_1<s_2<\cdots <s_k\le \frac{n}{2}.$ The graph $C_n(s_1,s_2,\ldots ,s_k)$ with n vertices $0,1,2,\ldots ,n-1$ is called circulant graph if the vertex $i,0\le i\le n-1$ is adjacent to the vertices $i\pm s_1, i\pm s_2,\ldots ,i\pm s_k$ (mod n). All vertices of the graph are of even degree 2k. If n is even and $s_k=\frac{n}{2}$, then the vertices i and $i+s_k$ are connected by two edges. In this paper, we also allow the empty circulant graph $C_n(\emptyset )$ consisting of n isolated vertices.

3 Circulant foliation over a graph

Let H be a connected finite graph on vertices $v_1,v_2,\ldots ,v_m$, allowed to have multiple edges but without loops. Denote the number of edges between vertices $v_i$ and $v_j$ by $a_{i\,j}$. Since H has no loops, we have $a_{i\,i}=0.$ To define the circulant foliation $H_n=H_n(G_1,\,G_2,\ldots ,G_m)$, we prescribe to each vertex $v_i$ a circulant graph $G_i=C_n(s_{i,1},\,s_{i,2},\ldots ,s_{i,k_i}).$ Then, the circulant foliation $H_n=H_n(G_1,\,G_2,\ldots ,G_m)$ over H with fibers $G_1,\,G_2,\ldots ,G_m$ is a graph with the vertex set $V(H_n)=\{(k,\,v_i)\ | \,k=1,2,\ldots n,\,i=1,2,\ldots ,m\},$ where for a fixed k the vertices $(k,\,v_i)$ and $(k,\,v_j)$ are connected by $a_{i\,j}$ edges, while for a fixed i,  the vertices $(k,\,v_i),\,k=1,2,\ldots n$ form a graph $C_n(s_{i,1},\,s_{i,2},\ldots ,s_{i,k_i})$ in which the vertex $(k,\,v_i)$ is adjacent to the vertices $(k\pm s_{i,1},v_i),(k\pm s_{i,2},\,v_i),\ldots ,(k\pm s_{i,k_i},\,v_i),(\text {mod}\ n).$

There is a projection $\varphi : H_n\rightarrow H$ sending the vertices $(k,\,v_i),\,k=1,\ldots ,n$ and edges between them to the vertex $v_i$ and for given k each edge between the vertices $(k,\,v_i)$ and $(k,\,v_j),\,i\ne j$ bijectively to an edge between $v_i$ and $v_j.$ For each vertex $v_i$ of graph H, we have $\varphi ^{-1}(v_i)=G_i,\,i=1,2,\ldots ,m.$

Consider an action of the cyclic group $\mathbb {Z}_n$ on the graph $H_n$ defined by the rule $(k,\,v_{i})\rightarrow (k+1,\,v_{i}),\,k\,\mod n.$ Then, the group $\mathbb {Z}_n$ acts free on the set of vertices and the set of edges and the factor graph $H_{n}/\mathbb {Z}_{n}$ is an equipped graph $\widehat{H}$ obtained from the graph H by attaching $k_i$ loops to each ith vertex of H.

By making use of the voltage technique [4], one can construct the graph $H_n$ in the following way. We put an orientation to all edges of $\widehat{H}$ including loops. Then, we prescribe the voltage 0 to all edges of subgraph H of $\widehat{H}$ and the voltage $s_{i,j},\,\mod n$ to the jth loop attached to ith vertex of H. The respective voltage covering is the graph $H_n.$ It is well known that the obtained graph $H_n$ is connected if and only if the voltages $\{s_{i,j},\,\mod n\}$ generate the full group $\mathbb {Z}_n.$ Equivalently, $H_n$ is connected if and only if $\gcd (n,s_{i,j},\,i=1,\ldots ,m,\,j=1,\ldots , k_i)=1.$ Moreover, if r is a unit in the ring $\mathbb {Z}_n$ (that is, there is an element $r^{\prime }$ in $\mathbb {Z}_n$ such that $r r^{\prime }=1$ ), then the graphs $H_n$ and $H_n^{\prime }$ obtained by the voltage assignments $\{s_{i,j},\,\mod n\}$ and $\{r\,s_{i,j},\,\mod n\}$ are isomorphic.

Recall that the adjacency matrix of the circulant graph $C_n(s_1,s_2,\ldots ,s_k)$ on the vertices $1,2,\ldots ,n$ has the form $\sum \nolimits _{p=1}^{k}(T_{n}^{s_p}+T_{n}^{-s_p}).$ Let the adjacency matrix of the graph H be

$$\begin{aligned} A(H)= \left( \begin{array}{ccccc} 0 &{}\quad a_{1,2} &{}\quad a_{1,3} &{}\quad \ldots &{}\quad a_{1,m} \\ a_{2,1} &{}\quad 0 &{}\quad a_{2,3} &{}\quad \ldots &{}\quad a_{2,m} \\ &{}\quad \vdots &{}\quad &{}\quad \ddots &{}\quad \vdots \\ a_{m,1} &{}\quad a_{m,2} &{}\quad a_{m,3} &{}\quad \ldots &{}\quad 0\\ \end{array}\right) . \end{aligned}$$

Then, the adjacency matrix of the circulant foliation $H_n=H_n(G_1,\,G_2,\ldots ,G_m)$ over a graph H with fibers $G_i=C_n(s_{i,1},\,s_{i,2},\ldots ,s_{i,k_i}),\,i=1,2,\ldots ,n$ is given by

$$\begin{aligned} A(H_n)= \left( \begin{array}{ccccc} \sum \limits _{p=1}^{k_1}(T_{n}^{s_{1,p}}+T_{n}^{-s_{1,p}}) &{}\quad a_{1,2}I_n &{}\quad a_{1,3}I_n &{}\quad \ldots &{}\quad a_{1,m}I_n \\ a_{2,1}I_n &{}\quad \sum \limits _{p=1}^{k_2}(T_{n}^{s_{2,p}}+T_{n}^{-s_{2,p}}) &{}\quad a_{2,3}I_n &{}\quad \ldots &{}\quad a_{2,m}I_n \\ &{}\quad \vdots &{}\quad &{}\quad \ddots &{}\quad \vdots \\ a_{m,1}I_n &{}\quad a_{m,2}I_n &{}\quad a_{m,3}I_n &{}\quad \ldots &{}\quad \sum \limits _{p=1}^{k_m}(T_{n}^{s_{m,p}}+T_{n}^{-s_{m,p}})\\ \end{array}\right) . \end{aligned}$$

As the first example, we consider the sandwich graph $SW_n=H_n(G_1,\,G_2,\ldots ,G_m)$ formed by the circulant graphs $G_i=C_n(s_{i,1},\,s_{i,2},\ldots ,s_{i,k_i}).$ To create $SW_n$, we take H to be the path graph on m vertices $v_1,v_2, \ldots ,v_m$ with the end points $v_1$ and $v_m.$ A very particular case of this construction, known as I-graph I(n, k, l),  occurs by taking $m=2,\,G_1=C_n(k)$ and $G_2=C_n(l).$ Also, the generalized Petersen graph [23] arises as $GP(n,k)=I(n,k,1).$ The sandwich of two circulant graphs $H_n(G_1,G_2)$ was investigated in [1].

As the second example, we consider the generalized Y-graph $Y_n=Y_n(G_1,G_2,G_3),$ where $G_1,G_2,G_3$ are given circulant graphs on n vertices. To construct $Y_n,$ we consider a Y-shape graph H consisting of four vertices $v_1,v_2,v_3,v_4$ and three edges $v_1v_4,v_2v_4,v_3v_4.$ Let $G_4=C_n(\emptyset )$ be the empty graph of n on vertices. Then, by definition, we put $Y_n=H_n(G_1,G_2,G_3,G_4).$ In a particular case, $G_1=C_n(k),G_2=C_n(l),$ and $G_3=C_n(m),$ the graph $Y_n$ coincides with the Y-graph Y(n; k, l, m) defined earlier in [5, 13].

The third example is the generalized H-graph $H_n(G_1,G_2,G_3,G_4,G_5,G_6),$ where $G_1,G_2,G_3,G_4$ are given circulant graphs and $G_5= G_6=C_n(\emptyset )$ are the empty graphs on n vertices. In this case, we take H to be the graph with vertices $v_1,v_2,v_3,v_4,v_5,v_6$ and edges $v_1v_5,v_5v_3,v_2v_6,v_6v_4,v_5v_6.$ In the case $G_1=C_n(i),G_2=C_n(j),G_3=C_n(k),G_4=C_n(l),$ we get the graph H(n; i, j, k, l) investigated in the paper [13]. Shortly, we will write $H_n(G_1,G_2,G_3,G_4)$ ignoring the last two empty graph entries.

4 Counting the number of spanning trees in the graph $H_n$

Let H be a finite connected graph with the vertex set $V(H)=\{v_1,\,v_2,\ldots ,v_m\}.$ Consider the circulant foliation $H_n=H_n(G_1,\,G_2,\ldots ,G_m),$ where $G_i=C_n(s_{i,1},\,s_{i,2},\ldots ,s_{i,k_i}),\,i=1,2,\ldots ,m.$ Let $L(H,\,X)$ be the generalized Laplacian of graph H with the set of variables $X=(x_1,x_2,\ldots ,x_m).$ We specify X by setting $x_{i}=2k_{i}+d_i-\sum \nolimits _{p=1}^{k_i}(z^{s_{i,p}}+z^{-s_{i,p}})$ and put $P(z)=\det (L(H,\,X))$, where $d_i$ is the degree of $v_i$ in H. We note that P(z) is an integer Laurent polynomial. Consider one more specification $L(H,\,W)$ for generalized Laplacian of H with the set $W=(w_1,w_2,\ldots ,w_m),$ where $w_{i}=2k_{i}+d_i-\sum \nolimits _{p=1}^{k_i}2\,T_{s_{i,p}}(w)$ and $T_k(w)=\cos (k\arccos w)$ is the Chebyshev polynomial of the first kind. See [17] for the basic properties of the Chebyshev polynomials. We set $Q(w)=\det (L(H,\,W)).$ Then, Q(w) is an integer polynomial of degree $s=s_{1,k_1}+s_{2,k_2}\cdots +s_{m,k_m}.$ For our convenience, we will call Q(w) a Chebyshev transform of P(z). The following lemma holds.

Lemma 4.1

We have $P(z)=Q(w)$ with $w=\frac{1}{2}(z+\frac{1}{z})$ and Q(w) is the order s polynomial with the leading coefficient $(-1)^{m}2^s,$ where $s=\sum \nolimits _{i=1}^{m}s_{i,k_i}.$ Moreover,

$$\begin{aligned} Q(1)=0,\,Q^{\prime }(1)=-2q\tau (H)\ne 0^{}, \end{aligned}$$

where $q=\sum \nolimits _{i=1}^{m}\sum \nolimits _{j=1}^{k_{i}}s_{i,j}^2$ and $\tau (H)$ is the number of spanning trees in the graph H. In particular, Q(w) has a simple root $w=1$ and P(z) has a double root $z=1.$

Proof

The equality $P(z)=Q(w)$ follows from the identity $T_{n}(\frac{1}{2}(z+\frac{1}{z}))=\frac{1}{2}(z^{n}+\frac{1}{z^{n}}).$ Recall that the leading term of $T_{n}(w)$ is $2^{n-1}w^{n}.$ The leading term of Q(w) is coming from the product $\prod \nolimits _{i=1}^{m}(-2T_{s_{i,k_{i}}}(w))$ and is equal to $(-1)^{m}2^{s}w^{s},$ where $s=\sum \nolimits _{i=1}^{m}s_{i,k_{i}}.$

Let $a_{i,j}$ be the number of edges between ith and jth vertices of the graph H. Then,

$$\begin{aligned} Q(w)=\det \left( \begin{array}{ccccc}x_1 &{}\quad -a_{1,2} &{}\quad -a_{1,3} &{}\quad \ldots &{}\quad -a_{1,m} \\ -a_{2,1} &{}\quad x_2 &{}\quad -a_{2,3} &{}\quad \ldots &{}\quad -a_{2,m} \\ &{}\quad \vdots &{}\quad &{}\quad \ddots &{}\quad \vdots \\ -a_{m,1} &{}\quad -a_{m,2} &{}\quad -a_{m,3} &{}\quad \ldots &{}\quad x_m\end{array}\right) , \end{aligned}$$

(1)

where $x_i=x_i(w)=2k_i+d_i-\sum \nolimits _{j=1}^{k_i}2T_{s_{i,j}}(w),\,i=1,2,\ldots ,m.$ In particular, for $w=1$ we have $x_i=d_i.$ Hence, $Q(1)=0$ because of valency of ith vertex is $d_i=\sum _j a_{i,j}.$ Let $x_i^{\prime }=x_i^{\prime }(w)$ be the derivative of $x_i$ with respect to w. Then,

$$\begin{aligned} Q^{\prime }(w)= & {} \det \left( \begin{array}{ccccc} x^{\prime }_1 &{}\quad -a_{1,2} &{}\quad -a_{1,3} &{}\quad \ldots &{}\quad -a_{1,m} \\ 0 &{}\quad x_2 &{}\quad -a_{2,3} &{}\quad \ldots &{}\quad -a_{2,m} \\ &{}\quad \vdots &{}\quad &{}\quad \ddots &{}\quad \vdots \\ 0 &{}\quad -a_{m,2} &{}\quad -a_{m,3} &{}\quad \ldots &{}\quad x_m\end{array}\right) \\&+ \det \left( \begin{array}{ccccc} x_1 &{}\quad 0 &{}\quad -a_{1,3} &{}\quad \ldots &{}\quad -a_{1,m} \\ -a_{2,1} &{}\quad x^{\prime }_2 &{}\quad -a_{2,3} &{}\quad \ldots &{}\quad -a_{2,m} \\ &{}\quad \vdots &{}\quad &{}\quad \ddots &{}\quad \vdots \\ -a_{m,1} &{}\quad 0 &{}\quad -a_{m,3} &{}\quad \ldots &{}\quad x_m\end{array}\right) \\&+\cdots + \det \left( \begin{array}{ccccc} x_1 &{}\quad -a_{1,2}&{}\quad -a_{1,3} &{}\quad \ldots &{}\quad 0 \\ -a_{2,1} &{}\quad x_2 &{}\quad -a_{2,3} &{}\quad \ldots &{}\quad 0 \\ &{}\quad \vdots &{}\quad &{}\quad \ddots &{}\quad \vdots \\ -a_{m,1} &{}\quad -a_{m,2} &{}\quad -a_{m,3} &{}\quad \ldots &{}\quad x^{\prime }_m\end{array}\right) \\= & {} x^{\prime }_1(w)Q_{1,1}(w)+x^{\prime }_2(w)Q_{2,2}(w)+\cdots +x^{\prime }_m(w) Q_{m,m}(w), \end{aligned}$$

where $Q_{i,i}(w)$ is the (i, i)th minor of the matrix in formula (1). For $w=1$, this matrix coincides with the Laplacian of H. By the Kirchhoff theorem, we have

$$\begin{aligned} Q_{1,1}(1)=Q_{2,2}(1)=\cdots =Q_{m,m}(1)=\tau (H), \end{aligned}$$

where $\tau (H)$ is the number of spanning trees in the graph H.

Since $T_s^{\prime }(w)=s\, U_{s}(w),$ where $U_s(w)$ is the Chebyshev polynomial of the second kind and $U_s(1)=s,$ we have $x^{\prime }_i(w)=-\sum \nolimits _{j=1}^{k_i}2s_{i,j}U_{s_{i,j}}(w)$ and $x^{\prime }_i(1)=-2\sum \nolimits _{j=1}^{k_i}s_{i,j}^2.$

As a result, $Q^{\prime }(1)=(x_1^{\prime }(1)+\cdots +x_m^{\prime }(1))\tau (H)= -2\sum \nolimits _{i=1}^{m}\sum \nolimits _{j=1}^{k_i}s_{i,j}^2\tau (H)=-2q\,\tau (H).$ $\square$

The main result of this section is the following theorem.

Theorem 4.2

The number of spanning trees $\tau (n)$ in the graph $H_{n}(G_1,\,G_2,\ldots ,G_m)$ is given by the formula

$$\begin{aligned} \tau (n)=\frac{n\,\tau (H)}{q}\prod _{p=1}^{s-1}|2T_n(w_p)-2|, \end{aligned}$$

where $s=s_{1,k_1}+s_{2,k_2}\cdots +s_{m,k_m},\,w_p\,(p=1,2,\ldots ,s-1)$ are all the roots different from 1 of the equation $Q(w)=0$, $\tau (H)$ is the number of spanning trees in the graph H and $q=\sum \nolimits _{i=1}^{m}\sum \nolimits _{j=1}^{k_i}s_{i,j}^2.$

Proof

By the classical Kirchhoff theorem, the number of spanning trees $\tau (n)$ is equal to the product of nonzero eigenvalues of the Laplacian of a graph $H_{n}(G_1,\,G_2,\ldots ,G_m)$ divided by the number of its vertices $m\times n.$ To investigate the spectrum of Laplacian matrix, we consider the shift operator $T_{n}=circ(0,1,\ldots ,0).$ The Laplacian $L=L(H_{n}(G_1,\,G_2,\ldots ,G_m))$ is given by the matrix

$$\begin{aligned} L= \left( \begin{array}{ccccc} A_{1}(T_{n}) &{}\quad -a_{1,2}I_n &{}\quad -a_{1,3}I_n &{}\quad \ldots &{}\quad -a_{1,m}I_n \\ -a_{2,1}I_n &{}\quad A_{2}(T_{n}) &{}\quad -a_{2,3}I_n &{}\quad \ldots &{}\quad -a_{2,m}I_n \\ &{}\quad \vdots &{}\quad &{}\quad \ddots &{}\quad \vdots \\ -a_{m,1}I_n &{}\quad -a_{m,2}I_n &{}\quad -a_{m,3}I_n &{}\quad \ldots &{}\quad A_{m}(T_{n})\\ \end{array}\right) , \end{aligned}$$

where $A_{i}(z)=2k_{i}+d_{i}-\sum \nolimits _{j=1}^{k_{i}}(z^{s_{i,j}}+z^{-s_{i,j}}),\,i=1,\ldots ,m.$

The eigenvalues of circulant matrix $T_{n}$ are $\varepsilon _{n}^{j},\,j=0,1,\ldots ,n-1,$ where $\varepsilon _n=e^\frac{2\pi i}{n}.$ Since all of them are distinct, the matrix $T_{n}$ is conjugate to the diagonal matrix $\mathbb {T}_{n}=diag(1,\varepsilon _{n},\ldots ,\varepsilon _{n}^{n-1})$ with diagonal entries $1,\varepsilon _{n},\ldots ,\varepsilon _{n}^{n-1}$. To find spectrum of L,  without loss of generality, one can assume that $T_{n}=\mathbb {T}_{n}.$ Then, all $n\times n$ blocks of L are diagonal matrices. This essentially simplifies the problem of finding eigenvalues of the block matrix L. Indeed, let $\lambda$ be an eigenvalue of L and let $(x_{1},x_{2},\ldots ,x_{m})$ with $x_{i}=(x_{i,1},x_{i,2}\ldots ,x_{i,n})^t,\,i=1,\ldots ,m$ be the respective eigenvector. Then, we have the following system of equations

$$\begin{aligned} \left( \begin{array}{ccccc} A_{1}(\mathbb {T}_{n}) -\lambda \,I_{n}&{}\quad -a_{1,2}I_{n} &{}\quad -a_{1,3}I_{n} &{}\quad \ldots &{}\quad -a_{1,m}I_{n} \\ -a_{2,1}I_{n} &{}\quad A_{2}(\mathbb {T}_{n})-\lambda \,I_{n} &{}\quad -a_{2,3}I_{n} &{}\quad \ldots &{}\quad -a_{2,m}I_{n} \\ &{}\quad \vdots &{}\quad &{}\quad \ddots &{}\quad \vdots \\ -a_{m,1}I_{n} &{}\quad -a_{m,2}I_{n} &{}\quad -a_{m,3}I_{n} &{}\quad \ldots &{}\quad A_{m}(\mathbb {T}_{n})-\lambda \,I_{n}\\ \end{array}\right) \left( \begin{array}{c} x_{1}\\ x_{2}\\ \vdots \\ x_{m}\\ \end{array}\right) =0.\end{aligned}$$

(2)

Recall that all blocks in the matrix under consideration are diagonal $n\times n$-matrices and the (j, j)th entry of $\mathbb {T}_{n}$ is equal to $\varepsilon _n^{j-1}.$

Hence, Eq. (2) splits into n equations

$$\begin{aligned} \left( \begin{array}{ccccc} A_{1}(\varepsilon _n^{j}) -\lambda &{}\quad -a_{1,2} &{}\quad -a_{1,3} &{}\quad \ldots &{}\quad -a_{1,m} \\ -a_{2,1} &{}\quad A_{2}(\varepsilon _n^{j})-\lambda &{}\quad -a_{2,3} &{}\quad \ldots &{}\quad -a_{2,m} \\ &{}\quad \vdots &{}\quad &{}\quad \ddots &{}\quad \vdots \\ -a_{m,1} &{}\quad -a_{m,2} &{}\quad -a_{m,3} &{}\quad \ldots &{}\quad A_{m}(\varepsilon _n^{j})-\lambda \\ \end{array}\right) \left( \begin{array}{c} x_{1,j+1}\\ x_{2,j+1}\\ \vdots \\ x_{m,j+1}\\ \end{array}\right) =0,\end{aligned}$$

(3)

$j=0,1,\ldots ,n-1$. Each equation gives m eigenvalues of L,  say $\lambda _{1,j},\lambda _{2,j},\ldots ,\lambda _{m,j}.$ To find these eigenvalues, we set

$$\begin{aligned} P(z,\lambda )=\det \left( \begin{array}{ccccc} A_{1}(z) -\lambda &{}\quad -a_{1,2} &{}\quad -a_{1,3} &{}\quad \ldots &{}\quad -a_{1,m} \\ -a_{2,1} &{}\quad A_{2}(z)-\lambda &{}\quad -a_{2,3} &{}\quad \ldots &{}\quad -a_{2,m} \\ &{}\quad \vdots &{}\quad &{}\quad \ddots &{}\quad \vdots \\ -a_{m,1} &{}\quad -a_{m,2} &{}\quad -a_{m,3} &{}\quad \ldots &{}\quad A_{m}(z)-\lambda \\ \end{array}\right) . \end{aligned}$$

(4)

Then, $\lambda _{1,j},\lambda _{1,j},\ldots ,\lambda _{m,j}$ are roots of the equation

$$\begin{aligned} P(\varepsilon _n^{j},\lambda )=0.\end{aligned}$$

(5)

In particular, by Vieta’s theorem, the product $p_{j}=\lambda _{1,j}\lambda _{2,j}\ldots \lambda _{m,j}$ is given by the formula $p_{j}=P(\varepsilon _n^j,0)=P(\varepsilon _n^j),$ where P(z) is the same as in Lemma 4.1.

Now, for any $j=0,\ldots , n-1,$ matrix L has m eigenvalues $\lambda _{1,j},\lambda _{2,j},\ldots ,\lambda _{m,j}$ satisfying the order m algebraic equation $P(\varepsilon _n^{j},\lambda )=0.$ In particular, for $j=0$ and $\lambda =\lambda _{i,0},\,i=1,2,\ldots ,m$ we have $P(1,\lambda )=0.$ In this case, $A_{i}(1)=d_{i},\,i=1,2,\ldots ,m.$ One can see that the polynomial $P(1,\lambda )$ is the characteristic polynomial for Laplace matrix of the graph H and its roots are eigenvalues of H.

Note that $\lambda _{1,0}=0$ and the product of nonzero eigenvalues $\lambda _{2,0}\lambda _{3,0}\ldots \lambda _{m,0}$ is equal to $m\,\tau (H),$ where $\tau (H)$ is the number of spanning trees in the graph H.

Now we have

$$\begin{aligned} \tau (n)= & {} \frac{1}{m\,n}\lambda _{2,0}\lambda _{3,0}\ldots \lambda _{m,0} \prod \limits _{j=1}^{n-1}\lambda _{1,j}\lambda _{2,j}\ldots \lambda _{m,j}\nonumber \\= & {} \frac{m\,\tau (H)}{m\,n}\prod \limits _{j=1}^{n-1}p_j=\frac{\tau (H)}{n}\prod \limits _{j=1}^{n-1}P(\varepsilon _n^j). \end{aligned}$$

(6)

To continue the proof, we replace the Laurent polynomial P(z) by $\widetilde{P}(z)=(-1)^{m}z^{s}P(z).$ Then, $\widetilde{P}(z)$ is a monic polynomial of the degree 2s with the same roots as P(z). We note that

$$\begin{aligned} \prod \limits _{j=1}^{n-1}\widetilde{P}(\varepsilon _{n}^{j})=(-1)^{m(n-1)}\varepsilon _{n}^{\frac{(n-1)n}{2}s} \prod \limits _{j=1}^{n-1}P(\varepsilon _{n}^{j})=(-1)^{(m+s)(n-1)}\prod \limits _{j=1}^{n-1}P(\varepsilon _{n}^{j}). \end{aligned}$$

(7)

By Lemma 4.1, all roots of polynomials $\widetilde{P}(z)$ and Q(w) are $1,1,z_{1},1/z_{1},\ldots ,z_{s-1},1/z_{s-1},\,z_{j}\ne 1\text { and }1\ne w_{j}=\frac{1}{2}(z_{j}+z_{j}^{-1}),\,j=1,\ldots ,s-1,$ respectively. Also, we can recognize the complex numbers $\varepsilon _{n}^{j},\,j=1,\ldots ,n-1$ as the roots of polynomial $\frac{z^n-1}{z-1}.$ By the basic properties of resultant ([21], Ch. 1.3), we have

$$\begin{aligned} \prod \limits _{j=1}^{n-1}\widetilde{P}(\varepsilon _{n}^{j})= & {} \text {Res}\left( \widetilde{P}(z),\frac{z^{n}-1}{z-1}\right) =\text {Res}\left( \frac{z^{n}-1}{z-1},\widetilde{P}(z)\right) \nonumber \\= & {} \prod \limits _{z:\widetilde{P}(z)=0}\frac{z^{n}-1}{z-1}= n^{2}\prod \limits _{j=1}^{s-1}\frac{z_{j}^{n}-1}{z_{j}-1}\frac{z_{j}^{-n}-1}{z_{j}^{-1}-1}= n^{2}\prod \limits _{j=1}^{s-1}\frac{z_{j}^{n}+z_{j}^{-n}-2}{z_{j}+z_{j}^{-1}-2}\nonumber \\= & {} n^{2}\prod \limits _{j=1}^{s-1}\frac{2T_{n}(w_{j})-2}{2w_{j}-2} =n^{2}\prod \limits _{j=1}^{s-1}\frac{T_{n}(w_{j})-1}{w_{j}-1}. \end{aligned}$$

(8)

Combine (6), (7) and (8), we have the following formula for the number of spanning trees

$$\begin{aligned} \tau (n)=(-1)^{(m+s)(n-1)}n\,\tau (H)\prod \limits _{j=1}^{s-1}\frac{T_{n}(w_{j})-1}{w_{j}-1}. \end{aligned}$$

(9)

We have the following important statement from formula (9).

Claim

The number of spanning trees $\tau (n)$ is a multiple of $n\,\tau (H).$

Proof of Claim:

To prove the lemma, we have to show that the number $R=\prod \nolimits _{j=1}^{s-1}\frac{T_{n}(w_{j})-1}{w_{j}-1}$ is an integer. Indeed, setting $w=\frac{\zeta +2}{2}$ one can represent R in the form

$$\begin{aligned} R=\prod \limits _{j=1}^{s-1}\frac{2T_{n}(\frac{\zeta _{j}+2}{2})-2}{\zeta _{j}}, \end{aligned}$$

where $\zeta _{j},\,j=1,2,\ldots ,s-1$ are nonzero root of the equation $Q(\frac{\zeta +2}{2})=0.$ We note that the function $j_n(\zeta )=2T_{n}(\frac{\zeta +2}{2})$ satisfies the recursive relation $j_{n+1}(\zeta )=(\zeta +2)j_{n}(\zeta )-j_{n-1}(\zeta )$ with initial data $j_0(\zeta )=2$ and $j_1(\zeta )=\zeta +2.$ Hence, $j_{n}(\zeta )$ is a monic polynomial of degree n with integer coefficients. Since $2T_n(1)=2,$ the same is true for the polynomial $f(\zeta )=\frac{2T_{n}(\frac{\zeta +2}{2})-2}{\zeta }.$ By definition, Q(w) is an integer polynomial in the variables $2T_{s_{i,j}}(w),\,i=1,2,\ldots ,m,\,j=1,2,\ldots ,k_i.$ By Lemma 4.1, we have $Q(1)=0,$ and $Q^{\prime }(1)\ne 0.$ Also, the leading coefficient of Q(w) is equal to $(-1)^{m}2^{s},$ where s is the degree of Q(w). Hence, $g(\zeta )=\frac{1}{\zeta }Q(\frac{\zeta +2}{2})$ with $g(0)\ne 0$ is also a monic polynomial with integer coefficients. Taking this into account, we get $R=\mathrm{Res}\,(f(\zeta ),g(\zeta )).$ Since both $f(\zeta )$ and $g(\zeta )$ are polynomials with integer coefficients, R is integer. $\square$

Since $\tau (n)$ is a positive number, by (9) we obtain

$$\begin{aligned} \tau (n)=n\,\tau (H)\prod _{j=1}^{s-1}\left| \frac{T_{n}(w_{j})-1}{w_{j}-1}\right| = n\,\tau (H)\prod _{j=1}^{s-1}|T_{n}(w_{j})-1|\big /\prod _{j=1}^{s-1}|w_{j}-1|. \end{aligned}$$

(10)

Now we evaluate the product $\prod _{j=1}^{s-1}|w_{j}-1|.$ We note that from Lemma 4.1, the polynomial Q(w) has the leading coefficient $a_{0}=(-1)^{m}2^{s},\,Q(1)=0$ and $Q^\prime (1)=-2q,$ where $q=\sum \nolimits _{i=1}^{m}\sum \nolimits _{j=1}^{k_{i}}s_{i,j}^2.$

As a result, we have

$$\begin{aligned} \prod _{j=1}^{s-1}|w_{j}-1|=|\frac{1}{a_0}Q^\prime (1)| =\frac{2q}{2^{s}}=\frac{q}{2^{s-1}}. \end{aligned}$$

(11)

Substituting Eq. (11) into Eq. (10), we finish the proof of the theorem. $\square$

5 Arithmetical properties of complexity for the graph $H_n$

Let H be a finite connected graph on m vertices. Consider the circulant foliation $H_n=H_n(G_1,\,G_2,\ldots ,G_m),$ where $G_i=C_n(s_{i,1},\,s_{i,2},\ldots ,s_{i,k_i}),\,i=1,2,\ldots ,m.$ Recall that any positive integer s can be uniquely represented in the form $s=p \,r^2,$ where p and r are positive integers and p is square-free. We will call p the square-free part of s.

Theorem 5.1

Let $\tau (n)$ be the number of spanning trees in the graph $H_n.$ Denote the square-free parts of $Q(-1)$ by p. Then, there exists an integer sequence a(n) such that

$1^{0}$:

$\tau (n)= n\,\tau (H)\,a(n)^{2},$ if n is odd,

$2^{0}$:

$\tau (n)=p\,n\,\tau (H)\,a(n)^{2},$ if n is even.

Proof

By formula (6), we have $n\,\tau (n)=\tau (H)\prod _{j=1}^{n-1}\lambda _{1,j}\lambda _{2,j}\ldots \lambda _{m,j}.$ Note that $\lambda _{1,j}\lambda _{2,j}\ldots \lambda _{m,j}=P(\varepsilon _{n}^{j})= P(\varepsilon _{n}^{n-j})=\lambda _{1,n-j}\lambda _{2,n-j}\ldots \lambda _{m,n-j}.$ Define $c(n)=\prod \nolimits _{j=1}^{\frac{n-1}{2}}\lambda _{1,j}\lambda _{2,j}\ldots \lambda _{m,j},$ if n is odd and $d(n)=\prod \nolimits _{j=1}^{\frac{n}{2}-1}\lambda _{1,j}\lambda _{2,j}\ldots \lambda _{m,j},$ if n is even. By [15], each algebraic number $\lambda _{i,j}$ comes into the products $\prod _{j=1}^{(n-1)/2}\lambda _{1,j}\lambda _{2,j}\ldots \lambda _{m,j}$ and $\prod _{j=1}^{n/2-1}\lambda _{1,j}\lambda _{2,j}\ldots \lambda _{m,j}$ with all of its Galois conjugate elements. Therefore, both products c(n) and d(n) are integers. Moreover, if n is even we get $\lambda _{1,\frac{n}{2}}\lambda _{2,\frac{n}{2}}\ldots \lambda _{m,\frac{n}{2}}=P(-1)=Q(-1).$ We note that $Q(-1)$ is always a positive integer. The precise formula for it is given in Remark 1.

Now, we have $n\tau (n)=\tau (H)\,c(n)^2$ if n is odd, and $n\tau (n)=\tau (H)\,Q(-1)\,d(n)^2$ if n is even. Let $Q(-1)=p\,r ^{2},$ where p is a square-free number. Then,

$1^{\circ }$:

$\displaystyle {\frac{\tau (n)}{n\,\tau (H)}=\left( \frac{c(n)}{n }\right) ^2}$ if n is odd,

$2^{\circ }$:

$\displaystyle {\frac{\tau (n)}{n\,\tau (H)}=p\left( \frac{r\,d(n)}{n}\right) ^2}$ if n is even.

By Claim in the proof of Theorem 4.2, the quotient $\frac{\tau (n)}{n\,\tau (H)}$ is an integer. Since p is square- free, the squared rational numbers in $1^{\circ }$ and $2^{\circ }$ are integer. Setting $a(n)=\frac{c(n)}{n}$ in the first case, and $\ a(n)=\frac{r\,d(n)}{n}$ in the second, we finish the proof of the theorem. $\square$

Remark 1

Denote the number of odd elements in the sequence $s_{i,1},\,s_{i,2},\ldots ,s_{i,k_i}$ by $t_i$. Now $Q(-1)=\det \,L(H,W),$ where $W=(d_{1}+4t_{1},d_{2}+4t_{2},\ldots ,d_{m}+4t_{m}).$ Indeed, $Q(w)=\det \,L(H,W),$ where $W=(w_{1},w_{2},\ldots ,w_{m})$ and $w_{i}=2k_{i}+d_{i}-\sum \nolimits _{j=1}^{k_{i}}2T_{s_{i,j}}(w).$ If $w=-1$ we have $T_{s_{i,j}}(-1)=\cos (s_{i,j}\arccos (-1))=\cos (s_{i,j}\pi )=(-1)^{s_{i,j}}$ and $w_i=d_{i}+4\sum \nolimits _{j=1}^{k_{i}}\frac{1-(-1)^{s_{i,j}}}{2}=d_{i}+4t_{i}.$

6 Asymptotic formulas for the number of spanning trees

In this section, we obtain the following result.

Theorem 6.1

The asymptotic behavior for the number of spanning trees $\tau (n)$ in the graph $H_n$ with $\gcd (s_{i,p},\,i=1,\ldots ,m,\,p=1,\ldots ,k_i)=1$ is given by the formula

$$\begin{aligned} \tau (n)\sim \frac{n}{q}A^n,\,n\rightarrow \infty , \end{aligned}$$

where $q=\sum \nolimits _{i=1}^{m}\sum \nolimits _{j=1}^{k_i}s_{i,j}^2$ and $A=\exp ({\int \limits _{0}^{1}\log |Q(\cos {2 \pi t})|\text {d}t}).$

To prove the theorem, we need the following preliminary lemmas.

Lemma 6.2

Let $a_{i,j},(a_{i,i}=0),\,i,j=1,2,\ldots ,m$ be nonnegative numbers. Let

$$\begin{aligned} D(x_1,x_2,\ldots ,x_m)=\det \left( \begin{array}{ccccc}x_1 &{}\quad -a_{1,2} &{}\quad -a_{1,3}&{}\quad \ldots &{}\quad -a_{1,m} \\ -a_{2,1} &{}\quad x_2 &{}\quad -a_{2,3} &{}\quad \ldots &{}\quad -a_{2,m} \\ &{}\quad \vdots &{}\quad &{}\quad \ddots &{}\quad \vdots \\ -a_{m,1} &{}\quad -a_{m,2} &{}\quad -a_{m,3} &{}\quad \ldots &{}\quad x_m\end{array}\right) . \end{aligned}$$

Then, for $x_i\ge d_i=\sum \nolimits _{j=1}^ma_{i,j},\,i =1,2,\ldots ,m$ we have $D(x_1,x_2,\ldots ,x_m)\ge 0.$ The equality $D(x_1,x_2,\ldots ,x_m)=0$ holds if and only if $x_i=d_i,\,i =1,2,\ldots ,m.$

Proof

We use induction on m to prove the lemma. For $m=1$, we have $D(x_1)=x_1\ge a_{1,1}=0$ and $D(x_1)=0$ iff $x_1=a_{1,1}.$ For $m=2$, one has $D(x_1,x_2)=x_1x_2-a_{1,2}a_{2,1}\ge 0$ with $D(x_1,x_2)=0$ if and only if $x_1=a_{1,2}$ and $x_2=a_{2,1}.$ Suppose that $m>2$ and lemma is true for all $D(x_1,x_2,\ldots ,x_k)$ with $k<m.$

The (i, i)th minor of the matrix in the statement of lemma is denoted by $D(x_1,\ldots ,\hat{x}_i,\ldots ,x_m),$ where $\hat{x}_i$ means that the variable $x_i$ is dropped. We note that $D^{\prime }_{x_1}(x_1,x_2,\ldots ,x_k)=D(x_2,\ldots ,x_k).$ Since

$$\begin{aligned} x_2\ge d_2= & {} \sum \limits _{j=1}^ma_{2,j}\ge \sum \limits _{j=2}^ma_{2,j},x_3\ge d_3\\= & {} \sum \limits _{j=1}^ma_{3,j}\ge \sum \limits _{j=2}^ma_{3,j}, \ldots ,x_m\ge d_m\\= & {} \sum \limits _{j=1}^ma_{m,j}\ge \sum \limits _{j=2}^ma_{m,j}, \end{aligned}$$

the function $D(x_2,\ldots ,x_m)$ satisfies the conditions of lemma. Hence, $D(x_2,\ldots ,x_m)\ge 0.$ In a similar way, for $i=2,\ldots ,m$ we have

$$\begin{aligned} D^{\prime }_{x_i}(x_1,x_2,\ldots ,x_m)=D(x_1,\ldots ,\hat{x}_i\ldots ,x_m)\ge 0. \end{aligned}$$

Since $D(d_1,d_2,\ldots ,d_m)=0,$ we obtain $D(x_1,x_2,\ldots ,x_m)\ge 0$ for all $x_i\ge d_i,\,i =1,2,\ldots ,m.$ If for some $i_0$ we have $x_{i_0}>d_{i_0},$ then, by induction, for all $i\ne i_{0}$ we get $D^{\prime }_{x_{i_0}}(x_1,x_2,\ldots ,x_m)=D(x_2,\ldots ,\hat{x}_{i_0}\ldots ,x_m)>0$ and $D(x_1,x_2,\ldots ,x_m)>0.$ $\square$

Lemma 6.3

Let $\gcd (s_{i,j},\,i=1,\ldots ,m,\,j=1,\ldots ,k_i)=1$ and $s=s_{1,k_1}+s_{2,k_2}\ldots +s_{m,k_m}$ Then, the roots of the Laurent polynomial P(z) counted with multiplicities are $1,\,1,\,z_{1},\,1/z_{1},\ldots ,\,z_{s-1},\,1/z_{s-1},$ where we have $|z_{p}|\ne 1,\,p=1,2,\ldots ,s-1.$ Polynomial Q(w) has the roots $1,\,w_{1},\ldots , w_{s-1},$ where $w_{p}=\frac{1}{2}(z_{p}+z_{p}^{-1})$ for all $p=1,\,2,\ldots ,s-1.$

Proof

By Lemma 4.1, we have $P(z)=Q(\frac{1}{2}(z+z^{-1}))$ and Q(w) has the simple root $w=1.$

Since the mapping $w=\frac{1}{2}(z+z^{-1})$ is two-to-one, the Laurent polynomial P(z) has the double root $z=1.$

To prove the lemma, we suppose that the Laurent polynomial P(z) has a root $z_{0}$ such that $|z_{0}|=1$ and $z_{0}\ne 1.$ Then, $z_{0}=e^{\text {i}\,\varphi _{0}},\,\varphi _{0}\in \mathbb {R}{\setminus }2\pi \mathbb {Z}.$ Now we have

$$\begin{aligned} P(e^{\text {i}\,\varphi _{0}})= \det \left( \begin{array}{ccccc}x_1 &{}\quad -a_{1,2} &{}\quad -a_{1,3}&{}\quad \ldots &{}\quad -a_{1,m} \\ -a_{2,1} &{}\quad x_2 &{}\quad -a_{2,3} &{}\quad \ldots &{}\quad -a_{2,m} \\ &{}\quad \vdots &{}\quad &{}\quad \ddots &{}\quad \vdots \\ -a_{m,1} &{}\quad -a_{m,2} &{}\quad -a_{m,3} &{}\quad \ldots &{}\quad x_m\end{array}\right) , \end{aligned}$$

where

$$\begin{aligned} x_{i}= & {} 2k_{i}+d_{i}-\sum \limits _{j=1}^{k_{i}}(z_{0}^{s_{i,j}}+z_{0}^{-s_{i,j}}) =2k_{i}+d_{i}-\sum \limits _{j=1}^{k_{i}}(e^{\text {i}\,s_{i,j}\,\varphi _{0}}+e^{-\text {i}\,s_{i,j}\,\varphi _{0}})\\= & {} d_{i}+\sum \limits _{j=1}^{k_{i}}(2-2\cos (s_{i,j}\,\varphi _{0})). \end{aligned}$$

Since $d_{i}=\sum \nolimits _{j=1}^{m}a_{i,j}$ and $x_{i}\ge d_{i},$ the conditions of Lemma 6.2 are satisfied. Hence, $P(e^{\text {i}\,\varphi _{0}})=0$ if and only if $x_{i}=d_{i},\,i=1,\ldots ,m.$ Then, $\cos (s_{i,j}\,\varphi _{0})=1$ for all $i=1,\ldots ,m,\,j=1,\ldots ,k_{i}.$ So $s_{i,j}\,\varphi _{0}=2\pi m_{i,j}$ for some integer $m_{i,j}.$ As $\gcd (s_{i,j},\,i=1,\ldots ,m,\,j=1,\ldots ,k_i)=1$ there exist integers $p_{i,j}$ such that $\sum \nolimits _{i=1}^{m}\sum \nolimits _{j=1}^{k_{i}}s_{i,j}p_{i,j}=1.$ See, for example, ([3], p. 21). Hence, $\varphi _0=\varphi _0\sum \nolimits _{i=1}^{m}\sum \nolimits _{j=1}^{k_{i}}s_{i,j}p_{i,j}= 2\pi \sum \nolimits _{i=1}^{m}\sum \nolimits _{j=1}^{k_{i}}m_{i,j}p_{i,j}\in 2\pi \mathbb {Z}.$ Contradiction. $\square$

Now we come to the proof of Theorem 6.1

Proof

By theorem 4.2, we have $\tau (n)=\frac{n\tau (H)}{q}\prod \nolimits _{j=1}^{s-1}|{2T_{n}(w_{j})-2}|,$ where $q=\sum \nolimits _{i=1}^{m}\sum \nolimits _{j=1}^{k_{i}}s_{i,j}^2$ and $w_{j},\,j=1,2,\ldots ,s-1$ are roots of the polynomial Q(w) different from 1.

By lemma 6.3, $T_{n}(w_{j})=\frac{1}{2}(z_{j}^{n}+z_{j}^{-n}),$ where the $z_{j}$ and $1/z_{j}$ are roots of the polynomial P(z) with the property $|z_{j}|\ne 1,\,j=1,2,\ldots ,s-1.$ Replacing $z_{j}$ by $1/z_{j},$ if it is necessary, we can assume that $|z_j|>1$ for all $j=1,2,\ldots ,s-1.$ Then, $T_{n}(w_{j})\sim \frac{1}{2}z_{j}^{n}$ and $|2T_{n}(w_{s})-2|\sim |z_{s}|^{n}$ as $n\rightarrow \infty .$ Hence,

$$\begin{aligned} \frac{n\tau (H)}{q}\prod _{j=1}^{s-1}|2T_{n}(w_{j})-2|\sim \frac{n\tau (H)}{q}\prod _{j=1}^{s-1} |z_{j}|^{n}=\frac{n\tau (H)}{q}\prod \limits _{P(z)=0,\,|z|>1}|z|^{n}=\frac{n A^n}{q}, \end{aligned}$$

where $A=\prod \nolimits _{P(z)=0,\,|z|>1}|z|$ is the Mahler measure of the polynomial P(z). By ([10], p. 67), we have $A=\exp \left( \int _{0}^{1}\log |P(e^{2 \pi i t })|\text {d}t\right) .$ Since $P(z)=Q(\frac{1}{2}(z+z^{-1})),$ we get $A=\exp ({\int \limits _{0}^{1}\log |Q(\cos {2 \pi t})|\text {d}t}).$ The theorem is proved. $\square$

Remark 2

We note that $Q(\cos (2\pi t))=\det \,L(H,W),$ where $W=(w_{1},w_{2},\ldots ,w_{m})$ and $w_{i}=2k_{i}+d_{i}-\sum \nolimits _{j=1}^{k_{i}}2T_{s_{i,j}}(\cos (2\pi t))=d_{i}+4\sum \nolimits _{j=1}^{k_{i}}\sin ^2(s_{i,j}\pi t),\,i=1,2,\ldots ,m.$

7 Examples

7.1 Circulant graph $C_{n}(s_{1},s_{2},\ldots ,s_{k})$

We consider the classical circulant graph $C_{n}(s_{1},s_{2},\ldots ,s_{k})$ as a foliation $H_{n}(G_{1})$ on the one vertex graph $H=\{v_{1}\}$ with the fiber $G_{1}=C_{n}(s_{1},s_{2},\ldots ,s_{k}).$ In this case $d_{1}=0,\,L(H,X)=(x_{1}),\,P(z)=2k-\sum \nolimits _{p=1}^{k}(z^{s_p}+z^{-s_p})$ and its Chebyshev transform is $Q(w)=2k-\sum \nolimits _{p=1}^{k}2T_{s_p}(w).$ Different aspects of complexity for circulant graphs were investigated in the papers [8, 11, 18, 19, 26].

7.2 I-graph I(n, k, l) and the generalized Petersen graph GP(n, k)

Let H be a path graph on two vertices, $G_{1}=C_{n}(k)$ and $G_{2}=C_{n}(l).$ Then, $I(n,k,l)=H_{n}(G_{1},G_{2})$ and $GP(n,k)=I(n,k,1).$ We get $P(z)=(3-z^{k}-z^{-k})(3-z^{l}-z^{-l})-1$ and $Q(w)=(3-2T_{k}(w))(3-2T_{l}(w))-1.$ The arithmetical and asymptotical properties of complexity for I-graphs were studied in [20].

7.3 Sandwich of m circulant graphs

Consider a path graph H on m vertices. Then, $H_{n}(G_{1},G_{2},\ldots ,G_{m})$ is a sandwich graph of circulant graphs $G_{1},G_{2},\ldots ,G_{m}.$ Here, $d_{1}=d_{m}=1$ and $d_{i}=2,\,i=2,\ldots ,m-1.$ We set

$$\begin{aligned} D(x_1,x_2,\ldots ,x_m)= \det \left( \begin{array}{ccccccc} x_1 &{}\quad -1 &{}\quad 0 &{}\quad \ldots &{}\quad 0 &{}\quad 0&{}\quad 0 \\ -1 &{}\quad x_2 &{}\quad -1 &{}\quad \ldots &{}\quad 0 &{}\quad 0&{}\quad 0 \\ &{}\quad \vdots &{}\quad &{}\quad \ddots &{}\quad &{}\quad \vdots &{}\quad \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots &{}\quad -1 &{}\quad x_{m-1} &{}\quad -1\\ 0 &{}\quad 0 &{}\quad 0 &{}\quad \ldots &{}\quad 0 &{}\quad -1 &{}\quad x_m\end{array}\right) . \end{aligned}$$

By direct calculation, we obtain

$$\begin{aligned} D(x_1,x_2,\ldots ,x_m)= & {} x_1D(x_2,\ldots ,x_m)-D(x_3,\ldots ,x_m),\\ D(x_1)= & {} x_1,\,D(x_1,x_2)=x_1x_2-1. \end{aligned}$$

Then, $Q(w)=D(w_1,w_2,\ldots ,w_m)$ and $Q(-1)=D(d_1+4t_1,d_2+4t_2,\ldots , d_m+4t_m),$ where $w_i$ and $t_i$ are the same as in Theorem 6.1.

7.4 Generalized Y-graph

Consider the generalized Y-graph $Y_{n}(G_{1},G_{2},G_{3})$ where $G_{i}=C_{n}(s_{i,1},\,s_{i,2},\ldots ,s_{i,k_i}),\,i=1,2,3.$ Here

$$\begin{aligned} Q(w)=3A_{1}(w)A_{2}(w)A_{3}(w)-A_{1}(w)A_{2}(w)-A_{1}(w)A_{3}(w)-A_{2}(w)A_{3}(w), \end{aligned}$$

where $A_{i}(w)=2k_{i}+1-\sum \nolimits _{j=1}^{k_{i}}2T_{s_{i,j}}(w).$

7.5 Generalized H-graph

Consider the generalized H-graph $H_{n}(G_{1},G_{2},G_{3},G_{4}),$ where $G_{i}=C_{n}(s_{i,1},s_{i,2},\ldots ,s_{i,k_i}),\,i=1,2,3,4.$ Now we have

$$\begin{aligned}&Q(w)=A_{1}(w)A_{2}(w)A_{3}(w)A_{4}(w)\\&\quad \left( (3-\frac{1}{A_{1}(w)}-\frac{1}{A_{2}(w)}) (3-\frac{1}{A_{3}(w)}-\frac{1}{A_{4}(w)})-1\right) \end{aligned}$$

where $A_{i}(w)$ are the same as above.

7.6 Discrete torus $T_{n,m}=C_n\times C_m$

We have $T_{n,m}=H_{n}(\underbrace{C_{n}(1),\ldots ,C_{n}(1)}_{m \text { times}}),$ where $H=C_{m}(1)$ is the cyclic graph on n vertices. So, the generalized Laplacian matrix with respect to the set of variables $X=(\underbrace{x,\ldots ,x}_{m \text { times}})$ has the form $L(H,X)= \left( \begin{array}{cccccc} x &{}\quad -1 &{}\quad 0 &{}\quad \ldots &{}\quad 0&{}\quad -1 \\ -1 &{}\quad x &{}\quad -1 &{}\quad \ldots &{}\quad 0&{}\quad 0 \\ &{}\quad \vdots &{}\quad &{}\quad \ddots &{}\quad &{}\quad \vdots \\ -1 &{}\quad 0 &{}\quad 0 &{}\quad \ldots &{}\quad -1 &{}\quad x\\ \end{array}\right) .$ Then, L(H, X) is an $m\times m$ circulant matrix with eigenvalues $\mu _j=x-e^{\frac{2\pi i j}{m}}-(e^{\frac{2\pi i j}{m}})^{m-1}=x-2\cos (\frac{2\pi j}{m}),j=0,\ldots ,m-1.$ Hence, $\det L(H,X)=\prod \nolimits _{j=0}^{m-1}\mu _j=2 T_{m}(x/2)-2.$ Substituting $x=4-z-z^{-1}$ and $w=\frac{1}{2}(z+z^{-1}),$ we get $Q(w)=2 T_{m}(2-w)-2.$

7.7 Direct product $C_n\times H$ where H is a regular graph

Let H be a connected d-regular graph. One can identify the direct product $C_n\times H$ with $H_{n}=H_{n}(\underbrace{C_{n}(1),\ldots ,C_{n}(1)}_{m \text { times}}).$ Let $X=(\underbrace{x,\ldots ,x}_{m \text { times}})$. Now $L(H,X)=x I_{m}-A(H).$ Hence, $\det \,L(H,X)$ coincides with the characteristic polynomial $\chi _{H}(x)$ of graph H. We have $Q(w)=\chi _{H}(2+d-2w).$ Then, $Q(-1)=\chi _{H}(4+d).$