
\documentclass[12pt]{article}
\usepackage{e-jc}
\specs{P27}{20(1)}{2013}

\usepackage{amsthm, amsmath, amssymb}

\title{A new determinant expression for the\\ weighted Bartholdi zeta function of a digraph}
\author{Iwao Sato\thanks{Supported by Grant-in-Aid for Science Research (C)}, 
\ Hideo Mitsuhasi\\
\small Oyama National College of Technology \\[-0.8ex]
\small Oyama, Tochigi 323-0806, Japan \\
\small\tt isato@oyama-ct.ac.jp\\
\and
Hideaki Morita\\
\small Division of System Engineering for Mathematics \\[-0.8ex]
\small Muroran Institute of Technology\\ [-0.8ex]
\small Muroran, Hokkaido 050-8585, Japan}

\date{\dateline{Sep 10, 2013}{Jan 26, 2013}{Feb 5, 2013}\\
\small Mathematical Subject Classifications: 05C50, 15A15}

 \begin{document}
 \maketitle


\begin{abstract}
We consider the weighted Bartholdi zeta function of a digraph $D$, and 
give a new determinant expression of it. 
Furthermore, we treat a weighted $L$-function of $D$, and give a new 
determinant expression of it. 
As a corollary, we present determinant expressions for 
the Bartholdi edge zeta functions of a graph and a digraph. 


\bigskip\noindent \textbf{Key words:} zeta function, digraph covering, $L$-function
\end{abstract}




\section{Introduction}

Zeta functions of graphs started from zeta functions of regular graphs 
by Ihara [7]. 
In [7], he showed that their reciprocals are explicit polynomials. 
A zeta function of a regular graph $G$ associated with a unitary 
representation of the fundamental group of $G$ was developed by 
Sunada [12,13]. 
Hashimoto [6] generalized Ihara's result on the zeta function of 
a regular graph to an irregular graph, and showed that its reciprocal is 
again a polynomial by a determinant containing the edge matrix. 
Bass [2] presented another determinant expression for the Ihara zeta function 
of an irregular graph by using its adjacency matrix. 

Stark and Terras [11] gave an elementary proof of Bass' Theorem, and 
discussed three different zeta functions of any graph. 
Furthermore, various proofs of Bass' Theorem were given by 
Foata and Zeilberger [4], Kotani and Sunada [8].

For two variable zeta function of a graph, Bartholdi [1] defined and gave 
a determinant expression of the Bartholdi zeta function of a graph. 
Mizuno and Sato [9] presented a decomposition formula for the Bartholdi 
zeta function of a regular covering of a graph. 

As a digraph version of the Bartholdi zeta function, Choe, Kwak, Park and Sato [3] 
defined the weighted Bartholdi zeta function of a digraph, and 
presented its determinant expression. 
 
As a multi-variable zeta function of a graph, Stark and Terras [11] defined 
the edge zeta function of a graph. 
Watanabe and Fukumizu [14] presented a determinant expression for the edge zeta function 
of a graph $G$ with $n$ vertices by $n \times n$ matrices.

In this paper, we present a new determinant expression of 
the weighted Bartholdi zeta function of a digraph $D$ by using 
the method of Watanabe and Fukumizu [14]: 

\vspace{1mm} 


{\bf Main Theorem}. 

Let $D$ be a connected digraph with $n$ vertices and $m$ arcs, and 
let ${\bf W} = {\bf W} (D)$ be a weighted matrix of $D$. 
Then the reciprocal of the weighted Bartholdi zeta function of $D$ 
is given by 
\[
\zeta (D,w,u,t )^{-1} = \det ( {\bf I}_n +(1-u) t^2 \tilde{{\bf D}} 
-t \tilde{{\bf A}}_1 -t \tilde{{\bf A}}_0 ) 
\prod^{m_1}_{i=1} ( 1- w(f_i ) w(f^{-1}_i ) (1-u )^2 t^2 ), 
\]
where $\tilde{{\bf D}} $, $\tilde{{\bf A}}_1 $ and $ \tilde{{\bf A}}_0 $ are 
defined in Section 3, and $f^{\pm 1}_1, \ldots, f^{\pm 1}_{m_1} $ are 
symmetric arcs of $D$. 


\vspace{1mm} 

Furthermore, we present a new decomposition formula for 
the weighted Bartholdi zeta function of a group covering of $D$, 
and a new determinant expression for the weighted Bartholdi $L$-function 
of $D$. 

\section{Preliminaries} 

Graphs and digraphs treated here are finite.
Let $G=(V(G),E(G))$ be a connected graph (possibly multiple edges and loops) 
with the set $V(G)$ of vertices and the set $E(G)$ of unoriented edges $uv$ 
joining two vertices $u$ and $v$. 
For $uv \in E(G)$, an arc $(u,v)$ is the oriented edge from $u$ to $v$. 
Set $D(G)= \{ (u,v),(v,u) \mid uv \in E(G) \} $. 
For $e=(u,v) \in D(G)$, set $u=o(e)$ and $v=t(e)$. 
Furthermore, let $e^{-1}=(v,u)$ be the {\em inverse} of $e=(u,v)$. 

A {\em path $P$ of length $n$} in $G$ is a sequence 
$P=(e_1, \cdots,e_n )$ of $n$ arcs such that $e_i \in D(G)$,
$t( e_i )=o( e_{i+1} )(1 \leq i \leq n-1)$, 
where indices are treated $mod \  n$. 
Set $ \mid P \mid =n$, $o(P)=o( e_1 )$ and $t(P)=t( e_n )$. 
Also, $P$ is called an {\em $(o(P),t(P))$-path}. 
We say that a path $P=(e_1, \cdots,e_n )$ has a {\em backtracking} 
or a {\em bump} at $t( e_i )$ if $ e^{-1}_{i+1} =e_i $ 
for some $i(1 \leq i \leq n-1)$. 
A $(v, w)$-path is called a {\em $v$-cycle} 
(or {\em $v$-closed path}) if $v=w$. 


We introduce an equivalence relation between cycles. 
Two cycles $C_1 =(e_1, \cdots,e_m )$ and 
$C_2 =(f_1, \cdots,f_m )$ are called {\em equivalent} if there exists 
$k$ such that $f_j =e_{j+k} $ for all $j$. 
The inverse cycle of $C$ is in general not equivalent to $C$. 
Let $[C]$ be the equivalence class which contains a cycle $C$. 
Let $B^r$ be the cycle obtained by going $r$ times around a cycle $B$. 
Such a cycle is called a {\em power} of $B$. 
A cycle $C$ is {\em reduced} if 
$C$ has no backtracking. 
Furthermore, a cycle $C$ is {\em prime} if it is not a power of 
a strictly smaller cycle. 


The {\em Ihara zeta function} of a graph $G$ is 
a function of $u \in {\bf C}$ with $|u|$ sufficiently small, defined by 
\[
{\bf Z} (G, t)= \prod_{[C]} (1- t^{ \mid C \mid } )^{-1},
\]
where $[C]$ runs over all equivalence classes of prime, reduced cycles 
of $G$(see [7]). 

Let $m$ be the number of edges of $G$. 
Furthermore, let two $m \times m$ matrices 
${\bf B} =( {\bf B}_{e,f} )_{e,f \in A(D)} $ and 
${\bf J}_0 =( {\bf J}_{e,f} )_{e,f \in A(D)} $ be defined as follows: 
\[
{\bf B}_{e,f} =\left\{
\begin{array}{ll}
1 & \mbox{if $t(e)=o(f)$, } \\
0 & \mbox{otherwise}
\end{array}
\right.
, 
{\bf J}_{e,f} =\left\{
\begin{array}{ll}
1 & \mbox{if $f= e^{-1} $, } \\
0 & \mbox{otherwise.}
\end{array}
\right.
\]
Then ${\bf B} -{\bf J}_0 $ is called the {\em edge matrix} of $G$. 
  

\newtheorem{theorem}{Theorem}
\begin{theorem}[Hashimoto; Bass]
Let $G$ be a connected graph with $n$ vertices and $m$ edges. 
Then the reciprocal of the Ihara zeta function of $G$ is given by 
\[
{\bf Z} (G, t)^{-1} = \det ( {\bf I}_{2m} - t ( {\bf B} - {\bf J}_0 ))
=(1- t^2 )^{m-n} \det ( {\bf I} -t {\bf A} (G)+ t^2 ({\bf D} -{\bf I} )), 
\]
where ${\bf A} (G)$ is the adjacency matrix of $G$, and ${\bf D} =( d_{ij} )$ is the diagonal matrix 
with $d_{ii} = \deg v_i $ where $V(G)= \{ v_1, \cdots, v_n \} $. 
\end{theorem}


Then the {\em Bartholdi zeta function} of $G$ is  defined by 
\[
\zeta {}_G (u,t)= \zeta (G,u,t)= 
\prod_{[C]} (1- u^{ cbc(C)} t^{ \mid C \mid } )^{-1}, 
\]
where $[C]$ runs over all equivalence classes of prime cycles 
of $G$(see [1]). 



\begin{theorem}[Bartholdi] 
Let $G$ be a connected graph with $n$ vertices and $m$ unoriented edges. 
Then the reciprocal of the Bartholdi zeta function of $G$ is given by 
\begin{align*}
\zeta(G,u,t )^{-1} &= \det ( {\bf I}_{2m} -t ( {\bf B} -(1-u) {\bf J}_0 )) \\ 
&= (1-(1-u )^2 t^2 )^{m-n} \det ( {\bf I} -t {\bf A} (G)+(1-u)( {\bf D} -(1-u) {\bf I} ) t^2 ). 
\end{align*}
\end{theorem}


In the case of $u=0$, Theorem 2 implies Theorem 1. 

We now state the weighted Bartholdi zeta function of a digraph.  
Let $D\!=(V(D),A(D))$ be a connected digraph with the set $V(D)$ of vertices 
and the set $A(D)$ of arcs. 
Furthermore, let $D$ have $n$ vertices $v_1,\ldots,v_n$ 
and $m$ arcs. 
Then we consider an $n \times n$ matrix 
${\bf W} ={\bf W} (D)=( w_{ij} )_{1 \leq i,j \leq n }$ with $ij$ entry 
nonzero complex number $w_{ij}$ if $( v_i, v_j ) \in A(D)$, 
and $w_{ij} =0$ otherwise. 
The matrix ${\bf W} = {\bf W} (D)$ is called the 
{\em weighted matrix} of $D$.
Furthermore, let $w( v_i, v_j )= w_{ij}, \  v_i, v_j \in V(D)$ and 
$w(e)= w_{ij}, e=( v_i, v_j ) \in A(D)$. 
For each path $P=( e_1, \cdots, e_r )$ of $G$, the {\em norm} 
$w(P)$ of $P$ is defined as follows: 
$w(P)= w (e_1 ) \cdots w ( e_r ) $. 


The {\em cyclic bump count} $cbc(C)$ of a cycle 
$C=( e_1, \cdots, e_n )$ of $G$ is 
\[
cbc(C)= \mid \{  i=1, \cdots, n \mid e_i = e^{-1}_{i+1} \} \mid, 
\]
where $e_{n+1} = e_1 $. 
Then the {\em weighted Bartholdi zeta function} of $D$ is 
a function of $u,t \in {\bf C}$ with $|u|,|t|$ sufficiently small, defined by 
\[
\zeta (D,w,u,t)= \prod_{[C]} (1-w(C) u^{cbc(C)} t^{ \mid C \mid } ) {}^{-1},
\]
where $[C]$ runs over all equivalence classes of prime cycles of $D$. 

If $w= {\bf 1} $, i.e., $w(v_i,v_j)=1$ for each $( v_i, v_j ) \in A(D)$, then the weighted 
Bartholdi zeta function of $D$ is the Bartholdi zeta function of $D$. 
If $D= D_G$ is the symmetric digraph corresponding to a graph $G$, and $w= {\bf 1} $, 
then the weighted Bartholdi zeta function of $D_G$ is the Bartholdi zeta 
function of $G$. 
If $D= D_G$, $w= {\bf 1} $ and $u=0$, then the weighted Bartholdi zeta function of $G$ is 
the Ihara zeta function of $G$. 

Two $m \times m$ matrices 
${\bf B}_w =( {\bf B}^w_{e,f} )_{e,f \in A(D)} $ and 
${\bf J}_w =( {\bf J}^w_{e,f} )_{e,f \in A(D)} $ are defined as follows: 
\[
{\bf B}^w_{e,f} =\left\{
\begin{array}{ll}
w(e) & \mbox{if $t(e)=o(f)$, } \\
0 & \mbox{otherwise}
\end{array}
\right.
, 
{\bf J}^w_{e,f} =\left\{
\begin{array}{ll}
w(e) & \mbox{if $f= e^{-1} $, } \\
0 & \mbox{otherwise.}
\end{array}
\right.
\]
Furthermore, we define two $n \times n$ matrices 
${\bf W}_1 ={\bf W}_1 (D)=(a_{uv} )$ and ${\bf W}_0$ as follows: 
\[
a_{uv} =\left\{
\begin{array}{ll}
w(u,v) & \mbox{if both $(u,v)$ and $(v,u) \in A(D)$, } \\
0 & \mbox{otherwise }
\end{array}
\right. 
\]
and 
\[
{\bf W}_0 = {\bf W}_0 (D)= {\bf W} (D)- {\bf W}_1. 
\]
Let an $n \times n$ matrix ${\bf S} =(s_{xy} )$ is the diagonal matrix 
defined by 
\[
s_{xx} = \mid \{ e \in A(D) \mid o(e)=x, e^{-1} \in A(D) \} \mid. 
\] 


\begin{theorem}[Choe, Kwak, Park and Sato]
Let $D$ be a connected digraph, and 
let ${\bf W} = {\bf W} (D)$ be a weighted matrix of $D$. 
Furthermore, let $m_1 =\mid \{ e \in A(D) \mid e^{-1} \in A(D) \} \mid /2$. 
Then the reciprocal of the weighted Bartholdi zeta function of $D$ 
is given by 
\[
\zeta (D,w,u,t )^{-1} 
= \det ({\bf I}_{m} -( {\bf B}_w -(1-u) {\bf J}_w )t),  
\]
where $n= \mid V(D) \mid $ and $m= \mid A(D) \mid $. 

Furthermore, if $w( e^{-1} )=w(e )^{-1}$ for each $e \in A(D)$ such that 
$e^{-1} \in A(D)$, then 
\[
\begin{array}{rcl}
\  &  & \zeta (D,w,u,t )^{-1} = (1-(1-u )^2 t^2 )^{m_1 -n} \\ 
\  &   &                \\ 
\  & \times & 
\det ({\bf I}_n -t {\bf W}_1 (D) -(1-(1-u )^2 t^2 )t {\bf W}_0 (D) 
+(1-u) t^2 ( {\bf S} -(1-u) {\bf I}_n )). 
\end{array}
\]
\end{theorem}


If $D=D_G$, $w= {\bf 1} $ and $u=0$, then Theorem 2 implies Theorem 1.

Now, we proceed to the edge zeta function of a graph $G$ with $m$ edges. 
Let $G$ be a connected graph and $D(G)= \{ e_1, \ldots, e_m, e_{m+1}, \ldots, 
e_{2m} \} (e_{m+i} = e^{-1}_i (1 \leq i \leq m ))$. 
We introduce $2m$ variables $z_1, \ldots, z_{2m} $, and set $g(C)=  z_{i_1 } \cdots z_{i_k} $ 
for each cycle $C=( e_{i_1 }, \ldots, e_{i_k} )$ of $G$. 
Set $z_{e_i} =z_i (1 \leq i \leq 2m)$ and ${\bf z} =(z_1, \ldots, z_{2m} )$. 
Then the {\em edge zeta function} $\zeta {}_G ( {\bf z} )$ of $G$ is defined by 
\[
\zeta {}_G ( {\bf z} )= \prod_{[C]} (1-g(C) )^{-1}, 
\]
where $[C]$ runs over all equivalence classes of prime, reduced cycles 
of $G$. 


\begin{theorem}[Stark and Terras]
Let $G$ be a connected graph with $m$ edges. 
Then 
\[
\zeta {}_G ( {\bf z} ) {}^{-1} = 
\det ( {\bf I}_{2m} - ( {\bf B} - {\bf J} {}_0) {\bf U} ),  
\]
where 
\[
{\bf U} =
\left[ 
\begin{array}{cccccc}
z_1 &   &   &   &   & 0 \\
   & \ddots &   &   &   &   \\
   &    & z_m  &   &   &  \\
   &    &   & z_{m+1}  &   &  \\ 
   &    &   &   & \ddots &  \\
0  &    &   &   &   & z_{2m} 
\end{array} 
\right] 
. 
\]
\end{theorem}


Let $G$ be a graph with $n$ vertices.  
Then we define an $n \times n$ matrix $\widehat{{\bf A}} =(a_{xy} )$ as follows: 
\[
a_{xy} =\left\{
\begin{array}{ll}
z_{(x,y)}/(1- z_{(x,y)} z_{(y,x)} ) & \mbox{if $(x,y) \in D(G)$, } \\
0 & \mbox{otherwise. }
\end{array}
\right. 
\]
Furthermore, an $n \times n$ matrix $\widehat{{\bf D}} =(d_{xy} )$ is the diagonal matrix 
defined by 
\[
d_{xx} = \sum_{o(e)=x} 
\frac{z_e z_{e^{-1}} }{1- z_e z_{ e^{-1}} }. 
\]


\begin{theorem}[Watanabe and Fukumizu] 
Let $G$ be a connected graph with $n$ vertices and $m$ edges.  
Then    
\[
\zeta {}_G ( {\bf z} ) {}^{-1} = \det ( {\bf I}_{n} + \widehat{{\bf D}} 
- \widehat{{\bf A}} ) \prod^{m}_{i=1} ( 1- z_{f_i } z_{f^{-1}_i } ),  
\]
where $D(G)= \{ f_1,f^{-1}_1, \ldots, f_m f^{-1}_m \} $. 
\end{theorem}

In Section 2, we present a new determinant expression of 
the weighted Bartholdi zeta function of a digraph $D$ by using 
the method of Watanabe and Fukumizu [14]. 
In Section 3, we present a new decomposition formula for 
the weighted Bartholdi zeta function of a group covering of $D$. 
In Section 4, we present a new determinant expression for 
the weighted Bartholdi $L$-function of $D$. 
In Section 5, we define the Bartholdi edge zeta functions of graphs and digraphs, 
and present their determinant expressions as corollaries of Theorem 6. 


\section{Weighted Bartholdi zeta functions of digraphs}

We present a new determinant expression of 
the weighted Bartholdi zeta function of a digraph. 

Let $D$ be a connected digraph with $n$ vertices $v_1, \cdots, v_n $ 
and $m$ arcs, and ${\bf W} = {\bf W} (D)$ a weighted matrix of $D$. 
Then we define two $n \times n$ matrices 
$\tilde{{\bf A}}_1 =\tilde{{\bf A}}_1 (D)=(a_{xy} )$ and 
$\tilde{{\bf A}}_0 =\tilde{{\bf A}}_0 (D)= (b_{xy} )$ as follows: 
\[
a_{xy} =\left\{
\begin{array}{ll}
w(x,y)/(1- w(x,y) w(y,x) (1-u )^2 t^2 ) & \mbox{if both $(x,y)$ and $(y,x) \in A(D)$, } \\
0 & \mbox{otherwise }
\end{array}
\right. 
\]
and 
\[
b_{xy} =\left\{
\begin{array}{ll}
w(x,y) & \mbox{if $(x,y) \in A(D)$ and $(y,x) \not\in A(D)$, } \\
0 & \mbox{otherwise }
\end{array}
\right. 
\]
Furthermore, an $n \times n$ matrix $\tilde{{\bf D}} = \tilde{{\bf D}} (D)= (d_{xy} )$ 
is the diagonal matrix defined by 
\[
d_{xx} = \sum_{o(e)=x, e^{-1} \in A(D)} 
\frac{w(e) w(e^{-1} )}{1- w(e) w( e^{-1} ) (1-u )^2 t^2 }. 
\] 

Let \( {\bf M}_{1} \oplus \cdots \oplus {\bf M}_{s} \) be the 
block diagonal sum of square matrices 
${\bf M}_{1}, \cdots, {\bf M}_{s}$. 
A new determinant expression for $ \zeta (D,w,u,t )$ is given as follows: 


\begin{theorem}
Let $D$ be a connected digraph, and 
let ${\bf W} = {\bf W} (D)$ be a weighted matrix of $D$. 
Then the reciprocal of the weighted Bartholdi zeta function of $D$ 
is given by 
\[
\zeta (D,w,u,t )^{-1} = \det ( {\bf I}_n +(1-u) t^2 \tilde{{\bf D}} 
-t \tilde{{\bf A}}_1 -t \tilde{{\bf A}}_0 ) 
\prod^{m_1}_{i=1} ( 1- w(f_i ) w(f^{-1}_i ) (1-u )^2 t^2 ), 
\]
where $n= \mid V(D) \mid $, $m= \mid A(D) \mid $ and 
$f^{\pm 1}_1, \ldots, f^{\pm 1}_{m_1} $ are symmetric arcs of $D$. 
\end{theorem}


{\bf Proof}.  Let $V(D)= \{ v_1, \cdots, v_n \} $ and, 
let $A(D)= \{ e_1, \cdots, e_{m_0 }, f_1, 
\cdots, f_{m_1}, f^{-1}_{1}, \cdots $, $ f^{-1}_{m_1} \} $ 
such that $e^{-1}_i \not\in A(D) (1 \leq i \leq m_0 )$. 
Note that  $m= m_0 +2 m_1 $. 

Arrange arcs of $D$ as follows: 
\[
e_1, \cdots, e_{m_0 }, f_{1}, f^{-1}_1, 
\cdots, f_{ m_1}, f^{-1}_{m_1 }. 
\]
Let 
\[
{\bf U} =
\left[ 
\begin{array}{cccccc}
w(e_1 ) &   &   &   &   & 0 \\
   & \ddots &   &   &   &   \\
   &    & w(e_{m_0} ) &   &   &  \\
   &    &   & w(f_1)  &   &  \\ 
   &    &   &   & w(f^{-1}_1 ) &  \\
0  &    &   &   &   &  \ddots 
\end{array} 
\right] 
. 
\]
Then we have 
\[
{\bf U} {\bf B} = {\bf B}_w \  and \  
{\bf U} {\bf J}_0 = {\bf J}_w. 
\]
Thus, 
\[
{\bf B}_w -(1-u) {\bf J}_w = {\bf U} ({\bf B} -(1-u) {\bf J}_0 ). 
\]
By Theorem 2, it follows that 
\[
 \zeta (D,w,u,t )^{-1} = \det ( {\bf I}_m 
-t {\bf U} ({\bf B} -(1-u) {\bf J}_0 )). 
\]


Now, let ${\bf K} =( {\bf K}_{ev} )$ ${}_{e \in A(D); v \in V(D)} $ be the $m \times n$ matrix defined 
as follows: 
\[
{\bf K}_{ev} :=\left\{
\begin{array}{ll}
1 & \mbox{if $o(e)=v$, } \\
0 & \mbox{otherwise. } 
\end{array}
\right.
\]
Furthermore, we define the $m \times n$ matrix 
${\bf L} =( {\bf L}_{ev} )_{e \in A(D); v \in V(D)} $ as follows: 
\[
{\bf L}_{ev} :=\left\{
\begin{array}{ll}
1 & \mbox{if $t(e)=v$, } \\
0 & \mbox{otherwise. } 
\end{array}
\right.
\] 
Then we have 
\[
{\bf L} {}^t {\bf K} = {\bf B}. 
\]
Thus, 
\[
\begin{array}{rcl}
\  &   & \det ( {\bf I}_m -t {\bf U} ({\bf B} -(1-u) {\bf J}_0 )) \\  
\  &   &                \\ 
\  & = & \det ( {\bf I}_m -t {\bf U} 
({\bf L} {}^t {\bf K} -(1-u) {\bf J}_0 )) 
= \det ( {\bf I}_m -t {\bf U} {\bf L} {}^t {\bf K} 
+(1-u)t {\bf U} {\bf J}_0 ). 
\end{array}
\]


But, we have 
\begin{equation}
{\bf I}_m +(1-u)t {\bf U} {\bf J}_0 
= {\bf I}_{m_0 } \oplus ( \oplus {}^{m_1 }_{j=1} 
\left[ 
\begin{array}{cc}
1  &  (1-u)t w(f_j ) \\
(1-u)t w(f^{-1}_j ) & 1 
\end{array} 
\right] 
) 
. 
\end{equation}
Since  $|u|,|t|$ are sufficiently small, we have 
\[
\det ( 
\left[ 
\begin{array}{cc}
1  &  (1-u)t w(f_j ) \\
(1-u)t w(f^{-1}_j ) & 1 
\end{array} 
\right] 
) 
=1-(1-u )^2 t^2 w(f_j ) w(f^{-1}_j ) \neq 0\  (1 \leq j \leq m_1 ). 
\]
Thus, ${\bf I}_m +(1-u)t {\bf U} {\bf J}_0 $ is invertible. 
Therefore,  
\[
\begin{array}{rcl}
\  &   & \det ( {\bf I}_m -t {\bf U} ({\bf B} -(1-u) {\bf J}_0 )) \\  
\  &   &                \\ 
\  & = & \det ( {\bf I}_m -t {\bf U} {\bf L} {}^t {\bf K} 
({\bf I}_m +(1-u)t {\bf U} {\bf J}_0 )^{-1} ) 
\det ( {\bf I}_m +(1-u)t {\bf U} {\bf J}_0 ). 
\end{array}
\]

But, if ${\bf A}$ and ${\bf B}$ are a $m \times n $ and $n \times m$ 
matrices, respectively, then we have 
\begin{equation}
\det ( {\bf I}_m - {\bf A} {\bf B} )= 
\det ( {\bf I}_n - {\bf B} {\bf A} ). 
\end{equation}
Thus, we have 
\[
\begin{array}{rcl}
\  &   & \det ( {\bf I}_m -t {\bf U} ({\bf B} -(1-u) {\bf J}_0 )) \\
\  &   &                \\ 
\  & = & \det ( {\bf I}_n -t \  {}^t {\bf K} 
({\bf I}_m +(1-u)t {\bf U} {\bf J}_0 )^{-1} {\bf U} {\bf L} ) 
\det ( {\bf I}_m +(1-u)t {\bf U} {\bf J}_0 ). 
\end{array}
\]

Next, we have 
\[
\det ( {\bf I}_m +(1-u)t {\bf U} {\bf J}_0 ) 
= \prod^{m_1}_{i=1} ( 1- w(f_i ) w(f^{-1}_i ) (1-u )^2 t^2 ). 
\]
Furthermore, the $m \times n$ matrix 
${\bf U} {\bf L} =( c_{ev} )_{e \in A(D); v \in V(D)} $ is given as follows: 
\[
c_{ev} :=\left\{
\begin{array}{ll}
w(e) & \mbox{if $t(e)=v$, } \\
0 & \mbox{otherwise. } 
\end{array}
\right.
\] 
But, we have 
\[
( {\bf I}_m +(1-u)t {\bf U} {\bf J}_0 )^{-1} =
{\bf I}_{m_0 } \oplus ( \oplus {}^{m_1 }_{j=1} 
\left[ 
\begin{array}{cc}
1/x_j & -(1-u)t  w(f_j )/x_j \\ 
-(1-u)t w(f^{-1}_j )/x_j & 1/x_j 
\end{array} 
\right] 
) 
,   
\]
where $x_i =1-w(f_i) w(f^{-1}_i ) (1-u )^2 t^2 \  (1 \leq i \leq m_1 )$.  

Now, for a symmetric arc $(x,y) \in A(D)$, 
\[
( {}^t {\bf K} ({\bf I}_m +(1-u)t {\bf U} {\bf J}_0 )^{-1} 
{\bf U} {\bf L} )_{xy} = w(x,y)/(1- w(x,y) w(y,x) (1-u )^2 t^2 ). 
\]
For a nonsymmetric arc $(x,y) \in A(D)$, 
\[
( {}^t {\bf K} ({\bf I}_m +(1-u)t {\bf U} {\bf J}_0 )^{-1} 
{\bf U} {\bf L} )_{xy} = w(x,y).  
\]
Furthermore, if $x=y$, then 
\[
( {}^t {\bf K} ({\bf I}_m +(1-u)t {\bf U} {\bf J}_0 )^{-1} 
{\bf U} {\bf L} )_{xx} 
=- \sum_{o(e)=x, e^{-1} \in A(D)} 
\frac{(1-u)t w(e) w(e^{-1} )}{1- w(e) w( e^{-1} ) (1-u )^2 t^2 }. 
\]

Thus, 
\[
\det ( {\bf I}_n -t \  {}^t {\bf K} 
({\bf I}_m +(1-u)t {\bf U} {\bf J}_0 )^{-1} {\bf U} {\bf L} ) 
= \det ( {\bf I}_n +(1-u) t^2 \tilde{{\bf D}} 
-t \tilde{{\bf A}}_1 -t \tilde{{\bf A}}_0 ). 
\]
Therefore, it follows that 
\[
\zeta (D,w,u,t )^{-1} = \det ( {\bf I}_n +(1-u) t^2 \tilde{{\bf D}} 
-t \tilde{{\bf A}}_1 -t \tilde{{\bf A}}_0 ) 
\prod^{m_1}_{i=1} ( 1- w(f_i ) w(f^{-1}_i ) (1-u )^2 t^2 ). 
\]
$\Box$

By Theorem 5, we obtain the second identity of Theorem 2. 


\newtheorem{corollary}{Corollary}
\begin{corollary}[Choe, Kwak, Park and Sato]
Let $D$ be a connected digraph, and 
let ${\bf W} = {\bf W} (D)$ be a weighted matrix of $D$. 
Furthermore, assume that $w( e^{-1} )=w(e )^{-1}$ for each $e \in A(D)$ such that 
$e^{-1} \in A(D)$. 
Then the reciprocal of the weighted Bartholdi zeta function of $D$ 
is given by 
\[
\begin{array}{rcl}
\  &  & \zeta (D,w,u,t )^{-1} 
= (1-(1-u )^2 t^2 )^{m_1 -n} \\
\  &   &                \\ 
\  & \times & 
\det ({\bf I}_n -t {\bf W}_1 (D) -(1-(1-u )^2 t^2 )t {\bf W}_0 (D) 
+(1-u) t^2 ( {\bf S} -(1-u) {\bf I}_n )). 
\end{array}
\]
where $n= \mid V(D) \mid $ and $m= \mid A(D) \mid $. 
\end{corollary}


{\bf Proof}.  Since $w( e^{-1} )=w(e )^{-1}$ for each symmetric arc $e \in A(D)$, 
we have $w( e^{-1} )$ $w(e )^{-1} =1$. 
Then we have 
\[
\tilde{{\bf D}} = \frac{1}{1-(1-u )^2 t^2} {\bf S}, \  
\tilde{{\bf A}}_1 = \frac{1}{1-(1-u )^2 t^2} {\bf W}_1 (D). 
\]
Furthermore, $\tilde{{\bf A}}_0 = {\bf W}_0 (D)$. 
Thus, 
\begin{align*}
\zeta (D,w,u,t )^{-1} 
&= (1-(1-u )^2 t^2 )^{m_1 } 
\det \Big({\bf I}_n -t/(1-(1-u )^2 t^2 ) {\bf W}_1 (D)\\ 
& \quad -t {\bf W}_0 (D)
+(1-u) t^2 /(1-(1-u )^2 t^2 ) {\bf S} \Big) \\
& = (1-(1-u )^2 t^2 )^{m_1 -n} 
\det\Big({\bf I}_n -t {\bf W}_1 (D) -(1-(1-u )^2 t^2 )t {\bf W}_0 (D) \\
&\quad + (1-u) t^2 ( {\bf S} -(1-u) {\bf I}_n )\Big).  
\end{align*}
$\Box$ 


\section{Weighted Bartholdi zeta functions of 
group coverings of digraphs}

We can generalize the notion of a $\Gamma$-covering of a graph to a simple 
digraph. 
Let $D$ be a connected digraph and $ \Gamma $ a finite group.
Then a mapping $ \alpha : A(D) \longrightarrow \Gamma $
is called a {\em pseudo ordinary voltage} {\em assignment}
if $ \alpha (v,u)= \alpha (u,v)^{-1} $ for each $(u,v) \in A(D)$ 
such that $(v,u) \in A(D)$.
The pair $(D, \alpha )$ is called an 
{\em ordinary voltage digraph}.
The {\em derived digraph} $D^{ \alpha } $ of the ordinary
voltage digraph $(D, \alpha )$ is defined as follows:
$V(D^{ \alpha } )=V(D) \times \Gamma $ and $((u,h),(v,k)) \in 
A(D^{ \alpha })$ if and only if $(u,v) \in A(D)$ and $k=h \alpha (u,v) $. 
The digraph $D^{ \alpha }$ is called a {\em $ \Gamma $-covering} of $D$.
Note that a $\Gamma$-covering of the symmetric digraph corresponding 
to a graph $G$ is a $\Gamma$-covering of $G$(see [5]). 


Let $D$ be a connected digraph, $ \Gamma $ a finite group and 
$ \alpha : A(D) \longrightarrow \Gamma $ a pseudo ordinary voltage assignment. 
In the $\Gamma $-covering $D^{ \alpha } $, set $v_g =(v,g)$ and $e_g =(e,g)$, 
where $v \in V(D), e \in A(D), g\in \Gamma $. 
For $e=(u,v) \in A(D)$, the arc $e_g$ emanates from $u_g$ and 
terminates at $v_{g \alpha (e)}$. 

Let ${\bf W} = {\bf W} (D)$ be a weighted matrix of $D$. 
Then we define the {\em weighted matrix} 
$\tilde{{\bf W}} = {\bf W} ( D^{ \alpha } )
=( \tilde{w} ( u_g, v_h ))$ of $ D^{ \alpha } $ 
{\em derived from} ${\bf W}$ as follows: 
\[
\tilde{w} ( u_g, v_h ) :=\left\{
\begin{array}{ll}
w(u,v) & \mbox{if $(u,v) \in A(D)$ and $h=g \alpha (u,v)$, } \\
0 & \mbox{otherwise.}
\end{array}
\right.
\]


If \( {\bf M}_{1} = {\bf M}_{2} = \cdots = {\bf M}_{s} = {\bf M} \),
then we write 
\( s \circ {\bf M} = {\bf M}_{1} \oplus \cdots \oplus {\bf M}_{s} \).
The {\em Kronecker product} $ {\bf A} \bigotimes {\bf B} $
of matrices {\bf A} and {\bf B} is considered as the matrix 
{\bf A} having the element $a_{ij}$ replaced by the matrix $a_{ij} {\bf B}$.


\begin{theorem} 
Let $D$ be a connected digraph with $n$ vertices and $m$ arcs, 
$ \Gamma $ a finite group, $ \alpha : A(D) \longrightarrow \Gamma  $ 
a pseudo ordinary voltage assignment and ${\bf W} = {\bf W} (D)$ a weighted 
matrix of $D$. 
Set $m_1 = \mid \{ e \in A(D) \mid e^{-1} \in A(D) \} \mid /2$ and 
$\mid \Gamma \mid =r$. 
Furthermore, let $ {\rho}_{1} =1, {\rho}_{2}, \cdots, {\rho}_{k} $
be the irreducible representations of $ \Gamma $, and 
$d_i$ the degree of $ {\rho}_{i} $ for each $i$, where 
$d_1=1$.
For $g \in \Gamma $, the matrix ${\bf A} {}_{1,g} =(a^{(g)}_{xy} )$ 
is defined as follows:
\[
a^{(g)}_{xy} :=\left\{
\begin{array}{ll}
w(x,y)/(1-w(x,y) w(y,x) (1-u )^2 t^2 ) & \mbox{if $(x,y),(y,x) \in A(D)$ 
and $ \alpha (x,y)= g $, } \\
0 & \mbox{otherwise.}
\end{array}
\right.
\]
Furthermore, the matrix ${\bf A} {}_{0,g} =(b^{(g)}_{xy} )$ 
is defined as follows:
\[
b^{(g)}_{xy} :=\left\{
\begin{array}{ll}
w(x,y) & \mbox{if $(x,y) \in A(D)$,$(y,x) \not\in A(D)$ and 
$ \alpha (x,y)= g $, } \\
0 & \mbox{otherwise.}
\end{array}
\right.
\]

Suppose that the $ \Gamma $-covering $D^{ \alpha } $ of $D$ is connected. 
Then the reciprocal of the weighted Bartholdi zeta function of $D {}^{ \alpha } $ is
\[
\zeta (D {}^{ \alpha }, \tilde{w},u,t )^{-1} 
= \prod^{m_1}_{i=1} ( 1- w(f_i ) w(f^{-1}_i )(1-u )^2 t^2 )^r  
\]
\[
\times 
\prod^{k}_{i=1} 
\{ \det ({\bf I}_{n d_i } -t \sum_{h \in \Gamma } 
{\rho}_{i} (h) \bigotimes {\bf A} {}_{1,h} 
-t \sum_{h \in \Gamma } 
{\rho}_{i} (h) \bigotimes {\bf A} {}_{0,h} 
+(1-u) t^2 
( {\bf I}_{d_i} \bigotimes \tilde{{\bf D}} (D))) \} {}^{d_i}, 
\] 
where $f^{\pm 1}_1, \ldots, f^{\pm 1}_{m_1} $ are symmetric arcs of $D$. 
\end{theorem}


{\bf Proof }. 
Let $V(D)= \{ v_1, \cdots, v_{n} \} $ and 
$ \Gamma = \{ 1=g_1, g_2, \cdots,g_r \} $.
Arrange vertices of $D^{ \alpha } $ in $n$ blocks:
$(v_1,1), \cdots, (v_{n},1);(v_1,g_2), \cdots, (v_{n},g_2); 
\cdots ; (v_1,g_r), \cdots,(v_{n},g_r). $
We consider the three matrices $\tilde{{\bf A}}_1 ( D {}^{ \alpha } ) $, 
$\tilde{{\bf W}}_0 ( D {}^{ \alpha } ) $ and $\tilde{{\bf D}} ( D {}^{ \alpha } ) $
under this order.
By Theorem 5, we have 
\[
\zeta (D {}^{ \alpha }, \tilde{w},u,t )^{-1} =
\det ( {\bf I}_{\nu m} -t \tilde{{\bf A}} {}_1 ( D^{\alpha} )
-t \tilde{{\bf A}} {}_0 ( D^{\alpha} )+
(1-u) t^2 \tilde{{\bf D}} ( D^{\alpha } )) 
\]
\[
\cdot \prod^{m_1}_{i=1} ( 1- w(f_i ) w(f^{-1}_i )(1-u )^2 t^2 )^r. 
\]

For $h \in \Gamma $, the matrix ${\bf P}_{h}=(p^{(h)}_{ij} )$ 
is defined as follows:
\[
p^{(h)}_{ij} = \left\{
\begin{array}{ll}
1 & \mbox{if $g_i h=g_j$,} \\
0 & \mbox{otherwise.}
\end{array}
\right.
\]
Suppose that $p^{(h)}_{ij} =1 $, i.e., $g_j=g_ih$.
Then $((u,g_i),(v,g_j)) \in A(D {}^{ \alpha } ) $
if and only if $(u,v) \in A(D)$ and 
$g_{j} = g_{i} \alpha (u,v)$,
i.e., $ \alpha (u,v)=g^{-1}_{i} g_j =g^{-1}_{i} g_i h=h$.
Thus we have
\[
\tilde{{\bf A}} {}_0 (D {}^{ \alpha } )= \sum_{h \in \Gamma } {\bf P}_{h} 
\bigotimes {\bf A} {}_{0,h} \  and \  
\tilde{{\bf A}} {}_1 (D {}^{ \alpha } )= \sum_{h \in \Gamma } {\bf P}_{h} 
\bigotimes {\bf A} {}_{1,h}. 
\]

Let $\rho$ be the right regular representation of $ \Gamma $.
Furthermore, let $ {\rho}_{1} =1, {\rho}_{2}, \cdots, {\rho}_{k} $
be the irreducible representations of $ \Gamma $, and 
$d_i$ the degree of $ {\rho}_{i} $ for each $i$, where 
$d_1=1$.
Then we have $\rho (h)= {\bf P}_{h} $ for $h \in \Gamma $.
Furthermore, there exists a nonsingular matrix ${\bf P}$ such that 
${\bf P}^{-1} \rho (h) {\bf P} = (1) \oplus d_2 \circ {\rho}_{2} (h) 
\oplus \cdots \oplus d_k \circ {\rho}_{k} (h)$ 
for each $h \in \Gamma $(see [10]). 
Putting 
${\bf B} =( {\bf P}^{-1} \bigotimes {\bf I}_{n} ) 
( \tilde{{\bf A}}_1 (D {}^{ \alpha } )+ \tilde{{\bf A}}_0 (D {}^{ \alpha } )) 
( {\bf P} \bigotimes {\bf I}_{n} )$,
we have 
\[
{\bf B}= \sum_{h \in \Gamma } 
\{ (1) \oplus d_2 \circ {\rho}_{2} (h) \oplus \cdots \oplus 
d_k \circ {\rho}_{k} (h) \} \bigotimes 
( {\bf A}_{1,h} + {\bf A}_{0,h} ). 
\]
Note that $\tilde{{\bf A}}_i (D) = \sum_{h \in \Gamma } {\bf A} {}_{i,h}\  (i=0,1)$ 
and $1+ d^2_2 + \cdots + d^2_k =r$.
Therefore it follows that 
\[
\zeta (D {}^{ \alpha }, \tilde{w},u,t )^{-1} 
= \prod^{m_1}_{j=1} ( 1- w(f_j ) w(f^{-1}_j )(1-u )^2 t^2 )^{r} 
\]
\[
\times 
\prod^{k}_{i=1} \det ({\bf I}_{n d_i } -t \sum_{h \in \Gamma } 
{\rho}_{i} (h) \bigotimes {\bf A} {}_{1,h} 
-t \sum_{h \in \Gamma } 
{\rho}_{i} (h) \bigotimes {\bf A} {}_{0,h} 
+(1-u) t^2 
( {\bf I}_{d_i} \bigotimes \tilde{{\bf D}} (D))) {}^{d_i}.
\]
$\Box$  


\section{$L$-functions of digraphs}

Let $D$ be a connected digraph with $m$ arcs, $ \Gamma $ a finite group, 
$ \alpha : A(D) \longrightarrow \Gamma $ a pseudo ordinary voltage 
assignment and ${\bf W} = {\bf W} (D)$ a weighted matrix of $D$. 
For each path $P=( e_1, \cdots, e_l)$ of $D$, set 
$ \alpha (P)= \alpha ( e_1 ) \cdots \alpha ( e_l )$ and 
$w(P)=w( e_1 ) \cdots w( e_l )$. 
Furthermore, let $ \rho $ be a representation of $ \Gamma $ 
and $d$ its degree. 

The {\em weighted Bartholdi $L$-function} of $D$ associated with 
$ \rho $ and $ \alpha $ is defined by 
\[
\zeta {}_D (w, u,t, \rho, \alpha )= 
\prod_{[C]} \det ( {\bf I}_d -w(C) \rho ( \alpha (C)) 
u^{ cbc(C) } t^{ \mid C \mid } )^{-1}, 
\]
where $[C]$ runs over all equivalence classes of prime cycles of $D$. 


Two $md \times md $ matrices 
${\bf B}^{\rho }_w =( {\bf B}_{e,f} )_{e,f \in A(D)} $ and 
${\bf J}^{\rho }_w =( {\bf J}_{e,f} )_{e,f \in A(D)} $ are defined as follows: 
\[
{\bf B}_{e,f} =\left\{
\begin{array}{ll}
w(e) \rho ( \alpha (e)) & \mbox{if $t(e)=o(f)$, } \\
{\bf 0}_d & \mbox{otherwise}
\end{array}
\right.
, 
{\bf J}_{e,f} =\left\{
\begin{array}{ll}
w(e) \rho ( \alpha (e)) & \mbox{if $f= e^{-1} $, } \\
{\bf 0}_d & \mbox{otherwise.}
\end{array}
\right.
\]

A determinant expression for the weighted Bartholdi $L$-function of $D$ associated with 
$ \rho $ and $ \alpha $ was given by Choe, Kwak, Park and Sato [3]. 
Let $1 \leq i,j \leq n$. 
Then the {\em $(i,j)$-block} ${\bf F}_{ij} $ of a $dn \times dn$ 
matrix ${\bf F} $ is the submatrix of ${\bf F} $ consisting of 
$d(i-1)+1, \ldots, di$ rows and $d(j-1)+1, \ldots, dj$ columns. 


\begin{theorem}[Choe, Kwak, Park and Sato]
Let $D$ be a connected digraph with $m$ arcs, 
$ \Gamma $ a finite group, $ \alpha : A(D) \longrightarrow \Gamma $ 
a pseudo ordinary voltage assignment and 
${\bf W} = {\bf W} (D)$ a weighted matrix of $D$. 
Furthermore, let $ \rho $ be a representation of $ \Gamma $, 
and $d$ the degree of $ \rho $. 
Then the reciprocal of the weighted Bartholdi $L$-function of $D$ associated with 
$ \rho $ and $ \alpha $ is
\[
\zeta {}_D (w,u,t, \rho, \alpha )^{-1} 
= \det ({\bf I}_{md} -( {\bf B}^{\rho }_w -(1-u) {\bf J}^{\rho }_w )t). 
\]
\end{theorem}


A new determinant expression for the weighted Bartholdi $L$-function of $D$ associated with 
$ \rho $ and $ \alpha $ is given as follows: 


\begin{theorem}
Let $D$ be a connected digraph, and 
let ${\bf W} = {\bf W} (D)$ be a weighted matrix of $D$. 
Then the reciprocal of the weighted Bartholdi $L$-function of $D$ 
is given by 
\[
\zeta {}_D (w,u,t, \rho, \alpha )^{-1} 
= \prod^{m_1}_{i=1} ( 1- w(f_i ) w(f^{-1}_i ) (1-u )^2 t^2 )^d 
\]
\[
\times 
\det ( {\bf I}_{nd} +(1-u) t^2 {\bf I}_d \bigotimes \tilde{{\bf D}} (D) 
-t \sum_{g \in \Gamma } \rho (g) \bigotimes {\bf A}_{1,g}  
-t \sum_{g \in \Gamma } \rho (g) \bigotimes {\bf A}_{0,g} ), 
\]
where $n= \mid V(D) \mid $, $m= \mid A(D) \mid $ and 
 $f^{\pm 1}_1, \ldots, f^{\pm 1}_{m_1} $ are symmetric arcs of $D$.  
\end{theorem}


{\bf Proof}.  Let $V(D)= \{ v_1, \cdots, v_{n} \} $ and, 
let $A(D)= \{ e_1, \cdots, e_{m {}_0 }, f_{1}, 
\cdots, f_{m_1}, f^{-1}_{1}, \cdots $, $f^{-1}_{m_1} \} $ 
such that $e^{-1}_i \not\in A(D) (1 \leq i \leq m_0 )$. 
Note that  $m= m_0 +2 m_1 $. 

Arrange arcs of $D$ as follows: 
\[
e_1, \cdots, e_{m_0 }, f_{1}, f^{-1}_1, 
\cdots, f_{ m_1}, f^{-1}_{m_1 }. 
\]
Let 
\[
{\bf U} =
\left[ 
\begin{array}{cccccc}
w(e_1 ) &   &   &   &   & 0 \\
   & \ddots &   &   &   &   \\
   &    & w(e_{m_0} ) &   &   &  \\
   &    &   & w(f_1)  &   &  \\ 
   &    &   &   & w(f^{-1}_1 ) &  \\
   &    &   &   &   &  \ddots 
\end{array} 
\right] 
. 
\]
Furthermore, let two $md \times md$ matrices 
${\bf B}_{\rho } =( {\bf B}^{\rho }_{e,f} )_{e,f \in A(D)} $ and 
${\bf J}_{\rho } =( {\bf J}^{\rho }_{e,f} )_{e,f \in A(D)} $ be defined as follows: 
\[
{\bf B}^{\rho }_{e,f} =\left\{
\begin{array}{ll}
\rho ( \alpha (e)) & \mbox{if $t(e)=o(f)$, } \\
{\bf 0}_d & \mbox{otherwise}
\end{array}
\right.
, 
{\bf J}^{\rho }_{e,f} =\left\{
\begin{array}{ll}
\rho ( \alpha (e)) & \mbox{if $f= e^{-1} $, } \\
{\bf 0}_d & \mbox{otherwise.}
\end{array}
\right.
\]
Then we have 
\[
( {\bf U} \bigotimes {\bf I}_d ) {\bf B}_{\rho } = {\bf B}^{\rho }_w \  and \  
( {\bf U} \bigotimes {\bf I}_d ) {\bf J}_{\rho } = {\bf J}^{\rho }_w. 
\]
Thus, 
\[
{\bf B}^{\rho }_w -(1-u) {\bf J}^{\rho }_w =( {\bf U} \bigotimes {\bf I}_d )
({\bf B}_{\rho } -(1-u) {\bf J}_{\rho } ). 
\]
By Theorem 7, it follows that 
\[
 \zeta {}_D (w,u,t, \rho, \alpha )^{-1} = \det ( {\bf I}_{md}  
-t ( {\bf U} \bigotimes {\bf I}_d )({\bf B}_{\rho } -(1-u) {\bf J}_{\rho } )). 
\]


Now, let ${\bf K} =( {\bf K}_{ev} )$ ${}_{e \in A(D); v \in V(D)} $ be the $md \times nd$ matrix defined 
as follows: 
\[
{\bf K}_{ev} :=\left\{
\begin{array}{ll}
{\bf I}_d & \mbox{if $o(e)=v$, } \\
{\bf 0}_d & \mbox{otherwise. } 
\end{array}
\right.
\]
Furthermore, we define the $md \times nd$ matrix 
${\bf L} =( {\bf L}_{ev} )_{e \in A(D); v \in V(D)} $ as follows: 
\[
{\bf L}_{ev} :=\left\{
\begin{array}{ll}
\rho ( \alpha (e)) & \mbox{if $t(e)=v$, } \\
{\bf 0}_d & \mbox{otherwise. } 
\end{array}
\right.
\] 

Set ${\bf U}_d = {\bf U} \bigotimes {\bf I}_d $. 
Then we have 
\[
{\bf L} {}^t {\bf K} = {\bf B}_{\rho }. 
\]
Thus, 
\[
\begin{array}{rcl}
\  &   & \det ( {\bf I}_{md} -t {\bf U}_d ({\bf B}_{\rho } -(1-u) {\bf J}_{\rho } )) \\  
\  &   &                \\ 
\  & = & \det ( {\bf I}_{md} -t {\bf U}_d 
({\bf L} {}^t {\bf K} -(1-u) {\bf J}_{\rho } )) 
= \det ( {\bf I}_{md} -t {\bf U}_d {\bf L} {}^t {\bf K} 
+(1-u)t {\bf U}_d {\bf J}_{\rho } ). 
\end{array}
\]


But, we have 
\[
{\bf I}_{md} +(1-u)t {\bf U}_d {\bf J}_{\rho } 
\]
\begin{equation}
= {\bf I}_{m_0 d} \oplus ( \oplus {}^{m_1 }_{j=1}
\left[ 
\begin{array}{cc}
{\bf I}_d &  (1-u)t w(f_j ) \rho ( \alpha (f_j )) \\
(1-u)t w(f^{-1}_j ) \rho ( \alpha (f^{-1}_j )) & {\bf I}_d  
\end{array} 
\right] 
) 
. 
\end{equation}
Since  $|u|,|t|$ are sufficiently small, we have 
\[
\det ( 
\left[ 
\begin{array}{cc}
{\bf I}_d &  (1-u)t w(f_j ) \rho ( \alpha (f_j )) \\
(1-u)t w(f^{-1}_j ) \rho ( \alpha (f^{-1}_j )) & {\bf I}_d 
\end{array} 
\right] 
) 
\]
\[
=(1-(1-u )^2 t^2 w(f_j ) w(f^{-1}_j ) )^d \neq 0\  (1 \leq j \leq m_1 ). 
\]
Thus, ${\bf I}_{md} +(1-u)t {\bf U}_d {\bf J}_{\rho } $ is invertible. 
Therefore,  
\[
\begin{array}{rcl}
\  &   & \det ( {\bf I}_{md} -t {\bf U}_d ({\bf B}_{\rho } -(1-u) {\bf J}_{\rho } )) \\  
\  &   &                \\ 
\  & = & \det ( {\bf I}_{md} -t {\bf U}_d {\bf L} {}^t {\bf K} 
({\bf I}_{md} +(1-u)t {\bf U}_d {\bf J}_{\rho } )^{-1} ) 
\det ( {\bf I}_{md} +(1-u)t {\bf U}_d {\bf J}_{\rho } ). 
\end{array}
\]

By (2), we have 
\[
\begin{array}{rcl}
\  &   & \det ( {\bf I}_{md} -t {\bf U}_d ({\bf B}_{\rho } -(1-u) {\bf J}_{\rho } )) \\
\  &   &                \\ 
\  & = & \det ( {\bf I}_{nd} -t \  {}^t {\bf K} 
({\bf I}_{nd} +(1-u)t {\bf U}_d {\bf J}_{\rho } )^{-1} {\bf U}_d {\bf L} ) 
\det ( {\bf I}_{md} +(1-u)t {\bf U}_d {\bf J}_{\rho } ). 
\end{array}
\]

Next, we have 
\[
\det ( {\bf I}_{md} +(1-u)t {\bf U}_d {\bf J}_{\rho } ) 
= \prod^{m_1}_{i=1} ( 1- w(f_i ) w(f^{-1}_i ) (1-u )^2 t^2 )^d. 
\]
Furthermore, the $md \times nd$ matrix 
${\bf U}_d {\bf L} =( c_{ev} )_{e \in A(D); v \in V(D)} $ is given as follows: 
\[
c_{ev} :=\left\{
\begin{array}{ll}
w(e) \rho ( \alpha (e)) & \mbox{if $t(e)=v$, } \\
0 & \mbox{otherwise. } 
\end{array}
\right.
\] 

But, we have 
\[
( {\bf I}_{md} +(1-u)t {\bf U}_d {\bf J}_{\rho } )^{-1} 
\]
\[
= {\bf I}_{m_0 d} \oplus ( \oplus {}^{m_1 }_{j=1} 
\left[ 
\begin{array}{cc}
1/ x_j {\bf I}_d &  -(1-u)t w(f_j )/ x_j  \rho ( \alpha (f_j )) \\
-(1-u)t w(f^{-1}_j )/ x_j  \rho ( \alpha (f^{-1}_j )) & 1/ x_j {\bf I}_d  
\end{array} 
\right] 
) 
. 
\]
where $x_i =1-w(f_i) w(f^{-1}_i ) (1-u )^2 t^2 \  (1 \leq i \leq m_1 )$.  

But, for a symmetric arc $(x,y) \in A(D)$, 
\[
( {}^t {\bf K} ({\bf I}_{md} +(1-u)t {\bf U}_d {\bf J}_{\rho } )^{-1} 
{\bf U}_d {\bf L} )_{xy} = w(x,y)/(1- w(x,y) w(y,x) (1-u )^2 t^2 ) \rho ( \alpha (x,y)). 
\]
For a nonsymmetric arc $(x,y) \in A(D)$, 
\[
( {}^t {\bf K} ({\bf I}_{md} +(1-u)t {\bf U}_d {\bf J}_{\rho } )^{-1} 
{\bf U}_d {\bf L} )_{xy} = w(x,y) \rho ( \alpha (x,y)).  
\]
Furthermore, if $x=y$, then 
\[
( {}^t {\bf K} ({\bf I}_{md} +(1-u)t {\bf U}_d {\bf J}_{\rho } )^{-1} 
{\bf U}_d {\bf L} )_{xx} 
=- \sum_{o(e)=x, e^{-1} \in A(D)} 
\frac{(1-u)t w(e) w(e^{-1} )}{1- w(e) w( e^{-1} ) (1-u )^2 t^2 } {\bf I}_d. 
\]

Thus, 
\[
\det ( {\bf I}_{nd} -t \  {}^t {\bf K} 
({\bf I}_{nd} +(1-u)t {\bf U}_d {\bf J}_{\rho } )^{-1} {\bf U}_d {\bf L} ) 
\]
\[
= \det ( {\bf I}_{nd} +(1-u) t^2 \tilde{{\bf D}} (D) \bigotimes {\bf I}_d 
-t \sum_{g \in \Gamma } {\bf A}_{1,g} \bigotimes \rho (g) 
-t \sum_{g \in \Gamma } {\bf A}_{0,g} \bigotimes \rho (g)), 
\]
Therefore, it follows that 
\[
\zeta {}_D (w,u,t, \rho, \alpha  )^{-1}  
= \prod^{m_1}_{i=1} ( 1- w(f_i ) w(f^{-1}_i ) (1-u )^2 t^2 )^d 
\]
\[
\times 
\det ( {\bf I}_{nd} +(1-u) t^2 {\bf I}_d \bigotimes \tilde{{\bf D}} (D) 
-t \sum_{g \in \Gamma } \rho (g) \bigotimes {\bf A}_{1,g} 
-t \sum_{g \in \Gamma } \rho (g) \bigotimes {\bf A}_{0,g} ), 
\] 
$\Box$


By Theorems 6,8, the following result holds. 


\begin{corollary}[Choe, Kwak, Park and Sato]
Let $D$ be a connected digraph, $ \Gamma $ a finite group, 
$ \alpha : A(D) \longrightarrow \Gamma $ a pseudo ordinary voltage 
assignment and ${\bf W} = {\bf W} (D)$ a weighted matrix of $D$. 
Then we have 
\[
\zeta (D {}^{ \alpha  }, \tilde{w},u,t) = \prod_{ \rho } 
\zeta {}_D (w,u,t, \rho, \alpha )^{ \deg \rho }, 
\]
where $ \rho $ runs over all inequivalent irreducible representations 
of $ \Gamma $. 
\end{corollary}


\section{Bartholdi edge zeta function of a digraph} 

Let $D$ be a connected digraph with $m$ arcs $e_1, \ldots, e_m $. 
Furthermore, let $z_1, \ldots z_{m} $ be $m$ variables. 
Set $z_{e_i } =z_i (1 \leq i \leq m)$ and ${\bf z} =( z_1, \ldots, z_m )$. 
Then the {\em Bartholdi edge zeta function} $\zeta (D,w,u)$ of $D$ is defined by 
\[
\zeta (D,{\bf z},u)= \prod_{[C]} (1-g(C) u^{cbc(C)} )^{-1}, 
\]
where $[C]$ runs over all equivalence classes of prime cycles of $D$. 
If $D=D_G$ is the symmetric digraph of a graph $G$, then 
the Bartholdi edge zeta function $\zeta (D_G,{\bf z},u)$ of $D_G$ is called the 
{\em Bartholdi edge zeta function} $\zeta (G,{\bf z},u)$ of $G$. 

Now, set $|V(D)|=n$. 
Then we define an $n \times n$ matrix  
${\bf A}^{\prime }_1 ={\bf A}^{\prime }_1 (D)=(a_{xy} )$ as follows: 
\[
a_{xy} =\left\{
\begin{array}{ll}
z_{(x,y)} /(1- z_{(x,y)} z_{(y,x)} (1-u )^2 ) & \mbox{if both $(x,y)$ and $(y,x) \in A(D)$, } \\
0 & \mbox{otherwise.  }
\end{array}
\right. 
\]
Furthermore, an $n \times n$ matrix ${\bf D}^{\prime } = {\bf D}^{\prime } (D)= (d_{xy} )$ 
is the diagonal matrix defined by 
\[
d_{xx} = \sum_{o(e)=x, e^{-1} \in A(D)} 
\frac{z_e z_{e^{-1} }}{1- z_e z_{ e^{-1} } (1-u )^2 }. 
\] 

Substituting $t=1$ in Theorem 5, we obtain the following result. 


\begin{corollary}
Let $D$ be a connected digraph with $m$ arcs and 
let ${\bf z} =( z_1, \ldots, z_m )$ be $m$ variables.
Then the reciprocal of the Bartholdi edge zeta function of $D$ 
is given by 
\[
\zeta (D,{\bf z},u )^{-1} = \det ( {\bf I}_n +(1-u) {\bf D}^{\prime }  
- {\bf A}^{\prime }_1 (D) - \tilde{{\bf A}}_0 )  
\prod^{m_1}_{i=1} ( 1- z_{f_i } z_{f^{-1}_i } (1-u )^2 ),  
\]
where  $n= \mid V(D) \mid $ and 
$f^{\pm 1}_1, \ldots, f^{\pm 1}_{m_1} $ are symmetric arcs of $D$. 
\end{corollary}


If $D=D_G$, then 

\begin{corollary}
Let $G$ be a connected graph with $m$ edges and 
let ${\bf z} =( z_1, \ldots, z_{2m} )$ be $2m$ variables.and 
let ${\bf W} = {\bf W} (G)$ be a weighted matrix of $G$. 
Then the reciprocal of the Bartholdi edge zeta function of $G$ 
is given by 
\[
\zeta (G,{\bf z},u )^{-1} = \det ( {\bf I}_n +(1-u) {\bf D}^{\prime }  
- {\bf A}^{\prime }_1 (G) - \tilde{{\bf A}}_0 )  
\prod^{m}_{i=1} ( 1- z_{f_i } z_{f^{-1}_i } (1-u )^2 ),  
\]
where  $n= \mid V(G) \mid $ and 
$D(G)= \{ f^{\pm 1}_1, \ldots, f^{\pm 1}_{m} \}$.  
\end{corollary}


\section{Example}

Finally, we give an example.
Let $D$ be the digraph with three vertices $v_1, v_2, v_3 $ and 
five arcs $( v_1,v_2 )$, $( v_2,v_1 ),( v_2,v_3 ),( v_3,v_2 ),
( v_3,v_1 )$. 
Furthermore, let 
\[
{\bf W} (D)=
\left[ 
\begin{array}{ccc}
0 & a & 0 \\
b & 0 & c \\
d & e & 0 
\end{array} 
\right] 
. 
\]
Then we have $n=3, m=5, m_1 =2$. 
By Theorem 5, we have 
\[
\begin{array}{rcl}
\  &  & {\zeta} (D,w,u,t )^{-1} \\
\  &   &                \\ 
\  & = & (1-ab(1-u )^2 t^2 )(1-ce(1-u )^2 t^2 ) 
\det ({\bf I}_3 -t \tilde{{\bf A}}_1 -t \tilde{{\bf A}}_0 
+(1- u) t^2 \tilde{{\bf D}} ) \\ 
\  &   &                \\ 
\  & = & AB 
\det \left(
\left[ 
\begin{array}{ccc}
1+abF/A & -at/A & 0 \\
-bt/A & 1+abF/A+ceF/B & -ct/B \\
-dt & -et/B & 1+ceF/B 
\end{array} 
\right] 
\right)
\\
\  &   &                \\ 
\  & = & 1-(ab+ce) u^2 t^2 +abce( u^4 -u^2 ) t^4 -acd t^3,  
\end{array}
\]
where $A=1-ab(1-u )^2 t^2 $, $B=1-ce(1-u )^2 t^2 $ and $F=(1-u) t^2 $. 

Let $\Gamma = Z_3 = \{ 1, \tau, { \tau }^2 \} ( { \tau }^3 =1)$ be 
the cyclic group of order 3, and let 
$ \alpha : A(D) \longrightarrow Z_3$ be the pseudo ordinary voltage 
assignment such that $ \alpha ( v_1, v_2 )= \tau $,
$ \alpha ( v_2, v_1 )= \tau {}^2 $ and 
$ \alpha ( v_2, v_3 )= \alpha ( v_3, v_2 )= \alpha ( v_3, v_1 )=1$. 
The characters of ${\bf Z}_3$ are given as follows: 
$\chi {}_i ( \tau {}^j )=( \xi {}^i )^j $, $ 0 \leq i, j \leq 2$, 
where $ \xi = \frac{-1+ \sqrt{-3} }{2} $. 

Now, we present the weighted Bartholdi $L$-function 
${\zeta}_D (w,u,t, \chi {}_1, \alpha )$ of $D$ 
associated with $ \chi {}_1 $ and $ \alpha $. 
Theorem 8 implies that 
\begin{align*}
%\  &  & {\zeta}_D (w,u,t, \chi {}_1, \alpha )^{-1} \\
%\  &   &                \\ 
{\zeta}_D (w,u,t, \chi {}_1, \alpha )^{-1}   & =  AB  
\det ({\bf I}_3 -t \sum^2_{i=0} \chi {}_1 ( \tau {}^i ) 
{\bf A}_{1, \tau {}^i } -t \sum^2_{i=0} \chi {}_1 ( \tau {}^i ) 
{\bf A}_{0, \tau {}^i } 
+(1-u) t^2 \tilde{{\bf D}} ) \\ 
%\  &   &                \\ 
\  & =  AB  
\det \left(
\left[ 
\begin{array}{ccc}
1+abF/A & -at \xi /A & 0 \\
-bt \xi {}^2 /A & 1+abF/A+ceF/B & -ct/B \\
-dt & -et/B & 1+ceF/B 
\end{array} 
\right] 
\right)
\\
%\  &   &                \\ 
\  & = 1-(ab+ce) u^2 t^2 +abce( u^4 -u^2 ) t^4 -acd t^3 \xi. 
\end{align*}

Similarly, we have 
\[
{\zeta}_D (w,u,t, \chi {}_2, \alpha )^{-1} = 
1-(ab+ce) u^2 t^2 +abce( u^4 -u^2 ) t^4 -acd t^3 \xi {}^2. 
\]
By Corollary 2, it follows that 
\begin{align*}
{\zeta} ( D^{\alpha}, \tilde{w}, u,t )^{-1} 
&= {\zeta} (D,w, u,t )^{-1} {\zeta}_D (w,u,t, \chi {}_1, \alpha )^{-1} 
{\zeta}_D (w,u,t, \chi {}_2, \alpha )^{-1} \\
&=(1-(ab+ce) u^2 t^2 +abce( u^4 -u^2 ) t^4 )^3 - a^3 c^3 d^3 t^9. 
\end{align*}

If $w( e^{-1} )=w(e )^{-1}$ for each symmetric arc $e \in A(D)$, then 
\[
{\zeta} (D,w,u,t )^{-1} =1-2 u^2 t^2 +( u^4 -u^2 ) t^4 -acd t^3,  
\]
\[
{\zeta}_D (w,u,t, \chi {}_i, \alpha )^{-1} 
=1-2 u^2 t^2 +( u^4 -u^2 ) t^4 -acd t^3 \xi {}^i \  (i=1,2) 
\]
and 
\[
{\zeta} ( D^{\alpha}, \tilde{w}, u,t )^{-1} 
=(1-2 u^2 t^2 +( u^4 -u^2 ) t^4 )^3 - a^3 c^3 d^3 t^9. 
\]

 
\subsection*{Acknowledgment} 

We would like to thank the referee for many valuable comments and 
helpful suggestions. 


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\end{document}