# On Spherical Slant Helices in Euclidean 3-space

Çetin Camcı, Levent Kula, Mesut Altınok  Çanakkale Onsekiz Mart University,
Faculty of Sciences and Arts
Department of Mathematics, Canakkale, Turkey<EMAIL_ADDRESS>Ahi Evran
University, Faculty of Sciences and Arts
Department of Mathematics, Kirsehir, Turkey<EMAIL_ADDRESS>Ahi Evran
University, Faculty of Sciences and Arts
Department of Mathematics, Kirsehir, Turkey<EMAIL_ADDRESS>
###### Abstract.

In this paper we consider the spherical slant helices in $\mathbb{R}^{3}$.
Moreover, we show how could be obtained to a spherical slant helix and we give
some spherical slant helix examples in Euclidean 3-space.

Key Words: Slant helix, spherical curve, geodesic curvature.

###### 2000 Mathematics Subject Classification:

53A04, 14H50.

##  1\. Introduction

In [2], A slant helix in Euclidean space $\mathbb{R}^{3}$ was defined by the
property that the principal normal makes a constant angle with a fixed
direction. Moreover, Izumiya and Takeuchi showed that $\alpha$ is a slant
helix in $\mathbb{R}^{3}$ if and only if the geodesic curvature of the
principal normal of a space curve $\alpha$ is a constant function.

In [6], Kula and Yayli have studied spherical images of tangent indicatrix and
binormal indicatrix of a slant helix and they showed that the spherical images
are spherical helix.

In [5], Kula, Ekmekci, Yayli and Ilarslan have studied the relationship
between the plane curves and slant helices in $\mathbb{R}^{3}$. They obtained
that the differential equations which are characterizations of a slant helix.

In this paper we consider the spherical slant helices in $\mathbb{R}^{3}$. We
also present the parametric slant helices, their curvatures and torsions.
Moreover, we give some slant helix examples in Euclidean 3-space.

##  2\. Preliminaries

We now recall some basic concept on classical geometry of space curves and the
definition of slant helix in $\mathbb{R}^{3}$. A curve
$\alpha:I\subset\mathbb{R}\to\mathbb{R}^{3}$ is a space curve.
$T(s)=\frac{\alpha^{\prime}(s)}{\|\alpha^{\prime}(s)\|}$ is a unit tangent
vector of $\alpha$ at $s$. If $\kappa(s)\neq 0$, then the unit principal
binormal vector $B(s)$ of the curve $\alpha$ at $s$ is given by
$\frac{\alpha^{\prime}(s)\wedge\alpha^{\prime\prime}(s)}{\|\alpha^{\prime}(s)\wedge\alpha^{\prime\prime}(s)\|}$.
The unit vector $N(s)=B(s)\wedge T(s)$ is called the unit normal vector of
$\alpha$ at $s$. For the derivatives of the Frenet frame the Serret-Frenet
formula hold:

$\displaystyle T^{\prime}(s)$ $\displaystyle=$
$\displaystyle\nu\kappa(s)N(s),$ (2.1) $\displaystyle N^{\prime}(s)$
$\displaystyle=$ $\displaystyle\nu(-\kappa(s)T(s)+\tau(s)B(s)),$
$\displaystyle B^{\prime}(s)$ $\displaystyle=$ $\displaystyle-\nu\tau(s)N(s),$

where $\kappa$ is the curvature of the curve $\alpha$ at $s$, $\tau$ is the
torsion of the curve $\alpha$ at $s$ and $\nu=\|\alpha^{\prime}\|$. For a
general parameter $s$ of a space curve $\alpha$, we can calculate $\kappa$ the
curvature and the torsion $\tau$ as follows:

$\kappa(s)=\frac{\|\alpha^{\prime}(s)\wedge\alpha^{\prime\prime}(s)\|}{\|\alpha^{\prime}(s)\|^{3}},~{}~{}~{}~{}\tau(s)=\frac{\det{\alpha^{\prime}(s),\alpha^{\prime\prime}(s),\alpha^{\prime\prime\prime}(s)}}{\|\alpha^{\prime}(s)\wedge\alpha^{\prime\prime}(s)\|^{2}}.$

######  Definition 2.1.

A curve $\alpha$ with $\kappa(s)\neq 0$ is called a slant helix if the
principal normal vector line of $\alpha$ make a constant angle with a fixed
direction [2].

######  Theorem 2.1.

$\alpha$ is a slant helix if and only if the geodesic curvature of the
spherical image of the principal normal indicatrix (N) of $\alpha$

(2.2)
$\sigma(s)=\left(\frac{\kappa^{2}}{\nu(\kappa^{2}+\tau^{2})^{\frac{3}{2}}}\left(\frac{\tau}{\kappa}\right)^{\prime}\right)(s)$

is a constant function [2].

######  Definition 2.2.

Specially, if $\kappa(s)=1$, a slant helix is called a Salkowski curve and if
$\tau(s)=1$, a slant helix is called an anti-Salkowski curve [4].

######  Theorem 2.2.

Let $\alpha$ be a space curve in $\mathbb{R}^{3}$. The following statements
are equivalent:

  1. (1)

$\alpha$ is a spherical, i.e., it is contained in a sphere of radius $r$

  2. (2)

$\left(\frac{1}{\kappa}\right)^{2}+\left(\frac{1}{\nu\tau}\left(\frac{1}{\kappa}\right)^{\prime}\right)^{2}=r^{2}.$

  3. (3)

$\frac{1}{\nu}\left(\frac{1}{\nu\tau}\left(\frac{1}{\kappa}\right)^{\prime}\right)^{\prime}+\frac{\tau}{\kappa}=0.$

  4. (4)

$\frac{1}{\kappa}=A\cos\left(\int\nu\tau ds\right)+B\sin\left(\int\nu\tau
ds\right)$

where, $A$, $B$ are constant and $\sqrt{A^{2}+B^{2}}=r$ [8].

##  3\. Slant helix and Its Projection

In this section, we investigate curvature of slant helix and projection of
slant helix in the plane.

######  Theorem 3.1.

Let $\alpha$ be a space curve in $\mathbb{R}^{3}$. The following statements
are equivalent:

  1. (1)

$\alpha$ is a slant helix.

  2. (2)

$\displaystyle\kappa$ $\displaystyle=$
$\displaystyle\frac{1}{a\nu}\theta^{\prime}\sin\theta$ $\displaystyle\tau$
$\displaystyle=$ $\displaystyle\frac{1}{a\nu}\theta^{\prime}\cos\theta.$

  3. (3)

$\vec{U}=\frac{\tau}{a\sqrt{\kappa^{2}+\tau^{2}}}~{}T+~{}N+\frac{\kappa}{a\sqrt{\kappa^{2}+\tau^{2}}}~{}B=constant$.

Where $a\neq 0$ is a constant and $\theta$ is a function of $s$ [3].

######  Theorem 3.2.

Let $\alpha$ be a slant helix with Frenet frame $\\{T,N,B\\}$, curvature
$\kappa$, torsion $\tau$ and
$\vec{a}=\cos\theta_{1}T+\cos\theta_{2}N+\cos\theta_{3}B$ be axis of slant
helix.

(3.1) $\alpha_{\pi}(s)=\alpha(s)-\langle\alpha(s),\vec{a}\rangle\vec{a}$

is a plane curve and its curvature $\kappa_{\pi}$ is

(3.2)
$\displaystyle\kappa_{\pi}=\frac{(1+a^{2})\sin\theta}{(a^{2}+\sin^{2}\theta)^{\frac{3}{2}}}\kappa.$

###### Proof.

Differentiating the eq. (3.1), we get

(3.3) $\displaystyle\alpha_{\pi}^{\prime}(s)=\nu(T-\cos\theta_{1}\vec{a}).$

Therefore

$\displaystyle\nu_{\pi}$ $\displaystyle=$
$\displaystyle\|\alpha_{\pi}^{\prime}(s)\|=\frac{ds_{\pi}}{ds}=\nu\|T-\cos\theta_{1}\vec{a}\|=\nu\sin\theta_{1}.$

If we derive eq. (3.3) again, we obtain

$\displaystyle\alpha_{\pi}^{\prime\prime}(s)=\nu^{\prime}T+\nu^{2}\kappa
N-\nu^{\prime}\cos\theta_{1}\vec{a}-\nu^{2}\kappa\cos\theta_{2}\vec{a}.$

Thus

$\displaystyle\alpha_{\pi}^{\prime}\times\alpha_{\pi}^{\prime\prime}$
$\displaystyle=$
$\displaystyle\nu^{3}\kappa(B-\cos\theta_{2}T\times\vec{a}-\cos\theta_{1}\vec{a}\times
N)$ $\displaystyle=$ $\displaystyle\nu^{3}\kappa\cos\theta_{3}\vec{a}.$

It follows

(3.4)
$\displaystyle\kappa_{\pi}=\frac{\|\alpha_{\pi}^{\prime}\times\alpha_{\pi}^{\prime\prime}\|}{\nu_{\pi}^{3}}=\frac{\cos\theta_{3}}{\sin^{3}\theta_{1}}\kappa$

and $\tau_{\pi}=0$. We can easily see that

$\displaystyle\cos\theta_{1}=-\frac{\cos\theta}{\sqrt{1+a^{2}}},~{}~{}~{}~{}~{}\cos\theta_{2}=\frac{a}{\sqrt{1+a^{2}}},~{}~{}~{}~{}~{}\cos\theta_{3}=\frac{\sin\theta}{\sqrt{1+a^{2}}}$

In this case,

$\displaystyle\sin\theta_{1}=\frac{\sqrt{a^{2}+\sin^{2}\theta}}{\sqrt{1+a^{2}}},$
(3.5)
$\displaystyle\frac{ds_{\pi}}{ds}=\nu\frac{\sqrt{a^{2}+\sin^{2}\theta}}{\sqrt{1+a^{2}}}$

and finally,

(3.6)
$\displaystyle\kappa_{\pi}=\frac{(1+a^{2})\sin\theta}{(a^{2}+\sin^{2}\theta)^{\frac{3}{2}}}\kappa.$

∎

##  4\. Spherical slant helices

In this section, we investigate parametric equation of spherical slant helix
and we give some theorem.

The axis $\vec{a}$ of slant helix could chosen axis Oz without loss of
generality. Now since the tangent and normal are orthogonal unit vectors, we
may write $T_{\pi}(s_{\pi})=(\cos\phi_{\pi},\sin\phi_{\pi})$ and
$N_{\pi}(s_{\pi})=(\sin\phi_{\pi},-\cos\phi_{\pi})$, $\phi_{\pi}(s_{\pi})$
being the angle between the $x$-axis and tangent. We observe that the
curvature has the interpretation

$\kappa_{\pi}=\phi_{\pi}^{\prime}$

i.e., it is derivative of the tangent-angle $\phi_{\pi}$ with respect to the
arc length $s_{\pi}$.

The function $\kappa_{\pi}(s_{\pi})$ specifying thr curvature in terms of arc
length along a plane curve is called the intrinsic equation of that curve. it
uniquely defines the curve. We have the explicit representation

(4.1)
$x(s)=\int\cos\phi_{\pi}ds_{\pi},~{}~{}~{}~{}~{}~{}y(s)=\int\sin\phi_{\pi}ds_{\pi}$

of the curve $\alpha(s)=(x(s),y(s),z(s))$, where

(4.2) $\phi_{\pi}=\int\kappa_{\pi}ds_{\pi}.$

[1].

Therefore, from eq. (4.2) and eq. (3.6)

(4.3) $\displaystyle\phi_{\pi}$ $\displaystyle=$
$\displaystyle-\arctan\left(\frac{\sqrt{1+a^{2}}}{a}\tan\theta\right)+\frac{\sqrt{1+a^{2}}}{a}\theta.$

and by using eq. (4.1), eq. (3.5),

$\displaystyle x(s)$ $\displaystyle=$
$\displaystyle\int\cos\phi_{\pi}ds_{\pi}$ $\displaystyle=$
$\displaystyle\int\left(\frac{a}{\sqrt{\sin^{2}\theta+a^{2}}}\cos\theta\cos\left[\frac{\sqrt{1+a^{2}}}{a}\theta\right]+\frac{\sqrt{1+a^{2}}}{\sqrt{\sin^{2}\theta+a^{2}}}\sin\theta\sin\left[\frac{\sqrt{1+a^{2}}}{a}\theta\right]\right)ds_{\pi}$
$\displaystyle=$
$\displaystyle\nu\int\left(\frac{a}{\sqrt{1+a^{2}}}\cos\theta\cos\left[\frac{\sqrt{1+a^{2}}}{a}\theta\right]+\sin\theta\sin\left[\frac{\sqrt{1+a^{2}}}{a}\theta\right]\right)ds.$

Then from the fourth equation of Theorem 2.2 and the second equation of
Theorem 3.1, we have

(4.4)
$ds=\frac{1}{a\nu}\left(A\sin\theta\cos\left[\frac{\sin\theta}{a}\right]+B\sin\theta\sin\left[\frac{\sin\theta}{a}\right]\right)d\theta.$

Therefore, by using eq. (4.4),

$\displaystyle x(s)$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{1+a^{2}}}(\sqrt{1+a^{2}}\cos\theta\sin\left[\frac{\sqrt{1+a^{2}}}{a}\theta\right]\left(B\cos\left[\frac{\sin\theta}{a}\right]-A\sin\left[\frac{\sin\theta}{a}\right]\right){}$
$\displaystyle{}-\cos\left[\frac{\sqrt{1+a^{2}}}{a}\theta\right]\left(\cos\left[\frac{\sin\theta}{a}\right](A+aB\sin\theta)+(B-aA\sin\theta)\sin\left[\frac{\sin\theta}{a}\right]\right)),$

And similarly,

$\displaystyle y(s)$ $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{1+a^{2}}}(\sqrt{1+a^{2}}\cos\theta\cos\left[\frac{\sqrt{1+a^{2}}}{a}\theta\right]\left(B\cos\left[\frac{\sin\theta}{a}\right]-A\sin\left[\frac{\sin\theta}{a}\right]\right){}$
$\displaystyle{}+\sin\left[\frac{\sqrt{1+a^{2}}}{a}\theta\right]\left(\cos\left[\frac{\sin\theta}{a}\right](A+aB\sin\theta)+(B-aA\sin\theta)\sin\left[\frac{\sin\theta}{a}\right]\right),$

Moreover, since

(4.5)
$\displaystyle\langle\alpha^{\prime\prime}(s),\vec{a}\rangle=z^{\prime\prime}(s)=-\nu^{\prime}\frac{\cos\theta}{\sqrt{1+a^{2}}}+\nu^{2}\frac{a}{\sqrt{1+a^{2}}}\kappa,$

we find that

$\displaystyle
z(s)=\frac{1}{\sqrt{1+a^{2}}}\cos\left[\frac{\sin\theta}{a}\right](-aA+B\sin\theta)-\sin\left[\frac{\sin\theta}{a}\right](aB+A\sin\theta).$

where $A^{2}+B^{2}=1$, $a$ is constant and $\theta=\theta(s)$.

As a result of the above findings, we can give the following theorem.

######  Theorem 4.1.

Under the above notation, $\alpha$ is a spherical slant helix. Moreover, all
spherical slant helix can be constructed by above method.

###### Proof.

Let $\alpha$ be a space curve with Frenet frame $\\{T,N,B\\}$, curvature
$\kappa$ and torsion $\tau$.

In this case, we will show that $\sigma$ is constant and $\alpha\in S^{2}$.

By simple calculation, spherical indicatricies $T(s)$, $N(s)$, $B(s)$ of the
curve $\alpha$, respectively, are

$\displaystyle T(s)$ $\displaystyle=$
$\displaystyle(\frac{a\cos\theta\cos\left[\frac{\sqrt{1+a^{2}}}{a}\theta\right]}{\sqrt{1+a^{2}}}+\sin\theta\sin\left[\frac{\sqrt{1+a^{2}}}{a}\theta\right],$
$\displaystyle{}\frac{a\cos\theta\sin\left[\frac{\sqrt{1+a^{2}}}{a}\theta\right]}{\sqrt{1+a^{2}}}-\sin\theta\cos\left[\frac{\sqrt{1+a^{2}}}{a}\theta\right],-\frac{\cos\theta}{\sqrt{1+a^{2}}}),$
$\displaystyle N(s)$ $\displaystyle=$
$\displaystyle(\frac{\cos\left[\frac{\sqrt{1+a^{2}}}{a}\theta\right]}{\sqrt{1+a^{2}}},\frac{\sin\left[\frac{\sqrt{1+a^{2}}}{a}\theta\right]}{\sqrt{1+a^{2}}},\frac{a}{\sqrt{1+a^{2}}}),$
$\displaystyle B(s)$ $\displaystyle=$
$\displaystyle(-\frac{a\sin\theta\cos\left[\frac{\sqrt{1+a^{2}}}{a}\theta\right]}{\sqrt{1+a^{2}}}+\cos\theta\sin\left[\frac{\sqrt{1+a^{2}}}{a}\theta\right],$
$\displaystyle{}-\frac{a\sin\theta\sin\left[\frac{\sqrt{1+a^{2}}}{a}\theta\right]}{\sqrt{1+a^{2}}}-\cos\theta\cos\left[\frac{\sqrt{1+a^{2}}}{a}\theta\right],\frac{\sin\theta}{\sqrt{1+a^{2}}}).$

And also, by the formulae of the curvature and the torsion for a general
parameter, we can calculate that

$\displaystyle\kappa(s)$ $\displaystyle=$
$\displaystyle\frac{1}{|A\cos\left[\frac{\sin\theta}{a}\right]+B\sin\left[\frac{\sin\theta}{a}\right]|},$
$\displaystyle\tau(s)$ $\displaystyle=$
$\displaystyle\frac{\cot\theta}{A\cos\left[\frac{\sin\theta}{a}\right]+B\sin\left[\frac{\sin\theta}{a}\right]}.$

Therefore, by using eq. (2.2)

$\sigma(s)=-a=constant$

which means that $\alpha$ is a slant helix.

Finally,

$\|\alpha\|=A^{2}+B^{2}=1.$

Then, $\alpha$ is spherical slant helix. Thus the proof of theorem is
completed. ∎

######  Proposition 4.1.

If $\frac{\sqrt{1+a^{2}}}{a}$ is a rational number, then spherical slant helix
$\alpha$ is closed curve. If $\frac{\sqrt{1+a^{2}}}{a}$ is not a rational
number, then the curve spherical slant helix $\alpha$ never closes.

######  Definition 4.1.

Let $\alpha$ be a spherical curve. Let us denote
$T(s)=\nu(s)\alpha^{\prime}(s)$, and we call $T$ a unit tangent vector of
$\alpha$ at $s$. We now set a vector $Y(s)=\alpha(s)\wedge T(s)$. By
definition, we have an orthonormal frame $\\{\alpha(s),T(s),Y(s)\\}$. This
frame is called the Sabban frame of $\alpha$ [7].

######  Theorem 4.2.

The indicatrix $Y$ of the spherical slant helix $\alpha$ is a spherical slant
helix.

###### Proof.

Let $Y$ be curve with Frenet frame $\\{T^{Y},N^{Y},B^{Y}\\}$, curvature
$\kappa^{Y}$, torsion $\tau^{Y}$ and the geodesic curvature of the spherical
image of the principal normal indicatrix $(N^{Y})$ of $Y$ be $\sigma^{Y}$.
Then we have

$\displaystyle Y^{\prime}(s)$ $\displaystyle=$
$\displaystyle\nu\kappa\alpha\wedge N,$ $\displaystyle Y^{\prime\prime}(s)$
$\displaystyle=$ $\displaystyle(\nu\kappa)^{\prime}\alpha\wedge
N+\nu^{2}\kappa B-\nu^{2}\kappa^{2}\alpha\wedge
T+\nu^{2}\kappa\tau\alpha\wedge B,$ $\displaystyle Y^{\prime\prime}(s)$
$\displaystyle=$ $\displaystyle(\nu\kappa)^{\prime\prime}\alpha\wedge
N+(\nu\kappa)^{\prime}\nu B+\nu\tau(\nu\kappa)^{\prime}\alpha\wedge
B-\nu\kappa(\nu\kappa)^{\prime}\alpha\wedge T$ $\displaystyle+$
$\displaystyle(\nu^{2}\kappa)^{\prime}B-\nu^{3}\kappa\tau
N-(\nu^{2}\kappa^{2})^{\prime}\alpha\wedge T-\nu^{3}\kappa^{3}\alpha\wedge N$
$\displaystyle+$ $\displaystyle(\nu^{2}\kappa\tau)^{\prime}\alpha\wedge
B-\nu^{3}\kappa\tau N-\nu^{3}\kappa\tau^{2}\alpha\wedge N.$

By the formulae the curvature and the torsion for a general parameter, we can
calculate that

$\displaystyle\kappa^{Y}$ $\displaystyle=$
$\displaystyle\frac{\kappa}{\sqrt{\kappa^{2}-1}},$ $\displaystyle\tau^{Y}$
$\displaystyle=$ $\displaystyle\frac{\tau}{\sqrt{\kappa^{2}-1}}.$

Moreover, $\sigma^{Y}(s)=a$, so that $Y(s)$ is a spherical slant helix. ∎

######  Remark 4.1.

There are not spherical Salkowski and anti-Salkowski curve.

###### Proof.

We choose that $\alpha$ is unit speed Salkowski curve without loss of
generality. Since $\kappa(s)=1$ and by using

$\left(\frac{1}{\tau}\left(\frac{1}{\kappa}\right)^{\prime}\right)^{\prime}+\frac{\tau}{\kappa}=0,$

we have $\tau(s)=0$. Therefore, $\alpha$ is plane curve and is not spherical
slant helix.

Similarly, we choose that $\alpha$ is unit speed Anti-Salkowski curve without
loss of generality. Then, by simple calculation, we have $\sigma(s)$ is not
constant which means that $\alpha$ is not spherical slant curve. ∎

##  5\. Example

In this section we give two example of spherical slant helices in Euclidean
$3$-space and draw its pictures and its tangent indicatrix, normal indicatrix,
and binormal indicatrix by using Mathematica.

######  Example 5.1.

For $a=1$, $A=1$, $B=0$, We consider a spherical slant helix $\alpha$ is
defined by

$\displaystyle x(s)$ $\displaystyle=$
$\displaystyle-\cos\theta\sin(\sqrt{2}\theta)\sin(\sin\theta){}$
$\displaystyle{}+\frac{1}{\sqrt{2}}\cos(\sqrt{2}\theta)(-\cos(\sin\theta))+\sin\theta\sin(\sin\theta))$
$\displaystyle y(s)$ $\displaystyle=$
$\displaystyle\cos\theta\cos(\sqrt{2}\theta)\sin(\sin\theta){}$
$\displaystyle{}+\frac{1}{\sqrt{2}}\sin(\sqrt{2}\theta)(-\cos(\sin\theta))+\sin\theta\sin(\sin\theta))$
$\displaystyle z(s)$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}(-\cos(\sin\theta)+\sin\theta\sin(\sin\theta)).$

The picture of the curve $\alpha$ is rendered in Figure 1.

The parametrization of the tangent indicatrix
$T=\left(T_{1},T_{2},T_{3}\right)$ of the spherical slant helix $\alpha$ is

$\displaystyle T_{1}(s)$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\cos\theta\cos(\sqrt{2}\theta)+\sin(\sqrt{2}\theta)\sin\theta,$
$\displaystyle T_{2}(s)$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\sin\theta\cos(\sqrt{2}\theta)-\sin(\sqrt{2}\theta)\cos\theta,$
$\displaystyle T_{3}(s)$ $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{2}}\cos\theta.$

The picture of the tangent indicatrix is rendered in Figure 2 (a).

The parametrization of the normal indicatrix
$N=\left(N_{1},N_{2},N_{3}\right)$ of the spherical slant helix $\alpha$ is

$\displaystyle N_{1}(s)$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\cos(\sqrt{2}\theta),$ $\displaystyle
N_{2}(s)$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\sin(\sqrt{2}\theta),$ $\displaystyle
N_{3}(s)$ $\displaystyle=$ $\displaystyle\frac{1}{\sqrt{2}}.$

The picture of the normal indicatrix is rendered in Figure 2 (b).

The parametrization of the binormal indicatrix
$B=\left(B_{1},B_{2},B_{3}\right)$ of the spherical slant helix $\alpha$ is

$\displaystyle B_{1}(s)$ $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{2}}\sin\theta\cos(\sqrt{2}\theta)+\sin(\sqrt{2}\theta)\cos\theta,$
$\displaystyle B_{2}(s)$ $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{2}}\sin\theta\sin(\sqrt{2}\theta)-\cos(\sqrt{2}\theta)\cos\theta,$
$\displaystyle B_{3}(s)$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\sin\theta.$

The picture of the binormal indicatrix is rendered in Figure 2 (c).

######  Example 5.2.

For $A=1$, $B=0$ and $a=2$, $a=3$, $a=4$, the picture of the spherical slant
helix $\alpha$, respectively, is rendered in Figure 4.

######  Example 5.3.

For $a=1$, $A=1$, $B=0$, We consider a spherical slant helix $Y$ is defined by

$\displaystyle x(s)$ $\displaystyle=$
$\displaystyle-\cos\theta\sin(\sqrt{2}\theta)\cos(\sin\theta){}$
$\displaystyle{}+\frac{1}{\sqrt{2}}\cos(\sqrt{2}\theta)(\cos(\sin\theta))\sin\theta+\sin(\sin\theta))$
$\displaystyle y(s)$ $\displaystyle=$
$\displaystyle\cos\theta\cos(\sqrt{2}\theta)\cos(\sin\theta){}$
$\displaystyle{}+\frac{1}{\sqrt{2}}\sin(\sqrt{2}\theta)(\cos(\sin\theta))\sin\theta+\sin(\sin\theta))$
$\displaystyle z(s)$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}(-\cos(\sin\theta)\sin\theta+\sin(\sin\theta)).$

The picture of the curve $\alpha$ is rendered in Figure 4.

The parametrization of the tangent indicatrix
$T^{Y}=\left(T^{Y}_{1},T^{Y}_{2},T^{Y}_{3}\right)$ of the spherical slant
helix $Y$ is

$\displaystyle T^{Y}_{1}(s)$ $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{2}}\cos\theta\cos(\sqrt{2}\theta)-\sin(\sqrt{2}\theta)\sin\theta,$
$\displaystyle T^{Y}_{2}(s)$ $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{2}}\cos\theta\sin(\sqrt{2}\theta)+\cos(\sqrt{2}\theta)\sin\theta,$
$\displaystyle T^{Y}_{3}(s)$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\cos\theta.$

The picture of the tangent indicatrix is rendered in Figure 5 (a).

The parametrization of the normal indicatrix
$N^{Y}=\left(N^{Y}_{1},N^{Y}_{2},N^{Y}_{3}\right)$ of the spherical slant
helix $Y$ is

$\displaystyle N^{Y}_{1}(s)$ $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{2}}\cos(\sqrt{2}\theta),$ $\displaystyle
N^{Y}_{2}(s)$ $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{2}}\sin(\sqrt{2}\theta),$ $\displaystyle
N^{Y}_{3}(s)$ $\displaystyle=$ $\displaystyle-\frac{1}{\sqrt{2}}.$

The picture of the normal indicatrix is rendered in Figure 5 (b).

The parametrization of the binormal indicatrix
$B^{Y}=\left(B^{Y}_{1},B^{Y}_{2},B^{Y}_{3}\right)$ of the spherical slant
helix $Y$ is

$\displaystyle B^{Y}_{1}(s)$ $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{2}}\sin\theta\cos(\sqrt{2}\theta)+\sin(\sqrt{2}\theta)\cos\theta,$
$\displaystyle B^{Y}_{2}(s)$ $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{2}}\sin\theta\sin(\sqrt{2}\theta)-\cos(\sqrt{2}\theta)\cos\theta,$
$\displaystyle B^{Y}_{3}(s)$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\sin\theta.$

The picture of the binormal indicatrix is rendered in Figure 5 (c).

## References

  * [1] Farouki, R.T., Pythagorean-Hodograph Curves: Algebra and Geometry Inseparable. Springer, Berlin (2008).
  * [2] Izumiya, S. and Tkeuchi, N., New special curves and developable surfaces, Turk J. Math 28 (2004), 153-163.
  * [3] Menninger, T., An explicit parametrization of the Frenet apparatus of the slant helix. http://arxiv.org/abs/1302.3175v3, (2013).
  * [4] Monterde, J., Salkowski curves revisited: A family of curves with constant curvature and non-constant torsion. Computer Aided Geometric Design, 26 (2009), 271 278.
  * [5] Kula, L., Ekmekci, N., Yayli, Y. and Ilarslan, K., Characterizations of slant helices in Euclidean 3-space. Turkish J. Math., 34 (2010), 261-274.
  * [6] Kula, L. and Yayli, Y., On slant helix and its spherical indicatrix. Applied Mathematics and Computation, 169 (2005), 600-607.
  * [7] Koenderink, J., Solid shape. MIT Press, Cambridge, MA (1990).
  * [8] Wong, Y. C., On an explicit characterization of spherical curves. Proceeding of the American Mathematical Society, 34 (1972), no.1, 239-242.

Figure 1. For $a=1$, $A=1$, $B=0$, spherical slant helix $\alpha$.

Figure 2. For $a=1$, $A=1$, $B=0$, tangent indicatrix of the spherical slant
helix $\alpha$ (a), normal indicatrix of the spherical slant helix $\alpha$
(b) and binormal indicatrix of the spherical slant helix $\alpha$ (c).

Figure 3. For $A=1$, $B=0$ and $a=2$, the spherical slant helix $\alpha$ (a),
For $A=1$, $B=0$ and $a=3$, the spherical slant helix $\alpha$ (b) and For
$A=1$, $B=0$ and $a=4$, the spherical slant helix $\alpha$ (c). Figure 4. For
$a=1$, $A=1$, $B=0$, spherical slant helix $Y$.

Figure 5. For $a=1$, $A=1$, $B=0$, tangent indicatrix of the spherical slant
helix $\alpha$ (a), normal indicatrix of the spherical slant helix $\alpha$
(b) and binormal indicatrix of the spherical slant helix $\alpha$ (c).


# THE SCATTERING MATRIX WITH RESPECT TO AN HERMITIAN MATRIX OF A GRAPH

Takashi KOMATSU
Department of Bioengineering, School of Engineering,
The University of Tokyo
Bunkyo, Tokyo, 113-8656, JAPAN
e-mail<EMAIL_ADDRESS>
Norio KONNO
Department of Applied Mathematics, Faculty of Engineering,
Yokohama National University
Hodogaya, Yokohama 240-8501, JAPAN
e-mail<EMAIL_ADDRESS>
Iwao SATO
Oyama National College of Technology
Oyama, Tochigi 323-0806, JAPAN
e-mail<EMAIL_ADDRESS>
###### Abstract

Recently, Gnutzmann and Smilansky [5] presented a formula for the bond
scattering matrix of a graph with respect to a Hermitian matrix. We present
another proof for this Gnutzmann and Smilansky’s formula by a technique used
in the zeta function of a graph. Furthermore, we generalize Gnutzmann and
Smilansky’s formula to a regular covering of a graph. Finally, we define an
$L$-fuction of a graph, and present a determinant expression. As a corollary,
we express the generalization of Gnutzmann and Smilansky’s formula to a
regular covering of a graph by using its $L$-functions.

2000 Mathematical Subject Classification: 05C50, 15A15.
Key words and phrases : Hermitian matrix, scattering matrix, zeta function

The contact author for correspondence:

Iwao Sato

Oyama National College of Technology, Oyama, Tochigi 323-0806, JAPAN

Tel: +81-285-20-2176

Fax: +81-285-20-2880

E-mail<EMAIL_ADDRESS>
##  1 Introduction

Ihara zeta functions of graphs started from Ihara zeta functions of regular
graphs by Ihara [9]. Originally, Ihara presented $p$-adic Selberg zeta
functions of discrete groups, and showed that its reciprocal is a explicit
polynomial. Serre [13] pointed out that the Ihara zeta function is the zeta
function of the quotient $T/\Gamma$ (a finite regular graph) of the one-
dimensional Bruhat-Tits building $T$ (an infinite regular tree) associated
with $GL(2,k_{p})$.

A zeta function of a regular graph $G$ associated with a unitary
representation of the fundamental group of $G$ was developed by Sunada [16,
17]. Hashimoto [8] treated multivariable zeta functions of bipartite graphs.
Bass [1] generalized Ihara’s result on the Ihara zeta function of a regular
graph to an irregular graph, and showed that its reciprocal is again a
polynomial. Various proofs of Bass’ Theorem were given by Stark and Terras
[15], Foata and Zeilberger [3], Kotani and Sunada [10]. Sato [12] defined the
second weighted zeta function of a graph by using not an infinite product but
a determinant.

The spectral determinant of the Laplacian on a quantum graph is closely
related to the Ihara zeta function of a graph(see [2, 4, 5, 14]). Smilansky
[14] considered spectral zeta functions and trace formulas for (discrete)
Laplacians on ordinary graphs, and expressed some determinant on the bond
scattering matrix of a graph $G$ by using the characteristic polynomial of its
Laplacian. Recently, Gnutzmann and Smilansky [5] presented a formula for the
bond scattering matrix of a graph with respect to a Hermitian matrix.

In this paper, we another proof for the Gnutzmann and Smilansky’s formula on
the bond scattering matrix of a graph with respect to a Hermitian matrix. by a
technique used in the zeta function of a graph, and treat some related topics.
In Section 2, we review the Ihara zeta function and the bond scattering matrix
of a graph $G$. In Section 3, we present another proof for the Gnutzmann and
Smilansky’s formula by a technique used in the zeta function of a graph. In
Section 4, we we express a new zeta function of $G$ on the bond scattering
matrix of $G$ with respect to a Hermitian matrix by using the Euler product.
In Section 5, we generalize the Gnutzmann and Smilansky’s formula to a regular
covering of $G$. In Section 6, we define an $L$-fuction of $G$, and present
its determinant expression. As a corollary, we express the generalization of
the Gnutzmann and Smilansky’s formula to a regular covering of $G$ by using
its $L$-functions.

##  2 The zeta functions and the bond scattering matrix of a graph

Graphs 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 $b=(u,v)\in D(G)$, set $u=o(b)$
and $v=t(b)$. Furthermore, let $b^{-1}=(v,u)$ be the inverse of $b=(u,v)$.

A path $P$ of length $n$ in $G$ is a sequence $P=(b_{1},\cdots,b_{n})$ of $n$
arcs such that $b_{i}\in D(G)$, $t(b_{i})=o(b_{i+1})(1\leq i\leq n-1)$, where
indices are treated $mod\ n$. Set $\mid P\mid=n$, $o(P)=o(b_{1})$ and
$t(P)=t(b_{n})$. Also, $P$ is called an $(o(P),t(P))$-path. We say that a path
$P=(b_{1},\cdots,b_{n})$ has a backtracking or back-scatter if
$b_{i+1}=b^{-1}_{i}$ for some $i(1\leq i\leq n-1)$. A $(v,w)$-path is called a
$v$-cycle (or $v$-closed path) if $v=w$. The inverse cycle of a cycle
$C=(b_{1},\cdots,b_{n})$ is the cycle
$C^{-1}=(\hat{b}_{n},\cdots,\hat{b}_{1})$.

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
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 power
of $B$. A cycle $C$ is reduced if both $C$ and $C^{2}$ have no backtracking.
Furthermore, a cycle $C$ is prime if it is not a power of a strictly smaller
cycle. Note that each equivalence class of prime, reduced cycles of a graph
$G$ corresponds to a unique conjugacy class of the fundamental group
$\pi{}_{1}(G,u)$ of $G$ at a vertex $u$ of $G$. Furthermore, an equivalence
class of prime cycles of a graph $G$ is called a primitive periodic orbit of
$G$(see [14]).

The Ihara zeta function of a graph $G$ is a function of a complex variable $t$
with $\mid t\mid$ sufficiently small, defined by

${\bf Z}(G,t)={\bf Z}_{G}(t)=\prod_{[p]}(1-t^{\mid p\mid})^{-1},$

where $[p]$ runs over all equivalence classes of prime, reduced cycles of
$G$(see [9]).

######  Theorem 1 (Ihara; Bass)

Let $G$ be a connected graph. Then the reciprocal of the Ihara zeta function
of $G$ is given by

${\bf Z}(G,t)^{-1}=(1-t^{2})^{r-1}\det({\bf I}-t{\bf A}(G)+t^{2}({\bf D}-{\bf
I})),$

where $r$ and ${\bf A}(G)$ are the Betti number and the adjacency matrix of
$G$, respectively, and ${\bf D}=(d_{ij})$ is the diagonal matrix with
$d_{ii}=v_{i}=\deg u_{i}$ where $V(G)=\\{u_{1},\cdots,u_{n}\\}$.

Let $G$ be a connected graph and $V(G)=\\{u_{1},\cdots,u_{n}\\}$. Then we
consider an $n\times n$ matrix ${\bf W}=(w_{ij})_{1\leq i,j\leq n}$ with $ij$
entry the complex variable $w_{ij}$ if $(u_{i},u_{j})\in D(G)$, and $w_{ij}=0$
otherwise. The matrix ${\bf W}={\bf W}(G)$ is called the weighted matrix of
$G$. Furthermore, let $w(u_{i},u_{j})=w_{ij},\ u_{i},u_{j}\in V(G)$ and
$w(b)=w_{ij},b=(u_{i},u_{j})\in D(G)$. For each path
$P=(e_{i_{1}},\cdots,e_{i_{r}})$ of $G$, the norm $w(P)$ of $P$ is defined as
follows: $w(P)=w(e_{i_{1}})w(e_{i_{2}})\cdots w(e_{i_{r}})$.

Let $G$ be a connected graph with $n$ vertices and $m$ unoriented edges, and
${\bf W}={\bf W}(G)$ a weighted matrix of $G$. Two $2m\times 2m$ matrices
${\bf B}={\bf B}(G)=({\bf B}_{e,f})_{e,f\in R(G)}$ and ${\bf J}_{0}={\bf
J}_{0}(G)=({\bf J}_{e,f})_{e,f\in R(G)}$ are defined as follows:

${\bf B}_{e,f}=\left\\{\begin{array}[]{ll}w(f)&\mbox{if $t(e)=o(f)$, }\\\
0&\mbox{otherwise, }\end{array}\right.\ {\bf
J}_{e,f}=\left\\{\begin{array}[]{ll}1&\mbox{if $f=\hat{e}$, }\\\
0&\mbox{otherwise.}\end{array}\right.$

Then the second weighted zeta function of $G$ is defined by

${\bf Z}_{1}(G,w,t)=\det({\bf I}_{n}-t({\bf B}-{\bf J}_{0}))^{-1}.$

If $w(e)=1$ for any $e\in D(G)$, then the zeta function of $G$ is the Ihara
zeta function of $G$.

######  Theorem 2 (Sato)

Let $G$ be a connected graph, and let ${\bf W}={\bf W}(G)$ be a weighted
matrix of $G$. Then the reciprocal of the second weighted zeta function of $G$
is given by

${\bf Z}_{1}(G,w,t)^{-1}=(1-t^{2})^{m-n}\det({\bf I}_{n}-t{\bf
W}(G)+t^{2}(\tilde{{\bf D}}-{\bf I}_{n})),$

where $n=\mid V(G)\mid$, $m=\mid E(G)\mid$ and $\tilde{{\bf D}}=(d_{ij})$ is
the diagonal matrix with $d_{ii}=\sum_{o(b)=u_{i}}w(e)$,
$V(G)=\\{u_{1},\cdots,u_{n}\\}$.

Next, we state the bond scattering matrix of a graph. Let $G$ be a connected
graph with $n$ vertices and $m$ edges, $V(G)=\\{u_{1},\ldots,u_{n}\\}$ and
$D(G)=\\{b_{1},\ldots,b_{m},b_{m+1},\ldots,b_{2m}\\}$ such that
$b_{m+j}=b^{-1}_{j}(1\leq j\leq m)$. The Laplacian (matrix) ${\bf L}={\bf
L}(G)$ of $G$ is defined by

${\bf L}={\bf L}(G)=-{\bf A}(G)+{\bf D}.$

Let $\lambda$ be a eigenvalue of ${\bf L}$ and
$\psi=(\psi{}_{1},\ldots,\psi{}_{n})$ the eigenvector corresponding to
$\lambda$. For each arc $b=(u_{j},u_{l})$, one associates a bond wave function

$\psi{}_{b}(x)=a_{b}{\rm e}^{i\pi x/4}+a_{b^{-1}}{\rm e}^{-i\pi x/4},\ x=\pm
1$

under the condition

$\psi{}_{b}(1)=\psi{}_{j},\psi{}_{b}(-1)=\psi{}_{l}.$

We consider the following three conditions:

  1. 1.

uniqueness: The value of the eigenvector at the vertex $u_{j}$, $\psi{}_{j}$,
computed in the terms of the bond wave functions is the same for all the arcs
emanating from $u_{j}$.

  2. 2.

$\psi$ is an eigenvector of ${\bf L}$;

  3. 3.

consistency: The linear relation between the incoming and the outgoing
coefficients (1) must be satisfied simultaneously at all vertices.

By the uniqueness, we have

$a_{b_{1}}{\rm e}^{i\pi/4}+a_{b^{-1}_{1}}{\rm e}^{-i\pi/4}=a_{b_{2}}{\rm
e}^{i\pi/4}+a_{b^{-1}_{2}}{\rm e}^{-i\pi/4}=\cdots=a_{b_{d_{j}}}{\rm
e}^{i\pi/4}+a_{b^{-1}_{d_{j}}}{\rm e}^{-i\pi/4},$

where $b_{1},b_{2},\ldots,b_{d_{j}}$ are arcs emanating from $u_{j}$, and
$d_{j}=\deg u_{j}$, $i=\sqrt{-1}$.

By the condition 2, we have

$-\sum^{d_{j}}_{k=1}(a_{b_{k}}{\rm e}^{-i\pi/4}+a_{b^{-1}_{k}}{\rm
e}^{i\pi/4})=(\lambda-v_{j})\frac{1}{v_{j}}\sum^{d_{j}}_{k=1}(a_{b_{k}}{\rm
e}^{i\pi/4}+a_{b^{-1}_{k}}{\rm e}^{-i\pi/4}).$

Thus, for each arc $b$ with $o(b)=u_{j}$,

$a_{b}=\sum_{t(c)=u_{j}}\sigma{}^{(u_{j})}_{b,c}(\lambda)a_{c},$ (1)

where

$\sigma{}^{(u_{j})}_{b,c}(\lambda)=i(\delta{}_{b^{-1},c}-\frac{2}{d_{j}}\frac{1}{1-i(1-\lambda/d_{j})}),$

and $\delta{}_{b^{-1},c}$ is the Kronecker delta. The bond scattering matrix
${\bf U}(\lambda)=(U_{ef})_{e,f\in D(G)}$ of $G$ is defined by

$U_{ef}=\left\\{\begin{array}[]{ll}\sigma^{(t(f))}_{e,f}&\mbox{if $t(f)=o(e)$,
}\\\ 0&\mbox{otherwise}\end{array}\right.$

By the consistency, we have

${\bf U}(\lambda){\bf a}={\bf a},$

where ${\bf a}={}^{t}(a_{b_{1}},a_{b_{2}},\ldots,a_{b_{2m}})$. This holds if
and only if

$\det({\bf I}_{2m}-{\bf U}(\lambda))=0.$

######  Theorem 3 (Smilansky)

Let $G$ be a connected graph with $n$ vertices and $m$ edges. Then the
characteristic polynomial of the bond scattering matrix of $G$ is given by

$\det({\bf I}_{2m}-{\bf U}(\lambda))=\frac{2^{m}i^{n}\det(\lambda{\bf
I}_{n}+{\bf A}(G)-{\bf D})}{\prod^{n}_{j=1}(d_{j}-id_{j}+\lambda
i)}=\prod_{[p]}(1-a_{p}(\lambda)),$

where $[p]$ runs over all primitive periodic orbits of $G$, and

$a_{p}(\lambda)=\sigma{}^{(t(b_{n}))}_{b_{1},b_{n}}\sigma{}^{(t(b_{n-1}))}_{b_{n},b_{n-1}}\cdots\sigma{}^{(t(b_{1}))}_{b_{2},b_{1}},\
p=(b_{1},b_{2},\ldots,b_{n})$

Mizuno and Sato [11] presented another proof for this Smilansky’s formula by
using the determinant expression of the second weighted zeta function of a
graph.

##  3 The scattering matrix of a graph with respect to a Hermitian matrix

Let $G$ be a connected graph with $n$ vertices and $m$ edges,
$V(G)=\\{1,\ldots,n\\}$ and
$D(G)=\\{e_{1},\ldots,e_{m},e_{m+1},\ldots,e_{2m}\\}$ such that
$e_{m+j}=e^{-1}_{j}(1\leq j\leq m)$. Furthermore, let an Hermitian matrix
${\bf H}={\bf H}(G)=(H_{uv})_{u,v\in V(G)}$ be given as follows:

$H_{uv}=\left\\{\begin{array}[]{ll}h_{f}{\rm e}^{2i\gamma{}_{f}}&\mbox{if
$f=(u,v)\in D(G)$, }\\\ 0&\mbox{otherwise, }\end{array}\right.$

where, for each $f\in D(G)$,

$h_{f}=h_{f^{-1}}\geq 0\ and\
\gamma{}_{f}=-\gamma{}_{f^{-1}}\in[-\pi/2,\pi/2].$

If $H_{uv}=H_{f}$ is real and negative, then we choose $\gamma{}_{f}=\pi/2$ if
$u\geq v$ and $\gamma{}_{f}=-\pi/2$ if $u<v$. Set

$h(u,v)=h_{uv}=h_{f}\ and\ \gamma(u,v)=\gamma{}_{uv}=\gamma{}_{f}\ for\
f=(u,v)\in D(G).$

Now, let $\lambda$ be an eigenvalue of ${\bf H}$ and
$\psi=(\psi{}_{1},\ldots,\psi{}_{n})$ the eigenvector corresponding to
$\lambda$. For each arc $b=(u,v)$, one associates a bond wave function

$\psi{}_{b}(x)=\frac{{\rm e}^{i\gamma{}_{b}}}{\sqrt{h_{b}}}(a_{b^{-1}}{\rm
e}^{i\pi x/4}+a_{b}{\rm e}^{-i\pi x/4}),\ x=\pm 1$

under the condition

$\psi{}_{b}(1)=\psi{}_{u},\psi{}_{b}(-1)=\psi{}_{v}.$

We consider the following three conditions:

  1. 1.

uniqueness: The value of the eigenvector at the vertex $u$, $\psi{}_{u}$,
computed in the terms of the bond wave functions is the same for all the arcs
emanating from $u$.

  2. 2.

$\psi$ is an eigenvector of ${\bf H}$;

  3. 3.

consistency: The linear relation between the incoming and the outgoing
coefficients (1) must be satisfied simultaneously at all vertices.

By the uniqueness 1, we have

$\frac{{\rm e}^{i\gamma{}_{b_{1}}}}{\sqrt{h_{b_{1}}}}(a_{b^{-1}_{1}}{\rm
e}^{i\pi/4}+a_{b_{1}}{\rm e}^{-i\pi/4})=\frac{{\rm
e}^{i\gamma{}_{b_{2}}}}{\sqrt{h_{b_{2}}}}(a_{b^{-1}_{2}}{\rm
e}^{i\pi/4}+a_{b_{2}}{\rm e}^{-i\pi/4})=\cdots$ $=\frac{{\rm
e}^{i\gamma{}_{b_{d}}}}{\sqrt{h_{b_{d}}}}(a_{b^{-1}_{d}}{\rm
e}^{i\pi/4}+a_{b_{d}}{\rm e}^{-i\pi/4})=\psi{}_{u},$

where $b_{1},b_{2},\ldots,b_{d}$ are arcs emanating from $u$, and $d=\deg u$,
$i=\sqrt{-1}$.

By the condition 2, we have

$(H_{uu}-\lambda)\psi{}_{u}+\sum_{v\in{\cal E}_{u}}H_{uv}\psi{}_{v}=0,$

and so,

$(H_{uu}-\lambda)\frac{{\rm
e}^{i\gamma{}_{b_{1}}}}{\sqrt{h_{b_{1}}}}(a_{b^{-1}_{1}}{\rm
e}^{i\pi/4}+a_{b_{1}}{\rm
e}^{-i\pi/4})=-\frac{1}{d}\sum^{d}_{k=1}H_{b_{j}}\frac{{\rm
e}^{i\gamma{}_{b_{k}}}}{\sqrt{h_{b_{k}}}}(a_{b^{-1}_{k}}{\rm
e}^{i\pi/4}+a_{b_{k}}{\rm e}^{-i\pi/4}),$

where ${\cal E}_{u}=\\{f\in D(G)\mid o(f)=u\\}$. Thus, for each arc $b$ with
$o(b)=u$,

$a^{-1}_{b}=ia_{b}-2\sum^{d}_{k=1}\frac{\sqrt{h_{b}}\sqrt{h_{b_{k}}}}{H_{uu}-\lambda-i\Gamma{}_{u}}{\rm
e}^{i(\gamma{}_{b_{k}}+\gamma{}_{b^{-1}})}a_{b_{k}},$

where

$\Gamma{}_{u}=\sum^{d}_{k=1}h_{b_{k}}.$

Let $e=b^{-1}$, $f=b_{k}$ and

$\sigma^{(u)}_{ef}(\lambda)=i\delta{}_{e^{-1}f}-2\frac{\sqrt{h_{e}}\sqrt{h_{f}}}{H_{uu}-\lambda-i\Gamma{}_{u}}{\rm
e}^{i(\gamma{}_{f}+\gamma{}_{e})},$

where $\delta{}_{e^{-1}f}$ is the Kronecker delta. Then we have

$a_{e}=\sum_{o(f)=u}\sigma^{(u)}_{ef}(\lambda)a_{f}$ (2)

for each arc $e$ such that $t(e)=u$. The bond scattering matrix ${\bf
U}(\lambda)=(U_{ef})_{e,f\in D(G)}$ of $G$ is defined by

$U_{ef}=\left\\{\begin{array}[]{ll}\sigma^{(t(e))}_{e,f}&\mbox{if $t(e)=o(f)$,
}\\\ 0&\mbox{otherwise}\end{array}\right.$

By the consistency 3, we have

${\bf U}(\lambda){\bf a}={\bf a},$

where ${\bf a}={}^{t}(a_{b_{1}},a_{b_{2}},\ldots,a_{b_{2m}})$. This holds if
and only if

$\det({\bf I}_{2m}-{\bf U}(\lambda))=0.$

We present another proof of Theorem 4 by using the technique on the Ihara zeta
function, which is different from a proof in [5].

######  Theorem 4 (Gnutzmann and Smilansky)

Let $G$ be a connected graph with $n$ vertices $1,\ldots,n$ and $m$ edges.
Then, for the bond scattering matrix of $G$,

$\det({\bf I}_{2m}-{\bf U}(\lambda))=\frac{(-1)^{n}2^{m}\det(\lambda{\bf
I}_{n}-{\bf H})}{\prod^{n}_{j=1}(H_{jj}-\lambda-i\Gamma{}_{j})}.$

Proof. The argument is an analogue of Watanabe and Fukumizu’s method [18].

Let $G$ be a connected graph with $n$ vertices and $m$ edges,
$V(G)=\\{1,\cdots,n\\}$ and
$D(G)=\\{b_{1},\ldots,b_{m},b^{-1}_{1},\ldots,b^{-1}_{m}\\}$. Set $d_{j}=\deg
j$ and

$x_{j}=\frac{2}{H_{jj}-\lambda-i\Gamma{}_{j}}$

for each $j=1,\ldots,n$. Furthermore, for $e\in D(G)$, let

$w(e)=\sqrt{h_{e}}{\rm e}^{i\gamma{}_{e}}.$

Them we have

$\sigma^{(t(e))}_{ef}(\lambda)=i\delta{}_{e^{-1}f}-x_{t(e)}w(e)w(f).$

Now, we consider a $2m\times 2m$ matrix ${\bf B}=(B_{ef})_{e,f\in D(G)}$ given
by

$B_{ef}=\left\\{\begin{array}[]{ll}x_{o(f)}w(e)w(f)&\mbox{if $t(e)=o(f)$, }\\\
0&\mbox{otherwise}\end{array}\right.$

Let ${\bf K}=({\bf K}_{i,j})$ 1≤i≤2m;1≤j≤n be the $2m\times n$ matrix defined
as follows:

${\bf K}_{i,j}:=\left\\{\begin{array}[]{ll}x_{j}w(b_{i})&\mbox{if
$o(b_{i})=j$, }\\\ 0&\mbox{otherwise. }\end{array}\right.$

Furthermore, we define two $2m\times n$ matrices ${\bf L}=({\bf
L}_{i,j})_{1\leq i\leq 2m;1\leq j\leq n}$ and ${\bf M}=({\bf M}_{i,j})_{1\leq
i\leq 2m;1\leq j\leq n}$ as follows:

${\bf L}_{i,j}:=\left\\{\begin{array}[]{ll}w(b_{i})&\mbox{if $t(b_{i})=j$,
}\\\ 0&\mbox{otherwise, }\end{array}\right.\ {\bf
M}_{i,j}:=\left\\{\begin{array}[]{ll}w(b_{i})&\mbox{if $o(b_{i})=j$, }\\\
0&\mbox{otherwise. }\end{array}\right.$

Note that

${\bf K}={\bf M}\left[\begin{array}[]{ccc}x_{1}&&0\\\ &\ddots&\\\
0&&x_{n}\end{array}\right]={\bf M}{\bf X}.$ (3)

Furthermore, we have

${\bf L}{}^{t}{\bf K}={\bf B}$ (4)

and

${}^{t}{\bf M}{\bf L}={\bf H}.$ (5)

Note that

$H_{uv}=w(u,v)^{2}\ if\ (u,v)\in D(G).$

But, since

$U_{ef}=\left\\{\begin{array}[]{ll}-x_{t(e)}w(e)w(f)&\mbox{if $t(e)=o(f)$ and
$f\neq e^{-1}$, }\\\ i-x_{t(e)}w(e)w(f)&\mbox{if $f=e^{-1}$, }\\\
0&\mbox{otherwise, }\end{array}\right.$

we have

${\bf U}(\lambda)=i{\bf J}_{0}-{\bf B}.$

Furthermore, if ${\bf A}$ and ${\bf B}$ are an $r\times s$ and an $s\times r$
matrix, respectively, then we have

$\det({\bf I}_{r}-{\bf A}{\bf B})=\det({\bf I}_{s}-{\bf B}{\bf A}).$

Thus,

$\begin{array}[]{rcl}\det({\bf I}_{2m}-u{\bf U}(\lambda))&=&\det({\bf
I}_{2m}-u(i{\bf J}_{0}-{\bf B}))\\\ &&\\\ &=&\det({\bf I}_{2m}-iu{\bf
J}_{0}+u{\bf L}\ {}^{t}{\bf K})\\\ &&\\\ &=&\det({\bf I}_{2m}+u{\bf L}\
{}^{t}{\bf K}({\bf I}_{2m}-iu{\bf J}_{0})^{-1})\det({\bf I}_{2m}-iu{\bf
J}_{0})\\\ &&\\\ &=&\det({\bf I}_{n}+u\ {}^{t}{\bf K}({\bf I}_{2m}-iu{\bf
J}_{0})^{-1}{\bf L})\det({\bf I}_{2m}-iu{\bf J}_{0}).\end{array}$

Arrange arcs of $D(G)$ as follows: $b_{1},b^{-1}_{1},\ldots,b_{m},b^{-1}_{m}$.
Then we have

$\det({\bf I}_{2m}-iu{\bf
J}_{0})=\det(\left[\begin{array}[]{cccc}1&-iu&\ldots&0\\\ -iu&1&&\\\
\vdots&&\ddots&\\\ 0&&&\end{array}\right])=(1+u^{2})^{m}.$

Furthermore,

$\begin{array}[]{rcl}({\bf I}_{2m}-iu{\bf
J}_{0})^{-1}&=&\left[\begin{array}[]{cccc}1&-iu&\ldots&0\\\ -iu&1&&\\\
\vdots&&\ddots&\\\ 0&&&\end{array}\right]^{-1}\\\ &&\\\
&=&\frac{1}{1+u^{2}}\left[\begin{array}[]{cccc}1&iu&\ldots&0\\\ iu&1&&\\\
\vdots&&\ddots&\\\ 0&&&\end{array}\right]\\\ &&\\\ &=&\frac{1}{1+u^{2}}({\bf
I}_{2m}+iu{\bf J}_{0}).\end{array}$

Therefore, it follows that

$\begin{array}[]{rcl}&&\det({\bf I}_{2m}-u{\bf U}(\lambda))\\\ &&\\\
&=&\det({\bf I}_{n}+\frac{u}{1+u^{2}}\ {}^{t}{\bf K}({\bf I}_{2m}+iu{\bf
J}_{0}){\bf L})(1+u^{2})^{m}\\\ &&\\\ &=&(1+u^{2})^{m-n}\det((1+u^{2}){\bf
I}_{n}+u\ {}^{t}{\bf K}{\bf L}+iu^{2}\ {}^{t}{\bf K}{\bf J}_{0}{\bf
L}).\end{array}$

But, we have

${}^{t}{\bf K}{\bf L}={\bf X}{}^{t}{\bf M}{\bf L}={\bf X}{\bf H}.$

Furthermore, we have

${}^{t}{\bf K}{\bf J}_{0}{\bf L}={\bf X}{}^{t}{\bf M}{\bf J}_{0}{\bf L}.$

Then, for $u,v\in V(G)$, we have

$\begin{array}[]{rcl}&&({}^{t}{\bf M}{\bf J}_{0}{\bf L})_{uv}\\\ &&\\\
&=&\delta{}_{uv}\sum_{o(e)=u}({}^{t}{\bf M})_{ue}({\bf J}_{0})_{ee^{-1}}({\bf
L})_{e^{-1}v}\\\ &&\\\ &=&\delta{}_{uv}\sum_{o(e)=u}w(e)\cdot 1\cdot
w(e^{-1})\\\ &&\\\ &=&\delta{}_{uv}\sum_{o(e)=u}\sqrt{h_{e}}{\rm
e}^{i\gamma{}_{e}}\sqrt{h_{e}}{\rm e}^{-i\gamma{}_{e}}\\\ &&\\\
&=&\delta{}_{uv}\sum_{o(e)=u}h_{e}=\delta{}_{uv}\Gamma{}_{u}.\end{array}$

Now, let

${\bf D}_{L}=\left[\begin{array}[]{ccc}\Gamma{}_{1}&&0\\\ &\ddots&\\\
0&&\Gamma{}_{n}\end{array}\right].$

Then

${}^{t}{\bf K}{\bf J}_{0}{\bf L}={\bf X}{\bf D}_{L}.$

Thus,

$\det({\bf I}_{2m}-u{\bf U}(\lambda))=(1+u^{2})^{m-n}\det((1+u^{2}){\bf
I}_{n}+u{\bf X}{\bf H}+iu^{2}{\bf X}{\bf D}_{L}).$

Substituting $u=1$, we obtain

$\begin{array}[]{rcl}&&\det({\bf I}_{2m}-{\bf U}(\lambda))\\\ &&\\\
&=&2^{m-n}\det(2{\bf I}_{n}+{\bf X}{\bf H}+i{\bf X}{\bf D}_{L})\\\ &&\\\
&=&2^{m-n}\det(\left[\begin{array}[]{ccccc}\ldots&2+i\Gamma{}_{u}\frac{2}{H_{uu}-\lambda-i\Gamma{}_{u}}&\ldots&\frac{2}{H_{uu}-\lambda-i\Gamma{}_{u}}h_{uv}{\rm
e}^{2i\gamma{}_{uv}}&\ldots\end{array}\right])\\\ &&\\\
&=&\frac{2^{m}}{\prod^{n}_{u=1}(H_{uu}-\lambda-i\Gamma{}_{u})}\det(-\lambda{\bf
I}_{n}+{\bf H})\\\ &&\\\
&=&\frac{(-1)^{n}2^{m}}{\prod^{n}_{u=1}(H_{uu}-\lambda-i\Gamma{}_{u})}\det(\lambda{\bf
I}_{n}-{\bf H}).\end{array}$

$\Box$

##  4 The Euler product with respect to the scattering matrix

We present the Euler product for the determinant formula of the scattering
matrix ${\bf U}(\lambda)$ of a graph.

###### Theorem 5

Let $G$ be a connected graph with $m$ edges, and ${\bf H}={\bf
H}(G)=(H_{uv})_{u,v\in V(G)}$ an Hermitian matrix defined in Section 2. Then
the characteristic polynomial of the bond scattering matrix of $G$ induced
from ${\bf H}$ is given by

$\det({\bf I}_{2m}-u{\bf U}(\lambda))=\prod_{[C]}(1-w_{C}u^{|C|}),$

let ${c}$ runs over all equivalence classes of prime cycles in $G$, and

$w_{C}=\sigma{}^{(t(e_{1}))}_{e_{1}e_{2}}\sigma{}^{(t(e_{2}))}_{e_{2}e_{3}}\cdots\sigma{}^{(t(e_{n}))}_{e_{n}e_{1}},\
C=(b_{1},b_{2},\ldots,b_{n})$

Proof. Let $D(G)=\\{b_{1},\cdots,b_{2m}\\}$ such that
$b_{m+j}=b^{-1}_{j}(1\leq j\leq m)$. Set ${\bf U}={\bf U}(\lambda)$. Since

$\log\det({\bf I}-u{\bf F})={\rm Tr}\log({\bf I}-u{\bf F}),$

for a square matrix ${\bf F}$, we have

$\log\det({\bf I}-u{\bf U})={\rm Tr}\log({\bf I}-u{\bf
U})=-\sum^{\infty}_{k=1}\frac{{\rm Tr}({\bf U}^{k})}{k}u^{k}.$

Here,

${\rm Tr}({\bf U}^{k})=\sum_{C}w_{C},$

where $C$ runs over all cycles of length $k$ in $G$, and

$w_{C}=\sigma{}^{(t(e_{1}))}_{e_{1}e_{2}}\sigma{}^{(t(e_{2}))}_{e_{2}e_{3}}\cdots\sigma{}^{(t(e_{k}))}_{e_{k}e_{1}},\
C=(b_{1},b_{2},\ldots,b_{k})$

Thus,

$\begin{array}[]{rcl}u\frac{d}{du}\log\det({\bf I}_{2m}-u{\bf
U})&=&\sum^{\infty}_{k=1}{\rm Tr}({\bf U}^{k})u^{k}\\\ &&\\\
&=&\sum_{C}w_{C}u^{|C|},\end{array}$

where $C$ runs over all cycles in $G$.

Now, let $C$ be any cycle in $G$. Then there exists exactly one prime cycle
$D$ such that

$C=D^{l}.$

Thus, we have

$u\frac{d}{du}\log\det({\bf I}_{2m}-u{\bf
U})=-\sum_{D}\sum^{\infty}_{k=1}w^{k}_{D}u^{k|D|},$

and so,

$\frac{d}{du}\log\det({\bf I}_{2m}-u{\bf
U})=-\sum_{D}\sum^{\infty}_{k=1}w^{k}_{D}u^{k|D|-1},$

where $D$ runs over all prime cycles in $G$. Therefore, it follows that

$\begin{array}[]{rcl}\log\det({\bf I}_{2m}-u{\bf
U})&=&-\sum_{D}\sum^{\infty}_{k=1}\frac{w^{k}_{D}}{k|D|}u^{k|D|}\\\ &&\\\
&=&-\sum_{[D]}\sum^{\infty}_{k=1}\frac{|D|}{k|D|}w^{k}_{D}u^{k|D|}\\\ &&\\\
&=&-\sum_{[D]}\sum^{\infty}_{k=1}\frac{1}{k}w^{k}_{D}u^{k|D|}\\\ &&\\\
&=&\sum_{[D]}\log(1-w_{D}u^{|D|}).\end{array}$

Hence,

$\det({\bf I}_{2m}-u{\bf U}(\lambda))=\prod_{[C]}(1-w_{C}u^{|C|}),$

$\Box$

##  5 Scattering matrix of a regular covering of a graph

Let $G$ be a connected graph, and let $N(v)=\\{w\in V(G)\mid(v,w)\in D(G)\\}$
denote the neighbourhood of a vertex $v$ in $G$. A graph $H$ is a covering of
$G$ with projection $\pi:H\longrightarrow G$ if there is a surjection
$\pi:V(H)\longrightarrow V(G)$ such that
$\pi{\mid}_{N(v^{\prime})}:N(v^{\prime})\longrightarrow N(v)$ is a bijection
for all vertices $v\in V(G)$ and $v^{\prime}\in{\pi}^{-1}(v)$. When a finite
group $\Pi$ acts on a graph $G$, the quotient graph $G/\Pi$ is a graph whose
vertices are the $\Pi$-orbits on $V(G)$, with two vertices adjacent in $G/\Pi$
if and only if some two of their representatives are adjacent in $G$. A
covering $\pi:H\longrightarrow G$ is regular if there is a subgroup B of the
automorphism group $Aut\ H$ of $H$ acting freely on $H$ such that the quotient
graph $H/{\it B}$ is isomorphic to $G$.

Let $G$ be a graph and $\Gamma$ a finite group. Then a mapping
$\alpha:D(G)\longrightarrow\Gamma$ is an ordinary voltage assignment if
$\alpha(v,u)=\alpha(u,v)^{-1}$ for each $(u,v)\in D(G)$. The pair $(G,\alpha)$
is an ordinary voltage graph. The derived graph $G^{\alpha}$ of the ordinary
voltage graph $(G,\alpha)$ is defined as follows:
$V(G^{\alpha})=V(G)\times\Gamma$ and $((u,h),(v,k))\in D(G^{\alpha})$ if and
only if $(u,v)\in D(G)$ and $k=h\alpha(u,v)$. The natural projection
$\pi:G^{\alpha}\longrightarrow G$ is defined by $\pi(u,h)=u$. The graph
$G^{\alpha}$ is a derived graph covering of $G$ with voltages in $\Gamma$ or a
$\Gamma$-covering of $G$. Note that $|{\cal E}_{(u,h)}|=|{\cal E}_{u}|$ for
each $(u,h)\in V(G^{\alpha})$. The natural projection $\pi$ commutes with the
right multiplication action of the $\alpha(e),e\in D(G)$ and the left action
of $\Gamma$ on the fibers: $g(u,h)=(u,gh),g\in\Gamma$, which is free and
transitive. Thus, the $\Gamma$-covering $G^{\alpha}$ is a
$\mid\Gamma\mid$-fold regular covering of $G$ with covering transformation
group $\Gamma$. Furthermore, every regular covering of a graph $G$ is a
$\Gamma$-covering of $G$ for some group $\Gamma$ (see [6]).

Let $G$ be a connected graph, $\Gamma$ be a finite group and
$\alpha:D(G)\longrightarrow\Gamma$ be an ordinary voltage assignment. In the
$\Gamma$-covering $G^{\alpha}$, set $v_{g}=(v,g)$ and $e_{g}=(e,g)$, where
$v\in V(G),e\in D(G),g\in\Gamma$. For $e=(u,v)\in D(G)$, the arc $e_{g}$
emanates from $u_{g}$ and terminates at $v_{g\alpha(e)}$. Note that
$e^{-1}_{g}=(e^{-1})_{g\alpha(e)}$.

Let $G$ be a connected graph, $\Gamma$ be a finite group and
$\alpha:D(G)\longrightarrow\Gamma$ be an ordinary voltage assignment.
Furthermore, let ${\bf H}={\bf H}(G)=(H_{uv})_{u,v\in V(G)}$ be an Hermitian
matrix such that

$H_{uv}=\left\\{\begin{array}[]{ll}h_{f}{\rm e}^{2i\gamma{}_{f}}&\mbox{if
$f=(u,v)\in D(G)$, }\\\ 0&\mbox{otherwise, }\end{array}\right.$

where, for each $f\in D(G)$,

$h_{f}=h_{f^{-1}}\geq 0\ and\
\gamma{}_{f}=-\gamma{}_{f^{-1}}\in[-\pi/2,\pi/2].$

We give the function $\tilde{h}:D(G^{\alpha})\longrightarrow\mathbb{R}$ and
$\tilde{\gamma}:D(G^{\alpha})\longrightarrow[-\pi/2,\pi/2]$ induced from $h$
and $\gamma$, respectively, as follows:

$\tilde{h}(u_{g},v_{k})=h_{uv}\ and\
\tilde{\gamma}(u_{g},v_{k})=\gamma{}_{uv}\ if\ (u,v)\in D(G)\ and\
k=g\alpha(u,v).$

Furthermore, we consider the Hermitian matrix $\tilde{{\bf H}}={\bf
H}(G^{\alpha})=(H_{u_{g}v_{k}})_{u_{g}v_{k}\in V(G^{\alpha})}$ of $G^{\alpha}$
induced from ${\bf H}$. At first, let

$H_{u_{g}u_{g}}=H_{uu}\ for\ each\ g\in\Gamma.$

For $(u_{g},v_{k})\in D(G^{\alpha})$, we have

$H_{u_{g}v_{k}}=\tilde{h}(u_{g},v_{k})e^{2i\tilde{\gamma}(u_{g},v_{k})}=h_{uv}e^{2i\gamma{}_{uv}}.$

Thus,

$H_{u_{g}v_{k}}=\left\\{\begin{array}[]{ll}h_{uv}e^{2i\gamma{}_{u_{g}v_{k}}}&\mbox{if
$(u,v)\in D(G)$ and $k=g\alpha(u,v)$, }\\\ H_{uu}&\mbox{if $u=v$ and $k=g$,
}\\\ 0&\mbox{otherwise. }\end{array}\right.$

Next, we consider the bond wave function of the regular covering $G^{\alpha}$
of $G$. Let $V(G)=\\{v_{1},\ldots,v_{n}\\}$,
$D(G)=\\{e_{1},\ldots,e_{m},e^{-1}_{1},\ldots,e^{-1}_{m}\\}$ and
$\Gamma=\\{g_{1}=1,g_{2},\ldots,g_{p}\\}$. Let $\lambda$ be a eigenvalue of
$\tilde{{\bf H}}={\bf H}(G^{\alpha})$, and let
$\tilde{\phi}=(\phi{}_{v_{1},g_{1}},\ldots,\phi{}_{v_{1},g_{p}},\ldots,\phi{}_{v_{n},g_{1}},\ldots,\phi{}_{v_{n},g_{p}})$
be the eigenvector corresponding to $\lambda$, where $\phi{}_{v_{i},g_{j}}$
corresponds to the vertex $(v_{i},g_{j})\ (1\leq i\leq n;1\leq j\leq p)$ of
$G^{\alpha}$. Furthermore let $b_{g}=(v_{g},z_{g\alpha(b)})$ be any arc of
$G^{\alpha}$, where $b=(v,z)\in D(G)$, $g\in\Gamma$. Then the bond wave
function of $G^{\alpha}$ is

$\phi{}_{b_{g}}(x)=\frac{e^{i\gamma{}_{b}}}{\sqrt{h_{b}}}(a_{b^{-1}_{g}}{\rm
e}^{i\pi x/4}+a_{b_{g}}{\rm e}^{-i\pi x/4}),\ x=\pm 1,\ i=\sqrt{-1}$

under the condition

$\phi{}_{b_{g}}(1)=\phi{}_{v_{g}}\ and\
\phi{}_{b_{g}}(-1)=\phi{}_{z_{g\alpha(b)}}.$

By (1), we have

$\begin{array}[]{rcl}a_{b^{-1}_{g}}&=&i\delta{}_{b^{-1}_{g}e_{g}}-2\sum_{o(e_{g})=v_{g}}\frac{\sqrt{\tilde{h}_{e_{g}}}\sqrt{\tilde{h}_{b_{g}}}}{H_{v_{g}v_{g}}-\lambda-i\Gamma{}_{v_{g}}}e^{i(\tilde{\gamma}_{b_{g}}+\tilde{\gamma}_{e_{g}})}a_{e_{g}}\\\
&&\\\
&=&\sum_{o(e_{g})=v_{g}}\sigma^{(v_{g})}_{b_{g}e_{g}}a_{e_{g}}\end{array}$

for each arc $b_{g}$ with $o(b_{g})=v_{g}$, where

$\sigma{}^{(v_{g})}_{b_{g}e_{g}}=i\delta{}_{b^{-1}_{g}e_{g}}-2\frac{\sqrt{\tilde{h}_{e_{g}}}\sqrt{\tilde{h}_{b_{g}}}}{H_{v_{g}v_{g}}-\lambda-i\Gamma{}_{v_{g}}}e^{i(\tilde{\gamma}_{b_{g}}+\tilde{\gamma}_{e_{g}})}a_{e_{g}}$

and

$\tilde{h}_{e_{g}}=\tilde{h}(e_{g}),\
\tilde{\gamma}_{e_{g}}=\tilde{\gamma}(e_{g}).$

By the definitions of $\tilde{h}$, $\tilde{\gamma}$ and $\tilde{{\bf H}}$, we
have

$\sigma{}^{(v_{g})}_{b_{g}e_{g}}=i\delta{}_{b^{-1}e}-2\frac{\sqrt{h_{e}}\sqrt{h_{b}}}{H_{vv}-\lambda-i\Gamma{}_{v}}e^{i(\gamma{}_{b}+\gamma{}_{e})}=\sigma^{(v)}_{be}=\sigma^{(t(b))}_{be}.$

Note that ${\cal E}_{(v,g)}={\cal E}_{v}$. Thus,

$a_{b^{-1}_{g}}=\sum_{o(e_{g})=v_{g}}\sigma^{(t(b))}_{be}a_{e_{g}}.$

Therefore, the bond scattering matrix $\tilde{{\bf
U}}(\lambda)=(U(e_{g},f_{h}))_{e_{g},f_{h}\in D(G^{\alpha})}$ of $G^{\alpha}$
is given by

$U(e_{g},f_{h})=\left\\{\begin{array}[]{ll}\sigma^{(t(e))}_{ef}&\mbox{if
$t(f_{h})=o(e_{g})$, }\\\ 0&\mbox{otherwise. }\end{array}\right.$

But, we have

$x_{v_{g}}=\frac{2}{H_{vv}-\lambda-i\Gamma{}_{v}}=x_{v}$

for $v_{g}\in V(G^{\alpha})$. Furthermore, let
$\tilde{w}:D(G^{\alpha})\longrightarrow{\bf C}$ be given as follows:

$\tilde{w}(e_{g})=\sqrt{\tilde{h}_{e_{g}}}e^{i\tilde{\gamma}_{e_{g}}}\ for\
each\ e_{g}\in D(G^{\alpha}).$

Then we have

$\tilde{w}(e_{g})=\sqrt{h_{e}}e^{i\gamma{}_{e}}=w(e),\ e_{g}\in
D(G^{\alpha}).$

For $g\in\Gamma$, let the matrix ${\bf H}{}_{g}=(H^{(g)}_{uv})$ be defined by

$H^{(g)}_{uv}=\left\\{\begin{array}[]{ll}h_{uv}e^{2i\gamma{}_{uv}}&\mbox{if
$\alpha(u,v)=g$ and $(u,v)\in D(G)$, }\\\
0&\mbox{otherwise.}\end{array}\right.$

Furthermore, let ${\bf U}_{g}=(U^{(g)}(e,f))$ be given by

$U^{(g)}(e,f)=\left\\{\begin{array}[]{ll}\sigma^{(t(e))}_{ef}&\mbox{if
$t(e)=o(f)$ and $\alpha(e)=g$, }\\\ 0&\mbox{otherwise,}\end{array}\right.$

Let ${\bf M}_{1}\oplus\cdots\oplus{\bf M}_{s}$ be the block diagonal sum of
square matrices ${\bf M}_{1},\ldots,{\bf M}_{s}$. 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 Kronecker product ${\bf
A}\bigotimes{\bf B}$ of matrices A and B is considered as the matrix A having
the element $a_{ij}$ replaced by the matrix $a_{ij}{\bf B}$.

###### Theorem 6

Let $G$ be a connected graph with $n$ vertices $v_{1},\ldots v_{n}$ and $m$
unoriented edges, $\Gamma$ be a finite group and
$\alpha:D(G)\longrightarrow\Gamma$ be an ordinary voltage assignment. Set
$\mid\Gamma\mid=p$. Furthermore, let
${\rho}_{1}=1,{\rho}_{2},\cdots,{\rho}_{k}$ be the irreducible representations
of $\Gamma$, and $f_{i}$ be the degree of ${\rho}_{i}$ for each $i$, where
$f_{1}=1$.

If the $\Gamma$-covering $G^{\alpha}$ of $G$ is connected, then, for the bond
scattering matrix of $G{}^{\alpha}$,

$\det({\bf I}_{2mp}-\tilde{{\bf U}}(\lambda))=\det({\bf I}_{2m}-{\bf
U}(\lambda))\prod^{k}_{i=2}\det({\bf
I}_{2mf_{i}}-\sum_{h}{\rho}_{i}(h)\bigotimes{\bf U}{}_{h})^{f_{i}}$
$=\frac{2^{mp}(-1)^{np}\det(\lambda{\bf I}_{n}-{\bf H})}{\prod_{u\in
V(G)}(H_{uu}-\lambda-i\Gamma{}_{u}}\prod^{k}_{i=2}\det(\lambda{\bf
I}_{nf_{i}}-\sum_{h\in\Gamma}{\rho}_{i}(h)\bigotimes{\bf H}{}_{h}-{\bf
I}_{f_{i}}\bigotimes{\rm diag}({\bf H}))^{f_{i}},$

where

${\rm diag}({\bf H})=\left[\begin{array}[]{ccc}H_{v_{1}v_{1}}&&{\bf 0}\\\
&\ddots&\\\ {\bf 0}&&H_{v_{n}v_{n}}\end{array}\right].$

Proof . Let $\mid\Gamma\mid=p$. By Theorem 4, for the bond scattering matrix
of $G^{\alpha}$, we have

$\det({\bf I}_{2mp}-\tilde{{\bf
U}}(\lambda))=\frac{2^{mp}(-1)^{np}\det(\lambda{\bf I}_{np}-{\bf
H}(G^{\alpha}))}{\prod_{u\in V(G)}(H_{uu}-\lambda-i\Gamma{}_{u})^{p}}.$

Let $D(G)=\\{e_{1},\ldots,e_{m},e_{m+1},\ldots,e_{2m}\\}$ such that
$e_{m+j}=e^{-1}_{j}(1\leq j\leq m)$ and
$\Gamma=\\{1=g_{1},g_{2},\ldots,g_{p}\\}$. Arrange arcs of $G^{\alpha}$ in $p$
blocks:
$(e_{1},1),\ldots,(e_{2m},1);(e_{1},g_{2}),\ldots,(e_{2m},g_{2});\ldots;$
$(e_{1},g_{p}),\ldots,(e_{2m},g_{p}).$ We consider the matrix $\tilde{{\bf
U}}(\lambda)$ under this order. 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., $g_{j}=g_{i}h$. Then
$U(e_{g_{i}},f_{g_{j}})\neq 0$ if and only if $t(e,g_{j})=o(f,g_{i})$.
Furthermore, $t(e,g_{j})=o(f,g_{i})$ if and only if
$(o(f),g_{j})=o(f,g_{j})=t(e,g_{i})=(t(e),g_{i}\alpha(e))$. Thus, $t(e)=o(f)$
and $\alpha(e)=g^{-1}_{i}g_{j}=g^{-1}_{i}g_{i}h=h$. Thus, we have

$\tilde{{\bf U}}(\lambda)=\sum_{h\in\Gamma}{\bf P}_{h}\bigotimes{\bf
U}{}_{h}.$

Furthermore, we have

${\rm diag}({\bf H}(G^{\alpha}))={\bf I}_{p}\bigotimes{\rm diag}({\bf H}).$

Let $\rho$ be the right regular representation of $\Gamma$. Furthermore, let
${\rho}_{1}=1,{\rho}_{2},\ldots,{\rho}_{k}$ be all inequivalent irreducible
representations of $\Gamma$, and $f_{i}$ the degree of ${\rho}_{i}$ for each
$i$, where $f_{1}=1$. Then we have $\rho(g)={\bf P}_{g}$ for $g\in\Gamma$.
Furthermore, there exists a nonsingular matrix ${\bf P}$ such that ${\bf
P}^{-1}\rho(g){\bf P}=(1)\oplus f_{2}\circ{\rho}_{2}(g)\oplus\cdots\oplus
f_{k}\circ{\rho}_{k}(g)$ for each $g\in\Gamma$(see [12]). Thus, we have

${\bf P}^{-1}{\bf P}_{g}{\bf P}=(1)\oplus
f_{2}\circ{\rho}_{2}(g)\oplus\cdots\oplus f_{k}\circ{\rho}_{k}(g).$

Putting ${\bf F}=({\bf P}^{-1}\bigotimes{\bf I}_{2m})\tilde{{\bf
U}}(\lambda)({\bf P}\bigotimes{\bf I}_{2m})$, we have

${\bf F}=\sum_{g\in\Gamma}\\{(1)\oplus
f_{2}\circ{\rho}_{2}(g)\oplus\cdots\oplus
f_{k}\circ{\rho}_{k}(g)\\}\bigotimes{\bf U}{}_{g}.$

Note that ${\bf U}(\lambda)=\sum_{g\in\Gamma}{\bf U}{}_{g}$ and
$1+f^{2}_{2}+\cdots+f^{2}_{k}=p$. Therefore it follows that

$\det({\bf I}_{2mp}-\tilde{{\bf U}}(\lambda))=\det({\bf I}_{2m}-{\bf
U}(\lambda))\prod^{k}_{i=2}\det({\bf
I}_{2mf_{i}}-\sum_{g}{\rho}_{i}(g)\bigotimes{\bf U}{}_{g})^{f_{i}}.$

Next, let $V(G)=\\{v_{1},\ldots,v_{n}\\}$. Arrange vertices of $G^{\alpha}$ in
$p$ blocks: $(v_{1},1),\ldots,(v_{n},1);$
$(v_{1},g_{2}),\ldots,(v_{n},g_{2});\ldots;(v_{1},g_{p}),\ldots,(v_{n},g_{p}).$
We consider the matrix ${\bf H}(G{}^{\alpha})$ under this order.

Suppose that $p^{(h)}_{ij}=1$, i.e., $g_{j}=g_{i}h$. Then
$((u,g_{i}),(v,g_{j}))\in D(G{}^{\alpha})$ if and only if $(u,v)\in D(G)$ and
$g_{j}=g_{i}\alpha(u,v)$. If $g_{j}=g_{i}\alpha(u,v)$, then
$\alpha(u,v)=g^{-1}_{i}g_{j}=g^{-1}_{i}g_{i}h=h$. Thus we have

${\bf H}(G{}^{\alpha})=\sum_{h\in\Gamma}{\bf P}_{h}\bigotimes{\bf
H}{}_{h}+{\bf I}_{p}\bigotimes{\rm diag}({\bf H}).$

Putting ${\bf E}=({\bf P}^{-1}\bigotimes{\bf I}_{n}){\bf H}(G{}^{\alpha})({\bf
P}\bigotimes{\bf I}_{n})$, we have

${\bf E}=\sum_{h\in\Gamma}\\{(1)\oplus
f_{2}\circ{\rho}_{2}(h)\oplus\cdots\oplus
f_{k}\circ{\rho}_{k}(h)\\}\bigotimes{\bf H}{}_{h}+{\bf I}_{p}\bigotimes{\rm
diag}({\bf H}).$

Note that ${\bf H}(G)=\sum_{h\in\Gamma}{\bf H}{}_{h}+{\rm diag}({\bf H})$.
Therefore it follows that

$\begin{array}[]{rcl}\det(\lambda{\bf I}_{np}-{\bf
H}(G^{\alpha}))&=&\det(\lambda{\bf I}_{n}-{\bf H}(G))\\\ &&\\\
&\times&\prod^{k}_{i=2}\det(\lambda{\bf
I}_{nf_{i}}-\sum_{h\in\Gamma}{\rho}_{i}(h)\bigotimes{\bf H}{}_{h}-{\bf
I}_{f_{i}}\bigotimes{\rm diag}{\bf H})^{f_{i}}.\end{array}$

Hence,

$\det({\bf I}_{2mp}-\tilde{{\bf U}}(\lambda))=\det({\bf I}_{2m}-{\bf
U}(\lambda))\prod^{k}_{i=2}\det({\bf
I}_{2mf_{i}}-\sum_{h}{\rho}_{i}(h)\bigotimes{\bf U}{}_{h})^{f_{i}}$
$=\frac{2^{mp}(-1)^{np}\det(\lambda{\bf I}_{n}-{\bf H}(G))}{\prod_{u\in
V(G)}(H_{uu}-\lambda-i\Gamma{}_{u})^{p}}\prod^{k}_{i=2}\det(\lambda{\bf
I}_{nf_{i}}-\sum_{h\in\Gamma}{\rho}_{i}(h)\bigotimes{\bf H}{}_{h}-{\bf
I}_{f_{i}}\bigotimes{\rm diag}{\bf H})^{f_{i}}.$

$\Box$

##  6 $L$-functions of graphs

Let $G$ be a connected graph with $n$ vertices and $m$ unoriented edges,
$\Gamma$ be a finite group and $\alpha:D(G)\longrightarrow\Gamma$ be an
ordinary voltage assignment. Furthermore, let ${\bf H}={\bf
H}(G)=(H_{uv})_{u,v\in V(G)}$ be an Hermitian matrix such that

$H_{uv}=\left\\{\begin{array}[]{ll}h_{f}{\rm e}^{2i\gamma{}_{f}}&\mbox{if
$f=(u,v)\in D(G)$, }\\\ 0&\mbox{otherwise, }\end{array}\right.$

where, for each $f\in D(G)$,

$h_{f}=h_{f^{-1}}\geq 0\ and\
\gamma{}_{f}=-\gamma{}_{f^{-1}}\in[-\pi/2,\pi/2].$

Let $\rho$ be a unitary representation of $\Gamma$ and $d$ its degree. The
$L$-function of $G$ associated with $\rho$ and $\alpha$ is defined by

${\bf Z}_{H}(G,\lambda,\rho,\alpha)=\det({\bf
I}_{2md}-\sum_{h\in\Gamma}{\rho}(h)\bigotimes{\bf U}{}_{h})^{-1}.$

If $\rho={\bf 1}$ is the identity representation of $\Gamma$, then

${\bf Z}_{H}(G,\lambda,{\bf 1},\alpha)=\det({\bf I}_{2m}-{\bf U})^{-1}.$

A determinant expression for the $L$-function of $G$ associated with $\rho$
and $\alpha$ is given as follows. For $1\leq i,j\leq n$, the $(i,j)$-block
${\bf F}_{i,j}$ of a $dn\times dn$ matrix ${\bf F}$ is the submatrix of ${\bf
K}$ consisting of $d(i-1)+1,\ldots,di$ rows and $d(j-1)+1,\ldots,dj$ columns.

###### Theorem 7

Let $G$ be a connected graph with $n$ vertices and $m$ unoriented edges,
$\Gamma$ be a finite group and $\alpha:D(G)\longrightarrow\Gamma$ be an
ordinary voltage assignment. If $\rho$ is a unitary representation of $\Gamma$
and $d$ is the degree of $\rho$, then the reciprocal of the $L$-function of
$G$ associated with $\rho$ and $\alpha$ is

${\bf Z}_{H}(G,\lambda,\rho,\alpha)^{-1}=\frac{2^{md}(-1)^{nd}}{\prod_{u\in
V(G)}(H_{uu}-\lambda-i\Gamma{}_{u})^{d}}\det(\lambda{\bf
I}_{np}-\sum_{g\in\Gamma}\rho(g)\bigotimes{\bf H}_{g}-{\bf
I}_{d}\bigotimes{\rm diag}({\bf H})).$

Proof. The argument is an analogue of Watanabe and Fukumizu’s method [18].

Let $V(G)=\\{v_{1},\ldots,v_{n}\\}$ and $D(G)=\\{e_{1},\ldots,e_{m},e_{m+1},$
$\ldots,e_{2m}\\}$ such that $e_{m+i}=e^{-1}_{i}(1\leq i\leq m)$. Note that
the $(e,f)$-block $(\sum_{g\in\Gamma}{\bf U}_{g}\bigotimes\rho(g))_{ef}$ of
$\sum_{g\in\Gamma}{\bf U}_{g}\bigotimes\rho(g)$ is given by

$(\sum_{g\in\Gamma}{\bf
U}_{g}\bigotimes\rho(g))_{ef}=\left\\{\begin{array}[]{ll}\rho(\alpha(e))\sigma{}^{(t(e))}_{ef}&\mbox{if
$t(e)=o(f)$, }\\\ {\bf 0}_{d}&\mbox{otherwise.}\end{array}\right.$

For $g\in\Gamma$, two $2m\times 2m$ matrices ${\bf B}_{g}=({\bf
B}^{(g)}_{ef})_{e,f\in D(G)}$ and ${\bf J}_{g}=({\bf J}^{(g)}_{ef})_{e,f\in
D(G)}$ are defined as follows:

${\bf B}^{(g)}_{ef}=\left\\{\begin{array}[]{ll}x_{o(f)}w(e)w(f)&\mbox{if
$t(e)=o(f)$ and $\alpha(e)=g$, }\\\ 0&\mbox{otherwise, }\end{array}\right.\
{\bf J}^{(g)}_{ef}=\left\\{\begin{array}[]{ll}1&\mbox{if $f=e^{-1}$ and
$\alpha(e)=g$, }\\\ 0&\mbox{otherwise.}\end{array}\right.$

Then we have

${\bf U}_{g}=i{\bf J}_{g}-{\bf B}_{g}\ for\ g\in\Gamma.$

Let ${\bf K}=({\bf K}_{ij})$ 1≤i≤2m;1≤j≤n be the $2md\times nd$ matrix defined
as follows:

${\bf K}_{ij}:=\left\\{\begin{array}[]{ll}x_{v_{j}}w(e_{i}){\bf
I}_{d}&\mbox{if $o(e_{i})=v_{j}$, }\\\ {\bf 0}_{d}&\mbox{otherwise.
}\end{array}\right.$

Furthermore, we define two $2md\times nd$ matrices ${\bf L}=({\bf
L}_{ij})_{1\leq i\leq 2m;1\leq j\leq n}$ and ${\bf M}=({\bf M}_{ij})_{1\leq
i\leq 2m;1\leq j\leq n}$ as follows:

${\bf L}_{ij}:=\left\\{\begin{array}[]{ll}w(e_{j})\rho(\alpha(e_{i}))&\mbox{if
$t(e_{i})=v_{j}$, }\\\ {\bf 0}_{d}&\mbox{otherwise, }\end{array}\right.\ {\bf
M}_{ij}:=\left\\{\begin{array}[]{ll}w(e_{i}){\bf I}_{d}&\mbox{if
$o(e_{i})=v_{j}$, }\\\ {\bf 0}_{d}&\mbox{otherwise. }\end{array}\right.$

Then we have

${\bf K}={\bf M}({\bf X}\bigotimes{\bf I}_{d})={\bf M}{\bf X}_{d},$

where

${\bf X}_{d}={\bf X}\bigotimes{\bf I}_{d}.$

Furthermore, we have

${\bf L}\ {}^{t}{\bf K}=\sum_{h\in\Gamma}{\bf B}{}_{h}\bigotimes\rho(h)={\bf
B}_{\rho}$ (6)

and

${}^{t}{\bf M}{\bf L}=\sum_{g\in\Gamma}{\bf H}_{g}\bigotimes\rho(g),$ (7)

where

${\bf B}_{\rho}=\sum_{g\in\Gamma}{\bf B}_{g}\bigotimes\rho(g).$

Thus,

$\begin{array}[]{rcl}&&\det({\bf
I}_{2md}-u\sum_{g\in\Gamma}\rho(g)\bigotimes{\bf U}_{g})=\det({\bf
I}_{2md}-u\sum_{g\in\Gamma}{\bf U}_{g}\bigotimes\rho(g))\\\ &&\\\ &=&\det({\bf
I}_{2md}-u\sum_{g\in\Gamma}(i{\bf J}_{g}-{\bf B}_{g})\bigotimes\rho(g))\\\
&&\\\ &=&\det({\bf I}_{2md}-iu\sum_{g\in\Gamma}{\bf
J}_{g}\bigotimes\rho(g)+u\sum_{g\in\Gamma}{\bf
B}_{g}\bigotimes\rho(g)).\end{array}$

Now, let

${\bf J}_{\rho}=\sum_{g\in\Gamma}{\bf J}_{g}\bigotimes\rho(g).$

Note that

${\bf J}^{2}_{\rho}={\bf I}_{2md}.$

Then we have

$\begin{array}[]{rcl}&&\det({\bf
I}_{2md}-u\sum_{g\in\Gamma}\rho(g)\bigotimes{\bf U}_{g})\\\ &&\\\ &=&\det({\bf
I}_{2md}-iu{\bf J}_{\rho}+u{\bf B}_{\rho})\\\ &&\\\ &=&\det({\bf
I}_{2md}+u{\bf B}_{\rho}({\bf I}_{2md}-iu{\bf J}_{\rho})^{-1})\det({\bf
I}_{2md}-iu{\bf J}_{\rho})\\\ &&\\\ &=&\det({\bf I}_{2md}+u{\bf L}\ {}^{t}{\bf
K}({\bf I}_{2md}-iu{\bf J}_{\rho})^{-1})\det({\bf I}_{2md}-iu{\bf
J}_{\rho})\\\ &&\\\ &=&\det({\bf I}_{nd}+u\ {}^{t}{\bf K}({\bf I}_{2md}-iu{\bf
J}_{\rho})^{-1}{\bf L})\det({\bf I}_{2md}-iu{\bf J}_{\rho}).\end{array}$

But, we have

$\det({\bf I}_{2md}-iu{\bf J}_{\rho})=\det(\left[\begin{array}[]{cccc}{\bf
I}_{d}&-iu\rho(\alpha(e_{1}))&&{\bf 0}\\\ -iu\rho(\alpha(e^{-1}_{1}))&{\bf
I}_{d}&&\\\ &&\ddots&\\\ {\bf 0}&&&\end{array}\right])=(1+u^{2})^{md}.$

Furthermore, we have

$\begin{array}[]{rcl}&&({\bf I}_{2md}-iu{\bf J}_{\rho})^{-1}\\\ &&\\\
&=&\left[\begin{array}[]{cccc}{\bf I}_{d}&-iu\rho(\alpha(e_{1}))&&{\bf 0}\\\
-iu\rho(\alpha(e^{-1}_{1}))&{\bf I}_{d}&&\\\ &&\ddots&\\\ {\bf
0}&&&\end{array}\right]^{-1}\\\ &&\\\
&=&\frac{1}{1+u^{2}}\left[\begin{array}[]{cccc}{\bf
I}_{d}&iu\rho(\alpha(e_{1}))&&{\bf 0}\\\ iu\rho(\alpha(e^{-1}_{1}))&{\bf
I}_{d}&&\\\ &&\ddots&\\\ {\bf 0}&&&\end{array}\right]\\\ &&\\\
&=&\frac{1}{1+u^{2}}({\bf I}_{2md}+iu{\bf J}_{\rho}).\end{array}$

Thus, we have

$\begin{array}[]{rcl}&&\det({\bf
I}_{2md}-u\sum_{g\in\Gamma}\rho(g)\bigotimes{\bf U}_{g})\\\ &&\\\
&=&(1+u^{2})^{md}\det({\bf I}_{nd}+u/(1+u^{2})\ {}^{t}{\bf K}({\bf
I}_{2md}+iu{\bf J}_{\rho}){\bf L})\\\ &&\\\ &=&(1+u^{2})^{md-
nd}\det((1+u^{2}){\bf I}_{nd}+u\ {}^{t}{\bf K}{\bf L}+iu^{2}\ {}^{t}{\bf
K}{\bf J}_{\rho}{\bf L}).\end{array}$

Now, we have

${}^{t}{\bf K}{\bf L}={\bf X}_{d}\ {}^{t}{\bf M}{\bf L}={\bf
X}_{d}\sum_{g\in\Gamma}{\bf H}_{g}\bigotimes\rho(g).$

Furthermore,

${}^{t}{\bf K}{\bf J}_{\rho}{\bf L}={\bf X}_{d}\ {}^{t}{\bf M}{\bf
J}_{\rho}{\bf L}.$

Then we have

$\begin{array}[]{rcl}&&({}^{t}{\bf M}{\bf J}_{\rho}{\bf L})_{uv}\\\ &&\\\
&=&\delta{}_{uv}\sum_{o(e)=u}({}^{t}{\bf M})_{ue}({\bf
J}_{\rho})_{ee^{-1}}({\bf L})_{e^{-1}v}\\\ &&\\\
&=&\delta{}_{uv}\sum_{o(e)=u}w(e){\bf
I}_{d}\rho(\alpha(e))w(e^{-1})\rho(\alpha(e^{-1}))\\\ &&\\\
&=&\delta{}_{uv}\sum_{o(e)=u}\sqrt{h_{e}}e^{i\gamma{}_{e}}\sqrt{h_{e}}e^{-i\gamma{}_{e}}{\bf
I}_{d}\\\ &&\\\ &=&\delta{}_{uv}\sum_{o(e)=u}h_{e}{\bf
I}_{d}=\delta{}_{uv}\Gamma{}_{u}{\bf I}_{d}.\end{array}$

Thus,

${}^{t}{\bf K}{\bf J}_{\rho}{\bf L}={\bf X}({\bf D}_{\Gamma}\bigotimes{\bf
I}_{d}),$

where

${\bf D}_{\Gamma}=\left[\begin{array}[]{ccc}\Gamma{}_{v_{1}}&&0\\\ &\ddots&\\\
0&&\Gamma{}_{v_{n}}\end{array}\right].$

Therefore, it follows that

$\begin{array}[]{rcl}&&\det({\bf
I}_{2md}-u\sum_{g\in\Gamma}\rho(g)\bigotimes{\bf U}_{g})\\\ &&\\\
&=&(1+u^{2})^{(m-n)d}\det((1+u^{2}){\bf I}_{nd}+u{\bf
X}_{d}\sum_{g\in\Gamma}{\bf H}_{g}\bigotimes\rho(g)+iu^{2}{\bf X}_{d}({\bf
D}_{\Gamma}\bigotimes{\bf I}_{d})).\end{array}$

Substituting $u=1$, we obtain

$\begin{array}[]{rcl}&&\det({\bf
I}_{2md}-\sum_{g\in\Gamma}\rho(g)\bigotimes{\bf U}_{g})\\\ &&\\\
&=&2^{(m-n)d}\det(2{\bf I}_{nd}+{\bf X}_{d}\sum_{g\in\Gamma}{\bf
H}_{g}\bigotimes\rho(g)+i{\bf X}_{d}({\bf D}_{\Gamma}\bigotimes{\bf
I}_{d}))\\\ &&\\\ &=&2^{(m-n)d}\det({\bf X}_{d})\det(2{\bf
X}^{-1}_{d}+\sum_{g\in\Gamma}{\bf H}_{g}\bigotimes\rho(g)+i{\bf
D}_{\Gamma}\bigotimes{\bf I}_{d}).\end{array}$

Then we have

$\det({\bf X}_{d})=\det({\bf X}\bigotimes{\bf I}_{d})=(\det({\bf
X}))^{d}=\frac{2^{nd}}{\prod_{u\in V(G)}(H_{uu}-\lambda-i\Gamma{}_{u})^{d}}.$

Furthermore, since

${\bf X}^{-1}_{d}={\bf X}^{-1}\bigotimes{\bf I}_{d},$

we have

$\begin{array}[]{rcl}(2{\bf X}^{-1}_{d}+i{\bf D}_{\Gamma}\bigotimes{\bf
I}_{d})_{uu}&=&(2\frac{H_{uu}-\lambda-i\Gamma{}_{u}}{2}+i\Gamma{}_{u})\bigotimes{\bf
I}_{d}\\\ &&\\\ &=&(H_{uu}-\lambda)\bigotimes{\bf I}_{d}.\end{array}$

That is,

$2{\bf X}^{-1}_{d}+i{\bf D}_{\Gamma}\bigotimes{\bf I}_{d}=-\lambda{\bf
I}_{nd}+{\rm diag}({\bf H})\bigotimes{\bf I}_{d}.$

Therefore, it follows that

$\begin{array}[]{rcl}&&\det({\bf
I}_{2md}-\sum_{g\in\Gamma}\rho(g)\bigotimes{\bf U}_{g})\\\ &&\\\
&=&\frac{2^{md}}{\prod_{u\in
V(G)}(H_{uu}-\lambda-i\Gamma{}_{u})^{d}}\det(-\lambda{\bf
I}_{nd}+\sum_{g\in\Gamma}{\bf H}_{g}\bigotimes\rho(g)+{\rm diag}({\bf
H})\bigotimes{\bf I}_{d})\\\ &&\\\ &=&\frac{(-1)^{nd}2^{md}}{\prod_{u\in
V(G)}(H_{uu}-\lambda-i\Gamma{}_{u})^{d}}\det(\lambda{\bf
I}_{nd}-\sum_{g\in\Gamma}\rho(g)\bigotimes{\bf H}_{g}-{\bf
I}_{d}\bigotimes{\rm diag}({\bf H})).\end{array}$

$\Box$

By Theorems 6 and 7 the following result holds.

###### Corollary 1

Let $G$ be a connected graph with $m$ edges, $\Gamma$ be a finite group and
$\alpha:D(G)\longrightarrow\Gamma$ be an ordinary voltage assignment. Then

$\det({\bf I}_{2mp}-\tilde{{\bf U}}(\lambda))=\prod_{\rho}{\bf
Z}_{H}(G,\lambda,\rho,\alpha)^{-\deg\rho},$

where $\rho$ runs over all inequivalent irreducible representations of
$\Gamma$ and $p=\mid\Gamma\mid$.

##  7 Example

We give an example. Let $G=K_{3}$ be the complete graph with three vertices
$1,2,3$ and six arcs $e_{1},e_{2},e_{3},e^{-1}_{1},e^{-1}_{2},e^{-1}_{3}$,
where $e_{1}=(v_{1},v_{2}),e_{2}=(v_{2},v_{3}),e_{3}=(v_{3},v_{1})$.
Furthermore, let

${\bf H}=\left[\begin{array}[]{ccc}a&be^{2i\alpha}&be^{2i\alpha}\\\
be^{-2i\alpha}&a&be^{2i\alpha}\\\
be^{-2i\alpha}&be^{-2i\alpha}&a\end{array}\right],$

where $a>0$, $b>0$ and $\alpha\in[-\frac{\pi}{2},\frac{\pi}{2})$. Then we have

$x_{1}=x_{2}=x_{3}=\frac{2}{a-\lambda-2ib}.$

Set $x=\frac{2}{a-\lambda-2ib}$. Considering ${\bf U}(\lambda)$ under the
order $e_{1},e_{2},e_{3},e^{-1}_{1},e^{-1}_{2},e^{-1}_{3}$, we have

${\bf
U}(\lambda)=\left[\begin{array}[]{cccccc}-xbe^{2i\alpha}&-xbe^{2i\alpha}&-xbe^{2i\alpha}&i-xb&-xb&-xb\\\
-xbe^{2i\alpha}&-xbe^{2i\alpha}&-xbe^{2i\alpha}&-xb&i-xb&-xb\\\
-xbe^{2i\alpha}&-xbe^{2i\alpha}&-xbe^{2i\alpha}&-xb&-xb&i-xb\\\
i-xb&-xb&-xb&-xbe^{-2i\alpha}&-xbe^{-2i\alpha}&-xbe^{-2i\alpha}\\\
-xb&i-xb&-xb&-xbe^{-2i\alpha}&-xbe^{-2i\alpha}&-xbe^{-2i\alpha}\\\
-xb&-xb&i-xb&-xbe^{-2i\alpha}&-xbe^{-2i\alpha}&-xbe^{-2i\alpha}\end{array}\right].$

By Theorem 4, we have

$\begin{array}[]{rcl}\det({\bf I}_{6}-{\bf
U}(\lambda))&=&\frac{2^{3}(-1)^{3}}{(a-\lambda-2ib)^{3}}\det(\lambda{\bf
I}_{3}-{\bf H})\\\ &&\\\
&=&\frac{-8}{(a-\lambda-2ib)^{3}}\left[\begin{array}[]{ccc}\lambda-a&-be^{2i\alpha}&-be^{2i\alpha}\\\
-be^{-2i\alpha}&\lambda-a&-be^{2i\alpha}\\\
-be^{-2i\alpha}&-be^{-2i\alpha}&\lambda-a\end{array}\right]\\\ &&\\\
&=&\frac{-8}{(a-\lambda-2ib)^{3}}\\{(\lambda-a)^{3}-3b^{2}(\lambda-a)-b^{3}(e^{2i\alpha}+e^{-2i\alpha})\\}\\\
&&\\\
&=&\frac{-8}{(a-\lambda-2ib)^{3}}\\{(\lambda-a)^{3}-3b^{2}(\lambda-a)-2b^{3}\cos
2\alpha\\}.\end{array}$

Next. let $\Gamma=\mathbb{Z}_{3}=\\{1,\tau,{\tau}^{2}\\}({\tau}^{3}=1)$ be the
cyclic group of order 3, and let
$\alpha:D(K_{3})\longrightarrow\mathbb{Z}_{3}$ be the ordinary voltage
assignment such that $\alpha(e_{1})=\tau$, $\alpha(e^{-1}_{1})=\tau{}^{2}$ and
$\alpha(e_{2})=\alpha(e^{-1}_{2})=\alpha(e_{3})=\alpha(e^{-1}_{3})=1$. Then
the $\mathbb{Z}_{3}$-coverng $K^{\alpha}_{3}$ of $K_{3}$ is the cycle graph of
length 9.

The characters of $\mathbb{Z}_{3}$ are given as follows:
$\chi{}_{i}(\tau{}^{j})=(\xi{}^{i})^{j}$, $0\leq i,j\leq 2$, where
$\xi=\frac{-1+\sqrt{-3}}{2}$. Then we have

${\bf H}_{1}=\left[\begin{array}[]{ccc}0&0&be^{2i\alpha}\\\
&0&be^{2i\alpha}\\\ be^{-2i\alpha}&be^{-2i\alpha}&0\end{array}\right],{\bf
H}_{\tau}=\left[\begin{array}[]{ccc}0&be^{2i\alpha}&0\\\ 0&0&0\\\
0&0&0\end{array}\right],{\bf
H}_{\tau{}^{2}}=\left[\begin{array}[]{ccc}0&0&0\\\ be^{-2i\alpha}&0&0\\\
0&0&0\end{array}\right].$

Now, by Theorem 7,

$\begin{array}[]{rcl}&&{\zeta}_{H}(K_{3},\lambda,\chi{}_{1},\alpha)^{-1}=\frac{2^{3}(-1)^{3}}{(a-\lambda-2ib)^{3}}\det(\lambda{\bf
I}_{3}-\sum^{2}_{j=0}\chi{}_{1}(\tau{}^{j}){\bf H}_{\tau{}^{j}}-{\rm
diag}({\bf H}))\\\ &&\\\
&=&\frac{-8}{(a-\lambda-2ib)^{3}}\left[\begin{array}[]{ccc}\lambda-a&-b\xi
e^{2i\alpha}&-be^{2i\alpha}\\\
-b\xi{}^{2}e^{-2i\alpha}&\lambda-a&-be^{2i\alpha}\\\
-be^{-2i\alpha}&-be^{-2i\alpha}&\lambda-a\end{array}\right]\\\ &&\\\
&=&\frac{-8}{(a-\lambda-2ib)^{3}}\\{(\lambda-a)^{3}-3b^{2}(\lambda-a)-b^{3}(\xi
e^{2i\alpha}+\xi{}^{2}e^{-2i\alpha})\\}\\\ &&\\\
&=&\frac{-8}{(a-\lambda-2ib)^{3}}\\{(\lambda-a)^{3}-3b^{2}(\lambda-a)-2b^{3}\cos
2(\alpha+\pi/3)\\}.\end{array}$

Similarly, we have

$\begin{array}[]{rcl}&&{\zeta}_{H}(K_{3},\lambda,\chi{}_{2},\alpha)^{-1}=\frac{2^{3}(-1)^{3}}{(a-\lambda-2ib)^{3}}\det(\lambda{\bf
I}_{3}-\sum^{2}_{j=0}\chi{}_{2}(\tau{}^{j}){\bf H}_{\tau{}^{j}}-{\rm
diag}({\bf H}))\\\ &&\\\
&=&\frac{-8}{(a-\lambda-2ib)^{3}}\left[\begin{array}[]{ccc}\lambda-a&-b\xi{}^{2}e^{2i\alpha}&-be^{2i\alpha}\\\
-b\xi e^{-2i\alpha}&\lambda-a&-be^{2i\alpha}\\\
-be^{-2i\alpha}&-be^{-2i\alpha}&\lambda-a\end{array}\right]\\\ &&\\\
&=&\frac{-8}{(a-\lambda-2ib)^{3}}\\{(\lambda-a)^{3}-3b^{2}(\lambda-a)-b^{3}(\xi{}^{2}e^{2i\alpha}+\xi
e^{-2i\alpha})\\}\\\ &&\\\
&=&\frac{-8}{(a-\lambda-2ib)^{3}}\\{(\lambda-a)^{3}-3b^{2}(\lambda-a)-2b^{3}\cos
2(\alpha+2\pi/3)\\}.\end{array}$

By Corollary 1, it follows that

$\begin{array}[]{rcl}&&\det({\bf I}_{18}-\tilde{{\bf U}}(\lambda))=\det({\bf
I}_{6}-{\bf
U}(\lambda))\zeta{}_{H}(K_{3},\lambda,\chi{}_{1},\alpha)^{-1}{\zeta}_{S}(K_{H},\lambda,\chi{}_{2},\alpha)^{-1}\\\
&&\\\
&=&\frac{-512}{(a-\lambda-2ib)^{9}}\\{(\lambda-a)^{3}-3b^{2}(\lambda-a)-2b^{3}\cos
2\alpha\\}\\\ &&\\\ &\times&\\{(\lambda-a)^{3}-3b^{2}(\lambda-a)-2b^{3}\cos
2(\alpha+\pi/3)\\}\\{(\lambda-a)^{3}-3b^{2}(\lambda-a)-2b^{3}\cos
2(\alpha+2\pi/3)\\}.\end{array}$

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