数理解析研究所講究録 1270 巻 2002 年 112-125

112

An expression of harmonic vector fieldsof hyperbolic 3-cone-manifolds,
in terms of the hypergeometric functions
Michihiko FUJII and Hiroyuki OCHIAI
落合啓之 (東京工業大学大学院・理工)藤井道彦 (京都大学・総合人間)\S 0. Introduction.
an orientable Riemannian 3-manifold of
singularity along simple closed geodesics .
withconstant sectional
To each component of the singularity , is associated acone angle . The subset $N:=C-\Sigma$
has asmooth, incomplete hyperbolic structure whose metric completion is identical to theof eachsingular hyperbolic structure on . Asufficiently small tubular neighborhood
component of in $N$ has the metric $dr^{2}+\sinh^{2}rd\theta^{2}+\cosh^{2}rd\phi^{2}$ , where is the distancefrom the singular locus, is the distance along the singular locus, is the angular measure
Let be the Laplacian of $N$ with this metric.
around the singular locus defined modulo
In this paper, we give an explicit expression of aharmonic vector field in , by means of
the hypergeometric functions. This expression can be obtain , since asimultaneous ordinary differential equation with variable which is aconsequence of separation of variables(this is equivalent to the equationof the partial differential equation $(\Delta+4)\tau=0$ on
on ), can be solved exactly by means of Riemann’s -equations, where denotes
By ahyperbolic 3cone-manifold, we will$\mathrm{c}\mathrm{u}\mathrm{r}\mathrm{v}\mathrm{a}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}-1$$C$$\mathrm{m}\mathrm{e}\mathrm{m}$$\Sigma$$\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{e}-\Psi \mathrm{p}\mathrm{e}$$\Sigma$$\alpha$$C$$U$$\Sigma$$r$$\theta$$\phi$$\alpha.$$\Delta$$v$$U$$\mathrm{e}\mathrm{d}$

$r,$

$U$

$\Delta v=\mathrm{O}$

$P$

$U$

$\tau$

a1-form dual to . In fact, single ordinary differential equations of higher order, which
are consequences of an elimination of functions from the simultaneous equation, are transinto linear differential equations of Fuchsian type
formed by the substitution
with three singular points. The differential operators of these equations are factorized into
operators which express Riemann’s $P$-equations(see Theorem 3.1). Also it is seen that somerelationships between differential operators of Riemann’s -equations holds (see Theorem3.2). Then the single ordinary differential equations can be solved without integration offunctions. Moreover, if aparameter, which is obtained at the procdure of the separation of
the variables, satisfies agenericity condition (Assumption 5.1), then fundamental systems of$v$$z=( \frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}\prime}{\infty \mathrm{s}\mathrm{h}\mathrm{r}})^{2}$

$P$

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solutions of the simultaneous differential equation can be concretely obtained (see \S 5.1 and\S 5.3). Then the functions consisting the fundamental systems are explicitly represented bymeans of the hypergeometric functions. In this paper, we will give the explicit expressionof (hence ), with some condition on parameters (see the begining of \S 3). See [2] for the$v$$\tau$

general case.

\S 1. Definition of hyperbolic 3-c0ne-manif0lds.
In this section, we give the definition of hyperbolic 3-c0ne-manif0lds (see [1]).
Consider an 3-dimensional manifold which can be triangulated so that the link of eachsimplex is piecewise linear homeomorphic to the standard sphere and give acomplete path
metric on such that the restriction of the metric to each simplex is isomorphic to ageodesic
simplex of constant sectional. The manifold together with the metric above is
caUed ahyperbolic 3-c0ne-manif0ld and denote it again by .
The singular locus Iof ahyperbolic 3-c0ne-manif0ld
consists of the points with noneighborhood isometric to aball in aRiemannian manifold. It is aunion of totally geodesic
closed simplices of dimension 1. At each point of Iin an open 1-simplex, there is acone angle
which is the sum of dihedral angles of 3-simplices containing the point. The subset $C-\Sigma$
has asmooth Riemannian metric of constant curvature -1, but this metric is incomplete
near I.
In this paper we consider hyperbolic 3-c0ne-manif0lds of the following type. Let $M$ be aclosed orientable 3-manifold and Ibe alink in $M$ of components. Let us denote bythe-th component of the link X. We assume that $M$ is the underlying space of ahyperbolic
3-c0ne-manif0ld with singular locus I. The subset $N:=C-\Sigma$ has asmooth Riemannian
metric with constant sectional curvature -1 which is incomplete near each componentofThe metric completion of the hyperbolic structure on $N$ gives rise to $C$ . Eachcomponentof is atotally geodesic submanifold, and in cylindrical coordinates around, the metric has the form
$C$

$C$

$\mathrm{c}\mathrm{u}\mathrm{r}\mathrm{v}\mathrm{a}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}-1$

$C$

$C$

$k$

$\Sigma^{j}$

$i$

$C$

$g$

$\Sigma^{j}$$\Sigma.$$\Sigma^{j}$$\Sigma$$\Sigma^{j}$$g$$dr^{2}+\sinh^{2}rd\theta^{2}+\cosh^{2}rd\phi^{2}$,

where is the distance from the singular locus, is the distance along the singular locus,
is the angular measure around the singular locus defined modulo ci for some.
The number
is acone angle at .
$r$$\phi$$\theta$$\alpha^{j}\in(0, \infty)$$\alpha^{j}$$\Sigma^{j}$114\S 2. Laplacian of hyperbolic 3-cone-manifol.ds.
In this section, we state a situation which we will argue in this paper, and make a preparation on differential geometry. (See Rosenberg [4] for general reference on Riemannian
geometry and Hodgson-Kerckhoff [3] for aspecial setting on hyperbolic 3-c0ne-manif0lds.)
be an orientable hyperbolic 3cone-manifold with singularity X. We assume thatLet
as in \S 1. The subset $N=C-\Sigma$ hae
the singular set forms a link
$C$$\Sigma=\Sigma^{1}\cup\ldots\cup\Sigma^{k}$$\Sigma$$\mathrm{a}$.
smooth Riemannian metric with constant sectional
denote the space of smooth, real-valued pforms of $N$ . Let
Letexterior derivative of smooth real-valued forms on $N$ :$\mathrm{c}\mathrm{u}\mathrm{r}\mathrm{v}\mathrm{a}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}-1$$g$$\Omega^{p}(N)$$d$$\mathrm{L}\mathrm{e}\mathrm{t}*\mathrm{b}\mathrm{e}$:$\Omega^{p}(N)arrow\Omega^{\mathrm{p}+1}(N)$for any real-valued p-form
be the adjoint of :$\phi$

anddN$=\phi\wedge\psi$$(3-p)- \mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}\psi,$be the usual.

the Hodge star operator defined by using the Riemannian metric$g(\phi, *\psi)$

$d$

$g$

on $N$ :,where $dN$ is the volume form of N. Let$\delta$$d$$\delta$:$\Omega^{\mathrm{p}}(N)arrow\Omega^{p-1}(N)$.

Let Abe the Laplacian on smooth real-valued forms for the Riemannian manifold $N$ :$\Delta=d\delta+\delta d$.

be asufficiently small neighborhood of a component of C. Let be the cone angle
the metric
If we use cyh.ndrical coordinatae,
along the component ofinas described in \S 1. We assume that the boundary ofin isconsists of the points whose distances ffom the component of are same. We adapt
thein . We denote by) for the( $dr$ ,sinhrffl,
.
and
Then
orthonormal frame in dual to
and $\phi=x^{3}$ . We express the metric on
For notational convenience, we set $r=x^{1},$

Let

$U$

$\alpha$$U$$(r,\theta,\phi),$$\Sigma.$$dr^{2}+\sinh^{2}r\theta^{2}+\mathrm{c}\mathrm{o}\mathrm{e}\mathrm{h}^{2}rd\phi^{2}$$U$$\Sigma$$U$$(\omega_{1},\{v_{2},\omega_{3}):=$$\mathrm{c}\mathrm{o}\mathrm{e}\mathrm{h}rd\phi$

$U$

$U$

$\mathrm{c}\mathrm{c}\succ \mathrm{r}\mathrm{a}\mathrm{m}\mathrm{e}$$(\omega_{1},\omega_{2},\omega_{3}).$$e_{1}= \frac{\partial}{\partial r},$$(e_{1},e_{2}, e_{3})$$e_{2}= \frac{1}{\epsilon \mathrm{i}\mathrm{n}\mathrm{h}r}\frac{\partial}{\partial\theta}$$\theta=x^{2},$as$\sum_{j}.,g:,jdx^{\dot{\iota}}$$dx^{j}$$g3,3=\mathrm{c}\mathrm{o}\mathrm{e}\mathrm{h}^{2}x^{1}$

and $g:,j=0(i\neq j)$ .

$\Gamma_{j,k}^{1}$$\mathrm{r}\mathrm{j}_{k},=\frac{1}{2}\sum_{l}\dot{g}.,l(\frac{\partial g_{j,l}}{\partial x^{k}}+\frac{\partial g_{k,l}}{\partial x^{j}}-\frac{\partial g_{j,k}}{\partial x^{l}})$where$(g^{k,l})=(g:\mathrm{j})^{-1}$. The Levi-Civita connection$\nabla$.

,can be calculated by$\Gamma_{j,k}^{1}.\frac{\partial}{\partial x^{1}}.$$\nabla_{\frac{\partial}{\delta_{l}J}}\frac{\partial}{\partial x^{k}}=\sum_{1}$$e_{3}= \frac{1\partial}{\mathrm{c}\mathrm{o}\mathrm{e}\mathrm{h}_{t}\partial\phi}$$g$. Then $g1,1=1,$ $g2,2=\sinh^{2}x^{1},$cican be calculated by using the formula
The Christoffel symbol
$U$

$g$

.

115
Adirect calculation shows that$( \nabla_{\frac{\partial}{\theta x^{J}}}\frac{\partial}{\partial x^{k}})$is equal to$(\begin{array}{lll}0 \frac{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}r}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}r}\frac{\partial}{\partial\theta} \frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}r}{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}r}\frac{\partial}{\partial\phi}\frac{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}_{t}\partial}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}\mathrm{r}\partial\theta} -\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}_{\Gamma}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}r_{T\mathrm{r}}^{\partial} 0\frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}\prime}{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}\mathrm{r}}\frac{\partial}{\partial\phi} 0 -\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}r\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}r\frac{\partial}{\theta r}\end{array})$

and moreover, that the connection 1-form$(\omega_{\lambda}^{\mu})$,

is equal to.

$(\begin{array}{lll}0 -\frac{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}_{\mathrm{f}}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}\mathrm{r}}\omega_{2} -\frac{\epsilon \mathrm{i}\mathrm{n}\mathrm{h}\mathrm{r}}{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}\mathrm{r}}\omega_{3}\frac{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}\tau}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}\mathrm{r}}\omega_{2} 0 0\frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}r}{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}\tau}\omega_{3} 0 0\end{array})$

Now let be avector field in $N$ which satisfies the differential equationin .
Namely, is harmonic in . Let be the 1-form dual to . Then, by Weitzenb\"ock formulaand the fact that the Ricci curvature ofsatisfies $(\Delta+4)\tau=0$ in .
If we express as$\Delta v=\mathrm{O}$

$v$

$U$

$v$

$U$

$v$

$\tau$$N\mathrm{i}\mathrm{s}-2,$$U$$\tau$$\tau$$\tau=f(r, \theta, \phi)\omega_{1}+g(r, \theta, \phi)\omega_{2}+h(r, \theta, \phi)\omega_{3}$

in

$U$

, then, by explicit calculation, we obtain the following (see [3] pp.26-27):$(\Delta+4)\tau=$$(-f_{\mathrm{r}r}-( \frac{s}{c}+\frac{c}{s})f_{f}+(\frac{s^{2}}{c^{2}}+\frac{c^{2}}{s^{2}}-2)f-\frac{1}{s^{2}}f_{\theta\theta}-\frac{1}{c^{2}}f_{\phi\phi}+\frac{2c}{s^{2}}g_{\theta}+\frac{2s}{c^{2}}h_{\phi})\omega_{1}$$+$$(-g_{\mathrm{r}r}-( \frac{s}{c}+\frac{c}{s})g_{f}+(\frac{c^{2}}{s^{2}}-2)g-\frac{1}{s^{2}}g_{\theta\theta}-\frac{1}{c^{2}}g_{\phi\phi}-\frac{2c}{s^{2}}f_{\theta})\omega_{2}$$+$$(-h_{ff}-( \frac{s}{c}+\frac{c}{s})h_{f}+(\frac{s^{2}}{c^{2}}-2)h-\frac{1}{s^{2}}h_{\theta\theta}-\frac{1}{c^{2}}h_{\phi\phi}-\frac{2s}{c^{2}}f_{\phi})\omega_{3}$,(1)where subscripts denote derivatives with respect to variables and $s:=\sinh r,$ $c:=\cosh r$ .
The 1-form in satifies equivariance properties depending on the shape of the neighborhood . Since the cone angle is equal to. If the componentof has length 1it further satisfies $\tau(r, \theta, \phi+l)=\tau(r,\theta+t,$ , where measures the twist
in the normal direction along the component of . The complex number $l+t\sqrt{-1}$ is socalled the complex length of the component of the singular locus X.
Because of the decomposition of the Laplacian in , we can use separation of variables,
assuming that $f(r, \theta,$ equals afunction $f(r)$ times afunction on the torus. Similarly for the other functions $g(r, \theta,$ and. It suffices further to decompose thefunctions on the torus into eigenfunctions of the Laplacian on the torus, which are of the$U$$\tau$$U$$\alpha,$$\tau(r, \theta+\alpha, \phi)=\tau(r, \theta, \phi)$$\Sigma$$t$$\Sigma$$U$$(=\partial U)$$h(r,\theta, \phi)$forms $\cos(a\theta+b\phi)$ and $\sin(a\theta+b\phi)$ , where
and $b:= \frac{(2\pi m+\alpha t)}{l}(n, m\in \mathrm{Z})$ . We
say such a1-form is an eigenform of the Laplacian. Then, from the expression (1), we see$a:= \frac{2\pi n}{\alpha}$$\tau$

116
that such a1-form$\tau=$$\tau$

must be of the folowing type:$f(r)\cos(a\theta+b\phi)\omega_{1}+g(r)\sin(a\theta+b\phi)\omega_{2}+h(r)\mathrm{s}$in$(a\theta+b\phi)\omega_{3}$(2),or $\tau=f(r)\sin(a\theta+b\phi)\omega_{1}+g(r)\cos(a\theta+b\phi)\omega_{2}+h(r)\cos(a\theta+b\phi)\omega_{3}$ .
Then, we can verify the following (see the equation (21) in [3]):$(\Delta+4)\tau=0$$\bullet f’’(r)+(\frac{s}{c}+\frac{\mathrm{c}}{s})f’(r)-(2+sp^{2}+\mathrm{c}^{2}+\urcorner a^{2}\pi+F)ssb^{2}f(r)-\frac{2a\mathrm{c}}{s^{2}}g(r)-2\tau hbs(r)=0$$\bullet g’’(r)+(\frac{s}{\mathrm{c}}+\frac{c}{s})\oint(r)-(2+\frac{\mathrm{c}^{2}}{s^{2}}+\frac{a^{2}}{s^{2}}+\frac{b^{2}}{c^{2}})g(r)-\frac{2a\mathrm{c}}{s^{2}}f(r)=0$$\Leftrightarrow\{$

,

,

.

$h”(r)+( \frac{s}{\mathrm{c}}+\frac{c}{s})h’(r)-(2+\frac{s^{2}}{\mathrm{c}^{2}}+\frac{a^{2}}{s^{2}}+p^{2}b)h(r)-2\tau fbs(r)=0$$\bullet$

Put

and

$z^{1/2}:= \frac{\sinh r}{\cosh r}$then we have that$z=( \frac{\sinh \mathrm{r}}{\infty \mathrm{s}\mathrm{h}r})^{2}$$(1-z)^{1/2}:= \frac{1}{\cosh r}$,

and that$(\Delta+4)\tau=0$$\bullet 4z^{2}f’’(z)+4zf’(z)-(\frac{2z}{(1-z)^{2}}+\frac{1}{(1-z)^{2}}+\frac{z^{2}}{(1-z)^{2}}+\frac{a^{2}}{1-z}+\frac{b^{2}z}{1-z})f(z)$$- \frac{2a}{(1-z)^{3/2}}g(z)-\frac{2bz^{3/2}}{(1-z)^{3/2}}h(z)=0$,$\Leftrightarrow\{$,(3)$\bullet 4z^{2}g’’(z)+4zd(z)-(\frac{2z}{(1-z)^{2}}+\frac{1}{(1-z)^{2}}+\frac{a^{2}}{1-z}+\frac{b^{2}z}{1-z})g(z)-\frac{2a}{(1-z)^{3/2}}f(z)=0$$\bullet 4z^{2}h’’(z)+4zh’(z)-(\frac{2z}{(1-z)^{2}}+z^{2}\overline{\overline{-z})}+\frac{a^{2}}{1-z}(1+\frac{b^{2}z}{1-z})h(z)-\frac{2bz^{3/2}}{(1-z)^{3/2}}f(z)=0$.

\S 3. How to solve asingle differential equation.
In this paper, we only consider the case where

and $b\neq 0$ (see [2] for all the othercases). Then we can transform the system of the simultaneous equation (3) to asingle
differential equation of the 6–th order with respect to the function $h(z)$ .
Let Abe asubset of defined by$a\neq \mathrm{O}$$\mathrm{C}$$\Lambda:=\{z\in \mathrm{R} ; z<0,1<z\}$In the rest of this paper, let us regard the variable

number in the domain -A. Thendomain -A.
By the third equation of (3), we have$\mathrm{C}$$z^{1/2}$$z$.

in the equations in (3) as acomplex

and $(1-z)^{1/2}$ are single valued functions on the$\mathrm{C}$$f(z)= \frac{2}{b}z^{\frac{1}{2}}(1-z)^{\frac{3}{2}}R_{1}(z, \frac{d}{dz},a, b)h(z)$,(4)

where we put$R_{1}(z, \frac{d}{dz},a,b):=\frac{d^{2}}{dz^{2}}+\frac{1}{z}\frac{d}{dz}+(\frac{a^{2}}{4z^{2}(z-1)}+\frac{b^{2}-1}{4z(z-1)}+\frac{-3}{4z(z-1)^{2}})$.

117
By eliminating the function $f(z)$ from the first equation in (3) and the relation (4) between$f(z)$ and $h(z)$ , we obtain arelation between $g(z)$ and $h(z)$ as follows:$g(z)= \frac{1}{ab}z^{\frac{-1}{2}}R_{2}(z, \frac{d}{dz}, a, b)h(z)$

,

(5)

where
$R_{2}(z, \frac{d}{dz}, a, b):=-4z^{3}(z-1)^{3}\frac{d^{4}}{dz^{4}}+12z^{2}(z-1)^{2}(1-2z)\frac{d^{3}}{dz^{3}}$$+2z(z-1)(a^{2}+15z-a^{2}z+b^{2}z-13z^{2}-b^{2}z^{2}) \frac{d^{2}}{dz^{2}}$$+(a^{2}-4z-a^{2}z-b^{2}z+3z^{2}+3b^{2}z^{2}-2z^{3}-2b^{2}z^{3}) \frac{d}{dz}$$+ \frac{1}{4z(z-1)}(8a^{2}-a^{4}-26a^{2}z+2a^{4}z-2a^{2}b^{2}z-8z^{2}+20a^{2}z^{2}-a^{4}z^{2}-6b^{2}z^{2}+4a^{2}b^{2}z^{2}$$-b^{4}z^{2}+6z^{3}-2a^{2}z^{3}+8b^{2}z^{3}-2a^{2}b^{2}z^{3}+2b^{4}z^{3}-z^{4}-2b^{2}z^{4}-b^{4}z^{4})$.

By eliminating $f(z)$ and $g(z)$ from the second equation of (3) and the relations (4) and(5), we obtain the following equation which $h(z)$ should satisfy:$h^{(6)}(z)+ \frac{9(-1+2z)}{z(z-1)}h^{(5)}(z)+\frac{72-3a^{2}-394z+3a^{2}z-3b^{2}z+387z^{2}+3b^{2}z^{2}}{4z^{2}(z-1)^{2}}h^{(4)}(z)$$+$$\frac{1}{2z^{3}(z-1)^{3}}(-12+3a^{2}+212z-12a^{2}z+6b^{2}z-543z^{2}+9a^{2}z^{2}-18b^{2}z^{2}+348z^{3}$$+$$12b^{2}z^{3})h^{(3)}(z)+ \frac{1}{16z^{4}(z-1)^{4}}(-12a^{2}+3a^{4}-272z+92a^{2}z-6a^{4}z-24b^{2}z+6a^{2}b^{2}z$$+$

1 $792z^{2}-158a^{2}z^{2}+3a^{4}z^{2}+212b^{2}z^{2}-12a^{2}b^{2}z^{2}+3b^{4}z^{2}-2828z^{3}+78a^{2}z^{3}-362b^{2}z^{3}$$+$$6a^{2}b^{2}z^{3}-6a^{4}z^{3}+1323z^{4}+174b^{2}z^{4}+3b^{4}z^{4})h’’(z)+ \frac{1}{16z^{5}(z-1)^{5}}(-12a^{2}+3a^{4}+24a^{2}z$$6a^{4}z-56z^{2}-6a^{2}z^{2}+3a^{4}z^{2}-44b^{2}z^{2}+6a^{2}b^{2}z^{2}-3b^{4}z^{2}+136z^{3}-12a^{2}z^{3}+148b^{2}z^{3}$-$12a^{2}b^{2}z^{3}+12b^{4}z^{3}-149z^{4}+6a^{2}z^{4}-164b^{2}z^{4}+6a^{2}b^{2}z^{4}-15b^{4}z^{4}+54z^{5}+60b^{2}z^{5}$$+$$6b^{4}z^{5})h’(z)+ \frac{1}{64z^{6}(z-1)^{6}}(-64a^{2}+20a^{4}-a^{6}+232a^{2}z-70a^{4}z+3a^{6}z+12a^{2}b^{2}z$-$3a^{4}b^{2}z-316a^{2}z^{2}+83a^{4}z^{2}-3a^{6}z^{2}-56a^{2}b^{2}z^{2}+9a^{4}b^{2}z^{2}-3a^{2}b^{4}z^{2}-16z^{3}+180a^{2}z^{3}$-$36a^{4}z^{3}+a^{6}z^{3}-48b^{2}z^{3}+82a^{2}b^{2}z^{3}-9a^{4}b^{2}z^{3}-18b^{4}z^{3}+9a^{2}b^{4}z^{3}-b^{6}z^{3}+56z^{4}-35a^{2}z^{4}$$+$$3a^{4}z^{4}+100b^{2}z^{4}-44a^{2}b^{2}z^{4}+3a^{4}b^{2}z^{4}+47b^{4}z^{4}-9a^{2}b^{4}z^{4}+3b^{6}z^{4}-34z^{5}+3a^{2}z^{5}$-$71b^{2}z^{5}+6a^{2}b^{2}z^{5}-40b^{4}z^{5}+3a^{2}b^{4}z^{5}-3b^{6}z^{5}+9z^{6}+19b^{2}z^{6}+11b^{4}z^{6}+b^{6}z^{6}$) $h(z)=0$ .

118This is adifferential equation of Fuchsian tyPe with regular singularities at z $=0,$ z $=1$
. The characteristic exponents are
and z$=\infty$$\pm\frac{a}{2},$$\frac{2\pm a}{2},$(z$\frac{4\pm a}{2}$$=0); \frac{-1}{2},$$\frac{3}{2},$$\frac{1}{2},$$\frac{5}{2},$$\frac{2\pm\sqrt{5}}{2}$(z$=1); \frac{-1\pm b\sqrt{-1}}{2},$$\frac{1\pm b\sqrt{-1}}{2},$$\frac{3\pm b\sqrt{-1}}{2}$(z $=\infty)$ .

Letdenote the differential operator which represents the equation above. Thenthe equation above is written as$X(z, \frac{d}{dz}, a, b)$$X(z, \frac{d}{dz}, a,b)h(z)=0$.

(6)

By direct computation, it can be verified that the theorem below holds:

Theorem 3.1. The differential operator$X(z, \frac{d}{dz}, a, b)$$X(z, \frac{d}{dz}, a, b)$

is factorised as below:$=P_{3}(z, \frac{d}{dz},a,b)P_{2}(z, \frac{d}{dz},a,b)P_{1}(z, \frac{d}{dz},a,b)$$=P_{3}(z, \frac{d}{dz}, -a,b)P_{2}(z, \frac{d}{dz}, -a, b)P_{1}(z, \frac{d}{dz},a,b)$

,

where
$P_{1}(z, \frac{d}{dz}, a, b)$$:=$$\frac{d^{2}}{dz^{2}}+(\frac{1}{z}-\frac{1}{z-1})\frac{d}{dz}+(\frac{a^{2}}{4z^{2}(z-1)}+\frac{b^{2}+1}{4z(z-1)}+\frac{-1}{4z(z-1)^{2}})$,$P_{2}(z, \frac{d}{dz}, a, b)$$:=$$\frac{d^{2}}{dz^{2}}+(\frac{2}{z}+\frac{4}{z-1})\frac{d}{dz}+(\frac{a(a+2)}{4z^{2}(z-1)}+\frac{b^{2}+25}{4z(z-1)}+\frac{5}{4z(z-1)^{2}})$,$P_{3}(z, \frac{d}{dz}, a, b)$$:=$$\frac{d^{2}}{dz^{2}}+(\frac{6}{z}+\frac{6}{z-1})\frac{d}{dz}+(\frac{(a-6)(a+4)}{4z^{2}(z-1)}+\frac{\theta+121}{4z(z-1)}+\frac{21}{4z(z-1)^{2}})$are ones which give Riemann’s
mental solutions are written by the Riemann P-function.
The operators$P_{i}(z, \frac{d}{dz}, a,b)’ \mathrm{s}$-equations and the funda-$P$

By direct computation, it can be checked that some relationship between

and$P_{2}(z, \frac{d}{dz}, a,b)$.

$P_{1}(z, \frac{d}{dz}, a, b)$holds:

Theorem 3.2. Put.

$P_{4}(z, \frac{d}{dz}, a, b):=\frac{d^{2}}{dz^{2}}+(\frac{3}{z}+\frac{3}{z-1})\frac{d}{dz}+(\frac{a^{2}-4}{4z^{2}(z-1)}+\frac{b^{2}+25}{4z(z-1)}+\frac{-1}{4z(z-1)^{2}})$

Then, the follorning equation as operators holds:$P_{1}(z, \frac{d}{dz}, a, b)z^{2}(z-1)^{4}P_{4}(z, \frac{d}{dz}, a, b)-z^{2}(z-1)^{4}P_{3}(z, \frac{d}{dz}, a, b)P_{2}(z, \frac{d}{dz}, a,b)=1$.

119
We obtain acorollary of Theorem 3.1 and Theorem 3.2:
Corollary 3.3. Solutins of the equation$X(z, \frac{d}{dz}, a, b)u(z)=0$

are written as follows:$u(z)=v(z)+z^{2}(z-1)^{4}P_{4}(z, \frac{d}{dz}, a, b)(w^{+}(z)+w^{-}(z))$,

where $v(z),$ $w^{+}(z)$ and $w^{-}(z)$ are solutions of the equations $P_{1}(z, \frac{d}{dz}, a, b)v(z)=0$ ,$P_{2}(z, \frac{d}{dz}, a, b)w^{+}(z)=0$ and $P_{2}(z, \frac{d}{dz}, -a, b)w^{-}(z)=0$ respectively.
To the contrary, if $v(z)$ ,$w^{+}(z)$ and $w^{-}(z)$ are solutionsof the equations $P_{1}(z, \frac{d}{dz}, a, b)v(z)=0$ ,$P_{2}(z, \frac{d}{dz}, a, b)w^{+}(z)=0$ and $P_{2}(z, \frac{d}{dz}, -a, b)w^{-}(z)=0$ respectively, then$u(z):=v(z)+z^{2}(z-1)^{4}P_{4}(z, \frac{d}{dz}, a, b)(w^{+}(z)+w^{-}(z))$

satisfies the equaiion $X(z, \frac{d}{dz}, a, b)u(z)=0$ .
\S 4. Fundamental systems of solutions of the simulataneous differential equation.
For asolution $h(z)$ which is given by Corollrary 3.3, the corresponding functions $f(z)$ and$g(z)$ are obtained by the relations (4) and (5) respectively. These relations are expressed by, the order of which are 2 and 4respectively.
the operators
andWe will see that each of these operators can be reduced to an operator of lower order and$R_{1}(z, \frac{d}{dz}, a, b)$$R_{2}(z, \frac{d}{dz}, a, b)$then give asimple expression of solutions.
By direct computation, we can verify the following lemma on the operator$R_{2}(z, \frac{d}{dz}, a, b)$:

Lemma 4.1. Put$Q(z, \frac{d}{dz}, a,b):=\frac{d^{2}}{dz^{2}}+(\frac{2}{z}+\frac{4}{z-1})\frac{d}{dz}+(\frac{a^{2}}{4z^{2}(z-1)}+\frac{b^{2}+25}{4z(z-1)}+\frac{5}{4z(z-1)^{2}})$.

Then the folloing equation holds:$R_{2}(z, \frac{d}{dz},a, b)-a^{2}=-4z^{3}(z-1)^{3}Q(z, \frac{d}{dz}, a, b)P_{1}(z, \frac{d}{dz}, a, b)$

Let the operators
andraepectively.
$R_{i}$$P_{1}.(z, \frac{d}{dz},$a, b),-a, b) and$P_{i}(z, \frac{d}{dz},$.

a, b) be abbreviated as$R_{:}(z, \frac{d}{dz},$$P_{\dot{l}}$.

$P_{\dot{l}}^{-}$

120
By (4) and (5), the components of each solution $(f(z),g(z),$ $h(z))$ of the simultaneousareequation (3) which corresponds to a solution $bv(z)$ of the$\alpha \mathrm{l}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}P_{1}v(z)=0$,$f(z)$$=$$\frac{2}{b}z^{\frac{1}{2}}(1-z)^{\frac{3}{2}}R_{1}bv(z)=-2z^{\frac{1}{2}}(1-z)^{\frac{1}{2}}(\frac{d}{dz}-\frac{1}{2(z-1)})v(z)$$g(z)$$=$$\frac{1}{ab}z^{\frac{-1}{2}}R_{2}bv(z)=az^{\frac{-1}{2}}v(z)$,$h(z)=bv(z)$ .

ffom the right and
The second equation on $f(z)$ is verified by dividing the operator by
the second equation on $g(z)$ is seen by Lemma 4.1.
Next, we will argue arepresentation of solutions which correspond to solutions of the$P_{1}$

equations $P_{2}w(z)=0$ and $P_{2}^{-}w^{-}(z)=0$ .
By Cor 3.3, the components $(f(z),g(z),$ $h(z))$ of each solution of the simultaneous equation(3) which corresponds to a solution $bw^{+}(z)$ of the equation $P_{2}w^{+}=0$ are$f(z)$$g(z)$$h(z)$$=$$\frac{2}{b}z^{\frac{1}{2}}(1-z)^{\frac{3}{2}}R_{1}z^{2}(z-1)^{4}P_{4}bw^{+}(z)$$=$$az^{f}(1-z)^{\tau}1 \tau(\frac{d}{dz}+\frac{a+2}{2z}+\frac{3}{2(z-1)}-\frac{a^{2}+b^{2}}{2a(z-1)})w^{+}(z)$$=$$\frac{1}{ab}z^{\overline{T}^{1}}R_{2}z^{2}(z-1)^{4}P_{4}hv^{+}(z)$$=$$az^{\frac{1}{2}}(1-z)^{3}( \frac{d}{dz}+\frac{a+2}{2z}+\frac{3}{2(z-1)}-\frac{2}{a(z-1)})w^{+}(z)$$=$$z^{2}(z-1)^{4}P_{4}bw^{+}(z)$$=$$bz(1-z)^{3}( \frac{d}{dz}+\frac{a+2}{2z}+\frac{3}{2(z-1)})w^{+}(z)$

,

,

.

An the second equations above can be verified by dividing the operators by the operator
and $T_{3}(a, b)$ denote the operators which correspond tofrom the right. Let $T_{1}(a,b),$$h(z),$ $g(z)$ and $f(z)$ respectively;$P_{2}$$T_{2}(a)$$T_{1}(a,b)$$:=$$T_{1}(z, \frac{d}{dz}, a, b):=z^{\frac{1}{2}}(1-z)^{\frac{7}{2}}(a\frac{d}{dz}+\frac{a(a+2)}{2z}+\frac{3a}{2(z-1)}-\frac{a^{2}+b^{2}}{2(z-1)})$$T_{2}(a)$$:=$$T_{2}(z, \frac{d}{dz}, a):=az^{f}(1-z)^{3}1(\frac{d}{dz}+\frac{a+2}{2z}+\frac{3}{2(z-1)}-\frac{2}{a(z-1)})$$T_{3}(a,b)$$:=$$T_{3}(z, \frac{d}{dz}, a, b):=bz(1-z)^{3}(\frac{d}{dz}+\frac{a+2}{2z}+\frac{3}{2(z-1)})$

,

,

.

Then the equations above are written ae$f(z)=T_{1}(a,b)w^{+}(z)$ ,$g(z)=T_{2}(a)w^{+}(z)$ ,$h(z)=T_{3}(a,b)w^{+}(z)$.

In the same manner, the components $(f(z),g(z),$ $h(z))$ of each solution of the simultaneous
equation (3) which corresponds to asolution $bw^{-}(z)$ of the equation $P_{2}^{-}w^{-}(z)=0$ are121
represented as follows, by using the operators$f(z)=T_{1}(-a, b)w^{-}(z)$ ,$T_{1},$$T_{2}$

and$T_{3}$$g(z)=-T_{2}(-a)w^{-}(z)$ ,,$h(z)=T_{3}(-a, b)w^{-}(z)$.

Summarizing above, we have the proposition below:
Proposition 4.2. Let $\{v_{1}(z), v_{2}(z)\},$
andbe fundamental
systems of solutions of the equations $P_{1}v(z)=0,$ $P_{2}w^{+}(z)=0$ and $P_{2}^{-}w^{-}(z)=0$ respectively. For each $i\in\{1,2\}$ , put$\{w_{1}^{+}(z), w_{2}^{+}(z)\}$$\{w_{1}^{-}(z), w_{2}^{-}(z)\}$$(f_{\dot{l}}(z),g_{i}(z),$ $h_{i}(z))$$:=$$(-2_{Z^{l}}^{1}(1-z) \tau 1(\frac{d}{dz}-\frac{1}{2(z-1)})v_{i}(z),az^{-\frac{1}{2}}v_{\dot{\iota}}(z),$$(f_{i+2}(z),g_{i+2}(z),$ $h_{i+2}(z))$$:=$$(T_{1}(a, b)w_{i}^{+}(z),T_{2}(a)w_{\mathrm{i}}^{+}(z),$ $T_{3}(a,b)w_{\dot{l}}^{+}(z))$$(f_{1+4}.(z),g_{i+4}(z),$ $h_{i+4}(z))$$:=$$(T_{1}(-a, b)w_{i}^{-}(z),$ $-T_{2}(-a)w_{i}^{-}(z),T_{3}(-a,b)w_{\dot{l}}^{-}(z))$$bv.\cdot(z))$

,

,

,

Then the triples $\{(fj(z),gj(z), h_{j}(z));j=1, \ldots,6\}$ forms a fundamental system of solutions of the simultaneous equations (3) on the domain.
$\theta$$\mathrm{C}-\Lambda$\S 5. Explicit expressions of the fundamental systems of solutions.
In this section, by imposing agenericity condition on the parameter , we will give an
explicit expression of the fundamental systems of solutions of the differential equation (3),$a$by means of the hypergeometric functions.

The characteristic exponents of the equations $P_{1}v(z)=0,$ $P_{2}w^{+}(z)=0$ and $P_{2}^{-}w^{-}(z)=0$

are$\bullet$$\frac{a}{2},$$\bullet$$\frac{a}{2},$$\frac{-a}{2}(z=0);\frac{2+\sqrt{5}}{2},$$\frac{2-\sqrt{5}}{2}(z=1);\frac{-1+b\sqrt{-1}}{2},$$\frac{-a-2}{2}(z=0);\frac{-1}{2},$$\frac{-5}{2}(z=1);\frac{5+b\sqrt{-1}}{2},$$\frac{-1-b\sqrt{-1}}{2}(z=\infty)$,$\frac{5-b\sqrt{-1}}{2}(z=\infty)$

and$\bullet$$\frac{a-2}{2},$$\frac{-a}{2}(z=0);\frac{-1}{2},$$\frac{-5}{2}(z=1);\frac{5+b\sqrt{-}}{2},$respectively. The differences of the exponents at$z=\mathrm{O}$are$\frac{5-b\sqrt{-1}}{2}(z=\infty)$$a,$$-a;a+1,$ $-a-1$ and $a-1$ ,$-a+1$ raepectively.

We will put the following assumption to impose the genericity condition on :$a$

Assumption 5.1. The parameter a is not an integer.
The condition of Assumption 5.1 is equivalent to that no one of a, -a, $a+1,$ -a-l, a-l,$-a+1$ is anegative integer.

122
Then we can and will choose fundamental systems of solutions explicitly as follows:$v_{1}(z)$$:=$$v_{2}(z)$$:=$$w_{1}^{+}(z)$$:=$$w_{2}^{+}(z)$$:=$$w_{1}^{-}(z)$$:=$$w_{2}^{-}(z)$

$:=$

where

1; $z)$ ,$F(\alpha,\beta;\gamma;$z) is the hypergeometric function.

Remark. If we do not put Assumption 5.1, we may employ the standard procedure in thetheory of hypergeometric functions, that is, we may have to take logarithmic terms to form
the fundamental systems of solutions.
Then, by Proposition 4.2 with using the formula$\frac{d}{dz}F(\alpha,\beta;\gamma;z)=\frac{\alpha\beta}{\gamma}F(\alpha+1,\beta+1;\gamma+1;z)$,

we obtain explicitly the fundamental system of solutions of the simultaneous equation (3):$f_{1}(z)$$z^{\frac{a-12z}{2}}(1-z)^{1} \mathit{4}_{5}^{\frac{d}{dz(Z}-\frac{1}{2(z-1)az+})\frac{d}{\ }1}+\sqrt{5}z-a)F(\frac{a+1+b\sqrt{-1}+\sqrt{5}\frac{1}{2}(1-z)^{\frac{1}{2}}(}{-2},\frac{a+1-+\sqrt{5}-\frac{1}{2(z-1)b\acute{-}1})v}{2},\cdot a+1,\cdot z)-\frac{1}{2}(1-z)^{\frac{1}{2}}v_{1}(z)=-2_{Z}(z,a,b)$$==$$- \frac{(a+1+b\sqrt{-1}+\sqrt{5})(a+1-b\frac{\sqrt-1+\sqrt{5}}{}}{-2(a+1)}z^{a1}+(1-z)^{2}\neq^{5}$$\cross F(\frac{a+3+b\sqrt{-1}+\sqrt{5}}{2},$ $\frac{a+3-b\sqrt{-1}+\sqrt{5}}{-2};a+2;z)$$=$$g_{1}(z)$$f_{1}(z, a,b)$,$=$$az^{\frac{-1}{2}}v_{1}(z)=az^{\frac{-1}{2}}v_{1}(z, a, b)$$=$$az^{\frac{a-1}{2}}(1-z)^{\frac{2+\sqrt{5}}{2}}F( \frac{a+1+b\sqrt{-1}+\sqrt{5}}{2},$$=$$h_{1}(z)$::$g_{1}(z, a, b)$$\frac{a+1-b\sqrt{-1}+\sqrt{5}}{-2};a+1;z)$,$=$$bv_{1}(z)=bv_{1}(z,a, b)$$=$$bz^{a}f(1-z)^{\underline{2}} \not\cong F(\frac{a+1+b\sqrt{-1}+\sqrt{5}}{-2},$ $\frac{a+1-b\sqrt{-1}+\sqrt{5}}{2};a+1;z)$$=$:$h_{1}(z, a, b)$,$=$$-2_{Z^{l}}^{1}(1-z)^{1} \mathrm{z}(\frac{d}{dz}-\frac{1}{2(z-1)})v_{2}(z)=-2z^{1}\tau(1-z)^{1}\tau(\frac{d}{dz}-\frac{1}{2(z-1)})v_{2}(z, -a,b)$$=$$f_{1}(z, -a, b)$$g_{2}(z)$$=$$az\overline{\tau}^{1}v_{2}(z)=az\overline{\tau}^{1}v_{1}(z, -a,b)=-g_{1}(z, -a,b)$$h_{2}(z)$$=$$bv_{2}(z)=bv_{1}(z, -a,b)=h_{1}(z, -a, b)$ ,$f_{2}(z)$

,
,

123
$f_{3}(z)$$=$$T_{1}(a, b)w_{1}^{+}(z)=T_{1}(a,b)w_{1}^{+}(z, a, b)$$=$$\frac{1}{2}z^{\frac{a-1}{2}}(1-z)^{2}(-4az-a^{2}z+b^{2}z+2a+2a^{2})F(\frac{a+4+b\sqrt{-1}}{2},$ $\frac{a+4-b\sqrt{-1}}{2};a+2;z)$$+ \frac{a(a+4+b\sqrt{-1})(a+4-b\sqrt{-1})}{4(a+2)}z^{\frac{a+1}{2}}(1-z)^{3}F(\frac{a+6+b\sqrt{-1}}{2},$:$=$$g_{3}(z)$$\frac{a+6-b\sqrt{-1}}{2};a+3;z)$,$f_{3}(z, a, b)$$=$$T_{2}(a)w_{1}^{+}(z)=T_{2}(a)w_{1}^{+}(z,a, b)$$=$$\frac{1}{2}z^{\frac{a-1}{2}}(1-z)^{\frac{3}{2}}(4z-4az-2a^{2}z+2a+2a^{2})F(\frac{a+4+b\sqrt{-1}}{2},$ $\frac{a+4-b\sqrt{-1}}{2};a+2;z)$$+ \frac{a(a+4+b\sqrt{-1})(a+4-b\sqrt{-1})}{4(a+2)}z^{\frac{a+1}{2}}(1-z)^{\frac{5}{2}}F(\frac{a+6+b\sqrt{-1}}{2},$ $\frac{a+6-b\sqrt{-1}}{2};a+3;z)$:$=$$h_{3}(z)$,$g_{3}(z, a,b)$$=$$T_{3}(a, b)w_{1}^{+}(z)=T_{3}(a,b)w_{1}^{+}(z, a, b)$$=$$\frac{b}{2}z^{\frac{a}{2}}(1-z)^{3}z(-4z-2az+2+2a)F(\frac{a+4+b\sqrt{-1}}{2},$ $\frac{a+4-b\sqrt{-1}}{2};a+2;z)$$+ \frac{b(a+4+b\sqrt{-1})(a+4-b\sqrt{-1})}{4(a+2)}z^{\frac{a+2}{2}}(1-z)^{\frac{5}{2}}F(\frac{a+6+b\sqrt{-1}}{2},$:$=$$f_{4}(z)$$\frac{a+6-b\sqrt{-1}}{2};a+3;z)$,$h_{3}(z, a, b)$$=$$T_{1}(a, b)w_{2}^{+}(z)=T_{1}(a,b)w_{1}^{+}(z, -a-2, b)$$=$$\frac{(a^{2}+b^{2}-2a)}{2}z^{\frac{-a-1}{2}}(1-z)^{2}F(\frac{-a+2+b\sqrt{-1}}{2},$ $\frac{-a+2-b\sqrt{-1}}{2};-a;z)$$- \frac{(-a+2+b\sqrt{-}(-a+2-b\sqrt{-1}}{4}z^{\frac{-a-1}{2}}(1-z)^{3}F(\frac{-a+4+b\sqrt{-1}}{2},$ $\frac{-a+4-b\sqrt{-1}}{2};-a+1;z)$:$=$$g_{4}(z)$$f_{4}(z, a, b)$,$=$$T_{2}(a)w_{2}^{+}(z)=T_{2}(a)w_{1}^{+}(z, -a-2, b)$$=$$(2-a)z^{\frac{-a-1}{2}}(1-z)^{\frac{3}{2}}F( \frac{-a+2+b\sqrt{-1}}{2},$ $\frac{-a+2-b\sqrt{-1}}{2};-a;z)$$- \frac{(-a+2+b\sqrt{-1})(-a+2-b\sqrt{-1})}{4}z^{\frac{-a-1}{2}}(1-z)^{\frac{5}{2}}F(_{\overline{\overline{2}}}^{-a+4+b\sqrt{-1}},$:$=$$h_{4}(z)$$g_{4}(z, a,b)$$\frac{-a+4-b\sqrt{-1}}{2};-a+1;z)$,$=$$T_{3}(a, b)w_{2}^{+}(z)=T_{3}(a,b)w_{1}^{+}(z, -a-2, b)$$=$$-bz^{\frac{-a}{2}}(1-z)^{\frac{3}{2}}F( \frac{-a+2+b\sqrt{-1}}{2},$$\frac{-a+2-b\sqrt{-1}}{2};-a;z)$$- \frac{b(-a+2+b\sqrt{-1})(-a+2-b\sqrt{-1})}{4a}z^{\frac{-a}{2}}(1-z)^{\frac{5}{2}}F(\frac{-a+4+b\sqrt{-1}}{2},$ $\frac{-a+4-b\sqrt{-1}}{\overline{2}};-a+1;z)$$=$:$h_{4}(z, a, b)$,,$f_{5}(z)$$=$$T_{1}(-a, b)w_{1}^{-}(z)=T_{1}(-a, b)w_{1}^{+}(z, -a, b)=f_{3}(z, -a, b)$$g_{5}(z)$$=$$-T_{2}(-a)w_{1}^{-}(z)=-T_{2}(-a)w_{1}^{+}(z, -a, b)=-g_{3}(z, -a, b)$$h_{5}(z)$$=$$T_{3}(-a, b)w_{1}^{-}(z)=T_{3}(-a, b)w_{1}^{+}(z, -a, b)=h_{3}(z, -a, b)$$f_{6}(z)$$=$$T_{1}(-a, b)w_{2}^{-}(z)=T_{1}(-a, b)w_{1}^{+}(z, a-2, b)=f_{4}(z, -a, b)$$g_{6}(z)$$=$$-T_{2}(-a)w_{2}^{-}(z)=-T_{2}(-a)w_{1}^{+}(z,a-2, b)=-g_{4}(z, -a, b)$ ,$h_{6}(z)$$=$$T_{3}(-a, b)w_{2}^{-}(z)=T_{3}(-a, b)w_{1}^{+}(z, a-2, b)=h_{4}(z, -a, b)$

,

,
,

.

124
Recall that the parameters a and b are real numbers and that z
to see that the following proposition holds:$\in \mathrm{C}$-A. Then, it is easy

Proposition 5.2. For each $i\in\{1, \ldots,6\}$ ,$f_{\dot{l}}(\overline{z})=\overline{f_{1}.(z)},$$g_{\dot{l}}(\overline{z})=\overline{g_{1}.(z)},$$h_{1}.(\overline{z})=\overline{f_{4}.(z)}$Especially, if $0<z<1$ , then for each $i\in\{1, \ldots,6\},$ $f_{}(z),$ $g:(z),$.
$h_{:}(z)\in \mathrm{R}$.

6. Eigenforms of the Laplacian.

Let $\{(f_{j}(z),g_{j}(z), h_{j}(z));j=1, \ldots,6\}$ be the fundamental system of solutions of the si, which is given in Proposition 4.2.
multaneous equation (3) on the domain
Let $fj(r),$ $gj(r)$ andbe functions of $r(>0)$ obtained by the substitution
respectively.
into the functions
andThen, by summarizing all the argument and calculations in the previous sections, we have:$\mathrm{C}-\Lambda$$h_{j}(r)$$f_{j}(z),$ $g_{j}(z)$$z=( \frac{\epsilon \mathrm{i}\mathrm{n}\mathrm{h}f}{\mathrm{c}\mathrm{o}\mathrm{e}\mathrm{h}f})^{2}$$h_{j}(z)$’s be the functions given as above. Then any harmonic
Theorem 6.1. Let
vector field on , whose dual 1-form is an eigenform (2) of the Laplacian withcondition thatand $b\neq 0$, is given by a linear combination as folloeus (or the sameinterchanged).
$fom$ withand$(f_{j}(r),g_{\mathrm{j}}(r),$

$v$

$U$

$h_{j}(r))$$\theta\iota e$$\tau$$a\not\in \mathrm{Z}$$\sin$$\cos$$v=\Sigma_{j=1}^{6}\{p_{j}f_{j}(r)\infty \mathrm{s}(a\theta+b\phi)e_{1}+q_{j}g_{j}(r)\sin(a\theta+b\phi)e_{2}+r_{\mathrm{j}}h_{j}(r)\sin(a\theta+b\phi)e_{3}\}$where$p_{j},$ $q_{j},r_{j}\in \mathrm{R}$,.

References

1. D. Cooper, C.D. Hodgson and S.P. Kerckhoff, Thrae-dimensional Orbifolds and ConeManifolds, MSJ Memories vol. 5, Mathematical Society of JaPan, 2000.
2. M. Fujii and H. Ochiai, Harmonic vector fields of hyperbolic 3cone-manifolds, Preprint,
2002.

3. C.D. Hodgson and S.P. Kerckhoff, Rigidity of hyperbolic cone-manifolds and hyperbolic
Dehn surgery, J. Diff. Geom. 48(1998), 1-59.

125
4. S. Rosenberg, The Laplacian on aRiemannian Manifold, London Mathematical Society
Student Texts 31, Cambridge, 1997.
5. E.T. Whittaker and G.N. Watson, ACourse of Modern Analysis, Cambridge, 1969.

Division of Mathematics
Faculty of Integrated Human Studies
Kyoto University
SakyO-ku
Kyoto 606-8501
JAPAN
e-mail address: mfujii@math.h.ky0t0-u.ac.jp
Department of Mathematics

Graduate School of Science and Engineering
Tokyo Institute of Technology
Oh-0kayama, Meguro ku
Tokyo 152-8551
JAPAN
e-mail address:ochiai@math.titech.ac.jp