An affine model of a Riemann surface associated to a Schwarz-Christoffel mapping

TL;DR

This paper constructs an affine model of a Riemann surface linked to Schwarz-Christoffel mappings, revealing connections between geodesics and billiard motions in complex polygons.

Contribution

It introduces a novel affine model of a Riemann surface associated with Schwarz-Christoffel mappings and explores its geometric and dynamical properties.

Findings

Established a flat Riemannian metric on the surface

Linked geodesics to billiard trajectories in polygons

Provided insights into the surface's geometric structure

Abstract

In this paper we construct an affine model of a Riemann surface with a flat Riemannian metric associated to a Schwarz-Christoffel mapping of the upper half plane onto a rational triangle. We explain the relation between the geodesics on this Riemann surface and billiard motions in a regular stellated $n$-gon in the complex plane.

Equations

$$F_T:{\mathbb{C}}^{+}\to T\subseteq \mathbb{C}:\xi \mapsto {\int }_0^{\xi }\frac{d\xi }{\eta }.$$

$${\eta }^n={\xi }^{n-n_0}(1-\xi {)}^{n-n_1}.$$

$$F_Q:\mathbb{C}\setminus \{0,1\}\to Q=T\cup \overline{T}$$

$$\mathcal{R}:{\mathcal{S}}_{\mathrm{reg}}\subseteq {\mathbb{C}}^2\to {\mathcal{S}}_{\mathrm{reg}}\subseteq {\mathbb{C}}^2:(\xi ,\eta )\mapsto (\xi ,{\mathrm{e}}^{2\pi i/n}\eta )$$

$$\hat{\pi }:{\mathcal{S}}_{\mathrm{reg}}\subseteq {\mathbb{C}}^2\to \mathbb{C}\setminus \{0,1\}:(\xi ,\eta )\mapsto \xi .$$

$${\delta }_Q:\mathcal{D}\subseteq {\mathcal{S}}_{\mathrm{reg}}\to Q\subseteq \mathbb{C}:(\xi ,\eta )\mapsto (F_Q\circ \hat{\pi })(\xi ,\eta ),$$

$$\displaystyle X(\xi,\eta)=\eta\frac{\partial}{\partial\xi}+\mbox{$\frac{{\scriptstyle n-n_{0}}}{{\scriptstyle n}}$}\,{\xi}^{n-n_{0}-1}(1-\xi)^{n-n_{1}-1}(1-\mbox{$\frac{{\scriptstyle 2n-n_{0}-n_{1}}}{{\scriptstyle n-n_{0}}}$}\xi)\frac{\partial}{\partial\eta}$$

$$\delta :\mathcal{D}\subseteq {\mathcal{S}}_{\mathrm{reg}}\to Q\subseteq \mathbb{C}:(\xi ,\eta )\mapsto (F_Q\circ \hat{\pi })(\xi ,\eta ),$$

$${\delta }_{K^{\ast }}:({\mathcal{S}}_{\mathrm{reg}},\Gamma )\to (K^{\ast },{\gamma }_{K^{\ast }}),$$

$$F_T:{\mathbb{C}}^{+}$$

$$\xi \mapsto {\int }_0^{\xi }\frac{dw}{w^{1-\frac{n_0}{n}}(1-w{)}^{1-\frac{n_1}{n}}}=z$$

$$\big{(}w^{n-n_{0}}(1-w)^{n-n_{1}}\big{)}^{\raisebox{-2.0pt}{$\scriptstyle{1/n}$}}=(r^{n-n_{0}}_{0}r^{n-n_{1}}_{1})^{1/n}{\mathrm{e}}^{i((n-n_{0}){\theta}_{0}+(n-n_{1}){\theta}_{1})/n}.$$

$$F_T(0)=0,F_T(1)=C,\mathrm{and}{F}_T(\infty )=D,$$

$$\displaystyle F_{Q}:\mathbb{C}\setminus\{0,1\}\rightarrow Q=T\cup\overline{T}\subseteq\mathbb{C}:\xi\mapsto z=\left\{\begin{array}[]{rl}F_{T}(\xi),&\mbox{if $\xi\in{\mathbb{C}}^{+}\setminus\{0,1\}$}\\ \rule{0.0pt}{14.0pt}\overline{F_{T}(\overline{\xi})},&\mbox{if $\xi\in\overline{{\mathbb{C}}^{+}\setminus\{0,1\}}$.}\end{array}\right.$$

$$g(\xi ,\eta )={\eta }^n-{\xi }^{n-n_0}(1-\xi {)}^{n-n_1}=0,$$

$$0$$

$$\displaystyle=-(n-n_{0}){\xi}^{n-n_{0}-1}(1-\xi)^{n-n_{1}-1}(1-\mbox{$\frac{{\scriptstyle 2n-n_{0}-n_{1}}}{{\scriptstyle n-n_{0}}}$}\xi)\mathop{\!\,\mathrm{d}\!}\nolimits\xi+n{\eta}^{n-1}\mathop{\!\,\mathrm{d}\!}\nolimits\eta$$

$$G(\xi ,\eta ,\zeta )={\zeta }^{n-n_0-n_1}{\eta }^n-{\xi }^{n-n_0}(\zeta -\xi {)}^{n-n_1}=0.$$

$$\overline{\pi }:\overline{\mathcal{S}}\subseteq {\mathbb{CP}}^2\to \mathbb{CP}:[\xi :\eta :\zeta ]\mapsto [\xi :\zeta ].$$

$$\pi =\overline{\pi }|\mathcal{S}:\mathcal{S}=\overline{\mathcal{S}}\cap \{\zeta =1\}\subseteq {\mathbb{C}}^2\to \mathbb{C}=\mathbb{CP}\cap \{\zeta =1\}:(\xi ,\eta )\mapsto \xi .$$

$$\eta$$

$$\displaystyle=(1-\xi)^{1-\frac{n_{1}}{n}}\big{(}1-(1-\mbox{$\frac{{\scriptstyle n_{0}}}{{\scriptstyle n}}$})(1-\xi)+\cdots\big{)},$$

$$\eta$$

$$\displaystyle=(-1)^{1-\frac{n_{1}}{n}}{\xi}^{1+\frac{n_{\infty}}{n}}\big{(}1-(1-\mbox{$\frac{{\scriptstyle n_{1}}}{{\scriptstyle n}}$})\mbox{$\frac{{\scriptstyle 1}}{{\scriptstyle\xi}}$}+\cdots\big{)},$$

$$2g=n+2-(d_0+d_1+d_{\infty }).$$

$$0=G(\xi ,\eta ,\zeta )={\zeta }^{n-(n_0+n_1)}{\eta }^n-{\xi }^{n-n_0}(\zeta -\xi {)}^{n-n_1}$$

$$(0,0,0)$$

$$\displaystyle=\big{(}-{\xi}^{n-n_{0}-1}(\zeta-\xi)^{n-n_{1}-1}\big{(}(n-n_{0})(\zeta-\xi)-(n-n_{1})\xi\big{)},$$

$$n{\eta }^{n-1}{\zeta }^{n-(n_0+n_1)},(n-(n_0+n_1)){\eta }^n{\zeta }^{n-n_0-n_1-1}$$

$$\displaystyle\hskip 54.2025pt-(n-n_{1}){\xi}^{n-n_{0}}(\zeta-\xi)^{n-n_{1}-1}\big{)}$$

$$\hat{\pi }=\pi |{\mathcal{S}}_{\mathrm{reg}}:{\mathcal{S}}_{\mathrm{reg}}\subseteq {\mathbb{C}}^2\to \mathbb{C}\setminus \{0,1\}:(\xi ,\eta )\mapsto \xi$$

$$X(\xi,\eta)=\eta\frac{\partial}{\partial\xi}+\mbox{$\frac{{\scriptstyle n-n_{0}}}{{\scriptstyle n}}$}\frac{{\xi}^{n-n_{0}-1}(1-\xi)^{n-n_{1}-1}(1-\mbox{$\frac{{\scriptstyle 2n-n_{0}-n_{1}}}{{\scriptstyle n-n_{0}}}$}\xi)}{{\eta}^{n-2}}\frac{\partial}{\partial\eta}.$$

$$T_{(\xi ,\eta )}\hat{\pi }:T_{(\xi ,\eta )}{\mathcal{S}}_{\mathrm{reg}}\to T_{\xi }(\mathbb{C}\setminus \{0,1\})=\mathbb{C}:X(\xi ,\eta )\mapsto \eta \frac{\partial }{\partial \xi }.$$

$$\mathcal{R}:{\mathbb{C}}^2\to {\mathbb{C}}^2:(\xi ,\eta )\mapsto (\xi ,{\mathrm{e}}^{2\pi i/n}\eta ).$$

Full text

1 Introduction

Here we give a detailed description of the contents of this paper.

Consider the conformal Schwarz-Christoffel mapping

[TABLE]

where

[TABLE]

The map $F_{T}$ sends the closed upper half plane ${\mathbb{C}}^{+}$ onto the rational triangle $T=T_{n_{0}n_{1}n_{\infty}}$, where $n_{0}+n_{1}+n_{\infty}=n$ and $1\leq n_{0}\leq n_{1}\leq n_{\infty}$. Because ${F_{T}}{|[0,1]}$ has real values, using the Schwarz reflection principal we extend $F_{T}$ to the conformal map

[TABLE]

of $\mathbb{C}\setminus\{0,1\}$ onto the quadrilateral $Q$.

Following Aurell and Itzykson [1] we associate to the map $F_{Q}$ the affine Riemann surface $\mathcal{S}\subseteq{\mathbb{C}}^{2}$ defined by ($\mathrm{I}\,2$). Then ${\mathcal{S}}_{\mathrm{reg}}=\mathcal{S}\setminus\{(0,0),(1,0)\}$ is a smooth submanifold of ${\mathbb{C}}^{2}\setminus\{\eta=0\}$. To determine the geometry of ${\mathcal{S}}_{\mathrm{reg}}$, we think of $\mathcal{S}$ as the $n$-fold branched covering $\pi:\mathcal{S}\subseteq{\mathbb{C}}^{2}\rightarrow\mathbb{C}:(\xi,\eta)\mapsto\xi$. The map $\pi$ has branch points at [math], $1$, and $\infty$ of degree $\frac{n}{d_{0}}$, $\frac{n}{d_{1}}$, and $\frac{n}{d_{\infty}}$, respectively, where $d_{j}=\gcd(n,n_{j})$ for $j=0,1,\infty$. Using the Riemann-Hurwitz formula, see McKean and Moll [6], it follows that the genus of the compact Riemann surface $\mathrm{cl}(\mathcal{S})\subseteq{\mathbb{CP}}^{2}$ is \mbox{\frac{\scriptstyle 1}{\scriptstyle 2},}(n+2-(d_{0}+d_{1}+d_{\infty})). Here $\mathrm{cl}$ denotes closure. Thus ${\mathcal{S}}_{\mathrm{reg}}$, which is $\mathrm{cl}(\mathcal{S})$ less three points, has the same genus as $\mathrm{cl}(\mathcal{S})$.

We now give a more geometric description of ${\mathcal{S}}_{\mathrm{reg}}$. The abelian group $\widehat{\mathcal{G}}$ generated by

[TABLE]

is the group of covering transformations of the holomorphic covering map

[TABLE]

Let $\mathcal{D}$ be a fundamental domain for the $\widehat{\mathcal{G}}$ action on ${\mathcal{S}}_{\mathrm{reg}}$, which is a “sheet” of the covering map $\widehat{\pi}$ ($\mathrm{I}\,3$). Its image under the map

[TABLE]

which is a holomorphic diffeomorphism of $\mathrm{int}\,\mathcal{D}$ onto $\mathrm{int}\,Q$ and a homeomorphism of $\partial\mathcal{D}$ onto $\partial Q$, is the quadrilateral $Q$.

Let $K^{\ast}=\coprod_{0\leq j\leq n-1}R^{j}\delta(Q)$, where $R:\mathbb{C}\rightarrow\mathbb{C}:z\mapsto{\mathrm{e}}^{2\pi i/n}z$. Then $K^{\ast}$ is a regular stellated $n$-gon, which is invariant under the action of the dihedral group $G$ generated by the rotation $R$ and the reflection $U:\mathbb{C}\rightarrow\mathbb{C}:z\mapsto\overline{z}$, that are subject to the relation $RU=UR^{-1}$. Using $\mathrm{cl}(K^{\ast})$ we build a model ${\widetilde{\mathcal{S}}}_{\mathrm{reg}}$ of the affine Riemann surface ${\mathcal{S}}_{\mathrm{reg}}$ following Richens and Berry [7]. We say that two closed edges $E$ and $E^{\prime}$ of $\mathrm{cl}(K^{\ast})$ are equivalent $\sim$ if they are not adjacent and $E^{\prime}$ is the reflection in the diagonal $R^{m}{\ell}^{j}$, where ${\ell}^{j}=R^{n_{j}}U\ell$ and $\ell$ is the edge of $Q$ contained in the ray $R^{\pi n_{0}/n}({\mathbb{R}}_{>0})$. The $G$ orbit space formed by first identifying equivalent points of $\mathrm{cl}(K^{\ast})$, which are on equivalent edges in $\partial K^{\ast}$ or are points in $\mathrm{int}\,\mathrm{cl}(K^{\ast})$, and then acting on the identification space $(\mathrm{cl}(K^{\ast})\setminus\{O\})^{\sim}$ by the induced action of the group $G$ gives ${\widetilde{\mathcal{S}}}_{\mathrm{reg}}$. Since the action of $G$ on the identification space is free and proper, ${\widetilde{\mathcal{S}}}_{\mathrm{reg}}$ is a smooth $1$-dimensional complex manifold. Its genus is \mbox{\frac{\scriptstyle 1}{\scriptstyle 2},}(n+2-(d_{0}+d_{1}+d_{\infty})). So ${\widetilde{\mathcal{S}}}_{\mathrm{reg}}$ is a model of the affine Riemann surface ${\mathcal{S}}_{\mathrm{reg}}$.

We construct an affine model of ${\widetilde{\mathcal{S}}}_{\mathrm{reg}}$ as follows. Reflecting in the edges of $K^{\ast}\setminus\{O\}$, which is $\mathrm{cl}(K^{\ast})$ less the vertices and center $O$, and then in the edges of the reflected $K^{\ast}\setminus\{O\}$ et cetera , gives $\mathbb{C}\setminus{\mathbb{V}}^{+}$, which is certain translations of $K^{\ast}\setminus\{O\}$ that generate the abelian group $\mathcal{T}$. Here ${\mathbb{V}}^{+}$ is the union of translations of the vertices of $\mathrm{cl}(K^{\ast})$ and its center $O$ by elements of $\mathcal{T}$. The group $\mathfrak{G}=G\ltimes\mathcal{T}$ acts freely, properly, and transitively on the identification space $(\mathbb{C}\setminus{\mathbb{V}}^{+})^{\sim}$ of equivalent points, which are either on equivalent edges of $\mathbb{C}\setminus{\mathbb{V}}^{+}$ or lie in the interior of some $\mathcal{T}$ translate of $K^{\ast}\setminus\{O\}$. The $\mathfrak{G}$ orbit space $(\mathbb{C}\setminus{\mathbb{V}}^{+})^{\sim}/\mathfrak{G}$ of the induced action of $\mathfrak{G}$ is holomorphically differomorphic to ${\widetilde{\mathcal{S}}}_{\mathrm{reg}}$. It is an affine model of ${\mathcal{S}}_{\mathrm{reg}}$ being the space of $\mathfrak{G}$ orbits on $\mathbb{C}\setminus{\mathbb{V}}^{+}$, where $\mathfrak{G}$ is a discrete subgroup of the $2$-dimensional Euclidean group.

We now look at dynamics on the affine Riemann surface ${\mathcal{S}}_{\mathrm{reg}}$. The vector

[TABLE]

is tangent to ${\mathcal{S}}_{\mathrm{reg}}$ at every $(\xi,\eta)\in\mathcal{D}$ and defines a nowhere vanishing holomorphic vector field on the fundamental domain $\mathcal{D}$. Since \frac{\partial}{\partial z}=T_{\xi}F_{Q}\big{(}\eta\frac{\partial}{\partial\xi}\big{)} for every $(\xi,\eta)\in\mathcal{D}$ we get T_{(\xi,\eta)}\delta\,X(\xi,\eta)=\frac{\partial}{\partial z}\rule[-6.0pt]{0.5pt}{14.0pt}\raisebox{-5.0pt}{,{\scriptscriptstyle z=\delta(\xi,\eta)} }, where

[TABLE]

the map $\delta$ ($\mathrm{I}\,6$) straightens the holomorphic vector field $X$ on $\mathcal{D}$. Since $\mathcal{D}$ is a connected open subset of ${\mathcal{S}}_{\mathrm{reg}}$, the map ${\delta}_{Q}$ ($\mathrm{I}\,3$) straightens the holomorphic vector field $X$ on ${\mathcal{S}}_{\mathrm{reg}}$ determined by $X$ on $\mathcal{D}$.

Let $u=\mathrm{Re}\,z$ and $v=\mathrm{Im}\,z$. Then \gamma=\mathop{\!\,\mathrm{d}\!}\nolimits u\raisebox{2.0pt}{\tiny,\bigodot}\mathop{\!\,\mathrm{d}\!}\nolimits u+\mathop{\!\,\mathrm{d}\!}\nolimits v\raisebox{2.0pt}{\tiny,\bigodot}\mathop{\!\,\mathrm{d}\!}\nolimits v=\mathop{\!\,\mathrm{d}\!}\nolimits z\raisebox{2.0pt}{\tiny,\bigodot}\mathop{\!\,\mathrm{d}\!}\nolimits\overline{z} is the Euclidean metric on $\mathbb{C}$. Pulling ${\gamma}_{Q}={\gamma}_{|Q}$ back by the map ${\delta}_{Q}$ ($\mathrm{I}\,3$) gives a Riemannian metric \Gamma=\frac{1}{\eta}\mathop{\!\,\mathrm{d}\!}\nolimits z\raisebox{2.0pt}{\tiny,\bigodot}\frac{1}{\overline{\eta}}\mathop{\!\,\mathrm{d}\!}\nolimits\overline{z} on ${\mathcal{S}}_{\mathrm{reg}}$. Since the metric ${\gamma}_{Q}$ is flat on $Q$, the metric $\Gamma$ on ${\mathcal{S}}_{\mathrm{reg}}$ is flat. In other words, the map ${\delta}_{Q}:({\mathcal{S}}_{\mathrm{reg}},\Gamma)\rightarrow(Q,{\gamma}_{Q})$ is an isometry. Thus ${\delta}_{Q}$ is a developing map in the sense of differential geometry, see Spivak [8, note 12, vol. 2] and Gauss [5]. Since the vector field $X$ on ${\mathcal{S}}_{\mathrm{reg}}$ preserves the metric $\Gamma$, the vector field $X$ ($\mathrm{I}\,5$) on ${\mathcal{S}}_{\mathrm{reg}}$ is the geodesic vector field for the metric $\Gamma$. However, $X$ is incomplete, since the image of a geodesic on ${\mathcal{S}}_{\mathrm{reg}}$ under the map ${\delta}_{Q}$ is a straight line on $Q$, which is parallel to the $u$ axis on $\mathbb{C}$, that runs off $Q$ in finite time. The group $\mathcal{G}$ generated by the mappings $\mathcal{R}:{\mathcal{S}}_{\mathrm{reg}}\rightarrow{\mathcal{S}}_{\mathrm{reg}}:(\xi,\eta)\mapsto(\xi,{\mathrm{e}}^{2\pi i/n}\eta)$ and $\mathcal{U}:{\mathcal{S}}_{\mathrm{reg}}\rightarrow{\mathcal{S}}_{\mathrm{reg}}:(\xi,\eta)\mapsto(\overline{\xi},\overline{\eta})$ preserves the metric $\Gamma$. The map ${\delta}_{Q}$ ($\mathrm{I}\,3$) extends to the developing map

[TABLE]

which is an isometry that intertwines the action of $\mathcal{G}$ on ${\mathcal{S}}_{\mathrm{reg}}$ with the action of $G$ on $K^{\ast}$. Since the geodesic vector field $X$ on ${\mathcal{S}}_{\mathrm{reg}}$ is invariant under the action of $\mathcal{G}$ and the vector field $\frac{\partial}{\partial z}$ on $K^{\ast}$ is invariant under the action of $G$, the map ${\delta}_{K^{\ast}}$ sends geodesics on ${\mathcal{S}}_{\mathrm{reg}}$ to geodesics on $K^{\ast}$. However, incompleteness of the vector field $X$ remains.

Following Richens and Berry [7] we impose the condition that when a geodesic, starting at a point in $\mathrm{int}(\mathrm{cl}(K^{\ast})\setminus\{O\})$, meets $\partial K^{\ast}$ it undergoes a reflection in the edge of $K^{\ast}$ that it meets. Such geodesics never meet a vertex of $\mathrm{cl}(K^{\ast})$. Thus this type of geodesic becomes a billiard motion in $\mathrm{cl}(K^{\ast})\setminus\{O\}$, which is defined for all time. Billiard motions in polygons have been extensively studied. For a nice overview see Berger [3, chpt. XI ] and references therein. An argument shows that $\widehat{\mathcal{G}}$ invariant geodesics on $({\mathcal{S}}_{\mathrm{reg}},\Gamma)$ correspond, under the map ${\delta}_{K^{\ast}\setminus\{O\}}$ ($\mathrm{I}\,7$), to billiard motions on $(\mathrm{cl}(K^{\ast})\setminus\{O\},{\gamma}_{\mathrm{cl}(K^{\ast})\setminus\{O\}})$.

Repeatedly reflecting a billiard motion in an edge of $\mathrm{cl}(K^{\ast})$ and suitable edges of suitable $\mathcal{T}$ translations of $\mathrm{cl}(K^{\ast})$ gives a straight line motion $\lambda$ on $\mathbb{C}\setminus{\mathbb{V}}^{+}$, which is invariant under the action of $\widehat{G}\ltimes\mathcal{T}$. Use the union of $\lambda$ and $U\lambda$, whose intersection with $\mathrm{cl}(K^{\ast})$ is a segment of an extended billiard motion. The image of this extended billiard motion in the orbit space $(\mathbb{C}\setminus{\mathbb{V}}^{+})^{\sim}/\mathfrak{G}={\widetilde{\mathcal{S}}}_{\mathrm{reg}}$ is a geodesic. Here we use the Riemannian metric $\widehat{\gamma}$, which is induced by the $\mathfrak{G}$ invariant Euclidean metric $\gamma$ on $\mathbb{C}\setminus{\mathbb{V}}^{+}$ restricted to $\mathrm{cl}(K^{\ast})\setminus\{O\}$. Consequently, $({\widetilde{\mathcal{S}}}_{\mathrm{reg}},\widehat{\gamma})$ is an affine analogue of the affine Riemann surface ${\mathcal{S}}_{\mathrm{reg}}$ thought of as the orbit space of a discrete subgroup of $\mathrm{PGl}(2,\mathbb{C})$ acting on $\mathbb{C}$ with the Poincaré metric, see Weyl [9].

2 A Schwarz-Christoffel mapping

Consider the conformal Schwarz-Christoffel mapping

[TABLE]

of the upper half plane ${\mathbb{C}}^{+}$ to the rational triangle $T=T_{n_{0}n_{1}n_{\infty}}$ with interior angles $\frac{n_{0}}{n}\pi$, $\frac{n_{1}}{n}\pi$, and $\frac{n_{\infty}}{n}\pi$, see figure $1$. Here $n_{0}+n_{1}+n_{\infty}=n$ and $n_{i}\in{\mathbb{Z}}_{\geq 1}$ for $i=0,1$ and $\infty$ with $1\leq n_{0}\leq n_{1}\leq n_{\infty}$. Because $n_{\infty}$ is greater than or equal to either $n_{0}$ or $n_{1}$, it follows that $OC$ is the longest side of the triangle

[TABLE]

$T=\bigtriangleup OCD$. In the integrand of (1) we use the following choice of complex $n^{\mathrm{th}}$ root. Suppose that $w\in\mathbb{C}\setminus\{0,1\}$. Let $w=r_{0}{\mathrm{e}}^{i{\theta}_{0}}$ and $1-w=r_{1}{\mathrm{e}}^{i{\theta}_{1}}$ where $r_{0},$ $r_{1}\in{\mathbb{R}}_{>0}$ and ${\theta}_{0}$, ${\theta}_{1}\in[0,2\pi)$. For $w\in(0,1)$ on the real axis we have ${\theta}_{0}={\theta}_{1}=0$, $w=r_{0}>0$, and $1-w=r_{1}>0$. So \big{(}w^{n-n_{0}}(1-w)^{n-n_{1}}\big{)}^{\raisebox{-2.0pt}{\scriptstyle{1/n}}}=(r^{n-n_{0}}_{0}r^{n-n_{1}}_{1})^{1/n}. In general for $w\in\mathbb{C}\setminus\{0,1\}$, we have

[TABLE]

From (1) we get

[TABLE]

where $C=\int^{1}_{0}\frac{\mathop{\!\,\mathrm{d}\!}\nolimits w}{w^{1-\frac{n_{0}}{n}}(1-w)^{1-\frac{n_{1}}{n}}}$ and D={\mathrm{e}}^{\frac{n_{0}}{n}\pi i}\big{(}\frac{\raisebox{4.0pt}{{\scriptscriptstyle\sin\frac{n_{1}}{n}\pi}}}{\rule{0.0pt}{7.0pt}\raisebox{0.0pt}{\scriptscriptstyle\sin\frac{n_{\infty}}{n}\pi}}\big{)}C. Consequently, the bijective holomorphic mapping $F_{T}$ sends $\mathrm{int}({\mathbb{C}}^{+}\setminus\{0,1\})$, the interior of the upper half plane less [math] and $1$, onto $\mathrm{int}\,T$, the interior of the rational triangle $T=T_{n_{0}n_{1}n_{\infty}}$, and sends the boundary of ${\mathbb{C}}^{+}\setminus\{0,1\}$ to the edges of $\partial T$ less their end points $O$, $C$ and $D$, see figure 1. Thus the image of ${\mathbb{C}}^{+}\setminus\{0,1\}$ under $F_{T}$ is $\mathrm{cl}(T)\setminus\{O,C,D\}$. Here $\mathrm{cl}(T)$ is the closure of $T$ in $\mathbb{C}$.

Because $F_{T}|_{[0,1]}$ is real valued, we may use the Schwarz reflection principle to extend $F_{T}$ to the holomorphic diffeomorphism

[TABLE]

Here $Q=Q_{n_{0}n_{1}n_{\infty}}$ is a quadrilateral with internal angles $2\pi\frac{n_{0}}{n}$, $\pi\frac{n_{\infty}}{n}$, $2\pi\frac{n_{1}}{n}$, and $\pi\frac{n_{\infty}}{n}$ and vertices at $O$, $D$, $C$, and $\overline{D}$, see figure 2. The conformal mapping $F_{Q}$

[TABLE]

sends $\mathbb{C}\setminus\{0,1\}$ onto $\mathrm{cl}(Q)\setminus\{O,D,C,\overline{D}\}$.

3 The geometry of an affine Riemann surface

Let $\xi$ and $\eta$ be coordinate functions on ${\mathbb{C}}^{2}$. Consider the affine Riemann surface $\mathcal{S}={\mathcal{S}}_{n_{0},n_{1},n_{\infty}}$ in ${\mathbb{C}}^{2}$, associated to the holomorphic mapping $F_{Q}$, defined by

[TABLE]

see [1].

We determine the singular points of $\mathcal{S}$ by solving

[TABLE]

For $(\xi,\eta)\in\mathcal{S}$, we have $\mathop{\!\,\mathrm{d}\!}\nolimits g(\xi,\eta)=0$ if and only if $(\xi,\eta)=(0,0)$ or $(1,0)$. Thus the set ${\mathcal{S}}_{\mathrm{sing}}$ of singular points of $\mathcal{S}$ is $\{(0,0),(1,0)\}$. So the affine Riemann surface ${\mathcal{S}}_{\mathrm{reg}}=\mathcal{S}\setminus{\mathcal{S}}_{\mathrm{sing}}$ is a complex submanifold of ${\mathbb{C}}^{2}$. Actually, ${\mathcal{S}}_{\mathrm{reg}}\subseteq{\mathbb{C}}^{2}\setminus\{\eta=0\}$, for if $(\xi,\eta)\in\mathcal{S}$ and $\eta=0$, then either $\xi=0$ or $\xi=1$.

Lemma 2.1 Topologically ${\mathcal{S}}_{\mathrm{reg}}$ is a compact Riemann surface $\overline{\mathcal{S}}\subseteq{\mathbb{C}\mathbb{P}}^{2}$ of genus $2g=n+2-(d_{0}+d_{1}+d_{\infty})$ less three points: $[0:0:1]$, $[1:0:1]$, and $[0:1:0]$. Here $d_{i}=\gcd(n_{i},n)$ for $i=0,1,\infty$,

Proof. Consider the (projective) Riemann surface $\overline{\mathcal{S}}\subseteq{\mathbb{C}\mathbb{P}}^{2}$ specified by the condition $[\xi:\eta:\zeta]\in\overline{\mathcal{S}}$ if and only if

[TABLE]

Thinking of $G$ as a polynomial in $\eta$ with coefficients which are polynomials in $\xi$ and $\zeta$, we may view $\overline{\mathcal{S}}$ as the branched covering

[TABLE]

When $\zeta=1$ we get the affine branched covering

[TABLE]

From (5) it follows that $\eta={\omega}_{k}({\xi}^{n-n_{0}}(1-\xi)^{n-n_{1}})^{1/n}$, where ${\omega}_{k}$ for $k=0,1,\ldots,n-1$ is an $n^{\mathrm{th}}$ root of unity with and $(\,\,\,\,)^{1/n}$ is the complex $n^{\mathrm{th}}$ root used in the definition of the mapping $F_{T}$ (1). Thus the branched covering mapping $\overline{\pi}$ (8) has $n$ “sheets” except at its branch points. Since

[TABLE]

$\infty$ is a branch point of the mapping $\overline{\pi}$ of degree $\frac{n}{d_{\infty}}$, where $d_{\infty}=\mathrm{gcd}(n,n_{\infty})$. Hence the ramification index of [math], $1$, $\infty$ is $d_{0}(\frac{n}{d_{0}}-1)=n-d_{0}$, $n-d_{1}$, and $n-d_{\infty}$, respectively. Thus the map $\overline{\pi}$ has $d_{0}$ fewer sheets at [math], $d_{1}$ fewer at $1$, and $d_{\infty}$ fewer at $\infty$ than an $n$-fold covering of $\mathbb{C}\mathbb{P}$. Thus the total ramification index $r$ of the mapping $\overline{\pi}$ is $r=(n-d_{0})+(n-d_{1})+(n-d_{\infty})$. By the Riemann-Hurwitz formula, the genus $g$ of $\overline{\mathcal{S}}$ is $r=2n+2g-2$. In other words,

[TABLE]

Consequently, the affine Riemann surface $\mathcal{S}$ is the compact connected surface $\overline{\mathcal{S}}$ less the point at $\infty$, namely, $\mathcal{S}=\overline{\mathcal{S}}\setminus\{[0:1:0]\}$. So ${\mathcal{S}}_{\mathrm{reg}}$ is the compact connected surface $\overline{\mathcal{S}}$ less three points: $[0:0:1]$, $[1:0:1]$, and $[0:1:0]$. $\square$

Examples of $\overline{\mathcal{S}}={\overline{\mathcal{S}}}_{n_{0},n_{1},n_{\infty}}$

$n_{0}=1$, $n_{1}=1$, $n_{\infty}=1$; $n=3$. So $d_{0}=d_{1}=d_{\infty}=1$. Hence $2g=5-3=2$. So $g=1$.

$n_{0}=1$, $n_{0}=1$, $n_{\infty}=4$; $n=6$. So $d_{0}=1$, $d_{1}=1$, $d_{\infty}=2$. Hence $2g=8-4=4$. So $g=2$.

$n_{0}=1$, $n_{1}=2$, $n_{\infty}=3$; $n=6$. So $d_{0}=1$, $d_{1}=2$, $d_{\infty}=3$. Hence $2g=8-6=2$. So $g=1$.

$n_{0}=2$, $n_{1}=2$, $n_{\infty}=3$; $n=7$. So $d_{0}=d_{1}=d_{\infty}=1$. Hence $2g=9-3=6$. So $g=3$. $\square$

Below is a table listing all the partitions $\{n_{1},n_{0},n_{\infty}\}$ of $n$, which give a low genus Riemann surface $\overline{\mathcal{S}}={\overline{\mathcal{S}}}_{n_{0},n_{1},n_{\infty}}$

[TABLE]

Table 1.Genus $g$ of $\overline{\mathcal{S}}={\overline{\mathcal{S}}}_{n_{0},n_{1},n_{\infty}}$. This table is based on the table in Aurell and Itzykson [1, p.193].

Corollary 2.1a If $n$ is an odd prime number and $\{n_{1},n_{0},n_{\infty}\}$ is a partition of $n$ into three parts, then the genus of $\overline{\mathcal{S}}$ is \mbox{\frac{\scriptstyle 1}{\scriptstyle 2},}(n-1).

Proof. Because $n$ is prime, we get $d_{0}=d_{1}=d_{\infty}=1$. Using (11) we obtain g=\mbox{\frac{\scriptstyle 1}{\scriptstyle 2},}(n-1). $\square$

Corollary 2.1b The singular points of the Riemann surface $\overline{\mathcal{S}}$ are $[0:0:1]$, $[1:0:1]$, and if $n_{\infty}>1$ then also $[0:1:0]$.

Proof. A point $[\xi:\eta:\zeta]\in{\overline{\mathcal{S}}}_{\mathrm{sing}}$ if and only if $[\xi:\eta:\zeta]\in\overline{\mathcal{S}}$, that is,

[TABLE]

We need only check the points $[0:0:1]$, $[1:0:1]$ and $[0:1:0]$. Since the first two points are singular points of $\mathcal{S}=\overline{\mathcal{S}}\setminus\{[0:1:0]\}$, they are singular points of $\overline{\mathcal{S}}$. Thus we need to see if $[0:1:0]$ is a singular point of $\overline{\mathcal{S}}$. Substituting $(0,1,0)$ into the right hand side of (12b) we get \left\{\!\!\!\!\begin{array}[]{l}(0,0,1),\mbox{\, if n_{\infty}=n-(n_{0}+n_{1})=1}\\ (0,0,0),\mbox{\, if n_{\infty}>1.}\end{array}\right. Thus $[0:1:0]$ is a singular point of $\overline{\mathcal{S}}$ only if $n_{\infty}>1$. $\square$

Lemma 2.2 The mapping

[TABLE]

is a surjective holomorphic local diffeomorphism.

Proof. Let $(\xi,\eta)\in{\mathcal{S}}_{\mathrm{reg}}$ and let

[TABLE]

By hypothesis $(\xi,\eta)\in{\mathcal{S}}_{\mathrm{reg}}$ implies that $\eta\neq 0$. The vector $X(\xi,\eta)$ is defined and is nonzero. From (X\mbox{,\rule{8.0pt}{0.5pt}\rule{0.5pt}{6.0pt},,}\mathop{\!\,\mathrm{d}\!}\nolimits g)(\xi,\eta)=0 and $T_{(\xi,\eta)}{\mathcal{S}}_{\mathrm{reg}}=\ker\mathop{\!\,\mathrm{d}\!}\nolimits g(\xi,\eta)$, it follows that $X(\xi,\eta)\in T_{(\xi,\eta)}{\mathcal{S}}_{\mathrm{reg}}$. Using the definition of $X(\xi,\eta)$ (14) and the definition of the mapping $\pi$ (9), we see that the tangent of the mapping $\widehat{\pi}$ (13) at $(\xi,\eta)\in{\mathcal{S}}_{\mathrm{reg}}$ is given by

[TABLE]

Since $X(\xi,\eta)$ and $\eta\frac{\partial}{\partial\xi}$ are nonzero vectors, they form a complex basis for $T_{(\xi,\eta)}{\mathcal{S}}_{\mathrm{reg}}$ and $T_{\xi}(\mathbb{C}\setminus\{0,1\})$, respectively. Thus the complex linear mapping $T_{(\xi,\eta)}\widehat{\pi}$ is an isomorphism. Hence $\widehat{\pi}$ is a local holomorphic diffeomorphism. $\square$

Corollary 2.2a $\widehat{\pi}$ (13) is a surjective holomorphic $n$ to $1$ covering map.

Proof. We only need to show that $\widehat{\pi}$ is a covering map. First we note that every fiber of $\widehat{\pi}$ is a finite set with $n$ elements, since for each fixed $\xi\in\mathbb{C}\setminus\{0,1\}$ we have ${\widehat{\pi}}^{-1}(\xi)=\{(\xi,\eta)\in{\mathcal{S}}_{\mathrm{reg}}\,\rule[-4.0pt]{0.5pt}{13.0pt}\,\eta={\omega}_{k}({\xi}^{n-n_{0}}(1-\xi)^{n-n_{1}})^{1/n}\}$. Here ${\omega}_{k}$ for $k=0,1,\ldots,n-1$, is an $n^{\mathrm{th}}$ root of $1$ and $(\,\,\,\,)^{1/n}$ is the complex $n^{\mathrm{th}}$ root used in the definition of the Schwarz-Christoffel map $F_{Q}$ (4). Hence the map $\widehat{\pi}$ is a proper surjective holomorphic submersion, because each fiber is compact. Thus the mapping $\widehat{\pi}$ is a presentation of a locally trivial fiber bundle with fiber consisting of $n$ distinct points. In other words, the map $\widehat{\pi}$ is a $n$ to $1$ covering mapping. $\square$

Consider the group $\widehat{\mathcal{G}}$ of linear transformations of ${\mathbb{C}}^{2}$ generated by

[TABLE]

Clearly ${\mathcal{R}}^{n}={\mathrm{id}}_{{\mathbb{C}}^{2}}=e$, the identity element of $\widehat{\mathcal{G}}$ and $\widehat{\mathcal{G}}=\{e,\mathcal{R},\ldots,{\mathcal{R}}^{n-1}\}$. For each $(\xi,\eta)\in\mathcal{S}$ we have

[TABLE]

So $\mathcal{R}(\xi,\eta)\in\mathcal{S}$. Thus we have an action of $\widehat{\mathcal{G}}$ on the affine Riemann surface $\mathcal{S}$ given by

[TABLE]

Since $\widehat{\mathcal{G}}$ is finite, and hence is compact, the action $\Phi$ is proper. For every $g\in\widehat{\mathcal{G}}$ we have ${\Phi}_{g}(0,0)=(0,0)$ and ${\Phi}_{g}(1,0)=(1,0)$. So ${\Phi}_{g}$ maps ${\mathcal{S}}_{\mathrm{reg}}$ into itself. At $(\xi,\eta)\in{\mathcal{S}}_{\mathrm{reg}}$ the isotropy group ${\widehat{\mathcal{G}}}_{(\xi,\eta)}$ is $\{e\}$, that is, the $\widehat{\mathcal{G}}$-action $\Phi$ on ${\mathcal{S}}_{\mathrm{reg}}$ is free. Thus the orbit space ${\mathcal{S}}_{\mathrm{reg}}/\widehat{\mathcal{G}}$ is a complex manifold.

Corollary 2.2b The holomorphic $\widehat{\mathcal{G}}$-principal bundle

[TABLE]

Here $[(\xi,\eta)]$ is the $\widehat{\mathcal{G}}$-orbit $\{{\Phi}_{g}(\xi,\eta)\in{\mathcal{S}}_{\mathrm{reg}}\,\rule[-4.0pt]{0.5pt}{13.0pt}\,g\in\widehat{\mathcal{G}}\}$ of $(\xi,\eta)$ in ${\mathcal{S}}_{\mathrm{reg}}$. The bundle presented by the mapping $\rho$ is isomorphic to the bundle presented by the mapping $\widehat{\pi}$ (13).

Proof. We use invariant theory to determine the orbit space $\mathcal{S}/\widehat{\mathcal{G}}$. The algebra of polynomials on ${\mathbb{C}}^{2}$, which are invariant under the $\widehat{\mathcal{G}}$-action $\Phi$, is generated by ${\pi}_{1}=\xi\,\,\,\mathrm{and}\,\,\,{\pi}_{2}={\eta}^{n}$. Since $(\xi,\eta)\in\mathcal{S}$, these polynomials are subject to the relation

[TABLE]

Equation (17) defines the orbit space $\mathcal{S}/\widehat{\mathcal{G}}$ as a complex subvariety of ${\mathbb{C}}^{2}$. This subvariety is homeomorphic to $\mathbb{C}$, because it is the graph of the function ${\pi}_{1}\mapsto{\pi}^{n-n_{0}}_{1}(1-{\pi}_{1})^{n-n_{1}}$. Consequently, the orbit space ${\mathcal{S}}_{\mathrm{reg}}/\widehat{\mathcal{G}}$ is holomorphically diffeomorphic to $\mathbb{C}\setminus\{0,1\}$.

It remains to show that $\widehat{\mathcal{G}}$ is the group of covering transformations of the bundle presented by the mapping $\widehat{\pi}$ (13). For each $\xi\in\mathbb{C}\setminus\{0,1\}$ look at the fiber ${\widehat{\pi}}^{-1}(\xi)$. If $(\xi,\eta)\in{\widehat{\pi}}^{-1}(\xi)$, then ${\mathcal{R}}^{\pm 1}(\xi,\eta)=(\xi,{\mathrm{e}}^{\pm 2\pi i/n}\eta)\in{\mathcal{S}}_{\mathrm{reg}}$, since $(\xi,{\mathrm{e}}^{\pm 2\pi i/n}\eta)\neq(0,0)$ or $(1,0)$ and $(\xi,{\mathrm{e}}^{\pm 2\pi i/n}\eta)\in\mathcal{S}$. Thus {\Phi}_{{\mathcal{R}}^{\pm 1}}\big{(}{\widehat{\pi}}^{-1}(\xi)\big{)}\subseteq{\widehat{\pi}}^{-1}(\xi). So {\widehat{\pi}}^{-1}(\xi)\subseteq{\Phi}_{\mathcal{R}}\big{(}{\widehat{\pi}}^{-1}(\xi)\big{)}\subseteq{\widehat{\pi}}^{-1}(\xi). Hence {\Phi}_{\mathcal{R}}\big{(}{\widehat{\pi}}^{-1}(\xi)\big{)}={\widehat{\pi}}^{-1}(\xi). Thus ${\Phi}_{\mathcal{R}}$ is a covering transformation for the bundle presented by the mapping $\widehat{\pi}$. So $\widehat{\mathcal{G}}$ is a subgroup of the group of covering transformations. These groups are equal because $\widehat{\mathcal{G}}$ acts transitively on each fiber of the mapping $\widehat{\pi}$. $\square$

4 Another model for ${\mathcal{S}}_{\mathrm{reg}}$

In this section we construct another model ${\widetilde{S}}_{\mathrm{reg}}$ for the smooth part ${\mathcal{S}}_{\mathrm{reg}}$ of the affine Riemann surface $\mathcal{S}$ (5).

Let $\mathcal{D}\subseteq{\mathcal{S}}_{\mathrm{reg}}$ be a fundamental domain for the $\widehat{\mathcal{G}}$ action $\Phi$ (16) on ${\mathcal{S}}_{\mathrm{reg}}$. So $(\xi,\eta)\in\mathcal{D}$ if and only if for $\xi\in\mathbb{C}\setminus\{0,1\}$ we have \eta=\big{(}{\xi}^{n-n_{0}}(1-\xi)^{n-n_{1}}\big{)}^{\raisebox{-2.0pt}{\scriptstyle 1/n}}. Here $(\,\,\,\,)^{1/n}$ is the $n^{\mathrm{th}}$ root used in the definition of the mapping $F_{Q}$ (4). The domain $\mathcal{D}$ is a connected subset of ${\mathcal{S}}_{\mathrm{reg}}$ with nonempty interior. Its image under the map $\widehat{\pi}$ (13) is $\mathbb{C}\setminus\{0,1\}$. Thus $\mathcal{D}$ is one “sheet” of the covering map $\widehat{\pi}$. So $\widehat{\pi}|_{\mathcal{D}}$ is one to one.

[TABLE]

Using the extended Schwarz-Christoffel mapping $F_{Q}$ (4), we give a more geometric description of the fundamental domain $\mathcal{D}$. Consider the mapping

[TABLE]

where the map $\widehat{\pi}$ is given by equation (13). The map $\delta$ is a holomorphic diffeomorphism of $\mathrm{int}\,\mathcal{D}$ onto $\mathrm{int}\,Q$, which sends $\partial\mathcal{D}$ homeomorphically onto $\partial Q$. Look

[TABLE]

at $\mathrm{cl}(Q)$, which is a closed quadrilateral with vertices $O$, $D$, $C$, and $\overline{D}$. The set $\delta(\mathcal{D})$ contains the open edges $OD$, $DC$, and $C\overline{D}$ but not the open edge $O\overline{D}$ of $\mathrm{cl}(Q)$, see figure 3 above. Let K^{\ast}=K^{\ast}_{n_{0},n_{1},n_{\infty}}={\amalg}_{0\leq j\leq n-1}R^{j}\big{(}\delta(\mathcal{D})\big{)} be the region in $\mathbb{C}$ formed by repeatedly rotating $Q=\delta(\mathcal{D})$ through an angle $2\pi/n$. Here $R$ is the rotation $\mathbb{C}\rightarrow\mathbb{C}:z\mapsto{\mathrm{e}}^{2\pi i/n}z$. We say that the quadrilateral $Q=Q_{2n_{0},n_{\infty},2n_{1},n_{\infty}}$ forms $K^{\ast}$ less its vertices, see figure 4 above.

Claim 3.1 The connected set $K^{\ast}$ is a regular stellated $n$-gon with its $2n$ vertices omitted, which is formed from the quadrilateral $Q^{\prime}=OD^{\prime}C\overline{D^{\prime}}$, see figure 5.

Proof. By construction the quadrilateral $Q^{\prime}=OD^{\prime}C\overline{D^{\prime}}$ is contained in the quadrilateral $Q=ODC\overline{D}$. Note that $Q\subseteq\bigcup^{[\frac{n_{1}+1}{2}]}_{j=[-\frac{n_{1}+1}{2}]}R^{j}(Q^{\prime})$. Thus

[TABLE]

So $K^{\ast}=\bigcup^{n}_{j=0}R^{j}(Q^{\prime})$. Thus $K^{\ast}$ is the regular stellated $n$-gon, one of whose sides is the diagonal $D^{\prime}\overline{D^{\prime}}$ of $Q^{\prime}$. $\square$

[TABLE]

We would like to extend the mapping $\delta$ (18) to a mapping of ${\mathcal{S}}_{\mathrm{reg}}$ onto $K^{\ast}$. Let

[TABLE]

where $\Phi$ is the $\widehat{\mathcal{G}}$ action defined in equation (16). So we have a mapping

[TABLE]

defined by $({\delta}_{K^{\ast}})|_{{\Phi}_{{\mathcal{R}}^{j}}(\mathcal{D})}={\delta}|_{{\Phi}_{{\mathcal{R}}^{j}}(\mathcal{D})}$. The mapping ${\delta}_{K^{\ast}}$ is defined on ${\mathcal{S}}_{\mathrm{reg}}$, because ${\mathcal{S}}_{\mathrm{reg}}={\amalg}_{0\leq j\leq n-1}{\Phi}_{{\mathcal{R}}^{j}}(\mathcal{D})$, since $\mathcal{D}$ is a fundamental domain for the $\widehat{\mathcal{G}}$-action $\Phi$ (16) on ${\mathcal{S}}_{\mathrm{reg}}$. Because K^{\ast}={\amalg}_{0\leq j\leq n-1}R^{j}\big{(}\delta(\mathcal{D})\big{)}, the mapping ${\delta}_{K^{\ast}}$ is surjective. Hence ${\delta}_{K^{\ast}}$ is holomorphic, since it is continuous on ${\mathcal{S}}_{\mathrm{reg}}$ and is holomorphic on the dense open subset ${\amalg}_{0\leq j\leq n-1}{\mathcal{R}}^{j}(\mathrm{int}\,\mathcal{D})$ of ${\mathcal{S}}_{\mathrm{reg}}$.

Let $U:\mathbb{C}\rightarrow\mathbb{C}:z\mapsto\overline{z}$ and let $G$ be the group generated by the rotation $R$ and the reflection $U$ subject to the relations $R^{n}=U^{2}=e$ and $RU=UR^{-1}$. Shorthand $G=\langle U,R\,\rule[-4.0pt]{0.5pt}{13.0pt}\,\,U^{2}=e=R^{n}\,\,\&\,\,RU=UR^{-1}\rangle$. Then $G=\{e;R^{p}U^{\ell},\,\ell=0,1\,\,\&\,\,p=0,1,\ldots,n-1\}$. The group $G$ is the dihedral group $D_{2n}$.

The closure $\mathrm{cl}(K^{\ast})$ of $K^{\ast}={\amalg}_{0\leq j\leq n-1}R^{j}(Q)$ is invariant under $\widehat{G}$, the subgroup of $G$ generated by the rotation $R$. Because the quadrilateral $Q$ is invariant under the reflection $U:z\mapsto\overline{z}$, and $UR^{j}=R^{-j}U$, it follows that $\mathrm{cl}(K^{\ast})$ is invariant under the reflection $U$. So $\mathrm{cl}(K^{\ast})$ is invariant under the group $G$.

We now look at some group theoretic properties of $K^{\ast}$.

Lemma 3.2 If $F$ is a closed edge of the polygon $\mathrm{cl}(K^{\ast})$ and $g|_{F}={\mathrm{id}}|_{F}$ for some $g\in G$, then $g=e$.

Proof. Suppose that $g\neq e$. Then $g=R^{p}U^{\ell}$ for some $\ell\in\{0,1\}$ and some $p\in\{0,1,\ldots,n-1\}$. Let $g=R^{p}U$ and suppose that $F$ is an edge of $\mathrm{cl}(K^{\ast})$ such that $\mathrm{int}(F)\cap\mathbb{R}\neq\varnothing$, where $\mathbb{R}=\{\mathrm{Re}\,z\,\rule[-4.0pt]{0.5pt}{13.0pt}\,\,z\in\mathbb{C}\}$. Then $U(F)=F$, but $U|_{F}\neq{\mathrm{id}}_{F}$. So $g|_{F}=R^{p}U|_{F}\neq{\mathrm{id}}_{F}$. Now suppose that $\mathrm{int}(F)\cap\mathbb{R}=\varnothing$. Then $U(F)\neq F$. So $U|_{F}\neq{\mathrm{id}}_{F}$. Hence $g|_{F}\neq{\mathrm{id}}_{F}$. Finally, suppose that $g=R^{p}$ with $p\neq 0$. Then $g(F)\neq F$. So $g|_{F}\neq{\mathrm{id}}|_{F}$. $\square$

Lemma 3.3 For $j=0,1,\infty$ put $S^{(j)}=R^{n_{j}}U$. Then $S^{(j)}$ is a reflection in the closed ray ${\ell}^{j}=\{t{\mathrm{e}}^{i\,\pi n_{j}/n}\in\mathbb{C}\,\rule[-4.0pt]{0.5pt}{13.0pt}\,\,t\in OD\}$. The closed ray ${\ell}^{0}$ is the closure of the side $OD$ of the quadrilateral $Q=ODC\overline{D}$ in figure $5$.

Proof. $S^{(j)}$ fixes every point on the closed ray ${\ell}^{j}$, because

[TABLE]

Since $(S^{(j)})^{2}=(R^{n_{j}}U)(R^{n_{j}}U)=R^{n_{j}}(UU)R^{-n_{j}}=e$, it follows that $S^{(j)}$ is a reflection in the closed ray ${\ell}^{j}$. $\square$

Corollary 3.3a For every $j=0,1,\infty$ and every $k\in\{0,1,\ldots,n-1\}$ let $S^{(j)}_{k}=R^{k}S^{(j)}R^{-k}$. Here $S^{(j)}_{n}=S^{(j)}_{0}=S^{(j)}$, because $R^{n}=e$. Then $S^{(j)}_{k}$ is a reflection in the closed ray $R^{k}{\ell}^{j}$.

Proof. This follows because $(S^{(j)}_{k})^{2}=R^{k}(S^{(j)})^{2}R^{-k}=e$ and $S^{(j)}_{k}$ fixes every point on the closed ray $R^{k}{\ell}^{j}$, for

[TABLE]

Corollary 3.3b For every $j=0,1,\infty$, every $S^{(j)}_{k}$ with $k=0,1,\ldots,n-1$, and every $g\in G$, we have $gS^{(j)}_{k}g^{-1}=S^{(j)}_{r}$ for a unique $r\in\{0,1,\ldots,n-1\}$.

Proof. We compute. For every $k=0,1,\ldots,n-1$ we have

[TABLE]

and

[TABLE]

where $t=-(k+2n_{j})\bmod n$. Since $R$ and $U$ generate the group $G$, the corollary follows. $\square$

Corollary 3.3c For $j=0,1,\infty$ let $G^{j}$ be the group generated by the reflections $S^{(j)}_{k}$ for $k=0,1,\ldots,n-1$. Then $G^{j}$ is a normal subgroup of $G$.

Proof. Clearly $G^{j}$ is a subgroup of $G$. From equations (20) and (21) it follows that $gS^{(j)}_{k}g^{-1}\in G^{j}$ for every $g\in G$ and every $k=0,1\ldots,n-1$, since $G$ is generated by $R$ and $U$. But $G^{j}$ is generated by the reflections $S^{(j)}_{k}$ for $k=0,1,\ldots,n-1$, that is, every $g^{\prime}\in G^{j}$ may be written as $S^{(j)}_{i_{1}}\cdots S^{(j)}_{i_{p}}$, where for $\ell\in\{1,\ldots p\}$ we have $i_{\ell}\in\{0,1,\ldots,n-1\}$. So $gg^{\prime}g^{-1}=g(S^{(j)}_{i_{1}}\cdots S^{(j)}_{i_{p}})g^{-1}=(gS^{(j)}_{i_{1}}g^{-1})\cdots(gS^{(j)}_{i_{p}}g^{-1})\in G^{j}$ for every $g\in G$, that is, $G^{j}$ is a normal subgroup of $G$. $\square$

As a first step in constructing ${\widetilde{S}}_{\mathrm{reg}}$ from the regular stellated $n$-gon $K^{\ast}$ we look at certain pairs of edges of $\mathrm{cl}(K^{\ast})$. We say two distinct closed edges $E$ and $E^{\prime}$ of $\mathrm{cl}(K^{\ast})$ are adjacent if and only if they intersect at a vertex of $\mathrm{cl}(K^{\ast})$. For $j=0,1,\infty$ let ${\mathcal{E}}^{j}$ be the set of unordered pairs of closed edges $E$ and $E^{\prime}$ of $\mathrm{cl}(K^{\ast})$, that is, the edges $E$ and $E^{\prime}$ are not adjacent and $E^{\prime}=S^{(j)}_{m}(E)$ for some generator $S^{(j)}_{m}$ of $G^{j}$. Recall that for $x$ and $y$ in some set, the unordered pair $[x,y]$ is precisely one of the ordered pairs $(x,y)$ or $(y,x)$. Geometrically, two nonadjacent closed edges $E^{\prime}$ and $E$ of $\mathrm{cl}(K^{\ast})$ are equivalent if and only if $E^{\prime}$ is obtained from $E$ by reflection in the line $R^{m}{\ell}^{j}$ for some $m\in\{0,1,\ldots,n-1\}$.

In figure 7, where $K^{\ast}=K^{\ast}_{1,1,4}$, parallel edges of $K^{\ast}$, which are labeled with the same letter, are $G^{0}$-equivalent. This is no coincidence.

[TABLE]

Lemma 3.4 Let $K^{\ast}$ be formed from the quadrilateral $Q=T\cup\overline{T}$, where $T$ is the isosceles rational triangle $T_{n_{0}n_{0}n_{\infty}}$ less its vertices. Then nonadjacent edges of $\partial\,\mathrm{cl}(K^{\ast})$ are $G^{0}$-equivalent if and only if they are parallel, see figure 6.

Proof. In figure 6 let $OAB$ be the triangle $T$ with $\angle AOB=\alpha$, $\angle OAB=\beta$, and $\angle ABO=\gamma$. Let $OABA^{\prime\prime}$ be the quadrilateral formed by reflecting the triangle $OAB$ in its edge $OB$. The quadrilateral $OABA^{\prime\prime}$ reflected it its edge $OA$ is the quadrilateral $OAB^{\prime}A^{\prime}$. Let $AC^{\prime}$ be perpendicular to $A^{\prime}B^{\prime}$ and $AC$ be perpendicular to $A^{\prime\prime}B$, see figure 6. Then $CAC^{\prime}$ is a straight line if and only if $\angle C^{\prime}AB^{\prime}+\angle B^{\prime}AB+\angle BAC=\pi$. By construction $\angle C^{\prime}AB^{\prime}=\angle BAC=\pi/2-2\gamma$ and $\angle B^{\prime}AB=2\pi-2\beta$. So

[TABLE]

if and only if $\alpha=\gamma$. Hence the edges $A^{\prime\prime}B$ and $A^{\prime}B^{\prime}$ are parallel if and only if the triangle $OAB$ is isosceles. $\square$

Theorem 3.5 Let $K^{\ast}$ be the regular stellated $n$-gon formed from the rational quadrilateral $Q_{n_{0}n_{1}n_{\infty}}$ with $d_{j}=\mathrm{gcd}(n_{j},n)$ for $j=0,1,\infty$. The $G$ orbit space formed by first identifiying equivalent edges of the regular stellated $n$-gon $K^{\ast}$ less $O$ and then acting on the identification space by the group $G$ is ${\widetilde{S}}_{\mathrm{reg}}$, which is a smooth $2$-sphere with $g$ handles, where $2g=n+2-(d_{0}+d_{1}+d_{\infty})$ less some points corresponding to the image of the vertices of $\mathrm{cl}(K^{\ast})$.

Before we begin proving theorem 3.5 we consider the following special case. Let $K^{\ast}=K^{\ast}_{1,1,4}$ be a regular stellated hexagon formed by repeatedly rotating the quadrilateral $Q^{\prime}=OD^{\prime}C\overline{D^{\prime}}$ by $R$ through an angle $2\pi/6$, see figure 7.

Let $G^{0}$ be the group generated by the reflections $S^{(0)}_{k}=R^{k}S^{(0)}R^{-k}=R^{2k+1}U$ for $k=0,1,\ldots,5$. Here $S^{(0)}=RU$ is the reflection which leaves the closed ray ${\ell}^{0}=\{t{\mathrm{e}}^{i\pi/6}\,\rule[-4.0pt]{0.5pt}{13.0pt}\,\,t\in OD^{\prime}\}$ fixed. Define an equivalence relation on $\mathrm{cl}(K^{\ast})$

[TABLE]

by saying that two points $x$ and $y$ in $\mathrm{cl}(K^{\ast})$ are equivalent, $x\sim y$, if and only if 1) $x$ and $y$ lie on $\partial\,\mathrm{cl}(K^{\ast})$ with $x$ on the closed edge $E$ and $y=S^{(0)}_{m}(x)\in S^{(0)}_{m}(E)$ for some reflection $S^{(0)}_{m}\in G^{0}$ or 2) if $x$ and $y$ lie in the interior of $\mathrm{cl}(K^{\ast})$ and $x=y$. Let $\mathrm{cl}(K^{\ast})^{\sim}$ be the space of equivalence classes and let

[TABLE]

be the identification map which sends a point $p\in\mathrm{cl}(K^{\ast})$ to the equivalence class $[p]$, which contains $p$. Give $\mathrm{cl}(K^{\ast})$ the topology induced from $\mathbb{C}$. Placing the quotient topology on $\mathrm{cl}(K^{\ast})^{\sim}$ turns it into a connected topological manifold without boundary. Let $K^{\ast}$ be $\mathrm{cl}(K^{\ast})$ less its vertices. The identification space $(K^{\ast}\setminus\{O\})^{\sim}=\pi(K^{\ast}\setminus\{O\})$ is a connected $2$-dimensional smooth manifold without boundary.

Let $G=\langle R,U\,\rule[-4.0pt]{0.5pt}{13.0pt}\,\,R^{6}=e=U^{2}\,\,\&\,\,RU=UR^{-1}\rangle$. The usual $G$-action

[TABLE]

preserves equivalent edges of $\mathrm{cl}(K^{\ast})$ and is free on $K^{\ast}\setminus\{O\}$. Hence it induces a $G$ action on $(K^{\ast}\setminus\{O\})^{\sim}$, which is free and proper. Thus its orbit map

[TABLE]

is surjective, smooth, and open. The orbit space ${\widetilde{S}}_{\mathrm{reg}}=\sigma((K^{\ast}\setminus\{O\})^{\sim})$ is a connected $2$-dimensional smooth manifold. The identification space $(K^{\ast}\setminus\{O\})^{\sim}$ has the orientation induced from an orientation of $K^{\ast}\setminus\{O\}$, which comes from $\mathbb{C}$. So ${\widetilde{S}}_{\mathrm{reg}}$ has a complex structure, since each element of $G$ is a conformal mapping of $\mathbb{C}$ into itself.

Our aim is to specify the topology of ${\widetilde{S}}_{\mathrm{reg}}$. The regular stellated hexagon $K^{\ast}\setminus\{O\}$ less the origin has a triangulation ${\mathcal{T}}_{K^{\ast}\setminus\{O\}}$ made up of $12$ open triangles $R^{j}(\bigtriangleup OCD^{\prime})$ and $R^{j}(\bigtriangleup OC{\overline{D}}^{\prime})$ for $j=0,1,\ldots,5$; $24$ open edges $R^{j}(OC)$, $R^{j}(O{\overline{D}}^{\prime})$, $R^{j}(C{\overline{D}}^{\prime})$, and $R^{j}(CD^{\prime})$ for $j=0,1,\ldots,5$; and $12$ vertices $R^{j}(D^{\prime})$ and $R^{j}(C)$ for $j=0,1,\ldots,5$, see figure 7.

Consider the set ${\mathcal{E}}^{0}$ of unordered pairs of equivalent closed edges of $\mathrm{cl}(K^{\ast})$, that is, ${\mathcal{E}}^{0}$ is the set $[E,S^{(0)}_{k}(E)]$ for $k=0,1,\ldots,5$, where $E$ is a closed edge of $\mathrm{cl}(K^{\ast})$. Table 1 lists the elements of ${\mathcal{E}}^{0}$.

[TABLE]

Table 1.

Elements of the set ${\mathcal{E}}^{0}$. Here $D^{\prime}_{k}=R^{k}(D^{\prime})$ and $\overline{D^{\prime}_{k}}=R^{k}(\overline{D^{\prime}})$ for $k=0,2,4$ and $C_{k}=R^{k}(C)$ for $k=\{0,1,\ldots,5\}$, see figure 7.

$G$ acts on ${\mathcal{E}}^{0}$, namely, g\cdot[E,S^{(0)}_{k}(E)]=[g(E),gS^{(0)}_{k}g^{-1}\big{(}g(E)\big{)}], for $g\in G$. Since $G^{0}$ is the group generated by the reflections $S^{(0)}_{k}$, $k=0,1,\ldots,5$, it is a normal subgroup of $G$. Hence the action of $G$ on ${\mathcal{E}}^{0}$ restricts to an action of $G^{0}$ on ${\mathcal{E}}^{0}$ and permutes $G^{0}$-orbits in ${\mathcal{E}}^{0}$. Thus the set of $G^{0}$-orbits in ${\mathcal{E}}^{0}$ is $G$-invariant.

We now look at the $G^{0}$-orbits on ${\mathcal{E}}^{0}$. We compute the $G^{0}$-orbit of $d\in{\mathcal{E}}^{0}$ as follows. We have

[TABLE]

Since

[TABLE]

and

[TABLE]

So the $G^{0}$ orbit $G^{0}\cdot d$ of $d\in{\mathcal{E}}^{0}$ is $(G^{0}/\langle UR|\,(UR)^{2}=e\rangle)\cdot d=H^{0}\cdot d=\{a,d,e\}$. Here $H^{0}=\langle V=R^{2}\,\rule[-4.0pt]{0.5pt}{13.0pt}\,\,V^{3}=e\rangle$, since $G^{0}=\langle V=R^{2},UR\,\rule[-4.0pt]{0.5pt}{13.0pt}\,\,V^{3}=e=(UR)^{2}\,\,\&\,\,V(UR)=(UR)V^{-1}\rangle$. Similarly, the $G^{0}$-orbit $G^{0}\cdot f$ of $f\in{\mathcal{E}}^{0}$ is $H^{0}\cdot f=\{b,c,f\}$. Since $G^{0}\cdot d\,\cup\,G^{0}\cdot f={\mathcal{E}}^{0}$, we have found all $G^{0}$-orbits on ${\mathcal{E}}^{0}$. The $G$-orbit of $OC$ is $R^{j}(OC)$ for $j=0,1,\ldots,5$, since $U(OC)=OC$; while the $G$-orbit of $OD^{\prime}$ is $R^{j}(OD^{\prime})$, $R^{j}(O\overline{D^{\prime}})$ for $j=0,1,\ldots,5$, since $U(OD^{\prime})=O\overline{D^{\prime}}$.

Suppose that $B$ is an end point of the closed edge $E$ of $\mathrm{cl}(K^{\ast})$. Then $E$ lies in a unique $[E,S^{(0)}_{m}(E)]$ of ${\mathcal{E}}^{0}$. Let $G^{0}\cdot[E,S^{(0)}_{m}(E)]$ be the $G^{0}$-orbit of $[E,S^{(0)}_{m}(E)]$. Then $g^{\prime}\cdot B$ is an end point of the closed edge $g^{\prime}(E)$ of $g^{\prime}\cdot[E,S^{(0)}_{m}(E)]\in{\mathcal{E}}^{0}$ for every $g^{\prime}\in G^{0}$. So $\mathcal{O}(B)=\{g^{\prime}\cdot B\,\rule[-4.0pt]{0.5pt}{13.0pt}\,\,g^{\prime}\in G^{0}\}$ the $G^{0}$-orbit of the vertex $B$. It follows from the classification of $G^{0}$-orbits on ${\mathcal{E}}^{0}$ that $\mathcal{O}(D^{\prime})=\{D^{\prime},D^{\prime}_{2},D^{\prime}_{4}\}$ and $\mathcal{O}(\overline{D^{\prime}})=\{\overline{D^{\prime}},{\overline{D^{\prime}}}_{2},{\overline{D^{\prime}}}_{4}\}$ are $G^{0}$-orbits of the vertices of $\mathrm{cl}(K^{\ast})$, which are permuted by the action of $G$ on ${\mathcal{E}}^{0}$. Also $\mathcal{O}(C)=\{C,C_{1},\ldots,C_{5}\}$ and $\mathcal{O}(D^{\prime}\&\overline{D^{\prime}})=\{D^{\prime},\overline{D^{\prime}},D^{\prime}_{2},{\overline{D^{\prime}}}_{2},D^{\prime}_{4},{\overline{D^{\prime}}}_{4}\}$ are $G$-orbits of vertices of $\mathrm{cl}(K^{\ast})$, which are end points of the $G$-orbit of the rays $OC$ and $OD^{\prime}$, respectively.

[TABLE]

To determine the topology of the $G$ orbit space ${\widetilde{S}}_{\mathrm{reg}}$ we find a triangulation of ${\widetilde{S}}_{\mathrm{reg}}$. Note that the triangulation ${\mathcal{T}}_{K^{\ast}\setminus\{O\}}$ of $K^{\ast}\setminus\{O\}$, illustrated in figure 7, is $G$-invariant. Its image under the identification map $\pi$ is a $G$-invariant triangulation ${\mathcal{T}}_{(K^{\ast}\setminus\{O\}}$ of $(K^{\ast}\setminus\{O\})^{\sim}$. After identification of equivalent edges, each vertex $\pi(v)$, each open edge $\pi(E)$, having $\pi(O)$ as an end point, or each open edge $\pi([F,F^{\prime}])$, where $[F,F^{\prime}]$ is a pair of equivalent edges of $\mathrm{cl}(K^{\ast})$, and each open triangle $\pi(T)$ in ${\mathcal{T}}_{(K^{\ast}\setminus\{O\})^{\sim}}$ lies in a unique $G$ orbit. It follows that $\sigma(\pi(v))$, $\sigma(\pi(E))$ or $\sigma(\pi([F,F^{\prime}]))$, and $\sigma(\pi(T))$ is a vertex, an open edge, and an open triangle, respectively, of a triangulation ${\mathcal{T}}_{{\widetilde{S}}_{\mathrm{reg}}}=\sigma({\mathcal{T}}_{(K^{\ast}\setminus\{O\})^{\sim}})$ of ${\widetilde{S}}_{\mathrm{reg}}$. The triangulation ${\mathcal{T}}_{{\widetilde{S}}_{\mathrm{reg}}}$ has $4$ vertices, corresponding to the $G$ orbits $\sigma(\pi(\mathcal{O}(D^{\prime})))$, $\sigma(\pi(\mathcal{O}(\overline{D^{\prime}})))$, $\sigma(\pi(\mathcal{O}(C)))$, and $\sigma(\pi(\mathcal{O}(D^{\prime}\&\overline{D^{\prime}})))$; $18$ open edges corresponding to $\sigma(\pi(R^{j}(OC)))$, $\sigma(\pi(R^{j}(OD^{\prime})))$, and $\sigma(\pi(R^{j}(CD^{\prime})))$ for $j=0,1,\ldots,5$; and $12$ open triangles $\sigma(\pi(R^{j}(\bigtriangleup OCD^{\prime})))$ and $\sigma(\pi(R^{j}(\bigtriangleup OC\overline{D^{\prime}})))$ for $j=0,1,\ldots,5$. Thus the Euler characteristic $\chi({\widetilde{S}}_{\mathrm{reg}})$ of ${\widetilde{S}}_{\mathrm{reg}}$ is $4-18+12=-2$. Since ${\widetilde{S}}_{\mathrm{reg}}$ is a $2$-dimensional smooth real manifold, $\chi({\widetilde{S}}_{\mathrm{reg}})=2-2g$, where $g$ is the genus of ${\widetilde{S}}_{\mathrm{reg}}$. Hence $g=2$. So ${\widetilde{S}}_{\mathrm{reg}}$ is a smooth $2$-sphere with $2$ handles, less a finite number of points, which lies in a compact topological space $\widetilde{S}=\mathrm{cl}(K^{\ast})^{\sim}/G$, that is its closure. $\square$

Proof of theorem 3.5 We now begin the construction of ${\widetilde{S}}_{\mathrm{reg}}$ by identifying equivalent edges of $\mathrm{cl}(K^{\ast})$. Let $[E,S^{(0)}_{m}(E)]$ be an unordered pair of equivalent closed edges of $\mathrm{cl}(K^{\ast})$. We say that $x$ and $y$ in $\mathrm{cl}(K^{\ast})$ are equivalent, $x\sim y$, if 1) $x$ and $y$ lie in $\partial\,\mathrm{cl}(K^{\ast})$ with $x\in E$ and $y=S^{(0)}_{m}(x)\in S^{(0)}_{m}(E)$ for some $m\in\{0,1,\ldots,n-1\}$ or 2) $x$ and $y$ lie in $\mathrm{int}\,\mathrm{cl}(K^{\ast})$ and $x=y$. The relation $\sim$ is an equivalence relation on $\mathrm{cl}(K^{\ast})$. Let $\mathrm{cl}(K^{\ast})^{\sim}$ be the set of equivalence classes and let

[TABLE]

be the map which sends $p$ to the equivalence class $[p]$, that contains $p$. Compare this argument with that of Richens and Berry [7]. Give $\mathrm{cl}(K^{\ast})$ the topology induced from $\mathbb{C}$ and put the quotient topology on $\mathrm{cl}(K^{\ast})^{\sim}$.

Claim 3.6 Let $K^{\ast}$ be $\mathrm{cl}(K^{\ast})$ less its vertices. Then $(K^{\ast}\setminus\{O\})^{\sim}=\pi(K^{\ast}\setminus\{O\})$ is a smooth manifold. Also $\mathrm{cl}(K^{\ast})^{\sim}$ is a topological manifold.

Proof. To show that $(K^{\ast}\setminus\{O\})^{\sim}$ is a smooth manifold, let $E_{+}$ be an open edge of $K^{\ast}$. For $p_{+}\in E_{+}$ let $D_{p_{+}}$ be a disk in $\mathbb{C}$ with center at $p_{+}$, which does not contain a vertex of $\mathrm{cl}(K^{\ast})$. Set $D^{+}_{p_{+}}=K^{\ast}\cap D_{p_{+}}$. Let $E_{-}$ be an open edge of $K^{\ast}$, which is equivalent to $E_{+}$ via the reflection $S^{(0)}_{m}$, that is, $[\mathrm{cl}(E_{+}),\mathrm{cl}(E_{-})=S^{(0)}_{m}(\mathrm{cl}(E_{+}))]\in{\mathcal{E}}^{0}$ is an unordered pair of $S^{(0)}_{m}$ equivalent edges. Let $p_{-}=S^{(0)}_{m}(p_{+})$ and set $D^{-}_{p_{-}}=S^{(0)}_{m}(D^{+}_{p_{+}})$. Then $V_{[p]}=\pi(D^{+}_{p_{+}}\cup D^{-}_{p_{-}})$ is an open neighborhood of $[p]=[p_{+}]=[p_{-}]$ in $(K^{\ast}\setminus\{O\})^{\sim}$, which is a smooth $2$-disk, since the identification mapping $\pi$ is the identity on $\mathrm{int}\,K^{\ast}$. It follows that $(K^{\ast}\setminus\{O\})^{\sim}$ is a smooth $2$-dimensional manifold without boundary.

We now handle the vertices of $\mathrm{cl}(K^{\ast})$. Let $v_{+}$ be a vertex of $\mathrm{cl}(K^{\ast})$ and set $D_{v_{+}}=\widetilde{D}\cap\mathrm{cl}(K^{\ast})$, where $\widetilde{D}$ is a disk in $\mathbb{C}$ with center at the vertex $v_{+}=r_{0}{\mathrm{e}}^{i\pi{\theta}_{0}}$. The map

[TABLE]

with $r\geq 0$ and $0\leq\theta\leq 1$ is a homeomorphism, which sends the wedge with angle $\pi$ to the wedge with angle $\pi s$. The latter wedge is formed by the closed edges $E^{\prime}_{+}$ and $E_{+}$ of $\mathrm{cl}(K^{\ast})$, which are adjacent at the vertex $v_{+}$ such that ${\mathrm{e}}^{i\pi s}E^{\prime}_{+}=E_{+}$ with the edge $E^{\prime}_{+}$ being swept out through $\mathrm{int}\,\mathrm{cl}(K^{\ast})$ during its rotation to the edge $E_{+}$. Because $\mathrm{cl}(K^{\ast})$ is a rational regular stellated $n$-gon, the value of $s$ is a rational number for each vertex of $\mathrm{cl}(K^{\ast})$. Let $E_{-}=S^{(0)}_{m}(E_{+})$ be an edge of $\mathrm{cl}(K^{\ast})$, which is equivalent to $E_{+}$ and set $v_{-}=S(v_{+})$. Then $v_{-}$ is a vertex of $\mathrm{cl}(K^{\ast})$, which is the center of the disk $D_{v_{-}}=S^{(0)}_{m}(D_{v_{+}})$. Set $D_{-}={\overline{D}}_{+}$. Then $D=D_{+}\cup D_{-}$ is a disk in $\mathbb{C}$. The map $W:D\rightarrow\pi(D_{v_{+}}\cup D_{v_{-}})$, where W|_{D_{+}}=\pi\raisebox{0.0pt}{\scriptstyle\circ,}W_{v_{+}} and W|_{D_{-}}=\pi\raisebox{0.0pt}{\scriptstyle\circ,}S^{(0)}_{m}\raisebox{0.0pt}{\scriptstyle\circ,}W_{v_{+}}\raisebox{0.0pt}{\scriptstyle\circ,}{\mbox{}}^{\overline{\rule{5.0pt}{0.0pt}}}, is a homeomorphism of $D$ into a neighborhood $\pi(D_{v_{+}}\cup D_{v_{-}})$ of $[v]=[v_{+}]=[v_{-}]$ in $\mathrm{cl}(K^{\ast})^{\sim}$. Consequently, the identification space $\mathrm{cl}(K^{\ast})^{\sim}$ is a topological manifold. $\square$

We now describe a triangulation of $K^{\ast}\setminus\{O\}$. Let $T^{\prime}=T_{1,n_{1},n-(1+n_{1})}$ be the open rational triangle $\bigtriangleup OCD^{\prime}$ with vertex at the origin $O$, longest side $OC$ on the real axis, and interior angles $\frac{1}{n}\pi$, $\frac{n_{1}}{n}\pi$, and $\frac{n-1-n_{1}}{n}\pi$. Let $Q^{\prime}$ be the quadrilateral $T^{\prime}\cup\overline{T^{\prime}}$. Then $Q^{\prime}$ is a subset of the quadrilateral $Q=ODC\overline{D}$, see figure 5. Moreover $K^{\ast}=\bigcup^{n-1}_{\ell=0}R^{\ell}(Q^{\prime})$. The $2n$ triangles $\mathrm{cl}(R^{j}(T^{\prime}))\setminus\{O\}$ and $\mathrm{cl}(R^{j}(\overline{T^{\prime}}))\setminus\{O\}$ with $j=0,1,\ldots,n-1$ form a triangulation ${\mathcal{T}}_{K^{\ast}\setminus\{O\}}$ of $K^{\ast}\setminus\{O\}$ with $2n$ vertices $R^{j}(C)$ and $R^{j}(D^{\prime})$ for $j=0,1,\ldots,n-1$; $4n$ open edges $R^{j}(OC)$, $R^{j}(OD^{\prime})$, $R^{j}(CD^{\prime})$, and $R^{j}(C\overline{D^{\prime}})$ for $j=0,1,\ldots,n-1$; and $2n$ open triangles $R^{j}(T^{\prime})$, $R^{j}(\overline{T^{\prime}})$ with $j=0,1,\ldots,n-1$. The image of the triangulation ${\mathcal{T}}_{K^{\ast}\setminus\{O\}}$ under the identification map $\pi$ (23) is a triangulation ${\mathcal{T}}_{K^{\ast}\setminus\{O\})^{\sim}}$ of the identification space $(K^{\ast}\setminus\{O\})^{\sim}$.

The action of $G$ on $\mathrm{cl}(K^{\ast})$ preserves the set of unordered pairs of $S^{(j)}_{m}$ equivalent edges of $\mathrm{cl}(K^{\ast})$ for $j=0,1,\infty$. Hence $G$ induces an action on $\mathrm{cl}(K^{\ast})^{\sim}$, which is proper, since $G$ is finite. The $G$ action is free on $K^{\ast}\setminus\{O\}$ and thus on $(K^{\ast}\setminus\{O\})^{\sim}$ by lemma A2. We have proved

Lemma 3.7 The $G$-orbit space $\widetilde{S}=\mathrm{cl}(K^{\ast})^{\sim}/G$ is a compact connected topological manifold with ${\widetilde{S}}_{\mathrm{reg}}=(K^{\ast}\setminus\{O\})^{\sim}/G$ being a smooth manifold. Let

[TABLE]

Then $\sigma$ is the $G$ orbit map, which is surjective, continuous, and open. The restriction of the map $\sigma$ to $K^{\ast}\setminus\{O\}$ has image ${\widetilde{S}}_{\mathrm{reg}}$ and is a smooth open mapping.

We now determine the topology of the orbit space ${\widetilde{S}}_{\mathrm{reg}}$. For $j=0,1,\infty$ and $\ell=0,1,\ldots,d_{j}-1$ let $A^{j}_{\ell}$ be an end point of a closed edge $E$ of $\mathrm{cl}(K^{\ast})$, which lies on the unordered pair $[E,S^{(j)}_{\ell}(E)]\in{\mathcal{E}}^{j}$. Then $H^{j}\cdot A^{(j)}_{\ell}$ is an end point of the edge $H^{j}\cdot E$ of the unordered pair $H^{j}\cdot[E,S^{(j)}_{\ell}(E)]$ of ${\mathcal{E}}^{j}$. See appendix A for the definition of the group $H_{j}$. Fix $j$. The sets $\mathcal{O}(A^{(j)}_{\ell})=\{H^{j}\cdot A^{(j)}_{\ell}\}$ with $\ell=0,1,\ldots,d_{j}-1$ are permuted by $G$. The action of $G$ on $K^{\ast}\setminus\{O\}$ preserves the set of open edges of the triangulation ${\mathcal{T}}_{K^{\ast}\setminus\{O\}}$. There are $3n$-orbits: $R^{j}(OC)$; $R^{j}(O\overline{D^{\prime}})$, since $OD^{\prime}=R(O\overline{D^{\prime}})$; and $R^{j}(CD)$, since $C\overline{D^{\prime}}=U(CD)$ for $j=0,1,\ldots,n-1$. So the image of the triangulation ${\mathcal{T}}_{K^{\ast}\setminus\{O\}}$ under the continuous open map

[TABLE]

is a triangulation ${\mathcal{T}}_{{\widetilde{S}}_{\mathrm{reg}}}$ of the $G$-orbit space ${\widetilde{S}}_{\mathrm{reg}}$ with $d_{0}+d_{1}+d_{\infty}$ vertices $\mu(\mathcal{O}(A^{(j)}_{\ell}))$, where $j=0,1,\infty$ and $\ell=0,1,\ldots,d_{j}-1$; $3n$ open edges $\mu(R^{j}(OC))$, $\mu(R^{j}(O\overline{D^{\prime}}))$, and $\mu(R^{j}(CD))$ for $j=0,1,\ldots,n-1$; and $2n$ open triangles $\mu(R^{j}(T^{\prime}))$ and $\mu(R^{j}(\overline{T^{\prime}}))$ for $j=0,1,\ldots n-1$. Thus the Euler characteristic $\chi({\widetilde{S}}_{\mathrm{reg}})$ of ${\widetilde{S}}_{\mathrm{reg}}$ is $d_{0}+d_{1}+d_{\infty}-3n+2n=d_{0}+d_{1}+d_{\infty}-n$. But ${\widetilde{S}}_{\mathrm{reg}}$ is a smooth manifold. So $\chi({\widetilde{S}}_{\mathrm{reg}})=2-2g$, where $g$ is the genus of ${\widetilde{S}}_{\mathrm{reg}}$. Hence $2g=n+2-(d_{0}+d_{1}+d_{\infty})$. Compare this argument with that of Weyl [9, p.174]. This proves theorem 3.5. $\square$

Since the quadrilateral $Q$ is a fundamental domain for the action of $G$ on $K^{\ast}$, the $G$ orbit map \overline{\mu}=\sigma\raisebox{0.0pt}{\scriptstyle\circ,}\pi:K^{\ast}\subseteq\mathbb{C}\rightarrow\widetilde{S} restricted to $Q$ is a bijective continuous open mapping. But ${\delta}_{Q}:\mathcal{D}\subseteq{\mathcal{S}}_{\mathrm{reg}}\rightarrow Q\subseteq\mathbb{C}$ is a bijective continous open mapping of the fundamental domain $\mathcal{D}$ of the $\mathcal{G}$ action on $\mathcal{S}$. Consequently, the $\mathcal{G}$ orbit space $\mathcal{S}$ is homeomorphic to the $G$ orbit space $\widetilde{S}$. The mapping $\overline{\mu}$ is holomorphic except possibly at [math] and the vertices of $\mathrm{cl}(K^{\ast})$. So the mapping \overline{\mu}\raisebox{0.0pt}{\scriptstyle\circ,}{\delta}_{K^{\ast}}:{\mathcal{S}}_{\mathrm{reg}}\rightarrow{\widetilde{S}}_{\mathrm{reg}} is a holomorphic diffeomorphism.

5 An affine model of ${\widetilde{S}}_{\mathrm{reg}}$

We construct an affine model of the Riemann surface ${\widetilde{S}}_{\mathrm{reg}}$.

We return to the regular stellated $n$-gon $K^{\ast}=K^{\ast}_{n_{0}n_{1}n_{\infty}}$, which is formed from the quadrilateral $Q=Q_{n_{0}n_{1}n_{\infty}}$ less its vertices. Repeatedly reflecting in the edges of ${K^{\ast}}$ and then in the edges of the resulting reflections of ${K^{\ast}}$ et cetera, we obtain a covering of $\mathbb{C}\setminus{\mathbb{V}}^{+}$ by certain translations of $K^{\ast}$. Here ${\mathbb{V}}^{+}$ is the union of the translates of the vertices of $\mathrm{cl}(K^{\ast})$ and its center $O$. Let $\mathfrak{T}$ be the group generated by these translations. The semidirect product $\mathfrak{G}=G\ltimes\mathfrak{T}$ acts freely, properly and transitively on $\mathbb{C}\setminus{\mathbb{V}}^{+}$. It preserves equivalent edges of $\mathbb{C}\setminus{\mathbb{V}}^{+}$ and it acts freely and properly on $(\mathbb{C}\setminus{\mathbb{V}}^{+})^{\sim}$, the space formed by identifying equivalent edges in $\mathbb{C}\setminus{\mathbb{V}}^{+}$. The orbit space $(\mathbb{C}\setminus{\mathbb{V}}^{+})^{\sim}/\mathfrak{G}$ is holomorphically diffeomorphic to ${\widetilde{S}}_{\mathrm{reg}}$ and is the desired affine model of ${\widetilde{S}}_{\mathrm{reg}}$. We now justify these assertions.

First we determine the group $\mathcal{T}$ of translations.

Lemma 4.1 Each of the $2n$ sides of the regular stellated $n$-gon $K^{\ast}$ is perpendicular to exactly one of the directions

[TABLE]

for $j=0,1,\ldots,n-1$.

Proof. From figure 9 we have $\angle D^{\prime}CO=\frac{n_{1}}{n}\pi$. So $\angle COH=\frac{1}{2}\pi-\frac{n_{1}}{n}\pi$. Hence the line ${\ell}_{0}$, containing the edge $CD^{\prime}$ of $K^{\ast}$, is perpendicular to the direction ${\mathrm{e}}^{[\frac{1}{2}-\frac{n_{1}}{n}]\pi}$. Since $\bigtriangleup CO\overline{D^{\prime}}$ is the reflection of $\bigtriangleup COD^{\prime}$ in the line segment $OC$, the line ${\ell}_{1}$, containing the edge $C\overline{D^{\prime}}$ of $K^{\ast}$, is perpendicular to the direction ${\mathrm{e}}^{[-\frac{1}{2}+\frac{n_{1}}{n}]\pi}$. Because the regular stellated $n$-gon ${K^{\ast}}$ is formed by repeatedly rotating the quadrilateral $Q^{\prime}=OD^{\prime}C\overline{D^{\prime}}$ through an angle $\frac{2\pi}{n}$, we find that equation (51) holds. $\square$

[TABLE]

Since $\angle COH=\frac{1}{2}\pi-\frac{n_{1}}{n}\pi$, it follows that $|H|=|C|\sin\pi\frac{n_{1}}{n}$ is the distance from the center $O$ of $K^{\ast}$ to the line ${\ell}_{0}$ containing the side $CD^{\prime}$, or to the line ${\ell}_{1}$ containing the side $C\overline{D^{\prime}}$. So $u_{0}=(|C|\sin\pi\frac{n_{1}}{n}){\mathrm{e}}^{[\frac{1}{2}-\frac{n_{1}}{n}]\pi i}$ is the closest point $H$ on ${\ell}_{0}$ to $O$ and $u_{1}=(|C|\sin\pi\frac{n_{1}}{n}){\mathrm{e}}^{[-\frac{1}{2}+\frac{n_{1}}{n}]\pi i}$ is the closest point $\overline{H}$ on ${\ell}_{1}$ to $O$. Since the regular stellated $n$-gon ${K^{\ast}}$ is formed by repeatedly rotating the quadrilateral $Q^{\prime}=OD^{\prime}C\overline{D^{\prime}}$ through an angle $\frac{2\pi}{n}$, the point

[TABLE]

lies on the line ${\ell}_{2j}=R^{j}{\ell}_{0}$, which contains the edge $R^{j}(CD^{\prime})$ of ${K^{\ast}}$; while

[TABLE]

lies on the line ${\ell}_{2j+1}=R^{j}{\ell}_{1}$, which contains the edge $R^{j}(C\overline{D^{\prime}})$ of ${K^{\ast}}$ for every $j\in\{0,1,\ldots,n-1\}$. Also the line segments $Ou_{2j}$ and $Ou_{2j+1}$ are perpendicular to the line ${\ell}_{2j}$ and ${\ell}_{2j+1}$, respectively, for $j\in\{0,1,\ldots,n-1\}$.

Corollary 4.1a For $j=0,1,\ldots,n-1$ we have

[TABLE]

Proof. We compute. From (26) it follows that

[TABLE]

while from (27) we get

[TABLE]

Corollary 4.1b For $\ell$, $j\in\{0,1,\ldots,2n-1\}$ we have

[TABLE]

Proof. If $\ell=2k$, then $u_{\ell}=R^{k}u_{0}$, by definition. So

[TABLE]

If $\ell=2k+1$, then $u_{\ell}=R^{k}u_{1}$, by definition. So

[TABLE]

For $j=0,1,\ldots,2n-1$ let ${\tau}_{j}$ be the translation

[TABLE]

Corollary 4.1c For $k$, $j\in\{0,1,\ldots,2n-1\}$ we have

[TABLE]

Proof. For every $z\in\mathbb{C}$, we have

[TABLE]

Reflecting the regular stellated $n$-gon $K^{\ast}$ in its edge $CD^{\prime}$ contained in ${\ell}_{0}$ gives a congruent regular stellated $n$-gon $K^{\ast}_{0}$ with the center $O$ of $K^{\ast}$ becoming the center $2u_{0}$ of $K^{\ast}_{0}$.

Lemma 4.2 The collection of all the centers of the regular stellated $n$-gons formed by reflecting ${K^{\ast}}$ in its edges and then reflecting in the edges of the reflected regular stellated $n$-gons et cetera is

[TABLE]

where for $j=0,1,\ldots,2n-1$ we have

[TABLE]

Proof. For each $k_{0}=0,1,\ldots,2n-1$ the center of the $2n$ regular stellated congruent $n$-gon $K^{\ast}_{k_{0}}$ formed by reflecting in an edge of $K^{\ast}$ contained in the line ${\ell}_{k_{0}}$ is ${\tau}_{k_{0}}(0)=2u_{k_{0}}$. Repeating the reflecting process in each edge of $K^{\ast}_{k_{0}}$ gives $2n$ congruent regular stellated $n$-gons $K^{\ast}_{k_{0}k_{1}}$ with center at {\tau}_{k_{1}}\big{(}{\tau}_{k_{0}}(0)\big{)}=2(u_{k_{1}}+u_{k_{0}}), where $k_{1}=0,1,\ldots 2n-1$. Repeating this construction proves the lemma. $\square$

The set $\mathbb{V}$ of vertices of the regular stellated $n$-gon $K^{\ast}$ is

[TABLE]

see figure 5. Clearly the set $\mathbb{V}$ is $G$ invariant.

Corollary 4.2a The set

[TABLE]

is the collection of vertices and centers of the congruent regular stellated $n$-gons $K^{\ast}$, $K^{\ast}_{k_{1}}$, $K^{\ast}_{k_{0}k_{1}},\ldots$.

Proof. This follows immediately from lemma 4.2. $\square$

Corollary 4.2b The union of $K^{\ast},K^{\ast}_{k_{0}},K^{\ast}_{k_{0}k_{1}},\ldots K^{\ast}_{k_{0}k_{1}\cdots k_{\ell}},\ldots$, where $\ell\geq 0$, $0\leq j\leq\ell$, and $0\leq k_{j}\leq 2n-1$, covers $\mathbb{C}\setminus{\mathbb{V}}^{+}$, that is,

[TABLE]

Proof. This follows immediately from K^{\ast}_{k_{0}k_{1}\cdots k_{\ell}}={\tau}_{k_{\ell}}\raisebox{0.0pt}{\scriptstyle\circ,}\cdots\raisebox{0.0pt}{\scriptstyle\circ,}{\tau}_{k_{0}}(K^{\ast}). $\square$

Let $\mathcal{T}$ be the abelian subgroup of the $2$-dimensional Eulcidean group $\mathrm{E}(2)$ generated by the translations ${\tau}_{j}$ (30) for $j=0,1,\ldots 2n-1$. It follows from corollary 4.2b that the regular stellated $n$-gon $K^{\ast}$ with its vertices and center removed is the fundamental domain for the action of the abelian group $\mathcal{T}$ on $\mathbb{C}\setminus{\mathbb{V}}^{+}$. The group $\mathcal{T}$ is isomorphic to the abelian subgroup $\mathfrak{T}$ of $(\mathbb{C},+)$ generated by ${\{2u_{j}\}}^{2n-1}_{j=0}$.

Next we define the group $\mathfrak{G}$ and show that it acts freely, properly, and transitively on $\mathbb{C}\setminus{\mathbb{V}}^{+}$.

Consider the group $\mathfrak{G}=G\ltimes\mathfrak{T}\subseteq G\times\mathfrak{T}$, which is the semidirect product of the dihedral group $G$, generated by the rotation $R$ through $2\pi/n$ and the reflection $U$ subject to the relations $R^{n}=e=U^{2}$ and $RU=UR^{-1}$, and the abelian group $\mathfrak{T}$. An element $(R^{j}U^{\ell},2u_{k})$ of $\mathfrak{G}$ is the affine linear map

[TABLE]

Multiplication in $\mathfrak{G}$ is defined by

[TABLE]

which is the composition of the affine linear map $(R^{j^{\prime}}U^{{\ell}^{\prime}},2u_{k^{\prime}})$ followed by $(R^{j}U^{\ell},2u_{k})$. The mappings $G\rightarrow\mathfrak{G}:R^{j}\mapsto(R^{j}U^{\ell},0)$ and $\mathfrak{T}\rightarrow\mathfrak{G}:2u_{k}\mapsto(e,2u_{k})$ are injective, which allows us to identify the groups $G$ and $\mathfrak{T}$ with their image in $\mathfrak{G}$. Using (33) we may write an element $(R^{j}U^{\ell},2u_{k})$ of $\mathfrak{G}$ as $(e,2u_{k})\cdot(R^{j}U^{\ell},0)$. So

[TABLE]

For every $z\in\mathbb{C}$ we have

[TABLE]

that is,

[TABLE]

Hence

[TABLE]

which is just equation (31). The group $\mathfrak{G}$ acts on $\mathbb{C}$ as $\mathrm{E}(2)$ does, namely, by affine linear orthogonal mappings. Denote this action by

[TABLE]

Lemma 4.3 The set of vertices ${\mathbb{V}}^{+}$ (32) is invariant under the $\mathfrak{G}$ action.

Proof. Let $v\in{\mathbb{V}}^{+}$. Then for some $({\ell}^{\prime}_{0},\ldots,{\ell}^{\prime}_{2n-1})\in{\mathbb{Z}}^{2n}_{\geq 0}$ and some $w\in\mathbb{V}\cup\{0\}$

[TABLE]

where $u^{\prime}=\sum^{2n-1}_{j=0}{\ell}^{\prime}_{j}u_{j}$. For $(R^{j}U^{\ell},2u)\in\mathfrak{G}$ with $j=0,1,\ldots,n-1$ and $\ell=0,1$ we have

[TABLE]

where $w^{\prime}={\psi}_{(R^{j}U^{\ell},0)}(w)=R^{j}U^{\ell}(w)\in\mathbb{V}\cup\{0\}$. If $\ell=0$, then

[TABLE]

while if $\ell=1$, then

[TABLE]

Here k^{\prime}(k)=$$\left\{\begin{array}[]{cl}2n-k+1,&\mbox{if kis even}\\ 2n-k-1,&\mbox{ifk is odd},\end{array}\right. see corollary 4.1a. So $(e,2(R^{j}U^{\ell}u^{\prime}+u))\in\mathfrak{T}$, which implies ${\psi}_{(e,2(R^{j}U^{\ell}u^{\prime}+u))}(w^{\prime})\in{\mathbb{V}}^{+}$, as desired. $\square$

Lemma 4.4 The action of $\mathfrak{G}$ on $\mathbb{C}\setminus{\mathbb{V}}^{+}$ is free.

Proof. Suppose that for some $v\in\mathbb{C}\setminus{\mathbb{V}}^{+}$ and some $(R^{j}U^{\ell},2u)\in\mathfrak{G}$ we have $v={\psi}_{(R^{j}U^{\ell},2u)}(v)$. Then $v$ lies in some $K^{\ast}_{k_{0}k_{1}\cdots k_{\ell}}$. So for some $v^{\prime}\in K^{\ast}$ we have

[TABLE]

where $u^{\prime}=\sum^{2n-1}_{j=0}{\ell}^{\prime}_{j}u_{j}$ for some $({\ell}^{\prime}_{0},\ldots,{\ell}^{\prime}_{2n-1})\in({\mathbb{Z}}_{\geq 0})^{2n}$. Thus

[TABLE]

This implies $R^{j}U^{\ell}=e$, that is, $j=\ell=0$. So $2u=2R^{j}u^{\prime}+2u=2u^{\prime}+2u$, that is, $u=0$. Hence $(R^{j}U^{\ell},u)=(e,0)$, which is the identity element of $\mathfrak{G}$. $\square$

Lemma 4.5 The action of $\mathcal{T}$ (and hence $\mathfrak{G}$) on $\mathbb{C}\setminus{\mathbb{V}}^{+}$ is transitive.

Proof. Let $K^{\ast}_{k_{0}\cdots k_{\ell}}$ and $K^{\ast}_{k^{\prime}_{0}\cdots k^{\prime}_{{\ell}^{\prime}}}$ lie in

[TABLE]

Since K^{\ast}_{k_{0}\cdots k_{\ell}}={\tau}_{k_{\ell}}\raisebox{0.0pt}{\scriptstyle\circ,}\cdots\raisebox{0.0pt}{\scriptstyle\circ,}{\tau}_{k_{0}}(K^{\ast}) and K^{\ast}_{k^{\prime}_{0}\cdots k^{\prime}_{{\ell}^{\prime}}}={\tau}_{k^{\prime}_{{\ell}^{\prime}}}\raisebox{0.0pt}{\scriptstyle\circ,}\cdots\raisebox{0.0pt}{\scriptstyle\circ,}{\tau}_{k^{\prime}_{0}}(K^{\ast}), it follows that ({\tau}_{k^{\prime}_{{\ell}^{\prime}}}\raisebox{0.0pt}{\scriptstyle\circ,}\cdots\raisebox{0.0pt}{\scriptstyle\circ,}{\tau}_{k^{\prime}_{0}})\raisebox{0.0pt}{\scriptstyle\circ,}({\tau}_{k_{\ell}}\raisebox{0.0pt}{\scriptstyle\circ,}\cdots\raisebox{0.0pt}{\scriptstyle\circ,}{\tau}_{k_{0}})^{-1}(K^{\ast}_{k_{0}\cdots k_{\ell}})=K^{\ast}_{k^{\prime}_{0}\cdots k^{\prime}_{{\ell}^{\prime}}}. $\square$

The action of $\mathfrak{G}$ on $\mathbb{C}\setminus{\mathbb{V}}^{+}$ is proper because $\mathfrak{G}$ is a discrete subgroup of $\mathrm{E}(2)$ with no accumulation points.

We now define an edge of $\mathbb{C}\setminus{\mathbb{V}}^{+}$ and what it means for an unordered pair of edges to be equivalent. We show that the group $\mathfrak{G}$ acts freely and properly on the identification space of equivalent edges.

Let $E$ be an open edge of $K^{\ast}$. Since $E_{k_{0}\cdots k_{\ell}}={\tau}_{k_{0}}\cdots{\tau}_{k_{\ell}}(E)\in K^{\ast}_{k_{0}\cdots k_{\ell}}$, it follows that $E_{k_{0}\cdots k_{\ell}}$ is an open edge of $K^{\ast}_{k_{0}\cdots k_{\ell}}$. Let

[TABLE]

Then $\mathfrak{E}$ is the set of open edges of $\mathbb{C}\setminus{\mathbb{V}}^{+}$ by lemma 4.2b. Since {\tau}_{k_{\ell}}\raisebox{0.0pt}{\scriptstyle\circ,}\cdots\raisebox{0.0pt}{\scriptstyle\circ,}{\tau}_{k_{0}}(0) is the center of $K^{\ast}_{k_{0}\cdots k_{\ell}}$, the element (e,{\tau}_{k_{\ell}}\raisebox{0.0pt}{\scriptstyle\circ,}\cdots\raisebox{0.0pt}{\scriptstyle\circ,}{\tau}_{k_{0}})\cdot(g,({\tau}_{k_{\ell}}\raisebox{0.0pt}{\scriptstyle\circ,}\cdots\raisebox{0.0pt}{\scriptstyle\circ,}{\tau}_{k_{0}})^{-1}) of $\mathfrak{G}$ is a rotation-reflection of $K^{\ast}_{k_{0}\cdots k_{\ell}}$, which sends an edge of $K^{\ast}_{k_{0}\cdots k_{\ell}}$ to another edge of $g\ast K^{\ast}_{k_{0}\cdots k_{\ell}}$. Thus $\mathfrak{G}$ sends $\mathfrak{E}$ into itself. For $j=0,1,\infty$ let ${\mathfrak{E}}^{j}_{k_{0}\cdots k_{\ell}}$ be the set of unordered pairs $[E_{k_{0}\cdots k_{\ell}},E^{\prime}_{k_{0}\cdots k_{\ell}}]$ of equivalent open edges of $K^{\ast}_{k_{0}\cdots k_{\ell}}$, that is, $E_{k_{0}\cdots k_{\ell}}\cap E^{\prime}_{k_{0}\cdots k_{\ell}}=\varnothing$, so the open edges $E_{k_{0}\cdots k_{\ell}}={\tau}_{k_{0}}\cdots{\tau}_{k_{\ell}}(E)$ and $E^{\prime}_{k_{0}\cdots k_{\ell}}={\tau}_{k_{0}}\cdots{\tau}_{k_{\ell}}(E^{\prime})$ of $\mathrm{cl}(K^{\ast}_{k_{0}\cdots k_{\ell}})$ are not adjacent, which implies that the open edges $E$ and $E^{\prime}$ of $K^{\ast}$ are not adjacent, and for some generator $S^{(j)}_{m}$ of the group $G^{j}$ of reflections we have

[TABLE]

Let ${\mathfrak{E}}^{j}=\cup_{\ell\geq 0}\cup_{0\leq j\leq\ell}\,\cup_{0\leq k_{j}\leq 2n-1}{\mathfrak{E}}^{j}_{k_{0}\cdots k_{\ell}}$. Then ${\mathfrak{E}}^{j}$ is the set of unordered pairs of equivalent edges of $\mathbb{C}\setminus{\mathbb{V}}^{+}$. Define an action $\ast$ of $\mathfrak{G}$ on ${\mathcal{E}}^{j}$ by

[TABLE]

where {\tau}^{\prime}={\tau}_{k_{\ell}}\raisebox{0.0pt}{\scriptstyle\circ,}\cdots{\tau}_{k_{0}}.

Define a relation $\sim$ on $\mathbb{C}\setminus{\mathbb{V}}^{+}$ as follows. We say that $x$ and $y\in\mathbb{C}\setminus{\mathbb{V}}^{+}$ are related, $x\sim y$, if

1. $x\in F=\tau(E)\in{\mathfrak{E}}^{0}$ and $y\in F^{\prime}=\tau(E^{\prime})\in{\mathfrak{E}}^{0}$ such that $[F,F^{\prime}]=[\tau(E),\tau(E^{\prime})]\in{\mathfrak{E}}^{0}$, where $[E,E^{\prime}]\in{\mathcal{E}}^{0}$ with $E^{\prime}=S^{(0)}_{m}(E)$ for some $S^{(0)}_{m}\in G^{0}$ and y=\tau\big{(}S^{(0)}_{m}({\tau}^{-1}(x))\big{)} or 2) $x$, y\in\big{(}\mathbb{C}\setminus{\mathbb{V}}^{+}\big{)}\setminus\mathfrak{E} and $x=y$. Then $\sim$ is an equivalence relation on $\mathbb{C}\setminus{\mathbb{V}}^{+}$. Let $(\mathbb{C}\setminus{\mathbb{V}}^{+})^{\sim}$ be the set of equivalence classes and let $\Pi$ be the map

[TABLE]

which assigns to every $p\in\mathbb{C}\setminus{\mathbb{V}}^{+}$ the equivalence class $[p]$ containing $p$.

Lemma 4.6 $\Pi|_{K^{\ast}}$ is the map $\pi$ (22).

Proof. This follows immediately from the definition of the maps $\Pi$ and $\pi$. $\square$

Lemma 4.7 The usual action of $\mathfrak{G}$ on $\mathbb{C}$, restricted to $\mathbb{C}\setminus{\mathbb{V}}^{+}$, is compatible with the equivalence relation $\sim$, that is, if $x$, $y\in\mathbb{C}\setminus\mathbb{V}$ and $x\sim y$, then $(g,\tau)(x)\sim(g,\tau)(y)$ for every $(g,\tau)\in\mathfrak{G}$.

Proof. Suppose that $x\in F={\tau}^{\prime}(E)$, where ${\tau}^{\prime}\in\mathcal{T}$. Then $y\in F^{\prime}={\tau}^{\prime}(E^{\prime})$, since $x\sim y$. So for some $S^{(0)}_{m}\in G^{\,0}$ we have $({\tau}^{\prime})^{-1}(y)=S^{(0)}_{m}({\tau}^{-1}(x))$. Let $(g,\tau)\in\mathfrak{G}$. Then

[TABLE]

So $(g,\tau)(y)\in(g,\tau)\ast F^{\prime}$. But $(g,\tau)(x)\in(g,\tau)\ast F$ and $[(g,\tau)\ast F,(g,\tau)\ast F^{\prime}]=(g,\tau)\ast[F,F^{\prime}]$. Hence $(g,\tau)(x)\sim(g,\tau)(y)$. $\square$

Because of lemma 4.7, the usual $\mathfrak{G}$-action on $\mathbb{C}\setminus{\mathbb{V}}^{+}$ induces an action of $\mathfrak{G}$ on $(\mathbb{C}\setminus{\mathbb{V}}^{+})^{\sim}$.

Lemma 4.8 The action of $\mathfrak{G}$ on $(\mathbb{C}\setminus{\mathbb{V}}^{+})^{\sim}$ is free and proper.

Proof. The following argument shows that it is free. Using lemma A2 we see that an element of $\mathfrak{G}$, which lies in the isotropy group ${\mathfrak{G}}_{[F,F^{\prime}]}$ for $[F,F^{\prime}]\in{\mathfrak{E}}^{0}$, interchanges the edge $F$ with the equivalent edge $F^{\prime}$ and thus fixes the equivalence class $[p]$ for every $p\in F$. Hence the $\mathfrak{G}$ action on $(\mathbb{C}\setminus{\mathbb{V}}^{+})^{\sim}$ is free. It is proper because $\mathfrak{G}$ is a discrete subgroup of the Euclidean group $\mathrm{E}(2)$ with no accumulation points. $\square$

Theorem 4.9 The $\mathfrak{G}$-orbit space $(\mathbb{C}\setminus{\mathbb{V}}^{+})^{\sim}/\mathfrak{G}$ is holomorphically diffeomorphic to the $G$-orbit space $(K^{\ast}\setminus\{O\})^{\sim}/G={\widetilde{S}}_{\mathrm{reg}}$.

Proof. This claim follows from the fact that the fundamental domain of the $\mathfrak{G}$-action on $\mathbb{C}\setminus{\mathbb{V}}^{+}$ is $K^{\ast}\setminus\{O\}$, which is the fundamental domain of the $G$-action on $K^{\ast}\setminus\{O\}$. Thus $\Pi(\mathbb{C}\setminus{\mathbb{V}}^{+})$ is a fundamental domain of the $\mathfrak{G}$-action on $(\mathbb{C}\setminus{\mathbb{V}}^{+})^{\sim}$, which is equal to $\pi(K^{\ast}\setminus\{O\})=(K^{\ast}\setminus\{O\})^{\sim}$ by lemma 4.6. Hence the $\mathfrak{G}$-orbit space $(\mathbb{C}\setminus{\mathbb{V}}^{+})^{\sim}/\mathfrak{G}$ is equal to the $G$-orbit space ${\widetilde{S}}_{\mathrm{reg}}$. So the identity map from $\Pi(\mathbb{C}\setminus{\mathbb{V}}^{+})$ to $(K^{\ast}\setminus\{O\})^{\sim}$ induces a holomorphic diffeomorphism of orbit spaces. $\square$

Because the group $\mathfrak{G}$ is a discrete subgroup of the $2$-dimensional Euclidean group $\mathrm{E}(2)$, the Riemann surface $(\mathbb{C}\setminus{\mathbb{V}}^{+})^{\sim}/\mathfrak{G}$ is an affine model of the affine Riemann surface ${\mathcal{S}}_{\mathrm{reg}}$.

6 The developing map and geodesics

In this section we show that the mapping

[TABLE]

straightens the holomorphic vector field $X$ (14) on the fundamental domain $\mathcal{D}\subseteq{\mathcal{S}}_{\mathrm{reg}}$, see Bates and Cushman [2] and Flaschka [4]. We verify that $X$ is the geodesic vector field for a flat Riemannian metric $\Gamma$ on $\mathcal{D}$.

First we rewrite equation (15) as

[TABLE]

From the definition of the mapping $F_{Q}$ (4) we get

[TABLE]

where we use the same complex $n^{\mathrm{th}}$ root as in the definition of $F_{Q}$. This implies

[TABLE]

For each $(\xi,\eta)\in\mathcal{D}$ using (38) and (39) we get

[TABLE]

So the holomorphic vector field $X$ (14) on $\mathcal{D}$ and the holomorphic vector field $\frac{\partial}{\partial z}$ on $Q$ are $\delta$-related. Hence $\delta$ sends an integral curve of the vector field $X$ starting at $(\xi,\eta)\in\mathcal{D}$ onto an integral curve of the vector field $\frac{\partial}{\partial z}$ starting at $z=\delta(\xi,\eta)\in Q$. Since an integral curve of $\frac{\partial}{\partial z}$ is a horizontal line segment in $Q$, we have proved

Claim 5.1 The holomorphic mapping $\delta$ (37) straightens the holomorphic vector field $X$ (14) on the fundamental domain $\mathcal{D}\subseteq{\mathcal{S}}_{\mathrm{reg}}$.

We can say more. Let $u=\mathrm{Re}\,z$ and $v=\mathrm{Im}\,z$. Then

[TABLE]

is the flat Euclidean metric on $\mathbb{C}$. Its restriction $\gamma|_{\mathbb{C}\setminus{\mathbb{V}}^{+}}$ to $\mathbb{C}\setminus{\mathbb{V}}^{+}$ is invariant under the group $\mathfrak{G}$, which is a subgroup of the Euclidean group $\mathrm{E}(2)$.

Consider the flat Riemannian metric $\gamma|_{Q}$ on $Q$, where $\gamma$ is the metric (40) on $\mathbb{C}$. Pulling back $\gamma|_{Q}$ by the mapping $F_{Q}$ (4) gives a metric

[TABLE]

on $\mathbb{C}\setminus\{0,1\}$. Pulling the metric $\widetilde{\gamma}$ back by the projection mapping $\widetilde{\pi}:{\mathbb{C}}^{2}\rightarrow\mathbb{C}:(\xi,\eta)\mapsto\xi$ gives

[TABLE]

on ${\mathbb{C}}^{2}$. Restricting $\widetilde{\Gamma}$ to the affine Riemann surface ${\mathcal{S}}_{\mathrm{reg}}$ gives \Gamma=\frac{1}{\eta}\mathop{\!\,\mathrm{d}\!}\nolimits\xi\,\raisebox{2.0pt}{\tiny,\bigodot}\,\frac{1}{\overline{\eta}}\overline{\mathop{\!\,\mathrm{d}\!}\nolimits\xi}.

Lemma 5.2 $\Gamma$ is a flat Riemannian metric on ${\mathcal{S}}_{\mathrm{reg}}$.

Proof. We compute. For every $(\xi,\eta)\in{\mathcal{S}}_{\mathrm{reg}}$ we have

[TABLE]

Thus $\Gamma$ is a Riemannian metric on ${\mathcal{S}}_{\mathrm{reg}}$. It is flat by construction. $\square$

Because $\mathcal{D}$ has nonempty interior and the map $\delta$ (37) is holomorphic, it can be analytically continued to the map

[TABLE]

since $\delta={\delta}_{Q}|_{\mathcal{D}}$. By construction ${\delta}^{\ast}_{Q}(\gamma|_{Q})=\Gamma$. So the mapping ${\delta}_{Q}$ is an isometry of $({\mathcal{S}}_{\mathrm{reg}},\Gamma)$ onto $(Q,\gamma|_{Q})$. In particular, the map $\delta$ is an isometry of $(\mathcal{D},\Gamma|_{\mathcal{D}})$ onto $(Q,\gamma|_{Q})$. Moreover, $\delta$ is a local holomorphic diffeomorphism, because for every $(\xi,\eta)\in\mathcal{D}$, the complex linear mapping $T_{(\xi,\eta)}\delta$ is an isomorphism, since it sends $X(\xi,\eta)$ to \frac{\partial}{\partial z}\rule[-6.0pt]{0.5pt}{15.0pt}\,\raisebox{-6.0pt}{\scriptscriptstyle z=\delta(\xi,\eta)} . Thus $\delta$ is a developing map in the sense of differential geometry, see Spivak [8, p.97] note on §12 of Gauss [5]. The map $\delta$ is local because the integral curves of $\frac{\partial}{\partial z}$ on $Q$ are only defined for a finite time, since they are horizontal line segments in $Q$. Thus the integral curves of $X$ (14) on $\mathcal{D}$ are defined for a finite time. Since the integral curves of $\frac{\partial}{\partial z}$ are geodesics on $(Q,\gamma|_{Q})$, the image of a local integral curve of $\frac{\partial}{\partial z}$ under the local inverse of the mapping $\delta$ is a local integral curve of $X$. This latter local integral curve is a geodesic on $(\mathcal{D},\Gamma|_{\mathcal{D}})$, since $\delta$ is an isometry. Thus we have proved

Claim 5.3 The holomorphic vector field $X$ (14) on the fundamental domain $\mathcal{D}$ is the geodesic vector field for the flat Riemannian metric $\Gamma|_{\mathcal{D}}$ on $\mathcal{D}$.

Corollary 5.3a The holomorphic vector field $X$ on the affine Riemann surface ${\mathcal{S}}_{\mathrm{reg}}$ is the geodesic vector field for the flat Riemannian metric $\Gamma$ on ${\mathcal{S}}_{\mathrm{reg}}$.

Proof. The corollary follows by analytic continuation from the conclusion of claim 5.3, since $\mathrm{int}\,\mathcal{D}$ is a nonempty open subset of ${\mathcal{S}}_{\mathrm{reg}}$ and both the vector field $X$ and the Riemannian metric $\Gamma$ are holomorphic on ${\mathcal{S}}_{\mathrm{reg}}$.

7 Discrete symmetries and billiard motions

Let $\mathcal{G}$ be the group of homeomorphisms of the affine Riemann surface $\mathcal{S}$ (5) generated by the mappings

[TABLE]

Clearly, the relations ${\mathcal{R}}^{n}={\mathcal{U}}^{2}=e$ hold. For every $(\xi,\eta)\in\mathcal{S}$ we have

[TABLE]

So the additional relation $\mathcal{U}{\mathcal{R}}^{-1}=\mathcal{R}\,\mathcal{U}$ holds. Thus $\mathcal{G}$ is isomorphic to the dihedral group $D_{2n}$.

Lemma 6.1 $\mathcal{G}$ is a group of isometries of $({\mathcal{S}}_{\mathrm{reg}},\Gamma)$.

Proof. For every $(\xi,\eta)\in{\mathcal{S}}_{\mathrm{reg}}$ we get

[TABLE]

and

[TABLE]

Recall that the group $G$, generated by the linear mappings

[TABLE]

is isomorphic to the dihedral group $D_{2n}$.

Lemma 6.2 $G$ is a group of isometries of $(\mathbb{C},\gamma)$.

Proof. This follows because $R$ and $U$ are Euclidean motions. $\square$

We would like the developing map ${\delta}_{Q}$ (41) to intertwine the actions of $\mathcal{G}$ and $G$ and the geodesic flows on $({\mathcal{S}}_{\mathrm{reg}},\Gamma)$ and $(Q,\gamma|_{Q})$. There are several difficulties. The first is: the group $G$ does not preserve the quadrilateral $Q$. To overcome this difficulty we extend the mapping ${\delta}_{Q}$ (41) to the mapping ${\delta}_{K^{\ast}}$ (19) of the affine Riemann surface ${\mathcal{S}}_{\mathrm{reg}}$ onto the regular stellated $n$-gon $K^{\ast}$.

Lemma 6.3 The mapping ${\delta}_{K^{\ast}}$ (19) intertwines the action $\Phi$ (16) of $\mathcal{G}$ on ${\mathcal{S}}_{\mathrm{reg}}$ with the action

[TABLE]

of $G$ on the regular stellated $n$-gon $K^{\ast}$.

Proof. From the definition of the mapping ${\delta}_{K^{\ast}}$ we see that for each $(\xi,\eta)\in\mathcal{D}$ we have {\delta}_{K^{\ast}}\big{(}{\mathcal{R}}^{j}(\xi,\eta)\big{)}=R^{j}{\delta}_{K^{\ast}}(\xi,\eta) for every $j\in\mathbb{Z}$. By analytic continuation we see that the preceding equation holds for every $(\xi,\eta)\in{\mathcal{S}}_{\mathrm{reg}}$. Since $F_{Q}(\overline{\xi})=\overline{F_{Q}(\xi)}$ by construction and $\widehat{\pi}(\overline{\xi},\overline{\eta})=\overline{\xi}$ (13), from the definition of the mapping $\delta$ (37) we get $\delta(\overline{\xi},\overline{\eta})=\overline{\delta(\xi,\eta)}$ for every $(\xi,\eta)\in\mathcal{D}$. In other words, {\delta}_{K^{\ast}}\big{(}\mathcal{U}(\xi,\eta)\big{)}=U{\delta}_{K^{\ast}}(\xi,\eta) for every $(\xi,\eta)\in\mathcal{D}$. By analytic continuation we see that the preceding equation holds for all $(\xi,\eta)\in{\mathcal{S}}_{\mathrm{reg}}$. Hence on ${\mathcal{S}}_{\mathrm{reg}}$ we have

[TABLE]

The mapping $\varphi:\mathcal{G}\rightarrow G$ sends the generators $\mathcal{R}$ and $\mathcal{U}$ of the group $\mathcal{G}$ to the generators $R$ and $U$ of the group $G$, respectively. So it is an isomorphism. $\square$

There is a second more serious difficulty: the integral curves of $\frac{\partial}{\partial z}$ run off the quadrilateral $Q$ in finite time. We fix this by requiring that when an integral curve reaches a point $P$ on the boundary $\partial Q$ of $Q$, which is not a vertex, it undergoes a specular reflection at $P$. (If the integral curve reaches a vertex of $Q$ in forward or backward time, then the motion ends). This motion can be continued as a straight line motion, which extends the motion on the original segment in $Q$ or $S(Q)$. To make this precise, we give $Q$ the orientation induced from $\mathbb{C}$ and suppose that the incoming (and hence outgoing) straight line motion has the same orientation as $\partial Q$. If the incoming motion makes an angle $\alpha$ with respect to the inward pointing normal $N$ to $\partial Q$ at $P$, then the outgoing motion makes an angle $\alpha$ with the normal $N$, see Richens and Berry [7]. Specifically, if the incoming motion to $P$ is an integral curve of $\frac{\partial}{\partial z}$, then the outgoing motion, after reflection at $P$, is an integral curve of $R^{-1}\frac{\partial}{\partial z}={\mathrm{e}}^{-2\pi i/n}\frac{\partial}{\partial z}$. Thus the outward motion makes a turn of $-2\pi/n$ at $P$ towards the interior of $Q$, see figure 10 (left). In figure 10 (right) the incoming motion has the opposite orientation from $\partial Q$.

[TABLE]

This extended motion on $Q$ is called a billiard motion. A billiard motion starting in the interior of $\mathrm{cl}(Q)$ is defined for all time and remains in $\mathrm{cl}(Q)$ less its vertices, since each of the segments of the billiard motion is a straight line parallel to an edge of $\mathrm{cl}(Q)$ and does not hit a vertex of $\mathrm{cl}(Q)$, see figure 12.

We can do more. If we apply a reflection $S$ in the edge of $Q$ in its boundary $\partial Q$, which contains the reflection point $P$, to the initial reflected motion at $P$, and then again to the extended straight line motion in $S(Q)$ when it reaches

[TABLE]

$\partial S(Q)$, et cetera, we see that the extended motion becomes a billiard motion in the regular stellated $n$-gon K^{\ast}=Q\cup{\amalg}_{0\leq k\leq n-1}SR^{k}(Q)\big{)}, see figure 12.

[TABLE]

So we have verified

Claim 6.4 A billiard motion in the regular stellated $n$-gon $K^{\ast}$, which starts at a point in the interior of $K^{\ast}\setminus\{O\}$ does not hit a vertex of $\mathrm{cl}(K^{\ast})$ and is invariant under the action of the isometry subgroup $\widehat{G}$ of the isometry group $G$ of $(K^{\ast},\gamma|_{K^{\ast}})$ generated by the rotation $R$.

Let $\widehat{\mathcal{G}}$ be the subgroup of $\mathcal{G}$ generated by the rotation $\mathcal{R}$. We now show

Lemma 6.5 The holomorphic vector field $X$ (14) on ${\mathcal{S}}_{\mathrm{reg}}$ is $\widehat{\mathcal{G}}$-invariant.

Proof. We compute. For every $(\xi,\eta)\in{\mathcal{S}}_{\mathrm{reg}}$ and for $\mathcal{R}\in\widehat{\mathcal{G}}$ we have

[TABLE]

Hence for every $j\in\mathbb{Z}$ we get

[TABLE]

for every $(\xi,\eta)\in{\mathcal{S}}_{\mathrm{reg}}$. In other words, the vector field $X$ is invariant under the action of $\widehat{\mathcal{G}}$ on ${\mathcal{S}}_{\mathrm{reg}}$. $\square$

Corollary 6.5a For every $(\xi,\eta)\in\mathcal{D}$ we have

[TABLE]

Proof. Equation (45) is a rewrite of equation (44). $\square$

Corollary 6.5b Every geodesic on $({\mathcal{S}}_{\mathrm{reg}},\Gamma)$ is $\widehat{\mathcal{G}}$-invariant.

Proof. This follows immediately from the lemma. $\square$

Lemma 6.6 For every $(\xi,\eta)\in{\mathcal{S}}_{\mathrm{reg}}$ and every $j\in\mathbb{Z}$ we have

[TABLE]

Proof. From equation (43) we get {\delta}_{K^{\ast}}\raisebox{0.0pt}{\scriptstyle\circ,}{\Phi}_{\mathcal{R}}={\Psi}_{R}\raisebox{0.0pt}{\scriptstyle\circ,}{\delta}_{K^{\ast}} on ${\mathcal{S}}_{\mathrm{reg}}$. Differentiating the preceding equation and then evaluating the result at $X(\xi,\eta)\in T_{(\xi,\eta)}{\mathcal{S}}_{\mathrm{reg}}$ gives

[TABLE]

for all $(\xi,\eta)\in{\mathcal{S}}_{\mathrm{reg}}$. When $(\xi,\eta)\in\mathcal{D}$, by definition ${\delta}_{K^{\ast}}(\xi,\eta)=\delta(\xi,\eta)$. So for every $(\xi,\eta)\in{\mathcal{S}}_{\mathrm{reg}}$

[TABLE]

Thus

[TABLE]

for every $(\xi,\eta)\in\mathcal{D}$. By analytic continuation (47) holds for every $(\xi,\eta)\in{\mathcal{S}}_{\mathrm{reg}}$. Now $T_{(\xi,\eta)}{\Phi}_{\mathcal{R}}$ sends $T_{(\xi,\eta)}{\mathcal{S}}_{\mathrm{reg}}$ to $T_{{\Phi}_{\mathcal{R}}(\xi,\eta)}{\mathcal{S}}_{\mathrm{reg}}$. Since $T_{(\xi,\eta)}{\Phi}_{\mathcal{R}}X(\xi,\eta)={\mathrm{e}}^{2\pi i/n}X(\xi,\eta)$ for every $(\xi,\eta)\in{\mathcal{S}}_{\mathrm{reg}}$, it follows that ${\mathrm{e}}^{2\pi i/n}X(\xi,\eta)$ lies in $T_{{\Phi}_{\mathcal{R}}(\xi,\eta)}{\mathcal{S}}_{\mathrm{reg}}$. Also since $T_{{\delta}_{K^{\ast}}(\xi,\eta)}{\Psi}_{R}$ sends $T_{{\delta}_{K^{\ast}}(\xi,\eta)}K^{\ast}$ to $T_{{\Psi}_{R}({\delta}_{K^{\ast}}(\xi,\eta)}K^{\ast}$, we get

[TABLE]

For every $(\xi,\eta)\in{\mathcal{S}}_{\mathrm{reg}}$ we obtain

[TABLE]

that is, equation (46) holds with $j=0$. A similar calculation shows that equation (48) holds with $\mathcal{R}$ replaces by ${\mathcal{R}}^{j}$. This verifies equation (46). $\square$

We now show

Theorem 6.7 The image of a $\widehat{\mathcal{G}}$ invariant geodesic on $({\mathcal{S}}_{\mathrm{reg}},\Gamma)$ under the developing map ${\delta}_{K^{\ast}}$ (19) is a billiard motion in $K^{\ast}$.

Proof. Because ${\Phi}_{{\mathcal{R}}^{j}}$ and ${\Psi}_{R^{j}}$ are isometries of $({\mathcal{S}}_{\mathrm{reg}},\Gamma)$ and $(K^{\ast},\gamma|_{K^{\ast}})$, respectively, it follows from equation (43) that the surjective map ${\delta}_{K^{\ast}}:({\mathcal{S}}_{\mathrm{reg}},\Gamma)\rightarrow(K^{\ast},\gamma|_{K^{\ast}})$ (19) is an isometry. Hence ${\delta}_{K^{\ast}}$ is a developing map. Using the local inverse of ${\delta}_{K^{\ast}}$ and equation (46), it follows that a billiard motion in $\mathrm{int}(K^{\ast}\setminus\{0\})$ is mapped onto a geodesic in $({\mathcal{S}}_{\mathrm{reg}},\Gamma)$, which is possibly broken at the points $({\xi}_{i},{\eta}_{i})={\delta}^{-1}_{K^{\ast}}(p_{i})$. Here $p_{i}\in\partial K^{\ast}$ are the points where the billiard motion undergoes a reflection. But the geodesic on ${\mathcal{S}}_{\mathrm{reg}}$ is smooth at $({\xi}_{i},{\eta}_{i})$ since the geodesic vector field $X$ is holomorphic on ${\mathcal{S}}_{\mathrm{reg}}$. Thus the image of the geodesic under the developing map ${\delta}_{K^{\ast}}$ is a billiard motion. $\square$

[TABLE]

Next we follow a $G$-invariant set of billiard motions in $(K^{\ast},\gamma|_{K^{\ast}})$, which is the union of an $R$-invariant billiard motion and its $U$ reflection. After identification of equivalent edges of $\mathrm{cl}(K^{\ast})$, see figure 13 (left) and (center) and then dividing out the induced $G$ action, we get a motion on the Riemann surface ${\widetilde{S}}_{\mathrm{reg}}$, which is a geodesic for the induced Riemannian metric $\widehat{\gamma}$ on the $\mathfrak{G}$-orbit space $(\mathbb{C}\setminus{\mathbb{V}}^{+})^{\sim}/\mathfrak{G}$, see figure 13 (right). We now justify these assertions.

A billiard motion ${\gamma}_{z}$ in the regular stellated $n$-gon $K^{\ast}$, which starts at a point $z$ in the interior of $\mathrm{cl}(K^{\ast})\setminus\{O\}$ and does not hit a vertex of $\mathrm{cl}(K^{\ast})$, is made up of line segments, each of which is parallel to an edge of $\mathrm{cl}(K^{\ast}$). It is invariant under the subgroup $\widehat{G}$ of $G$ generated by the rotation $R$. Let {\mathrm{Rfl}}^{{\gamma}_{z}}=\{p\in\partial\,\mathrm{cl}(K^{\ast})\,\rule[-4.0pt]{0.5pt}{13.0pt}\,\,p={\gamma}_{z}(T_{p})\,\,\mbox{for some T_{p}\in\mathbb{R}}\} be the set of reflection points in the boundary of $\mathrm{cl}(K^{\ast})$ of the billiard motion ${\gamma}_{z}$. Since ${\gamma}_{z}$ is invariant under the group $\widehat{G}$, the set ${\mathrm{Rfl}}^{{\gamma}_{z}}$ of reflection points is invariant under group $\widehat{G}$. Because ${\gamma}_{z}$ does not hit a vertex of $\mathrm{cl}(K^{\ast})$, $z$ is not fixed by the reflection $U$. The billiard motion ${\gamma}_{\,\overline{z}}$ starting at $\overline{z}=U(z)$ is invariant under the group $\widehat{G}$ and, by uniqueness of billiard motions with a given starting point, is equal to the billiard motion $U({\gamma}_{z})={\gamma}_{\,\overline{z}}$. So $U({\mathrm{Rfl}}^{{\gamma}_{z}})={\mathrm{Rfl}}^{{\gamma}_{\,\overline{z}}}$. From $U(z)\neq z$, it follows that ${\mathrm{Rfl}}^{{\gamma}_{z}}\cap{\mathrm{Rfl}}^{{\gamma}_{\,\overline{z}}}=\varnothing$. Let $E^{{\gamma}_{z}}$ be the set of closed edges of $\mathrm{cl}(K^{\ast})$, which the billiard motion ${\gamma}_{z}$ reflects off of. In other words, E^{{\gamma}_{z}}=\{\mbox{E$an edge of$\mathrm{cl}(K^{\ast})}\,\rule[-4.0pt]{0.5pt}{13.0pt}\,\,p\in E\,\,\mbox{for some p\in{\mathrm{Rfl}}^{{\gamma}_{z}}}\}.

Lemma 6.8 $E^{{\gamma}_{\overline{z}}}=U(E^{{\gamma}_{z}})$.

Proof. Suppose that $E\in E^{{\gamma}_{\overline{z}}}$. Then for some $p\in{\mathrm{Rfl}}^{{\gamma}_{\overline{z}}}$ we have $p\in E$. Since ${\mathrm{Rfl}}^{{\gamma}_{\overline{z}}}=U({\mathrm{Rfl}}^{{\gamma}_{z}})$, $U(p)\in U({\mathrm{Rfl}}^{{\gamma}_{\overline{z}}})={\mathrm{Rfl}}^{{\gamma}_{z}}$ and $U(p)\in U(E)$. Thus $U(E)\in E^{{\gamma}_{z}}$. So $U(E^{{\gamma}_{\overline{z}}})\subseteq E^{{\gamma}_{z}}$. A similar argument shows that $U(E^{{\gamma}_{z}})\subseteq E^{{\gamma}_{\overline{z}}}$. Hence $E^{{\gamma}_{\overline{z}}}=U(U(E^{{\gamma}_{\overline{z}}}))\subseteq U(E^{{\gamma}_{z}})\subseteq E^{{\gamma}_{\overline{z}}}$, which implies $E^{{\gamma}_{\overline{z}}}=U(E^{{\gamma}_{z}})$. $\square$

Lemma 6.9 The sets $E^{{\gamma}_{z}}$ and $E^{{\gamma}_{\overline{z}}}$ are $\widehat{G}$-invariant.

Proof. Let $E\in E^{{\gamma}_{z}}$ and $p\in E\cap{\mathrm{Rfl}}^{{\gamma}_{z}}$. Since ${\mathrm{Rfl}}^{{\gamma}_{z}}$ is $\widehat{G}$-invariant, it follows that $R(p)\in{\mathrm{Rfl}}^{{\gamma}_{z}}$ and $R(p)\in R(E)$. Hence $R(E)\in E^{{\gamma}_{z}}$. So $E^{{\gamma}_{z}}$ is $\widehat{G}$-invariant. Similarly, $E^{{\gamma}_{\overline{z}}}$ is $\widehat{G}$-invariant. $\square$

Lemma 6.10 Let $S_{0}$ be the reflection $R^{n_{0}}U$ and set $S_{m}=R^{m}S_{0}R^{-m}$ for $m\in\{0,1,\ldots,n-1\}$. Then $S_{m}({\mathrm{Rfl}}^{{\gamma}_{z}})=U({\mathrm{Rfl}}^{{\gamma}_{z}})$.

Proof. If $p\in{\mathrm{Rfl}}^{{\gamma}_{z}}$, then $S_{m}(p)\in U({\mathrm{Rfl}}^{{\gamma}_{z}})$, for $U(p)\in U({\mathrm{Rfl}}^{{\gamma}_{z}})$, which implies $R^{n_{0}}((U(p)))\in U({\mathrm{Rfl}}^{{\gamma}_{z}})$, since $U({\mathrm{Rfl}}^{{\gamma}_{z}})$ is $\widehat{G}$-invariant. Hence $S_{0}(p)\in U({\mathrm{Rfl}}^{{\gamma}_{z}})$. If $p\in{\mathrm{Rfl}}^{{\gamma}_{z}}$, then $R^{-m}(p)\in{\mathrm{Rfl}}^{{\gamma}_{z}}$, since ${\mathrm{Rfl}}^{{\gamma}_{z}}$ is $\widehat{G}$-invariant. So $S_{0}(R^{-m}(p))\in U({\mathrm{Rfl}}^{{\gamma}_{z}})$, which implies $R^{m}S_{0}(R^{-m}(p))\in U({\mathrm{Rfl}}^{{\gamma}_{z}})$, because $U({\mathrm{Rfl}}^{{\gamma}_{z}})$ is $\widehat{G}$-invariant. So $S_{m}({\mathrm{Rfl}}^{{\gamma}_{z}})\subseteq U({\mathrm{Rfl}}^{{\gamma}_{z}})$. A similar argument shows that $S_{m}(U({\mathrm{Rfl}}^{{\gamma}_{z}}))\subseteq{\mathrm{Rfl}}^{{\gamma}_{z}}$. Thus

[TABLE]

So $S_{m}(U({\mathrm{Rfl}}^{{\gamma}_{z}}))={\mathrm{Rfl}}^{{\gamma}_{z}}$, that is, $U({\mathrm{Rfl}}^{{\gamma}_{z}})=S_{m}({\mathrm{Rfl}}^{{\gamma}_{z}})$. $\square$

Lemma 6.11 Every reflection $S_{m}$ interchanges an edge in $E^{{\gamma}_{z}}$ with an edge in $E^{{\gamma}_{\,\overline{z}}}$, specifically, $S_{m}(E^{{\gamma}_{z}})=E^{{\gamma}_{\overline{z}}}$.

Proof. Let $E\in E^{{\gamma}_{z}}$. Then there is a $p\in{\mathrm{Rfl}}^{{\gamma}_{z}}$ such that $p\in E$. So $S_{m}(p)\in S_{m}(E)$. But $S_{m}(p)\in U({\mathrm{Rfl}}^{{\gamma}_{z}})$, which shows that $S_{m}(E)\in U(E^{{\gamma}_{z}})$. Hence $S_{m}(E^{{\gamma}_{z}})\subseteq U(E^{{\gamma}_{z}})$. A similar argument shows that $S_{m}(U(E^{{\gamma}_{z}}))\subseteq E^{{\gamma}_{z}}$. Thus $E^{{\gamma}_{z}}=S_{m}(S_{m}(E^{{\gamma}_{z}}))\subseteq S_{m}(U(E^{{\gamma}_{z}}))\subseteq E^{{\gamma}_{z}}$. So $S_{m}(U(E^{{\gamma}_{z}}))=E^{{\gamma}_{z}}$, which implies $S_{m}(E^{{\gamma}_{z}})=U(E^{{\gamma}_{z}})=E^{{\gamma}_{\overline{z}}}$. $\square$

An extended billiard motion ${\lambda}_{z}$ in $K^{\ast}$ starting at a point $z\in\mathrm{int}(K^{\ast}\setminus\{0\})$ is the union of a billiard motion ${\gamma}_{z}$ in $(\mathrm{int}\,K^{\ast})\setminus\{O\}$ starting at $z$ and a billiard motion ${\gamma}_{\,\overline{z}}$ in $(\mathrm{int}\,K^{\ast})\setminus\{O\}$ starting at $\overline{z}=Uz$. The motion ${\lambda}_{z}$ is invariant under the group generated by the rotation $R$ and the reflection $U$. So ${\lambda}_{z}$ is $G$-invariant. The set of points of an extended billiard motion in $K^{\ast}\setminus\{O\}$, which lie on $\partial K^{\ast}$ is $G$-invariant and is the disjoint union of reflection points ${\mathrm{Rfl}}^{{\gamma}_{z}}$ for the billiard motion ${\gamma}_{z}$ and ${\mathrm{Rfl}}^{{\gamma}_{\,\overline{z}}}=U({\mathrm{Rfl}}^{{\gamma}_{z}})$ for its $U$ reflection ${\gamma}_{\,\overline{z}}$. From lemma 6.10 it follows that the equivalence relation $\sim$ among the closed edges of $\mathrm{cl}(K^{\ast})$ interchanges these subsets. Identifying equivalent points in ${\mathrm{Rfl}}^{{\gamma}_{z}}$ and ${\mathrm{Rfl}}^{{\gamma}_{\,\overline{z}}}$ with the equivalent edges, in which they are contained, gives a continuous motion ${\lambda}^{\sim}_{z}=\Pi({\lambda}_{z})$ in the smooth space $(K^{\ast}\setminus\{O\})^{\sim}$, which is $G$-invariant. Here $\Pi$ is the map (36).

Theorem 6.12 Under the restriction of the mapping

[TABLE]

to $K^{\ast}\setminus\{O\}$ the image of an extended billiard motion ${\lambda}_{z}$ is a smooth geodesic ${\widehat{\lambda}}_{\nu(z)}$ on $({\widetilde{S}}_{\mathrm{reg}},\widehat{\gamma})$, where ${\nu}^{\ast}(\widehat{\gamma})=\gamma|_{\mathbb{C}\setminus{\mathbb{V}}^{+}}$.

Proof. Since the Riemannian metric $\gamma$ on $\mathbb{C}$ is invariant under the group of Euclidean motions, the Riemannian metric ${\gamma}|_{K^{\ast}\setminus\{O\}}$ on $K^{\ast}\setminus\{O\}$ is $G$-invariant. Hence ${\gamma}_{K^{\ast}\setminus\{O\}}$ is invariant under the reflection $S_{m}$ for $m\in\{0,1,\ldots,n-1\}$. So ${\gamma}|_{K^{\ast}\setminus\{O\}}$ pieces together to give a Riemannian metric ${\gamma}^{\sim}$ on the identification space $(K^{\ast}\setminus\{O\})^{\sim}$. In other words, the pull back of ${\gamma}^{\sim}$ under the map ${\Pi}|_{K^{\ast}\setminus\{O\}}:K^{\ast}\setminus\{O\}\rightarrow(K^{\ast}\setminus\{O\})^{\sim}$, which identifies equivalent edges of $K^{\ast}$, is the metric ${\gamma}|_{K^{\ast}\setminus\{O\}}$. Since ${\Pi}|_{K^{\ast}\setminus\{O\}}$ intertwines the $G$-action on $K^{\ast}\setminus\{O\}$ with the $G$-action on $(K^{\ast}\setminus\{O\})^{\sim}$, the metric ${\gamma}^{\sim}$ is $G$-invariant. It is flat because the metric $\gamma$ is flat. So ${\gamma}^{\sim}$ induces a flat Riemannian metric $\widehat{\gamma}$ on the orbit space $(K^{\ast}\setminus\{O\})^{\sim}/G={\widetilde{S}}_{\mathrm{reg}}$. Since the extended billiard motion ${\lambda}_{z}$ is a $G$-invariant broken geodesic on $(K^{\ast}\setminus\{O\},{\gamma}_{K^{\ast}\setminus\{O\}})$, which is made up of two continuous pieces, it gives rise to a continuous broken geodesic ${\lambda}^{\sim}_{\Pi(z)}$ on $((K^{\ast}\setminus\{O\})^{\sim},{\gamma}^{\sim})$, which is $G$-invariant. Thus ${\widehat{\lambda}}_{\nu(z)}=\nu({\lambda}_{z})$ is a piecewise smooth geodesic on the smooth $G$-orbit space $((K^{\ast}\setminus\{O\})^{\sim}/G={\widetilde{S}}_{\mathrm{reg}},\widehat{\gamma})$.

We need only show that ${\widehat{\lambda}}_{\nu(z)}$ is smooth. To see this we argue as follows. Let $s\subseteq K^{\ast}$ be a closed segment of a billiard motion ${\gamma}_{z}$, which is contained in the extended billiard motion ${\lambda}_{z}$ that does not meet a vertex of $\mathrm{cl}(K^{\ast})$. Then ${\gamma}_{z}$ is a horizontal straight line motion in $\mathrm{cl}(K^{\ast})$. Suppose that $E_{k_{0}}$ is the edge of $K^{\ast}$, perpendicular to the direction $u_{k_{0}}$, which is first met by ${\gamma}_{z}$ and let $P_{k_{0}}$ be the meeting point. Let $S_{k_{0}}$ be the reflection in $E_{k_{0}}$. The continuation of the motion ${\gamma}_{z}$ at $P_{k_{0}}$ is the horizontal line $RS_{k_{0}}({\gamma}_{z})$ in $K^{\ast}_{k_{0}}$. Recall that $K^{\ast}$ is the translation of $K^{\ast}$ by ${\tau}_{k_{0}}$. Since $O_{k_{0}}={\tau}_{k_{0}}(0)$ is the center of $K^{\ast}_{k_{0}}$, the extended motion is the same as the motion $U({\gamma}_{z})$ translated by ${\tau}_{k}$. Using a suitable sequence of

reflections in the edges of a suitable $K^{\ast}_{k_{0}\cdots k_{\ell}}$ followed by a rotation $R$, which gives rise to a reflection $U$ and a translation in $\mathcal{T}$ corresponding to their origins, we can extend $s$ to a smooth straight line $\lambda$ in $\mathbb{C}\setminus{\mathbb{V}}^{+}$, see figure 14. The line $\lambda$ is a geodesic in $(\mathbb{C}\setminus{\mathbb{V}}^{+},\gamma|_{\mathbb{C}\setminus{\mathbb{V}}^{+}})$, which in $K^{\ast}$ has image ${\widehat{\lambda}}_{\nu(z)}$ under the $\mathfrak{G}$-orbit map that is a smooth geodesic on $({\widehat{S}}_{\mathrm{reg}},\widehat{\gamma})$. The geodesic $\nu(\lambda)$ starts at $\nu(z)$. Thus the smooth geodesic ${\widehat{\lambda}}_{\nu(z)}$ and the possibly broken geodesic $\nu(\lambda)$ are equal. In other words, $\nu(\lambda)$ is a smooth geodesic. $\square$

Thus the affine orbit space ${\widetilde{S}}_{\mathrm{reg}}=(\mathbb{C}\setminus{\mathbb{V}}^{+})/\mathfrak{G}$ with flat Riemannian metric $\widehat{\gamma}$ is the affine analogue of the Poincaré model of the affine Riemann surface ${\mathcal{S}}_{\mathrm{reg}}$ as an orbit space of a discrete subgroup of $\mathrm{PGl}(2,\mathbb{C})$ acting on the unit disk in $\mathbb{C}$ with the Poincaré metric.

8 Appendix. Group theoretic properties

In this appendix we discuss some group theoretic properties of the set of equivalent edges of $\mathrm{cl}(K^{\ast})$, which we use to determine the topology of ${\widetilde{S}}_{\mathrm{reg}}$.

Let $\mathcal{E}$ be the set of unordered pairs $[E,E^{\prime}]$ of nonadjacent edges of $\mathrm{cl}(K^{\ast})$. Define an action $\bullet$ of $G$ on $\mathcal{E}$ by

[TABLE]

for every unordered pair $[E,E^{\prime}]$ of nonadjacent edges of $\mathrm{cl}(K^{\ast})$. For every $g\in G$ the edges $g(E)$ and $g(E^{\prime})$ are nonadjacent. This follows because the edges $E$ and $E^{\prime}$ are nonadjacent and the elements of $G$ are invertible mappings of $\mathbb{C}$ into itself. So $\varnothing=g(E\cap E^{\prime})=g(E)\cap g(E^{\prime})$. Thus the mapping $\bullet$ is well defined. It is an action because for every $g$ and $h\in G$ we have

[TABLE]

The action $\bullet$ of $G$ on $\mathcal{E}$ induces an action $\cdot$ of the group $G^{j}$ of reflections on the set ${\mathcal{E}}^{j}$ of equivalent edges of $\mathrm{cl}(K^{\ast})$, which is defined by

[TABLE]

for every $g_{j}\in G^{j}$, every edge $E$ of $\mathrm{cl}(K^{\ast})$, and every generator $S^{(j)}_{k}$ of $G^{j}$, where $k=0,1,\ldots,$ $n-1$. Since $g_{j}S^{(j)}_{k}g^{-1}_{j}=S^{(j)}_{r}$ by corollary 3.3b, the mapping $\cdot$ is well defined.

Lemma A1 The group $G$ action $\bullet$ sends a $G^{j}$-orbit on ${\mathcal{E}}^{j}$ to another $G^{j}$-orbit on ${\mathcal{E}}^{j}$.

Proof. Consider the $G^{j}$-orbit of $[E,S^{(j)}_{m}(E)]\in{\mathcal{E}}^{j}$. For every $g\in G$ we have

[TABLE]

because $G^{j}$ is a normal subgroup of $G$ by corollary 3.3c. Since

[TABLE]

and $gS^{(j)}_{m}g^{-1}=S^{(j)}_{r}$ by corollary 3.3b, it follows that g\,\,\mbox{\raisebox{1.0pt}{\tiny\bullet}}\,[E,S^{(j)}_{m}(E)]\in{\mathcal{E}}^{j}. $\square$

Lemma A2 For every $j=0,1,\infty$ and every $k=0,1,\ldots,n-1$ the isotropy group $G^{j}_{e^{j}_{k}}$ of the $G^{j}$ action on ${\mathcal{E}}^{j}$ at $e^{j}_{k}=[E,S^{(j)}_{k}(E)]$ is $\langle S^{(j)}_{k}\,\rule[-4.0pt]{0.5pt}{13.0pt}\,\,(S^{(j)}_{k})^{2}=e\rangle$.

Proof. Every $g\in G^{j}_{e^{j}_{k}}$ satisfies

[TABLE]

if and only if

[TABLE]

if and only if one of the statements 1) $g(E)=E$ & $S^{(j)}_{k}(E)=S^{(j)}_{r}(g(E))$ or 2) $E=g(S^{(j)}_{r}(E))$ & $g(E)=S^{(j)}_{k}(E)$ holds. From $g(E)=E$ in 1) we get $g=e$ using lemma 3.2. To see this we argue as follows. If $g\neq e$, then $g=R^{p}(S^{(j)})^{\ell}$ for some $\ell=0,1$ and some $p\in\{0,1,\ldots,n-1\}$, see equation (50). Suppose that $g=R^{p}$ with $p\neq 0$. Then $g(E)\neq E$, which contradicts our hypothesis. Now suppose that $g=R^{p}S^{(j)}$. Then $E=g(E)=R^{p}S^{(j)}(E)$, which gives $R^{-p}(E)=S^{(j)}(E)$. Let $A$ and $B$ be end points of the edge $E$. Then the reflection $S^{(j)}$ sends $A$ to $B$ and $B$ to $A$, while the rotation $R^{-p}$ sends $A$ to $A$ and $B$ to $B$. Thus $R^{-p}(E)\neq S^{(j)}(E)$, which is a contradiction. Hence $g=e$. If $g(E)=S^{(j)}_{k}(E)$ in 2), then $(S^{(j)}_{k}g)(E)=E$. So $S^{(j)}_{k}g=e$ by lemma 3.2, that is, $g=S^{(j)}_{k}$. $\square$

For every $j=0,1,\infty$ and every $m=0,1,\ldots,\frac{n}{d_{j}}-1$ let $G^{j}_{e^{j}_{md_{j}}}=\{g_{j}\in G^{j}\,\rule[-4.0pt]{0.5pt}{13.0pt}\,\,g_{j}\cdot e^{j}_{md_{j}}=e^{j}_{md_{j}}\}$ be the isotropy group of the $G^{j}$ action on ${\mathcal{E}}^{j}$ at $e^{j}_{md_{j}}=[E,S^{(j)}_{md_{j}}(E)]$. Since $G^{j}_{e^{j}_{md_{j}}}=\langle S^{(j)}_{md_{j}}\,\rule[-4.0pt]{0.5pt}{13.0pt}\,\,(S^{(j)}_{md_{j}})^{2}=e\rangle$ is an abelian subgroup of $G^{j}$, it is a normal subgroup. Thus $H^{j}=G^{j}/G^{j}_{e^{j}_{md_{j}}}$ is a subgroup of $G^{j}$ of order $(2n/d_{j})/2=n/d_{j}$. This proves

Lemma A3 For every $j=0,1,\infty$ and each $m=0,1,\ldots,\frac{n}{d_{j}}-1$ the $G^{j}$-orbit of $e^{j}_{md_{j}}$ in ${\mathcal{E}}^{j}$ is equal to the $H^{j}$-orbit of $e^{j}_{md_{j}}$ in ${\mathcal{E}}^{j}$.

Lemma A4 For $j=0,1,\infty$ we have $H^{j}=\langle V=R^{d_{j}}\,\rule[-4.0pt]{0.5pt}{13.0pt}\,\,V^{n/d_{j}}=e\rangle$.

Proof. Since

[TABLE]

we get $S^{(j)}_{md_{j}}=R^{(2m+\frac{n_{j}}{d_{j}})d_{j}}U=(R^{d_{j}})^{m}S^{(j)}$. Because the group $G^{j}$ is generated by the reflections $S^{(j)}_{k}$ for $k=0,1,\ldots,n-1$, it follows that

[TABLE]

$K_{j}$ is a subgroup of $G$ of order $2n/d_{j}$. Clearly the isotropy group $G^{j}_{e^{j}_{md_{j}}}=\langle S^{(j)}_{md_{j}}\,\rule[-4.0pt]{0.5pt}{13.0pt}\,\,(S^{(j)}_{md_{j}})^{2}=e\rangle$ is an abelian subgroup of $K^{j}$. Hence $H^{j}=G^{j}/G^{j}_{e^{j}_{md_{j}}}\subseteq K^{j}/G^{j}_{e^{j}_{md_{j}}}=L^{j}$, where $L^{j}$ is a subgroup of $K^{j}$ of order $(2n/d_{j})/2=n/d_{j}$. Thus the group $L^{j}$ has the same order as its subgroup $H^{j}$. So $H^{j}=L^{j}$. But $L^{j}=\langle V=R^{d_{j}}\,\rule[-4.0pt]{0.5pt}{13.0pt}\,V^{n/d_{j}}=e\rangle$. $\square$

Let $f^{j}_{\ell}=R^{\ell}\cdot e^{j}_{0}$. Then

[TABLE]

So

[TABLE]

This proves

[TABLE]

since every $k\in\{0,1,\ldots,n-1\}$ may be written uniquely as $md_{j}+\ell$ for some $m\in\{0,1,\ldots,\frac{n}{d_{j}}-1\}$ and some $\ell\in\{0,1,\ldots,d_{j}-1\}$.