TITLE: Drawing a directed graph in Maple
QUESTION [3 upvotes]: I have been playing around with some Maple today for a semigroup/ graph theory style project.
I want to draw a left cayley graph with vertices $\{1,....,12\}$ and edges $[5, 5]$, $\{[6, 2].[7, 1], [1, 3], [1, 5], [2, 4], [2, 6], [3, 5], [3, 7], [4, 5], [4, 8], [6, 9], [7, 5], [8, 5], [8, 10], [9, 5], [9, 11], [10, 4], [10, 5], [11, 5], [11, 12], [12, 5], [12, 9]\}$
This is a directed graph; e.g. $[6, 2]$ means I want $6\rightarrow2$ as an edge.
I tried the command:
S := Graph([5, 5], {[6, 2].[7, 1], [1, 3], [1, 5], [2, 4], [2, 6], [3, 5], [3, 7], [4, 5], [4, 8], [6, 9], [7, 5], [8, 5], [8, 10], [9, 5], [9, 11], [10, 4], [10, 5], [11, 5], [11, 12], [12, 5], [12, 9]})
However, I get the error message:
"Error, (in GraphTheory:-Graph) vertices must have distinct labels
"
How can I fix this? If someone could give me a code that would produce this graph, I would be forever grateful.
Thank you
Grant
EDIT:
So the problem was resolved; it was a period between the first 2 edges, silly me. However, I now have a second question: Is there a way it can be set up that Maple gives me the "neatest" looking layout of the vertices to make the graph look the most appealing? For example, with the code above it just lays out 1 to 12 in a circle and looks pretty messy.

REPLY [1 votes]: As discussed in the comments, remove the loop edge and correct the period that should be a comma. Then construct the graph S with your GraphTheory:-Graph command. Then issue the command
GraphTheory:-DrawGraph(S, style= spring, redraw, dimension= 3);
Then click on the graph. Enlarge it by stretching the bounding box in both directions. Right click on the plot. Go to the Manipulator submenu. Select Rotate. Now use your mouse to rotate the plot into the best view. You should be able to clearly see three tetrehedra that share a single vertex, 5. There are only two other vertices, 2 and 6, and they are on a path connecting vertices 4 and 9.
If you can't see all that, issue the DrawGraph command again: There is a small amount of randomness in the spring algorithm, and you may get a better plot.


TITLE: Proving limit of a sequence without Taylor series
QUESTION [0 upvotes]: I'm trying to prove that pi is irrational with Niven's proof. In the proof I need to show the following: $lim_{n\to\infty}\frac{\pi^{n+1}a^n}{n!}=0$ where $\pi=\frac{a}{b}$ to obtain the contradiction.
I know that this can be done by writing a Taylor series that converges meaning that the terms must go to 0, however, I would like to find a way around that using an $\epsilon - N$ proof. Any help will be appreciated.

REPLY [2 votes]: So your sequence is $\;a_n:=\cfrac{a^{2n+1}}{b^{n+1}n!}\;$ . Look at the numerical series with this general term, and since it is a positive series apply the $\;n\,-$th root test to it:
$$\sqrt[n]{\frac{a^{2n+1}}{b^{n+1}n!}}=\frac{a^2}{b}\sqrt[n]\frac ab\frac1{\sqrt[n]{n!}}\xrightarrow[n\to\infty]{}\frac{a^2}b\cdot1\cdot0=0$$
and thus the infinite series $\;\sum\limits_{n=1}^\infty a_n\;$ converges $\;\implies \lim\limits_{n\to\infty}a_n=0\;$


text:
\begin{document}

\title[The Euclidean Picard modular groups]{Generators for the Euclidean Picard Modular Groups}

\author{Tiehong zhao}
\address{Institut de Math\'{e}matiques\\Universit\'{e} Pierre et Marie Curie\\ 4, place Jussieu\\F-75252 Paris, France}
\email{zhao@math.jussieu.fr}

\date{}
\thanks{The author is supported by the China-funded Postgraduates Studying Aboard Program for Building Top
University.}

\begin{abstract}
The goal of this article is to show that five explicitly given transformations, a rotation, two screw Heisenberg rotations,
 a vertical translation and an involution generate the Euclidean Picard modular groups with coefficient in the Euclidean ring
  of integers of a quadratic imaginary number field. We also obtain the relations of the isotropy subgroup by analysis of the
  combinatorics of the fundamental domain in Heisenberg group.
\end{abstract}

\maketitle

\section{Introduction}
\label{sec-intro}

Let $\mathcal{K}=\mathbb{Q}(\sqrt{-d})$ be a quadratic imaginary number field. Let $\mathcal{O}_d$ be the ring
of algebraic integers of $\mathcal{K}$. The Bianchi groups $PSL_2(\mathcal{O}_d)$ are the simplest numerically defined discrete groups. In number theory they have been used to study the zeta-functions of binary Hermitian forms over the rings $\mathcal{O}_d$. As the isometric groups acting on the half-upper space, they are of interest in the theory of Fuchsian groups and the related theory of Riemann surfaces. The Bianchi groups can be considered as the natural algebraic generalization of the classical modular group $PSL_2(\mathbb{Z})$. A good general reference for the Bianchi groups and their relation to the modular group is \cite{Fi}. As a natural generalization of the Bianchi groups, the subgroups of $PU(2,1)$ with coefficients in $\mathcal{O}_d$ are called {\it Picard modular groups}, denoted by $PU(2,1;\mathcal{O}_d)$. These groups have attracted a great deal of attention both
for their intrinsic interest as discrete groups and also for their applications in complex hyperbolic geometry
 (as the holomorphic automorphism subgroups).

A general method to determine finite presentations for each Bianchi group $PSL_2(\mathcal{O}_d)$ was developed by Swan \cite{Sw} based on geometrical work of Bianchi, while a separate purely algebraic method was given by Cohn \cite{Co}. In general, fundamental domains for Lie groups were studied by \cite{GR}, but the complex hyperbolic space is a particularly challenging case since no existence of totally geodesic hypersurface. So far very few examples of
 complex hyperbolic lattices have been constructed explicitly. Due to the famous paper \cite{Mo} of Mostow, other explicit
  constructions of fundamental domains for lattices in $PU(2,1$) were obtained, see for example, the work of Goldman
  and Parker \cite{GP}, Deraux, Falbel and Paupert \cite{DFP}, Schwartz \cite{Sch}, the survey paper of Parker \cite{Pa}. In particular,
  the group $PU(2, 1;\mathcal{O}_3) = PU(2, 1;\mathbb{Z}[\omega]),$
   where $\omega$ is a cube root of unity was studied by Falbel and Parker in \cite{FP} and its sister was treated recently in \cite{Zh}.
   Analogously a fundamental domain of Gauss-Picard group $PU(2,1;\mathbb{Z}[i])$ was described in (\cite{FFP}, \cite{FL1}, \cite{FL2}) and
   analysis of the combinatorics of the fundamental domain gives rise to a presentation of the group in \cite{FFP}.

In this paper we give a description of generators for certain Picard modular groups $PU(2,1;\mathcal{O}_d)$ where the
ring $\mathcal{O}_d$ is Euclidean except for $d=1,3$ (these two exceptional cases have been studied in many aspects).
Among the quadratic imaginary number rings $\mathcal{O}_d$ only $\mathcal{O}_1,\mathcal{O}_2,\mathcal{O}_3,\mathcal{O}_7,\mathcal{O}_{11}$
have a Euclidean algorithm, see \cite{ST}, although there is a larger finite collection of $\mathcal{O}_d$'s
(such as $d=1,$ $2,$ $3,$ $7,$ $11,$ $19,$ $43$, $67,$ $163$, see \cite{Zi}) which have class number one.
For these values of $d$ the orbifold $\mathbf{H}^2_\mathbb{C}/PU(2, 1;\mathcal{O}_d)$ has only one cusp.
 The method is based on the construction of various shapes of precisely fundamental domains for the stabilisers of infinity
 of $PU(2,1;\mathcal{O}_d)$ and then on a determination of several neighboring isometric spheres such that the union of the boundaries of these isometric
 spheres contains the fundamental domain of the stabiliser, which was used in (\cite{FP}, \cite{FFP}, \cite{Zh}). Compared with other groups, the generators of these groups in (\cite{FP}, \cite{FFP}, \cite{Zh}) are easy to be obtained since the fundamental domain constructed lies completely inside the
  boundary of the isometric sphere centred at origin. Again this reflects the underlying number theory; $\mathcal{O}_1$
  and $\mathcal{O}_3$ have non-trivial units while the other three do not. A simple algorithm to decompose any
  transformation in the Picard group $PU(2,1;\mathcal{O}_1)$ as a product of the generators was given in \cite{FFLP},
   one would be interesting to extend their method to other Picard modular groups. However, it would also be important to find
    the generators in terms of geometric ways which will provide more informations that one continue to construct a fundamental
  domain explicitly for each of Picard modular groups.

  I would like to thank my advisor E. Falbel for his warm encouragements all along
this work and for a number of helpful comments.

\section{Complex hyperbolic space}
\label{sec-complex}
\subsection{The Siegel domain}

\medskip

In this section we give the necessary background material on complex hyperbolic space. To know more details of this material we refer the reader to \cite{Go}.

Let $\mathbb{C}^{2,1}$ denote the complex vector space equipped with the Hermitian form defined by
\begin{equation*}\langle \mathbf{z},\mathbf{w}\rangle=z_1\bar{w}_3+z_2\bar{w}_2+z_3\bar{w}_1,
\end{equation*}
where $\mathbf{z}$ and $\mathbf{w}$ be the column vectors $[z_1,z_2,z_3]^t$ and $[w_1,w_2,w_3]^t$ respectively.
The projective model of complex hyperbolic space $\mathbf{H}_{\mathbb{C}}^2$ is defined to be the collection of negative
lines in $\mathbb{C}^{2,1}$, namely, those points $\mathbf{z}$ satisfying $\langle \mathbf{z},\mathbf{z}\rangle<0$.

We mainly take the {\it Siegel domain} $\mathfrak{S}$ as a upper half-space model for the complex hyperbolic space, that is given by
$$\mathfrak{S}=\{(z_1,z_2)\in\mathbb{C}^2:2\Re ez_1+|z_2|^2<0\}.$$ The boundary of the {\it Siegel domain} $\mathfrak{S}$ is identified with the one-point compactification of the Heisenberg group. The Heisenberg group $\mathfrak{R}$ is $\mathbb{C}\times \mathbb{R}$ with the group law
\begin{equation*}(\zeta_1,t_1)\diamond(\zeta_2,t_2)=(\zeta_1+\zeta_2,t_1+t_2+2\Im m(\zeta_1\bar{\zeta}_2)).\end{equation*}
The Cygan metric on $\mathfrak{R}$ is given by
$$
\rho_0((\zeta_1,t_1),(\zeta_2,t_2))=\left||\zeta_1-\zeta_2|^2-it_1+it_2-2i\Im m(\zeta_1\bar{\zeta}_2)\right|,
$$
in terms of the operation of Heisenberg group, that is $\left|(\zeta_1,t_1)^{-1}\diamond(\zeta_2,t_2)\right|$.

We can extend the Cygan metric to an incomplete metric on
$\bar{\mathfrak{S}}-\{\infty\}$ as follows
$$\tilde{\rho}_0=\left||\zeta_1-\zeta_2|^2+|u_1-u_2|-it_1+it_2-2i\Im m(\zeta_1\bar{\zeta}_2)\right|.$$

The Siegel domain $\mathfrak{S}$ is parametrised in horospherical
coordinates by
\begin{equation}\label{eq:2-1}(\zeta,t,u)\longrightarrow\left[\begin{array}{c}(-|\zeta|^2-u+it)/2\\ \zeta\\1\end{array}\right]\end{equation}
and the point at infinity being
\begin{equation*}q_{\infty}=\left[\begin{array}{c}1\\0\\0\end{array}\right].\end{equation*}
Then $\mathfrak{S}=\mathfrak{R}\times\mathbb{R}_+$ and $\partial\mathfrak{S}=(\mathfrak{R}\times\{0\})\cup\{q_{\infty}\}$.

\medskip

\subsection{Complex hyperbolic isometries}

\medskip

Let $U(2,1)$ be the group of matrices that are unitary with respect to the form $\langle .,.\rangle$. The group of holomorphic isometries of complex hyperbolic space is the projective unitary group
$PU(2,1)=U(2,1)/U(1)$, with a natural identification $U(1)=\{e^{i\theta}I,\theta\in[0,2\pi)\}.$ We now describe the action of the stabiliser of $q_\infty$ on the Heisenberg group.

The Heisenberg group acts on itself by {\it Heisenberg translations}. For $(\tau,v)\in\mathfrak{R}$, this is
\begin{equation*}T_{(\tau,v)}:(\zeta,t)\mapsto(\zeta+\tau,t+v+2\Im m(\tau\bar{\zeta}))=(\tau,v)\diamond(\zeta,t).\end{equation*}
Heisenberg translation by $(0,v)$ for any $v\in\mathbb{R}$ is called {\it vertical translation} by $v$.

The unitary group $U(1)$ acts on the Heisenberg group by {\it Heisenberg rotations}. For $e^{i\theta}\in U(1)$, the rotation fixing $q_0=(0,0,0)$ is given by \begin{equation*}R_{\theta}:(\zeta,t)\mapsto(e^{i\theta}\zeta,t).\end{equation*} All other Heisenberg rotations may be obtained from these by conjugating by a Heisenberg translation.

For $\lambda\in\mathbb{R}_+$, {\it Heisenberg dilation} by $\lambda$ fixing $q_\infty$ and $q_0=(0,0,0)\in\partial\textbf{H}^2_{\mathbb{C}}$ is given by
\begin{equation*}D_\lambda:(\zeta,t)\mapsto(\lambda \zeta,\lambda^2 t).\end{equation*}
All other Heisenberg dilations fixing $q_\infty$ may be obtained by conjugating by a Heisenberg translation.

The stabiliser of $q_{\infty}$ in $PU(2,1)$ is generated by all Heisenberg translations, rotations and dilations. However, only Heisenberg translations and rotations are isometric with respect to various natural metrics on $\mathfrak{R}$. For this reason the group generated by all Heisenberg translations and rotations, which is the semidirect product $U(1)\ltimes\mathfrak{R}$, is called the {\it Heisenberg isometry group} $Isom(\mathfrak{R})$. The nontrivial central elements of the Heisenberg isometry group are precisely the vertical translations. In particular, each element of $Isom(\mathfrak{R})$ preserves every horosphere.

There is a canonical projection from $\mathfrak{R}$ to $\mathbb{C}$ called {\it vertical projection} and denoted by $\Pi$, given by $\Pi: (\zeta,t)\longmapsto \zeta.$
Using the exact sequence
\begin{equation*}0\longrightarrow\mathbb{R}\longrightarrow\mathfrak{R}\stackrel{\Pi}{\longrightarrow}\mathbb{C}\longrightarrow0,\end{equation*}
we obtain the exact sequence (see Scott \cite{Sc} page 467)
\begin{equation}\label{eq:2-2}0\longrightarrow\mathbb{R}\longrightarrow  Isom(\mathfrak{R})\stackrel{\Pi_{*}}{\longrightarrow}  Isom(\mathbb{C})\longrightarrow1.\end{equation}
Here $Isom(\mathbb{C})$ is the group of orientation preserving
Euclidean isometries of $\mathbb{C}$.

Observe the elements in $Isom(\mathbb{C})$ can be represented by
matrices in $GL(2,\mathbb{C})$ of the form
\begin{equation*}\left[\begin{array}{cc}
 e^{i\theta}&\zeta_0\\0&1
\end{array}\right]
\left[\begin{array}{c}
 \zeta\\1
 \end{array}\right]=\left[\begin{array}{c}
 e^{i\theta}\zeta+\zeta_0\\1
\end{array}\right]\end{equation*}
Therefore, the map $\Pi_{*}$ can be given by
\begin{equation}\label{eq:2-3}\Pi_{*}:\left[ \begin{array}{ccc}
 1&-\bar{\zeta_{0}}e^{i\theta}&(-|\zeta_{0}|^{2}+it_{0})/2\\0&e^{i\theta}&\zeta_{0}\\0&0&1
\end{array}\right]\longrightarrow\left[ \begin{array}{cc}
 e^{i\theta}&\zeta_0\\0&1
\end{array}\right].\end{equation}
It is clear that
\begin{equation*}Ker(\Pi_{*})=\left\lbrace \left[ \begin{array}{ccc}
 1&0&it_{0}/2\\0&1&0\\0&0&1
\end{array}\right]:t_{0}\in\mathbb{R}\right\rbrace\end{equation*}
is the group of vertical translations fixing $q_{\infty}$.

\medskip

\subsection{Isometric spheres}

\medskip

Given an element $G\in PU(2,1)$ with satisfying $G(q_{\infty})\neq
q_{\infty}$, we define the isometric sphere of $G$ to be the
hypersurface
$$\left\{\mathbf{z}\in\textbf{H}^2_{\mathbb{C}}:|\langle \mathbf{z},q_{\infty}\rangle|=|\langle \mathbf{z},G^{-1}(q_{\infty})\rangle|\right\}.$$
For example, the isometric sphere of
$$I_0=\left[\begin{array}{ccc}
0&0&1\\
0&-1&0\\
1&0&0
\end{array}\right]
$$
is \begin{equation}\label{eq:2-4}
\mathcal{B}_0=\left\{(\zeta,t,u)\in\mathfrak{S}:\left||\zeta|^2+u+it\right|=2\right\}\end{equation} in horospherical
coordinates or
\begin{equation}\label{eq:2-5}\mathcal
{B}_0=\left\{[z_1,z_2,z_3]\in\textbf{H}^2_{\mathbb{C}}:|z_1|=|z_3|\right\}\end{equation}
in homogeneous coordinates.

All other isometric spheres are images of $\mathcal
{B}_0$
by Heisenberg dilations, rotations and translations. Thus the
isometric sphere with radius $r$ and centre $(\zeta_0,t_0,0)$ is given
by
$$\left\{(\zeta,t,u):\left||\zeta-\zeta_0|^2+u+it-it_0+2i\Im m(\zeta\bar{\zeta}_0)\right|=r^2\right\}.$$

If $G$ has the matrix form \begin{equation}\label{eq:2-6}\left[\begin{array}{ccc}
a&b&c\\
d&e&f\\
g&h&j
\end{array}\right]
,\end{equation} then $G(q_{\infty})\neq q_{\infty}$ if and
only if $g\neq0$. The isometric sphere of $G$ has radius
$r=\sqrt{2/|g|}$ and centre $G^{-1}(q_{\infty})$, which in
horospherical coordinates is
$$(\zeta_0,t_0,0)=(\bar{h}/\bar{g},2\Im m(\bar{j}/\bar{g}),0).$$

Isometric spheres are examples of bisectors. Mostow \cite{Mo} showed that
a bisector is the preimage of a geodesic, called spine, under
orthogonal projection onto the unique complex line containing it.
The fibres of this projection are complex lines called the slices of
the bisector. Goldman \cite{Go} showed that a bisector is the union of
all totally real Larangian planes containing the spine. Such
Lagrangian planes are called the meridians.

\section{On the structure of the stabiliser}

In this section we will obtain the generators and relations of the stabiliser of Picard modular groups by analysis of the fundamental domain in Heisenberg group.

Let $\mathcal{O}_d$ be the ring of integers in the quadratic imaginary number field $\mathbb{Q}(i\sqrt{d})$, where $d$ is
a positive square-free integer. If $d\equiv1,2\ (mod\ 4)$, then $\mathcal{O}_d=\mathbb{Z}[i\sqrt{d}]$ and if
 $d\equiv3\ (mod\ 4)$, then $\mathcal{O}_d=\mathbb{Z}[\omega_d]$, where $\omega_d=(1+i\sqrt{d})/2   $. The group $\Gamma_d=PU(2,1;\mathcal{O}_d)$
 is called {\it Euclidean Picard modular group} if the ring $\mathcal{O}_d$ is Euclidean, namely, only the rings $\mathcal{O}_1,\mathcal{O}_2,\mathcal{O}_3,\mathcal{O}_7,\mathcal{O}_{11}$. Further relative to amalgamation property, these five groups can be subclassified into three groupings $\{\Gamma_1\},$ $\{\Gamma_3\},$ $\{\Gamma_2,\Gamma_7,\Gamma_{11}\}$. Since two classes $\{\Gamma_1\},\{\Gamma_3\}$ (\textit{c.f.} \cite{FP}, \cite{FFP}) have been studied in detail, we mainly describe the remaining class $\{\Gamma_2,\Gamma_7,\Gamma_{11}\}$.

\subsection{The stabiliser of $q_\infty$}

First we want to analyse $(\Gamma_d)_\infty$ with $d=2,7,11$, the stabiliser of $q_\infty$. Every element of $(\Gamma_d)_\infty$ is upper triangular and its diagonal entries are units in $\mathcal{O}_d$. Recall that the units of $\mathcal{O}_1$ are $\pm1,\pm i$, they are $\pm1,\pm\omega,\pm\omega^2$ for $\mathcal{O}_3$ and they are $\pm1$ for others. Therefore $(\Gamma_d)_\infty$ contains no dilations and so is a subgroup of $Isom(\mathfrak{R})$ and fits into the exact sequence as
$$0\longrightarrow\mathbb{R}\cap(\Gamma_d)_\infty\longrightarrow
(\Gamma_d)_\infty\stackrel{\Pi_{*}}{\longrightarrow}\Pi_{*}((\Gamma_d)_\infty)\longrightarrow1.$$

We can write the isometry group of the integer lattice as
\begin{equation*}
Isom(\mathcal{O}_d)=\left\{\left[\begin{array}{cc}\alpha&\beta\\0&1\end{array}\right]:\alpha,\beta\in\mathcal{O}_d, \alpha\ \text{is a unit}\right\}.
\end{equation*}

We now find the image and kernel in this exact sequence.

\begin{proposition}\label{eq:3-1}The stabiliser
$(\Gamma_d)_\infty$ of $q_{\infty}$ in $\Gamma_d$ satisfies
$$0\longrightarrow 2\sqrt{d}\mathbb{Z}\longrightarrow(\Gamma_d)_\infty\stackrel{\Pi_{*}}{\longrightarrow}\Delta\longrightarrow1,$$
where $\Delta\subset Isom(\mathcal{O}_d)$ is of index 2 if $d\equiv2(mod\ 4)$ and $\Delta=Isom(\mathcal{O}_d)$ if $d\equiv3(mod\ 4)$.
\end{proposition}

\begin{proof} Although we only consider the cases $d=2,7,11$, the ring $\mathcal{O}_2$ represents those for the values of $d$ with $d\equiv2(mod\ 4)$ and the rings $\mathcal{O}_7, \mathcal{O}_{11}$ represent those of the values $d\equiv3(mod\ 4)$, the remaining case is the same as $\mathcal{O}_1$ which has been done in \cite{FFP}. Observe that $Isom(\mathcal{O}_d)$ is generated by the subgroup of translations
$$\left\{\hat{T}_\beta=\left[\begin{array}{cc}1&\beta\\0&1\end{array}\right]:\beta\in\mathcal{O}_d\right\}$$
and the finite subgroup of order two
$$\left\{\hat{R}_\alpha=\left[\begin{array}{cc}\alpha&0\\0&1\end{array}\right]:\alpha\in\mathcal{O}_d,\text{$\alpha$ is a unit}\right\}.$$
Then, to understand $\Delta\subset Isom(\mathcal{O}_d)$, it suffices to determine which translations can be lifted. We divide into two cases to complete the proof.

\medskip

(i)\ {\bf The case $\mathcal{O}_d$ with $d\equiv2(mod\ 4)$}

 Suppose that $\beta\in\mathcal{O}_d=\mathbb{Z}[i\sqrt{d}]$ and consider the translation $\hat{T}_\beta$ by $\beta$ in $\mathbb{Z}[i\sqrt{d}]$ given above. The preimage of $\hat{T}_\beta$ under $\Pi_*$ has the
form
$$T_{\beta,t}=\left[\begin{array}{ccc}1&-\bar{\beta}&\frac{-|\beta|^2+it}{2}\\0&1&\beta\\0&0&1\end{array}\right].$$
This map is in $PU(2,1;\mathbb{Z}[i\sqrt{d}])$ if and only if $|\beta|^2$ is an even integer and $t\in2\sqrt{d}\mathbb{Z}$ . Writing $\beta=m+i\sqrt{d}n$ for $m,n\in\mathbb{Z}$, then we can obtain $m\equiv0\ (mod\ 2)$ from the conditions $|\beta|^2=m^2+dn^2\in2\mathbb{Z}$ and  $d\equiv2(mod\ 4)$. Therefore, we conclude that $\Delta\subset Isom(\mathbb{Z}[i\sqrt{d}])$ is of index 2. Also, the kernel of $\Pi_*$ is generated by
$$\left[\begin{array}{ccc}1&0&i\sqrt{d}\\0&1&0\\0&0&1\end{array}\right],$$ which is a vertical translation of $(0,2\sqrt{2})$.

\medskip

(ii)\ {\bf The case $\mathcal{O}_d$ with $d\equiv3(mod\ 4)$}

Suppose that $\beta=m+n\frac{1+i\sqrt{d}}{2}\in\mathcal{O}_d$ with $m,n\in\mathbb{Z}$ for $d\equiv3(mod\ 4)$. By the same argument of (i), it only suffices to determine $m,n$ such that $|\beta|^2$ is an integer. For $d\equiv3(mod\ 4)$, it is easy to show that $|\beta|^2=m^2+mn+n^2(d+1)/4\in\mathbb{Z}$ for any $m,n\in\mathbb{Z}$, which implies that $\Delta=Isom(\mathcal{O}_d)$. Obviously, the kernel of  $\Pi_*$ is generated by a vertical translation of $(0,2\sqrt{d})$.

\end{proof}

\subsection{Fundamental domain for the stabiliser}
As the first step toward the construction of a fundamental domain for the action of $(\Gamma_d)_\infty$ on $\mathfrak{R}$ for $d=2,7,11$, we shall find the suitable generators of $Isom(\mathcal{O}_d)$ to construct a fundamental domain in $\mathbb{C}$.

In the proof of Proposition \ref{eq:3-1} we saw that $\Delta=\Pi_*((\Gamma_2)_\infty)$ is a subgroup of index 2 in $Isom(\mathcal{O}_2)$ consisting of elements of $GL(2,\mathcal{O}_2)$ of the form
$$\left\{\left[\begin{array}{cc}(-1)^j&m+i\sqrt{2}n\\0&1\end{array}\right]:j=0,1,m,n\in\mathbb{Z},m\equiv0(mod\ 2)\right\}.$$
A fundamental domain for this group is the triangle in  $\mathbb{C}$ with vertices at $-1+\sqrt{2}i/2$ and $1\pm\sqrt{2}i/2$; see (a) in Figure 3.1.
Side paring maps are given by
$$r^{(2)}_1=\left[\begin{array}{cc}-1&0\\0&1\end{array}\right],\ r^{(2)}_2=\left[\begin{array}{cc}-1&2\\0&1\end{array}\right],\
r^{(2)}_3=\left[\begin{array}{cc}-1&\sqrt{2}i\\0&1\end{array}\right].
$$
The first of these is a rotation of order 2 fixing origin, the second is a rotation of order 2 fixing $1/2$ and the third is a rotation of order 2 fixing $\sqrt{2}i/2$. Indeed every element of $\Delta=GL(2,\mathcal{O}_2)$ is generated by $r^{(2)}_1,r^{(2)}_2,r^{(2)}_3$ as follows
\begin{align*}\left[\begin{array}{cc}(-1)^j&2m+\sqrt{2}ni\\0&1\end{array}\right]&=\left[\begin{array}{cc}1&2\\0&1\end{array}\right]^m
\left[\begin{array}{cc}1&\sqrt{2}i\\0&1\end{array}\right]^n\left[\begin{array}{cc}-1&0\\0&1\end{array}\right]^j\\
&=\left(r^{(2)}_2r^{(2)}_1\right)^m\left(r^{(2)}_3r^{(2)}_1\right)^n\left(r^{(2)}_1\right)^j.
\end{align*}

As the same argument, a fundamental domain for $Isom(\mathcal{O}_d)$ with $d=7$ or $11$ is the triangle in  $\mathbb{C}$ with vertices at $(-1+i\sqrt{d})/4$, $(1-i\sqrt{d})/4$ and $(3+i\sqrt{d})/4$; see (b) in Figure 3.1. Side paring maps are given by
$$r^{(d)}_1=\left[\begin{array}{cc}-1&0\\0&1\end{array}\right],\ r^{(d)}_2=\left[\begin{array}{cc}-1&1\\0&1\end{array}\right],\
r^{(d)}_3=\left[\begin{array}{cc}-1&(1+i\sqrt{d})/2\\0&1\end{array}\right].
$$
All these maps are rotations by $\pi$ fixing $0,1/2$ and $(1+i\sqrt{d})/4$ respectively.

\begin{figure}
\centering
  \subfigure[]{
  \psfrag{r1}{\Small$r_1^{(2)}$}\psfrag{r2}{\Small$r_2^{(2)}$}
\psfrag{r3}{\Small$r_3^{(2)}$}
\psfrag{0}{\Small$0$}
\psfrag{1}{\Small$1$}\psfrag{2}{\Small$\frac{i\sqrt{2}}{2}$}
    \label{fig:mini:subfig:1}

 \begin{minipage}[1]{0.45\textwidth} \centering
      \includegraphics[width=1.9in]{triangle1.eps}
  \end{minipage}
}
  \subfigure[]{
  \psfrag{r1}{\Small$r_1^{(d)}$}\psfrag{r2}{\Small$r_2^{(d)}$}
\psfrag{r3}{\Small$r_3^{(d)}$}
\psfrag{0}{\Small$0$}\psfrag{wd}{\Small$\frac{\omega_d}{2}$}
\psfrag{1/2}{\Small$\frac{1}{2}$}
    \label{fig:mini:subfig:2}
    \begin{minipage}[3]{0.45\textwidth}
      \centering
      \includegraphics[width=1.8in]{triangle2.eps}
    \end{minipage}
    }
\medskip
\makeatletter\def\@captype{figure}\makeatother
\bigskip
\caption{(a)\ Fundamental domain for a subgroup $\Delta$ of $Isom(\mathcal{O}_2)$ with index 2.
(b)\ Fundamental domain for $Isom(\mathcal{O}_d)$ with $d=7,11$. This is also true for all the values of $d$ with $d\equiv3(mod\ 4)$.}
  \label{fig:mini:subfig}
\end{figure}

\medskip

In order to produce a fundamental domain for $(\Gamma_d)_\infty$ we look at all the preimages of
the triangle (that is a fundamental domain of $\Pi_*((\Gamma_d)_\infty)$)  under vertical projection
$\Pi$ and we intersect this with a fundamental domain for $ker(\Pi_*)$. The inverse of image of the
triangle under $\Pi$ is an infinite prism. The kernel of $\Pi_*$ is the infinite cyclic group
generated by $T$, the vertical translation by $(0,2\sqrt{d})$.

\begin{figure}
\centering
\psfrag{T}{\Large$T^{(2)}$}\psfrag{R1}{\Large$R_1^{(2)}$}\psfrag{R2}{\Large$R_2^{(2)}$}\psfrag{R3}{\Large$R_3^{(2)}$}
\psfrag{TR2}{\Large$T^{(2)}R_2^{(2)}$}\psfrag{TR3}{\Large$T^{(2)}R_3^{(2)}$}
\psfrag{z1+}{\Large$v_1^+$}\psfrag{z2+}{\Large$v_2^+$}\psfrag{z3+}{\Large$v_3^+$}\psfrag{z4+}{\Large$v_4^+$}
\psfrag{z5+}{\Large$v_5^+$}\psfrag{z6+}{\Large$v_6^+$}
\psfrag{z1-}{\Large$v_1^-$}\psfrag{z2-}{\Large$v_2^-$}\psfrag{z3-}{\Large$v_3^-$}\psfrag{z4-}{\Large$v_4^-$}\psfrag{z5-}{\Large$v_5^-$}
\psfrag{z6-}{\Large$v_6^-$}
\resizebox{2.5in}{!}{\includegraphics{prism1.eps}}
\medskip
\makeatletter\def\@captype{figure}\makeatother
\caption{A fundamental domain $\mathbf{\Sigma}_2$ for $(\Gamma_2)_\infty$ in the Heisenberg group: the map $R^{(2)}_1$
 rotates through $\pi$ about $\zeta=0$, the map $R^{(2)}_2$ is a Heisenberg rotation through $\pi$
  about $\zeta=1$ and the map $R^{(2)}_3$ is a Heisenberg rotation through $\pi$ about $\zeta=\sqrt{2}i/2$.}
\end{figure}

\medskip

\begin{proposition} $(\Gamma_2)_\infty$ is generated by $$ R^{(2)}_1=\left[\begin{array}{ccc}1&0&0\\0&-1&0\\0&0&1\end{array}\right],
 R^{(2)}_2=\left[\begin{array}{ccc}1&2&-2\\0&-1&2\\0&0&1\end{array}\right],
 R^{(2)}_3=\left[\begin{array}{ccc}1&-i\sqrt{2}&-1\\0&-1&i\sqrt{2}\\0&0&1\end{array}\right]$$
and $$T^{(2)}=\left[\begin{array}{ccc}1&0&i\sqrt{2}\\0&1&0\\0&0&1\end{array}\right].$$

A presentation is given by
$$(\Gamma_2)_{\infty}=\langle R^{(2)}_j,T^{(2)}|{R^{(2)}_j}^2=[T^{(2)},R^{(2)}_j]=
\left({T^{(2)}}^2R^{(2)}_1R^{(2)}_3R^{(2)}_2\right)^2=id\rangle.
$$
\end{proposition}

\begin{proof}
Those matrices are constructed by lifting generators of the subgroup $\Delta\subset Isom(\mathcal{O}_2)$ with index 2
and also $T^{(2)}$ is a generator of the kernel of the map $\Pi_*$. A fundamental domain can be constructed with
side pairings as Figure 3.2, where the vertices of the prism are
$v_3^+=(-1+\sqrt{2}i/2,\sqrt{2}),$ $v_2^+=(1+\sqrt{2}i/2,\sqrt{2}),$ $v_1^+=(1-\sqrt{2}i/2,\sqrt{2})$ for the upper cap of the prism
and $v_3^-=(-1+\sqrt{2}i/2,-\sqrt{2}),$ $v_2^-=(1+\sqrt{2}i/2,-\sqrt{2}),$ $v_1^-=(1-\sqrt{2}i/2,-\sqrt{2})$ for the base. In particular, the points
$v_4^{\pm},$ $v_5^{\pm},$ $v_6^{\pm}$ are the middle points of the edges $(v_1^{\pm},v_2^{\pm}),$ $(v_2^{\pm},v_3^{\pm})$ and $(v_3^{\pm},v_1^{\pm})$, respectively.

The actions of side-pairing maps on $\mathfrak{R}$ are given by
\begin{align*}
R^{(2)}_1(\zeta,t)&=(-\zeta,t),\\
R^{(2)}_2(\zeta,t)&=(-\zeta+2,t+4\Im m{\zeta}),\\
R^{(2)}_3(\zeta,t)&=(-\zeta+i\sqrt{2},t-2\sqrt{2}\Re e{\zeta}),\\
T^{(2)}(\zeta,t)&=(\zeta,t+2\sqrt{2}).
\end{align*}
We describe the side pairing in terms of the action on the vertice:
\begin{eqnarray*}
R^{(2)}_1&:&(v_6^+,v_1^+,v_1^-,v_6^-)\longrightarrow(v_6^+,v_3^+,v_3^-,v_6^-),\\
R^{(2)}_2&:&(v_1^+,v_4^+,v_4^-)\longrightarrow(v_2^-,v_4^+,v_4^-),\\
T^{(2)}R^{(2)}_2&:&(v_1^+,v_1^-,v_4^-)\longrightarrow(v_2^+,v_2^-,v_4^+),\\
R^{(2)}_3&:&(v_2^+,v_5^+,v_5^-)\longrightarrow(v_3^-,v_5^+,v_5^-),\\
T^{(2)}R^{(2)}_3&:&(v_2^+,v_2^-,v_5^-)\longrightarrow(v_3^+,v_3^+,v_5^+),\\
T^{(2)}&:&(v_1^-,v_4^-,v_2^-,v_5^-,v_3^-,v_6^-)\longrightarrow(v_1^+,v_4^+,v_2^+,v_5^+,v_3^+,v_6^+).
\end{eqnarray*}
The presentation can be obtained following from the edge cycles of the fundamental domain.
\end{proof}

\medskip

\begin{proposition} $(\Gamma_7)_\infty$ is generated by $$ R^{(7)}_1=\left[\begin{array}{ccc}1&0&0\\0&-1&0\\0&0&1\end{array}\right],
 R^{(7)}_2=\left[\begin{array}{ccc}1&1&-\bar{\omega}_7\\0&-1&1\\0&0&1\end{array}\right],
 R^{(7)}_3=\left[\begin{array}{ccc}1&\bar{\omega}_7&-1\\0&-1&\omega_7\\0&0&1\end{array}\right]$$
and $$T^{(7)}=\left[\begin{array}{ccc}1&0&i\sqrt{7}\\0&1&0\\0&0&1\end{array}\right].$$

A presentation is given by
\begin{align*}(\Gamma_7)_{\infty}&=\langle R^{(7)}_j,T^{(7)}|{R^{(7)}_1}^2={R^{(7)}_3}^2=[T^{(7)},R^{(7)}_1]=[T^{(7)},R^{(7)}_3]\\
&\hspace{4cm}=T^{(7)}{R^{(7)}_2}^{-2}=\left(R^{(7)}_1R^{(7)}_3R^{(7)}_2\right)^2=id\rangle.
\end{align*}
\end{proposition}

\begin{proof}
Those matrices are constructed by lifting generators of $Isom(\mathcal{O}_7)$ and also $T^{(7)}$ is a generator of the kernel of the map $\Pi_*$.
A fundamental domain can be constructed with
side pairings as Figure 3.3, where the vertices of the prism are
$v_1^+=((1-i\sqrt{7})/4,\sqrt{7}),$ $v_2^+=((3+i\sqrt{7})/4,\sqrt{7}),$ $v_4^+=((-1+i\sqrt{7})/4,\sqrt{7})$ for the upper cap of the prism
and $v_1^-=((1-i\sqrt{7})/4,-\sqrt{7}),$ $v_2^-=((3+i\sqrt{7})/4,-\sqrt{7}),$ $v_4^-=((-1+i\sqrt{7})/4,-\sqrt{7})$ for the base.
The points $v_3^{\pm},v_5^{\pm}$ are the middle points of the edges $(v_2^{\pm},v^{\pm}_4)$ and $(v_4^{\pm},v^{\pm}_1)$. In particular,
we introduce more three points $w_1^+=((1-i\sqrt{7})/4,\sqrt{7}/2),$ $w_2^-=((3+i\sqrt{7})/4,-\sqrt{7}/2)$ and $w_3^+=((-1+i\sqrt{7})/4,\sqrt{7}/2)$.
The actions of side-pairing maps on $\mathfrak{R}$ are given by
\begin{align*}
R^{(7)}_1(\zeta,t)&=(-\zeta,t),\\
R^{(7)}_2(\zeta,t)&=(-\zeta+1,t+2\Im m{\zeta}+\sqrt{7}),\\
R^{(7)}_3(\zeta,t)&=(-\zeta+\omega_7,t+2\Im m{(\bar{\omega}_7\zeta})),\\
T^{(7)}(\zeta,t)&=(\zeta,t+2\sqrt{7}).
\end{align*}
We describe the side pairing in terms of the action on the vertice:
\begin{eqnarray*}
R^{(7)}_1&:&(v_5^+,v_1^+,v_1^-,v_5^-)\longrightarrow(v_5^+,v_4^+,v_4^-,v_5^-),\\
R^{(7)}_2&:&(v_1^-,v_2^-,w_1^-,w_1^+)\longrightarrow(w_1^-,w_1^+,v_1^+,v_2^+),\\
R^{(7)}_3&:&(v_2^+,w_1^-,v_3^-,v_3^+)\longrightarrow(w_2^+,v_4^-,v_3^-,v^+_3),\\
T^{(7)}R^{(7)}_3&:&(w_1^-,v_2^-,v_3^-)\longrightarrow(v_4^+,w_2^+,v_3^+),\\
T^{(7)}&:&(v_1^-,v_2^-,v_3^-,v_4^-,v_5^-)\longrightarrow(v_1^+,v_2^+,v_3^+,v_4^+,v_5^+).
\end{eqnarray*}
The presentation can be obtained following from the edge cycles of the fundamental domain.
\end{proof}

\begin{figure}
\centering
\psfrag{T}{\Large$T^{(7)}$}\psfrag{R1}{\Large$R_1^{(7)}$}\psfrag{R2}{\Large$R_2^{(7)}$}\psfrag{R3}{\Large$R_3^{(7)}$}
\psfrag{TR3}{\Large$T^{(7)}R_3^{(7)}$}
\psfrag{z1+}{\Large$v_1^+$}\psfrag{z2+}{\Large$v_2^+$}\psfrag{z3+}{\Large$v_3^+$}\psfrag{z4+}{\Large$v_4^+$}
\psfrag{z5+}{\Large$v_5^+$}
\psfrag{z1-}{\Large$v_1^-$}\psfrag{z2-}{\Large$v_2^-$}\psfrag{z3-}{\Large$v_3^-$}
\psfrag{z4-}{\Large$v_4^-$}\psfrag{z5-}{\Large$v_5^-$}
\psfrag{w1+}{\Large$w_1^+$}\psfrag{w2+}{\Large$w_2^+$}\psfrag{w1-}{\Large$w_1^-$}
\resizebox{2.5in}{!}{\includegraphics{prism2.eps}}
\medskip
\makeatletter\def\@captype{figure}\makeatother
\caption{A fundamental domain $\mathbf{\Sigma}_7$ for $(\Gamma_7)_\infty$ in the Heisenberg group: the map $R^{(7)}_1$ rotates through $\pi$ about $\zeta=0$,
the action of parabolic $R_2^{(7)}$ is a Heisenberg rotation through $\pi$ about $\zeta=1/2$ followed by an upward vertical
translation by $\sqrt{7}$ and the map $R_3^{(7)}$ is a Heisenberg rotation through $\pi$ about $\zeta=(1+i\sqrt{7})/4$.}
\end{figure}

\medskip

\begin{proposition} $(\Gamma_{11})_\infty$ is generated by $$ R^{(11)}_1=\left[\begin{array}{ccc}1&0&0\\0&-1&0\\0&0&1\end{array}\right],
 R^{(11)}_2=\left[\begin{array}{ccc}1&1&-\bar{\omega}_{11}\\0&-1&1\\0&0&1\end{array}\right],
 $$
$$R^{(11)}_3=\left[\begin{array}{ccc}1&\bar{\omega}_{11}&-1-\bar{\omega}_{11}\\0&-1&\omega_{11}\\0&0&1\end{array}\right]\quad \text{and}\quad T^{(11)}=\left[\begin{array}{ccc}1&0&i\sqrt{11}\\0&1&0\\0&0&1\end{array}\right].$$

A presentation is given by
\begin{align*}(\Gamma_{11})_{\infty}&=\langle R^{(11)}_j,T^{(11)}|{R^{(11)}_1}^2=[T^{(11)},R^{(11)}_1]=T^{(11)}{R^{(11)}_2}^{-2}\\
&\hspace{2cm}=T^{(11)}{R^{(11)}_3}^{-2}=T^{(11)}\left(R^{(11)}_1R^{(11)}_3R^{(11)}_2\right)^{-2}=id\rangle.
\end{align*}
\end{proposition}

\begin{proof}
Those matrices are constructed by lifting generators of $Isom(\mathcal{O}_{11})$ and also $T^{(11)}$ is a generator of the kernel of the map $\Pi_*$.
A fundamental domain can be constructed with
side pairings as Figure 3.4, where the vertices of the prism are
$v_1^+=((1-i\sqrt{11})/4,\sqrt{11}),$ $v_2^+=((3+i\sqrt{11})/4,3\sqrt{11}/2),$ $v_3^+=((-1+i\sqrt{11})/4,2\sqrt{11})$ for the upper cap of the prism
and $v_1^-=((1-i\sqrt{11})/4,-\sqrt{11}),$ $v_2^-=((3+i\sqrt{11})/4,-\sqrt{11}/2),$ $v_3^-=((-1+i\sqrt{11})/4,0)$ for the base.
The points $v_0^{\pm}$ are the middle points of the edges $(v_1^{\pm},v^{\pm}_3)$. In particular, we introduce more three points $w_1=((1-i\sqrt{11})/4,0),$
 $w_2=((3+i\sqrt{11})/4,\sqrt{11}/2)$ and $w_3=((-1+i\sqrt{11})/4,\sqrt{11})$.
The actions of side-pairing maps on $\mathfrak{R}$ are given by
\begin{align*}
R^{(11)}_1(\zeta,t)&=(-\zeta,t),\\
R^{(11)}_2(\zeta,t)&=(-\zeta+1,t+2\Im m{\zeta}+\sqrt{11}),\\
R^{(11)}_3(\zeta,t)&=(-\zeta+\omega_{11},t+2\Im m{(\bar{\omega}_{11}\zeta})+\sqrt{11}),\\
T^{(11)}(\zeta,t)&=(\zeta,t+2\sqrt{11}).
\end{align*}
We describe the side pairing in terms of the action on the vertice:
\begin{eqnarray*}
R^{(11)}_1&:&(v_0^+,v_1^+,w_1,v_0^-)\longrightarrow(v_0^+,w_3,v_3^-,v_0^-),\\
T^{(11)}R^{(11)}_1&:&(w_1,v_1^-,v_0^-)\longrightarrow(v_3^+,w_3,v_0^+),\\
R^{(11)}_2&:&(v_1^+,w_1,v_1^-,v_2^-)\longrightarrow(v_2^+,w_2,v_2^-,v_1^+),\\
R^{(11)}_3&:&(v_2^+,w_2,v_2^-,v_3^-)\longrightarrow(v_3^+,w_3,v_3^-,v_2^+),\\
T^{(11)}&:&(v_0^-,v_1^-,v_2^-,v_3^-)\longrightarrow(v_0^+,v_1^+,v_2^+,v_3^+).
\end{eqnarray*}
The presentation can be obtained following from the edge cycles of the fundamental domain.
\end{proof}

\begin{figure}
\centering
\psfrag{T}{\Large$T^{(11)}$}\psfrag{R1}{\Large$R_1^{(11)}$}\psfrag{R2}{\Large$R_2^{(11)}$}\psfrag{R3}{\Large$R_3^{(11)}$}
\psfrag{TR1}{\Large$T^{(11)}R_1^{(11)}$}
\psfrag{z1+}{\Large$v_1^+$}\psfrag{z2+}{\Large$v_2^+$}\psfrag{z3+}{\Large$v_3^+$}\psfrag{z0+}{\Large$v_0^+$}
\psfrag{z1-}{\Large$v_1^-$}\psfrag{z2-}{\Large$v_2^-$}\psfrag{z3-}{\Large$v_3^-$}
\psfrag{z0-}{\Large$v_0^-$}
\psfrag{w1}{\Large$w_1$}\psfrag{w2}{\Large$w_2$}\psfrag{w3}{\Large$w_3$}
\resizebox{2.5in}{!}{\includegraphics{prism3.eps}}
\medskip
\makeatletter\def\@captype{figure}\makeatother
\caption{A fundamental domain $\mathbf{\Sigma}_{11}$ for $(\Gamma_{11})_\infty$ in the Heisenberg group: the map $R^{(11)}_1$ rotates through $\pi$ about $\zeta=0$, the action of parabolic $R_2^{(11)}$ is a screw Heisenberg rotation through $\pi$ about $\zeta=1/2$ followed by an upward vertical translation by $\sqrt{11}$ and the map $R_3^{(11)}$ is a screw Heisenberg rotation through $\pi$ about $\zeta=(1+i\sqrt{11})/4$ followed by an upward vertical translation by $\sqrt{11}$.}
\end{figure}

\section{The statement of our method and results}

In this section, we introduce the method used in (\cite{FP}, \cite{FFP}) to determine the generators of the Euclidean Picard groups and then state our results.

Recall that the map
$$I_0=\left[\begin{array}{ccc}0&0&1\\0&-1&0\\1&0&0\end{array}\right],$$
defined in the Section 2.3. We consider the isometric sphere $\mathcal{B}_0$ of $I_0$ given by (\ref{eq:2-4}), which is a Cygan sphere centred $o=(0,0,0)$ with radius $\sqrt{2}$. Observe that $I_0$ maps $\mathcal{B}_0$ to itself and swaps the inside and the outside of $\mathcal{B}_0$. Given an element of $\Gamma_d$ as the form (\ref{eq:2-6}), we know that the radius of isometric sphere is $\sqrt{2/|g|}$. For each case $\mathcal{O}_d$, the radius of isometric sphere is not greater than $\sqrt{2}$ since the absolute of $g$ is not smaller than 1 for $g\in\mathcal{O}_d$. We show that the largest isometric spheres are all the images of $\mathcal{B}_0$ under the elements in $(\Gamma_d)_\infty$.

\begin{proposition}
An isometric sphere has the largest radius if and only if it is the image of $\mathcal{B}_0$ under an element in $(\Gamma_d)_\infty$.
\end{proposition}

\begin{proof}
Obviously, the image of $\mathcal{B}_0$ under an element in $(\Gamma_d)_\infty$ has the largest radius $\sqrt{2}$. On the contrary, given an element $G$ as the form (\ref{eq:2-6}) such that $G(q_\infty)\neq q_\infty$, then the isometric sphere of $G$ has the largest radius which leads to $g=1$. So the centre of isometric sphere of $G$ is $G^{-1}(\infty)=(\bar{h},2\Im m\bar{j},0)$ in horospherical coordinates. Since $\bar{h}$ and $2\Im m\bar{j}\in\mathcal{O}_d$, we can take a Heisenberg translation $T\in(\Gamma_d)_\infty$ mapping the origin to $(\bar{h},2\Im m\bar{j})$. Writing $T'=GTI_0$, we know that $T'$ fixes $\infty$. We conclude explicitly that the isometric sphere of $G$ is
$$\left\{\mathbf{z}\in\textbf{H}^2_{\mathbb{C}}:|\langle \mathbf{z},q_{\infty}\rangle|=|\langle \mathbf{z},G^{-1}(q_{\infty})\rangle|=|\langle \mathbf{z},TI_0(q_{\infty})\rangle|\right\},$$
which is the image of $\mathcal{B}_0$ under $T$.
\end{proof}

Our method is based on the special feature that the orbifold $\mathbf{H}^2_\mathbb{C}/\Gamma_d$ has only one cusp for $d=2,7,11$. For these types of orbifolds, one would like to construct a fundamental domain using the Ford domain (that is the exteriors of isometric spheres of all elements not fixing infinity), namely, the intersection of the Ford domain and a fundamental domain for the stabiliser of infinity. The Ford domain is canonical, but we can choose a fundamental domain for the stabiliser freely. As the first step toward the construction of a fundamental domain, we should always determine the generators of the group. In the previous section, we found suitable generators of the stabliser and constructed a fundamental domain. We will show that adjoining $I_0$ to $(\Gamma_d)_\infty$ gives the Euclidean Picard modular groups $\Gamma_d$. The basic idea of the proof can be described easily. Analogous to Theorem 3.5 of \cite{FP} we shall prove that $\langle R^{(d)}_1,R^{(d)}_2,R^{(d)}_3,T^{(d)},I_0\rangle $ has only one cusp. The fact that $PU(2,1;\mathcal{O}_d)$ has the same cusp and the stabiliser of infinity as the group generated by $R^{(d)}_1,R^{(d)}_2,R^{(d)}_3,T^{(d)},I_0$ shows that they are the same. The key step is to find a union of isometric spheres such that a fundamental domain for $\Gamma_d$ is contained in the intersection of their exteriors and a fundamental domain for the stabiliser, which implies that the group $\langle R^{(d)}_1,R^{(d)}_2,R^{(d)}_3,T^{(d)},I_0\rangle$ has only one cusp. In other words, we want to show the union of the boundaries of these isometric spheres in Heisenberg group contains each of the prisms we constructed above.  The problem of determining this will be discussed in the next section.

A simple lemma will be used in the proof of our theorems many times, we state it as follows.

\begin{lemma}(c.f. \cite{FFP})\ All Cygan balls are affinely convex.
\end{lemma}

Our aim is to prove the following results, we summarise them as three theorems.

\begin{theorem} Let $\mathcal {K}=\mathbb{Q}(\sqrt{-2})$ and let $\mathcal{O}_2=\mathbb{Z}[i\sqrt{2}]$. Then the group $PU(2,1,\mathcal{O}_2)$ is generated by the elements
$$I_0=\left[\begin{array}{ccc}0&0&1\\0&-1&0\\1&0&0\end{array}\right], R^{(2)}_1=\left[\begin{array}{ccc}1&0&0\\0&-1&0\\0&0&1\end{array}\right],
 R^{(2)}_2=\left[\begin{array}{ccc}1&2&-2\\0&-1&2\\0&0&1\end{array}\right],$$
$$ R^{(2)}_3=\left[\begin{array}{ccc}1&-i\sqrt{2}&-1\\0&-1&i\sqrt{2}\\0&0&1\end{array}\right]
\quad\text{and}\quad T^{(2)}=\left[\begin{array}{ccc}1&0&i\sqrt{2}\\0&1&0\\0&0&1\end{array}\right]
.$$
\end{theorem}

\medskip

\begin{theorem} Let $\mathcal {K}=\mathbb{Q}(\sqrt{-7})$ and let $\mathcal{O}_7=\mathbb{Z}[\omega_7]$, where $\omega_7=\frac{1}{2}(1+i\sqrt{7})$, be the ring of integers of $\mathcal{K}$. Then the group $PU(2,1,\mathcal{O}_7)$ is generated by the elements
$$I_0=\left[\begin{array}{ccc}0&0&1\\0&-1&0\\1&0&0\end{array}\right], R^{(7)}_1=\left[\begin{array}{ccc}1&0&0\\0&-1&0\\0&0&1\end{array}\right],
 R^{(7)}_2=\left[\begin{array}{ccc}1&1&-\bar{\omega}_7\\0&-1&1\\0&0&1\end{array}\right],$$
 $$
 R^{(7)}_3=\left[\begin{array}{ccc}1&\bar{\omega}_7&-1\\0&-1&\omega_7\\0&0&1\end{array}\right]\quad\text{and}\quad T^{(7)}=\left[\begin{array}{ccc}1&0&i\sqrt{7}\\0&1&0\\0&0&1\end{array}\right]
.$$
\end{theorem}

\medskip

\begin{theorem} Let $\mathcal {K}=\mathbb{Q}(\sqrt{-11})$ and let $\mathcal{O}_{11}=\mathbb{Z}[\omega_{11}]$, where $\omega_{11}=\frac{1}{2}(1+i\sqrt{11})$, be the ring of integers of $\mathcal{K}$. Then the group $PU(2,1,\mathcal{O}_{11})$ is generated by the elements
$$I_0=\left[\begin{array}{ccc}0&0&1\\0&-1&0\\1&0&0\end{array}\right], R^{(11)}_1=\left[\begin{array}{ccc}1&0&0\\0&-1&0\\0&0&1\end{array}\right],
 R^{(11)}_2=\left[\begin{array}{ccc}1&1&-\bar{\omega}_{11}\\0&-1&1\\0&0&1\end{array}\right],$$
 $$R^{(11)}_3=\left[\begin{array}{ccc}1&\bar{\omega}_{11}&-1-\bar{\omega}_{11}\\0&-1&\omega_{11}\\0&0&1\end{array}\right]\quad\text{and}\quad T^{(11)}=\left[\begin{array}{ccc}1&0&i\sqrt{11}\\0&1&0\\0&0&1\end{array}\right]
.$$
\end{theorem}

\begin{remark}For other values of $d$ such that $\mathcal{O}_d$ has class number one, namely $d=19,$ $43,$ $67,$ $163$, we can construct the same type of fundamental domain for $(\Gamma_d)_\infty$ in Heisenberg group as $(\Gamma_{11})_\infty$. Although all generators as the above types lie in $PU(2,1;\mathcal{O}_d)$, there is no reason why adjoining $I_0$ to $(\Gamma_d)_\infty$ should continue to generate the full group $PU(2,1;\mathcal{O}_d)$. From the point of view of geometric, the largest radius of isometric sphere is always $\sqrt{2}$ while the shape of the fundamental domain and the length of Heisenberg translations become large as $d$ getting large. In contrast the radius of isometric spheres other than the largest are going to be smaller and smaller. Consequently the mount of isometric spheres containing the fundamental domain increases so rapidly with the value of $d$ that it seem to be done by another way. Furthermore, the method of \cite{FFLP} could not be extended to non-Euclidean Picard modular groups.
\end{remark}

\bigskip

\section{The determination of isometric spheres}

Recall that the Cygan sphere $\mathcal{B}_0$ is the isometric sphere of $I_0$. The boundary of $\mathcal{B}_0$ is called a spinal sphere \cite{Go} in Heisenberg group, we denote by $\mathcal{S}_0$ which is defined by
\begin{equation}\label{eq:5-1}\mathcal{S}_0=\left\{(\zeta,t):\left||\zeta|^2+it\right|=2\right\}.\end{equation}

Indeed we only need to consider the boundaries of isometric spheres in Heisenberg group because two isometric spheres have a non-empty interior intersection if and only if the boundaries have a non-empty interior intersection.

\subsection{The case $\mathcal{O}_2$}

In the cases of $PU(2,1;\mathcal{O}_1)$ and $PU(2,1;\mathcal{O}_3)$, all the vertices of the fundamental domain for the stabiliser
of $q_\infty$ acting on $\partial\textbf{H}^2_\mathbb{C}$ lie inside $\mathcal{S}_0$.
For the group $PU(2,1;\mathcal{O}_2)$, it is not hard to show that six vertices of the prism $\mathbf{\Sigma}_2$ lie outside $\mathcal{S}_0$.
Therefore we need to find more isometric spheres whose boundaries together with $\mathcal{S}_0$ contain the prism $\mathbf{\Sigma}_2$.

We consider the map

\begin{equation*}
I_0R^{(2)}_2I_0=\left[\begin{array}{ccc}
1&0&0\\-2&-1&0\\-2&-2&1
\end{array}\right],
\end{equation*}
whose isometric sphere which we denote by $\mathcal{B}_1$ is a Cygan sphere centred at the point $(1,0,0)$ (in horospherical coordinates) with radius 1. The boundary of $\mathcal{B}_1$ is given by
\begin{equation}\label{eq:5-2}\mathcal{S}_1=\left\{(\zeta,t):\left||\zeta-1|^2+it+2i\Im m\zeta\right|=1\right\}.\end{equation}

Minimising the number of spinal spheres by the symmetry of $R^{(2)}_1$, it suffice to consider $\mathcal{S}_0$ and several images of $\mathcal{S}_1$ under some suitable elements in $(\Gamma_2)_\infty$, these are in Heisenberg coordinates given by
\begin{align*}
T^{(2)}(\mathcal{S}_1)&=\left\{(\zeta,t):\left||\zeta-1|^2+it-2i\sqrt{2}+2i\Im m\zeta\right|=1\right\},\\
{T^{(2)}}^{-1}(\mathcal{S}_1)&=\left\{(\zeta,t):\left||\zeta-1|^2+it+2i\sqrt{2}+2i\Im m\zeta\right|=1\right\},\\
R^{(2)}_1(\mathcal{S}_1)&=\left\{(\zeta,t):\left||\zeta+1|^2+it-2i\Im m\zeta\right|=1\right\},\\
{T^{(2)}}^{-1}R^{(2)}_1(\mathcal{S}_1)&=\left\{(\zeta,t):\left||\zeta+1|^2+it+2i\sqrt{2}-2i\Im m\zeta\right|=1\right\}.
\end{align*}
We claim that the prism $\mathbf{\Sigma}_2$ lies inside the union of $\mathcal{S}_0$ and these images of $\mathcal{S}_1$, see Figure 5.1 for viewing these spinal spheres.

\begin{figure}
\centering
  \subfigure[]{
    \label{fig:mini:subfig:1}

 \begin{minipage}[1]{0.42\textwidth} \centering
      \includegraphics[width=2in]{neighboring1.eps}
  \end{minipage}
}
  \subfigure[]{
    \label{fig:mini:subfig:2}
    \begin{minipage}[3]{0.42\textwidth}
      \centering
      \includegraphics[width=2in]{neighboring2.eps}
    \end{minipage}
}
\medskip
\makeatletter\def\@captype{figure}\makeatother
\bigskip
\caption{(a)\ The shading view of neighboring spinal spheres containing the fundamental domain for $(\Gamma_2)_\infty$.
(b)\ Another view for these spinal spheres.}
  \label{fig:mini:subfig}
\end{figure}

\begin{proposition}The prism $\mathbf{\Sigma}_2$ is contained in the union of the interiors of the spinal spheres $\mathcal{S}_0,$ $\mathcal{S}_1,$ $T^{(2)}(\mathcal{S}_1),$ ${T^{(2)}}^{-1}(\mathcal{S}_1),$ $R^{(2)}_1(\mathcal{S}_1)$
and ${T^{(2)}}^{-1}R^{(2)}_1(\mathcal{S}_1)$.
\end{proposition}

\begin{proof} It suffices to show there exists three points $(v_1^+)^{(j)}$ $(j=1,2,3)$ on the edges $(v_1^+,v_1^-),$ $(v_1^+,v_2^+)$ and $(v_1^+,v_3^+)$
which lie in the intersection of the interiors of $\mathcal{S}_0$ and $\mathcal{S}_1$ such that the tetrahedron $\mathbb{T}(v_1^+)$ with vertices
 $v_1^+,$ $(v_1^+)^{(1)},$ $(v_1^+)^{(2)},$ $(v_1^+)^{(3)}$ lies inside $\mathcal{S}_1$. By the same argument,
we can also obtain other five tetrahedra $\mathbb{T}(v_2^+),$ $\mathbb {T}(v_3^+),$ $\mathbb{T}(v_1^-),$ $\mathbb {T}(v_2^-),$ $\mathbb{T}(v_3^-)$
with apex $v_2^+,$ $v_3^+,$ $v_1^-,$ $v_2^-,$ $v_3^-$ respectively such that
 $\mathbb {T}(v_2^+)\in Int(T^{(2)}(\mathcal{S}_1)),$ $\mathbb {T}(v_3^+)\in Int(R^{(2)}_1(\mathcal{S}_1)),$ $\mathbb {T}(v_1^-)\in Int({T^{(2)}}^{-1}(\mathcal{S}_1)),$
  $\mathbb {T}(v_2^-)\in Int(\mathcal{S}_1)$ and $\mathbb {T}(v_3^-)\in Int({T^{(2)}}^{-1}R^{(2)}_1(\mathcal{S}_1))$. Moreover, the core part obtained by cutting off six the tetrahedra from the prism lies inside $\mathcal{S}_0$.

We shall prove the existence of the tetrahedron $\mathbb {T}(v_1^+)$ and the others follow similarly. The edge joining $v_1^+$ and $v_1^-$ is contained the complex line $\zeta=1-\sqrt{2}i/2$ which is given by points with Heisenberg coordinates
$$\zeta=1-\sqrt{2}i/2,\quad-\sqrt{2}\leq t\leq\sqrt{2}.$$
The edge joining $v_1^+$ and $v_2^+$ is given by points with Heisenberg coordinates
$$\Re e\zeta=1,\quad-\sqrt{2}/2\leq\Im m\zeta\leq\sqrt{2}/2,\quad t=\sqrt{2}.$$
The edge joining $v_1^+$ and $v_3^+$ is given by points with Heisenberg coordinates
$$\Re e\zeta=-\sqrt{2}\Im m\zeta,\quad t=\sqrt{2}.$$
From (\ref{eq:5-1}) and (\ref{eq:5-2}), the points on the edge $(v^+_1,v^-_1)$ lie in the intersection of the interiors of $\mathcal{S}_0$ and $\mathcal{S}_1$ if and only if
\begin{equation}\label{eq:5-3}
\left|3/2+it\right|<2\quad\text{and}\quad\left|1/2-(t-\sqrt{2})i\right|<1.
\end{equation}
By an easy calculation, the inequalities (\ref{eq:5-3}) are equivalent to $$\sqrt{2}-\sqrt{3}/2<t<\sqrt{7}/2.$$

Using the same argument as above, we obtain that the points on the edge $(v^+_1,v^+_2)$ lie in the intersection of the interiors of $\mathcal{S}_0$ and $\mathcal{S}_1$ if and only if $\Re e\zeta=1,$ and $-\sqrt{\sqrt{2}-1}<\Im m\zeta<\delta_1$, where $\delta_1\approx-0.208$ is the largest real root of the equation $x^4+4x^2+4\sqrt{2}x+1=0$. The points on the edge $(v^+_1,v^+_3)$ lie in the intersection of the interiors of $\mathcal{S}_0$ and $\mathcal{S}_1$ if and only if $\Re e\zeta=-\sqrt{2}\Im m\zeta$ and $-2^{1/4}/\sqrt{3}<\Im m\zeta<\delta_2$, where $\delta_2\approx-0.264$ is the largest real root of the equation $9x^4+12\sqrt{2}x^3+18x^2+8\sqrt{2}x+2=0$.

In term of these, we choose three points as $(v_1^+)^{(1)}=(1-\sqrt{2}i/2,1)$ on the edge $(v^+_1,v^-_1)$, $(v_1^+)^{(2)}=(1-i/2,\sqrt{2})$ on the edge $(v^+_1,v^+_2)$ and $(v_1^+)^{(3)}=(\sqrt{2}/2-i/2,\sqrt{2})$ on the edge $(v^+_1,v^+_3)$, which are inside the intersection of the interiors of $\mathcal{S}_0$, $\mathcal{S}_1$ and also the vertex $v_1^+$ lies inside $\mathcal{S}_1$. Therefore the tetrahedron $\mathbb {T}(v_1^+)$ with the vertices $v_1^+,$ $(v_1^+)^{(1)},$ $(v_1^+)^{(2)},$ $(v_1^+)^{(3)}$ lies inside $\mathcal{S}_1$ by Lemma 4.1.
\end{proof}

\subsection{The case $\mathcal{O}_7$}

In this case, the distance between the top and base of the fundamental domain for the
stabiliser $(\Gamma_{7})_\infty$ is greater than the diameter of $\mathcal{S}_0$, which implies that the prism $\mathbf{\Sigma}_7$ can not be
contained inside $\mathcal{S}_0$ completely. Due to increasing the length of Heisenberg translations, only the images of $\mathcal{S}_0$ under the elements in $(\Gamma_7)_\infty$ could not cover the whole prism. We show that there are also isometric spheres with Cygan radius smaller than $\sqrt{2}$ whose centres are near to origin.

Therefore we consider the map
$$Q = I_0R^{(7)}_2I_0=\left[\begin{array}{ccc}
1&0&0\\
1&1&0\\
\bar{\omega}_7&1&1
\end{array}\right].
$$
Consider the isometric spheres of $Q$ and $Q^{-1}$, which we denote by $\mathcal{B}_2$ and $\mathcal{B}_3$, respectively.
The centre of $\mathcal{B}_2$ is $Q^{-1}(\infty)$, which is the point with horospherical coordinates
$(1/4+i\sqrt{7}/4,\sqrt{7}/2,0)$ and the centre of $\mathcal{B}_3$, is $Q(\infty)$ which has horospherical coordinates $(1/4-i\sqrt{7}/4,\sqrt{7}/2,0)$. Both these isometric spheres have Cygan radius $\sqrt{2/|\omega_7|}=2^{1/4}$. The boundaries of these isometric spheres $\mathcal{B}_2$ and $\mathcal{B}_3$ are in Heisenberg coordinates given by
\begin{align}\label{eq:5-4}
\mathcal{S}_2&=\left\{(\zeta,t):\left|\left|\zeta-\omega_7/2\right|^2+it+i\sqrt{7}/2+i\Im m(\bar{\omega}_7\zeta)\right|=\sqrt{2}\right\},\\ \label{eq:5-5}
\mathcal{S}_3&=\left\{(\zeta,t):\left|\left|\zeta-\bar{\omega}_7/2\right|^2+it-i\sqrt{7}/2+i\Im m(\omega_7\zeta)\right|=\sqrt{2}\right\}.
\end{align}

In order to cover the prim $\mathbf{\Sigma}_7$ by the spinal spheres, we use the symmetry property of $R^{(7)}_1$, it suffice to consider $\mathcal{S}_0,\mathcal{S}_2$ and images of $\mathcal{S}_0$ and $\mathcal{S}_3$ under suitable elements in $(\Gamma_7)_\infty$, these are points with Heisenberg coordinates given by
\begin{align*}
R^{(7)}_2(\mathcal{S}_0)&=\left\{(\zeta,t):\left||\zeta-1|^2+it-i\sqrt{7}+2i\Im m\zeta\right|=4\right\},\\
{R^{(7)}_2}^{-1}(\mathcal{S}_0)&=\left\{(\zeta,t):\left||\zeta-1|^2+it+i\sqrt{7}+2i\Im m\zeta\right|=4\right\},\\
{R^{(7)}_2}^{-1}(\mathcal{S}_3)&=\left\{(\zeta,t):\left|\left|\zeta-(1+\omega_7)/2\right|^2+it+i\sqrt{7}\right.\right.\\
&\hspace{5cm}\left.\left.+i\Im m((1+\bar{\omega}_7)\zeta)\right|=\sqrt{2}\right\},\\
R^{(7)}_3R^{(7)}_2(\mathcal{S}_3)&=\left\{(\zeta,t):\left|\left|\zeta+\bar{\omega}_7/2\right|^2+it-i\sqrt{7}/2-i\Im m(\omega_7\zeta)\right|=\sqrt{2}\right\},\\
R^{(7)}_1R^{(7)}_3R^{(7)}_2(\mathcal{S}_3)&=\left\{(\zeta,t):\left|\left|\zeta-\bar{\omega}_7/2\right|^2+it-i\sqrt{7}/2+i\Im m(\omega_7\zeta)\right|=\sqrt{2}\right\}.
\end{align*}

We claim that the prism $\mathbf{\Sigma}_7$ lies inside the union of $\mathcal{S}_0,$ $\mathcal{S}_2$ and these
 images $R^{(7)}_2(\mathcal{S}_0),$ ${R^{(7)}_2}^{-1}(\mathcal{S}_0),$ ${R^{(7)}_2}^{-1}(\mathcal{S}_3)$, $R^{(7)}_3R^{(7)}_2(\mathcal{S}_3),$ $R^{(7)}_1R^{(7)}_3R^{(7)}_2(\mathcal{S}_3)$, see Figure 5.2 for viewing these spinal spheres.

\begin{figure}
\centering
  \subfigure[]{
    \label{fig:mini:subfig:1}

 \begin{minipage}[1]{0.42\textwidth} \centering
      \includegraphics[width=2in]{neighboring3.eps}
  \end{minipage}
}
  \subfigure[]{
    \label{fig:mini:subfig:2}
    \begin{minipage}[3]{0.42\textwidth}
      \centering
      \includegraphics[width=2in]{neighboring4.eps}
    \end{minipage}
}
\medskip
\makeatletter\def\@captype{figure}\makeatother
\bigskip
\caption{(a)\ The shading view of neighboring spinal spheres containing the fundamental domain for $(\Gamma_7)_\infty$. (b)\ Another view for these spinal spheres.}
  \label{fig:mini:subfig}
\end{figure}

\begin{proposition}\label{eq:5.2}The prism $\mathbf{\Sigma}_7$ is contained in the union of the interiors of the spinal spheres
 $\mathcal{S}_0,$ $\mathcal{S}_2,$ $R^{(7)}_2(\mathcal{S}_0),$ ${R^{(7)}_2}^{-1}(\mathcal{S}_0),$ ${R^{(7)}_2}^{-1}(\mathcal{S}_3),$ $R^{(7)}_3R^{(7)}_2(\mathcal{S}_3)$  and $R^{(7)}_1R^{(7)}_3R^{(7)}_2(\mathcal{S}_3)$.
\end{proposition}

\begin{proof}It suffice to show that the prism $\mathbf{\Sigma}_7$ can be decomposed into several pieces as polyhedra such that each polyhedron lies inside a spinal sphere which is described in the proposition and the common face of two adjacent polyhedra lie in the intersection of the interior of two spinal spheres which contain these two polyhedra.

We need to add sixteen points on the faces of the prism $\mathbf{\Sigma}_7$ in order to decompose the prim into seven polyhedra, in Heisenberg coordinates, these are given by
$$
\begin{array}{llll}
p_1=(1/4-i\sqrt{7}/4,3/2),&p_2=(0.11-i11\sqrt{7}/100,1.44+\sqrt{7}/50),\\
p_3=(1/2,8/5),&p_4=(-1/10+i\sqrt{7}/10,\sqrt{7}),\\
p_5=(-1/10+i\sqrt{7}/10,\sqrt{7}/2),&p_6=(3/4+i\sqrt{7}/4,1.7),\\
p_7=(-1/4+i\sqrt{7}/4,1),&p_8=(1/4-i\sqrt{7}/4,-1),\\
p_9=(1/60-i\sqrt{7}/60,-2\sqrt{7}/3),&p_{10}=(-1/20+i\sqrt{7}/20,-\sqrt{7}),\\
p_{11}=(3/5+i\sqrt{7}/10,-2\sqrt{7}/3),&p_{12}=(7/10+i\sqrt{7}/5,-\sqrt{7}),\\
p_{13}=(3/4+i\sqrt{7}/4,-2\sqrt{7}/3),&p_{14}=(5/12+i\sqrt{7}/4,-2\sqrt{7}/3),\\
p_{15}=(1/4+i\sqrt{7}/4,-\sqrt{7}),&p_{16}=(-1/4+i\sqrt{7}/4,-1).
\end{array}$$

\begin{figure}
\centering
\psfrag{T}{\Large$\mathbb{T}$}\psfrag{P1}{\Large$\mathbb{P}_1$}\psfrag{P2}{\Large$\mathbb{P}_2$}
\psfrag{P3}{\Large$\mathbb{P}_3$}\psfrag{H1}{\Large$\mathbb{H}_1$}\psfrag{H2}{\Large$\mathbb{H}_2$}
\psfrag{O}{\Large$\mathbb{O}$}
\psfrag{v1+}{\Large$v_1^+$}\psfrag{v2+}{\Large$v_2^+$}\psfrag{v4+}{\Large$v_4^+$}
\psfrag{v1-}{\Large$v_1^-$}\psfrag{v2-}{\Large$v_2^-$}\psfrag{v4-}{\Large$v_4^-$}
\psfrag{p1}{\Large$p_1$}\psfrag{p2}{\Large$p_2$}\psfrag{p3}{\Large$p_3$}\psfrag{p4}{\Large$p_4$}
\psfrag{p5}{\Large$p_5$}\psfrag{p6}{\Large$p_6$}\psfrag{p7}{\Large$p_7$}\psfrag{p8}{\Large$p_8$}
\psfrag{p9}{\Large$p_9$}\psfrag{p10}{\Large$p_{10}$}\psfrag{p11}{\Large$p_{11}$}\psfrag{p12}{\Large$p_{12}$}
\psfrag{p13}{\Large$p_{13}$}
\psfrag{p14}{\Large$p_{14}$}
\psfrag{p15}{\Large$p_{15}$}
\psfrag{p16}{\Large$p_{16}$}
\resizebox{2in}{!}{\includegraphics{division1.eps}}
\medskip
\makeatletter\def\@captype{figure}\makeatother
\caption{A view of the decomposition for the prism $\mathbf{\Sigma}_7$ as several polyhedra.}
\end{figure}

We describe these polyhedra as follows:

(i)\ The tetrahedron $\mathbb{T}$ with the vertice $v^+_1,$ $p_1,$ $p_2,$ $p_3$;

(ii)\ The hexahedron $\mathbb{H}_1$ with the vertice $v^+_1,$ $v_2^+,$ $p_2,$ $p_3,$ $p_4,$ $p_5,$ $p_6$;

(iii)\ The pentahedron $\mathbb{P}_1$ with the vertice $v_2^+,$ $p_4,$ $p_5,$ $p_6,$ $v^+_4,$ $p_7$;

(iv)\ The pentahedron $\mathbb{P}_2$ with the vertice  $v_1^-,$ $p_8,$ $p_9,$ $p_{10},$ $p_{11},$ $p_{12}$;

(v)\ The hexahedron $\mathbb{H}_2$ with the vertice $p_9,$ $p_{10},$ $p_{11},$ $p_{12},$ $p_{13},$ $v_2^-,$ $p_{14},$ $p_{15}$;

(vi)\ The pentahedron $\mathbb{P}_3$ with the vertice $p_9,$ $p_{10},$ $p_{14},$ $p_{15},$ $p_{16},$ $v_4^+$;

(vii)\ The octahedron $\mathbb{O}$ with the vertice $p_1,$ $p_2,$ $p_3,$ $p_5,$ $p_6,$ $p_7,$ $p_8,$ $p_9,$ $p_{11},$ $p_{13},$ $p_{14},$ $p_{16}$.

Note that the face $(p_1,p_2,p_3)$ of $\mathbb{T}$ and the face $(p_2,p_3,p_5,p_6)$ of $\mathbb{H}_1$ are on the face $(p_1,p_5,p_6)$ of $\mathbb{O}$; the common face $(v^+_2,p_4,p_5,p_6)$ of $\mathbb{H}_1$ and $\mathbb{P}_1$ is a vertical plane; the face $(p_9,p_{11},p_{13},p_{14})$ of $\mathbb{H}_2$ is parallel to the base of the prism. Furthermore, the faces $(p_9,p_{10},p_{11},p_{12})$ and $(p_9,p_{10},p_{14},p_{15})$ are the trapeziums since
the edge $(p_9,p_{11})$ is parallel to $(p_{10},p_{12})$ and the edge $(p_9,p_{14})$ is parallel to $(p_{10},p_{15})$.

By examining the location of the points and applying Lemma 4.1, we conclude that the tetrahedron $\mathbb{T}$ is inside the spinal sphere
 $R^{(7)}_1R^{(7)}_3R^{(7)}_2(\mathcal{S}_3)$; the hexahedron $\mathbb{H}_1$ is contained inside the spinal sphere $R^{(7)}_2(\mathcal{S}_0)$; the pentahedron $\mathbb{P}_1$ is inside  $R^{(7)}_3R^{(7)}_2(\mathcal{S}_3)$; the pentahedron $\mathbb{P}_2$ is contained inside
 ${R^{(7)}_2}^{-1}(\mathcal{S}_0)$; the hexahedron $\mathbb{H}_2$ is contained inside ${R^{(7)}_2}^{-1}(\mathcal{S}_3)$;
  the pentahedron $\mathbb{P}_3$ is inside $\mathcal{S}_2$;  the remaining octahedron $\mathbb{O}$ is inside $\mathcal{S}_0$; see Figure 5.3 for viewing the decomposition of the prism.
\end{proof}

\subsection{The case $\mathcal{O}_{11}$}

In this case, we know that the fundamental domain for the stabiliser
$(\Gamma_{11})_\infty$ cannot be still inside $\mathcal{S}_0$ completely. The radius of spinal spheres other than the largest are so small that these spinal spheres are not much contribution to covering the prism $\mathbf{\Sigma}_{11}$. Due to the different shape of the prism $\mathbf{\Sigma}_{11}$ with the case $\mathcal{O}_7$, we only need to consider the largest spinal spheres which are the images of $\mathcal{S}_0$ under the elements of $(\Gamma_{11})_\infty$. In order to determine a union of the spinal spheres which covering the prim $\mathbf{\Sigma}_{11}$,  we minimise their numbers by the symmetry of $R^{(11)}_1$, it suffice to consider $\mathcal{S}_0$ and the images of $\mathcal{S}_0$ under suitable elements in $(\Gamma_{11})_\infty$, these are in Heisenberg coordinates given by
\begin{eqnarray*}
T^{(11)}(\mathcal{S}_0)&=&\left\{(\zeta,t):\left||\zeta|^2+it-2i\sqrt{11}+2i\Im m\zeta\right|=4\right\},\\
R^{(11)}_2(\mathcal{S}_0)&=&\left\{(\zeta,t):\left||\zeta-1|^2+it-i\sqrt{11}+2i\Im m\zeta\right|=4\right\},\\
R^{(11)}_1R^{(11)}_2(\mathcal{S}_0)&=&\left\{(\zeta,t):\left||\zeta+1|^2+it-i\sqrt{11}-2i\Im m\zeta\right|=4\right\},\\
{R^{(11)}_2}^{-1}(\mathcal{S}_0)&=&\left\{(\zeta,t):\left||\zeta-1|^2+it+i\sqrt{11}+2i\Im m\zeta\right|=4\right\},\\
R^{(11)}_3(\mathcal{S}_0)&=&\left\{(\zeta,t):\left|\left|\zeta-\omega_{11}\right|^2+it-i\sqrt{11}-2i\Im m(\bar{\omega}_{11}\zeta)\right|=4\right\},\\
{R^{(11)}_3}^{-1}(\mathcal{S}_0)&=&\left\{(\zeta,t):\left|\left|\zeta-\omega_{11}\right|^2+it+i\sqrt{11}-2i\Im m(\bar{\omega}_{11}\zeta)\right|=4\right\},\\
R^{(11)}_3R^{(11)}_2(\mathcal{S}_0)&=&\left\{(\zeta,t):\left|\left|\zeta+\bar{\omega}_{11}\right|^2+it-i\sqrt{11}-2i\Im m(\omega_{11}\zeta)\right|=4\right\},\\
R^{(11)}_1R^{(11)}_3R^{(11)}_2(\mathcal{S}_0)&=&\left\{(\zeta,t):\left||\zeta-\bar{\omega}_{11}|^2+it-i\sqrt{11}+2i\Im m(\omega_{11}\zeta)\right|=4\right\},\\
R^{(11)}_1{R^{(11)}_3}^{-1}R^{(11)}_2(\mathcal{S}_0)&=&\left\{(\zeta,t):\left||\zeta-\bar{\omega}_{11}|^2+it+i\sqrt{11}+2i\Im m(\omega_{11}\zeta)\right|=4\right\}.
\end{eqnarray*}

\begin{definition}
Let $X$ be a closed polygonal chain (not necessarily in a plane), then a topological disk defined by the cone over $X$ with apex $v$ is called a \textbf{cone-polygon}, denoted by $\mathbb{D}_v(X)$.
\end{definition}

Note that a polygon in traditional sense can be interpreted as a cone-polygon, in that case, the boundary of cone-polygon and the apex lie in the same plane and moreover the apex is in the interior of the boundary. We claim that the prism $\mathbf{\Sigma}_{11}$ lies inside the union of $\mathcal{S}_0$ and its images as above, see Figure 5.4 for viewing these spinal spheres.

\begin{figure}
\centering
  \subfigure[]{
    \label{fig:mini:subfig:1}

 \begin{minipage}[1]{0.4\textwidth} \centering
      \includegraphics[width=2in]{neighboring5.eps}
  \end{minipage}
}
  \subfigure[]{
    \label{fig:mini:subfig:2}
    \begin{minipage}[3]{0.4\textwidth}
      \centering
      \includegraphics[width=2in]{neighboring6.eps}
    \end{minipage}
}
\medskip
\makeatletter\def\@captype{figure}\makeatother
\bigskip
\caption{(a)\ The shading view of neighboring spinal spheres containing the fundamental domain for $(\Gamma_{11})_\infty$. (b)\ Another view for these spinal spheres.}
  \label{fig:mini:subfig}
\end{figure}

\begin{proposition}The prism $\mathbf{\Sigma}_{11}$ is contained in the union of the interiors of the spinal spheres $\mathcal{S}_0,$ $T^{(11)}(\mathcal{S}_0),$  $R^{(11)}_2(\mathcal{S}_0),$ ${R^{(11)}_2}^{-1}(\mathcal{S}_0),$ $R^{(11)}_3(\mathcal{S}_0),$ ${R^{(11)}_3}^{-1}(\mathcal{S}_0),$  $R^{(11)}_1{R^{(11)}_3}^{-1}R^{(11)}_2(\mathcal{S}_0)$, $R^{(11)}_1R^{(11)}_3R^{(11)}_2(\mathcal{S}_0)$, $R^{(11)}_1R^{(11)}_2(\mathcal{S}_0)$ and $R^{(11)}_3R^{(11)}_2(\mathcal{S}_0)$.
\end{proposition}

\begin{proof}
Using the same argument of Proposition \ref{eq:5.2}, we want to decompose the prism $\mathbf{\Sigma}_{11}$ into several
polyhedral cells. The difference is the complicated intersection of the spinal spheres, which leads that the prism is difficultly decomposed into several polyhedral cells each of which is contained in one spinal sphere. Observe that a union of interiors of several spinal spheres is
a star-convex set if they have a non-empty interior intersection. We shall show that the collection of these spinal spheres can be separated into several parts such that each part contains certain polyhedral cell. All these polyhedral cells are defined by the star-disk as its boundary.

We first define a tetrahedron $\mathbb{T}$ with vertices $v_1^-,q_1,q_2,q_3$, where
$$\begin{array}{l}
q_1=\left(1/4-i\sqrt{11}/4,-2\sqrt{11}/3\right),\\
q_2=\left(3/20-3i\sqrt{11}/20,-4\sqrt{11}/5\right),\\
q_3=\left(7/20-3i\sqrt{11}/20,-9\sqrt{11}/10\right).
\end{array}$$
Observe that the points $q_1,q_2,q_3$ lie on the edges $(v^+_1,v^-_1),$ $(v^-_1,v_3^-)$ and $(v_1^-,v_2^-)$, respectively.
An easy calculation shows that this tetrahedron is contained inside $R^{(11)}_1{R^{(11)}_3}^{-1}R^{(11)}_2(\mathcal{S}_0)$ by Lemma 4.1.

Next, we define a hexahedron $\mathbb{H}_1$ with vertices $q_1,q_2,q_3,q_4,q_5,q_6,q_7,v^+_0$ and another hexahedron $\mathbb{H}_2$ with vertices $v_2^-,q_5,q_6,q_7,q_8,q_9$, where
$$
\begin{array}{llll}
q_4=\left(1/4-i\sqrt{11}/4,-1/2\right),&q_5=\left(0.42+0.26i,-0.71\sqrt{11}+0.39\right),\\
q_6=\left(0.6+i\sqrt{11}/10,-0.65\sqrt{11}\right),&q_7=\left(0.58+2i\sqrt{11}/25,-1.92\right),\\
q_8=\left(3/4+i\sqrt{11}/4,0\right),&q_9=(0.55+i\sqrt{11}/4,-2\sqrt{11}/5).
\end{array}$$

Observe that the points $q_4,q_6,q_8,q_9$ lie on the edges $(v^+_1,v^-_1),$ $(v^-_1,v_2^-),$ $(v_2^+,v_2^-)$ and $(v^-_2,v^-_3)$, respectively.
The points $q_5$ lies on the interior of the base of the prism and $q_7$ lies on the interior of the face $(v^+_1,v^-_1,v^-_2,v^+_2)$. Then we know the hexahedron $\mathbb{H}_1$ has the faces $(q_1,q_2,q_3),$ $(q_1,q_3,q_6,q_7,q_4),$ $(q_1,q_2,v^+_0,q_4),$ $(q_4,q_5,v^-_0),$ $(q_4,q_5,q_7)$ and $(q_5,q_6,q_7)$ and the hexahedron $\mathbb{H}_2$ has the faces $(q_5,q_6,q_7),$ $(q_5,q_7,q_8),$ $(v^-_2,q_8,q_9),$ $(q_5,q_8,q_9)$ $(q_6,q_7,q_8,v^-_2)$ and $(q_5,q_6,v^-_2,q_9)$. By examining the location of these points and Lemma 4.1, we conclude that the hexahedron $\mathbb{H}_1$ is contained inside ${R^{(11)}_2}^{-1}(\mathcal{S}_0)$ and the hexahedron $\mathbb{H}_2$ is lied inside ${R^{(11)}_3}^{-1}(\mathcal{S}_0)$.

We focus on describing other polyhedral cells in the decomposition of the prism $\mathbf{\Sigma}_{11}$. Let $\mathcal{U}_1$ denote the union of $R^{(11)}_2(\mathcal{S}_0),$ $ R^{(11)}_1R^{(11)}_2(\mathcal{S}_0)$ and $R^{(11)}_1R^{(11)}_3R^{(11)}_2(\mathcal{S}_0)$. We verify that $q_{10}=(0.2-0.4i,2.4)$ is in the intersection of the interiors of these three spinal spheres, which implies that $\mathcal{U}_1$ is a star-convex set about $q_{11}$. Analogously, we know $\mathcal{U}_2$, denoted by the union of $T^{(11)}(\mathcal{S}_0),$ $R^{(11)}_3(\mathcal{S}_0),$ $R^{(11)}_1R^{(11)}_2(\mathcal{S}_0)$ and $R^{(11)}_3R^{(11)}_2(\mathcal{S}_0)$, is a star-convex set about $q_{11}=(0.18+0.72i,4.8)$. This can be verified by examining the location of $q_{12}$ which is in the intersection of the interiors of these four spinal spheres.
We need to add the following points on the faces of the prism $\mathbf{\Sigma}_{11}$, each of which is in the intersection of the interiors of at least two spinal spheres.
$$ \begin{array}{ll}
q_{12}=(1/4-i\sqrt{11}/4,\sqrt{11}/2),&q_{13}=(0.21-0.21i\sqrt{11},\sqrt{11}/2),\\
q_{14}=(0,\sqrt{11}/2),&q_{15}=(-0.21+0.21i\sqrt{11},\sqrt{11}/2),\\
q_{16}=(i\sqrt{11}/4,1),&q_{17}=(3/4+i\sqrt{11}/4,1),\\
q_{18}=(0.42-2i\sqrt{11}/25,1.95),&q_{19}=(3/4+i\sqrt{11}/4,\sqrt{11}),\\
q_{20}=(0.6+i\sqrt{11}/10,27\sqrt{11}/20),&q_{21}=(0.42+0.26i,1.29\sqrt{11}+0.39),\\
q_{22}=(-1.4+1.4i\sqrt{11},4\sqrt{11}/5),&q_{23}=(-1/4+i\sqrt{11}/4,\sqrt{11}/2).
\end{array}
$$

Observe that the points $q_{12},q_{20},q_{23}$ lie on the edges $(v^+_1,v^-_1),$ $(v^+_1,v^+_2)$ and $(v^+_3,v^-_3)$ respectively and the points $q_{17},q_{19}$ lie on the edge $(v^+_2,v^-_2)$. Moreover, the points $q_{13},$ $q_{14},$ $q_{15},$ $q_{22}$ lie on the interior of the face $(v^+_1,v^-_1,v^-_3,v^+_3)$,
the point $q_{16}$ lies on the interior of the face $(v^+_2,v^-_2,v^-_3,v^+_3)$, the point $q_{18}$ lies on the interior of the face
$(v^+_1,v^-_1,v^-_2,v^+_2)$ and the points $q_{21}$ lies on the interior of the top $(v^+_1,v^+_2,v^+_3)$.
We need to add other three points in the interior of the prism $\mathbf{\Sigma}_{11}$ which are used to define the cone-polygon,
$$ \begin{array}{l}
q_{24}=(-0.16+0.74i,1.4),\\
q_{25}=(0.328-0.28i,1.99),\\
q_{26}=(0.325+0.29i,4.652).
\end{array}
$$

We verify the location of all these points as follows:

$\bullet$\ The point $q_{12}$ is in the intersection of the interiors of  $R^{(11)}_1R^{(11)}_3R^{(11)}_2(\mathcal{S}_0)$ and $\mathcal{S}_0$;

$\bullet$\ The point $q_{13}$ is in the intersection of the interiors of $\mathcal{S}_0$, $R^{(11)}_1R^{(11)}_2(\mathcal{S}_0)$ and $R^{(11)}_1R^{(11)}_3R^{(11)}_2(\mathcal{S}_0)$;

$\bullet$\ The point $q_{14}$ is in the intersection of the interiors of $\mathcal{S}_0$, $R^{(11)}_2(\mathcal{S}_0)$ and $R^{(11)}_1R^{(11)}_2(\mathcal{S}_0)$;

$\bullet$\ The point $q_{15}$ is in the intersection of the interiors of $\mathcal{S}_0$, $R^{(11)}_2(\mathcal{S}_0)$, $R^{(11)}_3(\mathcal{S}_0)$ and $R^{(11)}_3R^{(11)}_2(\mathcal{S}_0)$;

$\bullet$\ The points $q_{16},q_{19},q_{20}$ are in the intersection of the interiors of $R^{(11)}_2(\mathcal{S}_0)$ and $R^{(11)}_3(\mathcal{S}_0)$;

$\bullet$\ The point $q_{17}$ is in the intersection of the interiors of $\mathcal{S}_0$ and $R^{(11)}_2(\mathcal{S}_0)$;

$\bullet$\ The point $q_{18}$ is in the intersection of the interiors of $\mathcal{S}_0$, $R^{(11)}_2(\mathcal{S}_0)$ and $R^{(11)}_1R^{(11)}_3R^{(11)}_2(\mathcal{S}_0)$;

$\bullet$\ The point $q_{21}$ is in the intersection of the interiors of $T^{(11)}(\mathcal{S}_0)$, $R^{(11)}_2(\mathcal{S}_0)$ and $R^{(11)}_3(\mathcal{S}_0)$;

$\bullet$\ The point $q_{22}$ is in the intersection of the interiors of $R^{(11)}_1R^{(11)}_2(\mathcal{S}_0)$, $R^{(11)}_2(\mathcal{S}_0)$ and $R^{(11)}_3(\mathcal{S}_0)$;

$\bullet$\ The point $v^{+}_0$ is in the intersection of the interiors of $R^{(11)}_1R^{(11)}_2(\mathcal{S}_0)$, $T^{(11)}(\mathcal{S}_0)$ and $R^{(11)}_2(\mathcal{S}_0)$;

$\bullet$\ The point $q_{23}$ is in the intersection of the interiors of $\mathcal{S}_0$, $R^{(11)}_3(\mathcal{S}_0)$ and $R^{(11)}_3R^{(11)}_2(\mathcal{S}_0)$;

$\bullet$\ The point $q_{24}$ is in the intersection of the interiors of $\mathcal{S}_0,$ $R^{(11)}_2(\mathcal{S}_0),$ $R^{(11)}_3(\mathcal{S}_0)$ and $R^{(11)}_3R^{(11)}_2(\mathcal{S}_0)$;

$\bullet$\ The point $q_{25}$ is in the intersection of the interiors of $\mathcal{S}_0,$ $R^{(11)}_2(\mathcal{S}_0),$ $ R^{(11)}_1R^{(11)}_2(\mathcal{S}_0)$ and $R^{(11)}_1R^{(11)}_3R^{(11)}_2(\mathcal{S}_0)$;

$\bullet$\ The point $q_{26}$ is in the intersection of the interiors of $T^{(11)}(\mathcal{S}_0),$ $R^{(11)}_2(\mathcal{S}_0),$ $R^{(11)}_3(\mathcal{S}_0)$ and $R^{(11)}_1R^{(11)}_2(\mathcal{S}_0)$.

In term of these, we denote by $X_1$ a closed polygonal chain joining in order with the points $p_{12},$ $p_{13},$ $p_{14},$ $p_{15},$ $p_{16},$ $p_{17},$ $p_{18}$ and denote by $X_2$ a closed polygonal chain joining in order with the points $p_{16},$ $p_{19},$ $p_{20},$ $p_{21},$ $v^+_0$, $p_{22},$ $p_{24}$. So then we can define two cone-polygons $\mathbb{D}_{q_{25}}(X_1)$ and $\mathbb{D}_{q_{26}}(X_2)$. By examining the locations of these points, we show that $\mathbb{D}_{q_{25}}(X_1)$ is in the intersection of the interiors of $\mathcal{S}_0$, $\mathcal{U}_1$ and $\mathbb{D}_{q_{26}}(X_2)$ is in the intersection of the interiors of $R^{(11)}_2(\mathcal{S}_0)$ and
$T^{(11)}(\mathcal{S}_0),$ $R^{(11)}_1R^{(11)}_2(\mathcal{S}_0),$ $R^{(11)}_3(\mathcal{S}_0)$, namely, the intersection of the interiors of $\mathcal{U}_1$ and $\mathcal{U}_2$. The remaining faces can be easily verified which are contained inside
$\mathcal{S}_0,\mathcal{U}_1$ or $\mathcal{U}_2$ .

Finally, we define three polyhedral cells as follows:

(i)\ The polyhedral cell $\mathbb{P}_1$ is defined by the faces $\mathbb{D}_{q_{25}}(X_1),$ $\mathbb{D}_{q_{26}}(X_2),$ $(v^+_1,q_{12},q_{18},q_{17},q_{19},q_{20})$, $(v_1^+,q_{12},q_{13},q_{14},q_{15},q_{22},v^+_0)$ and $(v^+_1,q_{20},q_{21},v^+_0)$  as its boundary;

(ii)\ The polyhedral cell $\mathbb{P}_2$ is defined by the faces $\mathbb{D}_{q_{26}}(X_2),$ $(q_{15},q_{23},q_{24}),$ $(v^+_1,q_{20},q_{21},v^+_0)$ $(v^+_1,q_{12},q_{18},q_{17},q_{19},q_{20})$, $(v_1^+,q_{12},q_{13},q_{14},q_{15},q_{22},v^+_0)$ and  $(q_{23},q_{16},q_{24})$ as its boundary;

(iii)\ The polyhedral cell $\mathbb{P}_3$ is defined by the faces $\mathbb{D}_{q_{25}}(X_1),$ $(q_4,q_5,q_7),$  $(q_5,q_7,q_8),$ $(q_4,q_5,v^-_0),$ $(q_{15},q_{23},q_{24}),$ $(q_8,q_9,v^-_3,q_{23},q_{16},q_{17})$, $(q_{23},q_{16},q_{24})$, $(q_4,q_7,q_8,q_{17},q_{18},q_{12})$, $(v^-_3,v^-_0,q_5,q_9)$ and $(q_{12},q_{13},q_{14},q_{15},q_{23},v^-_3,v^-_0,q_4)$ as its boundary;

By Lemma 4.1 and the properties of star-convex of $\mathcal{U}_1$ and $\mathcal{U}_2$, we conclude that the polyhedral cell $\mathbb{P}_1$ contained inside $\mathcal{U}_1$; the polyhedral cell $\mathbb{P}_2$ contained inside $\mathcal{U}_2$; the polyhedral cell $\mathbb{P}_3$ is contained inside $\mathcal{S}_0$. This completes the proof.
\end{proof}

\medskip

config: arxiv


text:
\subparagraph*{Subsets of groups}
Regular languages are an extremely versatile tool in algorithmics on
sets of finite words. This is mainly due to two reasons. First, they
are robust in terms of representations and closure properties: They
can be described by finite automata, by recognizing morphisms, and
by monadic second-order logic and they are closed under Boolean and an
abundance of other operations. Second, many properties (such as emptiness) are easily
decidable using finite automata.

Given this success, there have been several attempts to develop an analogous
notion for subsets of (infinite, finitely generated) groups. Adapting the notion
of recognizing morphism yields \emph{recognizable subsets} of a group $G$. They
are closed under Boolean operations, and problems such as membership or
emptiness are decidable. However, since they are merely unions of cosets of
finite-index normal subgroups, their expressiveness is severely limited.

Another notion is that of \emph{rational subsets}, which transfer
(non-deterministic) finite automata to groups. Starting with
pioneering work by Benois~\cite{Benois1969} in 1969, they have matured
into an important tool in group theory. Rational subsets are quite
expressive: They include finitely generated submonoids and are closed
under (finite) union, pointwise product, and Kleene star. Moreover, they have
been applied successfully to solving
equations in groups~\cite{DIEKERT2005105,DBLP:conf/icalp/CiobanuE19},
as well as in
other settings~\cite{BaSi2010,silva2017automata}.

The high expressiveness of rational subsets comes at the cost of
undecidability of decision problems for many groups.  The most
fundamental one is the \emph{membership problem} for rational subsets:
Given a rational subset $R$ of a group $G$ and an element $g\in G$,
does $g$ belong to $R$?  Understanding for which groups this problem
is decidable received significant attention over the last two decades,
see~\cite{Lohrey2016survey} for a survey. Unfortunately, the rational
subsets do not quite reach the level of robustness of regular
languages. In general, the class of rational subsets of a group is not
closed under Boolean operations, and the
papers~\cite{DBLP:journals/ijac/LohreyS08,bazhenova2000rational} study
for which groups the rational subsets form a Boolean algebra.

\subparagraph*{Baumslag-Solitar groups} A prominent class of groups is
that of \emph{Baumslag-Solitar groups} $\BS(p,q)$. For each
$p,q\in\N$, the group is defined as
$\BS(p,q)=\langle a,t \mid ta^pt^{-1}=a^q\rangle$. They were
introduced in 1962 by Baumslag and Solitar to provide an example of a
two-generator one-relator group that is non-Hopfian. They recently
came into focus from the algorithmic perspective in a paper by
Kharlampovich, L\'opez, and
Miasnikov~\cite{KharlampovichLopezMiasnikov2019}, which shows that
solvability of equations is decidable in $\BS(1,q)$.  They have also
been studied from several other perspectives, such as the decidability
and complexity of the word
problem~\cite{Robinson1993,DBLP:conf/latin/DiekertMW14,Weiss2015}, the
conjugacy problem~\cite{DBLP:conf/latin/DiekertMW14,Weiss2015}, tiling problems~\cite{DBLP:journals/corr/AubrunK13}, and
computing normal
forms~\cite{DBLP:journals/ijac/DiekertL11,ELDER2013260,elder2010}.

More specifically to our setting, the Baumslag-Solitar groups have
recently been identified by Diekert, Potapov, and
Semukhin~\cite{DBLP:journals/corr/abs-1910-02302} as a stumbling block
in solving rational subset membership in the group $\GL(2,\Q)$, that is,
the group of invertible $2 \times 2$ matrices over $\Q$.
They show
that any subgroup of $\GL(2,\Q)$ containing $\GL(2,\Z)$ is either of the
form $\GL(2,\Z)\times\Z^k$ for $k\ge 1$ or contains $\BS(1,q)$ as a
subgroup for some $q\ge 2$.
Rational subset membership for $\GL(2,\Z)\times\Z^k$ is today a matter
of standard arguments~\cite{Lohrey2016survey},
because $\GL(2,\Z)$ is virtually free.
Therefore, making significant progress towards decidability in
larger subgroups requires understanding rational subsets of
$\BS(1,q)$.

One can represent the elements of $\BS(1,q)$ as pairs $(r,m)$, where
$r$ is a number in $\Z[\tfrac{1}{q}]$,
say $r=\pm\sum_{i=-n}^n a_iq^i$
for $a_{-n},a_{-n+1},\ldots,a_n\in\{0,\ldots,q-1\}$,
\footnote{
    $\Z[\tfrac{1}{q}]$ denotes (the additive group of)
    the smallest subring of $(\Q, +, \cdot)$ containing $\Z$
    and $1/q$; as a set, it consists of all rational numbers
    of the form $n \cdot q^j$, $n, j \in \Z$.
  }

and
$m\in\Z$. Here, one can think of $m$ as a \emph{cursor} pointing to a
position in the $q$-ary expansion
$a_{n}q^{n}+\cdots+a_{-n}q^{-n}$. Then the action of the generators of $\BS(1,q)$
is as follows.
Multiplication by $t$ or $t^{-1}$ moves
the cursor to the left or the right, respectively. Multiplication by $a$
adds $q^m$; likewise, multiplication by $a^{-1}$ subtracts $q^m$. Thus,
from an automata-theoretic perspective, one can view the rational
subset membership problem as the reachability problem for an extended
version of one-counter automata. Instead of storing a natural number,
such an automaton stores a number $r \in \Z[\frac{1}{q}]$. Moreover, instead of
instructions ``increment by 1'' and ``decrement by 1'', it has an
additional $\Z$-counter $m$ that determines the value to be
added in the next update. Then, performing ``increment'' on $r$ will add $q^m$ and
``decrement'' on $r$ will subtract $q^m$. The $\Z$-counter $m$ supports
the classical ``increment'' and ``decrement'' instructions.

\subparagraph*{Contribution} Our \emph{first main contribution} is to show is
that for each group $\BS(1,q)$, the rational subset membership problem is
decidable and $\PSPACE$-complete. To this end, we show that each rational subset can be
represented by a regular language of finite words that encode elements of
$\BS(1,q)$ in the natural way: For $(r,m)$ as above, we encode each digit $a_i$
by a letter; and we decorate the digits at position $0$ and at position
$m$.  We call this encoding the \emph{pointed expansion} (PE) of \((r,m)\).
This leads to a natural notion of subsets of $\BS(1,q)$, which we call
\emph{PE-regular}. We regard the introduction of this notion as the \emph{second
  main contribution} of this work.

The class of PE-regular subsets of $\BS(1,q)$ has several properties that
make them a promising tool for decision procedures for $\BS(1,q)$:
First, our proof shows that it effectively includes the large class of
rational subsets, in particular any finitely generated submonoid.
Second, they form an effective Boolean algebra. Third, due to them
being regular languages of words, they inherit many algorithmic tools
from the setting of free monoids.
We apply these properties to obtain \emph{three applications of our main results}.
\begin{enumerate}
\item Membership in each fixed rational subset can be decided in
  logarithmic space.

\item We show that it is decidable whether a given PE-regular subset
  (and thus a given rational subset) is recognizable. Recognizability
  of rational subsets is rarely known to be decidable for groups: The
  only examples known to the authors are free groups, for which
  decidability was shown by
  S\'{e}nizergues~\cite{DBLP:journals/acta/Senizergues96} (and
  simplified by Silva~\cite{DBLP:journals/ita/Silva04}) and free
  abelian groups (this follows from \cite[Theorem
  3.1]{GinsburgSpanier1966a}). Since (i)~finitely generated subgroups
  are rational subsets and (ii)~a subgroup of any group $G$ is
  recognizable if and only if it has finite index in $G$, our result
  implies that it is decidable whether a given finitely generated
  subgroup of $\BS(1,q)$ has finite index. Studying decidability of
  this finite index problem in groups was recently proposed by
  Kapovich~\cite[Section~4.3]{DBLP:journals/dagstuhl-reports/DiekertKLM19}.
\item Our results imply that emptiness of Boolean combinations (hence
  inclusion, equality, etc.) of rational subsets is decidable. (We
  also show that the rational subsets of $\BS(1,q)$ are not closed
  under intersection.)  This is a strong decidability property that
  already fails for groups as simple as $F_2\times \Z$ (this follows
  from \cite[Theorem~6.3]{Ibarra1978}), where $F_2$ is the free group
  over two generators, and hence for $\GL(2,\Z)\times\Z^k$, $k\ge 1$.
\end{enumerate}

Finally, we remark that
since $\BS(1,q)$ is isomorphic to the group of all matrices
$\begin{psmallmatrix} q^m & r \\ 0 & 1\end{psmallmatrix}$ for $m\in\Z$
and $r\in\Z[\tfrac{1}{q}]$, our results can be interpreted as solving
the rational subset membership problem for this subgroup of $\GL(2,\Q)$.

\subparagraph{Related work} It is well-known that membership in a
given finitely generated subgroup, called the \emph{generalized word
  problem} of $\BS(1,q)$, is decidable. This is due to a general result of
Romanovski\u{\i}, who showed in \cite{Romanovskii1974a} and
\cite{Romanovskii1980a} that solvable groups of derived length two
have a decidable generalized word problem (it is an easy exercise to
show that $\BS(1,q)$ is solvable of derived length two for each
$q\in\N$).

Another restricted version of rational subset membership is the
\emph{knapsack problem}, which was introduced by Myasnikov, Nikolaev,
and Ushakov~\cite{MyNiUs14}. Here, one is given group elements
$g_1,\ldots,g_k,g$ and is asked whether there exist
$x_1,\ldots,x_k\in\N$ with $g_1^{x_1}\cdots g_k^{x_k}=g$. A recent
paper on the knapsack problem in Baumslag-Solitar groups by Dudkin and
Treyer~\cite{DudkinTreyer2018} left open whether the knapsack problem
is decidable in $\BS(1,q)$ for $q\ge 2$. This was settled very
recently in \cite{LohreyZetzsche2020a}, where one expresses
solvability of $g_1^{x_1}\cdots g_k^{x_k}=g$ in a variant of B\"{u}chi
arithmetic. A slight extension of that proof yields a regular
language as above for the set
$S=\{g_1^{x_1}\cdots g_k^{x_k}\mid x_1,\ldots,x_k\in\N\}$. Note that
each element $g_i$ moves the cursor either to the left (i.e.\
increases $m$), to the right (i.e.\ decreases $m$), or not at all.
Thus, in a product $g_1^{x_1}\cdots g_k^{x_k}$, the cursor direction
is reversed at most $k-1$ times. The challenge of our translation from
rational subsets to PE-regular subsets is to capture products where the
cursor changes direction an unbounded number of times.

Finally, closely related to rational subsets, there is another
approach to group-theoretic problems via automata: One can represent
finitely generated subgroups of free groups using \emph{Stallings
  graphs}
. Due to the special setting of free groups, they behave in many ways
similar to automata over words and are thus useful for decision
procedures~\cite{KAPOVICH2002608}. Stallings graphs have recently been
extended to semidirect products of free groups and free abelian groups
by Delgado~\cite{Delgado2017a}. However, this does not include
products $\Z[\tfrac{1}{q}]\rtimes\Z$ and is restricted to subgroups.


TITLE: What will be the mathematical formula to fetch the position of a letter from this string?
QUESTION [1 upvotes]: The above full string is a combined form of two different strings named string-1 and string-2. Both strings has equal length of 32 characters. There is two other parameters interval and size. string-1 is broken into pieces of length = interval and string-2 is broken into pieces of length = size to form the full string. once all the pieces of string -2 is put in full string, pieces of string-1 is padded to the full string.
possible values for size variable are : 2, 4, 8, 16
possible values for interval variable are : any value from 1 to value of size variable
My requirement is to write a mathematical equation fetch index(in full string) of nth number of character of string-1 without separating the full string into its 2 original strings.
for e.g.
In above full string, if I want to fetch 5th character of string-1 it will be '2' whose index in full string is 12.
and for 6th character which is '1' has the index 15.
P.S. index of the character is counted from 0 to index, and the nth character is counted from 1 to n

REPLY [1 votes]: Given character $n$ of string $1$, it will be in the $\lfloor \frac{n}{interval}\rfloor +1$ block.  The reason I used floor instead of ceiling (which would get rid of the $+1$) is that integer divide gives the floor.  There will be $(\lfloor \frac{n}{interval}\rfloor )size$ characters from string $2$ preceeding it, so it will be at position $(\lfloor \frac{n}{interval}\rfloor )size+n$ in the final string.  Note that you are inconsistent in the referencing-you talk about the $5$th character as $2$, but it is in position $4$ if you index from $0$ and that you say both strings will have $32$ characters in contrast to the example.  This will be correct as long as you don't run out of string $2$ characters.  If you use all the string $2$ characters, the position will be $n+32$ if string $2$ is $32$ characters long.


TITLE: Find all subgroups
QUESTION [3 upvotes]: Find all subgroups of $\mathbb{Z}_{1001}$, where $1001=7\cdot 11\cdot 13$.
By Lagrange, we know the order of every subgroup must divide 1001, but I'm not sure how the subgroups should be listed/how to find them..

REPLY [1 votes]: Hints , since $\mathbb{Z_{1001}}$ is  finite cyclic group. So for every divisor  of $1001$ it has unique subgroup of that order ..


TITLE: Compact form of $\mathfrak{su}(2)$
QUESTION [4 upvotes]: Apologies for the completely elementary things, but I am stuck. Please help.
Consider $\mathfrak{g}=\mathfrak{sl}(2,\mathbb{C})$ and let $X_+,X_-,H$ be as in Theorem 6.6 of [1],
$$
[X_+,X_-]=H,\quad[X_\pm,H]=\pm X_\pm.
$$
Let
$$
J_0=\imath H,\quad J_1=X_+-X_-,\quad J_2=\imath(X_++X_-).
$$
According to Theorem 6.11 of [1], $\mathfrak{u}_0=\mathbb{R}\{J_0,J_1,J_2\}$ is a compact form of $\mathfrak{g}$. But a direct computation shows
$$
[J_0,J_1]=-J_2,\quad[J_1,J_2]=2J_0,\quad[J_2,J_0]=-J_1.
$$
Are these not the commutation relations of the non-compact form $\mathfrak{sl}(2,\mathbb{R})$?
Thanks.
[1] A. Knapp. Lie groups beyond an introduction. 2nd edition, Birkhaeuser 2002.

REPLY [3 votes]: Yes, you have an error in your original commutation relations. On the top of page 98, you have the equations
$$
[H_\alpha, E_\alpha] = 2E_\alpha, \quad [H_\alpha, E_{-\alpha}] = 2E_{-\alpha}, \quad [E_\alpha, E_{-\alpha}] = H_\alpha.
$$
Then on the bottom of page 295, $X_\alpha = a_\alpha E_\alpha$, with $a_\alpha a_{-\alpha} = 1$. This implies the commutation relations
$$
[H_\alpha, X_\alpha] = 2X_\alpha, \quad [H_\alpha, X_{-\alpha}] = 2X_{-\alpha}, \quad [X_\alpha, X_{-\alpha}] = H_\alpha.
$$
To map this to the usual $\mathfrak{su}(2)$ algebra, you can just take (for example)
$$
J_1 = \frac{1}{2}(X_\alpha-X_{-\alpha}), \quad J_2 = \frac{i}{2}(X_\alpha+X_{-\alpha}), \quad J_3 = \frac{i}{2}H_\alpha.
$$
These then satisfy $[J_i, J_j] = \epsilon_{ijk}J_k$.


TITLE: Do we sometimes have to go "each way" separately for iff proofs?
QUESTION [2 upvotes]: So, I often enjoy trying to prove "if and only if" statements by only using if and only if arguments.
i.e. RTP: $A \Leftrightarrow D$. Proof: $A \Leftrightarrow B \Leftrightarrow C \Leftrightarrow  D$
My question is whether or not this is always possible? I'm aware that it's often easier to go one way then the other way, but is this ever the only way to do it?
(Let me know if I've not explained myself properly - found it quite a hard question to word!)

REPLY [1 votes]: It is often that one of the directions is trivial, the other one hard.
If an "if and only if" argument is reasonably clean to obtain, it can be the nicest form of proof. But that isn't always easy to see how to do.


TITLE: Proof that the DE corresponds to a mapping of circle.
QUESTION [0 upvotes]: I'm struggling with the differential equation :
$${d^2r\over dt^2} = i{dr\over dt}$$
I know from physics that this is just circular motion(acceleration is normal to velocity.)but stated this way confuses me. How do i prove this is a circle?

REPLY [0 votes]: As the comment suggests:
$$u(t) =  Ce^{it}$$
By the euler formula we have :$$Ce^{it}=C(\cos(t)+i\sin(t))$$
Which is the polar form of the equation of the circle.


TITLE: Binominal distribution "fewer than"
QUESTION [0 upvotes]: I have this exercise:

We learned in Exercise 3.26 that about 90% of American adults had chickenpox before adulthood. We now consider a random sample of 120 American adults.
(a) What is the mean and standard deviation.
(c) What is the probability that 105 or fewer people in this sample have had chickenpox in their childhood?

I have used Z score to calculate it, since both $np$ and $n(1-p)$ is above 10, we can use normal distribution. First I calculated mean and SD from exercise (a) for a binominal distribution, then the Z score:
Mean:
$μ=np$
$μ=120*0.90$
$μ=108$
Then the Standard Deviation:
$σ=\sqrt{np(1-p)}$
$σ=\sqrt{120*0.90(1-0.90)}$
$σ=\sqrt{10.8}$
$σ=3.28$
Now I calculate the Z-score:
$Z=(observation-mean)/SD$
$Z=  (105-108)/3.28$
$Z=-0.91$
We will look up that number on the Z-table: 0.1814 = 18.14%
Is this correctly done? I tried looking it up on the internet, but I see people getting 21% when they use pbinom(105,120,.9) from the R language = 0.2181634 = 21%
The same website calculates it this way, without R and still gets 21%

REPLY [1 votes]: I think you've done the steps correctly, the discrepancy between the binomial and normal approximation seems believable in this case (if your $p$ is quite far from $0.5$ and your $n$ is not enormously large, then your answer can usually be a few percentage points off).
To get a more accurate answer using the normal approximation, some people advocate using a continuity correction. This means calculating the $z$-score as $\left(105.5 - 108\right)/3.28$ instead of $\left(105 - 108\right)/3.28$. Using this continuity correction gives an answer of $22.3\%$ which is a bit closer to the true probability compared to your answer.
Ultimately there are a few ways to go about solving this problem, some by using approximations which lead to slightly different answers. The steps you've taken appear sound, assuming your course has taught you to solve it that way.


TITLE: Factorizations versus integer solutions in single-variable quartics
QUESTION [2 upvotes]: If $x$ is a positive integer, and I have the factorization
$$(x-1)(79x^3+159x^2-513x+255)=0,$$
what is the easiest way to conclude that $x=1$ is the only integer solution?
Related question: If [in a separate circumstance] I have
$$79x^4+40x^3-168x^2+96x-15=0,$$
which doesn’t factor algebraically into any linear factors, can I immediately conclude that the equation has no positive integer solutions? If not, what is the easiest way to prove that?

REPLY [0 votes]: There is the
rational root theorem
which tests the combined factors of $x^3$ and $x^0\space$
but the easiest way to see that
$\space (79x^3+159x^2-513x+255)=0\space$
has no integer solutions is to use Wolfram Alpha as shown
here.
Or use the "cubic formula" to find one root that can be used to reduce the cubic to a quadratic.
$$\text{Let}\quad a=79\qquad b=159\qquad c=-513\qquad d=255$$
\begin{align*}
x&=\sqrt[3]{\biggl(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)+\sqrt{\biggl(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)^2+\biggl(\frac{c}{3a}-\frac{b^2}{9a^2}\biggr)^3}}\\
&+\sqrt[3]{\biggl(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)-\sqrt{\biggl(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)^2+\biggl(\frac{c}{3a}-\frac{b^2}{9a^2}\biggr)^3}}\\
&-\frac{b}{3a}   \end{align*}
This can often  be done in a spreadsheet where  it is common to put the  parentheses
$\bigg(\bigg)$
parts in separate cells  and have one or three other cells do the final computations. Sometimes there will be transient complex values from the radicals but there is robust software that can handle it, including Wolfram Alpha.
A third option is also tedious, a
Trigonometric solution for three real roots
here
or here.
It takes a long time to figure out and apply but it works better for finding all three roots if there are indeed three roots.


TITLE: Angular momentum in planetary disk formation
QUESTION [2 upvotes]: This question is actually more linked to astronomy and astrophysics than to pure physics. I tried posting it on the astronomy page, however it got no answers, so I though this page might help.
Reading about the formation of planetary disks, one of the major problems, it seems like, is the evacuation of angular momentum. Aparently planets can't form with the amount o angular momentum the system has in its early stages. I think I understand where that excess comes from, the collapse of the nebula onto itself and provoking a spin. Then there are many hypothesis on how it's evacuated, which are mostly pretty logical.
Now my question, as a beginner in the study of physics, is this: Why does the angular momentum even need to be evacuated? If the angular momentum is too big, why can't planets still form? Does this have something to do with too much kinetic energy in the system? Thank you!
edit: my main source is this video:  youtube.com/watch?v=tEgw0PXwkGE
Otherwise the information comes from books, notably L'exporation des planètes by Therese Encrenaz and James Lequeux
link to the astronomy page: https://astronomy.stackexchange.com/questions/10073/angular-momentum-in-planetary-disk-formation

REPLY [3 votes]: Consider a particle in a nice Keplerian orbit around a star, taken to be a circle for simplicity. If it is at a radius $r$ and the star has mass $M$, then its specific angular momentum will be
$$ l = vr = \sqrt{GMr} $$
and its total specific energy will be
$$ e = -\frac{GM}{2r}. $$
If you want to decrease $r$, you are going to have to decrease $l$ as well as $e$. If you try to place a particle at a certain radius but with too much angular momentum (i.e. too much tangential velocity), it will start to fly off away from the star -- it will be going too fast for gravity to curve its orbit into a circle. This doesn't necessarily mean that that particle will escape to infinity, but it certainly won't be in a circular orbit. Instead it would be in an ellipse, with the point you tried to place it becoming the periastron.
In a disk with lots of particles, collisions will tend to prevent eccentric orbits. If every particle were on an eccentric orbit with a random orientation, particles would be colliding into each other all the time. Collisions redistribute energy and angular momentum so as to make all the orbits match.
Thus sapping the energy out of particles (e.g. by having them occasionally collide, transfer kinetic energy into internal heat, and radiate that heat away into space) isn't sufficient. You need to get rid of angular momentum too, which is what that video is about. This is a very big subject in astrophysics, because disks are everywhere, from forming planetary systems to black holes to galaxies.
Of course, all this just answers the question "Why do we need to get rid of angular momentum in order for material to fall in?" The broader question of "Why does angular momentum matter for planet formation?" is due to the fact that planets need dense environments to form. A protoplanetary disk around a young star is many orders of magnitude denser than the giant molecular cloud that started to collapse.
In short, angular momentum transfer helps a disk collapse to a small size, where densities are high enough for planets to form. Even planets themselves might start to form from small accretion disks within the main disk, and these only work if they can get rid of angular momentum.
See also this related question on star formation.


text:
\pagebreak
\section{Summary and outlook}
In this work, we have presented a 3D-1D coupled multispecies model for tumor growth including the influence of nutrient transport in a vascular system that is located in the vicinity of a solid tumor. Flow and transport of nutrients within the vascular system are governed by one-dimensional partial differential equations. The corresponding flow and transport processes in the healthy and tumor tissue are based on Darcy's law as well as a standard convection-diffusion equation. Coupling of the three-dimensional equations with their one-dimensional counterparts is done via filtration laws and source terms. In the source terms of the three-dimensional partial differential equations, Dirac measures occur. They are concentrated on the vessel surfaces of the vascular system, since there the exchange processes between the tissue and the vascular system take place. The remaining three-dimensional equations governing the distribution of the tumor cells are of Cahn--Hilliard type. The evolution of matrix degrading enzymes and the tumor angiogensis factor are modeled by convection-diffusion equations. Lastly, the extracellular matrix density is described by an abstract ordinary differential equation.

The centerpiece of our work is a mathematical analysis of this model with a focus on the existence of solutions. We have shown the existence of weak solutions. Our proof is based on the Faedo--Galerkin method. Thereby, the system of partial differential equations is semi-discretized in space and reduced to a system of ordinary differential equations. Using the Cauchy--Peano theorem we show that the system of ordinary differential equations exhibits a solution. In a next step, the existence of weak solutions with respect to the partial differential equations is derived by means of the Banach--Alaoglu theorem. Finally, we present some simulation results for two different settings, illustrating the performance of our model. Our simulation results indicate that the tumor cells sense the vessels with an increased nutrient concentration and move towards them. Furthermore, the impact of several model parameters on the solution variables is discussed.

Among extensions and applications of the models described here are the simulation and optimal control of
chemotherapy drug and radiation as well as modeling of the onset of metastasis. Simulation of these protocols and phenomena represent challenging goals for future work.