text:
\begin{document}

\title[Generators of arithmetic groups]{On generators of arithmetic groups over function fields}

\author{Mihran Papikian}

\address{Department of Mathematics, Pennsylvania State University, University Park, PA 16802}

\email{papikian@math.psu.edu}

\thanks{The author was supported in part by NSF grant DMS-0801208 and Humboldt Research Fellowship.}

\begin{abstract} Let $F=\mathbb{F}_q(T)$ be the field of rational functions with
$\mathbb{F}_q$-coefficients, and $A=\mathbb{F}_q[T]$ be the subring
of polynomials. Let $D$ be a division quaternion algebra over $F$
which is split at $1/T$. Given an $A$-order $\La$ in $D$, we find an
explicit finite set generating $\La^\times$.
\end{abstract}

\maketitle

\section{Introduction} \noindent\textbf{Notation.}\hfill

$\F_q$ = the finite field with $q$ elements;

\hspace{0.3in} \textit{throughout the article $q$ is assumed to be
odd}.

$A=\F_q[T]$, $T$ indeterminate.

$F=\F_q(T)$ = the fraction field of $A$.

$|F|$ = the set of places of $F$.

For $x\in |F|$, $F_x$ = the completion of $F$ at $x$.

$\cO_x = \{z\in F_x\ |\ \ord_x(z)\geq 0\}$ = the ring of integers of
$F_x$.

$\F_x$ = the residue field of $\cO_x$; $\deg(x)=[\F_x:\F_q]$.

For $0\neq f\in A$, $\deg(f)$ = the degree of $f$ as a polynomial in
$T$, and $\deg(0)=+\infty$.

For $f/g\in F$, $\deg(f/g)=\deg(f)-\deg(g)$.

$\ord=-\deg$ defines a valuation on $F$; the corresponding place is
denoted by $\infty$.

$K=F_\infty$.

$\cO=\cO_\infty$.

$\pi=T^{-1}$ = uniformizer at infinity.

\vspace{0.1in}

Let $D$ be a quaternion division algebra over $F$ such that
$D\otimes_F K\cong \M_2(K)$. Let $\La$ be an $A$-order in $D$, i.e.,
$\La$, as a subring of $D$ containing $A$, is also an $A$-module
containing $4$ linearly independent generators over $F$. The
subgroup $\G:=\La^\times$ of units of $\La$ consists of elements
$\la\in \La$ with reduced norm $\Nr(\la)\in\F_q^\times$. It is known
that $\G$ is infinite finitely generated group. The problem of
finding explicit sets of generators for $\G$ naturally arises in the
study of these arithmetic groups. In this paper we develop a method
for finding such explicit sets.

Now we explain what we mean by ``explicit'' and state the main
result of the paper. Let $\fr\in A$ be the discriminant of $D$ -
this is the product of the monic generators of the primes in $A$
where $D$ ramifies; see Section \ref{Sec3}. There exists $\fa\in A$
such that $D$ is isomorphic to the algebra over $F$ with basis
$1,i,j, ij$ satisfying the relations
$$
i^2=\fa,\quad j^2=\fr,\quad ij=-ji
$$
(see Lemma \ref{lemHab}). Next, one needs to find an $A$-basis of
$\La$ in terms of $\{1,i,j,ij\}$. Although our method works for
general $A$-orders, to simplify the exposition, we restrict the
consideration to
$$
\La=A\oplus A i\oplus A j\oplus A ij
$$
which has the simplest presentation in terms of the basis
$\{1,i,j,ij\}$. Intrinsically, this is the level-$\fa$ Eichler order
in $D$. In this case,
$$
\G=\left\{a+bi+cj+dij\ \bigg|\ \begin{matrix} a,b,c,d\in A\\
a^2-\fa b^2-\fr c^2+\fa\fr d^2\in \F_q^\times\end{matrix}\right\}.
$$
Now the problem is to find finitely many quadruples $(a_n, b_n, c_n,
d_n)\in A^4$, $1\leq n\leq N$ for some $N$, such that
$\gamma_n=a_n+b_ni+c_nj+d_nij\in \La$ are in $\G$ and generate $\G$.
Finding minimal explicit sets of generators is very difficult. In
fact, for the analogous problem over $\Q$ only in finitely many
cases an explicit minimal set of generators is known, cf. \cite{AB}.
(See \cite{PapLocProp} for some examples in the function field
case.) Instead of trying to find minimal sets of generators for
$\G$, we look for possibly larger sets which are easy to describe.
\begin{defn}
Put $\deg(0)=0$ and for $\la=a+bi+cj+dij\in \La$ define
$$
\norm{\la}=\max(\deg(a), \deg(b), \deg(c),\deg(d)).
$$
\end{defn}

The main result of the paper is the following:

\begin{thm}\label{thm1} The finite subset
$$
S=\{\gamma\in \G\ |\ \norm{\gamma}\leq 4\deg(\fa\fr)+6\}
$$
generates $\G$.
\end{thm}

To prove Theorem \ref{thm1} we use the action of $\G$ on the
Bruhat-Tits tree $\cT$ of $\PGL_2(K)$. The key is to quantify the
discontinuity of the action of $\G$ on $\cT$. More precisely, for a
ball $\cT_B$ of radius $B$ around a fixed vertex and $\gamma\in \G$,
we show that $\gamma \cT_B\cap \cT_B=\emptyset$ once
$\norm{\gamma}>2B$. To deduce Theorem \ref{thm1} from this result,
one needs strong bounds on the diameter of the quotient graph $\G\bs
\cT$. Such a bound follows from the property of $\G\bs \cT$ being
covered by a Ramanujan graph.

A problem of similar nature over $\Q$ was considered by Chalk and
Kelly in \cite{CK}, although the bound they obtain is exponentially
worse than ours. The approach in \cite{CK} is analytic in nature and
relies on the study of isometric circles of $\G$ (see also
\cite{Johansson}).

\begin{rem}
Theorem \ref{thm1} raises the following interesting question: What
is the minimal $\sigma$ independent of $\G$ such that there exists a
constant $\delta$ (also independent of $\G$) with the property that
the set
$$
\{\gamma\in \G\ |\ \norm{\gamma}\leq \sigma \deg(\fa\fr)+\delta\}
$$
generates $\G$? The theorem gives $\sigma \leq 4$. We have a trivial
lower bound $1/4\leq \sigma$, since according to Proposition
\ref{prop2.5} the cardinality of a minimal set of generators for
$\G$ is approximately $q^{\deg(\fa\fr)}$.
\end{rem}

\begin{rem}
After this article was essentially completed, Ralf Butenuth informed
me that he obtained a result similar to Theorem \ref{thm1} by a
different method.
\end{rem}

\section{Arithmetic of quaternion algebras}\label{Sec3} In this section we
recall some facts about quaternion algebras. The standard reference
for this material is \cite{Vigneras}.

Let $D$ be a \textit{quaternion algebra} over $F$, i.e., a
$4$-dimensional $F$-algebra with center $F$ which does not possess
non-trivial two-sided ideals. A quaternion algebra is either a
division algebra or is isomorphic to the algebra of $2\times 2$
matrices. If $L$ is a field containing $F$, then $D\otimes_F L$ is a
quaternion algebra over $L$. Let $x\in |F|$ and denote
$D_x:=D\otimes_F F_x$. We say that $D$ \textit{ramifies} (resp.
\textit{splits}) at $x$ if $D_x$ is a division algebra (resp. is
isomorphic to $\M_2(F_x)$). Let $R\subset |F|$ be the set of places
where $D$ ramifies. It is known that $R$ is a finite set of even
cardinality, and conversely, for any choice of a finite set
$R\subset |F|$ of even cardinality there is a unique, up to an
isomorphism, quaternion algebra ramified exactly at the places in
$R$. In particular, $D\cong \M_2(F)$ if and only if $R=\emptyset$.

Explicitly quaternion algebras can be given as follows.  For $a,
b\in F^\times$, let $H(a,b)$ be the $F$-algebra with basis $1,i,j,
ij$ (as an $F$-vector space), where $i,j$ satisfy
$$
i^2=a,\quad j^2=b,\quad ij=-ji.
$$
$H(a,b)$ is a quaternion algebra, and any quaternion algebra $D$ is
isomorphic to $H(a,b)$ for some $a,b\in F^\times$ (although $a$ and
$b$ are not uniquely determined by $D$, e.g. $H(a,b)\cong H(b,a)$).

From now on we assume that $D$ is a division algebra (equiv. $R\neq
\emptyset$). Let $L\neq F$ be a non-trivial field extension of $F$.
Then $L$ embeds into $D$, i.e., there is an $F$-isomorphism of $L$
onto an $F$-subalgebra of $D$, if and only if $[L:F]=2$ and places
in $R$ do not split in $L$.

There is a canonical involution $\alpha\mapsto \alpha'$ on $D$ which
is the identity on $F$ and satisfies $(\alpha\beta)'=\beta'\alpha'$.
The \textit{reduced trace} of $\alpha$ is
$\Tr(\alpha)=\alpha+\alpha'$; the \textit{reduced norm} of $\alpha$
is $\Nr(\alpha)=\alpha\alpha'$; the \textit{reduced characteristic
polynomial} of $\alpha$ is
$$f(x)=(x-\alpha)(x-\alpha')=x^2-\Tr(\alpha)x+\Nr(\alpha).$$
If $\alpha\not \in F$, then the reduced trace and norm of $\alpha$
are the images of $\alpha$ under the trace and norm of the quadratic
field extension $F(\alpha)/F$.

From now on we also assume that $\infty\not \in R$. For $x\in
|F|-\infty$, denote by $(x)$ the prime ideal of $A$ corresponding to
$x$. Let $f_x\in A$ be the monic generator of $(x)$, and let
$\fr=\prod_{x\in R}f_x$. Given $a, b\in A$, with $b$ irreducible and
coprime to $a$, let
$$
\left(\frac{a}{b}\right)=\left\{
                           \begin{array}{ll}
                             1, & \hbox{if $a$ is a square mod $(b)$} \\
                             -1, & \hbox{otherwise}
                           \end{array}
                         \right.
$$
be the Legendre symbol. Let
$$
\Odd(R)=\left\{
         \begin{array}{ll}
           1, & \hbox{if $\deg(x)$ is odd for all $x\in R$;} \\
           0, & \hbox{otherwise.}
         \end{array}
       \right.
$$

\begin{lem}\label{lemHab}\hfill
\begin{enumerate}
\item Suppose $\Odd(R)=0$. There is a monic irreducible polynomial $\fa\in A$ of even degree
which is coprime to $\fr$ and satisfies
$$
\left(\frac{\fa}{f_x}\right)=-1\text{ for all }x\in R.
$$
For such $\fa$, $D\cong H(\fa, \fr)$.
\item Suppose $\Odd(R)=1$. Let $\xi\in \F_q$
be a non-square. Then $D\cong H(\xi, \fr)$.
\end{enumerate}
\end{lem}
\begin{proof} The proof of this lemma is quite standard; we give the
details for completeness.

(1) By the Chinese Remained Theorem \cite[Prop. 1.4]{Rosen}, there
exists $a\in A$ such that $(a/f_x)=-1$ for all $x\in R$. Consider
the set of polynomials $\{a+\fr b\ |\ b\in A\}$. By \cite[Thm.
4.8]{Rosen}, this set contains irreducible monic polynomials $\fa$
of even degrees. Fix such $\fa$. It is clear that $\fa$ satisfies
$(\fa/f_x)=-1$ for all $x\in R$. Next, by the Reciprocity Law
\cite[Thm. 3.3]{Rosen}
$$
\left(\frac{f_x}{\fa}\right)=(-1)^{\frac{q-1}{2}\deg(\fa)\deg(f_x)}\left(\frac{\fa}{f_x}\right).
$$
Since $\deg(\fa)$ is even, the right hand-side is equal to $-1$.
Hence
\begin{equation}\label{eq-frfa}
\left(\frac{\fr}{\fa}\right)=\prod_{x\in
R}\left(\frac{f_x}{\fa}\right)=(- 1)^{\# R}=1.
\end{equation}

Now we show that $D\cong H(\fa, \fr)$. It is enough to check that
$H(\fa,\fr)$ is ramified exactly at the places in $R$. For this we
need to show that the Hilbert symbol $(\fa,\fr)_x$ is $-1$ if and
only if $x\in R$, cf. \cite[p. 32]{Vigneras}. By \cite[p.
217]{SerreLF}, for $x\in |F|-(\fa)-\infty$, $(\fa,\fr)_x=1$ if and
only if $\fa^{\ord_x(\fr)}$ is a square modulo $(x)$. Now
$\ord_x(\fr)=0$ if $x\in |F|-R-\infty$, and $\ord_x(\fr)=1$ if $x\in
R$. Observe that the image of $\fa$ is not a square in $\F_x$ for
$x\in R$ by the choice of $\fa$. Therefore, $H(\fa, \fr)$ is
ramified at the places in $R$ and is unramified at
$|F|-R-\infty-(\fa)$. By the same argument, $H(\fa,\fr)$ is
unramified at $(\fa)$, since $\fr^{\ord_{(\fa)}(\fa)}=\fr$ is a
square modulo $(\fa)$ by (\ref{eq-frfa}). Finally, $H(\fa, \fr)$ is
unramified at $\infty$ since the number of places where a quaternion
algebra ramifies is even.

(2) Note that $\xi$ in not a square in $\F_x$, $x\in R$, since
$\deg(x)$ is odd. Now apply the argument in the previous paragraph.
\end{proof}

\begin{defn}
Let $\cR$ be a Dedekind domain with quotient field $L$ and let $B$
be a quaternion algebra over $L$. For any finite dimensional
$L$-vector space $V$, an \textit{$\cR$-lattice} in $V$ is a finitely
generated $\cR$-submodule $M$ in $V$ such that $L\otimes_\cR M\cong
V$. An \textit{$\cR$-order} in $B$ is a subring $\La$ of $B$, having
the same unity element as $B$, and such that $\La$ is an
$\cR$-lattice in $B$. A \textit{maximal $\cR$-order} in $B$ is an
$\cR$-order which is not contained in any other $\cR$-order in $B$.
\end{defn}

Let $\La$ be an $A$-order in $D$. It is known that $\La$ is maximal
if and only if $\La_x:=\La\otimes_A \cO_x$ is a maximal
$\cO_x$-order in $D_x$ for all $x\in |F|-\infty$. A maximal
$\cO_x$-order in $D_x$ is unique if $x\in R$ - it is the integral
closure of $\cO_x$ in $D_x$. On the other hand, for $x\not\in R$,
$\La_x$ is maximal if and only if there is an invertible element
$u\in \M_2(F_x)$ such that $u\La_x u^{-1}=\M_2(\cO_x)$.

\begin{defn}
Suppose $\fn\in A$ is square-free and coprime to $\fr$. $\La$ is an
\textit{Eichler order of level-$\fn$} if $\La_x$ is maximal for all
$x\in R$, and for $x\in |F|-R-\infty$ it is isomorphic to the
subring of $\M_2(\cO_x)$ given by the matrices
$$
\left\{ \begin{pmatrix} a & b \\ \fn c & d\end{pmatrix}\ |\
a,b,c,d\in \cO_x\right\}.
$$

Suppose $\La=Ae_1\oplus Ae_2\oplus Ae_3\oplus Ae_4$, where
$e_1,\dots, e_4$ is a basis of $D$ as an $F$-vector space. The
\textit{discriminant} of $\La$ is the ideal of $A$ generated by
$\det(\Tr(e_ie_j))_{i,j}$. It is known that the discriminant of any
order is divisible by $(\fr)^2$. Moreover, the maximal orders are
uniquely characterized by the fact that their discriminant is
$(\fr)^2$, and the level-$\fn$ Eichler orders are uniquely
characterized by the fact that their discriminants are equal to
$(\fn\fr)^2$. By a theorem of Eichler, since $D$ splits at $\infty$,
all maximal $A$-orders are conjugate in $D$; the same is true also
for the level-$\fn$ Eichler orders.
\end{defn}

\begin{defn}\label{defSO}
The order $\La=A\oplus A i\oplus Aj\oplus Aij$ in $H(\fa,\fr)$ will
be called the \textit{standard order}. By computing its
discriminant, we see that $\La$ is a level-$\fa$ Eichler order. In
particular, $\La$ is maximal if and only if $\fa\in \F_q^\times$.
\end{defn}

Given an $A$-order $\La$, the group $\La^\times$ of its invertible
elements consists of
$$
\{\la\in \La\ |\ \Nr(\la)\in \F_q^\times\}.
$$
If $\la\in \La^\times$ is a torsion element, then it is algebraic
over $\F_q$. This easily implies that $\la$ is torsion if and only
if $\Tr(\la)\in \F_q$; such elements will be called
\textit{elliptic}. An element $\la\in \La^\times$ which is not
elliptic will be called \textit{hyperbolic}.

\begin{lem}
If $\la$ is hyperbolic, then its image in $\GL_2(K)$ under an
embedding $D\hookrightarrow \M_2(K)$ has two distinct $K$-rational
eigenvalues.
\end{lem}
\begin{proof}
The reduced characteristic polynomial of $\la$ is
$h_\la:=x^2+\Tr(\la)x+\kappa$, where $\kappa\in \F_q^\times$. Since
$\la\not\in F^\times$, $F(\la)$ is quadratic, and therefore $h_\la$
is irreducible. Next, since $s:=\Tr(\la)\in A$ has degree $\geq 1$,
$s^2-4\kappa$ is a non-zero polynomial of even degree whose leading
coefficient is a square. This implies that $h_\la$ splits over $K$
and has distinct roots.
\end{proof}

Denote
$$
g(R)=1+\frac{1}{q^2-1}\prod_{x\in
R}(q^{\deg(x)}-1)-\frac{q}{q+1}\cdot 2^{\# R-1}\cdot \Odd(R).
$$

Let $\La$ be the standard order in $H(\fa,\fr)$, where $\fa$ is a
monic irreducible polynomial of even degree if $\Odd(R)=0$, and
$\fa=\xi$ is a non-square in $\F_q$ if $\Odd(R)=1$, cf. Lemma
\ref{lemHab}. Denote $\G:=\La^\times$. Note that $\F_q^\times$ is in
the center of $\G$.

\begin{prop}\label{prop2.5}\hfill
\begin{enumerate}
\item $\G/\F_q^\times$ has non-trivial torsion if and only
if $\Odd(R)=1$.
\item Suppose $\Odd(R)=1$. $\G$ can be generated by
$g(R)+2^{\#R-1}$ elements.
\item Suppose $\Odd(R)=0$. $\G/\F_q^\times$ is a free group of rank
$$
1+(q^{\deg(\fa)}+1)(g(R)-1).
$$
\end{enumerate}
\end{prop}
\begin{proof}
(1) and (2) follow from \cite[Thm. 5.7]{PapLocProp}. (3) Let
$\Upsilon$ be a maximal order containing $\La$. Denote
$\G'=\Upsilon^\times$. By \cite[Thm. 5.7]{PapLocProp},
$\G'/\F_q^\times$ is a free group of rank $g(R)$. It is not hard to
show that $[\G'/\F_q^\times:\G/\F_q^\times]=q^{\deg(\fa)}+1$, cf.
\cite[p. 212]{Miyake}. Now the claim that $\G/\F_q^\times$ is a free
group of given rank follows from Schreier's theorem \cite[p.
29]{SerreT}.
\end{proof}

\section{Geometry on the Bruhat-Tits tree}\label{sGBT}

We start by recalling some of the terminology from \cite{SerreT}.
Let $\cG$ be an (oriented) connected graph; see \cite[Def. 1, p.
13]{SerreT}. We denote by $\Ver(\cG)$ and $\Ed(\cG)$ the sets of
vertices and oriented edges of $\cG$, respectively. For $e\in
\Ed(\cG)$, $o(e), t(e)\in \Ver(\cG)$ and $\bar{e}\in \Ed(\cG)$
denote its origin, terminus and inversely oriented edge. We will
assume that for any $v\in \Ver(\cG)$ the number of edges $e$ with
$o(e)=v$ is finite; this number is the \textit{degree} of $v$. $\cG$
is \textit{$m$-regular} if every vertex in $\cG$ has degree $m$. The
\textit{distance} $d(v,w)$ between $v,w\in \Ver(\cG)$ in $\cG$ is
the obvious combinatorial distance, i.e., the number of edges in a
shortest path without backtracking connecting $v$ and $w$. The
\textit{diameter} $D(\cG)$ of a finite graph $\cG$ is the maximum of
the distances between its vertices. A graph in which a path without
backtracking connecting any two vertices $v$ and $w$ is unique is
called a \textit{tree}; the unique path between $v, w$ is called
\textit{geodesic}.

Let $\G$ be a group acting on a graph $\cG$, i.e., $\G$ acts via
automorphisms. We say that $v, w\in \Ver(\cG)$ are
$\G$-\textit{equivalent} if there is $\gamma \in \G$ such that
$\gamma v=w$. $\G$ acts with \textit{inversion} if there is
$\gamma\in \G$ and $e\in \Ed(\cG)$ such that $\gamma e=\bar{e}$. If
$\G$ acts without inversion, then we have a natural quotient graph
$\G\bs\cG$ such that $\Ver(\G\bs \cG)=\G\bs \Ver(\cG)$ and
$\Ed(\G\bs \cG)=\G\bs \Ed(\cG)$, cf. \cite[p. 25]{SerreT}.

Recall the notation $K:=F_\infty$ and $\cO:=\cO_\infty$. Let $V$ be
a $2$-dimensional $K$-vector space. Let $\La$ be an $\cO$-lattice in
$V$. For any $x\in K^\times$, $x\La$ is also a lattice. We call
$\La$ and $x\La$ equivalent lattices. The equivalence class of $\La$
is denoted by $[\La]$.

Let $\cT$ be the graph whose vertices $\Ver(\cT)=\{[\La]\}$ are the
equivalence classes of lattices in $V$, and two vertices $[\La]$ and
$[\La']$ are adjacent if we can choose representatives $L\in [\La]$
and $L'\in [\La']$ such that $L'\subset L$ and $L/L'\cong \F_q$. One
shows that $\cT$ is an infinite $(q+1)$-regular tree. This is the
\textit{Bruhat-Tits tree} of $\PGL_2(K)$.

Fix a vertex $[\La]$ of $\cT$. The set of vertices of $\cT$ at
distance $n$ from $[\La]$ is in natural bijection with
$\P^1(L/\pi^nL)$. An \textit{end} of $\cT$ is an equivalence class
of half-lines, two half-lines being equivalent if they differ in a
finite graph. Taking the projective limit over $n$, we get a
bijection
$$
\partial \cT:=\text{set of ends of }\cT\cong \P^1(\cO)=\P^1(K),
$$
which is independent of the choice of $[\La]$. Given two vectors
$f_1, f_2$ spanning $V$, we denote $[f_1, f_2]=[\cO f_1\oplus \cO
f_2]$. $\GL_2(K)$, as the group of $K$-automorphisms of $V$, acts on
the Bruhat-Tits tree $\cT$ via $g[f_1, f_2]=[gf_1, gf_2]$. It is
easy to check that this action preserves the distance between any
two vertices, and the induced action on $\partial\cT$ agrees with
the usual action of $\GL_2(K)$ on $\P^1(K)$ through fractional
linear transformations.

Fix the standard basis $e_1=\begin{pmatrix} 1\\ 0 \end{pmatrix}$,
$e_2=\begin{pmatrix} 0\\ 1 \end{pmatrix}$ of $V$, and let $O:=[e_1,
e_2]$. Since $\GL_2(K)$ acts transitively on the vertices of $\cT$
and the stabilizer of $O$ is $\GL_2(\cO)K^\times$, we have a
bijection
\begin{align}\label{eq-map}
\GL_2(K)/\GL_2(\cO)K^\times &\overset{\sim}{\To} \Ver(\cT)\\
g &\nonumber \mapsto g\cdot O.
\end{align}
Using the Iwasawa decomposition, one easily sees that the set of
matrices
\begin{equation}\label{eq-setM}
\left\{\begin{pmatrix} \pi^k & u \\ 0 & 1\end{pmatrix}\ \bigg|\
\begin{matrix} k\in \Z\\ u\in K,\ u\ \mod\ \pi^k\cO\end{matrix}\right\}
\end{equation}
is a set of representatives for $\Ver(\cT)$; see \cite[p.
370]{Improper}. The map (\ref{eq-map}) becomes
$$
\begin{pmatrix} \pi^k & u \\ 0 & 1\end{pmatrix}\mapsto [\pi^k e_1,
ue_1+e_2].
$$
Note that under this bijection the identity matrix corresponds to
$O$. We say that a matrix $M\in \GL_2(K)$ is in \textit{reduced
form} if it belongs to the set of matrices in (\ref{eq-setM}). For
two matrices $M,M'\in\GL_2(K)$, we write $M\sim M'$ if they
represent the same vertex in $\cT$.

\begin{lem}\label{lem2.1}
The distance between $\begin{pmatrix} \pi^k & u \\ 0 &
1\end{pmatrix}$ $($in reduced form$)$ and $\begin{pmatrix} 1 & 0 \\
0 & 1\end{pmatrix}$ is
$$
\left\{
  \begin{array}{ll}
    |k|, & \hbox{if $u=0$ or $\ord(u)\geq 0$;} \\
    k-2\cdot \ord(u), & \hbox{if $u\neq 0$ and $\ord(u)<0$.}
  \end{array}
\right.
$$
$($Note that for a matrix in reduced form $k>\ord(u)$.$)$
\end{lem}
\begin{proof}
This is an easy calculation.
\end{proof}

Given two distinct points $P,Q\in \P^1(K)$, there is a unique path
in $\cT$, without backtracking and infinite in both directions,
whose ends are $P$ and $Q$; this is the \textit{geodesic} $\cA(P,Q)$
connecting the two boundary points of $\cT$. For example, the
geodesic connecting $0=(0:1)$ and $\infty=(1:0)$ is the subgraph of
$\cT$ with vertices $\left\{\begin{pmatrix} \pi^k & 0 \\ 0 &
1\end{pmatrix}\ \bigg|\ k\in \Z\right\}$.

Assume $\gamma\in \GL_2(K)$ has two distinct $K$-rational
eigenvalues $a$ and $b$. The eigenvectors corresponding to $a$ and
$b$ can be regarded as two well-defined points on $\P^1(K)$ - if
$\begin{pmatrix} x\\ y
\end{pmatrix}$ is an eigenvector, then the corresponding point is $(x:y)$.
Let $\cA(\gamma)$ be the geodesic connecting these points. Suppose
$\begin{pmatrix} x_1\\ y_1 \end{pmatrix}$ and $\begin{pmatrix} x_2\\
y_2 \end{pmatrix}$ are eigenvectors corresponding to $a$ and $b$,
respectively. Since $\begin{pmatrix} x_1 & x_2 \\ y_1 &
y_2\end{pmatrix}$ maps $\infty$ to $(x_1:y_1)$ and $0$ to
$(x_2:y_2)$,
\begin{equation}\label{eq-geod}
\Ver(\cA(\gamma))=\left\{\begin{pmatrix} x_1 & x_2 \\ y_1 &
y_2\end{pmatrix}\begin{pmatrix} \pi^k & 0 \\ 0 & 1\end{pmatrix}\ |\
k\in \Z\right\}.
\end{equation}
The \textit{distance} from $\cA(\gamma)$ to $O$ is the minimum of
the distances from the vertices on $\cA(\gamma)$ to $O$.

\begin{lem}\label{lem-geod} The action of $\gamma$ on $\cT$ induces translation on
$\cA(\gamma)$ of amplitude $|\ord(a)-\ord(b)|$.
\end{lem}
\begin{proof}
This is easy to see after choosing the eigenvectors of $\gamma$ as a
basis of $V$.
\end{proof}

\begin{lem}\label{lem-BB}
Suppose $x_1y_1\neq 0$ and $x_2y_2\neq 0$. Denote $x:=x_1/y_1$ and
$y:=x_2/y_2$. Suppose $\ord(x)\geq B$ and $\ord(y)\geq B$, or
$\ord(x)\leq -B$ and $\ord(y)\leq -B$ for some $B\geq 0$. Then the
distance from $\cA(\gamma)$ to $O$ is at least $B$.
\end{lem}
\begin{proof} Note that $x\neq y$, as $\begin{pmatrix} x\\ 1
\end{pmatrix}$ and $\begin{pmatrix} y\\ 1 \end{pmatrix}$ are
eigenvectors for $a$ and $b$. Without loss of generality we can
assume $\ord(x)\geq \ord(y)$. The geodesic $\cA(\gamma)$ has
vertices
$$
\begin{pmatrix} x\pi^k & y \\ \pi^k & 1\end{pmatrix}, \quad k\in \Z.
$$
First, consider the case when $k>0$. Then $\begin{pmatrix} 1 & 0 \\
-\pi^k & 1\end{pmatrix}\in \GL_2(\cO)$, so by multiplying
$\begin{pmatrix} x\pi^k & y \\ \pi^k & 1\end{pmatrix}$ by this
matrix from the right we see that
$$
\begin{pmatrix} x\pi^k & y \\ \pi^k &
1\end{pmatrix} \sim \begin{pmatrix} (x-y)\pi^k & y \\ 0 &
1\end{pmatrix}.
$$
On the other hand, $\ord((x-y)\pi^k)\geq k+\ord(y)>\ord(y)$, so the
resulting matrix is in reduced form. The distance from this matrix
to $O$ is
$$
\ord((x-y)\pi^k)\geq k+\ord(y), \quad \text{if }\ord(y)\geq 0,
$$
and
$$
\ord((x-y)\pi^k)-2\ord(y)\geq k-\ord(y), \quad \text{if }\ord(y)< 0.
$$
In either case, we conclude that the distance is at least
$1+|\ord(y)|\geq 1+B$.

The matrix with $k=0$ is adjacent to the matrix with $k=1$, so from
the previous paragraph we conclude that the corresponding matrix and
$O$ are at a distance at least $(1+B)-1=B$.

Now suppose $k<0$. Then
$$
\begin{pmatrix} x\pi^k & y \\ \pi^k &
1\end{pmatrix} \sim \begin{pmatrix} x & \pi^{-k}y \\ 1 &
\pi^{-k}\end{pmatrix}\sim \begin{pmatrix} \pi^{-k}(y-x) & x \\ 0 &
1\end{pmatrix}.
$$

If $\ord(x)=\ord(y)$, then $\ord(x-y)\geq \ord(y)=\ord(x)$. Hence
$-k+\ord(x-y)>\ord(x)$, and the above matrix is in reduced form. The
distance from $O$ is
$$
-k+\ord(x-y)\geq -k+\ord(y)>B, \quad \text{if }\ord(y)\geq 0,
$$
or
$$
-k+\ord(x-y)-2\ord(y)\geq -k-\ord(y)>B, \quad \text{if }\ord(y)< 0.
$$

If $\ord(x)>\ord(y)$, then $\ord(x-y)= \ord(y)$. If
$\ord(\pi^{-k}(x-y))=-k+\ord(y)\leq \ord(x)$, then
$$
\begin{pmatrix} \pi^{-k}(y-x) & x \\ 0 &
1\end{pmatrix}\sim \begin{pmatrix} \pi^{-k}(y-x) & 0 \\ 0 &
1\end{pmatrix},
$$
so the distance is $|-k+\ord(y)|$. If $\ord(y)\geq B$, then this
last quantity is obviously $\geq 1+B$. On the other hand, if
$\ord(y)\leq -B$, then, due to the assumption of the lemma,
$-k+\ord(y)\leq \ord(x)\leq -B$. Thus, $|-k+\ord(y)|\geq B$ Finally,
if $-k+\ord(y)> \ord(x)$, then the distance is
$$
-k+\ord(x-y)= -k+\ord(y)>B, \quad \text{if }\ord(x)\geq 0,
$$
or
$$
-k+\ord(x-y)-2\ord(x)= -k+\ord(y)-2\ord(x)>-\ord(x)\geq B,
$$
if $\ord(x)< 0$.
\end{proof}

\begin{rem}
The inequalities $\ord(x)\gg 0$, $\ord(y)\gg 0$ (resp. $\ord(x)\ll
0$, $\ord(y)\ll 0$) essentially mean that both $x$ and $y$ are in a
small neighborhood of $0$ (resp. $\infty$). Now one can visualize
the previous lemma as follows: for a sufficiently small interval on
$\R$ the geodesic in the hyperbolic upper half-plane $\cH$
connecting any two distinct points in that interval is far from a
fixed point in $\cH$.
\end{rem}

\begin{notn} For $B\geq 0$, let $\cT_B$ be the finite subtree of $\cT$ with
set of vertices
$$
\Ver(\cT_B)=\{v\in \Ver(\cT)\ |\ d(v,O)\leq B\}.
$$
\end{notn}

\begin{lem}\label{lem-empty} Let $n$ be the amplitude of translation
with which $\gamma$ acts on $\cA(\gamma)$. Let $m$ be the distance
from $\cA(\gamma)$ to $O$. If $2m+n> 2B$, then $\gamma\cT_B\cap
\cT_B=\emptyset$.
\end{lem}
\begin{proof} Let $P\in \cA(\gamma)$ be the vertex closest to $O$.
Then $m=d(P,O)$ and
$$
d(O, \gamma O)=d(O,P)+d(P, \gamma P)+d(\gamma P, \gamma O)=2m+n,
$$
cf. \cite[Prop. 24 (iv), p. 63]{SerreT}. On the other hand, if
$\gamma\cT_B\cap \cT_B\neq \emptyset$, then $d(O, \gamma O)\leq 2B$.
Thus $2m+n\leq 2B$.
\end{proof}

\begin{prop}\label{lemDiam}
Assume $\G$ acts without inversion on $\cT$ and $\G\bs \cT$ is
finite. Let $B=D(\G\bs \cT)$. Let $S$ denote the set of $\gamma\in
\G$ such that $\gamma\cT_B\cap \cT_B\neq \emptyset$. Then $S$
generates $\G$.
\end{prop}
\begin{proof} Let $\bar{O}\in \G\bs \cT$ be the image of $O$. Let
$\bar{v}\in \Ver(\G\bs \cT)$. Consider a path $P$ of shortest length
connecting $\bar{O}$ and $\bar{v}$. Obviously $P$ is a subtree of
$\G\bs \cT$, hence by \cite[Prop. 14, p. 25]{SerreT} lifts to $\cT$.
This implies that there is a vertex $v$ in $\cT$ which maps to
$\bar{v}$ and $d(O,v)=d(\bar{O},\bar{v})$. In particular, $v\in
\cT_B$, so $\cT_B$ surjects onto $\G\bs \cT$ under the quotient map
$\cT\to \G\bs \cT$.

Let $\mathrm{real}(\cT)$ be the \textit{realization} of $\cT$; see
\cite[p. 14]{SerreT}. Recall that $\mathrm{real}(\cT)$ is a
CW-complex where each edge of $\cT$ is homeomorphic to the interval
$[0,1]\subset \R$. Let $U$ be the open subset of
$\mathrm{real}(\cT)$ consisting of points at distance $< 1/3$ from
$\mathrm{real}(\cT_B)$. Then $\gamma U\cap U\neq \emptyset$ if and
only if $\gamma\cT_B\cap \cT_B$, and $U\to \mathrm{real}(\G\bs \cT)$
is surjective. The claim of the proposition now follows from
\cite[(1), p. 30]{SerreT}.
\end{proof}

Let $\G$ be as in Proposition \ref{prop2.5}. $\G$ acts naturally on
$\cT$ (see Section \ref{SecHE}). The action is without inversion and
the quotient graph $\G\bs\cT$ is finite; see \cite[Lem.
5.1]{PapLocProp}.

\begin{lem}\label{lem3.7}
Let $V:=\#\Ver(\G\bs \cT)$.
\begin{enumerate}
\item If $\Odd(R)=1$, then
$$
V= \frac{2}{(q-1)(q^2-1)}\prod_{x\in
R}(q^{\deg(x)}-1)+\frac{q}{q+1}2^{\#R-1}.
$$
\item If $\Odd(R)=0$, then
$$
V= \frac{2(q^{\deg(\fa)}+1)}{(q-1)(q^2-1)}\prod_{x\in
R}(q^{\deg(x)}-1).
$$
\end{enumerate}
\end{lem}
\begin{proof}
(1) This follows from \cite[Thm. 5.5]{PapLocProp}. (2) Since
$\F_q^\times\lhd \G$ acts trivially on $\cT$ and $\G/\F_q^\times$ is
a free group, $\G\bs \cT$ is $(q+1)$-regular. Thus, if we denote by
$E$ the number of (non-oriented) edges of $\G\bs \cT$, then
$E=(q+1)V/2$. On the other hand, by \cite[Thm. $4'$, p. 27]{SerreT},
the rank of $\G/\F_q^\times$ is equal to $E+1-V$. Now the expression
for $V$ follows from Proposition \ref{prop2.5}.
\end{proof}

\begin{lem}\label{LPS} Let $\cG$ be an $m$-regular Ramanujan graph on $n$
vertices. Then $$D(\cG)\leq 2\log_{m-1}(n)+\log_{m-1}(4).$$
\end{lem}
\begin{proof}
For the definition of Ramanujan graphs see \cite{Lubotzky}. The
claim of the lemma is part of Proposition 7.3.11 in \textit{loc.
cit.}
\end{proof}

\begin{prop}\label{prop3.9}
$D(\G\bs\cT)\leq 2\deg(\fa\fr)+3$.
\end{prop}
\begin{proof}
Let $I=(T)\lhd A$ be the ideal generated by $T$. Let $\Upsilon$ be a
maximal order containing $\La$. Denote by $\G(T)$ the principal
level-$I$ congruence subgroup of $\Upsilon^\times$, cf. \cite{LSV}.
Let $\G':=\G\cap \G(I)$. $\G\bs\cT$ is naturally a quotient of
$\G'\bs\cT$, so $$D(\G\bs\cT)\leq D(\G'\bs\cT).$$ Moreover, since
$[\G:\G']<q^4$, we have $V'< q^4 V$, where $V':=\#\Ver(\G'\bs\cT)$.

By \cite[Thm. 1.2]{LSV}, $\G'\bs\cT$ is a $(q+1)$-regular Ramanujan
graph, so the previous paragraph and Lemma \ref{LPS} give the bound
$$
D(\G\bs\cT)\leq 2\log_q(q^4 V)+1.
$$
Now the proposition follows from the formulae in Lemma \ref{lem3.7}.
(Note that $\deg(\fr)=\sum_{x\in R}\deg(x)$, and $\deg(\fa)=0$ when
$\Odd(R)=1$.)
\end{proof}

\section{Proof of Theorem \ref{thm1}}\label{SecHE}

Let $(\fa, \fb)=(\fa', \fr)$ (resp. $(\fa, \fb)=(\fr, \xi)$) if
$\Odd(R)=0$ (resp. $\Odd(R)=1$), with $\fa',\xi$ chosen as in Lemma
\ref{lemHab}. Let $\La$ be the standard order in $H(\fa, \fb)$ and
$\G:=\La^\times$. By Proposition \ref{prop2.5}, this is a finitely
generated group, and we would like to find an explicit set of
generators. We start by embedding $H(\fa, \fb)$ into $\M_2(K)$. The
map
$$
i\mapsto \begin{pmatrix} \sqrt{\fa} & 0 \\ 0 &
-\sqrt{\fa}\end{pmatrix}, \quad j\mapsto \begin{pmatrix} 0 & 1\\ \fb
& 0\end{pmatrix}
$$
gives an embedding $H(\fa, \fb)\hookrightarrow \M_2(K)$. Indeed,
since $\fa\in A$ is monic and has even degree, the equation
$x^2=\fa$ has a solution in $K$. Thus, the above matrices are indeed
in $\M_2(K)$. It remains to observe that the given matrices satisfy
the same relations as $i$ and $j$. Under this embedding $\G$ is the
subgroup of $\GL_2(K)$ consisting of matrices
$$
\G=\left\{\begin{pmatrix} a + b\sqrt{\fa} & c + d\sqrt{\fa}
\\ \fb(c - d\sqrt{\fa}) & a - b\sqrt{\fa} \end{pmatrix}\ \bigg|\
\begin{matrix} a,b,c,d,\in A \\ \det=a^2-b^2\fa-c^2\fb+d^2\fa\fb
\in \F_q^\times\end{matrix}\right\}.
$$

\begin{prop}\label{prop5.3} Fix some $B\geq 0$.
If $\gamma\in \G$ is hyperbolic and satisfies $\norm{\gamma}>2B$,
then $\gamma \cT_B\cap \cT_B=\emptyset$.
\end{prop}
\begin{proof} To simplify the notation, we put $\deg(0)=0$.
The image of $\gamma$ in $\GL_2(K)$ is the matrix
$$
\begin{pmatrix} a + b\sqrt{\fa} & c + d\sqrt{\fa}
\\ \fb(c - d\sqrt{\fa}) & a - b\sqrt{\fa} \end{pmatrix}.
$$
Let $\alpha, \beta\in K$ be the eigenvalues of $\gamma$. We know
that $\Nr(\gamma)=\alpha\beta=:\kappa\in \F_q^\times$ and
$\Tr(\gamma)=\alpha+\beta=2a\not\in \F_q$. Thus,
$\beta=\kappa\alpha^{-1}$ and $\ord(\beta)=-\ord(\alpha)$. Without
loss of generality, we assume $\ord(\alpha)\geq 0$, so
$\ord(\beta)\leq 0$. Since $\deg(a)\geq 1$, $\ord(\beta)=\ord(a)\leq
-1$. Using Lemma \ref{lem-geod}, we conclude that $\gamma$ acts on
$\cA(\gamma)$ by translations with amplitude $2\deg(a)$. Let $m$ be
the distance from $\cA(\gamma)$ to $O$. Lemma \ref{lem-empty}
implies that if
\begin{equation}\label{eq-empty}
m+\deg(a)>B
\end{equation}
then $\gamma \cT_B\cap \cT_B=\emptyset$. In particular, if
$\deg(a)>B$, then $\gamma \cT_B\cap \cT_B=\emptyset$.

Suppose $\begin{pmatrix} 1\\
0\end{pmatrix}$ is an eigenvector.  Then $\fb(c - d\sqrt{\fa})=0$,
which forces $c=d=0$. Thus, we must have $a^2-b^2\fa\in
\F_q^\times$. If $\deg(b)>\deg(a)-\deg(\fa)/2$, then this is not
possible. It is easy to see that we
reach the same conclusion in the case when $\begin{pmatrix} 0\\
1\end{pmatrix}$ is an eigenvector.

Assume $\deg(b)> \deg(a)-\deg(\fa)/2$. Denote
$s:=\deg(b)+\deg(\fa)/2-\deg(a)>0$. From the previous paragraph we
know that there are eigenvectors for
$\alpha$ and $\beta$ of the form $\begin{pmatrix} x\\
1\end{pmatrix}$ and $\begin{pmatrix} y\\
1\end{pmatrix}$, with $x\neq 0$, $y\neq 0$. Now
\begin{align*}
& x(a + b\sqrt{\fa}) +(c + d\sqrt{\fa})=\alpha x \\
& x\fb(c - d\sqrt{\fa}) +(a - b\sqrt{\fa})=\alpha.
\end{align*}
From this we get
\begin{equation}\label{eq4.1}
x=-\frac{c + d\sqrt{\fa}}{a +
b\sqrt{\fa}-\alpha}=\frac{\alpha-a+b\sqrt{\fa}}{\fb(c -
d\sqrt{\fa})}.
\end{equation}
Similarly,
\begin{equation}\label{eq4.2}
y=-\frac{c + d\sqrt{\fa}}{a +
b\sqrt{\fa}-\kappa\alpha^{-1}}=\frac{\kappa\alpha^{-1}-a+b\sqrt{\fa}}{\fb(c
- d\sqrt{\fa})}.
\end{equation}
Hence
\begin{equation}\label{eq1Prop}
\frac{x}{y}=\frac{a + b\sqrt{\fa}-\kappa\alpha^{-1}}{a +
b\sqrt{\fa}-\alpha}
\end{equation}
and
\begin{equation}\label{eq2Prop}
x-y=\frac{(c + d\sqrt{\fa})(\kappa\alpha^{-1}-\alpha)}{(a +
b\sqrt{\fa}-\alpha)(a +
b\sqrt{\fa}-\kappa\alpha^{-1})}=x\frac{\alpha-\kappa\alpha^{-1}}{(a
+ b\sqrt{\fa}-\kappa\alpha^{-1})}.
\end{equation}
From our assumption $\deg(b)> \deg(a)-\deg(\fa)/2$ and
(\ref{eq1Prop}), we get
$$
\ord(x)=\ord(y)=:n.
$$
Similarly from (\ref{eq2Prop}), we get
$$
\ord(x-y)=n+s.
$$
By (\ref{eq-geod}), the geodesic $\cA(\gamma)$ has vertices
$$
\Ver(\cA(\gamma))=\left\{\begin{pmatrix} x\pi^k & y \\ \pi^k &
1\end{pmatrix}\ |\ k\in \Z\right\}.
$$
If $k\geq 0$, then
$$
\begin{pmatrix} x\pi^k & y \\ \pi^k &
1\end{pmatrix}\sim \begin{pmatrix} (x-y)\pi^k & y \\ 0 &
1\end{pmatrix}\sim \begin{pmatrix} \pi^{k+n+s} & y \\ 0 &
1\end{pmatrix}.
$$
Since $y\neq 0$ and $k+n+s>n=\ord(y)$, the last matrix is in reduced
form. By Lemma \ref{lem2.1}, the distance of this matrix from $O$ is
$k+|n|+s\geq s$. If $k<0$, then
$$
\begin{pmatrix} x\pi^k & y \\ \pi^k &
1\end{pmatrix}\sim \begin{pmatrix} x & y\pi^{-k} \\ 1 &
\pi^{-k}\end{pmatrix}\sim \begin{pmatrix} y\pi^{-k} & x \\
\pi^{-k} & 1\end{pmatrix}.
$$
A similar calculation shows that the distance from this matrix to
$O$ is again greater or equal to $s$. Now
$$
s+\deg(a)=\deg(b)+\deg(\fa)/2.
$$
If $\deg(b)>B-\deg(\fa)/2$, then this last quantity is greater than
$B$. Therefore, by (\ref{eq-empty}), $\gamma \cT_B\cap
\cT_B=\emptyset$. Note that if $\deg(a)\leq B$, then
$\deg(b)>B-\deg(\fa)/2$ implies $\deg(b)>\deg(a)-\deg(\fa)/2$.
Overall, we have shown that if $\deg(a)> B$ or
$\deg(b)>B-\deg(\fa)/2$, then $\gamma \cT_B\cap \cT_B=\emptyset$.

Now assume $\deg(a)\leq B$ and $\deg(b)\leq B-\deg(\fa)/2$, but
$\deg(c)> 2B$ or $\deg(d)> 2B$. From our earlier
discussion we know that $\begin{pmatrix} 1\\
0\end{pmatrix}$ or $\begin{pmatrix} 0\\
1\end{pmatrix}$ are not eigenvectors (since otherwise $c=d=0$). Now,
either $\ord(c+d\sqrt{\fa})<-2B$ or $\ord(c-d\sqrt{\fa})<-2B$ (and
only one of the inequalities holds due to $\Nr(\gamma)\in
\F_q^\times$). First, assume $\ord(c+d\sqrt{\fa})<-2B$. Since
$\ord(a + b\sqrt{\fa}-\alpha)\geq -B$ and $\ord(a +
b\sqrt{\fa}-\kappa\alpha^{-1})\geq -B$, from the first equality in
(\ref{eq4.1}) and (\ref{eq4.2}) we get $\ord(x)< -B$ and
$\ord(y)<-B$. Next, assume $\ord(c-d\sqrt{\fa})<-2B$. Since
$\ord(\alpha-a+b\sqrt{\fa})\geq -B$ and
$\ord(\kappa\alpha^{-1}-a+b\sqrt{\fa})\geq -B$, from the second
equality in (\ref{eq4.1}) and (\ref{eq4.2}) we get $\ord(x)> B$ and
$\ord(y)>B$. In either case, Lemma \ref{lem-BB} implies that the
distance from $\cA(\gamma)$ to $O$ is greater than $B$. As before,
from (\ref{eq-empty}) we conclude $\gamma \cT_B\cap
\cT_B=\emptyset$.
\end{proof}

\begin{prop}\label{prop4.5}
If $\gamma\in \G$ is elliptic and satisfies $\norm{\gamma}>B$, then
$\gamma \cT_B\cap \cT_B=\emptyset$.
\end{prop}
\begin{proof} To simplify the notation, we again put $\deg(0)=0$.
If $\gamma$ is elliptic and satisfies the inequality
of the proposition, then obviously $\gamma\not \in \F_q^\times$. By
Proposition \ref{prop2.5}, the existence of such $\gamma$ is
possible if and only if $\Odd(R)=1$. Hence, we can assume that
$$
\gamma=\begin{pmatrix} a + b\sqrt{\fr} & c + d\sqrt{\fr}
\\ \xi(c - d\sqrt{\fr}) & a - b\sqrt{\fr} \end{pmatrix}.
$$
Since $\gamma$ is elliptic, $a\in \F_q$. Assume $\deg(c)>B$.
Consider $\tau:=\gamma\cdot j\in \G$:
$$
\tau=\begin{pmatrix}  \xi(c + d\sqrt{\fr}) & a +
b\sqrt{\fr} \\
\xi(a - b\sqrt{\fr}) & \xi(c - d\sqrt{\fr}) \end{pmatrix}.
$$
Note that $\tau$ is hyperbolic since $\Tr(\tau)=2\xi c\not \in
\F_q$. Suppose $\gamma \cT_B\cap \cT_B\neq \emptyset$. Since $j$
fixes $O$, we get
$$
d(O,\tau O)=d(O, \gamma O)\leq 2B.
$$
On the other hand, $\tau$ acts on its geodesic by translation with
amplitude $2\deg(c)>2B$, so $d(O,\tau O)>2B$ - a contradiction.
Thus, from now on we can assume $\deg(c)\leq B$. If $\deg(b)> B$,
then $\deg(a^2-b^2\fr)>2B$. Since
$$
(a^2-b^2\fr)-\xi(c^2- d^2\fr) \in \F_q^\times,
$$
we must have $\deg(b)=\deg(d)=:s$. In this case we have
$$
\gamma=\begin{pmatrix} z_1\pi^{-s} & z_2 \pi^{-s}\\
z_3 \pi^{-s} & z_4\pi^{-s}
\end{pmatrix}\sim \begin{pmatrix} z_1 & z_2 \\
z_3  & z_4
\end{pmatrix},
$$
where $z_1,\dots, z_4\in \cO^\times$. Moreover, since
$\det(\gamma)\in \F_q^\times$,$$z_1z_4-z_2z_3\in
\pi^{2s}\cO^\times.$$ This implies
$$
\begin{pmatrix} z_1 & z_2 \\ z_3  & z_4 \end{pmatrix}
\sim \begin{pmatrix} \pi^{2s} & 0 \\ 0  & 1 \end{pmatrix}.
$$
The distance from this matrix to $O$ is obviously $2s$. On the other
hand, the matrix corresponding to the vertex $\gamma O$ under
(\ref{eq-setM}) is $\gamma$, so $d(O, \gamma O)=2s>2B$. The previous
argument works also under the assumption $\deg(d)> B$, and leads to
the same conclusion.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1}] Let $B=2\deg(\fa\fb)+3$. By Proposition \ref{prop3.9},
$D(\G\bs \cT)\leq B$. By Proposition \ref{prop5.3} and
\ref{prop4.5}, the set of elements $\gamma\in \G$ such that $\gamma
\cT_B\cap \cT_B\neq \emptyset$ is contained in $S$. Now the theorem
follows from Proposition \ref{lemDiam}.
\end{proof}

\subsection*{Acknowledgments} The work on this paper was started while I was visiting the
Department of Mathematics of Saarland University. I thank the
members of the department for their warm hospitality. I thank R.
Butenuth and E.-U. Gekeler for stimulating discussions.


TITLE: Frame-independent probability amplitude for Compton scattering in scalar QED
QUESTION [9 upvotes]: In scalar QED, the matrix element for compton scattering is the sum of three diagrams: the s-channel, t-channel and the seagull diagram. After simplifcation, making use of the orthogonality between photon polarization vectors and photon momenta, we get the following expression for the matrix element.
$$ \begin{align}
i\mathcal{M} &= -ie^{2} \left[ \frac{(2p+k)_\mu (2p'+k')_\nu}{(p+k)^2-m^2}+\frac{(2p'-k)_\mu (2p-k')_\nu}{(p-k')^2-m^2} -2\eta_{\mu\nu} \right]\epsilon^\mu \epsilon'^{\nu *} \\
&=-i2e^{2} \left[ \frac{p_\mu p'_\nu}{p.k}-\frac{p'_\mu p_\nu}{p.k'} -\eta_{\mu\nu} \right]\epsilon^\mu \epsilon'^{\nu *}
\end{align}
$$
where $p$ and $p'$ are the four momenta of the incoming and outgoing (scalar) electrons, while $k$ and $k'$ are the incoming and outgoing photons four momenta, which correspond with the polarization vectors $\epsilon $ and $ \epsilon'$. $\eta_{\mu \nu}$ is the metric tensor.
I want to find an expression for the probability amplitude, corresponding with this matrix element for unpolarized photons and which is frame independent. So I want to calculate
$$ \frac{1}{2}\sum_{\epsilon \epsilon'} \vert \mathcal{M} \vert^2=2e^{4}\left[ \frac{p_\mu p'_\nu}{p.k} - \frac{p'_\mu p_\nu}{p.k'} -\eta_{\mu \nu} \right]\left[ \frac{p_\alpha p'_\beta}{p.k}-\frac{p'_\alpha p_\beta}{p.k'} -\eta_{\alpha\beta} \right]\sum_{\epsilon \epsilon'}\epsilon^\mu \epsilon^{\alpha *} \epsilon'^{\nu *} \epsilon'^{\beta}
$$
The polarization vector sum
$
\sum_{\epsilon \epsilon}\epsilon^\mu \epsilon^{\alpha *}$ is equal to $-\eta^{\mu \alpha}$ (up to a term that does not contribute to the amplitude due to the Ward identity). Working out this expression and using
$$ \eta_{\mu \nu}\eta^{\mu \nu}=4
$$ and eliminating $p'$ using $$p'=p+k-k'$$ I get the following result ($m$ is the electron mass).
$$ \frac{1}{2}\sum_{\epsilon \epsilon'} \vert \mathcal{M} \vert^2=2e^{4} \left[ m^4 \left(\frac{1}{p.k}-\frac{1}{p.k'} \right)^2 +2m^2\left(\frac{1}{p.k}-\frac{1}{p.k'} \right)+4 \right]
$$
However, the last term of this expression, i.e. the 4, does not seem right to me. If I  apply this formula  in the LAB frame, I get
$$  \frac{1}{2}\sum_{\epsilon \epsilon'} \vert \mathcal{M} \vert^2=2e^{4} \left[ 3 +  \cos^2 \theta \right] $$
with $ \theta $ the deviation angle of the photon. However, there is another way to calculate this probability amplitude, i.e. by calculating  $\vert \mathcal M \vert ^2 $ in the LAB frame for each polarization state separately and taking the sum afterwards. This leads to
$$  \frac{1}{2}\sum_{\epsilon \epsilon'} \vert \mathcal{M} \vert^2=2e^{4} \left[ 1 +  \cos^2 \theta \right] $$
which is more credible, as it is proportional to the (classical) result for Thomson scattering. So, it seems that the factor between brackets should be $1+ \cos^2 \theta$ and that the term 4 in the frame independent expression should be 2.
Can somebody help me out and tell me where I made a mistake?
Details of the calculation below.

REPLY [1 votes]: Just to fill in a few of the blanks of TwoBs answer:
Using
\begin{align*}
\sum_{s} \epsilon^s(k)^{\mu}\epsilon^s(k)^{\nu *} = -\eta^{\mu\nu} + \frac{1}{2E^2}(k^{\mu}\bar{k}^{\nu}+k^{\nu}\bar{k}^{\mu}),
\end{align*}
from TwoBs notes, we would like to calculate
\begin{align*}
\sum_{s,s'} |\mathcal{M}|^2 = 4e^4 \left( \frac{p_{\mu}p'_{\nu}}{p\cdot k}-\frac{p'_{\mu}p_{\nu}}{p\cdot k'}-\eta_{\mu\nu}\right) \left( \frac{p_{\alpha}p'_{\beta}}{p\cdot k}-\frac{p'_{\alpha}p_{\beta}}{p\cdot k'}-\eta_{\alpha\beta}\right) \epsilon^{s}(k)^{\mu}\epsilon^{s}(k)^{\alpha *} \epsilon^{s'}(k')^{\beta } \epsilon^{s'}(k')^{\nu *}.
\end{align*}
We can see that the term in brackets in the spin sum involving $k$ always vanish. You can just do it once at a time: under  $\epsilon^{s}(k)^{\mu} \rightarrow k^\mu$ the expression is invariant, and similarly under $\epsilon^{s}(k)^{\nu *} \rightarrow k^\nu$. Both of these are thanks to $\epsilon(k)\cdot k=0$. So now we know that we can replace one of the spin sums with just the metric, as the momentum terms vanish. If we do the next spin sum, we don't have the luxury of dotting into the polarization vectors anymore - they're all gone. You can show that under $\epsilon(k') \rightarrow k'$, you are left with a term that is now $k' \cdot \bar{k}'$, which is not 0.
Doing one of these, say $\epsilon(k)^{\mu} \rightarrow k^{\mu}$, we have
\begin{align*}
(\frac{p\cdot k}{p\cdot k}p'_{\nu} - \frac{p'\cdot k}{p\cdot k'}p_{\nu}-k_{\nu})\epsilon(k')^{\nu*}... &= -k'\cdot \epsilon(k')^* = 0.
\end{align*}