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{\bf Question}

Set $G ={\rm stab}_{{\rm M\ddot{o}b}} (\infty) =\{ m\in {\rm
M\ddot{o}b}\: |\: m(\infty) =\infty\}$.  Write down a set of
generating elements for $G$.

\medskip
Let $\eta(z) \: |{\rm dz}|$ be an element of arc-length on ${\bf
C}$ which is invariant under $G$.  Prove that $\eta$ is constant.
\medskip

{\bf Answer}

If $m(z)=\ds\frac{az+b}{cz+d} \in $M\"{o}b$^+$ and
$m(\infty)=\infty$, then $c=0$, and so $m(z)=\alpha z+\beta$.
Generators of this are $\{P_{\beta}(z)=z+\beta|\beta \in {\bf{C}}\
\rm{and}\ L_{\alpha}(z)=\alpha z|\alpha \in {\bf C}-\{0\}\}$.

Note that $C(z)=\bar z$ also fixes $\infty$, and so if
$n(z)=\ds\frac{a\bar z +b}{c\bar z+d}$ fixes $\infty$, then $C
\circ n(z)$ fixes $\infty$, and so is as above.

So, generators for $G$ are:

$\begin{array} {l} P_{\beta}(z)=z+\beta;\ \beta \in {\bf C}\\
L_{\alpha}(z)=\alpha(z);\ \alpha \in {\bf C}-\{0\}\\ C(z)=\bar{z}
\end{array}$

\un{$\eta(z)|\,dz|$} invariant under $G$,\ \ $f:[a,b]
\longrightarrow {\bf{C}}$\ a path

\begin{eqnarray*} \rm{length}(f) & = & \ds\int_f \eta(z) |\,dz|\\ &
= & \ds\int_{P_{\beta} \circ f} \eta(z)|\,dz|\\ & = & \ds\int_a^b
\eta(P_{\beta} \circ f(t))|(P_{\beta}\circ f)'(t)| \,dt\\ & = &
\ds\int_a^b \eta(f(t)+\beta)|f'(t)| \,dt\\ & = & \ds\int_a^b
\eta(P_{\beta} \circ f(t))|f'(t)| \,dt\\ & = & \ds\int_f \eta
\circ P_{\beta}(z) |dz|.\end{eqnarray*}

and so $\eta(z)=\eta \circ P_{\beta}(z)=\eta(z+\beta)$ all $\beta
\in {\bf C}$. So, given $\omega,\ z \in {\bf C}$,set
$\beta=\omega-z$ and note that
$\eta(z)=\eta(z+\beta)=\eta(\omega)$ and to $\eta$ is constant.

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