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\title{The Uniformization Theorem}
\author{JWR}
\date{Tuesday December 11, 2001, 9:03 AM}

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\begin{document}

\maketitle

The  proof given here is a loose translation of~\cite{REYSSAT}.
There is another proof of the Uniformization Theorem in~\cite{FORSTER}
where it is called the {\em Riemann Mapping Theorem}.



\section{Harmonic functions}



\para
Throughout this section
$X$ denotes a connected Riemann surface, possibly noncompact.
The open unit disk in $\C$ is denoted by $\D$.
A \jdef{conformal disk} in $X$
centered at $p\in X$ is a pair $(z,D)$ where $z$ is a holomorphic coordinate
on $X$ whose image contains the
closed disk of radius $r$ about the origin in $\C$, $z(p)=0$, and
$$
    D=\{q\in X: |z(q)|<r\}.
$$
For a conformal disk $(z,D)$ we abbreviate the average value of a function $u$ on the boundary of  $D$ by
by
$$
    M(u,z,\p D):=\frac{1}{2\pi}\int_0^{2\pi} v(re^{i\theta})\,d\theta, \qquad v(z(q))=u(q).
$$
\arap

\para\label{def:Poisson}
The \jdef{Poisson kernel} is the function $P:\D\times\p\D\to\R$ defined by
$$
  P(z,\zeta):=\frac{1}{2\pi}\cdot\frac{|\zeta|^2-|z|^2}{|\zeta-z|^2}.
$$
The unique solution of Dirichlet's problem
$$
    \Delta u=0, \qquad u|\p \D=\phi
$$
where $\phi\in C^0(\p \D)$ on the unit disk $\D$ is given by
the \jdef{Poisson integral formula}
$$
    u(z)= \int_{\p \D} P(z,e^{i\theta}) \phi(e^{i\theta})\,d\theta.
$$
(See~\cite{ABR} page~13 for the proof.)
For a conformal disk $(z,D)$ in a Riemann surface $X$ and a continuous
function $u:X\to\R$ we denote by $u_D$ the unique continuous
function which agrees with $u$ on $X\setminus D$ and is harmonic in $D$.
(It is given by reading $u|\p D$ for  $\phi$ in the Poisson integral formula.)
\arap

\para The \jdef{Hodge star operator} on Riemann surface $X$ is the operation
which assigns to each 1-form $\omega$ the 1-form $*\omega$ defined by
$$
    (*\omega)(\xi) =-\omega(i\xi)
$$
for each tangent vector $\xi$. If $z$ is a  holomorphic coordinate on $X$
a real valued 1-form has the form
$$
       \omega = a\,dx+b\,dy
$$
where $a$ and $b$ are real valued functions, $x=\Re(z)$, and $y=\Im(z)$;
the form $*\omega$ is then given by
$$
     *\omega = -b\,dx+a\,dy.
$$
A function $u$ is called \jdef{harmonic} iff it is $C^2$ and $*du$ is closed, i.e.
$d*du=0$. In terms of the coordinate $z$ we have
$$
       d{*du}=(\Delta u)\,dx\wedge dy, \qquad
        \Delta u = \frac{\p^2u}{\p x^2}+\frac{\p^2u}{\p y^2}.
$$
Note that the operator $\Delta$ is not intrinsic, i.e. $\Delta u$ depends
on the choice of holomorphic coordinate. (However the operator $d{*d}$
is independent of the choice of coordinate  and hence
also the property of being harmonic.)
\arap

\begin{theorem}\label{thm:mvp}  Let $X$ a Riemann surface and $u:X\to\R$ be continuous.
Then the following are equivalent:
\begin{description}
\item[(i)] $u$ is  harmonic.
\item[(ii)]  $u$ satisfies \jdef{mean value property} i.e.
$
    u(p)=M(u,z,\p D)
$
for every conformal disk $(D,z)$ centered at $p$;
\item[(iii)] $u$ is locally the real part of a holomorphic function.
\end{description}
\end{theorem}


\begin{proof} It is enough to prove this for an open subset of $\C$.
That (ii)$\implies$(i) follows from the Poisson integral formula,
namely
$$
   u(0)=\int_{\p \D} P(0,e^{i\theta}) \phi(e^{i\theta})\,d\theta=M(u,\mathrm{id},\p\D).
$$
The general case follows by the change a change of variables $z\mapsto az+b$.
Note that the mean value property implies the \jdef{maximum principle}:
the function $u$ has no strict  maximum (or minimum) on any open set.
A function which is continuous on the closure of $D$ and satisfies
the mean value property in $D$ must therefore assume its maximum and minimum
on the boundary of $D$. Hence  (ii)$\implies$(i) because
$u-u_D$ (see~\ref{def:Poisson})
satisfies the mean value property in $D$ and vanishes on $\p D$
so $u-u_D=0$ on $D$ so $u$ is harmonic on $X$ as $D$ is arbitrary.
For (i)$\iff$(iii) choose a conformal disk $(z,D)$.
The function $u$ is harmonic if and only if the form $*du$ is closed.
The equation $*du=dv$ encodes the Cauchy Riemann equations; it holds
if and only if the function $u+iv$ is holomorphic.
\end{proof}



\begin{corollary} The form $*du$  is exact if and only there is a holomorphic
function $f:X\to\C$ with $u=\Re(f)$.
\end{corollary}

\begin{theorem}[Removable Singularity Theorem] \label{thm:removable}
If $u$ is harmonic and bounded on the punctured disk,
it extends to a harmonic function on the disk.
\end{theorem}

\begin{proof}
By shrinking the disk and subtracting the solution of the Dirichlet problem
we may assume that $u$ vanishes on $\p D$. For $\eps>0$
the harmonic function
$$
   v_\eps(z)= u(z)+\eps\log|z| -\eps
$$
is negative on $\p\D$ and near $0$ and thus negative on $\D\setminus\{0\}$.
Fix $z$ and let $\eps\to0$; we conclude that $u\le 0$ on $\D\setminus\{0\}$.
Similarly $-u\le 0$.
\end{proof}

\rmk B\^orcher's Theorem (see~\cite{ABR} page~50) says that a positive function
which harmonic on $\D\setminus 0$ has the form
$$
  u(z)=-b \log|z|+ h(z)
$$
where $b\ge 0$ and $h$ is harmonic on $\D$.
This implies the Removable Singularity Theorem (add a constant).
To prove B\^orcher's Theorem choose $b$ so that $*du+b \log|z|$ is exact.
Hence assume w.l.o.g. that $u$ is the real part of a holomorphic function
with a possible singularity at the origin. However if the Laurent expansion
for this function contains any negative powers of $z$ its real part $u$
will be unbounded in both directions.
\kmr






\begin{theorem}[\jdef{Harnack's Principle}]
A pointwise nondecreasing sequence of harmonic functions
converges uniformly on compact sets either to $\infty$ or to a harmonic function.
\end{theorem}

\begin{proof} If we assume that the sequence converges uniformly,
this follows from the characterization via the mean value property (Theorem~\ref{thm:mvp}).
For the general argument see~\cite{ABR} page~49.
\end{proof}


\begin{theorem} \label{thm:converge} If a sequence of harmonic functions
converges uniformly on compact subsets then the limit
is harmonic and for $k=1,2,\ldots$ the sequence
converges in $C^k$ (uniformly on compact subsets).
\end{theorem}

\begin{proof} In each holomorphic disk $(z,D)$ we have
$$
    \p^\kappa u =
  \int_0^{2\pi} \p^\kappa P_\zeta\cdot u(\zeta)d\theta
$$
for each multi-index $\kappa=(\kappa_1,\kappa_2)$
where $\zeta=re^{i\theta}=z(q)$, $q\in\p D$ (i.e. $|z(q)|=r$), and $P_\zeta(z)=P(z,\zeta)$
is the Poisson kernel. ( See~\cite{ABR} page~15.)
\end{proof}

\begin{theorem}[\jdef{Compactness Theorem}] \label{thm:compactness}
A uniformly bounded sequence
of harmonic functions contains a subsequence which converges uniformly
on compact sets.
\end{theorem}

\begin{proof} By the estimate in the proof of~\ref{thm:converge} the
the first derivatives of the sequence are uniformly bounded on any compact subset
of any open disk on which the functions $u_n$ are uniformly bounded.
Use Arzela Ascoli and diagonalize over compact sets.
(See~\cite{ABR} page~35.)
\end{proof}

\section{The Dirichlet Problem}


Compare the following lemma and definition with Theorem~\ref{thm:mvp}.

\begin{lemma} \label{def:subharmonic}
Let $u:X\to\R$ be continuous.
Then the following conditions are equivalent
\begin{description}
\item[(i)] The function $u$ satisfies the following form of the \jdef{maximum principle}:
For every connected open subset $W\subset X$ and every harmonic function $v$ on $W$
either $u-v$ is constant or else it does not assume its maximum in $W$;
 \item[(ii)] For every conformal disk $(z,D)$ we have $u\le u_D$;
 \item[(iii)] $u$ satisfies \jdef{mean value inequality}, i.e. $u(p)\le M(u,z,\p D)$
 for every conformal disk $(z,D)$ centered at $p$;
\end{description}
A function  $u$ which satisfies these conditions is called \jdef{subharmonic}.
A function $u$ is called \jdef{superharmonic} iff $-u$ is subharmonic.
\end{lemma}

\begin{proof} We prove (i)$\implies$(ii).  Choose $(z,D)$ and let $v=u_D$.
Then $u-u_D$ is zero on $\p D$ and thus either the constant $0$ on $D$
or else nowhere positive. In either case $u\le u_D$.

We prove (ii)$\implies$(iii). Choose $(z,D)$ centered at $p$.
Then by~(ii) we have $u(p)\le u_D(p)=M(u_D,z,\p D)=M(u,z,\p D)$.

We prove (iii)$\implies$(i). Suppose that $v$ is harmonic on a
connected open subset $W\subset X$
and that $u-v$ assumes its maximum $M$ at some point of $W$, i.e. that
the set
$$
 W_M:=\{p\in W:u(p)-v(p)=M\}
$$
is nonempty. We must show that $u-v=M$ on all of $W$, i.e. that $W_M=W$.
The set $W_M$ is closed in $W$ so it is enough to show that $W_M$ is open.
Choose $p\in W_M$ and let $(z,D)$ be a conformal disk centered at $p$. Then
$$
  M= (u-v)(p)\le M(u-v,z,\p D)\le M.
$$
But $u-v\le M$ so $u-v=M$ on $\p D$. By varying the radius of $D$
we get that $u-v=M$ near $p$.
\end{proof}

\begin{corollary} \label{cor:subharm}
Subharmonic functions satisfy the following properties.
\begin{description}
\item[(i)] The max and sum of two subharmonic functions is subharmonic
and a positive multiple of a subharmonic function is subharmonic.
\item[(ii)] The subharmonic property is local: if $X=X_1\cap X_2$
where $X_1$ and $X_2$ are open and $u\in C^0(X,\R)$
is subharmonic on $X_1$ and on $X_2$, then it is subharmonic
on $X$.
\item[(iii)] If $u$ is subharmonic so is $u_D$.
\item[(iv)] If $u:X\to\R$ is continuous, positive and harmonic on an open set $V$,
and vanishes on $X\setminus V$, then $u$ is subharmonic on $X$.
\end{description}
\end{corollary}

\begin{proof} Part~(i) is immediate and part~(ii) follows easily from part~(i)
of~\ref{def:subharmonic}. For~(iii) suppose that $v$ is harmonic and
$u_D-v$ assumes its maximum $M$ at $p$.
Since $u_D-v$ is harmonic in $D$ it follows by the maximum
principle that the maximum on $D$ is assumed on $\p D$ so we may
assume that  $p\in X\setminus D$. But $u=u_D$ on $X\setminus D$
and $u-v\le u_D-v\le M$ so  $u-v$ also assumes its maximum at $p$.
Hence $u-v=M$ and hence $u$ is harmonic. For~(iv) suppose $u-v$ assumes
its maximum at a point $p$ and $v$ is harmonic. We derive a contradiction.
After subtracting a constant we may assume that this maximum is zero.
Then $u\le v$ so $0\le v$ on $X\setminus V$ and $0<u\le v$
on $X$. If $p\notin X$ then $v(p)=u(p)=0$ and $v$ assumes its minimum
at $p$ which contradicts the fact that $v$ is harmonic.
If $p\in X$ then $u-v$ assumes its maximum at $p$ and this contradicts
the fact that $u$ is subharmonic on $V$.
\end{proof}

\rmk The theory of subharmonic functions works in all dimensions.
In dimension one,  condition~(ii) of lemma~\ref{def:subharmonic}
says that $u$ is a convex function.
\kmr

\xrcs A $C^2$ function $u$ defined on an open subset of $\C$ is subharmonic
iff and only if $\Delta u\ge 0$.
\scrx

\para\label{def:perron}
A \jdef{Perron family} on a Riemann surface $X$ is a  collection $\cF$ of
functions on $X$ such that
\begin{description}
\item[(P-1)] $\cF$ is nonempty;
\item[(P-2)] every $u\in\cF$ is subharmonic;
\item[(P-3)] if $u,v\in\cF$, then  there exists $w\in\cF$ with $w\ge\max(u,v)$;
\item[(P-4)] for every conformal disk $(z,D)$ in $X$ and every $u\in\cF$ we have $u_D\in\cF$;
\item[(P-5)] the function
$$
   u_\cF(q):=\sup_{u\in\cF}\; u(q)
$$
is everywhere finite.
\end{description}
\arap

\begin{theorem}[Perron's Method] \label{thm:perron} If $\cF$ is a Perron family,
then $u_\cF$ is harmonic.
\end{theorem}

\begin{proof}
Choose a conformal disk $(z,D)$.
Diagonalize on a countable dense subset
to construct a sequence $u_n$ of elements of $\cF$
which converges pointwise  to $u_\cF$ on $D$ on a dense set.
By~(P-3) choose $w_n\in\cF$ with $w_1=u_1$ and
$w_{n+1}\ge\max(w_1,\ldots,w_n)$. The new sequence is pointwise monotonically increasing.
By~(P-2) and~(P-4) we may assume that each $w_n$ is harmonic on $D$.
Then $u_\cF$ is harmonic in $D$ by Harnack's Principle and~(P-4).
\end{proof}



\para
Assume $Y\subseteq X$ open with $\p Y\ne\emptyset$.  Define
$$
\cF_\phi=\{u\in C^0(\bar Y,\R): u\mbox{ subharmonic on }Y,\;\;
                                u\le\sup_{\p Y}\phi,\; u|\p Y\le \phi\}
$$
and
$$
   u_\phi(q)=\sup_{u\in\cF_\phi} \; u(q).
$$
\arap


\begin{lemma} If $\phi:\p Y\to \R$ is continuous and bounded then
the family $\cF_\phi$ is a Perron family so $u_\phi$ is harmonic on $Y$.
\end{lemma}

\begin{proof} Maximum principle.
\end{proof}

\para
A \jdef{barrier function} at $p\in\p Y$ is a function $\beta$ defined
in a neighborhood $U$ of $p$ which is continuous on the closure
$\overline{Y\cap U}$ of $Y\cap U$,
superharmonic on $Y\cap U$, such that $\beta(p)=0$ and
$\beta>0$ on $\overline{Y\cap U}\setminus\{p\}$.
A point $p\in\p Y$ is called \jdef{regular} iff there is a barrier function at $p$
\arap

\begin{lemma}\label{lem:barrier}
If $\p Y$ is a $C^1$ submanifold,
it is regular at each of its points.
\end{lemma}

\begin{proof}
Suppose w.l.o.g that $Y\subset\C$ and that $\p Y$ is transverse to the real axis at $0$,
and that $Y$ lies to the right. Then
$$
   \beta(z)= \sqrt{r}\cos(\theta/2)= \Re(\sqrt{z}), \qquad z=re^{i\theta}
$$
is a barrier function at $0$.
\end{proof}

\begin{lemma}\label{lem:boundary}
If $p$ is regular, then $\lim_{y\to p}u_\phi(y)=\phi(p)$.
\end{lemma}

\begin{proof} The idea is that $u_\phi(p)\le\phi(p)$ and
if we had strict inequality we could make $u_\phi$ bigger
by adding $\eps-\beta$.  See~\cite{ABR} page~203.
\end{proof}


\begin{corollary}\label{cor:dirichlet}
If $\p Y$ is a $C^1$ submanifold of $X$ then $u_\phi$
solves the Dirichlet problem with boundary condition $\phi$, i.e.
it extends to
a continuous function on $Y\cup\p Y$ which agrees with $\phi$ on $\p Y$.
\end{corollary}

\section{Green Functions}


\dfn\label{def:green} Let $X$ be a Riemann surface and $p\in X$.
A \jdef{Green function} at $p$ is a function $g:X\setminus\{p\}\to\R$
such that
\begin{description}
\item[(G-1)] $g$ is harmonic;
\item[(G-2)] for some (and hence every) holomorphic coordinate $z$
centered at $p$ the function $g(z)+\log(z)$ is harmonic near $p$;
\item[(G-3)] $g>0$;
\item[(G-4)] if $g':X\setminus\{p\}\to\R$ satisfies~(G-1),(G-2),(G-3) then $g \le g'$.
\end{description}
Condition~(G-4) implies that the Green function at $p$ is unique
(if it exists) so we denote it by $g_p$.
Warning: When $X$ is the interior of a manifold with boundary,
the Green function defined here differs from the usual Green's
function by a factor of $-1/(2\pi)$.
\nfd

\dfn A Riemann surface $X$ is called
\jdef{elliptic} iff it is compact,
\jdef{hyperbolic} iff it admits a nonconstant negative subharmonic function,
and \jdef{parabolic} otherwise.
By the maximum principle for subharmonic functions (in the  elliptic case)
and definition (in the parabolic case) a nonhyperbolic surface admits
no nonconstant negative subharmonic function.
In particular, it admits no nonconstant negative harmonic function
and hence (add a constant) no  nonconstant bounded harmonic function.
\nfd


\begin{theorem} For a Riemann surface the following are equivalent.
\begin{description}
\item[(i)] there is a Green function at every point;
\item[(ii)] there is a Green function at some point;
\item[(iii)] $X$ is hyperbolic;
\item[(iv)] for each compact set $K\subset X$ such that $\p K$ smooth
and $W:=X\setminus K$ is connected,
there is a continuous function $\omega:W\cup\p W\to\R$ such that
$\omega\equiv 1$ on $\p K=\p W$ and  on $W$ we have both
that $0<\omega<1$ and that $\omega$ is harmonic on $W$.
\end{description}
\end{theorem}

\begin{proof} {\em That (i)$\implies$(ii) is obvious; we prove (ii)$\implies$(iii).}
Suppose $g_p$ is a Green function at $p$.
Then  $u=\max(-2,-g_p)$ is negative and subharmonic.
Now $u=-2$ near $p$,  so either $u$ is nonconstant or else
$-g_p\le-2$ everywhere. The latter case is excluded, since
otherwise $g'=g_p-1$ would satisfy~(G-1), (G-2), (G-3) but not~(G-4).

{\em We prove (iii)$\implies$(iv).}  Assume $X$ is hyperbolic.
Then there is a  superharmonic $u:X\to\R$
which is nonconstant and everywhere positive.
Choose a compact $K$ $K$ and let $W=X\setminus K$.
After rescaling we may assume that $\min_Ku=1$.
By the Maximum principle (for $-u$) and the fact that $u$ is not constant
there are points (necessarily in $W$)
where $u<1$ so after replacing $u$ by $\min(1,u)$ we may assume that $u \equiv 1$ on $K$.
The family
$$
\cF_K=\{v\in C^0(W\cup\p W,\R):v\le u  \mbox{ and }v \mbox{ subharmonic on }W\}
$$
is a Perron family:
(P-1)~$\cF_K\ne\emptyset$ as the restriction of $-u$ to $X\setminus K$ is in $\cF_K$;
(P-2)~$v\in\cF_K\implies v$ subharmonic by definition;
(P-3)~$v_1,v_2\in\cF_K\implies\max(v_1,v_2)\in\cF$;
(P-4)~$v\in\cF_K$ and $D$ a conformal disk
in $X\setminus K$ implies that $v_D\le u_D\le u$
as $v\le u$ on $\p D$ and $u$ is superharmonic; and
(P-5)~the function $\omega:=\sup_{v\in\cF}v$ is finite as $v\in\cF\implies v\le u$.
It remains to show that $0<\omega<1$ on $W$ and $\omega=1$ on $\p W$.
Suppose $Y\subset X$ is open, with $\p Y$ smooth and $Y\cup\p Y$ compact, and $\p K\subset \p Y$.
Let $w$ be the solution of the Dirichlet problem with $w=1$ on $\p K$ and
$w=0$ on $(\p Y)\setminus(\p K)$. Extend $w$ by zero on $W\setminus Y$.
The extended function $w$ is subharmonic by Corollary~\ref{cor:subharm} part~(iv).
Thus $w_Y-u$ is subharmonic and $\le 0$ on $X\setminus Y$ and on $\p Y$.
Hence $w_Y|W\in\cF$. As the sets $Y$ exhaust
$X$ and $w>0$ on $Y$ it follows that $\omega$ satisfies
$0<\omega$ on $W$ and $\omega=1$ on $\p W$. Since $\p W$ is smooth it follows
that $\omega$  is continuous on $\p W$. Since $\omega\le u$ and $u<1$ on $W$
we have that $\omega\le u<1$ on $Y$ so $\omega$ satisfies~(iv).

{\em We prove (iv)$\implies$(i).} Choose $p\in X$ and a
conformal disk $(z,D)$ centered at $p$.
Let $\cF$ be the set of all continuous functions
$v:X\setminus\{p\}\to\R$ satisfying the following conditions:
\begin{description}
\item[(a)] $\mathrm{supp}(v)\cup\{p\}$ is compact;
\item[(b)] $v$ is subharmonic on $X\setminus\{p\}$, and
\item[(c)] $v+\log|z|$ extends to a subharmonic on function on $D$.
\end{description}
We show that $\cF$ is a Perron family. The set $\cF$ is not empty
since it contains the function $-\ln|z|$ (extended by $0$).
Properties~(i-iii) in~\ref{def:perron} are immediate.
It remains to show~(iv), i.e. that $u_\cF$ is finite.
For $0<r\le 1$ let
$$
K_r=\{q\in D: z(q)\le r\},
$$
define $\omega_r$ as in~(iv) reading $K_r$ for $K$ and $\omega_r$ for $\omega$,
and let
$$
   \lambda_r= \max_{|z|=1}\omega_r.
$$
We will show that for $v\in\cF$ we have
$$
    v\le \frac{\log r}{\lambda_r-1}  \eqno(*)
$$
$X\setminus K_r$
and this shows that $u_\cF<\infty$ on $X\setminus\{p\}=\bigcup_{r>0}X\setminus K_r$.
Choose $v\in \cF$ and let $c_r=\max_{|z|=r}v$.
The function $v+\log|z|$ is subharmonic so its maximum on $K_1$ must occur
on $\p K_1$, i.e.
$$
c_r+\log r\le c_1.
$$
But $c_r\omega-v\ge 0$ on
$\p K_r$ and off the support of $v$ so $v\le c_r\omega$ on $X\setminus K_r$
and hence
$$
c_1\le c_r\lambda_r.
$$
It follows that
$$
    c_r\le \frac{\log r}{\lambda_r-1}
$$
i.e. that~$(*)$ holds on $\p K_r$. But $v$ has compact support so~$(*)$ holds
on $X\setminus K_r$.


The desired Green function is
$$
   g_p=u_\cF.
$$
From~\ref{thm:perron} we conclude that $g_p$ is harmonic on $X\setminus\{p\}$
and hence that $g_p+\ln|z|$ is harmonic on $D\setminus\{p\}$. From~$(*)$
we conclude that the inequality
$$
 v+\log|z|\le \frac{\log r}{\lambda_r-1}+\log r
$$
holds on $\p K_r$ and hence (as the left hand side is subharmonic)
on $K_r$. Thus the function $g_p+\ln|z|$ is bounded on $D$
and therefore harmonic on $D$ by the Removable Singularity Theorem~\ref{thm:removable}.
Moreover $g_p>0$ because $g_p\ge 0$ and $g_p$ is nonconstant.
Suppose $g'$ also satisfies these properties; we must show
$g_p\le g'$. If $v\in\cF$ then $v-g'$ is subharmonic on $X\setminus p$
(because $v$ is) and on $D$ (because it equals $(v+\ln|z|)-(g'+\ln|z|)$)
and hence on all of $X$. But $v-g'<0$ off the support of $v$ and
hence $v<g'$ everywhere. Thus $g=u_\cF\le g'$.
\end{proof}

\section{Nonhyperbolic surfaces}


\begin{theorem}[Extension Theorem] \label{thm:extend}
Assume $X$ is a nonhyperbolic connected Riemann surface.
Suppose $p\in X$ and that $f$ is a holomorphic function defined on
$D\setminus\{p\}$ where $(z,D)$ is a conformal disk centered at $p\in X$. Then there
is a unique harmonic function $u:X\setminus\{p\}\to\R$ bounded in the complement
of any neighborhood of $p$ such that $u-\Re(f)$ is harmonic in $D$ and vanishes at $p$.
\end{theorem}


\rmk
 Suppose that $X=\C$, that  $p=0$, and that the function $f$ has a Laurent expansion
$$
   f(z)= \sum_{n=-\infty}^\infty c_nz^n
$$
valid in $0<|z|<1$.  The function $u$ is given by $u=\Re(w)$ where
$$
    w(z) = \sum_{n=-\infty}^{-1} c_nz^n.
$$
The latter series converges for all $z\ne 0$.
\kmr

\begin{proof}[Proof of~\ref{thm:extend}]
The proof of uniqueness is easy. If $u_1$ and $u_2$ are two  functions as in the theorem,
then $u_1-u_2$ is bounded in the complement of every neighborhood of $p$ (as $u_1$ and $u_2$ are)
and near $p$ (as $u_1-u_2=(u_1-\Re(f))-(u_2-\Re(f))$ and is thus bounded,
hence constant (as $X$ is nonhyperbolic) hence zero (as it vanishes at $p$).
For existence we need two preliminary lemmas.  By the definition of conformal disk
the open set $z(D)$ contains the closed unit disk in $\C$;
for $r\le 1$ let
$$
       D_r=\{q\in D: |z(q)|<r\}.
$$
The following lemma is an immediate consequence of Stoke's Theorem
if $X$ is compact.


\begin{lemma}\label{lem:star}
If $r<1$ and $u$ is harmonic and bounded on $X\setminus D_r$, then
$$
        \int_{\p D} *du=0.
$$
\end{lemma}

\begin{proof} By adding a large positive constant we may assume w.l.o.g.
that $u$ is nonnegative on $\p D_r$.
Choose an increasing sequence of open subsets
$X_n\subset X$ such that $X_n\cup\p X_n$ is compact,
$\p X_n$ is smooth, and the closure of $D$ is a subset of $X_n$.
Let $u_n$ and $v_n$ be the solutions of the Dirichlet problem on $X_n\setminus D_r$
with boundary conditions $u_n=v_n=0$ on $\p X_n$, $u_n=u$ on $\p D_r$
and $v_n=1$ on $\p D_r$. By the maximum principle we have that
$0\le u_n\le u_m\le \max_{\p D} u$ and $0\le v_n\le v_m\le 1$ on $X_n$ for $m\ge n$.
Hence by Harnack and~\ref{thm:converge} $u_n$ and $v_n$ converge in $C^k$ uniformly on compact subsets
of $X\setminus D$ (in fact on compact subsets of the complement of the closure of $D_r$).
Moreover $\lim_nv_n=1$ on $\p D$ and  $\lim_nv_n\le 1$ on $X\setminus D$
so we must have $\lim_nv_n=1$ on $X\setminus D$ by~(iv) of Corollary~\ref{cor:subharm}
and the fact that $X$ is nonhyperbolic.
Hence
$$
     \int_{\p D} *du = \lim_{n\to\infty} \int_{\p D} v_n\, *du_n-u_n\, *dv_n
$$
But $u_n=v_n=0$ on $\p X_n$  on $\p D$ so this may be written
$$
     \int_{\p D} *du = -\lim_{n\to\infty} \int_{\p (X_n\setminus D)} v_n{*du_n}-u_n{*dv_n}.
$$
Now by Stokes
$$
\int_{\p (X_n\setminus D)} v_n{*du_n}-u_n{*dv_n}
=\int_{X_n\setminus D} v_n\, \Delta u_n- u_n\,\Delta v_n = 0.
$$
\end{proof}

\begin{lemma} For $0<\rho<1$ and let $u_\rho$ be the solution of the
Dirichlet problem on $X\setminus D_\rho$ with $u=\Re(f)$ on $\p D_\rho$.
Then for $0<r<1/20$ there is a constant $c(r)$ such that for $0<\rho<r$
we have
$$
    \max_{\p D_r} |u_\rho| \le c(r).
$$
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:star} the 1-form $*\,du_\rho$ is exact
on on the interior of $D\setminus D_\rho$ so there is a holomorphic
function $F_\rho$ with $u_\rho=\Re(F_\rho)$. The function
$F_\rho-f$ has a Laurent expansion about $0$; its real part is
$$
  (u_\rho-\Re f)(te^{i\theta} =\sum_{n=-\infty}^\infty (\alpha_n\cos(n\theta)+\beta_n\sin(n\theta)t^n
$$
valid for $\rho\le t\le 1$. (The coefficients $\alpha_n$ and $\beta_n$ depend on $\rho$.) Then
$$
\frac{1}{\pi}(u_\rho-\Re f)(te^{i\theta}\cos(k\theta)\,d\theta = \alpha_kt^{k}+\alpha_{-k}t^{-k}
$$
and
$$
\frac{1}{\pi}(u_\rho-\Re f)(te^{i\theta}\sin(k\theta)\,d\theta = \beta_kt^{k}+\beta_{-k}t^{-k}.
$$
For $t=\rho$ the integrand vanishes so
$$
\alpha_{-k}(\rho)=-\alpha_k(\rho)\rho^{2k},\qquad\beta_{-k}(\rho)=-\beta_k(\rho)\rho^{2k}.
\eqno(1)
$$
For $t=1$ we have
$$
|\alpha_k|(1-\rho^{2k})=|\alpha_k+\alpha_{-k}|\le 2 M_\rho, \qquad
|\beta_k|(1-\rho^{2k})=|\beta_k+\beta_{-k}|\le 2 M_\rho
$$
where
$$
  M_\rho = \max_{|z|=1}|u_\rho|+ \max_{|z|=1}|\Re(f)|;
$$
Hence for $\rho<1/2$ we have $|\alpha_k|,|\beta_k|\le 4M_\rho$ so
$$
  \max_{|z|=r}|u_\rho|\le\max_{|z|=r}|\Re(f)|+4M_\rho\sum_{n=0}^\infty r^n+\rho^{2n}r^{-n}.
$$
Since $\rho<r$ sum on the right is less than $2\sum r^n=1/(1-r)$ so we get
$$
 \max_{|z|=r}|u_\rho|\le\max_{|z|=r}|\Re(f)|+\frac{8M_\rho}{1-r}.
\eqno(2)
$$
The function $u_\rho$ is harmonic and bounded on $X\setminus D_\rho$ so
we have
$$
 \max_{|z|=1}|u_\rho|\le\max_{|z|=r}|\Re(f)|+\frac{8M_\rho}{1-r}.
$$
and hence
$$
   M_\rho\le\max_{|z|=1}|\Re(f)|+\max_{|z|=r}|\Re(f)|+\frac{8M_\rho}{1-r}.
$$
Since $8/(1-r)<1/2$ this gives the bound
$$
 M_\rho\le2\left(\max_{|z|=1}|\Re(f)|+\max_{|z|=r}|\Re(f)\right)
$$
on $M_\rho$ independent of $\rho<r$.
hence a bound of $|u_\rho|$ on $|z|=1$ (i.e. $\p D$), and hence a bound on $|u_\rho|$
on $X\setminus D$.
\end{proof}

\medskip

\noindent{\em We complete the proof of Theorem~\ref{thm:extend}.}
Let $r_n=1/(21n)$ so that $1/20 >r_1>r_2>\cdots$
and $\lim_{n\to\infty}r_n=0$. Let $u_k=u_{\rho_k}$ where $\rho_k=r_k/2$.
By the Compactness Theorem~\ref{thm:compactness} and the fact that $|u_k|<c(r_1)$
for $\rho_k<r_1$ there is a subsequence of the $u_k$ (still denoted
by $u_k$) converging uniformly to a harmonic function $u$ on $X\setminus D_{r_1}$.
For the same reason there is a subsequence converging uniformly
on $X\setminus D_{r_2}$, and a subsequence converging uniformly
on $X\setminus D_{r_3}$, etc. Diagonalize and we get a sequence
converging uniformly on compact subsets to a harmonic function
$u$ on $X\setminus\{p\}$.
\end{proof}

\section{Maps to $\P$}


The material in this section is not required for the proof of the
Uniformization Theorem.

\begin{theorem} Let $X$ be a Riemann surface,
$p_1,p_2,\ldots,p_n\in X$ be distinct, and $a_1,a_2,\ldots,a_n\in\P:=\C\cup\{\infty\}$.
Then there is a meromorphic function $f$
(i.e. a holomorphic map $f:X\to\P$) with $f(p_j)=a_j$ for $j=1,2,\ldots,n$.
\end{theorem}

\begin{proof}
First suppose $n=2$, $a_1=\infty$, and $a_2=0$ amd
Then choose holomorphic coordinates $z_j$  centered at $p_j$.
In case  that $X$ is nonhyperbolic there are functions
$u_j:X\setminus\{p_j\}\to\R$ with $u_j-\Re(1/z_j)$ bounded and harmonic.
In case  that $X$ is hyperbolic there are functions
$u_j:X\setminus\{p_j\}\to\R$ with $u_j-\log|z_j|$ bounded and harmonic.
In either case by the Cauchy Riemann equations
the function
$$
     f(z) = \frac{u_{1x}-iu_{1y}}{u_{2x}-iu_{2y}}
$$
locally a ratio of two holomorphic differentials and is independent
of the choice of local coordinates $z=x+iy$ used to defined it.
(Note: No need to assume $X$ is simply connected.) Now
$g_{12}=f/(f+1)$ takes the value $1$ at $p_1$ and $0$ at $p_2$.
For general $n$ the function $h_j=\prod_{k\ne j}g_{kj}$
satisfies $h_j(p_k)=\delta_{jk}$. Take $f=\sum_j a_jh_j$.
\end{proof}


\begin{theorem} Let $X$ be a compact Riemann surface and $p\in X$.
Then there is a meromorphic function $F:X\to\P$ having $p$ as its
only pole.
\end{theorem}

\begin{proof} Let $g$ be the genus of $X$ so that $\dim_\R\,H^1(X,\R)=2g$.
For $k=1,2,\ldots,2g+1$ let $u_k$ be the harmonic function on $X\setminus\{p\}$
given by Theorem~\ref{thm:extend} with $f=1/z^k$, i.e. $u_k$
is harmonic on $X\setminus\{p\}$ and $u_k-\Re(1/z_k)$ is harmonic near $p$.
Then $*du_k$ is a closed 1-form on $X$ so some nontrivial linear combination is exact;
i.e. $dv=\sum_k a_k\,{*du_k}$.
The function $v$ is the imaginary part  of a holomophic function $F=u+iv$ whose
real part $\Re(F)=\sum_ka_ku_k$ is bounded in the complement of every neighborhood of $p$.
Thus $p$ is the only pole of $F$.
\end{proof}


\section{The Uniformization Theorem}


\begin{theorem}[Uniformization Theorem]\label{thm:unif}
Suppose that $X$ is connected and simply connected. Then
\begin{enumerate}
\item if $X$ is elliptic, it is isomorphic to $\P^1$;
\item if $X$ hyperbolic, it is isomorphic to the unit disk $\D$;
\item if $X$ is parabolic, it is isomorphic to $\C$.
\end{enumerate}
\end{theorem}


\dfn A holomorphic function $F$ on $X$ is called a \jdef{holomorphic Green  function}
at the point $p\in X$ iff
$$
|F| = e^{-g_p}
$$
where $g_p$ is the Green function for $X$ at $p$.
\nfd

\begin{lemma}\label{lem:holoGreen}
Assume $X$ is simply connected and hyperbolic and $p\in X$. Then there is a
holomorphic Green function $F$ at $p$.
\end{lemma}

\begin{proof} Choose a holomorphic coordinate $z=x+iy$ centered at $p$
and let $h$ be a holomorphic function defined near $p$ with
$\Re(h)=g_p+\log|z|$. Let $F_p=e^{-h}z$. Then $F$ is holomorphic
and $\log|F|=-\Re(h)+\log|z|=-g_p$.
Now the condition $g_p=-\log|F|$ defines a holomorphic function $F$
(unique up to a multiplicative constant) in a neighborhood of any point other
than $p$ so $F$ extends to $X$ by analytic continuation.
\end{proof}

\begin{lemma} Let $F$ be holomorphic Green function  $p$. Then
\begin{description}
\item[(i)] $F$ is holomorphic;
\item[(ii)] $F$ has a simple zero at $p$;
\item[(iii)] $F$ has no other zero;
\item[(iv)] $F:X\to\D$;
\item[(v)] If $F'$ satisfies~(i-iv) then $|F'|\le|F|$.
\end{description}
By~(v) the holomorphic Green function  at $p$ is unique up to
a multiplicative constant of absolute value one.
\end{lemma}

\begin{proof} Since $F_p=e^{-h}z$ the function $F$
has a simple zero at $p$.  Since $g_p>0$ we have
that $F:X\to\D$ and $F$ has no other zero.
\end{proof}


\begin{lemma} A holomorphic Green function is injective.
\end{lemma}


\begin{proof}
Choose $q\in X$ and let
$$
        \phi(r)=\frac{F_p(q)-F_p(r)}{1-\bar{F}_p(q)F_p(r)}.
$$
Then $\phi$ is the composition of $F_p$ with an automorphism of $\D$
which maps $F_p(q)$ to $0$. Suppose that $\phi$ has a zero of order $n$ at a point $q$.
Let $u=-\log|\phi|/n$.
Let $z$ be a holomorphic coordinate at centered at $q$. Then $u+\log|z|$
is bounded near $p$ and hence (by B\^ocher) harmonic near $p$.
The Green function $g_q$ at $q$ is defined by $g_q=u_\cF$ where
$\cF$ is the set of all  $v$ of compact support, with $v$ subharmonic on
$X\setminus\{q\}$ and $v+\log|z|$ subharmonic near $q$. By the maximum principle,
and because $v$ has compact support we have $v\le u$ for $v\in\cF$.
Hence $g_q\le u$ so
$$
|F_q(r)|\ge|\phi(r)|^{1/n}\ge|\phi(r)|.\eqno(\#)
$$
Since  $F_p(p)=0$ we have $\phi(p)=F_p(q)$ so  $F_q(p)|\ge|F_p(q)|$.
Reversing $p$ and $q$ gives $|F_q(p)|=|F_p(q)|$. By~$(\#)$
$|F_q(r)/\phi(r)|\le1$ with equality at $r=p$. Hence $F_q=c\phi$ where
$c$ is a constant with $|c|=1$. But $f_q(r)\ne 0$ for $r\ne q$ so
$\phi(r)\ne 0$ for $r\ne q$ so $F_p(r)\ne F_p(q)$ for $r\ne q$,
i.e. $F_p$ is injective.
\end{proof}

The proof that $X$ is isomorphic to $\D$ now follows from the
Riemann Mapping Theorem. However we can also prove that $F_p$
is surjective as follows.

\begin{lemma} \label{lem:stretch} Suppose $W$ is  a simply connected open subset
of the unit disk $\D$ such that
$0\in W$ but $W\ne\D$. Then there is a injective holomorphic map
$H:W\to\D$ with $H(0)=0$ and $|H'(0)|>1$.
\end{lemma}

\begin{proof} Suppose $a^2\in\D\setminus W$.
Then the unction $(z-a^2)/(1-\bar{a}^2z)$ is holomorphic and nonzero on
$W$. Since $W$ is simply connected this function has a square root, i.e.
there is a holomorphic function $h$ on $W$ such that
$$
h(z)^2=\frac{z-a^2}{1-\bar{a}^2z}
$$
and $h(0)=ia$. Consider the function $H:W\to D$ defined by
$$
   H(z)=\frac{h-ia}{1+iah}
$$
Then $H'(0)=(1+|a|^2)/(2ia)$ so $|H'(0)|>1$.  This map is injective as
$\ds
H(z)=H(w)\implies h(z)=h(w) \implies  h(z)^2=h(w)^2\implies
\frac{z-a^2}{1-\bar{a}^2z}=\frac{w-a^2}{1-\bar{a}^2w}
\implies z=w.
$
\end{proof}

\begin{lemma}  A holomorphic Green function is surjective.
\end{lemma}

\begin{proof} Assume not.
Read $F_p(X)$ for $W$ in Lemma~\ref{lem:stretch}.
Note that both $F_p$ and $H\circ F_p$ has a simple zero at $p$.
The function $-\log|H\circ F_p|$ has  all the properties
of the Green function so
$$
   -\log|F_p|=g_p\le-\log|H\circ F_p|
$$
by the minimality of the Green function. Hence $|H\circ F_p|\le|F_p|$
so $|H|\le|z|$ near zero contradicting $|H'(0)|>1$.
\end{proof}



This proves the Uniformization Theorem in the hyperbolic case.
To prove the Uniformization Theorem in the
nonhyperbolic case we  introduce
a class of functions  to play
the role of the holomorphic Green function  of~\ref{lem:holoGreen}.



\dfn Let $X$ be a Riemann surface and $p\in X$.
A function $F:X\to\P:=\C\cup\{\infty\}$ is called \jdef{unipolar} at $p$ iff
it it is meromorphic, has a simple pole at $p$, and is bounded (hence holomorphic)
in the complement of every neighborhood of $p$.
In other words, a unipolar function
is a holomorphic map $F:X\to\P$ such that $\infty$ is a regular value,
$F^{-1}(\infty)$ consists of a single point, and $F$ is \jdef{proper at infinity}
in the sense that for any sequence
$q_n\in X$ we have $\lim_{n\to\infty} F(q_n)=\infty\implies\lim_{n\to\infty}q_n=p$.
By the Extension Theorem~\ref{thm:extend} for any point $p$ in a nonhyperbolic
Riemann surface $X$
there is a unique function  $u$ which is
unipolar at $p$ (take $f=1/z$).
\nfd

\begin{lemma}\label{lem:unipolar!}
Assume that $X$ is nonhyperbolic  and that
$F'$ and $F$ are both unipolar at $p$. Then $F'=aF+b$ for some $a,b\in\C$.
\end{lemma}

\begin{proof}
For some constant $a$, $F'-aF$ has no pole at $p$ and is hence bounded and
holomorphic on $X$. On a nonhyperbolic  surface the only bounded holomorphic
functions are the constant functions.
\end{proof}

\begin{lemma}\label{lem:nearby}
Assume that $X$ is nonhyperbolic, that $p\in X$,
and that  $F:X\to\P:=\C\cup\{\infty\}$ is meromorphic,
has a simple pole at $p$,
and that $\Re(F)$ bounded in the complement of every neighborhood of $p$.
Then for $q$ sufficiently near (but distinct from) $p$ the function
$G(r)=1/(F(r)-F(q))$ is unipolar at $q$.
In particular, $G$ is unipolar at $q$ if $F$ is unipolar at $p$.
\end{lemma}

\begin{proof}
Since $F$ has a simple pole at $p$ it maps  a neighborhood  $U$ of $p$
diffeomorphically to a neighborhood of infinity by the Inverse Function Theorem.
Let $M=\sup_{r\notin U} u(r)$.
For $q$ sufficiently near $p$ we have $|F(q)|>2M$.
For such $q$ we have that $q$ is the only pole
of $G$ in $U$ (as $F$ is injective on $U$) and that $|G(r)|<M$
for $r\notin U$ (since $|G(r)|=1/|F(r)-F(q)|\le1/|u(r)-F(q)|<1/M$).
Thus $q$ is the only pole of $G$.
Since $G=L\circ F$ where $L(w)=1/(w-F(q))$ we have that
$G$ maps $U$  diffeomorphically to a neighborhood of infinity
so $G$ is proper at infinity
so  $G$ is unipolar as required.
\end{proof}

\begin{lemma}\label{lem:unipolarExist}
Assume $X$ is simply connected and nonhyperbolic and that $p\in X$.
Then there is a  function $F$ unipolar at $p$.
\end{lemma}

\begin{proof} Use Theorem~\ref{thm:extend} with $f(z)=1/z$.
As $X$ is simply connected the resulting function $u$ is the real
part of a meromorphic function $F=u+iv$
with $u$ bounded in the complement of every neighborhood of $p$,
and $F-1/z$ holomorphic in a neighborhood of $p$ and vanishing at $p$.
We must show that $v$ is also bounded in the complement of every neighborhood of $p$.
Apply Theorem~\ref{thm:extend} with $f(z)=i/z$.
We get a meromorphic function $\tilde F=\tilde u+i\tilde v$ with
$\tilde u$ bounded in the complement of every neighborhood of $p$ and  $\tilde F-i/z$ holomorphic in a neighborhood of $p$.
Thus to prove that $v$ is bounded in the complement of every neighborhood of $p$ it is enough to show
that $\tilde F=iF$ for then $v=-\tilde u$.

By Lemma~\ref{lem:nearby} the functions $G(r)=1/(F(r)-F(q))$
$\tilde G(r)=1/(\tilde F(r)-\tilde F(q))$  are unipolar
for $q$ sufficiently near  $q$.
Then for suitable constants $a$ and $\tilde a$ the function
$aG(r)+\tilde a\tilde G(r)$ has no pole at $q$ (and hence no pole at all)
and hence, as $X$ is nonhyperbolic, must be constant.
solve the equation $aG(r)+\tilde a\tilde G(r)=c$ for $\tilde F$ in terms of $F$.
Then $\tilde F= (\alpha F+\beta)/(\gamma F+\delta)$. But
$F(z)=1/z+R(z)$ and $\tilde F(z)=i/z+\tilde R(z)$
where $R$ and $\tilde R$ vanish at $p$.
Hence $\tilde F=iF$ as claimed.
\end{proof}

\begin{lemma}
If $F$ is unipolar at $p$ and
$F'$ is unipolar at $q$, then $F'=L\circ F$ for
some automorphism $L$ of $\P$.
\end{lemma}

\begin{proof} Fix $p$ and let $S$ be the set of points $q$ where
the lemma is true. By Lemma~\ref{lem:unipolar!} $p\in S$ so it suffices to show that
$S$ is open and closed. Choose $q_0\in S$ and let $F_0$ be
unipolar at $q_0$. By Lemma~\ref{lem:nearby}
the function $F(r)=1/(F_0(r)-F_0(q))$ is unipolar at
$q$ for $q$ sufficiently near $q_0$. Now $F=L\circ F_0$ where $L(w)=1/(w-F_0(q))$
so by Lemma~\ref{lem:unipolar!} (and the fact that the automorphisms form a group)
the lemma holds for $q$ sufficiently near $q_0$,
i.e. $S$ is open. Now choose $q\in X$ and assume that $q=\lim_{n\to\infty}q_n$
where $q_n\in S$. By Lemma~\ref{lem:unipolarExist} let $F'$ be unipolar at $p$.
For $n$ sufficiently large $G'(r)=1/(F'(r)-F'(q_n))$ is unipolar at $q_n$
by Lemma~\ref{lem:nearby}. Hence $G'=L\circ F$ for some $L$
so $F'=F'(q_n)+1/(L\circ F)=L'\circ F$ so $q\in S$. Thus $S$ is closed.
\end{proof}



\begin{lemma} \label{lem:unipolar-injective}
Assume $X$ is simply connected  and nonhyperbolic.
Then a  unipolar function is injective.
\end{lemma}


\begin{proof} Suppose that $F$ is unipolar at some point $o\in X$
and assume that $F(p)=F(q)$. Let $F_p$ be unipolar at $p$.
Then there is an automorphism $L$ with $F_p=L\circ F$. Thus
$F_p$ has a pole at $q$ so $q=p$.
\end{proof}

\begin{proof}[Proof of the Uniformization Theorem continued]
By Lemma~\ref{lem:unipolar-injective} we may assume that
$X$ is an open subset of $\P=\C\cup\{\infty\}$.
If $X$ is elliptic we must have $X=\P$.
If $X=\P\setminus\{a\}$ then a suitable automorphism
of $\P$ maps $X$ to $\P\setminus\{\infty\}=\C$.
Hence it suffices to show that a simply connected open subset of $\P$ which omits
two points admits a bounded nonconstant holomorphic function and is hence hyperbolic.
By composing with an automorphism of $\P$ we may assume that
$X\subset\C\setminus\{0,\infty\}$. As $X$ is simply connected there is a square root
function $f$ defined on $X$,i.e. $f(z)^2=z$ for $z\in X$.
Hence $X\cap(-X)=\emptyset$
else $z=f(z)^2=f(-z)^2=-z$ for some $z\in X$ so either $0$ or $\infty$ is in $X$,
a contradiction.
As $-X$ is open the function $g(z)=1/(z-a)$ is bounded on $X$ for $a\in-X$.
\end{proof}

\section{Surfaces with abelian fundamental group}



\para
 The Uniformization Theorem classifies all connected Riemann surfaces $X$
whose fundamental group  $\pi_1(X)$ is trivial. In this section we
extend this classification to  surfaces $X$
whose fundamental group us abelian. We also determine the automorphism group
of each such $X$.
Note that the upper half plane $\H$ and
the unit disk $\D$ are isomorphic via the diffeomorphism $f:\H\to\D$
defined by $f(z)=(1+zi)/(1-zi)$.
\arap

\begin{theorem}\label{thm:abelian}
The connected Riemann surfaces
with abelian fundamental group are
\begin{description}
\item[(i)] the plane $\C$,
\item[(ii)] the upper half plane $\H$,
\item[(iii)] the Riemann sphere $\P=\C\cup\{\infty\}$,
\item[(iv)] the punctured plane $\C^*=\C\setminus\{0\}$,
\item[(v)] the punctured disk $\D^*=\D\setminus\{0\}$,
\item[(vi)] the annulus $\D_r=\{z\in\D: r<|z|\}$ where $0<r<1$,
\item[(vii)] the torus $\C/\Lambda_\tau$ where $\tau\in\H$ and $\Lambda_\tau=\Z+\Z\tau$.
\end{description}
\end{theorem}



\begin{theorem}\label{thm:distinct}
No  two of these are isomorphic except that $\C/\Lambda_\tau$ and $\C/\Lambda_{\tau'}$
are isomorphic if and only if $\Q(\tau)=\Q(\tau')$, i.e.
if and only if $\tau'=g(\tau)$ for some $g\in\mathrm{SL}_2(\Z)$.
\end{theorem}

\begin{theorem}\label{thm:Aut-abelian}
The automorphism groups of these surfaces $X$ are as follows.
\begin{description}
\item[(i)] The group $\Aut(\C)$of automorphisms of $\C$
is the group  consisting of  transformations $\phi$ of form
$$
    \phi(z)=az+b
$$
where $a,b\in\C$ and $a\ne 0$.
\item[(ii)] The group $\Aut(\P)$of automorphisms of the Riemann sphere $\P$
is the group $\mathrm{PGL}(2,\C)$ of all transformations $\phi$ of form
$$
    \phi(z)=\frac{az+b}{cz+d}
$$

where $a,b,c,d\in\C$ and $ad-bc\ne 0$.
\item[(iii)] The group $\Aut(\H)$ of automorphisms of the upper half plane $\H$
is the group $\mathrm{PGL}(2,\R)$ of all transformations $\phi$ of form
$$
    \phi(z)=\frac{az+b}{cz+d}
$$
where $a,b,c,d\in\R$ and $ad-bc\ne 0$.
\item[(iv)] The group $\Aut(\C^*)$ of automorphisms of the punctured plane $\C^*$
is the group of all transformations $\phi$ of one of the forms
$$
    \phi(z)=az \qquad \mbox{or} \qquad \phi(z)=\frac{a}{z}
$$
where $a\in\C$ and $a\ne 0$.
\item[(v)] The group $\Aut(\C/\Lambda_\tau)$ of automorphisms of the torus $\C/\Lambda_\tau$
is the group of all transformations $\phi$ of form
$$
    \phi(z+\Lambda_\tau)=az+b+\Lambda_\tau
$$
where $b\in\C$ and $a=1$ if $\tau\notin\Q(i)\cup\Q(j)$,
$a^4=1$ if $\tau\in\Q(i)$, and $a^6=1$ if $a\in\Q(j)$.
(Here $j$ is the intersection point in $\H$ of the two circles
$|z|=1$ and $|z-1|=1$.)
\item[(vi)] The group $\Aut(\D^*)$ of automorphisms of the punctured disk $\D^*$
is the group of all transformations $\phi$   of form
$$
    \phi(z)=az
$$
where $a\in\C$ and $|a|=1$.
\item[(vii)] The group $\Aut(\D_r)$ of automorphisms of the annulus $\D_r$
is the group of all transformations $\phi$ of one of the forms
$$
    \phi(z)=az \qquad \mbox{or} \qquad \phi(z)=\frac{ar}{z}
$$
where $a\in\C$ and $|a|=1$.
\end{description}
\end{theorem}

\begin{theorem} A Riemann surface has abelian fundamental group if and only
if its automorphism group is not discrete.
\end{theorem}


\para Fix a connected Riemann surface $X$.
By the Uniformization Theorem the universal cover $\tilde X$ of a (connected)
Riemann surface $X$ is one of $\P$, $\C$, or $\H\simeq\D$ and hence
$X$ is isomorphic to $\tilde X/G$ where $G\subset\Aut(\tilde X)$
is the group of deck transformations of the covering projection
$\pi:\tilde X\to X$, i.e.
$$
   G=\{g\in\Aut(\tilde X): \pi\circ g=\pi\}.
$$
Note that $G$ is discrete and acts freely.
\arap

\begin{lemma} The automorphism group of $X=\tilde X/G$ is isomorphic to the
quotient $N(G)/G$ where
$$
    N(G)=\{\phi\in\Aut(\tilde X):\phi\circ G\circ\phi^{-1}=G\}
$$
is the \jdef{normalizer} of $G$ in $\Aut(\tilde X)$.
\end{lemma}



\begin{proof}[Proof of Theorem~\ref{thm:Aut-abelian}(i)]
Let $\phi\in\Aut(\C)$. Then $\phi$ is an entire function.
It cannot have an essential singularity at infinity by
Casorati-Weierstrass and the pole at infinity must be simple  as
$\phi$ is injective. Hence $\phi(z)=az+b$.
\end{proof}




\begin{proof}[Proof of Theorem~\ref{thm:Aut-abelian}(ii)] Choose $\phi\in\Aut(\P)$.
After composing with an element of $\mathrm{PGL}(2,\C)$ we may
assume that infinity is fixed, i.e. that $\phi(z)=az+b$.
\end{proof}



\begin{proof}[Proof of Theorem~\ref{thm:Aut-abelian}(iii)] Choose $\phi\in\Aut(\H)$.
Let $f:\H\to\D$ be the isomorphism given by $f(z)=(1+zi)/(1-zi)$.
Then $\psi:=f^{-1}\circ\phi\circ f$ is an automorphism of the disk
$\D$. Composing with $\alpha(z)=(z-a)/(\bar{a}z-1)$ we may assume
that $\psi(0)=0$. Then $|\psi(z)|\le|z|$ by the Maximum Principle
($\psi(z)/z$ is holomorphic) and similarly $|\psi^{-1}(z)|\le|z|$.
Hence $|\psi(z)|=|z|$ so $\psi(z)=cz$ where $|c|=1$ by the
Schwartz lemma. Hence $\phi\in\mathrm{PGL}(2,\C)$. The
coefficients must be real as the real axis is preserved so
$\phi\in\mathrm{PGL}(2,\R)$.
\end{proof}


\begin{proof}[Proof of Theorem~\ref{thm:Aut-abelian}(iv)]
The universal cover of the punctured plane $\C^*$ is the map
$$
  \C\to \C^*: z\mapsto \exp(2\pi i z).
$$
The group $G$ of deck transformations is
the cyclic group generated by the translation $z\mapsto z+1$.
The normalizer $N(G)$ of $G$ in $\Aut(\C)$ is \ETC
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:Aut-abelian}(v)]
The universal cover of the torus $\C/\Lambda_\tau$ is the map
$$
  \C\to \C/\Lambda_\tau: z\mapsto z+\Lambda_\tau.
$$
The group $G$ of deck transformations is
the abelian group generated by the translations $z\mapsto z+1$
and $z\mapsto z+\tau$.
The normalizer $N(G)$ of $G$ in $\Aut(\C)$ is \ETC
\end{proof}


\begin{proof}[Proof of Theorem~\ref{thm:Aut-abelian}(vi)]
The universal cover of the punctured disk $\D^*$ is the map
$$
  \H\to \D^*: z\mapsto \exp(2\pi i z).
$$
The group $G$ of deck transformations is
the cyclic group generated by the translation $z\mapsto z+1$ .
The normalizer $N(G)$ of $G$ in $\Aut(\D)$ is \ETC
\end{proof}


\begin{proof}[Proof of Theorem~\ref{thm:Aut-abelian}(vii)]
The universal cover of the annulus $\D_r$ is the map
$$
  \H\to D_r: z\mapsto \exp\left(\frac{\log r\log z}{\pi i}\right).
$$
Here $\log z$ denotes the branch of the logarithm satisfying $0<\Im(\log z)<\pi$
so writing $z=\rho e^{i\theta}$ the cover takes the form
$$
  \H\to D_r: \rho e^{i\theta}\mapsto r^{\theta/\pi}\exp\left(\frac{\log r\log \rho}{\pi i}\right).
$$
The group $G$ of deck transformations is the cyclic group generated by $z\mapsto az$  where
$a=\exp(-2\pi^2/\log r)$.
The normalizer $N(G)$ of $G$ in $\Aut(\D)$ is \ETC
\end{proof}


\begin{lemma} If $\tilde X=\P$ then $G=\{1\}$ so $X=\P$.
\end{lemma}

\begin{proof} Any nontrivial element of $\mathrm{PSL}(2,\C)$ has a fixed point.
\end{proof}


\begin{lemma} If $\tilde X=\C$ then the group $G$ consists of a discrete abelian group
of translations. More precisely $G$ is the set of all transformations
$f(z)=z+b$ where $b\in\Gamma$ and
where the subgroup $\Gamma\subset\C$ is one of the following:
\begin{description}
\item[(i)] $\Gamma=\{0\}$ in which case $X=\tilde X=\C$;
\item[(ii)] $\Gamma=\omega\Z$ in which case $X\simeq\C^*$;
\item[(iii)] $\Gamma=\omega_1\Z+\omega_2\Z$
in which case $X\simeq \C/\Lambda_\tau$, $\tau=\omega_2/\omega_1$.
\end{description}
\end{lemma}

\begin{proof} Any automorphism of form $z\mapsto az+b$ where $a\ne 1$ has a fixed point
so $G$ is a discrete group of translations. Kronecker's theorem says that for
$\omega\in\R$ the group $\Z\omega+\Z$ is dense in $\R$ if and only if
$\omega\notin\Q$. (Proof: Consider a minimal positive element of $\Z\omega+\Z$.)
It follows easily that a discrete subgroup of the additive group
$\R^n$ has at most $n$ generators. Hence the three possibilities.
In case~(ii) the group $G$ is conjugate in $\Aut(\C)$ to the cyclic
group generated by the translations $z\mapsto z+\tau$.
In case~(iii) the group $G$ is conjugate in $\Aut(\C)$ to the free abelian
group generated by the translations $z\mapsto z+1$ and $z\mapsto z+\tau$.
\end{proof}


\begin{corollary}[Picard's Theorem] An entire function $f:\C\to\C$ which omits
two points is constant.
\end{corollary}



\begin{proof} $\C\setminus\{a,b\}$ has a nonabelian fundamental group
so its universal cover must be $\D$. A holomorphic map
$f:\C\to\C\setminus\{a,b\}$ lifts to a map $\tilde f:\C\to\D$ which
must be constant by Liouville.
\end{proof}

\begin{lemma} A fixed point free automorphism  $\phi$  of $\H$ is
conjugate in $\Aut(\H)$ either to a homothety   $z\mapsto az$
where $a>0$ or to the translation $z\mapsto z+1$.
\end{lemma}

\begin{proof} Let $A\in\mathrm{SL}(2,\R)$ be a matrix representing
the automorphism $\phi$. Since $\phi$ has no fixed points in $\H$
the eigenvalues  of $A$ must be real.
Since their product is one we may rescale so that they are positive.
If there are two eigenvalues $\lambda$ and $\lambda^{-1}$
then $A$ is conjugate in $\mathrm{SL}(2,\R)$ to a diagonal matrix and
so $\phi$ is conjugate in $\Aut(\H)$ to $z\mapsto\lambda^2 z$. Otherwise
the only eigenvalue is $1$  and $A$ is conjugate in $\mathrm{SL}(2,\R)$ to
$\Mat{cc}1&1\\0&1\Rix$ so $\phi$ is conjugate in $\Aut(\H)$ to $z\mapsto z+1$.
\end{proof}

\begin{corollary} If $\tilde X=\H$ and $G$ is abelian, then the group $G$
is conjugate in $\Aut(\H)$ to either a free abelian group
generated by a homothety $z\mapsto az$ where $a>0$ or the free
abelian group generated by the translation $z\mapsto z+1$. In the
former case $X\simeq\D^r$ for some $r$ and in the latter case
$X\simeq\D^*$.
\end{corollary}

\begin{proof} If $G$ contains a homethety it must be a subgroup
of the group of homotheties (as it is abelian) and hence cyclic as it is discrete.
Similarly, If $G$ contains a translation it must be a subgroup
of the group of translations and hence cyclic.
\end{proof}


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\bibitem{REYSSAT} E. Reyssat:
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\end{document}