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\begin{document}

\centerline{\Large Homework 8: M\"obius transformations, Riemann
  surfaces}

\vskip 1cm

Notation: $\mathbb D=\{z|\mid |z|<1\}$ and $\mathbb H=\{z\mid Im(z)>0\}$.
\vskip 1cm
{\bf M\"obius transformations.}
\begin{enumerate}[1.]
  \item Determine which of the following transformations are
    elliptic, parabolic, loxodromic. Recall that I don't distinguish
    between ``hyperbolic'' and ``loxodromic'' (if you are looking at Ahlfors).
    \begin{enumerate}[(i)]
    \item $z\mapsto z/(2z-1)$.
    \item $z\mapsto 2z/(3z-1)$.
    \item $z\mapsto (3z-4)/(z-1)$.
    \item $z\mapsto z/(2-z)$.
    \end{enumerate}
    
\item Let $f:\mathbb D\to\mathbb D$ be holomorphic. If $f(0)=a$ show
  that $f$ has no zeros in $\{|z|<|a|\}$. Hint: Find an automorphism
  of $\mathbb D$ that interchanges $0$ and $a$ and apply the Schwarz lemma.
\end{enumerate}
\vskip 1cm
{\bf Riemann surfaces}

\begin{enumerate}[resume]

\item In class we have seen that the unit disk $\mathbb D$ and the upper
  half-plane $\mathbb H$ are biholomorphically (or conformally)
  equivalent, and they are both equivalent to the strip $\{z\mid
  0<Im(z)<\pi\}$. Show that the following domains in $\C$ are also
  conformally equivalent to these three. Note: The Riemann mapping
  theorem says that every simply connected domain, not all of $\C$, is
  conformally equivalent to $\mathbb D$. The point here is that the
  conformal equivalence is realized by explicit maps.
  \begin{enumerate}[(i)]
  \item The sector $\{z\mid 0<arg(z)<\alpha\}$ for any $\alpha\in
    (0,2\pi]\}$.
    \item Upper half-plane with a slit $\mathbb H\smallsetminus
      [0,i]$. Hint: $z^2$.
      \item Upper half-plane with a circular slit $\mathbb
        H\smallsetminus \{z\mid |z|=1, Re(z)\geq 0\}$.
        \item Half-strip $\{z\mid Re(z)<0, 0<Im(z)<\pi\}$. Hint:
          $e^z$.
  \end{enumerate}

  \item Show that $\C\smallsetminus\{0\}$ and $\mathbb
    D\smallsetminus\{0\}$ are not conformally equivalent. Hint: Liouville.

  \item Recall from class that
    $$X(a,b,c,\infty)=\{(z,w)\in\hat\C\times\hat\C\mid
    w^2=(z-a)(z-b)(z-c)\}$$ is a Riemann surface for $a,b,c\in\C$
    distinct. It is a double branched cover over $\hat \C$ branched
    over $a,b,c,\infty$. Show that if the cross-ratios
    $(a,b,c,\infty)$ and $(a',b',c',\infty)$ are equal after possibly
    permuting the points, then $X(a,b,c,\infty)$ and $X(a,b,c,\infty)$
    are conformally equivalent. Start by observing that the
    permutation can be taken to fix $\infty$. Hint 1: This is really
    an exercise in covering spaces. Hint 2: It can also be an exercise
    in pull-backs. There is a diagram in the shape of a cube with the
    front and back faces pull-back diagrams that define
    $X(a,b,c,\infty)$ and $X(a',b',c',\infty)$, all faces are
    commutative, and maps relating front and back are (biholomorphic)
    homeomorphisms.  Note: The surface $X(a,b,c,\infty)$ is a torus,
    and we will see that if the cross-ratios are different then the
    tori are not biholomorphic. In fact, the ``moduli space'' of tori
    (or ``elliptic curves'' in algebraic geometry) is the space of
    quadruples of distinct points in the Riemann sphere with the
    cross-ratio equivalence. Tori, along with annuli, have simplest
    interesting moduli spaces.

    \comm{
    \item The cylinder $\C/2\pi i$ can be identified with
      $\C\smallsetminus\{0\}$ via $z\mapsto exp(z)$ and then
      compactified to the Riemann sphere by adding two points at
      infinity. The cylinder $X=\C/2\pi$ can similarly be compactified
      by first identifying it with the original cylinder via a
      rotation. The function $\sin:\C\to\C$ factors through $X$ and
      gives $S:X\to\C$. Show that $S$ extends to a holomorphic
      function $\hat S:\hat X\to\hat \C$ between the compactifications
      and then show that $\hat S$ is a double branched cover.
    }

    \item This problem is more involved than most. In the class we
      really considered $X=X(0,1,2,\infty)$. The projection to the
      second coordinate $X\to\hat\C$, $(z,w)\mapsto w$, is
      holomorphic, hence a branched cover. Compute branch points,
      their preimages, and the local degree at each. To start, compute
      all this for the map $\hat C\to \hat C$ given by the polynomial
      $z(z-1)(z-2)$.  Warning: Unlike the projection to $z$, after
      removing branch points and their primages the resulting covering
      map is not regular.
    
  \end{enumerate}

\subsection*{A ``well-known'' identity.}

The goal here is to prove that the ``principal'' sum

$$\sum_{n\in\Z} \frac 1{z-n} = \pi\cot(\pi z)$$

The sum doesn't converge in the usual sense, just like the harmonic
series doesn't, but if we group the terms for $n$ and $-n$ it does:


\begin{enumerate}[resume]

\item Show that $f(z)=\frac 1z + \sum_{n\geq 1} \frac {2z}{z^2-n^2}$
  converges absolutely for every $z\not\in\Z$, and $f$ is a
  meromorphic function on $\C$ with simple poles at $\Z$ and residue 1
  at each.

\item Show that $f(z+1)=f(z)$ for every $z$. Hint: $f(z)$ is the limit
  of $\sum_{n=-N}^N \frac 1{z-n}$.

\item Let $X$ be the Riemann surface $\C/z\sim z+1$, i.e. a
  cyclinder. Thus $f$ defines a holomorphic function $F:X\to\hat
  \C$. Let $\hat X$ be the compactification of $X$ biholomorphic to
  $\hat \C$, with two ideal points $\pm \infty$ (corresponding to
  imaginary part going to $\pm\infty$) as in Problem 6. Show that $F$
  extends to a holomorphic function $\hat F:\hat X\to\hat \C$ via
  $\hat F(\pm\infty)= \mp\pi i$. Hint: It suffices to extend
  continuously. When $z=Mi$ this amounts to computing $-2Mi\sum_{n\geq
    1} \frac 1{M^2+n^2}$. The sum can be estimated with $\int_0^\infty
  \frac 1{M^2+x^2}dx$, which can be computed via calculus or calculus
  of residues.

\item Show that $g(z)=\pi\cot(\pi z)$ also has the following
  properties: it is meromorphic with simple poles at $\Z$ and residue
  1 at each, $g(z+1)=g(z)$, and the induced map $G$ on the cylinder
  $X$ extends to a function $\hat G$ on the compactification with the
  same values at $\pm\infty$.

\item Show that this implies $f=g$. Hint: Show that $\hat F\hat
  G^{-1}$ is a M\"obius transformation that fixes three points.

\end{enumerate}

There is a more direct way to do this, but I think it involves more
calculations. Namely, using the method to solve Basel's problem, we
can calculate $f(z)$ for all $z$. The advantage of the method outlined
above is that the only real work is calculating $f(\pm\infty)$, the
rest is citing theorems about the Riemann sphere and M\"obius
transformations. 





  

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