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\begin{document}

\centerline{\Large Hyperbolic geometry}

\vskip 1cm

Notation: $\mathbb D=\{z|\mid |z|<1\}$, the Poincar\'e disk model, and
$\mathbb H=\{z\mid Im(z)>0\}$, the upper half-plane model of the
hyperbolic plane $\H^2$. Also, $S^1_\infty$ is the circle at infinity,
represented by $\partial\mathbb D=\{|z|=1\}$ or by
$\partial\H=\R\cup\{\infty\}$. 
\vskip 1cm
\subsection*{Synthetic geometry}

\begin{enumerate}[1.]
  \item
  In class we have seen that given a geodesic $\ell$ and a point $x$
  there is a unique geodesic through $x$ perpendicular to $\ell$. Show
  that if $\ell,\ell'$ are two disjoint geodesics whose endpoints on
  $S^1_\infty$ are all distinct, then there is a unique geodesic
  perpendicular to both. Show also that if $\ell,\ell'$ have a common
  endpoint (we say they are asymptotic) but are distinct, then there
  is no common perpendicular.
\end{enumerate}


       \subsection*{Distance computations.}
\vskip 1cm
       Very useful distances to remember are:
       \begin{itemize}
       \item In $\H$: $d(ai,bi)=\log(b/a)$ if $0<a<b$.
         \item In $\mathbb D$: $d(0,r)=\log\frac{1+r}{1-r}$, and of
           course $d(0,z)=d(0,|z|)$.
           \item $z\mapsto \frac{z-i}{z+i}$ sends $\H$ onto $\D$ and
             $i$ to 0.
       \end{itemize}
       

\begin{enumerate}[resume]

\item Consider the points $i$ and $x+i$ in $\H$, with $x>0$.
  \begin{enumerate}[(i)]
    \item Compute the hyperbolic distance between these two points and
      observe that for $x\to\infty$ it is approximately $2\log x$, up
      to a uniformly bounded error. Hint: There are many ways to do
      this, but I think the easiest is to map $x+i$ to $\mathbb D$ by
      a M\"obius transformation $\H\to\mathbb D$ that maps $i$ to 0
      and then take $\log\frac{1+r}{1-r}$ where $r$ is the norm of the
      image point.
      \item Compute the hyperbolic length of the path $t\mapsto t+i$
        for $t\in [0,x]$ and observe that it is exponentially longer
        for large $x$.
      \item Compute the length of the path $[i,xi]\cup [xi,x+xi]\cup
        [x+xi,x+i]$ and show that it is comparable to the hyperbolic
        distance between the endpoints (up to a uniformly bounded error).
  \end{enumerate}
  This comes up in ``real life''. If you are on google maps observing
  the math department at the U and you want to look at the Eiffel
  Tower, you would first zoom out, then go over to France, and then
  zoom in. The above is a model of this with zoomed in level represented
  by the horocycle $Im(z)=1$. For a 2-dimensional model one would look
  at the hyperbolic 3-space $\H^3$.

  There is also a combinatorial model for hyperbolic plane. It is
  pictured below, with all edges having length 1. It is good for rough
  calculations. For example, check the above calculations on the model.
  
    
      \begin{figure}[H]
        \centering
      \includegraphics[scale=0.5]{Baumslag-Solitar.png}
    \end{figure}
  


  
\item The goal of this exercise is to show that when $a,b\in\mathbb D$
  then the hyperbolic distance
  $$d(a,b)=\log\frac {1+\frac {|a-b|}{|1-a\overline b|}}
  {1-\frac {|a-b|}{|1-a\overline b|}}$$
  It is more complicated than our cross-ratio formula, but the
  advantage is that it doesn't involve the endpoints of the geodesic
  through $a$ and $b$.
  \begin{enumerate}[(i)]
  \item If $z_0\in\mathbb D, \theta\in\R$ then the transformation
    $$f:z\mapsto e^{i\theta}\frac {z-z_0}{1-z\overline z_0}$$
    is an automorphism of $\mathbb D$.
  \item Every automorphism of $\mathbb D$ has the form from (i).
  \item Denote $$\rho(a,b)=\bigg| \frac{a-b}{1-a\overline b}\bigg|$$
    for $a,b\in\mathbb D$.
    Note that $\rho(a,b)=\rho(b,a)$.
    The transformation $f$ from (i) sends $z_0\mapsto 0$. Show that
    $$\rho(z,z_0)=\rho(f(z),0)$$
    \item If $g$ is any automorphism of $\mathbb D$, show that
      $\rho(a,b)=\rho(g(a),g(b))$.
    \item Calculate directly that $\rho(0,r)=r$.
    \item Deduce the distance formula.
  \end{enumerate}
  
\item In $\H$ consider the geodesic $Re(z)=0$ and two straight lines $L_\pm$
  through the origin with slopes $\pm k$ for some $k>0$. Show that the
  distance from any point of $L_\pm$ to $Re(z)=0$ is some constant $C(k)$. You
  don't have to compute anything, just use the isometry group. These
  two lines are called ``equidistant lines''. Also
  show that if $|Re(w)/Im(w)|<1/k$ then the distance between $w$ and
  $Re(z)=0$ is less than $C(k)$ and if $|Re(w)/Im(w)|>1/k$ then the
  distance is $>C(k)$. Thus a metric neighborhood of $Re(z)=0$ is a
  sector. Sketch a metric neighborhood of a geodesic in $\H$
  represented by a semi-circle, and a metric neighborhood of a geodesic
  in $\mathbb D$. The name on the street for these is a ``banana
  neighborhood''. 

\end{enumerate}




\vskip 1cm
\subsection*{Isometries}

\begin{enumerate}[resume]

\item Suppose $f,g\in Isom_+(\H^2)$ are two commuting isometries and
  neither is identity. Show that one of the following holds:
  \begin{enumerate}[(i)]
  \item both $f,g$ are elliptic and fix the same point of $\H^2$, or
    \item both $f,g$ are parabolic and fix the same point on
      $S^1_\infty$, or
    \item both $f,g$ are hyperbolic and have the same axis.
  \end{enumerate}
  As an example, $z\mapsto z+1$ and $z\mapsto 2z$ do not commute, and
  in fact generate a solvable group called the Baumslag-Solitar group
  $BS(2,1)$. 
  Conclude that the rank 2 free abelian group $\Z^2$ doesn't act
  discretely and faithfully by orientation preserving isometries on
  $\H^2$.

\item Here we view $Isom_+(\H^2)=PSL_2(\R)$ as acting on $\H^2\cup
  S^1_\infty$. This is a Lie group.
  \begin{enumerate}[(i)]
    \item Show that the stabilizer of a point in $\H^2$ is a compact
      subgroup isomorphic to the circle $SO(2)$. (This is in fact a
      maximal compact subgroup of $PSL_2(\R)$.)
      \item Show that the stabilizer of a point in $S^1_\infty$ is a
        noncompact solvable Lie group.
  \end{enumerate}
  
    

\end{enumerate}


\subsection*{Hyperbolic Trigonometry}
\vskip 1cm
This is what a 1060 final in a hyperbolic universe might look like.

\begin{enumerate}[resume]
  \item Consider a geodesic triangle with one ideal vertex, one right
    angle, the finite side of hyperbolic length $d$, and the third
    angle $\alpha$. Prove the Lobachevsky formula
    $$\sin\alpha=\frac 1{\cosh d}$$
    As a guide, use the following figure. First show that $r=\frac
    1{\cos\alpha}-\tan\alpha$.

    
    \begin{figure}[H]
      \centering
      \includegraphics[scale=0.5]{lobachevsky.pdf}
    \end{figure}
    

    Thus if you lived in hyperbolic space, a straight line would
    appear to you at a visual angle $<\pi$ and moreover, from this
    visual angle you could compute the distance to the line!

  \item There are all kinds of trig formulas.
    Consider the following three:
    \begin{itemize}
    \item (Pythagorean theorem) If $a,b,c$ are the sides of a right
      triangle with $c$ opposite to the right angle, then
      $$\cosh c=\cosh a \cosh b$$
      \item (Law of Sines) In any triangle with sides $a,b,c$ and
        opposite angles $\alpha,\beta,\gamma$
        $$\frac{\sinh a}{\sin\alpha}=\frac{\sinh
          b}{\sin\beta}=\frac{\sinh c}{\sin\gamma}$$
      \item (Law of Cosines)
        $$\cosh c=\cosh a \cosh b - \sinh a \sinh b \cos\gamma$$
        (this generalizes the Pythagorean theorem)
    \end{itemize}
    I am not asking you to prove these, but consider the following
    thought experiment. Rescale the metric on $\H^2$ by larger and
    larger number. This amounts to ``zooming in''. In the limit you
    will get a metric space isometric to the Euclidean plane. By
    taking a sequence of suitably smaller and smaller triangles in
    $\H^2$ and rescaling, these triangles will converge to a Euclidean
    triangle. So one should be able to deduce Euclidean trig formulas
    from the hyperbolic ones! Here is a concrete question: Fix
    $a,b>0$ and consider a hyperbolic right triangle with sides
    $ta,tb$ and hypotenuse $tc(t)$, for $t>0$. Show that
    $$(c(t)^2-a^2-b^2)/t^2\to 0$$ as $t\to 0$. So in the limit
    $a,b,c(0)$ satisfy the Euclidean Pythagorean theorem.

    Do the same with the Law of Sines and deduce the Euclidean Law of
    Sines. 

  
\end{enumerate}



\subsection*{Gromov's $\delta$-hyperbolicity}
\vskip 1cm
Let $\delta\geq 0$. Unlike in calculus, this $\delta$ doesn't have to
be small. A triangle with vertices $A,B,C$ and sides $AB,AC,BC$ is
said to be $\delta$-thin if for every point $x$ on one of the sides,
there is a point $y$ on another side with $d(x,y)\leq\delta$. Vertices
belong to two sides, so you may as well assume $x$ is not a vertex,
otherwise $y=x$ works. If you take a large geodesic triangle in Euclidean
plane, it will be $\delta$-thin only for large $\delta$. A metric
space is said to be {\it $\delta$-hyperbolic} if there is some
$\delta\geq 0$ such that every geodesic triangle is $\delta$-thin. One
also says a metric space is Gromov hyperbolic if it is
$\delta$-hyperbolic for some $\delta$. This definition captures the
essential features of negative curvature. 
So
$\R^2$ with the usual metric is not Gromov hyperbolic.

\begin{enumerate}[resume]
  \item Show that $\H^2$ is Gromov hyperbolic. Hint: The ``worst''
    triangle is an ideal triangle, and you should show that it is
    $\delta$-thin for $\delta=\log(1+\sqrt 2)$.
  \item Use $\delta$-thinness of geodesic triangles to prove that
    every geodesic quadrilateral is $2\delta$-thin: every point on one
    of the sides is within distance $2\delta$ of another side.
  \item Consider a right angled triangle with sides $a,b,c$, with $c$
    being the longest (opposite the right angle). Show that $a+b-c$ is
    uniformly bounded, independently of the triangle. Hint: You can do
    this using the Pythagorean theorem, but a better proof is to use
    $\delta$-hyperbolicity. Namely, find a point on the long side
    that's within $\delta$ of the other two sides, and argue it's
    bounded distance from the vertex with the right angle.
  \item Use the previous problem to show the following. Let $\ell$ be
    a geodesic in $\H^2$, $x,y\in\H^2$ two points and $x',y'$ their
    nearest point (or orthogonal) projections to $\ell$. Show that
    there is a constant $C$ (some fixed multiple of $\delta$,
    e.g. $10\delta$) such that if
    $d(x',y')>4\delta$ (the hyperbolicity constant) then the length of
    $$[x,x']\cup[x',y']\cup [y',y]$$ is less than $d(x,y)+C$. So if
    $x',y'$ are sufficiently far apart, to go from $x$ to $y$ you may
    as well (up to uniformly bounded detour) go straight to $\ell$,
    then follow $\ell$ to $y'$ and then head straight to $y$.
  \item Gromov hyperbolicity is equivalent to the statement that the
    radius of an inscribed circle in every geodesic triangle is
    uniformly bounded. Calculate the radius of the inscribed circle in
    an ideal triangle. Answer: $(\log 3)/2$.

  \end{enumerate}
\end{document}
