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\title{Homework 6 --- Properly discontinuous groups}

\author{Danny Calegari}
\address{Department of Mathematics \\ Harvard \\ Cambridge, MA 02138}
\email{dannyc@math.harvard.edu}

\maketitle

This homework is due Friday December 8th at the start of class.

\begin{prob}
Suppose $\Gamma$ is an infinite 
properly discontinuous group in $\Isom^+(\E^2)$ which contains
some elements of order $3$. Show that the set $P$ of fixed points of
elements of order $3$ form the vertices of a tessellation by equilateral
triangles or the vertices of a tessellation by equilateral hexagons.
Deduce {\em Napoleon's theorem}: if $T$ is a triangle, and
three equilateral triangles $T_1,T_2,T_3$ are constructed on the three sides
of $T$, then the centers of the triangles $T_i$ themselves are the vertices
of an equilateral triangle.

Hint: find a tessellation of $\E^2$ by copies of $T$ and the $T_i$ and
let $\Gamma$ be its group of symmetries. 
\end{prob}

\begin{prob}
If $\Gamma \subset \Isom^+(\E^2)$ is a properly discontinuous group of symmetries,
then we have seen that $\Gamma$ can be written as a semi--direct product
$$\Gamma = T \rtimes R$$
where $T$ is a group of translations, and $R$ is the subgroup of rotations
fixing some point. Find an example of a properly discontinuous group of
symmetries $\Gamma \subset \Isom(\E^2)$ which cannot be written as a semi--direct
product, even though there is a short exact sequence
$$1 \to T \to \Gamma \to S \to 1$$
where $S$ is a finite subgroup of $O(2,\R)$.
\end{prob}

\begin{rmk}
Every finite subgroup of $SO(2,\R)$ is {\em cyclic} --- that is, generated by
a single element. Thus a short exact sequence $$1 \to T \to \Gamma \to S \to 1$$
where $S$ is a finite subgroup of $SO(2,\R)$, {\em splits} --- that is,
$\Gamma$ is a semi--direct product. But if $S$ is not cyclic, this sequence
does not necessarily split. This is the explanation for the phenomenon
in the previous problem.
\end{rmk}

\begin{prob}
What is the $(2,2,\infty)$--spherical orbifold? Show that it admits a
flat (i.e. Euclidean)
structure, and identify its orbifold fundamental group as the infinite
dihedral group. How does this group act in an orientation preserving
way on $\E^2$?
\end{prob}

\begin{prob}
The group $S_4$ acts on $\R^4$ by permuting the co--ordinates. This action
is linear and fixes the origin; that is, it is a homomorphism of the
group into $O(4,\R)$. This action fixes the vector $(1,1,1,1)$, and therefore
it fixes the three dimensional subspace $\pi \subset \R^4$ of vectors
whose co--ordinates sum to $0$. Show that the restriction
$S_4|_\pi$ is the group of (not necessarily orientation--preserving)
symmetries of the regular tetrahedron centered at the origin in $\pi$
whose vertices are $$(3,-1,-1,-1),(-1,3,-1,-1),(-1,-1,3,-1),(-1,-1,-1,3)$$
\end{prob}

\begin{prob}
Let $T$ be an arbitrary Euclidean triangle. Show that there is a tetrahedron
$\Delta$ whose four sides are congruent to $T$ which has a
$\Z/2\Z \oplus \Z/2\Z$ group of symmetries. This group is also known as
the {\em Klein $4$--group}. Here a {\em tetrahedron} is just a polygonal
sphere with four faces which are flat Euclidean triangles. It is not assumed
that the tetrahedron is the boundary of a solid object in $\R^3$ with flat
faces. Show the spherical polyhedron can be realized as the boundary of
a solid tetrahedron (in the usual sense) in a way which preserves the
group of symmetries, if $T$ is {\em acute}.
\end{prob}

\begin{prob}
Let $S$ be a genus $2$ surface. Show that $S$ can be built from regular right
angled pentagons with four pentagons around every vertex. If $\gamma$ is
a geodesic (a straight closed curve) in the $1$--skeleton of the surface so
obtained, we can {\em twist} the surface along this curve, and reglue the
pentagons on either side in a new configuration.
Enumerate the combinatorially distinct decompositions
of $S$ into right angled pentagons, and show how they are related by
twisting in a curve.
\end{prob}

\begin{prob}
Let $\Sigma$ be a hyperbolic orbifold (possibly obtained as a quotient of $\H^2$ by
a group containing some orientation--reversing elements). Using the Gauss--Bonnet theorem
for orbifolds, show that the area of $\Sigma$ is at least $\frac \pi {42}$.
(Hint: show $\Sigma$ has a double cover which is an orientable orbifold and
therefore has at worst point singularities.)
Deduce that if $S$ is any hyperbolic surface of genus $g$, the group $G$
of orientation--preserving symmetries of $S$ has at most $84(g-1)$ elements.
\end{prob}

\begin{prob}
Let $p>3$ be a prime number and let
$\Gamma(p)$ denote the kernel of the obvious homomorphism
$$\phi_p:PSL(2,\Z) \to PSL(2,\Z/p\Z)$$
\begin{itemize}
\item{What is the order of the group $PSL(2,\Z/p\Z)$?}
\item{Show that $\Gamma(p)$ acts on $\H^2$ freely. What is the
hyperbolic area of the quotient surface?}
\item{For $p=7$ identify the quotient $\H^2/\Gamma(7)$ explicitly, and
find a decomposition of it into ideal triangles.}
\end{itemize}
\end{prob}

\begin{prob}
Recall that a closed curve on a surface $\Sigma$ is {\em simple} if it is embedded, and
{\em essential} if it does not bound a disk in $\Sigma$.
If $\gamma,\delta$ are two essential simple closed curves on a hyperbolic surface which
do not intersect, show that their geodesic representatives do not intersect.
Deduce that a combinatorial pair--of--pants decomposition of a hyperbolic surface can be
realized by cutting along $3g-3$ geodesics.
\end{prob}

\begin{prob}[hard]
Let $\alpha_1,\alpha_2,\beta_1,\beta_2$ be four disjoint infinite geodesics in $\H^2$
which are not ``nested'', in the sense that there is a point $p$ which can
be joined by arcs to any one of
the $\alpha_i,\beta_i$ without crossing any others.
Let $\gamma,\delta$ be two hyperbolic translations, where $\gamma$ takes
$\alpha_1$ to $\alpha_2$ and $\delta$ takes $\beta_1$ to $\beta_2$. 
\begin{itemize}
\item{Show that $\Gamma = \langle \gamma,\delta \rangle$, the group generated
by $\gamma$ and $\delta$, is a free group on two elements $\Z * \Z$.}
\item{Suppose $\alpha_1,\beta_1,\alpha_2,\beta_2$ are the sides (in order)
of an ideal quadrilateral. Show that the quotient $\H^2/\Gamma$ has finite area
if and only if the commutator $[\gamma,\delta]=\gamma\delta\gamma^{-1}\delta^{-1}$
is parabolic.}
\end{itemize}
\end{prob}

Thanks to Andrew Cotton for pointing out some errors and omissions in the first
version of this problem set.
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