gauss-bonnet.tex

\chapter{Gauss--Bonnet formula}
\label{chap:gauss-bonnet}
\section{Formulation}

The following theorem was proved by Carl Friedrich Gauss \cite{gauss}
for geodesic triangles;
Pierre Bonnet and Jacques Binet independently 
generalized the statement for arbitrary curves.
\index{Gauss--Bonnet formula}

\begin{thm}{Theorem}\label{thm:gb}
Let $\Delta$ be a topological disc in a smooth oriented surface $\Sigma$ bounded by a piecewise smooth curve $\partial\Delta$.
Suppose $\partial \Delta$ is oriented so that $\Delta$ lies on its left.
Then 
\[\tgc{\partial\Delta}+\iint_\Delta K=2\cdot \pi,\eqlbl{eq:g-b}\]
where $K$ denotes the Gauss curvature.
\end{thm}

In this chapter,
we give an informal proof of this formula in a leading special case.
A formal computational proof will be given in Section~\ref{sec:gauss--bonnet:formal}.
Before going into the proofs, we suggest solving the following exercises using the Gauss--Bonnet formula.

\begin{thm}{Exercise}\label{ex:1=geodesic-curvature}
Assume $\gamma$ is a simple closed curve with constant geodesic curvature $1$ in a smooth closed surface $\Sigma$ with positive Gauss curvature.
Show that $\length\gamma< 2\cdot\pi$;
that is, $\gamma$ is shorter than the unit circle.  
\end{thm}

\begin{thm}{Exercise}\label{ex:GB-hat}
Suppose that a disc $\Delta$ lies in a graph $z=f(x,y)$ of a smooth function,
the boundary of $\Delta$ lies in the $(x,y)$-plane 
and $\Delta$ meets this plane at fix angle $\alpha$.
Show that $\iint_\Delta K=2\cdot\pi\cdot(1-\cos\alpha)$.
\end{thm}


\begin{thm}{Exercise}\label{ex:geodesic-half}
Let $\gamma$ be a simple closed geodesic on a smooth closed surface $\Sigma$ with positive Gauss curvature.
Denote by $\Norm$ the spherical map on $\Sigma$.
Show that the curve $\alpha=\Norm\circ\gamma$ divides the sphere into two regions of equal area.

Conclude that $\length \alpha\ge 2\cdot\pi.$
\end{thm}

\begin{thm}{Exercise}\label{ex:closed-geodesic}
Let $\gamma$ be a closed geodesic possibly with self-intersections on a smooth closed surface $\Sigma$ with positive Gauss curvature.
Suppose $R$ is one of the regions that $\gamma$ cuts from~$\Sigma$.
Show that $\iint_R K\le 2\cdot\pi$.

Conclude that any two closed geodesics on $\Sigma$ have a common point.
\end{thm}

\begin{thm}{Exercise}\label{ex:self-intersections}
Let $\Sigma$ be a closed smooth surface with positive Gauss curvature. 
Show that in a single coordinate chart, a closed geodesic in $\Sigma$ cannot look like one of the curves on the following pictures.

\begin{figure}[h]
\begin{minipage}{.32\textwidth}
\centering
\includegraphics{mppics/pic-46}
\end{minipage}
\hfill
\begin{minipage}{.32\textwidth}
\centering
\includegraphics{mppics/pic-47}
\end{minipage}
\hfill
\begin{minipage}{.32\textwidth}
\centering
\includegraphics{mppics/pic-471}
\end{minipage}

\medskip

\begin{minipage}{.32\textwidth}
\centering
\caption*{\textit{(easy)}}
\end{minipage}
\hfill
\begin{minipage}{.32\textwidth}
\centering
\caption*{\textit{(tricky)}}
\end{minipage}
\hfill
\begin{minipage}{.32\textwidth}
\centering
\caption*{\textit{(hopeless)}}
\end{minipage}
\vskip-5mm
\end{figure}

\end{thm}


The following exercise optimizes \ref{ex:ruf-bound-mountain}.

\begin{thm}{Exercise}\label{ex:sqrt(3)}
Suppose $f\:\mathbb{R}^2\to\mathbb{R}$ is a $\sqrt{3}$-Lipschitz smooth convex function.
Show that any geodesic in the graph $z=f(x,y)$ has no self-intersections.
\end{thm}

A surface $\Sigma$ is called \index{simply-connected surface}\emph{simply-connected} if any simple closed curve in $\Sigma$ bounds a disc.
Equivalently, any closed curve in $\Sigma$ can be continuously 
deformed into a \index{trivial curve}\emph{trivial curve}; that is, a curve that stands at one point all the time.

Planes and spheres are examples of simply-connected surfaces;
tori and the cylinders are not simply-connected.

\begin{thm}{Exercise}\label{ex:unique-geod}
Suppose $\Sigma$ is a simply-connected open surface with nonpositive Gauss curvature.

\begin{subthm}{ex:unique-geod:unique}
Show that any two points in $\Sigma$ are connected by a unique geodesic.
In particular, geodesics in $\Sigma$ do not have self-intersections. 
\end{subthm}

\begin{subthm}{ex:unique-geod:diffeomorphism}
Conclude that for any point $p\in \Sigma$,
the exponential map $\exp_p$ is a diffeomorphism from the tangent plane $\T_p$ to~$\Sigma$.
In particular, $\Sigma$ is diffeomorphic to the plane.
\end{subthm}

\end{thm}

\section{Additivity}

Let $\Delta$ be a topological disc in a smooth oriented surface $\Sigma$ bounded by a simple piecewise smooth curve $\partial \Delta$.
As before we suppose $\partial \Delta$ is oriented in such a way that $\Delta$ lies on its left.
Set \index{10gb@$\GB$ (Gauss--Bonnet formula)}
\[\GB(\Delta)
\df
\tgc{\partial\Delta}+\iint_\Delta K-2\cdot \pi,
\eqlbl{eq:GB}\]
where $K$ denotes the Gauss curvature.
The Gauss--Bonnet formula can be written as
\[\GB(\Delta)=0,\]
and $\GB$ stands for Gauss--Bonnet.

{

\begin{wrapfigure}{r}{40 mm}
\vskip-16mm
\centering
\includegraphics{mppics/pic-1750}
\end{wrapfigure}

\begin{thm}{Lemma}\label{lem:GB-sum}
Suppose the disc $\Delta$ is subdivided into two discs $\Delta_1$ and $\Delta_2$ by a curve $\delta$.
Then
\[
\GB(\Delta)=\GB(\Delta_1)+\GB(\Delta_2).
\]
\end{thm}

\parbf{Proof.}
Let us subdivide $\partial \Delta$ into two curves $\gamma_1$ and $\gamma_2$ that share endpoints with $\delta$, so that $\Delta_i$ is bounded by the arc $\gamma_i$  and~$\delta$ for $i=1,2$.

}

Denote by $\phi_1$, $\phi_2$, $\psi_1$, and $\psi_2$ the angles between $\delta$ and $\gamma_i$ marked on the picture.
Suppose the arcs $\gamma_1$, $\gamma_2$, and $\delta$ are oriented as on the picture. 
Then
\begin{align*}
\tgc{\partial \Delta}&= \tgc{\gamma_1}-\tgc{\gamma_2}+(\pi-\phi_1-\phi_2)+(\pi-\psi_1-\psi_2),
\\
\tgc{\partial \Delta_1}&= \tgc{\gamma_1}-\tgc\delta+(\pi-\phi_1)+(\pi-\psi_1),
\\
\tgc{\partial \Delta_2}&= \tgc\delta-\tgc{\gamma_2}+(\pi-\phi_2)+(\pi-\psi_2),
\\
\iint_\Delta K&=\iint_{\Delta_1} K+\iint_{\Delta_2} K.
\end{align*}
It remains to plug in the results in the formulas for $\GB(\Delta)$, $\GB(\Delta_1)$, and $\GB(\Delta_2)$.
\qeds

\section{Spherical case}

If $\Sigma$ is a plane, then its Gauss curvature vanishes;
therefore the Gauss--Bonnet formula \ref{eq:g-b} can be written as 
\[\tgc{\partial\Delta}=2\cdot \pi,\]
and it follows from \ref{prop:total-signed-curvature}.
In other words, $\GB(\Delta)=0$ for any plane disc~$\Delta$ with piecewise smooth boundary.

If $\Sigma$ is the unit sphere, then $K\equiv1$;
in this case, Theorem~\ref{thm:gb} can be formulated in the following way:

\begin{thm}{Proposition}\label{prop:area-of-spher-polygon}
Let $P$ be a spherical polygon bounded by a simple closed broken geodesic $\partial P$.
Assume $\partial P$ is oriented such that $P$ lies on the left of $\partial P$.
Then 
\[\GB(P)=\tgc{\partial P}+\area P-2\cdot \pi=0.\]

Moreover, the same formula holds for any spherical region bounded by a piecewise smooth simple closed curve.
\end{thm}

This proposition will be used in the next section.

\parbf{Sketch of proof.}
Suppose a spherical triangle $\Delta$ has angles 
$\alpha$, $\beta$, and~$\gamma$.
According to \ref{lem:area-spher-triangle},
\[\area\Delta=\alpha+\beta+\gamma-\pi.\]

Recall that $\partial\Delta$ is oriented so that $\Delta$ lies on its left. 
Then its oriented external angles are  $\pi-\alpha$, $\pi-\beta$, and $\pi-\gamma$.
Therefore,
\[\tgc{\partial\Delta}=3\cdot\pi-\alpha-\beta-\gamma.\]
It follows that $\tgc{\partial\Delta}+\area \Delta=2\cdot\pi$ or, equivalently, $\GB(\Delta)=0$.
 
We can subdivide a given spherical polygon $P$ into triangles by cutting a polygon in two along a broken geodesic on each step.
By the additivity lemma (\ref{lem:GB-sum}), 
\[\GB(P)=0\]
for any spherical polygon~$P$.

The second statement can be proved by approximation.
One has to show that the total geodesic curvature of  
a piecewise smooth simple curve can be approximated by 
the total geodesic curvature of inscribed broken geodesics.
We omit the detailed proof,
but it close to \ref{ex:total-curvature=}.
\qeds


\begin{thm}{Exercise}\label{ex:half-sphere-total-curvature}
Let $\gamma$ be a simple piecewise smooth loop on on the unit sphere $\mathbb{S}^2$.
Assume $\gamma$ divides $\mathbb{S}^2$ in two regions of equal area.
Denote by $p$ the base point of~$\gamma$.
Show that the parallel transport $\iota_\gamma\:\T_p\mathbb{S}^2\to\T_p\mathbb{S}^2$ is the identity map.
\end{thm}



\section{Intuitive proof}\label{sec:gb-intuitive-proof}

In this section, we prove a special case of the Gauss--Bonnet formula.
This case is leading --- the general case can be proved similarly, using the signed area counted with multiplicity, but will do something else \ref{sec:gauss--bonnet:formal}.

\parbf{Proof of \ref{thm:gb} for open and closed surfaces with positive Gauss curvature.}
Let $\Norm\:\Sigma\to\mathbb{S}^2$ be the spherical map.
By \ref{cor:intK}, we have
\[\GB(\Delta)=\tgc{\partial\Delta}+\area[\Norm(\Delta)]-2\cdot \pi.
\eqlbl{eq:gb-area}\]

Choose a loop $\alpha$ that runs along $\partial\Delta$ so that $\Delta$ lies on the left from it; suppose that $p\in \partial\Delta$ is its base.
Consider the parallel translation $\iota_\alpha\:\T_p\to\T_p$ along $\alpha$.
According to \ref{prop:pt+tgc}, $\iota_\alpha$ is the clockwise rotation by angle $\tgc{\alpha}_\Sigma$.

Set $\beta=\Norm\circ\alpha$.
By \ref{obs:parallel=}, we have $\iota_\alpha=\iota_\beta$, where $\beta$ is considered as a curve in the unit sphere.

Further, $\iota_\beta$ is a clockwise rotation by angle $\tgc{\beta}_{\mathbb{S}^2}$.
By \ref{prop:area-of-spher-polygon} 
\[\GB(\Norm(\Delta))=\tgc{\beta}_{\mathbb{S}^2}+\area[\Norm(\Delta)]-2\cdot \pi=0.\]
Therefore, 
$\iota_\beta$ is a counterclockwise rotation by $\area[\Norm(\Delta)]$

Summarizing, the clockwise rotation by $\tgc{\alpha}_\Sigma$ is identical to a counterclockwise rotation by $\area[\Norm(\Delta)]$.
The rotations are identical if the angles are equal modulo $2\cdot\pi$.
Therefore, 
\[
\begin{aligned}
\GB(\Delta)&=\tgc{\partial\Delta}_\Sigma+\area[\Norm(\Delta)]-2\cdot \pi=
2\cdot n \cdot \pi
\end{aligned}
\eqlbl{eq:sum=2pin}\]
for an integer~$n$.

It remains to show that $n=0$.
By \ref{prop:total-signed-curvature}, this is so for a topological disc in a plane. 
One can think of a general disc $\Delta$ as the result of a continuous deformation of a plane disc. 
The integer $n$ cannot change during the process since the left hand side in \ref{eq:sum=2pin} is continuous along the deformation; whence $n=0$ for the result of the deformation.
\qeds

\section{Simple geodesic}

The following theorem provides an interesting application of the Gauss--Bonnet formula; it is proved by Stephan Cohn-Vossen \cite[Satz 9]{convossen}.

\begin{thm}{Theorem}\label{thm:cohn-vossen}
Any open smooth surface with positive Gauss curvature has a simple two-sided infinite geodesic.
\end{thm}

\parbf{Proof.}
Let $\Sigma$ be an open surface with positive Gauss curvature and $\gamma$ a two-sided infinite geodesic in~$\Sigma$.

If $\gamma$ has a self-intersection, then it contains a simple loop;
that is, for some closed interval $[a,b]$,
the restriction $\ell=\gamma|_{[a,b]}$ is a simple loop.

By \ref{ex:convex-proper-plane}, $\Sigma$ is parametrized by an open convex region $\Omega$ in the plane.
By Jordan's theorem (\ref{thm:jordan}), $\ell$ bounds a topological disc in $\Sigma$; denote it by~$\Delta$.
If $\phi$ is the internal angle at the base of the loop, then by the Gauss--Bonnet formula,
\[\iint_\Delta K=\pi+\phi.\] 

Recall that
\[\iint_\Sigma K\le 2\cdot\pi;
\eqlbl{intK=<2pi+}\]
see \ref{ex:intK:2pi}.
Therefore, $0<\phi<\pi$; that is, $\gamma$ has no concave simple loops.

Assume $\gamma$ has two simple loops, say $\ell_1$ and $\ell_2$ that bound discs $\Delta_1$ and $\Delta_2$, respectively.
Then the discs $\Delta_1$ and $\Delta_2$ have to overlap;
otherwise, the curvature of $\Sigma$ would exceed $2\cdot\pi$  contradicting \ref{intK=<2pi+}.

It follows that after leaving $\Delta_1$, the geodesic $\gamma$ has to enter it again before creating another simple loop.
\begin{figure}[h!]
\vskip-0mm
\centering
\includegraphics{mppics/pic-1550}
\end{figure}
Consider the moment when $\gamma$ enters $\Delta_1$ again;
two possible scenarios are shown in the picture.
On the left picture, we get two nonoverlapping discs which, as we know, is impossible.
The right picture is impossible as well --- in this case, we get a concave simple loop.

It follows that $\gamma$ contains only one simple loop.
This loop cuts a disc from $\Sigma$ 
and goes around it either clockwise or counterclockwise.
This way we divide all the self-intersecting geodesics on $\Sigma$
into two types which we call {}\emph{clockwise} and {}\emph{counterclockwise}.

The geodesic $t\mapsto \gamma(t)$ is clockwise 
if and only if the same geodesic traveled backwards
$t\mapsto \gamma(-t)$
is counterclockwise.
Let us shoot a unit-speed geodesic in each direction from a given point $p=\gamma(0)$.
This gives a one-parameter family of geodesics $\gamma_s$ for $s\in[0,\pi]$ connecting the geodesic $t\mapsto \gamma(t)$ with $t\mapsto \gamma(-t)$; that is, $\gamma_0(t)\z=\gamma(t)$, and $\gamma_\pi(t)=\gamma(-t)$.

The subset of values $s\in [0,\pi]$ such that $\gamma_s$ is right (or left) is open in $[0,\pi]$.
That is, if $\gamma_s$ is right, then so is $\gamma_t$ for all $t$ sufficiently close to~$s$.%
\footnote{Informally speaking, this means that self-intersection cannot suddenly disappear. Try to convince yourself of this.}

Since $[0,\pi]$ is connected, it cannot be subdivided into two open sets.
It follows that for some $s$, the geodesic $\gamma_s$ is neither  clockwise nor counterclockwise;
that is, $\gamma_s$ has no self-intersections.
\qeds

{

\begin{wrapfigure}{r}{17 mm}
\vskip-4mm
\centering
\includegraphics{mppics/pic-1575}
\end{wrapfigure}

\begin{thm}{Exercise}\label{ex:cohn-vossen}
Let $\Sigma$ be an open smooth surface with positive Gauss curvature.
Suppose $\alpha\:[0,1]\z\to \Sigma$ is a smooth loop such that $\alpha'(0)+\alpha'(1)=0$.
Show that there is a simple two-sided infinite geodesic $\gamma$ that is tangent to $\alpha$.
\end{thm}

}

\section{General domains}
\index{Gauss--Bonnet formula}

The following generalization of the Gauss--Bonnet formula is due to Walther von Dyck \cite{dyck}.

\begin{thm}{Theorem}\label{thm:GB-generalized}
Let $\Lambda$ be a compact domain on a smooth surface.
Assume $\Lambda$ is bounded by a finite (possibly empty) collection of closed piecewise smooth curves $\gamma_1,\dots,\gamma_n$.
Suppose that each $\gamma_i$ 
is oriented in such a way that $\Lambda$ lies on its left.
Then
\[\iint_\Lambda K=2\cdot \pi\cdot \chi-\tgc{\gamma_1}-\dots-\tgc{\gamma_n}\eqlbl{eq:g-b++}\]
for an integer $\chi=\chi(\Lambda)$.

Moreover, if a graph with $v$ vertices and $e$ edges embedded in $\Lambda$ subdivides it into $f$ discs and contains all $\gamma_i$, then $\chi=v-e+f$.
\end{thm}

The number $\chi=\chi(\Lambda)$ is called the \index{Euler characteristic}\emph{Euler characteristic} of $\Lambda$. 
The value $\chi$ does not depend on the choice of the subdivision since the remaining terms in \ref{eq:g-b++} do not.
Formula \ref{eq:g-b++} is deduced from the standard Gauss--Bonnet (\ref{thm:gb}).
The argument is similar to \ref{lem:GB-sum}, but the combinatorics gets trickier.

\begin{wrapfigure}{o}{38 mm}
\vskip-0mm
\centering
\includegraphics{mppics/pic-1580}
\end{wrapfigure}

Before reading the proof, try to modify the argument in \ref{lem:GB-sum} to derive the formula for the subdivision of the annulus $A$ on the picture;
its graph has 4 vertices and 6 edges (one of them is a loop) and it subdivides $A$ into two discs $\Delta_1$ and $\Delta_2$.
Therefore, $\chi(A)=4-6+2=0$ and  
\[\iint_A K=-\tgc{\gamma_1}-\tgc{\gamma_2}.\]


\parbf{Proof.}
Suppose that a graph with $v$ vertices and $e$ edges subdivides $\Lambda$ into $f$ discs $\Delta_1,\dots,\Delta_f$.
Lets us apply the Gauss--Bonnet formula for each disc, and sum up the results:
\[
\begin{aligned}
\iint_\Lambda K&=\iint_{\Delta_1} K+\dots+\iint_{\Delta_f} K=
2\cdot f\cdot \pi-\tgc{\partial\Delta_1}-\dots-\tgc{\partial\Delta_f}.
\end{aligned}
\]
It remains to show that  
\[\tgc{\gamma_1}+\dots+\tgc{\gamma_n}-\tgc{\partial\Delta_1}-\dots-\tgc{\partial\Delta_f}
=
2\cdot\pi\cdot(v-e).
\eqlbl{eq:GB-sum}\]
To prove this identity, we compute the left-hand side adding up the contributions of each edge and each vertex separately.

Choose an edge $\sigma$ of the graph.
If $\sigma$ is not part of some $\gamma_i$,
then it appears twice in the boundary of the discs, say in  $\partial \Delta_i$ and $ \partial \Delta_j$. 
In the latter case, we may assume that $\Delta_i$ lies on the left from $\sigma$ and $\Delta_j$ is on its right, so 
 $\sigma$ contributes $\tgc\sigma$ to $\tgc{\partial\Delta_i}$ and $-\tgc\sigma$ to $\tgc{\partial\Delta_j}$; hence $\sigma$ contributes nothing to the left-hand side of \ref{eq:GB-sum}.
It might happen that $i=j$ (as for one of the edges on the picture above);
in this case, we have one disc on both sides of~$\sigma$, but still, it contributes nothing to \ref{eq:GB-sum}.
If $\sigma$ is part of some $\gamma_i$, then it also appears once in the boundary of some disc, say in $\partial \Delta_j$.
We may assume that both $\Lambda$ and $\Delta_j$ lie on the left from $\sigma$,
so it contributes $\tgc\sigma$ to both $\tgc{\gamma_i}$ and $\tgc{\partial\Delta_j}$.
These contributions cancel each other on the left-hand side of  \ref{eq:GB-sum}.

Summarizing, the contributions to the left-hand side of  \ref{eq:GB-sum} that come from the total geodesic curvatures of edges cancel out.

Let us now handle the external angles at the vertices.
Choose a vertex $p$.
Denote by $d$ the \index{degree of vertex}\emph{degree} of~$p$;
that is, the number of edges that end at~$p$.

\begin{wrapfigure}{r}{23 mm}
\vskip-6mm
\centering
\includegraphics{mppics/pic-1585}
\end{wrapfigure}

Suppose $p$ lies in the interior of $\Lambda$.
Denote by $\delta_1,\dots,\delta_d$ the internal angles at $p$ of the discs containing $p$ as a vertex. 
Let  $\phi_{i}=\pi-\delta_{i}$ be the corresponding external angles.
Then $p$ contributes  
$-\phi_1-\dots-\phi_d$ to the sum.
Note that $\delta_1+\dots+\delta_d=2\cdot\pi$;
so $p$ contributes
\[-\phi_1-\dots-\phi_d = (\delta_1+\dots+\delta_d) - d\cdot \pi=(2-d)\cdot \pi,\]
to the left-hand side of \ref{eq:GB-sum}.

\begin{wrapfigure}{r}{23 mm}
\vskip-0mm
\centering
\includegraphics{mppics/pic-1590}
\end{wrapfigure}

If $p$ lies on the boundary of $\Lambda$, then it is a vertex of $d-1$ internal angles
$\delta_1,\dots,\delta_{d-1}$, and $\phi_{i}\z=\pi-\delta_{i}$ are the corresponding external angles.
Note that
\[\delta_1+\dots+\delta_{d-1}\z=\pi-\theta,\]
where $\theta\in(-\pi,\pi)$ denotes the external angle of $\Lambda$ at $p$.
And again, $p$ contributes
\[\theta-\sum\phi_{i}=(2-d)\cdot \pi\]
to the left-hand side of \ref{eq:GB-sum}.

Summarizing, suppose $p_1,\dots,p_v$ are the vertices of the graph, and $d_1,\dots,d_v$ are their corresponding degrees.
Then the total contribution from external angles to the left-hand side of  \ref{eq:GB-sum} is 
\[2\cdot v\cdot \pi-(d_1+\dots+d_v)\cdot\pi.
\eqlbl{eq:GB-sum-d}\]
Since the edges contributed nothing, the left-hand side of  \ref{eq:GB-sum} equals \ref{eq:GB-sum-d}.

It remains to observe that $d_1+\dots+d_v=2\cdot e$.
Indeed, $d_1+\dots+d_v$ is the total number of ends of all edges in the graph, and each of $e$ edges has exactly two ends. 
This finishes the proof of \ref{eq:GB-sum} and \ref{eq:g-b++}.
\qeds



\begin{thm}{Exercise}\label{ex:g-b-chi}
Find the integral of the Gauss curvature on each of the following surfaces:

\setlength{\columnseprule}{0.4pt}
\begin{multicols}{2}

\begin{subthm}{ex:g-b-chi:torus}
A torus.
\end{subthm}

\begin{Figure}
\vskip-0mm
\centering
\includegraphics{mppics/pic-1595}
\end{Figure}

\begin{subthm}{ex:g-b-chi:moebius}
A Möbius band with geodesic boundary.
\end{subthm}

\begin{Figure}
\vskip-0mm
\centering
\includegraphics{mppics/pic-1605}
\end{Figure}

\begin{subthm}{ex:g-b-chi:pair-of-pants}
A pair of pants with geodesic boundary components.
\end{subthm}
\begin{Figure}
\vskip-0mm
\centering
\includegraphics{mppics/pic-1600}
\end{Figure}

\begin{subthm}{ex:g-b-chi:two-handles}
A sphere with two handles.
\end{subthm}

\begin{Figure}
\vskip-0mm
\centering
\includegraphics{mppics/pic-1610}
\end{Figure}

\end{multicols}

\begin{subthm}{ex:g-b-chi:cylinder}
A cylider such that its boundary curves have flat neighborhoods. 
\begin{Figure}
\vskip-0mm
\centering
\includegraphics{mppics/pic-1620}
\end{Figure}
\end{subthm}


\end{thm}