surfaces-def.tex

\chapter{Definitions}
\label{chap:surfaces-def}

\section{Topological surfaces}

We will be most interested in smooth surfaces defined in the following section.
The following general definition will be used only a few times.

A connected subset $\Sigma$ in the Euclidean space $\mathbb{R}^3$
is called a \index{surface}\index{topological!surface}\emph{topological surface} (more precisely an {}\emph{embedded topological surface without boundary}) 
if any point of $p\in \Sigma$ admits a neighborhood $W$ in $\Sigma$ 
that can be parametrized by an open subset in the Euclidean plane; 
that is, there is a homeomorphism $V\to W$ from an open set $V\subset \mathbb{R}^2$; see Section~\ref{sec:topology}.


\section{Smooth surfaces}\label{sec:def-smooth-surface}

Recall that a function $f$ of two variables $x$ and $y$ is called \index{smooth!function}\emph{smooth} if all its partial derivatives $\frac{\partial^{m+n}}{\partial x^m\partial y^n}f$ are defined and are continuous in the domain of definition of~$f$. 

A connected set $\Sigma \subset \mathbb{R}^3$ is called a \index{surface}\index{smooth!surface}\emph{smooth surface}%
\footnote{We use it as a shortcut for the more precise term {}\emph{smooth regular embedded surface}.} if it can be described locally as a graph of a smooth function in an appropriate coordinate system.

More precisely, for any point $p\in \Sigma$ one can choose a coordinate system $(x,y,z)$ and a neighborhood $U\ni p$ such that
the intersection $W=U\cap \Sigma$ is a graph $z=f(x,y)$ of a smooth function $f$ defined in an open domain $V$ of the $(x,y)$-plane.

Note that $(x,y)\mapsto (x,y,f(x,y))$ defines a homeomorphism $V\to W$.
Therefore, а smooth surface is an example of а topological surface.

\parbf{Examples.}
The simplest example of a smooth surface is the $(x,y)$-plane 
$\Pi=\set{(x,y,z)\in\mathbb{R}^3}{z=0}$.
The plane $\Pi$ is a surface since
it can be described as the graph of the function $f(x,y)=0$.

All other planes are smooth surfaces as well since one can choose a coordinate system so that it becomes the $(x,y)$-plane.
We may also present a plane as a graph of a linear function 
$f(x,y)=a\cdot x+b\cdot y+c$ for some constants $a$, $b$, and $c$
(assuming the plane is not perpendicular to the $(x,y)$-plane, in which case a different coordinate system is needed).

A more interesting example is the unit sphere 
\[\mathbb{S}^2=\set{(x,y,z)\in\mathbb{R}^3}{x^2+y^2+z^2=1}.\]
This set is not a graph,
but it is locally a graph ---
it can be covered by the following 6 graphs:
\begin{align*}
z&=f_\pm(x,y)=\pm \sqrt{1-x^2-y^2},
\\
y&=g_\pm(x,z)=\pm \sqrt{1-x^2-z^2},
\\
x&=h_\pm(y,z)=\pm \sqrt{1-y^2-z^2},
\end{align*}
where each function $f_+$, $f_-$, $g_+$, $g_-$, $h_+$, and $h_-$ is defined in an open unit disc.
Any point $p\in\mathbb{S}^2$ lies in one of these graphs therefore $\mathbb{S}^2$ is a smooth surface.

\section{Surfaces with boundary}
A connected subset in a surface that is bounded by one or more piecewise
smooth curves is called a \index{surface!with boundary}\emph{surface with boundary}; such curves form the \index{boundary!line}\emph{boundary line} of the surface.

When we say {}\emph{surface} we usually mean a {}\emph{smooth surface without boundary}.
If needed, one may use the term {}\emph{surface with possibly nonempty boundary}.

\section{Proper, closed, and open surfaces}
If the surface $\Sigma$ is formed by a closed set in $\mathbb{R}^3$, then it is called \index{proper!surface}\emph{proper}.
For example, for any smooth function $f$ defined on the whole plane, its graph $z=f(x,y)$ is a proper surface.
The sphere $\mathbb{S}^2$ gives another example of a proper surface.

On the other hand, the open disc 
\[\set{(x,y,z)\in\mathbb{R}^3}{x^2+y^2<1,\  z=0}\]
is not a proper surface; this set is neither open nor closed in $\mathbb{R}^3$.

A compact surface without boundary is called \index{closed!surface}\emph{closed}
(this term is closely related to the closed curve, but has nothing to do with closed set).

A proper noncompact surface without boundary is called \index{open!surface}\emph{open} (again, the term open curve is relevant, but open set is not).

For example, the paraboloid $z=x^2+y^2$
is an open surface; the 
sphere $\mathbb{S}^2$ is a closed surface.
Note that \textit{any proper surface without boundary is either closed or open}.

The following claim is a three-dimensional analog of the plane separation theorem (\ref{ex:proper-curve}).
It follows from the so-called {}\emph{Alexander's duality} \cite{hatcher}.
We omit its proof; it would take us far away from the main subject.

\begin{thm}{Claim}\label{clm:proper-divides}
The complement of any proper topological surface without boundary (or, equivalently any open or closed topological surface) has exactly two connected components. 
\end{thm}



\section{Implicitly defined surfaces}

\begin{thm}{Proposition}\label{prop:implicit-surface}
Let $f\:\mathbb{R}^3\to \mathbb{R}$ be a smooth function.
Suppose $0$ is a regular value of $f$;
that is, $\nabla_p f\ne 0$ at any point $p$ such that $f(p)=0$.
Then any connected component $\Sigma$ of the level set  $f(x,y,z)=0$ is a smooth surface.
\end{thm}

\parbf{Proof.}
Fix $p\in\Sigma$.
Since $\nabla_p f\ne 0$ we have 
$f_x(p)\ne 0$,
$f_y(p)\ne 0$, or
$f_z(p)\ne 0$.
We may assume that $f_z(p)\ne 0$;
otherwise, permute the coordinates $x,y,z$.

The implicit function theorem (\ref{thm:imlicit}) implies that a neighborhood of $p$ in $\Sigma$ is the graph $z=h(x,y)$ of a smooth function $h$ defined on an open domain in $\mathbb{R}^2$.
It remains to apply the definition of smooth surface (Section~\ref{sec:def-smooth-surface}).
\qeds

\begin{thm}{Exercise}\label{ex:hyperboloids}
For which constants $\ell$ is the level set $x^2+y^2-z^2=\ell$
a~smooth surface?
\end{thm}

\section{Local parametrizations}
\index{parametrization}

Let $U$ be an open domain in $\mathbb{R}^2$, and $s\:U\to \mathbb{R}^3$ be a smooth map.
We say that $s$ is \index{regular!parametrization}\emph{regular} if its Jacobian matrix has maximal rank;
in this case, it means that the derivatives $s_u$ and $s_v$ are linearly independent at any $(u,v)\in U$;
equivalently $s_u\times s_v\ne 0$, where $\times$ denotes the vector product.

\begin{thm}{Proposition}\label{prop:graph-chart}
If $s\:U\to \mathbb{R}^3$ is a smooth regular embedding of an open connected set $U\subset \mathbb{R}^2$, then its image $\Sigma=s(U)$ is a smooth surface.
\end{thm}

\parbf{Proof of \ref{prop:graph-chart}.}
Let $s(u,v)=(x(u,v),y(u,v),z(u,v))$.
Since $s$ is regular, its Jacobian matrix
\[\Jac s=
\renewcommand\arraystretch{1.3}
\begin{pmatrix}
x_u&x_v\\
y_u&y_v\\
z_u&z_v
\end{pmatrix}
\]
has rank two at any point $(u,v)\in U$.

Choose a point $p\in \Sigma$; by shifting the $(x,y,z)$ and $(u,v)$ coordinate systems, we may assume that $p = (0,0, 0) =s(0,0)$.
Permuting the coordinates $x,y,z$ if necessary, we may assume that 
the matrix $\left(\begin{smallmatrix}
x_u&x_v\\
y_u&y_v
\end{smallmatrix}\right)$
is invertible at the origin.
Note that this is the Jacobian matrix of the map $(u,v)\mapsto (x(u,v),y(u,v))$.

The inverse function theorem (\ref{thm:inverse}) implies that there is a smooth regular map
$w\:(x,y)\mapsto (u,v)$ defined on an open set $W\ni 0$ in the $(x,y)$-plane
such that $w(0,0)=(0,0)$ and  $s\circ w(x,y)=(x,y,f(x,y))$ where $f=z\circ w$.
That is, the subset $s\circ w(W)\subset \Sigma$ is 
the graph of $f$.

Again, by the inverse function theorem, $w(W)$ is an open subset of $U$.
Since $s$ is an embedding, our graph is open in~$\Sigma$;
that is, there is an open set $V\subset \mathbb{R}^3$ such that $s\circ w(W)=V\cap \Sigma$ is a graph of smooth function.

Since $p$ is arbitrary, we get that $\Sigma$ is a smooth surface.
\qeds

\begin{thm}{Exercise}\label{ex:9-surf}
Construct a smooth regular injective map $s\:\mathbb{R}^2\to\mathbb{R}^3$ such that its image is \textit{not} a surface.
\end{thm}

If we have $s$ and $\Sigma$ as in the proposition, then we say that $s$ is a \index{smooth!parametrization}\emph{smooth parametrization} of the surface~$\Sigma$. 

Not all smooth surfaces can be described by such a parametrization;
for example, the sphere $\mathbb{S}^2$ cannot.
However, \textit{any smooth surface $\Sigma$ admits a local parametrization at any point $p\in\Sigma$}; that is,  $p$ admits an open neighborhood $W\subset \Sigma$ with a smooth regular parametrization~$s$.
In this case, any point in $W$ can be described by two parameters, usually denoted by $u$ and $v$, 
which are called \index{local coordinates}\emph{local coordinates} at~$p$.
The map $s$ is called a \index{chart}\emph{chart} of~$\Sigma$.

If $W$ is a graph $z=h(x,y)$ of a smooth function $h$, then the map 
\[s\:(u,v)\mapsto (u,v,h(u,v))\] is a chart.
Indeed, $s$ has an inverse $(u,v,h(u,v))\mapsto (u,v)$ which is continuous;
that is, $s$ is an embedding.
Further,
$s_u=(1,0,h_u)$, and $s_v=(0,1,h_v)$. 
Whence the partial derivatives $s_u$ and $s_v$ are linearly independent;
that is, $s$ is a regular map.

\begin{thm}{Corollary}\label{cor:reg-parmeterization}
A connected set $\Sigma\subset \mathbb{R}^3$ is a smooth surface if and only if a neighborhood of any point in $\Sigma$ can be covered by a chart.
\end{thm}

A function $g\: \Sigma \to \mathbb{R}$ defined on a smooth surface $\Sigma$ is said to be \index{smooth!function}\emph{smooth} if for any chart $s \: U\to \Sigma$,
the composition $g\circ s$ is smooth; that is, all partial derivatives $\frac{\partial^{m+n}}{\partial u^m\partial v^n}(g\circ s)$ are defined and continuous in the domain of definition.

\begin{thm}{Exercise}\label{ex:smooth-fun(surf)}
Let $\Sigma\subset \mathbb{R}^3$ be a smooth surface.
Show that a function $g\:\Sigma\to\mathbb{R}$ is smooth if and only if for any point $p\in \Sigma$ there is a smooth function $h\:N\to\mathbb{R}$ defined in a neighborhood $N\subset \mathbb{R}^3$ of $p$ such that the equality $g(q)=h(q)$ holds for $q\in \Sigma\cap N$.

Construct a smooth surface $\Sigma$ and a smooth function $g\:\Sigma\to\mathbb{R}$ that cannot be extended to a smooth function $h\:\mathbb{R}^3\to\mathbb{R}$.
\end{thm}


\begin{thm}{Exercise}\label{ex:inversion-chart}
Consider the following map 
\[s(u,v)=(\tfrac{2\cdot u}{1+u^2+v^2},\tfrac{2\cdot v}{1+u^2+v^2},\tfrac{2}{1+u^2+v^2}).\]
Show that $s$ is a chart of the unit sphere centered at $(0,0,1)$; describe the image of~$s$.
\end{thm}

\begin{wrapfigure}{o}{31 mm}
\vskip-3mm
\centering
\includegraphics{asy/torus}
\vskip0mm
\end{wrapfigure}

Let $\gamma(t)=(x(t),y(t))$ be a plane curve.
Recall that the \index{surface!of revolution}\emph{surface of revolution} of the curve $\gamma$ around the $x$-axis can be described as the 
image of the map 
\[(t, s)\mapsto (x(t), y(t)\cdot\cos s,y(t)\cdot\sin s).\]
The parameters $t$ and $s$ are called \index{latitude}\emph{latitude} and \index{longitude}\emph{longitude} respectively;
for fixed $t$ or $s$ the obtained curves are called \index{parallel}\emph{parallels} or
\index{meridian}\emph{meridians} respectively. 
The parallels are formed by circles in the plane perpendicular to the axis of rotation.
The curve $\gamma$ is called the \index{generatrix}\emph{generatrix} of the surface.

\begin{thm}{Exercise}\label{ex:revolution}
Assume $\gamma$ is a simple closed smooth plane curve that does not intersect the $x$-axis.
Show that the surface of revolution around the $x$-axis with generatrix $\gamma$ is a smooth surface.
\end{thm}


\section{Global parametrizations}\label{sec:global-parametrizations}
\index{parametrization}

A surface can be described by an embedding from a known surface.

For example, consider the ellipsoid
\[\Theta=\set{(x,y,z)\in\mathbb{R}^3}{\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}+\tfrac{z^2}{c^2}=1}\]
for positive constants $a$, $b$, and~$c$.
By \ref{prop:implicit-surface}, $\Theta$ is a smooth surface.
Indeed, let $h(x,y,z)=\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}+\tfrac{z^2}{c^2}$,
then
\[\nabla h(x,y,z)=(\tfrac{2}{a^2}\cdot x,\tfrac{2}{b^2}\cdot y,\tfrac{2}{c^2}\cdot z).\]
Therefore, $\nabla h\ne0$ if $h=1$; that is, $1$ is a regular value of~$h$.
It remains to observe that $\Theta$ is connected.

The surface $\Theta$ can be defined as the image of the map $ \mathbb{S}^2\to\mathbb{R}^3$, defined as the restriction of the following map to the unit sphere $\mathbb{S}^2$:
\[(x,y,z)\z\mapsto (a\cdot x, b\cdot y,c\cdot z).\]


A map $f\:\Sigma \to \mathbb{R}^3$ (or $f\:\Sigma \to \mathbb{R}^2$) is smooth if each of its coordinate functions is smooth.
Further, a smooth map $f \: \Sigma \to \mathbb{R}^3$ is called a 
\emph{smooth parametrized surface} if it is an embedding, and for any chart $s \:U\to \Sigma$,
the composition $f\circ s$ is regular;
that is,  the two vectors 
$\frac{\partial}{\partial u}(f\circ s)$ and $\frac{\partial}{\partial v}(f\circ s)$ are linearly independent.
In this case, the image $\Sigma^{*}=f(\Sigma)$ is a smooth surface.
The latter follows since for any chart $s\:U\to \Sigma$ the composition $f\circ s\:U\to \Sigma^{*}$ is a chart of $\Sigma^{*}$. 

The map $f$ is called a \index{diffeomorphism}\emph{diffeomorphism} from $\Sigma$ to $\Sigma^{*}$;
the surfaces $\Sigma$ and $\Sigma^{*}$ are said to be {}\emph{diffeomorphic} if there is a diffeomorphism $f\:\Sigma\to\Sigma^{*}$.
The following exercise implies that \textit{being diffeomorphic} is an equivalence relation for surfaces.

\begin{thm}{Exercise}\label{ex:inv-diffeomorphism}
Show that the inverse of a diffeomorphism is a diffeomorphism.
\end{thm}

\begin{thm}{Advanced exercise}\label{ex:star-shaped-disc}
Show the following.

\begin{subthm}{ex:plane-n}
Complements of $n$-point sets in the plane are diffeomorphic to each other.
\end{subthm}


\begin{subthm}{ex:star-shaped-disc:smooth}
Open convex subsets of the plane bounded by smooth closed curves are diffeomorphic to each other.
\end{subthm}


\begin{subthm}{ex:star-shaped-disc:nonsmooth}
Any pair of open convex subsets of the plane are diffeomorphic to each other.
\end{subthm}

\begin{subthm}{ex:star-shaped-disc:star-shaped}
Open star-shaped subsets of the plane are diffeomorphic to each other.
\end{subthm}
\end{thm}