%!TEX root = 162AB Lecture Notes v2.tex

\chapter{Calculus on Surface}






\begin{example}
The (open) unit disk in $\R^2$ 
\[
D=\{(x,y)\mid x^2+y^2<1\}
\]
is an open set.
\end{example}
\begin{proof}
To prove this, we let $(a,b)\in D$. By definition, $a^2+b^2<1$. Let
\[
r=\frac12(1-\sqrt{a^2+b^2})>0.
\]
Let $(x,y)$ be a pair of real numbers such that
\[(x-a)^2+(y-b)^2<r^2.
\]

Let $d(\cdot,\cdot)$ be the distance function. Then
\[
\sqrt{x^2+y^2}=d((x,y), (0,0))\leq d((x,y), (a,b))+d((a,b),(0,0))<r+\sqrt{a^2+b^2}<1.
\]
Hence $(x,y)\in D$.  

\end{proof}

\begin{definition} \blue{A }  \red{Simple surface } \blue{is a mapping $\vec{\bf x}:\mathscr U\to\R^3$, where
$\mathscr U$ is an open set of $\R^2$, and
\begin{enumerate}
\item $\vec{\bf x}$ is a one-to-one mapping;
\item $\frac{\displaystyle\pa \vec {\bf x}}{\displaystyle\pa u^1}\times \frac{\displaystyle\pa \vec {\bf x}}{\displaystyle\pa u^2}\neq 0$.}
\end{enumerate}

\end{definition}

This can be compared to the definition of simple curve.

\begin{definition}{}{} A simple curve is a mapping $\vec{\bf \alpha}:\mathscr (a,b)\to\R^3$, where
$(a,b)$ is an open intervel, and

 $\vec{\bf \alpha}$ is a one-to-one mapping (not self-intersecting);
 




\end{definition}


The Jocobian of the mapping $\vec{\bf x}$ is given by 
\[
\left[\frac{\pa\vec{\bf x}}{\pa u^1}, \frac{\pa\vec{\bf x}}{\pa u^2}\right].
\]
That the surface is simple means that the Jacobian matrix of the mapping is of rank 2 at every point of $\mathscr U$. For a curve, the Jacobian is given by $\vec{\bf \alpha}$, a simple curve is of rank 1 at any point it defines.\\

Like in the curve theory, we \red{{\textsl {don't}} } study surfaces that are not \red{\textsl {simple}}.

A simple surface is sometimes called a \red{\textsl {regular surface}}. 



\begin{example}
Let $\mathscr U$ be an open ser of $\R^2$, and let $f$ be a $\mathcal C^k$ function on $\mathscr U$. Define
\[
\vec{\bf x}:\mathscr U\to\R^3,\qquad (u^1,u^2)\mapsto (u^1,u^2, f(u^1,u^2)).
\]
Then $\vec{\bf x}$ defines a simple surface. 


\end{example}

\begin{proof}
To prove this, we compute
\[
\frac{\pa\vec{\bf x}}{\pa u^1}=\begin{bmatrix}
1\\0\\\frac{\pa f}{\pa u^1}
\end{bmatrix},\qquad 
\frac{\pa\vec{\bf x}}{\pa u^2}=\begin{bmatrix}
0\\1\\\frac{\pa f}{\pa u^2}
\end{bmatrix}
\]
It follows that
\[
\frac{\pa\vec{\bf x}}{\pa u^1}\times \frac{\pa\vec{\bf x}}{\pa u^2}
=\begin{bmatrix}
1&0\\0&1\\\frac{\pa f}{\pa u^1}&\frac{\pa f}{\pa u^2}
\end{bmatrix},
\]
which is a rank 2 matrix. This proves that the surface is simple. 

\end{proof}



\begin{example}
Let
\[
\mathscr W=\{(w^1,w^2)\in\R\mid -\frac\pi 2<w^1<\frac \pi 2, -\pi<w^2<\pi\},
\]
and define a surface by
\[
z(w^1,w^2)=(\cos w^1\cos w^2,\cos w^1\sin w^2,\sin w^1).
\]
\end{example}


\begin{proof}
It is not hard to prove that the Jacobian is of rank 2, hence simple. We compute
\[
\begin{split}
& \frac{\pa z}{\pa w^1}=(-\sin w^1\cos w^2, -\sin w^1\sin w^2,\cos w^1);\\
&\frac{\pa z}{\pa w^2}=(-\cos w^1\cos w^2,\cos w^1\sin w^2,-\sin w^1\cos w^1).
\end{split}
\]
It follows that 
\[
\left\| \frac{\pa z}{\pa w^1}\times \frac{\pa z}{\pa w^2}\right\|
^2=\cos^4(w^1)+\sin^2(w^1)\cos^2(w^1)=\cos^2(w^1).
\]
\blue{Since $-\pi/2<w^1<\pi/2$, the above right side is not zero, this proves that the surface is simple. }\\

\red{It is a little bit complicated to prove that $z$ is one-to-one. } We assume that 
\[
z(w^1,w^2)=z(u^1,u^2).
\]
Therefore we have
\[
\begin{split}
&
\cos w^1\cos w^2=\cos u^1\cos u^2;\\&
\cos w^1\sin w^2=\cos u^1\sin u^2;\\&
\sin w^1=\sin u^1.
\end{split}
\]
From the third equation, we conclude that 
\[
w^1=u^1.
\]
Moreover, $\cos w^1,\cos u^1\neq 0$. Thus we have
\[
\cos w^2=\cos u^2,\qquad \sin w^2=\sin u^2.
\]
Hence we have $w^2=u^2$. 
\end{proof}


\begin{example}{A non-simple surface} Let
\[
\vec{x}\left(u^1, u^{2}\right)=\left((u^1)^{2},\left(u^{2}\right)^{2}, u^1 u^{2}\right)
\]
Then
\[
\frac{\partial \vec{x}}{\partial u^1}=\left(2 u^1, 0, u^{2}\right),\qquad
\frac{\partial \vec{x}}{\partial u^{2}}=\left(0,2 u^{2}, u^1\right).
\]
At the point $(u^1,u^2)=(0,0)$, $\vec x_1\times \vec x_2=0$, and hence the surface is not simple. 
\end{example}


\[
\frac{\partial f^{1}}{\partial V^{1}} \cdot \frac{\partial g^{1}}{\partial u^{1}}+\frac{\partial f^{1}}{\partial V^{2}} \cdot \frac{\partial g^{2}}{\partial y^{1}}=1 \quad \frac{\partial f^{2}}{\partial V^{1}} \cdot \frac{\partial g^{1}}{\partial u^{1}}+\frac{\partial f^{2}}{\partial V^{2}} \cdot \frac{\partial g^{2}}{\partial u^{1}}=0
\]


\begin{definition}{}{}
A $C^k$
\[
f: \mathscr V\to\mathscr U
\]
such that 
\begin{enumerate}
\item $f$ is a bijection;
\item $g=f^{-1}: \mathscr U\to\mathscr V$ is also $C^k$. 
\end{enumerate}
\end{definition}

\begin{problemset}
\item 
\end{problemset}


\section{}
\section{}
\section{}
\section{}
\section{}
\section{Integration of Forms}


The three fundamental properties for definite integral: \red{what is the geometric point of view to them?}
\begin{enumerate}
\item integration by substitution
\[
\int_a^b F(\phi(t))\phi'(t) dt=\int_{\phi(a)}^{\phi(b)} F(s) ds
\]
\item integration by parts
\[
\int_a^b f'(t)g(t) dt=\left.f(t)g(t)\right|^b_a-\int_a^b f(t)g'(t) dt.
\]
\item the Fundamental Theorem of Calculus
\[
\int_a^b F'(t) dt=F(b)-F(a).
\]
\end{enumerate}
First, we define the integration on curves.

\begin{definition}{}{}
Let $M$ be a surface, and let $\phi$ be a $1$-form on $M$. Let
\[
\alpha: [a,b]\to M
\]
be a curve segment of $M$. Then the \emph{integral of} $\phi$ \emph{over} $\alpha$ is
\[
\int_\alpha\phi=\int_{[a,b]}\alpha^*\phi=\int_a^b\phi(\alpha'(t)) dt.
\]
\end{definition}
$\int_\alpha\phi$ is called the \emph{line integral} of $\phi$ over the line $\alpha$.


The above definition needs some elaboration. \vspace{.2in}


Let's recall the notation $\alpha^*\phi$. Since $\phi$ is a $1$-form on $M$. Then by definition, $\alpha^*\phi$ is a $1$-form on the interval $[a,b]$. Now, $[a,b]$ is a $1$-dimensional space. The tangent space of $[a,b]\subset\R$ is spanned by
\[
\frac{\pa}{\pa t}
\]
and therefore the the cotangent space is spanned by $\{dt\}$. Question, what is $\alpha^*\phi$?
\blue{
\[
\alpha^*\phi(\frac{\pa}{\pa t})=\phi(\alpha_*\frac{\pa}{\pa t})=\phi(\alpha'(t))
\]
So $\alpha^*\phi=\phi(\alpha'(t))dt$.}


By the above point of view, the \emph{definite integral} can also be viewed as the line integral
\[
\int_a^b f(t) dt=\int_{[a,b]}\phi, 
\]
where 
\[
\phi=f(t) dt
\]
is a one form. 

\begin{theorem}{}{}
Let $f$ be a smooth function on $M$ and let $\alpha: [a,b]\to M$ be a line segment. Then
 \begin{tcolorbox}[colback=white,colframe=red!75!black, text width=4.5cm,top=-.15cm,center]
\[
\int_\alpha df=f(\alpha(b))-f(\alpha(a)).
\]
\end{tcolorbox}
\end{theorem}
\begin{proof}
\[
\int_\alpha df=\int_a^b df(\alpha'(t))dt=\int_a^b\frac{d}{dt}f(\alpha(t)) dt=f(\alpha(b))-f(\alpha(a)).
\]
\end{proof}

\begin{definition}{}{} A \emph{$2$-segment} is a smooth mapping ${\bf x}$ from  open set $R$ of $\R^2$ to M
\[
{\bf x}: [a,b]\times [c,d]\to M
\]
\end{definition}


\begin{definition}{}{}
Let $\eta$ be a $2$-form on $M$, and let ${\bf x}: R\to M$ be a 2-segment. Then
\[
\int_R\eta=\iint_R{\bf x}^*\eta=\int_a^b\int_c^d\eta({\bf x}_u,{\bf x}_v) dudv.
\]
\end{definition}

\begin{theorem}{}{}
Let $ U$ be an open set of $M$ with smooth boundary curve $\pa U$. Let $\omega$ be a $1$-form. Then
\[
\int_U d\omega=\int_{\pa U} \omega.
\]
\end{theorem}
We shall prove this theorem only under the case that $U$ is the image of a proper coordinate patch ${\bf x}$ which is a 2-segment.




\begin{proof}
We see that
\[
\int_Ud\omega=\int_R{\bf x}^*(d\omega).
\]
We also have
\[
\int_{\pa U} \omega=\int_{\pa R} {\bf x}^*\omega
\]
The key is to prove
 \begin{tcolorbox}[colback=white,colframe=red!75!black, text width=4.5cm,top=-.05cm,center]
 \[
{\bf x}^*(d\omega)=d({\bf x}^*\omega),
 \]
 \end{tcolorbox}
 that is, $d$ and ${\bf x}^*$ commutes. Once this is done, the theorem follows from the Green's formula. 
\end{proof}

\section{Topological properties of surfaces}

Topological property is a property of a topological space which is invariant under homeomorphisms. \vspace{.4in}
 
 \blue{Topology has been called rubber-sheet geometry. In a topology of two dimensions there is no difference between a circle and a square. A circle made out of a rubber band can be stretched into a square. There is a difference between a circle and a figure eight. A figure eight cannot be stretched into a circle without tearing.}
 
 



\begin{note}
\begin{enumerate}
\item \blue {Connectedness;}
\item \blue {Compactness;}
\item \blue {Orientability;}
\item \blue {Simply-connectedness.}
\end{enumerate}
\end{note}


%\includegraphics[width=0.5\linewidth, height=0.25\textheight]{CalabiYau5.jpg}








\red{We will talk about their relations to analysis.}

\begin{definition}{}{}
A surface is connected provided that for any two points $p,q$ of $M$, there is a curve segment in $M$ from $p$ to $q$ (path-connected?).
\end{definition}

\begin{example}
Let $f$ be a function on a connected surface $M$ such that $df=0$. Prove that $f$ is a constant. 
\end{example}

\begin{definition}{}{}
A surface is \emph{compact} provided that given any open covering of A admits a \emph{finite} sub-covering. 
\end{definition}
Example of a compact set:
\begin{example}{Heine-Borel Theorem}
Any open cover of the closed interval $[0,1]$ by open intervals admits a finite open cover.
\end{example}


\begin{theorem}{}{}
A continuous function on a compact surface $M$ is bounded, and moreover, the maximum is reached at some point.
\end{theorem}


\begin{definition}{}{}
{A surface $M$ is \emph{orientable} if there exists a differentiable (or
merely continuous) 2-form $\mu$  on $M$ that is nonzero at every point of $M$.}
\end{definition}
%\includegraphics[width=0.2\linewidth, height=0.15\textheight]{M_strip.jpg}


\begin{theorem}{}{}
A surface $M\subset  \R^3$ is orientable if and only if there exists
a unit normal vector field on $M$. If $M$ is connected as well as orientable, there
are exactly two unit normals, $\pm U$.
\end{theorem}





\begin{definition}{}{}
A surface $M$ is \emph{simply connected} provided it is connected
and every loop in $M$ is homotopic to a constant.
\end{definition}

\begin{lemma}{}{}
Let $\phi$ be a closed $1$-form on a surface $M$. If a loop $\alpha$ in $M$ is homotopic to a constant, then $\int_\alpha\phi=0$. 
\end{lemma}


On a simply connected surface, every closed $1$-form is exact.

\begin{example}
\[
\psi=\frac{xdy-ydx}{x^2+y^2}.
\]
\end{example}



\begin{theorem}{}{}
A compact surface in $\R^3$ is orientable.
\end{theorem}
\begin{theorem}{}{}
A simply connected surface is orientable.
\end{theorem}


\section{Manifolds}


In mathematics, a manifold is a topological space that locally resembles Euclidean space near each point. More precisely, each point of an $n$-dimensional manifold has a neighborhood that is homeomorphic to the Euclidean space of dimension $n$ -- \red{ from Wikipedia}


\begin{definition}{}{}
An \emph{abstract patch} in $M$ is just a one-to-one function ${\bf x}: D\to M$ from any open set $D$ of $\R^2$ into the \emph{set} $M$.
\end{definition}



\begin{definition}{}{}
A \emph{topological surface} is a set $M$ furnished with a collection $\mathscr P$ of
abstract patches in $M$ satisfying
\begin{enumerate}
\item
\emph{The covering axiom:} The images of the patches in the collection $\mathscr P$
cover $M$.

\item
\emph{The Hausdorff axiom:} For any points $p\neq q$ in $M$ there exist disjoint
(that is, nonoverlapping) patches ${\bf x}$ and ${\bf y}$ with $p$ in ${\bf x}(D)$ and $q$ in ${\bf y}(E)$.
\end{enumerate}
\end{definition}





\begin{definition}{}{}
A \emph{surface} is a set $M$ furnished with a collection $\mathscr P$ of
abstract patches in $M$ satisfying
\begin{enumerate}
\item
\emph{The covering axiom:} The images of the patches in the collection $\mathscr P$
cover $M$.
\item
\emph{The smooth overlap axiom:} For any patches ${\bf x}, {\bf y} \in\mathscr P$, the composite
functions ${\bf y}^{-1}{\bf x}$ and ${\bf x}^{-1} {\bf y}$ are Euclidean differentiable—and defined on open
sets of $\R^2$.
\item
\emph{The Hausdorff axiom:} For any points $p\neq q$ in $M$ there exist disjoint
(that is, non-overlapping) patches ${\bf x}$ and ${\bf y}$ with $p$ in ${\bf x}(D)$ and $q$ in ${\bf y}(E)$.
\end{enumerate}
\end{definition}

\begin{itemize}
\item
First, as a set $RP^2=S^2/\sim$, where $\sim$ is an equivalence relation $p\sim -p$.
\item let $U_{upper}, U_{lower}, U_{front},U_{rear},U_{left}, U_{right}$, be the upper, lower front, rear, left, right \emph{open}
hemisphere of $S^2$, and let $D$ be the open unit disk of $\R^2$. Let 
\[
{\bf x}_i: D\to U_i\to S^2
\]
be the coordinate patch, where $i\in\{upper, lower, front, rear, left, right\}$. 
\item Let 
\[
\tilde{\bf x}_i: D\to U_i\to S^2\to RP^2=S^2/\sim.
\]
Then $\tilde{\bf x}_i$ are one-to-one and are abstract patches. 
\end{itemize}



\begin{enumerate}
\item Let $E$ be any open subset of the open disk $D$. Then
\[
\tilde{\bf x}_{i,E}: E\to D\to S^2\to RP^2=S^2/\sim.
\] 
is an abstract coordinate patch. 
\item Using the above, we can prove that $RP^2$ is a topological surface. It is a surface because the transition functions are smooth. 
\end{enumerate}





One can generalize the above concept to \emph{$n$-dimensional manifold}.



How to define the tangent space of an abstract surface? Two ways: let $p\in M$
\begin{enumerate}
\item as the space of  velocity vector of any curve passing $p$.
\item as the dual space of the co-tangent space, where the co-tange nt space 
\[
T^*_PM=\frac{\mathscr I_p}{\mathscr I_p^2}
\]
\end{enumerate}

\begin{example}
The tangent bundle $TM$ of a non-abstract surface $M$ is a manifold of dimension $4$.
\end{example}
How to construct abstract coordinate patches?

If ${\bf x}: \mathscr D\to M$ be a coordinate patch. Then ${\bf y}: \mathscr D\times \R^2\to TM$ is given by 
\[
{\bf y}(a,b,u,v)=a\,{\bf x}_u(u,v)+b\,{\bf x}_{v}(u,v).
\]

