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


\chapter{Shape Operators}
\section{The shape operator of $M\subset \R^3$}

In $\R^3$, let $X,Y$ be two vector fields. Then in \S 2.7, we defined the covariant derivatives. Namely, we have
  \[
  \nabla_X Y=XY.
  \]
  
  The above notation is not very clear. What we really mean is that, if $Y=(a,b,c)$, or more precisely, if
  \[
  Y=a\frac{\pa}{\pa x}+b\frac{\pa}{\pa y}+c\frac{\pa}{\pa z},
  \]
  then we have 
  \[
  \nabla_XY=  X(a)\frac{\pa}{\pa x}+X(b)\frac{\pa}{\pa y}+X(c)\frac{\pa}{\pa z}.
  \]
  

  Now let's assume that $M\subset \R^3$ be a surface. Let $U$ be a unit normal vector of $M$. 
  
  \begin{definition}{}{}
  The shape operator is defined by
  \begin{tcolorbox}[colback=white,colframe=red!75!black, text width=6.5cm,top=-.0cm,center]
  \[
  S_p(v)=-\nabla_v U,
  \]
  \end{tcolorbox}
  where $v\in T_pM$.
  \end{definition}
  

  \begin{lemma}{}{}
 The shape operator is a linear operator  
 \[
 S_p: T_p M\to T_p M.
 \]
  \end{lemma}
 
  \begin{lemma}{}{}
 The shape operator is a symmetric  operator  
 \[
 S_p: T_p M\to T_p M.
 \]
  \end{lemma}

  \begin{example}
  What is the shape operator of $\R^2\subset \R^3$? It is zero, because $U$ is parallel. 
  \end{example}

  \begin{example}
  The shape operator of the sphere of radius $r$: $x^2+y^2+z^2=r^2$. 
  
  
  \end{example}
  The normal vector field is
  \[
  U=\frac 1r \sum_ix_i U_i.
  \]
 

  \begin{example}
  The shape operator of the sphere of radius $r$: $x^2+y^2=r^2$. 
  
  
  \end{example}
  
  \section{Normal Curvature}
  Prelude lemma:

\begin{lemma}{}{}
Let $\alpha$ be a curve in $M\subset \R^3$, then
\[
\langle \alpha'', U\rangle=\langle S(\alpha'), \alpha'\rangle.
\]
\end{lemma}
 
   \begin{definition}{}{}
   Let $u$ be a \emph{unit} vector field on a surface $M$. Then
    the normal curvature is defined to be
    \[
    k(u)=S(u)\cdot u.\qquad\qquad a.k.a.\quad \langle S(u), u\rangle
    \]
   \end{definition}
   
   One obvious observation is that 
   \[
   k(-u)=k(u),
   \]
  and it should be normal curvatures instead of normal curvature. 

  This is important. Let $S$ be the shape operator which is symmetric. Then in linear algebra, we study it by studying the corresponding the quadratic form $S(u)\cdot u$. 
  
  \begin{theorem}{}{}
  There is a one-to-one correspondence between a symmetric operator and its corresponding quadratic form. 
  \end{theorem}
  
  \begin{proof}
  Let $\iota$ be the map between the set of symmetric matrices and the corresponding quadratic form. Then
  \begin{enumerate}
  \item $\iota$ is an injective map;
  \item $\iota$ is a surjective map.
  \end{enumerate}
  \end{proof}
{Detailed proof here}
  
Comparing the normal curvature of a curve and it curvature as a space curve
  Let $\alpha$ be a unit-speed curve on $M$. When regarding it as a space curve,  its Frenet apparatus is given by $(T,N,B)$, where
  \[
  T=\alpha'(t),\qquad N=T'/\|T'\|.
  \]
  We therefore have
  \[
  k(\alpha'(t))=S(\alpha'(t))\cdot \alpha'(t)=\alpha''\cdot U=k(t)N\cdot U=k(t)\cos\theta,
  \]
  where $\theta$ is the angle between the normal vector of the surface and the principal normal of the space curve $\alpha$.
Principal curvatures
  \begin{definition}{}{}
The eigenvalues of the shape operator is called the principle curvatures of the surface. The corresponding eigenvectors are called principal vectors.
  \end{definition}
  This definition is different from the text book, where it gives a stupid definition. 
  
  \begin{theorem}{}{}
  The eigenvalues are the maximum and minimum of the normal curvatures $k(u)$.
  \end{theorem}
  The above theorem is used in the textbook as the definition of principle curvatures. But this kind of definition is only valid in 2-dimension.

 

 \begin{definition}{}{}
 A point $p$ is called \emph{umbilic}, if the two principal curvatures are the same. 
 \end{definition}
 
 \begin{theorem}{}{}
 A point is umbilic, if and only if the shape operator is a scalar multiplication. 
 \end{theorem}
 
 \begin{proof}
 The only if part is trivial. In order to prove the if part, we shall observe that if a symmetric matrix having the same eigenvalue, then it is a scalar multiplication. 
 \end{proof}

\section{Gauss Curvature}

\red{Prelude: invariants}

\blue{Let $T$ be a symmetric linear transformation from $\R^2$ to itself. What are the invariants of $T$?}

\red{Answer: ${\rm Tr}\, (T)$ and $\det (T)$.}

\red{Justification: Let $A,B$ be the matrix representations of $T$ under different bases. Then $A$ and $B$ are similar matrices, that is, 
\[
A=PBP^{-1}.
\]
Therefore $\det (A)=\det(B)$ and hence $\det(T)=\det(A)$ is an invariant. The trace ${\rm Tr}(A)$ is also an invariant, where 
\[
{\rm Tr}(A)=a+b,
\]
if 
\[
A=\begin{bmatrix}
a&c\\c&b\end{bmatrix}
\]
}

   \begin{definition}{}{}
The Gauss curvature $K$ is defined by
\[
K=\det(T)=\lambda_1+\lambda_2,
\]
where $\lambda_1,\lambda_2$ are the principal curvatures. The mean curvature is defined as
\[
H=\frac 12(\lambda_1+\lambda_2).
\]

   \end{definition}
   The Gauss curvature is independent of the choice of orientation. The mean curvature does depend on the orientation. 
  
{Geometric interpretation of the Gauss curvature}
   {\bf Case 1.} $K(p)>0$. 
   If the Gauss curvature at one point is positive, then   both of the principal curvatures are positive or negative. By changing the normal vector $U$ to $-U$ if necessary, we may assume that both principal curvatures are positive. Let $\alpha_1$ be a curve passing through $p$ with the direction as the principle direction. Then by the relationship between   the normal curvature and the principal normal vector of $\alpha_1$ as a curve in the 3-dimensional space, we know that the angle between them is $<\pi/2$. Thus in both eigenvector directions, the surface is bending away from  the normal vector. Thus the surface locally looks like a paraboloid
     \begin{tcolorbox}[colback=white,colframe=red!75!black, text width=6.5cm,top=-.0cm,center]
   \[
   z=\frac 12 \lambda_1x^2+\frac 12 \lambda_2 y^2.
   \]
   \end{tcolorbox}
{Geometric interpretation of the Gauss curvature-2}
       {\bf Case 2.} $K(p)<0$ -- similar discussion.
       
       %\includegraphics[width=0.9\linewidth, height=0.5\textheight]{gauss.pdf}
 {Geometric interpretation of the Gauss curvature-3}
            {\bf Case 3.} $K(p)=0$. 
            There are two case, either one principal curvature is zero and the other one is non-zero, in which the sahpe looks like Fig 5.17, or it looks close to a flat plane $\R^2$ in $\R^3$.
             % \includegraphics[width=0.9\linewidth, height=0.5\textheight]{flat.pdf}
       {How to compute the Gauss curvature (at $(0,0)$) of the surface?}
          \[
          z=\frac12\lambda_1u^2+\frac 12\lambda_2 v^2.
          \]
          We have
          \[
          {\bf x}_u=(1,0,\lambda_1 u),\qquad {\bf x}_v=(1,0,\lambda_2 v)
          \]
          The unit normal vector is given by 
          \[
          U=\left(\frac{-\lambda_1 u}{\sqrt{1+\lambda_1^2 u^2+\lambda_2^2 v^2}},\frac{-\lambda_2 v}{\sqrt{1+\lambda_1^2 u^2+\lambda_2^2 v^2}},\frac{1}{\sqrt{1+\lambda_1^2 u^2+\lambda_2^2 v^2}},\right)
          \]
          At the origin, ${\bf x}_u=(1,0,0)$ and ${\bf x}_v=(0,1,0)$, and we have
          \[
          S({\bf x}_u) =(\lambda_1,0,0)=\lambda_1{\bf x}_u,\qquad S({\bf x}_v) =(0, \lambda_2,0)=\lambda_2{\bf x}_v
          \]
          So the principal curvatures are given by $\lambda_1,\lambda_2$, and the Gauss curvature $K=\lambda_1\lambda_2$.




\section{Computational Techniques}


Let $M$ be a surface with coordiniate patch 
\[
{\bf x}: \mathscr D\to M.
\]
Let $U$ be the unit norm of $M$. Then the shape operator $S$ is defined by 
\[
S: T_pM\to T_pM,\qquad S(v)=-\nabla_vU.
\]
The shape operator is a symmetric operator. The corresponding quadratic form is called the normal curvatures of the surface. The eigenvalues of the shape operator are called principle curvatures, the average of the principle curvatures is called the mean curvature, and the multiplication of the principle curvatures is called the Gauss curvature. 

Define
\[
E={\bf x}_u\cdot {\bf x}_u,\quad F={\bf x}_u\cdot {\bf x}_v,\quad G={\bf x}_v\cdot {\bf x}_v.
\]
Then $\{E,F,G\}$ determines the dot product, namely, if we write
\[
{\bf v}=v_1 {\bf x}_u+v_2 {\bf x}_v,\quad {\bf w}=w_1 {\bf x}_u+w_2 {\bf x}_v,
\]
then
\[
{\bf v}\cdot {\bf w}=Ev_1w_1+F(v_1w_2+v_2w_1)+Gv_2w_2.
\]
We also have
\[
\|{\bf x}_u\times {\bf x}_v\|^2=EG-F^2>0.
\]

Now we define the second derivatives. Let
\[
{\bf x}_{uu}=\frac{\pa {\bf x}_u}{\pa u},\quad {\bf x}_{uv}=\frac{\pa {\bf x}_u}{\pa v},\quad
{\bf x}_{vv}=\frac{\pa {\bf x}_v}{\pa v}
\]

Let
\[
L=S({\bf x}_u)\cdot  {\bf x}_u,\quad M=S( {\bf x}_u)\cdot  {\bf x}_v=S( {\bf x}_v)\cdot  {\bf x}_u,\quad
N=S( {\bf x}_v)\cdot  {\bf x}_v.
\]
{Characterizing the principle curvatures}

\begin{theorem}{}{}
The principle curvatures are the solutions of the following quadratic equation
 \begin{tcolorbox}[colback=white,colframe=red!75!black, text width=6.5cm,top=-.25cm,center]
\[
\det\left(\begin{bmatrix}L&M\\M&N\end{bmatrix}-\lambda\begin{bmatrix}E&F\\F&G\end{bmatrix}\right)=0
\]
\end{tcolorbox}
\end{theorem}






{Continuation of the Proof.}

\begin{theorem}{}{}
We have
 \begin{tcolorbox}[colback=white,colframe=red!75!black, text width=7cm,top=-.25cm,center]
\[
K=\frac{LN-M^2}{EG-F^2},\qquad
H=\frac{GL+EN-2FM}{2(EG-F^2)}.
\]
\end{tcolorbox}
\end{theorem}


\begin{lemma}{}{}
We have
\[
L=U\cdot {\bf x}_{uu},\qquad M=U\cdot {\bf x}_{uv},\qquad N=U\cdot {\bf x}_{vv}
\]
\end{lemma}

\section{The implicit Case}

Let $V,W$ be two linearly independent vector fields. Let
\[
\begin{split}
&S(V)=aV+bW\\
&S(W)=cV+dW
\end{split}
\]

Then the Gauss curvature is given by $ad-bc$, and the mean curvature is given by $\frac 12 (a+d)$.

Let $M$ is defined by a single equation $g=g(x,y,z)=0$. By analytic geometry, the norm vector field is defined by
\[
Z=\nabla g,
\]
where $\nabla g$ is called \emph{gradient}, given by
\[
\nabla g=\sum_{i=1}^3 \frac{\pa g}{\pa x_i} U_i=g_i U_i.
\]


\begin{lemma}{}{} Assume that $V\times W=Z$. Then
     \begin{tcolorbox}[colback=white,colframe=red!75!black, text width=6.5cm,top=-.0cm,center]
\[
\begin{split}
&
K=\frac{Z\cdot \nabla_ VZ\times \nabla_W Z}{\|Z\|^4}\\&
H=-Z\cdot\frac{\nabla_V Z\times W+V\times \nabla_WZ}{2\|Z\|^3}
\end{split}
\]
\end{tcolorbox}
\end{lemma}


{Example}

\begin{example}[Ellipsoid]
\[
M:  g=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1.
\]
\end{example}


{Example}

\begin{example}[Scherk minimal surface, Problem 5, \S 5.5]
\[
M:  e^z\cos x=\cos y
\]
\end{example}

{A formula}
If a surface is defined by $F(x,y,z)=0$. 
\[
K=-\frac{\det\begin{bmatrix}F_{xx}& F_{xy}&F_{xz}& F_x\\
F_{xy} &F_{yy}&F_{yz}& F_y\\
F_{xz}& F_{yz}& F_{zz} &F_z\\
F_x& F_y & F_z& 0\end{bmatrix}}{|\nabla F|^4}
\]





\begin{definition}{}{}
A regular curve $\alpha$ in $M\subset \R^3$ is a \emph{principal curve} provided that the velocity $\alpha'$ of $\alpha$ always points in a principal direction. 
\end{definition}

\begin{lemma}{}{}
Let $\alpha$ be a regular curve in $M\subset \R^3$, and let $U$ be a unit normal vector field restricted to $\alpha$. Then
\begin{enumerate}
\item the curve $\alpha$ is principal if and only if $U'$ and $\alpha'$ are collinear at each point.
\item if $\alpha$ is a principal curve, then the principal curvature of $M$ in the direction of $\alpha'$ is $\alpha''\cdot U/\alpha'\cdot\alpha'$. 
\end{enumerate}
\end{lemma}

\begin{lemma}{}{}
Let $\alpha$ be a curve cut from a surface $M\subset \R^3$ by a plane $P$. If the angle between $M$ and $P$ is constant along $\alpha$, then $\alpha$ is a principal curve of $M$.
\end{lemma}

Directions tangent to $M\subset\R^3$ in which the normal curvature is zero are called \emph{asymptotic} directions. 
\begin{lemma}{}{}
Let $P\in M$.
\begin{enumerate}
\item if $K(P)>0$, there is no asymptotic directions at $P$;
\item if $K(P)<0$, there are exactly two asymptotic directions at $P$, and these are bissected by the principal directions at angle $\theta$ such that 
\[
\tan^2\theta=\frac{-\lambda_1(P)}{\lambda_2(P)}.
\]
\item if $K(P)=0$, then the two principal curvatures are zero, then every direction is asymptotic; otherwise, only one direction is asymptotic. 
\end{enumerate}
\end{lemma}
{Ruled Surface}
A \emph{ruled surface} is swept out by a line moving through $\R^3$. 
\begin{lemma}{}{}
A ruled surface $M$ has Gauss curvature $K\leq 0$. Moreover, $K=0$ if and only if  the unit normal $U$ is parallel along each ruling of $M$. 
\end{lemma}



\section{Surface of revolution}
Let $\alpha$ be a curve defined by the parametric equations
\[
x=g(u),\qquad y=h(u)>0.
\]
Or $\alpha(u)=(g(u),h(u),0)$.
Define a surface $M$ of revolution
\[
\x(u,v)=(g(u), h(u)\cos v, h(u)\sin v).
\]
We then have
\[
\begin{split}
& \x_u=(g',h'\cos v,h'\sin v)\\
& \x_v=(0,-h\sin v,h\cos v)
\end{split}
\]
Hence 
\[
E=(g')^2+(h')^2,\qquad F=0,\qquad G=h^2.
\]
{Continuing...}
\[
\begin{split}
&
\x_u\times \x_v=(hh', -hg'\cos v,-hg'\sin v)\\
&\|\x_u\times \x_v\|=\sqrt{EG-F^2}=h\sqrt{(g')^2+(h')^2}\\
& U=\frac{1}{\sqrt{(g')^2+(h')^2}}(h', -g'\cos v, -g'\sin v)
\end{split}
\]
We also have
\[
\begin{split}
& \x_{uu}=(g'',h''\cos v,h''\sin v)\\
&\x_{uv}=(0, -h'\sin v, h'\cos v)\\
&\x_{vv}=(0,-h\cos v, -h\sin v)
\end{split}
\]
Then
\[
L=\frac{-g'h''+g''h'}{\sqrt{(g')^2+(h')^2}},\qquad M=0,\qquad N=\frac{g' h}{\sqrt{(g')^2+(h')^2}}
\]
{Continuing...}
We thus have 
\[
S(\x_u)=\frac LE \x_u,\qquad S(\x_v)=\frac NG\x_v.
\]
Thus the two principal curvatures are given by 
\[
k_\mu=\frac LE=\frac{-(g'h''-h'g'')}{((g')^2+(h')^2)^{3/2}},\qquad
k_\pi=\frac NG=\frac{g'}{h((g')^2+(h')^2)^{1/2}}
\]
Thus
\[
K=k_\mu k_\pi=\frac{-g'(g'h''-h'g'')}{h((g')^2+(h')^2)^{2}}
\]
{Continuing...}
If we assume that $g(u)=u$, then 
\[
\begin{split}
&
k_\mu=\frac{-h''}{(1+(h')^2)^{3/2}},\quad k_\pi=\frac{1}{h(1+(h')^2)^{1/2}},\quad\\
&
K=\frac{-h''}{h(1+(h')^2)^{2}}
\end{split}
\]
{Example -- Torus}





\begin{example}
We rotate the the circle
\[
g(u)=r\sin u,\quad h(u)=R+r\cos u
\]
for constants $0<r<R$ to obtain
\[
\x(u,v)=(r\sin u, (R+r\cos u)\cos v, (R+r\cos u)\sin v)
\]
\end{example}


We then have
\[
\begin{split}
&
E=r^2,\quad F=0,\quad G=(R+r\cos u)^2\\
& L=r,\quad M=0,\quad N=(R+r\cos u)\cos u
\end{split}
\]Thus
\[
k_\mu=\frac 1r,\quad k_\pi=\frac{\cos u}{R+r\cos u},\quad K=\frac{\cos u}{r(R+r\cos u)}.
\]
{Example -- Catenoid}






\begin{example}
We rotate the the \emph{catenary}
\[
y=c\cosh(x/c)
\]

\end{example}



We then have 
\[
\x(u,v)=(u,c\cosh(u/c)\cos v, c\cosh(u/c)\sin v)
\]
We can compute 
\[
\begin{split}
&\x_u=(1,\sinh (u/c)\cos v,\sinh(u/c)\sin v)\\
&\x_v=(0,-c\cosh(u/c)\sin v,c\cosh (u/c)\cos v)
\end{split}
\]So
\[E= \cosh^2 (u/c),\quad F=0,\quad G=c^2\cosh^2(u/c)\]

{Example -- Catenoid, continued}
For the second derivatives, we have
\[
\begin{split}
&\x_{uu}=(0,\frac 1c\cosh(u/c)\cos v,\frac 1c\cosh(u/c)\sin v)\\
&\x_{uv}=(0,-\sinh (u/c)\sin v,\sinh(u/c)\cos v)\\
&\x_{vv}=(0,-c\cosh(u/c)\cos v, -c\cosh(u/c)\sin v)
\end{split}
\]
We also have
\[
U=\frac{\x_u\times \x_v}{\|\x_u\times \x_v\|}=(\tanh (u/c),-\sech (u/c)\cos v, -\sech(u/c)\sin v).
\]

Thus we have
\[
L=-\frac 1c,\qquad M=0,\qquad N=c
\]
We then have $H=0$ and $K=-\frac{1}{c^2\cosh^4(u/c)}$.

%\vspace{-1.4in}
\begin{theorem}{}{}
If a surface of revolution $M$ is a minimal surface, then $M$ is part of a plane or a catenoid.
\end{theorem}


We shall renormalizing the profile curve 
\[
\alpha(u)=(g(u),h(u))
\]to be of the unit speed. That is
\[
(g')^2+(h')^2=1.
\]

Under this renormalization, the curvature formulae, especially the formula for the Gauss curvature, is every simple. 

{Renormalization of the catenary}


{Surface of revolution of constant positive curvature}






\begin{example}
We rotate the unit speed curve $\alpha(u)=(g(u), h(u))$, where
\[
h(u)=a\cos(u/c)
\]
\end{example}



{Tractrix}

A tractrix (from the Latin verb trahere "pull, drag"; plural: tractrices) is the curve along which an object moves, under the influence of friction, when pulled on a horizontal plane by a line segment attached to a tractor (pulling) point that moves at a right angle to the initial line between the object and the puller at an infinitesimal speed.


{The differential equation of tractrix}
\[
h'=\frac{-h}{\sqrt{c^2-h^2}}.
\]
Parametric expression is given by (when $c=1$)
\[
x=t-\tanh t,\qquad y=\sech t.
\]
{Pseudo sphere}
Also called \emph{bugle surface}, \emph{tractroid}.




The Gauss curvature is given by 
\[
K=-\frac{h''}{h(1+(h')^2)^2}.
\]

