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\begin{document}
\centerline{Math 322 lecture for Monday, Week 11}
\bigskip

Our goal now is to formally define the index of a vector field on a surface
besides~$\R^2$.  Let~$S$ be a $2$-dimensional manifold.  So~$S=\cap_i U_i$ where
the~$U_i$ open sets and with homeomorphisms
\[
h_i\colon U_i\to V_i\subset\R^2
\]
which allow us to think of each~$U_i$ as an open subset~$V_i\subseteq\R^{2}$ of the plane.  On
overlaps~$U_i\cap U_j$, the \emph{change of coordinates} mapping~$h_j\circ h_i^{-1}$ is differentiable.  
Recall that the pair~$(U_i,h_i)$ is called a {\em chart} and the collection of
charts is an {\em atlas} .
Suppose that~$S\subseteq\R^n$. A {\em vector field} on~$S$ is
a~$C^1$-mapping~$f\colon S\to\R^n$ such that~$f(p)$ is tangent to~$S$, so there
is a curve~$\gamma\colon (-1,1)\to S$ such that~$\gamma(0)=p$ and~$\gamma'(0)=f(p)$.

To calculate the index of a critical point~$p$ of~$f$, i.e., at a point
where~$f(p)=0$, we first pick~$U_i$ such
that~$p\in U_i$, and we use~$h_i$ to identify~$f$ with a vector field
on~$\R^2$ with critical point~$h_i(p)$.  In detail, if~$q\in U_i$, we pick a
curve~$\gamma$ in~$S$ passing through~$q$ at time~$0$ and such
that~$\gamma'(0)=f(q)$.  To find the corresponding vector on~$\R^2$, we use the
composition
\[
  (-1,1)\xrightarrow{\gamma}U_i\xrightarrow{h_i}\R^2.
\]
The vector in~$\R^2$ at the point~$q$ will be~$(h_i\circ\gamma)'(0)$.

The next thing we need to define the index of a critical point of a vector field
on~$S$ is an orientation. An \emph{orientation} is a choice of charts so that
the Jacobian of each change of coordinates has positive determinant wherever
defined.  In other words, for all~$i,j$ and for all~$p\in U_{i}\cap U_{j}$, we
assume~$\det J(h_{j}\circ h_{i}^{-1})>0$.  Then, to define the
\emph{index} of a vector field at a critical point~$p$, first choose a
chart~$(U_{i},h_{i})$ with~$p\in U_{i}$.  Use~$h_i$ to translate the vector
field~$f$ to a vector field~$h_{i,*}(f)$ on~$V_{i}=h_{i}(U_{i})$ as described above.  Then
define the index~$I_{f}(p)$ to be the index of~$h_{i}(p)$ for the vector field
$h_{i,*}(f)$ i.e.,~$I_{f}(p):=I_{h_{i,*}(f)}(h_{i}(p))$.

Not all~$2$-dimensional manifolds are orientable, for instance, a M\"obius
strip is not orientable, nor is the projective plane~$\mathbb{P}^{2}$ (which
contains a M\"obius strip).

We now assume that~$S$ is compact and oriented.  It turns out that in this
case, we can take~$S$ to be a~$g$-hold donut embedded in~$\R^3$.  We will take
the orientation to be the one determined by the outward-pointing normal vector
for~$S$: for a chart at a point~$p$, just take a sufficiently small open piece
of the surface containing~$p$, and ``flatten is out''.  We take
``counterclockwise'' to be the direction for which the right-hand rule gives
the outward pointing normal vector:
\begin{center}
  \begin{tikzpicture}
    \draw[fill=blue!20] (0,0) ellipse (1.7cm and 1cm);
    \draw[->, thick] (0,0)--(0,2);
    \node at (0,2.4) {$\vec{\mathbf{n}}$};
    \draw[->] (-0.2,-0.5) to[bend right=40] (0.5,-0.2);
  \end{tikzpicture}
\end{center}
\medskip

\textbf{Theorem (Poincar\'e-Hopf index theorem).} Suppose the critical
points of the vector field~$f$ on~$S$ are~$p_1,\dots,p_k$.  Then
\[
  \sum_{i=1}^kI_f(S)=2-2g.
\]
\begin{proof}
 We first consider the case~$g=0$.  So~$S$ is an ordinary sphere
 in~$\R^3$.  Draw a tiny circle~$C$ around a regular point~$q$ (i.e.,
 non-critical point).  Call the side of~$C$ containing~$q$ the ``inside''
 of~$C$.  We take~$C$ tiny enough so that there are no critical points
 inside~$C$ and~$f$ is virtually constant around~$C$:
 \begin{center}
   \begin{tikzpicture}
     \draw (0,0) circle (1);
     \foreach \t in {0,...,11}{
       \draw[->]
       ({cos(\t*360/12)},{sin(\t*360/12)})--({cos(\t*360/12)+0.7},{sin(\t*360/12)});
     }
      \node at (-1.6,0) {$C$};
   \end{tikzpicture}
 \end{center}
 Now for the hard part: imagine cutting out the inside of~$C$,
 stretching~$C$ and the remaining part of~$S$ so that this remaining part
 (containing all of the critical points of~$f$) sits in~$\R^2$ with~$C$ as its
 boundary.  If you are careful with how the vector field morphs under this
 transformation, you should get the picture:
 \begin{center}
   \begin{tikzpicture}
     \draw (0,0) circle (1);
     \foreach \t in {0,...,11}{
       \draw[->]
       ({cos(\t*360/12)},{sin(\t*360/12)})--++({-0.4*cos(\t*360/6)},{-0.4*sin(\t*360/6)});
     }
	\node[xscale=-1] at (-0.4,0) {$C$};
   \end{tikzpicture}
 \end{center}
 The vector field now rotates clockwise twice as we go around~$C$.
 The critical points are now inside the circle in the picture above, so it looks
 like the sum of their indices should be~$2$.  However, if you are careful,
 you'll notice that the orientation has reversed (the normal vectors from the sphere are
  now pointing into the page).  So we should flip the picture, to get the usual
   orientation:
 \begin{center}
   \begin{tikzpicture}[xscale=-1]
     \draw (0,0) circle (1);
     \foreach \t in {0,...,11}{
       \draw[->]
       ({cos(\t*360/12)},{sin(\t*360/12)})--++({-0.4*cos(\t*360/6)},{-0.4*sin(\t*360/6)});
     }
	\node at (-0.4,0) {$C$};
   \end{tikzpicture}
 \end{center}
   As we travel around the boundary counter-clockwise, the vector field on the
   boundary rotates twice, again counterclockwise.  So the index is still~$2$
   (as opposed to~$-2$).

 Thus, the sum of the indices is
 \[
 2=2-2\cdot0=2-2g
 \]
 in the case of $S=S^{2}$.
 
 Now consider a~$g$-holed donut:
 \begin{center}
   \includegraphics[height=1.5in]{donut1.png}
 \end{center}
 The picture is taken from our text, where~$p$ is used instead of~$g$.  We will
 keep using~$g$ to denote the number of holes.
 The cylinders~$S_1,\dots,S_g$ are chosen so that they contain no critical
 points.  Next, contract the boundaries of the~$S_i$, morphing the~$S_i$ into
 spheres~$S_1',\dots,S_g'$:
 \begin{center}
   \includegraphics[height=1.5in]{donut2.png}
 \end{center}
 The vector field morphs, too, creating critical points~$Q_i$ and~$Q_i'$
 for~$i=1,\dots,g$ on the surface~$S'$ pictured above, and corresponding critical
 points~$P_i$ and~$P_i'$ on the spheres~$S_i'$.  Since the vector field at
 each~$P_i$ is the negative of that on~$Q_i$ and similarly for~$P_i'$
 and~$Q_i'$, we have
 \[
   I_f(P_i)=I_f(Q_i)\quad\text{and}\quad I_f(P_i')=I_f(Q_i').
 \]
 Since the~$S_i'$ are spheres, from our previous work it follows that
 \[
   I_f(P_i)+I_f(P_i')=2
 \]
 for all~$i$.  The surface~$S'$ is a sphere.  Therefore,
 \[
   I_f(S')=2.
 \]
 The result follows:
 \begin{align*}
   I_f(S)
   &= I_f(S')-\sum_{i=1}^g\left( I_f(Q_i)+I_f(Q_i') \right)\\[5pt]
   &= I_f(S')-\sum_{i=1}^p\left( I_f(P_i)+I_f(P_i') \right)\\[5pt]
   &= 2-2g.
 \end{align*}
\end{proof}
\bigskip

\textbf{Note:} The Poincar\'e-Hopf theorem also holds for non-orientable manifolds.  See our
text for a proof (that builds on the above proof).
\end{document}
