\classheader{2012-10-26}

As a reminder, the midterm is Friday, November 2, 11am - 12:30pm.



\subsection*{Maps of Spheres}

A map $f:\S^n\to\S^n$ induces a map $f_*:H_n(\S^n)\to H_n(\S^n)$. Recall that $H_n(\S^n)\cong\Z$, having $[\Delta_1-\Delta_2]$ as a generator, where $\Delta_1$ and $\Delta_2$ are $n$-simplices glued along their boundaries to form the ``equator'' of the sphere.

\begin{definition}
The degree of $f$ is the unique $\deg(f)\in\Z$ such that $f_*(z)=\deg(f)z$ for all $z\in H_n(\S^n)$.
\end{definition}
Note that
\[\deg(f\circ g)z=(f\circ g)_*(z)=f_*(g_*(z))=f_*(\deg(g)z)=\deg(g)f_*(z)=\deg(f)\deg(g)z,\]
and hence $\deg(f\circ g)=\deg(f)\deg(g)$. Also, if $f\sim g$, then $f_*=g_*$, and hence $\deg(f)=\deg(g)$. In fact, the converse is true; that is, considering the map
\[\{f:\S^n\to\S^n\}/{\sim}\;\;\longrightarrow\;\;\Z\]
sending $[f]$ to $\deg(f)$, this map is a bijection. This is an old theorem:
\begin{theorem}[Hopf]
For any $f,g:\S^n\to\S^n$, then $f\sim g\iff \deg(f)=\deg(g)$.
\end{theorem}

Here are some basic properties.

\begin{itemize}
\item If $f$ is not surjective, then $\deg(f)=0$. This is because there is an $x$ such that $f(\S^n)\subseteq\S^n-x$, and $\S^n-x\cong\R^n$, so the fact that we have a commutative diagram
\begin{center}
\begin{tikzcd}[column sep=large, row sep=large]
H_n(\S^n) \ar{r}{f_*} \ar{d}[swap]{f_*} & H_n(\S^n)\\
\genfrac{}{}{0pt}{0}{H_n(\S^n-x)}{=H_n(\R^n)=0} \ar{ru}
\end{tikzcd}
\end{center}
forces $f_*=0$.

\item We can consider the suspension of a map between spheres, $\Sigma f:\Sigma\S^n\to\Sigma\S^n$. Note that $\Sigma\S^n\cong\S^{n+1}$ (this is clear if you draw a picture). We get a commutative diagram
\begin{center}
\begin{tikzcd}[column sep=large, row sep=large]
H_n(\S^n) \ar{r}[swap]{i}{\cong} \ar{d}[swap]{f_*} & H_{n+1}(\S^{n+1}) \ar{d}{\Sigma f_*} \\
H_n(\S^n) \ar{r} & H_{n+1}(\S^{n+1})
\end{tikzcd}
\end{center}
implying that $\deg(f)=\deg(\Sigma f)$.\newpage

\item Let $d\geq 1$. Let $\psi_d:\S^1\to\S^1$ be the map $z\mapsto z^d$. We give $\S^1$ two different $\Delta$-complex structures,
\begin{center}
\begin{tikzpicture}[scale=1.3]
\node at (-1.65,0) {$X=$};
\draw[thick] (0,0) circle (1);
\draw[thick] (0:0.9) -- (0:1.1);
\draw[thick] (50:0.9) -- (50:1.1);
\draw[thick] (100:0.9) -- (100:1.1);
\draw[thick] (310:0.9) -- (310:1.1);
\node at (25:1.3) {$\tau_1$};
\node at (75:1.3) {$\tau_2$};
\node at (335:1.3) {$\tau_d$};
\node[rotate=35] at (120:1.2) {$\cdot$};
\node[rotate=35] at (127:1.2) {$\cdot$};
\node[rotate=35] at (134:1.2) {$\cdot$};
\begin{scope}[outer sep=2pt,decoration={markings,mark=at position 0.6 with {\arrow[scale=1.4]{angle 60}}}]
\draw[postaction={decorate}]
(1,0) arc (0:50:1) ;
\draw[postaction={decorate}]
(50:1) arc (50:100:1) ;
\draw[postaction={decorate}]
(310:1) arc (310:359:1) ;
\end{scope}
\end{tikzpicture}
\qquad
\begin{tikzpicture}[scale=1.3]
\node at (-1.65,0) {$Y=$};
\draw[thick] (0,0) circle (1);
\draw[thick] (0:0.9) -- (0:1.1);
\node at (90:1.3) {$\sigma$};
\begin{scope}[outer sep=2pt,decoration={markings,mark=at position 0.51 with {\arrow[scale=1.4]{angle 60}}}]
\draw[postaction={decorate}]
(1,0) arc (0:359:1) ;
\end{scope}
\end{tikzpicture}
\end{center}
Then $\psi_d(\tau_i)=\sigma$ for all $i$, so that the map
\[(\psi_d)_*:\underbrace{H_1(X)}_{\displaystyle\left\langle\big[\textstyle\sum_{i=1}^d \tau_i\big]\right\rangle}\to\;\;\; \underbrace{H_1(Y)}_{\displaystyle\langle[\sigma]\rangle}\]
satisfies
\begin{align*}
(\psi_d)_*([\textstyle\sum_{i=1}^d \tau_i])&=[(\psi_d)_\# \textstyle\sum_{i=1}^d \tau_i]\\
&=[\textstyle\sum_{i=1}^d\psi_d(\tau_i)]\\
&=[\textstyle\sum_{i=1}^d\sigma]\\
&=d[\sigma]
\end{align*}
and hence $\deg(\psi_d)=d$. Note that we implicitly used that this diagram commutes:
\begin{center}
\begin{tikzcd}[column sep=large, row sep=large]
H_1^\Delta(X) \ar{r}{f_*^\Delta}\ar{d}[swap]{\phi}{\cong} & H_1^\Delta(Y) \ar{d}{\phi}[swap]{\cong}\\
H_1^S(\S^1) \ar{r}[swap]{f_*^S} & H_1^S(\S^1)
\end{tikzcd}
\end{center}
but this is just because $\phi$ (the isomorphism between simplicial and singular homology) is natural, i.e. the composition of degrees doesn't depend on the chosen $\Delta$-complex structure.

Thus, for all $n\geq 1$ and $d\in\Z$ there is a map $f:\S^n\to\S^n$ of degree $d$, namely $\Sigma^{n-1}\psi_d$.


In differential topology, while you can run into trouble with a few bad points,  you can generically get the degree in a similar way. For example, consider the suspension $\Sigma\psi_d$, which is the map from $\S^2$ to $\S^2$ wrapping the sphere around itself so that a lune of angle $\frac{2\pi}{d}$ is wrapped over the entire sphere.

\begin{center}
$\text{}$\hspace{0.6in}\begin{tikzpicture}[scale=0.5,every node/.style={minimum size=1cm},>=latex]
%% some definitions
\def\R{4} % sphere radius
\def\angEl{25} % elevation angle
\def\angAz{-100} % azimuth angle
\def\angPhiOne{-70} % longitude of point P
\def\angPhiTwo{-35} % longitude of point Q
\def\angBeta{30} % latitude of point P and Q
%% working planes

\pgfmathsetmacro\H{\R*cos(\angEl)} % distance to north pole
\LongitudePlane[xzplane]{\angEl}{\angAz}
\LongitudePlane[pzplane]{\angEl}{\angPhiOne}
\LongitudePlane[qzplane]{\angEl}{\angPhiTwo}
\LatitudePlane[equator]{\angEl}{0}
\fill[ball color=white!60] (0,0) circle (\R); % 3D lighting effect
\coordinate (O) at (0,0);
\coordinate[mark coordinate] (N) at (0,\H);
\coordinate[mark coordinate] (S) at (0,-\H);
\path[xzplane] (\R,0) coordinate (XE);

%defining points outsided the area bounded by the sphere
\path[qzplane] (\angBeta:\R+5.2376) coordinate (XEd);
\path[pzplane] (\angBeta:\R) coordinate (P);
\path[pzplane] (\angBeta:\R+5.2376) coordinate (Pd);
\path[pzplane] (\angBeta:\R+5.2376) coordinate (Td);
\path[pzplane] (\R,0) coordinate (PE);
\path[pzplane] (\R+4,0) coordinate (PEd);
\path[qzplane] (\angBeta:\R) coordinate (Q);
\path[qzplane] (\angBeta:\R) coordinate (Qd);
\path[qzplane] (\R,0) coordinate (QE);
\path[qzplane] (\R+4,0) coordinate (QEd);
\DrawLongitudeCircle[\R]{\angPhiOne} 
\DrawLongitudeCircletwo[\R]{\angPhiTwo} 
\DrawLatitudeCircle[\R]{0} 

\path[pzplane] (0.5*\angBeta:\R) node[right] {$$};
\path[qzplane] (0.5*\angBeta:\R) node[right] {$$};
\draw[equator,->,very thick] (\angAz+60:1.5*\R) to[bend right=27]
node[pos=0.4,above] {} (\angPhiOne+80:1.5*\R);
\path[xzplane] (0:\R) node[below] {$$};
\path[xzplane] (\angBeta:\R) node[below left] {$$};
\end{tikzpicture}
\end{center}

Then the preimage of almost every point is a set of $d$ points, except the north and south poles, which have only one preimage (they are preserved).


\item Let $\S^n=\{(x_1,\ldots,x_{n+1})\mid \sum x_i^2=1\}\subset\R^{n+1}$. Let $r:\S^n\to\S^n$ be the reflection map \[r(x_1,x_2\ldots,x_{n+1})=(-x_1,x_2,\ldots,x_{n+1}).\] We claim that $\deg(r)=-1$. Here is the proof: consider the $\Delta$-complex $X=\Delta_1^n\sqcup \Delta_2^n$ where $\partial\Delta_1^n=\partial\Delta_2^n$. Note that $X\cong\S^n$. Then the induced map $r_\#:C_n(X)\to C_n(X)$ just swaps the simplices, $\Delta_1\mapsto\Delta_2$ and $\Delta_2\mapsto\Delta_1$. Also, note that $H_n(X)\cong\Z=\langle[\Delta_1-\Delta_2]\rangle$. Thus,
\[r_*([\Delta_1-\Delta_2])=[r_\#(\Delta_1-\Delta_2)]=[\Delta_2-\Delta_1]=-[\Delta_1-\Delta_2]\]
and hence $\deg(r)=-1$.

As a corollary, this implies that the antipodal map $A(x_1,\ldots,x_{n+1})=(-x_1,\ldots,-x_{n+1})$ has degree $(-1)^{n+1}$, because
\[\deg(A)=\deg(r_1\circ r_2\circ\cdots\circ r_{n+1})=\deg(r_1)\deg(r_2)\cdots\deg(r_{n+1})=(-1)^{n+1},\]
and hence $A\not\sim \id_{\S^n}$ when $n$ is even, and $A\sim \id_{\S^n}$ when $n$ is odd.

\end{itemize}


Which maps $f:\S^n\to\S^n$ have no fixed points? Well, the antipodal map $A$, obviously; any others?

The Lefschetz number of any map $f:\S^n\to\S^n$ is
\[\Lambda(f)=\underbrace{\tr(f_*:H_0(\S^n)\rcirclearrowleft)}_{=\,1}+(-1)^n\deg(f)\]
Note that if $f$ has no fixed points, then we must have $\Lambda(f)=0$, hence $\deg(f)=(-1)^{n+1}$, and therefore $f\sim A$ by Hopf's theorem. Thus,  $f:\S^n\to\S^n$ has no fixed points implies that $f\sim A$.

\begin{definition}
Consider $\S^{n-1}$ as a subset of $\R^n$. A continuous vector field on $\S^{n-1}$ is a continuous map $v:\S^{n-1}\to\R^n$ such that $v(z)\perp z$ for all $z\in\S^{n-1}$.
\end{definition}
\begin{theorem}
The $n$-sphere $\S^n$, for $n\geq 1$, admits a nowhere zero vector field $v$ $\iff$ $n$ is odd.
\end{theorem}
\begin{proof}
To see $\impliedby$, consider $v(x_1,\ldots,x_{n+1})=(-x_2,x_1,-x_4,x_3,\ldots,-x_{n+1},x_n)$. We needed $n$ odd to be able to pair off the coordinates.

To see $\implies$, use the vector field $v$ to prove $\id_{\S^{n-1}}\sim A$. If $v$ is our nowhere vanishing vector field, then consider the map $v_t:\S^{n-1}\times I\to\S^{n-1}$ defined by \[v_t(z)=-\cos(\pi t)z+\sin(\pi t)v(z).\]
As $t$ goes from 0 to 1, this map slides a point $z\in\S^n$ along the great circle connecting $z$ to $-z$ in the direction determined by $v(z)$. This is a homotopy from $\id_{\S^{n-1}}$ to $A$, so $\deg(\id_{\S^{n-1}})=1=(-1)^{n+1}=\deg(A)$, hence $n$ is odd.
\end{proof}

\begin{theorem}[Adams, 1962]
Let $n\geq 1$, and write $n+1=2^{4a+b}(2k+1)$ where $a,b,k\in\Z$, and $0\leq b\leq 3$. Then the maximum number of linearly independent vector fields on $\S^n$ is precisely $2^b+8a-1$.
\end{theorem}







