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\begin{document}
\parindent=0cm
\section*{VIII.2 Further Application to Sphere}

\begin{defn} $f : S^n \rightarrow S^n \Rightarrow f_*
: \hh (S^n ) \rightarrow \hh (S^n) $. \\Now let $\alp$ be a
generator of $\hh(S^n ) \cong \mathbb{Z}$ and let $f_* (\alp) = d
\alp.$\\ Then the integer $d$ is independent of choice of a
generator $\alp$, and is called the degree of $f$.
\end{defn}

\textbf{Remark.} $M^n$ :  connected orientable closed $n$-manifold
$\Rightarrow$ degree is defined.

\begin{thm} (Basic properties of degree)\\
(i) $f \simeq g \Rightarrow $ deg $f$ = deg $g$\\
(ii) deg ($f\circ g) \Rightarrow $(deg $f)$(deg $g$)\\
(iii) deg (id) = 1\\
(iv) deg ($c$) = 0 , where $c$ is a constant map.\end{thm}

\begin{pf}
(i) $\sim$ (iii) : clear\\
(iv) $c$ : constant map $\Rightarrow c_* = 0 : \widetilde{H} (X)
\rightarrow \widetilde{H} (X)$\\
($c : X \overset{\bar{c}}{\rightarrow} x_0
\overset{i}{\hookrightarrow} X \Rightarrow c_* : \h (X)
\overset{\bar{c_*}}{\rightarrow} \h (x_0 ) = 0 \overset{i_*
}{\rightarrow}\h (X)$)\end{pf}\\

\textbf{Note.} (1) $\alp_n : S^1 \rightarrow S^1 \Rightarrow $ deg
$(\alp_n ) = n$\\
$\hspace*{7em} z\rightarrow z^n$\\
\begin{pf}
\[
\xymatrix @M=1ex @C=3em @R=1em @*[c]{%
\pi_1(S^1) \ar[rr]^{(\alp_n )_\sharp} \ar[dd]^{\chi : \cong}& & \pi_1 (S^1 ) \ar[dd]^{\chi : \cong} \\
 &\curvearrowright & \\
H_1 (S^1 ) \ar[rr]^{(\alp_n )_*} & & H_1 (S^1 )}
\]
$\Rightarrow (\alp _n )_* \{\alp_1\} = \chi ((\alp_n)_\sharp
[\alp_1] ) = \chi ([\alp_n \circ \alp_1 ]) = \chi ([\alp_n]) \\
\hspace*{5.5em}= \chi (n [\alp_1 ]) = n \chi [\alp_1 ] = n
\{\alp_1\}$\end{pf}\\

(2) Brower fixed point theorem on $B^n$: Prove as follows.\\

(i) $S^n \overset{f}{\rightarrow} S^n \hspace{0.5em} \Rightarrow \hspace{0.5em}$deg$f = 0$ :
Applying homology functor to the diagram \\
$\hspace*{1.5em} \cup \hspace{0.2em} \curvearrowright
\hspace{0.2em}
\nearrow \bar{f}\hspace{7em}$and note that $\widetilde{H}(B^n )=0$\\
$\hspace*{1em} B^{n+1}$\\
(ii) $\nexists$ retraction $r : B^{n+1} \rightarrow S^n : \\id :
S^n \overset{i}{\rightarrow}
B^{n+1} \overset{r}{\rightarrow} S^n \overset{(i)}{\Rightarrow}$deg$(id) = 0$, a contradiction.\\

(iii) $\forall \phi : B^n \rightarrow B^n$ has a fixed point : The
proof is the same as before for the case of $B^2$. (If $\phi$ has
no fixed point, we can construct a retraction:$B^n \rightarrow
S^{n-1} .$)

\begin{thm} Let $S^n \hookrightarrow \rb^{n+1}$ be the standard
sphere and $f \in O(n+1,\rb )$ so that $f : S^n \rightarrow S^n .
$ Then  deg $f$ = det $f (= \pm$ 1) \end{thm}
\begin{pf}
\begin{lem} Let $r :S^n \rightarrow S^n$ be the reflection given
by \\ $r(x_1 , x_2 , \cdots , x_{n+1} ) = (-x_1 , x_2 ,\cdots ,
x_{n+1}).$ Then deg $r$ = -1 = det $r$.\end{lem}

\begin{pf} induction on $n$.\\
Let $S^0 = \{-1, 1\}$, and $x$ and $y$ be a 0-singular simplex
which has image -1 and 1 respectively. $\widetilde{H_0}(S^0 ) =
\{x-y\} $ and $r_* :
x-y \mapsto y-x\\
\therefore$ deg $r = - 1$.\\

$n > 0 :$  MV sequence$ \Rightarrow$\\
\[
\xymatrix @M=1ex @C=3.5em @R=1em @*[c]{%
\widetilde{H_n}(S^n ) \ar[rr]^{\bd_* : \cong} \ar[dd]^{r_*}& & \widetilde{H_{n-1}}(S^{n-1} )\ar[dd]^{r|_*}\\
& \circlearrowright &  \\
\widetilde{H_n}(S^n ) \ar[rr]^{\bd_* : \cong}& &
\widetilde{H_{n-1}}(S^{n-1} )}
\]
$\Rightarrow $ deg $r_* $ = deg $r|_*$\end{pf}
\newpage
\begin{lem} $f \in SO(n+1) \Rightarrow f \simeq id.$ \end{lem}

\begin{pf} Essentially this is to show $SO(n+1)$ is
path-connected.\\

Fact. $ A \in SO(n) \Rightarrow \exists B$ s.t.

\begin{displaymath}
BAB^{-1} = \left(\begin {array}{ccc}
\left(\begin{array}{cc}
cos \theta & -sin \theta\\
sin \theta & cos \theta
\end{array}\right)& & \cdots \\
   \vdots       &            \ddots & \\
           &         &         1
\end{array}\right)\overset{let}{=} D_{\theta}
\end{displaymath}

where $\theta=(\theta_1 , \cdots , \theta_i )$\footnote{ $<  , >$
: standard inner product on $\mathbb{R}^n \rightsquigarrow$ ( , )
: standard Hermitian inner product on $\mathbb{C}^n \\ A \in O(n) \Rightarrow A$ is an isometry for ( , )\\
Let $Ax = \lambda x \Rightarrow \lambda\bar{\lambda}(x,x) = (Ax ,
Ax) = (x,x) \Rightarrow |\lambda | =1 \\ A:$ real $\Rightarrow
e^{i\theta}, e^{-i\theta} :$ conjugate eigenvlaues in general
$\Rightarrow Az  = (cos\theta + i sin \theta )z \Rightarrow  x =\frac{z + \bar{z}}{2} \\
\hspace*{24.5em} A\bar{z} = (cos \theta - i sin
\theta)\bar{z}\hspace{1.5em} y =\frac{z - \bar{z}}{2} i $\\
Then $A$ can be written as $\left(\begin{array}{cc} cos \theta &
-sin \theta\\
sin \theta & cos \theta \end{array}\right)$ \\with repect to a
basis $\{x,y\}$ where $x =\frac{z + \bar{z}}{2}$ and $y =\frac{z -
\bar{z}}{2} i$. }

Using fact, $H_t = B^{-1}D_{t{\theta}} B : I \simeq A\hspace{1em}
(t = 0 $ 일 때 $I, t = 1$ 일 때 $A$)\end{pf}\\

정리 2 증명 계속\\
Now note that $O(n+1) = SO(n+1) \cup rSO(n+1)
\overset{det}{\rightarrow} \{ -1, 1\} = \mathbb{Z} / 2$.(보조정리
3, 4로부터)

$\therefore $ deg = det \end{pf}

\begin{cor} $a : S^n \rightarrow S^n$ the antipodal map
$\Rightarrow $ deg $a = (-1)^{n+1}$.\end{cor}

\begin{thm} $f, g : S^n \rightarrow S^n $ with $f(x) \neq
g(x)\hspace{1em} \forall x\in S^n \Rightarrow f \simeq ag$.\\ (or
equivalently, $f$ and $g$ are never antipodal $\Rightarrow f\simeq
g$)\end{thm}
\begin{pf} $f(x) \neq g(x) \hspace{1em} \forall x \Rightarrow
f(x)$ and $ag(x)$ are not antipodal. \\ $\Rightarrow (1-t)f(x) + tag(x) \neq 0$\\
$F : S^n \times I \Rightarrow S^n $ given by $ F(x,t) =
\frac{(1-t)f(x) + tag(x)}{| (1-t)f(x) + tag(x) |}$ is the desired
homotopy.\end{pf}

\begin{cor} (1) $f$ has no fixed point. $\Rightarrow f \simeq a$\\
(2) deg $f \neq (-1)^{n+1} \Rightarrow f$ has a fixed
point.\end{cor}
\begin{pf} (1) $f(x) \neq id(x) \overset{정리 6}{\Rightarrow}
f\simeq a$ \end{pf}\\

exc. Is the converse of (1) true?\\

\textbf{숙제 1.} 문제 16.11.

\begin{thm} $S^n$ has a non-vanishing vector field iff $n=$
odd.\end{thm}
\begin{pf} $(\Leftarrow ) n = 2m +1 \Rightarrow $ let $ v(x_0
,\cdots , x_{2m+1} ) = (-x_1 , x_0 , -x_3 , x_2 , \cdots ,
-x_{2m+1} , x_{2m } )$\\

$(\Rightarrow) $ Suppose $\exists$ a non-vanishing vector field
$v,$ we may assume $|v(x)| = 1 \\ \Rightarrow F(x,t) =
(cos\hspace{0.2em} t\pi
)x + (sin\hspace{0.2em} t\pi )v(x)$ gives a homotopy : $id \simeq a \\
\Rightarrow $ deg $(id)$ = deg $(a) = (-1)^{n+1} \\ \Rightarrow n
= $ odd.\end{pf}\\

\textbf{Remark.} vector field problem for $S^{odd}$: \\Adams
solution : Let $n+1 = (odd) 2^{4a+b},\hspace{1em} 0\leqslant
b\leqslant 3$. Then the number
of independent vector fields on $S^n$ = $ 2^b + 8a -1.$\\

\textbf{숙제 2. } 문제 16.10.
\end{document}
