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\define\GL{\operatorname{GL}}
\define\ORTH{\operatorname{O}}
\define\U{\operatorname{U}}
\define\Lie{\operatorname{Lie}}
\define\SO{\operatorname{SO}}
\define\SU{\operatorname{SU}}
\define\SPIN{\operatorname{Spin}}

\define\Hom{\operatorname{Hom}}
\define\Iso{\operatorname{Iso}}
\define\End{\operatorname{End}}
\define\Aut{\operatorname{Aut}}
\define\Ker{\operatorname{Ker}}
\define\Coker{\operatorname{Coker}}
\redefine\Im{\operatorname{Im}}

\define\Rank{\operatorname{Rank}}
\define\Tr{\operatorname{Tr}}
\define\Mat{\operatorname{Mat}}
\define\vect{\operatorname{Vect}}

\define\rp{\Bbb{RP}}
\define\cp{\Bbb{CP}}

\define\R{\Bbb R}
\define\C{\Bbb C}
\define\Z{\Bbb Z}
\redefine\P{\Bbb P}
\redefine\H{\Bbb H}


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\parindent 16 pt

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\title       The Geometry of Vector Bundles and an Introduction to Gauge
Theory \\
Lecture 38
\endtitle
\author      Professor Steven Bradlow \\ Class Notes From Math 433
\endauthor
\affil       University of Illinois at Urbana-Champaign
\endaffil
\address     273 Altgeld Hall, 1409 W. Green Street, Urbana, IL  61801
\endaddress
\email       bradlow\@math.uiuc.edu
\endemail
\date        May 1, 1998
\enddate

\endtopmatter

\document

\noindent
\underbar{Last time}: We saw that the obstruction to orientability of
a vector bundle $E\to B$ is the first
Stiefel-Whitney class $w_1(E)\in H^2(B,\Z_2)$.

Suppose that $w_1(E)=0$, so that we can write $E=\coprod
U_{\alpha}\times\R^n/g_{\alpha\beta}\ ,\ 
g_{\alpha\beta}\in\underline{\SO(n)}$.

In this case, we can look for a more subtle structure called a 
{\bf spin structure}.

\vskip.25in
\leftline{{\bf Spin Structures}}
\vskip.25in

Replace $E$ by principal $\SO(n)$-bundle:
$$P_{\SO(n)}=\coprod U_{\alpha}\times\SO(n)/g_{\alpha\beta}\,\,.$$

\vskip.12in
\noindent
{\bf Key Facts}:

\item{$\bullet$} For $n>2$, $\pi_1(\SO(n))=\Z_2$. 
 
\vskip.05in
\item{$\bullet$} There exists a simply connected Lie group
$\SPIN(n)$ such that 
$$1\to\Z_2\to\SPIN(n)\overset\rho\to\longrightarrow\SO(n)\to\ 1\ ,$$
that is, $\rho$ is a {\it double cover} and {\it group homomorphism}.

\vskip.1in
\example{Example 1} $\SPIN(2)=S^1$. ($S^1\to S^1$ via $z\mapsto z^2$.) 
\endexample

\example{Example 2}
$\SPIN(3)=\SU(2)=\text{$2\times 2$
unitary complex matrices with determinant $1$}$.

\vskip.05in
\underbar{To see $\SPIN(3)\to\SO(3)$}:

\vskip.08in
\itemitem{$\bullet$} $A\in\SU(2)$ can be written as 
$$A=\bmatrix z_1&z_2\\ -\overline{z_2}&\overline{z_1}\endbmatrix$$
with $|z_1|^2+|z_2|^2=1$. So 
$$\split
\SU(2)&\cong\{\ (z_1,z_1)\ :\ |z_1|^2+|z_2|^2=1\ \}\cong S^3\\
&\cong \{\ q=z_1+jz_2\in\H\ :\ |q|^2=1\ \}\,,
\endsplit\tag1$$ 
under $\C^2\cong\H=\R\oplus i\R\oplus j\R\oplus k\R$ and $q=z_1+jz_2$. 

\vskip.05in
\itemitem{$\bullet$} Define 
$$\alignat2
\SU(2)\times &\H && \to\H \\
(q_0\,,\, &q)  && \mapsto q_0\,q\,q_0^{-1}
\endalignat$$
Let $\R^3=\Im\H=i\R\oplus j\R\oplus k\R$.
Then the unit quaternians act on $\Im\H=\R^3$ by 
$$\alignat2
\SU(2)\times &\Im\H &&\to\Im\H\\
(q_0\,,\, &x) &&\mapsto
q_0\,x\,q_0^{-1}(\,=q_0\,x\,\overline{q}_0\quad\text{if $|q_0|^2=1$}\,)
\tag2\endalignat$$
This defines an action of $\SU(2)$ on $\R^3$.
In fact, if $|q_0|=1$ and $x\in\Im\H\cong\R^3$, then 
$x\overset {A_{q_0}}\to\mapsto q_0\,x\,q_0^{-1}$ is an $\SO(3)$ transformation.
\newline
\itemitem{}{\bf Claim}: 
$$
\alignat2
\SU(&2) &&\to\SO(3)\\
&q_0 &&\mapsto A_{q_0}
\endalignat
$$
is the $2:1$ double cover ($A_{q_0}=A_{-q_0}$).
Hence the map $q_0\mapsto A_{q_0}$ is the map we want\,\,!
\endexample

\example{Example 3} $\SPIN(4)=\SU(2)\times\SU(2)$\,.
\endexample

\example{Remark} Given $\SPIN(n)\overset\rho\to\longrightarrow\SO(n)$
and $g\in\SO(n)$, we can always find $\widetilde g\in\SPIN(n)$ such
that $\rho(\widetilde g)=g$.
\endexample

\vskip.12in
How about 
$$\sarrowlength=.9\harrowlength
\commdiag{
&&\SPIN(n)\cr
 &\arrow(1,1)\lft{?\ \exists\ \widetilde g}&\mapdown\rt{\rho}\cr
U&\mapright^{g} &\SO(n) }\qquad ?$$
If $U$ is contractible, then we can do it\,\,!

Apply this to $g_{\alpha\beta}:U_{\alpha}\cap U_{\beta}\to\SO(n)$ in
the given principal bundle $P_{\SO(n)}=\coprod
U_{\alpha}\times\SO(n)/g_{\alpha\beta}$ (assuming $\{U_{\alpha\beta}\}$
is ``nice''). Thus we can lift $g_{\alpha\beta}$ to 
$\widetilde g_{\alpha\beta}:U_{\alpha}\cap U_{\beta}\to\SPIN(n)$ if 
$U_{\alpha}\cap U_{\beta}$ is contractible. That is, the diagram commutes: 
$$\sarrowlength=.9\harrowlength
\commdiag{
&&\SPIN(n)\cr
 &\arrow(1,1)\lft{\widetilde g_{\alpha\beta}}\circ&\mapdown\rt{\rho}\cr
U_{\alpha}\cap U_{\beta}&\mapright^{g_{\alpha\beta}} &\SO(n) }$$

\vskip.12in
\noindent
{\bf Question}: Do the transition functions 
$\{\widetilde g_{\alpha\beta}:U_{\alpha}\cap U_{\beta}\to\SPIN(n)\}$
define a {\bf $\SPIN(n)$ bundle}, i.e., do we get 
$$P_{\SPIN(n)}=\coprod U_{\alpha}\times\SPIN(n)/\widetilde
g_{\alpha\beta}\,\,?$$ 
\vskip.12in

If {\bf YES}, then such a lift defines a {\bf Spin structure} on
$\rho$ and we have $P_{\SPIN(n)}\overset\rho\to\longrightarrow
P_{\SO(n)}$, $2:1$ on each lift. 

We require (the cocycle condition) 
$\widetilde g_{\alpha\beta}\ \widetilde g_{\beta\gamma}\ \widetilde
g_{\gamma\alpha}=I\ (\,\text{in $\SPIN(n)$}\,)$.

\vskip.12in
\noindent
{\bf How to Detect Obstruction}:

Given $\{g_{\alpha\beta}\}$, we can pick a system of lifts
$\{\widetilde g_{\alpha\beta}\}$. Define
$h_{\alpha\beta\gamma}:=\widetilde g_{\alpha\beta}\ \widetilde
g_{\beta\gamma}\ \widetilde g_{\gamma\alpha}$ for $U_{\alpha}\cap
U_{\beta}\cap U_{\gamma}\neq\emptyset$. 

\remark{Note}
$h_{\alpha\beta\gamma}\in\SPIN(n)$
but $\rho(h_{\alpha\beta\gamma})=g_{\alpha\beta}\ g_{\beta\gamma}\
g_{\gamma\alpha}=I\in\SO(n)$. So
$h_{\alpha\beta\gamma}\in\rho^{-1}(I)\cong\Z^2$
\endremark

\vskip.12in
So $h^{(2)}=\{h_{\alpha\beta\gamma}\}$ defines a \v Cech $2$-cochain.

\proclaim{Key Facts} (With $h_{\alpha\beta\gamma}=e^{i\pi
w_{\alpha\beta\gamma}}$ defining $w^{(2)}$.)
$$\delta w^{(2)}=0$$ so
$$[w^{(2)}]\in\check H^{(2)}(B,\Z_2)$$ is the obstruction\,\,!

If $[w^{(2)}]=0$, then there exists
$\{\lambda_{\alpha\beta}\}=\lambda^{(1)}$ a $1$-cochain such that
$w^{(2)}=\delta(\lambda^{(1)})$.
\endproclaim

\proclaim{FACT} The class $[w^{(2)}]=w_2(P_{\SO(n)})\in H^2(B,\Z_2)$
is the {\bf second Stiefel-Whitney class}. 
$$P_{\SO(n)}\ \text{admits a spin structure}\Longleftrightarrow
w_2(P_{\SO(n)})=0\,\,.$$
\endproclaim

\definition{Definition 1} An orientable manifold $M$ is {\it spin} if
a principal $\SO(n)$ frame bundle admits a lift to a principal
$\SPIN(n)$ bundle ($\Longleftrightarrow w_2(TM)\equiv w_2(M)=0$).
\enddefinition

\example{Examples}
\itemitem{(1)} $S^m$ is spin.

\itemitem{(2)} All orientable Riemann surfaces, $\Sigma^g$, are spin.
 
\itemitem{(3)} $\cp(k)$ is spin $\Longleftrightarrow k$ is
odd. ($\cp(2)$ is {\bf NOT} spin\,\,!)

\itemitem{(4)} All Lie groups are spin.
\endexample

\example{Remark} If $M$ is spin and $P_{\SPIN}$ is a lift of $\SO(n)$
(bundle of frames), we can get associated vector bundles 
$$S=P_{\SPIN}\times_{\rho}V$$ where $\rho:\SPIN(n)\to\Aut(V)$ defines 
{\bf representation of spin (``Spin Bundle'')}.
\endexample





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