
\input amstex

\documentstyle{amsppt}

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\define\GL{\operatorname{GL}}
\define\ORTH{\operatorname{O}}
\define\Hom{\operatorname{Hom}}
\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\R{\Bbb R}
\define\C{\Bbb C}



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\parindent 16 pt

\title       The Geometry of Vector Bundles and an Introduction to Gauge
Theory \\
Lecture 9
\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        February 11, 1998
\enddate

\endtopmatter

\document

\vskip.25in
\leftline{{\bf New Bundles From Old Bundles II}}
\vskip.25in
Given two bundles
$$E=(\coprod U_\alpha \times \R^n)/\{g_{\alpha \beta}\}$$ and

$$F=(\coprod U_\alpha \times \R^n)/\{h_{\alpha \beta}\}.$$
\proclaim{Claim}
There exists a bundle $\Hom(E,F)$ such that rank $=n\cdot m$ 
and $\Hom(E,F)_b=\Hom(E_b,F_b)$.
\endproclaim
 
{\bf Proof:} There are two points of view.

(1) For vector spaces $V$ and $W$, $\Hom(V,W)\cong W\otimes V^*$.
\remark{Exercise 1} Show that $\Hom(V,W)\cong W\otimes V^*$.
\endremark
We can define $$\Hom(E,F):= F\otimes E^*$$ which has the
properties: (i) rank $nm$, (ii) fiber $(F\otimes E^*)_b=F_b\otimes
E^*_b$, and (iii) transition functions $h_{\alpha \beta}\otimes 
{g_{\alpha \beta}^t}^{-1}$.

(2) More directly: We can identify $\Hom(E,F)$ by looking at its
sections.
\remark{Exercise 2} Show that
$s\in\Gamma(\Hom(E,F))\Longleftrightarrow s:E\to F$ is a bundle map.
\endremark

Using local trivializations, bundle maps can give local descriptions 
$$s_\alpha :U_\alpha \to \Hom(\R^n,\R^m)$$
with $$s_\beta =h_{\beta\alpha }s_\alpha g_{\beta\alpha}^{-1}.\tag
{$\ast$} $$

\underbar{Question:}  For which bundle is $(\ast)$ the description of a
section?

\underbar{Answer:}  Take principal bundles 
$$P_F=\coprod U_\alpha\times\GL(m)/\{h_{\alpha\beta}\} $$ and
$$P_E=\coprod U_\alpha\times\GL(n)/\{g_{\alpha\beta}\}.$$
Define $$P=P_F\times P_E=\coprod U_\alpha\times
(\GL(m)\times\GL(n))/\{h_{\alpha\beta}\times g_{\alpha\beta}\}.$$
Take the representation of $\GL(m)\times\GL(n)$ on
$\Hom(\R^n, \R^m)$ given by
$$\rho: \GL(m)\times\GL(n)\overset \text{Ad}\to\longrightarrow
\GL(\Hom(\R^n, \R^m)),$$ that is, 
such that $\rho(A,B)(U)=A\circ U\circ B^{-1}$.
Then the bundle we want is 
$$\Hom(E, F)\equiv P\times_{\text{Ad}}\Hom(\R^n, \R^m).$$
\remark{Exercise 3} Show that sections of this are given by ($\ast$).
\endremark
\remark{Exercise 4} Show that $\Hom(E, F)\cong E^*\otimes F$!
\endremark

\remark{Remark} The case $E=F$ is an important special case.
In this case, $\Hom(E, E)=\End(E)$.
\remark{Exercise 5} Show that $\End(E)=
P_E\times_{\text{Ad}} \text{Mat}_n$, where 
$$\split\text{Ad}: &\GL(n)\longrightarrow\GL(\text{Mat}_n)\\
&A\longmapsto (U\mapsto A\circ U\circ A^{-1}).\endsplit$$
\endremark

\remark{Note}  If we define $\Aut(E)$ to be the 
bundle of isomorphisms of $E$
such that $\Aut(E)|_p=\Aut(E_p)$,
then $\Aut(E)=P_E\times_{\text{Ad}} \GL(n)$ 
is a bundle of groups, but not quite a principal bundle
(transition functions are not quite by left action of structure group
on fibers). If $\frak G$ is the space of sections of $\Aut(E)$, then
$\frak G$ can be given a {\it group} structure using fiberwise
identifications $\Aut(E|_P)\simeq\GL(n)$. The group $\frak G$ is the
{\bf gauge group} of $E$. 
\endremark
\endremark

\vskip.25in
\leftline{\underbar{ Kernel, Image, and Cokernel Bundles}}
\vskip.25in

Given a bundle map

$$\sarrowlength=.42\harrowlength
\commdiag{&E \mapright^{f} &F\cr
&\quad\arrow(1,-1)\quad \arrow(-1,-1)\cr
&\quad B}$$

Define $$ (\Ker(f))_b:=\Ker(f_b: E_b\to F_b)$$

       $$ (\Im(f))_b:=\Im(f_b: E_b\to F_b)$$
 
       $$ (\Coker(f))_b:=F_b/(\Im(f)_b)$$ 

\underbar{Questions:}
\itemitem{(1)} Is $\cup_{b\in B}(\Ker(f))_b:=\Ker(f)$
               a bundle over $B$?(i.e. a subbundle of $E$.)
\itemitem{(2)} Is $\cup_{b\in B}(\Im(f))_b:=\Im(f)$ 
               a subbundle of $F$?
\itemitem{(3)} What about $\cup_{b\in
               B}(\Coker(f))_b:=\Coker(f)$?


\underbar{Answers:}  In general, the answers are \underbar{not} 
{\bf YES}. For example, take trivial bundles: 


$$\sarrowlength=.66\harrowlength
\commdiag{\R^n\times \R^m \mapright^{f} \R^n\times
\R^m\cr
\arrow(1,-1)\qquad \arrow(-1,-1)\cr
\R^n}$$ 



and define $f(x)=x\cdot I_n$. (It is certainly a bundle map.)
Then $$(\Ker(f))_x =\cases 0,&\text{for $x\neq 0$}\\
                         \R^n,&\text{for $x=0$}.\endcases$$
So fiber dimensions can jump!
\proclaim{Theorem 1}  If $\Rank f_b$ is \underbar{constant}, 
then $\Ker f$, $\Im f$, and $\Coker f$ are bundles.
\endproclaim
 
\remark{Note} The proof uses ``Constant Rank Theorem'' for bundle maps.
It amounts to showing that trivializations for $E$ and $F$ can be chosen
so on each $U_\alpha$,
$$E_b\overset\simeq^{\Psi_\alpha}\to
{\overrightarrow{\hphantom{\qquad}}}\R^n=\R^k\oplus\R^{n-k},$$
where $\R^k=\Ker(f_\alpha)$, and
$$F_b\overset\simeq^{\Phi_\alpha}\to
{\overrightarrow{\hphantom{\qquad}}}\R^m=\R^i\oplus\R^{m-i},$$
where $\R^i=\Im(f_\alpha)$.
Then, on $U_\alpha\cap U_\beta$, we have the commutative
diagram (w.r.t. given decompositions of $\R^n$):
 
$$\sarrowlength=.42\harrowlength
\commdiag{
\R^k\oplus\R^{n-k} &\mapright^{f_\alpha }&\R^i\oplus\R^{m-i} \cr
\mapdown\lft{g_{\beta \alpha}}&&\mapdown\rt{h_{\beta \alpha}} \cr
\R^k\oplus\R^{n-k} &\mapright_{f_\beta}&\R^i\oplus\R^{m-i}}$$

with        $$
            g_{\beta \alpha}=\left( \matrix
            k_{\beta \alpha} & \Lambda_{\beta \alpha} \\
            0 & C_{\beta \alpha}
            \endmatrix \right)
            $$
            
            $$
            h_{\beta \alpha}=\left( \matrix
            i_{\beta \alpha} & \Pi_{\beta \alpha} \\
            0 & D_{\beta \alpha}
            \endmatrix \right)
            $$
\endremark
\remark{Note}
$\{k_{\beta \alpha}\}$ are transition functions for $\Ker f$.
Similarly for $\Im f$ and $\Coker f$.
\endremark

\remark{Exercise 6}  Using \underbar{orthonormal frames} (assuming $E$
and $F$ have matrices), we can assume $g_{\beta \alpha}\in\ORTH(n)$ and
$k_{\beta\alpha}\in\ORTH(n)$.
Show that it follows that $\Lambda_{\beta\alpha}=0$ and
$\Pi_{\beta\alpha}=0$.
 
\endremark

\enddocument
\end



