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\title The Geometry of Vector Bundles and an Introduction to Gauge
Theory \\
Lecture 3
\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 January 26, 1998
\enddate
\endtopmatter
\document
Recall that a bundle is a quadruple, $(E,B,F,\pi)$, where $E,B,F$ are
topological spaces, $\pi:E \to B$ is a continuous map satisfying
$\pi^{-1}(x) \cong F$ for all $x \in B$ and there exists an open set
$U \subseteq B$ with $x \in U$ such that $\pi^{-1}(U) \cong U \times F$ in
a fiber preserving way. In the future, the bundle will just get denoted,
$\pi:E \to B$. If we are in the {\bf Smooth} category, $\pi$ will be
a smooth map.
\definition{Definition 1} For $\pi:E \to B$ a bundle, we say that a map $s:B \to E$ is a section if $\pi \circ s=\text{Id}_B$. That is, $s(b)$ is in the
fiber of $s$ over $B$.
\enddefinition
A special case of a section is given by the trivial bundle,
$$\commdiag{
E=B \times F \cr
\mapdown \rt{\pi=\text{projection}} \cr
B}$$
If $s$ is a section, then $s(b)=(b,\sigma(b))$, where $\sigma:B \to F$.
Conversely, if $\sigma:B \to F$, then the map $s:B \to B \times F$ given
by $s(b)=(b,\sigma(b))$ is a section. That is, there is a 1-1 correspondance
between sections of the bundle and $Map(B,F)$.
When $E$ is not the trivial bundle, it will help to think of sections as
a sort of twisted map from $B$ to $F$. Write $E_{|U}$ for $\pi^{-1}(U)$.
If $\psi:E_{|U} \to U \times F$ is an identification, then we get the
following diagram
$$\sarrowlength=.42\harrowlength
\commdiag{&E|U \mapright^{\fam6 \psi} &U \times F\cr
&\arrow(-1,1)\lft{\fam6 s}\qquad \arrow(1,1)\rt{\fam6 \psi \circ s=s_U}\cr
&U}$$
Using the local trivialization, the local desription of the section is
a map $U \to F$.
\underbar{Question:} Do sections always exist?
\underbar{Answer:}
\itemitem{(a)} For smooth vector bundles the answer is yes. ($C^\infty$ sections).
\itemitem{(b)} For holomorphic bundles the answer is no.
\itemitem{(c)} For principal bundles the answer is no.
We have defined the objects which we want to study, and now we defined the
maps between them.
\definition{Definition 2} If $\pi:E \to B$ and $\pi':E' \to B'$ are two
bundles, a bundle map (or map of bundles) is a pair of maps, $(u,f)$ such that
the diagram commutes:
$$\sarrowlength=.42\harrowlength
\commdiag{
E &\mapright^{\fam6 u}&E' \cr
\mapdown\lft{\fam6 \pi}&&\mapdown\rt{\fam6 \pi'} \cr
B &\mapright_{\fam6 f} &B'}$$
The definition says that $E_{|b} \mapsto E'_{|f(b)}$ under $u$. If
$E=E'$and $B=B'$, then $(u,f)$ is called a bundle endomorphism. If the maps are
invertible, then $(u,f)$ is a bundle automorphism. Observe that
if we are considering vector bundles, then on the fibers the maps must
be linear. If we are considering principal G-bundles, then we require
the map on the fibers to group homomorphism and be G equivariant. That is,
$f(p \cdot g)=f(p) \cdot g$.
\enddefinition
Note the special case of a bundle endomorphism where $f=\text{Id}_B$. Then,
we have that the following diagram commutes:
$$\sarrowlength=.42\harrowlength
\commdiag{&E \mapright^{\fam6 u} & E \cr
&\quad\arrow(1,-1)\lft{\fam6 \pi} \qquad\arrow(-1,-1)\lft{\fam6 \pi}\cr
&B}$$
So, $u$ just transforms points in each fiber. For the case of a vector bundle, on each fiber $u$ is a vector space isomorphism.
\vskip.25in
\leftline{{\bf Local Picture for Bundles}}
\vskip.25in
Suppose that $\pi:E \to B$ is a smooth vector bundle with $\pi^{-1}(b) \cong \Bbb R^n$. For each $b \in B$, take a neighborhood, say $U_b$, over which $E$ can be trivialized. The collection $\{U_b\}_{b \in B}$ then covers $B$. Let
$\psi_b:E_{|U_b} \to U_b \times \Bbb R^n$ be the trivialization. Suppose
that $b \in U_\alpha \cap U_\beta$. A natural question to ask is how
$\psi_\alpha$ and $\psi_\beta$ compare. We have the following diagram:
$$\sarrowlength=.42\harrowlength
\commdiag{&E|U_\alpha \cap U_\beta\cr &\arrow(-1,-1)\lft{\psi_\beta}\quad \arrow(1,-1)\rt{\psi_\alpha}\cr
(U_\alpha \cap U_\beta) \times \Bbb R^n&\mapleft_{g_{\beta \alpha}}&(U_\alpha \cap U_\beta) \times \Bbb R^n\cr}$$
Where $g_{\beta \alpha}=\psi_\beta \circ \psi^{-1}_\alpha$. Observe that
$g_{\beta \alpha}$, being the composite of linear maps, is itself a linear
map. Since $\psi_\alpha$ and $\psi_\beta$ are invertible, so is
$g_{\beta \alpha}$. That is we may regard $g_{\beta \alpha}:U_\alpha \cap U_\beta \to \text{GL}(n, \Bbb R)$. We get one of these for every pair, $U_\alpha \cap U_\beta \neq \emptyset$.
So, given a smooth vector bundle, we obtain a cover $\{U_\alpha\}$ of
$B$ by locally trivial neighborhoods and a set of transition functions,
$$\{g_{\beta \alpha}\}_{U_\alpha \cap U_\beta \neq \emptyset}$$
\underbar{Question:} When are the cover and the transition functions
equivalent to the information contained in the bundle? That is, when
does $\{U_\alpha\}$ and $\{g_{\beta \alpha}\}$ describe a bundle?
One observation can be made right away. If the cover and the transition functions do determine a cover, then we must have $g_{\alpha \alpha}=\text{Id}$, the
inverse of $g_{\beta \alpha}$ is $g_{\alpha \beta}$, and if
$U_\alpha \cap U_beta \cap U_\gamma \neq \emptyset$,
$g_{\alpha \beta}\circ g_{\beta \gamma} \circ g_{\gamma \alpha}=\text{Id}$
This last condition is known as the cocycle condition.
\enddocument
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