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\title The Geometry of Vector Bundles and an Introduction to Gauge
Theory \\
Lecture 4
\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 30, 1998
\enddate
\endtopmatter
\document
Recall from last time that given a vector bundle $\pi:E \to B$, say with fiber $\Bbb R^n$, we cover $B$ with trivializing neighborhoods, $U_\alpha$, with
trivializations $\psi_\alpha:E|U_\alpha \to U_\alpha \times \Bbb R^n$,
and obtain transition functions, $g_{\alpha \beta}:U_\alpha \cap U_\beta \to \text{GL}(n,\Bbb R)$ by $g_{\alpha \beta}(x)=\psi_\alpha(\psi_\beta^{-1}(x))$.
These satisfy three properties:
\itemitem{(1)} $g_{\alpha \alpha}=\text{Id} \in \text{GL}(n,\Bbb R)$
\itemitem{(2)} $g_{\alpha \beta} = g_{\beta \alpha}^{-1}$
\itemitem{(3)} (cocycle condition) $g_{\alpha \gamma} g_{\gamma \beta} g_{\beta \alpha}=\text{Id}$
\proclaim{Claim} Given a cover, $\{U_\alpha\}$ of $B$ and $\{g_{\alpha \beta}\}:U_\alpha \cap U_\beta \to \text{GL}(n,\Bbb R)$, one for every nonemtpy
intersection, such that the conditions 1,2,3 listed above hold, then there
exists a vector bundle (unique up to bundle isomorphism) for which $\{g_{\alpha \beta}\}$ are the transition functions.
\endproclaim
{\bf Proof:} The proof is a constructive one.
\itemitem{Step 1:} For each $U_\alpha$, define $$Z=\coprod_{\alpha \in A} U_\alpha \times \Bbb R^n$$ Put the (product?) topology on $Z$.
\itemitem{Step 2:} Define an equivalence relation on $Z$ as follows:
$$(x,v)_\alpha \sim (x',v')_\beta \text{ if and only if }x=x'\text{ and }
v'=g_{\beta \alpha}(x)(v)$$
\remark{Exercise 1} Using conditions 1,2, and 3 above, show that this defines an equivalence relation.
\endremark
\itemitem{Step 3:} Define $E=Z/~$. Put the quotient topology on $E$. We need
to establish several properties of $E$. First, we need to show that since
the $g_{\beta \alpha}$ are smooth maps, $E$ is a smooth manifold. Second,
the map $[(x,v)_\alpha] \mapsto x$ is a well defined map from $E$ to $B$ which
makes this a rank $n$ bundle. Third, we need to show that the set of transition functions for $E$ is precisely $\{g_{\alpha \beta}\}$.
First, $X$ is a smooth manifold if we can cover $X$ with coordinate charts which
are smoothly related. That is, whose transition functions are smooth. Assume
that the cover $\{U_\alpha\}$ of $B$ actually consists of coordinate patches with trivializations $\phi_\alpha:U_\alpha \to \Bbb R^n$.
Otherwise, we simply refine our cover so that it has this property. Assume that $B$ has rank $m$. We need to
define a collection, $\{V_\alpha\}$ of coordinate patches on $E$ and maps,
$\bar \phi_\alpha:V_\alpha \to \Bbb R^m \times \Bbb R^n$. Set $V_\alpha=[U_\alpha \times \Bbb R^n]$. Each $V_\alpha$ is open by the defintion of the quotient topology. Define $\bar \phi_\alpha$ by $[(x,v)_\alpha] \mapsto (\phi_\alpha(x),v)$. We have the following commutative diagram:
$$\sarrowlength=.42\harrowlength
\commdiag{&V_\alpha \cap V_\beta\cr &\arrow(-1,-1)\lft{\bar \phi_\beta}\quad \arrow(1,-1)\rt{\bar \phi_\alpha}\cr
(U_\alpha \cap U_\beta) \times \Bbb R^n&\mapleft_{\phi_\beta \phi_\alpha^{-1}}&(U_\alpha \cap U_\beta) \times \Bbb R^n}$$
On the overlap, $\bar \phi_\beta \bar \phi_\alpha^{-1}$, has the following result, $$(x,v) \mapsto \bar \phi_\beta([x,v]_\alpha)=\bar \phi_\beta([x,g_{\beta \alpha}(x)(v)]_\beta)=(x,g_{\beta \alpha}(x)(v))$$ Thus, on the overlap, the maps are smooth since the $g_{\beta \alpha}$ are. Since
we now actually have a smooth manifold, we continue.
We now need to examine the projection map, $\pi:E \to B$, defined above by
$[(x,v)_\alpha] \mapsto x$. Locally, we have
$$\sarrowlength=.42\harrowlength
\commdiag{&V_\alpha\cr &\arrow(-1,-1)\lft{\pi}\quad \arrow(1,-1)\rt{\bar \phi_\alpha}\cr
U_\alpha &\mapleft_{\pi_1}&U_\alpha \times \Bbb R^n\cr}$$
Where $\pi_1$ is the natural projection. We have that $\pi=\pi_1 \circ \bar \phi_\alpha$ and so $\pi$ is smooth (well defined, continuous, etc). By
definition, $$\bar \psi_\alpha^{-1}(\pi_1^{-1}(x)) = \{[(x,v)_\alpha]:v \in \Bbb R^n\} \subseteq \pi^{-1}(x)$$ Do we get anything else? Suppose that $[(x,v')_\beta]$ is in the fiber over $x$. Then $[(x,v')_\beta]=[(x,g_{\alpha \beta}(x)(v'))_\alpha]$. So, the full fiber over $x$ is $\Bbb R^n$.
The remainder of the proof is left as the exercise below. \qed
\remark{Exercise 2} Show that, in fact, the above statements also yield
the locally trivial condition and the transition properties
\endremark
Recall that a bundle, $\pi:P \to B$ is a principal bundle $G-$bundle if
$\pi^{-1}(x) \cong G$, where $G$ is a Lie group, $\pi$ is locally trivial, and
there is a free right $G-$action on $P$ which preserves the fibers. A map
of two $G-$bundles, $P$ and $P'$ is a bundle map, $h:P \to P'$ which is $G-$equivariant; that is, $h(p \cdot g)=h(p) \cdot g$. The locally trivial
condition says that the trivialization, $\psi_\alpha:P|U_\alpha \to U_\alpha \times G$ must be fiber preserving and $G-$equivariant; where the $G-$action
on $U_\alpha \times G$ is trivial on $U_\alpha$ and group multiplication on
the $G$ factor. So, on $U_\alpha \cap U_\beta$, $\psi_\beta \psi_\alpha^{-1}$ has the form, $$(x,g) = (x,1 g) \mapsto (\psi_\beta \psi_\alpha^{-1}(x,1))g$$
Define $g_{\beta \alpha}(x)=(\psi_\beta \psi_\alpha^{-1}(x,1))$. So, we must
have that $(x,g) \mapsto (x,g_{\beta \alpha}(x) \cdot g)$. Exactly as before,
we define $$P=(\coprod U_\alpha \times G)/\sim$$ where $\sim$ is defined, as
before, by the $\{g_{\beta \alpha}\}$.
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