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\title       The Geometry of Vector Bundles and an Introduction to Gauge
Theory \\
Lecture 5
\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 2, 1998
\enddate

\endtopmatter

\document
Recall that we can describe a vector bundle, $\pi:E \to B$, in terms of 
local trivializations over a cover of $B$ together with transition functions, 
$g_{\alpha \beta}:U_\alpha \cap U_\beta \to \text{GL}(n,\Bbb R)$, where $n$ 
is the ranks of the bundle.  We had that $$E=(\coprod U_\alpha \times \Bbb R^n)/\{g_{\alpha \beta}\}$$

Similarly, for principal bundles, $$P=(\coprod U_\alpha \times G)/\{g_{\alpha \beta}\}$$ where $g_{\alpha \beta}:U_\alpha \cap U_\beta \to G$ and the 
relation was $(b,v)_\alpha \sim (b,g_{\beta \alpha}(b) \cdot v)_\beta$.

An important observation can be made here.  Given a vector bundle, $\pi:E \to B$ given by $$\{g_{\alpha \beta}:U_\alpha \cap U_\beta \to \text{GL}(n,\Bbb R)\}$$ we can build a principal $\text{GL}(n,\Bbb R)-$bundle.  Set $$P=(\coprod U_\alpha \times \text{GL}(n,\Bbb R))/\{g_{\alpha \beta}\}$$  We say that 
$E$ is associated to $P$.  A natural question to ask is if the converse holds.
That is, given a principal $G-$bundle, $\pi:P \to B$, can we build an associated
vector bundle?  To build such a bundle, we need a group representation, 
$\rho:G \to \text{GL}(V)$ for some vector space $V$.

\example{Example 1} If $G=\text{GL}(n,\Bbb R)$, then just use the standard 
representation on $\Bbb R^n$.  If $G=\text{O}(n)$, then we may embed 
$\text{O}(n)$ in $\text{GL}(n,\Bbb R)$ and use the composite representation.
If $G=\text{U}(n)$, the unitary group, we get, via the embedding of  $\text{U}(n)$ into $\text{GL}(n,\Bbb C)$, a representation on $\Bbb C^n$.  We can get a real representation by thinking of $\text{GL}(n,\Bbb C)$ as a subgroup of $\text{GL}(2n,\Bbb R)$.
\endexample

We define $$E=(\coprod U_\alpha \times V)/\{\rho(g_{\alpha \beta})\},$$ where 
$$\rho(g_{\alpha \beta}):U_\alpha \cap U_\beta \to \text{GL}(n,\Bbb R)$$
The cocycle condition is easily verfied, for, 
$$\rho(g_{\alpha \beta}) \rho(g_{\beta \gamma}) \rho(g_{\gamma \alpha})=\rho(g_{\alpha \beta}g_{\beta \gamma}g_{\gamma \alpha})=\rho(\text{Id})=\text{Id}$$

Thus far, we have concentrated on local descriptions of associated bundles.
It is also useful to have a global description of such things. 
To do so, we need the following defintion.

\definition{Definition 1}  Given $\pi:P \to B$ a principal $G-$bundle and 
a representation $\rho:G \to \text{GL}(V)$, we define the {\it product of 
$P$ and $V$ over $G$} by $$E=P \times_G V,$$ where $E$ is a quotient of 
$P \times V$ under the relation $(p,v) \sim (p \cdot g,\rho(g^{-1})v)$ for all 
$g \in G$.
\enddefinition

\proclaim{Claim}  Under $[(p,v)] \mapsto \pi(p)$, we get a vector bundle, 
$E \to B$ with fiber $V$.  Furthermore, if $$P=(\coprod U_\alpha \times G)/\{g_{\alpha \beta}\}$$ then $$E=(\coprod U_\alpha \times V)/\{\rho(g_{\alpha \beta})\}$$  That is, the two constructions are the same.
\endproclaim
\remark{Exercise 1} Prove the above claim.  Hint: 
We can identify $$(U_\alpha \times G) \times_G V \to U_\alpha \times V$$
by $[((b,u)_\alpha,v)] \mapsto (b,\rho(u)v)$ with inverse map 
$(b,v) \mapsto [((b,e),v)]$, where $e \in G$ is the group identity.
\endremark

We now give a geometric interpretation of the principal $\text{GL}(n,\Bbb R)$ 
bundle associated to a rank $n$ vector bundle $\pi:E \to B$. 

\definition{Definition 2}  Given a real vector space $V$ of dimension $n$, a frame for $V$ is an identification of $\Bbb R^n$ with $V$, $f:\Bbb R^n \to V$.
This is equivalent to choosing a basis; that is, if $\{e_i\}$ is the standard
basis for $\Bbb R^n$, then $\{f(e_i)\}$ is the basis we have chosen for 
$V$.  Define $F(V)$ to be the set of all frames of $V$.
\enddefinition

Observe that $\text{GL}(n,\Bbb R)$ acts on $F(V)$ by:
$$\sarrowlength=.42\harrowlength
\commdiag{&\Bbb R^n\cr &\arrow(1,1)\lft{A}\quad \arrow(1,-1)\rt{f}\cr
\Bbb R^n&\mapright_{A(f)=f \circ A}&V\cr}$$

\remark{Exercise 2} Show that $F(V) \cong \text{GL}(n,\Bbb R)$.
\endremark

For a vector bundle, $\pi:E \to B$ and for each $b \in B$, we have that 
$E_{|b} \cong \text{GL}(n,\Bbb R)$.  So, we get a copy of $\text{GL}(n,\Bbb R)$
attached to each $b$ via $F(E_{|b})=P_b$.

\proclaim{Claim}  If $$\bigcup_{b \in B}P_b=P$$ then $P$ is a principal $\text{GL}(n,\Bbb R)-$bundle.  This is the same as the previous construction.
This is called the {\bf principal frame bundle}, or the 
{\bf principal bundle of frames}.
\endproclaim
{\bf Proof:}  The proof is left as an exercise, with the following hint.
Say $$E=(\coprod U_\alpha \times \Bbb R^n)/\{g_{\alpha \beta}\}$$ with 
trivializing maps, $\psi_\alpha:E|U_\alpha \to U_\alpha \times \Bbb R^n$.
Use $\psi_\alpha^{-1}(b)():\Bbb R^n \to E_b$ to define a frame, $f$.  We
then get a description of $P_b$ as $$P_b=\{f_\alpha \circ A:A \in \text{GL}(n,\Bbb R)\},$$  and hence, $P|U_\alpha \cong U_\alpha \times \text{GL}(n,\Bbb R)$ via the map $(b,A) \mapsto f_\alpha(b) A$.
Notice that the $P_b$ carries a right action of $\text{GL}(n,\Bbb R)$.  

\remark{Exercise 3}  Show that $P$ also carries a right action.
\endremark

\remark{Exercise 4}  Check the transition function conditions.
\endremark
\enddocument
\end

