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\title The Geometry of Vector Bundles and an Introduction to Gauge
Theory \\
Lecture 7
\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 6, 1998
\enddate
\endtopmatter
\document
Recall from last time that if $E$ and $E'$ are bundles and $h:E \to E'$ is
a bundle map, then we form a collection of local maps, $\{h_\alpha:U_\alpha \to \text{Hom}(\Bbb R^n,\Bbb R^m)\}$ such that $g'_{\beta \alpha} h_\alpha=h_\beta g_{\beta \alpha}$. If $h$ was, in fact, an automorphism, then we had $h_\beta=g_{\beta \alpha} h_\alpha g_{\alpha \beta}$.
It is important to know when a bundle, $E$, over $B$ is isomorphic to the
trivial bundle, $E \cong B \times \Bbb R^n$. Suppose $E$ is given by
$$E=(\coprod U_\alpha \times \Bbb R^n)/\{g_{\alpha \beta}\}$$ Observe that
the trivial bundle, $\underline{\Bbb R^n}$, is given by
$$\underline{\Bbb R^n}=B \times \Bbb R^n = (\coprod U_\alpha \times \Bbb R^n)/\{g'_{\alpha \beta}\}$$ where $g'_{\alpha \beta}$ is identically the
identity. A map $h:E \to \underline{\Bbb R^n}$is thus described by a collection of locally defined
maps, $\{h_\alpha:U_\alpha \to \text{GL}(n,\Bbb R)\}$ satisfying $g_{\beta \alpha}=h_\beta h_\alpha^{-1}$.
That is, if we can find a collection of such $h_\alpha$, then we can trivialize the bundle.
But, this says precisely that the associated principal $\text{GL}(n,\Bbb R)-$bundle is trivial if and only if it admits a section. It follows in the same way that any principal bundle is trivial if and only if it
admits a section. So, unlike the case for vector bundles, sections of
principal bundles completely detect whether or not a bundle is trivial.
\remark{Side bar} For those who know some homotopy. The zero section, $s:B \to E$
embeds $B$ diffeomorphically into $E$. Furthermore, this diffeomorphic copy
of $B$ in $E$ is a deformation retract of $E$. Hence, $B$ and $E$ have
the same homotopy type. So, to distinguish bundles, homotopy and homology
theory will be difficult to use.
\endremark
\vskip.25in
\leftline{{\bf New Bundles From Old Bundles}}
\vskip.25in
Constructions done on vector spaces will carry over almost directly to the
case for vector bundles. For example, given two vector spaces, $V_1,V_2$, we can form new vector spaces, $V_1 \oplus V_2,V_1 \otimes V_2,V_1^*,V_1 \wedge V_1$,etc. These all have bundle versions. If $E_1$ is a rank $r_1$ bundle
and $E_2$ is a rank $r_2$ bundle, then we construct a new bundle, $E_1 \oplus E_2$ with rank $r_1+r_2$ as follows:
Suppose that $E$ and $E'$ are represented by$$\aligned
E_1&=(\coprod U_\alpha \times \Bbb R^{n_1})/\{g^1_{\alpha \beta}\} \cr
E_2&=(\coprod U_\alpha \times \Bbb R^{n_2})/\{g^2_{\alpha \beta}\}
\endaligned$$
We define the sum of the bundles to be
$$E_1 \oplus E_2=(\coprod U_\alpha \times \Bbb R^{n_1+n_2})/\{g^\oplus_{\alpha \beta}\}$$
where $g^\oplus_{\alpha \beta}$ is the $n_1$ by $n_2$ matrix with $g^1_{\alpha \beta}$ in the upper left corner and $g^2_{\alpha \beta}$ in the lower right
corner. To check that does define a bundle, we need to check the cocycle
condition. This does, however, check out just by using that the two blocks
of the matrix satisfy it. $E_1 \otimes E_2$ is constructed similarly.
The construction of the dual bundle, $E^*$, is slightly more difficult.
We want the fibers of the new bundle to be $E_b^* \cong \text{Hom}(E_b,\Bbb R)$. If $E$ is given by
$$E=(\coprod U_\alpha \times \Bbb R^n)/\{g_{\alpha \beta}\}$$
then we really need to concentrate on what the $g^*_{\alpha \beta}$ should be.
Set $g^*_{\alpha \beta}=(g^t_{\alpha \beta})^{-1}$, the transpose-inverse of
the original transition functions. The $g^*_{\alpha \beta}$ will satisfy the
cocycle condition and so will define a bundle. But, is it the correct bundle?
That is, do we have $E^*_b \cong \text{Hom}(E_b,\Bbb R^n)$? Suppose that
$\psi_\alpha:E_b \to \Bbb R^n$ is an identification of the fiber and
$\lambda \in \text{Hom}(E_b,\Bbb R)$. Then, we get a map $\lambda_\alpha:\Bbb R^n \to \Bbb R$. This is expressed in the following diagram:
$$\harrowlength=.62\sarrowlength
\commdiag{E_b &\mapright^\lambda &\Bbb R \cr
\mapdown\lft{\psi_\alpha}&\arrow(1,1)\rt{\lambda_\alpha} \cr
\Bbb R^n}$$
We get a correspondance, $\lambda \leftrightarrow \lambda_\alpha$ and so
we get $E_b^* \cong \Bbb R^n \cong \text{Hom}(\Bbb R^n,\Bbb R)$. In fact,
we also get $E|U_\alpha \cong U_\alpha \times \Bbb R^n$. Over $U_\alpha \cap U_\beta$, we have $\lambda_\beta=\lambda \circ \psi_\beta^{-1}$ and $\lambda_\alpha=\lambda \circ \psi_\alpha^{-1}$. So, $$\lambda=\lambda_\beta \psi_\beta=\lambda_\alpha \psi_\alpha=\lambda_\alpha \circ(\psi_\alpha \circ \psi_\beta^{-1}) \circ \psi_\beta=\lambda_\alpha g_{\beta \alpha} \psi_\beta$$
The compatibility condition on the $\{\lambda_\alpha\}$ is that, as maps $\Bbb R^n \to \Bbb R$, they satisfy $\lambda_\beta=\lambda_\alpha \circ g_{\alpha \beta}$.
\remark{Exercise 1} Show that this is equivalent to the requirement that $$E=(\coprod U_\alpha \times \Bbb R^n)/\{(g_{\alpha \beta}^t)^{-1}\}$$
\endremark
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