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\title The Geometry of Vector Bundles and an Introduction to Gauge
Theory \\
Lecture 24
\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 March 18, 1998
\enddate
\endtopmatter
\document
Recall that given a bundle, $E \to B$, a bundle metric, $\langle \quad,\quad \rangle$, and a connection $D$, we say that the metric is compatible with
the connection if for all $s,t \in \Omega^0(E)$ the following product formula holds $$d \langle s,t \rangle =\langle Ds,t \rangle+\langle s,Dt \rangle$$ Such a connection is sometimes called an {\it orthogonal} connection.
If $E$ is a complex bundle and $\langle \quad , \quad \rangle$ is a Hermitian
metric, then the connections which satisfy the above compatibility condition
are called {\it unitary} connections.
Suppose that $\{e_i\}$ is a
local frame for $E$ with $\langle e_i,e_j \rangle=\delta_{ij}$ and that with
respect to this choice of a local frame, $D=d+A$. Then, we have that
$$\aligned 0&=d \langle e_i , e_j \rangle=\langle A_{ki} \otimes e_k,e_j \rangle+\langle e_i,A_{kj} \otimes e_k \rangle \cr
&=A_{ki}\delta_{ki}+A_{kj}\delta_{kj} \endaligned $$
Hence, $A_{ji}+A_{ij}=0$.
\vskip.25in
\remark{Note} In the complex case, using a Hermitian metric we get $A_{ji}+\bar A_{ij}=0$.
\endremark
\vskip.25in
This says that for an orthogonal connection, the connection 1-form, $A$, is skew symmetric with respect to
an orthogonal frame. (If Lie(O($n$)) denotes the Lie algebra on O($n$), then
we can identify the skew symmetric matrices with Lie(O($n$))).
Similarly, for unitary connections on complex bundles, the connection 1-forms
take their values in Lie(U($n$)) when computed in a unitary frame.
\vskip.25in
\remark{Note} We can realize $E$ as an associated principal O($k$)-bundle.
This turns out to be a general feature of connections on principal $G-$bundles:
The connection 1-forms take their values in Lie($G$).
\endremark
\vskip.25in
A direct computation show that $F_A=dA+A \wedge A$ satisfies $F_A+F_A^t=0$, again with respect to an orthogonal frame. Thus for any vector fields
$X,Y \in TB$, the bundle map $F(X,Y):E \to E$ is an isometry, i.e, with respect
to orthogonal frames, $F(X,Y)$ is O($n$)-valued.
\vskip.25in
\leftline{{\bf An Alternative Description of the Compatibility Condition}}
\vskip.25in
A bundle metric, $\langle \quad , \quad \rangle$, can be viewed as
a section of $E^* \otimes E^*$. On a vector space, $\langle \quad, \quad \rangle$ is a bilinear map, $V \times V \to \Bbb R$, which is equivalent
to a map $V \to V^*$. Thus, on $E$, $\langle \quad , \quad \rangle$ is equivalent to a map, $E \to E^*$; that is, to a section of $\text{Hom}(E,E^*) \cong E^* \otimes E^*$.
Given a local frame, $\{e_i\}$ of $E$, we define $$H=\sum H_{ij}e_i^* \otimes e_j^*$$ where $H_{ij}=\langle e_i,e_j \rangle$. So, $H(s,t)=\langle s,t \rangle$. But, given a connection, $D$, on $E$, we get an induced connection
on $E^* \otimes E^*$, denoted also by $D$. So, we can evaluate $D(H)$.
\vskip.25in
\remark{Exercise 1} Show that the following are equivalent:
\itemitem{(1)} $d \langle s,t \rangle = \langle Ds,t \rangle +\langle s,Dt \rangle$.
\itemitem{(2)} $D(H)=0$.
\endremark
\vskip.25in
Condition 2 above is sometimes called a covariant constant condition. That is,
called $H$ covariant constant if $D(H)=0$.
\vskip.25in
\leftline{{\bf Connections on TM - The Tangent Bundle}}
\vskip.25in
Recall that $TM \to M$ is a vector bundle with rank the dimension of $M$. We can define connections, $$D:\Omega^0(TM) \to \Omega^0(T^*M \otimes TM)=
\Omega^1(TM)$$ Sections of $TM$ are just vector fields. In this case,
$D_V(s)$ is the covariant derivative along the vector field $V$ of the section $s$. But, $V$ and $s$ are now the same sort of creature: Vector Fields. When
speaking of connections on a tangent bundle, we use the symbol $\nabla$, rather
than $D$. As before, $F_\nabla \in \Omega^2(\text{End}(TM))$ and given
$X,Y \in \Omega^0(TM)$, we get $F_\nabla(X,Y):TM \to TM$.
Recall that given a frame, $\{e_i\}$ and a connection $D=d+A$ with respect to
this frame, $A$ is a matrix of 1-forms. Write $$A=\sum_{\alpha=1}^n A_\alpha dx_\alpha$$ with respect to the coordinates $(x_1,\dots,x_n)$, where $A_\alpha$ is a matrix (map on the fibers of $E$). For $E=TM$, we can take $$\{\frac{\partial}{\partial x_i}\}_{i=1}^m,$$ where $m$ is the dimension of $M$, as a local frame for $TM$ over the coordinate patch where the coordinates
are $(x_1,\dots,x_n)$. If $\nabla=d+A$, then
$$\nabla(\frac{\partial}{\partial x_i})=A_{ji}(\frac{\partial}{\partial x_i})=A_{ji,k} dx_k \otimes \frac{\partial}{\partial x_j}$$ This is usually
written $\Gamma_{ji}^k dx_k \otimes \frac{\partial}{\partial x_j}$. The
$\Gamma_{ji}^k$ are called the {\it Cristofell} symbols. The curvature of
$\nabla$ is usually written, for maximal confusion, as $\Omega$.
\vskip.25in
\leftline{{\bf Parallel Transport and Horizontal Lifting}}
\vskip.25in
Given a curve, $\gamma(t)$, in $M$, a lift to $TM$ is a vector field along $\gamma(t)$. Given such a lift, says $\tilde \gamma(t)$, we can compute the
covariant derivative with respect to $\dot \gamma(t)$ along $\tilde \gamma(t)$, $\nabla_{\dot \gamma(t)} \tilde \gamma(t)$. But, for our lift, $\tilde \gamma(t)$, we can simply take $\dot \gamma(t)$.
\vskip.25in
\definition{Definition 1} Say that $\gamma(t)$ is {\it geodesic} if $\nabla_{\dot \gamma(t)} \dot \gamma(t)=0$.
\enddefinition
\vskip.25in
With respect to local coordinates, $(x_1,\dots,x_m)$, a curve $\gamma(t)$ has
component functions, $(x_1(t),\dots,x_m(t))$. Since $\nabla=d+\Gamma_{ij}^k dx_k$, we have the following exercise.
\vskip.25in
\remark{Exercise 2} Show that the geodesic condition yields a system of
ordinary differential equations,
$$\text{\" x}_k(t)+\Gamma_{ij}^k \gamma(t) \dot x_i(t)\dot x_j(t)=0$$
\endremark
\enddocument
\end