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\title The Geometry of Vector Bundles and an Introduction to Gauge
Theory \\
Lecture 25
\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 20, 1998
\enddate
\endtopmatter
\document
During the last lecture we began to examine connections on tangent bundles,
$TM \to M$. If $M$ has dimension $m$, then this is a rank $m$ bundle over
$M$. Given a connection $\nabla$ and local coordinates, $(x_1,\dots,x_m)$ on
$M$, we can produce a local frame, $\{\frac{\partial}{\partial x_i}\}$ for
$TM$ and $\{dx_i\}$ for $T^*M$. We saw that $$\nabla(\frac{\partial}{\partial x_i})=\Gamma_{ji}^k dx_k \otimes \frac{\partial}{\partial x_j}$$ This was
how we defined the Christoffel symbols. The geodesic equation was derived
from examining covariant constant vector fields. That is, given a curve
$\gamma(t)$ in $M$, we called $\gamma(t)$ geodesic if the velocity vector
field, $\dot \gamma(t)$ was covariant constant, $\nabla_{\dot \gamma} \dot \gamma=0$. In local coordinates, $\gamma(t)=(x_1(t),\dots,x_m(t))$, we get the following set of ordinary differential equations
$$\text{\"x}_k(t)+\Gamma_{ji}^k(\gamma(t)) \dot x_i(t) \dot x_j(t)=0,\qquad k=1,\dots ,m$$
A consequence of the existence and uniqueness of solutions to differential
equations implies that given any vector $\vec v \in T_xM$, there exists
a unique geodesic curve, $\gamma_{\vec v}(t)$ with $\gamma_{\vec v}(0)=x$ and
$\dot \gamma_{\vec v}(0)=\vec v$.
\vskip.25in
\remark{Exercise 1} Let $\gamma_{\vec v}(t)$ be the geodesic through $x$ in
the direction of $\vec v$. If $\lambda$ is some constant, show that
$\gamma_{\lambda \vec v}(t)=\gamma_{\vec v}(\lambda t)$.
\endremark
\vskip.25in
\proclaim{Corollary} Given a unit vector $\vec u \in T_xM$ (a unit vector with respect
to some Riemannian metric), for small enough $\lambda$, $\gamma_{\lambda \vec u}(t)$ will be defined at $t=1$. We define a map, $T_xM \to M$ defined on
a small neighborhood of $0 \in T_xM$ by $\vec v \mapsto \gamma_{\vec v}(1)$.
This map is called the exponential map and is denoted by exp$(\vec v)$.
\endproclaim
\vskip.25in
A fact from Riemannian geometry says that this is a diffeomorphism of the neighborhood of $0$ in $T_xM$ onto a neighborhood of $x$ in $M$. The proof
relies on the implicit function theorem. In order to set things up properly,
we must examine the derivative of this map, $D_0(\text{exp}):T_0(T_xM) \to T_xM$.
\vskip.25in
\remark{Exercise 2} Show that $D_0(\text{exp})=\text{Id}$.
\endremark
\vskip.25in
This defined geodesic coordinates.
\vskip.25in
\proclaim{Claim} With respect to these geodesic coordinates, $\Gamma_{ij}^k$
vanishes at $x$.
\endproclaim
\vskip.25in
{\bf Sketch of Proof:} Given any $\vec v \in T_xM$, evaluate $(\Gamma_{ij}^k dx_k)(\vec v)$ using the geodesic along $\vec v$, $\gamma_{\vec v}(t)$. Use
the geodesic equation and note that in geodesic coordinates, $\gamma_{\vec v}(t)=t \vec v=(x_1(t),\dots,x_m(t))$.
\vskip.25in
\leftline{{\bf Torsion}}
\vskip.25in
\definition{Definition 1} For a connection, $\nabla$, on the tangent bundle, $TM$, over $M$, the {\it torsion} of the connection is a tensor field in
$$\Omega^0(\wedge^2(T^*M) \otimes TM)$$ defined by
$$\tau(X,Y)=\nabla_XY-\nabla_YX-[X,Y]$$ for any vector fields, $X,Y \in TM$.
\enddefinition
\vskip.25in
It is not at all clear that $\tau$ is actually an element of $ \Omega^0(\wedge^2(T^*M) \otimes TM)$. The following exercise is in this direction.
\vskip.25in
\remark{Exercise 3} Show that $\tau(X,Y)_b$ depends only on the values of
$X_b$ and $Y_b$. This will show that $\tau_b \in \wedge^2 T_b^*M \otimes T_bM$. Show that if $f$ is a function, then $\tau(fX,Y)=f \tau(X,Y)$.
\endremark
\vskip.25in
What does the torsion measure? With respect to the frames $\{\frac{\partial}{\partial x_i}\}$ for $TM$ and $\{dx_i\}$ for $T^*M$,
we see that
$$\aligned
\tau(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j})&=\nabla_{\frac{\partial}{\partial x_i}}(\frac{\partial}{\partial x_j})-\nabla_{\frac{\partial}{\partial x_j}}(\frac{\partial}{\partial x_i}) \cr
&=(\Gamma_{ji}^k-\Gamma_{ij}^k) \frac{\partial}{\partial x_k} \endaligned$$
So, if $$\tau=\tau_{ij}^k dx_i \wedge dx_j \otimes \frac{\partial}{\partial x_k}$$ then $\tau_{ij}=\Gamma_{ij}^k-\Gamma_{ji}^k$. If $\tau=0$, then the connection has symmetric Christoffel symbols.
\vskip.25in
{\bf Fact:} Given $\nabla$, if $\tau \neq 0$, then we can modify $\nabla$ to
obtain a new connection, $\tilde \nabla$, with $\tilde \tau=0$. The modification is given by the following procedure. We have
$$\Omega^1(\text{End}(TM))=\Omega^0(T^*M \otimes \text{End}(TM))$$ and
$$T^*M \otimes \text{End}(TM) \cong T^*M \otimes (TM^* \otimes TM) \cong
(T^*M \otimes TM^*) \otimes TM$$ Since $\wedge^2(T^*M) \subseteq T^*M \otimes TM^*$, we have that $\tau \in \Omega^1(\text{End}(TM)$. We can use this to define $\tilde \nabla=\nabla-\frac{1}{2} \tau$.
\vskip.25in
\remark{Exercise 4} Show that the torsion of $\tilde \nabla$ is zero.
\endremark
\vskip.25in
\leftline{{\bf The Levi-Civita Connection}}
\vskip.25in
Recall that a metric, $g$, on a manifold $M$ if it is a bundle metric on the
tangent bundle of $M$. We can thus ask for connections on $TM \to M$ to be
compatible with the metric. By our discussion of orthogonal connections, this
can be expressed by the condition that $\nabla g=0$.
\proclaim{Claim} Given $g$, there is a unique connection, called the
Levi-Civita connection and denoted $\nabla^{lc}$, such that $\nabla^{lc}g=0$ and $\nabla^{lc}$ is torsion free.
\endproclaim
\vskip.25in
{\bf Proof:} Use local coordinates $\{x_1,\dots,x_n\}$. Write
$$g=\sum g_{ij}dx_i\otimes dx_j$$ with respect to the local frame $\{dx_i\}$ for $T^*M$. Set $$\Gamma_{ij}^k=\frac{1}{2} g^{-1}_{kl}[g_{jl,i}-g_{ij,l}+g_{li,j}]$$ where $g_{jl,i}=\frac{\partial}{\partial x_i}(g_{jl})$ to form the desired connection.
\qed
\vskip.25in
We now get induced connections on $T^*M,\otimes T^*M,\otimes TM$,etc. On
$\wedge^p T^*M$, we have $\nabla^{lc}:\Omega^p(M) \to \Omega^0(T^*M \otimes \wedge^0 T^*M)$ and $d:\Omega^p(M) \to \Omega^{p+1}(M)$. We would like
to relate these two maps.
\vskip.25in
\proclaim{Claim} $\text{Alt}(\nabla^{lc})=d$ where Alt is the alternation.
\endproclaim
\enddocument
\end