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\documentclass{article} 
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	\begin{document}   
 
	\begin{center}   
	{\bf Connections as Functors \\}
	{\small John\ C.\ Baez, October 21, 2004\\}   
	\vspace{0.3cm}   
        \end{center}

\noindent
{\it In this homework we'll see how 
a vector bundle $E$ equipped with a connection over a manifold
$X$ gives a functor 
\[            F \maps \P(X) \to \Vect  \]
where $\P(X)$ is the category of paths in $X$, defined in the
last homework.  Recall that objects
of $\P(X)$ are points of $X$, while morphisms are piecewise-smooth 
paths in $X$.  The functor $F$ maps each point $x \in X$ to the
the fiber of $E$ over the point $x$.  Similarly, $F$ maps each
path $\gamma \maps x \to y$ in $X$ to a linear operator
\[       F(\gamma) \maps F(x) \to F(y)  \]
defined using parallel transport along the path $\gamma$.

\vskip 1em
\noindent
To warm up, let's see how any linear ordinary differential equation
gives a functor.   I'll let you use this fact: 
}

\begin{thm} \et
Let $\End(\R^n)$ be the algebra of linear operators from $\R^n$
to itself.
Suppose $A \maps [a,b] \to \End(\R^n)$ is any smooth function
and $t_0 \in [a,b]$.  
Given any vector $\psi_0 \in V$, the differential equation
\begin{equation}            
\frac{d\psi(t)}{dt} = A(t) \psi(t)  
\label{diff}
\end{equation}
has a unique smooth solution $\psi \maps [a,b] \to \R^n$ 
with $\psi(t_0) = \psi_0$.  
\end{thm}

\vskip 1em
\noindent
{\it 
{\bf Sketch of Proof. }  With the given initial conditions, 
Equation {\rm (\ref{diff})} is equivalent to the integral equation
\[       \psi(t) = \psi_0 + \int_{t_0}^t A(s) \psi(s) ds  .\]
A solution of this equation is none other than a fixed
point of the map $T$ sending the function $\psi$ to the function
$T\psi$ given by
\[      (T\psi)(t) = \psi_0 + \int_{t_0}^t A(s) \psi(s) ds .\]
$T$ maps the Banach space 
of continuous $\R^n$-valued functions on $[a,b]$ 
to itself.  If $\int_a^b \|A(s)\| ds = M$ then
\[         \|T(\psi_1) - T(\psi_2)\| \le M \|\psi_1 - \psi_2 \|.    \]
We call a map with this property a {\bf contraction} if $M < 1$.
An easy argument shows that any contraction 
on a Banach space has a unique fixed point, so our equation has a 
unique solution.  If $M \not< 1$, we can chop the interval
$[a,b]$ into smaller intervals for which this bound does hold, 
and prove the theorem one piece at a time.  \qed
}

\vskip 1em
\noindent
1.  Let $K[a,b]$ be the category whose objects are points of the
interval $[a,b]$, with exactly one morphism from any object
to any other.   Given a function $A \maps [a,b] \to \End(\R^n)$
satisfying the conditions of Theorem 1, use the theorem to prove
there is a unique functor 
\[             F \maps K[a,b] \to \Vect   \]
such that:
\begin{itemize}
\item $F$ sends any object to $\R^n$.
\item $F$ sends any morphism $f \maps t_0 \to t_1$ to the 
linear operator 
\[                 \psi_0 \mapsto \psi(t_1)   \]
where $\psi \maps [a,b] \to \R^n$ is the unique solution of 
Equation {\rm (\ref{diff})} with $\psi(t_0) = \psi_0$.
\end{itemize}

\vskip 1em
\noindent 
{\it More poetically, $K[a,b]$ is the category whose
objects are {\bf moments of time} between time $a$ and time $b$.
The morphisms in this category are {\bf passages of time}.  Applied
to the passage of time from $t_0$ to $t_1$, the functor $F$ gives the
{\bf time evolution operator} mapping $\psi(t_0)$ to $\psi(t_1)$,
where $\psi$ is any solution of 
\[            \frac{d\psi(t)}{dt} = A(t) \psi(t)  .
\]

\vskip 1em
\noindent 
Next, suppose $X$ is a smooth manifold and $A$ is a smooth
$\End(\R^n)$-valued 1-form on $X$.  
For each point $x \in X$, such a thing gives a linear map
\[              A_x \maps T_x X \to \End(\R^n)  ,\]
and $A_x$ varies smoothly as a function of $x$.  If we take $n = 1$, 
$A$ becomes an ordinary 1-form and the following result reduces
to a problem in the last homework assignment:
}

\vskip 1em
\noindent
2.  Suppose $A$ is a smooth $\End(\R^n)$-valued 1-form on the manifold
$X$.  Show that there is a unique functor
\[    F \maps \P(X) \to \Vect  \]
such that
\begin{itemize}
\item $F$ sends any point of $X$ to $\R^n$.
\item $F$ sends any piecewise-smooth path $\gamma \maps [0,T] \to X$
to the linear operator 
\[                 \psi_0 \mapsto \psi(T)   \]
where $\psi \maps [0,T] \to \R^n$ is the unique solution of 
the equation
\begin{equation} 
       \frac{d\psi(t)}{dt} = A_{\gamma(t)}(\gamma'(t)) \; \psi(t)  
\label{parallel}
\end{equation}
with $\psi(0) = \psi_0$.
\end{itemize}

\vskip 1em
\noindent
{\it
An $\End(\R^n)$-valued 1-form $A$ is called a {\bf connection} on 
the trivial vector bundle 
\[         \pi \maps X \times \R^n \to X  .\]
If $\psi(t)$ satisfies Equation {\rm (\ref{parallel})}, we say the 
vector $\psi(t)$ is {\bf parallel transported} along the path
$\gamma$ using the connection $A$.  The linear operator 
$F(\gamma)$ is called the {\bf holonomy} of the connection $A$
along the path $\gamma$.

\vskip 1em
\noindent
All this stuff generalizes to the case of a connection on
a nontrivial vector bundle 
\[          \pi \maps E \to X \]
except that now the functor $F$ maps each point $x \in X$ to the
{\bf fiber of $E$ over $x$}, namely $E_x = \pi^{-1}(x)$.  To handle this
case, we choose an open cover of $X$ such that $E$ restricted
to each open set is trivializable, and reduce the problem 
to the case treated above.  After huffing and puffing, we get:
}


\begin{thm} \et 
Suppose $A$ is a smooth connection on a smooth vector bundle
$\pi \maps E \to X$ over a smooth manifold $X$.  Then there is a unique
functor 
\[             F \maps \P(X) \to \Vect \]
such that:
\begin{itemize}
\item For any object $x$ of $\P(X)$, $F(x)$ is the fiber of $E$ over $x$. 
\item For any morphism $\gamma \maps x \to y$ of $\P(X)$,
$F(\gamma)$ is the holonomy of $A$ along $\gamma$.  
\end{itemize}
\end{thm}

\noindent
{\it 
The converse is not true: there are functors $F \maps \P(X) \to \Vect$
that don't come from connections on vector bundles!
However, we can characterize the functors that do by means of
three conditions:
\begin{itemize}
\item $F(\gamma_1) = F(\gamma_2)$ when $\gamma_2$ is obtained
by reparametrizing $\gamma_1$:
\[   \gamma_2(t) = \gamma_1(f(t))  \]
for any monotone increasing function $f$.
\item $F(\gamma_2) = F(\gamma_1)^{-1}$  when $\gamma_2$ is a
reversed version of $\gamma_1$:
\[   \gamma_2(t) = \gamma_1(f(t))  \]
for any monotone decreasing function $f$.
\item $F(\gamma)$ depends smoothly on $\gamma$ (in a certain
precise sense).
\end{itemize}

\noindent
For some hints on how to prove this, try:
\vskip 1em
\begin{quote}
{\rm J.\ Barrett, Holonomy and path structures in general relativity and 
Yang--Mills theory, {\sl Int.\ J.\ Theor.\ Phys.}, {\bf 30} (1991), 
1171--1215.}
\end{quote}

\noindent
If we drop the smoothness condition, we call $F$ a {\bf generalized
connection}.  These play an important role in loop quantum gravity.
}
 


\vfill \eject

\end{document}


