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\title{Compactness in Finite-dimensional Morse Theory}
\author{Kevin Iga}
\date{August 1996}
\maketitle
Let $M$ be a compact Riemannian manifold without boundary, and $f:M\to \re$ a
$C^2$ Morse function. Let $M(a,b)$ be the moduli space of Morse flows between
critical points $a$ and $b$ (with parametrization). Note that by associating
each with their value at $0$, we can associate $M(a,b)$ with the set of points
of $M$ which lie on flows from $a$ to $b$. Let $\bar{M}(a,b)$ be the union
\[\bar{M}(a,b)=\bigcup_{c_1,c_2,...,c_k} M(a,c_1)\times\cdots\times
M(c_k,b)\]
topologized as a subset of $M$.
\begin{theorem}
The space $\bar{M}(a,b)$ is compact.
\end{theorem}
Proof:
We reparametrize the Morse flows $x(s)$ so that
\[\dot{x}(s)=-\frac{\nabla f}{|\nabla f|}\]
and $x(0)=a$. Away from critical points, this is an elliptic PDE of
first order. Note that the flow is of finite length, since it is
parametrized by arclength, and so the domain of $x$ is the interval
from zero to the length of the flow line.
A sequence of reparametrized Morse flows is equicontinuous,
since if $\epsilon>0$, we take $\delta=\epsilon$, and if $|s-s_0|<\delta$,
then by the Mean Value theorem, $|x_i(s)-x_i(s_0)|\le \max |\dot{x}_i||s-s_0|
< \delta=\epsilon$.
Since $M$ is compact, the sequence is uniformly bounded. By the Arzela-Ascoli
theorem, the sequence has a convergent subsequence, uniformly on compact
sets. Without loss of generality, we will assume that our sequence is such
a convergent subsequence. The only question is whether the limit is a
piecewise flow.
If $x_i(s_0)$ converges to a critical point, we will say nothing of
regularity. If it does not converge to a critical point, we can find
an interval $[a,b]$ around $s_0$ and a subsequence so that $x_i(s_0)$
is bounded away from a critical point.
Away from critical points, $V(x)=-\frac{\nabla f(x)}{|\nabla f(x)|}$ is
a continuous vector field. As such, if $x_i(s)$ converges uniformly on
$[a,b]$ to $x(s)$, then we consider the convergence of $V(x_i(s))$
on $[a,b]$.
Certainly, $V(x)$ is uniformly continuous on
the compact subset of $M$ containing $x_i([a,b])$ but excluding small
balls around the critical points of $f$. If $\epsilon>0$, we choose
the $\delta$ involved in uniform continuity of $V(x)$ in the domain just
described. Then by the uniform
convergence of $x_i$, we can choose an $N$ so that
$|x_i(s)-x(s)|<\delta$ for all $i>N$.
So
\[|V(x_i(s))-V(x(s))|<\epsilon.\]
In other words,
$\dot{x}_i(s)=-\frac{\nabla f}{|\nabla f|}\circ x_i(s)$ converges
uniformly to $-\frac{\nabla f}{|\nabla f|}\circ x(s)$.
The fact that $x_i(s)$ converges uniformly to $x(s)$ and $\dot{x}_i(s)$
converges uniformly implies that $\lim_{i\to \infty}\dot{x}_i(s)=\dot{x}(s)$.
Since $\dot{x}_i(s)$ converged uniformly to
$-\frac{\nabla f}{|\nabla f|}\circ x(s)$, we have that
\[\dot{x}(s)=-\frac{\nabla f}{|\nabla f|}\circ x(s),\]
in other words, that $x(s)$ is on a flow line except where $x(s)$ is a
critical point.
\qed
\end{document}