I'm working on a problem in do Carmo's Riemannian geometry book (chapter 7, problem 5). He states that a divergent curve on a noncompact Riemannian manifold $M$ is a curve $\alpha: [0, \infty) \to M$ where given any compact $K \subset M$ we have there exists $T>0$ such that $\alpha(t) \not\in K$ for $t>T$. Subsequently the length is defined in the usual sense for curves on a manifold:
$$
L(\alpha) \;\; =\;\; \int_0^\infty ||\alpha'(t)|| dt.
$$
The problem I'm trying to prove is
Prove that a noncompact manifold is complete if and only if every divergent curve has unbounded (i.e. infinite) length.
The "forwards" direction of the proof is easy; I'm having trouble with the other direction. If I assume that every divergent curve $\alpha$ has infinite length, how do I show completeness of the manifold? I'm borrowing one strategy from MIT OCW but I don't totally agree with their proof. They claim that we can start with some arbitrary geodesic $\gamma:[0,\epsilon) \to M$ and assume that such a geodesic cannot be extended to the interval $[0,\infty)$. The set $R$ of points $x$ on which $\gamma$ can be extended to $[0,x)$ is nonempty and bounded above, thus let $s = \sup R$. Define a re-parameterization $\alpha(t) = \gamma(s(1-e^{-t}))$, which is defined on $[0,\infty)$ but is of finite length. If we can show that $\alpha$ is a divergent curve we are done.
The problem I have with the proof in the link is that they claim that $\gamma$ approaches a point on the "boundary" of $M$, but I find this claim problematic since it's not clear how to define a boundary unless $M$ is embedded/immersed in some ambient space. Ultimately I would like to show that $\alpha$ escapes every compact subset of $M$, but I'm starting to think this proof might be problematic.
Can anyone suggest a way to amend the above proof as stated, or provide an alternate proof strategy?
Best Answer
Consider any geodesic of unit speed $ \gamma :[0,\epsilon ) \rightarrow M$ If $\gamma(t)$ goes to $q\in M$, then by considering $B_\delta (q)$ we can extend the geodesic $\gamma : [0,\epsilon +\delta )\rightarrow M$
If we can do same thing countably many and if we have $\gamma : [0, \infty) \rightarrow M$, then we have a completeness
And assume that $M$ is not complete So we have $\gamma : [0,\epsilon )\rightarrow M$ which can not be extended.
By reparametrization we have $\alpha(t) =\gamma (\epsilon (1-e^{-t})),\ t\in [0,\infty)$
If $ \alpha (n)$ are in some compact set $K$ so it has a limit in $K$. It is a contradiction since $\gamma$ can not be extended. So any compact set does not contain all $\alpha (n)$ That is $\alpha$ is divergent curve If $\alpha$ has infinite length then it is a contradiction since $ {\rm length}\ \alpha = {\rm length}\ \gamma =\epsilon$