I struggle with this exercise from Riemann Manifolds: Introduction to Curvature by Lee:

A curve $\gamma: [0, b) \to M (0 < b \leq \infty)$ is said to converge to infinity if for every compact set $K \subset M$, there is a time $T \in [0,b)$ such that $\gamma(t) \notin K$ for $t > T$. (This means that γ converges to the “point at infinity” in the one-point compactification of $M$.) Prove that a Riemannian manifold is complete if and only if every regular curve that converges to infinity has infinite length. (The length of a curve whose domain is not compact is just the supremum of the lengths of its restrictions to compact subintervals.)

I started by assuming there exists a curve that converges to infinity but with a finite length, and I'd like to show that this implies that $M$ is non complete (ie. geodesics are not defined at all time). However, I can't continue...


Take a curve $\gamma:[0,b)$ that converges to infinity but has finite length. Let $(t_k)$ be a strictly increasing sequence of times in $[0,b)$ so that $t_1=0$ and $\lim_{k\to\infty}t_k=b$.

Let $K\in\mathbb N$. If $m>k\geq K$, then the distance $d(\gamma(t_k),\gamma(t_m))$ is bounded by their distance along $\gamma$. Therefore $$ d(\gamma(t_k),\gamma(t_m)) \leq \int_{t_k}^{t_m}|\dot\gamma(t)|dt \leq \int_{t_K}^{b}|\dot\gamma(t)|dt. $$ The integral $\int_{t_K}^{b}|\dot\gamma(t)|dt$ can be made arbitrarily small by increasing $K$ since $t_K\to b$ as $K\to\infty$ and $\gamma$ has finite length. Therefore the sequence $(\gamma(t_k))$ is a Cauchy sequence. If $M$ were complete, it would have a limit, but it cannot because $\gamma$ leaves all compact sets.

  • $\begingroup$ Thanks for the post ! I don't know if it is obvious but I don't get what you mean by : "d(γ(tk),γ(tm)) is bounded by their distance along γ. " Why is it not an equality for the first inequality ? $\endgroup$ – Sasha Oct 13 '14 at 18:49
  • $\begingroup$ @Sasha, the distance between two points is the infimum of their distances over all possible curves. Our $\gamma$ is just one possible curve, and if it's not a geodesic, it will not minimize the distance. Think of a wiggly $\gamma$. (Even if it were a geodesic, it wouldn't need to minimize. Consider the longer part of a great circle joining two points on $S^2$.) $\endgroup$ – Joonas Ilmavirta Oct 14 '14 at 4:29
  • $\begingroup$ i got it, thanks! $\endgroup$ – Sasha Oct 14 '14 at 18:03

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