Hopf-Rinow theorem (Metric completeness implies Geodesic completeness)

Question

I use Petersen's book.

1. How do we know $\hat{K}$ takes exactly the form in the description? Specifically the condition $|v| \leq |\hat{\dot{c}} |$

2. Pretty sure there is a minor typo. They want to relabel first $\mu:(-\epsilon, \epsilon) \to M$ with $\mu(0) = q$ and $\dot{\mu}(0) = v$. Now how does $\mu$ patch with the $c(t_j) \in K$? If $b$ is arbitrary large and $\epsilon > 0$ is arbitrary small, there should be a gap right?

3. So the contradiction is that $[0,b)$ is maximal? Can someone spill out the details for me?

Answer 1

1. How do we know $\hat{K}$ takes exactly the form in the description? Specifically the condition $|v| \leq |\hat{\dot{c}} |$

The choice $\lvert v\rvert\leq \lvert\dot c\rvert$ is pretty arbitrary, we only need it to contain what we need for the argument that follows, but I suppose a $\leq$ looks more intuitive (e.g., corresponds to closed ball being the first example of compact sets).

2. Pretty sure there is a minor typo. They want to relabel first $\mu:(-\epsilon, \epsilon) \to M$ with $\mu(0) = q$ and $\dot{\mu}(0) = v$.

No, I don't see any typo here. It is an abuse of the letter $c$ for curve. :-) Let's change to $\mu$ as you suggested.

(cont'd) Now how does $\mu$ patch with the $c(t_j) \in K$?

You have $\hat K$ and now choose $\varepsilon$ so that for every tangent vector in $\hat K$ the exponential map is defined for time up to $\pm\varepsilon$ (noninclusive). Now we have earlier chosen a sequence $t_j\uparrow b<\infty$. At $t_j$, the geodesic was still defined, so we have $(c(t_j),\dot c(t_j))$, and this pair is in $\hat K$. So for some $0<\delta\leq\varepsilon$, we have two curves: $c(t_j+\tau)$ and $\mu_{c(t_j),\dot c(t_j)}(\tau)$, both defined for $\lvert\tau\rvert<\delta$ and has the same initial data at $\tau=0$. So they have to agree. Hence we may patch them together and extend the curve $c$ to at least time $\max(b,t_j+\varepsilon)>b$ (possibly excluding the end-point), contradicting the choice of $b$.

So we have proved:

Result: A geodesic $c$ that cannot be extended indefinitely to time $+\infty$ must leave every compact set, i.e., for each compact $K\subseteq M$ there is a time $t_K<\sup\operatorname{dom}c$ such that $c(t)\notin K$ for all $t>t_K$, $t\in\operatorname{dom} c$.

which will be used later.

(cont'd) If $b$ is arbitrary large and $\epsilon > 0$ is arbitrary small, there should be a gap right?

No, the order of choice is

• $b$ was chosen first (as the time the geodesic "falls off the manifold")
• A sequence $t_j\uparrow b$ chosen by "not leaving a compact $K\subseteq M$"
• compact $\hat{K}\subset TM$ chosen to contain $\dot{c}(t_j)$
• $\varepsilon>0$ chosen a bit later, depending on $K$,
• A specific $t_j\in (t_j)_{j\in\mathbb{N}}$ chosen within $\varepsilon$ of $b$

so you can't move $b$ somewhere else now.

3. So the contradiction is that $[0,b)$ is maximal? Can someone spill out the details for me?

Suppose $M$ is complete under metric $d$, and that, contrary to (1), there is a geodesic cannot be extended beyond $(-a,b)$, $b<\infty$, $0<a\leq\infty$. Then we have two conflicting conditions

• $c$ is a geodesic, so $d(c(s),c(t))\leq\lvert s-t\rvert\cdot\lvert\dot c\rvert$ for all $s,t\in[0,b)$. So there is a Cauchy sequence, say $x_n=c(b-2^{-n})$, which must therefore converges to some $x\in M$ since $M$ is metrically complete. So $K=\{c(t):t\in[0,b)\}\cup\{x\}$ is compact.
• But $c$ leaves every compact set before time $b$, so it must leave the compact set $K$, i.e., there exists a time $t_K$ such that $c(t)\notin K$ for $t_K<t<b$, which it can't do. conttradiction!