Maximal geodesics on compact manifolds

Question

I have two questions about the following passage in Taubes's book on differential geometry. I also quote the proposition it references.

9.1 The maximal extension of a geodesic Let $I \subset R$ denote a closed interval that is not the whole of $R$, and let $s$ denote an end point of $I$, for the sake of the discussion, the maximum point. Let $\gamma: I \rightarrow M$ denote a geodesic. It follows from Proposition 8.3 that there exists an extension of $\gamma$ to an interval with maximal endpoint $s + \epsilon$ for some $\epsilon > 0$. To see this, take a point $q =\gamma (s')$ for $s'$ nearly $s$. Then $q$ is very close to $p =\gamma (s)$. Proposition 8.3 tells us that the geodesic $t \rightarrow \gamma_q(t) = \gamma(s'+ t)$ is defined for $|t|<\epsilon$ for some $\epsilon$ that depends only on p. This extends $\gamma$. It follows as a consequence that each geodesic in $M$ has a maximum extension. In particular, if $M$ is compact, then each geodesic extends as a geodesic mapping $R$ into $M$.

Proposition 8.3 Let $M$ be a smooth manifold and let $g$ denote a Riemannian metric on $M$. Let $p\in M$ and let $v \in TM|_p$. There exists $\epsilon>0$ and a unique map from the interval $(-\epsilon, \epsilon)$ to $M$ that obeys the geodesic equation, sends $0$ to $p$, and whose differential at $0$ sends the vector $d/dt$ to $v$.

1. How is this extension argument valid? The argument claims that $\epsilon$ depends only on the endpoint of the interval $I$, but in proposition $8.3$, $\epsilon$ depends on the point $0$ is being mapped to. Here, that is $s'$. I'm lost on the details of the argument here and would appreciate any clarification you could provide.

2. Why does it follow from the fact that geodesics have maximal extensions that geodesics on compact manifolds have extensions to all of $\mathbb R$?

Answer 1

Proposition 8.3 does not seem to be very conveniently formulated for the purpose you need it for here. This is one way to see the required conclusion: You can take a coordinate chart around the point $p$ so that it contains the ball $B(p,2r)$ and choose $q\in B(p,r)$. (Use either the Riemannian distance or the Euclidean distance in a chart; it doesn't matter.) Then you can do the existence and uniqueness business for an ODE on this chart, and you know that any geodesic starting in the smaller ball has to exist at least up to the boundary of the bigger ball.

Suppose you had a geodesic $\gamma:(a,b)\to M$ that is maximal but $b<\infty$. (The case $a>-\infty$ is similar.) By the argument you have provided, you can in fact extend $\gamma$ to be defined on $(a,b+\delta)$ for some small $\delta>0$. This contradicts the maximality of the interval.

The key thing here is that if you have a geodesic defined on a bounded interval, the limit $\lim_{t\to b}\gamma(t)$ exists by compactness. (Uniqueness follows from arc length parametrization, which you should be using anyway when speaking about maximality.) On a noncompact manifold the limit may fail to exist, and on a manifold with boundary the limit may be a boundary point where things are a little different than in the interior.

Answer 2

Regarding item 1, in the context of 9.1 the function $\gamma : I \to M$ is given. Thus, the endpoint $s$ of $I$ determines the point $p = \gamma(s)$ and the vector $v = \partial I/dt |_{t=s} \in TM|_p$ which determines $\epsilon$.

Regarding item 2, let $J$ be the union of all subintervals of $R$ that contain $I$ such that $\gamma$ extends to a geodesic defined on $J$. If $J=\mathbb{R}$, you are done. If $J \ne \mathbb{R}$ then $J$ has an endpoint, and you can apply Proposition 8.3 to extend $\gamma$ beyond that endpoint, which contradicts how $J$ was defined.