Existence of only finitely many geodesics

Question

The following question arose from Theorem 16.3 (p.90) in the book Morse theory from John Milnor.

We are dealing with a complete Riemannian manifold $M$, an $a \in \mathbb{R_{>0}}$ and two points $p,q \in M$ which are not conjugate along any geodesic from $p$ to $q$ of length $\le \sqrt{a}$. Then the question is: Why are there only finitely many geodesics from $p$ to $q$ of length $\le \sqrt{a}$.

I have already tried arguing this way: Assume that there are infinitely many geodesics $(\gamma_n)_{n \in \mathbb{N}}$ of the kind spoken of. Since the closed ball $B:=\{ v \in T_pM : \Vert v \Vert \le \sqrt{a}\}$ is compact (by Heine-Borel), there exists a subsequence of the vectors $(\gamma_n(0))_{n \in \mathbb{N}} \in B$ which converges against an $v_\infty \in B$. The fact that solutions of DGLs depend continuously on the initial value provides that the curve $\gamma_{v_\infty} = exp(t \cdot v_\infty)$ also ends in $q$. Now the idea is, that this sequence yields a proper, geodesic variation $\alpha:(-\varepsilon, \varepsilon) \times [0,1] \to M$ of $\gamma_{v_\infty}=\alpha(0,-)$. This would yield a Jacobifield $J=\frac{D}{ds}\alpha(0,t)$ along $\gamma_{v_\infty}$, which vanishes at $p$ and $q$, so we get a contradiction to the premise that $p$ and $q$ are not conjugate along any geodesic.

Solution: I don't think there is a way to extract uncountably many geodesics of length $\le \sqrt{a}$, all of which connect $p$ and $q$. So first of all, to get a variation $\alpha:(-1,1) \times [0,1] \to M$ we may do the following: Let $(v_n)_{n \in \mathbb{N}}$ be the (sub)sequence converging to $v_\infty$. Set $v:(-1,1) \to B; v\left(\frac{1}{n}\right) := v_n$ for $n \ge 2$. Then extend $v$ differentiably (i.e. by connecting the $v_n$ through lines and then smoothing the edges out). Finally, set $\alpha(s,t) := exp(t \cdot v(s))$. As $M$ is complete, this makes sense and is well defined. This certainly is a geodesic variation of $\gamma_{v_\infty}$. What remains to show is that the vector field $\frac{D}{ds}\alpha(0,t)$ vanishes at $q$ (to get the contradiction from above). Notice, that the $\alpha(s,-)$ mustn't connect $p$ and $q$ for all $s$, but for $s=\frac{1}{n}$. Since $\alpha$ is differentiable, we can compute $\frac{D}{ds}\alpha(0,1)$ by choosing the sequence $(s_n=\frac{1}{n})_{n \in \mathbb{N}}$ which yields $\frac{D}{ds}\alpha(0,1) = 0$ since the difference quotient is 0 for all $s_n$ because the $\alpha(s_n,-)$ connect $p$ and $q$ by our assumption.

Answer 1

I have found another, more abstract solution:

The energy functional $E$ on $B \simeq \Omega^a$ (compare Milnor, Morse theory, p.88-90), whose critical points are exactly these geodesics, has, by the Lemma of Morse (p.6) and the fact that $B$ is compact, only finitely many critical points, i.e. only finitely many of these geodesics exist.

Answer 2

First recall theorem: Let $\gamma= [0, a] \to M$ be a geodesic and put $\gamma(0) =p$. The point $q = \gamma(t_0 )$, $t_0 \in (0, a]$, is conjugate to $p$ along $\gamma$ if and only if $v_0 = t_0 \gamma'(0)$ is a critical point of $exp_p$.

Now assume by contradiction that there are infinitely many geodesic of length $\le a$ from $p$ to $q$. Each geodesic $\gamma_i$ defines a vector $v_i \in B_a(0)$ by parameterizing the geodesic $\gamma_i:[0,1] \to M$ $\gamma_i(t)=exp_p(tv_i)$ where $\gamma_i(v_i)=q$. Let $v_\infty$ be the accumulation point of $\{v_i\}_{i \in \mathbb{N}}$ which exists as $B_a(0)$ is compact.

First observe that $\gamma_{\infty}:[0,1] \to M$ defined by $\gamma_\infty(t)=exp_p(tv_{\infty})$ is a geodesic from $p$ to $q$. This as you said rests on the fact that solutions of DGLs depend continuously on their initial value.

We will now show that $v_\infty$ is a critical point of $exp_p$. Assume by contradiction it isn't. By the inverse function theorem there is a neighborhood of $v_\infty$ such that $exp_p$ is a homeomorphism. But $exp_p$ cant be a homeomorphism around $v_\infty$ as we have a converging sequence $\{v_{i_k}\}_{k \in \mathbb{N}}$ to $v_\infty$ such that $exp(v_{i_k})=q$. In particular, there is no neighborhood of $v_{\infty}$ such that $exp_p$ is injective.

Using the theorem from above we can conclude that $q$ is conjugate to $p$ along $exp_p(t v_\infty)$.

The proof of the theorem I was using is not hard and you can read it in do Carmo's book Riemannian Geometry on page 117.