I'm currently working through some notes on surfaces in $\mathbb{R}^3$ and their geodesics, with the following definitions (it's kind of lengthy, I will highlight the key parts):
- For a sufficiently smooth surface $S$ parametrized by $\varphi : D \to \mathbb{R}^3$, a geodesic is a curve $\alpha : I \to S$ to have acceleration orthogonal to the tangent spaces, i.e. $\alpha$ is a geodesic if and only if
$$
\alpha''(t) \perp T_{\alpha(t)}S \quad (\forall t \in I).
$$
It is proven that fixing $p \in S$ and $v \in T_pS$ there is a unique geodesic $\gamma_{p,v} : I_{p,v} \to S$ such that $\gamma(0) = p$ and $\gamma'(0) = v$, with $I_{p,v}$ the maximal interval of definition. That is, geodesics that stem from $p$ with velocity $v$ are (locally) unique, because two different ones give the same solution to an ODE in a sufficiently small interval of $p$. It is also claimed that, since these curves and their definition intervals depend smoothly on $p$ and $v$ and $\gamma_{p,0}$ is defined on $\mathbb{R}$, by continuity there exists some open ball $B_R(0_p) \subset T_pS$ where $1 \in I_{p,v}$ for $v \in B_R(0_p)$. Hence it is possible to define the Riemannian exponential as
$$
\begin{align}
\exp_p : &B_R(0_p) \to S \\
& v \longmapsto \gamma_{p,v}(1)
\end{align}
$$
Shortly after it is proved that $\exp_p$ sends lines through $0$ to geodesics, since $\gamma_{s,v}(t) = \gamma_{s,tv}(1)$, and that $D(\exp_p)_0 = Id$ which says that in a neighbourhood of $0$, the exponential is a diffeomorphism. We assume from now on that $R$ is small enough for this to hold.
From the previous definitions and results, it is then claimed that one can see that if $q = \exp_p(v) \in \exp(B_R(0_p))$ then there is a unique geodesic that joins $p$ and $q$, namely
$$
\gamma(t) := \exp_p(tv)
$$
with $[0,1] \subset Dom(\gamma)$.
I do not see why local uniqueness is derived just from the previous results. How can this be proven?
I am aware that (given sufficient regularity hypotheses), any curve $\alpha \subset S$ in $\exp_p(B_R(0_p))$ is a lift $\alpha(t) = \exp_p(\beta(t))$ with $\beta$ a curve in $B_R(0_p)$ (that is, in it's identification with a ball of the plane) but I haven't been able to prove much with that. I also presume a strong usage of locality is needed, as for example in the sphere any two points there are two geodesics joining them.
Best Answer
You get local uniqueness only up to different reparametrization. More precisely fix $p\in S$ and an $R>0$ such that $\exp_p : B_R(0_p)\to U$ is a diffeomorphism onto an open subset $U$ of $S$. Then:
$\textbf{proof}$:
Set $w=\gamma '(0)$. Then $\gamma(t)=\exp_p(tw)$. Claim: $|Lw|<R$:
Assume not. Then there is $\rho\in [0,L]$ with $ |\rho w|=R$. Choose a nonnegative strictly increasing sequence $\rho_n\to\rho$. Then $\exp_p(\rho_n w)=\gamma(\rho_n)\to\gamma(\rho)$ in $U$ and hence by applying the local inverse of $\exp_p$, $\rho_n w$ converges to some point in $B_R(0_p)$ which is not possible as $|\rho_n w|\to |\rho w|=R$. This proves the claim.
Now since $\exp_p(Lw)=\gamma(L)=\exp_p(v)$ and $v, Lw\in B_R(0_p)$ we conclude $Lw=v$ by the injectivity of $\exp_p$ on $B_R(0_p)$. Hence for all $t\in [0,L]$ we have $\gamma(t)=\exp_p(tw)=\exp_p(t \frac v L)$.