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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.

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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:

If $\gamma :[0,L] \rightarrow U$ is a geodesic with $\gamma(0)=p$ and $\gamma(L)=\exp_p(v)$ for some $v\in B_R(0_p)$ then $\gamma(t)=\exp_p(t \frac v L)$ for all $t\in [0,L]$.

$\textbf{proof}$:

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 $\gamma(t)=\exp_p(tw)=\exp_p(t \frac v L)$.

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