A question about unique minimal geodesic

Question

Note that for a Riemannian manifold the minimal geodesics may not be unique, for example, there are infinitely many minimal geodesics connecting antipodal points in sphere.

However, I hope the following thing is true, since I do need it.

Let $(M,g)$ be a complete Riemannian manifold, $p,q\in M$ and $\gamma:[0,b]\to M$ is a unit-speed minimal geodesic connecting $p,q$. Then for any $0<\varepsilon<b$, $\gamma|_{[\varepsilon,b]}$ is the unique minimal geodesic connecting $\gamma(\varepsilon)$ and $q$.

Is there anyone can provide a proof? Thanks in advance!

Answer 1

Given a curve $c :(\alpha,\beta)\to M$, its arc-length from $p$ to $q$ is given by $$\ell(c):=\int_\alpha^\beta\sqrt{g(c’(t),c’(t)}\, dt. $$ Since $\gamma$ is unit speed and minimises the length between $p$ and $q$, $$ \inf \{ \ell(c) \text{ s.t. } c\in C^1([0,b]; M), c(0)=p,c(b)=q\}= \ell(\gamma)=b.$$ Suppose that $\gamma \vert_{[\varepsilon,b]}$ does not minimise the length between $\gamma(\varepsilon)$ and $b$. Then there exists a competitor $\tilde \gamma \in C^1([\varepsilon,b];M)$ with $\tilde \gamma (\varepsilon)=\gamma(\varepsilon)$ and $\tilde \gamma(b) =q$ such that $\ell(\tilde \gamma)<\ell (\gamma \vert_{[\varepsilon,b]})$. Now define $$\bar \gamma (t) = \begin{cases} \gamma(t), &\text{if } t\in [0,\varepsilon)\\ \tilde \gamma(t), &\text{if } t\in [\varepsilon,b]. \end{cases} $$ Formally, the arc-length from $p$ to $q$ of $\bar \gamma$ is \begin{align*}\ell (\bar \gamma) &= \int_0^\varepsilon\sqrt{g(\bar \gamma’(t),\bar \gamma’(t)}\, dt + \int_\varepsilon^b\sqrt{g(\bar \gamma’(t),\bar \gamma’(t)}\, dt \\ &< \int_0^\varepsilon\sqrt{g(\bar \gamma’(t),\bar \gamma’(t)}\, dt + \int_\varepsilon^b\sqrt{g(\tilde \gamma’(t),\tilde \gamma’(t)}\, dt \\ &=\ell(\gamma), \end{align*} which contradicts minimality of $\gamma$ from $p$ to $q$. The small technical issue I have with this argument is that $\bar \gamma$ may not longer by $C^1$, so is no longer admissible; however, taking a $C^1$ approximation, I would expect that you can fix this.

Answer 2

I will address the uniqueness statement in your question: the fact that $\gamma|_{[\varepsilon,b]}$ is minimizing has already been shown by @JackT. It will rely on the following facts:

• any two distinct points can be joined by a minimizing geodesic (from Hopf-Rinow's Theorem).
• geodesics are smooth, which implies that if $\varphi\colon I \to M$ is a rectifiable curve joining two points that is not smooth, there exists a geodesic joining its endpoints with length strictly less.

By contradiction, let $t_0 \in (0,b)$ be such that $\gamma|_{[t_0,b]}$ is not the unique length minimizing geodesic from $\gamma(t_0)$ to $\gamma(b)$. Let $\tilde{\gamma}\colon [t_0,b] \to M$ be a distinct minimizing geodesic, (arc-length parametrized). Consider the path $$ \varphi(t) = \begin{cases} \gamma(t) & \text{if} \quad t\in [0,t_0],\\ \tilde{\gamma}(t) & \text{if} \quad t\in [t_0,b]. \end{cases} $$ Then $\varphi$ is not smooth at $t_0$, and is therefore not a geodesic. Since $M$ is complete, by the Hopf-Rinow Theorem, there exists a minimizing geodesic $c\colon I\to M$ joining $\varphi(0)$ to $\varphi(b)$, which has length strictly less than that of $\varphi$. Finally, $$ \ell(c) < \ell(\varphi) = \ell(\varphi|_{[0,t_0]}) + \ell(\varphi|_{[t_0,b]}) = \ell(\gamma|_{[0,t_0]}) + \ell(\tilde{\gamma}|_{[t_0,b]}) = \ell(\gamma|_{[0,t_0]}) + \ell(\gamma|_{[t_0,b]}) = \ell(\gamma). $$ But $c$ joints $\varphi(0) = \gamma(0)$ to $\varphi(b)=\tilde{\gamma}(b) = \gamma(b)$. This contradicts the fact that $\gamma$ is a minimizing geodesic.