Let $\alpha, \beta:I\to M$, two geodesic. If exist a number $a\in I$ such that $\alpha'(a)=\beta'(a)$, then $\alpha=\beta$

Question

Let $\alpha, \beta:I\to M$, two geodesic define on diff. manifold $M$ and $I$ connected. If exist a number $a\in I$ such that $\alpha'(a)=\beta'(a)$, then $\alpha=\beta$.

My approach: Define $\gamma:I\to M$, by $\gamma(t)=\alpha(t)-\beta(t)$. So, $$0=\dfrac{D}{dt}\left(\dfrac{d\gamma}{dt}\right)=\dfrac{D}{dt}\left(\dfrac{d\alpha}{dt}-\dfrac{d\beta}{dt}\right)\implies \dfrac{D}{dt}\left(\dfrac{d\alpha}{dt}\right)=\dfrac{D}{dt}\left(\dfrac{d\beta}{dt}\right)$$ But How use the fact that $I$ is connected?. Thanks!

Answer 1

I think it may be the property that $\alpha, \beta$ are geodesics that you aren't using. Since $\alpha, \beta \subset (\varphi, U)$ where $U$ is some chart on $M$, you may as well assume that $\alpha, \beta: I \to \mathbb{R}^k$. Here I am just assuming that $M$ is a $k$-manifold. If we define $\gamma(t) = \alpha(t) - \beta(t)$ then $\gamma: I \to \mathbb{R}^k$; moreoverwe have

$$\|\gamma(x) - \gamma(y)\| = \|x-y\|, \forall x,y \in I$$

Now we have that if $\gamma$ is nonzero then;

$$\gamma'(t) = \lim_{t \to a} \frac{\|\gamma(t) - \gamma(a)\|}{\|t-a\|}= 1$$

However $\gamma'(a) = 0$; hence a contradiction. Therefore, $\gamma = 0 \Rightarrow \alpha(t) = \beta(t)$.