Autoparallel submanifolds and geodesics I have the following question in differential geometry. Any help is greatly appreciated. Let $M$ be an autoparallel submanifold of a manifold $S$ with respect to a connection $\nabla$. Let $\gamma$ be a $\nabla$-geodesic connecting two points $p$ and $q$ of $M$. Should then $\gamma$ completely lie in $M$? 
 A: From your definition, you can show that following:
Let $p\in M$ and $v \in T_pM$. Then the geodesic $\alpha$ in $S$ with $\alpha(0) = p$ and $\alpha'(0) = v$ lies completely in $M$. 
Proof: Let me use $\bar\nabla $ for the connection in $S$ and $\nabla = \bar\nabla ^\top$ be the connection on $M$ (Which is not the notation you used). Let $\alpha$ be a $\nabla$-geodesic in $M$. Then $\nabla _{\dot\gamma} \dot\gamma = 0$. But as $\gamma$ is in $M$ we have also $\bar\nabla_{\dot\gamma}\dot\gamma = 0$ since by definition 
$$\nabla_\dot\gamma \dot\gamma = \big(\bar \nabla_\dot\gamma \dot\gamma\big)^\top$$
and 
$$\big(\bar \nabla_\dot\gamma \dot\gamma\big)^\top = \bar \nabla_\dot\gamma \dot\gamma $$
as $M$ is autoparallel. So $\gamma$ is also a geodesic in $S$. By uniqueness of geodesic, $\gamma = \alpha$. Thus $\alpha$ lies completely in $M$. 
However, your assertion is not correct. There are two issues:


*

*Topological: Let $S= \mathbb R^2$ and $M$ be two parallel lines. Then for sure your assertion does not hold.

*Geometrical: Let $S = \mathbb S^2$ and $M$ be the equator in $\mathbb S^2$. Let $p, q\in M$ be two antipodal point in $M$. Then there are infinitely many geodesic containing $p$ and $q$, while only one of them is in $M$.
