Also asked here: <https://math.stackexchange.com/questions/1725787/when-are-the-minimizing-geodesics-of-a-totally-geodesic-submanifold-also-minimiz>


A reference on totally geodesic submanifold (TGS):


<http://www.map.mpim-bonn.mpg.de/Totally_geodesic_submanifold>


Let "|" stand for restricted metric. Let $(S,g|)$ be a TGS of a complete Riemannian manifold $(M,g)$. In this case, is it OK to assume that $S$ should be defined to be complete as well? For example, we can think of $S=\mathbb{R}^2-{0}$, as a TGS of $\mathbb{R}^3$, but what's the point of not having that $\{0\}$? (I agree this is not exactly a question!)


**Now onto the question!** 


Let $p,q\in S$ and let $c$ be a minimizing geodesic in $(S,g|)$ from $p$ to $q$. Now, since $S$ is TGS, $c$ lies fully in $M$, and a geodesic in $(M,g)$. But is $c$ a **minimizing geodesic** in $(M,g)$? if so, why? If not, what's a counterexample? Thanks in advance!



When are the minimizing geodesics of a totally geodesic submanifold also minimizing in the underlying manifold?
The totally geodesic submanifold might not be complete. Take the usual metric on the upper half of Euclidean 3-space, and some generic metric on the lower half, so that they agree where they meet to infinite order. There won't be any totally geodesic surfaces in a generic metric. So the flat planes in the upper half will not extend to totally geodesic submanifolds in the lower half. 


For a counterexample to your other problem, take a hemisphere and glue a flat disk to it along a circle. Smooth out a little. The circle is still totally geodesic, our totally geodesic submanifold. But the distance between antipodal points is smaller along the nearly flat plate than along that great circle.


