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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):


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!


marked as duplicate by Alex Degtyarev, Wolfgang, Stefan Kohl, Mikhail Katz, Deane Yang Apr 3 '16 at 18:56

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    $\begingroup$ Take a "principal" ellipse on an ellipsoid in 3-space. It is totally geodesic but it won't be minimizing unless it combines the two smallest half-axes. $\endgroup$ – Mikhail Katz Apr 3 '16 at 17:29

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.


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