When are the minimizing geodesics of a totally geodesic submanifold also minimizing in the underlying manifold? Asked here too: https://mathoverflow.net/questions/235178/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 $(,g|)S$ 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 miniµizing 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!
 A: For the completeness question, it all depends on your definition of a submanifold. If you require a submanifold to be properly embedded, a (connected) Riemannian submanifold $(N, h)\subset (M,g)$ of a complete manifold is necessarily complete. This follows from the fact the fact that the Riemannian distance function $d_h$ on $N$ and the Riemannian distance function $d_g$ on $M$ are related by the inequality
$$
d_g(p,q)\le d_h(p,q), p, q\in N. 
$$ 
Therefore, every Cauchy sequence $(x_i)$ in $(N, d_h)$ is also Cauchy in $(M, d_g)$. Since $(M,g)$ is complete, the sequence $x_i$ converges to some $x\in M$. But since $N$ is properly embedded in $M$, it is closed in $M$ and, hence, $x\in N$. From this, it is easy to see that $\lim_i x_i=x$ in the metric topology of $(N,d_h)$, hence, $(N,h)$ is complete. 
If you do not assume that $N$ is properly embedded, you already have a counter example. 
Now, to your main question, consider the flat 2-torus $(M,g)=T^2$, the quotient of $R^2$ by the standard integer group of translations. Now, take a line $L\subset R^2$. It projects to a totally geodesic submanifold $(N,h)\subset (M,g)$. If you want to the submanifold to be properly embedded, take $L$ to have rational slope. In any case, if every minimizing geodesic in $N$ were minimizing in $M$, it would follow that the inclusion map
$$
i: (N, d_h)\to (M, d_g)
$$
would be an isometry in the metric sense: $d_g(i(p), i(q))=d_h(p,q)$ for all $p, q\in N$. But this is clearly false once the length of $N$ is large enough (more than $\sqrt{2}$, say). 
