The distances on totally geodesic submanifold and manifold.

Question

There is a conlusion that the distances on a totally geodesic submanifold $M^{n} \subseteq N^{n+p}$ are the same with the distances on $N^{n+p}$(if we need we can add completeness). But I do not know how to prove it(Seems to be obvious)? Is there any proof or references about the question?

Answer 1

As I mentioned in a comment above, this is not true as stated.

Perhaps the simplest counterexample is the following. Let $N = S^1\times S^1$ be a flat torus, with each circle factor of total length $1$. Envisioning this as square with sides identified, it is easy to see that it has diameter $\frac{\sqrt{2}}{2}$.

Now, for $k\in \mathbb{Z}$, consider the geodesic $\gamma_k:[0,1]\rightarrow N$ with $\gamma_k(t) = (t, kt)$. Then $\|\gamma'(t)\| = \sqrt{1+k^2}$. By inspecting the first coordinate, it is clear that $\gamma_k$ is injective, except that $\gamma_k(0) = \gamma_k(1)$. In particular, the length of $\gamma_k$ is $\sqrt{1+k^2}$.

Take $N$ to be the image of $\gamma_k$, which is a totally geodesic submanifold of $M$. From the calcuation in the previous paragraph, we see that $d_N(\gamma_k(0),\gamma_k(1/2)) = \frac{\sqrt{1+k^2}}{2}$. For $|k| > 2$, it now follows that $d_N(\gamma_k(0),\gamma_k(1/2)) > d_M(\gamma_k(0), \gamma_k(1/2)).$