Background : If a compact Riemannian manifold $M$ with a no curvature condition has disjoint two submanifolds $N_i$, then the distance between them is attained by some minimizing geodesic $c$.

If $c'(0)$ is orthogonal to $T_{c(0)} N_1$, then $\exp_{c(0)}\ tv$ with $|v|=1,\ v\perp T_{c(0)}N_1$ goes to where ?

Question : I will add some curvature condition and shape condition of submanifold to the above since it may help. If $M$ has nonnegative sectional curvature and if $N$ is a totally geodesic submanifold, then assume that $c$ is a geodesic. Define $$d(\alpha (t), c(t))= d(N, c(t)) ,\ \alpha (t) \in N$$

Then image of $\alpha$ is an image of some geodesic ?

Here we give more condition for well-definedness of $\alpha$ : $c(0)\ \in N$ and $t\in [0,\epsilon]$

Thank you in anticipation

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    $\begingroup$ I do not see why the projection is well-defined. Say let $M$ be the round $n$-sphere, $N$ be the equator, and suppose $c$ passes through the north pole. What $\alpha(t)$ corresponds to the north pole? In general maps that take geodesics to geodesics are called totally geodesics. Many submersions aren't totally geodesic, see projecteuclid.org/euclid.jdg/1214429276 so I see no reason why this could work in your generality even if you assume the projection is well defined. $\endgroup$ Mar 17, 2016 at 16:05
  • $\begingroup$ I add some condition for well-definedness of $\alpha$ And this is just my curiosity. $\endgroup$ Mar 17, 2016 at 16:14
  • $\begingroup$ And thank you for your reply and reference $\endgroup$ Mar 17, 2016 at 16:14
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    $\begingroup$ In view of Peter Michor's answer, can you clarify whether you want $\alpha$ to be geodesic in the sense $\nabla_{\dot{\alpha}} \dot{\alpha} = 0$ or merely $\nabla_{\dot{\alpha}}\dot{\alpha} \propto \dot{\alpha}$? $\endgroup$ Mar 17, 2016 at 20:33
  • $\begingroup$ @Willie Wong Sorry. I did not give an detail. I fix the OP $\endgroup$ Mar 18, 2016 at 0:08

1 Answer 1


Counterexample: Let $M$ be the round sphere $S^2$, let $N$ be the equator. Let $c$ start from the equator, going nearly north, miss the north pole and come back again. The normal projection $\alpha(t)$ to the equator is along longitudinal lines. The projection moves first very slow along the equator. If $c(t)$ is near the north pole, $\alpha(t)$ moves fast, then it slows down again.

  • $\begingroup$ @ Peter Michor, I am Sorry. I did not give an detail. I fix the OP $\endgroup$ Mar 18, 2016 at 0:09

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