0
$\begingroup$

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

$\endgroup$
7
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ @ Peter Michor, I am Sorry. I did not give an detail. I fix the OP $\endgroup$ Mar 18, 2016 at 0:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.