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Hee Kwon Lee
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Projection of geodesic is geodesic

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 normal 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

Hee Kwon Lee
  • 1.1k
  • 1
  • 9
  • 25