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 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

• 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. Mar 17, 2016 at 16:05
• I add some condition for well-definedness of $\alpha$ And this is just my curiosity. Mar 17, 2016 at 16:14
• And thank you for your reply and reference Mar 17, 2016 at 16:14
• 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 $\nabla_{\dot{\alpha}}\dot{\alpha} \propto \dot{\alpha}$? Mar 17, 2016 at 20:33
• @Willie Wong Sorry. I did not give an detail. I fix the OP Mar 18, 2016 at 0:08

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.