**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



Projection of geodesic is geodesic
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. 


