Question

Suppose $M^n$ is an open complete nonnegatively curved Riemannian manifold. In Cheeger-Gromoll's proof of the soul theorem. They need an estimate on the cut radius of a totally convex set $C$. By a totally convex set $C\in M^n$, we mean for any two point $a, b\in C$ and any geodesic joining them must lie in $A$. In order to define a retraction from $^aC$ to $C$, where $^aC$ is defined by $$ ^a C=x\in M, d(x, C) < a.$$ Here $a$ must be small enough such that for any $x\in ^aC\setminus C$ there is a unique point $h(a)\in C$ such that $d(x, C)=d(x, h(x))$.

I am wondering whether there is a example showing that when $a$ is large, the projection $h$ is not well defined, i.e. there are two points $p, q\in C$ such that $d(x, p)=d(x, q)=d(x, C)$.

Answer 1

Let $M$ be the union of the unit northern hemisphere centered at the origin of $R^3$ with the cylinder $\{x^2+y^2=1\,,\, z\leq 0\}$ and $C$ be the geodesic segment $\{x^2+z^2=1\,,\, y=0\,,\, \vert x\vert\leq 1/2\}$. The point $P=(0,1,0)$ is at the same distance from every point of the segment $C$.
The metric here is only continuous but it seems to me that "cutting" the hemisphere a little "over" the equator and "gluing" a truncated cone you have a similar example, this time with some point $P$ on the cone which is at equal distance to the two ends of the segment $C$. Moreover, this property should be stable "smoothing" the metric in a thin strip around the gluing circle.

Answer 2

let M be the infinite cylinder ${(x,y,z) | x^2+y^2=1 }$, choose as $C$ a small geodesic disc around some point $p$ and $q$ the point on the opposite site of the circle with the same z coordinate. Then the distance between $q$ and $C$ is realized by two different paths going around the circle in different directions and these paths end in different points in $C$.