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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 $d(x, p)=d(x, q)=d(x, C)$.

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What about a small geodesic ball around the north pole and the south pole in a sphere? –  Carlo Mantegazza Jan 18 '13 at 12:45
    
@Carlo, Compact manifold cannot have totally convex subset. –  Ralph Jan 18 '13 at 13:14
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No such examples exist, it follows form the Perelman's proof of soul conjecture. –  Anton Petrunin Jan 18 '13 at 20:12
    
@Anton, Thanks for the confirmation. That is what I thought should be true. But, on the other hand my question seems has nothing to do with the dimension of the soul. It is about the uniqueness of the projection from $x\in ^aC$ to $\partial C$ which minimize the distance. I tried to find a direct proof without using any concepts of injective radius or convex radius. I played with triangle comparison but no luck. So do you know a direct proof? (it seems for the soul=pt case, there is no flat strip but the projection is still unique for large $a$, right? –  Ralph Jan 18 '13 at 22:17
    
@Ralph, It seems that your question is just as hard as the soul conjecture, so I do not think there is a more direct proof. On the other hand, the proof of soul conjecture is simple and short, so it is not clear what could be better. BTW, why are you talking about dimension of soul (?) –  Anton Petrunin Jan 19 '13 at 2:11
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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.

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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 $C$.

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@Ralf, The set $C$ in your construction is not totally convex, hence does not serve as a counter example. All compact totally convex sets in the cylinder are always look like a cylinder with finite height. –  Ralph Jan 18 '13 at 19:04
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