# On nontrapping manifolds

Suppose that $$(M,g)$$ is a compact connected smooth Riemannian manifold without boundary. Let $$U \subset M$$ be a smooth submanifold of codimension zero with smooth boundary and assume that $$U$$ is nontrapping, meaning that all inextendible unit speed geodesics eventually exit the set $$U$$. Does there exist a point $$p \in M\setminus U$$ such that $$A(p) \subset M\setminus U$$?

Here, $$A(p)$$ is the antipodal set of $$p$$ defined by $$A(p)=\{ q\in M\,:\, \textrm{dist}_g(q,p)\geq \textrm{dist}_g(y,p) \quad \forall\, y\in M\}.$$

• If I understood the question correctly, the manifold $M=\mathbb{S}^1$ with the geodesic metric, and the set $U$ equals $\mathbb{S}^1$ minus a small open interval, is a counterexample (I posted it as a comment in case the OP wanted to modify the question) Commented May 7 at 1:53
• @SaúlRM oh dang, that's much easier than the one I posted. Commented May 7 at 1:54

Start with the line segment $$\{(x,y) : x\in [1,1000000], y = \epsilon x\}$$ for some tiny $$\epsilon$$. Form the surface by rotating the segment about the $$x$$ axis. This makes a section of a very pointy cone. Cap it off smoothly on both ends to make it convex, while keeping it still a surface of revolution. Call this your $$M$$.
Take your $$U$$ to be all of the cone section plus the cap on the small end. As a surface of revolution it is easy to check that $$U$$ is non-trapping. But every point on the large cap, by virtue of the cone being very long and pointy, has its antipodal set entirely contained in $$U$$.