# is the closure of a totally convex set again totally convex?

Recall that a totally convex subset $C$ of a complete Riemannian manifold $M$ is a set which contains with any two points $p,q$ also all the geodesics between them.

We know that there is a totally geodesic, totally convex submanifold $N\subset M$ such that $N\subset C \subset \bar N$. So the question is: Is $\bar N$ totally convex?

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What about an open half sphere? – Piero D'Ancona Dec 3 '10 at 13:22
@Piero: an open half sphere is not totally convex. We want that all geodesics between points to be contained in the subset and not only the minimizing geodesics. – Luc Dec 3 '10 at 13:32
I'm sorry, I'm dumb. How do you get $N$? – Theo Buehler Dec 3 '10 at 14:04
@Theo: consider $\cal N$ the family of all the submanifolds of $M$ contained in $C$. $\cal N$ is nonempty (e.g. points in $C$ are submanifolds of dimension 0). And let $k$ be the maximal dimension of a submanifold in $\cal N$. Then $N$ is the union of all submanifolds in $\cal N$ of dimension $k$. The reference for this would be: J. Cheeger and D. Gromoll, On the structure of complete manifolds of nonnegative curvature. – Luc Dec 3 '10 at 14:22
The answer is definitely yes if $N$ has top dimension, i.e. $dim(N)=dim(M)$. Indeed, any geodesic outside $N$ that joins points of $\partial N$ can be extended to a geodesic joining points of $N$, and hence the geodesic must lie in $C$. – Igor Belegradek Dec 3 '10 at 17:32

No. Define a Riemannian metric tensor $g$ on $\mathbb R^2=\{(x,y)\}$ by $$g(x,y) = \begin{pmatrix} 1 & 0 \\ 0 & f^2(x) \end{pmatrix}$$ where $f:\mathbb R\to\mathbb R$ is a positive smooth even function such that $f(x) = \cos x$ for $|x|\le 1$ and $f''(x)/f(x)$ increases after $x=1$. Let $N$ be an open segment of length $\pi$ in the $y$-axis, e.g. the one between points $A=(0,0)$ and $B=(0,\pi)$.
(The plane with this metric is isometric to the universal cover of a surface of revolution that looks like a unit sphere with two infinite tubes attached near a pair of opposite points. Note that the Gaussian curvature $K$ is given by $K=-f''/f$, so $K\le 1$ everywhere.)
The strip $\{|x|\le 1\}$ is isometric to the universal cover of a neighbourhood of the equator of the standard sphere, so there are plenty of geodesics between $A$ and $B$. On the other hand, using Clairot integral and the fact that the Gaussian curvature does not exceed 1, it is easy to see that no geodesic can intersect the $y$-axis at two points with distance less than $\pi$ between them. Hence $N$ is totally convex but its closure is not.
I'm sorry to ask. My intuition tells me it should be clear. I really tried to get the computation but I didn't succeed. Could you give another hint in how to conclude that no geodesic can intersect the $y$-axis at two points with distance less than $\pi$? – Luc Dec 10 '10 at 14:29