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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
up vote 6 down vote accepted

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.

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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
Suppose it does. Then the intersection points are not conjugate along the vertical segment, so you can vary it (as a geodesic) by varying the endpoints. Let the endpoints go to each other along this supposed second geodesic. The family of geodesics defined by these endpoints (started from the vertical segment) has decreasing length, because they form sharp angles with that second geodesic. Hence this family will be free of conjugate points for all time, and you'll get an arbitrarily short pair of distinct geodesics with the same endpoints, a contradiction. – Sergei Ivanov Dec 10 '10 at 15:10

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