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Let $(M,g)$ be a Riemannian manifold and $S\subset M $ open and strongly convex, i.e. any two points $p,q\in \bar S$ are connected by a unique minimizing geodesic with interior completely lying in $S$.

Conjecture: There exists an open neighbourhood $S'$ of $\bar S$ which is also strongly convex.

New question (March 20): It seems like the conjecture is false in general and user Dap has suggested a counterexample as an answer. Can you find another counterexample, maybe a simpler one?

Some remarks and proof attempts:

For sets which are merely convex this is not true. For example an open hemisphere of $S^2$ cannot be enlarged without losing convexity. The strongly convex sets of $S^2$ on the other hand have their closure strictly contained in a hemisphere and thus can obviously be enlarged to a new strongly convex set. This picture (respectively the ones that I can image on other surfaces) leads me to the assumption that the question has a positive answer.

Strongly convex sets are totally normal, (i.e. for every point $p\in S$ there is an open set $V\subset T_pM$ such that $\exp_p\colon V\rightarrow S$ is a diffeomorphism). My first idea was to first find an $S''\supset \bar S$ totally normal and then try to restict it to a strictly convex set. But just working with total normality cannot work: $S^2 \backslash p_0$ is totally normal, but cannot be extended to a larger totally normal set. Unfortunately I don't know how strong convexity can be incorporated from the start.

Edit (March 17): Here is a false counterexample, that was posted as an (now deleted) answer. I find it interesting neverletheless: Let $M=\{(x,y)\vert y>-e^x\}\subset \mathbb{R}^2$ and $S$ the open upper half plane (which is only convex and not strongly convex). If $S'$ is a open set containing $\bar S$, then $(0,-\epsilon)\in S'$ for some $\epsilon >0$. Take $x> 0$ so large that $\epsilon> xe^{-x+1}$, then the straight line between $(-x,0)$ and $(0,-\epsilon)$ contains the point $(-x+1,-\epsilon/x)$, which lies outside of $M$. In particular $S'$ cannot be convex.

Jan Bohr
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I believe you can get an example from taking $M$ to be the elliptic paraboloid $z=x^2+y^2,$ with the Riemannian metric coming from $\mathbb R^3,$ and $S=\{p\in M\mid p_z<C\}$ for a particular constant $C.$

$C$ needs to be chosen so that there are conjugate points lying on the boundary of $S.$ There is a description of the conjugate points at "Example for conjugate points with only one connecting geodesic" and a description of the geodesics in do Carmo's "Differential Geometry of Curves and Surfaces", in terms of Clairaut's relation. The sets $\{p\in M\mid p_z\leq c\}$ are geodesically convex for any $c,$ with the minimizing geodesics going via $p_z<c,$ and the minimizers are unique for $c\leq C.$ As discussed in that MSE answer I linked to, this is an example where there are conjugate points not coming from a family of geodesics.

To show the strong convexity:

  • By Clairaut's relation, if we represent a non-meridian geodesic in co-ordinates $(r(t)\cos \theta(t),r(t)\sin\theta(t), r(t)^2),$ then $\theta(t)$ can be chosen to be an increasing function, and $dr/d\theta$ is some analytic function $f(r,\theta,c).$

  • By the identity theorem for analytic functions, there cannot be a continuous family of geodesics passing through two points.

  • A minimizing geodesic behaves sensibly in terms of longitudes. If a geodesic goes from polar co-ordinates $(r_1,\theta_1)$ to $(r_2,\theta_2)$ with $\theta_2>\theta_1+\pi,$ then by reflecting in the meridian going through longitudes $\theta_1,\theta_1+\pi,$ we can get a new curve that stay within the range of angle $[\theta_1,\theta_1+\pi].$ By smoothing this curve slightly we get a shorter curve. So the original geodesic could not have been minimizing. This means it makes sense to represent geodesics in polar co-ordinates.

  • Suppose for contradiction that there are distinct points $p,q$ in $\overline{S}$ joined by two minimizing geodesics. Pick $p_z<C$ if possible, and if the two points can be joined by a meridian then pick this as one of the geodesics - this allows a consistent choice of polar co-ordinates. Starting from $p,$ shoot a geodesic at an angle bisecting the two geodesics. This will hit one of the original two geodesics. This means we get a new pair of geodesics between some two points. The new endpoint must be $q,$ because otherwise we would get a pair of geodesics between two points contradicting the choice of $C,p,q.$ This kind of construction produces a family of geodesics between $p,q,$ contradicting the fact that there are no continuous families of geodesics joining $p,q.$

Dap
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  • Thank you for this example, it already gives me some insight on what can go wrong. Unfortunately the upper half plane is not strongly convex in the sense of the question: The interior of the straight line which connects two points on the boundary (which is the $x$-axis), does not lie in the open upper half plane. – Jan Bohr Mar 17 '18 at 12:03
  • @jabo: I've changed my answer to an example that should actually be strongly convex – Dap Mar 17 '18 at 16:50
  • This indeed feels like it should be correct. Unfortunately the paper does not give a precise description of conjugate points, but the picture on pg.3 suggests that lowering $p$ will increase $q$, which would mean that there is a unique height $C$ at which $p$ is level with $q$. – Jan Bohr Mar 18 '18 at 14:09
  • Then if I understand your reasoning correctly, you say that for this height $C$, the set $S$ will be strongly convex (because the conjugate points $p$ and $q$ do not come from two geodesics), but any larger set automatically contains two conjugate points which are connected by more than one geodesic. – Jan Bohr Mar 18 '18 at 14:09
  • From the information given in the paper respectively the MSE answer that you've linked to I cannot conclude whether this is actually true. I would now sit down and try to compute this height $C$ and see if I learn something from that. Or can you suggest a better approach to make this argument rigorous? – Jan Bohr Mar 18 '18 at 14:12
  • @jabo: I don't think you have to do any calculations. I've added a sketch proof that I think can be used to make the argument rigorous. I only know the discussion in do Carmo's book, but maybe there is some better literature. – Dap Mar 19 '18 at 18:28
  • Thanks already for your efforts! I have awarded the bounty, though I need some more time to meditate about your answer. I will accept it once I understand it and in the mean time give other people the chance to maybe provide a simpler example. – Jan Bohr Mar 20 '18 at 16:36