Let $X$ be a compact Alexandrov space with $curv\geq 1$ (and without boundary). Does $X$ always have a nontrivial compact convex subset without boundary?

Definition of a convex subset: $A\subseteq X$ is called convex if for every two points $p ,q\in A$, there exists a minimizing geodesic between them which is completely in $A$.


This is extremally rare, even if $X$ is a Riemannian manifold.

If the convex set $A$ has interior points, then any boundary point of the subset $A$ in $X$ lies on the boundary of Alexandrov space $A$. So if $X\ne A$ then $\dim A<\dim X$.

Note that $A$ has to be totally geodesic, otherwise an end of geodesic would be a boundary point of $A$. Generic Riemannian manifold do not have totally geodesic submanifolds of dimension at least 2.

So, we are left with the case $\dim A=1$. In other words, $A$ is a closed geodesic in $X$. Since $A$ is convex, it has to be length minimizing on each half. This is also extremally rare thing --- generic Riemannian manifold does not have such geodesics.

  • $\begingroup$ @AntonPetrunin.That is true, but by boundary, I mean the boundary of an Alexandrov space since the convex subset is an Alexandrov space. I, indeed, mean something like closed geodesics in the compact Riemanninan manifolds. $\endgroup$
    – Jayq
    Dec 27 '18 at 17:47
  • $\begingroup$ @Jayq Both types of boundaries (relative to $X$ and intrinsic boundary of $A$) coincide in this case (since $X$ has no boundary). This is indeed requires a proof, it can be done by induction on dimension; you need to show that space of directions $\Sigma_pA$ forms a closed convex set in $\Sigma_pX$. $\endgroup$ Dec 27 '18 at 18:39
  • $\begingroup$ This just seems to be true if $A$ is open, or what am I understanding wrong here? Take for example an equator as A inside a round sphere. $\endgroup$ Mar 5 '19 at 17:12
  • $\begingroup$ @BruceWayne yes sure, now the answer is updated. Thank you. $\endgroup$ Mar 5 '19 at 18:41

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