8
$\begingroup$

Let $X$ be a finite dimensional Alexandrov space with curvature bounded below and non-empty boundary. Let $\gamma$ be a shortest geodesic path in $X$ whose endpoints belong to the interior of $X$.

Is it true that $\gamma$ is contained in the interior of $X$? Is that true at least under the assumption that $X$ is smooth Riemannian manifold with smooth boundary (thus $X$ must be locally geodesically convex)?

$\endgroup$
1
  • $\begingroup$ The case when $X$ is a smooth compact domain in a Riemannian manifold follows from elementary arguments once you know that $X$ is geodesically convex (which follows from the assumption that $X$ is an Alexandrov space as I mention in the comments to Michor's answer). Namely, if $[x,y]$ is a segment in $X$ with $x,y$ in the interior, then the segments from $x$ to points of a small neighborhood of $y$ fill a neighborhood of $(x,y)$, and since by convexity of $X$ all such segments must lie in $X$, no point of $(x,y)$ lies in $\partial X$. $\endgroup$ Jul 14, 2020 at 2:29

1 Answer 1

1
$\begingroup$

No: Take a banana shaped domain with smooth boundary in the plane.

Edit: This answer is wrong. I leave it here for the very instructive comments by Igor Belegradek.

I think I see geodesic triangles near the concave boundary in the banana where Gauss-Bonet implies large negative curvature.

$\endgroup$
7
  • 3
    $\begingroup$ The banana is not an Alexandrov space. $\endgroup$ Jul 12, 2020 at 21:17
  • $\begingroup$ @IgorBelegradek Why is the banana not an Alexandrov space? It seems to me that by compactness, Toponogov's theorem should hold possibly with a very very negative lower curvature bound. $\endgroup$
    – Sak
    Jul 13, 2020 at 16:02
  • $\begingroup$ @Sak: see mathoverflow.net/questions/223934/…. $\endgroup$ Jul 13, 2020 at 17:34
  • $\begingroup$ @IgorBelegradek wow, the more you know! Thanks! $\endgroup$
    – Sak
    Jul 13, 2020 at 18:00
  • 3
    $\begingroup$ I wish to add that a smooth compact domain in a Riemannian manifold must be convex if the intrinsic metric on the domain is $CD(K,\infty)$, see Theorem 1.5 in arxiv.org/abs/1902.00942. This applies to every finite-dimensional Alexandrov space by arxiv.org/abs/1003.5948. $\endgroup$ Jul 14, 2020 at 1:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.