Let $M^n$ be the boundary of a convex compact set in $\mathbb{R}^{n+1}$ with non-empty interior.

Question 1. Is $M$ geodesically complete, i.e. is it true that every geodesic (= locally shortest path) can be extended infinitely in both directions? Will it still be true if the convex set is not necessarily compact, but only closed.

Question 2. Is it true that every shortest path on $M$ has both left and right first derivatives at every point?

UPDATE: The answer to Question 1 is NO, as explained by Igor Rivin below.

UPDATE: The answer to Question 2 is YES, as claimed by John Harvey below. Originally it was proved by I.M. Liberman in "Geodesic lines on convex surfaces", C. R. (Doklady) Acad. Sci. URSS (N.S.) 32, (1941). 310–313. The proof for 2-dimensional hypersurfaces is also reproduced in the book "Intrinsic geometry of convex surfaces" by A.D. Alexandrov, see Ch. IV, $\S$ 6, Theorem 1.


The answer to the question as stated is NO. Indeed, consider (the boundary of) a tetrahedron (or any convex polyhedron). Since the cone angles at the vertices are smaller than $2\pi,$ no geodesic can be continued past a vertex (since then it would subtend an angle of smaller than $\pi$ on at least one side, and so would not be locally shortest.)


As for question 2, the tangent cone of $M$ can be defined as the cone on the space of directions, and the space of directions is the completion of the space of geodesic directions. Therefore every geodesic is represented in the cone.

On the other hand, the cone could be defined as a limit object by rescaling $M$ around the point, and this is the definition which we use when talking about derivatives.

$M$ is an Alexandrov space, and so by Theorem 7.8.1 of the Burago--Gromov--Perelman paper, these two definitions coincide. So this theorem shows that shortest paths have one-sided derivatives.

Probably this was already known for convex hypersurfaces without having to use the full generality of Alexandrov geometry, but I'm not sure where you would go to find that.

  • $\begingroup$ Could you please elaborate a little bit. It seems that both approaches to the tangent cone you mentioned are purely intrinsic. But derivative depends on the imbedding to Euclidean space (unless you use some notion of covariant derivative which I am not aware of). $\endgroup$ – orbits Jun 9 '15 at 10:09
  • $\begingroup$ If you rescale all of $\mathbb{R}^{n+1}$, then you will find the tangent cone of $M$ appearing as a subcone of $\mathbb{R}^{n+1}$. This gives you a one-sided tangent vector for the geodesic as a function with range $\mathbb{R}^{n+1}$ $\endgroup$ – John Harvey Jun 9 '15 at 11:29

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