4
$\begingroup$

I just posted this question as a comment to the question Hypersurfaces and Elliptic Points but I don't know how many people will see it.

It's well known and easy to prove that a point on a closed hypersurface in $\mathbb{R}^n$ that has maximal distance from the origin has nonnegative second fundamental form.

However, this is apparently not necessarily true, if the hypersurface is only PL (piecewise linear). This raises the following question: What does the hypersurface look like near a point of maximal distance?

Improve this question
$\endgroup$
3
  • $\begingroup$ It occurs to me that I think I know the answer. I imagine that an extreme point would be a vertex just like the extreme vertices of Connelly's flexible polyhedron. One where there are several edges (which should be viewed as hinges) meeting a vertex and the faces "zig zag" around the vertex, which allows them to be flexible under bending along the hinges. The curvature, I would guess, is zero at the vertex. $\endgroup$
    – Deane Yang
    Jul 19, 2014 at 16:04
  • 1
    $\begingroup$ If curvature is defined as $2 \pi$ minus the sum of the incident face angles, I see no reason this need be zero. E.g., one could arrange $4 \pi$ incident angle with triangles in a ruff-like pattern. $\endgroup$
    – Joseph O'Rourke
    Jul 19, 2014 at 17:07
  • $\begingroup$ Oh, I see Anton made the same remark. $\endgroup$
    – Joseph O'Rourke
    Jul 19, 2014 at 17:08

1 Answer 1

Reset to default
3
$\begingroup$

It is easy to see that point of maximal distance has to be a vertex. Further, all the edges starting from this vertex point in the directions of open half-space.

That is it — any vertex with the described property can appear as a point of locally maximal distance. (For global maximum, in addition it has to be a vertex of the convex hull of the surface.)

The angle around such vertex can be bidder than $2{\cdot}\pi$, but in this case the surface can not be locally convex at this vertex.

Improve this answer
$\endgroup$

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.