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By Alexandrov's isometric embedding theorem, any locally convex metric prescribed on the sphere admits a realization as a convex surface in Euclidean 3-space, which, by Pogorelov's rigidity result, is unique up to an isometry of the ambient space.

Thus, the intrinsic geometry of a convex surface completely determines its extrinsic geometry, and in principle it should be possible to describe all geometric quantities in terms of the intrinsic metric. On the other hand, Alexandrov's proof is not constructive, and does not provide any hints as to how one might be able to do this. Hence my first question is:

Question 1: Is there any way to compute, recognize, or characterize intrinsically any of the geometric quantities or features of a convex surface which are not invariant under local isometric deformations, e.g., mean curvature, principal directions, principal curvatures, or umbilic points?

I recognize this might be too much to ask, because the procedures which the above question asks for would somehow need to take into account the whole metric, not just a neighborhood of a point. So let me try to make things a bit more concrete and maybe a little more accessible.

It is well known that any (smooth) convex surface must have at least one umbilic, i.e., a point where the principal curvatures are equal (if not then the direction of the larger principal curvature yields a nonvanishing line field on the surface which violates the Poincare-Hopf index theorem). Can one prove this without reference to the ambient space, or quantities that we only know how to compute extrinsically:

Question 2: Is it possible to prove the existence of an umbilic point of a smooth convex surface in a purely intrinsic way? If so, can one also find or approximate the location of the umbilics?

One motivation behind this questions is the famous conjecture of Caratheodory, which states that each convex surface must have at least two umbilics. The extrinsic or local approaches to this problem have always been problematic.

Adendum: As Robert Bryant correctly points out below, the topological argument for the existence of at least one umbilic described above is really intrinsic and independent of the notion of principal directions. So the main point of Question 2 is its second part, i.e., to somehow get a handle on the location of the umbilic. See also this related question.

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But the existence of an umbilic point on the sphere follows from topological considerations: The sum of the Hopf indices of the umbilics is 1 (by a theorem of Hopf) so there has to be at least one umbilic.

Put another way: If there were no umbilics, then union of the principal directions at each point would define a 4-fold covering space of the sphere, which would, because the sphere is simply connected, have a section, which would give a nowhere vanishing vector field on the sphere. Thus, there must be at least one umbilic. I don't see why this isn't 'intrinsic'.

As for locating the umbilics, I don't see how you could expect to do that in general.

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  • $\begingroup$ Yes, I referenced the topological argument above, but I guess that I was not including that as "intrinsic" because it refers to principal directions which are not intrinsic. The hope (and I understand that it could be a pretty long shot) would be to recognize the umbilic in terms of the metric. $\endgroup$ – Mohammad Ghomi Oct 16 '17 at 1:25
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    $\begingroup$ But isn't this really a matter of semantics? The umbilics are the points where the quadratic form $Q$ that satisfies the Gauss and Codazzi equations (which are 'intrinsic' since they depend only on the metric $g$, interpreted as the first fundamental form) is a multiple of $g$. (Globally on $S^2$, $Q$ is unique up to sign when the Gauss curvature of $g$ is positive.) You don't have to call the eigenvectors of $Q$ relative to $g$ 'principal directions' if you don't want to. Since there are many local solutions to the Gauss and Codazzi equations, there is no local argument forcing umbilics. $\endgroup$ – Robert Bryant Oct 16 '17 at 11:09
  • $\begingroup$ OK, fair enough. I agree that if we look at it this way, then one can establish the existence of at least one umbilic in a purely intrinsic way. $\endgroup$ – Mohammad Ghomi Oct 16 '17 at 11:18
  • $\begingroup$ Hi @Robert, would you know the answer to this (in the noncomplete case)? $\endgroup$ – Mikhail Katz Oct 19 '17 at 11:04
  • $\begingroup$ @MikhailKatz: Yes, I do know the answer. I'll enter it there. $\endgroup$ – Robert Bryant Oct 20 '17 at 11:34
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"Umbilic points" have no obvious meaning in Alexandrov's framework (since his theory is not restricted to smooth surfaces), but there has been a fair amount of progress on making the theory more effective (it was always constructive: you could approximate the metric by a polyhedral metric. A polyhedral metric could be embedded in finite time [slowly, true], and then results would converge to the embedding of the surface you started with. The speed of convergence was already addressed in the '50s by A. Volkov). The last word on effectivizing Alexandrov for polyhedra is:

Bobenko, Alexander I.; Izmestiev, Ivan, Alexandrov’s theorem, weighted Delaunay triangulations, and mixed volumes, Ann. Inst. Fourier 58, No. 2, 447-505 (2008). ZBL1154.52005.

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  • $\begingroup$ Maybe then the right question to ask is: what is an "umbilic point" of a convex polyhedron? $\endgroup$ – Mohammad Ghomi Oct 16 '17 at 2:26
  • $\begingroup$ @MohammadGhomi I am pretty sure there is no good definition of such, in particular, there is no definition that would be stable under Hausdorff convergence, which is what you need for this. $\endgroup$ – Igor Rivin Oct 16 '17 at 2:31
  • $\begingroup$ On the other hand, since there are four-vertex theorems for polygons, why there shouldn't be umbilic points theorems for polyhedra? $\endgroup$ – Ivan Izmestiev Oct 16 '17 at 7:23
  • $\begingroup$ A tentative definition: a vertex of a polyhedron is umbilic if its spherical link has more than four "curvature extrema" (which are called vertices for smooth curves). $\endgroup$ – Ivan Izmestiev Oct 16 '17 at 7:27
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    $\begingroup$ @Mohammad: Yes, you are right. Also, a vertex of degree four should be considered umbilic if the incident edges lie on a circular cone with the apex at the vertex. $\endgroup$ – Ivan Izmestiev Oct 16 '17 at 12:10

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