Timeline for Intrinsic vs Extrinsic geometry of convex surfaces
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 17, 2017 at 18:51 | comment | added | Ivan Izmestiev | @IgorRivin No, why boatloads? A generic spherical quadrilateral is not circular, so umbilic vertices of degree four are rare. But on the other hand, a generic octahedron will have no umbilics. So, at the moment I don't see anything promising. | |
| Oct 17, 2017 at 18:06 | comment | added | Igor Rivin | @IvanIzmestiev This seems to introduce boatloads of "umbilics", I am not sure this is the right definition.... | |
| Oct 16, 2017 at 12:10 | comment | added | Ivan Izmestiev | @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. | |
| Oct 16, 2017 at 11:39 | comment | added | Mohammad Ghomi | @Ivan: Or maybe more or less than four curvature extrema (so every vertex of degree 3 would also count as an "umbilic"). | |
| Oct 16, 2017 at 7:27 | comment | added | Ivan Izmestiev | 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). | |
| Oct 16, 2017 at 7:23 | comment | added | Ivan Izmestiev | On the other hand, since there are four-vertex theorems for polygons, why there shouldn't be umbilic points theorems for polyhedra? | |
| Oct 16, 2017 at 2:31 | comment | added | Igor Rivin | @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. | |
| Oct 16, 2017 at 2:26 | comment | added | Mohammad Ghomi | Maybe then the right question to ask is: what is an "umbilic point" of a convex polyhedron? | |
| Oct 16, 2017 at 2:20 | history | answered | Igor Rivin | CC BY-SA 3.0 |