Timeline for Convex caps with prescribed edges
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Oct 12, 2017 at 3:32 | history | edited | Mohammad Ghomi | CC BY-SA 3.0 |
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| Sep 27, 2017 at 18:36 | history | edited | Mohammad Ghomi | CC BY-SA 3.0 |
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| Sep 22, 2017 at 0:48 | history | edited | Mohammad Ghomi | CC BY-SA 3.0 |
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| Sep 21, 2017 at 22:29 | vote | accept | Mohammad Ghomi | ||
| Sep 21, 2017 at 18:41 | answer | added | Joseph Malkevitch | timeline score: 1 | |
| Sep 21, 2017 at 18:39 | history | edited | Mohammad Ghomi |
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| Sep 21, 2017 at 17:12 | comment | added | Mohammad Ghomi | OK, maybe then a better assumption to rule out the trivial or degenerate cases would be 3-connected. I have now edited the question accordingly. Thanks. | |
| Sep 21, 2017 at 17:10 | history | edited | Mohammad Ghomi | CC BY-SA 3.0 |
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| Sep 21, 2017 at 17:04 | comment | added | Ivan Izmestiev | The problem that Fedor has pointed at still remains. Consider now a pentagon with one interior vertex and three interior edges that subdivide it into two triangles and a convex quadrilateral. This subdivision cannot be lifted. | |
| Sep 21, 2017 at 16:55 | comment | added | Mohammad Ghomi | In the degenerate case there are really no edges. I edited the question to specifically stipulate the existence of an interior vertex in the subdivision, so that the degenerate case does not arise. Thanks for pointing this out. | |
| Sep 21, 2017 at 16:53 | history | edited | Mohammad Ghomi | CC BY-SA 3.0 |
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| Sep 21, 2017 at 16:40 | comment | added | Ivan Izmestiev | The graph of a convex cap is always 3-connected. So, I think the question is interesting only when the graph of the subdivision is 3-connected. | |
| Sep 21, 2017 at 16:13 | comment | added | Fedor Petrov | But for degenerated cup there is no difference between two diagonals. If it counts in this situation, why it does not count for inner vertices too? | |
| Sep 21, 2017 at 15:44 | comment | added | Mohammad Ghomi | In that case the function would be zero and the cap would be degenerate. The question is interesting only when the partition has some vertices in the interior of the polygon. | |
| Sep 21, 2017 at 15:37 | comment | added | Fedor Petrov | Consider a convex quadrilateral partitioned by a diagonal. What is a function? | |
| Sep 21, 2017 at 15:15 | comment | added | Mohammad Ghomi | We can think of the cap as the graph of a piecewise linear concave function over P with zero boundary values. So it looks like a dome over P. Further the cap is a topological disk, so its boundary is just the topological boundary of the disk. | |
| Sep 21, 2017 at 12:40 | comment | added | Fedor Petrov | What do you mean by a 'cap' and its 'boundary'? | |
| Sep 21, 2017 at 8:50 | answer | added | Ivan Izmestiev | timeline score: 6 | |
| Sep 21, 2017 at 2:30 | history | asked | Mohammad Ghomi | CC BY-SA 3.0 |