Timeline for If a triangle can be displaced without distortion, must the surface have constant curvature?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Oct 11, 2023 at 0:18 | answer | added | Vladimir Zolotov | timeline score: 2 | |
| Jun 21, 2018 at 4:35 | history | edited | Mohammad Ghomi | CC BY-SA 4.0 |
added 8 characters in body
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| Jun 19, 2018 at 1:04 | comment | added | Robert Bryant | @JosephO'Rourke: However, your question only states "Let $T$ be a geodesic triangle", not "Let $T$ be any geodesic triangle". If any geodesic triangle can be copied without 'distortion', then, sure, the Gauss curvature has to be constant. This is an easy consequence of geodesic normal coordinates. I thought you were asking about the harder problem of knowing only that you can copy a specific 'congruence class' of geodesic triangles without distortion. | |
| Jun 18, 2018 at 23:35 | vote | accept | Joseph O'Rourke | ||
| Jun 18, 2018 at 23:35 | |||||
| Jun 18, 2018 at 23:33 | comment | added | Joseph O'Rourke | @Acccumulation: I meant for $T$ to be an arbitrary triangle, not just one specific triangle. So $\forall T$. | |
| Jun 18, 2018 at 22:54 | answer | added | Acccumulation | timeline score: 0 | |
| Jun 18, 2018 at 22:45 | comment | added | Acccumulation | There is a slight ambiguity as to whether you are asking whether "$\exists T: T' \equiv T \rightarrow C$ is constant" or "$\forall T: T' \equiv T \rightarrow C$ is constant". | |
| Jun 18, 2018 at 15:16 | comment | added | Robert Bryant | If by 'moved around' you mean that there are intrinsic isometries of the surface $S$ that allow you to move a given vertex of $T$ to any other point of the surface, then, yes, the surface has constant Gauss curvature. This follows because the group of intrinsic isometries preserves the Gauss curvature, and your 'move around arbitrarily' hypothesis would then imply that the Gauss curvature must be the same at any two points. It would be better to put conditions on the set of triangles in $S$ congruent to $T$, i.e., that there should be 'enough' of them in an appropriate sense. | |
| Jun 18, 2018 at 14:17 | answer | added | Robert Bryant | timeline score: 7 | |
| Jun 18, 2018 at 11:03 | comment | added | Mikhail Katz | @Gro-Tsen, depending on the interpretation this may not be true in higher dimensions; see this answer. | |
| Jun 18, 2018 at 10:46 | answer | added | Mikhail Katz | timeline score: 5 | |
| Jun 18, 2018 at 10:18 | comment | added | Joseph O'Rourke | @Gro-Tsen: Yes, I believe you are correct: should be true in $\mathbb{R}^n$. | |
| Jun 18, 2018 at 9:51 | comment | added | Gro-Tsen | If this is true for surfaces, then it should be true in arbitrary dimension, right? Because it would imply constant sectional curvature. | |
| Jun 18, 2018 at 7:58 | comment | added | Ali Taghavi | Is isometryic congruent equivalent to "Edge lengths the same, vertex angle the same", as you wrote in your question?And does this equivalency characterize non negative constant curvature? | |
| Jun 18, 2018 at 3:07 | answer | added | Zurab Silagadze | timeline score: 14 | |
| Jun 18, 2018 at 2:19 | answer | added | j.c. | timeline score: 5 | |
| Jun 17, 2018 at 23:17 | history | asked | Joseph O'Rourke | CC BY-SA 4.0 |