Timeline for Symmetry of points on unit sphere determined by relation between triples of points
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Mar 3, 2021 at 10:12 | comment | added | Ted Burgess | With two equilateral triangles, each triangle obviously rotates onto itself. So two of the 6! triples of points satisfy the constraint. How about triples composed of two corners of one triangle and one corner of the other? | |
| Mar 3, 2021 at 10:05 | comment | added | user114668 | In this case, could you have an equilateral triangle at the north pole and one at the equator, in which case $X$ has no rotational symmetry but each triangle is related to itself in different order? | |
| Mar 3, 2021 at 8:01 | comment | added | Ted Burgess | Thanks for your question. The triple $j_{1}, j_{2}, j_{3}$ might differ from $l_{1}, l_{2}, l_{3}$ only in order. The ordered triples are distinct. I've updated the question to make this point clearer. | |
| Mar 3, 2021 at 7:59 | history | edited | Ted Burgess | CC BY-SA 4.0 |
Clarified points raised in the comments
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| Mar 2, 2021 at 22:44 | comment | added | user114668 | Is "at least one other triple" three different indices or at least one different index? | |
| Mar 2, 2021 at 16:59 | comment | added | LSpice | Oh, I see. Sorry, I thought you meant full rotational symmetry. Of course I now see that any rotational symmetry is good enough. | |
| Mar 2, 2021 at 15:59 | history | edited | Ted Burgess | CC BY-SA 4.0 |
edited title
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| Mar 2, 2021 at 15:12 | comment | added | Ted Burgess | Thanks for the comment. The hexagon corresponding to the six points you described has two-fold rotational symmetry about an axis passing through $r_{1}$. I would like to know whether satisfying the constraint always implies some sort of rotational (perhaps also reflection) symmetry of the points. | |
| Mar 2, 2021 at 15:04 | comment | added | LSpice | Take two points $r_1$ and $r_2$ on the sphere and in the $(x, y)$-plane. Reflect $r_2$ about the diametre (of the circle in the $(x, y)$-plane) through $r_1$ to get $r_3$. Reflect $r_1$, $r_2$, and $r_3$ about the diametre perpendicular to the one throguh $r_1$ to get $r_4$, $r_5$, and $r_6$. I think that this tuple satisfies your condition. | |
| Mar 2, 2021 at 14:49 | comment | added | Ted Burgess | I'm only interested in 3D and $n > 3$. A rotationally-symmetric polyhedron with vertices on the unit sphere obviously satisfies the constraint, but I would like to know what can be said in the opposite direction. | |
| Mar 2, 2021 at 14:31 | comment | added | LSpice | Do you have an example of such a tuple with not all coördinates equal? (Also, you haven't specified the dimension; are we working in the unit sphere in the same $\mathbb R^n$ as the number of points, or could the dimension vary?) | |
| Mar 2, 2021 at 14:03 | comment | added | Ted Burgess | That's right. For every $j_{1}, j_{2}, j_{3}$ there is some $l_{1}, l_{2}, l_{3}$ such that the equation is satisfied for some rotation. | |
| Mar 2, 2021 at 13:58 | comment | added | LSpice | And presumably $(l_1, l_2, l_3)$ is also allowed to depend on $(j_1, j_2, j_3)$? | |
| Mar 2, 2021 at 12:13 | history | asked | Ted Burgess | CC BY-SA 4.0 |