Timeline for Riemannian manifolds which admit a smooth free $\mathbb{Z}/3\mathbb{Z}$ action but do not admit an equilateral triangle action
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 21, 2020 at 18:54 | comment | added | Anton Petrunin | @YCor I thought that $|a(x)-x|=const$. | |
| Mar 21, 2020 at 13:07 | comment | added | Olivier Bégassat | @YCor Right,. ${}$ | |
| Mar 21, 2020 at 13:03 | answer | added | YCor | timeline score: 2 | |
| Mar 21, 2020 at 11:13 | answer | added | user44143 | timeline score: 5 | |
| Mar 21, 2020 at 7:17 | comment | added | YCor | @AntonPetrunin for any metric space an equilateral action makes sense (a free continuous action of $C_3$ in which every orbit is equilateral). So asking for a Riemannian manifold with no such action is perfectly meaningful. While your modified question is quite trivial: if there's a free $C_3$-action, you can average to get an invariant Riemannian metric. | |
| Mar 21, 2020 at 7:15 | comment | added | YCor | @OlivierBégassat Precisely, I'm giving a proof of your assertion (that there's no "equilateral action") for the plane minus 4 points, none of which form an equilateral triangle. | |
| Mar 21, 2020 at 5:25 | comment | added | Anton Petrunin | I guess you want a manifold with a free action of $\mathbb{Z}_3$ such that there is no equilateral action for ANY Riemanninan metric. (Otherwise the question has no sense.) | |
| Mar 21, 2020 at 1:06 | comment | added | Olivier Bégassat | @YCor I don't understand. The three points here don't form an equilateral triangle, and neither will the orbit of a point close to $p$. | |
| Mar 21, 2020 at 0:12 | comment | added | YCor | @OlivierBégassat for sure it works for the plane minus 4 points, none 3 of which form an equilateral triangle.If $f$ is a homeomorphism, it acts on the end compactification, which is a 2-sphere. If it fixes all end points, then this is a homeo of order 3 with exactly 5 fixed points and this doesn't exist. So it permutes 3 and fixes the last two. A limit argument shows that the point at infinity is fixed, and that the three moved ones form an equilateral triangle. | |
| Mar 21, 2020 at 0:00 | comment | added | Olivier Bégassat | $A$ is only conjugate to a rotation (i.e. satisfies $A^3=I_2$) but it's not an isometry. The Z/3Z action is that of $A$ on $M$. I'm removing two orbits from the action of $A$ on $\Bbb{R}^2$. | |
| Mar 20, 2020 at 23:52 | comment | added | Ali Taghavi | In this case what is the free action which already exist? | |
| Mar 20, 2020 at 23:50 | comment | added | Ali Taghavi | @OlivierBégassat I think you mean we do not remove $A^2p$, yes?otherwise the rotation irself give the required action right? | |
| Mar 20, 2020 at 23:42 | comment | added | Olivier Bégassat | Are you assuming compactness or completeness? If you don't I think you can find uninteresting counter examples of the form $M=\Bbb{R}^2\setminus\{0,p, Ap, A^2p\}$ where $A$ is a matrix that is conjugate to a rotation by $2\pi/3$, $p$ is any nonzero point and the Riemannian structure is induced from that of $\Bbb{R}^2$. I would expect you can make this into a complete example by tampering with the metric. So maybe you should assume compactness? | |
| Mar 20, 2020 at 22:25 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
added 3 characters in body; edited tags
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| Mar 20, 2020 at 22:07 | history | asked | Ali Taghavi | CC BY-SA 4.0 |