Timeline for Geometric (smooth) Rubik's cube
Current License: CC BY-SA 4.0
25 events
| when toggle format | what | by | license | comment | |
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| Mar 1, 2019 at 19:20 | comment | added | YCor | But most likely, dense or not dense, this topology is certainly not complete, so not what we like of a "Lie group". Another topology, closer to topological dynamics would be the one induced by the action on the Boolean algebra generated by spherical caps. | |
| Mar 1, 2019 at 18:48 | comment | added | Seewoo Lee | @Nathaniel Thank you for such interesting examples! | |
| Mar 1, 2019 at 18:42 | comment | added | Seewoo Lee | @PierrePC I think that is almost the same as Elkies' argument, isn't it? Anyway, that also show that the group should be very big. | |
| Mar 1, 2019 at 18:41 | comment | added | Seewoo Lee | @YCor Yes, your definition of $G$ is what I intended. Honestly, I didn't think about the topology seriously but I think the metric topology that you suggested is somehow the most natural way to give a topology. Also, I don't know why $G$ should be a dense subgroup of the measure-preserving automorphism group of $S^{2}$. | |
| Mar 1, 2019 at 17:21 | comment | added | Pierre PC | If I'm not mistaken, the action on the equator of the subgroup preserving the equator ‘contains’ the action of the interval exchange transformation group. In fact, I belive this stays true if look at the action on a (possibly infinite) collection of disjoint circles, provided each of them is at positive distance from the others. That group is big. | |
| Mar 1, 2019 at 10:49 | comment | added | user108998 | @YCor OK makes sense and actually is an interesting q, tho I'm inclined to disagree that it's dense at first glance. Nonetheless everything is so huge here my intuition isn't much to go by. In partic we can just cut out arbitrary positive measure chunks and stick them back on somewhere else... | |
| Mar 1, 2019 at 10:40 | comment | added | YCor | @EBz yes, I'm stupid. I should say the group of measure preserving automorphisms of the sphere. But I should be careful which topology one considers on this group. One such topology is the topology induced by the bi-invariant distance $d(f,g)$ = measure of $\{x:f(x)\neq g(x)\}$. | |
| Mar 1, 2019 at 10:38 | comment | added | user108998 | @YCor, aren't these partial rotations discontinous? I don't see a map to the homeo group... | |
| Mar 1, 2019 at 8:16 | comment | added | YCor | I tend to guess that this group is dense in the group of volume-preserving self-homeomorphisms of the round 2-sphere. | |
| Mar 1, 2019 at 8:14 | comment | added | YCor | Elkies didn't say it's an infinite dimensional Lie group, but that it contains an infinite-dimensional Lie group... anyway to be an infinite-dimensional Lie group requires defining some "good" group topology and proving that it's Lie, and both steps are unclear here. | |
| Mar 1, 2019 at 8:12 | comment | added | YCor | Let $r(L,H,\theta)$ be the partial rotation defined at the beginning. Do you implicitly define $G$ as the subgroup generated by all such partial rotations (for all $(L,H,\theta)$?). It's not clear to me but it seems the interpretation made by other users. | |
| Mar 1, 2019 at 6:46 | comment | added | user44191 | I'm pretty sure the "semi-sphere turn subgroup" is not finite-dimensional. For any convex polygon on the sphere, you can look at the subgroup generated by "semi-sphere turns" along its edges; infinitesimally, these should be linearly independent. | |
| Mar 1, 2019 at 5:35 | comment | added | N. Virgo | (I mention this because I think there are interesting unexplored mathematical possibilities hinted at in those threads.) | |
| Mar 1, 2019 at 5:33 | comment | added | N. Virgo | Note that there are finite subgroups that do not correspond to regular polyhedrons, such as the ones discussed here twistypuzzles.com/forum/viewtopic.php?t=31829. (You will have to sign up for an account to see most of the images.) Also note that there are plenty of infinite subgroups that are generated by only a finite set of the generators. (These are called "jumbling" puzzles.) See for example twistypuzzles.com/forum/viewtopic.php?f=1&t=25752 (the images are missing for the first few pages of that post, but if you keep reading you will be able to see what they are discussing.) | |
| Mar 1, 2019 at 5:29 | comment | added | N. Virgo | @SeewooLee I realised there are some counterexamples. For example, puzzles like the Bubbloids, whose axes don't intersect in a single point, can't be represented this way. But if we restrict ourselves to puzzles that can be made in the shape of a sphere, then it should work. As you say in the edited question, it's clear that all such puzzles correspond to finite subgroups of this group, so we only have to determine whether the group has any 'pathological' finite subgroups that don't correspond to physically implementable puzzles. I would be interested to know the answer to that. | |
| Mar 1, 2019 at 5:18 | comment | added | Seewoo Lee | @Nathaniel I agree with that and it was one of the motivations of the question. Is it possible to prove it? | |
| Mar 1, 2019 at 5:17 | comment | added | Seewoo Lee | @JimConant I just edited the question. In summary, I want to ignore such small movements. | |
| Mar 1, 2019 at 5:16 | comment | added | Seewoo Lee | @NoamD.Elkies Thank you for the comment. I added some questions that I forgot to ask. | |
| Mar 1, 2019 at 5:16 | history | edited | Seewoo Lee | CC BY-SA 4.0 |
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| Mar 1, 2019 at 5:05 | comment | added | N. Virgo | It presumably has the property that its finite subgroups are exactly the groups corresponding to doctrinaire twisty puzzles. ("Doctrinaire" being the term used by the members of twistypuzzles.com/forum to refer to a puzzle whose moves form a group, like the Rubik's cube.) I don't know what other interesting properties it might have. | |
| Mar 1, 2019 at 4:01 | comment | added | Phil Tosteson | en.m.wikipedia.org/wiki/Pseudogroup | |
| Mar 1, 2019 at 4:00 | comment | added | Phil Tosteson | Though this is large as a group, it is a reasonably natural piece of a groupoid / pseudogroup consisting of open subsets of the sphere and isometries between them. | |
| Mar 1, 2019 at 3:41 | comment | added | Jim Conant | What happens at $S^2\cap H$? Does that part move? | |
| Mar 1, 2019 at 3:18 | comment | added | Noam D. Elkies | That's a huge group $-$ certainly not finite-dimensional, because the transformations with $L$ fixed already generate an infinite-dimensional abelian subgroup. | |
| Mar 1, 2019 at 3:09 | history | asked | Seewoo Lee | CC BY-SA 4.0 |