# Geometric (smooth) Rubik's cube

I and my friend are thinking about a smooth analog of Rubik's cube. One idea is the following:

Consider the 2-dimensional sphere $$S^{2}$$. We choose three parameters: $$(L, H, \theta)$$. Here $$L$$ is a ray that passes through the origin, $$H$$ is a plane that intersects with $$L$$, the sphere, and orthogonal to $$L$$. $$\theta\in [0, 2\pi]$$ is an angle that we will rotate the part of the sphere. Here is the picture for the action:

Now I have some question about a group $$G$$ generated by this action on $$S^{2}$$.

1. Is $$G$$ has finite dimensional Lie group? If it is, what is a dimension of the group?

2. Is there any interesting and nontrivial relation among elements? For example, in $$\mathrm{SO}(3)$$, the composition of two rotations is also rotation. But our group clearly doesn't satisfy such things.

Edit. As Elkies said, this is an infinite dimensional group, since it contains an infinite dimensional abelian group (which is generated by elements with fixed $$L$$ and varying $$H\&\theta$$. But we may think about some subgroups that seem to be finite dimensional, or even finite.

Before we do that, there's some minor problem with the definition of $$G$$ that Jim Conant mentioned. I want to ignore the action that only moves measure 0 parts, such as an equator. To do this, for any two functions $$g_{1}, g_{2}:S^{2} \to S^{2}$$ we can define an equivalence relation as $$g_{1}\sim g_{2} \text{ iff } \{x\in S^{2}\,:\, g_{1}(x) \neq g_{2}(x)\} \text{ has a measure 0}$$ Now consider the quotient of the group by this equivalence relation, then we will get more reasonable group. There's some reason why I'm thinking this reasonable - there are some subgroups that I want to think about.

1. Assume that we only allow "semi-sphere turns". So we consider the subgroup of $$G$$ where the plane $$H$$ should pass the origin. Is this a finite-dimensional group? I think $$\mathrm{SO}(3)$$ is a codimension 1 subgroup of this semi-sphere-turn group, but I'm not sure about it.

2. What will be the finite subgroups of this $$G$$? There's a natural but nontrivial way to construct a finite subgroup of $$G$$: to think real Rubik's cube type puzzles. For example, put Rubik's cube in the center of the sphere and cur the sphere along the planes that correspond to the Rubik's cube's slice faces (I hope you understand what I mean), and turn the pieces along the axis of the cube. In other words, this is a subgroup that is isomorphic to the group of Rubik's cube. We can do the same thing with Pyraminx, Metaminx, Dogic, Skewb, or any regular polytope. I want to know if these covers all the possible finite subgroups of $$G$$.

3. How about higher dimensional spheres, or other highly symmetric manifolds (such as a torus or hyperbolic spaces)?

• That's a huge group $-$ certainly not finite-dimensional, because the transformations with $L$ fixed already generate an infinite-dimensional abelian subgroup. – Noam D. Elkies Mar 1 at 3:18
• 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. – Phil Tosteson Mar 1 at 4:00
• 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. – YCor Mar 1 at 8:12
• 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. – YCor Mar 1 at 8:14
• @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 $\{x:f(x)\neq g(x)\}$. – YCor Mar 1 at 10:40