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For $g\in\mathrm{SO}(3),S\subseteq \mathbb{R}^3,$ define $g\cdot S:=\{g\cdot p : p\in S\}.$ In words, if $g$ is a rotation of $\mathbb{R}^3$, $g\cdot S$ is the set of elements of $S$ rotated by $g$. For two elements $g,g'\in\mathrm{SO}(3)$, define an equivalence relation $\sim$ via rotations of the axis-aligned cube: $g\sim g'\iff (g\cdot [-1,1]^3 = g'\cdot [-1,1]^3)$.

What is the structure of $\mathrm{SO}(3)/\sim$? Does this space admit a simple parameterization, measure, or metric?

More broadly, is this a well-studied space with a name? Some interesting problems in discrete geometry seem to involve this structure.

Note: $\mathrm{SO}(3)$ is a simple group, so I expect this to a quotient space rather than a quotient group. This is a refined version of an earlier question.

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    $\begingroup$ This equivalence relation just says that two elements are in the same coset of $H$, where $H$ is the stabilizer of the cube inside $SO(3)$. An element which stabilizes the cube will also stabilize the points $\{0,1\}^3$. This set is finite, and it is possible to show that $H$ is also finiet, and to calculate it explicitly. $\endgroup$
    – Ehud Meir
    Commented Aug 23, 2015 at 0:25
  • $\begingroup$ Got it! I understand these ideas abstractly but am hoping for a fairly concrete characterization (parameterization, embedding, metric, or something similar) to help use this space in a computational system. The elements of $\mathrm{SO}(3)$ that stabilize the cube are the octahedral group, but as this is not a normal subgroup of $\mathrm{SO}(3)$ I'm not sure how to deal with the resulting quotient in a concrete way. $\endgroup$
    – Justin
    Commented Aug 23, 2015 at 0:28
  • $\begingroup$ Possibly(?) related: uniformly distributed random orthogonal matrices, used, e.g., in this MO question. $\endgroup$
    – Joseph O'Rourke
    Commented Aug 23, 2015 at 0:46
  • $\begingroup$ Interesting pointer! I don't immediately see the connection but will read closely. $\endgroup$
    – Justin
    Commented Aug 23, 2015 at 1:07
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    $\begingroup$ As a postdoc in a math department using this information for my research, I am puzzled and the tiniest bit offended that this topic is closed because it is not "research level math." But, many thanks to the folks who responded before this happened. $\endgroup$
    – Justin
    Commented Sep 7, 2015 at 14:31

3 Answers 3

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$\DeclareMathOperator{\SO}{SO}$$\DeclareMathOperator{\RP}{RP}$$\DeclareMathOperator{\Stab}{Stab}$There is a difference between the sets $C = [-1,1]^3$ and $E = [0,1]^3$. Note that $\Stab(C)$, the stabilizer of $C$ inside of $\SO(3)$, is called the cube group and it has 24 elements. Note that $\Stab(E)$ has three elements.

  • Are you sure you want $E$ and not $C$?

In either case, the manifold $X = X_E = \SO(3)/\Stab(E)$ is called a "spherical space form" and also an "elliptical manifold". This is because $X$ is a quotient of the three-sphere by a freely acting subgroup of $\SO(4)$ -- note that $\SO(3)$ is a copy of $\RP^3$ so it is also a spherical space form. Jeff Weeks' book "The shape of space" has an elementary treatment of the geometry and topology involved. Also, here is link to Wikipedia:

https://en.wikipedia.org/wiki/Spherical_3-manifold

The manifold $X$ is called a "prism manifold" there. Now, you ask

Does this space admit a simple parameterization, measure, or metric?

and you ask for

[e.g. an embedding into Euclidean space without having to identify points] What does the induced Riemannian metric look like?

You haven't said why you are looking for this, beyond mentioning a computational application. There are lots of embedding theorems (eg the Whitney embedding theorem) and there are ways to describe the induced metrics. These will (almost certainly) be useless for computational purposes.

Probably you will be much happier with the following much simpler approach. Encode elements of $X = \SO(3)/\Stab(E)$ as objects $[M]$ where

  • $M$ is an orthogonal matrix (of determinant one) and
  • $[M] = [N]$ if and only if $N^{-1} M \in \Stab(E)$.

I'll guess that these new objects (basically, cosets) will do everything you want, while avoiding dangerous mucking about with wacky embeddings. For example, computing distance in $X$ reduces to three distance computations in $\SO(3)$.

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  • $\begingroup$ Thanks for the catch -- I was looking for $C$ rather than $E$! This post is very helpful -- I'll take a look at the book by Weeks and some of the terms you mention. Parameterization is a tricky matter. Stab(C) is a large group (S_4), so those elements could be quite large sets! $\endgroup$
    – Justin
    Commented Aug 23, 2015 at 12:44
  • $\begingroup$ Hmm. Having to check 24 things in an innermost loop could be painful. (You'll have to do timings to check. Perhaps using the unit quaternions, instead of SO(3), will speed things up.) Nonetheless, a simple implementation that works slowly is better than a clever idea that doesn't work at all. :) $\endgroup$
    – Sam Nead
    Commented Aug 23, 2015 at 13:09
  • $\begingroup$ Fair enough! "Constant time" after all -- so a theoretical computer scientist would be OK with that check :-) . I'm hoping to generate smooth maps into this quotient space -- will have to think about how to represent such a thing...I wonder if its Laplacian is easy to write down $\endgroup$
    – Justin
    Commented Aug 23, 2015 at 17:46
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More useful than embedding it in some higher-dimensional space will be to give it as a manifold with identifications, I think. Topologically, $SO(3)/\!\sim$ is the same space as what you get by taking an octahedron and gluing the opposite faces with a $1/3$ turn. This is related to the 24-cell, one of the 6 different regular 4-dimensional polytopes (analogues of Platonic solids in 4 dimensions), using the fact that $SO(3)$ is topologically $RP^3$, the quotient of the 3-sphere $S^3$ by the antipodal map.

The space of course comes with a natural metric and measure (both inherited from the one on $SO(3)$. You can see this metric/measure on the octahedron by embedding it as a regular octahedron of an appropriate size subset of $SO(3)$. You can work out the size by looking at the volume: 12 of these octahedra tile $SO(3)$, or 24 tile the double-cover $S^3$. Alternatively, each corner of the octahedron in $SO(3)$ is a rotation by $\pi/4$ around one of the faces of the cube, and the center of each face of the octahedron is a rotation by $\pi/3$ around one of the vertices of the cube.

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  • $\begingroup$ Thanks for this description -- it's much easier to visualize! I guess you can think of this space sort of like a spherical triangle with circle "fibers," glued together the right way. This description may be simpler for generating a simplicial complex description of this space. $\endgroup$
    – Justin
    Commented Aug 24, 2015 at 13:35
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$SO(3)$ acts transitively on the space of all orthonormal oriented bases (these are ordered up to an even permutation). The stabilizer of the standard basis is exactly your subgroup $H$. So $X = SO(3)/\sim$ is the space of such bases. It inherits an $SO(3)$-invariant measure and Riemannian metric from $SO(3)$, which is actually a covering space of $X$ of degree $3$.

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  • $\begingroup$ Indeed, looks like you worked backward to the source of my question :-) Is there a more explicit parameterization of this space? [e.g. an embedding into Euclidean space without having to identify points] What does the induced Riemannian metric look like? $\endgroup$
    – Justin
    Commented Aug 23, 2015 at 2:16
  • $\begingroup$ @Justin You should be able to calculate the metric in terms of local coordinates using $SO(3)$. $\endgroup$
    – Amritanshu Prasad
    Commented Aug 23, 2015 at 3:31

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