# Space of arbitrary rotations of a cube

Suppose I have a cube $[-1,1]^3\subset\mathbb{R}^3$. I am allowed to rotate it about any angle/axis through the origin rather than just $90^\circ$ about the coordinate axes, e.g., by applying elements of $SO(3)$. To differentiate from $SO(3)$, however, I wish to identify any pair of rotations that align faces of the cube. In other words, I wish to identify any pair of rotations that make the cube "look" the same if it is painted uniformly.

If the cube is constrained to rotate $90^\circ$ along the coordinate axes, the resulting group of rotations is the octahedral group, isomorphic to $S_4$. This subgroup is not normal, however, so I can't construct a quotient.

Is there a name/structure/parameterization/representation of this quotient space?

[REVISION FROM ORIGINAL QUESTION: It appears $SO(3)$ is simple, meaning it is not possible to quotient by $S_4.$ Any guidance about how to characterize this quotient /space/, however, would be useful.]

• I think this is just the quotient group $SO(3)/S_4$. – Eric Tressler Aug 20 '15 at 1:09
• Is it obvious that $S_4$ is a normal subgroup that can be quotiented out? [My abstract algebra is rusty!] Also, does this group have a name or any associated structure? I'm hoping to find its irreducible representations. – Justin Solomon Aug 20 '15 at 1:10
• It's not obvious to me. After thinking about it some more, I'm trying to verify it for a general element of $SO(3)$ and $S_4$, but I can't visualize it well enough. I guess a random rotation matrix and some actual multiplication will be enough to decide. – Eric Tressler Aug 20 '15 at 1:21
• Is there a reason you actually expect this to be a group? In particular how are you defining the operation and why is it well defined? – Nate Aug 20 '15 at 2:29
• SO(3)/S4 is a quotient space. Its underlying set is the coset space of S4, and its topology is the quotient topology. It is not a quotient group because S4 is not normal, and certainly not a subgroup (quotients and subobjects are different - in a sense, they are "dual"). – whacka Aug 20 '15 at 7:53

Not sure this is what you mean by a parametrization, but $SO(3)$ is homeomorphic to a quotient of the ball $\{x\in R^3:|x|\le\pi\}$. The correspondence is, given $x$, there is the rotation of $R^3$ around $x$ by angle $|x|$. The quotient is to identify $x$ and $-x$ when $|x|=\pi$, since these give the same rotation. Then, since $SO(3)$ is a quotient of the ball, so is your space.
The further description in many topology texts is that $SO(3)$ is homeomorphic to the projective space $RP^3$. I'm guessing this won't be helpful, but just in case: it is realized by pushing the ball up into a hemisphere in $R^4$; then each line through the origin of $R^4$ is associated with a rotation of $R^3$.