MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the Lie group $Spin(3)$, which can be thought of geometrically as the 3-sphere (e.g., it can be represented by the collection of unit quaternions). The quotient $Spin(3)/\pm I$ yields the group $SO(3)$ of rotations of $\mathbb{R}^3$. The octahedral group $Oct$ is the symmetry group of the (regular) octahedron, i.e., it is the maximal subgroup of $SO(3)$ that maps the octahedron to itself. The binary octahedral group $2Oct$ is simply the double cover of the octahedral group in the spin group.

Question: what is the quotient $Spin(3)/2Oct$ as a topological space?

Equivalently, what is the topology of $SO(3)/Oct$? (And is there a natural geometry associated with this space?) Note that $Oct$ is not a normal subgroup of $SO(3)$, hence the quotient does not exist in the sense of groups---only in the sense of topological manifolds.

A closely related example is the quotient of $Spin(3)$ by the icosahedral group, i.e., the symmetries of the icosahedron in $\mathbb{R}^3$. In that case, the quotient is the Poincaré sphere. I am basically looking for the analogous object in the octahedral case.


share|cite|improve this question
When you say "what is the quotient", what do you mean? It's the quotient, you already have a definition as you gave it yourself. It has spherical geometry. Google "geometrization conjecture" and look up "spherical geometry". Your action of the binary octahedral group on $S^3$ is by isometries, so the quotient inherits the same local geometry as $S^3$. – Ryan Budney May 25 '13 at 23:03
What is it that you know about the Poincare sphere but don't know about this other manifold? It's homology? A description as a polytope with sides identified? – Tom Goodwillie May 25 '13 at 23:18
This 'octahedral group' is isomorphic to the symmetric group $S_4$. Its abelianization has order two, and the abelianization of the 'binary octahedral group' is the same. – Tom Goodwillie May 26 '13 at 1:06
It has different topological descriptions, as a Seifert-fibered space which is a bundle over a $(2,3,4)$ turnover (the unit tangent bundle to the $(2,3,4)$ turnover), or as a Dehn filling on the trefoil knot. It has a spherical geometry, and you can visualize flying through it with Jeff Weeks' program Curved Spaces: – Ian Agol May 26 '13 at 1:11
fuzzytron, for more information about spherical manifolds in general, you might be interested in Peter Scott's article 'The geometries of 3-manifolds', Bulletin of the LMS, 1983. – HJRW May 27 '13 at 4:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.