Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

How can I prove that the quotient group $S^3/\{+I,-I\}$ is isomorphic to $SO_3$ and that the group $S^3$ is not isomophic to $SO_3$?

Here $S^3$ is the subgroup of the quaternion group: $S^3=\{a+bi+cj+dk | a^2+b^2+c^2+d^2=1\}$

share|cite|improve this question
What do you mean by $S^3/\{\pm I\}$? Are you looking at the orbit space of $S^3$ under the action of $\Bbb{Z}/2\Bbb{Z}$ given by the antipodal map? – user38268 Nov 26 '12 at 0:22

To prove that $S^3/\pm 1$ is isomorphic to $SO(3)$, use the hypersphere of rotations. This gives a way of naturally thinking of $SO(3)$ as a quotient of $S^3$. Embedding everything inside the quaternions makes defining the mapping much cleaner.

To show that $SO(3)$ and $S^3$ are not isomorphic as lie groups, note that this gives a homeomorphism of the underlying topological spaces. But from the hypersphere of rotations, one can also see that $SO(3)$ and $\mathbb{RP}^3$ are homeomorphic, and $\pi_{1}(\mathbb{RP}^3) = \mathbb{Z}/2\mathbb{Z}$, whereas $\pi_{1}(S^3) = \{1\}$ since the $3$-sphere is simply connected.

share|cite|improve this answer
They are also not isomorphic as groups. $S^3$ has nontrivial center ($-1$ is in it), but the center of $SO(3)$ is trivial. – Jason DeVito Nov 25 '12 at 23:36

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.