## § Why quaternions work better

- We want to manipuate $SO(3)$. Imagine it like $SO(1)$.
- Unfortunately, $\pi_1(SO(3)) = \mathbb Z/2\mathbb Z$. This is a pain, much like rotations of a circle need to be concatenated with modulo, which is a pain.
- idea for why $\pi_1(SO(3))$ is $\mathbb Z/2\mathbb Z$: $SO(3)$ is sphere with antipodal points identified. So a path from the north pole to the south pole on the sphere is a "loop" in $SO(3)$. Concatenate this loop with itself (make another trip from the south pole to the north pole) to get a full loop around the sphere, which can be shrunk into nothing as $\pi_1(S^2)$ is trivial. So $ns^2 = e$, where $ns$ is the north-south path in $S^2$which is a loop in $SO(3)$).
- Key idea: deloop the space! How? find univesal cover. Lucikly, universal cover of $SO(3)$ is $SU(2)$ / quaternions, just as universal cover of $SO(1)$ is $\mathbb R$.
- Universal cover also explains why $SU(2)$ is a
*double * cover. Since $\pi_1(SO(3))$ is $\mathbb Z/2Z$, we need to deloop "once" to get the delooped space. - No more redundancy now! Just store a bloch sphere representation, or a quaternion (store $SU(2)$). Just like we can just store a real number for angle and add it.
- How to go back to $SO(3)$ or $SO(1)$? Move down the universal cover map $SU(2) \to SO(3)$ or $\mathbb R \to \mathbb SO(1)$.
- This is strange though. Why is $\mathbb R$ both the
*lie algebra * and the *covering space * of $SO(1)$ ? What about in general? - In general, the original lie group $SO(3)$ and the universal cover $SU(2)$ both have the same
*lie algebra *. It is only that the lie group has less or more fundamental group.