Describe the group of unit quaternions topologically and explain its relationship with \(\SO(3)\).
Solution
Write the quaternion algebra as \[ \HH=\{a+bi+cj+dk:a,b,c,d\in\RR\}. \] The norm is \[ |q|^2=a^2+b^2+c^2+d^2. \] Hence the unit quaternions \[ \{q\in\HH:|q|=1\} \] form the unit sphere \[ S^3\subset\RR^4. \] Under quaternion multiplication this is a Lie group, isomorphic to \[ \SU(2). \]
Let \(\operatorname{Im}\HH\cong\RR^3\) be the purely imaginary quaternions. For a unit quaternion \(q\), define \[ \rho(q)(v)=qvq^{-1}, \qquad v\in\operatorname{Im}\HH. \] Conjugation preserves the quaternion norm and fixes the real axis, so \(\rho(q)\) restricts to an orientation-preserving orthogonal transformation of \(\operatorname{Im}\HH\). Thus \[ \rho:S^3\to\SO(3) \] is a group homomorphism.
Every rotation of \(\RR^3\) is obtained in this way, so \(\rho\) is surjective. Its kernel consists of the unit quaternions commuting with every imaginary quaternion, namely the real unit quaternions \[ \{\pm1\}. \] Therefore \[ \SO(3)\cong S^3/\{\pm1\}, \] and \[ S^3\cong\SU(2)\longrightarrow\SO(3) \] is a double covering map.