Transformation in $SO(4)$

Question

I want to prove that for any unitary $q,q'\in \mathbb{H}$: $$T_{q,q'}:\mathbb{H}\rightarrow \mathbb{H}$$ $$x\mapsto qxq'$$ belongs to $SO(4)$. Is there any elegant way to show it without expanding this product in terms of coordinates?

Answer 1

One can define (a group isomorphic to) $O(4)$ as the $\mathbb{R}$-linear transformations $\mathbb{H}\to\mathbb{H}$ which preserve the inner product $\langle q,q'\rangle:=\operatorname{Re}(\overline{q}q')$. $SO(4)$ is then the subgroup of these transformations which are orientation preserving (or, equivalently, continuously connected to the identity). Presumably this is what you refer to when asking if a transformation is "in" $SO(4)$.

From there, one can show that the left multiplication maps $L_q(q'):=qq'$ and the right multiplication maps $R_q(q'):=q'q$ are $\mathbb{R}$-linear and preserve the inner product, and thus elements of $O(4)$. Since both the right translations and left translations are connected groups, they are in fact contained in $SO(4)$. Thus, any composition of these maps will also be an element of $SO(4)$.