Isoclinic rotations in $\mathbb{R}^4$ are what subgroup of $SO(4)$?

Question

According to Wikipedia, isoclinic rotations of $\mathbb{R}^4$ are normal subgroups of $SO(4)$. But which subgroup? $SU(2)$? $SO(3)$? Some other group?

Answer 1

If you see $SO(4)$ as acting on $\mathbb{H}\simeq \mathbb{R}^4$ and preserving the usual quaternion norm, then those subgroups of (left or right) isoclinic rotations $S^3_L$ and $S^3_R$ are given respectively by the $x\mapsto qx$ and $x\mapsto xq$ where $q$ is a unit quaternion. So both those groups are isomorphic to the unit sphere in $\mathbb{H}$, which is topologically the $3$-dimensional sphere, and as a group is $SL_1(\mathbb{H})$.

This group $SL_1(\mathbb{H})$ is isomorphic to $\operatorname{Spin}(3)$ and to $SU(2)$. You have a sequence of central isogenies $$SL_1(\mathbb{H})\times SL_1(\mathbb{H})\simeq S^3_L\times S^3_R\to SO(4)\to SO(3)\times SO(3)$$ where the first arrow glues together the centers of $S^3_L$ and $S^3_R$ to become the center of $SO(4)$, and the second one quotients out this center. The composition of those morphisms is then just the direct product of twice the quotient $SL_1(\mathbb{H})\to SO(3)$, which can be described as sending $q$ to the map $\mathbb{H}_0\to \mathbb{H}_0$ (where $\mathbb{H}_0\simeq \mathbb{R}^3$ are the pure quaternions) defined by $x\mapsto qxq^{-1}$.