# Geometric intuition for how quaternions correspond to isoclinic rotations

I've been reading the wikipedia for 4D rotations. I understand the algebraic argument for why any rotation matrix can be decomposed into a right-isoclinic and a left-isoclinic rotation, and why these correspond to right and left multiplication by unit quaternions. But how does one go from a unit quaternion to the planes and the angle of the corresponding isoclinic rotation?

In 3D, I have the nice and easy fact that the imaginary part of a unit quaternion specifies the axis (or plane) of rotation while the real part is the cosine of one half times the rotation angle. Is there an analogously simple way of thinking about it in 4D?

I tried reasoning this out but ran into a wall almost immediately: specifying a unit quaternion is the same as specifying a line through the origin of $\mathbb{R}^4$ as well as a hyperplane orthogonal to this line, but I have no idea how one could go from this information to a pair of orthogonal 2-planes plus an angle.

The key point you need to get your two orthogonal 2-planes is that the real numbers also form a distinguished line in $\Bbb{H}$.
Suppose first that we have a quaternion $\bf{q}$ which is of the form $\cos \theta + i \sin \theta$. Then multiplication by $\bf{q}$ (either left or right) clearly acts on the span of $\{1,i\}$ by rotating by $\theta$; it's not too hard to check that it also acts on the span of $\{j,k\}$ by rotating by $\theta$ (if you orient $\{j,k\}$ correctly; the correct choice of orientation will depend on whether you're multiplying on the left or on the right). So in this case your two orthogonal 2-planes are the span of $\{1,i\}$ and the span of $\{j,k\}$.
It's much the same for general unit quaternions. In general we have ${\bf q}=\cos \theta + {\bf r }\sin \theta$ where $\bf r$ is a purely imaginary unit quaternion. Then $\bf{q}$ will act on the span of $\{1,{\bf r}\}$ via rotation by $\theta$, and also on its orthogonal complement (but the direction of rotation on the orthogonal complement depends on whether you're left- or right-multiplying).