Quaternions and special orthogonal transformations

Question

I am asked to show that for $A,B\in\operatorname{SU}(2)$, the map $\phi\colon (A,B)\colon\mathbb H\to\mathbb H\colon h\mapsto AhB^{-1}$ is in $\operatorname{SO}(\mathbb H)$, where $\mathbb H$ are the quaternions (and we consider them as being embedded in $2$-by-$2$ complex matrices). Now, I know of a way to brute force this; I can consider a basis for $\mathbb H$, and then write out how $\phi(A,B)$ looks like. By means of Mathematica it can then be verified that indeed $\phi(A,B)\phi(A,B)^T=I$ and $\det(\phi(A,B))=1$. However... this method is unsatisfying. I'm hoping that if I were to choose a nice basis for $\mathbb H$, a quicker argument could be made. Would anyone have any idea?

Answer 1

You need only to realize that for any two quaternions $u,w$,

$$|uv| = |u| |w|.$$ Thus $$ | A hB^{-1} | = |A| |h| |B^{-1}|=|h|$$ and so $h\mapsto AhB^{-1}$ is in $O(4)$. Since $SU(2)$ is connected, we see that the determinant has to be one (by connecting to $A,B$ to $I$).