If $q \in \mathbb{H}$ is a quaternion, then one may write
$$q = a \, e^{\, i\frac{\psi}{2}}$$
where $a$ is pure imaginary (in the sense of quaternions), which means that $\bar{a} = -a$ and $\psi \in [0, 4\pi)$. Then let
$$ \omega = qid\bar{q} - dq i \bar{q} $$
which is a (real) $1$-form on $\mathbb{H}$, thought of as a smooth $4$-dimensional manifold, and let
$$ V = \frac{1}{r} $$
where $r = \lvert \mathbf{r} \rvert$ and $\mathbf{r} = qi\bar{q} = a i \bar{a}$. Then one may write
$$ \mathbf{r} = xi + yj + zk, $$
thus introducing coordinate functions $x$, $y$ and $z$ which, together with $\psi$, form a system of coordinates on $\mathbb{H} \setminus \{ 0 \}$.
A calculation reveals that the pair $(\omega, V)$ satisfies Bogomolny's equations on $\mathbb{R}^3 \setminus \{ \mathbf{0} \}$, where the latter is the quotient of $\mathbb{R}^4 \setminus \{ \mathbf{0} \}$ by the action of $U(1)$, where $e^{i \theta} \in U(1)$ acts by:
$$ q \mapsto q e^{-i \theta}. $$
More precisely, the Bogomolny's equations are:
$$ d \omega = * d V, $$
where the flat Euclidean metric is assumed on $\mathbb{R}^3 \setminus \{ \mathbf{0} \}$, $d$ is the $3$-dimensional Cartan exterior differential (with respect to $x$, $y$ and $z$) and $*$ is the Hodge star (with respect to the flat $3$-dimensional Euclidean metric).
I know how to check that $(\omega, V)$ as defined above indeed satisfy Bogomolny's equations, but the calculation was a bit tedious (for me at least). My question is whether or not there is a somewhat short quick way of doing the calculation or not.
An outline of my calculation is the following:
- I calculated $* d(1/r)$, first with respect to $x$, $y$ and $z$, then I found an expression for it in terms of $\mathbf{r}$ (thus essentially in terms of a more quaternionic notation). If I remember correctly, I think I got
$$ * d(1/r) = - \frac{1}{2r^3} d\mathbf{r} \mathbf{r} \wedge d\mathbf{r}.$$
- Another calculation shows that, after simplification, $d \omega$ turns out to be equal to the right-hand side of the previous formula.
This is the outline of my calculation. Is there perhaps a more direct way of doing the calculation?