Stuck on a computation with quaternions and moment maps I am trying to understand an article by Gibbons, Rychenkova and Goto, called "Hyperkähler quotient construction of BPS Monopole Moduli Spaces". I will paraphrase the relevant notions and formulas in order to get to my question.
Let $M = \mathbb{H}$ and $q$ be a quaternionic coordinate on $M$. Then the flat metric is
$$ g = dq \odot d\bar{q}.$$
Moreover, if we decompose
$$-\frac{1}{2} dq \wedge d\bar{q} = \omega_I i + \omega_J j + \omega_K k,$$
the tuple $(g,\omega_I,\omega_J,\omega_K)$ is a flat hyperkähler structure on $M$.
With respect to the action of $U(1)$ that maps $q$ to $qe^{it}$, the moment map is
$$ \mu = \frac{1}{2} q i \bar{q}.$$
It is natural to introduce other coordinates, which are adapted to the moment map $\mu$. One may write
$$q = ae^{i \psi/2}$$
where $a$ is a pure imaginary quaternion and $\psi \in (0,4\pi]$. Let
$$\mathbf{r} = qi\bar{q} = ai\bar{a} = -aia.$$
Then the authors write that a "short calculation" shows that, in the new coordinates $(\mathbf{r},\psi)$, the metric $g$ can be written as
$$ g = \frac{1}{4}\left(\frac{1}{r} d\mathbf{r}^2 + r(d\psi + \omega.d\mathbf{r})^2\right),$$
where $r = |\mathbf{r}|$ is the Euclidean norm of $\mathbf{r}$ and
$$\tag{1}\label{Bog}\nabla \times \omega = \nabla\left(\frac{1}{r}\right).$$
I attempted the calculation. After completing the square, I got that
$$ g = \frac{1}{4}\left(\frac{1}{r} d\mathbf{r}^2 + r(d\psi + \frac{1}{r}(aid\bar{a}-dai\bar{a}))^2\right),$$
which would agree with what the authors got, assuming that
$$ \omega.d\mathbf{r} = \frac{1}{r}(aid\bar{a}-dai\bar{a})$$
satisfies the Bogomolny equations \eqref{Bog}.
I tried to prove that, but somehow I got stuck. I obtained for instance that
$$*d\left(\frac{1}{r}\right) = - \frac{d\mathbf{r}\,\mathbf{r} \wedge d\mathbf{r}}{2r^3}$$
and that
$$d(\omega.d\mathbf{r}) = -\frac{dr}{r^2} \wedge (aid\bar{a} - dai\bar{a}) + \frac{2}{r} daid\bar{a},$$
but I could not yet show that $$*d(1/r) = d(\omega.d\mathbf{r}),$$
despite a few algebraic manipulations attempts.
 A: I was able to finally prove that
$$d(\omega.d\mathbf{r}) = - \frac{1}{2r^3}(d\mathbf{r}\,\mathbf{r} \wedge d\mathbf{r}).$$
In the process, I have learned a lot. The main issue for me was that I was dealing with differential forms with values in $\mathbb{H}$, the latter being of course non-commutative. Now I am much more comfortable with such differential forms.
Here is an outline. Note that $-\mathbf{r}^2 = r^2$, so that, upon differentiating, one obtains
$$ -(\mathbf{r}d\mathbf{r} + d\mathbf{r}\mathbf{r}) = 2rdr.$$
Let us also introduce the notation
$$d^c\mathbf{r} = aid\bar{a} - dai\bar{a}.$$
We then get
$$d(\omega.d\mathbf{r}) = \frac{1}{2r^3}(\mathbf{r}d\mathbf{r}+d\mathbf{r}\mathbf{r}) \wedge d^c\mathbf{r}-\frac{2}{r^3}dai\bar{a}\mathbf{r}aid\bar{a}.$$
Using the formulas
$$dai\bar{a} = \frac{1}{2}(d\mathbf{r} - d^c\mathbf{r})$$
$$aid\bar{a} = \frac{1}{2}(d\mathbf{r} + d^c\mathbf{r})$$
and after simplifying, one gets
$$d(\omega.d\mathbf{r}) = \frac{1}{2r^3}(-d\mathbf{r}\,\mathbf{r} \wedge d\mathbf{r} + d^c\mathbf{r}\,\mathbf{r} \wedge d^c\mathbf{r}).$$
But the second term inside the parentheses vanishes, so we obtained the desired formula.
