first-order equations, we first use the time-independence of the solutions

$$i\gamma^{m}\partial_{m} Z^{i}\epsilon^{I} 
-4\epsilon^{IJ}G^{i\, +}{}_{0m}\gamma^{m}\gamma^{0}\epsilon_{J}=0\, ,$$

and then the known constraint Eq. (\[eq:constraint\]) as to obtain

$$(\partial_{m} Z^{i} -4e^{i\alpha}G^{i\, +}{}_{0m})
\gamma^{m}\epsilon^{I}=0\, ,\,\,\,\,
\Rightarrow \partial_{m} Z^{i} =4e^{i\alpha}G^{i\, +}{}_{0m}\, .$$

Going over to curved indices, the equation takes the form

$$\frac{dZ^{i}}{dr} =2\sqrt{2}G^{i\, +}{}_{tr}/X^{*}\, .$$

The self-duality of $G^{i\, +}$ allows us to express the $G^{i\, +}{}_{tr}$ component in terms of the $G^{i\, +}{}_{\theta\phi}$:

$$G^{i\, +}{}_{tr}= i({}^{\star}G^{i\, +}){}_{\theta\phi}=
-i\frac{2|X|^{2}}{r^{2}\sin{\theta}} G^{i\, +}{}_{\theta\phi}\, ,$$

which combined with

$$G^{i\, +} =\mathcal{T}^{i}{}_{\Lambda}F^{\Lambda\, +}
= 
{\textstyle\frac{i}{2}}
\mathcal{G}^{ij^{*}}
\langle\, \mathfrak{D}_{j^{*}}\mathcal{V}^{*} \mid F\, \rangle
=
{\textstyle\frac{i}{2}}
\mathcal{G}^{ij^{*}}
\mathfrak{D}_{j^{*}}\langle\, \mathcal{V}^{*} \mid F\, \rangle\, ,$$

leads to

$$\label{eq:esaecuacion}
\frac{dZ^{i}}{dr} =2\sqrt{2} \frac{X}{r^{2}\sin{\theta}} 
\mathcal{G}^{ij^{*}}
\mathfrak{D}_{j^{*}}\langle\, \mathcal{V}^{*} \mid F_{\theta\phi}\, \rangle \, .$$

Since the form of all the fields in terms of $\mathcal{I}(r)$ is in principle known, we can try to find a more explicit form for this equation: using the general form of the vector fields Eq. (\[eq:esas\]) and of $\mathcal{I}(r)$, Eq. (\[eq:Iforspherical\]), we find

$$F_{\theta\phi} 
=
{\textstyle \frac{1}{\sqrt{2}}}r^{2}\sin{\theta}\frac{d\mathcal{I}}{dr}
=
-\frac{q}{\sqrt{2}}\sin{\theta}\, .$$

After substituting this into Eq. (\[eq:esaecuacion\]), one ends up with

$$\frac{dZ^{i}}{d\rho} =2 X \mathcal{G}^{ij^{*}}
\mathfrak{D}_{j^{*}}\mathcal{Z}^{*} \, ,$$

where $\rho\equiv 1/r$ and where

$$\mathcal{Z}(Z,q) \equiv \langle\, \mathcal{V} \mid q \, \rangle\, ,$$

is the 