Computing the curvature associated to "quaternionic" connection

Question

I am attempting to verify that for a matrix $A=\begin{pmatrix}\alpha &\beta\\\gamma & \delta\end{pmatrix}\in\text{Sp}(2)$ where $\text{Sp}(2)$ is the compact sympletic group and the matrix entries are quaternions, the connection defined by $\omega=\text{Im}(\bar{\alpha}d\alpha+\bar{\gamma}d\gamma)$, which takes values in $S^3\cong \text{SU}(2)$ satisfies the Yang-Mills equation (I think these solutions are known as BPST instantons in physics). Given a section of the principal bundle $S^7\xrightarrow{\pi}\mathbb{HP}^1$ for a local trivialisation, i.e. $\sigma:U_1\to \pi^{-1}(U_1)$, it is given that for $q\in\mathbb{H}$ we have $A_q=(\sigma^*\omega)_q=\text{Im}(\frac{\bar{q}}{1+|q|^2}dq)$. Let $q=x+iy+jv+kw$, and $X=X^1+iX^2+jX^3+kX^4$, likewise for $Y$. I computed that $$dA_q=\frac{1}{1+|q|^2}d\bar{q}\wedge dq \\ [A_q,A_q](X,Y)=\frac{2}{(1+|q|^2)^2}\text{Im}(\text{Im}(\bar{q}X)\text{Im}(\bar{q}Y)) $$ I feel like I am quite close to being able to express the "field strength" $F^A$ of the pulled back connection this way, since $F^A_q=A_q+\frac{1}{2}[A_q,A_q]$. After I do so, I think I can show that it is self-dual, and therefore satisfies the Yang-Mills equation. The issue is that I have not been able to write the expression I computed for the Lie bracket independent of the arguments $X$ and $Y$. Could anyone help me by giving a hint, or explaining how I can convert the given expression into an actual differential form? I considered $\text{Im}(\text{Im}(\bar{q}dq)\wedge\text{Im}(\bar{q}dr))$ for some $r$ but I did not quite see how to make it work. Thanks for your time.

Answer 1

I obtained a solution. I first showed that $\text{Im}(\bar{q}dq)\wedge\text{Im}(\bar{q}dq)=\text{Im}(\bar{q}dq\wedge\bar{q}dq)$, and using this fact along with $\bar{q}\wedge q$ being anti-self-dual,

$$F^A_q=dA_q+A_q\wedge A_q=\text{Im}(\frac{d\bar{q}\wedge dq}{1+|q|^2}+\bar{q}d(\frac{1}{1+\bar{q}q})\wedge dq+\frac{\bar{q}dq\wedge\bar{q}dq}{(1+|q|^2)^2})=\\ \text{Im}((1+|q|^2)\frac{d\bar{q}\wedge dq}{(1+|q|^2)^2}-\bar{q}\frac{qd\bar{q}\wedge dq}{(1+|q|^2)^2}-\bar{q}\frac{dq\wedge \bar{q} dq}{(1+|q|^2)^2}+\frac{\bar{q}dq\wedge\bar{q}dq}{(1+|q|^2)^2})=\\ \text{Im}((1+|q|^2)\frac{d\bar{q}\wedge dq}{(1+|q|^2)^2)}-|q|^2\frac{d\bar{q}\wedge dq}{(1+|q|^2)^2})=\\ \text{Im}(\frac{d\bar{q}\wedge dq}{(1+|q|^2)^2})$$ Which is anti-self-dual by the above remark.