= ADHM factorization identity
{c}
{title2=$\mathcal D_z\mathcal D_z^\dagger=1_2\otimes f^{-1}$}
With $z_1=x^1+ix^2$, $z_2=x^3+ix^4$, form
$$
\mathcal D_z=\begin{pmatrix}B_2-z_2&B_1-z_1&I\\-B_1^\dagger+\bar z_1&B_2^\dagger-\bar z_2&J^\dagger\end{pmatrix}.
$$
The two <ADHM construction> constraints say that $\mathcal D_z\mathcal D_z^\dagger$ has zero off-diagonal blocks and equal diagonal blocks, so it equals $1_2\otimes f^{-1}$. Assume it is invertible everywhere. Choose an orthonormal kernel <bundle frame> $\Psi$, so $\mathcal D_z\Psi=0$ and $\Psi^\dagger\Psi=1_N$, and set $A=\Psi^\dagger d\Psi$. The orthogonal complement projector is $1-\Psi\Psi^\dagger=\mathcal D_z^\dagger(1_2\otimes f)\mathcal D_z$. Differentiating the kernel equation gives
$$
F=\Psi^\dagger d\mathcal D_z^\dagger(1_2\otimes f)\wedge d\mathcal D_z\Psi.
$$
Its space-time <differential two-form> entries are combinations of $dz_1\wedge d\bar z_1-dz_2\wedge d\bar z_2$, $dz_1\wedge d\bar z_2$, and $d\bar z_1\wedge dz_2$, all anti-self-dual. Thus the constructed <gauge curvature> satisfies the <Anti-self-dual Yang-Mills equations>.
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