= Solution
Let $F_+=(F+*F)/2$ and $F_-=(F-*F)/2$, using the oriented <Euclidean metric>. For a <self-dual two-form> $\omega$,
$$
\omega\wedge F=\langle\omega,*F\rangle\,\mathrm{vol}
=\langle\omega,F_+\rangle\,\mathrm{vol}.
$$
Apply this componentwise in the <Lie algebra>. Since the three $\omega_i$ span the self-dual subspace, their vanishing wedges force $F_+=0$. Thus \b[$*F=-F$]. The <gauge-theory Bianchi identity> is $D_AF=0$, so $D_A*F=-D_AF=0$, the Euclidean <Yang-Mills equations>. This proves the implication locally; finite action would additionally be needed to call the solution an <instanton>.
Use the complex orientation for $w=x^1+ix^2$ and $z=x^3+ix^4$. The real and imaginary parts of $dw\wedge dz$, and $i(dw\wedge d\bar w+dz\wedge d\bar z)$, span the <self-dual two-forms>. Accordingly the three <gauge curvature> equations are
$$
F_{wz}=0,\qquad F_{\bar w\bar z}=0,\qquad F_{w\bar w}+F_{z\bar z}=0.
$$
For a <spectral parameter> $\lambda\in\mathbb{CP}^1$, consider the <Lax pair>
$$
\boxed{(D_w-\lambda D_{\bar z})\Psi=0,\qquad(D_z+\lambda D_{\bar w})\Psi=0.}
$$
Their <commutator> is
$$
[D_w-\lambda D_{\bar z},D_z+\lambda D_{\bar w}]
=F_{wz}+\lambda(F_{w\bar w}+F_{z\bar z})+\lambda^2F_{\bar w\bar z}.
$$
It vanishes for every $\lambda$ exactly when all three coefficients vanish. The derivative directions $\partial_w-\lambda\partial_{\bar z}$ and $\partial_z+\lambda\partial_{\bar w}$ are mutually orthogonal <null vectors> for the complexified metric. Flatness on the corresponding two-planes makes the linear parallel-transport equations locally compatible. Conversely compatibility for an invertible fundamental <matrix> $\Psi$ forces the <commutator> to vanish; a single zero solution would not establish it. This is the <Euclidean complex-coordinate anti-self-dual Lax pair>.
For an <Abelian> <U(1) connection>, all <commutators> of the connection components vanish, and the <gauge curvature> equations therefore read
$$
\boxed{\begin{aligned}
\partial_w A_z-\partial_z A_w&=0,\\
\partial_{\bar w}A_{\bar z}-\partial_{\bar z}A_{\bar w}&=0,\\
\partial_zA_{\bar z}-\partial_{\bar z}A_z
+\partial_wA_{\bar w}-\partial_{\bar w}A_w&=0.
\end{aligned}}
$$
Locally the first two equations say that $A^{1,0}$ is $\partial$-closed and $A^{0,1}$ is $\bar\partial$-closed. The <Dolbeault-Poincaré lemma>, and its <complex conjugate>, give scalar functions $u,v$ with
$$
A=\partial u+\bar\partial v.
$$
Substituting in the third equation gives
$$
(v-u)_{z\bar z}+(v-u)_{w\bar w}=0.
$$
In a complex <gauge transformation>, use $g=e^{-u}$ and the derivative-plus-connection convention $A\mapsto A+g^{-1}dg$. Then
$$
\boxed{A'=\bar\partial f,\qquad f=v-u,\qquad f_{z\bar z}+f_{w\bar w}=0.}
$$
This is the scalar <Laplace equation>, since for the stated metric $\Delta=2(\partial_z\partial_{\bar z}+\partial_w\partial_{\bar w})$. The complex <gauge transformation> need not remain in $U(1)$.
A reduction using only a <unitary> <gauge transformation> is also available. Since $A$ is <anti-Hermitian>, $A_{\bar w}=-\overline{A_w}$ and $A_{\bar z}=-\overline{A_z}$, so the potential for the barred components can be chosen as $v=-\bar u$. Write $u=p+iq$ with real $p,q$. Then $A=\partial p-\bar\partial p+i\,dq$, and $g=e^{-iq}$ removes the last term. With $f=-2p$ this gives
$$
\boxed{A'=\frac12(\bar\partial f-\partial f),\qquad f\text{ real},\qquad f_{z\bar z}+f_{w\bar w}=0.}
$$
Its <gauge curvature> is $\partial\bar\partial f$. Thus the <Abelian anti-self-dual connections from harmonic scalar potentials> reduction is local and does not require silently enlarging the real gauge group.
Back to article page
