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Locally a U(1) connection whose gauge curvature has no $(2,0)$ or $(0,2)$ part can be written $A=\partial u+\bar{\partial }v$ by the Dolbeault-Poincaré lemma. Unitarity permits $v=-\bar{u}$. Writing $u=p+iq$, a gauge transformation removes $idq$ and leaves $A=\partial p-\bar{\partial }p$. Setting $f=-2p$ gives the displayed unitary form, whose gauge curvature is $\partial \bar{\partial }f$. Its remaining anti-self-duality of gauge curvature condition is the scalar Laplace equation. Alternatively a complex gauge transformation by $e^{-u}$ gives $A=\bar{\partial }(v-u)$; this second gauge generally does not preserve unitarity.

// Destination: complex-geometry.bigb

Let $F_{+}=(F+\ast F)/2$ and $F_{-}=(F-\ast F)/2$, using the oriented Euclidean metric. For a self-dual two-form $\omega$,Apply this componentwise in the Lie algebra. Since the three ${\omega }_i$ span the self-dual subspace, their vanishing wedges force $F_{+}=0$. Thus $\ast F=-F$. The gauge-theory Bianchi identity is $D_{A}F=0$, so $D_A\ast F=-D_{A}F=0$, 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 $w=x^1+i{x}^2$ and $z=x^3+i{x}^4$. 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 areFor a spectral parameter $\lambda \in {\mathbb{CP}}^1$, consider the Lax pairTheir commutator isIt 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 readLocally 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 $u,v$ withSubstituting in the third equation givesIn a complex gauge transformation, use $g=e^{-u}$ and the derivative-plus-connection convention $A\mapsto A+g^{-1}dg$. ThenThis 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 $A=\partial p-\bar{\partial }p+idq$, and $g=e^{-iq}$ removes the last term. With $f=-2p$ this givesIts 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.

Let $\mathcal{J}=\star J$ be the current $(d-1)$-form on an oriented $d$-dimensional spacetime. For a slab bounded by two spacelike hypersurfaces ${\Sigma }_1,{\Sigma }_2$ and a side boundary $B$, the Stokes theorem givesThus $Q(\Sigma )={\int }_{\Sigma }\mathcal{J}$ is conserved if the side flux vanishes, for example for a spatially compact current or appropriate falloff at infinity. With side flux it instead obeys the corresponding charge-balance equation. This is the differential-form version of current conservation.

For the Abelian Chern--Simons theory, take $A\mapsto A+d\chi$ with a smooth compactly supported gauge parameter. The change in the gauge-field term is a boundary term because $d\chi \wedge dA=d(\chi dA)$. For the source,Discarding the boundary terms, the action variation isGauge invariance for every such $\chi$ and $c\ne 0$ therefore gives $d\star J=0$. If $c=0$, the current decouples and gauge invariance places no condition on it. On a spacetime with boundary, boundary conditions or boundary degrees of freedom are also needed to handle the discarded terms; on a nontrivial compact gauge bundle, large-gauge invariance is an additional global issue.

For the field equation, useThe one-form $\delta A$ commutes with the two-form $\mathcal{J}$ under the wedge product, soup to the boundary term. ConsequentlyApplying $d$ and using $d^2=0$ again gives current conservation for $c\ne 0$. The factor two comes from varying both occurrences of $A$.

Orient a spatial region $D$ so its boundary is $\gamma$, and define $Q_D={\int }_D\mathcal{J}$. Applying Stokes theorem and the field equation yieldsFor a U(1) connection, in a convention with unit minimal charge, the gauge-invariant quantity is its holonomy $e^{i\Phi }$, the Wilson-loop or Aharonov-Bohm phase. A large gauge transformation can change the chosen representative of $\Phi$ by $2\pi n$. Therefore $\Phi$ is a phase angle modulo $2\pi$, rather than an absolute gauge-invariant real number. A particle of charge $q$ has phase $e^{iq\Phi }$ in the corresponding normalization. The flux-charge relation thus attaches a gauge phase to enclosed charge.

The metric variation requires specifying the independent source. The pure Chern-Simons term $A\wedge dA$ has no metric dependence, hence contributes zero stress-energy tensor. If the one-forms named in the question have fixed covariant components $A_{\mu },J_{\mu }$, the Hodge star in the source does depend on the metric:Using $\delta \sqrt{-g}=-\sqrt{-g}{g}_{\mu \nu }\delta g^{\mu \nu }/2$ givesUnder that literal fixed-one-form convention,There is another common convention in the topological source theory: hold the conserved two-form $\mathcal{J}=\star J$, equivalently the vector current density, fixed as the metric varies. Then both $A\wedge dA$ and $\mathcal{J}\wedge A$ are metric-independent, and $T_{\mu \nu }=0$ for this action. These are different variations, not contradictory calculations. The Chern-Simons source stress convention explains why a source prescription is necessary; a dynamical matter source would contribute its own action and stress as well.