 Solution

Choose orientation ${ϵ}_{1234}=1$ and the anti-Hermitian basis in the question. The Pauli matrix commutator identity gives $[e_a,e_b]={ϵ}_{abc}{e}_c$. Define the gauge curvature and adjoint covariant derivative byFor a compactly supported variation $a_{\mu }=\delta A_{\mu }$,Antisymmetry and the invariant Lie-algebra inner product therefore giveCovariant integration by parts is valid because
$⟨X,[A,Y]⟩=-⟨[A,X],Y⟩$. Thus the Euler-Lagrange field equations are

The Euclidean Hodge star operator on two-forms isThe Anti-self-dual Yang-Mills equations mean $\boxed{\ast F=-F}$.

The gauge-theory Bianchi identity isTo prove it, act on the defining representation with ${\nabla }_{\mu }={\partial }_{\mu }+A_{\mu }$. Its commutator is multiplication by $F_{\mu \nu }$. The operator Jacobi identity for ${\nabla }_{\mu },{\nabla }_{\nu },{\nabla }_{\rho }$ becomes multiplication by the displayed cyclic sum, which must vanish. Equivalently, substitution of $F=dA+A\wedge A$ into $D_{A}F$ cancels the ordinary derivatives, leaving the Lie-algebra Jacobi identity. Contracting the cyclic identity with the alternating tensor gives $D_{\mu }(\ast F{)}_{\mu \nu }=0$. If $\ast F=-F$, this is exactly $D_{\mu }{F}_{\mu \nu }=0$, proving the first-order-to-second-order implication without assuming the field equations.

LetAll squared component sums below include both orders of each antisymmetric pair, as does the printed energy. Since the Hodge star preserves this norm,Consequently $V\ge |J|/2$. An anti-self-dual field has $J=-2V\le 0$ and saturates the first bound. Among all finite-energy potentials with the same value of $J$, it therefore has the absolute minimum $-J/2$. For positive $J$, the saturating sign is self-duality instead; there cannot be a nonzero anti-self-dual solution in that sector. This is the Yang-Mills instanton Bogomolny bound in the paper's normalization.

For the topological meaning, regard $F=\frac{1}{2}{F}_{\mu \nu }^a{e}_{a}d{x}^{\mu }\wedge d{x}^{\nu }$ as a matrix-valued two-form. Since $\mathrm{tr}(e_a{e}_b)=-{\delta }_{ab}/2$,In the Second Chern number convention used here,With the standard instanton boundary condition, the connection approaches a pure gauge at infinity and extends to a bundle over the compactified four-sphere. This makes $k$ an integer, the instanton number; reversing the orientation or using the opposite charge convention reverses its sign. The square-completion argument itself does not require quantization or a chosen framing.

One can see the boundary interpretation directly. The Chern-Simons 3-form obeysFor an asymptotic potential $A=-dg{g}^{-1}$, the flatness identity $dA=-A\wedge A$ givesThis is the winding number of $g:S_{\infty }^3\to SU(2)\cong S^3$, with the corresponding orientation convention. Smooth variations preserving that boundary class cannot change it. Thus the integral in the question labels the topological sector of the Yang-Mills instanton, and in an anti-self-dual sector the energy is $\boxed{V=32{\pi }^{2}k}$ with $k\ge 0$ in our convention.