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Take a smooth gauge transformation $g:{\mathbb{R}}^3\to SU(2)$. In the anti-Hermitian convention of the adjoint covariant derivative, setExpanding ${\partial }_j(g\Phi g^{-1})$ shows that its two differentiated-$g$ terms cancel with the extra terms in $[A_j^g,{\Phi }^g]$. Thus $D_j^g{\Phi }^g=g(D_j\Phi )g^{-1}$. The gauge field strength is $F_{jk}={\partial }_j{A}_k-{\partial }_k{A}_j+[A_j,A_k]$, and the same cancellation, or the commutator of covariant derivatives, gives $F_{jk}^g=g{F}_{jk}{g}^{-1}$.

The matrix trace is unchanged by conjugation, soEach field-strength norm, covariant-gradient norm and Higgs norm is therefore invariant pointwise. It follows that $V_{\lambda }(A^g,{\Phi }^g)=V_{\lambda }(A,\Phi )$. The displayed tensor made from these fields is also gauge invariant wherever $\Phi \ne 0$.

There is a normalization problem in the requested final identity. With the ordinary defining $2\times 2$ trace and the printed inner product, an explicit orthonormal basis isThe last identity follows from the Pauli matrix multiplication law. In particular it is not permissible to combine this orthonormality with unit structure constants of a Lie algebra.

Write $A_j=a_j^a{e}_a$ and $\Phi =\phi e_3$ with $\phi \ne 0$, and put $s=\mathrm{sgn}\phi$. Define $C_{jk}=a_j^1{a}_k^2-a_j^2{a}_k^1$. ThenOnly the bracket of the transverse terms contributes in the $e_3$ direction, givingHence the literal tensor printed in the question isThe transverse term does not cancel with this trace normalization of su(2).

A concrete counterexample to the printed closure assertion is $\Phi =e_3$, $A_1=e_1$, $A_2=x_3{e}_2$, $A_3=0$ on a ball. Directly, $F_{12}=\sqrt{2}{x}_3{e}_3$, $F_{23}=-e_2$ and $F_{31}=0$, while $D_1\Phi =-\sqrt{2}{e}_2$, $D_2\Phi =\sqrt{2}{x}_3{e}_1$ and $D_3\Phi =0$. It follows thatThe Higgs field never vanishes. If global finite energy is desired, multiply both nonzero gauge-potential components by a smooth compactly supported cutoff equal to one on this ball and keep $\Phi =e_3$ everywhere. The energy is finite and the same local counterexample remains. Thus the claimed closure is false with the ordinary trace normalization actually printed; it is not a consequence of gauge invariance alone.

The consistent 't Hooft electromagnetic tensor for the printed inner product is insteadIn the directional gauge, its commutator coefficient is $1/{\kappa }^2$, so the terms cancel exactly andOn a connected region where $\phi \ne 0$, $s$ is constant. Commutation of ordinary partial derivatives provesAt any nonzero-Higgs point, use the permitted local gauge transformation to align the field with a fixed internal direction. Since $\mathcal{F}$ is gauge invariant, the identity derived in that gauge holds in the original gauge on the same ball. These balls cover every nonzero-Higgs region, proving that the corrected tensor is a closed differential form there.

Alternatively, keeping the tensor's printed coefficient one but replacing the inner product by $-2\mathrm{tr}(XY)$ gives the orthonormal basis $e_a=-i{\sigma }_a/2$ with $\kappa =1$. The same calculation then gives $f_{jk}=s({\partial }_j{a}_k^3-{\partial }_k{a}_j^3)$ and the intended identity. Either repair makes the bracket and inner-product normalizations agree; they must not be silently mixed.

For a nonvanishing adjoint Higgs field, put $n=\Phi /|\Phi |$. If an oriented orthonormal internal basis satisfies $[e_a,e_b]=\kappa {\epsilon }_{abc}{e}_c$, the displayed tensor is gauge invariant. In a gauge with $n=e_3$, write $A_j=a_j^a{e}_a$; the transverse quadratic terms cancel, leaving ${\mathcal{F}}_{jk}={\partial }_j{a}_k^3-{\partial }_k{a}_j^3$. Hence it is a closed differential form wherever $\Phi \ne 0$. Under trace normalization of su(2) with $-\mathrm{tr}(XY)$, ${\kappa }^2=2$ and the commutator term has coefficient $1/2$. With $-2\mathrm{tr}(XY)$, its coefficient is one. Omitting this normalization factor generally destroys closure.