Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 52 / 3 / Solution

Covariance of the fundamental gauge covariant derivative requires $D_{\mu }^{f\prime }(g\chi )=g{D}_{\mu }^f\chi$ for every column $\chi$. Expanding the left side and cancelling $g{\partial }_{\mu }\chi$ gives $({\partial }_{\mu }g)+A_{\mu }^{\prime }g=g{A}_{\mu }$. Therefore the necessary and sufficient Yang-Mills gauge transformation isIn operator notation $D_{\mu }^{f\prime }=g{D}_{\mu }^f{g}^{-1}$. Expanding its commutator on an arbitrary column shows $[D_{\mu }^f,D_{\nu }^f]\chi =F_{\mu \nu }\chi$: the second derivatives cancel, as do the terms containing a single derivative of $\chi$. Conjugating this identity proves the gauge field strength transformationThus the gauge field strength transforms homogeneously even though the potential has an inhomogeneous derivative term.

For the adjoint scalar field, write $B_{\mu }=({\partial }_{\mu }g)g^{-1}$ and ${\Phi }^{\prime }=g\Phi g^{-1}$. Differentiating the inverse matrix givesThe extra terms cancel, proving $D_{\mu }^{\prime }{\Phi }^{\prime }=g(D_{\mu }\Phi )g^{-1}$ for the adjoint covariant derivative. Its curvature is obtained by another direct expansion:where the Jacobi identity combines the last two terms.

For the equations of motion take compactly supported variations, and assume $e\ne 0$, as required by the Lagrangian coefficient. Invariance of the Killing form impliesIndeed, the two connection terms sum to $\kappa ([A_{\mu },X],Y)+\kappa (X,[A_{\mu },Y])=0$. This is the gauge-covariant integration by parts identity. The needed variations areAntisymmetry of the gauge field strength combines its two variation terms. Applying gauge-covariant integration by parts to the action $S=\int \mathcal{L}{d}^{4}x$ givesHere $\kappa ([\delta A_{\nu },\Phi ],D^{\nu }\Phi )=\kappa (\delta A_{\nu },[\Phi ,D^{\nu }\Phi ])$ fixes the sign of the scalar current. Since the Killing form is nondegenerate on a semisimple Lie algebra, independent variations give the coupled Yang-Mills equations and scalar equationEquivalently, $D_{\mu }{F}^{\mu \nu }=e^2[D^{\nu }\Phi ,\Phi ]$. These signs use the stated Minkowski metric, with upper spatial derivatives $D^i=-D_i$.

It remains to verify every equation for the static Bogomolny equations. The hypotheses imply $F_{0i}=0$ and $D_0\Phi =0$, so the temporal gauge equation is automatically zero. The gauge-theory Bianchi identityfollows, for example, from the Jacobi identity for the fundamental gauge covariant derivatives, using their commutator $F_{ij}$. Define $B_i=\frac{1}{2}{\epsilon }_{ijk}{F}_{jk}$. Contracting the gauge-theory Bianchi identity with ${\epsilon }_{ijk}$ gives $D_i{B}_i=0$. The Bogomolny equations say $B_i=e{D}_i\Phi$, so $e\ne 0$ implies $D_i{D}_i\Phi =0$. In the static configuration $D_{\mu }{D}^{\mu }\Phi =-D_i{D}_i\Phi$, verifying the scalar equation.

For the spatial gauge equation, use $F^{ij}=F_{ij}$ and $D^j\Phi =-D_j\Phi$. The Bogomolny equations, the curvature commutator proved above, and the contraction of two Levi-Civita symbols givesince ${\epsilon }_{ijk}{\epsilon }_{ik\ell }=-2{\delta }_{j\ell }$. This is exactly $D_i{F}^{ij}+e^2[\Phi ,D^j\Phi ]=0$. Thus the static first-order Bogomolny system satisfies all four gauge equations and the scalar equation, with no extra ansatz or boundary assumption needed for this local implication. This is the mechanism by which Bogomolny monopole equations imply Yang-Mills-Higgs equations.