Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 302 / 4

Requirefor every $\varphi$. Expanding and cancelling $g{\partial }_{\mu }\varphi$ givesRight multiplication by $g^{-1}$ yields the gauge-field transformation law

The first term $g{A}_{\mu }{g}^{-1}$ lies in $\mathfrak{g}$ because the adjoint action preserves the Lie algebra. For the second, fix $x$ and consider the group curve $s\mapsto g(x+s{e}_{\mu })g(x{)}^{-1}$ through the identity. Its tangent at zero is $({\partial }_{\mu }g)g^{-1}$, so this is also in $\mathfrak{g}$. Since a Lie algebra is a vector space, $A_{\mu }^{\prime }\in \mathfrak{g}$.

Direct expansion givesSince $D_{\mu }^{\prime }(g\varphi )=g{D}_{\mu }\varphi$,Therefore the non-Abelian gauge field strength transforms covariantly:

The Killing form is invariant under the adjoint action:Together with $F_{\mu \nu }^{\prime }={\mathrm{Ad}}_g{F}_{\mu \nu }$, this immediately givesso every positive integral power ${\mathcal{L}}_1$ is gauge invariant.

The adjoint covariant derivative obeysThis follows either by substituting the transformation laws or by applying the covariance of $D_{\mu }^{\prime }$ to an adjoint-valued field. A second use of invariance of the Killing form giveswhich proves gauge invariance of ${\mathcal{L}}_2$.