Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 302 / 4 / Solution

Let $G$ be a simple Lie group, let $T^a$ be Hermitian matrices for a finite-dimensional unitary representation $R$, and normalizeA matter field $\psi$ transforms locally as ${\psi }^{\prime }(x)=U(x)\psi (x)$, where $U(x)=e^{ig{ϵ}^a(x)T^a}$. An ordinary derivative of $\psi$ does not transform covariantly because it differentiates $U$. Introduce a gauge field $A_{\mu }=A_{\mu }^a{T}^a$ and the gauge covariant derivativeDemanding $D_{\mu }^{\prime }{\psi }^{\prime }=U{D}_{\mu }\psi$ determines the Yang-Mills gauge transformationTo first order in $ϵ$,

The gauge field strength is defined by $[D_{\mu },D_{\nu }]=-ig{F}_{\mu \nu }$:or, in components,Covariance of the commutator givesThe commutator term distinguishes Yang-Mills theory from an Abelian gauge theory and produces cubic and quartic gauge-boson interactions.

For a Dirac field of mass $m$ in $R$, the Lagrangian isEquivalently, the gauge term is proportional to $-\mathrm{tr}(F_{\mu \nu }{F}^{\mu \nu })$. The cyclic property of the trace and $F_{\mu \nu }^{\prime }=U{F}_{\mu \nu }{U}^{-1}$ make it invariant. Unitarity gives ${\bar{\psi }}^{\prime }=\bar{\psi }{U}^{-1}$, while $D_{\mu }^{\prime }{\psi }^{\prime }=U{D}_{\mu }\psi$, so both the matter kinetic term and mass term are invariant. A complex scalar $\varphi$ in a unitary representation may instead be coupled throughprovided the scalar potential $V$ is $G$-invariant.

The simplicity assumption means that the Lie algebra $\mathfrak{g}$ is nonabelian and has no proper nonzero Ideal of a Lie algebra. Its Adjoint representation is therefore irreducible, and every invariant symmetric bilinear form is proportional to the Killing form. Consequently the pure gauge kinetic term has one overall gauge coupling for a simple factor. The theory has no independent Abelian gauge direction; if the gauge algebra were a direct sum of simple and Abelian ideals, each factor could instead carry its own coupling. A simple group may still have a discrete center, but this does not add a gauge boson because gauge bosons are indexed by the Lie algebra.