Let $G$ be a simple <Lie group>, let $T^a$ be Hermitian matrices for a finite-dimensional <unitary representation> $R$, and normalize
$$
[T^a,T^b]=if^{ab}{}_cT^c,
\qquad \operatorname{tr}(T^aT^b)=T(R)\delta^{ab}.
$$
A matter field $\psi$ transforms locally as $\psi'(x)=U(x)\psi(x)$, where $U(x)=e^{ig\epsilon^a(x)T^a}$. An ordinary derivative of $\psi$ does not transform covariantly because it differentiates $U$. Introduce a <gauge field> $A_\mu=A_\mu^aT^a$ and the <gauge covariant derivative>
$$
D_\mu=\partial_\mu-igA_\mu.
$$
Demanding $D_\mu'\psi'=U D_\mu\psi$ determines the <Yang-Mills gauge transformation>
$$
A_\mu'=UA_\mu U^{-1}-\frac{i}{g}(\partial_\mu U)U^{-1}.
$$
To first order in $\epsilon$,
$$
\boxed{\delta A_\mu=\partial_\mu\epsilon+ig[\epsilon,A_\mu],
\qquad
\delta A_\mu^a=\partial_\mu\epsilon^a+g f^{bc}{}_aA_\mu^b\epsilon^c.}
$$

The <gauge field strength> is defined by $[D_\mu,D_\nu]=-igF_{\mu\nu}$:
$$
F_{\mu\nu}
=\partial_\mu A_\nu-\partial_\nu A_\mu-ig[A_\mu,A_\nu],
$$
or, in components,
$$
F_{\mu\nu}^a
=\partial_\mu A_\nu^a-\partial_\nu A_\mu^a
+g f^{bc}{}_aA_\mu^bA_\nu^c.
$$
Covariance of the commutator gives
$$
F_{\mu\nu}'=UF_{\mu\nu}U^{-1},
\qquad
\delta F_{\mu\nu}=ig[\epsilon,F_{\mu\nu}].
$$
The 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 is
$$
\boxed{\mathcal L
=-\frac14F_{\mu\nu}^aF^{a\mu\nu}
+\bar\psi(i\gamma^\mu D_\mu-m)\psi.}
$$
Equivalently, the gauge term is proportional to $-\operatorname{tr}(F_{\mu\nu}F^{\mu\nu})$. The <cyclic property of the trace> and $F'_{\mu\nu}=UF_{\mu\nu}U^{-1}$ make it invariant. Unitarity gives $\bar\psi'=\bar\psi U^{-1}$, while $D_\mu'\psi'=UD_\mu\psi$, so both the matter kinetic term and mass term are invariant. A complex scalar $\phi$ in a unitary representation may instead be coupled through
$$
\mathcal L_\phi=(D_\mu\phi)^\dagger D^\mu\phi-V(\phi),
$$
provided 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.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-302/4/solution