Define
$$
F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu-ig[A_\mu,A_\nu].
$$
The gauge-field transformation is precisely the one for which the <gauge covariant derivative> $D_\mu=\partial_\mu-igA_\mu$ transforms by $D_\mu' =UD_\mu U^\dagger$. Since $[D_\mu,D_\nu]=-igF_{\mu\nu}$, it follows immediately that
$$
\boxed{F_{\mu\nu}'=UF_{\mu\nu}U^\dagger}.
$$
For $U=1+i\alpha^cT_c+O(\alpha^2)$,
$$
\delta F_{\mu\nu}=i[\alpha^cT_c,F_{\mu\nu}^bT_b].
$$
Define the <structure constant of a Lie algebra> by $[T_b,T_c]=if_{bc}{}^aT_a$. Antisymmetry then gives
$$
\boxed{\delta F_{\mu\nu}^a=f_{bc}{}^aF_{\mu\nu}^b\alpha^c}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 304 / 4 / a / Solution

Write $\Psi=\Psi^aT_a$. The infinitesimal form of $\Psi\mapsto U\Psi U^\dagger$ is
$$
\delta\Psi=i[\alpha^bT_b,\Psi^cT_c],
\qquad
\boxed{\delta\Psi^a=-f_{bc}{}^a\alpha^b\Psi^c}.
$$
The adjoint <gauge covariant derivative> is
$$
D_\mu\Psi=\partial_\mu\Psi-ig[A_\mu,\Psi],
$$
which transforms as $D_\mu\Psi\mapsto U(D_\mu\Psi)U^\dagger$. A gauge-invariant Lagrangian is therefore
$$
\boxed{\mathcal L=-\frac14\operatorname{Tr}(F_{\mu\nu}F^{\mu\nu})
+\operatorname{Tr}\!\left[\bar\Psi(i\gamma^\mu D_\mu-m)\Psi\right]}.
$$
The trace and cyclicity make each term invariant under conjugation.

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 304 / 4 / c / Solution

<Charge conjugation> reverses the gauge charge, so
$$
\boxed{A_\mu\mapsto-A_\mu}.
$$
Using the antisymmetric spinor metric $\epsilon=i\sigma^2$, a compatible action on the two Weyl fields is
$$
\boxed{
\psi_L\mapsto i\sigma^2\psi_R^*,
\qquad
\psi_R\mapsto-i\sigma^2\psi_L^*}.
$$
Complex conjugation reverses the sign of $i$ and of the gauge representation, while $A_\mu\mapsto-A_\mu$ restores the original <gauge covariant derivative>. The identities $\sigma^2(\sigma^\mu)^*\sigma^2=\bar\sigma^\mu$ and $\sigma^2(\bar\sigma^\mu)^*\sigma^2=\sigma^\mu$ then exchange the two equations of motion. Unit phases can be inserted in the two spinor transformations without changing the conclusion, subject to the mass-term relation.

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 305 / 1 / ii / Solution

The block-diagonal subgroup
$$
S(U(3)\times U(2))
=\{\operatorname{diag}(U_3,U_2):\det U_3\det U_2=1\}
\subset SU(5)
$$
is locally $SU(3)\times SU(2)\times U(1)$, with only a finite central quotient distinguishing the global groups. The hypercharge direction in the fundamental representation is
$$
Y=\operatorname{diag}\!\left(-\frac13,-\frac13,-\frac13,
\frac12,\frac12\right),
$$
which is traceless and commutes with the $SU(3)$ and $SU(2)$ blocks.

The canonically normalized $SU(5)$ generator must satisfy $\operatorname{Tr}(T_Y^2)=1/2$. Since $\operatorname{Tr}(Y^2)=5/6$,
$$
T_Y=\sqrt{\frac35}\,Y.
$$
The unified <gauge covariant derivative> contains $g_5T_YB_\mu$, whereas the Standard Model convention contains $g_YYB_\mu$. Hence $g_Y=\sqrt{3/5}\,g_5$. The non-Abelian generators already have the canonical normalization, so <gauge coupling unification> in the <SU(5) grand unified theory> predicts
$$
\boxed{g_s^2=g_w^2=g_5^2=\frac53g_Y^2}.
$$

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 305 / 2 / v / Solution

{title2=$D_\mu$}
{wiki}

A gauge covariant derivative combines an ordinary derivative with a <gauge field> so that differentiating a field preserves its gauge-transformation law. In a matrix representation one common convention is $D_\mu=\partial_\mu-igA_\mu$.

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