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Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/7/Physics%20Papers%202018.pdf?cookie_passthrough=1

The bundle filename says 2018, but the paper's cover establishes Trinity Term 2019, 24 April 2019, and formal code A15272W1. These are solutions to that 2019 paper. Use natural units, the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$, and the Feynman i-epsilon prescription. Write $/p={\gamma }^{\mu }{p}_{\mu }$. In question 1, $D=4-ϵ$, $g_D={\mu }^{ϵ/2}g$, and the amputated gluon self-energy insertion is $i{\Pi }^{\mu \nu ,ab}$. This fixes the overall sign and factor of $i$ in the results below; defining the insertion as $-i\Pi$ instead reverses all corresponding definitions of $\Pi$. The quadratic Casimir operator in the Adjoint representation is $C_A=N_c$ for $SU(N_c)$, hence $C_A=3$ in QCD.

Route a directed ghost loop so that its internal four-momenta are $l$ and $l+k$. At the external gluon with indices $(\mu ,a)$, let the incoming ghost carry colour $c$ and the outgoing ghost carry colour $d$; the other Feynman vertex has the reverse order. The stated ghost-gluon Feynman rules supply $g_D^2{f}^{acd}{f}^{bdc}(l+k{)}^{\mu }{l}^{\nu }$. Their colour factor isThe two ghost propagators supply $i^2=-1$, while the closed ghost loop supplies another $-1$. Thus, with the insertion convention stated above,This establishes every prefactor before integration. To isolate its ultraviolet pole, take generic off shell $k$, initially $k^2<0$, so the massless integral has no overlapping infrared divergence. Introduce a Feynman parameter, shift $r=l+xk$, and set $\Delta =-x(1-x)k^2>0$. Odd powers of $r$ vanish, givingThe needed dimensional regularization integrals follow from the supplied gamma function formula. In $D=4-ϵ$,Here $\Gamma (ϵ/2)=2/ϵ+O(1)$, rather than $1/ϵ$. Using $1/D=1/4+O(ϵ)$ and ${\int }_0^{1}x(1-x)dx=1/6$, obtainTherefore the ghost-loop pole of the gluon self-energy isThe factor ${\mu }^{ϵ}$ in $g_D^2$ affects only finite terms at this order. The isolated ghost loop is not transverse: contracting this result gives $g^2{C}_A{\delta }^{ab}{k}^2{k}^{\nu }/(32{\pi }^2ϵ)$. Transversality is a condition on the complete gluon self-energy, including the other loop graphs.

There are four topologies for the one-loop gluon self-energy, shown in the upper row of the figure:

The two internal-gluon graphs have symmetry factor $1/2$. The directed quark and ghost loops have symmetry factor one and a minus sign for a closed anticommuting loop. The four-gluon tadpole diagram must be included in the list even though its massless Feynman integral vanishes in dimensional regularization. Local counterterms are excluded here.

The Slavnov-Taylor identity implies transversality of the full perturbative gluon self-energy:The quark loop is separately transverse. This follows without evaluating its Feynman integral: contract its vector Feynman vertex with $k_{\mu }$ and use $/k=(/l+/k-m)-(/l-m)$ between its two Dirac propagators. One obtains a difference of two loop integrals that cancel after a shift of $l$ in dimensional regularization. The colour factor is proportional to $\mathrm{tr}(T^a{T}^b)$ and does not obstruct this argument.

The four-gluon tadpole diagram vanishes after removal of $m_g$. Thus the nontransverse parts of the two-three-gluon-vertex loop and the ghost loop must cancel. Contracting the preceding ghost-loop pole of the gluon self-energy givesFor QCD, set $C_A=3$. No evaluation of the remaining three-gluon Feynman diagram is required. Its uncontracted transverse part is not fixed by this identity.