 Solution

For the outgoing massless gauge bosons, take $k_1^2=k_2^2=0$. Work away from internal poles, with the Feynman i-epsilon prescription understood. The Dirac equations and four-momentum conservation giveMultiplying by the neighboring Dirac propagator numerator cancels its denominator. The two fermion-exchange diagrams therefore yield the matrix commutatorThe annihilation vector current obeys $q_{\lambda }{J}_c^{\lambda }=\bar{v}(/p_1+/p_2)t^{c}u=0$. Direct contraction of the three-boson tensor givesThusThe terms proportional to $J_c^{\mu }$ cancel, leaving the non-Abelian Ward identity for fermion annihilationThe open-index tensor is generally nonzero: neither ${ϵ}^{abc}$ nor $J_c\cdot k_2$ is constrained to vanish. For a concrete nonzero configuration, take centre-of-momentum frame kinematics $p_1=(E,0,0,p)$, $p_2=(E,0,0,-p)$, $p=\sqrt{E^2-m^2}$, $k_1=(E,E,0,0)$, $k_2=(E,-E,0,0)$. Standard Dirac spinors with two-component spin labels $\chi =(1,0{)}^T$ for $u$ and $\eta =(0,1{)}^T$ for $v$ give the spinor current $(0,2E,2iE,0)$ and its contraction with $k_2$ is $2E^2$. Choosing both colour spinors $(1,0{)}^T$ and $(a,b,c)=(1,2,3)$ multiplies this by $1/2$, still nonzero. This explicitly rules out an unconditional tensor identity $k_{2\nu }{\mathcal{M}}^{ab\mu \nu }=0$.

For a physical polarization vector ${\epsilon }_{1\mu }^{\ast }$, however, $k_1\cdot {\epsilon }_1^{\ast }=0$, and thereforeThis is the required decoupling of a pure-gauge polarization vector on leg 2 when the other leg is physical. Interchanging the legs gives the corresponding identity for leg 1. The role of the three-boson diagram is explicit: it cancels the noncommuting colour-current term left by the two fermion exchanges.