= Solution
The <double-longitudinal Ward identity for gluon fusion> uses the defining <Lie algebra> identity $[T^a,T^b]=if^{abc}T^c$ makes the boxed amplitude vanish:
$$
\boxed{k_1^\mu k_2^\nu\mathcal M_{\mu\nu}=0.}
$$
In <QED>, both external gauge bosons are <photons>, the charge matrices commute, and there is no three-photon vertex. The two fermion-exchange graphs then cancel without a third graph. In <QCD>, omitting the three-gluon interaction would leave a generally nonzero colour commutator: gauge-boson self-interaction is essential to the cancellation.
This is a tree-level manifestation of a <Ward identity> and, more generally in a gauge-fixed non-Abelian theory, the <Slavnov-Taylor identities>. They connect the matter representation, the structure constants and the gauge interactions, ensuring that auxiliary longitudinal polarizations do not enter physical probabilities. With all other external legs physical and on shell, replacing one external massless gauge-boson polarization by its <momentum> leaves zero. The double-contraction result proved here is one consequence, not by itself a proof that every open-index amplitude tensor is transverse off shell. At loop order the corresponding non-Abelian identities also involve <Faddeev-Popov ghosts> and a consistent symmetry-preserving renormalization.
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