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In Srednicki's textbook "Quantum Field Theory", section 75 discusses chiral gauge theories and anomalies. On page 447, it is written

We would now like to verify that $V^{\mu\nu\rho} (p, q, r)$ is gauge invariant. We should have \begin{equation} p_{\mu}V^{\mu\nu\rho} (p, q, r) = 0, \tag{75.19} \end{equation} \begin{equation} q_{\nu}V^{\mu\nu\rho} (p, q, r) = 0, \tag{75.20} \end{equation} \begin{equation} r_{\rho}V^{\mu\nu\rho} (p, q, r) = 0. \tag{75.21} \end{equation}

where $V^{\mu\nu\rho} (p, q, r)$ is a three-photon vertex, and $p_{\mu}, q_{\nu}, r_{\rho}$ are the four-momenta of the three photons. Why do eqs. (75.19) - (75.21) imply that $V^{\mu\nu\rho} (p, q, r)$ is gauge invariant? Is this the only way to verify a vertex is gauge invariant?

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To be honest, I'm not into anomaly stuff, so for me it is easier to think of gluons which have a 3-vertex without any anomalies. :) I might be wrong, but I think that when you plug in the gauge transformation $A^\mu\to A^\mu + \partial^\mu \Lambda$ into the 3-gluon interaction term of the Lagrangian, you will get additional terms, all including a $\partial^\mu \Lambda$ which is traced with something which is later identified as the vertex. In Fourier space the partial derivative becomes a momentum vector, so if the trace of the vertex with the momentum w.r.t. all indices is zero, all extra terms including $\partial^\mu \Lambda$ will vanish and therefore the 3-gluon term in the Lagrangian becomes gauge-invariant. Sorry, I'm in a hurry and cannot present the calculation working out the index stuff, but I hope that I made the conceptual point clear.

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