# How to verify a vertex is gauge invariant?

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 $$p_{\mu}V^{\mu\nu\rho} (p, q, r) = 0, \tag{75.19}$$ $$q_{\nu}V^{\mu\nu\rho} (p, q, r) = 0, \tag{75.20}$$ $$r_{\rho}V^{\mu\nu\rho} (p, q, r) = 0. \tag{75.21}$$

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?

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.