# Ultraviolet Power Counting in the Four Photon Vertex

In the textbook Quantum Field Theory From Basics to Modern Topics, by François Gelis, the genuine ultraviolet degree of divergence for the four-photon vertex, as illustrated below, is found to be $$-4$$, as oppose to $$0$$, as one gets from superficial power counting. This result has deep implication in the renormalizability of the theory, but I had difficulty understanding the proof of this statement. According to the book, if we denote the amputated four photon function by $$\Gamma^{\mu\nu\rho\sigma}(k_1,k_2,k_3,k_4)$$, then, by Ward-Takahashi identity, we have $$k_{1\mu}\Gamma^{\mu\nu\rho\sigma}=k_{2\nu}\Gamma^{\mu\nu\rho\sigma}=k_{3\rho}\Gamma^{\mu\nu\rho\sigma}=k_{4\sigma}\Gamma^{\mu\nu\rho\sigma}=0\tag{3.88}$$ which must also be true for the UV divergent part of the function (corresponding to the $$\epsilon^{-1}$$ pole in dimensional regularization), so that $$\Gamma_{\text{divergent}}^{\mu\nu\rho\sigma}A_{\mu}A_{\nu}A_{\rho}A_{\sigma}$$ is gauge invariant (I verified this myself). Now comes the confusing part, the book asserts that this result implies "it" (not sure what is represented by this) can be rewritten with four powers of the field strength tensor $$F_{\alpha\beta}$$. And the book says for this to be possible $$\Gamma_{\text{divergent}}^{\mu\nu\rho\sigma}$$ must contain four powers of the momenta, which implies the genuine UV divergence for the loop is $$-4$$ and the vertex is UV finite, granting the renormalizability of the theory.

Given that I know $$[A^\mu]=[M]$$, $$[F^{\mu\nu}]=[M^2]$$, and $$[\mathcal L]=[M^4]$$, I was not able to assemble these facts in a logically consistent way in order to understand the origin of real UV degree of divergence being $$-4$$. Any help is greatly appreciated!

1. On p. 319 it is argued that when we Taylor expand an amplitude in external momenta, the $$n$$th Taylor coefficient has its superficial degree of divergence (SDOD) lowered by $$n$$.