 Solution

Use $F_{\mu \nu }^a={\partial }_{\mu }{A}_{\nu }^a-{\partial }_{\nu }{A}_{\mu }^a+g{f}^{abc}{A}_{\mu }^b{A}_{\nu }^c$ and $\mathcal{L}=-F_{\mu \nu }^a{F}^{a\mu \nu }/4$. Its cubic and quartic Yang-Mills self-interactions areAntisymmetry of the Lie algebra structure constants lets the cubic expression also be written as $-g{f}^{abc}({\partial }_{\mu }{A}_{\nu }^a)A^{b\mu }{A}^{c\nu }$.

The cubic vertex has three fields and one derivative, corresponding to $n=2,r=1$ in part (b). The quartic vertex has four fields and no derivatives, corresponding to $n=4,r=0$. Their dimensions areWith both types present, the same Feynman diagram counting givesIn four dimensions both couplings are dimensionless, and the degree is bounded independently of vertex count. Gauge symmetry, implemented consistently after gauge fixing, restricts the ultraviolet counterterms to the allowed renormalizable gauge-invariant and gauge fixing/ghost structures. Thus the counting is consistent with perturbative renormalizability of ordinary four-dimensional Yang-Mills theory.

In six dimensions $[g]=-1$ and $[g^2]=-2$, giving $D=6-2E+V_3+2V_4$. Increasing the vertex count permits new divergent momentum structures and higher-dimensional operators without a finite closed list of renormalization parameters. Ordinary Yang-Mills theory is therefore power-counting renormalizable in four dimensions and nonrenormalizable in six dimensions. It can still be used in six dimensions as an effective field theory with a cutoff and higher-operator expansion; this conclusion concerns the usual perturbative theory.