# Where do the color indices come back in $SU(3)$ Yang-Mills Quantization?

Can the partition function of $$SU(3)$$ (the Generic Partition function for a yang-mills theory found on the linked wiki page below), be split into a sum of 8 functional integrals for each gauge field?

https://en.wikipedia.org/wiki/Yang-Mills_theory#Quantization

$$F_{\mu\nu}$$ is shorthand for $$F^a_{\mu\nu}T^a$$ where $$T^a$$ are eight SU(3) generator matrices satisfying $$[T^a,T^b]=i\,f^{abc}T^c$$ and $$\text{tr}(T^a T^b)=\frac{1}{2}\delta^{ab}$$. So the first term in $$Z$$ contains the expression
$$\text{Tr}(F^{\mu\nu}F_{\mu\nu})=\text{Tr}(F^{a\mu\nu}T^a F^b_{\mu\nu}T^b)=F^{a\mu\nu}F^b_{\mu\nu}\text{Tr}(T^a T^b)=\frac{1}{2}F^{a\mu\nu}F^a_{\mu\nu}$$
$$F^a_{\mu\nu}=\partial_\mu A^a_\nu-\partial_\nu A^a_\mu+gf^{abc}A^b_\mu A^c_\nu,$$
due to the fact that SU(3) is nonabelian, means that each of these terms contains potentials with other color indices. So there is no clean split into a $$Z$$ for each color index when expressed in terms of potentials. This is expressing the fact that gluons interact with gluons.
• how would we expand the $[dA]$ were integrating over? – Craig Nov 19 '18 at 7:02
• I’m rusty on the path integral approach, but it seems like it would be $\prod_{a=1}^{8}\prod_{\mu=0}^3 dA^{a\mu}$... all the different potentials. – G. Smith Nov 19 '18 at 7:10