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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

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$F_{\mu\nu}$ 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

$$\text{Tr}(F^{\mu\nu}F_{\mu\nu})=\text{Tr}(F^{a\mu\nu}T^a F^b_{\mu\nu}T^b)=F^{a\mu\nu}F^b_{\mu\nu}\text{Tr}(T^a T^b)=\frac{1}{2}F^{a\mu\nu}F^a_{\mu\nu}$$

Expressed in terms of the field strengths, this is eight terms, each one containing a different color index. But the third term in

$$F^a_{\mu\nu}=\partial_\mu A^a_\nu-\partial_\nu A^a_\mu+gf^{abc}A^b_\mu A^c_\nu,$$

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.

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  • $\begingroup$ how would we expand the $[dA]$ were integrating over? $\endgroup$ – Craig Nov 19 '18 at 7:02
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    $\begingroup$ 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. $\endgroup$ – G. Smith Nov 19 '18 at 7:10

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