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When we have an abelian group the gauge transformation of gauge field is given by

$$ A^\mu \rightarrow A^\mu + \partial^\mu \alpha \equiv A^{\prime \mu} $$

Here it's easy to see that the path integral is gauge invariant since

$$ DA^\mu \rightarrow \Big|\det\Big(\frac{DA}{DA^\prime}\Big)\Big|DA^{\prime \mu} = DA^{\prime \mu}, \quad \frac{DA^{\prime \mu}}{DA^\nu} = \delta^\mu_\nu + \frac{D\partial^\mu\alpha}{DA^\nu} = \delta^\mu_\nu, \quad \frac{D\partial^\mu\alpha}{DA^\nu} = 0 $$

Nevertheless, for the non-abelian case we have the field tranformation

$$ A^{\prime \mu} = UA^\mu U^\dagger + \frac{i}{g}U\partial^\mu U^\dagger $$

So,

$$ \frac{DU\partial^\mu U^\dagger}{DA^\nu} = 0,\ \mbox{but}\ \frac{DUA^\mu U^\dagger}{DA^\nu} = U\frac{DA^\mu }{DA^\nu}U^\dagger $$

The last equation shows that

$$ DA^{\prime \mu} = UDA^\mu U^\dagger \neq DA^\mu,\ \mbox{due to the group is non-abelian} $$

My problem comes form the fact that my Professor said in class that they had to be equal (as assumption), but I have proven contrary. What do you think?

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    $\begingroup$ Doesn't the fact that the determinant of $U$ is the inverse of the determinant of $U^{\dagger}$ solves the problem? $\endgroup$ – user245141 Oct 27 '19 at 8:10
  • $\begingroup$ @yu-v Right, I can't understand how I didn't see it. If you write an answer I can mark it as the right one in order to let this question closed $\endgroup$ – Vicky Oct 27 '19 at 15:59
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Since $\det(U) = \det(U^{\dagger})^{-1}$ and the determinant is multiplicative, you get $$\det \left[U \frac{DA^{\mu}}{DA^{\nu}} U^{\dagger}\right] = \det\left[\frac{DA^{\mu}}{DA^{\nu}}\right]$$

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