# Non-Abelian Gauge Field and Fermions Under Parity?

Under a discrete parity transformation, how does a non-abelian gauge field $$A^a_{\mu}(x)$$ transform? Is it possible to get mixing between the colors? How about the fermion $$\psi_n(x)$$ which is coupled to the gauge field? Let's say they transform under some representation of the gauged Lie group, with generators $$(t_a)_{nm}$$, does the fermion mix its $$n$$ index under a parity transformation?

The gauge field $$A_\mu$$ transforms as a covector (here $$A_\mu = T^a A^a_\mu$$ is the full connection matrix). This means that $$A_\mu$$ transforms in the same way that partial derivatives $$\partial_\mu$$ transform. This is most easily seen by looking at the covariant derivative $$D_\mu = \partial_\mu + A_\mu$$. The covariant derivative transforms under coordinate changes $$x \rightarrow y$$ as

$$\frac{D}{dy^\mu} = \frac{dx^\nu}{dy^\mu} \frac{D}{dx^\nu}$$

Or, written another way,

$$D_\mu \rightarrow \frac{dx^\nu}{dy^\mu} D_\nu$$

This implies that under a coordinate change, $$A_\mu$$ transforms the same way,

$$A_\mu \rightarrow \frac{dx^\nu}{dy^\mu} A_\nu$$

So under a reflection, there is one component $$x^i$$ that transforms to $$x^i \rightarrow -x^i$$ all others staying same. So this means that

$$A_i \rightarrow -A_i$$ (i.e $$A^a_i \rightarrow -A^a_i$$)

and all other components stay the same. Note that this a purely geometric statement having nothing to do with quantizing the theory, and comes from viewing $$A_\mu$$ as a connection on a vector bundle (for example, see this other StackExchange post).

Fermions transform as usual, $$\psi \rightarrow \gamma^0 \psi$$ under parity, which corresponds to switching the left and right components of the fermi field.