Yang-Mills CP violation

Question

Why does a term proportional to $\left(F,\,\tilde{F}\right)\propto Tr\left[ F_{\mu\nu}\tilde{F}^{\mu\nu}\right]$ in the Lagrangian of the pure Yang-Mills theory violate CP?

Answer 1

If we write the field strength in terms of "electric" and "magnetic" fields $\vec{E}$ and $\vec{B}$, the relevant expression can be written as

$$\text{Tr}F_{\mu\nu}\tilde{F}^{\mu\nu}=4\,\text{Tr}\vec{E}\cdot\vec{B}.$$.

Under parity transformations, $\vec{E}\rightarrow-\vec{E}$ and $\vec{B}\rightarrow\vec{B}$, while under charge conjugation, $\vec{E}\rightarrow-\vec{E}$ and $\vec{B}\rightarrow-\vec{B}$. Together, this leads to a transformation behaviour of the term in the Lagrangian under $CP$-transformations given by

$$\text{Tr}F_{\mu\nu}\tilde{F}\rightarrow-\text{Tr}F_{\mu\nu}\tilde{F}.$$

This clearly shows that the term is not invariant.

Answer 2

The usual action for Yang-Mills theory is, using differential forms $$S = \int \operatorname{tr} (F \wedge \star F)$$ where $\star$ is the Hodge dual. Now note that the integral of a differential form is always defined with respect to an orientation, and the Hodge dual is also defined with respect to an orientation. Parity is reversing the orientation, which changes the sign of both the integral and the Hodge dual, so the signs cancel.

However the term $$S_\theta = \int \theta \operatorname{tr} (F \wedge F)$$ only has one sign change when you reverse the orientation, so it is not $P$ invariant.