Consider, for instance, a fundamental massless three-form field $C_{\alpha\beta\gamma}$ in the Coulomb phase:

$$ \mathcal L = E_{\mu\alpha\beta\gamma}E^{\mu\alpha\beta\gamma} + C_{\alpha\beta\gamma}J^{\alpha\beta\gamma}\,, $$

where $E_{\mu\alpha\beta\gamma} = \partial_{[\mu}C_{\alpha\beta\gamma]}$ is the field strength and $J^{\alpha\beta\gamma}$ is a conserved external current.

In the absence of sources, the four-form electric field can take an arbitrary constant value, $E_{\mu\alpha\beta\gamma} = E_0\epsilon_{\mu\alpha\beta\gamma}$ with $E_0$ a constant number.

In a paper by Gia Dvali, it is claimed that

Any theory in which a three-form field is in the Coulomb phase ‘suffers’ from a generalized strong CP problem.

I would like to know why a constant electric field violates CP. Thanks.


1 Answer 1


This is because $E$ is a psuedo-scalar.

$E$ is a 4-form, which in 4-d is a volume form. Hence under a transformation $T^\mu_\nu$ it goes to $\det(T) E$. This makes it Lorenz invariant since Lorenz transformations have determinant 1, but not $P$ invariant since $\det(P)=-1$. Since we also expect $C$ to act trivially on $E$, we have $CP(E)=-E$.


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