How can I prove that the following Lagrangian is Lorentz-invariant iff the axionic field is a pseudoscalar? My professor has given me the following action stating that $a(x)$ is an axionic field and told us in class that for this action to be Lorentz invariant the field must be a pseudoscalar.
$$ S = -\int d^4x  \frac{1}{2} \partial^\mu a \partial_\mu a - \frac{1}{4} F_{\mu \nu} F^{\mu \nu} - \frac{1}{f}[aF_{\mu \nu} \left.^* F^{\mu \nu} - 2 \partial_\mu (a A_\nu \left.^*F^{\mu \nu}) \right. \right.]$$
This is the first time that I encounter such fields in my life. I tried googling around to find some information but I didn't even understand what such a field is.
I tried proving that $S$ is Lorentz invariant iff $a$ is a pseudoscalar using the definition of a Lorentz transformation
$$\delta x^b = \delta \omega^b_a x^a$$
But the computations seem to be too difficult to me. Is there any shortcut or alternative way to prove this?
 A: The nature of field $a$ - scalar or pseudoscalar - isn't relevant for the existence of invariance of action under continuous Lorentz transformations. Really, under continuous Lorentz transformation both scalar and pseudoscalar fields are transformed trivially,
$$
a(x) \to a'(x) = a(\Lambda^{-1} x),
$$
However, there is a big difference - scalar or pseudoscalar - when we want to check the invariance of action under discrete transformations of the Lorentz group. For example, $\tilde{F}_{\mu \nu} \equiv \frac{1}{2}\epsilon_{\mu \nu \alpha \beta}F^{\alpha \beta}$ is pseudotensor, since $\epsilon_{\mu \nu \alpha \beta}$ is pseudotensor:
$$
\epsilon_{\mu \nu \alpha \beta} \to D_{\mu}^{\ \mu{'}}D^{\ \nu{'}}_{\nu}D_{\alpha}^{\ \alpha{'}}D_{\beta}^{\ \beta{'}}\epsilon_{\mu{'}\nu{'}\alpha{'}\beta{'}} = \text{det}(D)\epsilon_{\mu \nu \alpha \beta}
$$
(here $D$ is the matrix of the transformation), while $F_{\mu \nu}$ is the usual tensor. So that $F_{\mu \nu}\tilde{F}^{\mu \nu}$ is pseudoscalar, so it isn't invariant under $P$ and $T$ discrete coordinate transformations of the Lorentz group. If $a$ is scalar, then $aF\tilde{F}$ term isn't invariant too. While if $a$ is pseudoscalar, then this term is $T-$ and $P-$invariant.
