What are the physical reasons for considering the Maxwell invariant ${\cal G}=E.B$ in the action? As I know, the Maxwell invariant ${\cal G}=E.B$ is the fundamental invariant such as ${\cal F}=\frac{1}{2}\left( {{{\bf{B}}^2} - {{\bf{E}}^2}} \right)$ that can be used to construct all possible invariants in electrodynamics. I was wondering what are the physical reasons to consider this invariant (in the form of ${\cal G}^2$) in the electrodynamics Lagrangian. Why the presence of ${\cal G}$ in essential in the Lagrangian? Which physical phenomena can not be explained in the absence of ${\cal G}$? Is it necessary to have ${\cal G}$ in order to construct Lagrangian in higher dimensions?
 A: $\int \theta(x) {\bf E} \cdot {\bf B}d^4x $ occurs in the action for axion electrodynamics. Where have you seen it to higher powers? Perhaps in the quantum corrections due to fermion loops?
In higher dimensions there are more invariants than just $|E|^2-|B|^2$ and ${\bf E} \cdot {\bf B}$.
A: The modern way of looking for things (effective field theory) is that you should include every term in the action consistent with the symmetries and containing the degrees of freedom you are working with. So from that point of view, you would need a reason not to include $\mathcal{G}$.
As you mentioned, in four dimensions the Lorentz invariant scalar combinations of $E$ and $B$ are $\mathcal{F}$ and $\mathcal{G}$. However, since $\mathcal{G}$ is a total derivative, it does not contribute to the equations of motion. So it's a bit like adding zero to the action; it has no effect. (In Yang-Mills theory the analogous total derivative terms can lead to subtle quantum mechanical effects like CP violation, but those effects don't occur in electromagnetism).
If you add an additional scalar degree of freedom (an axion) you can make this term more interesting, as discussed by mike stone in their answer.
