Is the Lagrangian density of electromagnetism half-blind? The Lagrangian density of electromagnetism is
$$
\mathcal{L}_{EM}=\frac{1}{4\mu_0}F^{ab}F_{ab}
$$
This represents one of two fundamental Lorentz invariants of electromagnetism. The second one is:
$$
\frac{1}{2}\epsilon_{abcd}F^{ab}F^{cd}
$$
Since $\mathcal{L}_{EM}$ contains only 1 out of 2 fundamental Lorentz invariant, how is it the case that $\mathcal{L}_{EM}$ not "half-blind"? Does the absence of the second fundamental Lorentz invariant from $\mathcal{L}_{EM}$ erases any features of electromagnetism from the solutions, that would otherwise be present in nature who obviously accounts for both invariants?
 A: You can add this to the Lagrangian if you want, but it will have no effect whatsoever. Try running the Lagrangian with the extra term through the Euler-Lagrange equation; it's a bit tedious, but you'll see it has no effect on the equations of motion. The reason why is that this term can be written as a total derivative (see this question), and two Lagrangians differing by the total derivative of a function will describe the same physical system (i.e. will return the same equations of motion).
A: The quantity you propose is a total derivative;  specifically,
$$
\frac{1}{2} \epsilon_{abcd} F^{ab} F^{cd} = \partial^a \left( \epsilon_{abcd} A^b F^{cd} \right).
$$
Since adding a total derivative to any Lagrangian doesn't change the classical equations of motion, it doesn't matter if this invariant is in the Lagrangian or not, and it's customary to just leave it out.
(At the quantum level there are interesting physically observable phenomena that can arise from total-derivative terms, but that's a separate question and one I'm not as qualified to answer.)
