electromagnetism - Fundamental invariants of the electromagnetic field

It is a standard exercise in relativistic electrodynamics to show that the electromagnetic field tensor $F_{\mu\nu}$, whose components equal the electric $E^i=cF^{i0}$ and magnetic $B_i=-\frac12\epsilon_{ijk}F^{jk}$ fields in the taken frame of reference, has two Lorentz invariant quantities, $$\frac12F^{\mu\nu}F_{\mu\nu}=\mathbf{B}^2-\mathbf{E}^2$$ and $$\frac14F_{\mu\nu} {}^\ast F^{\mu\nu}=\frac14\epsilon^{\mu\nu\alpha\beta }F_{\mu\nu}F_{\alpha\beta}=\mathbf{B}\cdot\mathbf{E}.$$

There is, however, a further Wikipedia article which states that these two quantities are fundamental, in the sense that any other invariant of this tensor must be a function of these two. While I find this plausible, I have never seen a proof of this fact, and it is absent from e.g. Jackson. Is there a simple proof of this fact? I'm particularly interested in higher-order invariants, but I would also like answers to include a proof that these are the only two bilinears.

To be more precise, I would like to see a proof that

Any function $I:F_{\mu\nu}\mapsto I(F)\in\mathbb{R}$ that takes electromagnetic field tensors to real scalars and which is Lorentz invariant (i.e. $I(\Lambda_{\mu}^\alpha \Lambda_{\nu}^\beta F_{\alpha\beta})= I(F_{\mu\nu})$ for all Lorentz transformations) must be a function $I(F)=I'(F^{\mu\nu}F_{\mu\nu},F_{\mu\nu}\ {}^\ast F^{\mu\nu})$ of the two fundamental invariants described above.

If there are multiple ways to arrive at this result, I would also appreciate comments on how they relate to each other.

Answer

Here is the proof taken from Landau & Lifshitz' "Classical Theory of Fields":

Take the complex (3)-vector: $$ \mathbf{F} = \mathbf{E}+i\, \mathbf{B}. $$ Now consider the behavior of this vector under Lorentz transformations. It is easy to show that Lorentz boosts correspond to rotations through the imaginary angles, for example boost in $(x,t)$ plane: \begin{gather} F_x=F'_x,\\ F_y = F'_y \cosh \psi - i F'_z \sinh \psi = F'_y \cos i \psi - F'_z \sin i \psi. \\ F_z = F'_z \cos i \psi + F'_y \sin i \psi, \end{gather} where $\tanh \psi = \frac{v}c$, correspond to rotation of $\mathbf{F}$ through imaginary angle $i \psi$ in the $(y,z)$ plane.

Overall, the set of all Lorentz transformations (including also the purely spacial rotations) is equivalent to the set of all possible rotations through complex angles in three-dimensional space (where the six angles of rotation in four-space correspond to the three complex angles of rotation of the three-dimensional system).

The only invariant of a vector with respect to rotation is its square: $\mathbf{F}^2 = E^2 - B^2 + 2 i (\mathbf{E}\cdot \mathbf{B})$ thus the real quantities $E^2-B^2$ and $(\mathbf{E}\cdot \mathbf{B})$ are the only two independent invariants of the tensor $F_{\mu\nu}$.

So in essence, we reduce the problem of invariant of $F_{\mu\nu}$ under Lorentz tranform to invariants of a 3-vector under rotations which is square of a vector (and only it). So any invariant $I(F)$ has to be the function of $\Re(F^2)$ and $\Im(F^2)$.