Faraday Tensor quadruple product I would like to compute the following:
$$F^{ab}F_{ac}F_{bd}F^{cd}$$
Is this equal to $4(E^2-B^2)^2?$
If so how can i quickly calculate it as such?
 A: There doesn't seem to be a really easy way to go about it.
Group the first and third factors together, as well as the second and fourth and your result is following squared Frobenius (trace) matrix norm:
$$\mathrm{tr}\left(M^T\,M\right)=\left\|M\right\|^2$$
where:
$$M = \left(\begin{array}{c|c}|\mathbf{E}|^2&0\\\hline\\0&\mathbf{E}\,\mathbf{E}^T+c^2\,\mathbf{B}^2\end{array}\right)$$
and $\mathbf{E}$ are the components of the electric field written as a $3\times 1$ column vector and $\mathbf{B}$ is the usual $3\times3$ magnetic induction bivector (the lower right $3\times 3$ partition of the doubly covariant/ contravariant Faraday). Now of course $\mathrm{tr}(M) = F_{a\,b} F^{a\,b} = 2(|\vec{E}|^2-c^2|\vec{B}|^2)$ whereas you should see that its squared Frobenius norm contains all kinds of complicated cross terms. The result is, in fact:
$$\begin{array}{cl}&4\,c^2 (\vec{E}\cdot\vec{B})^2 + 2 \,(|\vec{E}|^2-c^2|\vec{B}|^2)^2+2\, c^2\,|\vec{E}\times\vec{B}|^2\\=&2\,(|\vec{E}|^4+c^4\,|\vec{B}|^4+c^2\,(\vec{E}\cdot\vec{B})^2-c^2\,|\vec{E}|^2\,|\vec{B}|^2)\end{array}$$
