How to calculate $F^{\mu\nu}F_{\mu\nu}$ efficiently? The quantity $\frac{1}{2}F^{\mu\nu}F_{\mu\nu}$ is equal to $B^2-E^2$. I can show this pretty laboriously via matrix multiplication, and slightly faster via the index notation. Is there an elegant method of computing this, possibly using the fact that $F$ is a 2-form $F=dA$?
Note: I can do the brute force calculation in components, and I understand how to do that calculation efficiently. I'm looking for a geometric argument ideally because the component makes the result seem surprising.
 A: There are no escape, you need to single out time from space-time in order to define $E$ and $B$. If you want to work in terms of $d$ and $A$, you need separate time from space as:
$$
d=\tilde d+dt\partial_{t}\qquad A= \tilde A + A_t dt
$$
Now, the electric field and magnetic field can then be defined as:
$$
E dt=(\partial_{t}\tilde A+\tilde d A_{t})dt\qquad B=\tilde d \tilde A
$$
where $E$ is a $1$-form and $B$ is a $2$-form, both in a $3$-dimensional space. Now you can compute $|F|^{2}=F\wedge * F$ as
$$
|F|^{2}_{d=4}=|(\tilde d+dt\partial_{t})(\tilde A + A_t dt)|^{2}_{d=4}=|B+Edt|^{2}_{d=4} = (|B|^{2}_{d=4}+|Edt|^{2}_{d=4})
$$ 
where the crossing terms vanish since $B\wedge*Edt=0$.
The reason why  $B\wedge*Edt=0$  is because $*Edt$ will have at least one differential $dx^{i}$ that is also in $B$, so it will vanish since $dx^i\wedge dx^i =0$. Note that there are just 3 spatial differentials $dx^{i}$, $i=1$ to $3$, and both $B$ and $*(Edt)$ are $2$-forms that only depends on spatial differentials. 
Reducing $d=4$ to $d=3$, note that $(*B)_{d=4}=(*B)_{d=3}dt$ and $(*Edt)_{d=4}=-(*E)_{d=3}$, so we get:
$$
|B|^{2}_{d=4}=(B\wedge *B)_{d=4} = (B\wedge *B)_{d=3} dt = |B|^{2}_{d=3} dt
$$
$$
|Edt|^{2}_{d=4}= (E\wedge * E)_{d=4} = -(E\wedge * E)_{d=3} dt = -|E|^{2}_{d=3} dt
$$
Note that the minus sign comes from the star product when it hits a time-like differential: 
$$
*(dt\wedge dx^{i})=\eta^{t t}\eta^{i_1 j_1} \varepsilon_{t j_1 j_2 j_3} dx^{j_2}\wedge dx^{j_3} = - \varepsilon_{t i_1 j_2 j_3} dx^{j_2}\wedge dx^{j_3}
$$
since $\eta^{tt}=-1$.
