How does the bi-vector $\mathbf{F}=\mathbf{E}+i\mathbf{B}$ generalize to curved space? How does $\mathbf{F}=\mathbf{E}+i\mathbf{B}$ generalize to curved space? (where $\mathbf{F}$ is the bivector of electromagnetism).
Here is what I am struggling with:
On the one hand, I can expand $\mathbf{F}$ as follows:
$$
\mathbf{F}=E_x\gamma_0\wedge \gamma_1+E_y\gamma_0\wedge \gamma_2+E_z\gamma_0\wedge \gamma_3+B_x\gamma_2\wedge\gamma_3+B_y\gamma_1\wedge \gamma_3+B_z\gamma_1\wedge \gamma_2 \tag{1}
$$
Or I can do it as follows:
$$
\mathbf{F}=(E_x+IB_x)\gamma_0\wedge \gamma_1+(E_y+IB_y)\gamma_0\wedge \gamma_2+(E_z+IB_z)\gamma_0\wedge \gamma_3 \tag{2}
$$
Since $I=\gamma_0\wedge \gamma_1\wedge \gamma_2 \wedge \gamma_3$, (1) is equivalent to (2).
But if I convert (1) to curvilinear coordinates, I do not end up to the same end result as if I had started with (2). So what is the correct way to go to general curvilinear coordinates: $(\gamma_0,\gamma_1,\gamma_2,\gamma_3)\to(\mathbf{e}_0,\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3)$ ?
Replacing from (1) I get
$$
\mathbf{F}=E_x\mathbf{e}_0 \wedge \mathbf{e}_1+E_y\mathbf{e}_0 \wedge \mathbf{e}_2+E_z\mathbf{e}_1\wedge \mathbf{e}_3+B_x\mathbf{e}_2\wedge\mathbf{e}_3+B_y\mathbf{e}_1\wedge \mathbf{e}_3+B_z\mathbf{e}_1\wedge \mathbf{e}_2 \tag{3}
$$
whereas if I start with (2) I instead get:
$$
\mathbf{F}=(E_x+IB_x)\mathbf{e}_0\wedge \mathbf{e}_1+(E_y+IB_y)\mathbf{e}_0\wedge \mathbf{e}_2+(E_z+IB_z)\mathbf{e}_0\wedge \mathbf{e}_3  \tag{4}
$$
(3) is not the same as (4) because (4) expands as (just the x part):
$$
\mathbf{F}_x=E_x\mathbf{e}_0\wedge \mathbf{e}_1 + B_x \gamma_0\wedge \gamma_1\wedge \gamma_2 \wedge \gamma_3 \wedge \mathbf{e}_0\wedge \mathbf{e}_1
$$
In the case of (4) the $Es$ remain "orthogonal" to the $Bs$ in curved space, whereas for (3), the $Es$ are free to distort generally with respect to the $Bs$.
 A: Where does (2) come from? It seems to be wrong on many levels. For example, $(E_x + IB_x)$ is the addition between the base field with a 4-form; this is then "multiplied" by a base 2-form $\gamma_0\wedge\gamma_1$ and your last equality seems to suggest a mixed notion of multiplication. Observe that the exterior product of more than 4 1-forms in a space of dimension 4 is always 0.
I think one should treat the "complexification" of the Faraday tensor as just a mathematical convenience to express the complex structure on $\bigwedge^2 TM$ when $M$ is a Lorentzian manifold of dimension 4. For starters, we observe that, as vector spaces, $\bigwedge^2 TM\cong\mathbb C^3$. However, we can endow the space of 2-forms with a complex structure by observing that this is given by Hodge duality. Indeed, if we operate the usual decomposition of $F$ in terms of polar and axial components $(\mathbf E,\mathbf B)$, we have (modulo a sign) that $\star F = (\mathbf B,-\mathbf E)$. Therefore, if we evocatively write
$$F = \mathbf E + i\mathbf B,$$
Hodge duality is indeed multiplication/division (depending on sign convention) by $i$. By embedding $\bigwedge^3 TM\cong\mathbb R^4$ into $\mathbb C^4$ we can express Maxwell's equations in the unified form
$$\text diF +i\text dF+J=0.$$
More generally, if we distinguish between electric and magnetic currents $J_e$ and $J_m$, we can write the more general and symmetric equation
$$\text diF +i\text dF+J_e + iJ_m = 0.$$
This is just a convenient way of encoding a system of many equations, and in particular the pair
$$\text d F + J_m = 0,\qquad \text d\star F + J_e = 0$$
into a "single" complex equation.
So, back to the original question, the expression, when interpreted in this way, is already generalised to arbitrary Lorentzian manifolds.
