Electromagnetic tensor:

$F_{uv} = 2 A_{[v;u]}$

$F_{uv} = \downarrow u(i) \overset{\rightarrow v(j)}{\left[\matrix{ 0 & -E_j \\ E_i & {\epsilon_{ij}}^k B_k }\right]}$

Dual (of an antisymemtric $T_{uv}$):

$*T_{uv} = \frac{1}{2} T_{ab} {\epsilon^{ab}}_{uv} $

Self-dual:

$^+T_{uv} = \frac{1}{2}(1 - i*) T_{uv} = \frac{1}{2}( T_{uv} - i * T_{uv}) = \frac{1}{2}( T_{uv} - \frac{i}{2} {\epsilon^{ab}}_{uv} T_{ab})$

$^-T_{uv} = \frac{1}{2}(1 + i*) T_{uv} = \frac{1}{2}( T_{uv} + i * T_{uv}) = \frac{1}{2}( T_{uv} + \frac{i}{2} {\epsilon^{ab}}_{uv} T_{ab})$

For EM:
$*F_{uv} = \frac{1}{2} {\epsilon_{uv}}^{ab} F_{ab} = \downarrow u(i) \overset{\rightarrow v(j)}{\left[\matrix{ 0 & B_j \\ -B_i & {\epsilon_{ij}}^k E_k }\right]}$

$^+F_{uv} = \frac{1}{2}( T_{uv} - i * T_{uv} ) = \downarrow u(i) \overset{\rightarrow v(j)}{\left[\matrix{ 0 & \frac{1}{2} (E_j - i B_j) \\ \frac{1}{2} (-E_i + i B_i) & \frac{1}{2} {\epsilon_{ij}}^k (B_k + i E_k) }\right]}$

Then $^+{F^{uv}}_{;v} = \frac{1}{2} J^u$ is a combination of all four Maxwell laws, of both ${F^{uv}}_{;v} = J^u$ and ${*F^{uv}}_{;v} = 0$, for $J^u$ real

Stress-energy:
$8 \pi T_{uv} = 2 (F_{ua} {F_v}^a - \frac{1}{4} g_{uv} F_{ab} F^{ab})$

So how do you fit $\pm F_{uv}$ into the complex metric, to unify EM and gravity? And what does the metric look like?