Disclaimer:
This idea of $g = g_{matter} + i F$ does match up with Hermitian metrics,
but it will fail $\nabla g = 0$ simply because $\nabla F = J$
However, for entertainment, $g = g_{matter} + i \star F$ should clear that hurdle,
since $\nabla \star F = 0$.
But this still suffers from the fact that $\Gamma$ and $F$ separately describe the force.
I was wondering if there was a way to carry the $F$ through to the $\Gamma$...

Hmm, thinking more, $\nabla g = 0, g = g_{matter} + i F$ shouldn't depend on $\nabla F = 0$, because the rest of that information to cancel $\nabla g$ could be kept within $g_{matter}$. After all, $\nabla g = 0$ isn't a requirement of the geodesic equation -- it's a consequence of it.
So if I choose to add a metric for which $\nabla$ wrt the real components of the metric is zero, that might make things cohese, but it shouldn't ruin things to choose otherwise.

Let $g_{ab} \in \mathbb{C}$ be a Hermitian metric (on a Kahler manifold I think? I'm just now learning about them).

For entertainment I'm going to use a signature of $\eta_{ab} = diag(1, -1, -1, -1)$
For entertainment and for the hope of eventually mapping these to the quaternions.

Note, Hermitian means $g_{ab} = \bar{g}_{ba}$
Also note this implies that $a_a = g_{ab} a^b \ne a^b g_{ba} = \bar{g}_{ab} a^b$
So raising and lowering is different depending on whether it is a left- or right- matrix multiply.
I'm going to pick a left matrix multiply. It works best with the ADM decomposition below.

This works for raising/lowering indexes when neither (or both) are conjugated.
The opposite must be done when raising or lowering indexes when either the vector or the metric is conjugate:
$\bar{a}^b g_{ba} = \overline{( a^b \bar{g}_{ba} )} = \overline{( g_{ab} a^b )} = \bar{a_a}$
Notice this means that $\bar{a}_i \ne \overline{( a_i )}$
(Am I going to get into a situation where individual indexes will need to keep track of whether they are conjugated or not?)

What does this say about inner products?
Norms must be real:
$\bar{a}_i a^i = \bar{a}^i \gamma_{ij} a^j$
$= \bar{a}^i \bar{\gamma}_{ij} a^j$ by conjugation
$= \bar{a}^i \gamma_{ij} a^j$ by transpose and relabeling indexes
I'm picking this definition of the norm based on the must-be-real component of the ADM metric ...

ADM decomposition:
$g_{ab} = \overset{a(i)\downarrow b(j)\rightarrow}{ \left[ \matrix{ \alpha^2 - \bar{\beta}_k \beta^k & \bar{\beta}_j \\ \beta_i & -\gamma_{ij} } \right]}$
$g^{ab} = \overset{a(i)\downarrow b(j)\rightarrow}{ \left[ \matrix{ 1/\alpha^2 & \bar{\beta}^j /\alpha^2 \\ \beta^i /\alpha^2 & \beta^i \bar{\beta}^j /\alpha^2 - \gamma^{ij} } \right]}$

Orthogonality:
$g_{ac} g^{cb}$ $ = \left[ \matrix{ \alpha^2 - \bar{\beta}_l \beta^l & \bar{\beta}_k \\ \beta_i & -\gamma_{ik} } \right] \left[ \matrix{ 1/\alpha^2 & \bar{\beta}^j /\alpha^2 \\ \beta^k /\alpha^2 & \beta^k \bar{\beta}^j /\alpha^2 - \gamma^{kj} } \right]$

$g_{tc} g^{ct} = (\alpha^2 - \bar{\beta}_l \beta^l) (1/\alpha^2) + \bar{\beta}_k \beta^k /\alpha^2$
$= 1 - \bar{\beta}_l \beta^l /\alpha^2 + \bar{\beta}_k \beta^k /\alpha^2$
$= 1$

$g_{ic} g^{ct} = \beta_i (1/\alpha^2) - \gamma_{ik} \beta^k /\alpha^2$
$= \beta_i /\alpha^2 - \beta_i /\alpha^2$
$= 0$
Notice the lower using left multiply: $\beta_i = \gamma_{ik} \beta^k$

$g_{tc} g^{cj} = (\alpha^2 - \bar{\beta}_l \beta^l) (\bar{\beta}^j/\alpha^2) + \bar{\beta}_k (\beta^k \beta^j /\alpha^2 - \gamma^{kj})$
$= \alpha^2 \bar{\beta}^j/\alpha^2 - \bar{\beta}_l \beta^l \bar{\beta}^j/\alpha^2 + \bar{\beta}_k \beta^k \bar{\beta}^j /\alpha^2 - \bar{\beta}_k \gamma^{kj}$
$= \bar{\beta}^j - \bar{\beta}_k \gamma^{kj}$
$= \bar{\beta}^j - \bar{\beta}^j$
$= 0$
Notice raising a conjugate vector and non-conjugated metric requires a right-hand multiply: $\bar{\beta}_k \gamma^{kj} = \bar{\beta}^j$

$g_{ic} g^{cj} = \beta_i \bar{\beta}^j /\alpha^2 - \gamma_{ik} (\beta^k \bar{\beta}^j /\alpha^2 - \gamma^{kj})$
$= \beta_i \bar{\beta}^j /\alpha^2 - \gamma_{ik} \beta^k \bar{\beta}^j /\alpha^2 + \gamma_{ik} \gamma^{kj}$
$= \delta_i^j$
Once again, using a left-multiply to lower $\beta_i = \gamma_{ik} \beta^k$

Now for combining the gravitational metric (real) with the Faraday tensor (imaginary)
The Faraday tensor:
$F_{ab} = \left[\matrix{ 0 & -E_j \\ E_i & {\tilde{\epsilon}_{ij}}^k B_k }\right]$

Let the real components of $g_{ab}$ be represented as $\tilde{g}_{ab} \in \mathbb{R}$.
Let $g_{ab} \approx \tilde{g}_{ab} + i F_{ab}$
$\tilde{g}_{ab} \approx diag(1, -1, -1, -1)$
$g_{ab} \approx \left[\matrix{ 1 & -i E_x & -i E_y & -i E_z \\ i E_x & -1 & i B_z & -i B_y \\ i E_y & -i B_z & -1 & i B_x \\ i E_z & i B_y & -i B_x & -1 }\right]$
Full derivation is below.

Let $\tilde{\alpha}, \tilde{\beta}^i, \tilde{\gamma}_{ij} \in \mathbb{R}$ be the analogous components of $\tilde{g}_{ab}$

Let $\alpha = \tilde{\alpha} \approx 1$. This is mandated because $g_{ab}$ is Hermitian, so the diagonals (which include $1/\alpha^2$) must be real.
It is only $1/\alpha^2$ that must be real, so $\alpha$ could be a real, or purely imaginary.

Let $\beta^i = \tilde{\beta}^i + i E^i$ for $\tilde{\beta}^i \in \mathbb{R}, \tilde{\beta}^i \approx 0$

Let $\gamma_{ij} = \tilde{\gamma}_{ij} - i \tilde{\epsilon}_{ijk} B^k$ for $\tilde{\gamma}_{ij} \in \mathbb{R}, \tilde{\gamma}_{ij} \approx \delta_{ij}$
$\tilde{\epsilon}_{ijk} = \sqrt{det (\tilde{\gamma})} [ijk]$ is the Levi-Civita tensor on the spatial metric.
$[ijk]$ the sign of the permutation of $i,j,k$.

What would the spatial inverse metric $\gamma^{ij}$ be?
Check out the Maxima sheet for derivations:
$\gamma^{ij} = (\tilde{\gamma}^{ij} - B^i B^j + i \tilde{\epsilon}^{ijk} \tilde{\gamma}_{kl} B^l) / (1 - \tilde{\gamma}_{kl} B^k B^l)$
For $\tilde{\epsilon}^{ijk} = \frac{1}{\sqrt{\tilde{\gamma}}} [ijk]$
Notice that $\frac{\gamma}{\tilde{\gamma}} = 1 - \tilde{\gamma}_{kl} B^k B^l$
Let $\mu = \frac{\tilde{\gamma}}{\gamma}$
Now $\gamma^{ij} = \mu (\tilde{\gamma}^{ij} - B^i B^j + i \tilde{\epsilon}^{ijk} \tilde{\gamma}_{kl} B^l)$

Notice that lowering spatial vectors is done by $\gamma_{ij}$, which is a combination of the real spatial $\tilde{\gamma}_{ij}$ and the imaginary magnetic field dual $-i \tilde{\epsilon}_{ijk} B^k$.
$\beta_i = \gamma_{ij} \beta^j = \gamma_{ij} (\tilde{\beta}^j + i E^j)$
$= \tilde{\beta}_i + i E_i = (\tilde{\gamma}_{ij} - i \tilde{\epsilon}_{ijk} B^k) (\tilde{\beta}^j + i E^j)$
$\tilde{\beta}_i + i E_i = \tilde{\gamma}_{ij} \tilde{\beta}^j + \tilde{\epsilon}_{ijk} E^j B^k + i (\tilde{\gamma}_{ij} E^j - \tilde{\epsilon}_{ijk} \tilde{\beta}^j B^k)$
Compare real components:
$\tilde{\beta}_i = \gamma_{ij} \beta^j = \tilde{\gamma}_{ij} \tilde{\beta}^j + \tilde{\epsilon}_{ijk} E^j B^k$
This says that lowering the real shift vector with the Hermitian spatial metric is only equal to lowering the real shift vector with the real metric if the electric and magnetic fields are parallel.
Compare imaginary components:
$E_i = \gamma_{ij} E^j = \tilde{\gamma}_{ij} E^j - \tilde{\epsilon}_{ijk} \tilde{\beta}^j B^k$
Once again, lowering the $E$ field with the Hermitian metric only matches lowering with the real metric if the real shift vector and magnetic field vector are parallel.

Now to represent it in ADM metric form:
$g_{ab} = a(i)\downarrow \overset{b(j)\rightarrow}{ \left[ \matrix{ \alpha^2 - \bar{\beta}_k \beta^k & \bar{\beta}_j \\ \beta_i & -\gamma_{ij} } \right]}$
$= a(i)\downarrow \overset{b(j)\rightarrow}{ \left[ \matrix{ \alpha^2 - \overline{(\tilde{\beta}_k + i E_k)} (\tilde{\beta}^k + i E^k) & \tilde{\beta}_j - i E_j \\ \tilde{\beta}_i + i E_i & -\tilde{\gamma}_{ij} + i \tilde{\epsilon}_{ijk} B^k } \right]}$

How about that $g_{tt}$ term.
$g_{tt} = \alpha^2 - \bar{\beta}_k \beta^k$
$= \alpha^2 - \overline{(\gamma_{kl} \beta^l)} \beta^k$
$= \alpha^2 - \bar{\gamma}_{kl} \bar{\beta}^l \beta^k$
$= \alpha^2 - \bar{\gamma}_{kl} (\tilde{\beta}^l - i E^l) (\tilde{\beta}^k + i E^k)$
$= \alpha^2 - (\tilde{\gamma}_{kl} + i \tilde{\epsilon}_{klm} B^m) (\tilde{\beta}^l - i E^l) (\tilde{\beta}^k + i E^k)$
$= \alpha^2 - (\tilde{\gamma}_{kl} + i \tilde{\epsilon}_{klm} B^m) (\tilde{\beta}^l \tilde{\beta}^k + i \tilde{\beta}^l E^k - i E^l \tilde{\beta}^k + E^l E^k)$
$= \alpha^2 - ( \tilde{\gamma}_{kl} \tilde{\beta}^l \tilde{\beta}^k + i \tilde{\epsilon}_{klm} B^m \tilde{\beta}^l \tilde{\beta}^k + \tilde{\gamma}_{kl} E^l E^k + i \tilde{\epsilon}_{klm} B^m E^l E^k + i \tilde{\gamma}_{kl} E^k \tilde{\beta}^l - \tilde{\epsilon}_{klm} B^m E^k \tilde{\beta}^l - i \tilde{\gamma}_{kl} E^l \tilde{\beta}^k + \tilde{\epsilon}_{klm} B^m E^l \tilde{\beta}^k )$
$= \alpha^2 - \tilde{\gamma}_{kl} (\tilde{\beta}^l \tilde{\beta}^k + E^k E^l) - 2 \tilde{\epsilon}_{klm} \tilde{\beta}^k E^l B^m$
$= \tilde{g}_{tt} - \tilde{\gamma}_{kl} E^k E^l - 2 \tilde{\epsilon}_{klm} \tilde{\beta}^k E^l B^m$

Back to $g_{ab}$:
$g_{ab} = \left[\matrix{ \alpha^2 - \tilde{\gamma}_{kl} (\tilde{\beta}^l \tilde{\beta}^k + E^k E^l) - 2 \tilde{\epsilon}_{klm} \tilde{\beta}^k E^l B^m & \tilde{\beta}_j - i E_j \\ \tilde{\beta}_i + i E_i & -\tilde{\gamma}_{ij} + i \tilde{\epsilon}_{ijk} B^k }\right]$
$= \left[\matrix{ \alpha^2 - \tilde{\gamma}_{kl} (\tilde{\beta}^l \tilde{\beta}^k + E^k E^l) - 2 \tilde{\epsilon}_{klm} \tilde{\beta}^k E^l B^m & \tilde{\beta}_j \\ \tilde{\beta}_i & -\tilde{\gamma}_{ij} }\right] + \left[\matrix{ 0 & -i E_j \\ i E_i & i \tilde{\epsilon}_{ijk} B^k }\right]$
$= \left[\matrix{ \alpha^2 - \tilde{\gamma}_{kl} \tilde{\beta}^l \tilde{\beta}^k & \tilde{\beta}_j \\ \tilde{\beta}_i & -\tilde{\gamma}_{ij} }\right] + \left[\matrix{ 0 & -i E_j \\ i E_i & i \tilde{\epsilon}_{ijk} B^k }\right] + \left[\matrix{ -\tilde{\gamma}_{kl} E^k E^l - 2 \tilde{\epsilon}_{klm} \tilde{\beta}^k E^l B^m & 0 \\ 0 & 0 }\right]$
$= \tilde{g}_{ab} + i F_{ab} + (g^*)_{ab}$

For $\tilde{g}_{ab} = \left[\matrix{ \alpha^2 - \tilde{\gamma}_{kl} \tilde{\beta}^l \tilde{\beta}^k & \tilde{\beta}_j \\ \tilde{\beta}_i & -\tilde{\gamma}_{ij} }\right]$
is the metric formed from the real components of the lapse, shift, and spatial metric.
For $(g^*)_{ab} = \left[\matrix{ -\tilde{\gamma}_{kl} E^k E^l - 2 \tilde{\epsilon}_{klm} \tilde{\beta}^k E^l B^m & 0 \\ 0 & 0 }\right]$

Now for the inverse:
$g^{ab} = a(i)\downarrow \overset{b(j)\rightarrow}{ \left[ \matrix{ 1/\alpha^2 & \bar{\beta}^j /\alpha^2 \\ \beta^i /\alpha^2 & \beta^i \bar{\beta}^j /\alpha^2 - \gamma^{ij} } \right]}$
$= a(i)\downarrow \overset{b(j)\rightarrow}{ \left[ \matrix{ 1/\alpha^2 & (\tilde{\beta}^j - i E^j) /\alpha^2 \\ (\tilde{\beta}^i + i E^i) /\alpha^2 & (\tilde{\beta}^i + i E^i) (\tilde{\beta}^j - i E^j) / \alpha^2 - \mu (\tilde{\gamma}^{ij} - B^i B^j + i \tilde{\epsilon}^{ijk} \tilde{\gamma}_{kl} B^l) } \right]}$
$= a(i)\downarrow \overset{b(j)\rightarrow}{ \left[ \matrix{ 1/\alpha^2 & (\tilde{\beta}^j - i E^j) /\alpha^2 \\ (\tilde{\beta}^i + i E^i) /\alpha^2 & (\tilde{\beta}^i \tilde{\beta}^j + E^i E^j) /\alpha^2 - \mu (\tilde{\gamma}^{ij} - B^i B^j) + i (E^i \tilde{\beta}^j - \tilde{\beta}^i E^j) /\alpha^2 + i \mu \tilde{\epsilon}^{ijk} \tilde{\gamma}_{kl} B^l } \right]}$

$ = \left[\matrix{ 1/\alpha^2 & \tilde{\beta}^j / \alpha^2 \\ \tilde{\beta}^i / \alpha^2 & \tilde{\beta}^i \tilde{\beta}^j / \alpha^2 - \tilde{\gamma}^{ij} }\right] + i \left[\matrix{ 0 & -E^j / \alpha^2 \\ E^i / \alpha^2 & \mu \tilde{\epsilon}^{ijk} \tilde{\gamma}_{kl} B^l }\right] + \left[\matrix{ 0 & 0 \\ 0 & E^i E^j / \alpha^2 + \mu B^i B^j + (1 - \mu) \tilde{\gamma}^{ij} + i (E^i \tilde{\beta}^j - \tilde{\beta}^i E^j) }\right]$
...which doesn't quite exactly look like $\tilde{g}^{ab} + i F_{ab} + $ extra terms in the $x_i x_j$ components.
I'll call these three parts $\tilde{g}^{ab}$, $i (F')^{ab}$, and $(g')^{ab}$, respectively.
Notice that $(F')^{ab}$ is nearly $F_{ab}$, except for the factors of $\alpha$ (which are nearly 1 anyways) and $\mu$ (which is 1 so long as $B = 0$).

In fact, what would $F^{ab}$ look like, in terms of $E$, $B$, and the real metric ADM terms $\tilde{\alpha}, \tilde{\beta}^i, \tilde{\gamma}_{ij}$?
What would I be looking for in the structure of $g^{ab}$ if I wanted to extract $F^{ab}$? (A bit of a circular question.)
$F^{ab} = g^{ac} g^{bd} F_{cd} = g^{ac} F_{cd} \bar{g}^{db}$
$F^{ab} = \left[ \matrix{ 1/\alpha^2 & (\tilde{\beta}^k - i E^k) /\alpha^2 \\ (\tilde{\beta}^i + i E^i) /\alpha^2 & (\tilde{\beta}^i \tilde{\beta}^k + E^i E^k) /\alpha^2 - \mu (\tilde{\gamma}^{ik} - B^i B^k) + i (E^i \tilde{\beta}^k - \tilde{\beta}^i E^k) /\alpha^2 + i \mu \tilde{\epsilon}^{ikm} \tilde{\gamma}_{mn} B^n } \right] \cdot \left[ \matrix{ 0 & -E_l \\ E_k & \tilde{\epsilon}_{klp} B^p } \right] \cdot \left[ \matrix{ 1/\alpha^2 & (\tilde{\beta}^j + l E^j) /\alpha^2 \\ (\tilde{\beta}^l - i E^l) /\alpha^2 & (\tilde{\beta}^l \tilde{\beta}^j + E^l E^j) /\alpha^2 - \mu (\tilde{\gamma}^{lj} - B^l B^j) - i (E^l \tilde{\beta}^j - \tilde{\beta}^l E^j) /\alpha^2 - i \mu \tilde{\epsilon}^{ljq} \tilde{\gamma}_{qr} B^r } \right]$
$= \left[ \matrix{ (\tilde{\beta}^k - i E^k) / \alpha^2 E_k & - 1/\alpha^2 E_l + (\tilde{\beta}^k - i E^k) / \alpha^2 \tilde{\epsilon}_{klp} B^p \\ ((\tilde{\beta}^i \tilde{\beta}^k + E^i E^k) /\alpha^2 - \mu (\tilde{\gamma}^{ik} - B^i B^k) + i (E^i \tilde{\beta}^k - \tilde{\beta}^i E^k) /\alpha^2 + i \mu \tilde{\epsilon}^{ikm} \tilde{\gamma}_{mn} B^n) E_k & -(\tilde{\beta}^i + i E^i) / \alpha^2 E_l + ((\tilde{\beta}^i \tilde{\beta}^k + E^i E^k) /\alpha^2 - \mu (\tilde{\gamma}^{ik} - B^i B^k) + i (E^i \tilde{\beta}^k - \tilde{\beta}^i E^k) /\alpha^2 + i \mu \tilde{\epsilon}^{ikm} \tilde{\gamma}_{mn} B^n) \tilde{\epsilon}_{klp} B^p } \right] \cdot \left[ \matrix{ 1/\alpha^2 & (\tilde{\beta}^j + l E^j) /\alpha^2 \\ (\tilde{\beta}^l - i E^l) /\alpha^2 & (\tilde{\beta}^l \tilde{\beta}^j + E^l E^j) /\alpha^2 - \mu (\tilde{\gamma}^{lj} - B^l B^j) - i (E^l \tilde{\beta}^j - \tilde{\beta}^l E^j) /\alpha^2 - i \mu \tilde{\epsilon}^{ljq} \tilde{\gamma}_{qr} B^r } \right]$
$= \left[ \matrix{ (\tilde{\beta}^k E_k - i E^k E_k) / \alpha^2 & (-E_l + \tilde{\epsilon}_{klp} \tilde{\beta}^k B^p - i \tilde{\epsilon}_{klp} E^k B^p) / \alpha^2 \\ (\tilde{\beta}^i \tilde{\beta}^k + E^i E^k) E_k / \alpha^2 - \mu (\tilde{\gamma}^{ik} - B^i B^k) E_k + i (E^i \tilde{\beta}^k - \tilde{\beta}^i E^k) E_k / \alpha^2 + i \mu \tilde{\epsilon}^{ikm} \tilde{\gamma}_{mn} B^n E_k & - (\tilde{\beta}^i + i E^i) / \alpha^2 E_l + (\tilde{\beta}^i \tilde{\beta}^k + E^i E^k) \tilde{\epsilon}_{klp} B^p /\alpha^2 - \mu (\tilde{\gamma}^{ik} - B^i B^k) \tilde{\epsilon}_{klp} B^p + i (E^i \tilde{\beta}^k - \tilde{\beta}^i E^k) \tilde{\epsilon}_{klp} B^p /\alpha^2 + i \mu \tilde{\epsilon}^{ikm} \tilde{\gamma}_{mn} B^n \tilde{\epsilon}_{klp} B^p } \right] \cdot \left[ \matrix{ 1/\alpha^2 & (\tilde{\beta}^j + l E^j) /\alpha^2 \\ (\tilde{\beta}^l - i E^l) /\alpha^2 & (\tilde{\beta}^l \tilde{\beta}^j + E^l E^j) /\alpha^2 - \mu (\tilde{\gamma}^{lj} - B^l B^j) - i (E^l \tilde{\beta}^j - \tilde{\beta}^l E^j) /\alpha^2 - i \mu \tilde{\epsilon}^{ljq} \tilde{\gamma}_{qr} B^r } \right]$


Now for the geodesic:
Metric:
$\Gamma_{abc} = \frac{1}{2} (g_{ab,c} + g_{ac,b} - g_{bc,a})$
$= \frac{1}{2} (\tilde{g}_{ab,c} + \tilde{g}_{ac,b} - \tilde{g}_{bc,a} + (g^*)_{ab,c} + (g^*)_{ac,b} - (g^*)_{bc,a} + i (F_{ab,c} + F_{ac,b} - F_{bc,a}))$
$= \frac{1}{2} (\tilde{g}_{ab,c} + \tilde{g}_{ac,b} - \tilde{g}_{bc,a} + (g^*)_{ab,c} + (g^*)_{ac,b} - (g^*)_{bc,a} + i (2 F_{ab,c} - F_{ab,c} - F_{ca,b} - F_{bc,a}))$
$= \frac{1}{2} (\tilde{g}_{ab,c} + \tilde{g}_{ac,b} - \tilde{g}_{bc,a} + (g^*)_{ab,c} + (g^*)_{ac,b} - (g^*)_{bc,a} + i 2 F_{ab,c})$
$= \tilde{\Gamma}_{abc} + i F_{ab,c} + (\Gamma^*)_{abc}$
For $\tilde{\Gamma}_{abc}$ the connection of the real metric
For $(\Gamma^*)_{abc} = \frac{1}{2} ((g^*)_{ab,c} + (g^*)_{ac,b} - (g^*)_{bc,a})$, which is all square terms of E and B, so let's hope it's too small to notice.

Now for contravariant form:
${\Gamma^a}_{bc} = g^{ad} \Gamma_{dbc}$
$ = (\tilde{g}^{ad} + i (F')^{ad} + (g')^{ad}) (\tilde{\Gamma}_{dbc} + i F_{db,c} + (\Gamma^*)_{dbc})$
$ = \tilde{\Gamma^a}_{bc} - (F')^{ad} F_{db,c} + i (F')^{ad} \tilde{\Gamma}_{dbc} + i \tilde{g}^{ad} F_{db,c}$
$... + (g')^{ad} \tilde{\Gamma}_{dbc} + i (g')^{ad} F_{db,c} + \tilde{g}^{ad} (\Gamma^*)_{dbc} + (g')^{ad} (\Gamma^*)_{dbc} + i (F')^{ad} (\Gamma^*)_{dbc}$ ... but I'm hoping these terms are neglegible.
Likewise let's pretend the $\partial F$ terms are nearly zero as well.
That leaves us with
${\Gamma^a}_{bc} = \tilde{\Gamma^a}_{bc} + i (F')^{ad} \tilde{\Gamma}_{dbc}$
...which turns out to have an imaginary part of the acceleration put in place of the velocity of a charged particle in the Lorentz force law.
...which is probably what you would expect by putting the Faraday tensor alongside the metric instead of the electromagnetic potential...