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 $\eta_{ab} = diag(1, -1, -1, -1)$

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

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

The Faraday dual tensor:
$\star F_{ab} = \frac{1}{2} F^{uv} \epsilon_{uvab}$ is only going to make sense after we've defined $g_{ab}$, which influences raising, lowering, and $\epsilon_{abcd}$.
Until then, let's define $E$ and $B$ in terms of $\star F_{ab}$, as they would appear in the flat-space dual:
$\star F^{flat}_{ab} = a(i) \downarrow \overset{b(j) \rightarrow}{\left[\matrix{ 0 & B_j \\ -B_i & {\tilde{\epsilon}_{ij}}^k E_k }\right]}$

So our 'E's and 'B's are going to be defined in terms of their covariant form in the dual of F,
Which makes them slightly different from the 'E's and 'B's found in the covariant form of F itself.

Let the real components of $g_{ab}$ be represented as $\tilde{g}_{ab} \in \mathbb{R}$.
Let $g_{ab} \approx \tilde{g}_{ab} + i \star F_{ab}$
$\tilde{g}_{ab} \approx diag(1, -1, -1, -1)$
$g_{ab} \approx \left[\matrix{ 1 & i B_x & i B_y & i B_z \\ -i B_x & -1 & i E_z & -i E_y \\ -i B_y & -i E_z & -1 & i E_x \\ -i B_z & i E_y & -i E_x & -1 }\right]$
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 $\beta^i = \tilde{\beta}^i - i B^i$ for $\tilde{\beta}^i \in \mathbb{R}, \tilde{\beta}^i \approx 0$

Let $\gamma_{ij} = \tilde{\gamma}_{ij} - i \tilde{\epsilon}_{ijk} E^k$ 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:
$\gamma^{ij} = (\tilde{\gamma}^{ij} - E^i E^j + i \tilde{\epsilon}^{ijk} \tilde{\gamma}_{kl} E^l) / (1 - \tilde{\gamma}_{kl} E^k E^l)$
For $\tilde{\epsilon}^{ijk} = \frac{1}{\sqrt{\tilde{\gamma}}} [ijk]$
Notice that $\frac{\gamma}{\tilde{\gamma}} = 1 - \tilde{\gamma}_{kl} E^k E^l$
Let $\mu = \frac{\tilde{\gamma}}{\gamma}$
Now $\gamma^{ij} = \mu (\tilde{\gamma}^{ij} - E^i E^j + i \tilde{\epsilon}^{ijk} \tilde{\gamma}_{kl} E^l)$

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 $-i \tilde{\epsilon}_{ijk} E^k$.
$\beta_i = \gamma_{ij} \beta^j = \gamma_{ij} (\tilde{\beta}^j - i B^j)$
$= \tilde{\beta}_i - i B_i = (\tilde{\gamma}_{ij} - i \tilde{\epsilon}_{ijk} E^k) (\tilde{\beta}^j - i B^j)$
$\tilde{\beta}_i - i B_i = \tilde{\gamma}_{ij} \tilde{\beta}^j - \tilde{\epsilon}_{ijk} B^j E^k - i (\tilde{\gamma}_{ij} B^j + \tilde{\epsilon}_{ijk} \tilde{\beta}^j E^k)$
Compare real components:
$\tilde{\beta}_i = \gamma_{ij} \beta^j = \tilde{\gamma}_{ij} \tilde{\beta}^j - \tilde{\epsilon}_{ijk} B^j E^k$
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:
$B_i = \gamma_{ij} B^j = \tilde{\gamma}_{ij} B^j + \tilde{\epsilon}_{ijk} \tilde{\beta}^j E^k$
Once again, lowering the $B$ 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:
$g_{ab} = a(i)\downarrow \overset{b(j)\rightarrow}{ \left[ \matrix{ \alpha^2 - \bar{\beta}_k \beta^k & \bar{\beta}_j \\ \beta_i & -\gamma_{ij} } \right]}$
$= a(i)\downarrow \overset{b(j)\rightarrow}{ \left[ \matrix{ \alpha^2 - \overline{(\tilde{\beta}_k - i B_k)} (\tilde{\beta}^k - i B^k) & \tilde{\beta}_j + i B_j \\ \tilde{\beta}_i - i B_i & -\tilde{\gamma}_{ij} + i \tilde{\epsilon}_{ijk} E^k } \right]}$

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

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

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 $(g^*)_{ab} = \left[\matrix{ -\tilde{\gamma}_{kl} B^k B^l + 2 \tilde{\epsilon}_{klm} \tilde{\beta}^k B^l E^m & 0 \\ 0 & 0 }\right]$

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

$ = \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 & B^j / \alpha^2 \\ -B^i / \alpha^2 & \mu \tilde{\epsilon}^{ijk} \tilde{\gamma}_{kl} E^l }\right] + \left[\matrix{ 0 & 0 \\ 0 & B^i B^j / \alpha^2 + \mu E^i E^j + (1 - \mu) \tilde{\gamma}^{ij} + i (-B^i \tilde{\beta}^j + \tilde{\beta}^i B^j) }\right]$
...which doesn't quite exactly look like $\tilde{g}^{ab} + i \star F_{ab} + $ extra terms in the $x_i x_j$ components.

In fact, what would $\star F^{ab}$ look like, in terms of $B$, $E$, and the real metric ADM terms $\tilde{\alpha}, \tilde{\beta}^i, \tilde{\gamma}_{ij}$?
$\star F^{ab} = g^{ac} g^{bd} \star F_{cd} = g^{ac} \star F_{cd} \bar{g}^{db}$
$\star F^{ab} = \left[ \matrix{ 1/\alpha^2 & (\tilde{\beta}^k + i B^k) /\alpha^2 \\ (\tilde{\beta}^i - i B^i) /\alpha^2 & (\tilde{\beta}^i \tilde{\beta}^k + B^i B^k) /\alpha^2 - \mu (\tilde{\gamma}^{ik} - E^i E^k) + i (-B^i \tilde{\beta}^k + \tilde{\beta}^i B^k) /\alpha^2 + i \mu \tilde{\epsilon}^{ikm} \tilde{\gamma}_{mn} E^n } \right] \cdot \left[ \matrix{ 0 & B_l \\ -B_k & \tilde{\epsilon}_{klp} E^p } \right] \cdot \left[ \matrix{ 1/\alpha^2 & (\tilde{\beta}^j - i B^j) /\alpha^2 \\ (\tilde{\beta}^l + i B^l) /\alpha^2 & (\tilde{\beta}^l \tilde{\beta}^j + B^l B^j) /\alpha^2 - \mu (\tilde{\gamma}^{lj} - E^l E^j) - i (-B^l \tilde{\beta}^j + \tilde{\beta}^l B^j) /\alpha^2 - i \mu \tilde{\epsilon}^{ljq} \tilde{\gamma}_{qr} E^r } \right]$
$= \left[ \matrix{ -(\tilde{\beta}^k + i B^k) / \alpha^2 B_k & + 1/\alpha^2 B_l + (\tilde{\beta}^k + i B^k) / \alpha^2 \tilde{\epsilon}_{klp} E^p \\ -((\tilde{\beta}^i \tilde{\beta}^k + B^i B^k) /\alpha^2 - \mu (\tilde{\gamma}^{ik} - E^i E^k) + i (-B^i \tilde{\beta}^k + \tilde{\beta}^i B^k) /\alpha^2 + i \mu \tilde{\epsilon}^{ikm} \tilde{\gamma}_{mn} E^n) B_k & (\tilde{\beta}^i - i B^i) / \alpha^2 B_l + ((\tilde{\beta}^i \tilde{\beta}^k + B^i B^k) /\alpha^2 - \mu (\tilde{\gamma}^{ik} - E^i E^k) + i (-B^i \tilde{\beta}^k + \tilde{\beta}^i B^k) /\alpha^2 + i \mu \tilde{\epsilon}^{ikm} \tilde{\gamma}_{mn} E^n) \tilde{\epsilon}_{klp} E^p } \right] \cdot \left[ \matrix{ 1/\alpha^2 & (\tilde{\beta}^j - i B^j) /\alpha^2 \\ (\tilde{\beta}^l + i B^l) /\alpha^2 & (\tilde{\beta}^l \tilde{\beta}^j + B^l B^j) /\alpha^2 - \mu (\tilde{\gamma}^{lj} - E^l E^j) - i (-B^l \tilde{\beta}^j + \tilde{\beta}^l B^j) /\alpha^2 - i \mu \tilde{\epsilon}^{ljq} \tilde{\gamma}_{qr} E^r } \right]$
$= \left[ \matrix{ (-\tilde{\beta}^k B_k - i B^k B_k) / \alpha^2 & (B_l + \tilde{\epsilon}_{klp} \tilde{\beta}^k E^p + i \tilde{\epsilon}_{klp} B^k E^p) / \alpha^2 \\ -(\tilde{\beta}^i \tilde{\beta}^k + B^i B^k) B_k / \alpha^2 + \mu (\tilde{\gamma}^{ik} - E^i E^k) B_k - i (-B^i \tilde{\beta}^k + \tilde{\beta}^i B^k) B_k / \alpha^2 - i \mu \tilde{\epsilon}^{ikm} \tilde{\gamma}_{mn} E^n B_k & + (\tilde{\beta}^i - i B^i) / \alpha^2 B_l + (\tilde{\beta}^i \tilde{\beta}^k + B^i B^k) \tilde{\epsilon}_{klp} E^p /\alpha^2 - \mu (\tilde{\gamma}^{ik} - E^i E^k) \tilde{\epsilon}_{klp} E^p + i (-B^i \tilde{\beta}^k + \tilde{\beta}^i B^k) \tilde{\epsilon}_{klp} E^p /\alpha^2 + i \mu \tilde{\epsilon}^{ikm} \tilde{\gamma}_{mn} E^n \tilde{\epsilon}_{klp} E^p } \right] \cdot \left[ \matrix{ 1/\alpha^2 & (\tilde{\beta}^j - i B^j) /\alpha^2 \\ (\tilde{\beta}^l + i B^l) /\alpha^2 & (\tilde{\beta}^l \tilde{\beta}^j + B^l B^j) /\alpha^2 - \mu (\tilde{\gamma}^{lj} - E^l E^j) - i (-B^l \tilde{\beta}^j + \tilde{\beta}^l B^j) /\alpha^2 - i \mu \tilde{\epsilon}^{ljq} \tilde{\gamma}_{qr} E^r } \right]$


Now for the geodesic:
Metric:
$\Gamma_{abc} = \frac{1}{2} (g_{ab,c} + g_{ac,b} - g_{bc,a})$
$= \frac{1}{2} (\tilde{g}_{ab,c} + \tilde{g}_{ac,b} - \tilde{g}_{bc,a} + (g^*)_{ab,c} + (g^*)_{ac,b} - (g^*)_{bc,a} + i (\star F_{ab,c} + \star F_{ac,b} - \star F_{bc,a}))$
$= \frac{1}{2} (\tilde{g}_{ab,c} + \tilde{g}_{ac,b} - \tilde{g}_{bc,a} + (g^*)_{ab,c} + (g^*)_{ac,b} - (g^*)_{bc,a} + i (2 \star F_{ab,c} - \star F_{ab,c} - \star F_{ca,b} - \star F_{bc,a}))$
$= \frac{1}{2} (\tilde{g}_{ab,c} + \tilde{g}_{ac,b} - \tilde{g}_{bc,a} + (g^*)_{ab,c} + (g^*)_{ac,b} - (g^*)_{bc,a} + i 2 \star F_{ab,c})$
$= \tilde{\Gamma}_{abc} + (\Gamma^*)_{abc} + i \star F_{ab,c}$
For $\tilde{\Gamma}_{abc}$ the connection of the real metric
For $(\Gamma^*)_{abc} = \frac{1}{2} ((g^*)_{ab,c} + (g^*)_{ac,b} - (g^*)_{bc,a})$

${\Gamma^a}_{bc} = g^{ad} \Gamma_{dbc}$