For clarity I think is best to start with Minkowski spacetime.

The equation we are trying to solve to understand the radiation of a point particle is:
$$\square A^{b}=j^{b}$$

with the gauge $\nabla_{a}A^{a}=0$ and $j^{b}$ is the current density.

The potential
\begin{eqnarray}
A^{b}(t,x)&=&\int G^{b}_{a}(t,x,t'x')j^{a}(t',x')dx'^{3}dt\\
&=&\int\delta_{a}^{b}\delta(t−t′−|x − x′|)∕|x − x′|j^{a}(t',x')dx'^{3}dt'
\end{eqnarray}

where $G^{b}_{a}$ is the Green function with support in the past light cone. In fact, the potential $A^{b}(t,x)$ only depends in the single event $(t',x')$ in the past which is the intersection between the null cone from $(t,x)$ and the world line of the particle.

Now in curved spacetime the generalization
\begin{eqnarray}
A^{b}(t,x)&=&\int G^{b}_{a}(t,x;t'x')j^{a}(t,x)dV\\
&=&\int\delta_{a}^{b}\delta(\gamma(t,x,t'x'))∕\Gamma(x − x ′)|j^{a}(t',x')\sqrt(g)dx'^{3}dt'
\end{eqnarray}

where $\gamma$ is the null geodesic between the two points $(t,x),(t',x')$ and $\Gamma$ is the distance with respect the induced metric of a suitable spacelike surface that contains $x,x'$ does not work.

In general curved spacetimes the retarded Green function would depend on the whole causal past cone and not only the past light cone. This dependence comes from the interaction with the curvature and is related with the extra terms that you point out that vanish for Minkowski.

Therefore the potential is not only defined by the information that travel along the null geodesics but depends on the whole past of the particle. Nevertheless, singularities of the field travel along null geodesics globally. This is the content of the propagation of singularity theorems for linear hyperbolic systems and is related with the geometric optics limit.

As you required rigorous analysis I will point you to some papers with appropriate calculations:

Section 1.4 of
http://relativity.livingreviews.org/open?pubNo=lrr-2011-7&page=articlese1.html

http://arxiv.org/abs/1108.1825

http://arxiv.org/abs/gr-qc/0008047

Also notice that my answer is just about electromagnetism in curved spacetime. To talk about General Relativity we would need to solve also for the Einstein's Equations. The point particle will affect the metric as self force corrections to the background metric. These type of corrections are treated in depth in the first reference.

**Replacing covariants with partials:**

The source equation you cite involves a skew-symmetric tensor density.

You may know that if $\nabla$ is the Levi-Civita connection, you can calculate divergences of vector fields without using the connection coefficients/Christoffel symbols: $$ \nabla_\mu X^\mu=\frac{1}{\sqrt{-g}}\partial_\mu(X^\mu\sqrt{-g}). $$

Now, let $\mathcal{J}^\mu=X^\mu\sqrt{-g}$ be a vector density. Then $$ (\nabla_\mu X^\mu)\sqrt{-g}=\nabla_\mu(X^\mu\sqrt{-g})=\nabla_\mu\mathcal{J}^\mu=\partial_\mu(X^\mu\sqrt{-g})=\partial_\mu\mathcal{J}^\mu, $$ so vector density fields can be differentiated partially.

However, as it turns out the situation is the same for arbitrary (k,0)-type **antisymmetric** tensor fields: Let $F^{\mu\nu}$ be such a field. Then $$ \nabla_\nu F^{\mu\nu}=\partial_\nu F^{\mu\nu}+\Gamma^\mu_{\ \nu\sigma}F^{\sigma\nu}+\Gamma^{\nu}_{\ \nu\sigma}F^{\mu\sigma}, $$ but the second term here vanishes because $\Gamma$ is symmetric in the lower indices, but $F$ is skew-symmetric in the upper indices, so we're left with $$ \nabla_\nu F^{\mu\nu}=\partial_\nu F^{\mu\nu}+\Gamma^\nu_{\ \nu\sigma}F^{\mu\sigma}=\partial_\nu F^{\mu\nu}+\partial_\sigma\ln\sqrt{-g}F^{\mu\sigma} \\ =\frac{1}{\sqrt{-g}}\partial_\nu(F^{\mu\nu}\sqrt{-g}). $$

Defining $\mathcal{F}^{\mu\nu}=F^{\mu\nu}\sqrt{-g}$ gives $$ \nabla_\nu \mathcal{F}^{\mu\nu}=\partial_\nu \mathcal{F}^{\mu\nu}. $$

**Dependence of Maxwell's equations on the (pseudo-)Riemannian structure:**

Remember that the fundamental field here is $A_\mu$, which is a covector field. The only Maxwell equation that doesn't depend on the Riemannian structure is $\partial_{[\mu} F_{\nu\sigma]}=0$, because you can replace the covariants here with partials do to skew-symmetry.

Also do remember that $F_{\mu\nu}=\partial_\mu A_\nu -\partial_\nu A_\mu$ is also well-defined without covariants.

When we get to the source equation things change, however, because

- to define divergence we need upper indices, however $F$ is lower-indiced by nature, so we need the metric to raise indices;
- you can replace covariants with partials there
*only* if you multiply with $\sqrt{-g}$ to create densities. Which, of course, depends on the metric.

So you *can* weasel out of using the *connection* (which is great for computations), but you cannot weasel out of using the metric, therefore Maxwell's equations are absolutely *not* topological.

**An aside on differential forms:**

You should read about differential forms. I am trying to think of reference that should be quick-readable by physicist. Probably Flanders' book is a good one. Otherwise Anthony Zee's General Relativity and QFT books also contain differential forms but only in a heuristic manner. Sean Carroll's GR book also has an OK recount of them.

Basically, differential forms are totally antisymmetric covariant tensor fields. Instead of using index notation as in, say $\omega_{\mu_1,...,\mu_k}$ to denote them, usually 'abstract' notation is used as $\omega=\sum_{\mu_1<...<\mu_k}\omega_{\mu_1...\mu_k}\mathrm{d}x^{\mu_1}\wedge...\wedge\mathrm{d}x^{\mu_k}$, where the basis is written out explicitly. The wedge symbols are skew-symmetric tensor products.

Differential forms are good because they generalize vector calculus to higher dimensions, arbitrary manifolds and also to cases when you don't have a metric. Differential forms can be differentiated ($\omega\mapsto\mathrm{d}\omega$), where the "$\mathrm{d}$" operator, called the exterior derivative, turns a $k$-form into a $k+1$ form without the need for a metric or a connection, and generalizes grad, div and curl, all in one.

The integral theorems of Green, Gauss and Stokes are also generalized.

The point is, if you also have a metric, the theory of differential forms is enriched. You get an option to turn $k$-forms into $n-k$ forms ($n$ is the dimension of your manifold), and also to define a "dual" operation to the exterior derivative, called the codifferential. The codifferential essentially brings the concept of divergence to differential forms.

Written with differential forms, Maxwell's equations are given by $$ \mathrm{d}F=0 \\ \mathrm{d}^\dagger F=kJ, $$ where $\mathrm{d}^\dagger$ is the codifferential, and $k$ is some constant I care not about right now.

I am noting two things:

- The $F$ field strength 2-form is given by $F=\mathrm{d}A$, where $A$ is ofc the 4-potential. The exterior derivative satisfies $\mathrm{d}^2=0$ (think of $\text{div}\ \text{curl}=0$ and $\text{curl}\ \text{grad}=0$), so with potentials, the first equation is $\mathrm{d}F=\mathrm{d}^2A=0$, which is trivially true.
- The first equation contains only $\mathrm{d}$, which is well-defined without a metric. The second equation depends on the codifferential $\mathrm{d}^\dagger$, which
**does** depend on the metric. There is your metric dependance!

## Best Answer

It may be far too late for this to be of any value to you, but in this paper by Mora et al: http://arxiv.org/abs/1205.4468, it is shown that such a gauge choice does exist (for any $n$ in your notation).