casting electromagnetism for exterior differential calculus

Question

I am trying to understand curved spacetime Maxwell's equations in terms of exterior differential calculus. I am surfacing this topic due to working with a flat, but non-constant metric and I am actually interested in the resulting transformation laws. My idea was to take terms like

$\quad\partial_i F^{ij}$

(which is also found here) and interpret this with the codifferential $\delta$

$\quad\delta = (-1)^k\;{\star^{-1}}\circ{\mathrm d}\circ{\star}$

as

$\quad((\delta(F^{\flat}))^\sharp)^j = \big(\,(-1)^k\;{\sharp}\circ{\star^{-1}}\circ{\mathrm d}\circ{\star}\circ{\flat}\,\big)(F)\;^j = \partial_i\mathrm F^{ij}$

Unfortunately, this does not seem to hold as I obtain an additional term in the component derivation (see below).

In my current understanding, this might be, because in the hodge-duality $\star\circ\flat$ and $\sharp\circ\star^{-1}$ at least a volume form is involved, but terms like $\partial_i\mathrm F^{ij}$ are metric- and volume-form-independent. So there might be an additional assumption necessary for the term that I have derived.

However, there are formulations of Maxwell's equations using differential calculus using this approach

$\quad\mathrm d(\star F) = J$

Q1: (How) is it possible to identify both formulations with each other in presence of a non-constant metric?

(I found this other answer but it does also not talk about the metric involvement)

Update: I found the calculation performed in a more rigorous way where it is shown "that the differential form Maxwell's equations reduce to the covariant Maxwell's equations" and indeed they also arrive at

Lemma 2. For any rank-2 tensor $F_{\mu\nu}$, we have

$$\frac{1}{\sqrt{|g|}}\partial_{\lambda} \left(\sqrt{|g|}\, F^{\lambda \sigma} \right) = \nabla_{\lambda} F^{\lambda\sigma}.$$

Update 2: yet another post explains a bit more that it's the densities that translate to partial derivatives

if ∇ 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.

Q2: When the 2-covector $F^\flat$ corresponds to a covariant tensor and the 2-covector $\star(F^\flat)$ corresponds to a covariant tensor density; does this mean that the hodge star changes (normal) covectors into "density"-covectors then?

I am confused here, because I thought the transformation for (normal) k-vectors and k-covectors is given by their specific pullback only; and this is preserved for operations such as exterior product, interior product, exterior differential and also the hodge star. (is that the case?)

component calculations

I have assembled a rough sketch of my component calculations in a simplified manner. The index-up-down-positioning does not work out in this form, but I hope it shows where the additional term occurs. Suppose

$\quad[\mathrm F]^I$ are contravariant components of $\boxed{F}$

where $I$ is a multi-index $I=(I_1,I_2)$. Then

$\quad[\mathrm F]^I\;[g]_{IJ}$ are covariant components of $\boxed{(F^\flat)_J}$

where $[g]_{IJ}$ contains $[g]_{I_1J_1}\,[g]_{I_2J_2}$ and is the metric for index lowering. We have

$\quad[\mathrm F]^I\;[g]_{IJ}\;[g]^{JK}\;\sqrt{|\det{g}|}$ are covariant components of $\boxed{(\star(F^\flat))_{\text{not}(K)}}$

where $[g]^{JK}\;\sqrt{|\det{g}|}$ is from the component representation of the hodge star (signs need to be accounted for in $\text{not}(K)$).

Since $[g]^{JK}$ is the matrix inverse of $[g]_{IJ}$, we have

$\quad[\mathrm F]^K\sqrt{|\det{g}|}$ are covariant components of $\boxed{(\star(F^\flat))_{\text{not}(K)}}$

(This is where the hodge-duality $\star\circ\flat$ depends only on the volume form $\sqrt{|\det{g}|}$)

Now for $[\mathrm d]$ having values of $\{-1,0,+1\}$, the exterior derivative $\mathrm d$ can be represented in terms of combinations of partial derivatives:

$\quad[\mathrm d]^i_{\text{not}(K)\,L}\;\partial_i (\;[\mathrm F]^K\;\sqrt{|\det{g}|}\;)$ are covariant components of $\boxed{(\mathrm d(\star(F^\flat)))_{L}}$

the components of the inverse of the hodge star contain $[g]_{IK}\;\frac{1}{\sqrt{|\det{g}|}}$

$\quad\mathrm{id}=\star^{-1}\circ\star$

$\quad(\mathrm{id})^J_I=(\star^{-1}\circ\star)^J_I$

$\quad(\mathrm{id})^J_I=[\star^{-1}]_{IK}\;[\star]^{KJ}$

$\quad(\mathrm{id})^J_I=[\star^{-1}]_{IK}\;[g]^{KJ}\;\sqrt{|\det{g}|}$

$\quad(\mathrm{id})^J_I=[g]_{IK}\;\frac{1}{\sqrt{|\det{g}|}}\;[g]^{KJ}\;\sqrt{|\det{g}|}$

i.e.

$\quad\sum_K\;[\mathrm X]_J\;[g]_{JK}\;\frac{1}{\sqrt{|\det{g}|}}$ are covariant components of $\boxed{(\star^{-1}(X))_{\text{not}(K)}}$

Although we might lost a sign somewhere, with that we arrive at

$\quad[\mathrm d]^i_{\text{not}(K)\,L}\;\partial_i (\;[\mathrm F]^K\;\sqrt{|\det{g}|}\;)\;[g]_{LM}\;\frac{1}{\sqrt{|\det{g}|}}$

$\quad\quad$are covariant components of $\boxed{(\star^{-1}(\mathrm d(\star(F^\flat))))_{\text{not}(M)}}$

and finally

$\quad[\mathrm d]^i_{\text{not}(K)\,L}\;\partial_i (\;[\mathrm F]^K\;\sqrt{|\det{g}|}\;)\;[g]_{LM}\;\frac{1}{\sqrt{|\det{g}|}}\;[g]^{M\,\text{not}(N)}$

$\quad\quad$ are contravariant components of $\boxed{((\star^{-1}(\mathrm d(\star(F^\flat))))^{\sharp})^{N}}$

which, with the hodge duality $\sharp\circ\star^{-1}$, reduces to

$\quad[\mathrm d]^i_{\text{not}(K)\,\text{not}(N)}\;\partial_i (\;[\mathrm F]^K\;\sqrt{|\det{g}|}\;)\;\frac{1}{\sqrt{|\det{g}|}}$

$\quad\quad$ are contravariant components of $\boxed{((\star^{-1}(\mathrm d(\star(F^\flat))))^{\sharp})^{N}}$

here comes the issue: the partial derivative $\partial_i$ also acts on $\sqrt{|\det{g}|}$ so by product rule we get

$\quad[\mathrm d]^i_{\text{not}(K)\,\text{not}(N)}\Big(\;\partial_i ([\mathrm F]^K) + [\mathrm F]^K\;\partial_i(\sqrt{|\det{g}|})\;\frac{1}{\sqrt{|\det{g}|}}\;\Big)$

only the left term $[\mathrm d]^i_{\text{not}(K)\,\text{not}(N)}\;\partial_i ([\mathrm F]^K)$ produces the combination of partial derivatives as in $\partial_i\mathrm F^{ij}$ if one accounts for multi-indices and signs. But the right term depends on the partial derivatives of the determinant of the metric (so it depends on the volume form).