I have read about a generalization of Maxwell equation on manifolds that employs differential forms and Hodge duality that goes as follow: $$dF = 0\qquad \text{and}\qquad d \star F = J.\tag{1}$$ As I understand exterior derivative is defined to be the differential on 0-forms and to be an antiderivation (roughly speaking). The Hodge dual of a $p$-form $w$ is defined as the unique $(n-p)$-form $\star w$ that satisfy $\eta \wedge \star w = \langle \eta , w \rangle \mathrm{vol}$ for all $p$-forms $\eta$.
We can express the Faraday tensor in coordinates such that the metric tensor is trivialized at one point i.e. $g_{\mu \nu} = \eta_{\mu \nu}$. Then one should compute the hodge dual using the general formula in a basis:
\begin{equation}
\star \alpha = \frac{1}{k!(n-k)!} \epsilon_{i_1,\dots,i_n} \sqrt{|\det(g)|} \alpha_{j_1,\dots,j_k} g^{i_1,j_1} \cdots g^{i_k,j_k} e^{i_{k+1}} \wedge \cdots \wedge e^{i_n}
\end{equation}
That formula can be used to obtain $\star F$ in a single point if one put $g_{\mu \nu} = \eta_{\mu \nu}$. But due to the vanishing of first derivative of the metric tensor once differentiate with the exterior derivative one gets the correct expression in the point where the metric was trivialized. By decomposing the effect of the derivative as spatial and temporal parts one gets Maxwell equation. Here I'm following John Baez and Javier P. Muniain "Gauge fields, knots and gravity".
Thus one sees that in an local inertial coordinate reference usual Maxwell equation are obtained. This formulation uses the fact that there is a Lorenzian metric and that the manifold is orientable, but doesn't rely on any addiction structure like a connection.
However I am aware of another way to generalize maxwell equation: the minimal coupling which is the substitution of usual partial derivative with the covariant derivative leading to $$\nabla_a F^{ab} = J^{b}\qquad \text{and}\qquad\nabla_{[a} F_{bc]} = 0.\tag{2}$$ With the Levi-Civita connection.
I don't understand how this two are connected. Are those two generalization the same? How can it be possible if the differential form formulation is not aware of Levi-Civita connection? It will be remarkable if the "right" connection popped out from the differential forms version of the Maxwell equation!
Best Answer
Maxwell's equations in differential form notation reads $$ dF = 0~, \qquad d\ast F = \ast J~. $$ We now show that these equations are equivalent to $\nabla_{[a} F_{bc]} = 0$, $\nabla_a F^{ab} = J^b$.
First, by definition \begin{align} \nabla_{[a} F_{bc]} = \partial_{[a} F_{bc]} + \Gamma^d_{[ab}F_{c]d} - \Gamma^d_{[ac} F_{b]d} \end{align} If the connection is torsion-free (not necessarily the Levi-Civita connection) then $\Gamma^a_{[bc]} = 0$ so that \begin{align} \nabla_{[a} F_{bc]} = \partial_{[a} F_{bc]} = \frac{1}{3} (dF)_{abc} \end{align} The last equality is true by definition. Thus, $\nabla_{[a} F_{bc]} \implies dF = 0$.
Next, consider the second equation \begin{align} \nabla_a F^{ab} = \partial_a F^{ab} + \Gamma^a_{ac} F^{cb} + \Gamma^b_{ac} F^{ac} \end{align} Again, if the connection is torsion free, then the last term is zero. To simplify the second term, we have to assume that $\Gamma$ is the Levi-Civita connection so that $$ \Gamma^a_{ac} = \frac{1}{2} g^{ab} ( \partial_a g_{cb} + \partial_c g_{ab} - \partial_b g_{ac} ) = \frac{1}{2} g^{ab} \partial_c g_{ab} = \frac{1}{2} \partial_c \log \det g = \frac{1}{\sqrt{\det g}}\partial_c \sqrt{\det g} $$ Then, we have \begin{align} \nabla_a F^{ab} = \partial_a F^{ab} + \frac{1}{\sqrt{\det g}}\partial_c \sqrt{\det g} F^{cb} = \frac{1}{\sqrt{\det g}} \partial_a \left( \sqrt{\det g} F^{ab} \right)~. \end{align} So we can write Maxwell's equation as $$ \partial_e \left( \sqrt{\det g} F^{ed} \right) = \sqrt{\det g} J^d $$ Now, contract both sides with the Levi-Civita symbol (not tensor), ${\tilde \varepsilon}_{abcd}$ to get $$ \partial_e \left( \sqrt{\det g} {\tilde \varepsilon}_{abcd} F^{ed} \right) = \sqrt{\det g}{\tilde \varepsilon}_{abcd} J^d $$ Now, recall the the Levi-Civita tensor is ${\varepsilon}_{abcd} = \sqrt{\det g} {\tilde \varepsilon}_{abcd}$ $$ \partial_e \left( {\varepsilon}_{abcd} F^{ed} \right) = {\varepsilon}_{abcd} J^d = (\ast J)_{abc} \tag{1}$$ The last equality is true by definition. Finally, we wish to write the LHS in terms of $\ast F$. To do this, we write $$ F^{ed} = -\frac{1}{2} \varepsilon^{edmn} (\ast F)_{mn} $$ Then, $$ (\ast J)_{abc} = \frac{1}{2} \partial_e \left( {\varepsilon}_{abcd}\varepsilon^{demn} (\ast F)_{mn} \right) $$ Then, we use the property $$ {\varepsilon}_{abcd}\varepsilon^{demn} = 6 \delta^e_{[a} \delta^m_b \delta^n_{c]} $$ Finally, $$ (\ast J)_{abc} = 3 \delta^e_{[a} \delta^m_b \delta^n_{c]} \partial_e (\ast F)_{mn} = 3 \partial_{[a} (\ast F)_{bc]} = ( d \ast F )_{abc} $$ where again, the last equility is the definition of $d$. Thus, we see that $\nabla_a F^{ab} = J^b \implies d \ast F = \ast J$.
QED.
PS - To answer your last question. The differential form notation is aware of the connection through the Hodge dual in which $\sqrt{\det g}$ enters. Note also that in the divergence of any $p$-form, only the following component of the connection appears - $\Gamma^a_{ab}$ which depends entirely on $\sqrt{\det g}$. Other components never appear, i.e. in full generality $$ \nabla_a T^{[abc\cdots]} = \frac{1}{\sqrt{\det g}} \partial_a \left( \sqrt{\det g} T^{[abc\cdots]} \right)~. $$