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Use the metric Levi-Civita connection and hold the covariant potential $A_a$ fixed when varying the metric. Symmetry of the connection cancels its terms in the electromagnetic field tensor:The Electromagnetic Bianchi identity follows from commuting partial derivatives, or $F=dA$ and $d^2=0$. Written covariantly it is ${\nabla }_c{F}_{ab}+{\nabla }_a{F}_{bc}+{\nabla }_b{F}_{ca}=0$, since total antisymmetrization also cancels the symmetric connection terms. Hence $\boxed{F_{[ab;c]}=0}$ without using any field equation.

For the potential variation, $\delta F_{ab}={\nabla }_a\delta A_b-{\nabla }_b\delta A_a$. Antisymmetry in the Maxwell action giveswhere the second equality is integration by parts followed by relabeling; compactly supported variations, or vanishing boundary variations, remove the surface term. Arbitrary $\delta A_a$ therefore give the source-free Maxwell equationsNo gauge fixing of $A$ is needed for this variational calculation.

In signature $(+---)$ define the Hilbert stress-energy tensor by ${\delta }_{g}S=\frac{1}{2}\int \sqrt{-g}{T}_{ab}\delta g^{ab}{d}^{4}x$. Since the independent covariant potential makes $F_{ab}$ metric independent,Combining these two terms yields the electromagnetic stress-energy tensorEquivalently $\delta S=-\frac{1}{2}\int \sqrt{-g}{T}^{ab}\delta g_{ab}$. The overall stress-variation sign has been matched to the signature and action. As a check, in an orthonormal frame $F_{cd}{F}^{cd}=2(B^2-E^2)$ and $T^{00}=(E^2+B^2)/(8\pi )>0$ for a nonzero field; copying a mostly-plus formula without changing conventions would fail this check.

Let $J^a={\nabla }_b{F}^{ab}$. Contract the Bianchi identity with $F^{bc}$ to obtain $2F^{bc}{\nabla }_b{F}_{ac}=\frac{1}{2}{\nabla }_a(F_{cd}{F}^{cd})$, hence $F^{bc}{\nabla }_b{F}^a{}_c=\frac{1}{4}{\nabla }^a(F_{cd}{F}^{cd})$. Differentiating the stress tensor cancels these derivative terms and gives the electromagnetic stress-energy divergence identityThus Maxwell's equation implies stress conservation for any field. Conversely, where the matrix $F^a{}_c$ is nonsingular, conservation implies $J^c=0$ by matrix inversion. When it is singular, conservation only says that $J$ lies in its kernel, so the converse cannot be inferred without the printed nonsingularity assumption.

For the first-order formulation, temporarily call the independently constructed potential curvature $f_{ab}(A)={\nabla }_a{A}_b-{\nabla }_b{A}_a$. In the first-order Maxwell action, vary the independent antisymmetric $F^{ab}$, with $F_{ab}=g_{ac}{g}_{bd}{F}^{cd}$. Its variation isso $F_{ab}=f_{ab}(A)$. Varying $A$ independently gives exactly ${\nabla }_b{F}^{ab}=0$, by the same integration-by-parts calculation as before. Eliminating the auxiliary field algebraically makes the integrand $f^{ab}{f}_{ab}-2f^{ab}{f}_{ab}=-f^{ab}{f}_{ab}$, recovering the original Maxwell action. The Bianchi identity then follows from $f=dA$. Metric variation also agrees after elimination: the chain-rule term from the eliminated field multiplies its already vanishing field equation. Thus the two actions give the same source-free electromagnetic dynamics and stress tensor. The Palatini procedure here means independent first-order field variables, not an additional variation of the spacetime connection.