Variation of a differential form Physicists sometimes get the lagrangian $$\mathcal{L}=-\frac{1}{2}\mbox dA \wedge \star \mbox dA - A \wedge \star J$$ define a functional given by 
$$S(A)=\int_{N_4} \mathcal{L}= \int_{N_4}-\frac{1}{2}\mbox dA \wedge \star \mbox dA - A \wedge \star J$$ and calculate variation of such functional 
$$\delta S(A)(\delta A)=\delta \int_{N_4} \mathcal L = \int_{N_4} \delta \mathcal{L}=\int_{N_4}-\frac{1}{2}\delta(\mbox dA\wedge \star \mbox dA)-\delta A\wedge \star J$$ etc. and they derive Maxwell's equation $d\star F=-\star J$. The functional is well defined - it's simply an integral of a 4-differential form over an oriented manifold (submanifold). But how is the variation defined? If $M$ is Riemannian manifold we've got Riemannian metric and induced inner product $(\cdot, \cdot)$ given by $$(\omega, \eta):=\int_M \left<\omega, \eta \right>\mathrm{dvol}$$ Hence we've got induced norm of a differential form and can define variation of a functional and of a differential form as its Frechet derivative. Alas, in physics, manifold is psudoriemannian manifold hence  $$(\omega, \eta)=\int_M \left<\omega, \eta \right>\mathrm{dvol}$$ is no longer an inner product. We have no norm of a differential form, and without norm we cannot define variations as Frechet derivatives. Moreover! We cannot even define a local extremum of a functional. 
Can you tell me how is defined variation of such functional, and a differential form? Does it even make sence what physicists are doing? 
 A: To cut a long story short, you don't use a Fréchet derivative but a Gateaux derivative, which doesn't run into any of the issues you bring up.
Anyhow, if $A$ is a fixed $1$-form on $N$, then for any $1$-form $\delta A$ of compact support in the interior of $N^4$, you can still define the variation $\delta S(A)(\delta A)$ of $S$ about $A$ in the direction of $\delta A$ by the Gateaux derivative
$$
 \delta S(A)(\delta A) = \left.\frac{d}{d\epsilon}\right\rvert_{\epsilon=0}S(A+\epsilon\delta A) = \lim_{\epsilon \to 0}\frac{S(A+\epsilon\delta A) - S(A)}{\epsilon}.
$$
So, in this case, since
$$
 L(A+\epsilon\delta A) = -\frac{1}{2}d(A+\epsilon \delta A) \wedge \star d(A+\epsilon \delta A) -(A + \epsilon \delta A) \wedge \star J\\
= L(A) + \epsilon\left( -\tfrac{1}{2}dA \wedge \star d(\delta A) - \tfrac{1}{2}d(\delta A) \wedge \star dA -\delta A \wedge \star J\right) + O(\epsilon^2)\\
= L(A) + \epsilon\left(-d(\delta A) \wedge \star dA - \delta A \wedge \star J \right) + O(\epsilon^2)\\
= L(A) + \epsilon\left( -\delta A \wedge \left(d\!\star\!dA - \star J \right) -d\left(\delta A \wedge \star dA\right) \right) + O(\epsilon^2)
$$
it follows that
$$
 \delta S(A)(\delta A) = \int_{N^4} \delta A \wedge \left(-d\!\star\!dA - \star J\right) - \int_{N^4} d(\delta A \wedge \star dA)\\
= \int_{N^4} \delta A \wedge \left(-d\!\star\!dA - \star J\right) = \int_{N^4} g\left(\delta A,-\Delta A - J\right)\,dvol,
$$
where the Laplacian $\Delta$ on $1$-forms is defined (in Lorentzian signature) by $\Delta = \star d \! \star \! d$. Since the bilinear form $\langle A,B \rangle := \int_{N^4} g(A,B)\,dvol$ is still non-degenerate even in the Lorentzian case, it still follows that $\delta S(A)(\delta A) = 0$ for all suitable variations $\delta A$ if and only if the Euler–Lagrange equation $\Delta A = -J$ holds.
