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Let $M\cong\mathbb{R}^4_1$ be the usual Minkowski spacetime. Then we can formulate electrodynamics in a Lorentz invariant way by giving the EM-field $2$-form $\mathcal{F}\in\Omega^2(M)$ and reformuling the homogeneous Maxwell equations as $$d\mathcal{F} = 0$$ Then the Poincaré lemma tells us that the first of the two equations (i.e. $\mathcal{F}$ is closed) implies that $\mathcal{F} = d\mathcal{A}$ for some $\mathcal{A}\in\Omega^1(M)$ (i.e. $\mathcal{F}$ is exact). $\mathcal{A}$ is the usual potential for ED. This automatically gives us the Gauge symmetry $\mathcal{A}'=\mathcal{A}+d\chi$, for any $\chi\in C^\infty(M)$.

My question is: say we want to treat ED on ageneral spacetime, i.e. any $4$-semi-Rimannian manifold $(M,g)$ using the same Maxwell equations. Then if $H^2(M)\neq 0$ (the $2$nd cohomology group) we don't have anymore that $\mathcal{F} = d\mathcal{A}$, and we also lose the Gauge symmetry, which would make things harder. How is the problem approached? How do you treat ED in general spacetime?

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The key is Weyl's famous observation that electrodynamics is really (classical) $U(1)$-gauge theory. Concretely:

  1. You generalise the global $1$-form $\mathcal{A}$ on $M$ to a connection $\nabla$ on a Hermitian line bundle $\mathcal{L} \to M$, which can locally be written as $d + \mathcal{A}$ for $\mathcal{A}$ the so-called connection $1$-form.
  2. The differential $\mathcal{F} := d\mathcal{A}$ of the global $1$-form $\mathcal{A}$ is replaced by the curvature $$\mathcal{F} := d\mathcal{A} + \mathcal{A} \wedge \mathcal{A} = d\mathcal{A}$$ of the connection $\nabla$, which is still a global $2$-form and still satisfies $d\mathcal{F} = 0$ by the Bianchi identity as applied to a connection on a line bundle.
  3. Gauge symmetry in this context now still holds, for the curvature $2$-form $\mathcal{F}$ is unchanged if you replace $\nabla$ by $\nabla + df$ for $f \in C^\infty(M)$.

This all, of course, fits extremely nicely with your observation about $H^2(M)$, for the assignment $$ (\text{line bundle $\mathcal{L} \to M$}) \mapsto (\text{curvature $2$-form $\mathcal{F}$ of a connection $\nabla$ on $\mathcal{L}$}) $$ induces an isomorphism $$ \operatorname{Pic}(M) \cong H^2(M), $$ where the Picard group $\operatorname{Pic}(M)$ is the abelian group of isomorphism classes of line bundles on $M$, with $$ [\mathcal{L}] + [\mathcal{L}^\prime] := [\mathcal{L} \otimes \mathcal{L}^\prime], \quad -[\mathcal{L}] := [\mathcal{L}^\ast]; $$ then $H^2(M) = 0$ if and only if every closed $2$-form on $M$ is exact (i.e., $\mathcal{F} = d\mathcal{A}$ for some global $1$-form $\mathcal{A}$), if and only if every line bundle is trivial (so that, necessarily, $\nabla = d + \mathcal{A}$ for a global $1$-form $\mathcal{A}$). The moment that $H^2(M) \neq 0$, however, you do have non-trivial line bundles and non-exact closed $2$-forms, so that you really do need to consider your spacetime $M$ together with a potentially non-trivial line bundle $\mathcal{L} \to M$.

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  • $\begingroup$ Great answer, I would give it more than +1, if I could. Thank you. $\endgroup$ – Daniel Robert-Nicoud Jul 18 '13 at 16:07

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