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?


1 Answer 1


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 a homomorphism $$ \operatorname{Pic}(M) \to H^2_{\mathrm{dR}}(M) \cong H^2(M,\mathbb{R}), $$ 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 $H^2(M,\mathbb{R}) = 0$ 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 or torsion (so that, necessarily, $\nabla = d + \mathcal{A}$ for a global $1$-form $\mathcal{A}$). The moment that $H^2(M,\mathbb{R}) \neq 0$, 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 $\mathcal{L} \to M$.


  1. The cohomology group $H^2(M,\mathbb{R})$ contains an isomorphic copy of $H^2(M,\mathbb{Z})/\operatorname{Tor}(H^2(M,\mathbb{Z}))$ (via the UCT) as a lattice. It’s a non-trivial consequence of the Chern–Weil theory that our homomorphism $\operatorname{Pic}(M) \to H^2(M,\mathbb{R})$ not only maps into this lattice but actually recovers the Chern class $\operatorname{Pic}(M) \to H^2(M,\mathbb{Z})$ modulo torsion.

  2. Given a line bundle $\mathcal{L}$, a connection $\nabla$ on $\mathcal{L}$ is the gauge potential of an electromagnetic field in that topological sector, and the curvature $\mathcal{F}$ of $\nabla$ is the field strength of that electromagnetic field.

  3. The class of the line bundle $\mathcal{L}$ in $\operatorname{Pic}(M) \cong H^2(M,\mathbb{Z})$ is called the topological charge or topological defect. If $H^2(M,\mathbb{Z}) \cong \mathbb{Z}$, then, in suitable units, the integer corresponding to $[\mathcal{L}]$ can be interpreted as a monopole charge à la Dirac. Indeed, the Dirac monopole can be interpreted as a certain connection on a certain non-trivial line bundle on $M = \mathbb{R}^{1,3} \setminus \text{(timelike worldline)}$.

  • $\begingroup$ Great answer, I would give it more than +1, if I could. Thank you. $\endgroup$ Jul 18, 2013 at 16:07
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    $\begingroup$ Does $\mathcal{L}$ have some intutive interpretation like $M$, which is interpreted as the set of all events? $\endgroup$
    – Filippo
    Jun 22, 2021 at 17:20
  • $\begingroup$ I second the question of @Filippo, is $\mathcal L$ the electro-magnetic field maybe? $\endgroup$
    – Giafazio
    Jun 24, 2021 at 13:30
  • $\begingroup$ @Filippo please see the addenda $\endgroup$ Jun 24, 2021 at 17:46
  • $\begingroup$ @Giafazio, please see the addenda $\endgroup$ Jun 24, 2021 at 17:46

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