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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Daniel Robert-Nicoud
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1 Answers1


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 $M = \mathbb{R}^{1,3} \setminus \text{(timelike worldline)}$.

Branimir Ćaćić
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  • Great answer, I would give it more than +1, if I could. Thank you. – Daniel Robert-Nicoud Jul 18 '13 at 16:07
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    Does $\mathcal{L}$ have some intutive interpretation like $M$, which is interpreted as the set of all events? – Filippo Jun 22 '21 at 17:20
  • I second the question of @Filippo, is $\mathcal L$ the electro-magnetic field maybe? – Giafazio Jun 24 '21 at 13:30
  • @Filippo please see the addenda – Branimir Ćaćić Jun 24 '21 at 17:46
  • @Giafazio, please see the addenda – Branimir Ćaćić Jun 24 '21 at 17:46
  • Does this hold within general media, a polar fluid for instance? – A rural reader Jun 25 '21 at 15:03
  • I want to say that this is all in vacuo, but you might get an answer in Baez-Muniain or Nakahara? – Branimir Ćaćić Jun 25 '21 at 19:52
  • Thank you very much! But I have to admit, this is a bit too advanced for me. However, I heard that another approach is to postulate a set of transition charts and to refer to the [fiber bundle theorem](https://en.wikipedia.org/wiki/Fiber_bundle_construction_theorem). Do you know the interpretation of the transition charts by any chance? – Filippo Jun 27 '21 at 17:40
  • A line bundle is an example of a fibre bundle. Indeed, to do Yang–Mills gauge theory with any structure group more general than $U(1)$, you need to work with *principal* fibre bundles. In the special case of $U(1)$-gauge theory, however, it's enough to work with Hermitian line bundles (which correspond bijectively with principal $U(1)$-bundles via taking the unitary frame bundle), which yield the simplest computational framework. – Branimir Ćaćić Jun 27 '21 at 17:54
  • Thank you for the comment! "In the special case of U(1)-gauge theory, however, it's enough to work with Hermitian line bundles which correspond bijectively with principal U(1)-bundles via taking the unitary frame bundle" - I'd be interested to understand this in detail, could you please name a reference? – Filippo Jun 29 '21 at 06:21
  • By the way, I have studied fiber bundles (in particular, vector bundles and principal fiber bundles), but I have no knowledge of cohomology groups. – Filippo Jun 29 '21 at 06:31
  • I couldn’t think of a reference, but the main thing is that there’s an unusually straightforward bijection between [isomorphism classes of] Hermitian line bundles and [isomorphism classes of] principal $U(1)$-bundles. If $\mathcal{L}$ is a Hermitian line bundle over $X$, the corresponding principal $U(1)$-bundle $P$ is $\cup_{x\in X}\{v \in \mathcal{L}_x \mid \langle v,v \rangle_x = 1\}$ with principal $U(1)$-action given by fibrewise scalar multiplication. – Branimir Ćaćić Jun 29 '21 at 12:25
  • Given $P$, you can recover $\mathcal{L}$ up to isomorphism as the vector bundle associated to the standard representation of $U(1)$ on $\mathbb{C}$. – Branimir Ćaćić Jun 29 '21 at 12:26
  • @Filippo I’d recommend the book by Baez and Muniain. – Branimir Ćaćić Jun 29 '21 at 12:27
  • Thank you for the explanation and the reference! Will I find out what you mean by "isomorphism classes" by reading Baez and Muniain? – Filippo Jun 29 '21 at 14:31