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Electric charges on compact four-manifolds

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Textbook wisdom in electromagnetism tells you that there is no total electric charge on a compact manifold. For example, consider space-time of the form $\mathbb{R} \times M_3$ where the first factor is time. One defines the total charge via $Q(M_3) = \int_{M_3} \star j$ where $d\star F = \star j$ is the electric current. If $M_3$ has no boundary (e.g. if it is compact) one can use Stokes' theorem to argue that $$ Q(M_3) = \int_{M_3} d\star F = \int_{\partial M_3} \star F = 0.$$

I wonder what happens for general four-manifolds $M_4$, especially in the case that the third Betti number is zero (otherwise one can simply integrate over a three-cycle). Is there a sensible way to define charge in the above sense? Can one argue that it has to vanish if $\partial M_4 = 0$?

This post imported from StackExchange Physics at 2014-09-29 16:53 (UTC), posted by SE-user jws
asked Sep 29, 2014 in Theoretical Physics by jws (45 points) [ no revision ]

1 Answer

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The total charge is only defined on a spatial slice, but on any compact spatial slice your argument shows that the total charge is zero. A nice way of imagining this is to ask yourself "in the case of some charge excess, where would the extra field lines go?"

By the way, this fact can actually be violated in the case where there is some ABJ anomaly. For example in the background of a flat photon field one doesn't see any chiral current non-conservation, but if one considers the total charge on a compact spatial slice where the photon field has a non-trivial Chern-Simons invariant, then this invariant equals the total charge! This is a global manifestation of the chiral anomaly.

answered Sep 29, 2014 by Ryan Thorngren (1,605 points) [ revision history ]

To add some references to this statement: chiral anomaly.

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