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 $\partial M_4 = 0$?

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    $\begingroup$ in a sense charges express defects, non-linearities, spikes, non-compactness of a manifold (for lack of better word). $\endgroup$
    – Nikos M.
    Sep 29, 2014 at 18:48
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    $\begingroup$ Total charge is only defined on a spatial slices and there can be an exeptions from the fact mentioned in this question in case of ABJ anonaly as explained here $\endgroup$
    – Dilaton
    Sep 29, 2014 at 20:27
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    $\begingroup$ The case of compact manifolds without a boundary is extremely dull. If we allow boundaries, things become much more interesting as in Alvarez and Olive arxiv.org/abs/hep-th/0303229v1 $\endgroup$ Sep 12, 2017 at 14:47
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    $\begingroup$ Are you still intereseted in an answer? $\endgroup$
    – Creo
    Jul 4, 2018 at 15:09
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    $\begingroup$ @Creo, yes. If you have answer, please, post here $\endgroup$
    – Nikita
    Jun 12, 2020 at 7:45

1 Answer 1


Charge integral makes a point only in that the sum of all charges is zero. The difference to the open $R^3$ is, that there are no counter charges at spacelike $\infty$.

For all practical purposes one can distribute the excess charges outside a large sphere. Faraday and Maxwell could rely simply on the polarizabilty of the walls of their laboratory.


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