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I want to understand what is wrong with the following argument:

in a topologically compact spacetime, a closed 3D boundary separates the spacetime in two connected components, because of this continuous symmetry, either one of the two things must happen

1) the stokes theorem is not valid

2) the integral of the derivative of any form field in each connected component must be equal in magnitude and opposite in sign, since both must be equal to the surface integral of the form field, and they must be all identical to zero at all times

In layman terms, if the space is compact, a gauss surface does not separate the space neatly in a interior region and an external region, rather, both connected components are topologically equivalent, so one is led to the startling conclusion that the validity of the stokes theorem (which naively looks of a local, differential nature) depends for its validity on the asymptotic boundary conditions of the space

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The answer is just option 2--- the integral on the two sides are necessarily equal in a compact orientable space. The condition of orientability cannot be omitted, see this example: Dirac string on (periodic) compact space

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  • $\begingroup$ I was just going to say that. It's easier to visualize with a closed curve on the surface of a sphere, thus dividing it in two regions. You have to determine an orientation along which to integrate over the curve and if you mean to enclose the other region, then the orientation w.r.t. to your previous choice must reverse. $\endgroup$
    – Raskolnikov
    Commented Aug 9, 2012 at 9:08
  • $\begingroup$ Ron, so take a random form field $f$ for which the integral on both sides is equal. Now construct a smooth scalar function $\phi$ that is 0 in region A and 1 in region B, with all the variation happening in a $\epsilon$-thick surface around the surface. It seems to me that $\phi f$ will not satisfy that same property, unless integration by part of $\phi$ is magically accounting for the missing divergence near the surface. Is that what happens? $\endgroup$
    – user56771
    Commented Aug 9, 2012 at 13:50
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    $\begingroup$ @user56771: you are discovering cohomology. The integration by parts accounts for it, but not by magic, it's because you can write the surface integral as bulk integral. So divide the space into little simplices (tetrahedrons in 3d), and you learn that the integral of d-form over the bulk is equal to the integral of form over the boundary. If you draw some triangles and edges you can see how the magic works. The global topological idea is homology/cohomology, and it's by far the most important idea in topology (perhaps the only really truly significant idea there). It's in Hatcher's book. $\endgroup$
    – Ron Maimon
    Commented Aug 9, 2012 at 16:23

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