# what is wrong with the following argument about stokes law in compact universes?

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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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? –  user56771 Aug 9 '12 at 13:50