This is exactly right. (There is no problem with the "curl theorem" or how it was used in the example in my original post.) (And what I believe Gareth had roughly in mind when he used the phrase "not simply connected". In fact I guess a number of non-topologists just mean by this a space having some kind of holes, judging by Dan Kleitman's online MIT 18.022 notes, where he writes that if div(V) == 0, then V == curl(W) as long as the domain is "simply connected".) The interesting thing is that there are virtually no vector analysis notes that state this correctly: As long as the 2nd betti number (rank of the 2nd cohomology, or equivalently of the 2nd homology group, of the domain is equal to 0, then one always has that div(V) == 0 implies there is some W for which V == curl(W). (Except for notes aimed at topologists or very advanced calculus honors students.) So the resolution of the paradox is that MathWorld's claim in its article on "divergenceless" vector fields omits the conditions under which one can conclude that such a vector field is the curl of another one. --Dan P.S. Essentially, grad, curl, and div are respectively d: L^0 -> L^1, d: L^1 -> L^2, d: L^2 -> L^3, where L^k represents smooth k-forms on our 3-dimensional domain.* Functions are already 0-forms. To convert a 1-form to a vector field, you just take the dual with respect to the (Riemannian) metric at each point. To convert a 2-form to a vector field, you take the Hodge star** of it, which turns it into a 1-form, and then dualize as before. To convert a 3-form to a function, you just express it as a function times the standard volume form dx dy dx (the wedge of dx, dy, dz). ________________________________________________________________ * < http://en.wikipedia.org/wiki/Differential_form > ** < http://en.wikipedia.org/wiki/Hodge_star > On 2013-03-27, at 1:16 PM, Andy Latto wrote:
On Tue, Mar 26, 2013 at 7:09 PM, Dan Asimov <dasimov@earthlink.net> wrote:
No. Surfaces (in 3-space) can have boundaries, and that's what they're talking about.
OK. I thought you were comparing an integral on the interior of the sphere to an integral on its boundary (the sphere), using Stokes' theorem on a 3-manifold. You're comparing an integral on the sphere to an integral on its (null) boundary, using Stokes' theorem on a 2-manifold. In that case, I think the problem is that
(*) div(V) == 0 implies that there exists a vector field W such that V == curl(W).
Is true locally, but not globally. Locally, there are of course many vector fields W such that V == curl(W); you can add any curl-free field to W. But that doesn't mean there is a choice of W that works globally, on the entire 2-manifold. We can look at V as a 2-form, and not every closed 2-form is exact; quotient of the closed 2-forms by the exact 2-forms is the second cohomology group of the surface, which is non-trivial for an orientable compact manifold such as the sphere.