compact manifold question

redrzewski

I'm looking at prop 19.5 of Taylor's PDE book.

The theorem is:

If M is a compact, connected, oriented manifold of dimension n, and a is an n-form, then a=dB where B is an n-1 form iff the ∫a over M is 0.

I'm trying to understand why a=dB implies ∫a = 0.
If M has no boundary, than this follows from Stokes theorem.

However, if M has a boundary, then it seems like this is a counterexample:
a = dx^dy, B=xdy, M=unit square in R^2

Here, ∫a = 1, and a=dB.

The general definitions of compact manifold I've found don't assume no boundary.

What am I missing?
thanks

quasar987

Your analysis is impeccable. Usually, when ppl say manifold, they mean "without boundary". Find out what Taylor means by manifold.

kdbnlin78

I mean this with no patronisation intended, but you've answered your own question because in mathematics, a closed manifold is a special type of topological space, namely a compact manifold without boundary. In contexts where no boundary is possible, any compact manifold is a closed manifold.

Jamma

That's a very nicely constructed counter-example, you clearly understand the concepts reasonably well.

Indeed, this only applies to closed manifolds. However, I'm sure that you can fix it by asserting something about the behaviour of the n-1 form at the boundary. E.g - if you assert that it vanishes to zero around some open neighbourhood of the boundary.

There's a theorem about fixed points, I think it's called the Poincare-Hopf theorem, which says that for a vector field on some manifold with finitely many zero vectors, the sum of the indices of the fixed points is equal to the Euler characteristic of the surface, but if it has a boundary you need to make sure that all the vectors are outward pointing there. Perhaps there's something similar going on here. The suggestion I had above will definitely work though.

Jamma

Thinking about it, what I said is rubbish, just making the thing vanish near the boundary won't help at all, as an adjustment to your construction would show. So I imagine you do have to do something more inline with what I said in my second paragraph.

zhentil

This is not one of those cases. If Taylor's claim is correct, his terminology is non-standard.