The Unapologetic Mathematician

Mathematics for the interested outsider

Integrals over Manifolds (part 2)

Okay, so we can now integrate forms as long as they’re supported within the image of an orientation-preserving singular cube. But what if the form \omega is bigger than that?

Well, paradoxically, we start by getting smaller. Specifically, I say that we can always find an orientable open cover of M such that each set in the cover is contained within the image of a singular cube.

We start with any orientable atlas, which gives us a coordinate patch (U,x) around any point p we choose. Without loss of generality we can pick the coordinates such that x(p)=0. There must be some open ball around 0 whose closure is completely contained within x(U); this closure is itself the image of a singular cube, and the ball obviously contained in its closure. Hitting everything with x^{-1} we get an open set — the inverse image of the ball — contained in the image of a singular cube, all of which contains p. Since we can find such a set around any point p\in M we can throw them together to get an open cover of M.

So, what does this buy us? If \omega is any compactly-supported n form on an n-dimensional manifold M, we can cover its support with some open subsets of M, each of which is contained in the image of a singular n-cube. In fact, since the support is compact, we only need a finite number of the open sets to do the job, and throw in however many others we need to cover the rest of M.

We can then find a partition of unity \Phi=\{\phi\} subordinate to this cover of M. We can decompose \omega into a (finite) sum:

\displaystyle\omega=\sum\limits_{\phi\in\Phi}\phi\omega

which is great because now we can define

\displaystyle\int\limits_M\omega=\sum\limits_{\phi\in\Phi}\int\limits_M\phi\omega

But now we must be careful! What if this definition depends on our choice of a suitable partition of unity? Well, say that \Psi=\{\psi\} is another such partition. Then we can write

\displaystyle\sum\limits_{\phi\in\Phi}\int\limits_M\phi\omega=\sum\limits_{\phi\in\Phi}\int\limits_M\sum\limits_{\psi\in\Psi}\psi\phi\omega=\sum\limits_{\psi\in\Psi}\int\limits_M\sum\limits_{\phi\in\Phi}\phi\psi\omega=\sum\limits_{\psi\in\Psi}\int\limits_M\psi\omega

so we get the same answer no matter which partition we use.

September 7, 2011 Posted by | Differential Topology, Topology | 4 Comments

   

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