# The Unapologetic Mathematician

## 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