# On the integration of differential forms

In my notes the integration of a differential form on an oriented manifold $M$, with $\{(U_\alpha,\phi_\alpha) \}$ oriented atlas, is defined as: $\int_M \omega = \sum_{i \in \mathbb{N}}\int_M \omega_i$, where $\omega_i=\rho_i \omega$, where $\rho$ is a partition of unity subordinate to the cover $\{U_\alpha \}$. (For every $i$ there is $\alpha(i)$ s.t. supp $\rho_i$ is in $U_{\alpha (i)}$).

But why this sum is finite? I know that the set of supp $\omega_i$ is locally finite and $\omega$ has compact support but I think that I have however to sum on all $i$... Someone may help me?

• In general, this sum is, of course, infinite. In particular, the integral may or may not exist. Jan 29, 2014 at 11:18

Finite sum $\ne$ finite number of summands. If $M$ is compact, you have a finite cover, so a finite number of summands. And consider the 0-form (function) $\omega(x)=1\ \forall x\in M={\Bbb R}$.
The partition of unity is required to define the integral. In each "patch": $$\int_{patch}\omega_i =\int_{{\rm subset of } R^n}\omega_i{\rm -composed-with-bijection}$$ See http://www.math.cornell.edu/~sjamaar/papers/manifold.pdf, definition 12.1.
• To be clear, though, the above sum does have finitely many non-zero terms, even when $M$ is not compact, right? Because we're assuming that $\omega$ is compactly supported and we're using a partition of unity... It's late where I am, so maybe I'm missing something? Jan 29, 2014 at 11:27
• In my hypothesis M isn't compact, but as @JesseMadnick say the support of $\omega$ is compact and the support of each $\rho_i$ is compact by definition on partition of unity. This means that $\omega$ is define only in a finite number of $U_{\alpha}$? And in each of these only a finite number $\omega_i$ is non zero, by the locally finite property?? Jan 29, 2014 at 11:33
• @Martín-BlasPérezPinilla I know that the partition of unity is required to define the integral. My question in this post is how that sum are finite. there must be that only finitely many addenda are zero...But I don't know how conclude it with the property of partition of unity and $\omega$.. Jan 29, 2014 at 11:45
• @andreasvr, with compact support (or $M$ compact) you can cover $supp\omega$ with a finite number of $U_\alpha$ and the series has a finite number of nonzero summands. Jan 29, 2014 at 12:02
• Yes, I overlooked your second commentary until several minutes later. If supp compact you can choose a partition of unit s.t. $\omega$ is nonzero only in a finite number of $supp\ \rho_i$. Jan 29, 2014 at 12:11