Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The problem is about the proof of the following result.

If $\omega_1,\omega_2 \in \Omega^n_c(X)$ (where $X$ is smooth manifold) are such that $\int_X\omega_1=\int_X \omega_2$ then there is $\mu\in\Omega^{n-1}_c(X)$ such that $\omega_1-\omega_2=d\mu$.

I am using text at link. Part where I am stuck is on numerical page 202 which is page 30 in the file. It says that using partition of unity argument in Step 4 we can assume $\omega_1,\omega_2$ are supported in some parametrizable open sets.

Can someone explain to me why is this true?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Note that $\omega\in \Omega_c^n(X)$ means $\omega$ is a $\mathcal{C}^\infty$ $n$-form, compactly supported on the manifold $X$. The notation is introduced on page 106 of the document. Partitions of unity are discussed starting on page 188.

Regarding the beginning of step 4: By assumption, $\int_X \omega_i = \int_{U_i}\omega_i = 0$. Step 1 tells us that $\omega_i = d\mu_i$ for $\mu_i \in \Omega_c^{n-1}(U_i)$. This implies $\omega_1\sim \omega_2$.

Another way to look at this proof is to ignore the argument regarding the special case $c=0$. Assume $\int_X\omega_1 = \int_X\omega_2 = c$ where $c$ can be zero. Don't divide by $c$. Go through step 5 but instead choose $\int_X \alpha_i = c$. The rest of step 5 goes through as written, giving $(a)\Rightarrow (b)$.

share|improve this answer
    
I understand the notation, the problem is if we have $\int_X \omega_1=\int_X \omega_2$ and $\rho$ is any function from the partition of the unity we do not know that $\int_X \rho \omega_1=\int_X\rho\omega_2$? I feel they used this in the proof. Is there any explanation for this? –  dmm Jul 29 '12 at 8:27
    
@dmm: This is not used in the proof and is not true in general. I add some more detail above. Cheers. –  user26872 Jul 30 '12 at 3:26

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.