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In this question I asked about proving that a connection form $\alpha$ on a $\mathbb{C}^*$ bundle had to have $2\pi i(\alpha - \overline{\alpha})$ be exact. From the answer to that question I understand why on any local trivialization, this quantity will be exact. However, this would ordinarily only imply global exactness if the 1st cohomology is trivial. I am probably being very dense, but how do we show that the global condition follows?

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So if you're going in the direction "$\alpha$-invariant pairing exists$\Rightarrow$ $2\pi i(\alpha-\overline{\alpha})$ is exact", my previous answer demonstrates that $2\pi i(\alpha-\overline{\alpha}) = dg$ at each point. But the function $g$, which is defined as $g(p) = \ln\langle p,p\rangle$, has a global definition on $L-\textrm{zero section}$. You're not trying to patch together several local $g$'s. Does this answer your question?

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Why is $g$ globally defined? If the bundle is nontrivial there will be no global non-vanishing sections. –  Blake Sep 22 '11 at 0:29
    
On overlaps the sections change by a local $U(1)$ transformation, and hence the norm of the section relative to the hermitian inner product is invariant. –  José Figueroa-O'Farrill Sep 22 '11 at 0:32
    
The definition of $g$ doesn't involve local sections, it only involves the pairing $\langle\cdot,\cdot\rangle$, which is a globally defined structure. The function $h$ on the other hand does involve local sections, maybe that's what's confusing you. The relationship between $g$ and $h$ is $h=\sigma^* g$. –  user17945 Sep 22 '11 at 1:21
    
Just to be even more explicit, $g$ is globally defined, whereas $h=g\circ \sigma$ is defined only over the domain of $\sigma$ - it's kind of a `local representative' of $g$. –  user17945 Sep 22 '11 at 1:28

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