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I have a quick question regarding the proof of Proposition II.3.2 in Farkas & Kra (pg. 40). The proposition is that if $\alpha$ is a square-integrable, $C^1$ 1-form, then $\alpha$ lives in a certain subspace if and only if $\alpha$ is closed. The relevant portion of the proof goes as follows:

Suppose $\alpha$ is closed and that $f$ is a smooth function whose support lives inside a precompact set $D$. The inner product is then calculated as

$(\alpha, *df) = -\iint\limits_D \alpha \wedge d\bar{f}$

$\qquad = -\iint\limits_D [d(\alpha\bar{f}) - d\alpha \wedge \bar{f}]$

$\qquad = -\iint\limits_D d(\alpha\bar{f})$

$\qquad = -\int\limits_{\partial D} \alpha \bar{f} = 0$

I don't understand the very last "= 0" part. Maybe I'm being dense, but just because $\alpha$ is closed doesn't mean $\alpha \bar{f}$ is closed, does it? (I tried writing out $\alpha = g\, dx + h\, dy$, so $\bar{f}\alpha = \bar{f}g\,dx + \bar{f}h\,dy$, then $d(\bar{f}\alpha) = (\bar{f}_x h - \bar{f}_y g) dx\wedge dy$.)

Sorry if this is a stupid, obvious question, but I just don't see it. Many thanks.

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1 Answer

By assumption, $f = 0$ on $\partial D$. (The support is compact in $D$.)

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