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This is $\bar{\partial}$-Poincaré lemma: Given a holomorphic funtion $f:U\subset \mathbb{C} \to \mathbb{C}$ ,locally on $U$ there is a holomorphic function $g$ such that : $$\frac{\partial g}{\partial \bar z}=f$$

The author says that this is a local statement so we may assume $f$ with compact support and defined on the whole plane $\mathbb{C}$, my question is why she says that... thanks.

*Added*

$f,g$ are suppose to be $C^k$ not holomorphic, by definition $$\frac{\partial g}{\partial \bar z}=0$$ if $g$ were holomorphic...

Jr.
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    What book is this from? It would be helpful if you gave the title and page number so we could see the precise statement and surrounding discussion. – Potato Jun 09 '12 at 01:45
  • It is Voisin's book Hodge theory and complex algebraic geometry, p.35, theorem 1.28. – Jr. Jun 09 '12 at 01:49
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    Just as a general note, you should be aware the author is a "she," not a "he." – Potato Jun 09 '12 at 01:52
  • I fixed it , thanks. – Jr. Jun 09 '12 at 02:05
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    Dear Jr., There is something strange in your statement: if $g$ were truly holomorphic, then $\partial g/\partial \bar{z}$ would equal $0$ (this is the Cauchy--Riemann equations). So the $g$ you are looking for should probably not be holomorphic. And it seems likely to me that $f$ should not be required to be holomorphic either, since a compactly supported holomorphic function also necessarily vanishes. Regards, – Matt E Jun 09 '12 at 03:11
  • I fixed the wrong statement...now the question does make sense... – Jr. Jun 09 '12 at 12:16
  • @MattE $f,g$ are $C^k$, could you help me? – Jr. Jun 09 '12 at 16:35
  • Is my response still not clear? You can respond under my post to let me know what is still making you uncomfortable. – Chris Gerig Jun 09 '12 at 19:50

2 Answers2

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I don't have the book, and thus I can't check the statement. However, I believe that the statement holds for smooth $f$.

Basically we want to construct/find $g$ as the following integral:

$$g(z) = \frac{1}{2 \pi i}\int_{w\in \mathbb{C}} \frac{f(w)}{z-w} d\overline{w}\wedge dw$$

In order to do this, $f$ must be defined over the whole complex plane.

mkim
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The statement is defined on a local subset $U$... so we can make $f$ have compact support which vanishes outside $U$, thus trivially extending to the whole complex plane (defined to be zero outside the support).

Chris Gerig
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  • I don't understand why we get no loss of generality if we suppose $f$ with compact support. – Jr. Jun 09 '12 at 03:01
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    So $f: U\to \mathbb{C}$ and the statement applies to a local subset $V\subset U$... i.e. we don't care about outside $V$ (nor $U$ in $\mathbb{C}$)... so taking a compact set $N\subset U$ containing $V$, we haven't lost any information, and $f$ vanishes outside $N$ (hence can be extended to $\mathbb{C}$ trivially). In other words, the original function $f$ and the "new" function with compact support do not look any different in the local region that you care about. – Chris Gerig Jun 09 '12 at 04:59