Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


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

share|cite|improve this question
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 9 '12 at 1:45
It is Voisin's book Hodge theory and complex algebraic geometry, p.35, theorem 1.28. – Jr. Jun 9 '12 at 1:49
Just as a general note, you should be aware the author is a "she," not a "he." – Potato Jun 9 '12 at 1:52
I fixed it , thanks. – Jr. Jun 9 '12 at 2:05
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 9 '12 at 3:11

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.

share|cite|improve this answer

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).

share|cite|improve this answer
I don't understand why we get no loss of generality if we suppose $f$ with compact support. – Jr. Jun 9 '12 at 3:01
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 9 '12 at 4:59

Your Answer


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.