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Is every (p,p) ($p\geq1$) closed form in a contractible open set of $\mathbb{C}^n$, $\partial \bar{\partial}$ exact? We know that every d-exact (p,p) form on a compact Kahler manifold is $\partial \bar{\partial}$ exact (by the Hodge theorem), but unfortunately, that can't be applied here...

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An obvious point: A nonzero $1$-form cannot be $\partial \overline{\partial}$-exact, so you want to specify that you are thinking of degrees $2$ and higher. Although this seems nitpicky, in the compact Kahler case, the $\partial \overline{\partial}$ lemma has no such hypothesis: A $d$-exact $(1,0)$ form or $(0,1)$ on a compact Kahler manifold must be $0$. –  David Speyer Mar 25 '11 at 0:57
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Another nitpicky point: Take a holomorphic $(p-1, 0)$-form $\eta$ which isn't closed. Then $d \eta$ is a non-zero $d$-exact $(p,0)$ form, and it certainly isn't of the form $\partial \overline{\partial}$. So you'd better specify that both $p$ and $q \geq 1$. Again, in the compact Kahler case, all global holomorphic $(p-1,0)$ forms are closed, so this doesn't come up. –  David Speyer Mar 25 '11 at 0:59

2 Answers 2

up vote 10 down vote accepted

Okay, here is a counter-example. Let $X$ be the following open subset of $\mathbb{C}^2$: $$X:= \{ (z_1, z_2) : |z_1| < 2,\ |z_2| < 1 \} \cup \{(z_1, z_2): |z_1| < 1,\ |z_2| < 2 \}$$

This is the standard example of a contractible space for which $H^1(X, \mathcal{O})$ is nonzero. We know that $H^1(X, \mathcal{O})$ is given by the Dolbeault computation: $$H^1(X, \mathcal{O}) = \{ \bar{\partial}\mbox{-closed}\ (0,1)\mbox{-forms} \}/\{ \bar{\partial}\mbox{-exact}\ (0,1)\mbox{-forms} \}.$$ Let $\eta$ be a $(0,1)$ form which is $\bar{\partial}$-closed but not $\bar{\partial}$-exact.

Then there are several way we can use $\eta$ to make a counter example. Let $\psi = d \eta = \partial \eta$. This is a $(1,1)$-form and is manifestly $d$-closed. I claim that $\psi$ is not $\partial \bar{\partial} f$. If it were, then $\eta - \bar{\partial} f$ is $d$-closed. Since $X$ is contractible, this would mean that $\eta - \bar{\partial} f = (\partial + \bar{\partial}) g$ for some $g$. Extracting the $(0,1)$-forms from the formula, we get $\eta - \bar{\partial} f = \bar{\partial} g$ so $\eta = \bar{\partial}(f+g)$, contradicting that $\eta$ is not $\bar{\partial}$-exact.

Alternatively, take $\theta = (dz_1 dz_2) \eta$, so $d \theta = \bar{\partial}(\theta) = 0$. So $\theta$ is $d$-closed and, by topology, must be $d$-exact. If $\theta = \bar{\partial} \partial \beta$ then, in particular, $\theta = \bar{\partial} (f dz_1 dz_2)$, where $f dz_1 dz_2$ is the $(2,0)$-form $\partial \beta$. But then $\bar{\partial} f= \eta$, a contradiction.

So we have built counter-examples in degrees $(1,1)$ and $(2,1)$.


So, what the heck is $\eta$? This is closely modeled parts of my Problem set 4.

Define $$Y = \{ (z_1, z_2) : |z_1| < 2,\ |z_2| < 2,\ |z_1 z_2| < 2 \}$$ The key fact abut $X$ is that

$Y$ is strictly larger than $X$, but every holomorphic function on $X$ extends to a holomorphic function on $Y$.

The right way to do the next part would be to step by step run through the isomorphism between Cech and Dolbeault cohomology. I'll skip to the end and work back instead.

Write $w_i = z_i - 1.2$. The particular number $1.2$ doesn't matter, the point is that $(1.2, 1.2) \in Y \setminus X$.

Let $$\eta = \frac{\overline{w_2} d \overline{w_1} - \overline{w_1} d \overline{w_2}}{(w_1 \overline{w_1} + w_2 \overline{w_2})^2}$$ If I didn't make any typos, we should have $$\eta = \bar{\partial} \frac{\overline{w_1}}{w_2(w_1 \overline{w_1} + w_2 \overline{w_2})} = - \bar{\partial} \frac{\overline{w_2}}{w_1(w_1 \overline{w_1} + w_2 \overline{w_2})}.$$

If $\eta$ was $\bar{\partial} f$ for some function $f$, then let $$g_1 = \frac{\overline{w_1}}{w_2(w_1 \overline{w_1}+ w_2 \overline{w_2})} -f \quad g_2 = - \frac{\overline{w_2}}{w_1(w_1 \overline{w_1} + w_2 \overline{w_2})} -f$$

So $\bar{\partial} g_i = \eta - \bar{\partial} f =0$, and we see that $g_i$ is holomorphic on $X \setminus \{ w_i =0 \}$. Set $h_i = w_i g_i$. Note that $h_i$ is bounded on $X$. (We are using that $(w_1, w_2) = (0, 0) \not \in X$, so $w_1 \overline{w_1} + w_2 \overline{w_2}$ is bounded below on $X$.) By Riemman extension, we see that $h_i$ extends to a holomorphic function on $X$ and, thus, extends to a holomorphic function on $Y$.

Then look at $$w_2 h_1 - w_1 h_2 = w_1 w_2 \left( \frac{\overline{w_1}}{w_2(w_1 \overline{w_1}+ w_2 \overline{w_2})} + \frac{\overline{w_2}}{w_1(w_1 \overline{w_1} + w_2 \overline{w_2})} \right) = \frac{w_1 \overline{w_1} + w_2 \overline{w_2}}{w_1 \overline{w_1} + w_2 \overline{w_2}} =1$$

But the left hand side vanishes at the point $(w_1, w_2) = (0,0)$ of $Y$ and the right hand doesn't, a contradiction. So we conclude that $\eta$ is not $\bar{\partial}$-exact.

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Minor quibble.Don't you need your inequalities to be strict for X to be open? –  Mohan Ramachandran Mar 25 '11 at 17:43
    
Thanks, I'll fix. –  David Speyer Mar 25 '11 at 18:04

If in addition you assume that your domain is pseudoconvex then by a theorem of A.Aeppli what you want is true.The paper is titled :On the cohomology structure of Stein manifolds 1965 (Proc.Conf.Complex Analysis(Minneapolis Minn 1964) pages 58 to 70 Springer, Berlin. David's example indicates why one might need a hypothesis like pseudoconvexity.

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