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> **Question.** Is every $(p, \, p)$ closed form ($p\geq1$) in a contractible open set of $\mathbb{C}^n$ $\partial \bar{\partial}$ exact? 
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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...



$\partial \bar{\partial}$ lemma for contractible domains
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 $(2,1)$. 




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So, what the heck is $\eta$? This is closely modeled parts of my [Problem set 4](http://www.math.lsa.umich.edu/~speyer/632/PSet4.pdf). 


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 $Y$.
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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. 


