Dolbeault cohomology

Remarks

The class of a $\bar\partial$-closed form $\alpha$ is written $[\alpha]$; two closed forms define the same class when they differ by $\bar\partial \beta$. Example: on $X = \C$, every smooth $(0,1)$-form $g\, d\bar z$ with compact support is $\bar\partial$ of the function $u(z) = \frac{1}{2\pi i} \int_\C \frac{g(w)}{w - z}\, dw \wedge d\bar w$, so it gives the zero class in $H^{0,1}_{\bar\partial}(\C)$; in fact $H^{0,1}_{\bar\partial}(\C) = 0$. Example: $H^{p,0}_{\bar\partial}(X)$ is the space of holomorphic $p$-forms on $X$, since there are no $(p,-1)$-forms to divide by. A nonzero class: let $X = \C/\Lambda$ with $\Lambda = \Z + i\Z$. The form $d\bar z$ is $\bar\partial$-closed on $X$, but it is not $\bar\partial u$ for a smooth function $u$ on $X$. Indeed, if it were, the lift $\tilde u$ of $u$ to $\C$ would satisfy $\partial \tilde u / \partial \bar z = 1$, so $h = \tilde u - \bar z$ would be entire. Then $h(z + 1) = h(z) - 1$ and $h(z + i) = h(z) + i$, so $h'$ is a doubly periodic entire function; it is bounded, hence a constant $a$ by Liouville's theorem. The two relations give $a = -1$ and $a = 1$, a contradiction. So $[d\bar z] \ne 0$ in $H^{0,1}_{\bar\partial}(X)$. The groups are also written $H^{p,q}(X)$ when no confusion with other cohomologies arises.