# ddc lemma

Posted on May 4, 2016 by Dima

tags: kahler manifolds, ddc

In this note I will prove the easy local case of $$dd^c$$ lemma.

Let $$(M,I)$$ be a complex manifold. Extend the action of the complex structure on exterior powers of the complexified tangent bundle forms by $\mathbf{I}: \bigwedge{}^* M \to \bigwedge{}^* M, \mathbf{I} := \sum i^{p-q} \cdot \Pi^{p,q}$ where $$\Pi^{p,q}: \bigwedge^{p+q} \to \bigwedge^{p,q}$$ is the natural projection. Define a dwisted differential $$d^c = I \circ d \circ I^{-1}$$. On the cotangent bundle define it to be $$\mathbf{I}(\alpha)(v_1, \ldots, v_n) := \alpha(\mathbf{I}v_1, \ldots, \mathbf{I}v_n)$$.

Proposition. $$\partial = \dfrac{d + i \cdot d^c}{2}, \bar \partial = \dfrac{d - i \cdot d^c}{2}$$, where $$\partial$$ and $$\bar\partial$$ are projections of $$d$$ on $$(\cdot +1, \cdot)$$ and $$(\cdot, \cdot+1)$$ components.

Proof. Indeed $\partial + \bar\partial + i \mathbf{I}^{-1} (\partial + \bar\partial) \mathbf{I} = \partial + \bar\partial + i (i^{q-p-1} \partial + i^{q-p+1}\bar\partial)i^{q-p}=\\ = \partial + \bar\partial + \partial + i^2\bar\partial=2\partial$ Similarly for the second equality. $$\square$$

In particular, $$\partial\bar\partial = -\dfrac{i}{2}dd^c$$ on a complex manifold.

Lemma (Poincaré lemma). If $$\alpha$$ is a closed form on a polydisc then it is exact.

Lemma (Poincaré-Dolbeault-Grothendieck lemma). If $$\alpha$$ is a $$\partial$$-closed, not holomorphic (i.e. $$\alpha \notin A^{n,0} M$$), form on a polydisc then it is $$\bar\partial$$-exact.

(this lemma means in particular that Dolbault resoltions of sheaves of holomorphic forms are acyclic)

Note that since $$\bar(\partial\alpha) = \bar\partial \bar\alpha$$, the PDG lemma also holds for $$\partial$$.

Lemma (local $$dd^c$$ lemma). Let $$\eta$$ be a $$(1,1)$$-form on a polydisc. Assume either

• $$\eta$$ is exact, or
• $$\eta$$ is $$\partial$$-exact, $$\bar\partial$$-closed.

Then there exists a function $$\chi$$ such that $$\eta = dd^c \chi$$.

Proof. First establish the equivalence of the assumptions. Recall that $$d^2=\partial^2=\bar\partial^2=0$$, where $$d=\partial+\bar\partial$$, and therefore $$\bar\partial\partial=-\partial\bar\partial$$. If $$\eta = d\alpha$$ then there exist forms $$\beta \in A^{1,0}, \gamma\in A^{0,1}$$ such that $$\eta = \partial\beta + \bar\partial\gamma$$ and $$\bar\partial\beta=\partial\gamma=0$$. Then $$\partial(\eta) = \partial^2 \beta + \partial\bar\partial\gamma=-\bar\partial\partial\gamma=0$$. Similarly for $$\bar\partial\eta$$. Apply PDG and $$\partial$$-PDG to get $$\partial$$ and $$\bar\partial$$-exactness of $$\eta$$.

By Poincaré-Dolbeault-Grothendieck lemma, there exists a form $$\alpha \in A^{1,0}$$ such tha $$\bar\partial\alpha=0$$. Deduce $$\bar\partial\partial \alpha = -\partial\bar\partial \alpha=-\partial \eta = 0$$.

Therefore $$\partial\alpha$$ is a closed form: $$(\partial + \bar\partial)\alpha = \partial^2 \alpha + \bar\partial\alpha=0$$. By Poincar'e lemma, there exists a form $$\alpha'$$ such that $$d\alpha' = \partial\alpha' + \bar\partial\alpha = \partial\alpha$$. By grading considerations, $$\bar\partial\alpha = 0$$.

Consider the form $$\beta = \alpha - \alpha' \in A^{1,0}$$. Then $$\partial\beta = \partial\alpha - \partial\alpha' = 0$$ and $$\bar\partial\beta = \bar\partial\alpha - \bar\partial\alpha' = \eta$$. By $$\partial$$-PDG, there exists a function $$\psi$$ such that $$\partial\psi = \beta$$, so $$\bar\partial\partial\psi=\eta$$. Put $$\chi=-i\psi$$. $$\square$$

Interestingly, this statement is true globally, i.e. for global $$(p,q)$$-forms on a manifold $$M$$, if the manifold is Kähler.

Lemma (global $$dd^c$$ lemma). Let $$\alpha$$ be a form on a Kähler manifold, and assume that $$\eta$$ is either $$d$$-, $$\partial$$- or $$\bar\partial$$-exact. Then $$\alpha$$ is $$dd^c$$-exact (or, which is the same, $$\partial\bar\partial$$-exact).

This result is not as trivial, and requires Hodge theory.