*dd*^{c} lemma

^{c}

*Posted on May 4, 2016 by Dima*

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.