Holomorphic forms and harmonic forms

Question

Assume $M$ is a Kähler manifold. Is holomorphic $p$-form on $M$ necessarily a harmonic form?

Answer 1

Note that on a Kähler manifold there are two notions of a harmonic form. The first is the usual sense of harmonic forms, which satisfy $$\Delta \alpha = (dd^\ast + d^\ast d)\alpha = 0.$$ There are also $\bar{\partial}$-harmonic forms, which satisfy $$\Delta_{\bar{\partial}} \alpha = (\bar{\partial}\bar{\partial}^\ast + \bar{\partial}^\ast \bar{\partial})\alpha = 0.$$ Below we will show that a holomorphic $p$-form is always $\bar{\partial}$-harmonic, and then we will show that on a Kähler manifold the notions of harmonic and $\bar{\partial}$-harmonic forms coincide so that a holomorphic $p$-form is then harmonic in the usual sense as well.

It is easy to see that any $\alpha \in \Omega^{p,0}(M)$ is $\bar{\partial}$-harmonic. Such an $\alpha$ satisfies $\bar{\partial} \alpha = 0$ and $\bar{\partial}^\ast \alpha = 0$, so $$\Delta_{\bar{\partial}} \alpha = (\bar{\partial} \bar{\partial}^\ast \alpha + \bar{\partial}^\ast \bar{\partial}) \alpha = 0 \text{ for all } \alpha \in \Omega^{p,0}(M).$$

On a complex manifold $M$, $d = \partial + \bar{\partial}$ and $d^\ast = \partial^\ast + \bar{\partial}^\ast$. If the complex manifold is Kähler with Kähler form $\omega$, define $$L : \Omega^{p, q}(M) \longrightarrow \Omega^{p+1,q+1}(M),$$ $$L(\alpha) = \alpha \wedge \omega$$ and let $\Lambda = L^\ast$ be the formal adjoint of $L$ with respect to the Hermitian metric on $M$. Then one may prove the Kähler identities $$[\Lambda, \bar{\partial}] = -i\partial^\ast \text{ and } [\Lambda, \partial] = i\bar{\partial}^\ast.$$ Now we prove the following.

Theorem. On a Kähler manifold $M$, $$\Delta = 2\Delta_{\bar{\partial}}.$$

Proof. We start by defining the $\partial$-Laplacian $$\Delta_\partial = \partial\partial^\ast + \partial^\ast \partial.$$ Then by the Kähler identities we have that \begin{align} \Delta_\partial & = i(\partial[\Lambda, \bar{\partial}] + [\Lambda, \bar{\partial}]\partial) \\ & = i(\partial \Lambda \bar{\partial} - \partial\bar{\partial}\Lambda + \Lambda\bar{\partial}\partial - \bar{\partial}\Lambda\partial) \\ & = i(\partial\Lambda\bar{\partial} + \bar{\partial}\partial\Lambda -\Lambda\partial\bar{\partial} - \bar{\partial}\Lambda\partial) \\ & = -i(\bar{\partial}[\Lambda, \partial] + [\Lambda, \partial]\bar{\partial}) \\ & = \Delta_{\bar{\partial}}, \end{align}

Additionally, we have that $$\partial \bar{\partial}^\ast + \bar{\partial}^\ast \partial = -i(\partial(\Lambda\partial - \partial\Lambda) + (\Lambda\partial -\partial\Lambda)\partial) = 0$$ and similarly $$\bar{\partial}\partial^\ast + \partial^\ast\bar{\partial} = 0.$$

Therefore \begin{align} \Delta & = dd^\ast + d^\ast d \\ & = (\partial + \bar{\partial})(\partial^\ast + \bar{\partial}^\ast) + (\partial^\ast + \bar{\partial}^\ast)(\partial + \bar{\partial}) \\ & = \partial\bar{\partial}^\ast + \partial\partial^\ast + \bar{\partial}\partial^\ast + \bar{\partial}\bar{\partial}^\ast + \partial^\ast \partial + \partial^\ast \bar{\partial} + \bar{\partial}^\ast \partial + \bar{\partial}^\ast \bar{\partial} \\ & = \Delta_\partial + \Delta_{\bar{\partial}} \\ & = 2\Delta_{\bar{\partial}}. \end{align} Hence $\Delta = 2\Delta_{\bar{\partial}}$. $\Box$

By the above theorem, it is clear that on a Kähler manifold, $$\Delta \alpha = 0 \iff \Delta_{\bar{\partial}}\alpha = 0.$$ Since we already determined that a holomorphic $p$-form is always $\bar{\partial}$-harmonic, we have that a holomorphic $p$-form is also always harmonic in the usual sense as well (at least on a Kähler manifold; for a general Hermitian manifold the notions of harmonicity may be distinct).

Answer 2

Yes. In general a $p,q$ form is harmonic if and only if it vanishes both under $\overline{\partial}$ and $\overline{\partial}^*$, where the latter is defined by the adjoint formula $$ (\overline{\partial}^* \alpha,\beta) = (\alpha, \overline{\partial}\beta). $$ In particular it's defined as a map from $(p,q)$ forms to $(p,q-1)$ forms. In other words, if $\alpha$ is a $(p,0)$ form, $\overline{\partial}^* \alpha = 0$. Hence if $\alpha$ is holomorphic $(\overline{\partial}\alpha = 0$), it is harmonic, and vice versa.