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

3 Answers 3

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No - consider $dz \wedge dw$ on a three-dimensional complex torus. (A form on a compact Kähler manifold is harmonic iff it's closed and annihilated by the adjoint of $d$.)

This is true on Riemann surfaces, where "harmonic" is the same as "closed."

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    Oh yes, I'm totally wrong! Will delete answer once OP reads this.2013-01-07
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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.

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    @User24601 Hey there : ) I'm sorry for the late reply, I only saw your comment by "coincidence" (I visited your profile to see if you had written any new nice answers). Users don't get notified of your comments unless you precede the username by an `@`. Thank you for putting back your nice answer!2013-01-09
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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).

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    @User24601 Thanks for reminding me about the Kähler identities. I fixed my post to use them in showing that the two notions of harmonicity on a Kähler manifold coincide.2013-01-02