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

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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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Thank you for the detailed answer. I totally forgot that there were two notions of harmonic forms. I meant the former one. –  Pooya Dec 30 '12 at 23:20
    
Henry, actually, I think the two notions of harmonic form agree on a Kahler manifold, because on a Kahler manifold the two Laplacians differ by a constant. (This is one of the main consequences of the Hodge identities of the operators $L$ and $\Lambda$.) It's on a general Hermitian manifold that the two notions of harmonic differ. –  user54535 Jan 2 '13 at 5:57
    
@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. –  Henry T. Horton Jan 2 '13 at 22:37
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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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I left a comment for you. It was upvoted so at least one other person seems to agree. But it's up to you of course. (will remove this comment later) –  Matt N. Dec 30 '12 at 22:28
    
Thank you for the answer. I now have better understanding on the problem. –  Pooya Dec 30 '12 at 23:30
    
Thanks, Matt! It's up again. –  user54535 Jan 4 '13 at 4:58
    
@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! –  Matt N. Jan 9 '13 at 15:58
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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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$dz\wedge dw$ is clearly holomorphic (and closed), but what is $\overline{\partial}^*dz\wedge dw$ (or $d^*dz\wedge dw$) in this case? Of course it depends on a metric on torus, but I don't quite see what it is. –  Pooya Dec 30 '12 at 23:28
    
Presumably he means the Kahler structure inherited from standard $\mathbb{C}^3 \supset (\mathbb{C}^*)^3$. In which case, Pooya, as you point out -- $\overline{\partial}^* (dz \wedge dw) = 0$. This isn't really a counter-example as far as I can tell. –  user54535 Jan 2 '13 at 15:44
    
By complex torus, I mean $\mathbb{C}^3$ mod a lattice (with the kahler structure coming from the universal cover), not torus in the sense of algebraic groups. Anyway, the point is $dz \wedge dw$ is not closed because there are three variables. –  user29743 Jan 3 '13 at 11:16
    
Maybe I'm not understanding your notation. If $z, w$ are two of the three coordinates on $\mathbb{C}^3$, then $dz \wedge dw$ seems closed to me. –  user54535 Jan 3 '13 at 19:00
    
Oh yes, I'm totally wrong! Will delete answer once OP reads this. –  user29743 Jan 7 '13 at 4:33
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