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Let $X$ be a Kähler manifold and consider the dual operator $\Lambda$ of the Lefschetz operator $L$. Is the following a true statement?

A closed $(1,1)$-form $\eta$ is harmonic if and only if $\Lambda\eta = \text{constant}$.

This is supposedly proven in Siu - Lectures on Hermitian–Einstein metrics for stable bundles and Kähler–Einstein metrics at Remark (1.4) part (ii), but the author states in the proof that "because the product of two harmonic forms is harmonic" which certainly isn't always true. I'm wondering if this is a consequence of some of the Kähler identities, but could not come up with a complete proof. If $\eta$ is harmonic, then $\Delta\eta = 0$. Since $\Lambda$ commutes with $\Delta$ we have $$\Delta(\Lambda \eta)=0$$ i.e. $\Lambda \eta$ is harmonic. Since $\Lambda \eta$ is a $0$-form it must be constant. The converse seems harder.

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  • $\begingroup$ Your argument for $\Delta \eta = 0$ implies $\Lambda \eta$ constant seems to assume that $X$ is compact and connected? If $X$ is not compact, then $\Delta(\phi)=0$ doesn't imply $\phi$ locally constant (EG $X = \mathbb{C}$ and $\phi$ any holomorphic function) and, of course, if $X$ is disconnected, than locally constant doesn't imply constant. $\endgroup$
    – David E Speyer
    Commented May 4 at 12:51
  • $\begingroup$ You're right, that's a bit problematic. Using the parallelity of $\omega$ as Jeffrey mentioned, would not need these assumptions on compactness and connectedness? @DavidESpeyer $\endgroup$
    – Nikolai
    Commented May 4 at 13:04

1 Answer 1

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Let $\eta$ be a closed $(1,1)$-form. Then $\partial\eta=0$ and $\overline{\partial}\eta=0$. Recall the Kähler identity \begin{equation*} [\partial,\Lambda] = -i\overline{\partial}^\ast . \end{equation*} Suppose that $\Lambda\eta$ is closed (this follows if it is constant). Then \begin{equation*} 0 = \partial\Lambda\eta = [\partial,\Lambda]\eta + \Lambda\partial\eta = i\overline{\partial}^\ast\eta . \end{equation*} Thus $\overline{\partial}^\ast\eta=0$. Therefore $\eta$ is harmonic.

You already gave the argument showing that if $\eta$ is a harmonic $(1,1)$-form, then $\Lambda\eta$ is a harmonic function. If your Kähler manifold is compact, this implies that $\Lambda\eta$ is locally constant. If $X$ is also connected, then $\Lambda\eta$ is constant.

Without assuming that the Kähler manifold is compact and connected, one cannot say that if $\eta$ is harmonic, then $\Lambda\eta$ is constant. For example, if $(X,\omega)$ is a one-dimensional Kähler manifold and $\eta$ is a $(1,1)$-form, then $\eta = f\omega$ for some function $f$. Moreover, $f = \Lambda\eta$ and $\Delta\eta = (\Delta f)\omega$. Hence $\eta$ is harmonic if and only if $\Lambda\eta$ is harmonic. But there are nonconstant harmonic functions when $X$ is noncompact (e.g. $f=z$) and when $X$ is compact but not connected (e.g. $f=1$ on one component and zero on the others).

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    $\begingroup$ TeX note: You probably want the first, rather than the second, of $\smash{\bar\partial}^* \bar\partial^*$ \smash{\bar\partial}^* \bar\partial^* (note the height of the $*$), but I didn't edit in case you prefer it the latter way. $\endgroup$
    – LSpice
    Commented May 4 at 14:39
  • $\begingroup$ Thank you for this. About your last comment, in general is the wedge product of a harmonic form with a parallel form harmonic again? @jeffrey-case $\endgroup$
    – Nikolai
    Commented May 4 at 20:04
  • $\begingroup$ No, the wedge product of harmonic forms need not be harmonic. But it is if one of those forms is parallel. $\endgroup$
    – Jeffrey Case
    Commented May 4 at 21:34
  • $\begingroup$ Sorry, I made a mistake in my last comment. The wedge product of two harmonic forms need not be harmonic, even if one of the factors is harmonic. I’ve also edited my answer to address this and to make sure my answer isn’t implicitly assuming $X$ is compact and connected, as in David Speyer’s comment to your question. $\endgroup$
    – Jeffrey Case
    Commented May 4 at 23:28
  • $\begingroup$ Not sure I follow anymore. Did you meant to say that "The wedge product of two harmonic forms need not be harmonic, even if one of the factors is parallel"? Also the argument I gave relies on compactness and connectedness, but I suppose the other direction which you gave did not, even in the first place. There should be a way to prove the converse without resorting to compactness and connectedness and I thought this came from parallelity of $\omega$, but guess not? @JeffreyCase $\endgroup$
    – Nikolai
    Commented May 5 at 19:11

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