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Let $M$ be a Kähler manifold with fundamental form $\omega(X,Y) = h(JX, Y)$. I am trying to show that $\omega$ is harmonic. The Kähler condition implies that $\omega$ is closed with respect to $d$, so it suffices to show that $\delta \omega = *d*\omega = 0$. However, I have been unable to do so. Any suggestions?

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3 Answers 3

up vote 4 down vote accepted

Choose a local orthonormal frame $X_1, Y_1, ..., X_m, Y_m$ such that $J(X_i) = Y_i$ and denote by $(\alpha_1, \beta_1, ..., \alpha_m, \beta_m)$ the dual frame ($\alpha_i = X_i^{\flat}, \beta_i = Y_i^{\flat})$. In such a frame, $\omega$ is given by $\omega = \sum \alpha_i \wedge \beta_i$. The orthonormal frame $(X_1, Y_1, ..., X_m, Y_m)$ is positively oriented and so $(\alpha_1, \beta_1, ..., \alpha_m, \beta_m)$ is positively oriented frame of covectors with respect the induced orientation on $T^{*}M$. Hence, it is easy to describe how the Hodge star acts on $k$-forms which are wedge products of members of the frame.

For $\alpha_i \wedge \beta_i$, we have $$ *(\alpha_i \wedge \beta_i) = \pm \alpha_1 \wedge \beta_1 \wedge \ldots \hat{\alpha_i} \wedge \hat{\beta_i} \ldots \wedge \alpha_m \wedge \beta_m, $$ where the sign is determined by orientation of the basis $$ (\alpha_i, \beta_i, \alpha_1, \beta_1, \ldots, \alpha_m, \beta_m). $$ Since this is also positively oriented, the sign is $+1$ and $$ *\omega = \sum \alpha_1 \wedge \beta_1 \wedge \ldots \hat{\alpha_i} \wedge \hat{\beta_i} \ldots \wedge \alpha_m \wedge \beta_m = \frac{1}{(m-1)!} \omega^{m-1}. $$

This immediately implies that $*\omega$ is closed and so $\omega$ is co-closed and harmonic.

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I have another idea to solve this problem. We konw that $\delta$ can be described by the connection and the contraction as follow: $$\delta (\omega)=-\sum_{k=1}^{n}i(V_k)D_{\bar{V}_k}\omega.$$ Specially, we choose the Chern-connection, then the $D^{(0,1)}=\bar{\partial}$. And $$D_{\bar{V}_k}\omega=\bar{\partial}\omega(\bar{V}_k)=0.$$ So $\omega$ is harmonic.

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Does that formula for $\delta$ really hold for any connection? (I genuinely don't know, but I suspect not.) –  Jesse Madnick Dec 3 '13 at 9:33
    
Of couse not, but it is true for the Chern-connection. –  Daniel Jan 15 at 6:38

That the Kahler form is harmonic also follows from the identity $$ [L,\Delta_{d}] = 0, $$ where $L$ is the Lefschetz operator, and acts according to $L(\nu) := \nu \wedge \omega$, where $\nu$ is any form. This identity in turn follows from the Kahler identities. In fact, this identity shows that wedging any harmonic form with the Kahler form will again produce a harmonic form.

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