7
$\begingroup$

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?

$\endgroup$
9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ In fact, this argument shows that on any symplectic manifold, equipped with a metric and almost complex structure compatible with the symplectic structure (these always exist), the symplectic form is always co-closed. In other words, it also works in the almost-Kähler case. $\endgroup$ – Danu Sep 15 '18 at 19:01
7
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.