For a compact Kähler manifold, we say that a form is primitive if it is contaned in the kernel of the dual Lefschetz operator, or the co-Lefschetz operator. For all examples I know, a primitive form $\omega$ is closed with respect to the $d$ de Rham exterior derivative if and only if $\omega$ is harmonic. I suspect that this is true for any compact Kähler manifold, but I don't know how one would prove it.

  • $\begingroup$ @Lars: Just to be clear, $\omega$ in your question is an arbitrary primitive form right? The reason I ask is that $\omega$ is usually reserved for the Kähler form. $\endgroup$ – Michael Albanese Nov 21 '17 at 22:31

If $\omega$ is harmonic, then it is $d$-closed. For the other direction, suppose $\omega$ is $d$-closed and primitive, i.e. $d\omega = 0$ and $\Lambda\omega = 0$. Then, by a Kähler identity,

$$i\bar{\partial}^*\omega = [\Lambda, \partial]\omega = \Lambda(\partial\omega) - \partial(\Lambda\omega) = \Lambda(\partial\omega)$$

and similarly $-i\partial^*\omega = \Lambda(\bar{\partial}\omega)$. So

$$d^*\omega = \partial^*\omega + \bar{\partial}^*\omega = \Lambda(i\bar{\partial}\omega - i\partial\omega) = \Lambda(d^c\omega)$$

where $d^c = i(\bar{\partial}-\partial)$. So $\omega$ is harmonic if and only if if $d^c\omega$ is primitive.

If $\omega$ is a $(p, q)$-form, then $d\omega = 0$ is equivalent to $\partial\omega = 0$ and $\bar{\partial}\omega = 0$, in which case it follows that $\partial^*\omega = 0$ and $\bar{\partial}^*\omega = 0$. Therefore $d^*\omega = 0$ and hence $\omega$ is harmonic.

I'm not sure if the statement still holds if $\omega$ is an arbitrary form. The issue is that $d\omega = 0$ does not necessarily imply $\partial\omega = 0$ and $\bar{\partial}\omega = 0$ in this case.

| cite | improve this answer | |

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.