Question about the proof of $\Delta_d = 2\Delta_{\bar{\partial}} = 2\Delta_{\partial}$ in Principles of Algebraic Geometry by Griffiths and Harris.

Question

On page $115$ of Principles of Algebraic Geometry by Griffiths and Harris, in the proof of $\Delta_d = 2\Delta_{\bar{\partial}} = 2\Delta_{\partial}$, they state that

$$\sqrt{-1}(\partial\bar{\partial}^*+\bar{\partial}^*\partial) = \partial( \Lambda \partial -\partial \Lambda )+ (\Lambda \partial -\partial \Lambda)\partial = \partial \Lambda \partial - \partial \Lambda \partial.$$

I don't think the last line of equality is right, am I right?

I think it should be equal to: $\Lambda \partial \partial -\partial \partial \Lambda$, in which case why does it vanish?

Answer 1

Note that

\begin{align*} \sqrt{-1}(\partial\bar{\partial}^* + \bar{\partial}^*\partial) &= \partial(\Lambda\partial - \partial\Lambda) + (\Lambda\partial - \partial\Lambda)\partial\\ &= \partial\Lambda\partial - \partial\partial\Lambda + \Lambda\partial\partial - \partial\Lambda\partial\\ &= \Lambda\partial\partial - \partial\partial\Lambda. \end{align*}

As pointed out in the comments, on any complex manifold $\partial^2 = \partial\partial = 0$, so

$$\sqrt{-1}(\partial\bar{\partial}^* + \bar{\partial}^*\partial) = 0.$$

Note, the Kähler hypothesis was needed in order to use the Kähler identity

$$\sqrt{-1}\bar{\partial}^* = [\Lambda, \partial] = \Lambda\partial - \partial\Lambda.$$