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Let $M$ be a complex manifold and $\Delta^{\bar \partial} = \bar\partial^* \bar\partial + \bar\partial\bar\partial^*$ the complex laplacian. Is it true that $\Delta^{\bar\partial} f = \Delta f$ (the ordinary Hodge laplacian of $f$) for $f\in C^\infty(M)$ a $0$-form?

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The answer to your question is no.

If $M$ is Kähler, $\Delta^{\bar{\partial}} = \frac{1}{2}\Delta$. Therefore, if $\Delta f \neq 0$, we have $\Delta f = \frac{1}{2}\Delta^{\bar{\partial}}f \neq \Delta^{\bar{\partial}}f$. For example, if $M = \mathbb{C}$ and $f(z) = |z|^2$, then $\Delta f = \pm 4$ (depending on your sign convention for $\Delta$) but $\Delta^{\bar{\partial}}f = \pm 2$.