For a Kahler manifold $M$ we have two well-known Laplacians: the de Rham Laplacian $\Delta_{\text{d}} = ($d$ + $d$^\ast)^2$, and the Dolbeault Laplacian $\Delta_{\overline{\partial}} = (\overline{\partial} + \overline{\partial}^\ast)^2$. Now on smooth functions, these two operators are related by the well-known formula $$ \Delta_{\text{d}} = 2\Delta_{\overline{\partial}} $$ Now both these operators act on the exterior algebra. Does there exist a similar formula in this more general setting?
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If $(X,\omega)$ is Kähler, then it is always true that $$ \Delta'=\Delta''=\frac 12\Delta, $$ where these three Laplacians are with respect, in order, to $\partial$, $\bar\partial$ and $d$. This is valid when they act on any space of complex-valued differential forms. More generally, you can look to differential forms with values in a hermitian vector bundle $E\to X$. In this case, take $D_E$ to be the (unique) Chern connection of $E$ and let $D_E=D'_E+D''_E$ its decomposition in the $(1,0)$ and $(0,1)$ part (then, by definition $D''_E=\bar\partial$). In this case, you can again compare $\Delta'_E$ and $\Delta''_E$. They no longer coincide, but differ by a order zero operator which is expressed in terms of the curvature $\Theta(E)=D^2_E$ and the (formal) adjoint $\Lambda_\omega$ of the operator $L_\omega=\omega\wedge\bullet$ of wedge product with $\omega$. The relation is
On the other hand, coming back to complex-valued differential forms, if you merely suppose your manifold to be hermitian than the relation between the three Laplacians is a little bit more complicated. You have to introduce the torsion operator
$$
\tau=[\Lambda_\omega,\partial\omega]
$$
which is of type $(1,0)$ and order zero (observe that if $\omega$ is Kähler then $\partial\omega=0$). With these notations, you have
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