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?



Relation between the de Rham and Hodge Laplacians on the Exterior Algebra
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
$$
\Delta''\_E=\Delta'\_E+[i\Theta(E),\Lambda\_\omega],
$$
where $[\bullet,\bullet]$ is the (graded) commutator.


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
$$
\Delta''=\Delta'+[\partial,\tau^\*]-[\bar\partial,\bar\tau^\*]
$$
and
$$
\Delta=\Delta'+\Delta''-[\partial,\bar\tau^\*]-[\bar\partial,\tau^\*],
$$
so that $\Delta'$, $\Delta''$ and $\frac 12\Delta$ no longer coincide but they differ by linear differential operators of order $1$ only. 


