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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asked Mar 24, 2011 at 18:03
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3$\begingroup$ Yes, the identity holds for differential forms. This is what's behind the Hodge decomposition. See your favourite book on Kahler manifolds (Griffiths-Harris, Wells...) for an explanation. $\endgroup$– Donu ArapuraCommented Mar 24, 2011 at 18:20
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2$\begingroup$ (I don't think you want the superscript $2$ in your displayed formula.) On a Kähler manifold, the three laplacians associated with $d$, $\partial$ and $\bar\partial$ satisfy $$\Delta_d = 2 \Delta_{\partial} = 2 \Delta_{\bar\partial}$$ not just on functions, but also on forms. This is why on a compact Kähler manifold, one has the decomposition of the de Rham cohomology in terms of the Dolbeault cohomology groups. This can be found, e.g., in Well's Differential analysis on complex manifolds. $\endgroup$– José Figueroa-O'FarrillCommented Mar 24, 2011 at 18:24
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$\begingroup$ He does want superscripts, since what he calls Laplacian is not... He is writing the Dirac operator instead... $\endgroup$– diveriettiCommented Mar 24, 2011 at 18:28
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$\begingroup$ Sorry, about that, superscripts in the wrong place. Yes, it still works out, but best to stick to convention. $\endgroup$– Jean DelinezCommented Mar 24, 2011 at 18:31
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1 Answer
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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 $$ \Delta''_E=\Delta'_E+[i\Theta(E),\Lambda_\omega], $$ 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.
