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For a Kahler manifold $M$, let us denote its Dirac operator $\overline{\partial} + \overline{\partial}^\ast$, with respect to a metric $g$, by $D$. Moreover, let us dentoe the Levi-Civita connection wrt $g$ by $\nabla$, and the corresponding connection Laplacian by $\nabla^*\nabla$.

If I am not mistaken, then the Lichnerowicz formula says that $$ D^2 = \nabla^{\ast} \nabla + O, $$ where $O$ is a zero order operator.

I have two questions:

(i) Does there exist a global version of this proof? The two versions I've looked at are in Andre Moroianu's notes, and Thomas Friedrich's book. Both are given in local terms, but both seem realtively algebraic, causing me to suspect that there may exist a global version. I am right here?

(ii) I know that the Lichnerowicz formula can be used to prove that $D$ has compact resolvent, wrt the standard Hilbert space completion of the exterior algebra. However, I'm finding it difficult to see the wood for the trees in proofs I have to hand. Could someone explain why a relation between the Dirac-Laplacian and the connection Laplacian tells me some about $(1+D)^{-1}$? Is it some clear that $\nabla^{\ast} \nabla$ has nice properties as Hilbert space operator?

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    $\begingroup$ I have asked a lot of knowledgeable people about (i), and nobody could point me to a satisfying global or conceptual argument. I think the physicists have some sort of heuristic explanation, but that's about it. $\endgroup$
    – Paul Siegel
    Commented Mar 3, 2012 at 16:58
  • $\begingroup$ As for (ii), Garding's inequality and Rellich's lemma together imply that any elliptic first order differential operator has compact resolvent. So the issue isn't really the Hilbert space properties of $\nabla^* \nabla$ but rather the way you intend to compute the symbol of $D$. $\endgroup$
    – Paul Siegel
    Commented Mar 3, 2012 at 17:13
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    $\begingroup$ Could one of you explain what you mean by a global proof? Isn't this a pointwise statement? $\endgroup$
    – Deane Yang
    Commented Mar 3, 2012 at 18:44
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    $\begingroup$ The operator $\nabla^*\nabla$ is a self-adjoint positive elliptic operator operator, so its inverse is smoothing. Those are rather strong and useful properties. $\endgroup$
    – Deane Yang
    Commented Mar 3, 2012 at 18:46
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    $\begingroup$ Jean, the formula you give is for the spin Dirac operator on a spin manifold. However, the Dolbeault operator is an example of a spin-c Dirac operator $D_A$, associated with a connection $A$ in the determinant line bundle - in this case, the canonical line bundle for the complex manifold. The Lichnerowicz formula for such operators has another term, a multiple of $F_A$, viewed through Clifford multiplication as an endomorphism of the spinor bundle. $\endgroup$
    – Tim Perutz
    Commented Mar 3, 2012 at 22:42

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I seem not to be able to comment, so this is not an answer to the question:

@Paul Maybe it gives some intuition if you regard the flat case first. (I think this was the historical beginning of Dirac-Operators): In the flat case the difference $D^2-\nabla^*\nabla $ ist just $0$. If you now regard a bundle $S$ with curvature instead it is not too surprising that curvature comes into play. Why the curvature term decomposes into twisting- and scalar curvature might become a bit clearer if you calculate the clifford action of the induced curvature two form. Roe does this calculation in "elliplic operators, topology and asymtotic methods"(p.46 Lemma 3.13 and 3.15)

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