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In the Euclidean setting, the Dirac operator was constructed so as to give the square of the Laplacian. Now for a K\"{a}hler manifold with a spin$^c$ structure we have the a corresponding Dirac operator $D$. Moreover, we have a Laplacian $(d+d^{\ast})$, where $d^{\ast}$ is the coadjoint $\ast d \ast $, for $\ast$ the Hodge $\ast$-mapping. Now in the case where the manifold is also symmetric we get a relationship between the square of the Dirac and the Laplacian that involves an extra curvature term. Does this extend to all K\"{a}hler manifolds, and if it does, what is the exact relationship?

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I highly suspect you are just talking about the Weitzenbock identities? en.wikipedia.org/wiki/Weitzenbock_identity It extends to any spin manifold and all Riemannian manifolds. –  Willie Wong Nov 19 '10 at 20:57
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(also, $d + \delta$ is not the Laplacian. THe Hodge Laplacian is the "square" of that.) –  Willie Wong Nov 19 '10 at 21:03
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As Willie says, I think that you are talking about the Lichnerowicz formula for the Dirac operator. –  José Figueroa-O'Farrill Nov 19 '10 at 21:15

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This is a very general statement valid for any first order differential operator such that its square has the same principal symbol as a Laplacian. One can then prove that the square of that first order operator differs from the covariant Laplacian $\nabla^*\nabla$ by a zeroth order term. For more details, see the monograph of Berligne, Getzler, Vergne.

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