In the Euclidean setting, the Dirac operator was constructed so as to give the square of the Laplacian. Now for a Kähler 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 Kähler manifolds, and if it does, what is the exact relationship?
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asked Nov 19, 2010 at 19:58
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4$\begingroup$ 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. $\endgroup$– Willie WongCommented Nov 19, 2010 at 20:57
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2$\begingroup$ (also, $d + \delta$ is not the Laplacian. THe Hodge Laplacian is the "square" of that.) $\endgroup$– Willie WongCommented Nov 19, 2010 at 21:03
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1$\begingroup$ As Willie says, I think that you are talking about the Lichnerowicz formula for the Dirac operator. $\endgroup$– José Figueroa-O'FarrillCommented Nov 19, 2010 at 21:15
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1 Answer
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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.
answered Mar 14, 2012 at 11:27