most recent 30 from http://mathoverflow.net 2013-05-25T06:14:33Z http://mathoverflow.net/feeds/question/46674 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46674/square-of-the-dirac-and-the-laplacian-on-a-k-ahler-manifold Jean Delinez 2010-11-19T19:58:25Z 2012-03-14T11:27:48Z

<p>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?</p>

http://mathoverflow.net/questions/46674/square-of-the-dirac-and-the-laplacian-on-a-k-ahler-manifold/91162#91162 Liviu Nicolaescu 2012-03-14T11:27:48Z 2012-03-14T11:27:48Z

<p>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.</p>