Square of the Dirac and the Laplacian on a K\"{a}hler Manifold - MathOverflow 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 Square of the Dirac and the Laplacian on a K\"{a}hler 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 Answer by Liviu Nicolaescu for Square of the Dirac and the Laplacian on a K\"{a}hler Manifold 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>