If we consider a spin manifold $M$, we can define the Ricci curvature $Ricc (X,Y)$ which is a symmetric tensor, moreover the spinors are defined, so that we can define a Dirac-Ricci operator: $$DR(\psi)=\sum_{ij}Ricc(e_i,e_j)[ e_i. \nabla_{e_j}\psi+e_j .\nabla_{e_i}\psi]$$ with $(e_i)$ an orthonormal basis.

If the manifold $M$ is an Einstein manifold ($Ricc=\lambda g$), then $DR=\lambda D$, the Dirac-Ricci operator is the Dirac operator $D$. Have we nice properties of such an operator in the general case?

  • $\begingroup$ The Dirac operator has an invariant definition as the composition $$ C^\infty(S)\stackrel{\nabla}{\to} C^\infty(T^*M\otimes S)\stackrel{c}{\to} C^\infty(S),$$ where $S$ denotes the spinor bundle and $c$ denotes the Clifford multiplication. I do not understand why the operator $DR$ is invariantly defined, If we replace the orthonormal moving frame $(e_i)$ by another, do we get the same thing? $\endgroup$ – Liviu Nicolaescu Sep 7 '18 at 18:12
  • $\begingroup$ If we have an expression like $\sum_i f(e_i,e_i)$ with $f$ bilinear, then it is always invariantly defined. We can indeed replace by another basis with an orthogonal automorphism and use $\sum_i a_{ij} a_{ik}=\delta_j^k$. $\endgroup$ – Antoine Balan Sep 7 '18 at 19:11

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