# The Einstein-Killing spinors

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Let $(M,g)$ be a spin manifold with Ricci curvature $Ric$. An Einstein-Killing spinor $\psi$ is defined by the following equation:

$$X.{\cal D}\psi =\mu Ric(X).\psi$$

where ${\cal D}$ is the Dirac operator and $X$ is a variable tangent vector, $\mu$ is a constant.

If the manifold $M$ is Einstein, then the spinor $\psi$ is a proper vector of the Dirac operator.

What is the space of Einstein-Killing spinors?

asked Jun 15, 2022
edited Jun 15, 2022

## 1 Answer

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Taking the norm, we have that:

$$||Ric(X)||=\lambda ||X||$$

so that the manifold is Einstein and the spinor $\psi$ is a proper vector of the Dirac operator.

Another interesting equation would be:

$$\nabla_X {\cal D} \psi = \mu Ric(X).\psi$$

with $\nabla$ the spinorial connection.

answered Jun 15, 2022 by (-80 points)
edited Jun 16, 2022

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