# The Einstein-Killing spinors

+ 0 like - 0 dislike
440 views

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?

edited Jun 15, 2022

+ 0 like - 0 dislike

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 15, 2022

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar\varnothing$sicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). Please complete the anti-spam verification