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  The Clifford-Einstein equations

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Let $(M,g)$ be a riemannian manifold. I define an element in the Clifford algebra:

$$\tilde R (X)= \sum_{i,j} e_i .e_j . R(X,e_i)e_j$$

where $R\in \Lambda^2(TM) \otimes End(TM)$ is the Riemannian curvature. The Clifford-Einstein equations are:

$$\tilde R (X)=\lambda X$$

Can we solve the Clifford-Einstein equations for black holes?

asked Jun 6, 2022 in Mathematics by Antoine Balan (-80 points) [ no revision ]

Trivial answer is: Yes, of course, since every Einstein manifold (see A. Besse's mongraph) is a solution to the equation you defined as "Clifford-Einstein." Since a Clifford bundle contains as the spin bundle as a natural sub-bundle, stationary spherically symmetric solutions to Dirac-Einstein equations are also solutions. If you gauge to N=4 SUGRA, supersymmetric black-holes (coupled to gravitino) are also another class of solutions.

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