# The Clifford-Einstein equations

+ 0 like - 0 dislike
433 views

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?

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.

 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$\hbar$ysicsOv$\varnothing$rflowThen 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