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  What are Killing spinors?

+ 6 like - 0 dislike
1613 views

What are Killing spinors? How can they be motivated? Are they directly related to Killing vectors and Killing tensors and is there an overarching motivation for all three objects? Any answer is greatly appreciated but a less formal one would be preferred.

This post imported from StackExchange Physics at 2014-09-10 17:12 (UCT), posted by SE-user theriddler
asked Sep 10, 2014 in Theoretical Physics by theriddler (30 points) [ no revision ]
retagged Sep 10, 2014

1 Answer

+ 2 like - 0 dislike

There is an interesting relation, as the \(n\)lab says. I'll try to explain it as informally as possible.

Let \(\mathcal{M}\) be a pseudo-Riemannian manifold. Then, a Killing vector field on \(\mathcal{M}\) is a covariantly constant vector field on \(\mathcal{M}\), and "pairing two covariant constant spinors (parallel spinors, i.e., Killing spinors with \(\lambda=0\)) to a vector yields a Killing vector". Similarly, a Killing tensor on \(\mathcal{M}\) is a covariantly constant section of \(\mathrm{Sym}^k(\Gamma(\mathrm{T}(\mathcal{M})))\). Therefore, you may interpret ``Killing'' as being synonymous with ``covariantly constant'' (at least in these three cases).

answered Sep 11, 2014 by SDevalapurkar (285 points) [ revision history ]
edited Sep 12, 2014 by SDevalapurkar

This is not the definition given by most authors.  Killing spinors and parallel spinors verify \(\nabla_X \psi = \lambda X.\psi, \lambda \in \mathbb C\) . Parallel spinors (covariantly constant spinors) correspond to the case \(\lambda = 0\), while Killing spinors correspond to the case \(\lambda \neq 0\)

True. I've fixed that nLab entry.

I, too, have edited my answer accordingly. 

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