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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.

Qmechanic
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theriddler
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    Can you give us some idea of what research you've already done? For example you provide links to a couple of Wikipedia articles but no indication that you've read the Wikipedia article on Killing spinors. If you have read that article can you give some idea of what the article leaves unclear? – John Rennie Sep 10 '14 at 17:41
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    The links were not added by me but by the editor. My knowledge of general relativity is introductory (taken a graduate course in it). I know that Killing vectors lead to quantities that are conserved for geodesics. I have been introduced to the Lie derivative definition of Killing vectors (not just via the Killing equation). Killing tensors I have not used nor Killing spinors. – theriddler Sep 10 '14 at 17:57
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    As explained here something being a "killing object" is related to it being covariantly invariant. – Dilaton Sep 12 '14 at 08:13

1 Answers1

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The term is used slightly differently in mathematics and physics, in particular in supergravity. Supergravity extends general relativity with local supersymmetry and includes a spin-3/2 particle called the gravitino.

In GR the parameter of an infinitesimal diffeomorphism is a vector. If the diffeomorphism leaves the metric invariant it is an isometry and the parameter is a Killing vector.

In supergravity the parameter of a supersymmetry is a spinor. If it leaves all the fields invariant it is called a Killing spinor.

Here's an excerpt from https://arxiv.org/abs/hep-th/9803116 by K.S Stelle.

sugra transform

Here $\epsilon$ is the supersymmetry parameter, $\psi$ is the gravitino, $F$ is an antisymmetric 4th rank tensor, $e$ is a vielbein, $\omega$ is the spin connection and the $\Gamma^{AB...}$ are products of Dirac gamma matrices. This makes $D_A$ the covariant derivative of a spinor.

The condition for a Killing spinor is that the right hand side of (4.17) vanishes. Note that when $F=0$ this reduces to $D_A\epsilon=0$, i.e. $\epsilon$ is covariantly constant. Mathematicians call this a parallel spinor. For them a Killing spinor satisfies $D_A\epsilon = \lambda\Gamma_A\epsilon$ for some $\lambda$.

Subhaneil Lahiri
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  • Is there a geometric interpretation for a non-parallel spinor? – TheQuantumMan Mar 31 '19 at 06:29
  • Sorry, I have no idea. – Subhaneil Lahiri Mar 31 '19 at 10:07
  • Also, why do we impose that the gravitino vanishes? – TheQuantumMan Mar 31 '19 at 15:02
  • The gravitino is a fermionic field, so it cannot take on classical values (see [https://physics.stackexchange.com/questions/108112/why-cannot-fermions-have-non-zero-vacuum-expectation-value]) Anyway, the important thing is setting $\delta\psi=0$, so that it is invariant under the transformation. – Subhaneil Lahiri Mar 31 '19 at 17:14
  • Great answer and sorry for insisting on this point, but the top answer in the link your provided argues that only scalars can have non zero vaccuum expectation value. That means that even the graviton and photon would be trivial, which is not true since we can have classical solutions with those two being non-zero. – TheQuantumMan Mar 31 '19 at 18:16
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    Yeah, I think the second answer there was a lot better. I should have mentioned that in my last comment. – Subhaneil Lahiri Mar 31 '19 at 21:37