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From https://en.wikipedia.org/wiki/Killing_spinor we see that a Killing Spinor $\epsilon$ is defined as a solution of the following equation:

$$\nabla_{X}\epsilon = \lambda X * \epsilon \quad , \quad \lambda \in \mathbb{C}$$

Where $X$ is a tangent vector and $*$ is known as 'Clifford Multiplication' (however no definition is specified).

Then, we choose the connection to be the spin connection and consider the following equation:

$$ \nabla_{a} \epsilon = \frac{i}{3} (b_{a} - \sigma_{ac}b^{c})\epsilon$$

I am attempting to show that this takes the general form of a Killing Spinor equation, however the lack of a definition for $*$ seems to be causing difficulties. I am able to write the equation in the form:

$$ \nabla_{X} \epsilon = \frac{i}{3} (b_{a}X^{a} \sigma_{0}- X^{a}\sigma_{ac}b^{c}) \epsilon$$

However, this does not seem to be correct.

Another approach I was considering was to adjust the connection as follows:

$$ \mathcal{D}_{X} := \nabla_{X} - \frac{i}{3} X^{a}b_{a} \sigma_{0} $$

Which would yield:

$$ \mathcal{D}_{X} \epsilon = - \frac{i}{3} X^{a} \sigma_{ac} b^{c} \epsilon$$

I think that the confusion lies mostly in the lack of a formal definition of the Clifford Multiplication available both online and in texts as most assume this as prior knowledge. After knowing the definition is it trivial to then show that the originial equation is indeed a Killing spinor equation?

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  • $\begingroup$ How familar are you with the concept of a spin-structure of a given manifold? I think the problem is that you try to compare local quantities with global ones, a reference might be ''Spin Geometry'' by H.Blaine Lawson, JR. and Marie-Louise Michelsohn (specifically Chapter II, $\S$ 3), though this is not an easy read if your are not familar with Differential Geometry and Topology. $\endgroup$ – Creo Oct 16 '18 at 9:31
  • $\begingroup$ Maybe an easier approach (or if you haven't got a copy of Blaine et al.) could be to click the link to Clifford multiplication in your Killing Spinor link? It goes straight to en.wikipedia.org/wiki/Clifford_algebra . If that doesn't satisfy, ask via the Talk page of your Killing spinor link? (Please forgive me if I'm "teaching granny to suck eggs"; No offence intended.) $\endgroup$ – iSeeker Oct 21 '18 at 18:25

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