Note: For this question I am using the conventions of "Ideas and Methods of Supersymmetry and Supergravity" by Ioseph Buchbinder and Sergei Kuzenko (mostly p16 & p44).

Let our space be equipped with the spin connection, which defines the covariant derivative:

$$\nabla_a = e_a \, ^m \partial_m + \tfrac{1}{2} \omega_{abc} M^{bc}$$

Where $e_{a} \,^{m}$ is the vielbein, $\omega_{abc}$ are the connection coefficients and $M^{bc}$ are the Lorentz generators. (Here the $m$ refers to a world index while $a , b , c$ refer to flat space indicies).

On this space we have a (2 component) spinor field $\epsilon_{\beta}$ which we assume is covariantly constant with respect to this connection i.e. :

$$ \nabla_{a} \epsilon_{\beta} = 0 $$

This obviously implies that $\big[ \nabla_a , \nabla_b \big] \epsilon_{\beta} = 0$. We can then convert the flat space indicies in our deviative to spinor form by some basic manipulation:

$$ \big[\nabla_{\alpha \dot{\alpha}} , \nabla_{\beta \dot{\beta} } \big] = R_{\alpha \dot{\alpha},\beta \dot{\beta}, \gamma \delta} M^{\gamma \delta} + \bar{R}_{\alpha \dot{\alpha},\beta \dot{\beta}, \dot{\gamma} \dot{\delta}} \bar{M}^{\dot{\gamma} \dot{\delta}}$$

Which then implies that:

$$ \big[\nabla_{\alpha \dot{\alpha}} , \nabla_{\beta \dot{\beta} } \big] \epsilon_{\gamma} = R_{\alpha \dot{\alpha},\beta \dot{\beta}, \gamma \delta} \epsilon^{\delta} = 0$$

If we then impose that $R_{abcd}$, the flat space index Riemann curvature tensor is non-degenerate, does this then imply that the only possible solution to $\nabla_{a} \epsilon_{\beta} = 0$ is $\epsilon_{\beta} = 0$? Are there other implications that I have missed?

Edit: After some further research, in particular from http://www.igt.uni-stuttgart.de/LstGeo/Semmelmann/Publikationen/2000jmp.pdf , which describes these objects as 'parallel spinors', it seems that in order for a spin manifold to admit such objects, it must be that the Ricci tensor vanishes.


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