Here I transfered the question from the comment The relationship between spin and spinor curvature

How $\mathcal{R}_{ab} = \frac{1}{4}R_{abst}\gamma^s \gamma^t$ is from $\Psi \mapsto \Psi + \frac{1}{4} \epsilon_{\mu\nu}\gamma^\mu\gamma^\nu \Psi$ arise?

When $\mathcal{R}_{ab}$ is defined by

$\newcommand{\Rcal}{\mathcal{R}} \Rcal_{ab} \Psi = [D_a, D_b] \Psi$

and $R_{abst}$ is Riemann tensor constructed by metric.

$D_a$ is the covariant derivative in the Dirac equation in the curved spacetime
Dirac Equation in General Relativity

This post imported from StackExchange Physics at 2014-06-25 20:56 (UCT), posted by SE-user user48875