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


Let me try to explain a vox populi secret.

If you have a two sphere, $S^2$, and define a vector on a point $p$ (you can imagine a vector as an arrow), the vector does not lie on the sphere!. In fact the set of all possible vectors at that point generate (or live) in a flat and tangent space to the sphere at $p$, denoted as $T_p S^2$.

First note: On $T_p S^2$ you can define an Euclidean action. If your manifold is not $S^2$ but a spacetime, you can define a Lorentz action on the tangent space. This action of the Euclidean (Lorentzian) group is local, because is defined for each point $p\in\mathcal{M}$.

When people uses the parallel transport, the vector moves from a tangent space to another, these motion is achieved by defining a relation (or connection) among the tangent spaces... if you want this relation tells you how the tangent spaces are glued together.

Second note: The connection tells you how to move from a tangent space to another, and relates all of the Euclidean (Lorentzian) actions on them.

The curvature measures whether or not it is possible to define a global Euclidean (Lorentzian) structure on your manifold. If there is no curvature then you can!... if the curvature does not vanishes, you are out of luck. So, somehow the curvature tells you if the local structure of your manifold can be "extended" to be global.

Third note: Curvature is related with your local Euclidean or Lorentzian structure.

Now, some fun!!! (I'll restrict to Lorentz... got tired of writing Euclidean -Lorentz- )

The Lorentz group is defined by a set of generators. It is possible to find inequivalent sets of generators (this is related with the different representations of a -Lie- group).

A particular set of generators of the Lorentz group is given by the commutator of gamma matrices $$\mathbb{J}^{ab} \simeq \left[\gamma^a,\gamma^b\right].$$

You can show explicitly that those $\mathbb{J}$'s satisfy the algebra of Lorentz.

Finally, as in any local (or gauge) theory, one defines a covariant derivative $$\mathcal{D}_\mu = \partial_\mu + \Omega_\mu,$$with $\Omega = \frac{1}{2}(\omega_\mu)_{ab}\mathbb{J}^{ab}$.

Calculating the commutator you'll get the desired result!!!

That is the relation between the curvature and Lorentz group ;-)


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