Covariant derivative of a Dirac spinor and Kosmann lift

Question

In [1] I have found a definition of the covariant derivative of a Dirac field with a general connection $\omega_{\mu a}{}^{b}$ (with torsion and non-metricity) [see eq. (29)]:

$$\nabla_{\mu}\psi=\partial_{\mu}\psi-\frac{1}{4}\omega_{\mu ab}\gamma^{a}\gamma^{b}\psi$$

They use the so-called Kosmann lift, to build the spinor connection $$\Gamma_{\mu}=-\frac{1}{4}\omega_{\mu ab}\gamma^{a}\gamma^{b}$$ from $\omega_{\mu a}{}^{b}$. I have two questions,

(Question 1) Is there a simple way to explain that “Kosmann lift”? I have basic ideas about fiber bundle theory, but I get completely lost if we go deeper in that formalism. I am not looking for a strictly rigorous explanation.

(Question 2) If you expand that covariant derivative you obtain the standard Lorentzian one plus a new term that does not transform well under Lorentz transformations:

$$\nabla_{\mu}\psi=\partial_{\mu}\psi-\frac{1}{4}\omega_{\mu ab}\gamma^{[a}\gamma^{b]}\psi- \frac{1}{4}\omega_{\mu a}{}^{a}\psi =\nabla_{\mu}^{Lor}\psi-\frac{1}{4}\omega_{\mu a}{}^{a}\psi$$

I would think that this expression cannot be a “good covariant derivative” because you want to write something Diff and (local) Lorentz invariant. Am I wrong?

[1] M. Adak, T. Dereli, L.H. Ryder, Dirac equation in spacetimes with torsion and non-metricity. arXiv:gr-qc/0208042

Answer 1

Kosmann lift

Lie derivatives are defined naturally on vector and tensor fields, because we know how to lift diffeomorphisms of the manifold to its tangent and cotangent bundle. Kosmann was able to define a Lie derivative of a spinor field along a Killing vector. In contrast to Lie derivatives of vector fields, the Kosmann Lie derivative of a spinor field depends on the metric, but otherwise satisfies all the properties of a Lie derivative. Her idea is as follows:

The Lie algebra of the group $SO(n)$ is generated by antisymmetric matrices $E_{\alpha \beta}$. This algebra acts on spinors via the spinor representation: $$\rho(E) = E_{\alpha \beta} \gamma^{\alpha} \gamma^{\beta}$$ ($\gamma$ are the Dirac matrices). It is not difficult to show that the matrices $\rho(E)$ satisfy the Lie algebra of $\mathrm{Spin}(n)$ which is the same as $SO(n)$.

On a Riemannian spin manifold $(M, g)$, a Killing vector $\xi$ satisfies the Killing equation: $$\nabla_{\alpha }\xi_{\beta } + \nabla_{\beta} \xi_{\alpha }=0$$ thus the tensor field $\nabla_{\alpha }\xi_{\beta }$ is antisymmetric (only in the case of a Killing vector), therefore can be represented on spinors via the same relation above. This enables to define a Lie derivative on spinors: $$\mathcal{L}_{\xi} \psi = \xi^{\alpha} \nabla_{\alpha} \psi - \nabla_{\alpha }\xi_{\beta } \gamma^{\alpha} \gamma^{\beta} \psi$$ The motivation for the above definition is that the Lie derivative on vectors can written in an analogous way in terms of covariant derivatives, but in the case of vectors all dependence on the metric cancels out unlike the case of spinors: $$\mathcal{L}_{\xi} \zeta = [ \xi, \zeta] = \xi^{\alpha} \nabla_{\alpha} \zeta - \nabla_{\alpha} \xi \zeta^{\alpha}$$

Second question

I think that the additional term is a true tensor and all properties of the covariant derivative are preserved.