Spin connection vs. Cartan connection

Question

I am studying the tetradic Palatini formalism of general relativity. In this formalism, one usually considers a manifold $M$, which is either non-compact or compact with Euler-characteristic $\chi(M)=0$ so that there exist Lorentzian metrics on $M$. Choosing such a metric $g$ on $M$, one defines the bundle of orthonormal co-frames on $M$, denotes $\mathcal{F}_{\mathrm{ort}}(TM)$. The two variables of interest in the Palatini formalism are the "frame field", which is a bundle isomorphism $e:\mathcal{T}\to TM$, where $\mathcal{T}:=\mathcal{F}_{\mathrm{ort}}(TM)\times_{\rho}\mathbb{R}^{1,3}$ denotes the associated vector bundle using the fundamental representation $\rho$ of $\operatorname{SO}(1,3)$ and the "Cartan connection" $\omega$, which is a connection $1$-form of $M$. Now, in many texts, $\omega$ is also referred to as "spin connection", a terminology which is usually also used for connections of a spinor bundle. So, a natural question is the following:

How is the Cartan connection appearing in the tetradic formulation of gravity related to a connection of a spinor bundle?

Answer 1

The possible spin structures of the Lorentz metric are the possible choices of principal $\operatorname{Spin}(3,1)$-bundle and equivariant bundle morphism to the frame bundle. By discreteness of $\operatorname{Spin}(3,1)\to SO(3,1)$, the Lie algebras are the same. The possible connections on a principal $G$-bundle $\pi\colon B\to M$ are the splittings of the sequence of vector bundles $0\to B\times^G\mathfrak{g}\to TB^G \to TM\to 0$, the $G$-quotient of the obvious inclusion of vertical vectors into all tangent vectors: $0\to B\times\mathfrak{g}\to TB \to \pi^*TM\to 0$ (see Atiyah, 1957, Complex analytic connections in fiber bundles, for details). But if the Lie algebras are the same, say we have groups $G\to\bar{G}$ and associated bundles $B\to\bar{B}$, these quotients turn out to be the same, since quotienting by the kernel of $G\to\bar{G}$ (for us, $\operatorname{Spin}(3,1)\to SO(3,1)$) acts trivially on $\mathfrak{g}=\mathfrak{so}(3,1)$.