Poincaré bundle, a way of understanding it

So I will call the Poincaré bundle $(FM,pi, M)$ the principal fiber bundle that has the Poincaré group as a structure group, the space of linear frames as total space and $M$ as the Riemann-Cartan space.

I didn’t find a book or a paper that treats this kind of principal bundle (I would be happy if you share resources that you found) so I did my own point of view on this typical example and I want you to correct me if I dismiss some things in my line of reasoning.

So the idea is that we have a Poincaré group (the restricted one) $mathcal{P}^{uparrow}_{+}$ and we find that this group is too big because the goal (for me) is the representation of the restricted Lorentz group (RLG) in a spinor space so we restrict the $mathcal{P}^{uparrow}_{+}$ to the RLG, so to do that we know that the RLG is a Lie subgroup of $mathcal{P}^{uparrow}_{+}$ (which is also a Lie group) and with the Cartan theorem on closed subgroups, we have as a result: the RLG is a closed subgroup. Nevertheless, we still need to search for a global section for the $SO(1,3)^{uparrow}_{+}$-principal fiber bundle, the obvious one is the section that associates $forall x in M $ a section $ s: Mlongrightarrow OFM$, $OFM$ being the Orthogonal frame space.
Consequently, we restricted the Poincaré bundle to the $SO(1,3)^{uparrow}_{+}$-principal fiber bundle.

Searching for a representation to this group, we can use the principal bundle morphism that is $rho$-equivariant with $rho: SO(1,3)^{uparrow}_{+} longrightarrow Spin(1,3)$ and with the representation space being the spin frame bundle, after that, we associate naturally a spin bundle with typical fiber $F cong mathbb{R^4}$ by virtue of a linear left action of the Spin group that we can construct from the Clifford algebra or just simply by a linear representation of the $SO(1,3)^{uparrow}_{+}$ into the space of 4-by-4 matrices $S(Lambda)$ that acts on spinor fields.

Then we can define the one-form spin connection and the covariant derivative of a spinor field (section of the spinor bundle).

Thank you for your attention and just for you to know that am not trying to be extremely focused on rigorous step-by-step reasoning, but you are free to give me critics.