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In the course of QFT i learnt that the gauge field emerges from the need of a gauge invariance in the action, as we use the covariant derivative in minimal coupling. Now i'm studying how spin fields couple on curved space and it seems that is instead the gauge invariance that emerges from this attempt to couple spin field with a curved space, since the tetrad formalism has more degrees of freedom than the metric one, which correspond to the dimension of the proper orthochronous Lorentz group:

$$\text{dim}(e^{\mu}_{a})-\text{dim}(g_{\mu\nu})=d^2-\frac{d(d+1)}{2}=\text{dim}(\tilde{L}^+).$$

Anyway i aspect that this gauge invariance is not the same as the one we require in QFT since QFT is made on flat space. What is wrong and what is true in this reasoning? Is there a physical meaning of the additional degrees of freedom in tetrad formalism?

EDIT: I thought about this question and i believe that i was a bit confused when asked it because the teacher made the exemple of covariant electromagnetism when obtained the covariant derivative of a spin vector (with spin connection). Now, i think that the question were:

  1. Is spin connection a complete different object than connection due to some symmetry of the theory?

  2. Could we have both spin connection and gauge connection in the covariant derivative?

  3. As the connection of (i.e.) $U(1)$ gauge symmetry give rise to electromagnetism, what kind of interaction give rise the coupling with spin connection?

  4. As the generator of gauge $U(1)$ is the electric charge (so an observable), is the generator of the orthochronous Lorentz group an observable too?

EDIT

As a result of conversation with another user it turns out that the physical meaning of the 6 extra dof is that they describe the coupling with gravity, and they are not present for scalar fields because those don't interact with gravity.

See this: Spin field on curved space: meaning of coupling with spin connection

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    1. No, a connection is a connection. In the case of spin connections they arise from the symmetry group that double cover the Lorentz group. At each point in spacetime you have the freedom to choose a reference frame, so you have a $SO(1,3)$ gauge invariance. Lifting this to its double cover you obtain the spin connection in the same way you obtain the EM field connection from the $U(1)$ gauge group.
    1. Of course, if you want a complete action describing all the forces you need to insert both the spin connection and the other connections from the standard model groups into the covariant derivative.
    1. The coupling with the spin connection give rise to the gravitational interaction. The curvature tensor given by the spin connection is in fact the curvature tensor of general relativity. So you could frame general relativity in terms of a connection inside couplings and a self interacting part made with the curvature tensor of such a connection.
    1. The generators that give rise to such a connection are observables: they are the particle spins! So each spin can be seen as the "charge" of the spin connection interaction. (Note: for scalars in general relativity you don't use the covariant derivative but the ordinary partial derivative: they have 0 "spin charge", in fact scalar fields are spin-0 bosonic fields)
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  • $\begingroup$ Thank you for the answer. So, if I understood: given a spinorial field in 4D flat space we have SO(1,3)+traslation=10 dof, so we can use metric formalism; in 4D curved space beyond those 10 dof we have an additional SO(1,3)=6 dof that gives rise to gravitational interaction, and we use tetrad formalism to describe it. To what those additional dof are due? To some freedom in choosing a frame that fixes the spin of the particle? And is maybe the graviton the spin connection? $\endgroup$
    – polology
    Commented Aug 31, 2023 at 12:08
  • $\begingroup$ No wait, you don't have extra degrees of freedom. In flat spacetime you have 0 dof because you can define a global frame, you have a global $SO(1,3)$ and 4 global translations, like when you have a global phase in the wavefunction, you have no $U(1)$ interaction. When you gauge the $SO(1,3)$ and the 4 translations you obtain 10 dof: 6 spin connection fields and 4 vielbeins. This is analogue to gauging the global phase obtaining the single $U(1)$ connection: the EM field. Those new 10 gauged dof are in fact your 10 metric dof you have in general relativity. $\endgroup$
    – LolloBoldo
    Commented Aug 31, 2023 at 12:45
  • $\begingroup$ Okay now I understand, but what about the computation of ndof I wrote in the question (and that my teacher showed me)? From that calculation in tetrad formalism we have 6 dof more than metric's one, and since it's the dimension of $\tilde{L}^+$ I thought that the 6 spin connection fields came from that $\endgroup$
    – polology
    Commented Sep 1, 2023 at 11:25
  • $\begingroup$ I think your problem is that you say that $dim(e^a _\mu)$ has dimension $d^2$ since it's a square matrix, but don't use the same criteria for the metric tensor, which is simmetric. $e^a_\mu$ is not an arbitrary matrix so it can't have $d^2$ degrees of freedom $\endgroup$
    – LolloBoldo
    Commented Sep 1, 2023 at 11:39
  • $\begingroup$ You say that is not an arbitrary matrix beacuse we impose the frame basis is orthonormal (although it doesn't necessarily have to be so)? If so, how to show that this condition gives us the right ndof for the vielbein? $\endgroup$
    – polology
    Commented Sep 1, 2023 at 11:56

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