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I'm studying qft on curved space and i'm a bit confused as the teacher made the example of covariant electromagnetism when obtained the covariant derivative of a spin vector (with spin connection). So i have some questions:

  1. Is spin connection a complete different object than connection due to some symmetry of the theory (EDIT: other than the group of diffeomorphism, i mean). So could we have both spin connection and gauge connection in the covariant derivative?
  2. As the connection of (i.e.) $U(1)$ gauge symmetry gives rise to electromagnetism, what kind of interaction gives rise the coupling with spin connection? And is the generator of the orthochronous Lorentz group an observable as it is the electric charge?
  3. Why for bosonic fields the coupling is with the metric and we don't need tetrads?
polology
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1 Answers1

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  1. No, the spin connection is not different from the usual gauge connection. They are connections.

You can (and MUST, for spinors) indeed have both the spin connection and the gauge connection inside your covariant derivative. For example, for Dirac spinors with EM, you have:

$$D_\mu = \partial_\mu + 1/2 (\omega_{\alpha \beta})_\mu \sigma^{\alpha \beta} -ieA_\mu$$

  1. The interactions that give rise to spin connections are gravitational interactions. In the language of gauge theory, in curved space at each point you are free to choose a locally inertial frame, you can interpret this as a sort of gauge theory for the Lorentz group.

It is more complicated than this and more mathematical involved than it is physically relevant, tell me in the comments if I need to edit this to make it more precise.

  1. The reasons why for bosonic fields is sufficient to use the metric and not the tetrad is mainly for mathematical reasons.

I will work with Euclidean signature just for convenience, the same is true for the full Lorentz group. The Lorentz group, here $SO(4)$, is used to build what is called an Orthonormal Frame Bundle. The fields are built using Associated Bundles. The covariant derivative lives in such a bundle. The connection always acts on representations of your gauge group. For bosonic fields you can use representations of $SO(4)$, so the connection is of vector-type, the usual Levi-Civita connection. For fermionic fields you can't use $SO(4)$ representations but you need to go to representations of its double cover. This gives you what is called a spin connection.

To build such a structure you will need also a Spin Structure.

It's important to note that you can think of the spin connection as the fundamental connection, remembering that on bosonic fields the representations of the $SO(4)$ group are equivalent to the ones of its double cover. In other words, you can, if you want to, express the Levi-Civita connection with tetrad instead of using the metric, because those gives you equivalent structures in vector-type representations

LolloBoldo
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