# How do fermions explicitly interact with curvature via the tetrad?

I am aware of the basics of the tetrad formalism and am clear on why bosonic fields do not have couplings to curvature via their covariant derivatives in a curved space Lagrangian i.e. why $$\nabla_\mu\phi\nabla^\mu\phi = \partial_\mu\phi\partial^\mu\phi$$ and $$\nabla_\mu A_\nu - \nabla_\nu A_\mu = \partial_\mu A_\nu - \partial_\nu A_\nu$$. What is not clear however, is how a free fermion is affected by curved space.

For a fermion kinetic term that looks like $$|e|\frac{i}{2}\bar{\psi}\gamma^a e^\mu_a\overset{\leftrightarrow}{\nabla}_\mu\psi \quad \quad \nabla_\mu\psi = (\partial_\mu + \Gamma_\mu)\psi \quad ,$$ I have read that we can expand the tetrad around flat space and write $$e^\mu_a(x) \approx \delta^\mu_a + k^\mu_a(x) \quad \Rightarrow \quad \Gamma_\mu = \frac{1}{2}e^\beta_l(\partial_\mu e_{\beta h})\Sigma^{hl} \approx \partial_\mu(\frac{1}{2}k_{\beta h}\Sigma^{h\beta}) \quad$$ [Yepez, arXiv: 1106.2037].

I have also read that by going to the Lorentz gauge, "all vierbein fluctuations can be written in terms of the metric fluctuations" [O. Zanusso et al., Physics Letters B 689 (2010) 90–94]. What do these statements mean for perturbative QFT calculations? Explicitly, what kind of interaction terms appear in the Lagrangian because of the tetrads presence here?

To phrase my question another way, what kind of tetrad/graviton-fermion Feynman diagrams can I draw based on how the tetrad appears with the gamma matrix and as part of the spin connection?