# How to prove explicitly that by including Dirac fermions into the Einstein-Hilbert action we make torsion to be non-zero?

Recently I've heard the statement that by including Dirac fermions into the Einstein-Hilbert action we make torsion be non-zero, so that is one of problem of quantum gravity. How to prove that explicitly? Intuitively it's somehow connected with the form of Dirac action in curved spacetime (which includes vierbein), but I don't know how to demonstrate it directly.

Maybe it can be done by assuming Christoffel symbols and metric as independent quantities and then by variation of action by Christoffel symbol? In result I'll get some equation for Christoffel symbol, then I'll add to one equation the another one with indices permutation for getting equation on torsion tensor. If the tensor-free part of equation will Benin-zero, then torsion isn't zero. Is this thinking right?

It's convenient to rewrite EH action in terms of spin connection. From its definition $$\omega^{\mu}_{ab} = e^{\nu}_{a}\partial^{\mu}e_{\nu b} - \Gamma^{\mu}_{\lambda \sigma} e_{a}^{\lambda}e_{b}^{\sigma}$$ you can (by using orthogonality relation for tetrads) get $$\tag 1 \Gamma^{\mu}_{\kappa \delta} = e^{b}_{\delta}\partial^{\mu}e_{\kappa b} - e^{a}_{\kappa}e^{b}_{\delta}\omega^{\mu}_{ab}.$$ Eq. $(1)$ makes possible to rewrite geometrical quantities, such as curvature, torsion, in terms of tetrads and spin connection: $$R_{\mu \nu}^{ab} = e^{\kappa a}e_{\sigma}^{b}R^{\sigma}_{\ \kappa \mu \nu} = \partial_{\mu}\omega_{\nu}^{ab} - \partial_{\nu}\omega_{\mu}^{ab} + \omega_{\mu}^{ac}\omega^{\ b}_{c\nu } - \omega_{\nu}^{ac}\omega^{\ b}_{c\mu },$$ $$S^{a}_{\mu \nu} = \partial_{\mu}e^{a}_{\nu} - \partial_{\nu}e^{a}_{\mu} + \omega^{a}_{\ \mu b}e^{b}_{\nu} - \omega^{a}_{\ \nu b}e^{b}_{\mu}.$$ Then the EH action can be rewritten as $$\tag 2 S_{EH} = \int d^{4}x\sqrt{-g} e^{\mu}_{a}e^{\nu}_{b}R_{\mu \nu}^{ab} = \int d^{4}x e e^{\mu}_{a}e^{\nu}_{b}R_{\mu \nu}^{ab}, \quad e = det (e^{\mu}_{a}).$$ Let's add Dirac action to $(2)$: $$\tag 3 S = S_{EH} + S_{D} = \int d^{4}x e e^{\mu}_{a}e^{\nu}_{b}R_{\mu \nu}^{ab} + \kappa \int d^{4}x e \bar{\psi}\left(i\gamma^{\mu}e_{\mu}^{a}D_{a}(\omega ) - m \right)\psi .$$ Formally we may consider $\omega , e$ as independent quanties. Then variation of $(3)$ with respect to $e$ will give Einstein equation (leaving tetrad degree of freedom), while variation with respect to spin connection will give, after set of simplifications, something like $$S^{a}_{\mu \nu} = i\kappa \bar{\psi}\gamma^{a}\sigma_{\mu \nu}\psi .$$ So formally fermionic field will produce torsion.