Derivation of the torsion in the Einstein–Cartan formulation The EC action is given by:
$ S_{EC}= \int \epsilon_{abcd}e^a \wedge e^b \wedge R^{cd}$ where $R^{cd}=d\omega^{cd}+\omega^c_{ e}\wedge\omega^{ed}$
The Dirac action for the fermionic field in the EC formulation is given by:
$ S_D=i \kappa \int \epsilon_{abcd}e^a \wedge e^b \wedge e^c \wedge \bar \psi (\gamma^d \overleftrightarrow d + \{\gamma^d, \omega \} + \frac{im}{2}e^d)\psi  $
where $\omega=\omega_{ab}[\gamma^a,\gamma^b]/8$.
The $\delta(S_{EC}+S_D)/\delta\omega$ eguation should give the torsion $T_a \equiv \nabla e_a = \frac{3\kappa}{4} i\epsilon_{abcd}e^b\wedge e^c \bar \psi \gamma^5 \gamma^d \psi $, but I have trouble deriving it. Where does $ \gamma^5$ come from?
 A: So, I used the identity (which can be proved by expanding this anticommutator in terms of basis $\Gamma$)
$\{\gamma_\mu , \sigma_{\nu\rho} \}=-2\epsilon_{\alpha\mu\nu\rho}\gamma^5\gamma^{\alpha} $
$\frac{\delta S}{\delta \omega_{mn}}=-\epsilon_{abcd} de^a \wedge e^b + \epsilon_{abcd} e^a \wedge de^b+ \epsilon_{abcp} e^a \wedge e^b \wedge {\omega_d}^p - \epsilon_{abpd} e^a \wedge e^b \wedge {\omega_c}^p $
$\frac{\delta S_D}{\delta \omega_{mn}}=-\frac{\kappa}{4} \epsilon_{mnzd} \epsilon^{dabc}e_a \wedge e_b \wedge e_c \bar \psi (\gamma^5 \gamma^z) \psi$
Multiplying $\frac{\delta S+ S_D}{\delta \omega_{mn}}$with $\epsilon^{cdmn}$
$4(\nabla e^m)\wedge e^n-4(\nabla e^n)\wedge e^m+\frac{\kappa_1}{2}2
\begin{vmatrix}
\delta^m_z & \delta^m_d \\
\delta^n_z & \delta^n_d
\end{vmatrix}
\epsilon ^{dabc}e_a \wedge e_b \wedge e_c \bar \psi (\gamma^3 \gamma^z)\psi$
$4 \nabla e^{[m} \wedge e^{n]}+\kappa _1 \epsilon ^{abc[n}e_a \wedge e_b \wedge e_c \bar \psi (\gamma^5 \gamma^{m]}) \psi =0$
$4 \nabla _\mu e^m \nu dx^ \mu \wedge dx^\nu e^n_\rho dx^\rho - 4\nabla _\mu e^n_\nu dx^ \mu \wedge dx^\nu e^m_\rho dx^\rho + \kappa_1 {\epsilon_{abc}}^m e^a_\mu dx^\mu \wedge e^b_\nu dx^\nu \wedge e^c_\rho dx^\rho \bar \psi (\gamma^5 \gamma^n)\psi - \kappa_1 {\epsilon_{abc}}^n e^a_\mu dx^\mu \wedge e^b_\nu dx^\nu \wedge e^c_\rho dx^\rho \bar \psi (\gamma^5 \gamma^m)\psi =0$
Multiplying with $dx^\sigma e^\alpha_m e^\beta_n$
$\epsilon^{\mu \nu \rho \sigma}(4 \nabla _\mu e^m_\nu e^\alpha _m \delta^\beta _\rho - 4\nabla_\mu e^n_\nu e^\beta _n \delta^\alpha_\rho + \kappa_1 {\epsilon_{abc}}^m e^a_\mu e^b_\nu e^c_\rho e^\alpha_m e^\beta_n \bar \psi (\gamma^5 \gamma^n)\psi - \kappa_1 {\epsilon_{abc}}^n e^a_\mu e^b_\nu e^c_\rho e^\beta_n e^\alpha_m  \bar \psi (\gamma^5 \gamma^n)\psi) =0$
$\epsilon^{\mu \nu \rho \sigma}(4 \nabla _\mu e^m_\nu e^\alpha _m \delta^\beta _\rho - 4\nabla_\mu e^n_\nu e^\beta _n \delta^\alpha_\rho + \kappa_1 {\epsilon_{\mu \nu \rho}}^\alpha e^\beta_n \bar \psi (\gamma^5 \gamma^n)\psi - \kappa_1 {\epsilon_{\mu \nu \rho}}^\beta e^\alpha_m \bar \psi (\gamma^5 \gamma^n)\psi) =0$
$\epsilon^{\mu \nu \rho \sigma}(4 \nabla _\mu e^m_\nu e^\alpha _m \delta^\beta _\rho - 4\nabla_\mu e^n_\nu e^\beta _n \delta^\alpha_\rho) + \kappa_1(-6g^{\beta \sigma} e^\alpha_m \bar \psi (\gamma^5 \gamma^n)\psi +6g^{\beta \sigma}\bar \psi (\gamma^5 \gamma^n)\psi) =0$
Multiplying everthing with $\epsilon_{\gamma \delta \beta \sigma}$ and we get
$4(\nabla e^\alpha_{ \delta \gamma} -\nabla e^\nu _{\gamma \nu} \delta ^\alpha _\sigma + \nabla e^\nu _{\delta \nu} \delta ^\alpha _\gamma) - 6\kappa_1 g^{\alpha \tau}\epsilon_{\beta \tau \gamma \delta} \bar \psi (\gamma^5 \gamma^\beta)\psi + 6\kappa_1 g^{\beta \tau}\epsilon_{\beta \tau \gamma \delta} \bar \psi (\gamma^5 \gamma^\alpha)\psi =0$
Where $(\nabla e)^\alpha_{\mu \nu} = \nabla_\mu e^m_\nu e^\alpha_m$
For the contraction $\alpha = \delta$ we have $4(\nabla e^\alpha_{\alpha \gamma} -\nabla e^\nu _{\gamma \nu} 4 + \nabla e^\nu _{\gamma \nu}) = 0$
That means that $\nabla e^\nu_{\gamma \nu} =0$
Inserting this result into the previous equation, we get $4\nabla e^\alpha_{\delta \gamma} -6\kappa_1 g^{\alpha \tau}\epsilon_{\beta \tau \gamma \delta} \bar \psi (\gamma^5 \gamma^\beta)\psi =0 $
Using $\gamma^\beta=e^\beta_a \gamma_a$ , $\epsilon_{\beta \mu \gamma \delta} e_a^\beta =\epsilon_{abcd}e^b_\mu e^c_\gamma e^d_\delta$ and $\nabla e^\alpha_{\delta \gamma}=e^\alpha_b \nabla e^b_{\delta \gamma} $ we obtain wanted expression.
