For field, theory what i know $i.e$,complex scalar QED \begin{align} D_\mu \phi = \partial_{\mu} \phi - i Q A_{\mu} \phi \end{align} and \begin{align} D_\mu \phi^{\dagger} = \partial_{\mu} \phi^{\dagger} + i Q A_{\mu} \phi^{\dagger} \end{align}
In Geometry, by adapting vielbein formulation for spinor $\psi$, \begin{align} D_\nu \psi = \partial_\nu \psi + \frac{1}{4} w_{\nu ab} \gamma^{[a} \gamma^{b]} \psi, \quad D_\nu \bar{\psi} = \partial_\nu \bar{\psi} - \frac{1}{4} w_{\nu ab} \bar{\psi} \gamma^{[a}\gamma^{b]} \end{align}
Consider charged spinor, then the covariant derivatives
\begin{align} D_a \psi= e^{\mu}{}_a (\partial_\mu - i eQ A_\mu + \frac{1}{2} \sigma^{bc} e_b{}^\nu \partial_\mu e_{c\nu})\psi, \quad \sigma^{ab} = \frac{\gamma^a \gamma^b - \gamma^b \gamma^a}{4} \end{align} Then how about $D_a \bar{\psi}$?
My guess is \begin{align} D_a \bar{\psi}= e^{\mu}{}_a (\partial_\mu\bar{\psi} + i eQ A_\mu \bar{\psi} - \frac{1}{2} \bar{\psi}\sigma^{bc} e_b{}^\nu \partial_\mu e_{c\nu}), \quad \sigma^{ab} = \frac{\gamma^a \gamma^b - \gamma^b \gamma^a}{4} \end{align} Is this right?
