# Covariant derivatives of the frame field, are they the same?

There is a exterior covariant derivative $$D \alpha_a = d \alpha_a -\omega^b{}_a \wedge \alpha_b\;, \tag 1$$ and another $D$ in the vierbein postulate $$D_\mu e_{I\alpha}=\partial_\mu e_{I\alpha}-\omega_\mu{}^J{}_Ie_{J\alpha}-\Gamma_\mu{}^{\lambda}{}_\alpha e_{I\lambda}=0\;.\tag 2$$ The question is: Do they the same thing?

• Would Mathematics be a better home for this question? Feb 22 '17 at 16:33
• No. The clue is in the term "exterior differential" - which is a skew differential. Regarding the orthogonal moving frame, $e_{I\alpha}$ - which I presume is the frame basis vector and not a $2$-form - which typically should be written as $e_{a}(x)=e^{i}_{a}(x)\frac{\partial}{\partial x_{i}}$ where $a$ is the frame index and $i$ is the tangent space index. Jan 3 '20 at 20:37

You first statement looks like the definition of the torsion form $$D{\bf e}^{*a}\equiv d{\bf e}^{*a}+ {\omega^a}_b \wedge {\bf e}^{*b}= T^a.$$ The second is a confused aspect of the ordinary covariant derivative and has nothing to do torsion or metricity.
The $$e_a^\mu$$ and so on are just an array of numbers. Their covariant derivative is just the ordinary partial derivative. The expression you have is therefore not the covaraint derivative of $$e^\mu_a$$. It is the geometric basis vectors $${\bf e}_a$$ or $${\boldsymbol \partial}_\mu$$ for the tangent space, or $${\bf e}^{*a}$$, or $$dx^\mu$$ for the cotangent space that need the connection forms $${\omega^b}_{a\mu}dx^\mu$$ or $${\Gamma^\lambda}_{\nu\mu}dx^\mu$$.
In particular the "vierbein postulate" is not a postulate. It is an identy that always holds. It is in fact an appallingly bad and confusing name for the formula for converting the covariant derivative $$\nabla_X\equiv X^\mu \nabla_\mu$$ from its definition in terms of a frame-field $$\nabla_X {\bf e}_a= {\bf e}_b{\omega^b}_{a\mu}X^\mu$$ to its definition in terms of the coordinate basis
$$\nabla_X {\boldsymbol \partial }_\nu= {\boldsymbol \partial }_\lambda {\Gamma^\lambda}_{\nu\mu} X^\mu.$$ In each case $$\nabla_X$$ is the same object. So by expanding
$${\bf e}_a = e^\nu_a {\boldsymbol \partial}_\nu$$ and using the derivation property (Leibnitz rule) of $$\nabla_X$$, we evaluate $$\nabla_X {\bf e}_a = e_b^\nu {\omega^b}_{a\mu}X^\mu{\boldsymbol \partial}_\nu$$ in the equivalent form $$\nabla_X {\bf e}_a = X^\mu (\partial_ \mu e^\nu_a){\boldsymbol \partial}_\nu + X^\mu e^\nu_a (\nabla_\mu {\boldsymbol \partial}_\nu)\\ =X^\mu (\partial_\mu e_a^\nu+ e^\lambda_a {\Gamma^\nu}_{\lambda \mu}){\boldsymbol \partial}_\nu.$$ By comparing coefficients of $$X^\mu {\boldsymbol \partial}_\nu$$, we read off that
$$\partial_\mu e_a^\nu+ e^\lambda_a {\Gamma^\nu}_{\lambda \mu}- e_b^\lambda {\omega^b}_{a\mu}=0.$$ Your equation (2) is just an example of this manipulation for the co-frames $${\bf e}^{*a}= e^{*a}_\mu dx^\mu$$.