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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?

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  • $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Feb 22 '17 at 16:33
  • $\begingroup$ 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. $\endgroup$ Jan 3 '20 at 20:37
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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 ${\bf e}^{*a}= e^{*a}_\mu dx^\mu$.

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As far as I known the anti-symmetric part of (2) equals to (1).

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