Covariant derivative of the spin connection I wish to compute $$[\nabla_{\mu}, \nabla_{\nu}]e^{\lambda}_{~~a}. $$
To do so, I make use of $\nabla_{\nu}e^{\lambda}_{~~a} = \omega_{a~~~\nu}^{~~b}e^{\lambda}_{~~b}$, so that I may write
$$\nabla_{\mu}\nabla_{\nu}e^{\lambda}_{~~~a} = \nabla_{\mu}(\omega_{a~~~\nu}^{~~~b})e^{\lambda}_{~~b}+\omega_{a~~\nu}^{~~b}~\omega_{b~~~\mu}^{~~c}e^{\lambda}_{~~c}$$In the end, I intend to anti-symmterize in $\mu, \nu$ to get the desired object. Therefore,  I would like to know what is the covariant derivative of the spin-connection $\omega$ in order to finish my computation. Is $\omega$ a scalar, a vector or what? How do you decide? Can someone help?
 A: It's imporant o keep track of what is a vector, and and what are just numbers. The components of vectors, tensors etc are numbers, and the covariant derivative of a number-valued function is just the ordinary derivative.  In particular the array of numbers ${\omega^a}_{b\mu}(x)$ are just number-valued functions, so
$$
\nabla_\nu {\omega^a}_{b\mu} =\partial_\nu {\omega^a}_{b\mu}.
$$
Let's use the definition $\nabla_\nu{{\bf e}_a} = {\bf e}_b {\omega^b}_{a\nu} $ together with Liebnitz' rule to work out
$$
\nabla_\mu \nabla_\nu {{\bf e}_a}  = \nabla_\mu ({\bf e}_b {\omega^b}_{a\nu})\\
= (\nabla_\mu {\bf e}_b) {\omega^b}_{a\nu}+ {\bf e}_b(\nabla_\mu {\omega^b}_{a\nu})\\ 
={\bf e}_c {\omega^c}_{b\mu} {\omega^b}_{a\nu}+ {\bf e}_b\partial_\mu {\omega^b}_{a\nu} \\
={\bf e}_c ({\omega^c}_{b\mu} {\omega^b}_{a\nu}+ \partial_\mu {\omega^c}_{a\nu}).
$$
So
$$
(\nabla_\mu \nabla_\nu {{\bf e}_a})^c = {\omega^c}_{b\mu} {\omega^b}_{a\nu}+ \partial_\mu {\omega^c}_{a\nu}
$$
is the $c$-th compoent of $\nabla_\mu \nabla_\nu {{\bf e}_a}$.
Thus
$$
[\nabla_\mu ,\nabla_\nu]{\bf e}_a =  {\bf e}_c(\partial_\mu {\omega^c}_{a\nu}-\partial_\nu {\omega^c}_{a\mu}+ {\omega^c}_{b\mu} {\omega^b}_{a\nu}-{\omega^c}_{b\nu} {\omega^b}_{a\mu}).
$$
or
$$
([\nabla_\mu ,\nabla_\nu]{\bf e}_a)^c =  \partial_\mu {\omega^c}_{a\nu}-\partial_\nu {\omega^c}_{a\mu}+ {\omega^c}_{b\mu} {\omega^b}_{a\nu}-{\omega^c}_{b\nu} {\omega^b}_{a\mu}.
$$
We can also write the components in the coordinate frame as ${\bf e}_a = {e_a}^\lambda {\boldsymbol \partial}_\lambda$ and then
$$
([\nabla_\mu ,\nabla_\nu]{\bf e}_a)^\lambda =  (\partial_\mu {\omega^c}_{a\nu}-\partial_\nu {\omega^c}_{a\mu}+ {\omega^c}_{b\mu} {\omega^b}_{a\nu}-{\omega^c}_{b\nu} {\omega^b}_{a\mu}){e_c}^\lambda
$$
It's a bad, but common, habit  to  write things like $\nabla_\mu X^\nu= \partial_\mu X^\nu+ X^\alpha {\Gamma^\nu}_{\alpha\mu}$ when you mean
$$
(\nabla_\mu {\bf X})^\nu= \partial_\mu X^\nu+ X^\alpha {\Gamma^\nu}_{\alpha\mu}, \quad {\bf X}= X^\nu {\boldsymbol \partial}_\nu .
$$
A: Note that from the relation $e^\lambda{}_{a;\nu} = \omega_a{}^b{}_\nu e^\lambda{}_b$ you give you can deduce by contracting with $e_{\lambda c}$
$$e^\lambda{}_{a;\nu}e_{\lambda c} = \omega_{ac\nu}$$
Note, however, that I am using the definition of the covariant derivative that takes tetrad indices $a,b,c$ as mere labels and thus the covariant derivative of the tetrad leg vector field is
$$e^\lambda{}_{a;\nu} = e^\lambda{}_{a,\nu} + \Gamma^\lambda{}_{\kappa\nu} e^\kappa{}_{a}\,.$$
That is, the covariant derivative is covariant with respect to coordinate transforms but not wrt vielbein transforms.
In other words, $\omega_{ab\nu}$ transforms as a tensor in the $\nu$ index and as such it will have the corresponding Christoffel connection term in the covariant derivative. This should help you derive the commutator (relation between the Riemann tensor and $\omega$) as desired.
