Differential forms manipulation in Cartan's formalism this may be a silly question but I'm not seeing any easy answer.
Right now I'm dealing with the condition of zero-torsion of the spin connection in General Relativity:
$$
0=de^a +\omega{^a_b}\wedge e^b 
$$
and I wonder if there is a way to explicit $\omega^a_b$, without exploiting the tensor formalism, but only the differential forms formalism.
 A: If the connection is metric compatible, then $\omega_{ab}=-\omega_{ba}$. Thus, 
$$
e_b\rfloor d\vartheta_a = \omega_{cab}\vartheta^c+\omega_{ab}\\
e_a\rfloor d\vartheta_b = \omega_{cba}\vartheta^c+\omega_{ba}\\
e_a\rfloor e_b\rfloor d\vartheta_c\wedge\vartheta^c=(\omega_{cba}-\omega_{cab})\vartheta^c
$$
where $\rfloor$ denotes the interior product. Then, you can easily see that
$$
\omega_{ab}=\tfrac{1}{2}(e_b\rfloor d\vartheta_a-e_a\rfloor d\vartheta_b+e_a\rfloor e_b\rfloor d\vartheta_c\wedge\vartheta^c)
$$
Notation
$\vartheta^a$ are the form components of the coframe $\vartheta=\vartheta^a\otimes e_a$. $e_a$ are the components of the frame $e$, such that $e_a\rfloor \vartheta^b=\delta^b_a$ (duality). $\omega^a{}_{bc}$ are the components of $\omega^a{}_b$. 
A: Assuming $<e^a,e_{b}>=\sum_{i}e^{a}_{i}e^{i}_{b}=\delta^{a}_{b}$, then I presume you may be looking for this form $$\omega^{ab}_{i}=\frac{1}{2}[e^{ak}(\partial_{i} e^{b}_{k}-\partial_{k} e^{b}_{i})-e^{bk}(\partial_{i} e^{a}_{k}-\partial_{k} e^{a}_{i})-e^{aj}e^{bl}(\partial_{j} e_{cl}-\partial_{i} e_{cj})e^{c}_{i}]$$
where $a,b,$ and $c$ are the frame indices, and $i,j$ and $k$ are the tangent indices. 
And that you're not looking for this
$$\omega^{a}_{\;b}\wedge e^b=e^{a}_{jb}e^{b}_{i}dx^{j}dx^{i}$$
