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    $\begingroup$ I couldn't understand exactly your question, but any connection on a principal bundle can be written locally as $d + A$ where $A$ is a Lie algebra valued form. Furthermore you can extend the connection to higher forms by Leibniz rule in order to get a "complex" (not exactly a complex if your connection is not flat). $\endgroup$ Commented Oct 7, 2016 at 4:51
  • $\begingroup$ By reading better your question, it seems that the term $- \alpha ([X, Y])$ is missing in your $\omega_{\alpha}$. The general expression should be $d_{\nabla} \alpha (X_1, …, X_{k + 1}) = \sum_i (-1)^{i + 1} \alpha (X_1, …, \hat{X_i}, …, X_{k+ 1}) + \sum_{i > j}(-1)^{i + j} \alpha ([X_i, X_j], …, \hat{X_i}, …, \hat{X_j}, …, X_{k + 1})$ . This is actually the differential for the tangent Lie algebroid with values in a vector bundle (i.e, a representation of this Lie algebroid on a vector bundle). $\endgroup$ Commented Oct 7, 2016 at 6:27
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    $\begingroup$ No, there is nothing missing. By Leibniz we have $(\nabla_X\alpha)(Y) = \nabla_X(\alpha(Y)) - \alpha(\nabla_XY)$. Hence $$(\nabla_X\alpha)(Y) - (\nabla_Y\alpha)(X) = X(\alpha(Y)) - Y(\alpha(X)) - \alpha(\nabla_XY - \nabla_YX).$$ The last term is $\alpha$ evaluated on the commutator, provided that $\nabla$ is torsion free. This also answers the question what is going wrong if there is a torsion... $\endgroup$ Commented Oct 7, 2016 at 8:54
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    $\begingroup$ @dorebell What you are writing makes perfect sense. You may consult Besse, Einstein manifolds. Section 1.12 defines the exterior derivative of forms with values in a vector bundle by antisymmetrizing the covariant derivative. Section 1.19 provides a connection between $\nabla$ and $d$ in your context. It is a good exercise to check that the $\nabla$-definition of $d$ coincides with a more usual one. $\endgroup$ Commented Oct 7, 2016 at 9:06
  • $\begingroup$ @IvanIzmestiev Whoops! I misinterpreted the question. The OP actually wants to recover $d$ itself. $\endgroup$ Commented Oct 7, 2016 at 9:45