The point is that, with your second equation, you are dealing with in a local coordinate patch, say an open set $U\subset M$ equipped with coordinates $x^\mu \equiv x^1,x^2,x^3,x^4$. Therefore, if $p\in U$, you can handle two bases of the tangent space $T_pM$. One is made of (pseudo)orthonormal vectors $e_a$, $a=1,2,3,4$ and the other is the one associated with the coordinates $\frac{\partial }{\partial x^\mu}|_p$, $\mu= 1,2,3,4$. The metric at $p$ reads:
$$g_p = \eta_{ab} \omega^a \otimes \omega^b$$
where, by definition, the co-vectors $\omega^a \in T^*_pM$ (defining another pseudo-orthonormal tetrad but in the cotangent space at $p$) satisfy
$$\omega^a (e_b) = \delta^a_b\:\:.\qquad(0)$$

Correspondingly you have:
$$\omega^a = \omega^a_\mu dx^\mu|_p \:, \qquad (1)$$
and $\Omega :=[\omega^a_\mu]$ is a $4\times 4$ invertible matrix. Invertible because it is the transformation matrix between two bases of the same vector space ($T_p^*M$).
Similarly
$$e_a = e_a^\mu \frac{\partial}{\partial x^\mu}|_p\:,\qquad (2)$$
where
$E:= [e^a_\mu]$ is a $4\times 4$ invertible matrix. It is an elementary exercise to prove that, in view of (0): $$E= \Omega^{-1t}\:,\qquad (3)$$
so you can **equivalently** write
$$e_a^\mu = (\omega^{-1})_a^\mu$$
where the **transposition** operation in (3) is now apparent from the fact that we have swapped the positions of Greek and Latin indices (compare with (1)).

An identity as yours:
$$R^{\mu}_{\nu \lambda \tau} = (\omega^{-1})^{\mu}_a \, \omega^b_\nu \, \omega^c_\lambda \, \omega^d_\tau \, R^{a}_{bcd}$$
is understood as a trivial change of basis relying upon (1) and (2). It could equivalently be written down as:
$$R^{\mu}_{\nu \lambda \tau} = e^{\mu}_a \, \omega^b_\nu \, \omega^c_\lambda \, \omega^d_\tau \, R^{a}_{bcd}\:.$$

This post imported from StackExchange Physics at 2015-03-04 16:09 (UTC), posted by SE-user Valter Moretti