Cartan equations versus Einstein equations in classical gravity Are Cartan structural equations equivalent to Einstein's equations
 $$G_{\mu\nu}=T_{\mu\nu}$$
and why (in the case of torsionless geometries, of course)? Does it also apply with a non-null cosmological constant?
 A: No.  The structure equations give expressions for torsion and curvature in terms of the spin connection.  In contrast, Einstein's equations tell you how the curvature responds to energy-momentum and vice versa.
Addendum: 4 April 2013.
The structure equations can be written in a non-coordinate basis (vielbein) $e^a = e_\mu^a dx^\mu$ as (see Carroll's Spacetime and Geometry (J.28) and (J.29))
\begin{align}
  T^a &= de^a + \omega^a_{\phantom a b}\wedge e^b \tag{1}\\
  R^a_{\phantom ab} &= d\omega^a_{\phantom a b}+\omega^a_{\phantom ac}\wedge \omega^c_{\phantom cb}
\end{align}
where $\omega^a_{\phantom ab}$ is the spin connection, $T^a$ is the torsion, and $R^a_{\phantom ab}$ is basically the Riemann tensor in the form of a bunch of two-forms.  
In practice, if the torsion vanishes, then the first structure equation often allows one to solve for the spin connection.  Once this is done, the second structure equation can be used to determine the Riemann curvature.  As you can see from the first structure equation with zero torsion, the vielbein and spin connection are related, but not the same.  
So basically, if the metric is known, then the structure equations can be used to determine the curvature.  This in turn is related to the energy-momentum tensor via Einstein's equations.
A: The covariant exterior derivative of Cartan structure equations imply the Bainchi identity: By starting with
\begin{align}
  T^a &= de^a + \omega^a_{\phantom a b}\wedge e^b \tag{1}\\
  R^a_{\phantom ab} &= d\omega^a_{\phantom a b}+\omega^a_{\phantom ac}\wedge \omega^c_{\phantom cb}
\end{align}
Then operate them with the covariant exterior derivative provide us the Bianchi identities
\begin{align}
  D_\omega T^a &= R^a{}_b \wedge e^b\\
   D_\omega R^a_{\phantom ab} &= 0
\end{align}
The last equation is the second Bianchi identity. The usual local coordinate form can be derived from it as shown below (roughly done)
\begin{eqnarray}
 D_\Gamma R^{\mu\nu} &=& 0 \;,\\
 d R^{\mu\nu} + \Gamma^\mu{}_\lambda \wedge R^{\lambda  \nu} + \Gamma^\nu{}_\gamma  \wedge R^{\mu \gamma } &=&0\;,\\
 \partial_{[\beta} R^{\mu\nu}{}_{{\rho\sigma}]} +\Gamma_{[\beta|}{}^\mu{}_\lambda R^{\lambda  \nu}{}_{|{\rho\sigma}]}+ \Gamma_{[\beta|}{}^\nu{}_\lambda  R^{\mu \lambda  }{}_{|{\rho\sigma}} -\Gamma_{[\beta}{}^\gamma {}_{\rho |} R^{\mu\nu}{}_{\gamma  | \sigma]} - \Gamma_{[\beta}{}^\gamma {}_\rho R^{\mu\nu}{}_{\rho ] \gamma} &=&0\;,\\
 \therefore \nabla_{[\beta} R^{\mu\nu}{}_{\rho\sigma]} &=&0\;.
\end{eqnarray} The contracted second Bianchi identity is equivalents to the divergence-free condition of Einstein tensor
$$\nabla_\mu \Big( R_{\mu\nu} -\frac 1 2 R g_{\mu\nu} \Big)=0$$
not exactly the vacuum EFE.
