Constraint on vierbein vectors Is it reasonable to choose the vierbein frame $e_{a}^{\mu}$, with the following constraint being imposed: $e_{a\mu}^{\quad;\mu} = 0$? If yes, how one can find such vierbein vectors for the Kerr metric.
 A: I think the answer by AGML is a bit misleading. If $X=X^\mu\partial_\mu$ is a tangent-vector field, the covariant derivative of the vierbein frame field with respect to $X$ defines the connection one-form  ${\omega^b}_{a\mu} dx^\mu$  through 
$$
\nabla_X{\bf e}_a = {\bf e}_b {\omega^b}_{a\mu} X^\mu.
$$
When we choose the  frame field  to coincide with  the coordinate basis vectors $\partial_\mu$ 
 we usually write
$$
\nabla_X \partial_\lambda = {\Gamma^\nu}_{\lambda\mu} X^\mu \partial_\nu.
$$
where  ${\Gamma^\nu}_{\lambda\mu}$ is the traditional notation for the connection coefficients in a coordinate basis. 
If we write the vierbein in terms of its coordinate components as ${\bf e}_a= e_a^\mu \partial_\mu$ then we can  use  the fact that $\nabla_X$ obeys Leibnitz rule  to compute $\nabla_X {\bf e}_a$ by two  different routes. 
We must, however, bear in mind that on functions $f$ we have $\nabla_Xf = X^\mu\partial_\mu f$ and that  real-number components $e_a^\mu$  are functions not vectors or tensors. The two routes give
$$
{\bf e}_b \,{\omega^{b}}_{a\mu}X^\mu= \nabla_X {\bf e}_a = (\nabla_X e_a^\nu)\partial_\nu + e^\lambda_a (\nabla_X \partial_\lambda)
$$
$$
=X^\mu( \partial_\mu e^\nu_a + e^\lambda_a {\Gamma^\nu}_{\lambda\mu}){ \partial}_\nu.
$$
Writing
$$
{\bf e}_b \,{\omega^{b}}_{a\mu}X^\mu = {\omega^{b}}_{a\mu}X^\mu e^\nu_b\partial_\nu
$$
in the equality 
$$ 
 {\bf e}_b \,{\omega^{b}}_{a\mu}X^\mu= X^\mu( \partial_\mu e^\nu_a + e^\lambda_a {\Gamma^\nu}_{\lambda\mu}){ \partial}_\nu
$$
and comparing coefficients of the the basis vector $\partial_\nu$ leads to 
$$
\partial_\mu  e_{a}^\nu- e_{b}^{\nu}{\omega^b}_{a\mu} + {\Gamma^\nu}_{\lambda\mu}e_{a}^\lambda=0
$$
 which is the so-called  "tetrad postulate'' . This is very bad name. I have no idea who invented this term. A postulate is something that we are free to accept or reject, but this "postuate"   is simply the   relation expressing  the general vierbein connection coefficients 
 ${\omega^a}_{b\mu}$ in terms of  coordinate-frame  connection coefficients ${\Gamma^\lambda}_{\nu\mu}$ (or vice versa) and must always hold. In particular, and despite its appearence, it is not the statement that the covariant derivative $\nabla_X {\bf e}_a$  of the vierbein is zero.
A: The covariant derivatives of the vielbeins are usually assumed to vanish by the appropriately named "vielbein postulate", since otherwise the modifications to the covariant derivative necessary to accomodate the Lorentz indices make it metric-incompatible. Given that your "constraint" provides no new information.
If you just want an admissible set of vielbeins for a given metric, the most striaghtforward approach is to write the metric out as a matrix and then diagonalize it. If you write out the definitions of the vielbeins as matrix, rather than indexed, equations you will see this guarantees they satisfy their assumed properties.
