Why is covariant derivative of Null-tetrads not zero? I think the null tetrads are just re-writing of the vierbiens. However, the vierbiens satisfy $\nabla_\mu e^{a}_{~~\nu} = 0$, but the same is not true for the null tetrads because the spin coefficients are defined by taking covariant derivatives of these null tetrads. Can anyone explain what is wrong with my thinking?
 A: It is deeply misleading to say that  $\nabla_\mu {e^a}_\nu=0$.
What is true is that $\nabla_\mu {\bf e}^a = - {\omega^a}_{b\mu}{\bf e}^b$ defines the affine connection coefficients in the veilbein basis fof the cotangent bundle. If we write expand the coframe ${\bf e}^a$ in terms of the coordinate dual basis as
$$
{\bf e}^a= {e^a}_\nu {dx}^\nu
$$
and use the coordinate basis definition of the Christoffel symbols
$$
\nabla_\mu (dx^\nu)= - {\Gamma^\nu}_{\lambda\mu} dx^\lambda
$$
together with  Leibnitz rule
$$
\nabla_\mu({e^a}_\nu dx^\nu)= (\nabla_\mu  {e^a}_\nu)dx^\nu + {e^a}_\nu \nabla_\mu (dx^\nu),
$$
while remembering that the ${e^a}_\nu$ are just functions so that
$$
\nabla_\mu {e^a}_\nu\equiv \partial_\mu {e^a}_\nu,
$$
you   get
$$
(\partial_\mu {e^a}_\nu+  {\omega^a}_{b\mu}{e^b}_\nu - {e^a}_\lambda {\Gamma^\lambda}_{\nu\mu}) dx^\nu =0.
$$
The expression in parentheses  does  look rather look  like something you might call "$\nabla_\mu {e^a}_\nu$" if the ${e^a}_\nu$ were the components of some sort of  a tensor.  The  array of numbers  ${e^a}_\nu$, considered as two-index objects, are not the component of a normal tensor, however.  They  are the  change-of-basis matrices from the ${\bf e}^a$ basis of the cotangent bundle to the $dx^\mu$ basis of the same bundle. The formula
$$
\partial_\mu {e^a}_\nu+  {\omega^a}_{b\mu}{e^b}_\nu -  {e^a}_\lambda {\Gamma^\lambda}_{\nu\mu}=0
$$
which is sometimes (I don't know who started this missnaming) called the "tetrad postulate" is not a postulate. A postulate is something that you are free to accept or reject. This formula cannot be rejected because it is always true. It  is simply the transformation rule that tells us  how the connection components ${\omega^a}_{b\mu}$ written in the vielbein  basis are related to the components ${\Gamma^\lambda}_{\nu\mu}$ of the same connection written in the coordinate basis.
I must confess,  however, that I remember how to get the signs in the formula by pretending in my mind that it is a covariant derivative of ${e^a}_\mu$, but I know that it is not such a thing.
It is true that if we define
$$
{\bf I}=  {\bf e}_a   \otimes ({e^a}_\mu dx^\mu) \equiv   {\bf e}_a \otimes {\bf e}^a 
$$
then Leibnitz rule gives
$$
\nabla_\mu {\bf I}= (\partial_\mu {e^a}_\nu+  {\omega^a}_{b\mu}{e^b}_\nu -  {e^a}_\lambda {\Gamma^\lambda}_{\nu\mu}) {\bf e}_a\otimes dx^\nu
$$
but ${\bf I}$ is easly seen to be the identity operator ${\bf I}:TM\to TM$, and is not the same as ${\bf e}^a$.
Some details and further rants about this can be found in my lecture notes here.
