# 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?

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 $${\bf e}^a$$.

• In your lecture note, I noticed that you write $X^{\mu}\nabla_{\mu}\partial_{\nu} = X^{\mu}\Gamma^{\lambda}_{\mu\nu}\partial_{\lambda}$. Why is there not $X^{\mu}\partial_{\mu}\partial_{\nu}$ there? Commented Apr 13, 2021 at 5:35
• the $\partial_\mu$ are just coordinate-fram basis vectors and could have been written as ${\bf e}_\mu$. They are not acting on anything. I tried writing ${\boldsymbol \partial}_\mu$ and ${\bf dx}^\mu$ make the distinction clear, but it did not look nice. You have, therefore, to go by context to figure out. Commented Apr 13, 2021 at 11:39