Finding the transformation for the Newman-Penrose formalism I am working with the Kerr metric (suppose it is $g_{\mu \nu}$) where I want to understand how the Teukolsky master equation is derived using the Newman-Penrose formalism. The original article can be found here. Given the Kerr metric, he considers a tetrad of four vectors given by $ l^{\mu}, n^{\mu}, m^{\mu} \text{ and } \bar{m^\mu}$ where the bar represents complex conjugation. Consider the explicit form of $l^\mu$ in the coordinate basis (as I understand it).
$$ l^{\mu} = [(r^2 + a^2)/\Delta, 1, 0, a/\Delta] $$
where $\Delta = r^2 - 2Mr + a^2$. These are supposed to form null vectors and hence when I transform them to the tetrad basis, I should have a null tetrad set. Taking the conditions on l,n,m one can show that the metric to lower tetrad indices can be given by:
$$\eta _{ab} = \left[\matrix{0 &1 &0 &0 \\
                       1 &0 &0 & 0 \\
                       0 &0 & 0&-1 \\
                       0&0 &-1 &0} \right]$$
For the rest of the question I take greek indices to mean coordinate indices and roman indices to mean tetrad indices. I want to find out the components of the vector l in the tetrad basis. I guess this can be given as $$ l^a = E_\mu ^{ \text{  }a}\text{ }l^{\mu} $$ where $E_{\mu}^{\text{ } a}$ represents the transformation from the coordinate to the tetrad basis.

*

*How do I find out this $E_\mu ^{ \text{  }a}$?
I know using the knowledge of the metric, I can write $$\eta _{ab} = E_{a}^{\text{ }\mu} E_{b}^{\text{ }\nu} g_{\mu \nu} $$
and since I know both $\eta_{ab}$ and $g_{\mu \nu}$ I can in principle, find the transformation matrix (at least modulo coordinate transformations). However, I am unable to find out this matrix.


*I want to  explicitly show that $l_a l^a = 0$  (tetrad basis inner product - lowering is done using $\eta$)

This is one of the conditions for choosing the Newman Penrose tetrad vectors. I have also tried to compute $l_\mu l^{\mu}$ (coordinate basis inner product - lowering done using $g$) but I am not getting this to be zero. Shouldn't this also be zero, since the norm of the vector must be invariant?
EDIT: Had a typo in one of the equations.
 A: I think you are slightly confused about the overall picture. Let's review it again:
All is about the construction of suitable frames. By "frame", I mean a set of four basis vectors at each point of spacetime. Given a coordinate system over the spacetime, we already have one: the "coordinate basis" given by partial derivatives $\partial_\mu$ along each coordinate (e.g. $\{\partial_t,\partial_r,\partial_\theta,\partial_\phi \}$). However, the problem with this frame is that it is not orthonormal in a curved spacetime, since $\partial_\mu\cdot \partial_\nu=g_{\mu\nu}$. For various reasons, we prefer to work with orthonormal frames. Therefore we transform from the coordinate basis $\partial_\mu$ to another set of four basis vectors $e_{(a)}=e_{(a)}{}^\mu\partial_\mu$ for $a=\{0,1,2,3\}$ such that
\begin{align}
e_{(a)}\cdot e_{(b)}=\eta_{(a)(b)}
\end{align}
I put parantheses around "$a,b$" indices to stress that they are "label indices" to be distinguished from coordinate indices $\mu$.
For a Rimannian geometry it is natural to take $\eta_{(a)(b)}=\text{diag}(1,1,1,1)$, which implies that the four vectors are orthonormal, i.e. of unit norm and orthogonal to each other. For a Lorentzian spacetime, we instead typically consider $\eta_{(a)(b)}=\text{diag}(-1,1,1,1)$ so that $e_{(0)}$ is timelike while the other three are spacelike.
However, in some situations (including black hole perturbation theory), we prefer to work with null basis vectors, i.e. $|e_{(0)}|^2=e_{(0)}\cdot e_{(0)}=0$ and similarly for the other three basis vectors. To form a complete basis, then one typically takes $e_{(0)}\cdot e_{(1)}=-1, e_{(2)}\cdot e_{(3)}=1$ and the rest of the inner products to be zero. It is also conventional to give these basis vectors separate names: $e_{(0)}^\mu=l^\mu,e_{(1)}^\mu=n^\mu,e_{(2)}^\mu=m^\mu,e_{(3)}^\mu=\overline{m}^{\mu}$, where overline means complex conjugation. Therefore the matrix $\eta_{(a)(b)}$ looks like what you wrote above [except that you have replaced (0,1) with (2,3)]. In Boyer-Lindquist coordinates for Kerr spacetime, a choice for the basis vectors is
\begin{aligned}
l^{\mu} &=\frac{1}{\Delta}\left(r^{2}+a^{2}, \Delta, 0, a\right), \\
q^{\mu} &=\frac{1}{2 \rho^{2}}\left(r^{2}+a^{2},-\Delta, 0, a\right), \\
m^{\mu} &=\frac{1}{\sqrt{2}} \frac{1}{r+i a \cos \theta}\left(i a \sin \theta, 0,1, \frac{i}{\sin \theta}\right),
\end{aligned}
and $e_{(a)}{}^\mu$ is a 4$\times$4 matrix whose columns are the basis vectors, i.e. $e_{(a)}{}^\mu=\Big(l^\mu\,|\,n^\mu\,|\,m^{\mu}\,|\,\overline{m}^{\mu}\Big)$.
The inverse of $e_{(a)}{}^\mu$ is instead a set of four one-forms $e^{(a)}=e^{(a)}{}_\mu dx^\mu$ whose components are given by
$$e^{(a)}{}_\mu=\eta^{(a)(b)}g_{\mu\nu}e_{(b)}{}^\nu$$
From this we see that as a column matrix  $e^{(a)}{}_\mu=\Big(-n_\mu\,|\,-l_\mu\,|\,\overline{m}_{\mu}\,|\,{m}_{\mu}\Big)$
Now it is very clear that $e^{(a)}{}_\mu l^\mu=(-n_\mu l^\mu, -l_\mu l^\mu,\overline{m}_\mu l^\mu,m_\mu l^\mu,)=(1,0,0,0)$. So as you see $l^{(a)}=e^{(a)}{}_\mu l^\mu$ the vector $l$ is the zeroth basis vector and hence contains no particularly useful information. Also obviously $l^{(a)}l_{(a)}=\eta_{(a)(b)}l^{(a)}l^{(b)}=\eta_{(0)(0)}=0$. this is another way to say that $l^\mu$ is by construction a null vector.
I hope this helps.
