In a set of notes, I came across the following few lines involving the covariant derivative, and just wanted to make sure I understood the notation correctly:


Let $\lbrace F\_{1},F\_{2},F\_{3},F\_{4}\rbrace$ be a null tetrad. Then from Newman Penrose formalism we have an equation


$\nabla\_{F\_{4}}F\_{3}=(\alpha-\overline{\beta})F\_{3}+\overline{\mu}F\_{1}-\rho F\_{2}$


Then we have


$F\_{4}\hskip .5pt ^{l}\nabla\_{l}F\_{3}\hskip .5pt^{k}=(\alpha-\overline{\beta})F\_{3}\hskip .5pt ^{k}+\overline{\mu}F\_{1}\hskip .5pt ^{k}-\rho F\_{2}\hskip .5pt ^{k}$


So, do I understand correctly that in the second equation, we simply take the components of the original Newman Penrose equation?



Clarification on tetrad indices
$$F\_4 = F\_4^l \frac{\partial}{\partial x^l}$$
$$\nabla\_{F\_4}F\_3 = \nabla\_{\left(F\_4^l \frac{\partial}{\partial x^l}\right)}F\_3 = F\_4^l \nabla\_{\frac{\partial}{\partial x^l}}F\_3 = F\_4^l \nabla\_lF\_3 = F\_4^l \Big(\nabla\_lF\_3\Big)^k\frac{\partial}{\partial x^k}$$
$$ F\_4^l \Big(\nabla\_lF\_3\Big)^k\frac{\partial}{\partial x^k} = F\_4^l \nabla\_l F\_3^{\,k}\frac{\partial}{\partial x^k} = \Big((\alpha - \bar{\beta}) F\_3^k + \bar{\mu}F\_1^k - \rho F\_2^k\Big) \frac{\partial}{\partial x^k}$$


Basically, they have omitted the parenthesis that specify that they have taken the $k$-th component of the covariant derivative of the vector field $F\_3$ in the $l$-th coordinate direction. 


