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?

• It looks to me more like abstract index notation than taking components. – Willie Wong Aug 11 '16 at 13:27
• @WillieWong could you please elaborate a little bit? – GregVoit Aug 11 '16 at 16:50
• See Sec 2.4 in Wald's book on GR. – Igor Khavkine Aug 11 '16 at 19:57

$$F_4 = F_4^l \frac{\partial}{\partial x^l}$$ $$\nabla_{F_4}F_3 = \nabla_{\left(F_4^l \frac{\partial}{\partial x^l}\right)}F_3 = F_4^l \nabla_{\frac{\partial}{\partial x^l}}F_3 = F_4^l \nabla_lF_3 = F_4^l \Big(\nabla_lF_3\Big)^k\frac{\partial}{\partial x^k}$$ $$F_4^l \Big(\nabla_lF_3\Big)^k\frac{\partial}{\partial x^k} = F_4^l \nabla_l F_3^{\,k}\frac{\partial}{\partial x^k} = \Big((\alpha - \bar{\beta}) F_3^k + \bar{\mu}F_1^k - \rho F_2^k\Big) \frac{\partial}{\partial x^k}$$
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.
• so are you saying that in this notation, $\nabla_{l}(v^{k})=\left(\nabla_{l}v\right)^{k}$? @Futurologist – GregVoit Aug 12 '16 at 5:04
• No, I think such index "commutativity" does not hold in general. I simply mean that the notation is, as you said it yourself, taking the $k$-th component of the covariant derivative $(\nabla_{F_4}F_3)^k = (F_4^l \nabla_l F_3)^k = F_4^l (\nabla_l F_3)^k$ is written as $F_4^l \nabla_l F_3^{\, k}$. Simply the parentheses are omitted. In general $$(\nabla_l v)^k \partial_k = \nabla_l ( v^k ) \partial_k + v^j \Gamma_{ l j }^k \partial_k$$ – Futurologist Aug 12 '16 at 5:40