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I have an issue with Eq. 3.3.6 of Wald's General Relativity. There he would like to prove that for Gaussian normal coordinates, the geodesic tangent field remains orthogonal to all coordinate basis fields which generate hypersurfaces indexed by parameter $t$ (page 43).

The step is as follows: Let $n^a$ be tangent vector to the geodesic and $X^a$ be any one of the coordinate basis fields. Then \begin{equation} \begin{aligned} n^b\nabla_b(n_aX^a)&=n_an^b\nabla_bX^a\\&=n_aX^b\nabla_bn^a\\&=\frac{1}{2}X^b\nabla_b(n^an_a)\\&=0 \end{aligned} \end{equation} First equality is due to product rule and also the definition of geodesics (parallel transport its own tangent vector). Third equality follows by reversing product rule. Fourth equality follows because $n_an^a$ is a constant (norm of $n^a$). But I do not quite get second equality. Wald mentioned that it follows because as coordinate basis for the tangent space at $p$, $[n^a,X^b]=0$. But I cannot see how that enters the second equality.

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The commutator of two vector fields $n^a$ and $X^b$ is $[n,X]^a = n^b \nabla_b X^a - X^b \nabla_b n^a$. Since this vanishes, it follows that $n^b \nabla_b X^a = X^b \nabla_b n^a$, and the second step follows from there.

I don't have my copy of Wald in front of me, but I'm 99% sure that the commutator of two vector fields is defined in terms of the covariant derivative (or vice versa) somewhere in Chapter 3.

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  • $\begingroup$ It's on page 31 of Wald's book, equations (3.1.1)-(3.1.2) $\endgroup$
    – user_35
    Commented May 15, 2016 at 21:26
  • $\begingroup$ Thanks to both of you, I completely forgot about the expression for commutator. It's actually easy (: @user35736 $\endgroup$ Commented May 15, 2016 at 21:33

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