Deriving a unique derivative operator with torsion such that $\triangledown g = 0$

Question

From Wald's General Relativity, Chapter 3 problem 1c), in the solution from http://www.physics.drexel.edu/~dcross/papers/wald.pdf:

$\triangledown g = 0$ becomes again

$$\triangledown_a g_{bc} = \bar{\triangledown}_ag_{bc} - C^d_{ab}g_{dc}- C^d_{ac}g_{bd} = \bar{\triangledown}_ag_{bc} - C_{cab}-C_{bac}$$

If we add the permutation (ab) and subtract the permutation (cab) as before we get

$$\bar{\triangledown}_a g_{bc} + \bar{\triangledown}_ag_{bc} - \bar{\triangledown}_ag_{bc} = T_{bac}-T_{abc}+2C_{c(ab)}$$

How have we gone form C's to T's?

Answer 1

Remember that the torsion in basically the antisymmetric part of $C$

$$ T_{abc} = C_{a[bc]} \tag{1} $$

So that

\begin{eqnarray} \color{blue}{C_{bac} - C_{bca}} &=& \color{blue}{C_{b[ac]} = T_{bac}} \\ \color{red}{C_{abc} - C_{acb}} &=& \color{red}{C_{a[bc]} = T_{abc}} \\ \color{magenta}{C_{cab} + C_{cba}} &=& \color{magenta}{2C_{c(ba)}} \tag{2} \end{eqnarray}

With this in mind ($\nabla g = 0$)

\begin{eqnarray} \tilde{\nabla}_ag_{bc} + \tilde{\nabla}_bg_{ac} - \tilde{\nabla}_cg_{ab} &=& [\nabla_a g_{bc} + C_{cab} + C_{bac}] + [\nabla_{b}g_{ac} + C_{cba} + C_{abc}] \\ && -[\nabla_cg_{ab} + C_{bca} + C_{abc}] \\ &=& (\color{magenta}{C_{cab} + C_{cba}}) + \color{blue}{C_{bac} - C_{bca}} + (\color{red}{C_{abc} - C_{acb}}) \\ &\stackrel{(2)}{=}& \color{magenta}{2C_{c(ba)}} + \color{blue}{C_{b[ac]}} + \color{red}{C_{a[bc]}} \\ &\stackrel{(1)}{=}& \color{magenta}{2C_{c(ab)}} + \color{blue}{T_{bac}} + \color{red}{T_{abc}} \end{eqnarray}