- #1

- 394

- 15

## Main Question or Discussion Point

**1. The problem statement, all variables and given/known**

I don't understand the proof of the following theorem:

**Let ##g_{ab}## be a metric. Then there exists a unique derivative operator ##\nabla_a## satisfying ##\nabla_a\,g_{bc}=0##**

Theorem 3.1.1

Theorem 3.1.1

2. Homework Equations

2. Homework Equations

After some manipulations it is found that:

[tex]C^c_{ab}=\frac{1}{2}g^{cd}\{\overline{\nabla }_a\,g_{bd}+\overline{\nabla }_b\,g_{ad}-\overline{\nabla }_d\,g_{ab} \} [/tex]

At this point Wald asserts: " This choice of ##C^c_{ab}## solves equation ##\nabla_a\,g_{bc}=0## and it is manifestly unique, which completes the proof.""

## The Attempt at a Solution

The problem is that ##\overline{\nabla}_a## is an arbitrary derivative operator and the expresion

[tex]\overline{\nabla }_a\,g_{bd}+\overline{\nabla }_b\,g_{ad}-\overline{\nabla }_d\,g_{ab}\ldots (1)[/tex]

is not "manifestly unique".

In fact, replacing ##\overline{\nabla}_a## by another ##\nabla_a## using

[tex]\overline{\nabla}_a\,g_{bc}=\nabla_a g_{bc}-C'^d_{ab}\,g_{dc}-C'^d_{ac}\,g_{bd}[/tex]

it is found that eq. (1) is not invariant with respect the derivative operator, i.e.

[tex]\overline{\nabla }_a\,g_{bd}+\overline{\nabla }_b\,g_{ad}-\overline{\nabla }_d\,g_{ab} \neq

\nabla_a\,g_{bd}+\nabla_b\,g_{ad}-\nabla_d\,g_{ab}[/tex]