# Problem with a basic theorem in Wald's GR book

## Main Question or Discussion Point

1. The problem statement, all variables and given/known
I don't understand the proof of the following theorem:

Theorem 3.1.1
Let $g_{ab}$ be a metric. Then there exists a unique derivative operator $\nabla_a$ satisfying $\nabla_a\,g_{bc}=0$

2. Homework Equations

After some manipulations it is found that:
$$C^c_{ab}=\frac{1}{2}g^{cd}\{\overline{\nabla }_a\,g_{bd}+\overline{\nabla }_b\,g_{ad}-\overline{\nabla }_d\,g_{ab} \}$$
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
$$\overline{\nabla }_a\,g_{bd}+\overline{\nabla }_b\,g_{ad}-\overline{\nabla }_d\,g_{ab}\ldots (1)$$
is not "manifestly unique".
In fact, replacing $\overline{\nabla}_a$ by another $\nabla_a$ using
$$\overline{\nabla}_a\,g_{bc}=\nabla_a g_{bc}-C'^d_{ab}\,g_{dc}-C'^d_{ac}\,g_{bd}$$
it is found that eq. (1) is not invariant with respect the derivative operator, i.e.
$$\overline{\nabla }_a\,g_{bd}+\overline{\nabla }_b\,g_{ad}-\overline{\nabla }_d\,g_{ab} \neq \nabla_a\,g_{bd}+\nabla_b\,g_{ad}-\nabla_d\,g_{ab}$$

Related Special and General Relativity News on Phys.org
Orodruin
Staff Emeritus
Homework Helper
Gold Member
While $\bar \nabla$ is arbitrary, it should be considered a fixed derivative operator throughout. You fixed it in the beginning. Then you found out what the C coefficients were given that operator. Had you had a different operator from the beginning, you clearly would have had different C coefficients because it would have a different difference to the Levi-Civita connection.

Also, the Levi-Civita connection is not a unique metric compatible connection. It is a unique metric compatible and torsion free connection.

PeterDonis
Mentor
2019 Award
Moderator's note: moved to relativity forum as this isn't really a homework problem.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Put another way, the equation for C is clearly not invariant under change of connection but it should not be. It is the expression for the Levi-Civita connection that should be invariant.

While $\bar \nabla$ is arbitrary, it should be considered a fixed derivative operator throughout. You fixed it in the beginning. Then you found out what the C coefficients were given that operator. Had you had a different operator from the beginning, you clearly would have had different C coefficients because it would have a different difference to the Levi-Civita connection.
That's right, now I see it. A different deriative operator should give a different C. However I don't see how this solves the uniqueness problem. Let's say we have two different derivative operators $\nabla$ and $\overline{\nabla}$ ,then according to the theorem and using $C$ and $\overline{C}$ we can go to operators $\nabla'$ and $\overline{\nabla}'$ such that $\nabla'_c\,g_{ab}=\overline{\nabla}'_c\,g_{ab}=0$, but this does not mean $\nabla'=\overline{\nabla}'$

Also, the Levi-Civita connection is not a unique metric compatible connection. It is a unique metric compatible and torsion free connection.
In the context of the theorem $\nabla$ is supposed to be torsion free.
As I see it the uniquenes issue is essential for the usefullness of the theorem because after the theorem is proved Wald uses the ordinary derivative operator $\partial_c$ to obtain the ussual expression for the Levi-Civita conection, however since the operator $\partial_c$ is not chart independ we do not know if this procedure yields a chart independent derivative operator

Orodruin
Staff Emeritus
Homework Helper
Gold Member
That's right, now I see it. A different deriative operator should give a different C. However I don't see how this solves the uniqueness problem. Let's say we have two different derivative operators ∇∇\nabla and ¯¯¯¯¯∇∇¯\overline{\nabla} ,then according to the theorem and using CCC and ¯¯¯¯CC¯\overline{C} we can go to operators ∇′∇′\nabla' and ¯¯¯¯¯∇′∇¯′\overline{\nabla}' such that ∇′cgab=¯¯¯¯¯∇′cgab=0∇c′gab=∇¯c′gab=0\nabla'_c\,g_{ab}=\overline{\nabla}'_c\,g_{ab}=0, but this does not mean ∇′=¯¯¯¯¯∇′∇′=∇¯′\nabla'=\overline{\nabla}'
Sorry, but this is not correct. Given a connection there is only one set of Cs that will change it to a metric compatible one and therefore there is only one metric compatible connection. It does not matter that you get different Cs starting from different connections. You will always recover the same unique connection. You can easily check this by considering two different connections by comparing the resulting metric compatible connection. You will find that the Cs change in exactly the correct way to cancel out in the resulting metric compatible connection.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
In the context of the theorem ∇∇\nabla is supposed to be torsion free.
As I see it the uniquenes issue is essential for the usefullness of the theorem because after the theorem is proved Wald uses the ordinary derivative operator ∂c∂c\partial_c to obtain the ussual expression for the Levi-Civita conection, however since the operator ∂c∂c\partial_c is not chart independ we do not know if this procedure yields a chart independent derivative operator
You can just as well derive the connection coefficients directly from the metric compatibility and see that it transforms in exactly the correct way under coordinate transformations.

You can just as well derive the connection coefficients directly from the metric compatibility and see that it transforms in exactly the correct way under coordinate transformations.
Yes, Know that. The problem here is that I'm trying to understand it in the context of Wald's approach. For intance he does not consider the transformation properties. His appraoch is coordinate independend.
Anyway now I understand why in his approach it is unique and your observations help me to achieve that. Thank you.

Orodruin
Staff Emeritus