Covariant derivative definition in Wald

[Orodruin]

For a (pseudo-)Riemannian manifold we have some additional requirements. One is that the manifold is torsion free and the other that the gradient of the (pseudo-)metric vanishes. From these requirements you get the well-known unique definition of the Christoffel symbols in terms of partial derivatives of the (pseudo-)metric.

To clarify a bit here, these are not requirements that you have to impose specifically for a manifold with a metric - although we often do so. Any connection which is a connection without a metric tensor is still a connection when there is a metric tensor. You could easily imagine connections that fulfil none, one, or both of these conditions. An example is the connection on a sphere (without the poles) which preserves compass directions during parallel transport. This is a metric compatible, but not torsion-free, connection.

The point is that it is only when you require both that the connection is compatible with the metric and that it is torsion free that you obtain the unique Levi-Civita connection. You are also free to require that a connection is torsion free even if you do not have access to a metric, although it will not uniquely fix the connection.

 

[stevendaryl]

Indeed, it is very important to distinguish between tensors (including scalars and vectors as tensors of rank zero and one, respectively), which are independent of the choice of coordinates and their components with respect to a basis. The Christoffel symbols are NO tensor components

I just demonstrated exactly the opposite. Let me go through it again:

1. Let $V$ be any vector field.
2. Then $\nabla V$ is a $(1,1)$ tensor field.
3. A basis vector $e_\mu$ for a coordinate system is, in fact, a vector field.
4. Therefore, for any basis vector $e_\mu$, $\nabla e_\mu$ is a $(1,1)$ tensor field.
5. The components of $\nabla e_\mu$ in that basis are, in fact $(\nabla e_\mu)^\lambda_\nu = \Gamma^\lambda_{\mu \nu}$
6. So $\Gamma^\lambda_{\mu \nu}$ is in fact, the components of a tensor field.

What is perhaps confusing is that, in spite of appearances, the roles of $\mu$ and $\nu$ in $\Gamma^\lambda_{\mu \nu}$ are not analogous: $\lambda$ and $\nu$ are component indices of the tensor field $\nabla e_\mu$, while $\mu$ is not a component index, but tells us which tensor field we are talking about: $\nabla e_\mu \neq \nabla e_\nu$ if $\mu \neq \nu$

 

[Orodruin]

I just demonstrated exactly the opposite. Let me go through it again:

1. Let $V$ be any vector field.
2. Then $\nabla V$ is a $(1,1)$ tensor field.
3. A basis vector $e_\mu$ for a coordinate system is, in fact, a vector field.
4. Therefore, for any basis vector $e_\mu$, $\nabla e_\mu$ is a $(1,1)$ tensor field.
5. The components of $\nabla e_\mu$ in that basis are, in fact $(\nabla e_\mu)^\lambda_\nu = \Gamma^\lambda_{\mu \nu}$
6. So $\Gamma^\lambda_{\mu \nu}$ is in fact, the components of a tensor field.

What is perhaps confusing is that, in spite of appearances, the roles of $\mu$ and $\nu$ in $\Gamma^\lambda_{\mu \nu}$ are not analogous: $\lambda$ and $\nu$ are component indices of the tensor field $\nabla e_\mu$, while $\mu$ is not a component index, but tells us which tensor field we are talking about: $\nabla e_\mu \neq \nabla e_\nu$ if $\mu \neq \nu$

Just to underline the structure in a coordinate representation, the mixed tensor $\nabla e_\mu$ is given by
$$
\nabla e_\mu = \Gamma_{\mu\nu}^\sigma e_\sigma \otimes dx^\nu.
$$
It is here quite obvious that if you use a coordinate basis $e_\mu = \partial_\mu$, then $\partial_\mu$ is a very specific vector field and clearly $\nabla V$ depends on which vector field $V$ you use. Using the vector field $\partial'_\mu$ is clearly going to result in a different tensor. The important part is that the $\mu$ is labelling which vector field we are talking about and is not denoting a tensor component, which I think is the key point in both yours and vanhees' statements. The Christoffel symbols are not the components of a type (1,2) tensor, they are the components of $n$ different (1,1) tensors. Which $n$ tensors we are talking about depends on what vector basis we have chosen.

 

[stevendaryl]

It is here quite obvious that if you use a coordinate basis $e_\mu = \partial_\mu$, then $\partial_\mu$ is a very specific vector field and clearly $\nabla V$ depends on which vector field $V$ you use. Using the vector field $\partial'_\mu$ is clearly going to result in a different tensor. The important part is that the $\mu$ is labelling which vector field we are talking about and is not denoting a tensor component, which I think is the key point in both yours and vanhees' statements. The Christoffel symbols are not the components of a type (1,2) tensor, they are the components of $n$ different (1,1) tensors. Which $n$ tensors we are talking about depends on what vector basis we have chosen.

Exactly. The point of $\nabla$ versus $\partial$ is this: If $V$ is a vector field, then $\nabla V$ is a unique tensor field, which you can evaluate in any coordinate system you like. In contrast, $\partial V$ is not a unique tensor field; you have to know which basis was used in order to know which tensor field you mean. It's still a tensor field (but probably not the one you meant).