Notation of General Relativity

[Mike2]

I'm reading, General Relativity, by Robert M. Wald. On page 39 he has the following notation:


\displaystyle{
{\rm R}_{{\rm [abc]}}^{\,\,\,\,\,\,\,\,\,\,\,\,{\rm d}} = 0
}



and


\displaystyle{
\nabla _{{\rm [a}} {\rm R}_{{\rm bc]d}}^{\,\,\,\,\,\,\,\,\,\,{\rm e}} = 0
}


What do the square brakets [ ] mean? How can the subscript of the \nabla be included in these brakets? I've not seen this notation in the other books I have, so I could use a little help.

Thanks.

[chroot]

The brackets denote that tensor is symmetric in the indices inside the brackets.

The reason why the index of the \nabla can be included in the brackets is simple: the entire construction \nabla_a R_{bcd}{}^{e} is itself another tensor. It has five indices, under four of which it is symmetric.

You could think of it as a new tensor T:

T_{abcd}{}^{e} = \nabla_a R_{bcd}{}^{e}

where all of the lower indices of T are symmetric:

T_{[abcd]}{}^{e} = 0

- Warren

[Mike2]

Originally posted by chroot
The brackets denote that tensor is symmetric in the indices inside the brackets.

The reason why the index of the \nabla can be included in the brackets is simple: the entire construction \nabla_a R_{bcd}{}^{e} is itself another tensor. It has five indices, under four of which it is symmetric.

You could think of it as a new tensor T:

T_{abcd}{}^{e} = \nabla_a R_{bcd}{}^{e}

where all of the lower indices of T are symmetric:

T_{[abcd]}{}^{e} = 0

- Warren

So this bracket notation is the same as the sum of all permutations with a coefficient of +1 for even permutations and -1 for odd permutations?

The subscript on the \nabla confuses me because on page 31, Wald writes, "It is often notationally convenient to attach an index directly to the derivative operator and write it as \nablaa, although this is to some extent an abuse of the index notation since \nablaa is not a dual vector."

So what is he doing using it like this?

[jeff]

Originally posted by Mike2
What do the square brakets [ ] mean?

This notation is defined on p26.

Originally posted by chroot
The brackets denote that tensor is symmetric in the indices inside the brackets.

It wasn't stated explicitly by either of you so, just in case, I'll point out that it's the vanishing of the antisymmetric part that means it's completely symmetric in those indices. In other words

R_{[abc]}{}^d=0 \Leftrightarrow
R_{abc}{}^d=R_{(abc)}{}^d.

Originally posted by chroot
The reason why the index of the \nabla can be included in the brackets is simple: the entire construction \nabla_a R_{bcd}{}^{e} is itself another tensor.

It doesn't matter that it's a tensor.

Originally posted by Mike2
...since \nablaa is not a dual vector."

So what is he doing using it like this?

To make operations on the components of \nabla explicit.

[Mike2]

Originally posted by jeff

To make operations on the components of \nabla explicit.

Does the subscript on \nabla stand for differentiation with respect to the a'th coordinate curve of some as yet unspecified system?

Is that system always orthogonal?

Thanks.

[jeff]

Originally posted by Mike2
Does the subscript on \nabla stand for differentiation with respect to the a'th coordinate curve of some as yet unspecified system?

If you mean covariant differentiation with respect to the ath coordinate of some as yet unspecified coordinate system, then yes.

Originally posted by Mike2
Is that system always orthogonal?

Again, by "system" I'll assume you mean coordinate system. A coordinate system S is orthogonal with respect to some inner product ( , ) if S's basis vectors are mutually orthogonal with respect to ( , ). In GR, ( , ) is the spacetime metric, and S needn't be chosen orthogonal with respect to it. On the other hand, given any point p on a manifold, one may always choose a coordinate system which is orthogonal at p.