# Tensor Algebra in Wald Exercise

1. Sep 29, 2009

### Quiablo

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

Problem 2, chapter 3 of Wald's General Relativity. The details don't matter much, but it is given a totally anti-symetric tensor field Eab such that EabEab=2(-1)^(s), s being the signature of the metric. I have checked a solution to the exercise, and somewhere during the development there is the following reasoning:

2. Relevant equations

$$\nabla$$cEabEab=0 (because EabEab is contant);

This implies that:

2Eab$$\nabla$$cEab = 0 (applying Leibniz rule and noting that $$\nabla$$gab = 0);

And then, the reasoning goes on saying that, because Eab is totally anti-symetric and non-vanishing, we can conclude that:

$$\nabla$$cEab = 0

Which doesn't make sense to me. Can anyone explain to me the last line, please?

3. The attempt at a solution

Last edited: Sep 30, 2009
2. Sep 30, 2009

### javierR

Recall the simpler case of an nxn matrix, A, acting on an n-vector, v, such that A*v=0, If v is not the zero vector, then either (1) A=0; that is, all components vanish, or (2) there is a vector of *constraints* and the basis is linearly dependent. Well if we know we have a linearly independent basis, like that of our spacetime, then (1) has to be true.

Next, upgrade the matrix to rank-3 tensor ("nxnxn"), A, and the vector to a rank-2 tensor ("nxn") matrix, V, such that A*V=0. *Let's assume again that V is non-zero.* The equation again represents a vector-valued object. But now there is another possibility. If (1) A is symmetric but V is antisymmetric in the contracting indices, you can write out components and see that the equation is satisfied trivially. So no constraint equations are required. But if A and V are both antisymmetric or both symmetric in the contracted indices, you can again write out components to see that there is either (a) a non-trivial constraint equation among components or (b) A=0 (recall the assumption that V is not zero). We know that the spacetime basis is linearly independent, so (a) cannot be true once again. We are left with the requirement that A=0.

To compare with your problem, let the tensor $$A_{cab}=\nabla_{c} E_{ab}$$ and the $$V^{ab}=E^{ab}$$, where the matrices are anti-symmetric in the contracted (ab) indices.

Last edited: Oct 1, 2009
3. Oct 1, 2009

### Quiablo

Hey JavierR, thanks for answering. I am not sure if I understand what you say exactly, but I got kind of lost in your reasoning. First, in the second paragraph, I believe instead of meaning 3x3 matrix and 2x2 matrix, you probably mean rank 3 tensor and rank 2 tensor, respectively, because you cannot multiply tensors of diferent dimensions (3x3 matrixes cannot be multiplied by 2x2 matrices, not in the ordinary sense of multiplication). In oder words, you can only contract two indices if they have the same number of components (same dimension).

Secondly, you say that A*v=0, A a matrix, v a vector, means that either A = 0 or there is a set of constraints in the basis vectors. Well, as far as I can tell, A*v=0, for matrices A and vector v of any dimension, v not equal to zero, is equivalent to saying that A is singular, which means det(A) = 0. Thats all you can say about A. I am not sure if you can say that if the basis is linearly independent that means that A = 0. But I would be glad if you could cite a reference with that information, because I really wanted to check the reasoning myself and get really clear about that point.

Thanks a lot.

4. Oct 1, 2009

### javierR

Oops. Yeah, clearly I meant to say a rank-3 tensor and rank-2 tensor (of the same dimension indices). I changed it in the above. What I had in mind is that they are nxn and nxnxn objects, so that the situation is analogous to the familiar A*v case.

I'll try to go through it again (there's no specific reference I have in mind...just linear algebra and tensor algebra resources). I'm going to change the names of my objects from the last post because it’s more convenient. In the analogous case of A*V=0, where A is still a rank-2 tensor but V is now a rank-3 tensor, we are contracting over two indices. The following imply each other:
A is non-singular
<-> A is of matrix rank n <-> A has n linearly independent row vectors etc.
<--> its inverse exists (which we know is true since it appears in your equation, and more generally, since we are allowed to casually raise and lower indices with the metric)
<--> V=0 is the solution

In your case A is $$E^{ab}$$ and V is $$\nabla_{c}E_{ab}$$.

Regarding the point about linear independence and spacetime: A spacetime tensor lives on a local tangent space and can be written in the form
$$T=T^{ij\ldots }_{lm\ldots } e_{i}\otimes e_{j}\otimes\cdots e^{l}\otimes e^{m}$$
where the e’s are basis vectors and co-vectors (resp.) and the T’s in front are coefficients. In particular,
$$E=E^{ab}e_{a}\otimes e_{b}$$. The coefficients are just numbers, but the tensor structure is encoded in the *tensor products of the basis vectors*, which are those of spacetime itself.

But normal spacetime cannot be degenerate…if there are n dimensions, these *spacetime tensors* had better be “rank n” meaning nxnxnx….objects seen as collections of linearly row or column independent vectors. That's why we can raise and lower indices of spacetime tensors (i.e., invert) so casually using the metric tensor.