# I Metric tensor

1. Nov 25, 2016

### kent davidge

I wonder if it is possible to write the components of the metric tensor (or any other tensor) as a summ of functions of the coordinates? Like this:

$g^{\mu\nu} = \sum_{\mu}^{D}\sum_{\nu}^{D} g_{_1}(x^{\mu}) g_{_2}(x^{\nu})$

where g1 and g2 are functions of one variable alone and D is the dimension of the Manifold. I hope you understand my poor English. Thanks in advance.

If no, then what would be a way of writing the components of a tensor? I dont like just gμν.... It would be better if there were a deeper way of representing that.

Last edited: Nov 25, 2016
2. Nov 25, 2016

### andrewkirk

It is always possible, at least in GR, to find, for any given point P on the manifold, a coordinate system in which the metric is diagonal at P. That metric will trivially satisfy your requirement [EDIT: on reflection, I think that's wrong. A diagonal metric does not satisfy the condition], but only in that coordinate system, and only at that point.

Conversely, for any coordinate system in which your condition holds at a point P, there will be other coordinate systems and/or other points for which it does not hold - in fact probably most of them.

For your condition to be true for all points in all coordinate systems, it would have to be the case that the metric tensor is everywhere the tensor product of two vectors, that is
$$\mathbf g=\vec u\otimes\vec v=(u^\alpha\vec e_\alpha)\otimes(v^\beta\vec e_\beta)=u^\alpha v^\beta(\vec e_\alpha\otimes\vec e_\beta)=:u^\alpha v^\beta\mathbf e_{\alpha\beta}$$
The set of tensors that are tensor products of two vectors forms a generating set for the space of order 2 tensors. But they are only a small part of that space, which is formed from all linear combinations of tensor products. It might help to read this page explaining tensor products and how tensor spaces are constructed from them.

Metric tensors are a somewhat restricted subset of the set of all order 2 tensors, because they are required to be symmetric. But the class of symmetric tensors is much larger than the class of tensors that are tensor products of vectors.

Last edited: Nov 26, 2016
3. Nov 25, 2016

### kent davidge

Thanks andrewkirk for your detailed answer.

Why must the metric be diagonal at P to my condition be satisfied?

4. Nov 25, 2016

### Staff: Mentor

It doesn't have to. He's saying that if it is diagonal it will satisfy your condition, not that that's the only way of satisfying it.

5. Nov 26, 2016

### kent davidge

Thank you.

6. Nov 26, 2016

### kent davidge

Can you show me with equations why a diagonal metric will satisfy that condition?

7. Nov 26, 2016

### andrewkirk

Actually, on second thoughts, I don't think a diagonal metric does satisfy that condition. The rest of the post is correct though.

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