Tensors create form metric

1. Jul 4, 2010

paweld

Is it true that the only combination of second order derivative of metric which
transforms tensorially is Riemann tensor (and its traces)?

2. Jul 4, 2010

Yes.

AB

3. Jul 4, 2010

paweld

Do you know any proof of it.

4. Jul 4, 2010

Altabeh

Actually this is based on the fact that until today there we have only Riemann tensor created from the combination of the first and second order derivatives of metric tensor and the metric tensor itself. But for an informal proof, see

GRAVITATION AND COSMOLOGY: PRINCIPLES AND APPLICATIONS OF THE GNERAL RELATIVITY by S. Weinberg. John Wiley & Sons, Inc., 1972, pp 133-34.

AB

5. Jul 4, 2010

atyy

How about the Einstein tensor?

6. Jul 4, 2010

paweld

It's lieanr combination of traces of Riemann tensor.

7. Jul 4, 2010

Of course!

AB

8. Jul 4, 2010

Thanks guys!

9. Jul 4, 2010

atyy

I followed Altabeh's suggestion to look at Weinberg, and there he states uniqueness with all the conditions previously mentioned in this thread, plus the requirement that it be linear in second derivatives.

There is an interesting comment in Berger's http://books.google.com/books?id=d_...&resnum=3&ved=0CCAQ6AEwAg#v=onepage&q&f=false "A important remark is in order: many people think that the curvature and its derivatives are the only Riemannian invariants. This is true and classical when looking for algebraic invariants which stem from the connection, see page 165 of Schouten 1954 [1109] and the references there. But things are dramatically different if one asks only for tensors which are invariant under isometries (called natural ). Then there is no hope to get any kind of classification, as explained in Epstein 1975 [491]. For more see Munoz & Valdes 1996 [952]."