# Ricci Tensor and Trace

1. Aug 23, 2006

### waht

Just wondering if Traces can be applied to tensors.

If the Ricci tensor is Rii then is sums over diagonal elements.

So technically, can you say the trace of the Riemann tensor is the Ricci tensor?

Last edited: Aug 23, 2006
2. Aug 23, 2006

### dmoravec

well what you are referring to is Tensor Contraction. A tensor contraction sums much like the trace of a matrix and gets rid of two indicies. Think of a matrix being a rank 2 tensor being contracted to a rank 0 (scalar) tensor. That is what a trace is, a specific form of a tensor contraction (if the matrix is made of tensor elements).
Here is the mathworld site:
http://mathworld.wolfram.com/TensorContraction.html
So in that case the Ricci tensor is contracted from the Riemann Tensor.

3. Aug 23, 2006

### waht

Thanks alot,

For a trace, the matrix has to be n x n, right. So, can all tensors with a rank greater or equal to 2 be written as n x n matrices?

Sorry i'm asking a silly question, I didn't work with tensors in long time.

Last edited: Aug 23, 2006
4. Aug 23, 2006

### dmoravec

not exactly. Any tensor with an even rank can be contracted down to a nxn matrix (effectivly a rank 2 tensor), but odd rank tensors will be contracted to other odd rank tensors and ultimately a vector (n x 1 matrix).
Picture a rank 3 tensor as a n x n x n 'cubic matrix'. A contraction would basically take the trace of each level of this 'cubic matrix' and make a vector out of it.
I don't think there is a way to get rid of a single rank in the way of contraction. But all even rank tensors could be contracted (multiple times) down to a n x n matrix.

5. Aug 23, 2006

### waht

Thanks, that makes alot more sense now.

6. Sep 9, 2006

### gvk

The contraction can not be done with covariant or contravariant tensors , but only with mixed tensors. So, the Ricci tensor $$R_{ij}=R^k_{k,ij}$$ is the contraction of Riemann's tensor $$R^l_{k,ij}$$ (you can not contract $$R_{lk,ij}$$, one index should be rised).
Further, if you want to contract the Ricci tensor $$R_{ij}$$, you need first to rise the index 'i' and then contract $$g^{ki}R_{ij}$$ by k=j. The result will be the scalar curvature R.

Last edited: Sep 9, 2006
7. Jan 10, 2010

### Omega137

Weyl tensor $$C_{abcd}$$ is traceless in the sense $$g^{ab} g^{cd} C_{abcd} = 0$$? Am I right?

8. Jan 11, 2010

### Omega137

Sorry; I've made a mistake in the indexing...

I should have said

$$C_{abcd}$$ is traceless in the sense $$g^{ac} C_{abcd} = 0$$ or eqvivalently $$C^a_{bad} = 0$$ for arbitrary $$b , d$$

Sorry for the mistake...