# Quick Ricci tensor/scalar contraction manipulation question

1. Apr 17, 2015

### binbagsss

So my textbook definitions of the ricci tensor and ricci scalar are:
$R_{ab}=R_{acbd}g^{cd}$ - I note the contraction is over the 2nd and 4th index. and the 1st and 3rd.
$R=R_{ab}g^{ab}$

Now, I'm trying to show that $g^{ad}g^{ce}(\bigtriangledown_{a}R_{bcde}+\bigtriangledown_{b}R_{cade}+\bigtriangledown_{c}R_{abde})=2\bigtriangledown^{a}R_{ab}-\bigtriangledown_{b}R$

It's obvious that the $R$ term must come from $g^{ad}g^{ce}\bigtriangledown_{b}R_{cade}$ . But, I'm not seeing this properly, if the ricci vector is defined for the contraction to only be over specific indicies -or can it be any 2? As here the summation is over indicies consecutive to each other, but in the definition given above its over the 1st and 3rd, and 2nd and 4th.

Also, I'm totally clueless as to where a minus sign is coming from?

Thanks very much !!

Last edited: Apr 17, 2015
2. Apr 17, 2015

### Orodruin

Staff Emeritus
This does not make sense, you have three ds in your right hand side. Even if you replace the $g^{ad}$ by $g^{ac}$, the right hand side does not have any free indices at all and the left hand side has a and b as free.

3. Apr 17, 2015

### binbagsss

Apologies. edited.

4. Apr 17, 2015

### Orodruin

Staff Emeritus
Do you perhaps know some symmetries that the Riemann tensor has?

5. Apr 17, 2015

### Orodruin

Staff Emeritus
It still cannot be true. The left hand side has two free indices and the right hand side none.

6. Apr 17, 2015

### binbagsss

Edited, apologies again ! ta.

7. Apr 17, 2015

### Orodruin

Staff Emeritus
So what symmetries of the Riemann tensor are you familiar with?

8. Apr 17, 2015

### binbagsss

Ah yes. antisymmetric in the last/first two indices. the only thing i can see to explain a minus sign. thank you !

9. Apr 17, 2015

### binbagsss

Apologies, another q, is the ricci vector defined to be over the 2nd and 4th only?

And then by use of $R_{abcd}=R_{badc}$, which is from swapping both the first 2 indices and the last 2, double negative, this is equivalent to contracting over the 1st and 3rd, or by definition is it either the 2nd and 4th or 1st and 3rd.

Thanks.

10. Apr 17, 2015

### Orodruin

Staff Emeritus
As you noticed, the two choices are equivalent due to the anti-symmetries of the tensor. It does not matter which one you select.