General Relativity - Riemann Tensor and Killing Vector Identity

[Tangent87]

Homework Statement

I am trying to show that for a vector field V_a which satisfies $$V_{a;b}+V_{b;a}$$ that $$V_{a;b;c}=V_eR^e_{cba}$$ using just the below identities.

Homework Equations

$$V_{a;b;c}-V_{a;c;b}=V_eR^e_{abc}$$(0)

$$R^e_{abc}+R^e_{bca}+R^e_{cab}=0$$ (*)

$$V_{a;b}+V_{b;a}=0$$ (**)

The Attempt at a Solution

So far I have:

$$V_eR^e_{cba}=-V_eR^e_{acb}-V_eR^e_{bac}=V_{a;b;c}-V_{a;c;b}+V_{b;c;a}-V_{b;a;c}=2V_{a;b;c}+V_{b;c;a}-V_{a;c;b}$$

Which gives:

$$V_eR^e_{cba}-V_{a;b;c}=V_eR^e_{abc}-V_{c;b;a}$$

I want to say that this implies $$V_{a;b;c}=V_eR^e_{cba}$$ but I can't justify why both sides of the above equation must be zero. Can anyone see what else I can do? Thanks

 

[Tangent87]

The problem is part of a larger question (see page 59 here http://www.maths.cam.ac.uk/undergrad/pastpapers/2004/Part_2/list_II.pdf ) and I'm having trouble with the last bit as well where it goes into the EM stuff, I know it must obviously somehow relate to everything we've done above but I just don't see the relevance.

 

[Tangent87]

Can anybody help me out with deriving the identity $$V_{a;b;c}=V_eR^e_{cba}?$$

Forget about the EM stuff I don't care so much about that but I'd be very grateful for some help in deriving that identity. Thanks

 

[fzero]

There's a couple of ways to show that. One way is to start from (*) and use (0) to write 6 terms involving the 2nd derivatives of the V's. Then you can use (**) to show that

$$V_{a;bc} + V_{b;ca} + V_{c;ab} =0.$$

You can then use (**) and (0) to show that

$$V_{b;ca} + V_{c;ab} = V_e {R^e}_{cab}.$$

 

[Tangent87]

There's a couple of ways to show that. One way is to start from (*) and use (0) to write 6 terms involving the 2nd derivatives of the V's. Then you can use (**) to show that

$$V_{a;bc} + V_{b;ca} + V_{c;ab} =0.$$

You can then use (**) and (0) to show that

$$V_{b;ca} + V_{c;ab} = V_e {R^e}_{cab}.$$

Ahhh thanks, done it now. It seems I was definitely going about it the wrong way.