Raising and lowering Ricci Tensor

[unscientific]

Taken from Hobson's book:

How is this done? Starting from:

$$R_{abcd} = -R_{bacd}$$

Apply $g^{aa}$ followed by $g^{ab}$

$$g^{aa}g^{aa} R_{abcd} = -g^{ab}g^{aa}R_{bacd}$$
$$g^{ab}R^a_{bcd} = -g^{ab}g^{aa}R_{bacd}$$
$$R^{aa}_{cd} = - g^{ab}g^{aa} R_{bacd}$$

Applying $g_{aa}$ to both sides:
$$g_{aa}R^{aa}_{cd} = - g^{ab}g^{aa} g_{aa} R_{bacd}$$
$$R^a_{acd} = -g^{ab} R_{bacd}$$
$$R^a_{acd} = - R^a_{acd} = 0$$

Is there a quicker way?

 

[samalkhaiat]

Taken from Hobson's book:

How is this done? Starting from:

$$R_{abcd} = -R_{bacd}$$

Apply $g^{aa}$ followed by $g^{ab}$

$$g^{aa}g^{aa} R_{abcd} = -g^{ab}g^{aa}R_{bacd}$$
$$g^{ab}R^a_{bcd} = -g^{ab}g^{aa}R_{bacd}$$
$$R^{aa}_{cd} = - g^{ab}g^{aa} R_{bacd}$$

Applying $g_{aa}$ to both sides:
$$g_{aa}R^{aa}_{cd} = - g^{ab}g^{aa} g_{aa} R_{bacd}$$
$$R^a_{acd} = -g^{ab} R_{bacd}$$
$$R^a_{acd} = - R^a_{acd} = 0$$

Is there a quicker way?

$g^{aa}$ has no meaning. The metric is symmetric tensor, $g^{a b} = g^{b a}$, so when you contract it with some antisymmetric tensor like $R_{a b c d e f g} = - R_{b a c d e f g}$, you get zero:
$$g^{ab} R_{a b c d} = - g^{b a} R_{b a c d}, \ \Rightarrow \ R^{b}{}_{b c d} = - R^{a}{}_{a c d} .$$
So, you have something equal to minus itself. It must be zero.

 

[unscientific]

$g^{aa}$ has no meaning. The metric is symmetric tensor, $g^{a b} = g^{b a}$, so when you contract it with some antisymmetric tensor like $R_{a b c d e f g} = - R_{b a c d e f g}$, you get zero:
$$g^{ab} R_{a b c d} = - g^{b a} R_{b a c d}, \ \Rightarrow \ R^{b}{}_{b c d} = - R^{a}{}_{a c d} .$$
So, you have something equal to minus itself. It must be zero.

Thanks for replying, I think I almost got the hang of it.
Quick but unrelated question:
Suppose I start off with $R_{abcd}$, how do I get to $R_{bc}$?

I'm thinking $g^{aa}R_{abcd} = R^a _{bcd}$ then $-R^a_{bdc}$ then $-g_{de} g^{ea} R^a_{bdc} = -\delta_d^a R^a_{bdc} =- R^a_{bac} =- R_{bc}$.

 

[samalkhaiat]

Quick but unrelated question:
Suppose I start off with $R_{abcd}$, how do I get to $R_{bc}$?

This depends on the book you use. I contract the first with the last index $R_{a b} = g^{c d} R_{c b a d} = R^{d}{}_{b a d}$ others do it between the first and the 3rd.

 

[unscientific]

This depends on the book you use. I contract the first with the last index $R_{a b} = g^{c d} R_{c b a d} = R^{d}{}_{b a d}$ others do it between the first and the 3rd.

I don't understand how you can do that... I'm confused.

 

[unscientific]

This depends on the book you use. I contract the first with the last index $R_{a b} = g^{c d} R_{c b a d} = R^{d}{}_{b a d}$ others do it between the first and the 3rd.

Starting with the Bianchi Identity:

$$\nabla_e R_{abcd} + \nabla_c R_{abde} + \nabla_d R_{abec} = 0$$

I raise index $a$ by multiplying $g^{aa}$:

$$\nabla_e R^a_{bcd} + \nabla_c R^a_{bde} + \nabla_d R^a_{bec} = 0$$

I'm trying to understand their steps:

 

[unscientific]

This depends on the book you use. I contract the first with the last index $R_{a b} = g^{c d} R_{c b a d} = R^{d}{}_{b a d}$ others do it between the first and the 3rd.

It's alright, I got it - It can be done by contracting $a$ with $c$ then multiplying $g^{be}$.

 

[samalkhaiat]

Starting with the Bianchi Identity:

$$\nabla_e R_{abcd} + \nabla_c R_{abde} + \nabla_d R_{abec} = 0$$

I raise index $a$ by multiplying $g^{aa}$:

$$\nabla_e R^a_{bcd} + \nabla_c R^a_{bde} + \nabla_d R^a_{bec} = 0$$

I told you, we don't have an object like $g^{aa}$. If you have a tensor $T_{abc}$ and you want to raise the first index, you multiply your tensor with $g^{a d}$, i.e. $T^{d}{}_{b c} = g^{a d} T_{a b c}$.

 

[samalkhaiat]

Multiply the BI with $g^{ad}$. Let us do it term by term:
The first term $$g^{ad} \nabla_{e} R_{abcd} = \nabla_{e} g^{ad} R_{abcd} = \nabla_{e} R^{d}{}_{bcd} = \nabla_{e} R_{bc} .$$ The second term $$g^{ad} \nabla_{c} R_{abde} = \nabla_{c} R^{d}{}_{bde} = - \nabla_{c} R_{be} .$$ And the third term is easy, just raise the index on the covariant derivative $$g^{ad} \nabla_{d} R_{abec} = \nabla^{a} R_{abec} = \nabla_{a} R^{a}{}_{bec} .$$

 

[unscientific]

Multiply the BI with $g^{ad}$. Let us do it term by term:
The first term $$g^{ad} \nabla_{e} R_{abcd} = \nabla_{e} g^{ad} R_{abcd} = \nabla_{e} R^{d}{}_{bcd} = \nabla_{e} R_{bc} .$$ The second term $$g^{ad} \nabla_{c} R_{abde} = \nabla_{c} R^{d}{}_{bde} = - \nabla_{c} R_{be} .$$ And the third term is easy, just raise the index on the covariant derivative $$g^{ad} \nabla_{d} R_{abec} = \nabla^{a} R_{abec} = \nabla_{a} R^{a}{}_{bec} .$$

The ricci tensor is defined as $R_{bc} = R^a_{bac}$ in my notes. I'm trying to work with their notation.