A Difficulty in understanding contracted Bianchi identities

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1. Jul 7, 2016

lichen1983312

I am confused about the contraction in the proof of the contracted Bianchi identities in

https://en.wikipedia.org/wiki/Proofs_involving_covariant_derivatives

from the step
$${g^{bn}}(R_{bmn;l}^m - R_{bml;n}^m + R_{bnl;m}^m) = 0$$
it seems that the following two quantities are equal
$${g^{bn}}R_{bml;n}^m = R_{l;n}^n$$
$$- {g^{bn}}R_{bnl;m}^m = R_{l;m}^m$$
but I don't understand how is this done if I write them explicitly
$${g^{bn}}({\nabla _n}R)_{bml}^m$$
$$- {g^{bn}}({\nabla _m}R)_{bnl}^m$$
Can anybody help me? I am new to this field and I feel there is something missing. Please help to point out.

Last edited: Jul 7, 2016
2. Jul 7, 2016

Orodruin

Staff Emeritus
The only difference between the terms is the name of the summation index. Sums do not depend on what you call the summation indices.

3. Jul 7, 2016

lichen1983312

Thanks for replying. you are right it doesn't matter what dummy indices are used for summation. But I feel that here is a game of cancling indecis, which does not make sense. does the proof imply
$${g^{bn}}R_{bml}^m = R_l^n$$
and
$${g^{bn}}R_{bnl}^m = -R_l^m$$
in this case
$$- {g^{bn}}R_{bnl}^m = R_l^m = {g^{bm}}R_{bkl}^k$$
I just don't see how this is true

4. Jul 7, 2016

Orodruin

Staff Emeritus
There is no cancelling of indices, just the definition of the Ricci tensor and use of the symmetries of the Riemann tensor.

5. Jul 8, 2016

lichen1983312

Can you be more specific? is this relationship right?

$$- {g^{bn}}R_{bnl}^m = R_l^m = {g^{bm}}R_{bkl}^k$$

sorry for haunting you. I have been stuck here for 2 days.

6. Jul 8, 2016

Orodruin

Staff Emeritus
I suggest using the symmetries and antisymmetries of the Riemann tensor with all indices lowered and you should be able to figure it out.