Attempted proof of the Contracted Bianchi Identity

In summary, the conversation discusses the attempted proof of the identity ##R^{mn}_{;n} - \frac {1} {2} g^{mn} R_{;n} = 0## using the second Bianchi Identity and the inverse metric tensor. The steps involved include manipulating indices, contracting, summing, and using the symmetric property of the metric tensor and its inverse. The result obtained is ##2 R^{mn}_{;n} = g^{mn} R_{;n} ##, which is the desired identity.
  • #1
TL;DR Summary
I constructed an attempted proof of the Contracted Bianchi Identity, and cannot identify any errors myself. However, obviously, this does not mean that the proof is watertight. For any errors present, I am hoping to get details and explanations on where I went wrong, how to avoid the mistake in the future, etc.
Criticism welcome!
My Attempted Proof
##R^{mn}_{;n} - \frac {1} {2} g^{mn} R_{;n} = 0##
##R^{mn}_{;n} = \frac {1} {2} g^{mn} R_{;n}##
So, we want ##2 R^{mn}_{;n} = g^{mn} R_{;n} ##

Start w/ 2nd Bianchi Identity ##R_{abmn;l} + R_{ablm;n} + R_{abnl;m} = 0##
Sum w/ inverse metric tensor twice ##g^{bn} g^{am} (R_{abmn;l} + R_{ablm;n} + R_{abnl;m}) = g^{bn} g^{am} (0)##
Distributing ##g^{bn} (g^{am} R_{abmn;l} + g^{am} R_{ablm;n} + g^{am} R_{abnl;m}) = 0##
Index manipulating ##g^{bn} (R^{m}_{~~~bmn;l} + R^{m}_{~~~blm;n} + R^{m}_{~~~bnl;m}) = 0##
Contracting ##g^{bn} (R_{bn;l} - R_{bl;n} + R^{m}_{~~~bnl;m}) = 0##
Distributing and manipulating indices ## R^n_{n;l} - R^n_{l;n} + g^{bn} R^{m}_{~~~bnl;m} = 0##
Summing ##R_{;l} - R^n_{l;n} + g^{bn} R^{m}_{~~~bnl;m} = 0##
Sum W/ ##g^{lp}##: ##g^{lp} (R_{;l} - R^n_{l;n} + g^{bn} R^{m}_{~~~bnl;m} = 0##
Distributing ##g^{lp} R_{;l} - g^{lp} R^n_{l;n} + g^{lp} g^{bn} R^{m}_{~~~bnl;m} = 0##
Summing ##g^{lp} R_{;l} - R^{np}_{;n} + g^{lp} g^{bn} R^{m}_{~~~bnl;m} = 0##
Ricci Tensor (downstairs indices is symmetric, so I THINK it is valid to flip the 2 top indices here ##g^{lp} R_{;l} - R^{pn}_{;n} + g^{lp} g^{bn} R^{m}_{~~~bnl;m} = 0##
##g^{lp} R_{;l} + g^{lp} g^{bn} R^{m}_{~~~bnl;m} = R^{pn}_{;n}## becomes ##R^{pn}_{;n} = g^{lp} R_{;l} + g^{lp} g^{bn} R^{m}_{~~~bnl;m}##
##R^{pn}_{;n} = g^{lp} R_{;l} - g^{lp} g^{bn} R^{m}_{~~~bln;m}##
##R^{pn}_{;n} = g^{lp} R_{;l} - R^{mnp}_{~~~~~~~~n;m}##
##R^{pn}_{;n} + R^{mnp}_{~~~~~~~~n;m} = g^{lp} R_{;l} ##
The metric tensor and its inverse are symmetric ##R^{pn}_{;n} + R^{mnp}_{~~~~~~~~n;m} = g^{pl} R_{;l} ##
Summing over ##n##, ##R^{pn}_{;n} + R^{mp}_{;m} = g^{pl} R_{;l} ##
I've seen others just replace indices like I am about to, so I am about 60% sure the following operation is valid. However, I am not sure. Any insight and corrections would be much appreciated here especially! ##R^{mn}_{;n} + R^{mn}_{;m} = g^{mn} R_{;n} ##
Combining like terms ##2 R^{mn}_{;n} = g^{mn} R_{;n} ##, which is just the result we're looking for, assuming I did it right.

Any help is much appreciated!
P.S., I'm not always great at articulating my thoughts, so my apologies if this question isn't clear. Also, I know this isn't high school material, but I am currently in high school, which is why I made my level "Basic/high school level."
Lastly, I know that this isn't the typical way one would go about proving this identity, but I was wondering if this was still a valid avenue to do so (I did it from scratch).
 
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  • #2
Moderator's note: Thread moved to the Differential Geometry forum.
 

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