# Bianchi Identity-where did I go wrong?

zn5252
hello ,
In the Bianchi Identities of the second kind, we have ∇a Rbcde + ∇b Rcade + ∇c Rabde ≡ 0
but since ∇c Rabde = - ∇c Rbade
we get :
in the last term, we exchange the c and the b indices and we would arrive at:
∇a Rbcde = 0 .
but this incorrect ? did I do something wrong ?
I would like actually to arrive at the formula :
2 ∇b Racde − ∇a Rbcde ≡ 0.

see formula 9 here from original Bel's article from 1938:

http://gallica.bnf.fr/ark:/12148/bp...ction+d'un+tenseur+du+quatrième+ordre;.langEN

and
(see formula 5 here : https://docs.google.com/viewer?a=v&...9JJLhB&sig=AHIEtbRUONCiZUSV_8erdxK9YSMiouUrjA)

Thanks,
cheers,

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in the last term, we exchange the c and the b indices
Only "dummy" indices can be freely changed like this, i.e. indices that are being summed over.

zn5252
Thanks I see , I must have been on a rush to get to the formula.
My problem is then : how do i get rid of the last variable on which the divergence is acting I mean c ? is there a formula I must have have overlooked perhaps ?

I don't think what you're trying to prove is true. Bel's result follows only after he's multiplied by another factor of the Riemann tensor.

zn5252
I see, now I think I found it . I have tried it out indeed with the additional tensor factor. it goes like this :
we have :
a Rbcde + ∇b Rcade + ∇c Rabde ≡ 0
multiply out by Rbcdf , we get :
Rbcdfa Rbcde + Rbcdfb Rcade + Rbcdfc Rabde ≡ 0

=>in the last term , we exchange the b and the c since now they are dummy indices I presume and in the second term, we use the antisymmetry of a and c:

Rbcdfa Rbcde - Rbcdfb Racde + Rcbdfb Racde = Rbcdfa Rbcde - 2 Rbcdfb Racde

Thank you.

PS : Sorry the Bel's article above is from 1958. I must have confused it with the article of Lanczos .

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