Bianchi Identity-where did I go wrong?

[zn5252]

hello ,
In the Bianchi Identities of the second kind, we have ∇a R_bcde + ∇b R_cade + ∇c R_abde ≡ 0
but since ∇c R_abde = - ∇c R_bade
we get :
∇a R_bcde + ∇b R_cade - ∇c R_bade = 0
in the last term, we exchange the c and the b indices and we would arrive at:
∇a R_bcde + ∇b R_cade - ∇b R_cade = 0
which leads to :
∇a R_bcde = 0 .
but this incorrect ? did I do something wrong ?
I would like actually to arrive at the formula :
2 ∇b R_acde − ∇a R_bcde ≡ 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,

 

[Bill_K]

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 ?

 

[Bill_K]

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 R_bcde + ∇_b R_cade + ∇_c R_abde ≡ 0
multiply out by R^bcd_f , we get :
R^bcd_f ∇_a R_bcde + R^bcd_f ∇_b R_cade + R^bcd_f ∇_c R_abde ≡ 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:

R^bcd_f ∇_a R_bcde - R^bcd_f ∇_b R_acde + R^cbd_f ∇_b R_acde = R^bcd_f ∇_a R_bcde - 2 R^bcd_f ∇_b R_acde

Thank you.

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