How to Prove contracted Bianchi Identity

[jojoo]

How to prove
g^{im}<br /> \nabla_{\partial_j}R_{ilkm}=\nabla_{\partial_j}R_{lk}.

of cause g^{im}R_{ilkm}=R_{lk}, but I don't know how the contraction can pass through the covariant derivative?

 

[Hurkyl]

How to prove
g^{im}<br /> \nabla_{\partial_j}R_{ilkm}=\nabla_{\partial_j}R_{lk}.

of cause g^{im}R_{ilkm}=R_{lk}, but I don't know how the contraction can pass through the covariant derivative?

Because derivatives and (finite) sums commute. e.g.

<br /> \frac{d}{dx} \sum_{i = a}^b f_i(x) = \sum_{i = a}^b \frac{df_i}{dx}(x)<br />

 

[robphy]

...but I don't know how the contraction can pass through the covariant derivative?

The derivative operator is usually taken to be "metric compatible", i.e.,
\nabla_a g_{bc}=0.

 

[jojoo]

The derivative operator is usually taken to be "metric compatible", i.e.,
\nabla_a g_{bc}=0.

Yes, I know that. But \nabla_{\partial_j}R_{ilkm}=(\nabla_{\partial_j}R)(\partial_i,\partial_l,\partial_k,\partial_m) How can the contraction really happen?(since \nabla_a g_{bc}=0 means (\nabla_a g)(\partial_b,\partial_c)=0)
Would you like to give me more detail? Thank you!

 

[Chris Hillman]

This is OT, but George Jones just pointed out that today's issue of the daily paper in Toronto featured a picture of the uncontracted Bianchi identities. For some reason a politician is in the foreground.

Industrious students can look for papers pointing out that Bianchi himself credited these identities to someone else.

 

[neorayner]

g^{im}\nabla_{\partial_j}R_{ilkm}=\nabla_{\partial_j}(g^{im}R_{ilkm}) - R_{ilkm}\nabla_{\partial_j}g^{im}=\nabla_{\partial_j}R_{ lk}

because \nabla_a g_{bc}=0

it respects the Leibniz derivation