Riemann tensor cyclic identity (first Bianchi) and noncoordinate basis

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SUMMARY

The discussion focuses on the cyclic sum identity of the Riemann curvature tensor, specifically the first Bianchi identity, expressed as {R^\alpha}_{[ \beta \gamma \delta ]}=0. The user references established proofs from textbooks such as Wald, Padmanabhan, and Weinberg, which utilize a coordinate basis where commutation coefficients are zero. The challenge arises when attempting to prove the identity in a noncoordinate basis, where additional terms involving commutation coefficients complicate the proof. The user concludes that the identity should hold in noncoordinate bases, emphasizing the tensorial nature of {R^\alpha}_{[ \beta \gamma \delta ]}.

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miracu113
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I got trouble to understand the cyclic sum identity (the first Bianchi identity) of the Riemann curvature tensor:

{R^\alpha}_{[ \beta \gamma \delta ]}=0

or equivalently,

{R^\alpha}_{\beta \gamma \delta}+{R^\alpha}_{\gamma \delta \beta}+{R^\alpha}_{\delta \beta \gamma}=0.

I can understand the proofs from the most general textbooks, for example:
Wald P.39 (3.2.18), Padmanabhan P.200 (5.41), Weinberg P.141 (6.6.5) and so on.

But I found that all of them do the proof under a coordinate basis, so commutation coefficients are zero, {c_{\mu\nu}}^\alpha=0. In this case, the Riemann tensors are
{R^\alpha}_{\beta \gamma \delta}<br /> =<br /> {\Gamma^\alpha}_{\beta \gamma , \delta}<br /> -{\Gamma^\alpha}_{\beta \gamma , \delta}<br /> +{\Gamma^\alpha}_{\mu \delta}{\Gamma^\alpha}_{\beta \gamma}<br /> -{\Gamma^\alpha}_{\mu \gamma}{\Gamma^\alpha}_{\beta \delta}<br />
and
[\nabla_\gamma, \nabla_\delta]\,v^\alpha<br /> ={R^\alpha}_{\beta \gamma \delta}\,v^\beta
for a vector v=v^\alpha \,e_\alpha. I've checked that we can easily prove the above cyclic sum identity by starting from either of these.

But in a noncoordinate basis case, the Riemann tensors are
{R^\alpha}_{\beta \gamma \delta}<br /> =<br /> {\Gamma^\alpha}_{\beta \gamma , \delta}<br /> -{\Gamma^\alpha}_{\beta \gamma , \delta}<br /> +{\Gamma^\alpha}_{\mu \delta}{\Gamma^\alpha}_{\beta \gamma}<br /> -{\Gamma^\alpha}_{\mu \gamma}{\Gamma^\alpha}_{\beta \delta}<br /> -{\Gamma^\alpha}_{\beta \mu} \,{c_{\delta \gamma}}^\mu<br />
and
([\nabla_\gamma, \nabla_\delta]-{c_{\gamma \delta}}^\mu\nabla_\mu)\,v^\alpha<br /> ={R^\alpha}_{\beta \gamma \delta}\,v^\beta.

I've tried the proof again for a noncoordinate basis and stuck:
3{R^\alpha}_{[ \beta \gamma \delta ]}<br /> =<br /> {c_{\delta \beta}}^\alpha{}_{,\gamma}<br /> +{c_{\beta \gamma}}^\alpha{}_{,\delta}<br /> +{c_{\gamma \delta}}^\alpha{}_{,\beta}<br /> +{c_{\gamma \mu}}^\alpha{c_{\delta \beta}}^\mu<br /> +{c_{\delta \mu}}^\alpha{c_{\beta \gamma}}^\mu<br /> +{c_{\beta \mu}}^\alpha{c_{\gamma \delta}}^\mu<br />

I think the identity will be true in a noncoordinate basis too. But how can I proove or understand it in a noncoordinate basis?
 
Last edited:
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I think all you need is to note that {R^\alpha}_{[ \beta \gamma \delta ]} is a tensor because it's the sum of three tensors, and that if a tensor is zero in one basis it's zero in all.
 

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