# Riemann tensor cyclic identity (first Bianchi) and noncoordinate basis

1. Jul 12, 2012

### miracu113

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
${R^\alpha}_{\beta \gamma \delta} = {\Gamma^\alpha}_{\beta \gamma , \delta} -{\Gamma^\alpha}_{\beta \gamma , \delta} +{\Gamma^\alpha}_{\mu \delta}{\Gamma^\alpha}_{\beta \gamma} -{\Gamma^\alpha}_{\mu \gamma}{\Gamma^\alpha}_{\beta \delta}$
and
$[\nabla_\gamma, \nabla_\delta]\,v^\alpha ={R^\alpha}_{\beta \gamma \delta}\,v^\beta$
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
${R^\alpha}_{\beta \gamma \delta} = {\Gamma^\alpha}_{\beta \gamma , \delta} -{\Gamma^\alpha}_{\beta \gamma , \delta} +{\Gamma^\alpha}_{\mu \delta}{\Gamma^\alpha}_{\beta \gamma} -{\Gamma^\alpha}_{\mu \gamma}{\Gamma^\alpha}_{\beta \delta} -{\Gamma^\alpha}_{\beta \mu} \,{c_{\delta \gamma}}^\mu$
and
$([\nabla_\gamma, \nabla_\delta]-{c_{\gamma \delta}}^\mu\nabla_\mu)\,v^\alpha ={R^\alpha}_{\beta \gamma \delta}\,v^\beta$.

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

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: Jul 12, 2012
2. Jul 13, 2012

### pervect

Staff Emeritus
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.