I must to compute value $$ [[D_{\mu}, D_{\nu}],D_{\lambda}]A^{\rho}. $$ It is equal to $$ [D_{\mu}, D_{\nu}]D_{\lambda}A^{\rho} - D_{\lambda} ([D_{\mu}, D_{\nu}]])A^{\rho} - [D_{\mu}, D_{\nu}]D_{\lambda}A^{\rho} = -D_{\lambda} ([D_{\mu}, D_{\nu}])A^{\rho}. $$ So, the question: can I formally take $A^{\rho}$ under the sign of the derivative for using the identity $[D_{\mu}, D_{\nu}]A^{\rho} = R^{\rho}_{\quad \sigma \mu \nu}A^{\sigma}$ and, afer that, take $A^{\sigma}$ outside the derivative? I'm afraid that no, but I hope that it is possible.


$[[D_{\mu}, D_{\nu}],D_{\lambda}]A^{\rho} = [D_{\mu}, D_{\nu}]D_{\lambda}A^{\rho}-D_{\lambda}[D_{\mu}, D_{\nu}]A^{\rho}$

$=-R^{\tau}_{{\lambda}\mu \nu}D_{\tau}A^{\rho}+R^{\rho}_{\sigma \mu \nu}D_{\lambda}A^{\sigma}- D_{\lambda}(R^{\rho}_{\sigma \mu \nu}A^{\sigma})$

$=-R^{\tau}_{{\lambda}\mu \nu}D_{\tau}A^{\rho}+ R^{\rho}_{\sigma \mu \nu ; \lambda}A^{\sigma}$

When you cycle over $\mu, \nu, \lambda$ you will need/get the first and second bianchi identities

1st BI: $R^{\tau}_{ \mu \nu \lambda}+ R^{\tau}_{\lambda \mu \nu }+ R^{\tau}_{\nu \lambda \mu } = 0$

2nd BI: $R^{\rho}_{\sigma \mu \nu ; \lambda}+R^{\rho}_{\sigma \lambda \mu ; \nu}+R^{\rho}_{\sigma \nu \lambda ; \mu}=0$


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