Given a metric $g_{\mu \nu}$, one can select an orthonormal basis $\omega^{\hat{a}}$ such that, $$ds^2= \omega^{\hat{t}}\otimes\omega^{\hat{t}} - \omega^{\hat{x}} \otimes \omega^{\hat{x}} - ...$$

By employing the Cartan structural equations, one can 'read off' the components of the Ricci tensor and Riemann curvature tensor in the orthonormal basis. My question is: how do I take my curvature tensors in the orthonormal basis back to the coordinate basis after?


1 Answer 1


I) The vielbein $e^a{}_{\mu}$ in the Cartan formalism is an intertwiner

$$\tag{1} g_{\mu\nu}~=~e^a{}_{\mu} ~\eta_{ab} ~e^b{}_{\nu} $$

between the curved (pseudo) Riemannian metric $g_{\mu\nu}$ and the corresponding flat metric $\eta_{ab}$. Here $\mu,\nu,\lambda, \ldots,$ are so-called curved indices, while $a,b,c, \ldots,$ are so-called flat indices.

II) The Riemann curvature tensor $R^d{}_{abc}$ with flat indices is related to the Riemann curvature tensor

$$\tag{2} R^{\sigma}{}_{\mu\nu\lambda} ~=~(e^{-1})^{\sigma}{}_{d}~ R^d{}_{abc}~e^a{}_{\mu}~e^b{}_{\nu}~e^c{}_{\lambda} $$

with curved indices by multiplying with the appropriate factors of the vielbein and its inverse in the obvious way. Similarly for the Ricci tensor.


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