1
$\begingroup$

Let $\omega^{\hat{a}}$ be an orthonormal basis, and $\theta^{\hat{a}}_{\hat{b}}$ be the associated connections. From Cartan's second structure equation, we may compute the curvature 2-form, i.e.

$$R^{\hat{a}}_{\hat{b}} = \mathrm{d}\theta^{\hat{a}}_{\hat{b}} + \theta^{\hat{a}}_{\hat{c}} \wedge \theta^{\hat{c}}_{\hat{b}}.$$

We may also relate the curvature 2-form to the Riemann curvature, via the relation,

$$R^{\hat{a}}_{\hat{b}} = \frac{1}{2}R^{\hat{a}}_{\hat{b} \hat{c} \hat{d}} \omega^{\hat{c}} \wedge \omega^{\hat{d}}$$

from which we may 'read off' the terms. However, both the Ricci and Riemann curvature are in an orthonormal basis. How can I 'convert' $R^{\hat{a}_{\hat{b}}}$ and $R^{\hat{a}}_{\hat{b}\hat{c}\hat{d}}$ back to an ordinary coordinate basis?

| cite | improve this question | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.