Curvature tensor of 2-sphere using exterior differential forms (tetrads)

Question

$ds^2= r^2 (d\theta^2 + \sin^2{\theta}d\phi^2)$

The following is the tetrad basis

$e^{\theta}=r d{\theta} \,\,\,\,\,\,\,\,\,\, e^{\phi}=r \sin{\theta} d{\phi}$

Hence, $de^{\theta}=0 \,\,\,\,\,\, de^{\phi}=r\cos{\theta} d{\theta} \wedge d\phi = \frac{\cot{\theta}}{r} e^{\theta}\wedge e^{\phi}$

Setting the torsion tensor to zero gives: $de^a + \omega^a _b \wedge e^b =0$.

This equation for $a=\theta$ gives $\omega^{\theta}_{\phi}=0$. (I have used $\omega^{\theta}_{\theta}=\omega^{\phi}_{\phi}=0$)

The equation for $\phi$: $\omega^{\phi}_{\theta} \wedge e^{\theta}=\frac{\cot{\theta}}{r} e^{\phi} \wedge e^{\theta} \implies \omega^{\phi}_{\theta}=\frac{\cot{\theta}}{r} e^{\phi}=\cos{\theta} d{\phi}$

$d\omega^{\phi}_{\theta}=-\sin{\theta} d\theta \wedge d\phi$

Hence $R^i_j = d\omega^i_j+ \omega^i_b \wedge \omega^b_j$ gives $R^{\phi}_{\theta}=-\sin{\theta} d\theta \wedge d\phi$ and $R^{\theta}_{\phi}=0$

Writing in terms of components gives $R^{\phi}_{\theta \theta \phi}=-\sin{\theta}$, and $R^{\theta}_{\phi \phi \theta}=0$

However this is wrong. I have done the same problem using Christoffel connections, and the answer which I know to be correct is

$R^{\phi}_{\theta \theta \phi}=-1$, and $R^{\theta}_{\phi \phi \theta}=\sin^2{\theta}$

Please could anyone tell mw what I am doing wrong? Any help will be appreciated?

Answer 1

$R^{\phi}_{\theta}=-\sin{\theta} d\theta \wedge d\phi$

So:

$R^{\phi}_{\theta}= - \frac{ e^\theta \wedge e^\phi}{r^2}$

So:

$R^{\phi}_{\theta \theta \phi}= - \frac{1}{r^2}$

Then $R_{\phi\theta \theta \phi}= g_{\phi\phi}R^{\phi}_{\theta \theta \phi} = -sin^2\theta$

Then $R_{\theta \phi \theta \phi} = - R_{\phi\theta \theta \phi} = sin^2\theta$

Then $R^\theta _{\phi \theta \phi} = g^{\theta\theta}R_{\theta \phi \theta \phi} = \frac {sin^2\theta}{r^2}$

Also, you could not have $\omega^\theta_\phi=0$, it is because $\omega^{ab}=−\omega^{ba}$ and the metrics is diagonal