Curvature tensor of 2-sphere using exterior differential forms (tetrads) $ds^2= r^2 (d\theta^2 + \sin^2{\theta}d\phi^2)$
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 $\phi$:  $\omega^{\phi}_{\theta} \wedge e^{\theta}=\frac{\cot{\theta}}{r} e^{\phi} \wedge e^{\theta} \implies \omega^{\phi}_{\theta}=\frac{\cot{\theta}}{r}=\cos{\theta} d{\phi}$
$d\omega^{\phi}_{\theta}=-\sin{\theta} d\theta \wedge d\phi$ 
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?
 A: A few comments for you. First of all, for the Riemannian connection in an orthonormal frame field, we should have $\omega_\phi^\theta = -\omega_\theta^\phi$. What you do know is that $\omega_\phi^\theta\wedge e^\theta = 0$. Indeed, $\omega_\phi^\theta = -\cos\theta\,d\phi$. Similarly, $R_\phi^\theta = -R_\theta^\phi$. 
Now, you need to write the curvature tensor in terms of the frame field, i.e.,
$$R_\theta^\phi = -\sin\theta\,d\theta\wedge d\phi = -\frac1{r^2} e^\theta\wedge e^\phi\,,$$
which tells us that, in fact, the sphere has constant curvature $1/r^2$. Similarly, you need
$R_\theta^\phi = R_{\theta\theta\phi}^\phi e^\theta\wedge e^\phi$, so $R_{\theta\theta\phi}^\phi = -1/r^2$.
Last, there's no reason the curvature tensor components should agree with a different computation. When you do the classic computation in coordinates $(\theta,\phi)$, the corresponding vector fields $\partial/\partial\theta$, $\partial/\partial\phi$ do not give an orthonormal frame field. The actual curvature itself (sectional curvature) is, of course, independent of how you calculate, as is the curvature tensor (not its components alone).
