Explicit calculation of spin connection through Cartan's first structure equation Given the metric 
$$
ds^2 = F(r)^2dr^2 + r^2d\theta^2 + r^2 \sin^2(\theta)\,  d\phi^2, 
$$
I'm trying to find the corresponding spin connections $\omega^a_{\ b}$ using the first structure equation:
$$
de + \omega e = 0. 
$$
I found the vielbeins $e$ and their exterior derivatives $de$: 
$$ 
de^1 = 0, \quad de^2 = drd\theta, \quad de^3 = sin(\theta)drd\phi + r\cos(\theta)d\theta d\phi,
$$
but I am stuck on actually working out the $\omega$. 
I read through Zee's 'GR in a nutshell', and he does the same calculation but just says: "In general, write $\omega^a_{\,b} = \omega^a_{\,bc}e^c = \omega^a_{\,b\mu}dx^\mu$. Plug this into the first structure equation and match terms. How do I actually go about calculating $\omega^1_{\, 2}$, $\omega^1_{\, 3}$, and $\omega^2_{\, 3}$ at this point?  
 A: As you have, the first step is to identify $e^r = F(r)\mathrm{d}r, e^\theta = r\mathrm{d}\theta$ and $e^\phi = r\sin \theta \mathrm{d}\phi$. The trick is to then take the derivatives but re-express them in terms of $e$ again. Thus,
$$\mathrm{d}e^r = 0, \quad \mathrm{d}e^\theta = -\mathrm{d\theta} \wedge \mathrm{d}r = -\frac{1}{rF(r)}e^\theta \wedge e^r$$
and, 
$$\mathrm{de^\phi} = -\sin\theta \mathrm{d\phi} \wedge \mathrm{d}r - r\cos\theta \mathrm{d}\phi \wedge \mathrm{d}\theta = -\frac{1}{rF(r)} e^\phi \wedge e^r - \frac{\cot \theta}{r^2} e^\phi \wedge e^\theta.$$
Now let's take an example of using Cartan's first equation. We have $\mathrm{d}e^a + \omega^a_b \wedge e^b = 0$ and if we choose $a=\theta$ the equations read,
$$\frac{1}{rF(r)}e^\theta \wedge e^r = \omega^\theta_r \wedge e^r + \omega^\theta_\theta \wedge e^\theta + \omega^\theta_\phi \wedge e^\phi.$$
We have $\omega^\theta_\theta = 0$ by anti-symmetry. We can identify now $\omega^\theta_r = -\omega^r_\theta = \frac{1}{rF(r)}e^\theta$. Notice the last term we could choose $\omega^\theta_\phi = 0$ however Cartan's equations are a system of equations, so we are not free to make this choice yet without considering the other equations. We can at best say $\omega^\theta_\phi$ is proportional to $\mathrm{d}\phi$ to ensure $\omega^\theta_\phi \wedge e^\phi = 0$. As it turns out, we don't have $\omega^\theta_\phi = 0$ because of the $a = \phi$ equation, which will give you $\omega^\theta_\phi = -r^{-2}\cot\theta\, e^\phi$.
I hope this elucidates how to use Cartan's structure equation. Computing the Ricci tensor is then much simpler, as rather than solving for components you're just plugging in and computing.
A: There is also an explicit procedure that is often better if the vielbein is simple.
We have $$ \mathrm d e^a=-\frac{1}{2}C^a_{bc}e^b\wedge e^c, $$ where the $C^a_{bc}$ are the vielbein commutators. We can invert the first structure equation explicitly as $$ 0=\mathrm de^a+\omega^a_{\ b}\wedge e^b \\ =-\frac{1}{2}C^a_{bc}e^b\wedge e^c+\omega^a_{c\ b}e^c\wedge e^b \\ \frac{1}{2}C^a_{bc}e^b\wedge e^c=\frac{1}{2}\left( \omega^a_{b\ c}-\omega^a_{c\ b} \right)e^b\wedge e^c, $$ so $$ C^a_{bc}=\omega^a_{b\ c}-\omega^a_{c\ b}. $$
Lowering the index, we get $$ C_{a,bc}=\omega_{b,ac}-\omega_{c,ab} \\ C_{b,ca}=\omega_{c,ba}-\omega_{a,bc} \\ -C_{c,ab}=-\omega_{a,cb}+\omega_{b,ca}, $$ now sum these up: $$ C_{a,bc}+C_{b,ca}-C_{c,ab}=2\omega_{c,ba} \\ \omega_{c,ab}=\frac{1}{2}\left(C_{c,ab}-C_{a,bc}-C_{b,ca}\right) \\ \omega_{ab}=\frac{1}{2}\left(C_{c,ab}-C_{a,bc}-C_{b,ca}\right)e^c. $$
If the veilbein is simple, then the $\mathrm de^a=-\frac{1}{2}C^a_{bc}e^b\wedge e^c$ will only involve a few terms at most, and the spin connection is very easy to calculate from this.
