I'm trying to compute the curvature tensor from the following metric:
$$g = -(\frac{r}{M})^2 dt^2 + (\frac{M}{r})^2 dr^2$$
To compute the curvature I'm employing the first Cartan structure equation which reads:
$$d\theta^a + \omega^a_b \wedge \theta^a = 0$$
where I assumed that the manifold I'm working with is torsion free and $\theta^a$ define an orthonormal local frame given by: $$\theta^0 = \frac{r}{M}dt, \theta^1 = \frac{M}{r}dr$$
Then from the first Cartan equation I obtain the two equations:
$$d\theta^0 + w^0_1 \wedge \theta^1 = 0$$ $$d\theta^1 + w^1_0 \wedge \theta^0 = 0$$
The first one gives: $$\frac{1}{M} dr \wedge dt + w^0_1 \wedge \frac{r}{M}dr = 0$$ $$0 + w^1_0 \wedge \frac{r}{M}dt = 0$$
From the first equation I can see that $w^0_1$ should be given by $\frac{r}{M^2}dt$ but the second equation seems give $w^1_0=0$ which also would imply that $w^0_1$ by antisymmetry. Is there a fault in my reasoning?
First of all, because you're working with a metric of signature $(1,1)$, you will have $\omega^0_1 = \omega^1_0$ rather than skew-symmetry. Second, your first line tells you that $$\omega^0_1 = \frac{dt}r + A(r,t)dr$$ and your second line tells you that $\omega_0^1=\omega_1^0$ is a multiple of $dt$ (not necessarily $0$). Thus, we conclude that $A(r,t)=0$ and there's no problem.