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The connection one-forms can be defined as follows: let $(M,g)$ be spacetime, $U\subset M$ an open set and $\{e_a\}$ an orthonormal basis of vector fields on $U$ with dual basis $\{\omega^a\}$ of one-forms. We define the connection one-forms as follows. First define the connection coefficients by

$$\nabla e_a=\Gamma^b_{ca}e_b\otimes \omega^c,$$

then define the connection one-forms

$$\theta^b_a=\Gamma^b_{ca}\omega^c.$$

Thus the connection one-forms are defined so that

$$\nabla_X e_a=\theta^b_a(X)e_b.$$

We can thus show that if the basis is orthonormal then $\theta^a_b=-\theta^b_a$.

We then have Cartan's first structural equation:

$$d\omega^a=\omega^b\wedge \theta^a_b.$$

I've heard then that this can be used to give a method to compute the connection one-forms. Find $\omega^a$, then take the exterior derivatives, expand in terms of $\{\omega^a\}$ and read of the terms as above.

Now in the case of Schwarzschild I'm running into a inconsistency. The metric is of the form

$$ds^2=f^2 dt^2-g^2 dr^2-h^2 (d\theta^2+\sin^2\theta d\phi^2)$$

We thus have

$$\omega^0=fdt,\quad \omega^1=gdr,\quad \omega^2=hd\theta,\quad \omega^3=h\sin\theta d\phi.$$

On the other hand we can compute

$$d\omega^0=-\frac{f'}{fg}\omega^0\wedge \omega^1, \quad d\omega^1=0$$

Now it would seen from $d\omega^0$ that

$$\theta^0_1=-\frac{f'}{fg}\omega^0$$

but from $d\omega^1$ it seems $\theta^1_0=0$, however $\theta^0_1=-\theta^1_0$ which leads to one inconsistency.

So what am I missing here? Why am I running into one inconsistency?

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  • $\begingroup$ This lecture youtube.com/watch?v=Ypu4l_ONhKY explains how to get Schwarzschild from one-forms very nicely, how to read everything off in a unified way. $\endgroup$ – bolbteppa Mar 27 '18 at 23:12
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First off, note that it is $\theta_{ab} = -\theta_{ba}$ with both indices in covariant position, and we have let the metric act to lower the connection form indices: $\theta_{ab} = g_{ac}\theta^c{}_b$. This follows, as you may know, from the metricity of the connection and by choosing a frame such that the metric components are constant functions (rigid frame). The distinction is important, although irrelevant to this particular mistake, because even in an orthonormal frame, using e.g. the $(+---)$ sign convention, we have $\theta^0{}_1 = \theta_{01} = -\theta_{10} = \theta^1{}_0$.

Your mistake is that $\theta^1{}_0 = 0$ does not follow from $d\omega^1 = 0$. What does follow is that $\theta^1{}_0 = A\omega^0$ for some function $A$, which is, as you can see, precisely what you get from your expression for $d\omega^0$. In effect, you neglect the fact that the wedge product between two parallell one-forms vanishes.

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