Suppose we have a Lorentzian metric of the form
\begin{align}
g&=-f(r)^2\,dt^2+ h(r)^2(dr^2+r^2\,d\theta^2+r^2\sin^2\theta\,d\phi^2)
\end{align}
Where $f,h$ are say strictly positive functions. We use the Levi-Civita connection. I introduced the 1-forms
\begin{align}
e^0=f(r)\,dt,\quad e^1=h(r)\,dr,\quad e^2=rh(r)\,d\theta,\quad e^3=rh(r)\sin\theta\,d\phi
\end{align}
which diagonalize the metric, and now I'm trying to use these to calculate the connection 1-forms $\omega^a\_{\,b}$ using Cartan's structural equation $de=-\omega\wedge e$ (since Levi-Civita connection is torsion free).

---

**Question 1.**

The issue I'm facing is that once I calculate $de$, I'm not sure how to identify $\omega$ from those equations: initially I tried the most naive thing by just looking at the appropriate coefficient and calling that the appropriate component of $\omega$, but I think this naive approach is wrong, probably because the wedge-product of non-zero forms can still be zero (so "cancelling" terms won't work).

To be more explicit, I calculated
\begin{align}
\begin{cases}
de^0= f'(r)\,dr\wedge dt\\
de^1= 0\\
de^2=(h(r)+rh'(r))\,dr\wedge d\theta\\
de^3= (h(r)+rh'(r))\sin\theta\,dr\wedge d\phi+ rh(r)\cos\theta\,d\theta\wedge d\phi
\end{cases}
\end{align}
When I first did the calculation, I naively concluded that
\begin{align}
de^0=f'(r)\,dr\wedge dt=-\left[-\frac{f'(r)}{f(r)}\,dr\right]\wedge e^0,
\end{align}
and thus that $\omega^0\_0=-\frac{f'(r)}{f(r)}\,dr, \omega^0\_1=\omega^0\_2=\omega^0\_3=0$. Next, from $de^1=0$ I naively concluded that $\omega^1\_{\,b}=0$ for all $b=0,1,2,3$. I did a similar thing with the other equations. But now I realize this is wrong, because for example, we can also write
\begin{align}
de^0=f'(r)\,dr\wedge dt=
-\left[\frac{f'(r)}{h(r)}\,dt\right]\wedge e^1,
\end{align}
so if I were to use my above logic, I would have $\omega^0\_0=0, \omega^0\_1=\frac{f'(r)}{h(r)}\,dt, \omega^0\_2=\omega^0\_3=0$. So clearly my mistake stems from the fact that the wedge of non-zero forms can be zero. But now I'm not sure what the correct approach is.

I have read [this answer](https://math.stackexchange.com/a/3725231/568204) by @Ted Shifrin, and it seems like the correct answer is the second approach, but I'm not sure why. Also, I can't really understand that answer because it's not clear to me why certain certain $\omega^a\_b$ are equal to certain functions and why others are multiples of some $e^i$, and why some others are zero.

---

**Question 2.**

The equation $de=-\omega\wedge e$ consists of four equations relating $2$-forms. However, $\omega$ being a $4\times 4$ matrix (in this case) of $1$-forms, consists a-priori of 16 unknowns. I believe in this case due to the Lorentzian signature and the diagonalizability of the metric, there is some relationship between $\omega^a\_b$ and $\omega^b\_a$, so that it can be written as
\begin{align}
[\omega^a\_b]&=
\begin{pmatrix}
0&\alpha\_1&\alpha\_2&\alpha\_3\\
\alpha\_1&0&\beta\_1&\beta\_2\\
\alpha\_2&-\beta\_1&0&\beta\_3\\
\alpha\_3& -\beta\_2&-\beta\_3&0
\end{pmatrix}
\end{align}
for some 1-forms $\alpha\_1,\beta\_i$. So, now there are only 6-unknowns, but this is still too many unknowns for the number of equations.

So my question is whether we can always use this structural equation to determine $\omega$ completely? I believe the answer is yes because for the case of Christoffel symbols $\Gamma^i\_{jk}$ we have explicit formulas for it in terms of the metric, and now since $\omega$ are related to $\Gamma$ in some fashion, the same ought to hold true; but now I'm not sure how to reconcile this with the above counting argument (6 unknowns vs 4 equations).