I'm trying to follow this example from my class lecture notes, but am unable too.
It says we look at plane polar coordinates: $ds^2 = dr^2 + r^2 d\theta^2$.
Then, we introduce an orthonormal basis: $\omega^r = dr$ and $\omega^\theta = r d\theta$.
I get the first exterior derivative $dw^r = 0$ because this is just $d(dr) = d^2 r=0$.
But, I don't understand how they got: $d \omega^\theta = \frac{1}{r}w^r \wedge w^\theta$.
That is, I don't see how the factor of $1/r$ appears.
Thank you.
By direct computation, you have $$d \omega^{\theta}=d(r d\theta)=dr\wedge d\theta=\frac{1}{r}dr\wedge rd\theta=\frac{1}{r}\omega^{r}\wedge\omega^{\theta},$$ so the factor $1/r$ appears when comparing $d\theta$ to $\omega^{\theta}$.