Let E be the cylindrical frame field $E_1 = \cos\theta U_1 + \sin\theta U_2, E_2 = − \sin\theta U_1 + \cos\theta U_2, E_3 = U_3$
(a) Starting from the basic cylindrical equations $x = r \cos\theta, y = r \sin\theta, z = z$, show that the dual 1-forms are $θ_1 = dr, θ_2 = rd\theta, θ_3 = dz$.
I started by taking the derivatives of E however, I'm not sure if that's the right first step or not. Any guidance on this please? Thanks!
One way to do it is just recalling from linear algebra that the transformation between dual basis uses the inverse of the transition matrix between the original bases. Using columns whose entries are the vectors themselves requires an extra transposition. Meaning that if $$\begin{pmatrix} E_1 \\ E_2 \\ E_3 \end{pmatrix} = \begin{pmatrix} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix} U_1 \\ U_2 \\ U_3 \end{pmatrix}$$then $$\begin{pmatrix} \theta_1 \\ \theta_2 \\ \theta_3 \end{pmatrix} = \left(\begin{pmatrix} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{pmatrix}^{-1}\right)^\top\begin{pmatrix} dx\\ dy \\ dz \end{pmatrix} = \begin{pmatrix} \cos \theta \,dx + \sin\theta\,dy \\ -\sin \theta\,dx + \cos \theta\,dy \\ dz \end{pmatrix}.$$For free, $\theta_3 = dz$. And sure enough, using $x = r\cos \theta$ and $y=r \sin \theta$ you get $$\begin{align} \theta_1 &= \cos\theta(\cos\theta \,dr - r\sin\theta\,d\theta) + \sin\theta(\sin\theta\,dr + r\cos\theta\,d\theta) = dr, \\ \theta_2 &= -\sin\theta(\cos\theta\,dr-r\sin\theta\,d\theta)+\cos\theta(\sin\theta\,dr+r\cos\theta\,d\theta) = r\,d\theta. \end{align}$$