I Connection forms and dual 1-forms for cylindrical coordinate

[Gene Naden]

I ran across exercise 2.8.4 in Oneill's Elementary Differential Geometry. It says "Given a frame field $E_1$ and $E_2$ on $R^2$ there is an angle function $\psi$ such that $E_1=\cos(\psi)U_1+\sin(\psi)U_2$, $E_2=-\sin(\psi)U_1+\cos(\psi)U2$

(where $U_1$, $U_2$, $U_3$ are the natural (rectangular) unit vectors)

express the connection form and dual 1-forms in terms of $\psi$ and the natural coordinates x,y."

For the connection forms I get $\omega_{12}=-\omega_{21}=d\psi$ with the other components zero. For the dual 1-forms I had difficulty. The definition of dual 1-forms states they are the 1-forms $\theta_i$ such that $\theta_i(v)=v\cdot E_i(p)$ for each tangent vector v. I have difficulty applying this definition. One reason is that it is independent of any coordinate system.

I noticed that the given frame field matches the frame field of cylindrical coordinates with the angle $\psi$. So that was my answer, the dual 1-forms of cylindrical coordinates: $\theta_1=dr$ where $r=\sqrt{x^2+y^2}$ and $\theta_2=rd\psi$.

So my question is, is my answer correct? Can I just pick a coordinate system and use it to compute the dual 1-forms?

 

[Gene Naden]

This is still an open question for me: how do you generate the 1-forms for a frame field when you don't have a coordinate system associated with that field. Perhaps the correct tactic is to somehow derive a coordinate system and then use that set of coordinates to get the 1-forms.

 

[martinbn]

So my question is, is my answer correct? Can I just pick a coordinate system and use it to compute the dual 1-forms?

Yes, except the question asks you to do it in the natural coordinates $x,y$.

 

[Gene Naden]

So I get $\theta_1=\frac{xdx+ydy}{\sqrt{x^2+y^2}}$ and $\theta_2=\frac{-ydx+xdy}{\sqrt{x^2+y^2}}$

I got this from the cylindrical coordinate transformation. I am not sure this is the right approach; it says to use natural coordinates.