Connection forms and dual 1-forms for cylindrical coordinate

  • #1
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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?
 

Answers and Replies

  • #2
319
64
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.
 
  • #3
martinbn
Science Advisor
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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##.
 
  • #4
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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.
 

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