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How to calculate the Poincare dual of a ray on R2-{0}?

  1. Oct 4, 2011 #1
    S={(x,0)|x>0} on R^2-{0},I need to calculate the closed Poincare dual of S.

    Assume [tex] \omega=f(x,y)dx+g(x,y)dy [/tex] on [tex]R^2[/tex]-{0} have compact support.Then we need to find a form [tex] \eta [/tex]in [tex]H^1 (R^2 - {0} )[/tex] satisfying [tex] \int\limits_S {i^* \omega = \int\limits_M {\omega \wedge \eta } } [/tex],

    The book let me prove [tex] d\theta /2\pi[/tex]([tex] \theta [/tex]is the angle function) is the poincare dual.

    But in my calculation,[tex]\int\limits_S {i^* \omega = \int_0^{ + \infty } {f(x,0)dx} } [/tex] and [tex]
    1/2\pi \int\limits_M {\omega \wedge d\theta } = 1/2 \pi \int_0^{ + \infty } {dr} \int_0^{2\pi } {(f(r\cos \theta ,r\sin \theta )\cos \theta + g(r\cos \theta ,r\sin \theta )\sin \theta )d\theta } [/tex] ,how to prove they are equal?
     
    Last edited: Oct 4, 2011
  2. jcsd
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