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

[kakarotyjn]

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?

 

[jvicens]

First, let's clarify what the closed Poincare dual of S is. The closed Poincare dual of S is a differential form \eta such that \int\limits_S {i^* \omega = \int\limits_M {\omega \wedge \eta } } for any differential form \omega on R^2-{0} with compact support. In other words, it is a form that "pairs" with \omega to give the integral over S.

Now, in order to prove that d\theta /2\pi is the closed Poincare dual of S, we need to show that \int_0^{ + \infty } {f(x,0)dx} = 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 }.

To do this, we can use the change of variables formula for integrals. Let u = r\cos \theta and v = r\sin \theta. Then, our integral becomes:

1/2 \pi \int_0^{ + \infty } {dr} \int_0^{2\pi } {(f(u,v)\cos \theta + g(u,v)\sin \theta )d\theta } = 1/2 \pi \int_0^{ + \infty } {dr} \int_0^{2\pi } {(f(u,v)u + g(u,v)v)d\theta }

Note that we can change the order of integration here because all of the functions involved have compact support. Now, using the fact that r = \sqrt{u^2 + v^2}, we can rewrite the integral as:

1/2 \pi \int_0^{ + \infty } {dr} \int_0^{2\pi } {(f(u,v)\sqrt{u^2 + v^2}\cos \theta + g(u,v)\sqrt{u^2 + v^2}\sin \theta )d\theta }

Then, using the fact that \cos \theta = u/\sqrt{u^2 + v^2} and \sin \theta = v/\sqrt