Maps B2->S2 Define F+ & F-: Show Int w S2 = Int F+*(w) B2 - Int F-*(w)

[huyichen]

Define F+:B2->S2 by
F+=(u,v)=(u,v,sqrt(1-u^2-v^2))
and F-:B2->S2 by
F-=(u,v)=(u,v,-sqrt(1-u^2-v^2))

Then to show that Integral of w on S2=Integral of F+^*(w) on B2-Integral of F-^*(w), why do we need to justify the limits(As the integral on the right hand side are defined as limits as R->1 of the integrals over B_R(0))

 

[quasar987]

What do you mean by "justify the limits"?

 

[huyichen]

Justify means that we can indeed replace Integral F^*w with the limit integral as R--> 1

 

[quasar987]

To my mind,

$$\int_{S^2}\omega=\int_{B_1(0)}F_+^*\omega - \int_{B_1(0)}F_-^*\omega$$

and that's all there is to it. We're missing the equator but this has measure 0 so nobody's going to complain.

Again, I guess I just don't understand the nature of your question. You'll have to provide a lot more information on the context behind where this question is coming from.

 

[huyichen]

Actually, this is problem 14-6 from Introduction to Smooth Manifolds, Lee's book, we can not directly talk about the boundary because F+, F- is continuous but not smooth up to the equator. And the integrand is unbounded, but if we interpret in an appropriate limiting sense, then we can show Integral w on S^2=Integral F+^*w on B^2-Integral F-^*w.

 

[quasar987]

Ok, I see.

So what you have is for every R<1,

$$\int_{S^2-\rho(R)}\omega=\int_{B_R(0)}F_+^*\omega - \int_{B_R(0)}F_-^*\omega$$

You must show that

$$\lim_{R\rightarrow 1^-}\int_{S^2-\rho(R)}\omega=\int_{S^2}\omega$$

Where $\rho(R)=S^2 - F_-(B_R(0)) - F_+(B_R(0))$ is some band\ring\cylinder nbhd of the equator.

 

[huyichen]

But as F- and F+ agrees on equator, then the limit indeed hold, right?

 

[quasar987]

I argue that the form in bounded in the band, and the band has area converging to 0 as R-->1.