Poopsilon
Jun5-11, 10:55 PM
So I'm going over Rudin's chapter on differential forms in his Principles of Mathematical Analysis and I'm looking at Example 10.36 which gives the 1 form \eta = \frac{xdy-ydx}{x^2+y^2} on the set \mathbb{R}^2-{0} and then parametrizes the circle \gamma(t)=(rcos(t),rsin(t)) for fixed r>0 and 0≤t≤2pi.
Nowd\eta=0yet direct computation shows that \int_{\gamma}\eta=2\pi.
Thus by Stokes' Theorem we can then conclude that gamma is not the boundary of any 2-chain of class C'' in the punctured plane.
Now I understand that because the origin is not included we can't just use the disk of radius r as our 2-surface with boundary equal to gamma (I think this is related to the Cauchy Residue Theorem) but what if we extended ourselves to \mathbb{R}^3-{0} than we could parametrize some sort of cone-like 2-surface with boundary equal to gamma which we could probably make C'' which would then by Stokes' Theorem force the integral to be \int_{\gamma}\eta=0 thus contradicting the integral given above.
Also what is up with this C'' requirement? I can't figure out why its important.
Nowd\eta=0yet direct computation shows that \int_{\gamma}\eta=2\pi.
Thus by Stokes' Theorem we can then conclude that gamma is not the boundary of any 2-chain of class C'' in the punctured plane.
Now I understand that because the origin is not included we can't just use the disk of radius r as our 2-surface with boundary equal to gamma (I think this is related to the Cauchy Residue Theorem) but what if we extended ourselves to \mathbb{R}^3-{0} than we could parametrize some sort of cone-like 2-surface with boundary equal to gamma which we could probably make C'' which would then by Stokes' Theorem force the integral to be \int_{\gamma}\eta=0 thus contradicting the integral given above.
Also what is up with this C'' requirement? I can't figure out why its important.