Question about Stokes' Thm and Boundaries of Surfaces

In summary, Rudin's chapter on differential forms in his Principles of Mathematical Analysis provides an example of 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. However, by Stokes' Theorem we can then conclude that gamma is not the boundary of any 2-chain of class C'' in the punctured plane. To solve the problem, Rudin suggests pulling a 1-form back to the unit
  • #1
Poopsilon
294
1
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 [tex]\eta = \frac{xdy-ydx}{x^2+y^2}[/tex] on the set [tex]\mathbb{R}^2-{0}[/tex] and then parametrizes the circle [tex]\gamma(t)=(rcos(t),rsin(t))[/tex] for fixed r>0 and 0≤t≤2pi.

Now[tex]d\eta=0[/tex]yet direct computation shows that [tex]\int_{\gamma}\eta=2\pi[/tex].

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 [tex]\mathbb{R}^3-{0}[/tex] 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 [tex]\int_{\gamma}\eta=0[/tex] thus contradicting the integral given above.

Also what is up with this C'' requirement? I can't figure out why its important.
 
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  • #2
The formula for eta as it stands can be interpreted as a 1-form on R² - 0 or R³ - "z axis". If you want a 1-form on R³ - 0, you need to specify in addition what eta is to be on "z-axis" - 0. But you're going to have trouble doing that in a continuous fashion!
 
  • #3
Can't I just leave it is as so it doesn't even involve the z axis just sits inside the xy-plane?
 
  • #4
We if you're going to talk about a 1-form on R³ - 0, you better tell what its value is on each point of R³ - 0. Otherwise, that's not a 1-form on R³ - 0.

It's like if you say "consider the function f: R-->R defined by f(0)=0". That's nonsense: the formula f(0)=0 only defines a map f:{0}-->R.
 
  • #5
Ah Sorry you are completely right the interpretation I will have to have is R³ - 'z-axis' which would disallow any type of cone structure with boundary gamma. Could you tell me why the C'' condition is important for my surface, I mean C' I understand because you need to be able to take the Jacobian upon changing variables, but why C''?
 
  • #6
I don't know about the C² thing.
 
  • #7
I think that technically you would need to show that eta has no extension to a closed 1 form in R^3 - 0.

Try pulling such a form back to the unit sphere via the inclusion map and then reducing the problem to the planar case.
 

Related to Question about Stokes' Thm and Boundaries of Surfaces

1. What is Stokes' Theorem?

Stokes' Theorem is a fundamental theorem in vector calculus that relates the surface integral of a vector field over a surface to the line integral of the same vector field along the boundary of the surface.

2. How is Stokes' Theorem used in science?

Stokes' Theorem is used in various fields of science, such as physics, engineering, and mathematics, to calculate the flux of a vector field through a surface or to solve problems involving circulation and curl of vector fields.

3. What is the significance of boundaries of surfaces in Stokes' Theorem?

The boundaries of surfaces play a critical role in Stokes' Theorem as they determine the limits of integration for the line integral. The theorem states that the line integral along the boundary is equal to the surface integral over the entire surface.

4. Can Stokes' Theorem be applied to any surface?

No, Stokes' Theorem can only be applied to surfaces that are smooth and have a well-defined boundary. This means that the surface must have a continuous tangent plane at every point and the boundary must also be continuous.

5. Are there any limitations to using Stokes' Theorem?

One limitation of Stokes' Theorem is that it only applies to vector fields in three-dimensional space. It also cannot be used for surfaces with self-intersections or for surfaces that are not orientable, such as a Mobius strip.

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