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Integration on chains in Spivak's calculus on manifolds

by quasar987
Tags: calculus, chains, integration, manifolds, spivak
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quasar987
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Aug17-08, 11:58 PM
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I would like to discuss this chapter with someone who has read the book.

From looking at other books, I realize that Spivak does things a little differently. He seems to be putting less structure on his chains (for instance, no mention of orientation, no 1-1 requirement and so on), and as a result, I find that things get a little weird.

For instance, the first thing I asked myself after reading the definition of the integral of a k-form over a k-chain is whether or not the result is independent of the chain. More precisely, if c and d are two k-chains with identical images, does

[tex]\int_c\omega=\int_d\omega[/tex]

as intuition demands??

I found a little guidance in answering this in the person of problem 4-25 (Independence of parametrization), but that's not entirely satisfying, because after all, coudn't it be that there is no 1-1 p such that c o p = d? If c and d are not injective for instance, the obvious p(t) := c^-1(d(t)) fails. And that det p'(x) >= 0 condition... what does it say about p? What characterize reparametrizations p with det p'(x) >= 0? (Does injectivity implies that the determinant does not chance sign? locally okay, but globally?!)
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Hurkyl
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Aug18-08, 12:14 AM
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Quote Quote by quasar987 View Post
From looking at other books, I realize that Spivak does things a little differently. He seems to be putting less structure on his chains (for instance, no mention of orientation, no 1-1 requirement and so on), and as a result, I find that things get a little weird.
Isn't a chain a formal linear combination of parametrized regions? The parametrization on a region gives it an orientation.

More precisely, if c and d are two k-chains with identical images, does

[tex]\int_c\omega=\int_d\omega[/tex]

as intuition demands??
If we set d = 2c, then wouldn't they have the same image and different integrals?
quasar987
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Aug18-08, 01:37 AM
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Oops, I wrote k-chain everywhere where I should have written k-cube.

His k-cube on A (subset of R^n) is a smooth map c:[0,1]^k-->A.

Hurkyl
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Aug18-08, 02:37 AM
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Integration on chains in Spivak's calculus on manifolds

Well, it's easy enough to construct counterexamples in the same spirit. For example, c could be a curve tracing out a circle on the Euclidean plane, and d could be another curve that traces out the same circle twice.

(Of course, I doubt your intuition ever really demanded that these be the same....)
quasar987
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Aug18-08, 11:09 AM
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Mmmh, true.

And about problem 4-25? It reads,

"Let c be a k-cube and p:[0,1]^k-->[0,1]^k a bijection with det p'(x) >= 0 everywhere. If w is a k-form, then

[tex]\int_c\omega=\int_{c\circ p}\omega[/tex]"

The proof is direct... what I'm wondering is say I want to reparametrize c with a p as in the exercise. What does a p with det p'(x) >= 0 looks like? What does det p'(x) >= 0 says about p geometrically or otherwise?
alexn49
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Nov23-09, 03:34 PM
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Hello everybody!
I am an exchange student in Canada and one of the course I choose is Calculus on Manifolds. This is a intersting as it is difficult but I deal with it! Anyway, I try to compute an aera over the chain but I can't find the right chain!

I have to compute the area on R2 of a square with a semi-circle on its top (I hope it is easy to understand). If you could give me any to start with or hint, I'd be glad because for now I don't even have an idea.

Hope to hear from you,
Alex


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