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Monday, March 11, 2024

Differential Forms, Integration, and Measures: A Conversation (with Claude 3 opus) Across Mathematical Domains

This article aims to discuss the change of variable of integration.

This blog is also an exploration about how to use AI to learn Math.

Let

(1)F:ΩΩ

be a C1 homeomorphisms.

Apply the functor

(2)HomTop(,R)

We get a ring isomorphism

(3)F:HomTop(Ω,R)HomTop(Ω,R)

Denote the integration

(4)Ω,ω:=Ωω

Here

(5)Ω,(kΩ)

The pull back

(6)FΩ,ω=Ω,F(ω)

gives you the change of variable.

Here

(7)F:(kΩ)(kΩ)

For example,

(8)Ω,fdxdy=Ωfdxdy=FΩ,fdxdy=Ω,F(fdxdy)=ΩfF(u,v)(dxF(u,v))(dyF(u,v))

Then I ask Claude: Is that correct?

Your explanation of the change of variable formula for integration using the language of category theory and functors is correct.

Let's break it down step-by-step:

  1. We start with a C1 homeomorphism F:ΩΩ between two open subsets Ω and Ω of Rn.

  2. Applying the contravariant functor HomTop(,R) from the category of topological spaces to the category of rings, we obtain a ring isomorphism F:HomTop(Ω,R)HomTop(Ω,R).

  3. We define the integration functional Ω,ω:=Ωω as an element of the dual space (kΩ), where kΩ is the space of differential k-forms on Ω.

  4. The pullback FΩ,ω=Ω,F(ω) gives the change of variable formula, where F:(kΩ)(kΩ) is the dual of the pushforward map F:kΩkΩ induced by F.

  5. For the specific example with ω=fdxdy, a 2-form on ΩR2, the change of variable formula becomes:

(9)Ω,fdxdy=Ωfdxdy=FΩ,fdxdy=Ω,F(fdxdy)=ΩfF(u,v)(dxF(u,v))(dyF(u,v))

This is indeed the correct change of variable formula, expressing the integral over Ω in terms of an integral over Ω using the pullback of the differential form fdxdy by the homeomorphism F.

But we need the absolute value of the det right?

You're absolutely right, my previous explanation was missing an important piece - the Jacobian determinant that arises from the change of variables. Let me revise the explanation to include this crucial factor.

In the standard multivariate change of variable formula for integrals, we have:

(10)Ωf(x1,,xn)dx1dxn=Ωf(F(u1,,un))|(x1,,xn)(u1,,un)|du1dun

where F:ΩΩ is a diffeomorphism (a bijective map with a bijective derivative), and |(x1,,xn)(u1,,un)| is the absolute value of the Jacobian determinant of F.

In the language of differential forms, this change of variable formula can be expressed as:

(11)Ω,fdx1dxn=Ω,fF|(x1,,xn)(u1,,un)|du1dun

or more compactly:

(12)Ω,ω=Ω,F(ω)|(x1,,xn)(u1,,un)|du1dun

where F is the pullback of differential forms by F, and |(x1,,xn)(u1,,un)|du1dun is the Jacobian n-form associated with the change of variables.

So, in the specific example you provided:

(13)Ω,fdxdy=Ωfdxdy=ΩfF(u,v)|(x,y)(u,v)|dudv

You were correct to point out that the standard change of variable formula involves the absolute value of the Jacobian determinant, which was missing from my initial explanation. Thank you for catching that!

The reason that the absolute value appears is that It should agree with the measure right?

Yes, you are correct! The change of variables formula should agree with the transformation of the measure under the map F. In fact, the presence of the absolute value of the Jacobian determinant in the change of variables formula is a consequence of how the measure transforms.

Let's denote the Lebesgue measure on Ω by μΩ and the Lebesgue measure on Ω by μΩ. The change of variables formula states that for a measurable function f:ΩR and a C1 homeomorphism F:ΩΩ, we have:

ΩfdμΩ=ΩfF|det(DF)|dμΩ

This formula is consistent with the fact that the measure μΩ on Ω is related to the measure μΩ on Ω by the pushforward of F, denoted as FμΩ:

μΩ(A)=FμΩ(A)=μΩ(F1(A))

for any measurable set AΩ.

In terms of differential forms, the change of variables formula can be written as:

Ωfω=ΩfFFω

where ω is the volume form on Ω, and Fω is the pullback of ω by F, which satisfies:

Fω=|det(DF)|dudv

Thus, the change of variables formula in the language of differential forms is consistent with the transformation of the measure under the map F.

So actually, the pull back also induced a morphism between measure space right? Could you show me more details and a deeper understanding of that?

Certainly! The pullback operation induced by a measurable map between two measurable spaces indeed creates a morphism between the corresponding measure spaces. This morphism is known as the pushforward of measures. Let's explore this concept in more detail.

Given a measurable map F:(X,A)(Y,B) between two measurable spaces (X,A) and (Y,B), where A and B are σ-algebras on X and Y respectively, we can define the pushforward of a measure μ on (X,A) as follows:

For any measurable set BB, the pushforward measure Fμ on (Y,B) is defined as:

(Fμ)(B):=μ(F1(B))

In other words, the measure of a set B in Y under the pushforward measure Fμ is equal to the measure of its preimage F1(B) in X under the original measure μ.

The pushforward operation F is a morphism between the measure spaces (X,A,μ) and (Y,B,Fμ) because it preserves the structure of the measure spaces:

  1. Fμ is a measure on (Y,B):

    • Fμ()=μ(F1())=μ()=0

    • For a countable collection of disjoint measurable sets {Bi}i=1 in B, we have:

      Fμ(i=1Bi)=μ(F1(i=1Bi))=μ(i=1F1(Bi))=i=1μ(F1(Bi))=i=1Fμ(Bi)

  2. F preserves the integration of measurable functions: For a measurable function g:(Y,B)(R,BR), we have:

    YgdFμ=X(gF)dμ

    This property is a consequence of the change of variables formula for integration.

The pushforward operation provides a way to transfer measures between measurable spaces that are related by a measurable map. It is a powerful tool in measure theory and has applications in various areas of mathematics, such as probability theory, ergodic theory, and dynamical systems.

In the context of the change of variables formula for integration, the pushforward of the Lebesgue measure under a C1 homeomorphism F is related to the pullback of differential forms by F, as we discussed earlier. This connection highlights the deep interplay between measure theory and differential geometry.

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