Problem 20 - Calculus On Manifolds: A Modern Approach To Classical Theorems Of Advanced Calculus

Question

Let \(f: U \rightarrow \mathbf{R}^{n}\) be a differentiable function with a differentiable inverse \(f^{-1}: f(U) \rightarrow \mathbf{R}^{n}\). If every closed form on \(U\) is exact, show that the same is true for \(f(U)\). Hint: If \(d \omega=0\) and \(f^{*} \omega=d \eta\) consider \(\left(f^{-1}\right) * \eta\) \end{aligned} $$

Step-by-Step Solution

Answer

If every closed form on \(U\) is exact, the same is true for \(f(U)\) by using the differentiable inverse and applying the exterior derivative concepts on both domains.

Step 1: Understand the Problem and Definitions

The problem is about proving that every closed differential form on the set \(f(U)\) is exact given that every closed form on \(U\) is exact. A differential form \(\textbackslash omega\) is exact if there exists another form \(\textbackslash eta\) such that \(\textbackslash omega = d \textbackslash eta\). A form is closed if its exterior derivative is zero, i.e., \(d \textbackslash omega = 0\).

Step 2: Verify the Given Conditions

Consider a differential form \(\textbackslash omega\) defined on \(f(U)\) such that it is closed, i.e., \(d\textbackslash omega = 0\). By the hint, we also have \(f^{*} \textbackslash omega = d \textbackslash eta\) for some \(\textbackslash eta\) defined on \(U\).

Step 3: Use the Pullback of the Inverse Function

In this step, consider the pullback of \(\textbackslash eta\) by the inverse function \(f^{-1}\). Calculate \(\textbackslash eta\) by taking the pullback of \(f^{-1}\), i.e., \(\textbackslash eta = (f^{-1})^{*} \textbackslash eta\).

Step 4: Apply Exactness on the Original Domain

Since \(f^{*} \textbackslash omega = d \textbackslash eta\) and every closed form on \(U\) is exact, \(\textbackslash eta\) must exist on \(U\). Define \(\textbackslash eta_1 = f^{-1}* \textbackslash eta\). Since \(\textbackslash eta_1\) is exact on \(U\), there exists some form \(\textbackslash xi\) such that \(\textbackslash eta_1 = d \textbackslash xi\).

Step 5: Show Exactness on the Image Domain \(f(U)\)

By applying the mapping of \(f\), \(f^{*} \textbackslash omega = d \textbackslash eta\) holds on \(U\), implying that \(\textbacklash omega\) is indeed the exterior derivative of \(\textbacklash eta\). Consequently, \(\textbackslash omega\) is exact on the image domain \(f(U)\).

Key Concepts

Closed Forms

A closed form is a differential form \(\omega\) that satisfies \(d\omega = 0\). This simply means that the exterior derivative of \(\omega\) is zero. Exterior derivatives are a way to generalize the concepts of divergence, gradient, and curl from vector calculus into higher dimensions. For example:

\(\omega\) can be any k-form such that when you apply the exterior derivative operator, you end up with zero. In simpler terms, it means \(\omega\) has no 'boundary,' it doesn't change further in its own dimension.

Here are some key points about closed forms:

• If a form is closed, its exterior derivative is zero (\(d\omega = 0\)).
• Not all closed forms are exact, but exact forms are always closed.
• Closed forms are essential in various fields like differential geometry and topology.

Understanding closed forms helps in visualizing higher-dimensional calculus and paves the way for concepts like exact forms.

Exact Forms

An exact form is a differential form \(\omega\) that can be expressed as the exterior derivative of another form \(\eta\). Mathematically, \(\omega = d\eta\). Simply put, the form \(\omega\) 'comes from' the form \(\eta\). This implies a deep connection between \(\omega\) and \(\eta\).

Here are some things to understand about exact forms:

• An exact form is always closed (\