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Chapter 2: Problem 7

Let \(M\) be a nonempty smooth \(n\)-manifold with or without boundary, and suppose \(n \geq 1\). Show that the vector space \(C^{\infty}(M)\) is infinite- dimensional. [Hint: show that if \(f_{1}, \ldots, f_{k}\) are elements of \(C^{\infty}(M)\) with nonempty disjoint supports, then they are linearly independent.]

Short Answer

Expert verified
The vector space \(C^{\infty}(M)\) is infinite-dimensional due to the existence of infinitely many smooth functions with disjoint supports.

Step by step solution

01

Understanding the Problem

We need to prove that the vector space of smooth functions, denoted by \(C^{\infty}(M)\), on an \(n\)-dimensional manifold \(M\), is infinite-dimensional for \(n \geq 1\). A smooth manifold is a topological manifold with a globally defined differential structure. The hint provided suggests using functions with disjoint supports to show linear independence.
02

Define Disjoint Support

Two functions \(f, g \in C^{\infty}(M)\) are said to have disjoint supports if there are no points in \(M\) where both \(f\) and \(g\) are non-zero simultaneously. In mathematical terms, \(\text{supp}(f) \cap \text{supp}(g) = \emptyset\), where \(\text{supp}(f)\) denotes the support of \(f\).
03

Show Linear Independence

To show the linear independence of functions \(f_1, \ldots, f_k\) with nonempty disjoint supports, consider a linear combination \(c_1f_1 + c_2f_2 + \ldots + c_kf_k = 0\). At any point \(x \in M\), if \(f_i(x) eq 0\), then \(f_j(x) = 0\) for all \(j eq i\) due to the disjoint support condition. Consequently, each coefficient \(c_i\) must be zero for the linear combination to equal zero at every point \(x\). This implies that such functions are linearly independent.
04

Existence of Functions with Disjoint Support

On a manifold, it is always possible to construct infinitely many smooth functions that have disjoint supports. For instance, on \(\mathbb{R}^n\), bump functions鈥攕mooth functions that are 1 on a compact set and 0 elsewhere鈥攁re classic examples. By translating or scaling bump functions, we can ensure their supports do not overlap.
05

Conclude Infinite Dimensionality

Since for any \(k\), we can construct \(k\) linearly independent functions with disjoint supports, the space \(C^{\infty}(M)\) cannot be of finite dimension, as finite-dimensional spaces can only have a finite number of linearly independent vectors.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Smooth Manifold
A smooth manifold is a fundamental concept in modern mathematics that combines topology and calculus. To understand it comprehensively, think of it as a space that locally resembles Euclidean space, but may have a more complex global structure. It can be seen as a flexible surface with no creases or sharp corners.

  • **Local Resemblance:** By locally resembling Euclidean space, manifolds can be covered with charts. These charts are seamless pieces that lay flat and allow calculus operations.
  • **Differential Structure:** A smooth manifold comes equipped with a set of smooth transitions between charts. This means we can differentiate functions on these manifolds just like in regular calculus.
By allowing calculus to work on them, smooth manifolds are crucial in fields like physics and engineering, where they appear naturally in the study of space and surfaces.
Linearly Independent Functions
Linearly independent functions form a core part of vector space theory. In the context of functions on manifolds, linear independence means no function in a set can be written as a linear combination of the others.

When you have functions like \(f_1, f_2, \ldots, f_k\), they are considered linearly independent if:
  • The equation \(c_1f_1 + c_2f_2 + \ldots + c_kf_k = 0\) holds only when all coefficients \(c_i\) are zero.
  • This condition ensures their uniqueness in forming combinations, hence they span a larger space.
Establishing this independence using functions with disjoint supports, as suggested in the exercise, serves to demonstrate that the space is infinite, utilizing that no function can cancel the "action" of others in non-overlapping regions.
Disjoint Support
The idea of disjoint support is crucial when working with linearly independent functions. Simply put, two functions have disjoint supports if there's no overlap in the region where they are non-zero.
  • **Support of a Function:** This is the closure of the set where the function doesn't take the value zero.
  • **Disjoint Condition:** For functions \(f\) and \(g\), they have disjoint supports if their supports, denoted \(\text{supp}(f)\) and \(\text{supp}(g)\), do not intersect: \(\text{supp}(f) \cap \text{supp}(g) = \emptyset\).
This condition simplifies our work immensely since it allows each function to be "active" in its own domain, with no interference from others. This is key to establishing their linear independence in infinite-dimensional spaces.
Bump Functions
Bump functions are fascinating tools in mathematical analysis. They are smooth functions with very controlled and compact behavior, making them ideal for constructing examples in manifolds.
  • **Characteristics:** A bump function is smooth (infinitely differentiable), supported on a compact region, and zero outside a certain range.
  • **Construction:** On \(\mathbb{R}^n\), one commonly used bump function is defined by using a smooth, non-negative function that is one on a smaller compact set and drops to zero outside a larger compact set.
  • **Utility in Manifolds:** By tweaking their position and size, bump functions can be scaled and translated to work on different regions of a manifold without overlapping.
Bump functions are key in proving that an infinite number of linearly independent functions exist on smooth manifolds, as each bump function can be crafted to have disjoint support from others.

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Most popular questions from this chapter

Let \(P: \mathbb{R}^{n+1} \backslash\\{0\\} \rightarrow \mathbb{R}^{k+1} \backslash\\{0\\}\) be a smooth function, and suppose that for some \(d \in \mathbb{Z}, P(\lambda x)=\lambda^{d} P(x)\) for all \(\lambda \in \mathbb{R} \backslash\\{0\\}\) and \(x \in \mathbb{R}^{n+1} \backslash\\{0\\}\). (Such a function is said to be homogeneous of degree \(d\).) Show that the map \(\widetilde{P}: \mathbb{R} \mathbb{P}^{n} \rightarrow \mathbb{R P}^{k}\) defined by \(\widetilde{P}([x])=[P(x)]\) is well defined and smooth.

Suppose \(A\) and \(B\) are disjoint closed subsets of a smooth manifold \(M\). Show that there exists \(f \in C^{\infty}(M)\) such that \(0 \leq f(x) \leq 1\) for all \(x \in M\), \(f^{-1}(0)=A\), and \(f^{-1}(1)=B\).

For any topological space \(M\), let \(C(M)\) denote the algebra of continuous functions \(f: M \rightarrow \mathbb{R}\). Given a continuous map \(F: M \rightarrow N\), define \(F^{*}: C(N) \rightarrow C(M)\) by \(F^{*}(f)=f \circ F\). (a) Show that \(F^{*}\) is a linear map. (b) Suppose \(M\) and \(N\) are smooth manifolds. Show that \(F: M \rightarrow N\) is smooth if and only if \(F^{*}\left(C^{\infty}(N)\right) \subseteq C^{\infty}(M)\). (c) Suppose \(F: M \rightarrow N\) is a homeomorphism between smooth manifolds. Show that it is a diffeomorphism if and only if \(F^{*}\) restricts to an isomorphism from \(C^{\infty}(N)\) to \(C^{\infty}(M)\). [Remark: this result shows that in a certain sense, the entire smooth structure of \(M\) is encoded in the subset \(C^{\infty}(M) \subseteq C(M)\). In fact, some authors define a smooth structure on a topological manifold \(M\) to be a subalgebra of \(C(M)\) with certain properties; see, e.g., [Nes03].] (Used on p. 75.)

For any topological space \(M\), let \(C(M)\) denote the algebra of continuous functions \(f: M \rightarrow \mathbb{R}\). Given a continuous map \(F: M \rightarrow N\), define \(F^{*}: C(N) \rightarrow C(M)\) by \(F^{*}(f)=f \circ F\). (a) Show that \(F^{*}\) is a linear map. (b) Suppose \(M\) and \(N\) are smooth manifolds. Show that \(F: M \rightarrow N\) is smooth if and only if \(F^{*}\left(C^{\infty}(N)\right) \subseteq C^{\infty}(M)\). (c) Suppose \(F: M \rightarrow N\) is a homeomorphism between smooth manifolds. Show that it is a diffeomorphism if and only if \(F^{*}\) restricts to an isomorphism from \(C^{\infty}(N)\) to \(C^{\infty}(M)\). [Remark: this result shows that in a certain sense, the entire smooth structure of \(M\) is encoded in the subset \(C^{\infty}(M) \subseteq C(M)\). In fact, some authors define a smooth structure on a topological manifold \(M\) to be a subalgebra of \(C(M)\) with certain properties; see, e.g., [Nes03].] (Used on p. 75.)

Show that the inclusion map \(\bar{B}^{n} \hookrightarrow \mathbb{R}^{n}\) is smooth when \(\bar{B}^{n}\) is regarded as a smooth manifold with boundary.
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