Problem 22 - Calculus On Manifolds: A Modern Approach To Classical Theorems Of Advanced Calculus

Question

Let \(S\) be the set of all singular \(n\)-cubes, and \(\mathbf{Z}\) the integers. An \(n\)-chain is a function \(f: S \rightarrow \mathbf{Z}\) such that \(f(c)=0\) for all but finitely many c. Define \(f+g\) and \(n f\) by \((f+g)(c)=\) \(f(c)+g(c)\) and \(n f(c)=n \cdot f(c) .\) Show that \(f+g\) and \(n f\) are \(n\)-chains if \(f\) and \(g\) are. If \(c \in S\), let \(c\) also denote the function \(f\) such that \(f(c)=1\) and \(f\left(c^{\prime}\right)=0\) for \(c^{\prime} \neq c .\) Show that every \(n\)-chain \(f\) can be written \(a_{1} c_{1}+\ldots+a_{k} c_{k}\) for some integers \(a_{1}, \ldots, a_{k}\) and singular \(n\)-cubes \(c_{1}, \ldots, c_{k}\)

Step-by-Step Solution

Answer

f+g and n f are n-chains because their non-zero values remain finite. Any n-chain can be written as a_1 c_1 + ... + a_k c_k.

Step 1 - Understand n-chains

An n-chain is a function f: S → ℤ such that f(c)=0 for all but finitely many c. This means most elements in the set S will map to 0, making the function f finite.

Step 2 - Define Operations

For two n-chains f and g: (f+g)(c) = f(c) + g(c) and for an integer n: n f(c) = n ⋅ f(c).

Step 3 - Prove f+g is an n-chain

Since f and g are both n-chains, they are non-zero for only finitely many values. Adding two such functions will result in another function that is non-zero for only finitely many values: Let c_1, c_2, ..., c_k be the points where f is non-zero and let d_1, d_2, ..., d_m be the points where g is non-zero. The set of points where (f+g) is non-zero is a subset of {c_1, c_2, ..., c_k, d_1, d_2, ..., d_m}. This is still a finite set, thus (f+g) is an n-chain.

Step 4 - Prove n f is an n-chain

Given f is an n-chain, f is non-zero for only finitely many values. Multiplying f by any integer n does not change the number of non-zero values. Let c_1, c_2, ..., c_k be the points where f is non-zero. Then n f is non-zero only at these points as well. Thus, n f is an n-chain.

Step 5 - Expand an n-chain function as a linear combination

Any n-chain f maps only a finite number of singular n-cubes to non-zero integers. Let these singular n-cubes be c_1, c_2, ..., c_k with f(c_i) = a_i for i=1, 2, ..., k. Define c_i: S → ℤ such that c_i(c_j) = 1 if i = j, and 0 otherwise. Then f can be written as a linear combination of these c_i functions: f = a_1 c_1 + a_2 c_2 + ... + a_k c_k.

Key Concepts

Singular n-cubes

In the context of n-chains, a singular n-cube is an essential concept. It refers to a mapping of an n-dimensional cube into a space, often represented as a function. These functions are denoted as \(c: [0, 1]^n \rightarrow \mathbb{R}^m\). This idea is crucial because it sets the stage for understanding higher-dimensional geometric objects in calculus and topology. Singular n-cubes serve as the building blocks for constructing more complex structures called chains.

Finite mappings

When we talk about finite mappings in the realm of n-chains, we are referring to functions that map elements from one set to another and are non-zero for only a finite number of elements. An n-chain itself is a finite mapping from the set of singular n-cubes \(S\) to integers \(\mathbf{Z}\). This ensures that while there could be infinitely many singular n-cubes, the function values are zero for all but a finite number. This finiteness makes computations and theoretical developments much more manageable. For two n-chains f and g, we define their sum and scalar multiplication by finite mappings:

• \((f + g)(c) = f(c) + g(c)\)
• \(n f(c) = n \cdot f(c)\)

These operations maintain the finiteness of the mappings, ensuring that f+g and n f are also n-chains.

Linear Combination

Another significant concept in understanding n-chains is the notion of a linear combination. A linear combination refers to an expression made up of a set of terms (like singular n-cubes) multiplied by constants and then summed. In the context of n-chains, every n-chain can be represented as a linear combination of singular n-cubes. For any n-chain \(f\), which maps most elements to zero, it can be expanded as: \[f = a_1 c_1 + a_2 c_2 + ... + a_k c_k\] Here, \(c_1, c_2, ..., c_k\) are the singular n-cubes where f is non-zero, and \(a_1, a_2, ..., a_k\) are integers. This linear combination is finite and encapsulates the idea that while there can be infinitely many singular n-cubes, only a select few contribute to any specific n-chain. Understanding this foundation assists greatly in advanced calculus and algebraic topology.