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Chapter 1: Problem 4

(The Bohr compactification.) Show that there is a compact, abelian topological group \(\mathbb{R}\) and a continuous, in jective group homomorphism \(t: \mathbb{R} \rightarrow \widetilde{\mathbb{R}}\), such that \(t(\mathbb{R})\) is dense in \(\widetilde{\mathbb{R}}\). Hint: Let \(\mathbb{R}\) denote \(\mathbb{R}\) as a discrete group and put \(\widetilde{\mathbb{R}}=\) hom \((\mathbb{R}, \mathbb{T})\), cf. E 1.6.15. Define \(t(x)\) to be the homomorphism \(y \rightarrow \exp (\mathrm{i} x y), y \in \mathbb{R}\) for each \(x\) in \(\mathbb{R}\).

Short Answer

Expert verified
The mapping \\(t:\mathbb{R} \to \widetilde{\mathbb{R}}\\) is a continuous, injective homomorphism, and \\(t(\mathbb{R})\\) is dense in \\(\widetilde{\mathbb{R}}\\).

Step by step solution

01

Understand the Concept of Bohr Compactification

The Bohr compactification of a locally compact abelian group, in this case \(\mathbb{R}\), is a way to embed it into a compact abelian topological group. Here, the group \(\widetilde{\mathbb{R}}\) is the set of homomorphisms from the discrete group \(\mathbb{R}\) to the circle group \(\mathbb{T}\), making \(\widetilde{\mathbb{R}}\) compact.
02

Define the Homomorphism

For each \(x \in \mathbb{R}\), define the map \(t(x)\) as a homomorphism \(y \to \exp(\mathrm{i}xy)\), where \(y \in \mathbb{R}\). This maps \(x\) into \(\widetilde{\mathbb{R}}\), respecting the group operation.
03

Show Continuity of the Homomorphism

Since \(\mathbb{R}\) with the standard topology and group operation is continuous, and the mapping \(y \to \exp(\mathrm{i}xy)\) is continuous for fixed \(x\), \(t:\mathbb{R} \to \widetilde{\mathbb{R}}\) is continuous as each homomorphism is continuous with respect to pointwise convergence.
04

Prove Injectivity

To show \(t\) is injective, assume \(t(x_1) = t(x_2)\), which means \(\exp(\mathrm{i}x_1y) = \exp(\mathrm{i}x_2y)\) for all \(y \in \mathbb{R}\). Considering specific values of \(y\), we can deduce that \(x_1 = x_2\), proving injectivity.
05

Demonstrate Density of Image

To establish that \(t(\mathbb{R})\) is dense in \(\widetilde{\mathbb{R}}\), consider any non-zero homomorphism \(f \in \widetilde{\mathbb{R}}\). For \(f\) to be continuous and approach every point in the circle group \(\mathbb{T}\), we can approximate \(f\) arbitrarily closely by elements form \(t(\mathbb{R})\), thereby demonstrating density.

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Key Concepts

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

abelian topological group
An abelian topological group is a group that is both abelian, meaning the group operation is commutative (i.e., the order of operation does not matter), and a topological group, meaning that its group operation and inversion are continuous with respect to the topology of the group. In simpler terms:
  • **Abelian**: For any two elements, say \(a\) and \(b\), in the group, \(a \cdot b = b \cdot a\).
  • **Topological Group**: The way elements are combined respects the structure of the space the group is defined on, such as open or closed sets, convergence, and continuity.

In this context, \( \widetilde{\mathbb{R}} \), the Bohr compactification of \( \mathbb{R} \), forms a compact abelian topological group by considering the homomorphisms from \( \mathbb{R} \) to \( \mathbb{T} \), the circle group. This creates a structure where both the algebraic and topological properties come together beautifully.
continuous homomorphism
Continuous homomorphism is a function between two topological groups that is both a homomorphism (preserving the group structure) and continuous (preserving the topological structure).
  • **Homomorphism**: A map \( t : G \to H \) is a homomorphism if it preserves the group operation, meaning \( t(a \cdot b) = t(a) \cdot t(b) \) for all elements \(a, b\) in \(G\).
  • **Continuous**: The behavior of the map doesn't "jump" between values; it smoothly transitions and behaves predictively, respecting the topology of the group.

In the Bohr compactification exercise, the homomorphism \( t : \mathbb{R} \rightarrow \widetilde{\mathbb{R}} \) is defined by \( t(x)(y) = \exp(i x y) \). This map respects the group operation and is continuous, making the function useful for embedding structures in a compact way.
dense subset
A dense subset of a topological space is one that comes arbitrarily close to every point in that space. If a subset is dense in a space, it means:
  • Every point in the space is either in the subset or can be approximated as closely as desired by points in the subset.
  • The closure of the subset in the topology of the space is the entire space.

For the Bohr compactification, \( t(\mathbb{R}) \) being dense in \( \widetilde{\mathbb{R}} \) implies that for any point in \( \widetilde{\mathbb{R}} \), you can find a point in \( t(\mathbb{R}) \) that is as close to the original as you want. This density ensures that the small-scale, local behavior in the compact group mimics that of the larger space.
injective map
An injective map is a function between two sets that preserves distinctness. This means:
  • **One-to-One**: If \( f(a) = f(b) \), then it must be that \( a = b \).
  • This property ensures the absence of collision; no two distinct inputs produce the same output.

In the problem, showing that \( t \) is injective indicates that the continuous homomorphism \( t: \mathbb{R} \rightarrow \widetilde{\mathbb{R}} \) respects distinctness. Suppose \( t(x_1) = t(x_2) \) implies that for any \( y \in \mathbb{R} \), \( \exp(i x_1 y) = \exp(i x_2 y) \). By analyzing this equation, it becomes evident that this implies \( x_1 = x_2 \), ensuring injectivity.

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Most popular questions from this chapter

(The Sorgenfrey line.) Give the set \(\mathbb{R}\) the topology \(\tau\) for which a basis consists of the half-open intervals \([y, z[\), where \(y\) and \(z\) range over R. Show that every basis set is closed in \(\tau\). Show that \((\mathbb{R}, \tau)\) is a separable space that satisfies the first but not the second axiom of countability. Hint: If \(\rho\) is some basis for \(\tau\) and \(x \in \mathbb{R}\), then \(\rho\) must contain a set \(A\) such that \(x=\operatorname{Inf}\\{y \in A\\} .\)

(The fundamental group.) Let \((X, \tau)\) be a nonempty arcwise connected (cf. E 1.4.14) topological space, and choose a base point \(x_{0}\) in \(X\). A loop in \(X\) is a continuous function (curve) \(f:[0,1] \rightarrow X\) such that \(f(0)=f(1)=x_{0}\). On the space \(L(X)\) of loops we define a composition \(f g\) (product) by $$ f g(t)=g(2 t), \quad 0 \leq t \leq \frac{1}{2} ; \quad f g(t)=f(2 t-1), \quad \frac{1}{2} \leq t \leq 1, $$ for \(f\) and \(g\) in \(L(X)\). We define homotopy of loops, written \(f \sim g\), if there is a continuous function \(F:[0,1] \times[0,1] \rightarrow X\) such that \(F(s, 0)=F(s, 1)=x_{0}\) for every \(s\) and \(F(0, t)=f(t), F(1, t)=g(t)\) for every \(t\). Show that the set \(\pi(X)\) of equivalence classes (under homotopy) of loops is a group under the product \(\pi(f) \pi(g)=\pi(f g)\), where \(\pi: L(X) \rightarrow \pi(X)\) is the quotient map. Hint: If \(F\) is a homotopy between the loops \(f_{1}\) and \(f_{2}\), and \(G\) is a homotopy between the loops \(g_{1}\) and \(g_{2}\), set $$ \begin{array}{cll} H(s, t)=F(s, 2 t) & \text { for } & 0 \leq s \leq 1, & 0 \leq t \leq \frac{1}{2} \\ (s, t)=G(s, 2 t-1) & \text { for } & 0 \leq s \leq 1, & \frac{1}{2} \leq t \leq 1 \end{array} $$ and check that \(H\) is a homotopy between \(f_{1} g_{1}\) and \(f_{2} g_{2}\). The product in \(\pi(X)\) is therefore well-defined. If \(f \in L(X)\), define \(f^{-1}\) in \(L(X)\) by \(f^{-1}(t)=f(1-t)\) and check that \(f^{-1} f \sim e\), where \(e(t)=x_{0}\) for all \(t\). The relevant homotopy is $$ \begin{array}{lll} F(s, t)=f(2 s t) & \text { for } & 0 \leq s \leq 1, & 0 \leq t \leq \frac{1}{2} \\ F(s, t)=f(2 s(1-t)) & \text { for } & 0 \leq s \leq 1, & \frac{1}{2} \leq t \leq 1 \end{array} $$ Similarly \(f f^{-1} \sim e, f e \sim e f \sim f\), so that \(\pi(e)\) is the identity in \(\pi(X)\). Given \(f, g, h\) in \(L(X)\) we have $$ \begin{aligned} &f(g h)(t)= \begin{cases}h(4 t) & \text { for } 0 \leq t \leq \frac{1}{4} \\ g(4 t-1) & \text { for } \frac{1}{4} \leq t \leq \frac{1}{2} \\ f(2 t-1) & \text { for } & \frac{1}{2} \leq t \leq 1\end{cases} \\ &(f g) h(t)= \begin{cases}h(2 t) & \text { for } 0 \leq t \leq \frac{1}{2} \\ g(4 t-2) & \text { for } \frac{1}{2} \leq t \leq \frac{3}{4} \\ f(4 t-3) & \text { for } \frac{3}{4} \leq t \leq 1\end{cases} \end{aligned} $$ To show that \(f(g h) \sim(f g) h\), use the homotopy $$ F(s, t)= \begin{cases}h\left(4 t(1+s)^{-1}\right) & \text { for } 4 t-1 \leq s \\\ g(4 t-s-1) & \text { for } 4 t-2 \leq s \leq 4 t-1 \\ f\left((4 t-s-2)(2-s)^{-1}\right) & \text { for } s \leq 4 t-2\end{cases} $$

Let \(X\) be the subset of points \((x, y)\) in \(\mathbb{R}^{2}\) such that either \(x=y=0\) or \(x y=1\), and give \(X\) the relative topology. Let \(f: X \rightarrow \mathbb{R}\) be the restriction to \(X\) of the projection of \(\mathbb{R}^{2}\) to the \(x\)-axis. Is \(f\) a continuous map? Is \(f\) an open map?

Let \(C\) and \(D\) be compact subsets of topological spaces \(X\) and \(Y\), respectively. Show that if \(G\) is an open subset of \(X \times Y\) containing \(C \times D\), there are open sets \(A\) and \(B\) in \(X\) and \(Y\), respectively, such that \(C \subset A, D \subset B\), and \(A \times B \subset G\).

(Topology according to Hausdorff.) Suppose that to every point \(x\) in a set \(X\) we have assigned a nonempty family \(\mathscr{U}(x)\) of subsets of \(X\) satisfying the following conditions: (i) \(x \in A\) for every \(A\) in \(\mathcal{U}(x)\). (ii) If \(A \in \mathscr{U}(x)\) and \(B \in \mathscr{U}(x)\), then there is a \(C\) in \(\mathscr{U}(x)\) with \(C \subset A \cap B\). (iii) If \(A \in \mathscr{U}(x)\), then for each \(y\) in \(A\) there is a \(B\) in \(\mathscr{\ell}(y)\) with \(B \subset A\). Show that if \(\tau\) is the weakest topology containing all \(\mathscr{U}(x), x \in X\), then \(\mathscr{U}(x)\) is a neighborhood basis for \(x\) in \(\tau\) for every \(x\) in \(X\).
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