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Chapter 10: Problem 3

Let \(G\) be a topological group and \(H \subset G\) a subgroup. (a) If \(H\) is open, then so is every coset \(g H\). (b) If \(H\) is open, then \(H\) is closed.

Short Answer

Expert verified
If H is open, all cosets gH are open. If H is open, H is also closed.

Step by step solution

01

Understand the Coset Definition

Recall that a coset of a subgroup H of G is a set of the form gH = {gh : h 鈭� H} for some g in G. We want to show that if H is open, then each coset gH is also open.
02

Utilize the Group's Topological Properties

Since G is a topological group, the map \(\theta_g: G \rightarrow G \) given by \(\theta_g(x) = gx\) is continuous for each g in G. If H is open in G, consider the map \(\theta_g\).
03

Analyze the Image under Continuity

The preimage of any open set under a continuous map is open. Therefore, the image \(gH = \theta_g(H)\) of the open set H under the continuous map \(\theta_g\) is open in G.
04

Conclude Part (a)

Since we have shown that gH = \(\theta_g(H)\) is open for any g in G, we conclude that every coset of H is open in G.
05

Analyze the Closure Properties of H

We need to show that if H is open, then H is closed in G. Suppose, for the sake of contradiction, that H is not closed and the closure of H, \(\bar{H}\), has an element not in H, say x 鈭� \(\bar{H}\) \ H.
06

Use the Density Argument

Since H is open, consider the coset xH. If x 鈭� \(\bar{H}\), then for any neighborhood V of x, V 鈭� H 鈮� 鈭�. This implies xH intersects H non-trivially.
07

Contradiction and Conclusion for Part (b)

However, if x 鈭� H, then xH 鈮� H, contradicting the fact that H is a subgroup (and thus a coset of itself). This contradiction implies that \(\bar{H}\) does not have any element outside H, thus H is closed.

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Key Concepts

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

Understanding Cosets in a Topological Group
A coset is a fundamental concept when dealing with subgroups in a topological group. If you have a subgroup 饾懐 of a topological group 饾懏 and an element 饾拡 in 饾懏, the coset of 饾懐 by 饾拡 is denoted as 饾拡饾懐. This is the set of all elements formed by multiplying 饾拡 with each element of 饾懐, that is 饾拡饾懐 = {饾拡饾拤 : 饾拤 鈭� 饾懐}. Cosets help us understand how subgroups partition the group into distinct and non-overlapping collections. Understanding cosets and their properties can be beneficial when dealing with problems of open and closed structures in topological groups.
What are Open Sets?
In topology, an open set is a concept that helps describe the 'openness' or closeness of sets in topological spaces. A set 饾懠 in a topological space 饾懣 is called open if, for every point 饾樄 in 饾懠, there is a neighborhood of 饾樄 that is still entirely contained within 饾懠. Essentially, if you pick any point within the open set, you can find a 'bubble' around that point which lies completely within the open set. Open sets are the building blocks of a topology and are used to define continuous functions, convergence, and more.
Defining Closed Sets
Closed sets are another fundamental concept in topology, complementary to open sets. A set 饾懎 in a topological space 饾懣 is defined as closed if its complement, 饾懣 - 饾懎, is open in 饾挋. This means that if you take every point not in 饾懎 and form an open set, the original set 饾懎 itself will be considered closed. Closed sets contain all their limit points, meaning any sequence or net converging into the set will have its limit also within the set. Recognizing and working with closed sets can be crucial in topological proofs and concepts.
Continuous Maps in Topological Groups
A map (or function) between two topological spaces is called continuous if the preimage of every open set is also an open set. This means that small changes in the input result in small changes in the output. In topological groups, continuity plays a vital role. Consider the map 胃鈧嶐潚堚値: 饾懏 鈫� 饾懏 given by 胃鈧嶐潚堚値(饾樄) = 饾拡饾樄 for each 饾拡 in 饾懏. Since 饾懏 is a topological group, this map is continuous. Using the continuity of this map ensures that if 饾懐 is an open set, its coset 饾拡饾懐 will also be open. This is crucial in solving the problems related to the openness and closeness of subgroups in topological groups.

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Most popular questions from this chapter

(a) Let \((x, V)\) be a coordinate system around \(e \in G\) with \(x^{i}(e)=0\). Let $$ x^{i}(a b)=f^{i}\left(x^{1}(a), \ldots, x^{n}(a), x^{1}(b), \ldots, x^{n}(b)\right) $$ for \(C^{\infty}\) functions \(f^{i}\). Show that $$ D_{j} f^{i}(0)=D_{n+j} f^{i}(0)=\delta_{j}^{i} $$ (b) If \(\alpha, \beta:(-\varepsilon, \varepsilon) \rightarrow G\) are differentiable, show that $$ (\alpha \cdot \beta)^{\prime}(0)=\alpha^{\prime}(0)+\beta^{\prime}(0) $$ (c) Also deduce this result from Theorem \(14(1)\). (Not even the full strength of \((1)\) is needed; it suffices to know that \(\exp t X \exp t Y=\exp \\{t(X+Y)+O(t)\\}\). The argument of part (a) is essentially equivalent to the initial part of the deduction of \((1) .\) )

(a) Show that $$ \exp \left(\begin{array}{cc} 0 & a \\ -a & 0 \end{array}\right)=\left(\begin{array}{cc} \cos a & \sin a \\ -\sin a & \cos a \end{array}\right) $$ (b) Use the matrices \(A\) and \(B\) below to show that \(\exp (A+B)\) is not generatly equal to \((\exp A)(\exp B)\). $$ A=\left(\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right) \quad B=\left(\begin{array}{cc} 0 & 0 \\ -1 & 0 \end{array}\right) $$

For \(a \in G\), consider the map \(b \mapsto a b a^{-1}=L_{a} R_{a}^{-1}(b) .\) The map $$ \left(L_{a} R_{a}^{-1}\right)_{*}: \mathfrak{g} \rightarrow \mathfrak{g} $$ is denoted by \(\operatorname{Ad}(a)\); usually \(\operatorname{Ad}(a)(X)\) is denoted simply by \(\operatorname{Ad}(a) X\). (a) \(\operatorname{Ad}(a b)=\operatorname{Ad}(a) \circ \mathrm{Ad}(b)\). Thus we have a homomorphism Ad: \(G \rightarrow A u t(\mathrm{~g})\) where \(\operatorname{Aul}(\mathrm{g})\), the automorphism group of \(\mathrm{a}\), is the set of all non-singular linear transformations of the vector space \(\mathrm{g}\) onto itself (thus, isomorphic to \(\mathrm{GL}(n, \mathbb{R})\) if \(\mathfrak{s}\) has dimension \(n\) ). The map Ad is called the adjoint representation. (b) Show that $$ \exp (\operatorname{Ad}(a) X)=a(\exp X) a^{-1} $$(c) For \(A \in G L(n, \mathbb{R})\) and \(M \in \mathrm{g} I(n, \mathbb{R})\) show that $$ \operatorname{Ad}(A) M=A M A^{-1} $$ (It suffices to show this for \(M\) in a neighborhood of \(0 .\) ) (d) Show that $$ \operatorname{Ad}(\exp t X) Y=Y+t[X, Y]+O\left(t^{2}\right) $$ (e) Since Ad: \(G \rightarrow g\), we have the map $$ \mathrm{Ad}_{* e}: \mathrm{a}\left(=G_{e}\right) \rightarrow \begin{aligned} &\text { tangent space of } \operatorname{Aul}(\mathrm{g}) \text { at the } \\\ &\text { identity map } \mathrm{I}_{\mathrm{a}} \text { of } \mathrm{a} \text { to itself. } \end{aligned} $$ This tangent space is isomorphic to \(E n d(\beta)\), where \(E n d(q)\) is the vector space of all hinear transformations of \(\mathrm{q}\) into itself: If \(c\) is a curve in \(\operatorname{Aul}(\mathrm{a})\) with \(c(0)=1_{\mathrm{g}}\), then to regard \(c^{\prime}(0)\) as an element of \(A u t(\mathrm{~g})\), we let it operate on \(Y \in \mathrm{g}\) by $$ c^{\prime}(0)(Y)=\left.\frac{d}{d t}\right|_{t=0} c(Y) $$ (Compare with the case \(g=\mathbb{R}^{n}, \operatorname{Aul}(\mathfrak{a})=\mathrm{GL}(n, \mathbb{R}), \operatorname{End}(\mathrm{g})=n \times n\) matrices.) Use (d) to show that $$ \operatorname{Ad}_{* e}(X)(Y)=[X, Y] $$ (A proof may atso be given using the fact that \(\left.[\tilde{X}, \tilde{Y}]=L_{\tilde{x}} \tilde{Y}_{.}\right)\)The map \(Y \mapsto[X, Y]\) is denoted by ad \(X \in \operatorname{End}(\mathrm{g})\) (f) Conclude that $$ \operatorname{Ad}(\exp X)=\exp (\operatorname{ad} X)=1_{9}+\operatorname{ad} X+\frac{(\operatorname{ad} X)^{2}}{2 !}+\cdots $$ (g) Let \(G\) be a connected Lie group and \(H \subset G\) a Lie subgroup. Show that \(H\) is a normal subgroup of \(G\) if and onty if \(\mathfrak{l}=\mathcal{L}(H)\) is an ideal of \(\mathfrak{=} \mathcal{L}(G)\), that is, if and only if \([X, Y] \in \mathrm{b}\) for atl \(X \in \mathfrak{q}, Y \in \mathrm{l})\).

Let \(f: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}\) be distance preserving, with \(f(0)=0\). (a) Show that \(f\) takes straight lines to straight lines. (b) Show that \(f\) takes planes to planes. (c) Show that \(f\) is a linear transformation, and hence an element of \(O(n)\). (d) Show that any element of \(E(n)\) can be written \(A \cdot \tau\) for \(A \in O(n)\) and \(\tau\) a translation.

Let \(G\) and \(H\) be topological groups and \(\phi: U \rightarrow H\) a map on a connected open neighborhood \(U\) of \(e \in G\) such that \(\phi(a b)=\phi(a) \phi(b)\) when \(a, b, a b \in U\) (a) For each \(c \in G\), consider pairs \((V, \psi)\), where \(V \subset G\) is an open neighborhood of \(c\) with \(V \cdot V^{-1} \subset U\), and where \(\psi: V \rightarrow H\) satisfies \(\psi(a) \cdot \psi(b)^{-1}=\) \(\phi\left(a b^{-1}\right)\) for \(a, b \in V\). Define \(\left(V_{1}, \psi_{1}\right) \sim\left(V_{2}, \psi_{2}\right)\) if \(\psi_{1}=\psi_{2}\) on some smaller neighborhood of \(c\). Show that the set of all \(\widetilde{c}\) equivalence classes, for all \(c \in G\), can be made into a covering space of \(G\). (b) Conclude that if \(G\) is simply-connected, then \(\phi\) can be extended uniquet: to a homomorphism of \(G\) into \(H\).
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