10.8 Topological groups

中文

There are a number of useful ways in which topological and algebraic structure can be combined. The principal requirement connecting the two types of structure is that the functions representing the algebraic laws of composition be continuous with respect to the topology imposed on the underlying set. In this section we combine group theory with topology.

A topological group is a set $G$ that is both a group and a Hausdorff topological space such that the map $\psi :G\times G\to G$ defined by $\psi (g,h)=g{h}^{-1}$ is continuous. The topological group $G$ is called discrete if the underlying topology is discrete.

The maps $\varphi :G\to G$ and $\tau :G\to G$ defined by $\varphi (g,h)=gh$ and $\tau (g)=g^{-1}$ are both continuous. For, by Theorem 10.3 the injection map $i:G\to G\times G$ defined by $i(h)={\iota }_e(h)=(e,h)$ is continuous. The map $\tau$ is therefore continuous since it is a composition of continuous maps, $\tau =\psi \circ i$ . Since $\varphi (g,h)=\psi (g,\tau (h))$ ) it follows immediately that $\varphi$ is also a continuous map.

Exercise

Show that $\tau$ is a homeomorphism of $G$ .

Exercise

If $\varphi$ and $\tau$ are continuous maps, show that $\psi$ is continuous.

Example 10.20

The additive group ${\mathbb{R}}^n$ , where the ‘product’ is vector addition

$$\varphi (\mathbf{x},\mathbf{y})=(x^1,\dots ,x^n)+(y^1,\dots ,y^n)=(x^1+y^1,\dots ,x^n+y^n)$$

and the inverse map is

$$\tau (\mathbf{x})=-\mathbf{x}=(-x^1,\dots ,-x^n),$$

is an abelian topological group with respect to the Euclidean topology on ${\mathbb{R}}^n$ . The $n$-torus $T^n={\mathbb{R}}^n/{\mathbb{Z}}^n$ is also an abelian topological group, where group composition is addition modulo 1.

Example 10.21

The set $M_n(\mathbb{R})$ of $n\times n$ real matrices has a topology homeomorphic to the Euclidean topology on ${\mathbb{R}}^{n^2}$ . The determinant map $\det$ : $M_n(\mathbb{R})\to \mathbb{R}$ is clearly continuous since $\det A$ is a polynomial function of the components of $A$. Hence the general linear group $GL(n,\mathbb{R})$ is an open subset of $M_n(\mathbb{R})$ since it is the inverse image of the open set $\dot{\mathbb{R}}=\mathbb{R}-\{0\}$ under the determinant map. If $GL(n,\mathbb{R})$ is given the induced relative topology in $M_n(\mathbb{R})$ then the map $\psi$ reads in components,

$$(\psi (\mathsf{A},\mathsf{B}){)}_{ij}=\sum _{k=1}^n{A}_{ik}(B^{-1}{)}_{kj}.$$

These are continuous functions since ${\left(B^{-1}\right)}_{ij}$ are rational polynomial functions of the components $B_{ij}$ with non-vanishing denominator $\det B$.

A subgroup $H$ of $G$ together with its relative topology is called a topological subgroup of $G$. To show that any subgroup $H$ becomes a topological subgroup with respect to the relative topology, let $U^{\prime }=H\cap U$ where $U$ is an arbitrary open subset of $G.$ . By continuity of the map $\varphi$ , for any pair of points $g,h\in H$ such that $gh\in U^{\prime }\subset U$ there exist open sets $A$ and $B$ of $G$ such that $A\times B\subset {\varphi }^{-1}(U)$ . It follows that $\varphi (A^{\prime }\times B^{\prime })\subset H\cap U$ , where $A^{\prime }=A\cap H$ , $B^{\prime }=B\cap H$ , and the continuity of ${\varphi |}_H$ is immediate. Similarly the inverse map $\tau$ is continuous when restricted to $H$. If $H$ is a closed set in $G$ , it is called a closed subgroup of $G$.

For each $g\in G$ let the left translation $L_g:G\to G$ be the map

$$L_g(h)\equiv L_{g}h=gh,$$

as defined in Example 2.25. The map $L_g$ is continuous since it is the composition of two continuous maps, $L_g=\varphi \circ {\iota }_g$ , where ${\iota }_g:G\to G\times G$ is the injection map ${\iota }_g(h)=(g,h)$

(see Theorem 10.3). It is clearly one-to-one, for $gh=g{h}^{\prime }⟹h=g^{-1}g{h}^{\prime }=h^{\prime }$ , and its inverse is the continuous map $L_{g^{-}1}$ . Hence $L_g$ is a homeomorphism. Similarly, every right translation $R_g:G\to G$ defined by $R_{g}h=hg$ is a homeomorphism of $G$ , as is the inner automorphism $C_g:G\to G$ defined by $C_{g}h=gh{g}^{-1}=L_h\circ R_{h^{-1}}(g)$

Connected component of the identity

If $G$ is a topological group we will denote by $G_0$ the connected component containing the identity element $e,$ simply referred to as the component of the identity.

Theorem 10.20 · The identity component is a closed normal subgroup

Let $G$ be a topological group, and $G_0$ the component of the identity. Then $G_0$ is a closed normal subgroup of $G$.

Proof

By Theorem 10.17 the set $G_0{g}^{-1}$ is connected, since it is a continuous image under right translation by $g^{-1}$ of a connected set. If $g\in G_0$ then $e=g{g}^{-1}\in G_0{g}^{-1}$ . Hence $G_0{g}^{-1}$ is a closed connected subset containing the identity $e,$ , and must therefore be a subset of $G_0$ . We have therefore $G_0{G}_0^{-1}\subseteq G_0$ , showing that $G_0$ is a subgroup of $G$. Since it is a connected component of $G$ it is a closed set. Thus, $G_0$ is a closed subgroup of $G$.

For any $g\in G$ , the set $g{G}_0{g}^{-1}$ is connected as it is the image of $G_0$ under the inner automorphism $h\mapsto C_g(h)$ ). Since this set contains the identity $e,$ we have $g{G}_0{g}^{-1}\subseteq G_0.$ and $G_0$ is a normal subgroup.

A topological space $X$ is said to be locally connected if every neighbourhood of every point of $X$ contains a connected open neighbourhood. A topological group $G$ is locally connected if it is locally connected at the identity $e,$ for if $V$ is a connected open neighbourhood of $e$ then $gV=L_{g}V$ is a connected open neighbourhood of any selected point $g\in G$ . If $K$ is any subset of a group $G$ , we call the smallest subgroup of $G$ that contains $K$ the subgroup generated by $K$. It is the intersection of all subgroups of $G$ that contain $K$.

Theorem 10.21 · Connected identity neighbourhoods generate the identity component

In any locally connected group $G$ the component of the identity $G_0$ is generated by any connected neighbourhood of the identity $e$.

Proof

Let $V$ be any connected neighbourhood of $e,$ , and $H$ the subgroup generated by $V$. For any $g\in H$ , the left coset $gV=L_{g}V\subset H$ is a neighbourhood of $g$ since $L_g$ is a homeomorphism. Hence $H$ is an open subset of $G$. On the other hand, if $H$ is an open subgroup of $G$ it is also closed since it is the complement in $G$ of the union of all cosets of $H$ that differ from $H$ itself. Thus $H$ is both open and closed. It is therefore the connected component of the identity, $G_0$

Let $H$ be a closed subgroup of a topological group $G$ , we can give the factor space $G/H$ the natural topology induced by the canonical projection map $\pi :g\mapsto gH$ . This is the finest topology on $G/H$ such that $\pi$ is a continuous map. In this topology a collection of cosets $U\subseteq G/H$ is open if and only if their union is an open subset of $G$. Clearly $\pi$ is an open map with respect to this topology, meaning that $\pi (V)$ is open for all open sets $V\subseteq G$

Theorem 10.22 · Connectedness from a subgroup and its quotient

If $G$ is a topological group and $H$ a closed connected subgroup such that the factor space $G/H$ is connected, then $G$ is connected.

Proof

Suppose $G$ is not connected. There then exist open sets $U$ and $V$ such that $G=U\cup V,$ , with $U\cap V=\varnothing$ . Since $\pi$ is an open map the sets $\pi (U)$ and $\pi (V)$ are open in $G/H$ and $G/H=\pi (U)\cup \pi (V)$ . But $G/H$ is connected, $\pi (U)\cap \pi (V)\ne \varnothing .$ , so there exists a coset $gH\in \pi (U)\cap \pi (V)$ . As a subset of $G$ this coset clearly meets both $U$ and $V$, and $gH=(gH\cap U)\cup (gH\cap V)$ , contradicting the fact that $gH$ is connected (since it is the image under the continuous map $L_g$ of a connected set $H$). Hence $G$ is connected.

Example 10.22

The general linear group $GL(n,\mathbb{R})$ is not connected since the determinant map $\det$ : $GL(n,\mathbb{R})\to \mathbb{R}$ has image $\dot{\mathbb{R}}=\mathbb{R}-\{0\}$ , which is a disconnected set. The component $G_0$ of the identity $I$ is the set of $n\times n$ matrices with determinant $>0$ , and the group of components is discrete

$$GL(n,\mathbb{R})/G_0\cong \{1,-1\}=Z_2.$$

Note, however, that the complex general linear group $GL(n,\mathbb{C})$ is connected, as may be surmised from the fact that the Jordan canonical form of any non-singular complex matrix can be continuously deformed to the identity matrix $I$.

The special orthogonal groups $SO(n)$ are all connected. This can be shown by induction on the dimension $n$. Evidently $SO(1)=\{1\}$ is connected. Assume that $SO(n)$ is connected. It will be shown in Chapter 19, Example 19.10, that $SO(n+1)/SO(n)$ is homeomorphic to the $n$-sphere $S^n$ . As this is a connected set (see Example 10.19) it follows from Theorem 10.22 that $SO(n+1)$ is connected. By induction, $SO(n)$ is a connected group for all $n=1,2,\dots$ . However the orthogonal groups $O(n)$ are not connected, the component of the identity being $SO(n)$ while the remaining orthogonal matrices have determinant $-1$.

Similarly, $SU(1)=\{1\}$ and $SU(n+1)/SU(n)\cong S^{2n-1}$ , from which it follows that all special unitary groups $SU(n)$ are connected. By Theorem 10.22 the unitary groups $U(n)$ are also all connected, since $U(n)/SU(n)=S^1$ is connected.

Problem

Problem 10.22

If $G_0$ is the component of the identity of a locally connected topological group $G$, the factor group $G/G_0$ is called the group of components of $G$. Show that the group of components is a discrete topological group with respect to the topology induced by the natural projection map $\pi :g\mapsto g{G}_0$