group theory

Contents

Idea

A topological group is a topological space with a continuous group structure: a group object internal to the category Top of topological spaces and continuous functions between them.

Definition

Definition

A topological group is

1. a group, hence

1. a set $G$,

2. a neutral element $e \in G$,

3. $(-)\cdot (-) \;\colon\; G \times G \to G$,

4. a function $(-)^{-1} \;\colon\; G \to G$ such that $g \cdot g^{-1} = e = g^{-1} \cdot g$ for all $g \in G$;

2. a topology $\tau_G \subset P(G)$ giving $G$ the structure of a topological space

such that the operations $(-)^{-1}$ and $(-)\cdot (-)$ are continuous functions (the latter with respect to the product topology).

Sometimes topological groups are understood to be Hausdorff (e.g. Bredon 72, p. 1).

Properties

General

Lemma

(open subgroups of topological groups are closed)

Every open subgroup $H \subset G$ of a topological group is closed, hence a closed subgroup.

Proof

The set of $H$-cosets is a cover of $G$ by disjoint open subsets. One of these cosets is $H$ itself and hence it is the complement of the union of the other cosets, hence the complement of an open subspace, hence closed.

Proposition

(connected locally compact topological groups are sigma-compact)

Every connected locally compact topological group is sigma-compact.

Every locally compact topological group is paracompact.

Proof

By assumption of local compactness, there exists a compact neighbourhood $C_e \subset G$ of the neutral element. We may assume without restriction of generality that with $g \in C_e$ any element, then also the inverse element $g^{-1} \in C_e$.

For if this is not the case, then we may enlarge $C_e$ by including its inverse elements, and the result is still a compact neighbourhood of the neutral element: Since taking inverse elements $(-)^{-1} \colon G \to G$ is a continuous function, and since continuous images of compact spaces are compact, it follows that also the set of inverse elements to elements in $C_e$ is compact, and the union of two compact subspaces is still compact (obviously, otherwise see this prop).

Now for $n \in \mathbb{N}$, write $C_e^n \subset G$ for the image of $\underset{k \in \{1, \cdots n\}}{\prod} C_e \subset \underset{k \in \{1, \cdots, n\}}{\prod} G$ under the iterated group product operation $\underset{k \in \{1, \cdots, n\}}{\prod} G \longrightarrow G$.

Then

$H \coloneqq \underset {n \in \mathbb{N}} {\cup} C_e^n \;\subset\; G$

is clearly a topological subgroup of $G$.

Observe that each $C_e^n$ is compact. This is because $\underset{k \in \{1, \cdots, n\}}{\prod}C_e$ is compact by the Tychonoff theorem, and since continuous images of compact spaces are compact. Thus

$H = \underset{n \in \mathbb{N}}{\cup} C_e^n$

is a countable union of compact subspaces, making it sigma-compact. Since locally compact and sigma-compact spaces are paracompact, this implies that $H$ is paracompact.

Observe also that the subgroup $H$ is open, because it contains with the interior of $C_e$ a non-empty open subset $Int(C_e) \subset H$ and we may hence write $H$ as a union of open subsets

$H = \underset{h \in H}{\cup} Int(C_e) \cdot h \,.$

Finally, as indicated in the proof of Lemma , the cosets of the open subgroup $H$ are all open and partition $G$ as a disjoint union space (coproduct in Top) of these open cosets. From this we may draw the following conclusions.

• In the particular case where $G$ is connected, then there is just one such coset, namely $H$ itself. The argument above thus shows that a connected locally compact topological group is $\sigma$-compact and (by local compactness) also paracompact.

• In the general case, all the cosets are homeomorphic to $H$ which we have just shown to be a paracompact group. Thus $G$ is a disjoint union space of paracompact spaces. This is again paracompact (by this prop.).

Uniform structure

A topological group $G$ carries two canonical uniformities: a right and left uniformity. The right uniformity consists of entourages $\sim_{l, U}$ where $x \sim_{l, U} y$ if $x y^{-1} \in U$; here $U$ ranges over neighborhoods of the identity that are symmetric: $g \in U \Leftrightarrow g^{-1} \in U$. The left uniformity similarly consists of entourages $\sim_{r, U}$ where $x \sim_{r, U} y$ if $x^{-1} y \in U$. The uniform topology for either coincides with the topology of $G$.

Obviously when $G$ is commutative, the left and right uniformities coincide. They also coincide if $G$ is compact Hausdorff, since in that case there is only one uniformity whose uniform topology reproduces the given topology.

Let $G$, $H$ be topological groups, and equip each with their left uniformities. Let $f: G \to H$ be a group homomorphism.

Proposition

The following are equivalent:

• The map $f$ is continuous at a point of $G$;

• The map $f$ is continuous;

• The map $f$ is uniformly continuous.

Proof

Suppose $f$ is continuous at $g \in G$. Since neighborhoods of a point $x$ are $x$-translates of neighborhoods of the identity $e$, continuity at $g$ means that for all neighborhoods $V$ of $e \in H$, there exists a neighborhood $U$ of $e \in G$ such that

$f(g U) \subseteq f(g) V$

Since $f$ is a homomorphism, it follows immediately from cancellation that $f(U) \subseteq V$. Therefore, for every neighborhood $V$ of $e \in H$, there exists a neighborhood $U$ of $e \in G$ such that

$x y^{-1} \in U \Rightarrow f(x) f(y)^{-1} = f(x y^{-1}) \in V$

in other words such that $x \sim_U y \Rightarrow f(x) \sim_V f(y)$. Hence $f$ is uniformly continuous with respect to the right uniformity. By similar reasoning, $f$ is uniformly continuous with respect to the right uniformity.

Unitary representation on Hilbert spaces

Definition

A unitary representation $R$ of a topological group $G$ in a Hilbert space $\mathcal{H}$ is a continuous group homomorphism

$R \colon G \to \mathcal{U}(\mathcal{H})$

where $\mathcal{U}(\mathcal{H})$ is the group of unitary operators on $\mathcal{H}$ with respect to the strong topology.

Remark

Here $\mathcal{U}(\mathcal{H})$ is a complete, metrizable topological group in the strong topology, see (Schottenloher, prop. 3.11).

Remark

In physics, when a classical mechanical system is symmetric, i.e. invariant in a proper sense, with respect to the action of a topological group $G$, then an unitary representation of $G$ is sometimes called a quantization of $G$. See at geometric quantization and orbit method for more on this.

Why the strong topology is used

The reason that in the definition of a unitary representation, the strong operator topology on $\mathcal{U}(\mathcal{H})$ is used and not the norm topology, is that only few homomorphisms turn out to be continuous in the norm topology.

Example: let $G$ be a compact Lie group and $L^2(G)$ be the Hilbert space of square integrable measurable functions with respect to its Haar measure. The right regular representation of $G$ on $L^2(G)$ is defined as

$R: G \to \mathcal{U}(L^2(G))$
$g \mapsto (R_g: f(x) \mapsto f(x g))$

and this will generally not be continuous in the norm topology, but is always continuous in the strong topology.

Protomodularity

Proposition

The category TopGrp of topological groups and continuous group homomorphisms between them is a protomodular category.

A proof is spelled out by Todd Trimble here on MO.

Examples

The following monograph is not particulary about group representations, but some content of this page is based on it: