Contents

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: