Functional Analysis/Harmonic Analysis/Topological Group

Introduction

The main algebraic structure studied in harmonic analysis is the topological group. In summary, a topological group is a group whose underlying set possesses a topology compatible with the group structure.

Preliminaries

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Definition 9.1.1: A topological group is a triple ${\displaystyle (G,*,\tau )}$ , where ${\displaystyle (G,*)}$  is a group, ${\displaystyle (G,\tau )}$  is a topological space, such that:

1. The product map ${\displaystyle *:G\times G\rightarrow G}$  is continuous where ${\displaystyle G\times G}$  is equipped with the canonical product topology.
2. The inverse map ${\displaystyle \iota :G\rightarrow G}$  is continuous.

We abuse notation slightly and write ${\displaystyle G}$  for a topological group when the product and topologies are understood from context, unless we need to be careful about a situation, for example, when talking about two different topologies on the same group.

Examples:

1. Any group equipped with the discrete topology becomes a topological group.
2. ${\displaystyle \mathbb {R} }$ , with the addition of numbers as product and the usual line topology. More generally, if ${\displaystyle V}$  is a finite dimensional ${\displaystyle \mathbb {K} }$ -vector space, then ${\displaystyle V}$  equipped with the canonical product topology and addition of vectors is a topological group.
3. If ${\displaystyle V}$  is a ${\displaystyle \mathbb {K} }$ -vector space, then the set ${\displaystyle GL(V)=\{T:V\rightarrow V|\ T}$  is linear and invertible ${\displaystyle \}}$  is a topological group equipped with map composition as product and the subspace topology inherited from the vector space ${\displaystyle {\text{End}}(V)}$ .

The following proposition gives an equivalent definition of topological group.

Proposition 9.1.2: Let ${\displaystyle (G,*)}$  be a group and ${\displaystyle (G,\tau )}$  a topological space with the same underlying set. Then ${\displaystyle (G,*,\tau )}$  is a topological group if and only if the map ${\displaystyle \phi :G\times G\rightarrow G}$ , given by ${\displaystyle \phi (x,y)=xy^{-1}}$  is continuous.

proof: First notice that we can write the map ${\displaystyle \phi }$  as ${\displaystyle \phi =*\circ id\times \iota :G\times G\rightarrow G}$ . Suppose ${\displaystyle (G,*,\tau )}$  is a topological group. Then, by definition 9.1.1, 1 and 2 , ${\displaystyle \phi =\phi =*\circ (id\times \iota )}$  is a composition of continuous maps, and is therefore continuous.

Conversely, assume ${\displaystyle \phi =*\circ (id\times \iota )}$  is continuous. Since the inclusion ${\displaystyle i_{2}:G\rightarrow G\times G}$  given by ${\displaystyle i_{2}(y)=(e,y)}$  is continuous. We can then conclude that the composition ${\displaystyle \iota =\phi \circ i_{2}:G\rightarrow G}$  is continuous. Finally, by a similar line of reason the product map ${\displaystyle *=\phi \circ (id\times \iota )}$  is continuous. QED

Definition 9.1.3: Let ${\displaystyle (G_{1},*_{1},\tau _{1})}$  and ${\displaystyle (G_{2},*_{2},\tau _{2})}$  be topological groups. A topological group homomorphism, or simply a homomorphism between ${\displaystyle G_{1}}$  and ${\displaystyle G_{2}}$  is a continuous group homomorphism ${\displaystyle \phi :G_{1}\rightarrow G_{2}}$ . To be more precise, a homomorphism of topological groups is a ${\displaystyle \phi :G_{1}\rightarrow G_{2}}$  such that:

1. ${\displaystyle \phi (xy)=\phi (x)\phi (y)}$  for all ${\displaystyle x,y\in G_{1}}$ .
2. ${\displaystyle \phi }$  is a continuous map between the topological spaces ${\displaystyle (G_{1},\tau _{1})}$  and ${\displaystyle (G_{1},\tau _{2})}$ .

An isomorphism between topological groups is a bijective continuous map whose inverse is also continuous.

As with purely algebraic groups, isomorphic topological groups are seen as being the same topological group, except for very specific contexts.

Definition 9.1.5: Let ${\displaystyle G}$  be a topological group and ${\displaystyle H}$  a topological group such that ${\displaystyle H}$  considered as a pure algebraic group is a subgroup of ${\displaystyle G}$ . We call ${\displaystyle H}$  a topological subgroup of ${\displaystyle G}$  if the inclusion map is continuous.

Proposition 9.1.6: Let ${\displaystyle \phi :G_{1}\rightarrow G_{2}}$  be a homomorphism. Then ${\displaystyle \phi (G_{1})\subset G_{2}}$  is a topological subgroup and ${\displaystyle \ker(\phi )\subset G_{1}}$  is a normal topological subgroup. Furthermore

Proof: If ${\displaystyle \phi }$  is a homomorphism, we know from group theory that the image ${\displaystyle \phi (G_{1})\subset G_{2}}$  is a subgroup. But we also recall from topology that the image of a continuous map is canonically equipped with the subspace topology. But the restriction of the product and inverse maps to ${\displaystyle \phi (G_{1})}$  are continuous in the subspace topology and thus ${\displaystyle \phi (G_{1})}$  is a topological group. Lastly, we know from topology that the subspace topology makes the inclusion map continuous and therefore ${\displaystyle \phi (G_{1})}$  is a topological subgroup of ${\displaystyle G_{2}}$ . The second assertion follows from the same line of reasoning.

We use the first isomorphism theorem for purely algebraic groups to conclude that ${\displaystyle {\frac {G_{1}}{\ker(\phi )}}\simeq \phi (G_{1})}$  as groups, with isomorphism given by ${\displaystyle {\tilde {\phi }}(x\ker(\phi ))=\phi (x)}$ . But since the map ${\displaystyle {\tilde {\phi }}}$  is the quotient map of ${\displaystyle \phi }$ , it is continuous and open. These properties together with surjectivity show that ${\displaystyle {\tilde {\phi }}}$  is an isomorphism of topological groups. QED.

Lemma: The left and right translations (ref) (def of Lx, Rx, group theory) by a given element are homeomorphisms of the group with itself. More precisely, the maps ${\displaystyle L_{x}(y)=xy,R_{x}(y)=yx}$  are homeomorphisms of ${\displaystyle G}$ .

Proof: The product map is jointly continuous by assumption and therefore separately continuous. The inverses of these maps are the maps ${\displaystyle L_{x^{-1}},R_{x^{-1}}}$  which are continuous by the same reason. QED.

Since we shall almost exclusively deal with topological groups, we shall say homomorphism instead of homomorphism of topological groups, and if we mean pure group homomorphism we say algebraic homomorphism.

Neighborhood of the neutral element are particularly important for a topological group.

Definition: For ${\displaystyle x\in G}$ , denote the set of all neighborhoods of ${\displaystyle x}$  in ${\displaystyle G}$  by ${\displaystyle {\mathcal {N}}_{G}(x)}$ .

Lemma: For any ${\displaystyle y\in G}$  we have ${\displaystyle {\mathcal {N}}_{G}(y)=\{yN|\ N\in {\mathcal {N}}_{G}(e)\}=\{Ny|\ N\in {\mathcal {N}}_{G}(e)\}}$ . In other words, the neighborhoods of a point in are the translations of the neighborhoods of the neutral element by that point.

Proof: If ${\displaystyle N\in {\mathcal {N}}_{G}(e)}$ , then by lemma (ref) (translations are homeos), ${\displaystyle yN,Ny}$  are neighborhoods of ${\displaystyle y}$ . Similarly, if ${\displaystyle N\in {\mathcal {N}}_{G}(y)}$ , then ${\displaystyle Ny^{-1},y^{-1}N}$  are neighborhoods of ${\displaystyle e}$  such that ${\displaystyle N=yy^{-1}N=Ny^{-1}y}$ . QED.

This suggests that the neighborhoods of the neutral element are sufficient for the description of the topology of the group. Indeed, some topological properties of maps, groups, etc... depend only on their behaviour at the neutral element. For example we have:

Lemma: Let ${\displaystyle \phi :G_{1}\rightarrow G_{2}}$ , be an algebraic homomorphism. In order for ${\displaystyle \phi }$  to be a homomorphism, it is necessary and sufficient for ${\displaystyle \phi }$  to be continuous at ${\displaystyle e}$ .

Proof: Necessity is clear. To show sufficiency, let ${\displaystyle N\subset G_{2}}$  be a nonempty open set, and ${\displaystyle x\in N}$ . Then ${\displaystyle x^{-1}N}$  is a neighborhood of the neutral element ${\displaystyle e_{2}\in G_{2}}$ , and by assumption ${\displaystyle \phi ^{-1}(x^{-1}N)}$  is an open neighborhood of ${\displaystyle e_{1}\in G_{1}}$ . For each ${\displaystyle y\in \phi ^{-1}(x)}$  we have the open set ${\displaystyle y\phi ^{-1}(x^{-1}N)}$  satisfying ${\displaystyle \phi (y\phi ^{-1}(x^{-1}N))\subset N}$ . We claim that:

${\displaystyle \phi ^{-1}(N)=\bigcup _{x\in N,y\in \phi ^{-1}(x)}y\phi ^{-1}(x^{-1}N)}$ .

Indeed if ${\displaystyle z\in \phi ^{-1}(N)}$  then ${\displaystyle z\in z\phi ^{-1}(\phi (z)^{-1}N)}$  since ${\displaystyle e_{1}\in \phi ^{-1}(\phi (z)^{-1}N)}$ . Consequently ${\displaystyle \phi ^{-1}(N)}$  is an open set and ${\displaystyle \phi }$  is continuous. QED.

Proposition: For every ${\displaystyle x}$  contained in the topological group ${\displaystyle G}$ , we have ${\displaystyle {\mathcal {N}}_{G}(x)=x{\mathcal {N}}_{G}(e)=\{xN|\ N\in {\mathcal {N}}_{G}(e)\}}$  and ${\displaystyle {\mathcal {N}}_{G}(x)={\mathcal {N}}_{G}(e)x=\{Nx|\ N\in {\mathcal {N}}_{G}(e)\}}$ .

Proof: Let ${\displaystyle x\in G}$ . Then for each each ${\displaystyle N\in {\mathcal {N}}_{G}(e)}$ , by proposition (ref) (translations are homeos) we have ${\displaystyle xN\in {\mathcal {N}}_{G}(x)}$ . Conversely, if ${\displaystyle N\in {\mathcal {N}}_{G}(x)}$ , then ${\displaystyle N_{1}=:x^{-1}N\in {\mathcal {N}}_{G}(e)}$ . But then we can write ${\displaystyle N=xN_{1}}$ , ${\displaystyle N_{1}\in {\mathcal {N}}_{G}(e)}$ . QED.

This lemma suggests that in order to find topologies in a group that make it into a topological group it suffices to find a "nice" base of neighborhoods for the neutral element. This is indeed true, and we have:

Theorem: Let ${\displaystyle G}$  be a topological group and ${\displaystyle {\mathcal {N}}}$  be a class of subsets of ${\displaystyle G}$  containing ${\displaystyle e}$ . Then the class ${\displaystyle G{\mathcal {N}}:=\{xN|\ x\in G,N\in {\mathcal {N}}\}}$  is the basis for a topology making ${\displaystyle G}$  a topological group if and only it satisfies the following properties:

1. If ${\displaystyle N_{1},N_{2}\in {\mathcal {N}}}$  and if ${\displaystyle x\in N_{1}\cap N_{2}}$  then there exists ${\displaystyle N_{3}\in {\mathcal {N}}}$  such that ${\displaystyle xN_{3}andN_{3}x\subset N_{1}\cap N_{2}}$

Appendices

Here, you will find a list of unsorted chapters. Some of them listed here are highly advanced topics, while others are tools to aid you on your mathematical journey. Since this is the last heading for the wikibook, the necessary book endings are also located here.