Functional Analysis/Harmonic Analysis/Topological Group

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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.


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Definition 9.1.1: A topological group is a triple (G,*,τ), where (G,*) is a group, (G,τ) is a topological space, such that:

  1. The product map *:G×GG is continuous where G×G is equipped with the canonical product topology.
  2. The inverse map ι:GG is continuous.

We abuse notation slightly and write 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. ℝ, with the addition of numbers as product and the usual line topology. More generally, if V is a finite dimensional 𝕂-vector space, then V equipped with the canonical product topology and addition of vectors is a topological group.
  3. If V is a 𝕂-vector space, then the set GL(V)={T:VV| T is linear and invertible } is a topological group equipped with map composition as product and the subspace topology inherited from the vector space End(V).


The following proposition gives an equivalent definition of topological group.


Proposition 9.1.2: Let (G,*) be a group and (G,τ) a topological space with the same underlying set. Then (G,*,τ) is a topological group if and only if the map ϕ:G×GG, given by ϕ(x,y)=xy1 is continuous.

proof: First notice that we can write the map ϕ as ϕ=*id×ι:G×GG. Suppose (G,*,τ) is a topological group. Then, by definition 9.1.1, 1 and 2 , ϕ=ϕ=*(id×ι) is a composition of continuous maps, and is therefore continuous.

Conversely, assume ϕ=*(id×ι) is continuous. Since the inclusion i2:GG×G given by i2(y)=(e,y) is continuous. We can then conclude that the composition ι=ϕi2:GG is continuous. Finally, by a similar line of reason the product map *=ϕ(id×ι) is continuous. QED

Definition 9.1.3: Let (G1,*1,τ1) and (G2,*2,τ2) be topological groups. A topological group homomorphism, or simply a homomorphism between G1 and G2 is a continuous group homomorphism ϕ:G1G2. To be more precise, a homomorphism of topological groups is a ϕ:G1G2 such that:

  1. ϕ(xy)=ϕ(x)ϕ(y) for all x,yG1.
  2. ϕ is a continuous map between the topological spaces (G1,τ1) and (G1,τ2).

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 G be a topological group and H a topological group such that H considered as a pure algebraic group is a subgroup of G. We call H a topological subgroup of G if the inclusion map is continuous.


Proposition 9.1.6: Let ϕ:G1G2 be a homomorphism. Then ϕ(G1)G2 is a topological subgroup and ker(ϕ)G1 is a normal topological subgroup. Furthermore

Proof: If ϕ is a homomorphism, we know from group theory that the image ϕ(G1)G2 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 ϕ(G1) are continuous in the subspace topology and thus ϕ(G1) is a topological group. Lastly, we know from topology that the subspace topology makes the inclusion map continuous and therefore ϕ(G1) is a topological subgroup of G2. The second assertion follows from the same line of reasoning.

We use the first isomorphism theorem for purely algebraic groups to conclude that G1ker(ϕ)ϕ(G1) as groups, with isomorphism given by ϕ~(xker(ϕ))=ϕ(x). But since the map ϕ~ is the quotient map of ϕ, it is continuous and open. These properties together with surjectivity show that ϕ~ 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 Lx(y)=xy,Rx(y)=yx are homeomorphisms of G.

Proof: The product map is jointly continuous by assumption and therefore separately continuous. The inverses of these maps are the maps Lx1,Rx1 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 xG, denote the set of all neighborhoods of x in G by 𝒩G(x).

Lemma: For any yG we have 𝒩G(y)={yN| N𝒩G(e)}={Ny| N𝒩G(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 N𝒩G(e), then by lemma (ref) (translations are homeos), yN,Ny are neighborhoods of y. Similarly, if N𝒩G(y), then Ny1,y1N are neighborhoods of e such that N=yy1N=Ny1y. 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 ϕ:G1G2, be an algebraic homomorphism. In order for ϕ to be a homomorphism, it is necessary and sufficient for ϕ to be continuous at e.

Proof: Necessity is clear. To show sufficiency, let NG2 be a nonempty open set, and xN. Then x1N is a neighborhood of the neutral element e2G2, and by assumption ϕ1(x1N) is an open neighborhood of e1G1. For each yϕ1(x) we have the open set yϕ1(x1N) satisfying ϕ(yϕ1(x1N))N. We claim that:

ϕ1(N)=xN,yϕ1(x)yϕ1(x1N).

Indeed if zϕ1(N) then zzϕ1(ϕ(z)1N) since e1ϕ1(ϕ(z)1N). Consequently ϕ1(N) is an open set and ϕ is continuous. QED.

Proposition: For every x contained in the topological group G, we have 𝒩G(x)=x𝒩G(e)={xN| N𝒩G(e)} and 𝒩G(x)=𝒩G(e)x={Nx| N𝒩G(e)}.

Proof: Let xG. Then for each each N𝒩G(e), by proposition (ref) (translations are homeos) we have xN𝒩G(x). Conversely, if N𝒩G(x), then N1=:x1N𝒩G(e). But then we can write N=xN1, N1𝒩G(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 G be a topological group and 𝒩 be a class of subsets of G containing e. Then the class G𝒩:={xN| xG,N𝒩} is the basis for a topology making G a topological group if and only it satisfies the following properties:

  1. If N1,N2𝒩 and if xN1N2 then there exists N3𝒩 such that xN3andN3xN1N2


Appendices Template:Stage

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.



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