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Introduction

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Definition:

Proposition:

Proof: QED.

Theorem:

Proof: QED.

Lemma:

Proof: QED.

Notation

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.

Notation And Previous Definitions

Definition: A subset $X$ of a group $G$ is called symmetric if $X=X^{-1}$.

Definition: Let $f:X\rightarrow Y$ be a map between two sets. For any subset $Z\subset X$, we define $f(Z)=\{f(z)|\ z\in Z\}\subset Y$. In particular, if $A,B$ are subsets of a group $G$ we have:

1. $AB=\{ab|\ a\in A,b\in B\}$
2. $A^{-1}=\{a^{-1}|\ a\in A\}$
3. If $A=\{a\}$ is a singleton, we denote $AB=:aB$ and $BA=:Ba$.

Preliminaries

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

1. The product map $*:G\times G\rightarrow G$ is (jointly) continuous where $G\times G$ is equipped with the canonical product topology.
2. The inversion map $\iota :G\rightarrow G$ 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. $\mathbb {R}$, with the addition of numbers as product and the usual line topology. More generally, if $V$ is a finite dimensional $\mathbb {K}$-vector space, then $V$ equipped with the canonical product topology and addition of vectors is a topological group.
3. If $V$ is a $\mathbb {K}$-vector space, then the set $GL(V)=\{T:V\rightarrow V|\ T$ is linear and invertible $\}$ is a topological group equipped with map composition as product and the subspace topology inherited from the vector space ${\text{End}}(V)$.
4. If $(G,*,\tau )$ is a topological group, the opposite or reversed topological group is the group $G^{op}=(G,*^{op},\tau )$, where $x*^{op}y=y*x$.
5. Let $S$ be a nonempty set and consider $Bij(S)$ the set of all bijections from $S$ to $S$. If equipped with the product $f*g(x)=f(g(x))$, i.e., the composition of maps, then $Bij(S)$ becomes a group.

From now on, when we use a topological property or charateristic to describe a group, one should understand that we are talking about a topological group. Therefore instead of saying connected topological group or locally compact topological group we say connected group or locally compact group.

The following proposition gives an equivalent definition of topological group.

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

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

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

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

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

An isomorphism between topological groups is a bijective continuous map whose inverse is also continuous. In other words, it is a group homomorphism which is also a homeomorphism of topological spaces

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

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 $L_{x}(y)=xy,R_{x}(y)=yx$ 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 $L_{x^{-1}},R_{x^{-1}}$ which are continuous by the same reason. QED.

Corollary: Any topological group $G$ is a homogeneous topological space.

Proof: If $x,y\in G$, then by lemma (ref) (translations are homeos) the map $L_{xy^{-1}}:G\rightarrow G$ is an homeomorphism of $G$ sending $y$ to $x$.

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. Until the end of this chapter fix a topological group $G$.

Subgroups

Proposition: If $H$ is an algebraic subgroup of $G$ equipped with the subspace topology, then $H$ is a topological group.

Proof: This follows from the fact that the product and inversion, which are continuous maps, of $G$ restricted to $H$ remain continuous QED.

Definition: Let $G$ be a topological group and $H$ an algebraic subgroup, which is also a topological group by (ref) . In this case $H$ is called a subgroup of $G$. We denote $H\leq G$ if $H$ is a subgroup of $G$.

Recall from topology (ref) that the inclusion of subspaces are continuous, and from abstract algebra that the inclusion of groups (ref) is a group homomorphism. Combining this information we conclude that the inclusion of a subgroup is a homomorphism of topological groups.

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

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

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

Proposition: Suppose $G$ is a $T_{1}$ topological space. If $H\leq G$, then ${\overline {H}}\leq G$. Furthermore:

1. $H$ is abelian if and only if ${\overline {H}}$ is abelian.
2. If $H$ is normal, then ${\overline {H}}$ is normal.

Proof: Indeed, let $x,y\in {\overline {H}}$, and let $N_{1}\in {\mathcal {N}}_{G}(xy),N_{2}\in {\mathcal {N}}_{G}(x^{-1})$ be neighborhoods. Then $x^{-1}N_{1}y^{-1}\in {\mathcal {N}}_{G}(e)$. Using proposition (ref) we find a symmetric $M_{1}\in {\mathcal {N}}_{G}(e)$ such that $M_{1}^{2}\subset x^{-1}N_{1}y^{-1}$. Then $xM_{1}$ is a neighborhood of $x$ and therefore there exists $h_{1}\in xM_{1}\cap H$ since $x\in {\overline {H}}$. Similarly we find $h_{2}\in M_{1}y\cap H$. But then $h_{1}h_{2}\in xMMy\subset xx^{-1}N_{1}y^{-1}y=N_{1}\cap$. Thus $xy\in {\overline {H}}$.

If M_2 is a symmetric neighborhood of $e$ such that $M_{2}^{2}\subset xN_{2}$, then $M_{2}x\in {\mathcal {N}}_{G}(x)$. Similarly, there exists $h\in H\cap M_{2}x$ which means that $h^{-1}\in x^{-1}M_{2}^{-1}=x^{-1}M_{2}\subset x^{-1}xN_{2}=N_{2}$. Thus $x^{-1}\in {\overline {H}}$.

To prove (1.) let $H$ be abelian, and let $x,y\in {\overline {H}}$. Take $(x_{\alpha })_{A},(y_{\beta })_{B}$ nets in $H$ converging to $x,y$ respectively. Then the net $(x_{\alpha }y_{\beta })_{A\times B}=(y_{\beta }x_{\alpha })_{A\times B}$ converges to $xy$ and to $yx$. Since $G$ is $T_{1}$, $xy=yx$. The reverse implication is clear.

To prove (2.) let $x\in G$, $h\in {\overline {H}}$, and let $(h_{\alpha })_{A}$ be a net in $H$ converging to $h$. For every $\alpha \in A$ $xh_{\alpha }x^{-1}=h'_{\alpha }\in H$. But $\lim _{\alpha \in A}xh_{\alpha }x^{-1}=xhx^{-1}=\lim _{\alpha \in A}h'_{\alpha }\in {\overline {H}}$. QED.

Lemma: Let ${\mathcal {H}}$ be any collection of subgroups of a given group $G$. Then the intersection of all the subgroups in ${\mathcal {H}}$ is a subgroup of $G$. Symbolically, $\bigcap _{H\in {\mathcal {H}}}H\leq G$.

Proof: Since all $H\in {\mathcal {H}}$ are subgroups, the neutral element is contained in all of them, and therefore in their intersection. If $x,y\in H$ for each $H\in {\mathcal {H}}$ then $xy\in H$ and $x^{-1}\in H$ for each $H\in {\mathcal {H}}$, and therefore $x^{-1},\ xy\in \bigcap _{H\in {\mathcal {H}}}H.$ QED.

Proposition: Given any subset $S\subset G$, there exists a unique subgroup $\langle S\rangle \leq G$ containing $S$ which is minimal with this property. This unique subgroup is called the subgroup generated by $S$.

Proof: To show existence we use the previous lemma on the collection ${\mathcal {H}}(S):=\{H\leq G|\ S\subset H\}$ of all subgroups containing $S$, which is nonempty since $G\in {\mathcal {H}}$. To prove uniqueness, denote $\langle S\rangle :=\bigcap _{H\in {\mathcal {H}}(S)}H$. If $H_{0}$ is another subgroup containing $S$ with the minimality property then, by the minimality property $\langle S\rangle \subset H_{0}$ and $H_{0}\subset \langle S\rangle$. QED

Topological Quotient Spaces

Throughout this subsection, fix $H\leq G$ a closed subgroup, and denote by $\tau$ the topology of $G$. We shall study the quotient space $G/H$ of left equivalence classes modulo $H$. The canonical topology on this set is the quotient topology, i.e., the topology $\tau _{G/H}=\{X\in G/H|\ \cup _{xH\in X}xH\in \tau \}$.

Proposition: A subset $X\subset G/H$ is open if and only if $\pi ^{-1}(X)\subset G$ is open. The quotient map $P_{H}:G\rightarrow G/H;P_{H}(x)=xH$ is open and closed.

HA-TOPGP-TQS-001

Proof: QED.

Proposition: If $H$ is a closed subgroup of $G$, then the space $G/H$ is Hausdorff (satisfies the $T_{2}$ axiom of separability).

HA-TOPGP-TQS-002

Proof: Let $xH\neq yH\in G/H$. Since $H$ is closed, $G\setminus yH$ is an open neighborhood of $x$. QED.

From this proposition we may conclude that if $G$ possesses a topological property $P$ preserved on continuous images, then $G/H$ also has this property. For example, if $G$ is compact or connected, then so is $G/H$. Unfortunately, local compactness is not preserved by continuous images. However, the quotient map is also open, so we have:

Proposition: If $G$ is locally compact, then so is $G/H$.

HA-TOPGP-TQS-

Proof: If $xH\in G/H$, let $K$ be a compact neighborhood of $x$. Since $P_{H}$ is open and continuous, $P_{H}(K)$ is a compact neighborhood of $XH$. QED.

Proposition: '

HA-TOPGP-TQS-

Proof: QED.

Proposition: '

HA-TOPGP-TQS-

Proof: QED.

Proposition: '

HA-TOPGP-TQS-

Proof: QED.

Neighborhoods of the Neutral Element

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

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

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

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

Proposition: For any $A\subset G$, we have

${\overline {A}}=\bigcap _{N\in {\mathcal {N}}_{G}(e)}NA$

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

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

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

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

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

Proof: Let $x\in G$. Then for each each $N\in {\mathcal {N}}_{G}(e)$, by proposition (ref) (translations are homeos) we have $xN\in {\mathcal {N}}_{G}(x)$. Conversely, if $N\in {\mathcal {N}}_{G}(x)$, then $N_{1}=:x^{-1}N\in {\mathcal {N}}_{G}(e)$. But then we can write $N=xN_{1}$, $N_{1}\in {\mathcal {N}}_{G}(e)$. 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 $G$ be a topological group and ${\mathcal {N}}$ be a class of subsets of $G$ containing $e$. Then the class $G{\mathcal {N}}:=\{xN|\ x\in G,N\in {\mathcal {N}}\}$ is the basis for a topology making $G$ a topological group if and only it satisfies the following properties:

1. For every $N_{1},N_{2}\in {\mathcal {N}}$ there exists $N_{3}\in {\mathcal {N}}$ such that $N_{3}\subset N_{1}\cap N_{2}$
2. For every $N\in {\mathcal {N}}$ there exists $N_{0}\in {\mathcal {N}}$ such that $N_{0}^{2}\subset N$.
3. For every $N\in {\mathcal {N}}$ there exists $N_{0}\in {\mathcal {N}}$ such that $N_{0}^{-1}\subset N$.
4. For every $N\in {\mathcal {N}}$ and every $x\in N$ there exists $N_{0}\in {\mathcal {N}}$ such that $N_{0}x\subset N$.
5. For every $N\in {\mathcal {N}}$ and every $x\in G$ there exists $N_{0}\in {\mathcal {N}}$ such that $xN_{0}x^{-1}\subset N$.

Furthermore, this topology is $T_{1}$ (ref) (def of t1 space) if and only if $\{e\}=\bigcap _{\mathcal {N}}N$.

Proof: QED.

Proposition: Every topological group possesses a base of neighborhoods of the neutral element composed of symmetric open sets.

Proof: QED.

Proposition Every topological group is a Tychonoff space, i.e., completely regular and Hausdorff

Proof QED.

Proposition: If $N$ is a neighborhood of a subset $C$ of $G$, then there exists $M\in {\mathcal {N}}_{G}(e)$ such that $MC\subset N$.

Proof: QED.

Functions on Topological Groups

Proposition: Let $G$ be a locally connected group. Then the connected component $G_{0}$ containing the neutral element is a subgroup of $G$.

Proof: By assumption, $G_{0}$ contains the neutral element.

Proposition: An open subgroup is also closed.

Operations With Subsets

Proposition: Let $A,B$ be subsets of $G$.

1. If $A$ is open, then $AB$ is open
2. If $A,B$ are connected, then $A^{-1}$ and $AB$ are connected
3. If $A$ is closed and $B$ is compact, then both $AB$ and $BA$ are closed.
4. If $A,B$ are compact, then $AB$ is compact.

Proof: Exercise (ref). QED.

Proposition: If $H$ is a closed subgroup of the LCG $G$, then the topological space $G/H$ is locally compact.

Proof: QED.

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.

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Actions of Topological Groups

Definition: An action of a topological group $G$ is a pair $(\pi ,X)$ where $X$ is a topological space (ref) and $\pi :G\times X\rightarrow X$ is a continuous group action (ref). In other words it satisfies, for all $x,y\in G$ and $z\in X$  :

1. $\pi (x,\pi (y,z))=\pi (xy,z)$ .
2. $\pi (e,z)=z$.
3. The map $\pi$ is continuous.

_____________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Locally Compact Groups

In this section we define the most well-known class of topological groups, namely locally compact groups. This class includes compact groups which in turn includes all finite groups, finite-dimensional Lie groups, etc.

Preliminaries

Definition 9.2.1: A locally compact group is a topological group whose underlying topological space is locally compact.

Examples:

1. All compact, and therefore all finite groups are locally compact.
2. A discrete group is always locally compact.
3. Any finite-dimensional vector space is a locally compact group (equipped with addition).

The Hilbert space $l^{2}(\mathbb {N} )$ is not locally compact in the norm topology.

Proposition: An open subgroup of a locally compact group is always closed. A closed subgroup of a locally compact group is locally compact.

Proof: Indeed, let $H\subset G$ be an open subgroup of $G$. Choose a set $\{y_{j}\in G|\ j\in J\}$, one $y_{j}$ for each class in $G/H$, but choosing $e$ for the class of $H$. We then have the disjoint union $G=\sqcup _{j\in J}y_{j}H$. Since left multiplication by a given element $y_{j}$ is a homeomorphism between $H$ and $y_{j}H$, we have that each such set is open in $G$. Therefore the complement of $H=eH$ is open in $G$ and therefore $H$ is also closed.

If now $H\subset G$ be an closed subgroup of $G$, let $x\in H$. There exists a compact neighborhood $X$ of $x$ in $G$. But then the intersection $X\cap H$ is a compact neighborhood of $x$ in $H$. QED.

Combining the statements in the last proposition we conclude that an open subgroup of a locally compact group is also locally compact.

Proposition: Let $G$ be a topological group. In order for $G$ to be locally compact it is necessary and sufficient that the neutral element $e$ possesses a compact neighborhood.

Proof: Indeed, if $X$ is a compact neighborhood of $e$, then $Xx$ is a compact neighborhood of $x$ for any $x\in X$, since $Xx=R_{x}(X)$ is the image of a continuous map by lemma (ref) (left and right multiplication maps). QED.

Examples

1) Sn and symmetries of polygons

2) $GL(n,K)$ and its subgroups

3) $a-$adic numbers

4) Isometries of a metric space.

5) Complex functions on a topological group, finite, N, Z, R, C, Q, Qua, GL,

6)

1) Finite Groups.

The canonical topology on finite groups is the discrete topology. Consider, then, the symmetric groups $S_{n}={\text{Perm}}(X_{n})$, where $X_{n}$ is any set with $n$ elements.

2) The General Linear Group.

Let $\mathbb {K}$ be a field and $n\in \mathbb {N}$. Consider the topological vector space ${\text{End}}(\mathbb {K} ^{n})$ of linear maps from $\mathbb {K}$ into itself with the operator norm topology, i.e., for each $T\in {\text{End}}(\mathbb {K} ^{n})$, $||T||:={\text{sup}}_{||x||=1}||Tx||$.

Definition: The general linear group of $\mathbb {K}$ and $n$ is the group of invertible linear operators $GL(n,\mathbb {K} ):=\{T\in {\text{End}}(\mathbb {K} ^{n})|\ \ker(T)\neq 0\ {\text{and}}\ T(\mathbb {K} ^{n})=\mathbb {K} ^{n}\}$.

Proposition: Equipped with the subspace topology inherited from the topological vector space ${\text{End}}(\mathbb {K} ^{n})$, $GL(n,\mathbb {K} )$ is a topological group.

Proof: QED.

3)

Exercises

1) If A is open and B is any set, then AB is open

2) find an example in which A and B are closed but AB is not closed.

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.

• List of Mathematical Symbols
• List of Theorems
• References
• Index