# Topological Spaces and Groups: Part 2: Fast Review

Reference: Chapter 1, Topological Transformation Groups by Deane Montgomerry and Leo Zippin

1.10 Sequential Convergence:

The proof of the principal theorem of this section illustrates a standard technique. It will involve choosing an infinite sequence, then an infinite subsequence, then again an infinite subsequence and so on repeating this construction a countably infinite number of times. A special form of this method is called the Cantor diagonalization procedure. To facilitate the working of this technique we shall sometimes use the following notation:

1.10.1

The letter I will denote the sequence of natural numbers 1, 2, 3, …When subsequences of I need to be chosen, they will be labelled in some systematic way: $I_{1}, I_{2}, \ldots$ or $I^{'}, I^{''}, \ldots$ or $I^{*}, I^{**}, \ldots$ and so on. Then given a sequence of elements $x_{n}$, $n \in I$, we can refer to a subsequence as : $x_{n}$, $n \in I_{1}$, or : $x_{n}, n \in I^{*}$ and so on.

DEFINITION:

Let S denote a metric space and let $K_{n}, n \in I$ be a sequence of subsets of S. The sequence $K_{n}$ is said to converge to a set K if for every $\epsilon >0$

1.10.3 $K_{n} \subset S_{\epsilon}(K)$ and $K \subset S_{\epsilon}(K_{n})$

for n sufficiently large (depending only on $\epsilon$)

If $K_{n}$ is given and if a K exists satisfying relation 1.10.3 then $\overline{K}$ also satisfies 1.10.3. If two closed sets $K^{'}$ and $K^{''}$ satisfy 1.10.3 for the same sequence $K_{n}$ then $K^{'}=K^{''}$. In the special case that the sets $K_{n}$ are single points, the set K if it exists is a point and is unique.

1.10.4 THEOREM

Every sequence of non-empty subsets of a compact metric space S has a convergent subsequence.

Proof:

Let $K_{n}, n \in I$ be an arbitrary sequence of subsets of S and let $W_{n}, n\in I$ be a basis for open sets in S. (Theorem 1.9)

Let I be called $I_{0}$ and suppose a sequence $I_{m-1}$ has been defined. Consider $W_{m} \bigcap K_{n}$, $n \in I_{m-1}$ m fixed. Then either $W_{m} \bigcap K_{n}$ is not empty for an infinite subsequence of integers $n \in I_{m-1}$ or on the contrary $W_{m} \bigcap K_{n}$ is empty for almost all $n \in I_{n-1}$. In the first of these cases define $I_{m}$ as the set of indices n in $I_{m-1}$. In the first of these cases define $I_{m}$ as the set of indices n in $I_{m-1}$ such that $W_{m} \bigcap K_{n}$ is not empty for $n \in I_{m}$. Then in all possible cases $I_{m} \subset I_{m-1}$ is uniquely defined. We now consider $I_{m}$ to be defined by induction for all $m \in I$.

We can now specify what subsequence of $K_{n}$ we may take as convergent. Let $I^{*} \subset I$ denote the diagonal sequence of the sequences $I_{m}$, that is $I^{*}$ contains the m-th element of $I_{m}$ for each m. We shall show that $K_{n}, n \in I^{*}$ is convergent. It follows from the definition of I^{*} that for each $m \in I$, if we except at most the first m integers in I^{*},

$W_{m} \bigcap K_{n}, n \in I^{*}$,

is always empty or is never empty depending on m.

We next define the set K to which the sets $K_{n}, n \in I^{*}$ will be shown to converge. Let W denote the union of those sets $W_{m}, m \in I$ for which $W_{m} \bigcap K_{n}, n \in I^{*}$ is almost empty. Let $K = S - W$ Since each $W_{m}$ forming W meets at most a finite number of the sets $K_{n}$ no finite number of these $W_{m}$ can cover S. Therefore W cannot cover S, since S is compact, and hence K is not empty. The set K is closed and therefore compact.

Let $\epsilon >0$ be given. There is a covering of K by sets $W_{k_{1}}, W_{k_{2}}, \ldots, W_{k_{s}}, k_{i} \in I$

each meeting K and each of diameter less than $\epsilon$. None of the sets $W_{k_{i}}$ can belong to W and each must intersect almost all the $K_{n}, n \in I^{*}$. Therefore for sufficiently large n, $n \in I^{*}$

$W_{k_{i}} \bigcap K_{n} \neq \Phi$

It follows that $K \subset S_{\epsilon}(K_{n})$, $n \in I^{*}$ for all n sufficiently large. Finally it can be seen that the closed set $S - S_{\epsilon}(K) \subset W$. It follows that the complement of $S_{\epsilon}(K)$ is covered by sets $W_{j_{1}}, W_{j_{2}}, \ldots, W_{j_{t}}$ each of which is an element in the union defining W. Therefore $W_{j_{k}} \bigcap K_{n}$, $n \in I^{*}$ $k=1,2, \ldots t$

is almost always empty. Therefore for sufficiently large $n \in I^{*}$

$K_{n} \subset S_{\epsilon}(K)$.

This completes the proof of the theorem. QED.

1.10.5

Let X and Y be closed subsets of a compact metric space S. Define Hausdorff metric: $d(X,Y)$ as the greatest lower bound of all $\epsilon$ such that symmetrically $X \subset S_{\epsilon}(Y), Y \subset S_{\epsilon(Y)}$

This is a metric for the collection of closed subsets of S.

THEOREM:

The set F of all closed subsets of a compact metric space S is a compact metric space in the metric defined above.

Proof.

The set F is a metric space in the metric defined above. It $A_{n}$ is in F then $A_{n}$ has a subsequence converging to a set A in the sense of convergence defined above and the set A may be assumed closed. It follows that the subsequence also converges to A in the sense of the metric of F. Hence, every sequence in F has a convergent subsequence.

Let $W_{n}$ where $n \in \mathcal{N}$ be a basis for open sets in S as in the preceding Theorem. For each n and each choice of integers $k_{1}, k_{2}, \ldots, k_{n}$ let $W(k_{1}, k_{2}, \ldots, k_{n})$ denote the union of the sets $W_{k_{1}}, W_{k_{2}}, \ldots, W_{k_{n}}$. Now in F let $W^{*}(k_{1}, \ldots, W_{k_{n}})$ consist of all the compact sets in S which belong to $W(k_{1}, \ldots, k_{n})$ and meet each $W_{k_{i}}$. This gives a countable collection of subsets of F. The proof of the preceding theorem shows that this collection is a basis for open sets in F.

Finally, let $\{ O_{n}\}$ where n=1, 2, …be a countable collection of open sets of F which cover F. We have to find an integer m such that $\bigcup_{1}^{m}O_{i}$ covers F. If no such integer existed we could find a sequence of points $x_{n}$ where $x_{n} \in F - \bigcup_{1}^{n}U_{i}$. This sequence would have to have a susbsequence converging to some point x. Since $x \in O_{m}$ for some m, it follows that infinitely many of the $x_{n}$ belong to $O_{m}$; this contradiction proves the theorem.

QED.

1.10.6

Note that separability implies that every collection of covering sets has a countable covering subcollection. It also implies that there exists a countable set of points which is everywhere dense in the space.

THEOREM.

A metric space S is compact if and only if every infinite sequence of points has a convergent subsequence.

Proof:

Suppose that every infinite subsequence of points of S has a convergent subsequence. We shall prove that S is separable. The last paragraph of the preceding section then shows that S is compact. The converse is shown in 1.10.4

For each positive integer n construct a set $P_{n}$ such that (1) every point of $P_{n}$ is at a distance at least $\frac{1}{n}$ from every other point of $P_{n}$ (2) every point of S not in $P_{n}$ is at a distance less than 1/n from some point of $P_{n}$. It is easy to see that no sequence of points in any one $P_{n}$ can be convergent and it follows that $P_{n}$ is a finite point set. Let $P= \bigcup P_{n}$. Then P is countable and every point of S is a limit point of P. For each rational $r >0$ and each point of P construct the “sphere” with that point as centre and radius r. The set of these spheres is countable and is a basis for open sets. This concludes the proof.

QED.

EXAMPLE.

Let S be a compact metric space let H be the space of real continuous functions defined on S with values in $R_{1}$. Each continuous function $f(x)$ determines a closed subset of $S \times R_{1}$ namely the graph consisting of the pairs $(x, f(x))$, $x \in S$. Hence H is a subset of a compact metric space (see examples 2 and 3 in Sec 1.9).

Example 2 of Section 1.9 reproduced below:

The set F of continuous functions defined on a compact space S with values in a metric space M becomes a metric space by defining for f, g in F $d(f,g) = lub (x \in S) [d_{M}(f(x), g(x))]$ where $d_{M}$ is the metric in M. Recall also the following here in this example: Theorem: Let S be a compact space and $\{ D_{a}\}$ a collection of closed subsets such that $\bigcap_{a}D_{a}$ is empty. Then there is some finite set $D_{a_{1}}, \ldots, D_{a_{n}}$ such that $\bigcap_{i}D_{a_{i}}$ is empty. From this theorem it follows that: A lower semi-continuous (upper semi-continuous) real valued function on a compact space has finite g.l.b. (and L.u.b) and always attains these bounds at some points of space.

Example 3 of Section 1.9 reproduced below:

If S is a compact metric space and $E_{1}$ denotes the real line, then $S \times E_{1}$ is a metrizable locally compact space with a countable basis for open sets.

Remarks: Notice that the metric defined for H in example 2 of 1.9 and the metric which it gets from $S \times H_{1}$ are topologically equivalent. This proves that H itself is a separable metric space.

QED.

1.11 TOPOLOGICAL GROUPS

Topological groups were first considered by Lie, who was concerned with groups defined by analytic relations (to be discussed later). Around 1900-1910 Hilbert and others were interested in more general topological groups. Brouwer showed that the Cantor middle third set can be made into an abelian topological group. Later Schreier and Leja gave a definition in terms of topological spaces whose theory had been developed in the intervening time.

A topological group is a topological space whose points are elements of an abstract group, the operations of the group being continuous in the topology of the space. A detailed definition containing some redundancies is as follows:

DEFINITION:

A topological group G is a space in which for $x, y \in G$ there is a unique product $xy \in G$, and

i) there is a unique identity element e in G such that $xe = ex =x$ for all $x \in G$

ii) to each $x \in G$ there is an inverse $x^{-1} \in G$ such that $xx^{-1} = x^{-1}x=e$

iii) $x(yz) = (xy)z$ for $x, y, z \in G$

iv) the function $x^{-1}$ is continuous on G and $xy$ is continuous on $G \times G$

Familiar examples are the real or complex numbers under addition with the usual topologies for $E_{1}$ and $E_{2}$ respectively, and the complex numbers of absolute value one under multiplication with their usual topology as a subset of $E_{2}$. The space of this last group is homeomorphic to the circumference of a circle.

1.11.1

Of course, properties 1, 2 and 3 define a group in the customary sense and a topological group may be thought of as a set of elements which is both an abstract group and a space, the two concepts being united through 4. Wnen a subset H of G is itself a group we shall call H a subgroup of G, but we shall understand that H is to be given the relative topology. It is easy to see that then H becomes a topological group.

If H is a subgroup of G and if x and y are points of G belonging to the closure of H, then every neighbourhood of the product xy contains points of H. For, let U be a neighbourhood of xy. Then by property 4, there exists neighbourhoods V of x and W of y such that every product of an element of V and an element of W is contained in U. We see from this that xy belongs to the closure of H. Similarly, $x^{-1}$ belongs to the closure. Thus, $\overline{H}$ is a group, and we shall call it a closed subgroup.

1.11.2

If $G_{a}$ where $a \in \{ a\}$ is a collection of topological groups and if G denotes the product space defined in 1.6, (previous blog), then G can be regarded as a group (the product of two elements of G being defined by the product of their components in each factor $G_{a}$). Because each neighbourhood of the PROD $G_{a}$ depends on only a finite number of the factors it is easy to see that G becomes a topological group. We shall call it the topological group of the factors $G_{a}$.

By way of examples, note that the product of an arbitrary number of groups each isomorphic to $C_{1}$: the group of reals modulo one (isomorphic to the complex numbers of modulus one under multiplication) is a compact topological group. The product of an arbitrary number of factors each isomorphic to the group of reals is a topological group which is locally compact if the number of factors is finite.

A finite group with the discrete topology is compact and the topological product of any collection of finite groups is therefore compact.

1.11.3

If x and y are in topological group, $x \neq y$, then as will be seen (section 1.16) we may choose a neighbourhood W of e such that

$y \notin WW^{-1}$

Hence, yW and xW are disjoined, and thus two distinct points of a topological group are in disjoined open sets. This is called the Hausdorff property (see section 1.10); it implies that a point is a closed set.

It is easy to see that if H is an abelian subgroup of a topological group G then $\overline{H}$ is also abelian. Thus, if x, y $\in \overline{H}$ and $xy \neq yx$ there are neighbourhoods $U_{1}$ of xy and $U_{2}$ of yx with $U_{1} \bigcap U_{2} = \Phi$. There exist neighbourhoods V of x and W of y such that for every $v \in V$ and $w \in W$ $vw \in U_{1}$ and $wv \in U_{2}$. However, if v, w $\in H$ then $wv = vw$ and we are led to a contradiction proving that the closure of H is abelian.

QED.

1.11.4

Important examples of topological groups are given below:

EXAMPLE 1.

The sets $M_{n}(R)$ and $M_{n}(C)$ of all $n \times n$ matrices of real and complex elements under addition with the distance of $A = (a_{ij})$, $B = (b_{ij})$ defined by

$d(A, B) = max_{i,j}|a_{ij} - b_{ij}|$

The spaces of these two groups are homeomorphic to $E_{n^{2}}$ and $E_{2n^{2}}$. They are in fact the sets of real or complex vectors with $n^{2}$ co-ordinates, and hence are vector spaces as well as groups. Another example is the set H of continuous real valued functions on a metric space under addition.

EXAMPLE 2.

The sets of non-singular real or complex $n \times n$ matrices $GL(n,R), GL(n,C)$ under multiplication; these are subsets of $M_{n}(R)$ and $M_{n}(C)$ respectively and are given the induced topology. They are open subsets and are therefore locally compact and locally euclidean. (see 1.27 later)

EXAMPLE 3.

Let S be a compact metric space and let G be the group of all homeomorphisms of S onto itself topologized as a subspace of the space of continuous maps of S into itself. (Section 1.9 previous blog, example 2)

1.12 ISOMORPHISM of TOPOLOGICAL GROUPS

The spaces associated with two topological groups may be homeomorphic but the groups essentially different; for example, one abelian and the other not.

EXAMPLE 1.

The matrices $\left |\begin{array}{cc}a & 0 \\ 0 & b \end{array}\right |$ where a and b are real numbers under addition. This is an abelian group with $E_{2}$ as space.

EXAMPLE 2.

The matrices $\left| \begin{array}{cc} e^{a} & b \\ 0 & e^{-a} \end{array} \right |$ where a and b are real under multiplication. This is a non-abelian group with $E_{2}$ as space.

If we give the space in this example (or example 1) the discrete topology we obtain a new topological group with the same algebraic structure.

EXAMPLE 3.

In the additive group of integers, for each pair of integers h and k, $k \neq 0$ let the set $\{ h \pm nk\}$ where n=0, 1,2, …, be called an open set and let the collection of all these sets be taken as a basis for open sets.

EXAMPLE 4.

Introduce a metric into the additive gorup of integers, depending on the prime number p, defined thus:

$d(a,b) = \frac{1}{p^{n}}$

if $a \neq b$ and $p^{n}$ is the highest power of p which is a factor of $a-b$.

EXAMPLE 5.

Let G be the integers under addition with any set called open if it is the complement of a finite set (or is the whole space or the null set). Algebraically G is a group and it is also a space. However, it is not a topological group because addition is now not simultaneously continuous. It is true however that addition is continuous in each variable separately.

For some types of group spaces separate continuity implies simultaneous continuity. It is not known whether this is true for a compact Hausdorff group space.

DEFINITION.

Two topological groups will be called isomorphic if there is a one-one correspondence between their elements which is a group isomorphism (preserves products and inverses) and a space homeomorphism (preserves open sets).

An isomorphic map of G onto G is called an automorphism.

In examples (3) and (4) the abstract group structure of the additve group of integers is embodied in infinitely many non-isomorphic topological groups.

1.13 SET PRODUCTS

If G is a group $A \subset G$ let $A^{-1}$ denote the inverse set $\{ a^{-1}\}$ where $a \in A$. Clearly, $(A^{-1})^{-1}=A$. If $B \subset G$ let AB denote the set $\{ ab\}$ where $a \in A, b \in B$. It is understood that the product set is empty if either factor is empty. We shall write $AA = A^{2}$ and so on. It can be seen that $(AB)C = A (BC)$ and $(AB)^{-1} = B^{-1}A^{-1}$. The set $AA^{-1}$ satisfies

$(AA^{-1})^{-1}=AA^{-1}$

that is, it is symmetric. Similarly, the set intersection $A \bigcap A^{-1}$ is symmetric. The intersection of symmetric sets is symmetric.

A set H in G is called invariant if $gH = Hg$ for every $g \in G$ equivalently if $gHg^{-1}=H$.

THEOREM 1.

Let G be a topological group and let $A \subset G$ be an open set. Then $A^{-1}$ is open.

Proof:

Let $a^{-1}$ be in A. By the continuity of the inverse there exists an open set B containing a such that $b \in B$ implies $b^{-1} \in A$. This means that $B^{-1} \subset A$ and therefore $B \subset A^{-1}$. Thus $A^{-1}$ is a union of open sets and is open.

COROLLARY:

The map $x \rightarrow x^{-1}$ is a homeomorphism.

LEMMA.

Let G be a topological group, A an open subset, b an element. Then Ab and bA are open.

Proof:

Let $a \in A$ and let $c=ab$. Then $a = cb^{-1}$. Because “a” regarded as a product is continuous in c there must exist an open set C containing c such that if $c^{'} \in C$ then $c^{'}b^{-1} \in A$. But then $c^{'} \in Ab$ and it follows that $C \subset Ab$. Therefore Ab is a union of open sets and is open. The proof that bA is open is similar.

QED.

COROLLARY.

For each $a \in G$, the left and right translations : $a \rightarrow ax$ and $x \rightarrow xa$ are homeomorphisms.

THEOREM 2.

Let G be a topological group and let A and B be subsets. If A or B is open then $AB$ is open.

Proof.

Since AB is a union of sets of the form Ab, $b \in B$, it is open if A is open. SimilarlyAB is a union of sets aB, $a \in A$ and is open if B is open.

COROLLARY.

Let A be a closed subset of a topological group. Then Ab and bA are closed.

Proof:

This is true because left and right translations by the constant b are homeomorphisms of G onto G.

Now, a function $f(x)$ taking a group $G_{1}$ into another group $G_{2}$ will be called a homomorphism if

*) $f(x)f(y) = f(xy)$ where $x, y \in G_{1}$

When $G_{1}$ and $G_{2}$ are topological we shall ordinarily require that f be continuous. The most useful case is where f is open as well as continuous. In many situations (see 1.26.4 and 2.13 in later blogs) continuity implies openness but this is not true in general. The set of elements going into e is a subgroup and if f is continuous it is a closed subgroup (a point in a topological group is a closed set). This is the kernel of the homomorphism.

EXAMPLE.

Let $V_{1}$ be the additive group of real numbers and G be a topological group. A continuous homomorphism $h(t)$ of $V_{1}$ into G is called a one-parameter group in G. If h is defined only on an open interval around zero satisyfying the definition of homomorphism so far as it has meaning, then $h(t)$ is called a local one-parameter group in G. If $h(t)$ is a one-parameter group, the image of $V_{1}$ may consist of e alone and then $h(t)$ is a trivial one-parameter group. If this is not the case and if for some $t_{1} \neq 0$ and $h(t_{1})=e$, then the image of $V_{1}$ is homeomorphic to a circumference. In case $h(t)=e$ only for $t =0$ the image of $V_{1}$ is a one-one image of the line which may be homeomorphism of the line or a very complicated imbedding of the line. To illustrate this let G be a torus which we obtain from the plane vector group $V_{2}$ by reducing mod one in both the x and y directions. In $V_{2}$ any line through the origin is a subgroup isomorphic to $V_{1}$ and after reduction the line $y=ax$ is mapped onto the torus G thus giving a one-parameter group in G. If a is rational the image is a simple closed curve but if a is irrational the image is everywhere dense on the torus.

1.14 PRODUCTS of Closed Sets

If A and B are subgroups of a group G, AB is not necessarily a subgroup. However if A is an invariant subgroup (that is, $g^{-1}Ag = A$ where $g \in G$) and B is a subgroup then $AB$ is a subgroup.

The product of closed subsets, even if they are subgroups, need not be closed. As an example let G be the additive group of real numbers, $H_{1}$ the subgroup of integers $\{ \pm n\}$ and $H_{2}$ the subgroup $\{ \pm n\sqrt{2}\}$. The product $H_{1}H_{2}$ is countable and a subgroup but it is not a closed set.

It will be shown later that if A is a compact invariant subgroup and B is a closed subgroup then AB is a closed subgroup (corollary of 2.1 in later blogs). (Remark: I think in I N Herstein’s language of Topics in Algebra, an “invariant subgroup” is a normal subgroup. Kindly correct me if I am mistaken).

1.15 Neighbourhoods of the identity

Let G be a topological group and U an open subset containing the identity e. We showed in 1.13 that $xU$ is open and clearly $x \in xU$. Conversely, if $x \in O$, O is open, then $U = x^{-1}O$ is an open set containing e.

If a collection of open sets $\{ U_{a}\}$ is a basis for open sets at e then every open set of G is a union of open sets of the form $x_{a}U_{a}$, where $x_{a} \in G$, $U_{a} \in \{ U_{a}\}$ and the topology of G is completely determined by the basis at e. In particular the collection $\{ xU_{a}\}$ is a basis for open sets at x so also is $\{ U_{a}x\}$

If U is a neighbourhood of e, $U^{-1}$ is a neighbourhood of e and $U \bigcap U^{-1}$ is a symmetric neighbourhood of e.

THEOREM

Let G be a topological group and U a neighbourhood of e. There exists a symmetric neighbourhood W of e such that $W^{2} \subset U$.

Proof:

Since $e.e = e$ and the product is simultaneously continuous in x and y there must exist neighbourhoods $V_{1}$ and $V_{2}$ of e such that $V_{1}V_{2} \subset U$. Define $W = V_{1} \bigcap V_{1}^{-1} \bigcap V_{2} \bigcap V_{2}^{-1}$. Then $W^{2} \subset U$ and this completes the proof.

QED.

COROLLARY.

Let G be a topological group. If $x \neq e$ there exists a neighbourhood W of e such that $W \bigcap xW$ is empty.

Proof.

Since G is a topological space there is a neighbourhood $V_{1}$ of e not containing x or there is a neighbourhood $V_{2}$ of e not containing x. In either case there is a neighbourhood of e not containing x. Let W be a symmetric neighbourhood of e, $W^{2} \subset U$. If $W \bigcap xW$ were not vacuous there would exist $w, w^{'} \in W$ with $w^{'}=w$. But this gives $x = w^{'}w^{-1} \in W^{2} \subset U$ which is false.

QED.

1.15.1

If G is a topological group then a set S of open neighbourhoods $\{ V\}$ which forms a basis at the identity e has the following properties:

(a) the intersection of all V in S is $\{ e\}$

(b) the intersection of two sets of S contains a third set of S

(c) given U in S there is a V in S such that $VV^{-1} \subset U$

(d) if U is in S and $a \in U$ then there is a V in S such that $Va \subset U$

(e) If U is in S and a is in G there is a V in S such that $aVa^{-1} \subset U$.

Conversely, a system of subsets of an abstract group having these properties may be used to determine a topology in G as will be formulated in the following theorem, the proof of which is contained in main part in remarks already made.

THEOREM.

Let G be an abstract group in which there is given a system S of subsets satisfying (a) to (e) above. If open sets in G are defined as unions of sets of the form Va, $a \in G$ then G becomes topological with S a basis for open sets at e. This is the only topology making G a topological group with S a basis at e.

1.16 COSET Spaces

Let G be a group and H a subgroup. The sets $xH$ and $yH$ where $x, y \in G$ either coincide or are mutually exclusive; and $xH = yH$ if and only if $x^{-1}y \in H$. Each set xH is called a coset of H, more specifically a left coset. Right cosets Hx, Hy will be used infrequently. We use the notation G/H for the set of all left cosets. When G is a (topological space) G/H will be made into a space (see below 1.16.2) but we speak of it as a space in the present case also, although it carries no topology at present.

If H is an invariant subgroup, that is if $x^{-1}Hx=H$ equivalently if $xH=Hx$ for any $x \in G$, then

$xHyH = xyH$ where $x, y \in G$

is a true equation in sets of elements of G. In this case, the coset space becomes a group, the factor group $G/H$.

1.16.1

DEFINITION

By the natural map T of a group G onto the coset space G/H, H being a subgroup of G, we mean the map

$T : x \rightarrow xH$, where $x \in G, xH \in G/H$

For any subset $U \subset G$ we have

$T^{-1}(T(U)) = UH \subset G$

Let G be a topological group, H a subgroup. It is useful to topologize the coset space and to do this so that the natural map T is continuous. It will become clear as we proceed that unless the group H is a closed subgroup of G, it will not be possible in general to have T continuous and G/H a topological space; for this reason only the case where H is closed will be considered.

1.16.2

DEFINITION.

Let G be a topological group and let H be a closed subgroup of G, that is a subgroup which is a closed set. By an open set in G/H we mean a set whose inverse under the natural map I is an open set in G/H.

THEOREM.

With open sets defined as above, G/H is a topological space and the map T is continuous and open. If $xH \neq yH$, there exist neighbourhoods $W_{1}$ and $W_{2}$ of xH and yH respectively such that $W_{1} \bigcap W_{2}$ is empty.

Proof:

Let U denote an arbitrary open set in G. Then U/H is open (1.13) and is the inverse of $T(U)$. Since T(U) is open in G/H, T is an open map.

Let $x, y \in G$ and $xH \neq yH$. Then $x \in yH$ and yH is closed because H is closed. There exists a neighbourhood U of e such that $Ux \bigcap yH$ is empty. Let W be open, $e \in W$ and $W^{*} \subset U$ (Theorem in 1.15 above). Then if \$latex $WxH \bigcap WyH$ is not empty we can find $w, w_{1} \in W$ and $h, h_{1} \in H$ so that $w_{1}xh = wy h_{1}$. But this leads to $w^{-1}w_{1}x = yh_{1}h^{-1}$ and implies that Ux meets yH. Therefore $WxH \bigcap WyH$ is empty. The sets Wx and Wy are open. Therefore $W_{1}=T(Wx)$ and $W_{2} = T(Wy)$ are open in G/H; $xH \in W_{1}$ $yH \in W_{2}$ and $W_{1} \bigcap W_{2}$ is empty. This is the main part of the proof and depends on the fact that H is closed. We have proved more than condition (4) of 1.1 (The T_{0} separation axiom) The remaining three conditions of 1.1 are easy to verify. The fact that T is a continuous map is stated in the definition of open set in G/H. The fact that T is open was proved in the last paragraph.

QED.

COROLLARY.

If G is a topological group, $x, y \in G$ where $x \neq y$, then there exist neighbourhoods $W_{1}$ of x and $W_{2}$ of y such that $W_{1} \bigcap W_{2}$ is empty.

Proof:

To see this it is only necessary to take $H=e$. A space in which every pair of distinct points belong to mutually exclusive open sets is called a Hausdorff space. Therefore it has been shown that a topological group G and a coset space G/H, H closed in G, are Hausdorff spaces.

COROLLARY.

Suppose that G is a topological group and H a closed invariant subgroup. Then with the customary definition of product : $(xH)(yH) = xyH$,, G/H becomes a topological group. The natural map of G onto G/H is a continuous and open homomorphism.

1.17 A FAMILY OF NEIGHBOURHOODS

Suppose that we are given a topological group G and a sequence of neighbourhoods of e: $Q_{0}, Q_{1}, \ldots$ By repeated of the following Theorem 1.15: ( Let G be a topological group and U a neighbourhood of e. There exists a symmetric neighbourhood W of e such that $W^{2} \subset U$) .We can choose a sequence of symmetric open neighbourhoods of e : $U_{0}, U_{1}, \ldots$ with $U_{0}=Q_{0}$ such that

(1) $U_{n+1}^{2} \subset U_{n}\bigcap Q_{n}$ where n=0, 1, ….

In this section we shall show how to imbed the sets $U_{n}$ in a larger family of neighbourhoods possessing a multiplicative property which generalizes (1). We shall use this family in the next section to construct a real and non-constant function which is continuous on G. In 1.22 we shall use a similar family in order to construct a metric in a metrizable group. We remark in passing that in groups which do not satisfy the first countability axiom the set: $\bigcap U_{n}$ may be of considerable interest (to be discussed in a future blog section 2.6); it is a closed group (if $x, y \in \bigcap U_{n}$ then for every n, $xy \in U_{n+1}^{2} \subset U_{n}$).

Now, for each dyadic rational $r = \frac{k}{2^{n}}$, for n=0, 1, …and $k=1, \ldots, 2^{n}$ we define an open neighbourhood $V_{r}$ of e as follows:

(2) $V_{1/2^{n}}=U_{n}$ for all n

and then using (3) and (4) alternately by induction on k.

(3) $V_{2k/2^{n+1}}=V_{k/2^{n}}$

(4) $V_{(2k+1)/2^{n+1}} = V_{1/2^{n+1}}V_{k/2^{n}}$

Each $V_{n}$ depends on the dyadic rational r, and not on the particular representation by $k/2^{n}$. The entire family has the property:

(5) $V_{1/2^{n}}V_{m/2^{n}} \subset V_{m+1/2^{n}}$ where $m+1 \leq 2^{n}$

For $m=2k$, (5) is an immediate consequence of (3) and (4). For $m=2k+1$ the left side of (5) becomes: $V_{1/2^{n}}(V_{1/2^{n}}V_{k/2^{n-1}}) \subset V_{1/2^{n-1}}V_{k/2^{n-1}}$

The right side of (5) becomes $V_{(k+1)/2^{n-1}}$. This sets up an induction on n and since (5) holds for $n=1$, we have proved that (5) is true for all n. It follows from (5) and also more directly that:

(6) $V_{r} \subset V_{r^{'}}$ if $r

1.18 COMPLETE REGULARITY

THEOREM.

Suppose that G is a topological group and that F is a closed subset of G not containing e. Then one can define on G a continuous real function f, $0 \leq f(x) \leq 1$, $f(e) =0$, $f(x)=1$ if $x \in F$.

Proof:

In virtue of the property described in the theorem, G is called a completely regular space at the point e. The theorem is due to Pontrjagin.

Set $Q_{a}=G-F$ and setting $Q_{n}=Q_{0}$ for every n, construct a family of sets $V_{r}$ as in the preceding section.

Define f(x), $x \in G$ as follows:

(1) f(x)=0 if $x \in V_{r}$ for every r

(2) f(x)=1 if $x \in V_{1}$

and in all other cases

(3) $f(x)=lub_{r}$ where $\{ r \leq 1, x \in V_{r}\}$

It is clear that e belongs to every $V_{r}$ and that $F = G - V_{1}$ and does not meet $V_{1}$ so that there remains only to prove that f is continuous; let $\epsilon>0$ be given and let n be a positive integer such that $1/2^{n}<\epsilon$.

Now suppose that $f(x) <1$ at some point $x \in G$. Then there is a pair of integers m and k such that $k >n$ (same n as above), $m < 2^{n}$ and (interpreting $V_{0}$, if it occurs, as the null set),

$x \in V_{m/2^{k}}-V_{m-1/2^{k}}$

Let y be an arbitrary element in the neighbourhood $V_{1/2^{k}}x$. Then,

$y \in V_{1/2^{k}}V_{m/2^{k}} \subset V_{(m+1)/2^{k}}$

By the choice of y, $yx^{-1} \in V_{1/2^{k}}$; therefore $xy^{-1} \in V_{1/2^{k}}$ and $x \in V_{1/2^{k}}y$. It follows from this that $y$ cannot belong to $V_{(m-2)/2^{k}}$ and this shows that

$(m-2)/2^{k} \leq f(y) \leq (m+1)2^{k}$

Concerning x we know that

$(m-1)/2^{2^{k}} \leq f(x) \leq m/2^{k}$ and we may conclude from both inequalities that

$|f(x)-f(y)| \leq 2/2^{k} \leq 1/2^{n}<\epsilon$

Suppose next that $f(y)=1$ and choose $k>n$ as before. Let y be an arbitrary element in $V_{1/2^{k}}x$. Now y cannot belong to $V_{m/2^{k}}$ with $m < 2^{k}-2$ without implying that $f(x)<1$. It follows that

$1-2/2^{k} \leq f(y) \leq 1$

and again we get $|f(x) - f(y)| \leq e$

This concludes the proof and $f(x)$ is continuous on G.

QED.

COROLLARY.

A topological group is completely regular at every point.

1.19 HOMOGENEOUS SPACES

THEOREM.

Let G be a topological group, H a closed subgroup. Each element of G determines a homeomorphism of the coset space G/H onto itself, and G becomes a group of homeomorphisms of this space (topological space); furthermore G is transitive on the space: that is, each point may be carried to any other by an element of G.

Proof

Let $a \in G$. Associate to a the mapping $T_{a}: xH \rightarrow axH$.

This is a one-one transformation of G/H onto itself with $T_{a^{-1}}$ as inverse. These transformations are open and each transformation is a homeomorphism.

Since $T_{b}T_{a}$ is given by

$T_{b}T_{a}(xH) = T_{b}(axH)=baxH = T_{ba}(xH)$.

the association of $a \in G$ and $T_{a}$ makes a group of transformations of G/H. It is clear that xH is carried to yH by $T_{a}$ with $a = yx^{-1}$. This completes the proof.

QED.

A space is called HOMOGENEOUS when a group of homeomorphisms is transitive on it. We have shown that G/H is homogeneous with G being the transitive group of homeomorphisms. This implies that the unit segment $R_{1}$, for example, cannot be the underlying space of a group or even a coset space G/H since an end point of $R_{1}$ cannot go to an interior point to a homeomorphism of $R_{1}$.

It follows from the simultaneous continuity of $ax \in G$ in the factors a and x that the image point $axH$ of $xH$ under $T_{a}$ is continuous simultaneously in the counter point xH and the element $T_{a}$ of G. This makes G an instance of what is called a topological transformation group of a space M which will be defined below.

However, we shall be principally concerned with transformation groups which are locally compact and separable, acting on spaces which are topologically locally euclidean.

1.20 LOCAL GROUPS

An open neighbourhood of the identity of a topological group when it is regarded as a space in the relative topology has some of the properties of a group. There will usually be pairs of elements for which no product element exists in the neighbourhood. A structure of this kind is called a local group and will be defined below. Local groups often arise in a natural way, especially in the case of analytic group (Lie groups of transformations) and they have been intensively studied in that form

DEFINITION.

A space G is called a local group if a product xy is defined as an element in G for some pairs x, y in G and the following conditions are satisfied:

i) there is a unique element e in G such that ex and xe are defined for each x in G and $ex = xe=x$.

ii) If x, y are in G and xy exists then there is a neighbourhood U of x and a neighbourhood V of y such that if $x^{'} \in U$ and $y^{'} \in V$ then $x^{'}y^{'}$ exists. The product xy is continuous wherever defined.

iii) The associative law holds whenever it has meaning.

iv) If $ab=e$ then $ba=e$. An element b satisfying this relation is called an inverse and is denoted by $a^{-1}$. We assume that $a^{-1}$ is unique and continuous where defined and that if it exists for an element a it exists for all elements in some neighbourhood of a. Note that $a^{-1}$ always exists in some neighbourhood of e. In fact, there exists a symmetric open neighbourhood U of e such that $U^{2}$ is defined.

The above definition is somewhat redundant.

EXAMPLE.

Any neighbourhood O of the identity oa topological group is a local group if the neighbourhood is open.

We shall call two local groups isomorphic if there is a homeomorphism between their elements which carries inverse to inverse and product to product in so far as they are defined. However, in some applications, it is natural to regard two local groups as equivalent if they belong to the same local equivalence class, that is, a neighbourhood of e in one is isomorphic to a neighbourhood of e in the other. In this book an isomorphism and preserves group operations so far as they are defined.

LEMMA.

Let G be a local group with U the symmetric open neighbourhood of e described in the definition. Given any neighbourhood V of e, $V \subset U$, there is a symmetric neighbourhood W of e, $W^{3} \subset V$. The product sets AB, BC, (AB)C, A(BC) exist for $A, B, C \subset W$ and $A(BC) = (AB)C$. The set AB is open if either A or B is open. The sets A, bA, and Ab are homeomorphic for $b \in W$. Any two points of W have homeomorphic neighbourhoods.

The proof is omitted, the details being as in 1.13, 1.14 and 1.15.

Regards,

Nalin Pithwa

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