Reference: A Course on Topological Groups by K Chandrashekharan, Hindustan Book Agency, available Amazon Indiia.
Chapter 1. Topological Preliminaries:
Section 1.1 Topological Space:
A topology T in a set X is a class of subsets of X (called open sets) satisfying the following axioms:
(1) the union of any number of open sets is open
(2) the intersection of any two (or finitely many) open sets is open
(3) X and the emtpy set of null set are both open.
A topological space is a set X with a topology T in X. T is called the trivial topology if latex \phi$) and is called the discrete topology if T= (A: ). (PS: the trivial topology has too many limit points; it is like a lumped mass) whereas the discrete topology has no limit points, yet this latter topology is quite useful ! )
Let X be a topological space and . We define an induced topology in A as follows: the class of open sets in A is the class of sets of the form where U runs through all open sets in X.
In a topological space X, a closed set is any set whose complement is an open set.
The closure of any set E is the intersection of all closed sets containing E. By axioms (1) and (3) above, the closure is a closed set.
A neighbourhood of is any open set containing x. Let and be two topological spaces and f a mapping
A mapping is an assignment to each of an element f(x) in . Then f is continuous if and only if for every open set , the set is an open set in . Here, .
Let . We say that f is continuous at x, if to every neighbourhood U of f(x) there exists a neighbourhood V of x such that . We say that f is continuous, if f is continuous at every point of . This definition is equivalent to the preceding one.
The mapping f is said to be open if for every open set , the set is an open set in .
Let and be two continuous mappings. Then the composite mapping is continuous.
Examples
(i) If X is a discrete topological space, Y is a topological space, then every mapping is continuous. (PS: this is vacuously true).
(ii) If X is any topological space, and Y a set wtih the trivial topology, then is continuous.
(iv) is the identity mapping.
The topological product of two topological spaces and is the topological space , whose set is the cartesian product of and , namely, with the open sets being all unions of sets of the form where is an open set in and is an open set in . Sets of the form form a basis of open sets in
Note that if and are the projections defined by and respectively then and are continuous mappings where and .
COVERING. A family where of subsets of a set X is a covering of X, if . That is to say, each point of X belongs to at least one . If further, X is a topological space, and each is an open subset of X, we say that is an open covering of X.
A covering is a finite covering if I is a finite set.
COMPACT SETS. A topological space X is said to be compact if every covering of X contains a finite subcover. That is to say, where is open implies that there exists all belonging to I such that . A subset is compact if A is compact in the induced topology.
Examples:
(i) The is not compact.
(ii) Let X be a topological space and / If A consists of finitely many elements of X, then A is compact. (PS: A finite point set has no limit points).
(iii) Let a, b belong to reals. Let . Then the set is compact.
An equivalent definition of compactness is the following:
If is a family of closed sets such that every finite subfamily has a non-empty intersection, then . (If is a family of closed sets with the finite intersection property, then the intersection of the whole class is non-empty).
A closed subset of a compact set is compact. A continuous image of a compact set is compact. (If is compact, is Hausforff, and is continuous, then is closed in ).
A product of compact spaces is compact (Tychonoff theorem).
A space X is locally compact if for every , there is a neighbourhood of x such that the closure of is compact.
Every compact space is locally compact but not vice-versa.
HOMEOMORPHISMS.
Let X and Y be topological spaces and f a mapping . We say that f is a homeomorphism if f is one-to-one, onto and both f and are continuous.
Two topological spaces X and Y are said to be homeomorphic if there exists a homeomorphism f from X to Y.
Examples:
(i) and for .
(ii) (a) and . X and Y are homeomorphic under the mapping
(ii) (b) and the open unit ball
(iii) If and then X and Y are not homeomorphic. Note that Y is compact while is not compact.
(iv) If a mapping is one to one, onto and continuous, it does not follow that it is a homeomorphism. For example, let X = a set with more than one element and T is the discrete topology in X, and is the trivial topology in X. Then the identity mapping is continuous, but not a homeomorphism. (PS: is not a bijection).
SEPARATION AXIOMS:
Two sets and can be separated by open sets if there exist open sets and such that and and .
and can be separated by a real function if there exists a continuous real function f on X such that for and for and for .
separation axiom:
For any two disjoint points, there exists a neighbourhood of either point not containing the other. (This implies that the complement of each point is an open set or that each point is a closed set).
separation axiom:
Any two distinct points can be separated by open sets (or any two distinct points have disjoint neighbourhoods). A topological space is Hausdorff if it satisfies ).
Examples:
(i) A set X with the discrete topology (PS: If a topological space is , then it is also but the converse is not true in general).
(ii) If X is a set with more than one element, then X with the trivial topology is not Hausdorff !
(iii) If X is Hausdorff and , then A is Hausdorff with the induced topology.
(iv) Let be compact, be Hausdorff and be continuous. Then is closed in .
(v) Let X be Hausdorff with and let A be compact. Then A is closed.
AXIOM :
A closed set F and a point can be separated by open sets.
A topological space is regular if it satisfies separation axioms and / A completely regular topological space is one in which a closed set F and a point can be separated by a real function.
AXIOM :
Two disjoint closed sets can be separated by open sets. A topological space is normal if it satisfies seperation axioms and .
(Solomon Lefshetz, Algebraic Topology): In a normal space, every two disjoint closed sets can be separated by a real function.
Section 1.2
A topological group is a set G which is both a group and a space with the topology and group structure related by the assumption that the functions and are continuous. Here, and is the group inverse element of x in G.
Equivalently, the function (from to G) is continuous.
Examples:
(i) The additive group of real numbers is the underlying group of a topological group whose underlying topological space is the usual space of real numbers. More generally, the Euclidean n-space under the usual topology.
(ii) Any group with the usual discrete topology (every set is its own closure).
(iii) The n-dimensional torus (that is to say, the product of n circles). Here a circle is the topological group under multiplication.
(iv) , the general linear group, that is, the group of all non-singular matrices with complex coefficients. For the topology use that “induced” by considering the matrices as a subset of .
(v) Any product of topological groups.
Let e be the identity in the topological group G. If then (this is the definition, by the way)
(1) If , and O is a neighbourhood of x, there exist neighbourhoods P of x and Q of y such that .
Let denote the mapping and . Since is continuous and O is open, it follows that is open. Now contains so it also contains a set of the form where is a neighbourhood of x, and is a neighbourhood of y, and . Take and .
(1′) If , then for every neighbourhood V of there exists a neighbourhood U of a, such that . This follows again from the fact that is continuous. (If g is a continuous mapping of a topological space R into , then for every point and every neighbourhood of there exists a neighbourhood U of a such that )
(2) For each , the mappings and are homeomorphisms or topological mappings.
The mapping is one-to-one (because implies that ). The inverse mapping is of the same form. Hence, it suffices to prove that f is continuous. Let O be any neighbourhood of xy. By (1) there exists a neighbourhood of x, and a neighbourhood of y such that . Hence, , that is to say . This shows that f is continuous (cf. (1′)).
(3) The mapping is a homeomorphism. For is one-to-one, and is its own inverse (by group law). It is continuous by definition.
(4) If O is open in G, then are also open, where and .
By (2) xO and Ox are open so is by (3). Now EO is equal to , hence, open, similarly also OE.
(5) If V is any neighbourhood of e, it contains a neighbourhood W of e, such that .
Proof:
Since , there exist neighbourhoods of e, and of , such that by (1).
Let . Then is open and so that is a neighbourhood of e.
Next . For implies and , and implies that , which in turn implies that . Hence, implies that . Choose .
(6) A neighbourhood V of e is defined to be symmetric if and only if .
Every neighbourhood W of e contains a symmetric neighbourhood (for example, ). By (4) is open. Hence, is a symmetric neighbourhood of e.
(7) Every neighbourhood of is of the form xV as well as of the form Wx, where V and W are neighbourhoods of the identity.
Let U be any neighbourhood of x. Then, , and is open by (4). Hence, V is a neighbourhood of e. Similarly, is a neighbourhood of e. Hence, .
(8) The continuity of is equivalent to the continuity of together with the continuity of .
(i) If and are continuous, then and are continuous, hence the composite is continuous.
(ii) Conversely, if is continuous, then is continuous. Further, is continuous (obviously). Hence, is continuous. Therefore, is continuous. But by hypothesis, is continuous. Hence, is continuous.
The topological space of a topological group G is Hausdorff.
Let x,y belong to a group G such that x and y are distinct elements. By the property (Given two points of S, each of them lies in an open set not containing the other) we can find a neighbourhood V of e not containing . By (5) there exists a neighbourhood of e, such that . Then , are neighbourhoods of x and of y respectively. And , are neighbourhoods of x and of y respectively. And we claim that . This is because of the following: (indirect proof) if then there exist points (or elements of G or T) such that so that contradicting the choice of V. Hence, the Hausdorff axiom is satisfied because .
If , then where V extends over all neighbourhoods of e. (Note that E could be any, some or run through all subsets of G)
First we prove that (i)
If any arbitrary element for all V. we shall see/prove that every neighbourhood of x intersects E and so we would have .
Now, let and let O be any neighbourhood of x. Then by (7) above, , where V is a neighbourhood of e. By hypothesis, , which implies that , where , , or . Hence xV intersects E (that is, has a non-empty intersection with E). Therefore, O intersects E, hence, .
(ii) .
If , then every neighbourhood of x intersects E. By (7), is a neighbourhood of x (where V is a neighbourhood of e). Hence, intersects E. This implies that (for there exists y , such that , or . Hence,
A topological group is homogeneous. Given any two elements —- , there exists a topological mapping f of G onto itself which takes p into q.
Take and take .
It is sufficient for many purposes therefore to verify local properties for a single element only. For example, to show that G is locally compact, it is suffieient to show that its identity e has a neighbourhood U whose closure is compact — so also with regularity.
The topological space of a topological group G is regular (Kolmogorov).
Note: A topological space is if the following is true: Given two points of a topological space S, each of them is contained in an open set not containing the other. Axiom : If C is a closed set in the space S, and if p is a point not contained in C, then there are disjoint open sets in S, one containing C and the other containing p.3
A space that satisfies both and is called regular.
We can separate e and any closed set F not containing not containing e. Now let . Then O is a neighbourhood of e. Now there exists a neighbourhood V of e, such that (because of (5) and (6) above). (Remark I think the author has assumed here that V is symmetric)V and are disjoint open sets. We shall see that .
Since V is a neighbourhood of e, that will prove this lemma. Now,
, where W is a neighbourhood of e (by Lemma 2)
(since V is a W, a neighbourhood of e_
(by the choice of V and O).
(So (by definition of O)
Hence, or or .
Atopological space which is the underlying space of a topological group G is completely regular.
It is sufficient to show that if F is a closed set not containing e, then F and e can be separated by a continuous real function. (cf. the remark above)
Note: Definition: A topological spacce is said to be (also called
) if for every point p of S and for any open set U containing p, there is a continuous function of S to such that and for all points x in . I is a closed interval .
Let . Then V is a neighbourhood of e. Choose a sequence of neighbourhoods of e, such that , where . [This is possible, see the proof of Lemma 3 in which (5) and (6) were used.
Let be finite dyadic real number (a dyadic rational or binary rational number is a number that can be expressed as a fraction with denominator whose fraction is a power of two)
where
Define (group product ) where is e.
We will show that implies that
If then first j digits agree for some so that
and for some
Now define
note that here is the right most 1.
Then
Clearly, (since
We shall see that
(which will imply that
Let
Then
and
Hence by Lemma 2,
(since ) by definition
Hence, and therefore
Now detine
if
if x is in some
Then we have for all . Furtheer
for x = e (since Slatex V_{i}^{0}=e$)
for (for ) (as ` (see above)
We shall see that f is continuous. The sets and , (where k is a real number) are both open. For,
This is equal to the following:
So we get the following now:
(this is a closed set)
Note that hence
for if x is such that , and such that , then for all (since ), hence , a contradiction. The opposite inclusion is trivial. Thus,
But since on the one hand, and on the other so that implies that (as is a dyadic rational), hence is open.
Similarly, is also open. For if the set contains the point x, it contains a whole neighbourhood of it; for, if x is such that , then there exists , such that , which is neighbourhood of x all of which is contained in note ()
Similarly, is also open. For if the set contains the point x, it contains a whole neighbourhood of it; for, if x is such that , then there exists such that which is a neighbourhood of x all of which is contained in , .
If are compact subsets of G, then is compact
Proof: Consider the mapping which takes into xy. This is continuous. The product is compact (Tychonoff). The proof follows from the fact that the continuous image of a compact set is compact.
If are closed, it does not follow that is closed. In let and . Then . Here by Q we mean the set of rational numbers greater than or equal to zero.
A locally compact group is normal.
Note: we recall here the definition of normal: Axiom : Given two points of a topological space S, each of them lies in an open set not containing the other. Axiom : If H and K are disjoint closed sets in the topological space S, then there exist disjoint open sets, one containing H and the other containing K. A space or a normal space is one that satisfies both Axiom and Axiom .
If the group is compact, the proof is easy since compactness together with regularity (or Hausdorff ) implies normality.
Otherwise take a symmetric neighbourhood U of e with compact closure, and consider . Then is an open and closed subgroup of G, and it is sufficient to prove normality for . To do that, use the fact that is compact. That is, where is compact. .
Note: A locally compact group is para-compact, hence normal. (Reference: Hewitt and Ross, Abstract Harmonic Analysis. Volume I. page 76, Theorem 8.13
Let G be a topological group, and H a subset of G. Then H is, by definition, a subgroup of the topological group G if and only if H is a subgroup of the abstract group G, and H is a closed set in the topological space G.
Let G be a topological group, and H a subset of it which is a subgroup of G considered as an abstract group. Then H is also a topological subgroup with the induced topology. In particular, a subgroup of an abstract group which is a topological group is itself a topological group.
A subgroup N of the topological group G is defined to be a normal subgroup if N is the normal subgroup of the abstract group G.
Let G be a topological group, and let H be a subgroup of the abstract group G. Then is a subgroup of the topological group G. If H is a normal subgroup of the abstract group G, then is also normal, that is, a normal subgroup of the topological group G.
Let us recall that if G is any group, and H a subgroup of G, a left coset of H is a subset of G of the form xH, where x is an element of G. The left coset set is the set of all left cosets of H, denoted by G/H. We have a natural map or projection where defined by equal to the left coset of H which contains x.
If G is a topological group, we shall topologize G/H, assuming that H is closed. A set is open if and only if is open in G. This means that we require to be a continuous map.
G/H is a space, and is open.
Again we first recall definitions of and open: Axiom says: Given two points of a topological space S, each of them lies in an open set not containing the other. Also, a mapping f is said to be open if for every open set in the set $latex $f(O_{1}) is an open set in .. Note: .
Since H is closed, xH is also closed. (homeomorphism) so that is open in G. Write , coset containing x. Now , which is open. Therefore, G/H is (the complement of each point is an open set). We know that is continuous, we have to show that it is also open. Let O be open, where . Then is in the coset space; it is open if and only if is open. But is OH, which is open since , where Ox is open. Thus, O open is also open. QED.
G/H is a
Let us recall definition of or Hausdorff space: Given two distinct points of a topological space S, there are disjoint open sets each containing just one of the two points.
Let and be two distinct points of and such that and . Choose a neighbourhood v of e such that . This is possible because (H is closed as yH is closed. Vx is a neighbourhood of x, use the definition of .) It follows that . For if , then , say,, where , . Hence , which contradicts .
Let be a neighbourhood of e such that . Then , hence . Therefore,
Now since and and , are open (by Lemma 1, and are open, and by Lemma 5 is open mapping. Hence, are separated by disjoint open sets. QED.
If G is any topological group, H a closed subgroup, we have topologized the coset set G/H in such a way that it is a space. We may call G/H the quotient space, and the given topology the quotient space topology. On the other hand, if G is any group, and H a normal subgroup (that is, , , or ), then the coset G/H is, in fact, a group known as the quotient group. The next lemma shows that the quotient group, with the quotient space topology is a topological group, if, to start with, G is a topological group.
If G is a topological group, and H a normal subgroup, then the quotient group G/H with the quotient space topology is a topological group.
We have only to show that the mapping given by where is continuous.
Let be given by where x, y are elements of G. We then have the following “loop”:
Trivially, we have
Since and are continuous, is also continuous. Hence, is continuous.
Let be an open set in G/H. Then, is open in . But is open. Hence,
is open in ; that is to say, is open, hence is continuous. QED.
Let G be a topological group, and H a subgroup.
(a) If G is compact, then H and G/H are both compact.
(b) If G is locally compact, then both H and G/H are locally compact.
(a) H is a closed subset of a compact set, hence, compact, G/H is the continuous image of a compact set, hence, compact.
(b) H is locally compact since it is a closed subset of a locally compact space. [Let S be a locally compact space, . Then T is locally compact. For, so that and is compact. Now, is a compact neighbourhood of p.]
To prove that G/H is locally compact, let , let , and a neighbourhood of q, that is, an open set containing q.) Let
and let . so that . Since G is locally compact, and U is open, there exists a neighbourhood O of x, such that , with compact. [Let G be any topological group. To every neighbourhood U of e, there exists a neighbourhood V of e, such that . For let V be a symmetric neighbourhood of e, such that ]. (See (5) and (6) above). Now . Hence, , where . Therefore . Hence, . Then we have
.
Since a neighbourhood of q(O is a neighbourhood of x). And, is compact. (since ). Since is a space, is closed.
We have …..see above which implies that (compact).
Hence, is compact (closed subset of a compact set of a compact set). Thus, is a neighbourhood of q with compact closure. It follows that G/H is locally compact.
Let denote the n-dimensional complex cartesian space. It is a vector space of dimension n over the field of complex numbers.
Let ( where 1 is in the ith position. Then, form a basis of over .
An endomorphism of is defined when the elements is equal to are given. To corresponds the matrix of degree n, and conversely.
We use the same letter for the matrix as well as the endomorphism.
We define a multiplication of two endos, with matrics respectively by defining the corresponding matrix as the product of the matrices, namely
Now let denote the set of all matrices of degree n with coefficients in . If , put (as j goes from 1 to n, j-1 goes from 0 to n-1, and (j-1)n from zero to in steps of n, while i goes from 1 to n; so that i+(j-1)n goes from one to )
To we associate the point with the coordinates in . In this way, we get a one to one correspondence between and . Since is a topological space, we define a topology in by requiring the correspondence to be a homeomorphism.
Let T be any topological space, and let map T into ; . If , is a matrix with coefficients , say. Clearly, is continuous if and only if each function is continuous. Note that:
(), where (In general, the following situation holds: If , then ” f is continuous iff is continuous for *every* . On the one hand, it is trivial that : f continuous implies is continuous”. On the other, if is continuous and *open*, with , then is open, that is, is open. But is open, since sets of the form , U open, form a sub-basis of open sets of . So f is continuous).
It follows from this remark, and (4.1) that the product of two matrices and is continuous funcction of the pair considered as a point of the space .
We denote by the transpose of the matrix ; . We denote by the complex conjugate of ; .
Clearly, and are homeomorphisms, of order 2, of onto itself.
If and are any two matrices, then and
An matrix is regular (or non-singular), if it has an inverse, that is, if there exists a matrix , such that , where is the unit matrix of degree n.
A necessary and sufficient condition for to be regular is that its determinant is non zero.
If an endomorphism of maps onto itself (and not some subspace of lower dimension) the corresponding matrix is regular, and has a reciprocal endomorphism .
If is a regular matrix, we have
= (^{t}\sigma)^{-1} and
If and are regular matrices, is also regular, and we have
.
Hence, the regular matrices od degree n form a group with respect to multiplication, which is called the general linear group .
Since the determinant of a matrix is obviously a continuous function (being a polynomial) of the matrix, is an open subset of (), det is a continuous function. ) The elements of may be considered as points of a topological space which is a subspace of the topological space .
If is a regular matrix, the coefficients of are given by , where the are polynomials in the coefficients . Hence, the mapping of onto itself is continuous. Since the mapping coincides with its reciprocal mapping, it is a homeomorphism of with itself.
The mappings and are also homeomorphisms of with itself. The first is an automorphism of the group, not the second (preserves sums and inverts the order of the products)
If define
Then, we have and
Thus, we have: and . Hence, is a homeomorphism, and an automorphism of order 2 of .
namely,
Let . We say that is orthogomal if . The set of all orthogonal matrices of degree n we denote by . If only , then is called complex orthogonal, and the set of all such we denote by . If , then is called unitary, and we denote by the set of all such .
Since and are continuous, the sets are closed subsets of . [Note that is closed if f is real and continuous. is continuouz for all i,j and is closed]. Because these mappings are automorphisms, are subgroups of .
[Note, in parenthesis, that if X is any topological space, and Y a Hausdorff space, and f,g are continuous mappings of X into Y, then the set is closed. One can see then the set E – is open in X. Let . Since is open in X and …, and Y is Hausdorff, there exist neighbourhoods of respectively so that ). Since f and g are are continuous, there exist neighbourhoods of in X such that . Let . Then V is a neighbourhood of , and for , hence . It follows that F is open.]
Clearly, ….call this (i)
is real, if its coefficients are real, that is, if . The set of all real matrices of degree n we denote by , and we define . Hence, .
Since the determinant of the product of two matrices is the product of their determinants, the matrices of determinant I form a subgroup of . The group of all matrices with determinant I in is called the special linear group .
We set
….call this (ii)
….call this (iii)
Clearly, are subgroups and closed subsets of . They may be considered as subspaces of .
are compact.
We have only to show that is compact, since are closed subsets of . We shall see that is homeomorphis to a bounded, closed subset of .
A matrix is unitary if and only if , where is the unit matrix. []
If . then
[ is regular, that is, ]
The left hand side of the last equations are continuous functions of . is not only a closed subset of but also of . [For is an intersection of closed sets].
Further,
for .
Therefore the coefficients of the matrix are bounded. Since is a homeomorphism, is closed and bounded, and a subset of and hence compact.
Cheere,
Nalin Pithwa
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