Peter Weyl theorem – MScMathematics

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 $\phi$ are both open.

A topological space is a set X with a topology T in X. T is called the trivial topology if $T = (X,$ latex \phi$) and is called the discrete topology if T= (A: $A \subset X$ ). (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 $A \subset X$ . We define an induced topology in A as follows: the class of open sets in A is the class of sets of the form $U \bigcap A$ 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 $\overline{E}$ 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 $x \in X$ is any open set containing x. Let $X_{1}$ and $X_{2}$ be two topological spaces and f a mapping

$f : X_{1} \rightarrow X_{2}$

A mapping is an assignment to each $x \in X_{1}$ of an element f(x) in $X_{2}$ . Then f is continuous if and only if for every open set $O_{2} \in X_{2}$ , the set $f^{-1}(O_{2})$ is an open set in $O_{1}$ . Here, $f^{-1}(O_{2}) = \{ x: x \in X_{1}, f(x) \in O_{2}\}$ .

Let $x \in X_{1}$ . 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 $f(V) \subset U$ . We say that f is continuous, if f is continuous at every point of $X_{1}$ . This definition is equivalent to the preceding one.

The mapping f is said to be open if for every open set $O_{1} \in X_{1}$ , the set $f(O_{1})$ is an open set in $X_{2}$ .

Let $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ be two continuous mappings. Then the composite mapping $g \circ f: X \rightarrow Z$ is continuous.

Examples

(i) If X is a discrete topological space, Y is a topological space, then every mapping $f: X \rightarrow Y$ is continuous. (PS: this is vacuously true).

(ii) If X is any topological space, and Y a set wtih the trivial topology, then $f: X \rightarrow Y$ is continuous.

(iv) $f_{X}: X \rightarrow X$ is the identity mapping.

The topological product of two topological spaces $X_{1}$ and $X_{2}$ is the topological space $X = X_{1} \times X_{2}$ , whose set is the cartesian product of $X_{1}$ and $X_{2}$ , namely, $\{ (x_{1}, x_{2}) : x_{1} \in X_{1}, x_{2} \in X_{2}\}$ with the open sets being all unions of sets of the form $O_{1} \times O_{2}$ where $O_{1}$ is an open set in $X_{1}$ and $O_{2}$ is an open set in $X_{2}$ . Sets of the form $O_{1} \times O_{2}$ form a basis of open sets in $X_{1} \times X_{2}$

Note that if $p_{1}: X_{1} \times X_{2} \rightarrow X_{1}$ and $p_{2}: X_{1} \times X_{2} \rightarrow X_{2}$ are the projections defined by $p_{1}(x_{1}, x_{2}) = x_{1}$ and $p_{2}(x_{1}, x_{2}) = x_{2}$ respectively then $p_{1}$ and $p_{2}$ are continuous mappings where $x_{1} \in X_{1}$ and $x_{2} \in X_{2}$ .

COVERING. A family $\{ V_{\alpha} \}$ where $\alpha \in I$ of subsets of a set X is a covering of X, if $\bigcup_{\alpha \in I}V_{\alpha} = X$ . That is to say, each point of X belongs to at least one $V_{\alpha}$ . If further, X is a topological space, and each $V_{\alpha}$ is an open subset of X, we say that $\{ V_{\alpha}\}$ is an open covering of X.

A covering $\{V_{\alpha}\}_{\alpha \in I}$ 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, $X = \bigcup_{\alpha \in I}U_{\alpha}$ where $U_{\alpha}$ is open implies that there exists $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ all belonging to I such that $\bigcup_{1 \leq i \leq n}U_{\alpha_{i}}=X$ . A subset $A \subset X$ is compact if A is compact in the induced topology.

Examples:

(i) The $\Re$ is not compact.

(ii) Let X be a topological space and $A \subset X$ / 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 $a \leq b$ . Then the set $a \leq x \leq b$ is compact.

An equivalent definition of compactness is the following:

If $\{ F_{\alpha}\}$ is a family of closed sets such that every finite subfamily has a non-empty intersection, then $\bigcap_{\alpha}F_{\alpha} \neq \phi$ . (If $\{ F_{\alpha}\}$ 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 $X_{1}$ is compact, $X_{2}$ is Hausforff, and $f: X_{1} \rightarrow X_{2}$ is continuous, then $f(X_{1})$ is closed in $X_{2}$ ).

A product of compact spaces is compact (Tychonoff theorem).

A space X is locally compact if for every $x \in X$ , there is a neighbourhood $O_{1}$ of x such that the closure of $O_{1}$ is compact.

Every compact space is locally compact but not vice-versa.

HOMEOMORPHISMS.

Let X and Y be topological spaces and f a mapping $f: X \rightarrow Y$ . We say that f is a homeomorphism if f is one-to-one, onto and both f and $f^{-1}$ 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) $X = Y = \Re$ and $f(x) = x^{3}-x$ for $x \in X$ .

(ii) (a) $X = \Re$ and $Y = \{ x \in \Re -1 < x < 1 \}$ . X and Y are homeomorphic under the mapping $f(x) = \frac{x}{1+|x|}$

(ii) (b) $\Re^{n}$ and the open unit ball $B = \{ x \in \Re^{n}: ||x||<1\}$

(iii) If $X = \Re$ and $Y = \{ -1 < x \leq 1\}$ then X and Y are not homeomorphic. Note that Y is compact while $\Re$ 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 $T^{'}$ is the trivial topology in X. Then the identity mapping $I_{X}: (X, T) \rightarrow (X, T^{'})$ is continuous, but not a homeomorphism. (PS: $f^{-1}$ is not a bijection).

SEPARATION AXIOMS:

Two sets $M_{1}$ and $M_{2}$ can be separated by open sets if there exist open sets $O_{1}$ and $O_{2}$ such that $M_{1} \subset O_{1}$ and $M_{2} \subset O_{2}$ and $O_{1} \bigcap O_{2} = \phi$ .

$M_{1}$ and $M_{2}$ can be separated by a real function if there exists a continuous real function f on X such that $0 \leq f(x) \leq 1$ for $x \in X$ and $f(x)=0$ for $x \in M_{1}$ and $f(x)=1$ for $x \in M_{2}$ .

$T_{1}$ 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).

$T_{2}$ 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 $T_{2}$ ).

Examples:

(i) A set X with the discrete topology (PS: If a topological space is $T_{2}$ , then it is also $T_{1}$ 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 $A \subset X$ , then A is Hausdorff with the induced topology.

(iv) Let $X_{1}$ be compact, $X_{2}$ be Hausdorff and $f: X_{1} \rightarrow X_{2}$ be continuous. Then $f(X_{1})$ is closed in $X_{2}$ .

(v) Let X be Hausdorff with $A \subset X$ and let A be compact. Then A is closed.

AXIOM $T_{3}$ :

A closed set F and a point $x \notin F$ can be separated by open sets.

A topological space is regular if it satisfies separation axioms $T_{1}$ and $T_{3}$ / A completely regular topological space is one in which a closed set F and a point $x \notin F$ can be separated by a real function.

AXIOM $T_{4}$ :

Two disjoint closed sets can be separated by open sets. A topological space is normal if it satisfies seperation axioms $T_{1}$ and $T_{4}$ .

$\bf {Urysohns \hspace{0.1in} \bf {Lemma}}$

(Solomon Lefshetz, Algebraic Topology): In a normal space, every two disjoint closed sets can be separated by a real function.

Section 1.2 $\bf {Topological \hspace{0.1in} \bf {Groups}}$

A topological group is a set G which is both a group and a $T_{1}$ space with the topology and group structure related by the assumption that the functions $(x, y) \rightarrow x.y$ and $x \rightarrow x^{-1}$ are continuous. Here, $x, y \in G$ and $x^{-1}$ is the group inverse element of x in G.

Equivalently, the function $(x,y) \rightarrow xy^{-1}$ (from $G \times G$ 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 $\{ x \in \mathcal{C}: |x| =1\}$ under multiplication.

(iv) $GL(n,C)$ , the general linear group, that is, the group of all non-singular $n \times n$ matrices with complex coefficients. For the topology use that “induced” by considering the $n \times n$ matrices as a subset of $\mathcal{C^{n}}$ .

(v) Any product of topological groups.

$\bf{Trivial \hspace{0.1in} properties}$

Let e be the identity in the topological group G. If $E, F \subset G$ then $EF \equiv \{ xy : x \in E, y \in F\}$ (this is the definition, by the way)

(1) If $z=xy$ , and O is a neighbourhood of x, there exist neighbourhoods P of x and Q of y such that $PQ \subset O$ .

Let $\phi$ denote the mapping $(x,y) \rightarrow xy$ and $\overline{O} = \phi^{-1}(O)$ . Since $\phi$ is continuous and O is open, it follows that $\overline{O}$ is open. Now $\overline{O}$ contains $(x,y)$ so it also contains a set of the form $O_{1} \times O_{2}$ where $O_{1}$ is a neighbourhood of x, and $O_{2}$ is a neighbourhood of y, and $\phi(O_{1} \times O_{2}) = O_{1}.O_{2} \subset O$ . Take $P = O_{1}$ and $Q = O_{2}$ .

(1′) If $a \in G$ , then for every neighbourhood V of $a^{-1}$ there exists a neighbourhood U of a, such that $U^{-1} \subset V$ . This follows again from the fact that $a \rightarrow a^{-1}$ is continuous. (If g is a continuous mapping of a topological space R into $R^{'}$ , then for every point $a \in R$ and every neighbourhood $U^{'}$ of $a^{'} = g(a) \in R^{'}$ there exists a neighbourhood U of a such that $g(U) \subset U^{'}$ )

(2) For each $x \in G$ , the mappings $y \rightarrow yx$ and $y \rightarrow xy$ are homeomorphisms or topological mappings.

The mapping $f: y \rightarrow xy$ is one-to-one (because $xy = x^{'}y$ implies that $x = x^{'}$ ). The inverse mapping $f^{-1}: y \rightarrow x^{-1}y$ 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 $O_{1}$ of x, and a neighbourhood $O_{2}$ of y such that $O_{1}O_{2} \subset O$ . Hence, $xO_{2} \subset O$ , that is to say $f(O_{2}) \subset O$ . This shows that f is continuous (cf. (1′)).

(3) The mapping $x \rightarrow x^{-1}$ is a homeomorphism. For $x \rightarrow x^{-1}$ 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 $O^{-1}, xO, EO, Ox, OE$ are also open, where $x \in G$ and $E \subset G$ .

By (2) xO and Ox are open so is $O^{-1}$ by (3). Now EO is equal to $\bigcup_{x \in E}xO$ , hence, open, similarly also OE.

(5) If V is any neighbourhood of e, it contains a neighbourhood W of e, such that $W.W^{-1} \subset V$ .

Proof:

Since $ee^{-1}=e$ , there exist neighbourhoods $V_{1}$ of e, and $V_{2}$ of $e^{-1}=e$ , such that $W.W^{-1} \subset V$ by (1).

Let $V_{3} = V_{1} \bigcap V_{2}^{-1}$ . Then $V_{3}$ is open and $e \in V_{3}$ so that $V_{3}$ is a neighbourhood of e.

Next $V_{3}V_{3}^{-1} \subset V$ . For $x \in V_{3}$ implies $x \in V_{1}$ and $x \in V_{2}^{-1}$ , and $y^{-1} \in V_{3}^{-1}$ implies that $y \in V_{3}$ , which in turn implies that $y^{-1} \in V_{2}$ . Hence, $xy^{-1} \in V_{2}V_{2}^{-1}$ implies that $xy^{-1} \in V_{1}V_{2} \subset V$ . Choose $W = V_{3}$ .

(6) A neighbourhood V of e is defined to be symmetric if and only if $V=V^{-1}$ .

Every neighbourhood W of e contains a symmetric neighbourhood (for example, $W \bigcap W^{-1}$ ). By (4) $W^{-1}$ is open. Hence, $W \bigcap W^{-1}$ is a symmetric neighbourhood of e.

(7) Every neighbourhood of $x \in G$ 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, $V = x^{-1}U \ni e$ , and $x^{-1}U$ is open by (4). Hence, V is a neighbourhood of e. Similarly, $W = Ux^{-1}$ is a neighbourhood of e. Hence, $U=xV=Wx$ .

(8) The continuity of $(x,y) \rightarrow xy^{-1}$ is equivalent to the continuity of $(x,y) \rightarrow xy$ together with the continuity of $x \rightarrow x^{-1}$ .

(i) If $(x,y) \rightarrow xy$ and $x \rightarrow x^{-1}$ are continuous, then $(x,y^{-1}) \rightarrow xy^{-1}$ and $(x,y) \rightarrow (x,y^{-1})$ are continuous, hence the composite $(x,y) \rightarrow xy^{-1}$ is continuous.

(ii) Conversely, if $(x,y) \rightarrow xy^{-1}$ is continuous, then $(e,y) \rightarrow ey^{-1}=y^{-1}$ is continuous. Further, $y \rightarrow (e,y)$ is continuous (obviously). Hence, $y \rightarrow y^{-1}$ is continuous. Therefore, $(x,y) \rightarrow (x,y^{-1})$ is continuous. But by hypothesis, $(x,y) \rightarrow xy^{-1}$ is continuous. Hence, $(x,y) \rightarrow xy$ is continuous.

$\bf{Separation \hspace{0.1in} properties}$

$\bf{Lemma \hspace{0.1in}1}$

The topological space of a topological group G is Hausdorff.

$\bf{Proof}$

Let x,y belong to a group G such that x and y are distinct elements. By the $T_{1}$ 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 $y^{-1}x$ . By (5) there exists a neighbourhood $V_{1}$ of e, such that $V_{1}V_{1}^{-1} \subset V$ . Then $xV_{1}$ , $yV_{1}$ are neighbourhoods of x and of y respectively. And $xV_{1}$ , $yV_{1}$ are neighbourhoods of x and of y respectively. And we claim that $xV_{1} \subset yV_{1} = \phi$ . This is because of the following: (indirect proof) if $xV_{1} \bigcap yV_{1} \neq \phi$ then there exist points (or elements of G or T) $v^{'}, v^{''}$ such that $xv^{'}=yv^{''}$ so that $y^{-1}x$ contradicting the choice of V. Hence, the Hausdorff axiom is satisfied because $xV_{1} \bigcap yV_{1} = \phi$ .

$\bf{Lemma \hspace{0.1in}2}$

If $E \subset G$ , then $\overline{E} = \bigcap {EV} =\bigcap{VE}$ 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) $\overline{E} \supset \bigcap {EV}$

If any arbitrary element $x \in EV$ for all V. we shall see/prove that every neighbourhood of x intersects E and so we would have $x \in \overline {E}$ .

Now, let $x \in \bigcap EV$ and let O be any neighbourhood of x. Then by (7) above, $O \in xV$ , where V is a neighbourhood of e. By hypothesis, $x \in EV^{-1}$ , which implies that $x = ay^{-1}$ , where $a \in E$ , $y \in V$ , or $xy=a$ . Hence xV intersects E (that is, has a non-empty intersection with E). Therefore, O intersects E, hence, $x \in \overline{E}$ .

(ii) $\overline{E} \subset \bigcap{EV}$ .

If $x \in \overline{E}$ , then every neighbourhood of x intersects E. By (7), $xV^{-1}$ is a neighbourhood of x (where V is a neighbourhood of e). Hence, $xV^{-1}$ intersects E. This implies that $x \in EV$ (for there exists y $\in V$ , such that $xy^{-1} \in E$ , or $x \in EY \subset EV$ . Hence, $x \in \bigcap EV$

$\bf{Remark}$ A topological group is homogeneous. Given any two elements —- $p. q \in G$ , there exists a topological mapping f of G onto itself which takes p into q.

Take $a= p^{-1}q$ and take $f(x)-xa$ .

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.

$\bf{Lemma \hspace{0.1in}}$

The topological space of a topological group G is regular (Kolmogorov).

Note: A topological space is $T_{1}$ 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 $T_{3}$ : 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 $T_{1}$ and $T_{3}$ is called regular.

$\bf{Proof}$

We can separate e and any closed set F not containing not containing e. Now let $O = F^{c}$ . Then O is a neighbourhood of e. Now there exists a neighbourhood V of e, such that $V^{2} \subset O$ (because of (5) and (6) above). (Remark I think the author has assumed here that V is symmetric)V and $\overline{F^{c}}$ are disjoint open sets. We shall see that $F \subset \overline{V^{c}}$ .

Since V is a neighbourhood of e, that will prove this lemma. Now,

$\overline{V} = \bigcap{VW}$ , where W is a neighbourhood of e (by Lemma 2)

$\overline{V} \subset V^{2}$ (since V is a W, a neighbourhood of e_

$\subset \overline{V} \subset O$ (by the choice of V and O).

(So $\overline{V} = F^{c}$ (by definition of O)

Hence, $V \subset F^{c}$ or $\overline{V^{c}}$ or $\overline{V^{c}} \supset F$ .

$\bf{Theorem \hspace{0.1in}}$ Atopological space which is the underlying space of a topological group G is completely regular.

$\bf{Proof}$ 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 $\bf{completely} \hspace{0.1in} \bf{regular}$ (also called

$\bf{Tychonoff} \hspace{0.1in}\bf{space}$ ) if for every point p of S and for any open set U containing p, there is a continuous function of S to $I^{2}$ such that $f(p)=0$ and $f(x)=1$ for all points x in $S-U$ . I is a closed interval $[a,b]$ .

Let $V=F^{c}$ . Then V is a neighbourhood of e. Choose a sequence $V_{1}, V_{2}, \ldots$ of neighbourhoods of e, such that $V_{1}^{2}\subset V$ , $V_{k+1}^{2}\subset V_{k}$ where $k \geq 1$ . [This is possible, see the proof of Lemma 3 in which (5) and (6) were used.

Let $\alpha$ 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)

$\alpha = 0.\alpha_{1}\alpha_{2}\ldots 000$ where $\alpha_{i}=0,1$

Define $O_{\alpha}=V_{1}^{\alpha_{1}} V_{2}^{\alpha_{2}}\ldots V_{k}^{\alpha_{k}}$ (group product ) where $V_{i}^{0}$ is e.

We will show that $\alpha < \beta$ implies that $\overline{O}_{\alpha}\subset O_{\beta}$

If $\alpha < \beta$ then first j digits agree for some $j \geq 0$ so that

$\alpha = 0.\alpha_{1}\alpha_{2}\ldots \alpha_{j}0\alpha_{j+2}\ldots \alpha_{k}000$

$\beta = 0.\beta_{1}\beta_{2}\ldots \beta_{j}1\beta_{j+2}\ldots \beta{m}000$

and $\alpha_{l} = \beta_{l}$ for some $l \leq j$

Now define

$\alpha^{'}=0.\alpha_{1}\ldots 0 111 \ldots 1000$ note that here $\alpha_{k}$ is the right most 1.

$\beta^{'} = 0.\alpha_{1}\ldots \alpha_{j}1000 \ldots$

Then

$\alpha \leq \alpha^{'}< \beta^{'} \leq \beta$

Clearly, $O_{\alpha} \subset O_{\alpha^{'}} \subset O_{\beta^{'}} \leq \O_{\beta}$ (since $V_{i}^{0}=e$

We shall see that

${\overline{O}}_{\alpha^{'}} \subset O_{\beta^{'}}$ (which will imply that ${\overline{O}}_{\alpha} \subset O_{\beta}$

Let $O = V_{1}^{\alpha_{1}}\ldots V_{j}^{\alpha_{j}}$

Then

$O_{\alpha^{'}} = OV_{j+2}\ldots V_{k}$ and $O_{\beta^{'}} = OV_{j+1}$

Hence by Lemma 2,

${\overline{O}}_{\alpha^{'}}V_{k} = OV_{j+2}\ldots V_{k-1}V_{k}^{2}$

$\subset OV_{j+2}\ldots V_{k-2}V_{k-1}^{2}$ (since $V_{k}^{2} \subset V_{k-1}$ ) $\subset OV_{j+2}^{2} \subset OV_{j+1} = O_{\beta^{'}}$ by definition

Hence, ${\overline{O}}_{\alpha^{'}} \subset O_{\beta^{'}}$ and therefore $(\overline{O})_{\alpha} \subset O_{\beta}$

Now detine

$f(x) = 1$ if $x \notin O_{\alpha}$

$f(x) = \inf \{\alpha : x \in O_{\alpha}\}$ if x is in some $O_{\alpha}$

Then we have $0 \leq f(x) \leq 1$ for all $x \in G$ . Furtheer

$f(x) = 0$ for x = e (since Slatex V_{i}^{0}=e$)

$f(x) - 1$ for $x \in F$ (for $x \in F$ ) (as ` $V = F^{c}$ (see above)

We shall see that f is continuous. The sets $\{ x: f(x)>k\}$ and $\{ x : f(x) <k\}$ , (where k is a real number) are both open. For,

$\{ x : f(x) >k\}$ This is equal to the following: $(\bigcap_{\alpha>k}(\overline{O})_{\alpha})^{c}$

$\Longleftrightarrow \{ x: f(x) \leq k\}$ So we get the following now:

$\{ x: f(x) >k\} = (\bigcap_{\alpha>k}{\overline{O}}_{\alpha})$ (this is a closed set)

Note that $\alpha < \beta \Longrightarrow O_{\alpha} \subset O_{\beta}$ hence

$\{ x: f(x)\leq k\} \subset \{ x: x \in \bigcap_{\alpha >k} O_{\alpha}\}$

for if x is such that $f(x) \leq k$ , and $\beta>k$ such that $x \notin O_{\beta}$ , then $x \notin O_{\alpha}$ for all $\alpha < \beta$ (since $O_{\alpha} \subset O_{\beta}$ ), hence $f(x) \geq \beta > k$ , a contradiction. The opposite inclusion is trivial. Thus,

$\{ x: f(x) \leq k\} = \{ x: x \in \bigcap_{\alpha >k}O_{\alpha}\}$

But $\bigcap_{\alpha >k}{\overline{O}}_{\alpha}$ since on the one hand, $O_{\alpha} \subset{\overline{O}}_{\alpha}$ and on the other $\alpha < \beta \Longrightarrow {\overline{O}}_{\alpha} \subset O_{\beta}$ so that $x \in \bigcap_{\alpha>k}{\overline{O}}_{\alpha}$ implies that $x \in \bigcap_{\alpha>k}O_{\alpha}$ (as $\alpha$ is a dyadic rational), hence $\{ x: f(x)>k\}$ is open.

Similarly, $\{ x: f(x)<k\}$ is also open. For if the set contains the point x, it contains a whole neighbourhood of it; for, if x is such that $f(x)<k$ , then there exists $\alpha < k$ , such that $x \in O_{\alpha}$ , which is neighbourhood of x all of which is contained in $\{ x : f(x)<k\}$ note ( $y \in {O_{\alpha}} \Longrightarrow f(y) \leq \alpha <k$ )

Similarly, $\{ x: f(x) < k\}$ is also open. For if the set contains the point x, it contains a whole neighbourhood of it; for, if x is such that $f(x)<k$ , then there exists $\alpha <k$ such that $x \in O_{\alpha}$ which is a neighbourhood of x all of which is contained in $\{ x: f(x)<k\}$ , $y \in O_{\alpha} \Longrightarrow f(y) \leq \alpha <k$ .

$\bf{Lemma \hspace{0.1in}4}$

If $C_{1}, C_{2}$ are compact subsets of G, then $C_{1}C_{2}$ is compact

Proof: Consider the mapping $G \times G \rightarrow G$ which takes $(x,y)$ into xy. This is continuous. The product $C_{1} \times C_{2}$ is compact (Tychonoff). The proof follows from the fact that the continuous image of a compact set is compact.

$\bf{Remark}$

If $F_{1}, F_{2}$ are closed, it does not follow that $F_{1}.F_{2}$ is closed. In $\Re^{1}$ let $F_{1}= \{ n \in \mathcal{Z}: n \geq 1\}$ and $F_{2} = \{ 0\} \bigcup \{ \frac{1}{n}: n \in F_{1}\}$ . Then $F_{1}.F_{2}=Q \subset \Re^{1}$ . Here by Q we mean the set of rational numbers greater than or equal to zero.

$\bf{Theorem \hspace{0.1in} 2}$

A locally compact group is normal.

Note: we recall here the definition of normal: Axiom $T_{1}$ : Given two points of a topological space S, each of them lies in an open set not containing the other. Axiom $T_{4}$ : 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 $T_{4}$ space or a normal space is one that satisfies both Axiom $T_{1}$ and Axiom $T_{4}$ .

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 $G^{'} = \bigcup_{n=1}^{\infty} U^{n}$ . Then $G^{'}$ is an open and closed subgroup of G, and it is sufficient to prove normality for $G^{'}$ . To do that, use the fact that $G^{'}$ is $\sigma$ compact. That is, $G^{'} = \bigcup_{n=1}^{\infty}K_{n}$ where $K_{n}$ is compact. $K_{n}=\bigcup_{l=1}^{n}{\overline{U}}^{l}$ .

Note: A locally compact $T_{1}$ group is para-compact, hence normal. (Reference: Hewitt and Ross, Abstract Harmonic Analysis. Volume I. page 76, Theorem 8.13

$\bf{1.3}$ $\bf{Subgroups}$ $\bf{Quotient}$ $\bf{groups}$

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 $\overline{H}$ is a subgroup of the topological group G. If H is a normal subgroup of the abstract group G, then $\overline{H}$ 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 $\pi: G \rightarrow G/H$ where $x \mapsto xH$ defined by $\pi (x)$ 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 $O \subset G/H$ is open if and only if $\pi^{-1}(O)$ is open in G. This means that we require $\pi$ to be a continuous map.

$\bf{Lemma \hspace{0.1in}5}$ G/H is a $T_{1}$ space, and $\pi$ is open.

Again we first recall definitions of $T_{1}$ and open: Axiom $T_{1}$ 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 $O_{1}$ in $X_{1}$ the set $latex $f(O_{1}) is an open set in $X_{2}$ .. Note: $f: X_{1} \rightarrow X_{2}$ .

$\bf{Proof}$

Since H is closed, xH is also closed. (homeomorphism) so that $(xH)^{c}$ is open in G. Write $\overline{x}=xH$ , coset containing x. Now $\pi^{-1}(G/H - \overline{x}) =(xH)^{c}$ , which is open. Therefore, G/H is $T_{1}$ (the complement of each point is an open set). We know that $\pi$ is continuous, we have to show that it is also open. Let O be open, where $O \subset G$ . Then $\pi O$ is in the coset space; it is open if and only if $\pi^{-1}(\pi O)$ is open. But $\pi^{-1}(\pi O)$ is OH, which is open since $OH = \bigcup_{x \in H}Ox$ , where Ox is open. Thus, O open $\Longrightarrow \pi O$ is also open. QED.

$\bf{Lemma \hspace{0.1in}6}$ G/H is a $T_{2}$

Let us recall definition of $T_{2}$ 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.

$\bf{Proof}$

Let $x_{1}$ and $y_{1}$ be two distinct points of $G/H = Q$ and $x,y \in G$ such that $\pi(x)=x_{1}$ and $\pi(y)=y_{1}$ . Choose a neighbourhood v of e such that $Vx \bigcap yH = \phi$ . This is possible because $x \notin yH = \overline{yH}$ (H is closed as yH is closed. Vx is a neighbourhood of x, use the definition of $\overline{yH}$ .) It follows that $VxH \bigcap yH=\phi$ . For if $VxH \bigcap yh \neq \phi$ , then $vxh_{1}=yh_{2}$ , say,, where $v \in V$ , $h_{1}, h_{2} \in H$ . Hence $vx = yh_{2}h_{1}^{-1}=yh_{3}$ , which contradicts $Vx \bigcap yH = \phi$ .

Let $V_{1}$ be a neighbourhood of e such that $V_{1}^{-1}V_{1} \subset V$ . Then $V_{1}^{-1}V_{1}xH \bigcap yH = \phi$ , hence $V_{1}xH \bigcap V_{1}yH= \phi$ . Therefore,

$\pi(V_{1}x) \pi(V_{1}y) = \phi$

Now $x_{1} \in \pi (V_{1}x)$ since $e \in V_{1}$ and $y_{1} \in \pi (V_{1}y)$ and $\pi (V_{1}x)$ , $\pi (V_{1}y)$ are open (by Lemma 1, $V_{1}x$ and $V_{1}y$ are open, and by Lemma 5 $\pi$ is open mapping. Hence, $x_{1}, y_{1}$ are separated by disjoint open sets. QED.

$\bf{Remark}$

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 $T_{2}$ 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, $\forall {x} \in H$ , $\forall a \in G$ , $axa^{-1} \in H$ or $aHa^{-1} \subset H$ ), 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.

$\bf{Lemma \hspace{0.1in}7}$ 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.

$\bf{Proof}$

We have only to show that the mapping $\psi_{1}: G/H \times G/H \rightarrow G/H$ given by $\psi_{1}(x_{1}, y_{1}=x_{1}y_{1}^{-1})$ where $x_{1}, y_{1} \in G/H$ is continuous.

Let $\psi: G \times G \rightarrow G$ be given by $(x,y) \mapsto xy^{-1}$ where x, y are elements of G. We then have the following “loop”:

$\pi \times \pi : G \times G \rightarrow G/H \times G/H$

$\psi_{1}: G/H \times G/H \rightarrow G/H$

$\psi : G \times G \rightarrow G$

$\pi G \rightarrow G/H$

Trivially, we have $\pi \psi = \psi_{1}(\psi \times \psi)$

Since $\pi$ and $\psi$ are continuous, $\pi\psi$ is also continuous. Hence, $\psi_{1}(\pi \times \pi)$ is continuous.

Let $O_{1}$ be an open set in G/H. Then, $[\psi_{1}(\pi \times \pi)]^{-1}O_{1}$ is open in $G \times G$ . But $\pi \times \pi$ is open. Hence,

$(\pi \times \pi)(\pi \times \pi)^{-1}\psi_{1}^{-1}(O_{1})$ is open in $G/H \times G/H$ ; that is to say, $\psi_{1}^{-1}(O_{1})$ is open, hence $\psi_{1}$ is continuous. QED.

$\bf{Lemma \hspace{0.1in}8}$

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.

$\bf{Proof}$

(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, $T \subset S, T = \overline{T}$ . Then T is locally compact. For, $p \in T \Longrightarrow p \in S \Longrightarrow \exists U_{p} \subset S$ so that $p \in U_{p}$ and $(\overline{U})_{p}$ is compact. Now, $T \bigcap (\overline{U})_{p}$ is a compact neighbourhood of p.]

To prove that G/H is locally compact, let $q$ , let $q \in G/H$ , and $U_{1}$ a neighbourhood of q, that is, an open set containing q.) Let

$U = \pi^{-1}(U_{1})$

and let $x \in U$ . so that $\pi(x)=q$ . Since G is locally compact, and U is open, there exists a neighbourhood O of x, such that $\overline{O} \subset U$ , with $\overline{O}$ compact. [Let G be any topological group. To every neighbourhood U of e, there exists a neighbourhood V of e, such that $\overline{V} \subset U$ . For let V be a symmetric neighbourhood of e, such that $V^{2} \subset U$ ]. (See (5) and (6) above). Now $x \in \overline{V} \Longrightarrow (xV) \bigcap V \neq \phi$ . Hence, $xv_{1}=v_{2}$ , where $v_{1}, v_{2} \in V$ . Therefore $x=v_{2}v_{1}^{-1} \in V.V^{-1}\subset V^{2}\subset U$ . Hence, $\overline{V} \subset U$ . Then we have

$\pi(\overline{O}) \subset \pi(U) = U_{1}$ .

Since $O_{1} = \pi{O}$ a neighbourhood of q(O is a neighbourhood of x). And, $\pi(\overline{O})$ is compact. (since $\overline{O}$ ). Since $G/H$ is a $T_{2}$ space, $\pi(\overline{O})$ is closed.

We have $O_{1} \subset \pi(\overline{O})$ …..see above $O_{1} = \pi(O)$ which implies that $(\overline{O})_{1} \subset \overline{\pi{O}} = \pi(\overline(O))$ (compact).

Hence, $O_{1}$ is compact (closed subset of a compact set of a compact set). Thus, ${O}_{1}$ is a neighbourhood of q with compact closure. It follows that G/H is locally compact.

$\bf{I.4 \hspace{0.1in} Examples}$

Let $\mathcal{C}^{n}$ denote the n-dimensional complex cartesian space. It is a vector space of dimension n over the field $\mathcal{C}$ of complex numbers.

Let ( $\overline{e})_{i}=\{ \}0,0,0, \ldots ,1,0,0,0\}$ where 1 is in the ith position. Then, $(\overline{e})_{1}, (\overline{e})_{2}, \ldots, (\overline{e})_{n}$ form a basis of $\mathcal{C}^{n}$ over $\mathcal{C}$ .

An endomorphism $\alpha$ of $\mathcal{C}^{n}$ is defined when the elements $\alpha \overline{e}_{i}$ is equal to $\Sigma_{j=1}^{n}a_{ji}(\overline{e})_{j}$ are given. To $\alpha$ corresponds the matrix $\{ a_{ij}\}$ of degree n, and conversely.

We use the same letter $\alpha$ for the matrix as well as the endomorphism.

We define a multiplication $\alpha \circ \beta$ of two endos, $\alpha, \beta$ with matrics $\{ a_{ij}\}, \{ b_{ij} tr \}$ respectively by defining the corresponding matrix $\{ e_{ij}\}$ as the product of the matrices, namely

$e_{ij} = \Sigma_{k=1}^{n}a_{ik}b_{kj}$

Now let $M_{n}(\mathcal{C})$ denote the set of all matrices of degree n with coefficients in $\mathcal{C}$ . If $\{a_{ij} \} \in M_{n}(\mathcal{C})$ , put $b_{i+(j-1)n}=a_{ij}$ (as j goes from 1 to n, j-1 goes from 0 to n-1, and (j-1)n from zero to $n^{2}-n$ in steps of n, while i goes from 1 to n; so that i+(j-1)n goes from one to $n^{2}$ )

To $(a_{ij})$ we associate the point with the coordinates $b_{1}, b_{2}, \ldots, b_{n^{2}}$ in $\mathcal{C}^{n^{2}}$ . In this way, we get a one to one correspondence between $M_{n}(\mathcal{C})$ and $\mathcal{C}^{n^{2}}$ . Since $\mathcal{C}^{n^{2}}$ is a topological space, we define a topology in $M_{n}(\mathcal{C})$ by requiring the correspondence to be a homeomorphism.

Let T be any topological space, and let $\phi$ map T into $M_{n}(\mathcal{C})$ ; $\phi: T \rightarrow M_{n}(\mathcal{C})$ . If $t \in T$ , $\phi(t)$ is a matrix with coefficients $a_{ij}$ , say. Clearly, $\phi$ is continuous if and only if each function $a_{ij}(t)$ is continuous. Note that:

( $\underbrace{T \stackrel{\phi}\rightarrow M_{n}{(C)}\stackrel{\psi}\rightarrow \mathcal{C}^{n} \stackrel{\pi_{ij}}\rightarrow \mathcal{C}}_{\phi_{ij}}$ ), where $\phi_{ij}(t)=a_{ij}(t)$ (In general, the following situation holds: If $T \stackrel{I}\rightarrow \prod_{a \in A}X_{a} \stackrel{\pi_{a}}\rightarrow X_{a}$ , then ” f is continuous iff $\pi_{a} \circ f$ is continuous for *every* $a \in A^{n}$ . On the one hand, it is trivial that : f continuous implies $\pi_{a} \circ f$ is continuous”. On the other, if $\pi_{a} \circ f$ is continuous and *open*, with $U \subset X_{a}$ , then $(\pi_{a} \circ f)^{-1}U$ is open, that is, $f^{-1}(\pi_{a}^{-1}(U))$ is open. But $\pi_{a}^{-1}(U)$ is open, since sets of the form $\pi_{a}^{-1}(U)$ , U open, form a sub-basis of open sets of $\prod_{a \in A}X_{a}$ . So f is continuous).

It follows from this remark, and (4.1) that the product $\sigma\tau$ of two matrices $\sigma$ and $\tau$ is continuous funcction of the pair $(\sigma, \tau)$ considered as a point of the space $M_{n}(\mathcal{C}) \times M_{n}(\mathcal{C})$ .

$\bf{Notation}$ We denote by $^{t}a$ the transpose of the matrix $\alpha = (a_{ij})$ ; $^{t}a = (a^{'}_{ij})=a_{ji}$ . We denote by $\overline{\alpha}$ the complex conjugate of $\alpha$ ; $\overline{\alpha}= (\overline{a}_{ij})$ .

Clearly, $\alpha \mapsto ^{t}\alpha$ and $\alpha \mapsto \overline{\alpha}$ are homeomorphisms, of order 2, of $M_{n}(\mathcal{C})$ onto itself.

If $\alpha$ and $\beta$ are any two matrices, then $^{t}(\alpha\beta) = ^{t}\alpha^{t}\beta$ and $\overline{\alpha\beta} = \overline{\alpha}\cdot \overline{\beta}$

An $n \times n$ matrix $\sigma$ is regular (or non-singular), if it has an inverse, that is, if there exists a matrix $\sigma^{-1}$ , such that $\sigma\sigma^{-1}=\sigma^{-1}\sigma = \varepsilon$ , where $\varepsilon$ is the unit matrix of degree n.

A necessary and sufficient condition for $\sigma$ to be regular is that its determinant is non zero.

If an endomorphism $\sigma$ of $\mathcal{C}^{n}$ maps $\mathcal{C}^{n}$ onto itself (and not some subspace of lower dimension) the corresponding matrix $\sigma$ is regular, and $\sigma$ has a reciprocal endomorphism $\sigma^{-1}$ .

If $\sigma$ is a regular matrix, we have

$^{t}(\sigma^{-1})$ = (^{t}\sigma)^{-1} and $(\overline{\sigma})^{-1}=\overline{(\sigma^{-1})}$

If $\sigma$ and $\tau$ are regular matrices, $\sigma\tau$ is also regular, and we have

$(\sigma\tau)^{-1}=(\tau)^{-1}\sigma^{-1}$ .

Hence, the regular matrices od degree n form a group with respect to multiplication, which is called the general linear group $GL(n, \mathcal{C})$ .

Since the determinant of a matrix is obviously a continuous function (being a polynomial) of the matrix, $GL_{n, \mathcal{C}}$ is an open subset of $M_{n}(\mathcal{C})$ ( $GL(n, \mathcal{C}) = \{ \sigma: det \sigma \neq 0\}$ ), det is a continuous function. ) The elements of $GL{(n, \mathcal{C})}$ may be considered as points of a topological space which is a subspace of the topological space $M_{n}(\mathcal{C})$ .

If $\sigma = (a_{ij})$ is a regular matrix, the coefficients $b_{ij}$ of $\sigma^{-1}$ are given by $b_{ij}=A_{ij}(\det \sigma)^{-1}$ , where the $A_{ij}$ are polynomials in the coefficients $f \sigma$ . Hence, the mapping of $GL(n, \mathcal{C})$ onto itself is continuous. Since the mapping coincides with its reciprocal mapping, it is a homeomorphism of $GL(n, \mathcal{C})$ with itself.

The mappings $\sigma \mapsto \overline{\sigma}$ and $\sigma \mapsto ^{t}\sigma$ are also homeomorphisms of $GL(n, \mathcal{C})$ with itself. The first is an automorphism of the group, not the second (preserves sums and inverts the order of the products)

If $\sigma \in GL(n, \mathcal{C})$ define $\sigma^{*} = (^{t}\sigma)^{-1}$

Then, we have $(\sigma\tau)^{*}= \sigma^{*}\tau^{*}$ and $\sigma_{}$

Thus, we have: $(\sigma\tau)^{*}=(\sigma\tau)^{*}=\sigma^{*}\tau^{*}$ and $(\sigma^{*})^{-1}$ . Hence, $\sigma \mapsto \sigma^{*}$ is a homeomorphism, and an automorphism of order 2 of $GL(n, \mathcal{C})$ .

$\bf{The \hspace{0.1in} subgroups}$ namely, $O(n), O(n, \mathcal{C}, U(n) \hspace{0.1in} of GL(n, \mathcal{C}))$

Let $\sigma \in GL(n, \mathcal{C})$ . We say that $\sigma$ is orthogomal if $\sigma = \overline{\sigma}=\sigma^{*}$ . The set of all orthogonal matrices of degree n we denote by $O(n)$ . If only $\sigma= \sigma^{*}$ , then $\sigma$ is called complex orthogonal, and the set of all such $\sigma$ we denote by $O(n, \mathcal{C})$ . If $\overline{\sigma}=\sigma^{*}$ , then $\sigma$ is called unitary, and we denote by $U(n)$ the set of all such $\sigma$ .

Since $\sigma \mapsto \overline{\sigma}$ and $\sigma \mapsto \sigma^{*}$ are continuous, the sets $O(n), O(n, \mathcal{C}, U(n))$ are closed subsets of $GL(n, \mathcal{C})$ . [Note that $\{ x: f(x)=\sigma\}$ is closed if f is real and continuous. $\phi_{ij}(\sigma)=(\overline{a})_{ij}-a_{ij}^{*}$ is continuouz for all i,j and $\{ \sigma: \phi_{ij}(\sigma)=0\}$ is closed]. Because these mappings are automorphisms, $O(n), O(n, \mathcal{C}, U(n))$ are subgroups of $GL(n,\mathcal{C})$ .

[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 $E=\{ x: x \in X, f(x)=g(x0\}$ is closed. One can see then the set E – $\{ x : x \in X, f(x) \neq g(x)\}$ is open in X. Let $x_{0}\in F$ . Since $F = \{ x : x \in X, f(x) \neq g(x)\}$ is open in X and … $f(x_{0}\neq g(x_{0})$ , and Y is Hausdorff, there exist neighbourhoods $U_{1}, U_{2}$ of $f(x_{0}, g(x_{0}))$ respectively so that $U_{1}\bigcap U_{2}=\phi$ ). Since f and g are are continuous, there exist neighbourhoods $V_{1}, V_{2}$ of $x_{0}$ in X such that $f(V_{1}) \subset U_{1}, g(V_{2} \subset U_{2})$ . Let $V = V_{1} \bigcap V_{2}$ . Then V is a neighbourhood of $x_{0}$ , and $f(x) \neq g(x)$ for $x \in V$ , hence $V \subset F$ . It follows that F is open.]

Clearly, $O(n) = O(n, \mathcal{C}) \bigcap U(n)$ ….call this (i)

$\sigma$ is real, if its coefficients are real, that is, if $\sigma = \overline{\sigma}$ . The set of all real matrices of degree n we denote by $M_{n}(\Re)$ , and we define $GL(n, \Re) = M_{n}(\Re) \bigcap GL (n, \mathcal{C})$ . Hence, $O(n) = GL(n, \Re) \bigcap O(n, \mathcal{C})$ .

Since the determinant of the product of two matrices is the product of their determinants, the matrices of determinant I form a subgroup of $GL(n, \mathcal{C})$ . The group of all matrices with determinant I in $GL(n, \mathcal{C})$ is called the special linear group $SL(n, \mathcal{C})$ .

We set

$SL(n, \Re) = SL(n,\mathcal{C}) \bigcap GL(n, \Re)$

$SU(\overline{n}) = SL(\overline{n}, \mathcal{C}) \bigcap U(n)$ ….call this (ii)

$O(n) = SL(n, \mathcal{C}) \bigcap O(n)$ ….call this (iii)

Clearly, $SL(n, \mathcal{C}), SL(n, \Re), SU(n), SO(n)$ are subgroups and closed subsets of $GL(n, \mathcal{C})$ . They may be considered as subspaces of $GL(n, \mathcal{C})$ .

$\bf{Theorem 3}$ $U(n), O(n), SU(n), SO(n)$ are compact.

$\bf{Proof}$ We have only to show that $U(n)$ is compact, since $O(n), SU(n), SO(n)$ are closed subsets of $U(n)$ . We shall see that $U(n)$ is homeomorphis to a bounded, closed subset of $\mathcal{C}^{n^{2}}$ .

A matrix $\sigma$ is unitary if and only if $^{t}\sigma\overline{\sigma}=\varepsilon$ , where $\varepsilon$ is the unit matrix. [ $\overline{\sigma}= \sigma^{*}=(^{t}\sigma)^{-1}$ ]

If $\sigma = \{ a_{ij}\}$ . then

$(^{t}\sigma)\overline{\sigma}=e \Longleftrightarrow \Sigma_{j}(a_{ji}).(\overline{a_{jk}})=\delta_{ik}$

[ $\sigma$ is regular, that is, $\sigma\sigma^{-1}=\varepsilon$ ]

The left hand side of the last equations are continuous functions of $\sigma$ . $U(n)$ is not only a closed subset of $GL(n, \mathcal{C})$ but also of $M_{n}(\mathcal{C})$ . [For $\{ \sigma: \Sigma a_{ji}\overline{a}_{jk}=0, i \neq k\} \bigcap \{ \sigma: \Sigma a_{jk}\cdot \overline{a}_{ji} = I\}$ is an intersection of closed sets].

Further,

$\Sigma_{j}a_{ji}\overline{a}_{ji}=1 \Longleftarrow |a_{ij}| \leq 1$ for $1 \leq i,j \leq n$ .

Therefore the coefficients of the matrix $\sigma \in U(n)$ are bounded. Since $f: M_{n}(C) \mapsto \mathcal{C}^{n^{2}}$ is a homeomorphism, $f(U(n))$ is closed and bounded, and a subset of $\mathcal{C}^{n^{2}}$ and hence compact.