Let H and J be as in the above lemma. The result of the lemma shows that the integral

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1 Let and be as in the above lemma. The result of the lemma shows that the integral ( f(x, y)dy) dx is well defined; we denote it by f(x, y)dydx. By symmetry, also the integral ( f(x, y)dx) dy is well defined; we denote it by f(x, y)dxdy Theorem Let and be compact topogroups and f K( ). Then f(x, y)dxdy = f = f(x, y)dydx. Proof. We show that by setting M (f) = f(x, y)dxdy = f. Define a mapping M : K( ) C f(x, y)dxdy for every f K( ). We show that M is linear mapping and M satisfies Conditions a) l, b) and c) of Theorem We show first that M is linear. Let f, f K( ) and c C. Then ( M (cf + f) ( = cf(x, y) + f(x, ) ) ( y) dx dy = cf(x, y)dx + = c f(x, y)dxdy + f(x, y)dxdy = c f(x, y)dxdy + f(x, y)dxdy = cm (f) + M ( f). a ) f(x, y)dx dy We have shown that M is linear. Condition c) of Theorem 7.13 holds, since M (1) = ( 1dx) dy = 1dy = 1. To show that also Conditions a) l and b) are satisfied, we choose f K( ) and we define a function l : C by the formula l(y) = f(x, y)dx. Then M (f) = l(y)dy. Let (h, j). Then M ( (h,j) f) = = ( ) (h,j)f(x, y)dxdy = f(hx, jy)dx dy ( ) f(x, jy)dx dy = l(jy)dy = l(y)dy = M (f). We have shown such that Condition a) l is satisfied. Next we assume that f K + ( ). Then, for every y, the function x f(x, y) belongs to K + (), and it follows that l(y) = f(x, y)dx 0. ence we have l K +() and therefore M (f) = l 0. We have shown that Condition b) is satisfied. By Theorem 7.13, M is the invariant integral of. As a consequence, we have f(x, y)dxdy = f for every j K( ). Similarly, we see that f(x, y)dydx = f. 69

2 8. Linear groups and representations of topological groups. We use the symbol GL(n) to denote the collection of all n n regular (i.e., invertible ) C valued matrices. Then GL(n) is a group with respect to matrix multiplication. Let τ be the topology which GL(n) inherits from the topology of C n2 via the identification a 1,1 a 1,n.. (a 11,..., a 1n, a 21,..., a n1,..., a nn ). a n,1 a n,n We shall show later that the triple (GL(n),,τ) is a topogroup. The topogroups GL(n) are called general linear topogroups. The topogroups GL(n) and some of their closed subgroups are called classical topogroups and they are of central importance in investigations of the structure of compact topogroups. The usefulness of linear topogroups in the theory of compact groups is a consequence of the famous Peter Weyl Theorem, which states that every compact T 1 topogroup has enough continuous finite-dimensional representations. A (continuous, finite-dimensional) representation of a topogroup G is a continuous homomorphism G GL(n) for some n = 1, 2,... The Peter Weyl Theorem is the following result: if G is a compact T 1 topogroup with neutral element e, then for every g G \ {0}, there exists a representation φ of G such that φ(g) φ(e). To prove the Peter Weyl Theorem, it is useful to represent linear topogroups in a more abstract form, as groups of operators of finite-dimensional normed spaces. We start by reviewing some basic concepts and results dealing with normed spaces. In the following, a linear space always means a linear space over the complex numbers. A norm of a linear space V is a mapping v v from V to R such that the following conditions are satisfied for all u, v V and c C: 1 u 0, and u = 0 only if u = 0. 2 u + v u + v. 3 cu = c u 8.1 Lemma Let be a norm of the linear space V. Then is an invariant length function of the group (V, +). For every v V, the sets {u V : u v < ε}, ε > 0, form a neighborhood base of v in the group topology determined by. 70

3 Proof. Condition 3 above shows that v = v for every v V. It follows, by Conditions 1 and 2, that is a length function of the group (V, +). Since the group (V, +) is commutative, the length function is invariant. The sets V,ε = {w V : w < ε}, ε > 0, form a neighborhood base for the neutral element 0 of the group (V, +) in the group topology τ determined by the length function. For all v V and ε > 0, we have v + V,ε = {v + w : W V and w < ε} = {u V : u v < ε}. The last assertion of the lemma follows, since the sets v + V,ε, ε > 0, form a neighborhood base for v in the topology τ. A normed space is a couple (V, ), where V is a linear space and is a norm of V. The norm topology of a normed space (V, ) is the group topology determined by the length function of the group (V, +). We show that, in a finite-dimensional linear space, all norms determine the same topology. 8.2 Theorem In a finite-dimensional linear space, all norms determine the same topology, and this topology is locally compact. Proof. Let V be a finite-dimensional linear space and let {v 1,..., v n } be a basis of V. Denote by π the Euclidean topology of C n. We define a mapping f : C n V by the formula f(c 1,...c n ) = c 1 v 1 + c n v n. The mapping f is a group isomorphism from (C n, +) onto (V, +). We show that for every norm topology τ of V, the mapping f is a homeomorphism (C n,π) (V,τ). The stated result then follows, since the topology π is locally compact. Let be a norm of V and let τ be the topology of V determined by. We show that f is a continuous mapping (C n,π) (V,τ). By Theorem 2.29, it suffices to to show that f is continuous at the point (0,..., 0) C n. Let O be a neighborhood of the point f(0,..., 0) = 0 in the topology τ. Then there exists ε > 0 such that {v V : v < ε} O. Let M = max( v 1,..., v n ). For every (c 1,..., c n ) C, if c c n < ε M, then f(c 1,..., c n ) = c 1 v 1 + +c n v n c 1 v c n v n M ( c c n ) < ε. 71

4 As a consequence, { (c1,..., c n ) C : c c n < ε } f 1 ( {v V : v < ε} ) f 1 (O). M Since the leftmost set above is a neighborhood of the point (0,..., 0) in the topology π, we have shown that f is continuous at the point (0,..., 0). By the foregoing, f is continuous mapping (C n,π) (V,τ). We show that f is an open mapping at the point (0,..., 0) C n. Let ε positive number. The subset K = {(c 1,..., c n ) : c c n = ε} of (C n,π) is compact. It follows, by continuity of the mapping f, that the subset f(k) of the space (V,τ) is compact. Since 0 / f(k), we have v > 0 for every v f(k). By the remark made after Definition 4.10, the function is continuous topology with respect to the topology τ. It follows that the number δ = inf{ w : w f(k)} is positive. For every v = c 1 v c n v n V, if 0 < v < δ, then the following holds for the point ε c c n (c 1,..., c n ) of K: ( ) v < δ f ε c c n (c 1,..., c n ) = ε c c n v. ence we have c c n < ε. From the foregoing it follows that {v V : v < δ} f({(c 1,..., c n ) C n : c c n < ε}). We have shown that f is an open mapping C n at the point (0,..., 0). By Theorem 2.29, f is and open mapping. As a consequence, f is a homeomorphism. We now use the above result to show that linear mappings between finite-dimensional vector spaces are automatically continuous. 8.3 Corollary Every linear mapping between finite-dimensional normed spaces is continuous with respect to the norm topologies. Proof. Let (V, ) and (W, ) be finite-dimensional normed spaces and let L : V W be a linear mapping. We define a function : V R by the formula v = v + Lv. It is easy to see that is a norm of V. Let τ, τ and τ be the topologies determined by the norms, and,respectively. For every v V we have Lv v. As a consequence, L is a continuous mapping (V,τ ) (W,τ ). By Theorem 8.2, we have τ = τ. It follows that L is a continuous mapping (V,τ) (W,τ ). 72

5 In particular, a one-to-one onto linear mapping between finite-dimensional normed spaces is a homeomorphism. We move on to consider operators. Let V be a normed space with norm. An operator of V is a mapping L : V V, which satisfies the following conditions: 1 L(cu + dv) = clu + dlv for all u, v V and c, d C. 2 sup{ Lu : u V and u 1} <. Condition 1 just requires that L is a linear mapping V V. Linear mappings satisfying Condition 2 are called bounded linear mappings. We see easily that a linear mapping L : V V is bounded if, and only if, L is continuous with respect to the norm topology of V. We use the symbol L(V ) to denote the collection of all operators of the normed space V. The set L(V ) is made into a linear space by defining, for all L, T L(V ) and c C, mappings L + T and cl by the formulas (L + T)v = Lv + Tv and (cl)v = c(lv). It is easy to see that L + T L(V ) and cl L(V ). We define a norm in the linear space L(V ) by setting, for every L L(V ), L = sup{ Lv : v V and v 1}. 8.4 Lemma The function is a norm of L(V ). Proof. We have L 0 for every L L(V ). The zero-element of L(V ) is the constant mapping 0 : 0(v) = 0 for every v V. Clearly, 0 = 0. If L 0, then there exists u V such that Lu 0. Since L0 = 0, we have u 0. Let v = u u. Then v = 1 and Lv = Lu 0. It follows that L 0. u Let L, T L(V ) and c C. For every v V, we have (L + T)v = Lv + Tv Lv + Tv and (cl)v = c Lu. As a consequence, we have L + T L + T and cl = c L. The function is called the operator norm of L(V ). It is easy to see an operator L L(V ) has norm L = sup{ Lu : u V and u = 1}, and that Lv L v for every v V. Besides addition, we can define another binary operation in the set L(V ). For all L, T L(V ), we denote in the usual way L T the composition of the mappings L and T. We show that is, indeed, an operation in L(V ). 73

6 8.5 Lemma Let V be a normed pace. For all L, T L(V ), we have L T L(V ) and L T L T. Proof. Clearly, a composition of linear mappings is linear. For every v V, we have Lv L v and Tv T v. It folllows that, for every u V, we have (L T)u = L(Tu) L Tu L T u. From this it follows that L T is bounded and its norm satisfies the inequality L T L T. We move on to consider the finite-dimensional case. First we note that, by Corollary 8.3, the following result obtains. 8.6 Lemma Let V be a finite-dimensional linear space. Then every linear mapping V V is V operator. Up to isomorphism, operator spaces of finite-dimensional linear spaces can be characterized as spaces formed by complex matrices. For every n N, denote by M(n) the set of all complex n n matrices. The set M(n) is made into a linear space by defining addition and scalar multiplication by the formulas (a ij ) + (b ij ) = (a ij + b ij ) and c(a ij ) = (ca ij ). Besides addition, we can also define multiplication in M(n). We use the formula (a ij )(b ij ) = ( n k=1 a ikb kj ). The recall the following result from linear algebra. 8.7 Lemma Let V be an n dimensional normed space, 0 < n <. Then there exists a linear one-to-one onto mapping ϕ : L(V ) M(n) such that, for all L, T L(V ), we have ϕ(l T) = ϕ(l)ϕ(t). Clearly, the elements (c kl ij ) of M(n), where ckl kl = 1 and ckl ij = 0 for every (i, j) (k, l), form a basis of the linear space M(n). ence it follows from the above lemma that the linear space L(V ) is n 2 dimensional when V is an n dimensional normed space. An operator L of a normed space V is regular, if L is a one-to-one mapping from V onto V and the inverse mapping L 1 is an operator of V. The inverse mapping L 1 of a linear one-to-one onto mapping L : V V is linear. ence L 1 is an operator of V if, and only if, L 1 is bounded. Denote by L s (V ) the set of all regular operators of V. The topology, which L s (V ) inherits from the operator norm topology of L(V ), is called the norm topology of L s (V ) 74

7 8.8 Theorem Let V be a normed space and let τ be the norm topology of L s (V ). Then the triple (L s (V ),,τ) is a topogroup. Proof. We see easily that the pair (L s (V ), ) is a group. Denote by the norm of V (and that of L(V )). For every L L s (V ), the sets {T L s (V ) : T L < ε}, where ε > 0, form a neighborhood base of L in the topology τ of L s (V ). We use Theorem 1.31 to show that τ is a group topology of the group (L s (V ), ). We show that the translations L T L and L L T are continuous with respect to τ. Let T L s (V ). By Lemma 8.5 we have, for all L, L L s (V ), T L T L = T (L L ) T L L and It follows that if L L < L T L T = (L L ) T L L T. ε T +1, then T L T L < ε and L T L T < ε. By the foregoing, the translations L T L and L l T are continuous. Let I be the neutral element of L s (V ), that is, the identity mapping of V. We show that the mapping (L, T) L T is continuous at the point (I, I) of the product space L s (V ) L s (V ). For all L, T L s (V ), we have L T I = (L I) (T I) + (L I) + (T I) and hence L T I (L I) (T I) + (L I) + (T I). It follows from the foregoing that if 0 < ε < 1 and L I < ε 3 and T I < ε 3, then L T I < ε. As a consequence, the mapping (L, T) L T is continuous at the point (I, I). It remains to show that the mapping L L 1 is continuous at the point I of L s (V ). For every L L s (V ), we have L 1 I = L 1 (I L) = (L 1 I + I) (I L) = (L 1 I) (I L) + (I L) 75

8 and hence, further, L 1 I L 1 I L I + L I. From the foregoing it follows that L 1 I L I 1 L I Therefore, if L I < 1 2 ε < 1 2, then L 1 I < 1 2 ε 1 2 L L 1 is continuous at the point I. whenever L I 1. = ε. It follows that the mapping We have shown that the topology τ satisfies the conditions of Theorem By the theorem, τ is a group topology of the group (L s (V ), ). We now present the promised abstract characterization of general linear topogroups. 8.9 Theorem Let n be a positive integer. Then GL(n) is a topogroup. If V is an n dimensional normed space, then L s (V ) = GL(n). Proof. By Lemma 8.7, there exists linear one-to-one mapping ϕ from L(V ) onto M(n) such that, for all T, L L(V ), we have ϕ(t L) = ϕ(t)ϕ(l). Denote by ϕ the restriction of the mapping ϕ to the set L s (V ). Results of linear algebra show that ϕ isomorphism (L s (V ), ) (M(n), ). is a group We show that ϕ is a homeomorphism. We define a function : M(n) R by the formula (a ij ) = max{ a ij : 1 i, j n}. We see easily that is a norm of M(n). Denote by τ the topology of M(n) determined by the normn. Clearly, the topology of GL(n), as defined in the beginning of this chapter, is the same as the relative topology of GL(n) in the space (M(n),τ). Denote by π the operator norm topology of the linear space L(V ). Then the norm topology of L s (V ) is the same as the relative topology of L s (V ) in the space (L(V ),π). By Corollary 8.3, the linear one-to-one onto mapping ϕ is a homeomorphism L(V ) M(n), with respect to the topologies π and τ. Since ϕ(l s (V )) = GL(n), the restriction ϕ of ϕ to the set L s (V ) is a homeomorphism L s (V ) GL(n). From this it follows, since ϕ is a group isomorphism and L s (V ) is a topogroup (by Theorem 8.8), that GL(n) is a topogroup and L s (V ) = GL(n). 76

9 Besides L s (V ), we shall consider another subset of the set L(V ) of all operators of a normed space V. Let be the norm of V. A mapping L : V V is a linear isometry of V if L is a linear mapping and we have Lv = v for every v V. Denote by L i (V ) the set of all linear isometries of V. Clearly, L i (V ) L(V ) and L = 1 for every L L i (V ). We show that if V is finite-dimensional, then L i (V ) is a subgroup of the group L s (V ). First we recall a result from linear algebra Lemma The following conditions are mutually equivalent when V is a finite-dimensional linear space and L is a linear mapping V V : a) L is one to one. b) L is onto Theorem Let V be a finite-dimensional normed space. Then the topogroup L s (V ) is locally compact and L i (V ) is a compact subgroup of L s (V ). Proof. We show first that if A is a subset of a finite-dimensional normed space (W, ) such that sup a A a <, then the closure A of A in the topology of W determined by the norm is compact. Let M R be such that sup a A a M. Using the triangle inequality satisfied by a norm (Condition 2 ), we see that sup a A a M. By Lemma 8.2, the space W is locally compact. ence 0 has a compact neighborhood K. Let ε > 0 be such that {w W : w ε} K, and let n N satisfy the inequality nε M. The closed subset C = {w W : w ε} of the compact set K is compact. It follows by Theorem 1.27 that the set nc is compact. We see easily that nc = {w W : w nε} and it follows, since nε M, that the set {w W : w M} is compact. Since A {w W : w M}, it follows that A is compact. Next we show that L s (V ) is locally compact. By Theorem 8.2 and the remark made after Theorem 8.7, the norm topology of L(V ) is locally compact. Since L s (V ) is a subspace of L(V ), local compactness of L s (V ) follows once we show that L s (V ) is an open subset of L(V ). Let L L s (V ). Set s = inf{ Lv : v V and v = 1}. Since the mappings v Lv and u u are continuous and the set {v V : v = 1} 77

10 is compact, there exists an element a of V such that a = 1 and La = s. Since L is regular and a 0, we have La 0 and hence s = La > 0. Let ε = 1 2 s and O = {T L(V ) : T L < ε}. Then O η L (L(V )). We show that O L s (V ). Let T O. For every u V, if u = 1, then Tu Lu Tu Lu = Lu (T L)u Lu T L 2ε ε > 0. From the foregoing it follows that T is one to one mapping and from this it follows, by Lemmas 8.10 and 8.6, that T L s (V ). We have shown that O L s (V ). By the foregoing, L s (V ) is an open subset of L(V ). Since every L L i (V ) is one to one mapping, Lemmas 8.10 and 8.6 show that L i (V ) L s (V ). We see easily that, for all L, T L i (V ), we have L 1 L i (V ) and L T L i (V ); hence L i (V ) is a subgroup of the group (L s (V ), ). We show that L i (V ) is a compact subset of L s (V ). Since L s (V ) has the relative topology from L(V ) and since L i (V ) is contained in the compact subset {L L(V ) : L = 1} of L(V ), we see that compactness of L i (V ) follows once we show that L i (V ) is closed in L(V ). Let L n L i (V ) for n = 1, 2,..., and let L L(V ) be such that L n L 0 when n. We show that L L i (V ). Let v V. If v = 0, then Lv = 0 = v. We assume that v 0, and we show that also then the equation Lv = v holds. For every n N, we have 0 L n v Lv L n L v. It follows that L n v Lv 0 when n. Moreover, for every n N, we have Ln v Lv Ln v Lv. It follows from the foregoing that L n v Lv when n. Since {L 1, L 2,...} L i (V ), we have L n v = v for every n = 1, 2,..... As a consequence, Lv = v. We have shown that L L i (V ). We have shown that L i (V ) is closed in L(V ). We now give a new definition for the concept of a representation. By Theorem 8.9, the new definition is, essentially, equivalent with the previous definition Definition Let G be a topogroup. A representation of G is a continuous homomorphism G L s (V ), where V is some finite-dimensional normed space. 78