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asp.net aspx barcode F.5 Direct integrals of unitary representations in .NET Insert bar code 39 in .NET F.5 Direct integrals of unitary representations

F.5 Direct integrals of unitary representations using barcode generation for none control to generate, create none image in none applications.asp.net+qr+code+generator As previously mentioned (see none for none remark after Corollary C.4.7), direct sums are not suf cient in order to decompose a given unitary representation into irreducible ones.

One needs the notion of a direct integral of unitary representations. Let (Z, ) be a measure space, where is a - nite positive measure on Z. A eld of Hilbert spaces over Z is a family (H(z))z Z , where H(z) is a Hilbert.

USPS Intelligent Mail Weak containment and Fell s topology space for each z Z. Elemen none for none ts of the vector space z Z H(z) are called vector elds over Z. To de ne a measurable eld of Hilbert spaces over Z, we have to specify the measurable vector elds.

This depends on the choice of a fundamental family of measurable vector elds; by de nition, this is a sequence (xn )n N of vector elds over Z with the following properties: (i) for any m, n N, the function z xm (z), xn (z) is measurable; (ii) for every z Z, the linear span of {xn (z) : n N} is dense in H(z). Fix a fundamental family of measurable vector elds. A vector eld x z Z H(z) is said to be a measurable vector eld if all the functions z x(z), xn (z) , n N.

are measurable. As is easily shown (Exercise F.6.

6), the set M of measurable vector elds is a linear subspace of z Z H(z). Moreover, if x, y M , then the function z x(z), y(z) is measurable. The pair ((H(z))z Z , M ) , simply denoted by z H(z), is called a measurable eld of Hilbert spaces over Z.

In the sequel, we identify two measurable vector elds which are equal -almost everywhere. A measurable vector eld x is a square-integrable vector eld if x(z) 2 d (z) < ..

The set H of all square-inte none for none grable vector elds is a Hilbert space for the inner product x(z), y(z) d (z) ,. x, y H. We write H= H(z)d (z). and call H the direct integr none for none al of the eld (H(z))z Z of Hilbert spaces over Z. Example F.5.

1 (i) Let Z be a countable set and let be a measure on Z such that (z) = 0 for all z Z. Then every vector eld is measurable and. H(z)d (z) =. H(z). is the direct sum of the Hilbert spaces H(z), z Z. F.5 Direct integrals of unitary representations (ii) Let (Z, ) be a - nit e measure space and let H(z) = C for all z Z. We can choose a fundamental family of measurable vector elds such that the measurable vector elds are the measurable complex-valued functions on Z. Then.

H(z)d (z) = L2 (Z, ).. (iii) Let (Z, ) be a - ni te measure space and let K be a xed separable Hilbert space. Set H(z) = K for all z Z. We can choose a fundamental family of measurable vector elds such that the measurable vector elds are the measurable mappings Z K, with respect to the Borel structure on K given by the weak topology.

Then. H(z)d (z) = L2 (Z, K),. the Hilbert space of all square-integrable measurable mappings Z K. Z H(z)d (z). For every z none none Z, let T (z) be a bounded operator on H(z). We say that (T (z))z Z is a measurable eld of operators on Z if all the functions.

Let z H(z) be a measurable eld of Hilbert spaces over Z. Let H = z T (z)x(z), y(z) ,. x, y H,. are measurable. If, moreover none for none , z T (z) is -essentially bounded, then we can de ne a bounded operator T : H H by (Tx)(z) = T (z)x(z),. x H, z Z. In this case, we write T = Z none for none T (z)d (z). Let now G be a second countable locally compact group. Let (Z, ) be a - nite measure space, let z H(z) be a measurable eld of Hilbert spaces over Z, and let H = Z H(z)d (z).

A measurable eld of unitary representations of G on the H(z) s is a family z (z), where (z) is a unitary representation of G on H (z), such that ( (z)(g))z Z is a measurable eld of operators on H for every g G. Let g G. Since (z)(g) is a unitary operator on H(z), we can de ne a unitary operator (g) on H by (g) =.

(z)(g)d (z).. It is clear that g (g) is none for none a homomorphism from G to the unitary group of H. It can be shown [Dixmi 69, Proposition 18.7.

4] that the mapping g (g) is.
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