Talk:Direct integral
Latest comment: 17 years ago by Mct mht in topic Center of algebra of bounded operators
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For the author
editFTA: "A Borel space is standard if and only if it is isomorphic to the underlying Borel space a Polish space."
This makes no sense to me.
- No it doesn't. I fixed it.--CSTAR 03:32, 15 July 2006 (UTC)
Center of algebra of bounded operators
editIf H is an infinite dimensional space, I don't see how you prove the center of L(H) consists of scalars without using the spectral theorem for self adjoint operators. --CSTAR 19:06, 9 January 2007 (UTC)
- the following argument looks like it works: Suppose T ∈ Z(L(H)) is not of the form T = λ · I. Then there exist h, such that h and Th are linearly independent. Let P be the rank-1 projection onto the linear span of h. Then TPh = Th. But PTh ≠ Th. Thus P and T does not commute.
- also, how does one prove it using the spectral theorem? if T is in Z(L(H)) is self adjoint, take two projections P_1 and P_2 from its resolution of the identity. if Q is the partial isometry from ran(P_1) to ran(P_2), then TQ ≠ QT. the general case can then be shown by writing T as a Cartesian decomposition T = S_1 + i S_2. that sound right? Mct mht 04:18, 10 January 2007 (UTC)
- Yeah you're right that argument with 1 dimensional projectors seems to work. I was thinking of the more general fact that a C*-algebra has no invariant subspaces iff its commutator algebra consists of scalars and had blotted out forever simpler arguments for irreducibility of L(H). That general fact uses the spectral theorem as per your argument.--CSTAR 04:46, 10 January 2007 (UTC)
- Wait a minute.. some sort of argument is needed to show there is an h, such that h and Th are linearly independent. You are assuming that T is not a scalar to arrive at a contradiction. Note that the fact T is not a multiple of the identity merely says that Th is not the same scalar multiple of h. It does not exclude the possibility that Th is a scalar multiple of h which depends on h. Spectral theorem works here of course to exclude this possibility, but is there an easier ragument.--CSTAR 15:11, 10 January 2007 (UTC)
- Suppose there is a scalar function λ of h such that T h is λ(h) h, for any vector h ∈ H, any orthonormal basis for H diagonalizes T. In particular, conisder a pair of vectors e, f in this basis with different eigenvalues λ(e), λ(f). It's now an argument in two-dimensional Hilbert space to show the existence of such an h such that h, Th are independent. This is a computation with 2 x 2 matrices. --CSTAR 16:27, 10 January 2007 (UTC)
- Wait a minute.. some sort of argument is needed to show there is an h, such that h and Th are linearly independent. You are assuming that T is not a scalar to arrive at a contradiction. Note that the fact T is not a multiple of the identity merely says that Th is not the same scalar multiple of h. It does not exclude the possibility that Th is a scalar multiple of h which depends on h. Spectral theorem works here of course to exclude this possibility, but is there an easier ragument.--CSTAR 15:11, 10 January 2007 (UTC)
- Yeah you're right that argument with 1 dimensional projectors seems to work. I was thinking of the more general fact that a C*-algebra has no invariant subspaces iff its commutator algebra consists of scalars and had blotted out forever simpler arguments for irreducibility of L(H). That general fact uses the spectral theorem as per your argument.--CSTAR 04:46, 10 January 2007 (UTC)
- ok, suppose T ∈ Z(L(H)) is of the form T h = λ(h) h then λ(h) must be independent of h, i.e. T = λ · I. for if this is not the case, take e, f in H with λ(e) ≠ λ(f). let S be the rank one operator that takes e to f. so TS e = λ(f)f ≠ ST e = λ(e) f, contradicting the assumption that T lies in the center. Mct mht 07:32, 11 January 2007 (UTC)