Timeline for Hilbert space automorphisms realized as induced by transformations of some base-spaces
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Oct 29, 2010 at 21:09 | history | edited | Andrey Rekalo | CC BY-SA 2.5 |
edited title
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| Oct 29, 2010 at 21:04 | history | edited | Bad English | CC BY-SA 2.5 |
added 1218 characters in body
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| Oct 28, 2010 at 21:54 | history | edited | Bad English | CC BY-SA 2.5 |
added 181 characters in body
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| Oct 28, 2010 at 21:20 | comment | added | Bad English | anon, in following answer i will try to describe common construction. | |
| Oct 23, 2010 at 9:28 | answer | added | Costabel | timeline score: 2 | |
| Oct 22, 2010 at 0:45 | comment | added | Yemon Choi | My feeling is that this question may be related to the following one: is the set of normal, invertible operators on (separable) Hilbert space a group? There the answer is no, even for the finite-dimensional case, which makes me suspect that the answer to BA's original question is also no. | |
| Oct 22, 2010 at 0:04 | comment | added | anon | Is A a map from H to H? By "automorphism in Banach-spaces sense," do you mean that A is a bounded and bijective linear map from H to H? In the definition of "realizable," is the "isomorphism" from H to the L^2 space a unitary from one Hilbert space to the other, or just an invertible linear map? What is the "direct image of A"? What does "realized" mean? Maybe your definition has the form: A is realizable if there is a map f [of some kind] from H to some L^2 space and a map g [of some kind: maybe a composition operator?] on the L^2 space with A = f^{-1} g f. Is this true? Please clarify. | |
| Oct 21, 2010 at 21:39 | history | asked | Bad English | CC BY-SA 2.5 |