# Hilbert space automorphisms realized as induced by transformations of some base-spaces

Following question may be soft. Fix abstract hilbert space H and consider any automorphism A in banach-spaces sence (i.e. no conditions on metric). Call A is realizable if exist measure space $(X,\mu)$ and isomorphism from H onto (or into, depending on the preferred definition) $L_2(X,\mu)$ such that direct image of A realized by some homeomorphism (or other type of transformations) of X i.e. in "new coordinates" A induced by automorphism of X. Set of all realizable automorphisms is conjugate-invariant. My main question: Is it group? i.e. can i construct measure space $(X,\mu)$ from $(X_1,\mu_1)$ and $(X_2,\mu_2)$ with operators A,B induced by transformations of $X_1,X_2$ respectively, such that AB realizible on $(X,\mu)$? Sorry for bad english.

Added: simple argument that it haven't group structure on usual multiplicity structure gived by Costabel. But questions related to common construction described bellow are stayes.

Yemon Choi, your point of view is proper. In fact we working on some discrete category of functional spaces (L_2) on certain-type measure-spaces (probability spaces, topological groups and so on). Morphisms in such category called by "transformations", and in my first message they are obtained by transformations on themselves measure-spaces. We can consider our category with new object - abstract hilbert space H with morphisms from H to each other object - just bounded and left-invertible linear maps (or isometries, as you want). Inverse arrows just bounded maps. Obtained morphisms from H to itself called realizable. For instance, consider such category where transformations is multiplication on square-integrable function and all maps from H is isometries, so spectral theorem gives that self-adjoint operators in precisely are realizible in such category. I really upset by the fact, that obtained sets of realizable operators do not have any natural algebraic structure. So that on it moment question may be closed, but any comments are welcome. May be certain category of measure-spaces (lie groups with haar measure) will gives something interesting? May be this sets need to be considering in total?

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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. –  anon Oct 22 '10 at 0:04
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. –  Yemon Choi Oct 22 '10 at 0:45
anon, in following answer i will try to describe common construction. –  Bad English Oct 28 '10 at 21:20

Note first that the only multiples of the identity that are realizable are zero and the identity itself (Proof: Whatever your definition of "transformation of $X$" is, it will map characteristic functions of sets of finite measure to functions that take the values $0$ and $1$ only).
Now take $H=\mathbb{R}^2$ and let $A$ be defined by the matrix $0\ \ \ 1\choose 1\ \ \ 0$. This is realizable, by transposition of indices in $H\equiv \ell^2(\{0,1\})$. Let $U$ be the automorphism of $H$ defined by the matrix $1\ \ \ \ 0\choose \ \ 0\ \ -1$. Then $B=UAU^{-1}$ is also realizable. But $B=-A$ and $AB=-1$, which is not realizable.