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Consider $L^2(A)$ and $L^2(B)$. If $\{a_i\}$ is an o.n basis of $L^2(A)$, how many linear homeomorphisms $F:L^2(A) \to L^2(B)$ do there exist such that $Fa_i$ is an orthonormal basis of $L^2(B)$?

Is this a very restrictive assumption on the maps, if I wanted to discuss something about homeomorphism between the spaces?

aere
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1 Answers1

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If you fix one such $F$, the rest are found by composition with elements of the unitary group of $L^2(A)$. The unitary group of a Hilbert space (sometimes called the Hilbert group, $\mathrm{Hilb}\,(L^2(A))$) is very large: it contains a copy of every compact group. At the same time, it is contractible, by Kuiper's theorem.

You may also be interested in MO discussions on the subject:

Community
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40 votes
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  • Thank you for your answer. I am not familiar with groups at all so it is hard for me to understand the mathoverflow threads, but I will read some more. You say it's very large which is good news (I think). Are you aware of some result like: for every linear homemorphism $G$ between $L^2(A)$ and $L^2(B)$ there exists a $F$ mapping o.n basis to o.n basis that is arbitrarily close to $G$ in some norm. – aere Jul 15 '13 at 16:00
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    @aere Of course, that would be false. If a map sends ONB to ONB, it is an isometric isomorphism between Hilbert spaces; it preserves the norm of every vector. In particular, its operator norm is $1$. If you take any linear bijective map between Hilbert spaces, it need not be close to such a thing. Try to test your conjectures on the example with both spaces being two-dimensional... – 40 votes Jul 15 '13 at 16:04