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Let $\mathcal{H}$ be an infinite dimensional separable (complex) Hilbert space. What is a natural space which parameterizes the choices of orthonormal bases for $\mathcal{H}$?

It seems like one option, in analogy to the finite-dimensional Stiefel manifold (something I know nothing about beyond what I just glanced at on the wikipedia page), would be to consider the set of all unitary operators from $\ell^2(\mathbb{N})\to\mathcal{H}$, with some topology. What kind of topology would one consider? Ideally, I'd like it to be Polish.

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Since you're looking for a basis for the full space, you're actually looking at the unitary group, not the Stiefel manifold. For topological considerations, look for example at Kuiper's theorem:'s_theorem – Yoav Kallus May 14 '13 at 1:36
@Yoav For some reason, I was trying to avoid identifying my space with $\ell^2$ (i.e., fixing a basis to begin with), but if I do so, then you're right, then unitary group does the job, and is Polish in the strong operator topology. – Iian Smythe May 14 '13 at 1:47
That said, I'm open to other parameterizing spaces. – Iian Smythe May 14 '13 at 1:49
If one wants to use the unitary group of $\ell^2$ for this, then one would have to quotient out by the permutation operators. $\:$ Otherwise, one could take the topology generated by the sets of bases that intersect an open subset of $\mathcal{H}$. $\;\;$ – Ricky Demer May 14 '13 at 2:03
up vote 3 down vote accepted

You could look at the set of operators in $B(H)$ which are diagonalized by a given basis. For each choice of orthonormal basis, this gives you an atomic maximal abelian von Neumann algebra (atomic masa). Now there is a natural topology on the space of all von Neumann algebras sitting inside $B(H)$ called the Effros-Maréchal topology, so you could just restrict this topology to the set of atomic masas.

Here is a link that may be helpful.

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