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I like writing things for this blog, and so far I’ve tended to go in for long-ish comments or explanations of things. I’d definitely like to keep doing that, since it’s helpful to put some thoughts in order and get them out there. I also would like to try writing some shorter update-style entries about little things. One reason is that it might be useful to sit down more often with the plan of writing something here.

I’m also thinking it might be interesting to broaden out the range of things I write about, since my original theme is pretty open-ended. We’ll see how that goes.

Update

I started sitting in on Masoud Khalkhali’s Noncommutative Geometry seminar for this term, which is mostly attended by his graduate students. This term, at least to begin with, he’s planning to fill them in on some basics of operator algebras, (C*-algebras, von Neumann algebras, classification theorems, etc.) I’d like to learn some of this in any case – not least because it’s used in a lot of the standard work on groupoids. The first seminar was a refresher on the basics of Hilbert spaces – also well worth going through from the beginning again since I’ve been thinking about what’s involved in categorification of infinite-dimensional Hilbert spaces.

So we looked at the basic definitions and a little of the history of how von Neumann came up with the idea of a Hilbert space (hence the name) in the late 1920’s under the influence of Hilbert’s work on classical analysis, and the
then-recent work on quantum mechanics. (The reason Hilbert spaces seem to do just what quantum mechanics needs them to do (at the beginning) is that they were specifically designed for it). Some basic tools, like the (complex) polarization identity, the Cauchy-Schwartz inequality and its corollaries like the triangle inequality, and the important theorem that closed convex sets in Hilbert space have unique minimum-norm elements.

There was one question that came up which I’m not sure of the answer to. Remember that there’s a difference between a “Hilbert basis” and a vector-space basis for a Hilbert space \mathcal{H}. A Hilbert basis for \mathcal{H} is just a maximal orthonormal set of vectors (which always exists, by an argument involving Zorn’s Lemma). A vector-space basis has to span \mathcal{H}: in other words, any vector must be expressible as a finite linear combination of basis vectors. So, in particular, in the infinite-dimensional case, the linear span of a Hilbert basis B will only be a subspace of \mathcal{H}, since a vector might only be in the completion of the set of finite combinations of vectors in B.

So, for example, the space of square-summable sequences, l^2(\mathbb{N}), has a Hilbert basis consisting of sequences like (0, \dots, 0, 1, 0 \dots) with just a single nonzero entry. This is a countably infinite set. However, not all square-summable sequences can be written as a finite linear combination of these. Using the axiom of choice (!) it’s possible to prove that there is some vector-space basis, but it will be a uncountably infinite. In fact, no Hilbert space has countably-infinite vector-space dimension.

Now, Hilbert spaces are determined up to isomorphism (which in this context means isometry) by their Hilbert dimension – the cardinality of a Hilbert basis. For finite dimensional spaces, this is the same as the vector space dimension, so there’s no trouble here. The question is: are Hilbert spaces determined up to isometry by their vector space dimension?

In other words: is the relationship between the cardinalities of a vector-space basis and a Hilbert basis for \mathcal{H} a 1-1 relationship? Or can there be Hilbert spaces with different Hilbert dimension (therefore non-isometric) which have the same vector-space dimension? If there is such a counterexample, it will have at least one of the Hilbert spaces non-separable (i.e. no countable Hilbert basis).

In fact, since this question involves large infinite cardinals, I wouldn’t be terribly surprised to learn that the answer is indeterminate in ZFC, but neither do I see a good argument that it should be. A quick web-search doesn’t reveal the answer at once. Does anyone know?

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