**Meta**

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 . A Hilbert basis for 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 : 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 will only be a subspace of , since a vector might only be in the completion of the set of finite combinations of vectors in .

So, for example, the space of square-summable sequences, , has a Hilbert basis consisting of sequences like 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 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?

September 12, 2008 at 3:20 am

To do good a web search for this question, you need to use the phrase “Hamel basis” or “Hamel dimension”. “Hamel basis” is a a commonly used name for what you’re calling a “vector-space basis”, mainly in situations when there’s some other sort of basis, like an orthonormal basis for a Hilbert space. “Hamel dimension” is the cardinality of a Hamel basis.

But, I haven’t yet found the answer to your question!

I’m embarrassed, because I was going to try to solve it, and then I realized I don’t even know the Hamel dimension of your garden-variety countably-infinite-dimensional Hilbert space. Clearly the Hamel dimension is bigger than alelph-nought. And clearly it’s less than or equal to aleph-nought to the (2 to the aleph_nought), since that’s the

cardinalityof this Hilbert space. But that leaves a lot of wiggle room.Is there some sort of “formula” for the Hamel dimension in this case?

If so, can we generalize this formula to fancier cases, by replacing aleph-nought with bigger cardinals?

September 12, 2008 at 5:30 pm

After a good night’s sleep I came to my senses a bit and realized that “aleph-nought to the (2 to the aleph-nought)” is just a long-winded way of saying “2 to the aleph-nought”. The cardinality of a countable-dimensional Hilbert space is the cardinality of the continuum. Why? Since the coordinates of a vector form a square-summable countable list of real numbers, and the set of countable lists of real numbers has cardinality equal that of the real numbers.

So, I hope I convinced you that the Hamel dimension of a countable-dimensional Hilbert space is greater than aleph-nought and less than or equal to 2 to the aleph-nought.

So, if you’re willing to buy the Continuum Hypothesis, it has to be 2 to the aleph-nought!

What if we don’t assume the Continuum Hypothesis?

September 15, 2008 at 5:34 pm

If we don’t assume CH, I would guess that there are models with many different Hamel dimensions for a countable-dimension Hilbert space. In any case, I’m prepared to assume CH as long as we do it explicitly. I’m not sure I have an opinion on whether it’s “true” (whatever that means in this context – I’m not sure how you would check).

But perhaps we don’t need to assume CH to answer the question. It seems that the Hamel and Hilbert dimensions are related by:

So this turns my original question into one about large-cardinal arithmetic: is this function 1-1? I’m no expert, but I’m thinking this might depend on some other axioms about large cardinals (though if anyone knows otherwise I’d like to know for sure). For the kind that are liable to come up in any “real” application, such as or even , my vague recollections of cardinal arithmetic imply that it is 1-1.

Non-separable Hilbert spaces may come up in “real life” (e.g. in LQG), but probably not ones with dimensions up among the exotic large cardinals – inaccessibles or hyper-inaccessibles etc.

September 17, 2008 at 12:37 am

“For the kind that are liable to come up in any “real” application, such as d = \aleph_0 or even d = 2^{\aleph_0}, my vague recollections of cardinal arithmetic imply that it is 1-1.”

For these choices of d, one can actually “force” things like d^\aleph_0 = d^\aleph_1 = aleph_2 and a tremendous number of other things. A very general result was obtained by Easton that you can force 2^c for all

regularcardinals c [cardinals equal to their own cofinality] to be just about anything you like, provided that(1) c \leq c’ implies 2^c \leq 2^c’, and

(2) cof(2^c) > c.

So to settle this type of question, you are pretty much forced to add axioms to ZFC.

September 17, 2008 at 2:19 am

Thanks for clearing that up, Todd! I suspected that might be the case, but I wasn’t aware there was that much leeway.

September 7, 2011 at 6:07 pm

How do you prove that the vector-space basis is uncountably infinite ?

I’ve been unsuccessfully searching for study material on this . The closest thing I came up with is that in works for spaces over Q , so it “should” work for spaces over R , but that’s not an argument .

September 8, 2011 at 1:45 pm

Dan:

A little looking around revealed a claim (under “Hamel dimension” in the Wikipedia article on Banach spaces) that this follows from the Baire Category Theorem. This is the one that says that for certain topological spaces (in particular, complete metric spaces with the metric topology, such as a Banach space ), the intersection of any countable collection of dense subsets is dense. The way I would use this to give the argument is like this:

Say we’ve already used the axiom of choice to argue that there is a Hamel basis, then suppose it’s countable, of the form , and get a contradiction. Then let be the subspace spanned by the first vectors in the basis, so that is a dense open subset of . Thus, the intersection of the over all is also a dense open subset of . But this is exactly the complement of the union of all the , which, if we really had a Hamel basis, would have been all of .

Presumably there are some details to sort out there, but that seems about right to me.

September 9, 2011 at 2:50 pm

Thanks for the reply :) . Subsequently I found this

http://data.imf.au.dk/kurser/advanalyse/F06/lecture1.pdf

, which was of help . (it also has a proof of the Baire theorem ) .

May 25, 2012 at 11:32 am

Hello,

Found your blog while searching for the Hamel dimension of Hilbert space. Your formula $\latex d_{\text{Hamel}} = d^{\aleph_0}$ was exactly what I was looking for.

By the way, $\latex d\mapsto d^{\aleph_0}$ is certainly not injective, even in the simplest cases. For instance, if $\latex d\geq 2$ then $d^\aleph_0 \geq 2^\aleph_0$. On the other hand, $\latex (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0^2} = 2^\aleph_0$.