An Innocent Problem

Introduction

In meditating on my axiomatic approach to life I recalled a problem from Papa Rudin:

All Hilbert spaces are basically the same.

More precisely, given two Hilbert spaces, one of them is isometrically isomorphic to a subspace of the other. This seems pretty clear once you have the definitions straight, and you can easily prove it by looking at orthonormal bases for the two spaces.

But hang on, how do you know that every Hilbert space has an orthonormal basis? That is a theorem!

Every Hilbert space has an orthonormal basis.

You know that this feels right and even have an idea for proving it. Namely, Gram-Schmidt: you just keep subtracting projections away from stuff not in your span and appending the results to your growing basis. In finite dimensions this terminates. In separable spaces, this converges. How?

We have elided important details; we did not say what is meant by basis, even. A basis is a linearly independent set whose span is dense in your Hilbert space, henceforth $H$. An orthonormal basis is a basis that is orthonormal 😉

So, phew, we do not have to produce a set of vectors whose span gives us exactly $H$. In particular, the span is not closed under limits and we would not have completeness of $H$ if that were the definition. In fact, the span of a basis of $H$ is a proper subspace of an infinite-dimensional $H$.

If you wanted to ‘construct’ a basis for a separable $H$, then, you could take from the definition a countable dense subset and hit it with Gram-Schmidt. It might be more delicate than that, but bear with me because we’ve got bigger fish to fry 🐟

What happens in a non-separable $H$? You must gift yourself the construction. This wish is granted by the Hausdorff maximal principle, a variant of Zorn’s lemma.

Both are equivalent to the axiom of choice.

That Problem

We are now on solid ground, aren’t you glad? We have settled our debts and nothing can go wrong from here on out.

Suppose that $H_1$ and $H_2$ are Hilbert spaces and $B_1$ and $B_2$, respectively, orthonormal bases for them. This is going to be a cinch, but first we have to find out which basis has the larger cardinality - its $H$ will have a subspace isometrically isomorphic to the other $H$ and we will be done.

OK, great so which one is it? Surely one of the following three conditions must hold:

1. $B_2$ is bigger than $B_1$
2. $B_1$ is bigger than $B_2$
3. $B_1$ and $B_2$ have the same size

Yes, but it is not “immediate” and it does not “fall out” from the definition of (comparison of) cardinality. It is called the law of trichotomy and it is equivalent to the axiom of choice 😬

Harbor any delusion long enough and its attendant challenges will stalk you with predatory eyes, leaping into view just when you are least able to contend with them.¹

And A Wicked One

All this Hilbert space talk got me thinking: is the invariant subspace problem solved yet? Back in May, Per Enflo claimed a proof and today I learned of it. Bandwidth limitations preclude a deep dive, so I wrote an ancient friend and he wrote back:

Good to hear from you Martin.

The proof did not last a couple of months. There is a gap there. Very technical stuff, I did not even try to untangle.

Warm regards,

[REDACTED]

Such is life 🌎🌍🌏