## Friday, June 25, 2021

### On Choice

This is a follow-up to a Twitter discussion with John Baez that did not fit into the 260 character limit. And before starting, I should warn you that I have never studied set theory in any seriousness and everything I am about to write here is only based on hearsay and is probably wrong.

I am currently teaching "Mathematical Statistical Physics" once more, large part of which is to explain the operator algebraic approach to quantum statistical physics, KMS states and all that. Part of this is that I cover states as continuous linear functionals on the observables (positive and normalised) and in the example of B(H), the bounded linear operators on a Hilbert space H, I mention that trace class operators $$\rho\in {\cal S}^1({\cal H})$$ give rise to those via $$\omega(a) = {\hbox tr}(\rho a).$$

And "pretty much all" states arise in this way meaning that the bounded operators are the (topological) dual to trance class operators but, for infinite dimensional H, not the other way around as the bi-dual is larger. There are bounded linear functionals on B(H) that don't come from trace class operators. But (without too much effort), I have never seen one of those extra states being "presented". I have very low standards here for "presented" meaning that I suspect you need to invoke the axiom of choice to produce them (and this is also what John said) and for class this would be sufficient. We invoke choice in other places as well, like (via Hahn-Banach) every C*-algebra having faithful representations (or realising it as a closed Subalgebra of some huge B(H)).

So much for background. I wanted to tell you about my attitude towards choice. When I was a student, I never had any doubt about it. Sure, every vector space has a basis, there are sets that are not Lebesque measurable. A little abstract, but who cares. It was a blog post by Terrence Tao that made me reconsider that (turns out, I cannot find that post anymore, bummer). It goes like this: On one of these islands, there is this prison where it is announced to the prisoners that the following morning, they are all placed in a row and everybody is given a hat that is either black or white. No prisoner can see his own hat but all of those in front of him. They can guess their color (and all other prisoners hear the guess). Those who guess right get free while those who guess wrong get executed. How many can go free?

The answer is: All but one half. The last one says white if he sees an even number of white hats in front of him. Then all the others can deduce their color from this parity plus the answers of the other prisoners behind them. So all but the last prisoner get out and the last has a 50% chance.

But that was too easy. Now, this is a really big prison and there are countably infinitely many prisoners. How many go free? When they are placed in a row, the row extends to infinity only in one direction and they are looking in this infinite direction. The last prisoner sees all the other prisoners.