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?

Think about it.

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.

Think about it.

In this case, the answer is: Almost all, all but finitely many. Out of all infinite sequences of black/white form equivalence classes where two sequences are equivalent if they differ at at most finitely many places, you could say they are asymptotically equal. Out of these equivalence classes, the axiom of choice allows us to pick one representative of each. The prisoners memorise these representatives (there are only aleph1 many, they have big brains). The next morning in the court of the prison, all prisoners can see almost all other prisoners, so they know which equivalence class of sequences was chosen by the prison's director. Now, every prisoner announces the color of the memorised representative at his position and by construction, only finitely many are wrong.

This argument as raised some doubts in me if I really want choice to be true. I came to terms with it and would describe my position as agnostic. I mainly try to avoid it and better not rely too much on constructions that invoke it. And for my physics applications this is usually fine.

But it can also be useful in everyday's life: My favourite example of it is in the context of distributions. Those are, as you know, continuous linear functionals on test functions. The topology on the test functions, however, is a bit inconvenient, as you have to check all (infinitely many) partial derivates to converge. So you might try to do the opposite thing: Let's study a linear functional on test functions that is not continuous. Turns out, those are harder to get hold of than you might think: You might think this is like linear operators where continuity is equivalent to boundedness. But this case is different: You need to invoke choice to find one. But this is good, since this implies that every concrete linear functional that you can construct (write down explicitly) is automatically continuous, you don't have to prove anything!

This type of argument is a little dangerous: You really need more than "the usual way to get one is to invoke choice". You really need that it is equivalent to choice. And choice says that for every collection of sets, the Cartesian product is non-empty. It is the "every" that is important. The collection that consists of copies of the set {apple} trivially has an element in the Cartesian product, it is (apple, apple, apple, ...). And this element is concrete, I just constructed it for you.

This caveat is a bit reminiscent of a false argument that you can read far too often: You show that some class of problems is NP-complete (recent incarnations: deciding if an isolated string theory vacuum as a small cosmological constant, deciding if a spin-chain model is gapped, determining the phase structure of some spin chain model, ...) and then arguing that these problems are "hard to solve". But this does not imply that a concrete problem in this class is difficult. It is only that solving all problems in this class of problems is difficult. Every single instance of practical relevance could be easy (for example because you had additional information that trivialises the problem). It could well be that you are only interested in spin chain Hamiltonians of some specific form and that you can find a proof that all of them are gapped (or not gapped for that matter). It only means that your original class of problems was too big, it contained too many problems that don't have relevance in your case. This could for example well be for the string theory vacua: In the paper I have in mind, that was modelled (of course actually fixing all moduli and computing the value of the potential in all vacua cannot be done with today's technology) by saying there are N moduli fields and each can have at least two values with different values of its potential (positive or negative) and we assume you simply have to add all those values. Is there one choice of the values of all moduli fields such that the sum of the energies is epsilon-close to 0? This turns out to be equivalent to the knapsack-problem which is known to be NP-complete. But for this you need to allow for all possible values of the potential for the individual moduli. If, for example, you knew the values for all moduli are the same, that particular incarnation of the problem is trivial. So just knowing that the concrete problem you are interested in is a member of a class of problems that is NP-complete does not make that concrete problem hard by itself.

What is you attitude towards choice? When is the argument "Here is a concrete, constructed example of a thing x. I am interested in some property P of it. To show there are y that don't have property P, I need to invoke choice. Does this prove x has property P?" to be believed?

2 comments:

Anonymous said...

The reason you can't find it is because it wasn't a Terry Tao post (although he did comment on it). It's at the Everything Seminar. Also, after being very offended when I saw an economist use Hahn-Banach to prove a theorem in economics (indicative of what's very wrong in econ theory, but that's another rant), I did a little digging, and you don't actually need choice (or the BPIT, more precisely) in the case of a separable space, where there's a constructive version.

Aaron

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