Tuesday, August 25, 2020

When can we have exact symmetries of hyperreal probabilities?

In many interesting cases, there is no way to define a regular hyperreal-valued probability that is invariant under symmetries, where “regular” means that every non-empty set has non-zero probability. For instance, there is no such measure for all subsets of the circle with respect to rotations: the best we can do is approximate invariance, where P(A)−P(rA) is infinitesimal for every rotation. On the other hand, I have recently shown that there is such a measure for infinite sequences of fair coin tosses where the symmetries are reversals at a set of locations.

So, here’s an interesting question: Given a space Ω and a group G of symmetries acting on Ω, under what exact conditions is there a hyperreal finitely-additive probability measure P defined for all subsets of Ω that satisfies the regularity condition P(A)>0 for all non-empty A and yet is fully (and not merely approximately) invariant under G, so that P(gA)=P(A) for all g ∈ G and A ⊆ Ω?

Theorem: Such a measure exists if and only if the action of G on Ω is locally finite. (Assuming the Axiom of Choice.)

The action of G on Ω is locally finite iff for every x ∈ Ω and every finitely-generated subgroup H of G, the orbit Hx = {hx : h ∈ H} of x under H is finite. In other words, we have such a measure provided that applying the symmetries to any point of the space only generates finitely many points.

This mathematical fact leads to a philosophical question: Is there anything philosophically interesting about those symmetries whose action is locally finite? But I’ve spent so much of the day thinking about the mathematical question that I am too tired to think very hard about the philosophical question.

Sketch of Proof of Theorem: If some subset A of Ω is equidecomposable with a proper subset A′, then a G-invariant measure P will assign equal measure to both A and A′, and hence will assign zero measure to the non-empty set A − A′, violating the regularity condition. So, if the requisite measure exists, no subset is equidecomposable with a proper subset of itself, which by a theorem of Scarparo implies that the action of G is locally finite.

Now for the converse. If we could show the result for all finitely-generated groups G, by using ultraproduct along an ultrafilter on the partially ordered set of all finitely generated subgroups of G we could show this for a general G.

So, suppose that G is finitely generated and the orbit of x under G is finite for all x ∈ Ω. A subset A of G is said to be G-invariant provided that gA = A for all g ∈ G. The orbit of x under G is always G-invariant, and hence every finite subset of A is contained in a finite G-invariant subset, namely the union of the orbits of all the points in A.

Consider the set F of all finite G-invariant subsets of Ω. It’s worth noting that every finite subset of G is contained in a finite G-closed subset: just take the union of the orbits under G. For A ∈ F, let PA be uniform measure on A. Let F* = {{B ∈ F : A ⊆ B}:A ∈ F}. This is a non-empty set with the finite intersection property. Let U be an ultrafilter extending F*. Let *R be the ultraproduct of the reals over F with respect to U, and let P(C) be the equivalence class of the function A ↦ PA(A ∩ C) on F. Note that C ↦ PA(A ∩ C) is G-invariant for any G-invariant set A, so P is G-invariant. Moreover, P(C)>0 if C ≠ ∅. For let C′ be the orbit of some element of C. Then {B ∈ F : C′⊆B} is in F*, and PA(A ∩ C′) > 0 for all A such that C′⊆A, so the set of all A such that PA(A ∩ C′) > 0 is in U. It follows that P(C′) > 0. But C′ is the orbit of some element x of C, so every singleton subset of C′ has the same P-measure as {x} by the G-invariance of P. So P({x}) = P(C′)/|C′| > 0, and hence P(C)≥P({x}) > 0.

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