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Ergodic systems with regular orbits concentrate on one orbit

Statement

Assume AC. Let (U,P) be an ergodic system of imprimitivity for a Borel action of a group G on a standard Borel space X, acting on a nonzero separable Hilbert space, and suppose the orbit equivalence relation of the action is regular: there is a countable family E1,E2,… of G-invariant Borel subsets of X such that every orbit is the intersection of the sets En that contain it. Equivalently, some countable family of invariant Borel sets separates distinct orbits, which is the condition that the orbit space is countably separated. Then there is an orbit C⊆X with P(X∖C)=0.

Facts & Assumptions

Given: AC, the Borel G-space X, the ergodic system of imprimitivity (U,P) on a nonzero separable Hilbert space, and a countable family (En) of invariant Borel sets as in the statement.

[F1]

(U,P) is a strongly continuous unitary representation together with a PVM with UgP(E)Ug−1=P(gE), and ergodicity means that every Borel E with UgP(E)Ug−1=P(E) for all g satisfies P(E)=0 or P(E)=I; for invariant Borel E one has P(E) invariant (Systems of imprimitivity for a Borel G-space, Left group actions, transitive actions, and faithful actions, A measurable function between measurable spaces).

[F2]

Projections in the range of a PVM satisfy P(A)P(B)=P(A∩B); a projection P(A) is zero exactly when the scalar measures Ex(A)=⟨P(A)x,x⟩ vanish for all x; and P is strongly countably additive (Scalar and complex measures from a pvm, Bounded borel pvm integral).

[F3]

X is standard Borel, so Borel sets are closed under countable unions and intersections; invariance of a Borel set E means gE=E for all g and implies invariance of its complement (Standard Borel spaces).

[F4]

AC is the standing hypothesis; it is inherited from the ambient system (The Axiom of Choice, Multiplicity model of a projection-valued measure over a standard Borel base).

Proof

technique · direct

Given: AC, the ergodic system (U,P) and the countable invariant family (En).

1.1F3

The two regularity formulations are equivalent. If every orbit is the intersection of the invariant Borel sets containing it, the family (En) separates distinct orbits: if x,y lie in different orbits and y belonged to every En containing x, then y would lie in the intersection defining the orbit of x. Conversely, adjoin the complements to a countable separating family and reenumerate it as (En). Then for every x the intersection Ix=⋂{En:x∈En} is contained in the orbit of x: a point y∉Gx is separated from x by some En, and replacing En by its complement if necessary (also invariant Borel by [F3]) gives En∋x, En∌y; the reverse inclusion holds because each En is invariant.

1.2F1

Each P(En) is 0 or I: since En is invariant, UgP(En)Ug−1=P(gEn)=P(En) for every g, so ergodicity applies.

2.1F2F3step 1.2

Define Fn=En if P(En)=I and Fn=X∖En if P(En)=0; each Fn is invariant Borel and P(Fn)=I. For C:=⋂nFn one has X∖C=⋃n(X∖Fn) with P(X∖Fn)=0; strong countable additivity gives P(⋃n(X∖Fn))=lim⁡NP(⋃n≤N(X∖Fn)) and the scalar measures of each finite union vanish, so P(X∖C)=0.

3.1step 1.1step 2.1F3

C is nonempty because P(C)=I≠0 on the nonzero Hilbert space, and C is invariant. Choose x∈C. Its orbit satisfies Gx⊆C since C is invariant. Conversely, if y∈C and En∋x, then either Fn=En, in which case y∈C⊆En; or Fn=X∖En and x∉En, a contradiction. Hence y belongs to every En containing x, so by regularity y∈Gx. Therefore C=Gx is a single orbit and is Borel as a countable intersection.

4.1step 2.1step 3.1F4∎

Combining [step 2.1] and [step 3.1]: the invariant Borel set C is exactly one orbit and P(X∖C)=0, which is the concentration claim.

Depends on

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