Hopf decomposition

• Wandering sets and dissipative actions. A measurable subset W of X is wandering if its characteristic function q = χ_W in A = L^∞(X) satisfies qτⁿ(q) = 0 for all n; thus, up to null sets, the translates Tⁿ(W) are disjoint. An action is called dissipative if X = ∐ Tⁿ(W) for some wandering set W.
• Conservative actions. If X has no wandering subsets of positive measure, the action is said to be conservative.
• Incompressible actions. An action is said to be incompressible if whenever a measurable subset Z satisfies T(Z) ⊊ Z then Z \ TZ has measure zero. Thus if q = χ_Z and τ(q) ≤ q, then τ(q) = q.
• Recurrent actions. An action T is said to be recurrent if q ≤ τ(q) ∨ τ²(q) ∨ τ³(q) ∨ ... for any q = χ_Y.
• Infinitely recurrent actions. An action T is said to be infinitely recurrent if q ≤ τ^m (q) ∨ τ^{m + 1}(q) ∨ τ^m+2(q) ∨ ... for any q = χ_Y and any m ≥ 1.

Theorem. If T is an invertible transformation on a measure space (X,μ) preserving null sets, then the following conditions are equivalent on T (or its inverse):[1]

1. T is conservative;
2. T is recurrent;
3. T is infinitely recurrent;
4. T is incompressible.

Since T is dissipative if and only if T⁻¹ is dissipative, it follows that T is conservative if and only if T⁻¹ is conservative.

If T is conservative, then r = q ∧ (τ(q) ∨ τ²(q) ∨ τ³(q) ∨ ⋅⋅⋅)^⊥ = q ∧ τ(1 - q) ∧ τ²(1 -q) ∧ τ³(q) ∧ ... is wandering so that if q < 1, necessarily r = 0. Hence q ≤ τ(q) ∨ τ²(q) ∨ τ³(q) ∨ ⋅⋅⋅, so that T is recurrent.

If T is recurrent, then q ≤ τ(q) ∨ τ²(q) ∨ τ³(q) ∨ ⋅⋅⋅ Now assume by induction that q ≤ τ^k(q) ∨ τ^k+1(q) ∨ ⋅⋅⋅. Then τ^k(q) ≤ τ^k+1(q) ∨ τ^k+2(q) ∨ ⋅⋅⋅ ≤ . Hence q ≤ τ^k+1(q) ∨ τ^k+2(q) ∨ ⋅⋅⋅. So the result holds for k+1 and thus T is infinitely recurrent. Conversely by definition an infinitely recurrent transformation is recurrent.

Now suppose that T is recurrent. To show that T is incompressible it must be shown that, if τ(q) ≤ q, then τ(q) ≤ q. In fact in this case τⁿ(q) is a decreasing sequence. But by recurrence, q ≤ τ(q) ∨ τ²(q) ∨ τ³(q) ∨ ⋅⋅⋅ , so q ≤ τ(q) and hence q = τ(q).

Finally suppose that T is incompressible. If T is not conservative there is a p ≠ 0 in A with the τⁿ(p) disjoint (orthogonal). But then q = p ⊕ τ(p) ⊕ τ²(p) ⊕ ⋅⋅⋅ satisfies τ(q) < q with q − τ(q) = p ≠ 0, contradicting incompressibility. So T is conservative.

Theorem. If T is an invertible transformation on a measure space (X,μ) preserving null sets and inducing an automorphism τ of A = L^∞(X), then there is a unique τ-invariant p = χ_C in A such that τ is conservative on pA = L^∞(C) and dissipative on (1 − p)A = L^∞(D) where D = X \ C.[2]

Without loss of generality it can be assumed that μ is a probability measure. If T is conservative there is nothing to prove, since in that case C = X. Otherwise there is a wandering set W for T. Let r = χ_W and q = ⊕ τⁿ(r). Thus q is τ-invariant and dissipative. Moreover μ(q) > 0. Clearly an orthogonal direct sum of such τ-invariant dissipative q′s is also τ-invariant and dissipative; and if q is τ-invariant and dissipative and r < q is τ-invariant, then r is dissipative. Hence if q₁ and q₂ are τ-invariant and dissipative, then q₁ ∨ q₂ is τ-invariant and dissipative, since q₁ ∨ q₂ = q₁ ⊕ q₂(1 − q₁). Now let M be the supremum of all μ(q) wirh q τ-invariant and dissipative. Take q_n τ-invariant and dissipative such that μ(q_n) increases to M. Replacing q_n by q₁ ∨ ⋅⋅⋅ ∨ q_n, t can be assumed that q_n is increasing to q say. By continuity q is τ-invariant and μ(q) = M. By maximality p = I − q is conservative. Uniqueness is clear since no τ-invariant r < p is dissipative and every τ-invariant r < q is dissipative.

Corollary. The Hopf decomposition for T coincides with the Hopf decomposition for T⁻¹.

Since a transformation is dissipative on a measure space if and only if its inverse is dissipitive, the dissipative parts of T and T⁻¹ coincide. Hence so do the conservative parts.

Corollary. The Hopf decomposition for T coincides with the Hopf decomposition for Tⁿ for n > 1.

If W is a wandering set for T then it is a wandering set for Tⁿ. So the dissipative part of T is contained in the dissipative part of Tⁿ. Let σ = τⁿ. To prove the converse, it suffices to show that if σ is dissipative, then τ is dissipative. If not, using the Hopf decomposition, it can be assumed that σ is dissipative and τ conservative. Suppose that p is a non-zero wandering projection for σ. Then τ^a(p) and τ^b(p) are orthogonal for different a and b in the same congruence class modulo n. Take a set of τ^a(p) with non-zero product and maximal size. Thus |S| ≤ n. By maximality, r is wandering for τ, a contradiction.

Corollary. If an invertible transformation T acts ergodically but non-transitively on the measure space (X,μ) preserving null sets and B is a subset with μ(B) > 0, then the complement of B ∪ TB ∪ T²B ∪ ⋅⋅⋅ has measure zero.

Note that ergodicity and non-transitivity imply that the action of T is conservative and hence infinitely recurrent. But then B ≤ T^m (B) ∨ T^{m + 1}(B) ∨ T^m+2(B) ∨ ... for any m ≥ 1. Applying T^−m, it follows that T^−m(B) lies in Y = B ∪ TB ∪ T²B ∪ ⋅⋅⋅ for every m > 0. By ergodicity μ(X \ Y) = 0.

Let (X,μ) be a measure space and S_t a non-sngular flow on X inducing a 1-parameter group of automorphisms σ_t of A = L^∞(X). It will be assumed that the action is faithful, so that σ_t is the identity only for t = 0. For each S_t or equivalently σ_t with t ≠ 0 there is a Hopf decomposition, so a p_t fixed by σ_t such that the action is conservative on p_tA and dissipative on (1−p_t)A.

• For s, t ≠ 0 the conservative and dissipative parts of S_s and S_t coincide if s/t is rational.[3]

This follows from the fact that for any non-singular invertible transformation the conservative and dissipative parts of T and Tⁿ coincide for n ≠ 0.

• If S₁ is dissipative on A = L^∞(X), then there is an invariant measure λ on A and p in A such that

1. p > σ_t(p) for all t > 0
2. λ(p – σ_t(p)) = t for all t > 0
3. σ_t(p) ${\displaystyle \uparrow }$ 1 as t tends to −∞ and σ_t(p) ${\displaystyle \downarrow }$ 0 as t tends to +∞.

Let T = S₁. Take q a wandering set for T so that ⊕ τⁿ(q) = 1. Changing μ to an equivalent measure, it can be assumed that μ(q) = 1, so that μ restricts to a probability measure on qA. Transporting this measure to τⁿ(q)A, it can further be assumed that μ is τ-invariant on A. But then λ = ∫¹
₀ μ ∘ σ_t dt is an equivalent σ-invariant measure on A which can be rescaled if necessary so that λ(q) = 1. The r in A that are wandering for Τ (or τ) with ⊕ τⁿ(r) = 1 are easily described: they are given by r = ⊕ τⁿ(q_n) where q = ⊕ q_n is a decomposition of q. In particular λ(r) =1. Moreover if p satisfies p > τ(p) and τ^–n(p) ${\displaystyle \uparrow }$ 1, then λ(p– τ(p)) = 1, applying the result to r = p – τ(p). The same arguments show that conversely, if r is wandering for τ and λ(r) = 1, then ⊕ τⁿ(r) = 1.

Let Q = q ⊕ τ(q) ⊕ τ² (q) ⊕ ⋅⋅⋅ so that τ^k (Q) < Q for k ≥ 1. Then a = ∫^∞
₀ σ_t(q) dt = ∑_k≥0 ∫¹
₀ σ_k+t(q) dt = ∫¹
₀ σ_t(Q) dt so that 0 ≤ a ≤ 1 in A. By definition σ_s(a) ≤ a for s ≥ 0, since a − σ_s(a) = ∫^∞
_s σ_t(q) dt. The same formulas show that σ_s(a) tends 0 or 1 as s tends to +∞ or −∞. Set p = χ_[ε,1](a) for 0 < ε < 1. Then σ_s(p) = χ_[ε,1](σ_s(a)). It follows immediately that σ_s(p) ≤ p for s ≥ 0. Moreover σ_s(p) ${\displaystyle \downarrow }$ 0 as s tends to +∞ and σ_s(p) ${\displaystyle \uparrow }$ 1 as s tends to − ∞. The first limit formula follows because 0 ≤ ε ⋅ σ_s(p) ≤ σ_s(a). Now the same reasoning can be applied to τ⁻¹, σ_−t, τ⁻¹(q) and 1 – ε in place of τ, σ_t, q and ε. Then it is easily checked that the quantities corresponding to a and p are 1 − a and 1 − p. Consequently σ_−t(1−p) ${\displaystyle \downarrow }$ 0 as t tends to ∞. Hence σ_s(p) ${\displaystyle \uparrow }$ 1 as s tends to − ∞. In particular p ≠ 0 , 1.

So r = p − τ(p) is wandering for τ and ⊕ τ^k(r) = 1. Hence λ(r) = 1. It follows that λ(p −σ_s(p) ) = s for s = 1/n and therefore for all rational s > 0. Since the family σ_s(p) is continuous and decreasing, by continuity the same formula also holds for all real s > 0. Hence p satisfies all the asserted conditions.

• The conservative and dissipative parts of S_t for t ≠ 0 are independent of t.[4]

The previous result shows that if S_t is dissipative on X for t ≠ 0 then so is every S_s for s ≠ 0. By uniqueness, S_t and S_s preserve the dissipative parts of the other. Hence each is dissipative on the dissipative part of the other, so the dissipative parts agree. Hence the conservative parts agree.

• Ergodic flow

• Aaronson, Jon (1997), An introduction to infinite ergodic theory, Mathematical Surveys and Monographs, 50, American Mathematical Society, ISBN 0-8218-0494-4
• Hopf, Eberhard (1937), Ergodentheorie (in German), Springer
• Krengel, Ulrich (1968), "Darstellungssätze für Strömungen und Halbströmungen I", Math. Annalen (in German), 176: 181−190
• Krengel, Ulrich (1985), Ergodic theorems, De Gruyter Studies in Mathematics, 6, de Gruyter, ISBN 3-11-008478-3