Hopf decomposition explained

In mathematics, the Hopf decomposition, named after Eberhard Hopf, gives a canonical decomposition of a measure space (X, μ) with respect to an invertible non-singular transformation T:X→X, i.e. a transformation which with its inverse is measurable and carries null sets onto null sets. Up to null sets, X can be written as a disjoint union C ∐ D of T-invariant sets where the action of T on C is conservative and the action of T on D is dissipative. Thus, if τ is the automorphism of A = L^∞(X) induced by T, there is a unique τ-invariant projection p in A such that pA is conservative and (I–p)A is dissipative.

Definitions

• 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 pairwise disjoint. An action is called dissipative if X = ∐ Tⁿ(W) a.e. 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 has measure zero. Thus if q = χ_Z and τ(q) ≤ q, then τ(q) = q a.e.
• Recurrent actions. An action T is said to be recurrent if q ≤ τ(q) ∨ τ²(q) ∨ τ³(q) ∨ ... a.e. 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) ∨ ... a.e. for any q = χ_Y and any m ≥ 1.

Recurrence theorem

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. 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, contradicting incompressibility. So T is conservative.

Hopf decomposition

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.

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) with 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 dissipative, 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.

Hopf decomposition for a non-singular flow

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.

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

1. λ(p – σ_t(p)) = t for all t > 0

1. σ_t(p)

\uparrow 1 as t tends to −∞ and σ_t(p) \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 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)

\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 .

Let Q = q ⊕ τ(q) ⊕ τ² (q) ⊕ ⋅⋅⋅ so that τ^k (Q) < Q for k ≥ 1. Then so that 0 ≤ a ≤ 1 in A. By definition σ_s(a) ≤ a for s ≥ 0, since . 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)

\downarrow 0 as s tends to +∞ and σ_s(p) \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) \downarrow 0 as t tends to ∞. Hence σ_s(p) \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.

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.

See also

• Ergodic flow