Hopf decomposition

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 = χW 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 $Z \ TZ$ has measure zero. Thus if q = χZ 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 = χY.
 * Infinitely recurrent actions. An action T is said to be infinitely recurrent if q ≤ τm (q) ∨ τm + 1(q) ∨ τm+2(q) ∨ ... a.e. for any q = χY 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 ≤ τk(q) ∨ τk+1(q) ∨ ⋅⋅⋅. Then τk(q) ≤ τk+1(q) ∨ τk+2(q) ∨ ⋅⋅⋅ ≤. Hence q ≤ τk+1(q) ∨ τk+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.

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 = χC 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 = χW 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&prime;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 qn τ-invariant and dissipative such that μ(qn) increases to M. Replacing qn by q1 ∨ ⋅⋅⋅ ∨ qn, t can be assumed that qn 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 τa(p) and  τb(p) are orthogonal for different a and b in the same congruence class modulo n. Take a set of τa(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 ≤ Tm (B) ∨ Tm + 1(B) ∨ Tm+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 St a non-sngular flow on X inducing a 1-parameter group of automorphisms σt of A = L∞(X). It will be assumed that the action is faithful, so that σt is the identity only for t = 0. For each St or equivalently σt with t ≠ 0 there is a Hopf decomposition, so a pt fixed by σt such that the action is conservative on ptA and dissipative on (1−pt)A.


 * For s, t ≠ 0 the conservative and dissipative parts of Ss and St 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
 * p > σt(p) for all t > 0
 * λ(p – σt(p)) = t for all t > 0
 * σt(p) $$\uparrow$$ 1 as t tends to −∞ and σt(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 $λ = ∫1 0 μ ∘ σ_{t} dt$ 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(qn) where q = ⊕ qn 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 $⊕ τ^{n}(r) = 1$.


 * Let Q = q ⊕ τ(q) ⊕ τ2 (q) ⊕ ⋅⋅⋅ so that τk (Q) < Q for k ≥ 1. Then $a = ∫∞ 0 σ_{t}(q) dt = Σ_{k≥0} ∫1 0 σ_{k+t}(q) dt = ∫1 0 σ_{t}(Q) dt$ so that 0 ≤ a ≤ 1 in A. By definition σs(a) ≤ a for s ≥ 0, since $a − σ_{s}(a) = ∫∞ s σ_{t}(q) dt$. The same formulas show that σs(a) tends 0 or 1 as s tends to +∞ or −∞. Set p = χ[ε,1](a) for 0 < ε < 1. Then σs(p) = χ[ε,1](σs(a)). It follows immediately that σs(p) ≤ p for s ≥ 0. Moreover σs(p) $$\downarrow$$ 0 as s tends to +∞ and  σs(p) $$\uparrow$$ 1 as s tends to − ∞. The first limit formula follows because 0 ≤ ε ⋅ σs(p) ≤ σs(a). Now the same reasoning can be applied to τ−1, σ−t, τ−1(q) and 1 – ε in place of τ, σt, 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 σs(p) $$\uparrow$$ 1 as s tends to − ∞. In particular p ≠ 0, 1.


 * So r = p − τ(p) is wandering for τ and ⊕ τk(r) = 1. Hence λ(r) = 1. It follows that λ(p −σs(p) ) = s for s = 1/n and therefore for all rational s > 0. Since the family  σs(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 St for t ≠ 0 are independent of t.


 * The previous result shows that if St is dissipative on X for t ≠ 0 then so is every Ss for s ≠ 0. By uniqueness, St and Ss 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.