# 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 = χ$_W$ in A = L$^∞$(X) satisfies qτ$^n$(q) = 0 for all n; thus, up to null sets, the translates T$^n$(W) are pairwise disjoint. An action is called dissipative if X = ∐ T$^n$(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 = 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) ∨ τ$^2$(q) ∨ τ$^3$(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):
* T is conservative;
* T is recurrent;
* T is infinitely recurrent;
* 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) ∨ τ$^2$(q) ∨ τ$^3$(q) ∨ ⋅⋅⋅)$^⊥$ = q ∧ τ(1 - q) ∧ τ$^2$(1 -q) ∧ τ$^3$(q) ∧ ... is wandering so that if q < 1, necessarily r = 0. Hence q ≤ τ(q) ∨ τ$^2$(q) ∨ τ$^3$(q) ∨ ⋅⋅⋅, so that T is recurrent.
If T is recurrent, then q ≤ τ(q) ∨ τ$^2$(q) ∨ τ$^3$(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 τ$^n$(q) is a decreasing sequence. But by recurrence, q ≤ τ(q) ∨ τ$^2$(q) ∨ τ$^3$(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 τ$^n$(p) disjoint (orthogonal). But then q = p ⊕ τ(p) ⊕ τ$^2$(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 = 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 = ⊕ τ$^n$(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 q$_1$ and q$_2$ are τ-invariant and dissipative, then q$_1$ ∨ q$_2$ is τ-invariant and dissipative, since q$_1$ ∨ q$_2$ = q$_1$ ⊕ q$_2$(1 − q$_1$). 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 q$_1$ ∨ ⋅⋅⋅ ∨ 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 T$^n$ for n > 1.
    If W is a wandering set for T then it is a wandering set for T$^n$. So the dissipative part of T is contained in the dissipative part of T$^n$. Let σ = τ$^n$. 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 ∪ T$^2$B ∪ ⋅⋅⋅ 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 ∪ T$^2$B ∪ ⋅⋅⋅ 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$_t$A 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 T$^n$ coincide for n ≠ 0.
* If S$_1$ is dissipative on A = L$^∞$(X), then there is an invariant measure λ on A and p in A such that
   * p > σ$_t$(p) for all t > 0
   * λ(p – σ$_t$(p)) = t for all t > 0
   * σ$_t$(p) $\uparrow$ 1 as t tends to −∞ and σ$_t$(p) $\downarrow$ 0 as t tends to +∞.
    Let T = S$_1$. Take q a wandering set for T so that ⊕ τ$^n$(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 τ$^n$(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 ⊕ τ$^n$(r) = 1 are easily described: they are given by r = ⊕ τ$^n$(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 ⊕ τ$^n$(r) = 1.
    Let Q = q ⊕ τ(q) ⊕ τ$^2$ (q) ⊕ ⋅⋅⋅ so that τ$^k$ (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 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) $\downarrow$ 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.