Equivariant thinning over a free group

TL;DR

This paper constructs entropy-increasing monotone factors within Bernoulli shifts over free groups of rank two or more, advancing understanding of entropy dynamics in non-abelian group actions.

Contribution

It introduces a novel method for creating entropy-increasing monotone factors in Bernoulli shifts over free groups, a previously unexplored area.

Findings

Established existence of entropy-increasing factors

Extended entropy theory to free group actions

Provided new tools for studying non-abelian group dynamics

Abstract

We construct entropy increasing monotone factors in the context of a Bernoulli shift over the free group of rank at least two.

Equations

$$H(\kappa ):=-\sum _{i\in K}\kappa (i)\log \kappa (i).$$

$$\rho (0,0)=1-p,\rho (0,1)=0,\rho (1,0)=rp,\text{ and }\rho (1,1)=(1-r)p$$

$$\gamma (100^{n-2},0^n)={\kappa }^n(100^{n-2})$$

$$\gamma (010^{n-2},0^n)={\kappa }^n(010^{n-2}).$$

$$\varrho (100^{n-2},0^n)=\varrho (010^{n-2},0^n)=p(1-p{)}^{n-1}$$

$$\varrho (100^{n-2},100^{n-2})=\varrho (010^{n-2},010^{n-2})=0.$$

$$\begin{array}{rl}\sum _{x\in {\left\{0,1\right\}}^n}\varrho (x,0^n)& =\sum _{x\in {\left\{0,1\right\}}^n}{\rho }^n(x,0^n)-{\rho }^n(100^{n-2},0^n)-{\rho }^n(010^{n-2},0^n)\\ & +\varrho (100^{n-2},0^n)+\varrho (010^{n-2},0^n)\\ & =(1-q{)}^n+2p(1-p{)}^{n-1}(1-r),\end{array}$$

$$\begin{array}{rl}\sum _{x\in {\left\{0,1\right\}}^n}\varrho (x,100^{n-2})& =\sum _{x\in {\left\{0,1\right\}}^n}{\rho }^n(x,100^{n-2})-{\rho }^n(100^{n-2},100^{n-2})\\ & +\varrho (100^{n-2},100^{n-2})\\ & =q(1-q{)}^{n-1}-p(1-p{)}^{n-1}(1-r)+0\\ & =\sum _{x\in {\left\{0,1\right\}}^n}\varrho (x,010^{n-2}),\end{array}$$

$$\varrho (x,0^n)={\rho }^n(x,0^n)=p^2(1-p{)}^{n-2}{r}^2,$$

$$\varrho (x,100^{n-2})={\rho }^n(x,100^{n-2})=p^2(1-p{)}^{n-2}r(1-r),$$

$$\varrho (x,010^{n-2})={\rho }^n(x,010^{n-2})=p^2(1-p{)}^{n-2}r(1-r).$$

$$\sum _{x\in B_1\cup B_2}\varrho (x,0^n)=2(n-2)p^2(1-p{)}^{n-2}{r}^2>2p(1-p{)}^{n-1}(1-r).$$

$$\left\{(x,0^n):x\in B_1\cup B_2\right\}\cup \left\{(x,100^{n-2}):x\in B_1\right\}\cup \left\{(x,010^{n-2}):x\in B_2\right\},$$

$$\gamma (x,0^n)=p^2(1-p{)}^{n-2}{r}^2-\frac{p(1-p{)}^{n-1}(1-r)}{n-2}>0,$$

$$\gamma (x,100^{n-2})=p^2(1-p{)}^{n-2}(1-r)r+\frac{p(1-p{)}^{n-1}(1-r)}{n-2},$$

$$\gamma (x,010^{n-2})=p^2(1-p{)}^{n-2}(1-r)r+\frac{p(1-p{)}^{n-1}(1-r)}{n-2}.$$

$$q_a(j):=\mathbb{P}(Y\in \left\{b_1,\dots ,b_j\right\}|X=a)=\frac{\mathbb{P}(Y\in \left\{b_1,\dots ,b_j\right\},X=a)}{\mathbb{P}(X=a)}$$

$$\mathbb{P}(q_a(j-1)\le U<q_a(j))=\frac{\mathbb{P}(Y=b_j,X=a)}{\mathbb{P}(X=a)}.$$

$$\Gamma (a,u):=b_j\text{ if }{q}_a(j-1)\le u<q_a(j).\text{qed}$$

$$X^j:=(X_{jn},\dots ,X_{(j+1)n-1}),$$

$$\big{(}X,(\Gamma(X^{i},U_{i}))_{i\in{\mathbb{N}}}\big{)}$$

$$(\Gamma (X^i,U_i){)}_{i\in \mathbb{N}}=(\Gamma ({\bar{X}}^i,U_i){)}_{i\in \mathbb{N}}.$$

$$b(X):=(1[X^{s_i}=100^{n-2}]{)}_{i\in \mathbb{N}}=(X_{s_{i}n}{)}_{i\in \mathbb{N}}$$

$$[\Gamma (X^i,U_i){]}_{i\in \mathbb{N}}$$

$$\phi(x):=\big{[}\Gamma({x^{i}},c(b(x))_{i})\big{]}_{i\in{\mathbb{N}}}$$

$$\big{(}(X_{1},\ldots,X_{n}),\Gamma(X_{1},\ldots,X_{n},U_{1})\big{)}$$

$$(\Phi (x,u{)}_j,\dots ,\Phi (x,u{)}_{j+n})=\Gamma (x_j,\dots ,x_{j+n},u_j).$$

$$X_{jn+2}^{\prime }=X_{jn+3}^{\prime }=\cdots =X_{(j+1)n-1}^{\prime }=0,$$

$$\varphi (x)(g)=(x(g)\oplus x(ga),x(g)\oplus x(gb))$$

$$\varphi :{\left\{0,1\right\}}^{{\mathbb{F}}_r}\to (\left\{0,1\right\}\times \left\{0,1\right\}{)}^{{\mathbb{F}}_r}\equiv {\left\{00,01,10,11\right\}}^{{\mathbb{F}}_r}$$

Full text

Funded in part by a New Faculty General Research Fund

1. Introduction

Let $\kappa$ be a probability measure on a finite set $K$. We will mainly be concerned with the simple case where $K=\left\{{0,1}\right\}$, where we call $\kappa(1):=\kappa(\left\{{1}\right\})\in(0,1)$ the intensity of $\kappa$. Let $G$ be a group. A Bernoulli shift over $G$ with base $(K,\kappa)$ is the measure-preserving system $(G,K^{G},\kappa^{G})$, where $G$ acts on $K^{G}$ via $(gx)(f)=x(g^{-1}f)$ for $x\in K^{G}$ and $g,f\in G$. Let $\iota$ be a probability measure of lower intensity. We say that a measurable map $\phi:K^{G}\to K^{G}$ is an equivariant thinning from $\kappa$ to $\iota$ if $\phi(x)(g)\leq x(g)$ for all $x\in K^{G}$ and $g\in G$, the push-forward of $\kappa^{G}$ under $\phi$ is $\iota^{G}$, and $\phi$ is equivariant $\kappa^{G}$-almost-surely; that is, on a set of full-measure, $\phi\circ g=g\circ\phi$ for all $g\in G$.

Theorem 1**.**

Let $\kappa$ and $\iota$ be probability measures on $\left\{{0,1}\right\}$ and $\iota$ be of lower intensity. For Bernoulli shifts over the free group of rank at least two, there exists an equivariant thinning from $\kappa$ to $\iota$.

Theorem 1 does not hold with such generality in the case of a Bernoulli shift over an amenable group like the integers. Recall that the entropy of a probability measure $\kappa$ on a finite set $K$ is given by

[TABLE]

Theorem 2** (Ball [3], Soo [16]).**

Let $\kappa$ and $\iota$ be probability measures on $\left\{{0,1}\right\}$ and $\iota$ be of lower intensity. For Bernoulli shifts over the integers, there exists an equivariant thinning from $\kappa$ to $\iota$ if and only if $H(\kappa)\geq H(\iota)$.

In Theorem 2, the necessity of $H(\kappa)\geq H(\iota)$ follows easily from the classical theory of Kolmogorov-Sinai entropy [8, 19], which we now recall. Let $G$ be a group and let $\kappa$ and $\iota$ be probability measures on a finite set $K$. An equivariant map $\phi$ is a factor from $\kappa$ to $\iota$ if the push-forward of $\kappa^{G}$ under $\phi$ is $\iota^{G}$, and is an isomorphism if $\phi$ is a bijection and its inverse also serves as a factor from $\iota$ to $\kappa$. In the case $G={\mathbb{Z}}$, Kolmogorov proved that entropy is non-increasing under factor maps; this implies the necessity of $H(\kappa)\geq H(\iota)$ in Theorem 2. Furthermore, Sinai [15] proved that there is a factor from $\kappa$ to $\iota$ if $H(\kappa)\geq H(\iota)$, and Ornstein [12] proved there is an isomorphism from $\kappa$ to $\iota$ if and only if $H(\kappa)=H(\iota)$. Thus entropy is a complete invariant for Bernoulli shifts over ${\mathbb{Z}}$. Ornstein and Weiss [13] generalized these results to the case where $G$ is an amenable group. See also Keane and Smorodinsky for concrete constructions of factor maps and isomorphisms [9, 10].

The sufficiency of $H(\kappa)>H(\iota)$ in Theorem 2 was first proved by Ball [3]. The existence of an isomorphism that is also an equivariant thinning in the equal entropy case was proved by Soo [16]. Let us remark that the factor maps given in standard proofs of the Sinai and Ornstein theorems will not in general be monotone; that is, they may not satisfy $\phi(x)(i)\leq x(i)$ for all $x\in\left\{{0,1}\right\}^{{\mathbb{Z}}}$ and $i\in{\mathbb{Z}}$.

Towards the end of their 1987 paper, Ornstein and Weiss [13] give a simple but remarkable example of an entropy increasing factor in the case where $G$ is the free group of rank at least two, which is further elaborated upon by Ball [2]. It was an open question until recently whether all Bernoulli shifts over a free group of rank at least two are isomorphic. This question was answered negatively by Lewis Bowen [5] in 2010, who proved that although entropy can increase under factor maps, in the context of a free group with rank at least two, it is still a complete isomorphism invariant. Recently, there has been much interest in studying factors in the non-amenable setting; see Russell Lyons [11] for more information.

Our proof of Theorem 1 will make use of a variation of the Ornstein and Weiss example in Ball [2] and a primitive version of a marker-filler type construction, in the sense of Keane and Smorodinsky [9, 10]. Our construction uses randomness already present in the process in a careful way as to mimic a construction that one would make if additional independent randomization were available. This approach was taken by Holroyd, Lyons, and Soo [7], Angel, Holroyd, and Soo [1], and Ball [4] for defining equivariant thinning in the context of Poisson point processes.

2. Tools

2.1. Coupling

Let $(A,\alpha)$ and $(B,\beta)$ be probability spaces. A coupling of $\alpha$ and $\beta$ is a probability measure on the product space $A\times B$ which has $\alpha$ and $\beta$ as its marginals. For a random variable $X$, we will refer to the measure ${\mathbb{P}}(X\in\cdot)$ as the law or the distribution of $X$. If two random variables $X$ and $Y$ have the same law, we write $X\stackrel{{\scriptstyle d}}{{=}}Y$. Similarly, a coupling of random variables $X$ and $Y$ is a pair of random variables $(X^{\prime},Y^{\prime})$, where $X^{\prime}$ and $Y^{\prime}$ are defined on the same probability space and have the same law as $X$ and $Y$, respectively. Thus a coupling of random variables gives a coupling of the laws of the random variables. Often we will refer to the law of a pair of random variables as the joint distribution of the random variables. In the case that $A=B$ and $A$ is a partially ordered by the relation $\preceq$, we say that a coupling $\gamma$ is monotone if $\gamma\left\{{(a,b)\subset A\times A:b\preceq a)}\right\}=1$. We will always endow the space of binary sequences $\left\{{0,1}\right\}^{I}$ indexed by a set $I$ with the partial order $x\preceq y$ if and only if $x_{i}\leq y_{i}$ for $i\in I$.

Example 3** (Independent thinning).**

Let $\kappa$ and $\iota$ be probability measures on $\left\{{0,1}\right\}$, where $\kappa(1):=p\geq\iota(1):=q$. Let $r:=\tfrac{p-q}{p}.$ Then the measure $\rho$ on $\left\{{0,1}\right\}^{2}$ given by

[TABLE]

is a monotone coupling of $\kappa$ and $\iota$. Thus under $\rho$, a $1$ is thinned to a [math] with probability $r$ and kept with probability $1-r$. Clearly, the product measure $\rho^{n}$ is a monotone coupling of $\kappa^{n}$ and $\iota^{n}$. We will refer to the coupling $\rho^{n}$ as the independent thinning of $\kappa^{n}$ to $\iota^{n}$. $\Diamond$

The following simple lemma is one of the main ingredients in the proof of Theorem 1. In it we construct a coupling of $\kappa^{n}$ and $\iota^{n}$ for $n$ sufficiently large which will allow us to extract spare randomness from a related coupling of $\kappa^{G}$ and $\iota^{G}$. We will write $0^{n}1^{m}$ to indicate the binary sequence of length $n+m$ of $n$ zeros followed by $m$ ones.

Lemma 4** (Key coupling).**

Let $\kappa$ and $\iota$ be probability measures on $\left\{{0,1}\right\}$, where $\kappa$ is of greater intensity. For $n$ sufficiently large, there exists a monotone coupling $\gamma$ of $\kappa^{n}$ and $\iota^{n}$ such that

[TABLE]

and

[TABLE]

Proof.

Let $p=\kappa(1)$, $q=\iota(1)$, and $\rho^{n}$ be the independent thinning of $\kappa^{n}$ to $\iota^{n}$ as in Example 3. We will perturb $\rho^{n}$ to give the required coupling. We specify a probability measure $\varrho$ on $\left\{{0,1}\right\}^{n}\times\left\{{0,1}\right\}^{n}$ by stating that it agrees with $\rho^{n}$ except on the points $(100^{n-2},0^{n})$, $(010^{n-2},0^{n})$, $(100^{n-2},100^{n-2})$, and $(010^{n-2},010^{n-2})$, where we specify that

[TABLE]

and

[TABLE]

Thus $\varrho$ is almost a monotone coupling of $\kappa^{n}$ and $\iota^{n}$, except that from our changes to $\rho^{n}$ we have

[TABLE]

and

[TABLE]

where $r=\tfrac{p-q}{p}$.

We perturb $\varrho$ to obtain the desired coupling $\gamma$. Consider the set $B_{1}$ of all binary sequences of length $n$, where $x\in B_{1}$ if and only if $x_{1}=1$, $x_{2}=0$, and $\sum_{i=3}^{n}x_{i}=1$. Similarly, let $B_{2}$ be the set of all binary sequences of length $n$, where $x\in B_{2}$ if and only if $x_{1}=0$, $x_{2}=1$, and $\sum_{i=3}^{n}x_{i}=1$. The sets $B_{1}$ and $B_{2}$ are disjoint, and each have cardinality $n-2$.

For $x\in B_{1}\cup B_{2}$,

[TABLE]

for $x\in B_{1}$,

[TABLE]

and for $x\in B_{2}$,

[TABLE]

Note that for $n$ sufficiently large

[TABLE]

Let $\gamma$ be equal to $\varrho$ except on the set of points

[TABLE]

where we make the following adjustments. For $x\in B_{1}\cup B_{2}$, set

[TABLE]

for $x\in B_{1}$, set

[TABLE]

and for $x\in B_{2}$, set

[TABLE]

That $\gamma$ has the required properties follows from its construction. ∎

To illustrate the utility of Lemma 4, we will give a different proof of the following result of Peled and Gurel-Gurevich [6]. Let ${\mathbb{N}}=\left\{{0,1,2,\ldots}\right\}$.

Theorem 5** (Peled and Gurel-Gurevich [6]).**

Let $\kappa$ and $\iota$ be probability measures on $\left\{{0,1}\right\}$, where $\kappa$ is of greater intensity. There exists a measurable map $\phi:\left\{{0,1}\right\}^{{\mathbb{N}}}\to\left\{{0,1}\right\}^{{\mathbb{N}}}$ such that the push-forward of $\kappa^{{\mathbb{N}}}$ under $\phi$ is $\iota^{{\mathbb{N}}}$ and $\phi(x)(i)\leq x(i)$ for all $x\in\left\{{0,1}\right\}^{{\mathbb{N}}}$ and all $i\in{\mathbb{N}}$.

We note that in [6, Theorem 1.3], they use the dual terminology of thickenings; their equivalent theorem states that for probability measures $\iota$ and $\kappa$ on $\left\{{0,1}\right\}$, where $\iota$ is of lesser intensity, there is a measurable map $\phi:\left\{{0,1}\right\}^{{\mathbb{N}}}\to\left\{{0,1}\right\}^{{\mathbb{N}}}$ such that the push-forward of $\iota^{{\mathbb{N}}}$ under $\phi$ is $\kappa^{{\mathbb{N}}}$ and $\phi(x)(i)\geq x(i)$ for all $x\in\left\{{0,1}\right\}^{{\mathbb{N}}}$ and all $i\in{\mathbb{N}}$.

In the proof of Theorem 5, we will make use of the following two lemmas. We say that a random variable $U$ is uniformly distributed in $[0,1]$ if the probability that $U$ lies in a Borel subset of the unit interval is given by the Lebesgue measure of the set.

Lemma 6**.**

Let $(X,Y)$ be a pair of discrete random variables taking values on the finite set $A\times B$ with joint distribution $\gamma$. There exists a measurable function $\Gamma:A\times[0,1]\to B$ such that if $U$ is uniformly distributed in $[0,1]$ and independent of $X$, then $(X,\Gamma(X,U))$ has joint distribution $\gamma$.

Proof.

Assume that ${\mathbb{P}}(X=a)>0$, for all $a\in A$. Let $B=\left\{{b_{1},\ldots,b_{n}}\right\}$. For each $a\in A$, let

[TABLE]

for all $1\leq j\leq n$. Set $q_{a}(0)=0$ and note that $q_{a}(n)=1$, so that

[TABLE]

For each $1\leq j\leq n$, let

[TABLE]

We call a $\left\{{0,1}\right\}$-valued random variable a Bernoulli random variable. The following lemma allows us to code sequences of independent coin-flips into sequences of uniformly distributed random variables.

Lemma 7**.**

There exists a measurable function $c:\left\{{0,1}\right\}^{{\mathbb{N}}}\to[0,1]^{{\mathbb{N}}}$ such that if $B=(B_{i})_{i\in{\mathbb{N}}}$ is a sequence of i.i.d. Bernoulli random variables with mean $\tfrac{1}{2}$, then $(c(B)_{i})_{i\in{\mathbb{N}}}$ is a sequence of i.i.d. random variables that are uniformly distributed in $[0,1]$.

Proof.

The result follows from the Borel isomorphism theorem. See [17, Theorem 3.4.23] for more details. ∎

Proof of Theorem 5.

Let $\gamma$ be the monotone coupling of $\kappa^{n}$ and $\iota^{n}$ given by Lemma 4, so that $\gamma$ is a measure on $\left\{{0,1}\right\}^{n}\times\left\{{0,1}\right\}^{n}\equiv(\left\{{0,1}\right\}\times\left\{{0,1}\right\})^{n}$. Thus the product measure $\gamma^{2}$ is a monotone coupling of $\kappa^{2n}$ and $\iota^{2n}$ and $\gamma^{{\mathbb{N}}}$ gives a monotone coupling of $\kappa^{{\mathbb{N}}}$ and $\iota^{{\mathbb{N}}}$. We will modify the coupling $\gamma^{{\mathbb{N}}}$ to become the required map $\phi$. In order to do this, it will be easier to think in terms of random variables rather than measures.

Let $X=(X_{i})_{i\in{\mathbb{N}}}$ be an i.i.d. sequence of Bernoulli random variables with mean $\kappa(1)$. For each $j\geq 0$, let

[TABLE]

so that the random variables are partitioned into blocks of size $n$. Let $U=(U_{i})_{i\in{\mathbb{N}}}$ be an i.i.d. sequence of random variables that are uniformly distributed in $[0,1]$. Also assume that $U$ is independent of $X$, and let $Y=(Y_{i})_{i\in{\mathbb{N}}}$ be an i.i.d. sequence of Bernoulli random variables with mean $\iota(1)$.

By Lemmas 4 and 6, let $\Gamma:\left\{{0,1}\right\}^{n}\times[0,1]\to\left\{{0,1}\right\}^{n}$ be a measurable map such that $(X^{1},\Gamma(X^{1},U_{1}))$ has joint law $\gamma$ and $\Gamma(w,v)=0^{n}$ for all $v\in[0,1]$ if $w\in\left\{{100^{n-2},010^{n-2}}\right\}$. We have that

[TABLE]

gives a monotone coupling of $X$ and $Y$ with law $\gamma^{{\mathbb{N}}}$.

For each $j\in{\mathbb{N}}$, call $X^{j}$ special if $X^{j}\in\left\{{100^{n-2},010^{n-2}}\right\}$ and let $S\subset{\mathbb{N}}$ be the random set of $j\in{\mathbb{N}}$ for which $X^{j}$ are special. Note that almost surely, $S$ is an infinite set. Let $\bar{X}=(\bar{X}_{i})_{i\in{\mathbb{N}}}$ be the sequence of binary digits such that $\bar{X}^{j}=X^{j}$ if $j\not\in S$ and $\bar{X}^{j}=0^{n}$ if $j\in S$. We have that

[TABLE]

Let $(s_{i})_{i\in{\mathbb{N}}}$ be the enumeration of $S$, where $s_{0}<s_{1}<s_{2}<s_{3}\cdots$. Consider the sequence of random variables given by

[TABLE]

Since $100^{n-2}$ and $010^{n-2}$ occur with equal probability, we have that $b(X)$ is an i.i.d. sequence of Bernoulli random variables with mean $\tfrac{1}{2}$. Furthermore, we have that $b(X)$ is independent of $\bar{X}$, since $b(X)$ only depends on the values of $X$ on the special blocks. Let $c$ be the function from Lemma 7, so that $c(b(X))\stackrel{{\scriptstyle d}}{{=}}U$. Since $b(X)$ is independent of $\bar{X}$,

[TABLE]

Thus \Big{(}X,\big{[}\Gamma({X}^{i},c(b(X))_{i})\big{]}_{i\in{\mathbb{N}}}\Big{)} is another monotone coupling of $X$ and $Y$. Hence, we define

[TABLE]

for all $x\in\left\{{0,1}\right\}^{{\mathbb{N}}}$ when the set $S$ is infinite, and set $\phi(x)=0^{{\mathbb{N}}}$ when $S$ is finite–an event that occurs with probability zero. ∎

2.2. Joinings

Let $T:\left\{{0,1}\right\}^{{\mathbb{Z}}}\to\left\{{0,1}\right\}^{{\mathbb{Z}}}$ be the left-shift given by $(Tx)_{i}=x_{i+1}$ for all $x\in\left\{{0,1}\right\}^{{\mathbb{Z}}}$ and all $i\in{\mathbb{Z}}$. Let $\kappa$ and $\iota$ be probability measures on $\left\{{0,1}\right\}$. A joining of $\kappa^{{\mathbb{Z}}}$ and $\iota^{{\mathbb{Z}}}$ is a coupling $\varrho$ of the two measures with the additional property that $\varrho\circ(T\times T)=\varrho$. We will make use of the following joining in the proof of Theorem 1.

Example 8**.**

Let $\kappa$ and $\iota$ be probability measures on $\left\{{0,1}\right\}$. Assume that the intensity of $\kappa$ is greater than the intensity of $\iota$. Let $x\in\left\{{0,1}\right\}^{{\mathbb{Z}}}$, and let $n$ be sufficiently large as in Lemma 4. Call the subset $[j,j+2n+1]\subset{\mathbb{Z}}$ a marker if $x_{i}=0$ for all $i\in[j,j+2n]$ and $x_{j+2n+1}=1$. Notice that two distinct markers have an empty intersection. Call an interval a filler if it is nonempty and lies between two markers. Thus each $x\in\left\{{0,1}\right\}^{{\mathbb{Z}}}$ partitions ${\mathbb{Z}}$ into intervals of markers and fillers. Call a filler fitted if it is of size $n$, and call a filler special if it is both fitted and of the form $100^{n-2}$ or $010^{n-2}$.

Let $X$ have law $\kappa^{{\mathbb{Z}}}$ and $Y$ have law $\iota^{{\mathbb{Z}}}$. In what follows we describe explicitly how to obtain a monotone joining of $X$ and $Y$, where the independent thinning is used everywhere, except at the fitted fillers, where the coupling from Lemma 4 is used. Let $U=(U_{i})_{i\in{\mathbb{Z}}}$ be an i.i.d. sequence of random variables that are uniformly distributed in $[0,1]$ and independent of $X$. By Example 3 and Lemma 6, let $R:\left\{{0,1}\right\}\times[0,1]\to\left\{{0,1}\right\}$ be a measurable function such that $R(X_{1},U_{1})\leq X_{1}$ is a Bernoulli random variable with mean $\iota(1)$. Let $\Gamma$ and $\gamma$ be as in the proof of Theorem 5, so that

[TABLE]

has law $\gamma$. Consider the function $\Phi:\left\{{0,1}\right\}^{{\mathbb{Z}}}\times[0,1]^{{\mathbb{Z}}}\to\left\{{0,1}\right\}^{{\mathbb{Z}}}$ defined by $\Phi(x,u)_{i}=R(x_{i},u_{i})$ if $i$ is not in a fitted filler. For $(j,j+1,\ldots,j+n)$ in a fitted filler, we set

[TABLE]

The law of $X$ restricted to a filler interval is the law of a finite sequence of i.i.d. Bernoulli random variables with mean $\kappa(1)$, conditioned not to contain a marker. Note that since a fitted interval is of size $n$, and a marker is of size $2n+1$, the law of $X$ restricted to a fitted interval is just the law of a finite sequence of i.i.d. Bernoulli random variables with mean $\kappa(1)$. Furthermore, conditioned on the locations of the markers, the restrictions of $X$ to each filler interval are independent (see for example Keane and Smorodinsky [9, Lemma 4] for a detailed proof). Hence, $\Phi(X,U)\stackrel{{\scriptstyle d}}{{=}}Y$. In addition, since all the couplings involved are monotone, we easily have that $\Phi(X,U)_{i}\leq X_{i}$ for all $i\in{\mathbb{Z}}$. $\Diamond$

Remark 9**.**

To emphasize the strong form of independence in Example 8, we note that if $A=(A_{i})_{i\in\mathbb{Z}}$ are independent Bernoulli random variables with mean $\tfrac{1}{2}$ that are independent of $X$, then $(A_{jn})_{j\in S}$ has the same law as $(X_{jn})_{j\in S}$. Recall if $j\in S$ then $X^{j}=(X_{jn},\ldots,X_{(j+1)n-1})$ is special. In addition, if $X^{\prime}$ is such that $X^{\prime}_{i}=X_{i}$ for every $i$ not in a special filler of $X$ and on each special filler of $X$ we set $X^{\prime}_{jn}=A_{jn}$, $X^{\prime}_{jn+1}=1-A_{jn}$, and

[TABLE]

then $X^{\prime}\stackrel{{\scriptstyle d}}{{=}}X$. Thus we can independently resample on the special fillers without affecting the distribution of $X$. $\Diamond$

2.3. The example of Ornstein and Weiss

Let $\mathbb{F}_{r}$ be the free group of rank $r\geq 2$. Let $a$ and $b$ be two of its generators. The Ornstein and Weiss [13] entropy increasing factor map is given by

[TABLE]

for all $x\in\left\{{0,1}\right\}^{\mathbb{F}_{r}}$ and all $g\in\mathbb{F}_{2}$, where

[TABLE]

pushes the uniform product measure $(\tfrac{1}{2},\tfrac{1}{2})^{\mathbb{F}_{r}}$ forward to the uniform product measure $(\tfrac{1}{4},\tfrac{1}{4},\tfrac{1}{4},\tfrac{1}{4})^{\mathbb{F}_{r}}$; the required independence follows from the observation that if $m\oplus n:=m+n\bmod 2$, if $X$, $X^{\prime}$, and $Y$ are independent Bernoulli random variables with mean $\tfrac{1}{2}$, and if $Z:=X\oplus Y$ and $Z^{\prime}:=X^{\prime}\oplus Y$, then $Z$ and $Z^{\prime}$ are independent, even though they both depend on $Y$.

Ornstein and Weiss’s example can be iterated to produce an infinite number of bits at each vertex in the following way. As in Ball [2, Proposition 2.1], we will define $\phi_{k}:\left\{{0,1}\right\}^{\mathbb{F}_{r}}\to(\left\{{0,1}\right\}^{k})^{\mathbb{F}_{r}}$ inductively for $k\geq 2$. Let $\tilde{\phi_{k}}:\left\{{0,1}\right\}^{\mathbb{F}_{r}}\to\left\{{0,1}\right\}^{\mathbb{F}_{r}}$ be the last coordinate of $\phi_{k}$ so that $\tilde{\phi_{k}}(x)(g)=[\phi_{k}(x)(g)]_{k}$ for all $x\in\left\{{0,1}\right\}^{\mathbb{F}_{r}}$ and all $g\in\mathbb{F}_{2}$. Set $\phi_{2}=\phi$ . For $k\geq 3$, let $\phi_{k}$ be given by

[TABLE]

for all $x\in\left\{{0,1}\right\}^{\mathbb{F}_{r}}$ and all $g\in\mathbb{F}_{2}$. At each step we are saving one bit to generate two new bits using the original map $\phi$. The map $\phi_{k}$ pushes the uniform product measure $(\tfrac{1}{2},\tfrac{1}{2})^{\mathbb{F}_{r}}$ forward to the uniform product measure on $(\left\{{0,1}\right\}^{k})^{\mathbb{F}_{r}}$. By taking the limit, we obtain the mapping

[TABLE]

which yields a sequence of i.i.d. fair bits at each coordinate $g\in\mathbb{F}_{2}$, independently. Note that $\phi_{\infty}(x)(g)_{k}=\phi_{n}(x)(g)_{k}$ for all $n>k$. In our proof of Theorem 1 we will use this iteration, which Ball attributes to Timár.

3. Proof of the main theorem

Proof of Theorem 1.

Let $r\geq 2$. We begin by extending the same monotone joining defined in Example 8 to a monotone joining of $\kappa^{\mathbb{F}_{r}}$ and $\iota^{\mathbb{F}_{r}}$. Let $X$ have law $\kappa^{\mathbb{F}_{r}}$ and $Y$ have law $\iota^{\mathbb{F}_{r}}$; then $X=(X_{g})_{g\in\mathbb{F}_{r}}=(X(g))_{g\in\mathbb{F}_{r}}$ are i.i.d. Bernoulli random variables with mean $\kappa(1)$. As in the Ornstein and Weiss example, it will be sufficient to use only two generators $a$ and $b$ in the expression of our equivariant thinning. We refer to the string of generators and their inverses that make up the representation of an element in $\mathbb{F}_{r}$ as a word, and the individual generators and inverses as letters. We call a word reduced if its string of letters has no possible cancellations.

Consider $\mathbb{F}_{r}$ as being partitioned into infinitely many $\mathbb{Z}$ copies $Z(w)$ in the following way. Let $\mathbb{F}_{r}^{\prime}$ be the set of reduced words in $\mathbb{F}_{r}$ that do not end in either $b$ or $b^{-1}$. For each $w\in\mathbb{F}_{r}^{\prime}$, set $Z(w):=\{wb^{i}\}_{i\in{\mathbb{Z}}}$. Indeed, any element in $\mathbb{F}_{r}$ may be written as $wb^{i}$ for unique reduced $w\in\mathbb{F}_{r}^{\prime}$ and $i\in{\mathbb{Z}}$.

Let $n$ be sufficiently large for the purposes of Lemma 4. We define markers, fillers, fitted fillers, and special fillers on each of the ${\mathbb{Z}}$ copies in the obvious way. For example, if $x\in\left\{{0,1}\right\}^{\mathbb{F}_{r}}$ and $w\in\mathbb{F}_{r}^{\prime}$, then the set $\left\{{wb^{j},\ldots,wb^{j+2n+1}}\right\}$ is a marker if $x(wb^{i})=0$ for all $i\in[j,2n]$ and $x(wb^{2n+1})=1$.

Let $U^{\prime}=(U^{\prime}_{g})_{g\in{\mathbb{F}_{r}}}$ be i.i.d. uniform random variables independent of $X$. Let $\Phi$ be as in Example 8. Define $\hat{\Phi}:\{0,1\}^{\mathbb{F}_{r}}\times[0,1]^{\mathbb{F}_{r}}\rightarrow\{0,1\}^{\mathbb{F}_{r}}$ by

[TABLE]

for all $w\in\mathbb{F}_{r}^{\prime}$ and all $i\in{\mathbb{Z}}$, where $x(Z(w)):=(x(wb^{j}))_{j\in{\mathbb{Z}}}$ and $u^{\prime}(Z(w)):=(u^{\prime}(wb^{j}))_{j\in{\mathbb{Z}}}$. Thus we have the monotone joining $\Phi$ on each $\mathbb{Z}$ copy $Z(w)$ in $\mathbb{F}_{r}$, so that

[TABLE]

and $\hat{\Phi}(X,U^{\prime})_{g}\leq X_{g}$ for all $g\in\mathbb{F}_{r}$. Additionally, since $\Phi$ is a joining, the joint law of $(X,\hat{\Phi}(X,U^{\prime}))$ is invariant under $\mathbb{F}_{r}$-actions.

Recall that a special filler has length exactly $n$, and the filler has two choices of values $010^{n-2}$ or $100^{n-2}$, which occur with equal probability. We define an initial vertex of a special filler in $Z(w)$ to be an element $wb^{n_{0}}\in Z(w)$ where the entire special filler takes values sequentially at vertices on the minimal path from $wb^{n_{0}}$ to $wb^{n_{0}+n}$. For each $x\in\left\{{0,1}\right\}^{\mathbb{F}_{r}}$, let $V=V(x)$ be the set of initial vertices in $\mathbb{F}_{r}$. Note that as in Example 8, the law of $X$ restricted to a fitted interval is just the law of a finite sequence of i.i.d. Bernoulli random variables with mean $\kappa(1)$. Furthermore, conditioned on the locations of the markers, the restrictions of $X$ to each filler interval are independent. Thus for all $v\in V(X)$, $X(v)$ is a Bernoulli random variable with mean $\frac{1}{2}$, and conditioned on $V(X)$, the random variables $(X(v))_{v\in V}$ are independent.

We have the same strong form of independence here as emphasized in Remark 9 for Example 8, again by Keane and Smorodinsky [9, Lemma 4]. This is key in our construction: we will use the Bernoulli random variables $(X(v))_{v\in V}$ to build deterministic substitutes for $U^{\prime}$.

Now we adapt the iteration of the Ornstein and Weiss example to assign a sequence of i.i.d. Bernoulli random variables to each $v\in V$. For each $v\in V$, let $k$ be the smallest positive integer such that $va^{k}\in V$; set $\alpha(v)=va^{k}$. Similarly, let $k^{\prime}$ be the smallest positive integer such that $vb^{k^{\prime}}\in V$ and set $\beta(v)=vb^{k^{\prime}}.$ For each $v\in V$, define

[TABLE]

Conditioned on $V$, we have that $(\psi(X))_{v\in V}$ is a family of independent random variables uniformly distributed on $\left\{{00,01,10,11}\right\}$. We iterate the map $\psi$ as we did with the Ornstein and Weiss map $\phi$. Set $\psi_{2}=\psi$. For $k\geq 3$, let

[TABLE]

where $\tilde{\psi}_{k-1}(x)(v)=[\psi_{k-1}(x)(v)]_{k-1}$ is the last coordinate of $\psi_{k}$. Let $\psi_{\infty}$ be the limit, and let $B_{v}=\psi_{\infty}(X)(v)$, so that conditioned on $V$, the random variables $(B_{v})_{v\in V}$ are independent, and each $B_{v}$ is an i.i.d. sequence of Bernoulli random variables with mean $\frac{1}{2}$.

For all $x\in\{0,1\}^{\mathbb{F}_{r}}$, let $\bar{x}(g)=x(g)$ for all $g$ not in a special filler, and let $\bar{x}(g)=0$ if $g$ belongs to a special filler. It follows from Remark 9 that if $B^{\prime}=(B^{\prime}_{g})_{g\in\mathbb{F}_{r}}$ are independent Bernoulli random variables with mean $\tfrac{1}{2}$ independent of $X$, then $(B^{\prime}_{v})_{v\in V(X)}$ has the same law as $(B_{v})_{v\in V(X)}$. Moreover,

[TABLE]

We assign, in an equivariant way, one uniform random variable to each element in $\mathbb{F}_{r}$ using the randomness provided by $(B_{v})_{v\in V}$. Let $c:\left\{{0,1}\right\}^{{\mathbb{N}}}\to[0,1]^{{\mathbb{N}}}$ be the function from Lemma 7, and let $g\in\mathbb{F}_{r}$. Then almost surely there exist $v\in V$ and a minimal $j>0$ such that $gb^{j}=v$; set $U_{g}=c(B_{v})_{j}$. Define $\mathbf{u}:\left\{{0,1}\right\}^{\mathbb{F}_{r}}\to[0,1]^{\mathbb{F}_{r}}$ by setting $\mathbf{u}(X):=(U_{g})_{g\in\mathbb{F}_{r}}$. Recall that $U^{\prime}=(U_{g}^{\prime})_{g\in\mathbb{F}_{r}}$ are independent random variables uniformly distributed in $[0,1]$ independent of $X$. From (2),

[TABLE]

Let $R:\left\{{0,1}\right\}\times[0,1]\to\left\{{0,1}\right\}$ and $\Gamma:\left\{{0,1}\right\}^{n}\times[0,1]\to\left\{{0,1}\right\}^{n}$ be the functions that appear in the definition of $\Phi$ in Example 8. Recall that $R$ facilitated independent thinning and $\Gamma$ the key monotone coupling of Lemma 4. Also recall $\Gamma(100^{n-2},t)=0=\Gamma(010^{n-2},t)$ for all $t\in[0,1]$.

Now define $\phi:\{0,1\}^{\mathbb{F}_{r}}\rightarrow\{0,1\}^{\mathbb{F}_{r}}$ by

[TABLE]

for $g$ not in a fitted filler; if $\left\{{wb^{i},\ldots,wb^{i+n-1}}\right\}$ is a fitted filler, then set

[TABLE]

Note $\phi$ is defined so that $\phi(x)=\hat{\Phi}(x,\mathbf{u}(x))$. The map $\phi$ is equivariant and satisfies $\phi(x)(g)\leq x(g)$ by construction. It remains to verify that $\phi(X)\stackrel{{\scriptstyle d}}{{=}}Y$.

By the definition of $\Gamma$, we have $\phi(X)=\phi(\bar{X})$; that is, all special fillers are sent to $0^{n}$. A similar remark applies to the map $\hat{\Phi}$. From (1) and (3),

[TABLE]

4. Generalizations and questions

4.1. Stochastic domination

Let $[N]=\left\{{0,1,\ldots,N-1}\right\}$ be endowed with the usual total ordering. Let $\kappa$ and $\iota$ be probability measures on $[N]$. We say that $\kappa$ stochastically dominates $\iota$ if $\sum_{i=0}^{j}\kappa_{i}\leq\sum_{i=0}^{j}\iota_{i}$ for all $j\in[N]$. An elementary version of Strassen’s theorem [18, Theorem 11] gives that $\kappa$ stochastically dominates $\iota$ if and only if there exists a monotone coupling of $\kappa$ and $\iota$. Notice that in the case $N=2$, we have that $\kappa$ stochastically dominates $\iota$ if and only if $\iota$ is not of higher intensity than $\kappa$. Thus Theorem 1 gives a positive answer to a special case of the following question.

Question 1**.**

Let $\kappa$ and $\iota$ be probability measures on $[N]$, where $\kappa$ stochastically dominates $\iota$, and $\kappa$ gives positive measure to at least two elements of $[N]$. Let $G$ be the free group of rank at least two. Does there exist a measurable equivariant map $\phi:[N]^{G}\to[N]^{G}$ such that the push-forward of $\kappa^{G}$ is $\iota^{G}$ and $\phi(x)(g)\leq x(g)$ for all $x\in[N]^{G}$ and $g\in G$?

In Question 1, we call the map $\phi$ a monotone factor from $\kappa$ to $\iota$. A necessary condition for the existence of a monotone factor from $\kappa$ to $\iota$ is that $\kappa$ stochastically dominates $\iota$. In the case $G={\mathbb{Z}}$, Ball [3] proved that there exists a monotone factor from $\kappa$ to $\iota$ provided that $\kappa$ stochastically dominates $\iota$, $H(\kappa)>H(\iota)$, and $\iota$ is supported on two symbols; Quas and Soo [14] removed the two symbol condition on $\iota$.

In the non-amenable case, where $G$ is a free group of rank at least two, one can hope that Question 1 can be answered positively, without any entropy restriction. However, the analogue of Lemma 4 that was key to the proof of Theorem 1 does not apply in the simple case where $\kappa=(0,\tfrac{1}{2},\tfrac{1}{2})$ and $\iota=(\tfrac{1}{3},\tfrac{1}{3},\tfrac{1}{3})$. In particular, for all $n\geq 1$, there is no coupling $\rho$ of $\kappa^{n}$ and $\iota^{n}$ for which there exists $x\in\left\{{1,2}\right\}^{n}$ and $y\in\left\{{0,1,2}\right\}^{n}$ such that $\rho(x,y)=\kappa^{n}(x)=(\tfrac{1}{2})^{n}$, since $\rho(x,y)\leq\iota^{n}(y)=(\tfrac{1}{3})^{n}$.

4.2. Automorphism-equivariant factors

The Cayley graph of $\mathbb{F}_{n}$ is the regular tree $\mathbb{T}_{2n}$ of degree $2n$. We note that $\mathbb{F}_{n}$ is a strict subset of the group of graph automorphisms of $\mathbb{T}_{2n}$. The map that we constructed in Theorem 1 is not equivariant with respect to the full automorphism group of $\mathbb{T}_{2n}$. In particular, our definition of a marker is not equivariant with respect to the automorphism which exchanges $a$-edges and $b$-edges in $\mathbb{T}_{2n}$. However, Ball generalizes the Ornstein and Weiss example to the full automorphism group in [2, Theorem 3.3] by proving that for any $d\geq 3$, there exists a measurable mapping $\phi:\left\{{0,1}\right\}^{\mathbb{T}_{d}}\to[0,1]^{\mathbb{T}_{d}}$ which pushes the uniform product measure on two symbols forward to the product measure of Lebesgue measure on the unit interval, equivariant with respect to the group of automorphisms of $\mathbb{T}_{d}$. Moreover, she proved the analogous result for any tree with bounded degree, no leaves, and at least three ends.

Question 2**.**

Let $T$ be a tree with bounded degree, no leaves, and at least three ends. Let $\kappa$ and $\iota$ be probability measures on $\left\{{0,1}\right\}$ and $\iota$ be of lower intensity. Does there exists a thinning from $\kappa$ to $\iota$ that is equivariant with respect to the full automorphism group of $T$?

Acknowledgements

We thank the referee for carefully reviewing the paper and providing various helpful comments and suggestions.