Monod, Problem 12 — H(ℤ) is not amenable (open)

What the statement asserts

The statement has no hypotheses and no free variables. It asserts that one specific group, called $H$ below, does not admit a left-invariant, finitely additive probability measure defined on all of its subsets. Everything below spells out what $H$ is and what that measure condition means.

The space and its homeomorphism group

Let $X = \mathbb{R} \cup \{\infty\}$ be the one-point compactification of the real line. A set $U \subseteq X$ is open exactly when $U \cap \mathbb{R}$ is open in $\mathbb{R}$ and, if $\infty \in U$, the complement $\mathbb{R} \setminus U$ is compact in $\mathbb{R}$. So a neighbourhood of $\infty$ contains $\{\infty\} \cup \{t \in \mathbb{R} : |t| > R\}$ for some $R$.

$\mathrm{Homeo}(X)$ is the group of all homeomorphisms $X \to X$, with product given by composition,

$$(f g)(x) = f(g(x)),$$

identity the identity map, and inverse the inverse map.

Möbius maps

For a subring $A \subseteq \mathbb{R}$, $\mathrm{SL}_2(A)$ is the set of $2 \times 2$ matrices $g = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ with $a, b, c, d \in A$ and $ad - bc = 1$. Such a $g$ acts on $X$ by

$$g \cdot t = \begin{cases} \dfrac{at + b}{ct + d} & \text{if } t \in \mathbb{R},\ ct + d \neq 0,\\[4pt] \infty & \text{if } t \in \mathbb{R},\ ct + d = 0, \end{cases} \qquad g \cdot \infty = \begin{cases} a / c & \text{if } c \neq 0,\\ \infty & \text{if } c = 0. \end{cases}$$

Two rings are used below: $A = \mathbb{R}$ itself, and $A = \mathbb{Z}$ (the smallest subring of $\mathbb{R}$, consisting of the integer multiples of $1$). So $\mathrm{SL}_2(\mathbb{Z})$ is the set of integer matrices of determinant $1$.

A matrix $g \in \mathrm{SL}_2(\mathbb{Z})$ is called hyperbolic when its trace satisfies $|a + d| > 2$ (strict inequality).

Let

$$P = \{\, p \in X : g \cdot p = p \text{ for some hyperbolic } g \in \mathrm{SL}_2(\mathbb{Z}) \,\}.$$

(For instance $\infty \notin P$: $g \cdot \infty = \infty$ forces $c = 0$, hence $ad = 1$ with $a,d$ integers, hence $a + d = \pm 2$, which is not hyperbolic.)

Piecewise-projective homeomorphisms

Given a subring $A \subseteq \mathbb{R}$ and a set $E \subseteq X$, say that $f \in \mathrm{Homeo}(X)$ is piecewise $A$-projective with breaks in $E$ if there is a finite set $B \subseteq E$ (possibly empty) such that for every point $x \in X \setminus B$ — including $x = \infty$ if $\infty \notin B$ — there exist a matrix $g \in \mathrm{SL}_2(A)$ and a neighbourhood $U$ of $x$ in $X$ with $f(y) = g \cdot y$ for all $y \in U$. The matrix $g$ may depend on $x$. Nothing is required of $f$ at points of $B$ beyond $f$ being a homeomorphism.

Define:

• $G_{pp} \le \mathrm{Homeo}(X)$: the subgroup generated by all homeomorphisms that are piecewise $\mathbb{R}$-projective with breaks in $X$ (that is, with an arbitrary finite exceptional set, the local matrices taken from $\mathrm{SL}_2(\mathbb{R})$).
• $S$: the set of those $f \in \mathrm{Homeo}(X)$ that both lie in $G_{pp}$ and are piecewise $\mathbb{Z}$-projective with breaks in $P$ (the finite exceptional set lies inside $P$, and the local matrices come from $\mathrm{SL}_2(\mathbb{Z})$).
• $G$: the subgroup of $\mathrm{Homeo}(X)$ generated by $S$ (all finite products of elements of $S$ and their inverses).
• $H = \{\, f \in G : f(\infty) = \infty \,\}$, the stabiliser of $\infty$ inside $G$.

Note the order of operations: $H$ is obtained by first generating $G$ from $S$ and then intersecting with the stabiliser of $\infty$; it is not the group generated by those elements of $S$ that fix $\infty$.

$H$ is regarded as an abstract group with the composition product above; no topology on $H$ plays any role.

The measure condition, and the claim

A function $m$ assigning to every subset $Y \subseteq H$ a value $m(Y) \in [0, \infty]$ (the extended non-negative reals, with $\infty + u = \infty$) is a left-invariant finitely additive probability measure on $H$ if all of the following hold:

1. $m(\varnothing) = 0$;
2. $m(Y \cup Z) = m(Y) + m(Z)$ for all disjoint $Y, Z \subseteq H$;
3. $m(H) = 1$;
4. $m(hY) = m(Y)$ for every $h \in H$ and every $Y \subseteq H$, where $hY = \{\, h y : y \in Y \,\} = \{\, h \circ y : y \in Y \,\}$ (left translation).

Every subset of $H$ is in the domain; there is no measurability restriction. Conditions 2 and 3 together force $m(Y) + m(H \setminus Y) = 1$, so in fact every value of $m$ lies in $[0, 1]$. Only invariance under left translation is required; nothing is said about right translation or conjugation.

The statement asserts: there is no function $m$ on the subsets of $H$ satisfying conditions 1–4.